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600 | https://cage.ugent.be/~hs/billiards/billiards.html | ELLIPTICAL BILLIARD TABLES
Normally a billiards game is played on a rectangular table. On a "mathematical billiard table" we consider only one ball. Moreover, this ball is reduced to a point that moves in a straight line until it hits an edge and bounces back in such a way that the angle of reflection equals the angle of incidence. Both angles can be considered as the angles formed by the side of the table involved and the incoming and outgoing lines. The mathematician Charles L. Dodgson, better known as Lewis Carroll, the author of "Alice in Wonderland" thought and wrote about circular billiard tables. We'll consider a billiard table that takes the shape of an ellipse. I started this page in 2002, but only in 2007 I discovered the book Geometry and Billiards by Serge Tabachnikov. In this book (186 pages) the author treats different billiard table shapes and the (often not so elementary) mathematics behind them. The bouncing law is linked to the reflection law in optics. As to elliptical billiards this leads in a short and elegant way to the property that a shot passing through one focus is reflected through the other focus. A nice geometrical proof of this property is mentioned further. The book also contains an extensive bibliography.
The animations on this page have been made using Maple.
First we remind some facts about ellipses.
Definition and a useful property of an ellipse
Given two fixed point F1 and F2 in a plane, an ellipse is the locus of all points P in that plane for which the sum of the distances to F1 and F2 equals a given value 2a. The given points F1 and F2 are called the foci or focal points of the ellipse, PF1 and PF2 are called the focal radii of the point P. For all points outside the ellipse the sum of the distances to the focal points is larger than 2a, for all points inside the ellipse it is smaller than 2a.
Given the value 2a and calling 2c the distance between the foci F1 and F2 we are easily led to the length of the axes of the ellipse. The length of the major axis AA' is 2a and the length of the minor axis BB' is 2b. The relation between a, b and c is illustrated on the next picture. The property that will be most useful in the further treatment of our subject is illustrated on the next picture: for any point P on the ellipse, the bisectors b and b'of FP and F'P are the tangent and normal in P to the ellipse.
The orthogonal line from F to b meets b in Q and F'P in F". For any point R (different from P) on b we have: RF = RF". Furthermore RF'+ RF" > F'F" = F'P + PF = 2a.
From the definition of the ellipse it follows that R is outside of the ellipse, and so are all points on b different from P. This means that the bisectors b and n are the tangent and normal in P to the ellipse.
A special case: the ball passes over a focus
The ball is positioned in P1 and shot in the direction of the focus F1. If the ball hits the border of the table in P2 it will be be bounced according to the law of reflection: the angle between the incoming path and the normal in P2 equals the angle between this normal and the outgoing path. As a consequence of the property last mentioned the ball will bounce back along P2F2 ! After a new hit the ball will bounce through the other focus F1. In the following illustration we consider a sequence of consecutive bounces.
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After a rather small number of bounces the ball seems to travel along the major axis!
The general case: the first shot doesn't pass over a focus
If the first shot doesn't pass over a focus none of all subsequent reflected paths will pass over a focus. But also in this case the mathematical elliptical billiard game leads to interesting patterns!
First case: the first shot doesn't pass between the foci of the ellipse.
General situation
It follows from the mentioned property of the focal radii in relation with the normal or tangent in the reflection point that none of all subsequent reflected segments will pass between the foci of the ellipse. If we consider a large number of reflected segments there seems to arise a new ellipse! Some knowledge of calculus is needed to find out that this curve, known as the evolute of the reflected segments indeed is an ellipse and moreover the foci of this ellipse are the same as those of the original ellipse!
In Tabachnikovs book one can find a proof of the property that (in general) a billiard trajectory inside an ellipse remains tangent to a confocal conic.
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In the following animated picture 50 reflections are shown.
Some exceptions: periodical paths.
One easily can see that there exist particular situations: in some cases the paths are periodical and don't lead to a new ellipse inside the given one. We give two examples: a quadrangular path and a hexagonal path.
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The circle as a special ellipse.
A circle can be considered as a special ellipse: the foci F1 and F2 coincide, a = b = radius circle ; c = 0. If a path is non periodical the evolute of the reflected shots is a new circle with the same centre. Perhaps the existence of periodical paths is more easy to be seen.
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Second case: the first shot passes between the foci of the ellipse.
General situation
It follows from the mentioned property of the focal radii in relation with the normal or tangent in the reflection point that all subsequent reflected segments will pass between the foci of the ellipse. If we consider a large number of reflected segments there seems to arise a new curve! Some knowledge of calculus is needed to find out that this curve, known as the evolute of the reflected segments is a hyperbola and moreover the foci of this hyperbola are the same as those of the original ellipse! The definition of a hyperbola resembles the definition of an ellipse: given two fixed point F1 and F2 in a plane, a hyperbola is the locus of all points P in that plane for which the difference of the distances to F1 and F2 equals a given value 2a. The given points F1 and F2 are called the foci of the hyperbola and PF1 and PF2 are called the focal radii of the point P.
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In the following animated picture 50 reflections are shown.
Also in this case one can imagine that exceptions exist. An example: if we push the ball along the minor axis of the ellipse it will continuously bounce along the same axis.
Herman Serras, Last updated February 2018 return to the homepage |
601 | https://www.geeksforgeeks.org/maths/solving-multi-step-linear-equations-with-fractions/ | Published Time: 2024-07-30 07:11:56
Solving Multi-Step Linear Equations with Fractions - GeeksforGeeks
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Next Article: What are Numbers?
Solving Multi-Step Linear Equations with Fractions
Last Updated : 23 Jul, 2025
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Solving multi-step linear equations with fractions requires a clear understanding of algebraic principles and fraction operations. This process involves systematically isolating the variable by performing operations such as addition, subtraction, multiplication, and division on both sides of the equation.
In this article, we'll break down the process of solving multi-step linear equations with fractions into simple, manageable steps. My goal is to make sure you feel confident and capable when tackling these problems. We'll cover the essential techniques, highlight common mistakes to avoid, and provide practical examples to show how these equations are used in real life. Plus, we'll include FAQs and solved problems to help reinforce your understanding.
What is Multi-Step Linear Equations with Fractions?
Multi-step linear equations with fractions are algebraic equations that involve multiple operations (addition, subtraction, multiplication, division) to isolate the variable. These equations also contain fractions, which add an extra layer of complexity. The goal is to simplify the equation step-by-step until the variable is isolated on one side of the equation.
Fractions: Parts of a whole, represented as a ratio of two integers, such as a b\frac{a}{b}b a, Where a is the numerator and bnumerator and b is the denominator.
Operations: The sequence of mathematical actions performed to simplify and solve the equation, including addition, subtraction, multiplication, and division.
Isolation of the Variable:The process of manipulating the equation to get the variable (usually represented as x) alone on one side.
Example Multi-Step Linear Equations with Fractions
Consider the equation 3 x 5−2 3=1 2\frac{3x}{5} - \frac{2}{3} = \frac{1}{2}5 3 x−3 2=2 1. To solve this equation, follow these steps :
Step 1:Find theleast common denominator (LCD)of fractions, i.e. 30.
Step 2:Multiply each term by 30 to clear the fractions:
30(3x/5) - 30(2/3) = 30(1/2)
Step 3: Simplify: 18x - 20 = 15
Step 4: Isolate the variable xby adding 20 to both sides: 18x = 35
Step 5:Finally, divide by 18 to solve for x: x = 35/18
Important Formulas Multi-Step Linear Equations with Fractions
Understanding key formulas and concepts is crucial for solving multi-step linear equations with fractions. Here are the important ones:
Distributive Property
The distributive property allows you to eliminate parentheses by distributing multiplication over addition or subtraction within the parentheses.
Formula: a(b+c) = ab + ac
Combining Like Terms
Combine terms with the same variable to simplify the equation.
Formula: ax + bx = (a+b)x
Addition and Subtraction Properties of Equality
Add or subtract the same value from both sides of the equation to help isolate the variable.
Formula:if a = b,than a + c = b+cand a-c = b-c
Multiplication and Division Properties of Equality
Multiply or divide both sides of the equation by the same nonzero number to isolate the variable.
Formula:If a=b, than ac = bc(for c≠0 c \neq 0 c=0) and a c=b c\frac{a}{c} = \frac{b}{c} c a=c b(for c≠0 c \neq 0 c=0).
Reciprocal of a Fraction
When dividing by a fraction, multiply by its reciprocal.
Formula:a b c=a⋅c b\frac{a}{\frac{b}{c}} = a\cdot \frac{c}{b}c ba=a⋅b c
Cross-Multiplication (for Equations with Two Fractions)
For equations involving two fractions set equal to each other, cross-multiply to clear the fractions.
Formula:If a b=c d,\frac{a}{b} = \frac{c}{d},b a=d c, then ad = bc.
Linear Equations with Fractions
Concept related to Linear equation with fractions is added below:
Definition: Linear equations are mathematical statements of equality involving variables and constants. When fractions are included, these equations can be written in the form , where a,b, c,d, e are integers, and c and e are not zero.
Examples of Linear Equations with Fractions
3 x+2 4=5 6\bold{\frac{3x+2}{4} = \frac{5}{6}}4 3x+2=6 5
2 x−7 5+3 8=4 9\bold{\frac{2x-7}{5} + \frac{3}{8} = \frac{4}{9}}5 2x−7+8 3=9 4
Steps to Solve Multi-Step Linear Equations with Fractions
Follow the steps added below to Solve Multi-Step Linear Equations with Fractions
Step 1: Simplify the Fractions
Ensure all fractions are in their simplest form. For example, reduce 4 8\frac{4}{8}8 4 to 1 2\frac{1}{2}2 1
Step 2: Find theLeast Common Denominator (LCD)
Identify the LCD of all the fractions in the equation. The LCD is the smallest number that all denominators can divide into evenly.
Step 3: Clear the Fractions
Multiply every term in the equation by the LCD to eliminate the fractions. For example, in the equation 2 x 3+1 4=5 6\bold{\frac{2x}{3} + \frac{1}{4} = \frac{5}{6}}3 2x+4 1=6 5, the LCD is 12. Multiply every term by 12:
12(2 x 3)+12(1 4)=12(5 6)\bold{12(\frac{2x}{3}) + 12(\frac{1}{4}) = 12(\frac{5}{6})} 12(3 2x)+12(4 1)=12(6 5)
8 x+3=10\bold{8x + 3 = 10}8x+3=10
Step 4: Solve the Simplified Equation
Solve the resulting linear equation using standard algebraic methods. For example:
8 x+3=10\bold{8x + 3 = 10}8x+3=10
8 x=7\bold{8x = 7}8x=7
x=7 8\bold{x = \frac{7}{8}}x=8 7
Common Mistakes and How to Avoid Them
Misinterpreting Fractions : Always double-check the fractions to ensure they are correctly understood.
Incorrectly Simplifying Fractions :Verify that all fractions are simplified before proceeding.
Errors in Arithmetic Operations :Carefully perform all arithmetic operations and recheck your work to avoid mistakes.
Examples Related to Solving Multi-Step Linear Equations with Fractions
Example 1: Solve3 x 5−2 3=1 2\bold{\frac{3x}{5} - \frac{2}{3} = \frac{1}{2}}5 3x−3 2=2 1
Solution:
Find the Least Common Denominator (LCD) of 5, 3, and 2, which is equal to 30.
Multiply through by 30:
30(3 x 5)−30(2 3)=30(1 2)\bold{30(\frac{3x}{5}) - 30(\frac{2}{3}) = 30(\frac{1}{2})} 30(5 3x)−30(3 2)=30(2 1)
18 x−20=15\bold{18x − 20 = 15} 18x−20=15
18 x=35\bold{18x=35}18x=35
x=35 18\bold{x = \frac{35}{18}}x=18 35
Example 2: Solve4 x 7+5 6=3 2\bold{\frac{4x}{7} + \frac{5}{6} = \frac{3}{2}}7 4x+6 5=2 3
Solution:
Find the LCD of 7, 6, and 2, which is equal to 42.
Multiply through by 42:
42(4 x 7)+42(5 6)=42(3 2)\bold{42(\frac{4x}{7}) + 42(\frac{5}{6}) = 42(\frac{3}{2})} 42(7 4x)+42(6 5)=42(2 3)
24 x+35=63\bold{24x+35=63}24x+35=63
24 x=28\bold{24x=28}24x=28
x=28 24=7 6\bold{x = \frac{28}{24} = \frac{7}{6}}x=24 28=6 7
Example 3: Solve2 x 3+1 4=5 6\frac{2x}{3} + \frac{1}{4} = \frac{5}{6}3 2 x+4 1=6 5
Solution:
Find Least Common Denominator (LCD): The LCD of 3, 4, and 6 is 12
Multiply every term by 12:
12⋅2 x 3+12⋅1 4=12⋅5 6 12 \cdot \frac{2x}{3} + 12 \cdot \frac{1}{4} = 12 \cdot \frac{5}{6}12⋅3 2 x+12⋅4 1=12⋅6 5
Simplifies to:
8x + 3 = 10
Subtract 3 from both sides:
8x = 7
x = 7/8
Example 4: Solve4 x 7−3 5=1 2\frac{4x}{7} - \frac{3}{5} = \frac{1}{2}7 4 x−5 3=2 1
Solution:
LCD of 7, 5, and 2 is 70.
Multiply every term by 70
70⋅4 x 7−70⋅3 5=70⋅1 2 70 \cdot \frac{4x}{7} - 70 \cdot \frac{3}{5} = 70 \cdot \frac{1}{2}70⋅7 4 x−70⋅5 3=70⋅2 1
Simplifies to:
40x - 42 = 35
Add 42 to both sides:
40x = 77
x = 77/40
Example 5: Solve2 x−3 5=4 3\frac{2x-3}{5} = \frac{4}{3}5 2 x−3=3 4
Solution:
Cross-Multiply:
3(2 x−3)=4⋅5 3(2x-3) = 4 \cdot 5 3(2 x−3)=4⋅5
Simplifies to:
6x - 9 = 20
Add 9 to both sides
6x = 29
x = 29/6
Example 6: Solve7 x−1 3=2 x+5 4\frac{7x-1}{3} = \frac{2x+5}{4}3 7 x−1=4 2 x+5
Solution:
Cross-Multiply:
4(7 x−1)=3(2 x+5)4(7x - 1) = 3(2x+5)4(7 x−1)=3(2 x+5)
Simplifies to:
28x - 4 = 6x + 15
Combine Like Terms: Subtract 6x from both sides:
22x - 4 = 15
Add 4 to both sides:
22x = 19
x = 19/22
Example 7: Solve3 x+2 4=5 6\frac{3x+2}{4} = \frac{5}{6}4 3 x+2=6 5
Solution:
Cross-Multiply:
6(3 x+2)=5⋅4 6(3x + 2) = 5 \cdot 4 6(3 x+2)=5⋅4
Simplifies to:
18x + 12 = 20
Subtract 12 from both sides:
18x = 8
x = 8/18 = 4/9
Example 8: Solve6 x+1 2=3 x−4 4\frac{6x+1}{2} = \frac{3x-4}{4}2 6 x+1=4 3 x−4
Solution:
Cross-Multiply:
4(6 x+1)=2(3 x−4)4(6x+1) = 2(3x-4)4(6 x+1)=2(3 x−4)
Simplifies to:
24x+4 = 6x-8
Combine Like Terms: Subtract 6x from both side:
18x+4 = -8
Subtract 4 from both sides:
18x = -12
x = -12/18 = -2/3
Example 9: Solve3 x−2 5=7 3\frac{3x-2}{5} = \frac{7}{3}5 3 x−2=3 7
Solution:
Cross-Multiply:
3(3 x−2)=7⋅5 3(3x-2) = 7\cdot5 3(3 x−2)=7⋅5
Simplifies to:
9x-6 = 35
Add 6 to both sides:
9x = 41
x = 41/9
Example 10: Solve4 x+3 2=5 x−1 3\frac{4x+3}{2} = \frac{5x-1}{3}2 4 x+3=3 5 x−1
Solution:
Cross-Multiply:
3(4 x+3)=2(5 x−1)3(4x + 3) = 2(5x - 1)3(4 x+3)=2(5 x−1)
Simplifies to:
12x + 9 = 10x -2
Combine Like Terms: Subtract 10x from both side :
2x + 9 = -11
Subtract 9 from both sides:
2x = -11
x = -11/2
Multi-Step Linear Equations with Fractions - Practice Problems
Problem 1: Solve:5 x−3 7=2 x+4 3\frac{5x-3}{7} = \frac{2x+4}{3}7 5 x−3=3 2 x+4
Problem 2: Solve:2 x+1 5=x−3 2\frac{2x+1}{5} = \frac{x-3}{2}5 2 x+1=2 x−3
Problem 3: Solve:3 x−2 4=x−3 2\frac{3x-2}{4} = \frac{x-3}{2}4 3 x−2=2 x−3
Problem 4: Solve:4 x+3 8=5 x−2 7\frac{4x+3}{8} = \frac{5x-2}{7}8 4 x+3=7 5 x−2
Problem 5: Solve:7 x−5 9=2 x+6 7\frac{7x-5}{9} = \frac{2x+6}{7}9 7 x−5=7 2 x+6
Problem 6: Solve:6 x+2 3=4 x−1 5\frac{6x+2}{3} = \frac{4x-1}{5}3 6 x+2=5 4 x−1
Problem 7:Solve:5 x−4 6=3 x+2 8\frac{5x-4}{6} = \frac{3x+2}{8}6 5 x−4=8 3 x+2
Problem 8: Solve:8 x+1 7=2 x−3 5\frac{8x+1}{7} = \frac{2x-3}{5}7 8 x+1=5 2 x−3
Problem 9: Solve:9 x−2 4=3 x+5 6\frac{9x-2}{4} = \frac{3x+5}{6}4 9 x−2=6 3 x+5
Problem 10: Solve:10 x+3 5=4 x−7 9\frac{10x+3}{5} = \frac{4x-7}{9}5 10 x+3=9 4 x−7
Conclusion
Understanding and applying the steps to solve these equations not only enhances your mathematical skills but also prepares you for practical problem-solving in numerous fields.
In this article, we’ve broken down the process into clear, manageable steps, highlighted common mistakes to avoid, and provided practical examples to demonstrate real-world applications. By practicing these methods and using the tips provided, you can approach multi-step linear equations with confidence and accuracy.
Read More:
Simplifying Fractions
Least Common Denominator (LCD)
Solving Linear Equations
Linear Equations in One Variable
Frequently Aksed Questions
Why is the use of Solving Linear Equations?
Solving fractions simplifies the equation, making it easier to solve.
What is the Least Common Denominator (LCD)?
LCD is the smallest number that all denominators in the equation can divide into evenly.
What are Simple Fractions?
Simple fractions, often just called "fractions," are a way of representing parts of a whole or a division of quantities.
What are Examples of Simple Fractions?
Various examples of simple fractions are: 1/2, 3/4, 7/10, etc.
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603 | https://chengzhaoxi.xyz/7a900672.html | 分层抽样 | 潮汐朝夕
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数统计
分层抽样
字数统计: 2.1k字 | 阅读时长: 7分
307
2021-12-12
数据分析 16, 智能风控 10, 统计 20
摘要: 分层抽样
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分层抽样的概念
抽样时，将总体分成互不交叉的层，然后按照一定的比例，从各层独立地抽取一定数量的个体，将各层取出的个体合在一起作为样本，这种抽样方法叫分层抽样。
有几个关键要点
总体个体差异明显，每层的差异比较大，层内个体间的差异比较小
每层可以抽取多少样本，常见的有以下这些方案
如果根据它在总体中占的比例来抽取，就是等比例抽样
也可以对不同的层赋予不同的权重，手动控制各层的抽样规模。
对每一层都分配同样的个体数
各层抽得的样本数与所抽得的总样本数之比等于该层方差与各类方差之和的比
每层抽取样本时，采用简单随机抽样
分层抽样的步骤
(1) 确定总体与样本容量抽取的比例
抽取比例 = 样本容量 / 总体个数
(2) 由分层情况，确定各层抽取的样本数
如果是等比例抽样，各层抽取个数 = 抽取比例 各层个数
如果是加权抽样，各层抽取个数自定义
(3) 各层抽取数之和应等于样本容量
等比例分层抽样的例子
一单位有 500 职工，不到 35 岁的有 125 人，35~49 有 280 人，50 以上有 95 人。
为了解单位职工与身体状况有关的某指标。抽取一个容量为 100 的样本。
思路
身体状况与年龄的关系非常大，因此适合分层抽样。
step1:j 确定比例
样本容量 / 总体个数 = 100 / 500 = 1/5
step2: 各层抽取的样本数
每一个层都应该按照 1/5 抽取，各层抽取的个数为
125/5, 280/5, 95/5 = 25, 56, 19
step3: 在各层用简单随机抽样
分层抽样的代码模板 (Pandas)
假设数据中有一个分类型变量 var0，以该变量 df[“var0”] 为依据分层抽样，各层的抽样个数均为 10
1
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3
4
5
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7# 分层抽样
each_sample_count = 10 # 各层抽样个数
label_unique = np.unique(df['var0']) # 分层依据
df_sample = pd.DataFrame(np.zeros((1, len(df.columns))), columns=df.columns ,dtype=int)
for label in label_unique:
sample = pd.DataFrame.sample(df[df['var0']==label], each_sample_count)
df_sample = pd.concat([df_sample,sample])
分层抽样的优缺点
优点
如果层内的具有较低的标准偏差（与总体中的总体标准偏差相比），则分层会产生较小的估计误差。
对于许多应用，当人口被分组到层中时，测量变得更易于管理。
分层抽样特别适用于既要对总体参数进行推断，也要对各子总体(层)的参数进行推断的情形
评分卡建模数据集切分的例子
在评分卡建模项目中，时间窗口确定之后，数据集也就定了下来，这份数据集相当于是总体。
在建模前，数据集一般分为 3 个子集：开发样本(dev)，验证样本(val)，时间外样本(OOT)。
一般情况下，时间外样本通常使用整个建模样本中最后一段时间的样本。而开发样本与验证样本使用分层抽样进行划分，目的是保证两个数据集中的负样本占比相同。
但是评分卡建模中，从经验上看正负样本数量都应该 >= 1500，且总样本量最好不超过 50000，因为超过 5 万后，模型的效果就不在随着样本量增加而有显著变化了。此时一般需要对正样本做欠采样处理。
用分层抽样对好样本欠采样
分层抽样是这里的欠采样的常用方法，保证抽样后，开发样本，验证样本，时间外样本的正负比例相同。
具体的做法是首先根据总体的负样本数，确定一个正样本和负样本想要保留的样本数，正负样本比例也顺便确定了。然后将最后一段时间范围的样本作为时间外样本的候选样本，其余的为训练样本和测试样本的候选样本。
然后按照正负样本比例和两份候选样本的负样本个数，以及训练集验证集的比例，得到训练集、验证集、时间外样本集需要抽样的正样本数，然后进行分层抽样即可。
然后按照正负样本比例和时间内样本的负样本个数，得到需要抽样的正样本数，对时间外样本的候选样本分层抽样后，得到时间外样本；
例子
例如我们有 200000 样本，其中有 5000 负样本(2.5%)，195000 正样本。
首先我们确定一个比例，就是负样本占比 10%(2.5% 的四倍)。基于这个比例我们确定负样本取 5000，正样本取 45000，这样负样本占比就是 10% 了。
然后我们选定最后一段时间内的样本作为时间外样本的候选样本，假设这部分样本有 50000，其中 1000 负样本，占比 2%。那么开发样本和测试样本的候选样本就是 150000，其中 4000 负样本，占比 2.67%。
接下来抽样获取时间外样本。由于时间外样本有 1000 负样本，负样本比例 10%，因此我们需要抽样 9000 正样本。
开发样本和测试样本的候选样本我们称为时间内总体，共 150000，其中 4000 负样本。由于负样本比例 10%，因此我们需要从 146000 中抽样 36000。然后将 4000 负样本和 36000 正样本分别按某个测试集比例(例如 0.3) 切割即可。
抽样完成后，数据记录表大致是下面这样
| | 好样本 | 坏样本 | 样本合计 |
--- --- |
| 数量(总体) | 195000 | 5000 | 200000 |
| 占比(总体) | 97.5% | 2.5% | 100% |
| 数量(时间内总体) | 146000 | 4000 | 150000 |
| 占比(时间内总体) | 97.33% | 2.67% | 100% |
| 数量(训练集抽样) | 25200 | 2800 | 28000 |
| 占比(训练集抽样) | 90% | 10% | 100% |
| 权重(训练集抽样) | 4.0556 | 1
| 加权(训练集抽样) | 102200 | 2800 | 105000(训练集占比0.7) |
| 数量(测试集抽样) | 10800 | 1200 | 12000 |
| 占比(训练集抽样) | 90% | 10% | 100% |
| 权重(测试集抽样) | 4.0556 | 1
| 加权(测试集抽样) | 43800 | 1200 | 45000(测试集占比0.3) |
| 数量(时间外总体) | 49000 | 1000 | 50000 |
| 占比(时间外总体) | 98.0% | 2.0% | 100% |
| 数量(时间外抽样) | 9000 | 1000 | 10000 |
| 占比(时间外总体) | 90% | 10% | 100% |
| 权重(时间外抽样) | 5.44 | 1
| 加权(时间外抽样) | 49000 | 1000 | 50000 |
例子2
上面的例子中为了看的清楚，数据取的都很整。这里把数据做的乱一点再看一下。
总体数据为 230257，负样本 8139(占比 3.53%)，分出时间外样本后：
时间外候选样本 33115，负样本 1034(占比 3.12%)
开发测试候选样本 197142，负样本 7105(占比 3.60%)
依然取负样本比例 10%，测试集比例 0.3，分层抽样后，数据记录表大致如下(注意负样本也做了一定采样，因此抽样后，负样本也是有权重的)
| | 好样本 | 坏样本 | 样本合计 |
--- --- |
| 数量(总体) | 222118 | 8139 | 230257 |
| 占比(总体) | 96.47% | 3.53% | 100% |
| 数量(时间内总体) | 190037 | 7105 | 197142 |
| 占比(时间内总体) | 96.40% | 3.60% | 100% |
| 数量(训练集抽样) | 26945 | 3045 | 29990 |
| 占比(训练集抽样) | 89.85% | 10.15% | 100% |
| 权重(训练集抽样) | 4.93743 | 1.631856
| 加权(训练集抽样) | 133039 | 4969 | 138008(训练集占比0.7) |
| 数量(测试集抽样) | 11544 | 1309 | 12853 |
| 占比(训练集抽样) | 89.82% | 10.18% | 100% |
| 权重(测试集抽样) | 4.93746 | 1.63178
| 加权(测试集抽样) | 56998 | 2136 | 59134(测试集占比0.3) |
| 数量(时间外总体) | 32081 | 1034 | 33115 |
| 占比(时间外总体) | 96.88% | 3.12% | 100% |
| 数量(时间外抽样) | 6511 | 646 | 7157 |
| 占比(时间外总体) | 90.97% | 9.03% | 100% |
| 权重(时间外抽样) | 4.9272 | 1.600619
| 加权(时间外抽样) | 32081 | 1034 | 33115 |
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604 | https://allen.in/dn/qna/646470158 | A string is wrapped around the rim of a wheel of moment of inertia 0.20 kg m^2 and radius 20 cm.the wheel is free to rotate about its axis .Initially,the wheel is at rest.The string is now pulled by a force of 20 N.Find the angular velocity of the wheel after 5.0 seconds.
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HomeClass 11PHYSICSA string is wrapped around the rim of a ...
A string is wrapped around the rim of a wheel of moment of inertia 0.20 k g m 2 and radius 20 cm.the wheel is free to rotate about its axis .Initially,the wheel is at rest.The string is now pulled by a force of 20 N.Find the angular velocity of the wheel after 5.0 seconds.
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Home » Law » The Doctrine of Ultra Vires in Company Law: From Strict Application to Statutory and Judicial Dilution
The Doctrine of Ultra Vires in Company Law: From Strict Application to Statutory and Judicial Dilution
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Introduction
The doctrine of ultra vires, meaning ‘beyond the powers,’ has historically played a pivotal role in company law by restricting companies to activities within the scope of their objects clause as defined in their memorandum of association. This principle was crystallised in the landmark case of Ashbury Railway Carriage & Iron Co. Ltd v Riche (1875), where the House of Lords held that acts outside a company’s stated objects were void and incapable of ratification. However, over time, statutory reforms and judicial interpretations, both in the United Kingdom and in other common law jurisdictions such as Uganda, have significantly diluted the strict application of this doctrine. This essay critically examines the origins and initial strict application of ultra vires, before analysing how subsequent legislative changes, including provisions in Uganda’s Companies Act, Cap 106, and evolving judicial attitudes have softened its impact. By exploring key case law and statutory provisions, the essay evaluates the extent to which the doctrine retains relevance in modern company law.
The Origins and Strict Application of Ultra Vires
The doctrine of ultra vires emerged as a mechanism to protect shareholders and creditors by ensuring that a company did not deviate from the purposes for which it was incorporated. In Ashbury Railway Carriage & Iron Co. Ltd v Riche (1875), the company entered into a contract to finance the construction of a railway, an activity outside its objects clause, which was limited to manufacturing and selling railway carriages. The House of Lords ruled that the contract was void, as the company lacked the capacity to undertake such activities, and even shareholder ratification could not validate it (Hannigan, 2018). This decision underscored the strict interpretation of ultra vires, premised on the notion that a company, as an artificial legal entity, only possessed powers explicitly granted by its memorandum of association.
The strict application of ultra vires served a dual purpose: it protected investors by ensuring funds were used for intended purposes and safeguarded creditors by limiting the risk of companies engaging in speculative ventures. However, this rigidity often led to harsh consequences, particularly for third parties dealing with companies in good faith, as contracts deemed ultra vires were void and unenforceable. Indeed, the doctrine’s inflexibility arguably stifled commercial innovation, as companies could not adapt to changing economic circumstances without amending their objects clause, a cumbersome and costly process (Griffin, 2020).
Statutory Reforms and the Dilution of Ultra Vires in the UK
Recognising the practical limitations of the ultra vires doctrine, legislative reforms in the UK progressively undermined its strict application. The Companies Act 1989, for instance, introduced significant changes by allowing companies to adopt broad objects clauses or register with the general object of carrying on business as a ‘general commercial company,’ effectively bypassing ultra vires restrictions (Davies, 2020). Furthermore, provisions in the Companies Act 2006 (UK) largely abolished the doctrine’s external application by protecting third parties acting in good faith. Section 39 of the 2006 Act states that 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. This statutory shift prioritised commercial certainty over the traditional protective function of ultra vires (Sealy & Worthington, 2013).
While remnants of the doctrine persist internally—such as in disputes between shareholders and directors regarding misuse of company powers—the external constraints on third-party transactions have been significantly eroded. Thus, the legislative trend in the UK reflects a deliberate move away from the strict principles established in Ashbury Railway, aiming to balance flexibility with accountability.
Statutory and Judicial Developments in Uganda
In Uganda, a common law jurisdiction influenced by British legal principles, the evolution of the ultra vires doctrine mirrors, to some extent, developments in the UK, though with contextual differences. The Companies Act, Cap 106 (now repealed and replaced by the Companies Act, 2012), historically incorporated provisions that required companies to adhere to their objects clause, reflecting the traditional ultra vires doctrine. However, similar to reforms in the UK, the Companies Act, 2012 introduced reforms to mitigate the harshness of ultra vires. Section 34 of the Act provides that a company has the capacity and powers of an individual, thereby allowing it to engage in activities beyond a strictly defined objects clause unless otherwise restricted by its articles (Uganda Law Reform Commission, 2012).
Judicial approaches in Uganda have also contributed to the dilution of ultra vires. While there is a scarcity of high-profile Ugandan case law directly challenging the application of ultra vires in the manner of Ashbury Railway, courts have generally adopted a pragmatic stance. For instance, in Kampala Capital City Authority v Kabagambe (2015), a case indirectly touching on corporate powers, the Ugandan judiciary demonstrated a willingness to prioritised transactional security over strict adherence to formalistic interpretations of a company’s capacity (Musoke, 2016). Although this case does not explicitly address ultra vires, it reflects a judicial trend towards flexibility, aligning with the statutory intent of the Companies Act, 2012. Regrettably, limited access to comprehensive Ugandan case law on ultra vires restricts a more detailed analysis of judicial attitudes in this context, but the available evidence suggests a departure from rigid application.
Comparative Analysis and Broader Implications
Across common law jurisdictions, the trajectory of ultra vires illustrates a broader tension between legal formalism and commercial pragmatism. In jurisdictions such as Australia and Canada, similar statutory reforms have curtailed the doctrine’s impact. For example, Canada’s Canada Business Corporations Act (1985) grants companies the capacity of a natural person, rendering ultra vires largely obsolete in corporate transactions (MacMillan, 2019). These comparative developments highlight a global shift towards facilitating business flexibility, paralleling changes in Uganda and the UK.
However, the dilution of ultra vires raises concerns about accountability. While third-party protection is a laudable objective, the reduced emphasis on objects clauses may allow directors to engage in risky or unrelated ventures without sufficient checks, potentially prejudicing minority shareholders. Therefore, while statutory reforms address the doctrine’s historical harshness, they arguably shift the burden of oversight onto internal governance mechanisms, such as shareholder resolutions and directors’ duties (Hannigan, 2018).
Conclusion
In conclusion, the doctrine of ultra vires, as solidified in Ashbury Railway Carriage & Iron Co. Ltd v Riche (1875), originally served as a strict boundary on corporate capacity, safeguarding stakeholders by confining companies to their stated objects. However, its rigidity proved impractical, prompting legislative and judicial interventions in the UK, Uganda, and other common law jurisdictions. In the UK, reforms under the Companies Act 2006 have largely abolished the doctrine’s external effect, while Uganda’s Companies Act, 2012 and judicial trends reflect a similar move towards flexibility. While these changes enhance commercial certainty, they also underscore unresolved issues of accountability and internal governance. Ultimately, the evolution of ultra vires demonstrates the law’s adaptability to economic realities, though it raises critical questions about how best to balance corporate freedom with stakeholder protection in modern company law.
References
Davies, P. L. (2020) Gower and Davies’ Principles of Modern Company Law. 11th ed. Sweet & Maxwell.
Griffin, S. (2020) Company Law: Fundamental Principles. 6th ed. Pearson Education.
Hannigan, B. (2018) Company Law. 5th ed. Oxford University Press.
MacMillan, C. (2019) Corporate Law in Common Law Jurisdictions. Routledge.
Musoke, R. (2016) Commercial Law Developments in Uganda. Kampala: LawAfrica Publishing.
Sealy, L. & Worthington, S. (2013) Sealy & Worthington’s Cases and Materials in Company Law. 10th ed. Oxford University Press.
Uganda Law Reform Commission (2012) Report on the Reform of Company Law in Uganda. Government Printer, Kampala.
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606 | https://www.cdc.gov/gonorrhea/hcp/clinical-care/index.html | A .gov website belongs to an official government organization in the United States.
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Related Topics:
Clinical Treatment of Gonorrhea
At a glance
Treatment options
Treatment regimens for gonorrhea can be found in STI Treatment Guidelines, 2021.
Use of EPT in the Treatment of Gonorrhea
Because of EPT’s effectiveness in reducing gonorrhea reinfection rates,1 CDC has recommended its use since 2006 for the heterosexual partners of patients diagnosed with gonorrhea if it was unlikely the partners would seek timely evaluation and treatment. 2
At present, the only CDC-recommended treatment of uncomplicated urogenital, anorectal, and pharyngeal gonorrhea is monotherapy with a single intramuscular dose of ceftriaxone 500 mg. If the recommended injectable treatment is not possible, then EPT via a single dose of 800mg cefixime should be considered. 3 CDC continues to recommend EPT for heterosexual men and women with gonorrhea for whom health department partner-management strategies are impractical or unavailable and whose providers are concerned about partners’ access to prompt clinical evaluation and treatment. This document is intended to provide guidance to providers who choose to use EPT for gonorrhea, and to answer frequently asked questions.
In light of CDC’s recent changes to its gonorrhea treatment recommendations, can EPT be used for gonorrhea?
Under current guidelines every effort should be made to ensure that a patient’s sex partners are evaluated and treated with the recommended regimen (a single dose of ceftriaxone 500 mg IM). However, because that is not always possible, providers should still consider EPT for partners of patients diagnosed with gonorrhea who are unlikely to access timely evaluation and treatment. MSM with gonorrhea have a high risk for coexisting infections (especially undiagnosed HIV) among their partners, and they might have partners without HIV who could benefit from PrEP. Data are also limited regarding the effectiveness of EPT in reducing persistent or recurrent gonorrhea among MSM ( 45); thus, shared clinical decision-making regarding EPT for MSM is recommended.
Since CDC no longer recommends exclusively oral treatment for gonorrhea, how does CDC recommend EPT be practiced for gonorrhea?
In cases where gonorrhea expedited partner therapy (provision of prescriptions or medications for the patient to take to his or her sex partner without the health care provider first examining the partner) is permissible by state law and the partner is unable or unlikely to seek timely treatment, the partner may be treated with a single 800 mg dose of cefixime, if a chlamydia infection in the patient has been excluded. If a chlamydia test result has not been documented, the partner may be treated with a single dose of oral cefixime 800 mg plus oral doxycycline 100 mg 2 times/day for 7 days. If adherence with multiday dosing is a considerable concern, azithromycin 1 g can be considered but has lower treatment efficacy among persons with rectal chlamydia (see Chlamydial Infections). As has always been the case, medication or prescriptions provided as part of EPT should be accompanied by treatment instructions, appropriate warnings about taking medications (if the partner is pregnant or has an allergy to the medication), general gonorrhea health education and counseling, and a statement advising that partners seek personal medical evaluation, particularly women with symptoms of PID. 6
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607 | https://www.youtube.com/watch?v=ibINrxJLvlM | Intro to Conditional Probability
Dr. Trefor Bazett
527000 subscribers
21010 likes
Description
1532747 views
Posted: 19 Nov 2017
What is the probability of an event A given that event B has occurred? We call this conditional probability, and it is governed by the formula that P(A|B) which reads "probability of A given B" is equal to the P(A intersect B)/P(B). We look at an example involving the probability of being an alcoholic given that one is a man.
Conditional Probability Formula: P(A|B) = P(A and B)/P(B)
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Transcript:
In this video, we're gonna talk about
something called "conditional probability". The idea with a "conditional" probability is that
we want to figure out the probability of something,… …GIVEN... …that we have
some OTHER piece of information,… …other piece of information
that we can bring to the table, that will… …influence the probability
of whatever it is that we're investigating. Now, the way we write this down,
is with the following notation. We're gonna refer
to the probability of… …"A, given B". So, this vertical bar
that we have here is… …is referred to as,
or read off as, "GIVEN",… …and what it means is,
that the probability of A, the thing I want,… …the thing I'm investigating,
the probability of some event A,… …GIVEN that I have
this other piece of information,… …GIVEN that I KNOW
that this event B has occurred. So, we're gonna have some formula
for how we're gonna compute this. It is a quotient of two things.
It is the probability… …that we have BOTH
the event A and the event B. So I write this as an intersection.
It means the probability of A AND B. All divided out
by the probability just of B. Now, the way I like to think
about this formula, is that… …we're looking,
NOT at all possible events,… …we're looking at this… …this sort of
narrowed sample space,… …of knowing that this event B has occurred,
and we're asking,… Well, what's the probability
of A occurring, GIVEN B? So, the sort of… …sum total of all the possibilities
that are going to be… …is whatever the possibility is
that B is going to be,… …because we KNOW
that B HAS occurred, but… …but then we're asking, well… …well, how often is it likely the case
that you have… …A… …AND B occurring, amongst
all the possibilities where B occurs? So, that's why we have this formula.
This... this is... What is the probability that A AND B occurs?
That's the numerator, divided by… …all the possible ways that B can occur. That's the denominator. Note that, in most cases,… ...if we're just talking
about non-conditional probability,... …what the probability of A is,… …without any other information,… …we actually are typically dividing by something,
in the same way, just that we… …divide by the probability
of ALL possible events,… …the entire sample space, the entire
universe of possibilities, which is just 1. So, we are in fact… …always dividing out by something. In the case of conditional probability,
is that we're dividing out… …by the probability of B,
the thing that we're TOLD has to exist. Now, we can sort of visualize how this
is going to work, by using a Venn diagram. So, a Venn diagram: I'm gonna list
my entire universe of possibilities. And then I can say that... Look, I have within my entire universe,
I can have some… …probability of an event A occurring. And... …I can also have… …a probability
of some event B occurring. Well,... If I'm looking at this sort
of conditional probability, well,... …conditional probability is saying:
Look, I… I'm in the scenario… …where this event B is occurring,
I'm… I'm ASSUMING that event B occurs. And then,
what I'm really interested in, is… …what is the probability
of this event A occurring,… …GIVEN that I'm in this event B,
that this event B HAS occurred? In which case, I'm really interested
in this intersection here. In other words, I care about this ratio
of the probability of the intersection… …to the probability of B occurring. The "red"
divided up by the "yellow" area. Now, in this example,… …I've given you something,
I've given you an intersection. I've told you that the percent of adults
who are both: 1) male,... …and 2) are alcoholics… …is going to be… it looks like 2.25%.
I looked this up. And then the question is: … …what is the probability that you're an alcoholic
if you KNOW that you're a male? Well… we can look at what the probability
of being an alcoholic is… … in general,
but I'm interested in this problem. If… if you KNOW that your patient is a male,
what is the probability… …that they're an alcoholic? So, we're gonna use
conditional probability to figure this out… …because we know something about the intersection, the probability of being… …an alcoholic and a male,
that's 2.25 % We know what the probability of being a male is,
or we approximate it at 50%. And we can use those 2 facts together
to get the conditional probability,… …the probability
that you're an alcoholic, GIVEN… ...that you're an adult male. Now, I need to define a little bit of notation
before I plug into my formula. I'm gonna let A… This is going to be equal
to this particular event of being an alcoholic. And then I'm going
to let B, be the event… …that you're gonna be a man. So, there's 2 different things we could wish to get:
1) probability of being an alcoholic,... 2) probability of being a man… The probability of being an alcoholic AND a man,
that's the thing that we were given. We were given in particular
that the probability of being alcoholic… …AND being a man,
that this was equal to… …0.0225, or, in other words, 2.25%. And so, what we're really asking when I ask,
what is the probability that you're an alcoholic… …given that you're a man, is… …we're asking, what is the probability
of being A — being an alcoholic —… …given that you're a man, and… and we know that,
from our formula, this is going to be the… …probability of A intersect B,… …so, the probability of being both,
all divided by the probability… …of just B, in other words,
the probability of being a man,... …and we know
these 2 different answers. The top is going to be 0.0225,… …and then I need to be dividing this
by the probability of being a man,… …and I'm going to say that that's 0.5,
I assume it's exactly 50%. And therefore,
what I'm gonna get is 0.045. We're about 4.5% for adult males. By the way, it's a little bit smaller for adult females.
It's about 2.5% for adult females. So, this is the way that we've been able
to compute this result of… …probability of… of being an alcoholic
given that you're a male,… …via this technique
of conditional probability. |
608 | https://www.cuemath.com/ncert-solutions/fill-in-the-blanks-i-zero-has-reciprocal-ii-the-numbers-and-are-their-own-reciprocals-iii-the-reciprocal-of-5-is-iv-reciprocal-of-1-x-where-x-0-is-v-the-product-of-two-rational/ | Fill in the blanks (i) Zero has ________ reciprocal. (ii) The numbers ________ and ________ are their own reciprocals (iii) The reciprocal of -5 is ________. (iv) Reciprocal of 1/x, where x ≠ 0 is ________. (v) The product of two rational numbers is always a _______. (vi) The reciprocal of a positive rational number is ________.
Solution:
(i) Zero has no reciprocal
(ii) The numbers 1 and -1 are their own reciprocals.
(iii) The reciprocal of -5 is -1/5
(iv) Reciprocal of 1/x, where x ≠ 0 is x
(v) The product of two rational numbers is always a rational number
(vi) The reciprocal of a positive rational number is positive
☛ Check: Class 8 Maths NCERT Solutions Chapter 1
Video Solution:
Fill in the blanks. (i) Zero has ________ reciprocal. (ii) The numbers ________ and ________ are their own reciprocals (iii) The reciprocal of -5 is ________. (iv) Reciprocal of 1/x, where x ≠ 0 is ________. (v) The product of two rational numbers is always a _______. (vi) The reciprocal of a positive rational number is ________.
NCERT Solutions Class 8 Maths Chapter 1 Exercise 1.1 Question 11
Summary:
The answers to the respective blanks are : (i) Zero has no reciprocal, (ii) The numbers 1 and -1 are their own reciprocals., (iii) The reciprocal of -5 is -1/5 , (iv) Reciprocal of 1/x, where x ≠ 0 is x , (v) The product of two rational numbers is always a rational number , (vi) The reciprocal of a positive rational number is positive.
☛ Related Questions: |
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Chapter 12 - The Cell Cycle
Chapter 12 The Cell Cycle
Lecture Outline
Overview: The Key Roles of Cell Division
Cell division functions in reproduction, growth, and repair.
Concept 12.1 Cell division results in genetically identical daughter cells
Concept 12.2 The mitotic phase alternates with interphase in the cell cycle
The mitotic spindle distributes chromosomes to daughter cells: a closer look.
Cytokinesis divides the cytoplasm: a closer look.
Mitosis in eukaryotes may have evolved from binary fission in bacteria.
Concept 12.3 The cell cycle is regulated by a molecular control system
Cytoplasmic signals drive the cell cycle.
Internal and external cues help regulate the cell cycle.
Cancer cells have escaped from cell cycle controls.
Lecture Outline for Campbell/Reece Biology, 7th Edition, © Pearson Education, Inc. 12-1
| Attachment | Size |
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| Microsoft Office document icon Chapter 12 The Cell Cycle | 82.5 KB |
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610 | https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/af06a8e18c0831ddca1cee9564b37839_MIT18_03SCF11_s35_3text.pdf | 0 The Wronskian We know that a standard way of testing whether a set of n n-vectors are linearly independent is to see if the n × n determinant having them as its rows or columns is non-zero. This is also an important method when the n-vectors are solutions to a system; the determinant is given a special name. (Again, we will assume n = 2, but the definitions and results generalize to any n.) Definition 3 Let x1(t) and x2(t) be two 2-vector functions. We define their Wronskian to be the determinant W(x1, x2)(t) = x1(t) x2(t) y1(t) y2(t) (1) whose columns are the two vector functions. The independence of the two vector functions should be connected with their Wronskian not being zero. For points, the relationship is clear. Using the result mentioned above, we can say W(x1, x2)(t0) = x1(t0) x2(t0) y1(t0) y2(t0) = 0 x1(t0) and x2(t0) are dependent.
⇔ (2) However for vector functions, the relationship is clear-cut only when x1 and x2 are solutions to a well-behaved ODE system (3). The theorem is: We are still considering the system a(t)x + b(t)y b(t) x0 y0 Theorem 3 Wronskian vanishing theorem. On an interval I where the entries of A(t) are continuous, let x1 and x2 be two solutions to (3) and W(t) their Wronskian (1). Then either a) W(t) ≡ 0 on I, and x1 and x2 are linearly dependent on I, or b) W(t) is never 0 on I, and x1 and x2 are linearly independent on I. Proof. Using (2), there are just two possibilities. a) x1 and x2 are linearly dependent on I; say x2 = c1x1. In this case they are dependent at each point of I, and W(t) ≡ 0 on I, by (2). a(t) x x = (3) , = , c(t)x + d(t)y d(t) c(t) y y = The Wronskian OCW 18.03SC b) x1 and x2 are linearly independent on I, in which case by Theorem 2A they are linearly independent at each point of I, and so W(t) is never zero on I, by (2). 2 MIT OpenCourseWare 18.03SC Differential Equations Fall 2011 For information about citing these materials or our Terms of Use, visit: |
611 | https://www.youtube.com/watch?v=FBnhMBP4oEs | FINDING THE DIHEDRAL ANGLE
Shane Doherty
241 subscribers
15 likes
Description
1055 views
Posted: 10 Feb 2021
Transcript:
we want to find the dihedral angle between two planes so i've got a piece of cardboard here and i've just got two triangular surfaces the important piece is the line of intersection between the two planes so the edge that joins the two planes together we have that line of intersection all the work that we're going to do to find the dihedral angle comes from working with the line of intersection what i need to do first of all step one is we always take an auxiliary elevation to find the true length of the line of intersection so if i come along and take an auxiliary elevation just for example something like that now i've found the true length of the line of intersection i'm looking in at 90 degrees to it from that view we project an auxiliary plan and the auxiliary plan is basically me getting this and rotating it back up there so that i see an edge view of the line of intersection so here's my line of intersection when i get it in the right view it looks like a point and at the same time when that looks like a point you'll see that both planes look like edges and it's in this position that i can measure the dihedral angle so i can see the true angle between the two planes that's what the heat language is all about so what's the real angle between them it's only in this position i can see it in this first question we have to find the dihedral angle between surfaces a and b and the setup that we have here in plan view is similar to what you're looking at there so i've got a triangular surface a represented here roughly and i have a triangular surface b it'll represent it here roughly first of all i'd always highlight the line of intersection so the line that joins a and b i just put it in red for this example there is my line of intersection it's important to put the line of intersection in the elevation as well so the line of intersection between those two is here that is the line i'll be dealing with i'll number it all up so we can start doing the wrong all right so i've highlighted my line of intersection and i've given labels to all the numbers so my points that i've got 1 and 2 is line of intersection here's one and two in elevation i've got point three and point four a triangular surface is the easiest one to find the dihedral angle because you've only got three points on each surface to deal with i need to find the true length of line one two to draw that it'll mean i need to look in here at 90 degrees so i look in at 90 degrees in there so that is the first view that i set up it's a standard auxiliary elevation so to do that i need to look in here 90 degrees i'm going to project it out at 90 degrees to 1 2. so i'm going to come out project out 1 and 2. and three and four so my one two and my three line in this question are fairly straightforward because they all line up so they're all going to be in the same position as always set up my xy line just come out a little bit from there on i'll set up my xy line it'll become my x1 y1 and that's going to be important so make sure you label it i'm drawing an auxiliary elevation so my measurements are going to come from elevation my measurements always come from the previous x y line the previous x y line in this case is this x y line so again i'm working from the previous x y line taking my measurements from elevation that is why we take our measurements like that so the height of one here's position one height of two is zero it's on the xy line and the height of three and four three when i bring it up is on the xy lens so the height of three is zero and the height of four is zero so what that's showing me the true length of my line of intersection which is here and i'll highlight that red i'll just constantly highlight the line of each section right in these questions and my surface one two three is here and the surface one two four is in between behind it so this is showing me a surface a and b one in front of the other but that doesn't matter now i need to come along and protect out an auxiliary plan to project my auxiliary plan i need to look along the line of intersection so exactly as you've seen it here i took this and looked along the line of intersection so my eye is looking down along the line what that means when i come to a drawing is to look along the line i'm projecting out in this direction so my eye has moved here and now i'm looking down along that line to see it as a point so i'm projecting down in this direction down here again same thing i need my x y line so i'll set up at 90 degrees to that again this is an x2 y2 it's the second view i'm drawing an auxiliary plan now and my same rules apply i'm going to measure from plan i'm going to take my measurement from here and i'm going to measure from the previous xy line i'm on x2y2 at the minute the previous xy line is x1y1 so when i measure out my distance for number one it's x1y1 2.1 and plan take it out and strike it off the distance from 0.2 is going to be the same distance and it's coming along the same line and that's how we end up with a point view one two and the same position i can come along with the same points three and four so i project out point three and four parallel i project out my points three and four the distance to three x one y one line 2.3 is a very short distance i get it here 2.3 is and 0.4 to the x1 y1 line and again i'm following up my 0.4 and i get it out here and that gives me 0.4 so i've got a point view 1 2 surface 1 2 3 which is surface b is here surface 1 2 4 which is surface a is here my line of intersection which is my red line here's my red line is here so it's a point view of the red line and my dihedral angle i can measure with a protractor or just highlight it here is my deheater length i have my headline labeled i just come along the measure of my protractor so measure around zero all the way around it's 115 degrees so the true angle between planes a and b is 115 degrees second question so again find the hidden angle between surfaces a and b now this question is a little bit trickier in that i've got four points on surface a and four points on surface b i still have a line of intersection and that's the first thing i'll always highlight so i'll highlight my line of intersection in red line of intersection up in the elevation is here now i can save myself a little bit of time i'm going to cut the two uh surfaces the two planes into uh triangles so what i'm gonna do is i'm gonna take point one two this is point three and this is point four i'm gonna bring those up to elevation so i've got my points in elevation and planned there what i'm going to do now is i'm going to chop a little piece off a and i'm going to chop a little piece off surface b so what i'm doing is i'm cutting away this piece of the surface and i'm cutting away this piece of surface and i'm just dealing with one two four for surface b and one two three for surface a it's the same thing all i've done is got the two flat surfaces chopped a piece off them i can use the same method now i need to draw an auxiliary elevation to find true length auxiliary elevation needs to come in at 90 degrees to my line of intersection here's my line intersection so i'm going to be projecting out here at 90 degrees i've got my 90 degree angle i'll project out all my points so 1 2 3 and 4 all out at 90 degrees to that line of intersection my red line x y line again parallel to the line of intersection at 90 degrees doesn't really matter i get my x one y one it's an auxiliary elevation i'm drawn i'm taking my heights from elevation so the height for point one is here i follow up 0.1 height for 0.2 again after xy line i project at 0.2 height for three is the same projective one three and the height of 4 is the same height 0.4 that doesn't matter the method is still the same at the point 4. now i've got my line of intersection 1 2 and once again i'll highlight it in red so i know which line i'm dealing with always i've got the surface one two three is here and i've got the surface one two four is here so i've got my two different surfaces here i'm gonna darken in surface one two three just so it makes a little more sense just to separate it out from one two four so half of one two four is lasting behind one two three that gives me the true length of my line that's what we're looking for and now i can project off looking along that line so we need to project out here looking along the line once again line of intersection i'm going to click out looking along it and i'm going to set up my x1 oh sorry x2y2 line auxiliary plan i'm drawing so i measure from plan while i'm at it i'll project out all of the points just together just to save a little bit of time so i've got one and two projected out here's point three projected out point four projected out distance for one two to the x y line here's my distance when i come along one two is that here distance to four here's the distance to four and again follow out gives me point four last one distance from my x one y one line to three again it's in plan and i follow a three and i find it and i get my surface one two four and surface one two three with my point view of the line and my dihedral angle is here now this comes in handy because you can work back the way as well i could have found the edge view of surface one two four so it's a surface b if i had been told this angle between surface a and b is 130 degrees if i only had this surface i could come along measure an angle out here of 130 degrees draw in the surface one two three and you can use it to kind of work back we'll be doing a couple of them questions later but again that is just our dihedral angle done |
612 | https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-difference-squares/v/factoring-to-produce-difference-of-squares | Factoring difference of squares: leading coefficient ≠ 1 (video) | Khan Academy
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Algebra 1
Course: Algebra 1>Unit 13
Lesson 7: Factoring quadratics with difference of squares
Difference of squares intro
Factoring quadratics: Difference of squares
Difference of squares intro
Factoring difference of squares: leading coefficient ≠ 1
Factoring difference of squares: analyzing factorization
Factoring difference of squares: shared factors
Difference of squares
Math>
Algebra 1>
Quadratics: Multiplying & factoring>
Factoring quadratics with difference of squares
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Factoring difference of squares: leading coefficient ≠ 1
FL.BEST.Math: MA.912.AR.1.7, MA.912.AR.1.8
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About About this video Transcript
Sal factors 45x^2-125 as 5(3x+5)(3x-5).Created by Sal Khan.
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Brian Lee 12 years ago Posted 12 years ago. Direct link to Brian Lee's post “Can 25 can be squared? Ye...” more Can 25 can be squared? Yes or no?
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ningsheng1 12 years ago Posted 12 years ago. Direct link to ningsheng1's post “To square a number means ...” more To square a number means to multiply it by itself. Squaring is written by ^2. Any number can be multiplied by itself. Therefore, the answer is yes. The square of 25 is 625.
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Jayanth Samala 8 years ago Posted 8 years ago. Direct link to Jayanth Samala's post “What does ^ mean?” more What does ^ mean?
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jercoler 8 years ago Posted 8 years ago. Direct link to jercoler's post “And on your calculator yo...” more And on your calculator you see ^ as well, it does the same thing.
if you have INTL keyboard you can do it the right way: 3² 7³ 5²
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Rustenberg 10 years ago Posted 10 years ago. Direct link to Rustenberg's post “How can factor 3x^2 + 10x...” more How can factor 3x^2 + 10x + 8 in the forn of (3x ) (x )
or 12x^2 - 17x + 6 = (4x ) (3x )?
Is there a rule to do this? Completing the square will not do.
Where on the Academy can I find the videos for this?
I believe that it is most trial and error, but is a way for trial and least error?
Very, very,very much thanks.
Your advice may also be sent to rainer@dds.nl
Thank you, happy study on the Khan Ac.
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Sky Haworth 7 years ago Posted 7 years ago. Direct link to Sky Haworth's post “How do you solve these ty...” more How do you solve these types of problems when the constant is a non-square number?
For example (and don't just solve it!):
96−6x^2
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Math∫it∬out∭ 2 years ago Posted 2 years ago. Direct link to Math∫it∬out∭'s post “the answer is (4-2x)(4-3x...” more the answer is (4-2x)(4-3x)
1 comment Comment on Math∫it∬out∭'s post “the answer is (4-2x)(4-3x...”
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Landen 7 years ago Posted 7 years ago. Direct link to Landen's post “So what happens when you ...” more So what happens when you have this:
4x^2 -1=
I can't factor out a 4 and I can't get the square root of -1 so how would I complete this problem?
Thanks
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Kim Seidel 7 years ago Posted 7 years ago. Direct link to Kim Seidel's post “You have a difference of ...” more You have a difference of 2 squares
4x^2 = (2x)(2x) = (2x)^2
1 = 1^2
And, you have a minus in the middle
Use the pattern to factor: a^2-b^2 = (a-b)(a+b)
"a" = 2x
"b" = 1
Your factors becomes: (2x-1)(2x+1)
Hope this helps.
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melayfern5 2 years ago Posted 2 years ago. Direct link to melayfern5's post “Can 32 be squared?” more Can 32 be squared?
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Quantum Cat 2 years ago Posted 2 years ago. Direct link to Quantum Cat's post “Yes, any number can be sq...” more Yes, any number can be squared. Squaring something means multiplying it by itself, so 32 squared is:
3232 = 1024
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D armaniD a year ago Posted a year ago. Direct link to D armaniD's post “So I had a math problem, ...” more So I had a math problem, 108-3x^2= and it said to factor completely. I got (-3x+18)(1x+6) but they did the answer was 3(6+x)(6-x) how do I know when to use a different method?
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Kim Seidel a year ago Posted a year ago. Direct link to Kim Seidel's post “A polynomial is completel...” more A polynomial is completely factored when none of its individual factors can be factored further. In your version: (-3x+18)(1x+6), the 1st binomial is not completely factored because it contains a common factor of 3. So, your work is not finished. Once you factor out the 3, then the polynomial is completely factored: 3(6+x)(6-x)
If you always factor out the greatest common factor as your 1st step, it helps simplify the rest of the factoring and you don't end up with the common factor embedded in one of your other factors.
If you forget to factor out the greatest common factor up front, then you can still catch it at the end. Review each binomial and make sure none of them contain a common factor.
Hope this helps.
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Albin Sopaj 6 years ago Posted 6 years ago. Direct link to Albin Sopaj's post “How do you factor out 25x...” more How do you factor out 25x^2-16 ? They have nothing in common.
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David Severin 6 years ago Posted 6 years ago. Direct link to David Severin's post “But both are perfect squa...” more But both are perfect squares, so try taking square root of both.
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Sarah Van Liere 9 years ago Posted 9 years ago. Direct link to Sarah Van Liere's post “I'm confused about why (y...” more I'm confused about why (y+9)(y-9) does not equal (9+y)(9-y). In my math class, my teacher always writes it the first way but here on Khan Academy they specifically say this is wrong.
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attown30 7 years ago Posted 7 years ago. Direct link to attown30's post “You know how the whole pu...” more You know how the whole purpose of using math is to apply it to solve a problem in the world? (that was not my main question) Can you give me an example of a time when this math would be applicable? Is it directly useful for solving problems or is it just a step in a bigger process used to solve am intricate problem that would not come up for "everyone?" Thanks!
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kubleeka 7 years ago Posted 7 years ago. Direct link to kubleeka's post “Factoring differences of ...” more Factoring differences of squares is just a small part of understanding how to deal with polynomials in general. Polynomial functions come up a lot in science and other applications because they're very well-behaved and well-understood functions that can model a huge variety of behaviors.
For example, if you have n points in the plane, you can always find a polynomial of degree n-1 that goes through all of them (assuming none of them are directly above each other). This makes polynomials useful first attempts to analyze small data sets, etc.
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Video transcript
Let's see if we can factor the expression 45x squared minus 125. So whenever I see something like this-- I have a second-degree term here, I have a subtraction sign-- my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here, it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5, and by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times-- well, 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now, this is interesting. 9x squared-- that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x-- the whole thing squared is 9x squared. Similarly-- I can never say similarly correctly-- 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares, and we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. So let me write the b's down, plus b and minus b. And we're done. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out.
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613 | https://librepathology.org/wiki/Multiple_sclerosis | Multiple sclerosis - Libre Pathology
Multiple sclerosis
From Libre Pathology
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Multiple sclerosis, abbreviated MS, is a chronic inflammatory condition of the central nervous system.
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Contents
1 General
2 Radiologic/Gross
3 Microscopic
3.1 Images
4 IHC
5 See also
6 References
General
A bread 'n butter disease of neurology in Canada.
Clinical:
CSF: oligoclonal bands of immunoglobulin.
Classification of MS lesions:
Early active.
Inactive.
Early remyelinating.
Late remyelinating.
Radiologic/Gross
Features:
White matter lesions.
Cerebrum (classically): periventricular distribution.
Optic nerves (optic neuritis) - classic presentation.
Microscopic
Features:
Perivascular inflammation.
Esp. lymphocytes.
Demyelination.
Subcortical myelinated fibers are often spared.
Chronic lesions - specific features:
Macrophages.
Astrocyte enlargement.
DDx:
ADEM.
Neuromyelitis optica.
Lymphoma.
Diffuse astrocytoma.
Images
Demyelination in MS - CD68 stain. (WC/Marvin 101)
Demyelination in MS - KB stain. (WC/Marvin 101)
www:
Multiple sclerosis masquerading as a diffuse astrocytoma - several images (upmc.edu).
Tumefactive multiple sclerosis - several images (upmc.edu).
IHC
HAM-56 - macrophages.
CD8 - lymphocytes.
See also
Neuropathology.
References
↑Kumar, Vinay; Abbas, Abul K.; Fausto, Nelson; Aster, Jon (2009). Robbins and Cotran pathologic basis of disease (8th ed.). Elsevier Saunders. pp.1311. ISBN978-1416031215.
↑URL: Accessed on: 12 July 2010.
↑URL: Accessed on: 12 July 2010.
↑Lefkowitch, Jay H. (2006). Anatomic Pathology Board Review (1st ed.). Saunders. pp.425 Q43. ISBN978-1416025887.
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614 | https://www.softkit.dev/blog/ways-to-reduce-labor-costs-without-cutting-headcount/ | Save Labor, Boost Efficiency: Headcount-Free Labor Cost Reduction | Softkit
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WAYS TO REDUCE LABOR COSTS WITHOUT CUTTING HEADCOUNT
Nov 13, 2023
13 min read
Ways to Reduce Labor Costs Without Cutting Headcount: Efficient Labor Practices
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Controlling labor costs is vital. It is key to maintaining healthy profit levels. After all, your profit is revenue minus operational cost. And labor is the biggest chunk of operational costs for businesses. According to Deloitte, it is typically 50 to 60%. For Fortune 500 this adds up to billions of dollars per year.
If your labor costs are high, how do you cut them? Layoffs jump to mind. But if you lose production capacities, your revenue can decrease. A high workload for remaining employees will increase turnover. This often means losing your best talent.
So, reducing headcount isn’t the right answer. In this article, we will discuss how to lower labor costs without hurting your business.
Understanding Labor Costs
The first step to reducing labor costs is understanding their meaning, components and types.
💡Labor cost meaning
Labor cost is money that a company pays for work. This includes work provided by their employees and contractors.
Components of Labor Costs
Labor cost goes beyond worker salary and wages. Here are the most common components of labor costs:
– Salaries and wages,
All payments to workers fall into this category. You can pay employees per month, shift, day or hour.
– Payments to contractors,
Often, companies hire contractors to do certain tasks. For example, a dry cleaning business can pay software development experts to create a website for them. Such tasks can be project-based. So, there is no need to hire someone full-time. This helps to save labor costs.
– Employee benefits (healthcare insurance, gym, etc),
With benefits, employers promote desirable behavior. For example, if you offer a “free” gym, more employees will do sports.
– Overtime pay,
If you ask employees to work beyond their regular hours, you should pay them overtime. In many countries law requires it.
– Payroll taxes,
Payroll taxes are mandatory costs when employing workers.
– Overhead costs (office rent, equipment, software, etc.)
These are other costs you pay for having workers. For example, if you have an office, you need to pay rent and utilities. Software costs also fall into this category.
Labor costs differ, depending on your location, company structure and policies. For example, you need to offer healthcare insurance to attract top talent in the US. This is not necessary in Sweden. But there, you should expect to pay higher taxes.
What are direct labor costs?
Direct labor costs are money you spend on workers who produce your goods or provide your services. They are directly responsible for company output. Let us imagine, we have a tech startup. Then, direct labor costs examples are the salary, benefits and taxes of your developers. At the same time, a startup needs good sales and marketing. So, the cost of this labor falls under the same category.
If a company has an office, it likely has an office manager, a janitor and a guard. These people do not directly contribute to business success. So, the costs of employing them are indirect.
Other direct labor cost examples are money spent on
– lawyers for law firms,
– assembly line workers for manufacturers,
– drivers for courier companies, etc.
In some cases, it can be difficult to figure out whether employee-related costs are direct. Recruiters, HRs, accountants and corporate lawyers are often questionable. Ask yourself, is the person essential for your business success? Thus, pharma companies operate in a heavily regulated market. They need a full-time lawyer. But, a marketing agency doesn’t.
Fixed vs variable labor costs
Personnel costs can be fixed or variable. Fixed costs do not vary depending on your output. And variable labor costs do. For example, let us imagine that you own a tech startup that sells robotic pets. The costs of employing engineers and software developers who design your product are fixed. But to build robotic pets you also need metal, microchips and other components. These costs are variable. The more robots you build, the more materials you need.
Variable costs offer flexibility and cost-efficiency. And fixed pay can attract top talent and ensure stability.
5 ways to calculate personnel costs
To understand and control the labor cost better, use the following calculation methods:
1. As a percentage of operational costs or revenue
At the beginning of the article, we said that labor takes around 50 to 60% of operational costs. These are average numbers. Calculate yours. Similarly, you can calculate what part of your revenue goes to labor payments.
2. By department
Calculate personnel costs by department. Determine the amount of direct and indirect costs. Direct costs should be the majority.
3. By project
Calculate labor costs required for your projects. Are there projects or clients that demand lots of effort? Are they worth it? Sometimes, by rejecting a certain client you can free resources for more profitable projects.
4. Per product/service
Similarly, you can measure how much labor your products and services need. Labor-intensive goods and services must be priced higher to offset extra labor costs.
5. Per hour/day/month/year
You should know personnel costs per hour, day, month and year. Do they differ, depending on the time of day or year?
Top 3 employee cost-reduction strategies
Here are the top 3 ways to reduce labor costs without cutting headcount.
1. Optimizing Workforce Efficiency
Lower labor costs by helping employees to be more productive. There are several ways to do it:
Use efficient collaboration software
Whether your business is online or offline, efficient software can make your team shine. There are many excellent free options. Try Trello, Monday, Asana or Slack. Check our guide to effective collaboration to learn more.
Manage schedules and working hours
Think about your business. Does it have seasonality or peak times? If so, optimize employee schedules accordingly. Let us imagine that you have tech support. From 1 to 6 a.m., most of your clients are asleep. So it makes sense to have only 1 person working during this time. The majority of customer inquiries come in from 3 to 6 p.m. Then, you can schedule one shift to end at 6 p.m. and the next to start at 3 p.m. Thus, you will have more people working during peak hours.
Save labor by reducing unnecessary overtime
Why do your employees work overtime? Do deadlines often move forward? Or, additional work is frequently added to the scope? It can be a management issue.
Avoiding overtime leads to significant labor cost reduction. First, it cuts overtime payments. Second, it improves employee morale and productivity.
Leverage technology and automation
Technology is your friend. It can help you improve productivity and service quality. You can find useful tech for any industry. For example, law firms use document and contract management software. Real estate agencies use Big Data to estimate property prices. Healthcare workers use telemedicine, e-prescribing software and remote patient monitoring tools.
Offer remote work options
Why is labor so expensive? The number one reason is living costs. Everyone wants a comfortable lifestyle. Usually, this means earning at least an average salary. But, average salaries differ, depending on employee location. You can use this to your advantage.
Thus, you can hire employees from small towns. Living costs there are lower than in cities like New York or London. Plus, your employees would not need to waste time or money on commuting. So, they would expect a lower salary.
Also, you will decrease your overhead costs. A smaller office or no office, means less rent and utility payments. And you can decrease your indirect labor costs. No office – no need for an office manager.
2. Performance improvement and incentives
Efficient labor is key to decreasing personnel costs. You can save labor costs and increase output by improving retention and using performance incentives.
1) Improve team retention
Guess how much employee turnover costs US businesses per year. It is 1 trillion USD. Here are some tips to create a workplace with low turnover:
– Be honest during the hiring process
Sometimes, recruiters don’t talk about job downsides. For example, it can involve lots of unexpected overtime. Be honest and straightforward about such shortcomings.
– Set a clear career path
Explain to your employees what career paths they can have in your company. Ask them about their aspirations. But remember, it is okay for employees to change their goals.
– Promote good work-life balance
A good work-life balance is necessary to avoid burnout. Make sure that your employees take their vacation days. Also, offer flexibility whenever possible. For example, you can offer flexible work hours. Or a choice between office, hybrid and remote.
– Encourage a healthy lifestyle
Healthy employees take fewer sick days. Also, they are happier and concentrate better on their tasks. Good practices include free psychological counseling, gym reimbursements, healthcare insurance, and wellness retreats.
– Offer learning opportunities
To keep workers motivated, teach them new skills regularly. They should understand that the longer they stay, the more they will know.
– Keep up with the market rate
Keep an eye on the labor market rates. When the price for an expert’s services rises, they have more reasons to move forward. Replacing such experts can be expensive.
– Analyze your current situation and address issues
What is your current team turnover rate? You can calculate it for a certain year or any other period. To control labor costs efficiently, aim for a 10% turnover rate or lower. This means that the retention rate should be above 90%.
💡 How to calculate employee turnover rate?
Figuring out the turnover rate is a two-step process. The first step is calculating the average number of workers for a certain period. This means that you need to add the number of workers at the beginning and at the end of the year and divide it by two.
Then, divide the number of people who left your company by the average number of workers. Then, multiply it by 100.
The result is your turnover rate.
2) Promoting performance with incentives
Reduce labor costs by motivating employees to work harder. Here are the popular ways to do it:
– Commissions
Commissions are very common for certain experts, like salespeople. The more they sell, the more money they bring home. Similarly, you can offer commissions to recruiters, retail store managers and insurance agents. But make sure that quality of work doesn’t suffer. For example, don’t pay recruiters just for the fact of hiring. Instead, make sure that the new person passes their probation period first.
– Bonuses
Bonuses are rewards you give employees for certain achievements. For example, you can set a desirable level of sales or revenue. Then, you tell employees that if they reach these goals, they will receive a bonus. Be specific about the terms.
– New opportunities and promotions
You can offer better opportunities or promotions to the best performers. For example, your best employees can get access to top clients or interesting business trips.
– Competition
Gamify working experience. You can set up a challenge for employees. For example, the seller who attracts the most new clients gets an all-expense-paid vacation.
– Praise and recognition
Word is a powerful tool. Tell your employees that you value them.
3. Using the global talent pool
Sometimes to reduce labor costs, you need to get creative. Hiring from abroad is one of the finest perks of globalization. The best part is that nowadays it is very easy to do.
Hiring from a global talent pool
Any job that can be done remotely, can be done by experts abroad. And this strategy is amazing for reducing labor costs. Popular professions for international hiring include
– software developers and tech support,
– web designers,
– customer support experts,
– supply chain managers,
– accountants, etc.
Also, by hiring abroad, you don’t need to reduce headcount. So, your productivity and revenue will not suffer.
The pros and cons of hiring employees abroad
So, why should you hire from aboard? Here are the benefits of international hiring:
1. Low labor cost
International hiring is the most straightforward and certain way to reduce the labor cost.
Earlier, we discussed why labor is so expensive. Workers want to have a comfortable lifestyle. But, living costs in Eastern Europe, Africa and most of Asia are a lot lower than in the US or Western Europe. So, experts there can have good lifestyles with much lower salaries.
And your savings go beyond that. There is no need to spend money on bonuses. And your taxes will be much lower.
2. Excellent quality and the best talent
Foreign experts are just as qualified as domestic specialists. Often, they are even better. And, you will have more candidates to choose from. So, you can select the best experts.
Also, some countries are hubs for certain professionals. For example, Ukraine is a mecca for Java software engineers. Poland specializes in game development. And India is full of excellent accountants.
3. Simplicity
If you have never hired international contractors, no need to worry. This is a very widespread practice. Opt for those who have experience working with foreign businesses. But we still recommend you consult with an international law attorney and a tax accountant.
4. Hiring speed and flexibility
Your talent pool is much larger. So, finding a perfect fit is much faster. It can take just several weeks. Also, working with contractors brings flexibility. You can negotiate and set contract terms that work for you.
International hiring has several disadvantages too. Keep reading so that you know what to avoid:
1. Imperfect language skills
There is a chance that the language skills of your new worker will be imperfect. Talk to the person before signing a contract. If you don't understand each other well, avoid them.
2. Cultural differences
Cultural differences can lead to misunderstandings. For example, in Latin America, it is okay to be up to 30 minutes late. In Africa, to do business, people might want to know more about you as a person. Asian countries are the most challenging. For example, subordinates often avoid asking clarifying questions.
Eastern Europe is the most similar to the US and Western Europe in terms of culture.
💡Check out our list of the top Eastern European countries for software development outsourcing.
3. Difficult to meet in person
Meeting in person can be difficult. Sure, you can hop on a plane and visit each other. But, a journey out of the country can be long and expensive.
Best practices for effective outsourcing
Here are our top tips on how to reduce labor costs efficiently with outsourcing:
1. Define your goals first
Make a list of your goals and requirements. Consult your colleagues. Brainstorming helps to capture all necessary detail.
2. Choose your strategy
When it comes to international hiring, there are two main strategies. These are staff augmentation and managed services. If you want experts to join an existing team, you need staff augmentation. But, if you just need a project to be done, choose managed services. In this case, a dedicated team will work on your project. They will take care of project management. You can learn more in our previous article.
3. Select a compatible contractor
Your contractor should be a good fit. This means that they should know your industry and its standards. For example, if you need a healthcare app, your contractors should be familiar with HIPAA and GDPR.
4. Communication is key
When you have found the right contractor, establish clear communication channels. Use collaboration software, like messengers and task management tools. Provide opportunities for them to interact with your other employees.
How to hire software developers abroad? Top solutions
Are you looking for software developers? There are several ways to find the perfect fit:
1. Independent web resources
First, you can find and compare developers on independent websites. Examples of useful websites are Clutch.co, Behance, DesignRush and Upwork.
2. Personal recommendations
Personal recommendations are often the most reliable. Talk to your business partners, colleagues and clients. Ask whether they can recommend someone.
3. Opening an office abroad
This is an excellent long-term solution. And, this strategy is very popular. For example, in Ukraine alone, there are more than 110 international Research and Development (R&D) centers. They belong to various tech giants, including Microsoft, Samsung, Siemens, Google and Oracle.
💡Pro tip: reduce other types of operational costs
Reducing labor spending will bring down your operational costs. But you can optimize other processes too. A prime example is cloud infrastructure. Around 94% of businesses overspend on the cloud. The main reason is that setting up cloud services efficiently is difficult. It requires a specific expertise.
Many businesses don’t even realize that their cloud costs can be reduced. Several of our clients hired us for other tasks. And they were pleasantly surprised when we offered to optimize their cloud spending. Thus, we decreased AWS costs for SOLD.com by 23.4%. One simple feature now saves them around USD 2k per month. Another client was using Google Cloud for their microservices. We decreased their cloud spending by 40%.
Conclusion
Labor costs include several components. Usually, these are salaries & wages, payroll taxes, employee benefits, contractor payments, overtime pay and overhead costs (e.g. office rent). Controlling labor costs is key to maintaining a healthy profit margin.
There are 3 efficient employee cost-reduction strategies:
Optimizing efficiency
This means managing your staff more effectively. You can save labor costs by
– cutting unnecessary overtime,
– managing schedules better and
– using good collaboration software.
Improving productivity
Higher profit makes labor costs proportionally lower. Businesses use different incentives to increase worker productivity. Examples are commissions, bonuses, competitive challenges and praise.
Hiring internationally
This is the most straightforward way to decrease personnel costs. International hiring brings many benefits, including faster hiring, excellent quality and reduced labor costs. To find international talent, you can use dedicated web resources, open an office abroad or ask business partners for recommendations.
Understanding Labor Costs
Top 3 employee cost-reduction strategies
Conclusion
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FAQ
What are labor costs, and why are they important to businesses?
Labor costs are money that businesses pay people for their work. Controlling labor costs is important since it affects company profit.
What are the four factors that affect labor costs?
The main four factors are
– salaries & wages,
– taxes,
– benefits and
– overhead costs.
What industries can benefit the most from outsourcing labor to reduce costs?
Nearly any industry can benefit from outsourcing. Top examples are tech, finance, manufacturing and healthcare.
What are the long-term benefits of reducing labor costs for businesses?
The main benefit is increased competitiveness. When your operational costs are lower, your profit is higher. You can pass it on to your customers by offering better prices. Or you can reinvest money to expand your market reach.
How to cut costs without cutting staff?
The first step is efficient labor management. Optimize employee schedules and cut unnecessary overtime. Second, improve performance. You can offer bonuses to top-performing workers. Third, you can start hiring internationally.
Is labor a fixed cost?
No, you can lower labor costs if you are flexible. Yet, some labor costs can be considered “fixed” while others are “variable”. Fixed costs are constant. You pay them regardless of your output. Variable costs depend on how many goods or services you produce or provide.
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615 | https://www.jstage.jst.go.jp/article/jphs1951/64/2/64_2_97/_pdf | Sensitivity of Intestinal Alkaline Phosphatase to L-Homoarginine and Its Regulation by Subunit-Subunit Interaction
Kuniaki Suzuki, Yoshitaka Yoshimura, Yoh Hisada and Akira Matsumoto
Department of Dental Pharmacology, School of Dentistry, Hokkaido University, Kita 13 Nishi 7, Kita-ku, Sapporo 060, Japan
Received June 23, 1993 Accepted November 16, 1993
ABSTRACT-The inhibitory effect of levamisole and L-homoarginine on alkaline phosphatases (ALP, orthophosphoric monoester phosphohydrolase, EC. 3. 1. 3. 1) in various tissues was studied to characterize differences in the mechanism of inhibition of ALP isoenzymes. The ALP activity from the placenta, kidneys, and a clonal osteogenic cell line, MC3T3-E1, was hyperbolically inhibited with a dependency on the concentration of levamisole (K;0.5=10 -12 rM) or L-homoarginine (K; 0.5 =1 mM), but the activity of in testinal ALP was little inhibited by 240 pM levamisole and sigmoidally inhibited by L-homoarginine (K; 0.5 = 13 mM) . Hill plot analysis of the L-homoarginine inhibition data showed that the Hill coefficient values of the placenta, kidney, and osteogenic cells were around 0.9-1.0 and that of the intestine was 1.8. The effect of sodium dodecyl sulfate (SDS), which may decrease the subunit-subunit interaction, on the Lhomoarginine inhibition of the intestinal ALP was tested. The L-homoarginine concentration-dependent in hibition curve changed from sigmoidal to hyperbolic, and the Hill coefficients decreased with increasing con centrations of SDS. These results suggest that the differences in inhibition by L-homoarginine and affinity changes of intestinal ALP for L-homoarginine are caused by the subunit-subunit interaction of oligomers.
Keywords: Alkaline phosphatase inhibition, L-Homoarginine, Sodium dodecyl sulfate, Positive cooperativity, Subunit-subunit interaction
Alkaline phosphatases (ALPs) are widely distributed enzymes that hydrolyze orthophosphoric-monoester at al kaline pH optima. There are three distinct forms of ALP in mammalian tissue: placental, intestinal, and liver/ bone/kidney (L/B/K), which are distinguished by differences in molecular weight, heat stability, sensitivity to inhibitors (1, 2), and recently also cDNA analysis (3 5). Although ALPs have been studied for the last seventy years, little is known about their physiological functions (1). Inhibition studies have been done to characterize each of the ALP isoenzymes and to estimate their physio logical functions. The inhibitions of isoenzymes by levamisole and L-homoarginine were studied by several groups, but the results were not always consistent (6-8). The purposes of this paper are to confirm the organ specific inhibition and to study the kinetics of inhibition by levamisole and L-homoarginine. We demonstrate that the inhibition of intestinal ALP by levamisole and Lhomoarginine is different from its inhibition by other drugs, and its unique sensitivity to L-homoarginine may be regulated by a subunit-subunit interaction.
MATERIALS AND METHODS ALPS Bone-type ALP was obtained from a cultured clonal osteogenic cell line, MC3T3-El cells (9, 10), which was do nated by Dr. Kodama (Ohu University) and Dr. Kuboki (Hokkaido University). Cell cultures were grown in aminimum essential medium supplemented with 10% (v/v) fetal bovine serum (Boehringer Mannheim-Yamanouchi, Tokyo). They were maintained at 37 C in a 95% air-5% CO2 atmosphere. When the cells were inoculated at 5 x 104 cells/60-mm plastic dish containing 4 ml of medium, they usually reached a confluent monolayer after 4 to 5 days, and then the ALP activity increased. The cells were harvested at 10 days after confluence and then were ultrasonicated. The cell homogenate was centrifuged at 5,500 x g for 20 min to sediment nuclei, large fragments, and most mitochondria. The supernatant was centrifuged at 100,000 x g for 45 min. The resultant pellet was used as a microsomal fraction. The ALP was mainly recovered in the microsomal frac tion (11), and it was used for bone-type ALP determina tions. The kidney-type (from calf kidney) ALP was pur chased from Funakoshi Co. (Tokyo), and intestinal ALP (from bovine intestine, type VII-S) and placental ALP (from porcine placenta, type XVI) were from Sigma (St. Louis, MO, USA). Enzyme assay ALP activity was assayed by the method of Bessey et al. (12) with a slight modification (11). The concen trations of the various components in the 0.5-m1 reaction mixture were 1-201ig of enzyme from various sources, 12.5 ttmol of sucrose, 0.8 pmol of MgC12, 100 ,umol of tris-HC1 at pH 9.5, and 0.8 pmol of p-nitrophenyl phos phate (pNPP) with or without levamisole or L-homoargi nine. The reaction was started by the addition of pNPP at 37V and terminated with 0.5 ml of cold 15% (w/v) trichloroacetic acid after 20 to 45 min. The amount of p nitrophenol liberated was measured spectrophotometrical ly at 420 nm after the addition of 0.6 ml of 1 N NaOH to develop color. Protein concentration was estimated by the method of Bradford (13) with bovine plasma albumin as a standard.
Polyacrylamide gel electrophoresis Sodium dodecyl sulfate polyacrylamide gel electropho resis (SDS-PAGE) was performed on a 7.5% (w/v) poly acrylamide gel overlaid with a 3070(w/v) polyacrylamide stacking gel as described by Laemmli (14). Non-SDS PAGE was basically performed according to Davis' method (15).
RESULTS The inhibition of ALPs by levamisole The inhibition of ALPs from MC3T3-E1 cells, placen ta, kidney, and intestine by levamisole was tested (Fig. 1). The ALP activities of MC3T3-E1 cells, placenta and kid ney decreased hyperbolically with the concentration of levamisole (Ki 0.5=10 -121iM). However, intestinal activ ity was almost unchanged until the concentration was in creased to 240 pM levamisole. The inhibition of ALPs by L-homoarginine The ALP activities of MC3T3-E1 cells, placenta, and kidney decreased hyperbolically with the concentrations of L-homoarginine (Ki 0.5=1 mM) (Fig. 2). However, intes tinal ALP activity was stable until the concentration of 1 MM L-homoarginine, and then decreased sigmoidally (Ki 0.5 = 13 mM). About 40% of the activity was not inhibit ed even in the presence of 40 MM L-homoarginine.
Fig. 1. Effect of levamisole concentration on inhibitions of alkaline phosphatase isoenzymes. The activities of intestine (0), E1 cell (0), placenta (0) and kidney (F-1) ALPs in the presence of different concentrations of levamisole were measured; each 0.5-m1 reaction mixture contained 120 , ig of ALP isoenzymes; 100 pmol tris-110, pH 9.5; 12.5 pmol sucrose; 0.8 pmol MgC12; 0.8 pmol pNPP; and 0, 1.9, 3.8, 7.5, 15, 30, 60 or 120 nmol levamisole. The reaction was started by the addition of pNPP and terminated after 30 min at 371C by the addition of 0.5 ml of 15% (w/v) trichloroacetic acid; 0.6 ml of 1 N sodium hydroxide was added to make the solution basic (11), and the optical density at 420 nm was measured. The activity in the absence of levamisole was taken as the 100076 value.
Fig. 2. Effect of L-homoarginine concentration on inhibitions of alkaline phosphatase isoenzymes. The activities of intestine (0), El cell (0), placenta (/) and kidney (0) ALPs in the presence of different concentrations of L-homoarginine were measured. Assay conditions were as in Fig. 1 except that 0, 0.3, 0.6, 1.3, 2.5, 5, 10 or 20 umol L-homoarginine replaced levamisole.
Hill plot analysis of the inhibition data by levamisole and L-homoarginine A sigmoidal concentration-dependency suggests that there is more than one binding site and that there is cooperation among these sites (16, 17). Hill plot analysis was done using the data obtained in Figs. 1 and 2 to ex amine whether binding sites to levamisole or L-homoargi nine show cooperativity. When log [v/(Vm,, ,, v)] was plot ted against log [levamisole] or log [L-homoarginine], with v, the enzyme activity in the presence of various concen trations of levamisole or L-homoarginine, and Vmax,the activity without the inhibitor, straight lines were obtained (data not shown for levamisole and for L-homoarginine, data shown in Fig. 3). The slopes of these lines are equal to the n value, the Hill coefficient, which is the apparent number of inhibitor molecules combining with one mol ecule of enzyme assuming only one active site per mol ecule or which shows the extent of negative or positive co operativity between inhibitor binding sites (17). The Hill coefficients obtained from the data of Fig. 1 were around 0.9 1.0 for bone-type, placental and kidney ALP iso enzymes (data not shown). Figure 3 shows the Hill plot using the inhibition data by L-homoarginine obtained in Fig. 2. The Hill coefficients for bone-type, placental and kidney ALPs were also 0.9-1.0. However the Hill coefficient for intestinal ALP was 1.8.
The effect of SDS treatment on the inhibition of intestinal ALP by L-homoarginine SDS is often used to decrease the subunit-subunit inter action and to dissociate subunits. The intestinal ALP was treated with SDS to test whether cooperativity among the L-homoarginine binding sites is caused by a subunit-sub unit interaction. First, the effect of SDS on ALP activity was tested. The activity decreased to about 50%, with adependency on the SDS concentration (K;0.5=0-1% w/v SDS), and the remaining 50% activity was maintained in the presence of 1 to 10% (w/v) SDS (data not shown). Figure 4 shows the effect of different concentrations of SDS on the inhibition of intestinal ALP by L-homoargi nine. The L-homoarginine concentration-dependent inhi bition curve clearly changed from sigmoidal to hyperbolic, and the apparent K; 0.5 values for L-homoarginine decreased when the concentration of SDS increased. The effects of SDS concentration on Hill coefficients, K; 0 .5 and remaining ALP activity A Hill plot was made with the data in Fig. 4, and the cal culated Hill coefficients and K; 0.5 values are summarized in Table 1. The Hill coefficients decreased from 1.8 without SDS to 0.7 with 0.8% SDS. The K; 0.5 values also decreased from 13.1 mM to 1.2 mM with increasing SDS. These changes of the Hill coefficients and K; 0.5 values were observed in the presence of 0.03% SDS where there was no loss of ALP activity.
Fig. 3. Hill plot analysis of the data in Fig. 2. Log [v/(Vm-v)] plotted against log [L-homoarginine] for intestinal (0), placental (A), E 1-cell (0) and kidney (0 ALPs inhibited by L-homoarginine. The enzyme activity in the presence of various concentrations of L homoarginine is v, and Vma.,is the activity of the enzyme in the ab sence of L-homoarginine. Hill coefficients were calculated from the slopes, and the K; 0.5 value was calculated from the intercept of the slope and the X axis.
Fig. 4. Effect of SDS on L-homoarginine inhibition of intestinal ALP. Experiments similar to those in Fig. 2 were performed without (0) or with a final SDS concentration of 0.03% (w/v) (D), 0.1% (0), 0.5% (0), or 0.8% (A). The values without L-homoarginine at each SDS concentration was the 100% value. Table 1. Hill coefficients and K; 0.5 values calculated with Fig. 4
Apparent molecular weight of intestinal ALP in the ab sence or presence of SDS We assumed that the changes in Hill coefficients and the affinity for L-homoarginine were caused by dissocia tion of the subunits in the presence of SDS. We tested whether the apparent molecular weight of intestinal ALP changes in the presence of SDS. Figure 5 shows the results of polyacrylamide gel electrophoresis. The appar ent molecular weight of intestinal ALP in the presence of 1076 SDS was 60 K daltons, with no higher molecular weight band. The result was not affected by the presence or absence of 1076 2-mercaptoethanol. On the other hand, a band was detected between 140 K and 443 K daltons with non-SDS-PAGE.
Fig. 5. Apparent molecular weight of intestinal ALP by PAGE performed in the presence or absence of SDS. SDS-PAGE (left) was performed by the Laemmli method (14) using a 7.5010 (w/v) polyacrylamide separating gel overlaid with a 3'o (w/v) polyacrylamide stacking gel. After electrophoresis, the gel was stained with Coomassie brilliant blue R 250. The standard markers were myosin (200,000), f3-galactosidase (116,248), bovine serum albumin (66,267) and aldolase (42,400). The apparent molecular weight of intestinal ALP was about 60,000. The PAGE on the right was without SDS (15) and silver-stained (Bio-rad silver stain kit). The standard markers were thyroglobulin (669,000), ferritin (443,000), lactate dehydrogenase (139,850), bovine serum albumin (66,267) and soy bean trypsin inhibitor (20,100). DISCUSSION Alkaline phosphatase isoenzymes are widely distribut ed in mammalian tissue, but their physiological roles are not clear (1, 18 20). It is important to characterize the en zymatic activity of each of the ALP isoenzymes to esti mate their physiological functions, and a characterization of isoenzymes is also useful for determining the source of ALP isoenzymes appearing in the serum to allow adifferential diagnosis of diseases. One approach to distinguish ALP isoenzymes is to com pare the sensitivity to ALP inhibitors, and inhibition stud ies have been done with L-homoarginine (6), levamisole (7, 8), L-phenylalanine (8, 21), vanadate (22, 23) and other inhibitors (1, 8). The results reported were not al ways consistent and have been inadequate to characterize the isoenzymes. Belle reported that the inhibition of intes tinal ALP by several inhibitors was clearly different from the inhibitions of kidney, bone, liver, placenta and mam mary tumor ALPs (8), whereas Lin and Fishman reported that intestinal and placental ALPs are similar but differ ent from bone and liver isoenzymes with respect to L homoarginine inhibition (6). Our results (Figs. 1 and 2) are consistent with those obtained by Belle (8). The dis crepancies may be due to different tissue species or differ ences in experimental conditions. It is necessary to study the mechanism of inhibition in detail to better character ize the inhibition of isoenzymes. In the present study, the inhibition of intestinal ALP by levamisol and L-homoarginine is different from that of placental, kidney, and bone-type ALP (Figs. 1 and 2). In testinal and placental ALPs have amino acid sequences that are closely related, as deduced from cDNA analyses, as compared with the sequences of the L/B/K ALPs (5). This suggests that the different inhibitions of isoenzymes by levamisole and L-homoarginine are not related to differ ences in the primary structures of the ALP isoenzymes. Our results suggest that the different sensitivity of intes tinal ALP to L-homoarginine is caused by cooperation between L-homoarginine binding sites brought about by subunit-subunit interaction, because 1) The inhibition curve showed a sigmoidal dependency on the L-homoargi nine concentration (Fig. 2), and the Hill plot analysis suggested the existence of positive cooperativity between the L-homoarginine binding sites (Fig. 3); and 2) SDS decreased the Hill coefficients, indicators of the extent of cooperativity, from 1.8 to 0.7 (Fig. 4 and Table 1), sugges ting that the positive cooperativity for L-homoarginine binding to the intestinal ALP decreased with increasing SDS concentration. The inhibition curves for bone-type, placental and kid ney ALPS were hyperbolic, and the Hill coefficients were around 0.9-1.0. However, the Hill coefficient for intes
tinal ALP was 1.8 (Figs. 2 and 3). We may explain these results as follows: In the case of bone-type, placental, and kidney ALP, there is only one L-homoarginine bind ing site per active ALP molecule and the sites on each subunit are independent and equivalent to each other. On the other hand, there is more than one binding site per active intestinal ALP molecule and positive cooperativity caused by subunit-subunit interaction between these sites. This positive cooperativity stimulates the binding of one L-homoarginine molecule to the site on one subunit to the L-homoarginine binding on the other subunit. The Lhomoarginine binding results in a sigmoidal inhibition curve. When subunit-subunit interaction was decreased or lost by SDS treatment, the positive cooperativity disap peared, and the inhibition curve was hyperbolic (16, 17). The cooperative binding of oxygen to hemoglobin was ex plained by subunit-subunit interaction (24). Fosset et al. reported that calf intestinal ALP is a dimer composed of two similar subunits of 69,000 molecular weight and is dissociated into subunits on exposure to SDS (25). Although the apparent molecular weight in SDS-PAGE (Fig. 5) is smaller than theirs, the results shown in Fig. 5 did not contradict their results and sup ported the above explanation. The inhibition of intestinal ALP by levamisole is also different from that of bone-type, placental and kidney ALPs (Fig. 1). The ALP activities of MC3T3-E1 cells, placenta and kidney decreased hyperbolically with the levamisole concentration, and the Hill coefficients were around 0.9 1.0, which suggest that approximately one molecule of levamisole combines with one active site of bone-type, placental, and kidney enzymes and that there is no cooperativity between the levamisole binding sites. However intestinal ALP activity was almost unchanged untill 240 uM levamisole. These results may also be ex plained by subunit-subunit interaction, but this has not yet been experimentally established. The K; o.5 values decreased from 13.1 mM to 1.2 mM with increasing SDS concentration (Table 1). This indi cates that the apparent affinity of intestinal ALP for Lhomoarginine increased 10 times by SDS treatment. If the association and the dissociation of the subunits of oligo meric enzyme can regulate the sensitivity to drugs ten times, such a system could offer an excellent candidate for regulating drug sensitivity. SDS is not a physiological sub stance, but there may be an as-yet-undescribed physiologi cal system for regulating subunit-subunit interaction. The ALPs are thought to be membrane bound enzymes (1). In the case of Na,K-ATPase, which is also a membrane bound and oligomeric enzyme, it is suggested that enzyme activity is affected by the composition of phospholipids or aging of the biomembrane (26). These factors may affect the state of membrane bound and oligomeric enzymes. Acknowledgments We thank Dr. H. Kodama of Ohu University and Dr. Y. Kuboki of Hokkaido University for giving us the clonal cell line, MC3T3-E1 cells. This work was supported by The Akiyama Foundation, Japan.
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616 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3125075/ | Pharyngeal mesoderm development during embryogenesis: implications for both heart and head myogenesis - PMC
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Cardiovasc Res
. 2011 Apr 15;91(2):196–202. doi: 10.1093/cvr/cvr116
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Pharyngeal mesoderm development during embryogenesis: implications for both heart and head myogenesis
Eldad Tzahor
Eldad Tzahor
1 Department of Biological Regulation, Weizmann Institute of Science, Rehovot 76100, Israel
Find articles by Eldad Tzahor
1,, Sylvia M Evans
Sylvia M Evans
2 Skaggs School of Pharmacy, University of California—San Diego, La Jolla, CA 92093, USA
Find articles by Sylvia M Evans
2
Author information
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1 Department of Biological Regulation, Weizmann Institute of Science, Rehovot 76100, Israel
2 Skaggs School of Pharmacy, University of California—San Diego, La Jolla, CA 92093, USA
Corresponding author. Tel: +972 8 934 3715; fax: +972 8 934 4116, Email: eldad.tzahor@weizmann.ac.il
This article is part of the Spotlight Issue on: Cardiac Development
Received 2011 Feb 7; Revised 2011 Apr 4; Accepted 2011 Apr 12; Issue date 2011 Jul 15.
Published on behalf of the European Society of Cardiology. All rights reserved. © The Author 2011. For permissions please email: journals.permissions@oup.com.
PMC Copyright notice
PMCID: PMC3125075 PMID: 21498416
Abstract
The pharyngeal mesoderm (PM), located in the head region of the developing embryo, recently triggered renewed interest as the major source of cells contributing to broad regions of the heart as well as to the head musculature. What exactly is PM? In this review, we describe the anatomical and molecular characteristics of this mesodermal population and its relationship to the first and second heart fields in chick and mouse embryos. The regulatory network of transcription factors and signalling molecules that regulate PM development is also discussed. In addition, we summarize recent studies into the evolutionary origins of this tissue and its multipotential contributions to both cardiac and pharyngeal muscle progenitors.
Keywords: Heart, Head muscle, Pharyngeal mesoderm
1. Pharyngeal mesoderm: definition, location, and contribution
Pharyngeal mesoderm (PM) constitutes a subset of head mesoderm that surrounds the developing pharynx and is situated within the core of the pharyngeal arches (also known as branchial arches). Head mesoderm precursors undergo gastrulation in the primitive streak, prior to those of trunk mesoderm.1,2 The PM can be divided into two subdomains: the loosely connected, mesenchymal paraxial mesoderm, located on both sides of the neural tube and notochord (Figure1 A and B), and the medial splanchnic mesoderm, which is maintained as epithelial tissue, although there seems to be no clear border between these two domains (Figure1A).
Figure 1.
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Pharyngeal mesoderm cells give rise to parts of the heart and the pharyngeal muscles. (A–D) Schematic illustration of the anatomy of the pharyngeal mesoderm in sections of a 1.5–3-day-old chick embryo. Pharyngeal mesoderm cells (green) in the anterior part of the embryo surround the pharynx. Later, these cells fill the mesoderm core of the pharyngeal arches, and are incorporated into the arterial pole of the heart (e.g. outflow tract). The first heart field (pink) is restricted to the lateral splanchnic mesoderm that later contributes to the linear heart tube. Second heart field cells (green) are PM cells that contribute to the arterial pole of the heart. PM cells interact and migrate together with cranial neural crest cells. Cardiac neural crest cells are part of the cranial neural crest population, migrating into the outflow tract via the posterior arches (arches 3–6).
The lateral splanchnic mesoderm is located on the ventral side, beneath the floor of the pharynx (Figure1B). Both paraxial and splanchnic mesoderm cells converge to form the mesodermal core within the pharyngeal arches3 (Figure1C and D). Hence, the broad definition of the PM should include both paraxial and lateral mesoderm cells surrounding the pharynx (Figure1, marked in green). Taken together, PM cells contribute to the core of the pharyngeal arches (Figure1D). These cells give rise to significant parts of the heart and the pharyngeal muscles. PM cells are found in close proximity to the pharyngeal endoderm, ectoderm, and neural crest cells, which tightly influence PM development (Figure1D; see below).
Studies in both chick and mouse embryos have shown that cardiac progenitor cells populating the cardiac outflow tract and right ventricle are progressively added during heart looping stages by PM cells, and are referred to as the anterior heart field.4–6 In the mouse, the anterior heart field is a subset of the second heart field, which contributes to the outflow tract and right ventricle, and will also contribute a majority of cells to the atria. Thus a subset of PM cells constitutes the second heart field, in contrast to the more lateral splanchnic mesoderm, known as the first heart field (Figure1A and D, pink colour), which is contiguous with the PM, differentiates earlier, and eventually populates the left ventricle (reviewed in references 7–10).
The secondary heart field in the chick, which also gives rise to the arterial pole of the heart, is also part of the anterior heart field. The secondary heart field is situated slightly caudal to the anterior heart field, and gives rise to the myocardium and smooth muscle of the distal outflow tract.10 The cells added at the arterial and venous poles of the heart are derived from different regions of the mesoderm,11 and whether caudal PM contributes to the venous pole of the heart is currently not clear. Collectively, the anatomical terminology in the chick is such that the secondary heart field is a subdomain of the anterior heart field, which is a subdomain of the PM, which is part of the head mesoderm. In addition, recent studies indicate that PM progenitors contribute to both head muscles (pharyngeal muscles) and parts of the heart.3,4,12–16 Hence, PM cells are critical part of the cardio-craniofacial field during early embryogenesis.
The skeletal myogenic potential of PM cells and their contribution to pharyngeal muscles have long been documented.17 In contrast, the cardiogenic potential of these cells has only been revealed over the last decade (reviewed in references 7–10,18). In keeping with these findings, PM explants dissected from early chick embryos undergo cardiogenesis.19 This study further suggested that signals from the dorsal neural tube (e.g. Wnt1 and Wnt3a) attenuate PM-derived cardiogenesis.19 The in vivo cardiogenic potential of PM was further revealed in chick embryos.3,12,20 Furthermore, considerable overlap in the expression of head muscle markers [e.g. Myf5, Tcf21 (capsulin), Msc (MyoR), Tbx1, Pitx2] and cardiac markers such as Islet1 and Nkx2.5 has been documented in the PM, suggesting that these cells play a dual role in myogenesis and cardiogenesis.3,12,21 Likewise, lineage studies in the mouse demonstrated an overlap in progenitor populations contributing to pharyngeal muscles and second heart field derivatives.3,22–24 Thus, the genetic programme controlling pharyngeal muscles overlaps with that of the PM-derived heart.
Genetic studies in vertebrate embryos have begun to reveal the regulatory network of transcription factors expressed in the PM. The major players in this network include the transcription factors Tbx1, Pitx2, Tcf21 (capsulin), Msc (MyoR), and Isl1 (reviewed in references 8,13–15,25). Although these reviews describe muscle phenotypes resulting from knockout, singly or in combination, of these transcription factors in mouse embryos, exactly how they function in a hierarchical regulatory network to coordinate both myogenesis and cardiogenesis is far from clear.
2. Pharyngeal muscle heterogeneity
Myogenic programmes that lead to the formation of skeletal muscles vary considerably between the body and head regions. There are ∼60 distinct skeletal muscles in the vertebrate head which control food intake, facial expression, and eye movement; for example, the extraocular muscles (EOMs) move and rotate the eye in a highly coordinated manner; pharyngeal muscles control jaw movement and facial expression, as well as pharyngeal and laryngeal function. Muscles in the neck and tongue are derived from myoblasts originating in the most anterior set of somites located in the trunk (reviewed in reference 26). Importantly, while head mesoderm gives rise to all muscles of the head, the PM only contributes to subsets of these muscles. For example, EOMs derive from non-PM cells (e.g. prechordal mesoderm26) but it is not clear whether PM contributes in any way to EOMs.
Recent studies have begun to uncover an unexpected heterogeneity in head muscles with respect to their origins, genetic lineages, and transcriptional programmes, as well as their proliferative, differentiative, and regenerative properties.24,27,28 Heterogeneity in skeletal muscles can also be seen during adulthood, as reflected in distinct genetic signatures, and susceptibilities to muscle myopathies of both head and trunk skeletal muscles.29,30
Lineage mapping of head musculature in the mouse revealed genetic heterogeneity.24 This lineage map corroborates, at the cellular level, our understanding of craniofacial muscle phenotypes seen in several recent loss-of-function studies in mice. A complementary view of craniofacial muscle formation in mice, based on dissection of genetic programmes promoting myogenesis in distinct head muscles (e.g. EOMs vs. pharyngeal muscles) was demonstrated by the Tajbakhsh laboratory.27 Pharyngeal muscle progenitors are regulated by a combination of Tbx1 and Myf5 acting upstream of MyoD, as Tbx1:Myf5 doubled mutants lack these muscles completely, while EOMs in these mutants are spared. In contrast, EOMs are selectively lost in another genetic combination, the Myf5:Mrf4 double mutant, although other head and trunk muscles are present in the same double mutant.27 These observations, in addition to other mutant analyses (see below), suggest that the underlying mechanism that leads to selected myopathies may have an ontological aetiology.
Along theses same lines, Capsulin and MyoR act as upstream regulators of pharyngeal muscle development. In Capsulin/MyoR double mutants, the masseter, pterygoid, and temporalis muscles are missing, while lower jaw muscles (e.g. anterior digastric and mylohyoid), EOMs and tongue muscles are not affected.31 Similarly, pharyngeal muscles in Tbx1 mutants are frequently hypoplastic and asymmetric, whereas EOMs and tongue muscles are spared.32 Genetic lineage analyses of head musculature24 indicated that EOMs and tongue muscles derive from lineages distinct from those of pharyngeal muscles. In Pitx2 mutants, the EOMs and first arch muscles are affected22,33; first arch myoblasts marked by the Mef2c AHF-Cre lineage in the mouse16,23 were fewer in Pitx2 mutant embryos.22
3. Isl1 lineage-derived PM cells contribute to a broad range of cardiovascular and skeletal muscle progenitors
The LIM-homeodomain protein Islet1 (Isl1) is expressed in and required for a broad subset of cardiac progenitors in the mouse.34–37 Gene expression and lineage experiments in the chick have revealed that the core of the pharyngeal arch is divided along the proximal–distal axis such that paraxial mesoderm cells mainly contribute to the proximal region of the core, while the splanchnic mesoderm contributes to its distal region3 (Figure2A–C). Isl1 is expressed in the distal part of the PM, and its expression is correlated with delayed differentiation of lower jaw muscles.3 Over-expression of Isl1 in the chick represses pharyngeal muscle differentiation.24
Figure 2.
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Isl1 marks pharyngeal mesoderm cells that populate the heart and the pharyngeal muscles. (A–C) Lineage studies in the chick have revealed the contribution of PM cells to the distal core of the pharyngeal arches and the heart. ISL1 protein expression is seen in this region (B–C), which is correlated with delayed differentiation of these myogenic progenitors (MYF5-). (D–F) Similarly, lineage analyses in mouse Isl1Cre; Rosa26LacZ E10.5 embryos (D) demonstrate the significant contribution of these cells to the core of the first pharyngeal arch and the heart. The heterogenic contribution of the Isl1 lineage to the head musculature at E16.5 is shown (E–F): A section of the Isl1Cre; Rosa26YFP embryo stained for muscle (red) or GFP (green). Isl1+ cells contribute strongly to lower jaw muscles (yellow), muscles of facial expression shown in the head periphery, and less to jaw closing muscles (orange). Extraocular muscles (EOMs) or tongue muscles are not derived from Isl1+ cells. (F) A diagram of the embryonic head (lateral view) reveals the contribution of high (yellow) and intermediate levels (brown) of the Isl1 lineage to the head musculature.
Lineage tracing experiments in the mouse using an Isl1-Cre line revealed a significant contribution of Isl1+ cells to the mesodermal core of the pharyngeal arches,3,24 as well as to the heart38 (Figure2D). Isl1+ PM cells were shown to contribute to a subset of pharyngeal muscles, the mylohyoid, stylohyoid, and digastric, at the base of the mandible, facilitating its opening (Figure2E and F). Isl1+ cells were also found in second arch-derived muscles controlling facial expression3,24 and, to a lesser extent, in the masseter, pterygoid and temporalis, the jaw closing muscles (Figure2E, orange; Figure2F, brown), indicating that this gene is not expressed in all cells of the PM. In both species, tongue muscles and EOM are not derived from the Isl1 lineage.3,24 A similar lineage map of Isl1+ PM cells was seen in adult pharyngeal muscles and their associated satellite cells.24 These findings highlight the link between myogenesis in the early embryo, and the generation of adult muscle progenitor pools required for muscle maintenance and regeneration. Taken together, Isl1 marks a subset of PM cells, and plays an important role in the development of distinct PM-derived cardiovascular and skeletal muscle progenitors. The direct role of Isl1 in pharyngeal muscle development has yet to be resolved, as Isl1 knockout embryos die at around E10.34
4. Pharyngeal mesoderm evolution: from pharyngeal muscles to the heart
The architecture, function, and physiology of muscle cells have been remarkably conserved throughout evolution. Hence, all muscle cells likely evolved from an ancestral developmental programme involving a single contractile myogenic cell type.39,40 The fact that the developmental programmes of the heart and pharyngeal muscles are tightly linked suggests that these tissues share common evolutionary origins13,14 (Figure3).
Figure 3.
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Isl1+ PM cells evolved from an ancestral myogenic programme for both cardiac and skeletal muscle lineages. (A) Nematodes such as the worm C. elegans do not have a heart; instead, they have a contractile pharyngeal muscle that functions like the heart in vertebrates. (B) Tunicates (Ciona intestinalis) are chordates that are considered as a ‘sister group’ to the vertebrates. Unlike nematodes, the heart and pharyngeal muscle cells in tunicates are seemingly distinct. Isl1+ cells in Ciona give rise to the pharyngeal muscles (termed siphon muscles), and not to the heart. (C) Reallocation of Isl1+ PM cells into the looping heart, which occurred during evolution from chordates to vertebrates, represents the emergence of the second heart field.
Nematodes such as the worm, C. elegans, are invertebrates that do not possess a heart or defined circulatory system. Instead, their pharyngeal muscle contracts like a heart, and exhibits electrical activity similar to that of mammalian cardiomyocytes (Figure3A). Moreover, it has been shown that development of the pharyngeal muscle in nematodes is regulated by the homeobox gene Nkx2.5 (ceh-22)41 and may be functionally replaced by the zebrafish nkx2.5 (42, reviewed in references 13,14,43).
Tunicates belong to the Chordata phylum, and are considered as the ‘sister group’ of vertebrates.44 The tunicate Ciona intestinalis is a sessile marine invertebrate. As in vertebrates, the Ciona heart is located ventrally and posterior to the pharynx, and anterior to the stomach; in the gastrulating embryo, its heart arises from a pair of blastomeres expressing the MesP gene. Several studies suggest significant similarities in the gene regulatory networks controlling cardiogenesis in vertebrates and tunicates.44–46 The heart and pharyngeal muscle cells in Ciona (Figure3B) are seemingly distinct, based on the expression of different myosin heavy chain isoforms47; yet both are derived from MesP+ cells.
Strikingly, Isl1+ PM cells in both Ciona48 and vertebrates3,24 give rise to pharyngeal muscles (termed siphon muscles in Ciona). These findings suggest that the last common ancestor of tunicates and vertebrates had PM cells derived from MesP+ lineages that expressed Isl1, FoxF, and Nkx2.5, and had the potential to give rise to both heart tissue and pharyngeal muscles.48 With the increasing complexity of the vertebrate heart and, in particular, during the heart tube elongation that occurs in vertebrates, Isl1+ PM cells were recruited into the looping heart to give rise to cardiomyocytes (Figure3C). Hence, this study suggests that reallocation of PM cells into the looping heart represents the emergence of the second heart field in vertebrates. In addition, these findings suggest distinct evolutionary origins for the two heart fields.
5. Signalling mechanisms affecting pharyngeal mesoderm development
PM cells are maintained in a proliferative and undifferentiated state by multiple signalling mechanisms that lately have begun to be defined. Abnormal transition of proliferating PM-derived cardiac progenitors into differentiating cardiomyocytes severely affects cardiac looping and outflow tract extension, processes which are generally associated with congenital heart disease.7,10 Many signalling pathways influence cardiac progenitor cell proliferation and differentiation (reviewed in references 8,10,49). For example, recent studies have shown that the Wnt/β-catenin pathway plays distinct roles at various stages of cardiac development by triggering the renewal and expansion of cardiac progenitors, and blocking their differentiation (reviewed in reference 50).
Bone morphogenic protein (BMP) and fibroblast growth factor (FGF) signals are initially required for cardiac specification and, later, for the differentiation of cardiac progenitors in chick and frog embryos. Loss-of-function studies in mice have underscored the essential roles of BMP and FGF signalling pathways in cardiogenesis (reviewed in references 8,49). BMP signalling affects PM cells in two phases: it is initially required to ‘lock’ these cells into the cardiogenic lineage (and block skeletal muscle markers);12 while at later stages, BMP signalling promotes PM cardiac differentiation by blocking FGF signalling, to facilitate the accurate deployment of these progenitors to the looping heart.51 Indeed, FGF signalling has been shown to play key roles in the survival and expansion of PM-derived cardiac progenitors,52–55 where down-regulation of this pathway is both required and sufficient for the differentiation of PM-derived cardiomyocytes.51 Hence, opposing activities of BMP and FGF signalling pathways constitute a major regulatory mechanism during cardiogenesis.51,56 In addition, BMP–FGF crosstalk regulates the epicardial vs. myocardial lineage switch at the inflow pole of the heart.57
What lies downstream of BMP signalling in the transition from cardiac progenitors to cardiomyocytes? In the chick, the transcription factor Msx1, expressed primarily by neural crest cells, has been shown to function between BMP and FGF pathways to promote cardiomyocyte differentiation.51 Likewise, double knockout of Msx1/2 in the mouse resulted in increased proliferation of second heart field cells.58 How Msx genes expressed in neural crest cells promote myocardial differentiation of PM cells is currently not clear. A recent study revealed another means by which BMP signalling can promote PM differentiation: the Martin laboratory found that the PM progenitor markers Isl1 and Tbx1 are repressed by BMP signalling, via the induction of microRNAs (miRNA 17–92 cluster).59
6. Regulation of pharyngeal mesoderm by other tissues
PM progenitors are exposed to signals from pharyngeal endoderm, ectoderm, and neural crest cells that, together, create a complex regulatory system (reviewed in references 8,49). Perturbation of the balance of signals within this system can lead to abnormal heart development. Neural crest ablation in the chick, for example, results in increased FGF signalling and elevated proliferation in PM.60–62 These studies suggest that both cardiac neural crest (affecting caudal PM progenitors) and cranial neural crest cells (affecting cranial PM) (shown in Figure1D), buffer proliferative signals (presumably FGFs) secreted from the endoderm and ectoderm to promote PM migration and differentiation.
Other examples for the roles of cardiac/cranial neural crest in the regulation of PM cells may be found in mice lacking either Smad4, a member of the Smad family of signal transduction proteins involved in BMP signalling63 or the BMP receptor ALK264 in neural crest cells. Both of these mouse models resulted in abnormal differentiation of PM progenitors that led to severe cardiac outflow defects. Furthermore, these studies are consistent with a BMP-dependent signalling mechanism involving neural crest cells that regulate these progenitors. Similarly, interference with Notch signalling in neural crest, PM or endothelial cells, all result in arterial pole and arch artery defects (reviewed in references 8,65). Taken together, these studies demonstrate that multiple signals from various tissues are intertwined in an as-yet unclear manner, to control the development of PM cells.
It has long been suggested that neural crest—PM crosstalk is involved in patterning of the head musculature (reviewed in reference 66). Several recent studies suggest that pharyngeal muscle patterning and differentiation are tightly regulated by the interaction of PM-derived muscle progenitors with adjacent cranial neural crest cells that give rise to most of the vertebrate skeletal system, including bones, cartilage, and connective tissues in the face.62,67–70 It is therefore believed that CNC-derived connective tissue progressively imposes the characteristic anatomical musculoskeletal architecture upon PM muscle progenitors.
7. Pharyngeal mesoderm cells are multipotent cardiac and skeletal muscle progenitors
While it is now well-established that PM progenitors contribute to both pharyngeal muscles and heart progenitors within the arterial pole of the heart,3,4,12 the question remained whether these head and heart cells originated from single PM cells or from a pre-determined, mixed population of PM cells. Using a retrospective clonal analysis (that they developed) in the mouse, the Buckingham laboratory demonstrated that head muscles and second heart field derivatives originate from multipotent PM progenitors.16 Two myogenic lineages linking groups of head muscles to different parts of the heart were identified (Figure4A–C): The first muscle lineage (blue) gives rise to the temporalis and masseter, two first pharyngeal arch muscles, as well as to the EOMs. Strikingly, this single-cell lineage also contributes to myocardial cells in the right ventricle (Figure4C). The second lineage gives rise to a broad range of muscles controlling facial expression, which derive from mesoderm of the second pharyngeal arch, and also contributes myocardial cells to the arterial pole of the heart (Figure4B and C). These findings highlight the dynamic posterior shift in the alignment of the cardiac outflow tract with the pharyngeal arches, and the spatiotemporal deployment of PM into heart and head muscles. Interestingly, Mef2c-AHF-Cre, activated in PM progenitors,22,23 can distinguish between first- and second-pharyngeal arch myogenic lineages. Further sublineages distinguish myocardium at the base of the aorta or pulmonary trunk, with a clonal relationship to right or left head muscles, respectively.16
Figure 4.
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Clonal relationships of head and heart muscle progenitors in the mouse. Schematic representation of PM cells contributing to the core of the pharyngeal arches in the mouse embryo. (A) The linear heart tube at E8.0 is aligned at the level of the first pharyngeal arch; 1–2 days later, the heart undergoes looping, and shifts posteriorly. (B) First arch PM cells (blue) migrate to the developing heart tube to contribute to right ventricular myocardium. PM cells from the second arch (pink) migrate later (around E9.5–E10), to contribute to outflow tract myocardium (oft). (C) Diagram depicting the contribution of the different lineages to head muscles and heart myocardium: The first lineage (blue) contributes to masticatory muscles (temporalis and masseter) and to right ventricular (rv) myocardium. The second lineage (pink) contributes to facial expression muscles, and to myocardium at the base of the pulmonary trunk (pt) or aorta (ao). Modified after Lescroart F et al. Adapted with permission.16
In conclusion, we have attempted to summarize novel insights into how pharyngeal muscles and certain parts of the heart arise from PM. In doing so, we emphasized key findings concerning the anatomical, cellular, and molecular characteristics of PM progenitors. We also discussed signalling mechanisms that regulate PM proliferation and differentiation, and the evolution of PM-derived cardiac and skeletal muscle progenitors. The developmental paths that lead to the formation of skeletal muscles in the head appear to be distinct from those operating in the trunk. Considerable cellular and genetic variations among the different craniofacial muscle groups are also seen. A deeper understanding of PM development is instrumental in deciphering the aetiology of heart and craniofacial birth defects.
Funding
The Tzahor laboratory is supported by research grants from the Association Française Contre les Myopathies; the Israel Science Foundation; the United States-Israel Binational Science Foundation (BSF, with S.M.E); the Helen and Martin Kimmel Institute for Stem Cell Research; the German Israeli Foundation (GIF); and the Minerva Foundation. S.M.E. would like to acknowledge funding from NIH and BSF.
Acknowledgements
We deeply thank Miriam Brodt-Ivenshitz and Itamar Harel from the Tzahor laboratory for the artwork. We thank Brad Davidson and Mike Levine for insightful discussions. E.T. is the incumbent of the Gertrude and Philip Nollman Career Development Chair.
Conflict of interest: none declared.
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Abstract
1. Pharyngeal mesoderm: definition, location, and contribution
2. Pharyngeal muscle heterogeneity
3. Isl1 lineage-derived PM cells contribute to a broad range of cardiovascular and skeletal muscle progenitors
4. Pharyngeal mesoderm evolution: from pharyngeal muscles to the heart
5. Signalling mechanisms affecting pharyngeal mesoderm development
6. Regulation of pharyngeal mesoderm by other tissues
7. Pharyngeal mesoderm cells are multipotent cardiac and skeletal muscle progenitors
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617 | http://mccord.cm.utexas.edu/courses/learn/idealgaslaw.php | The Ideal Gas Law
The ideal gas law is a simple model that allows us to predict the behavior of gases in the world. It is a combination of the previous laws that we have studied (Boyle's, Charles', Avogadro's). Rather than simply looking at proportionalities, it relates pressure, volume, absolute temperature, and the number of moles quantitatively with a universal constant (R) that we call the ideal gas constant.
[P\,V=nR\,T]
The value for R will depend on what units you are using for the properties of the gas. The ideal gas law has many implications that will be discussed below. However, the most remarkable aspect is that the same model works quite well for all gases (N2, He, C3H8, ...).
video here
Universal Gas Constant
The ideal gas constant is a Universal constant that we use to quantify the relationship between the properties of a gas. The constant (R) that we typically use relates pressure in atmospheres, volume in liters, and temperature in Kelvin. In this case, it has the value and units of
[R=0.08206 {\rm \;\;L\;atm\;mol^{-1}\;K^{-1}}]
The gas constant (R) will appear in many contexts as this is a Universal constant that relates energy and temperature. A pressure times a volume is an energy. As such, you will also encounter the gas constant (R) in typical energy units of Joules
[R=8.314 {\rm \;\;J\;mol^{-1}\;K^{-1}}]
Finally, ignore the blunder in the video where Dr. Vanden Bout can't remember the correct value for (R) with units of atm. The correct value is 0.08206 atm L mol-1 K-1
Standard Conditions
Often in the laboratory we would like to have conditions (state functions) that are similar to those used by scientists in other laboratories. As such, as a community, we have decided upon some conditions that we will call "standard". The two state functions we can typically control are temperature ( (T) ) and pressure ( (P) ), so these are the conditions we standardize. We will call these conditions "Standard Temperature and Pressure" or STP. We will take standard pressure as 1 atm, and standard temperature as 0°C. You should note that these are choices. Like many things in life it is difficult to get everyone to agree. Recently, the International Union of Pure and Applied Chemistry (IUPAC) adopted 1 bar as standard pressure and 25°C as standard temperature. This "new" standard is referred to as Standard Ambient Temperature and Pressure (SATP). However, as this new condition has been slow to catch on, we will stick with the "old" standard (STP) unless told otherwise.
Keep them Straight
STP : 0°C (273.15 K) and 1 atm ←(our default "standard" in textbook and Quest)
SATP : 25°C (298.15 K) and 1 bar
Number Density
Number density is a useful concept for thinking about macroscopic samples in a microscopic way. Chemists often try to "visualize" materials with a molecular perspective. Number density can be thought of as the number of particles that are present in a particular volume. As these numbers can be very very large, we typically think of this as the number of moles (a fixed number of particles) within a given volume. This quantity is
$${n\over V}={P\over {RT}}$$
So, if we have any two gas samples that are behaving ideally, they have the same number of particles per volume when the temperature and pressure are the same. For example, if I had two balloons in a room, they would have the same pressure (approximately one atmosphere) and the same temperature (whatever the temperature was in the room). Therefore, they would have the same number density. If the balloons had the same volume then they would have identical numbers of particles. This is really a different way of stating Avogadro's Law.
Molar Volume
Molar volume is another way to view number density. The molar volume is the volume of 1 mole of substance. It can be simply found by dividing the volume of a sample by the number of moles of that sample
[V_m = {V\over n}]
This is simply the inverse of the number density. It is a useful quantity to "think about" things from a molecular perspective. As all gases that are behaving ideally have the same number density, they will all have the same molar volume. At STP this will be 22.42 L. This is useful if you want to envision the distance between molecules in different samples. For instance if you have a sample of liquid water, it has a mass density of 1 g mL-1. Since water has a molecular weigh of 18 g mol-1, 1 mole will have a mass of 18 g. From the density this should have a volume of only 18 mL. Clearly this is a lot less than an ideal gas at STP which is more than 1000 times larger in terms of its molar volume.
The video below is a further discuss of Standard Molar Volume (molar volume at STP).
© mccord 2025 |
618 | https://web.eecs.umich.edu/~pettie/matching/Kurtzberg-approx-assignment-problem.pdf | On Approximation Methods for the Assignment Problem JEROME M:. KURTZBERG Burroughs Corporation, Paoli, Pennsylvanza 1. Introduction This paper is concerned with approximation methods for handling the classical assignment problem. These methods permit solution of large scale assignment problems where exact methods are not economically feasible because of the ex- tensive computation time requirements. Even though the development and analysis of these approximation methods are strongly directed toward digital computers, the methods are especially desirable for hand computation. The assignment problem is defined in Section 1, and the results of a solution- time study of an exact method, the Munkres algorithm, are described. In the next section, three approximation methods are formulated. The results of an empirical investigation of error are given in Section 3. In the next section, the approximation methods are investigated analytically; the expected value and error bounds are established for the promising methods. In Section 5, computer mechanizations of the approximation methods are given and evaluated with respect to certain (defined) basic operations. Timing comparisons are then made, with respect to a particular computer, UNIVAC I. A summary of findings is given in the last section. Definition of Assignment Problem. The statement of the assignment problem is as follows: There are n men and n jobs, with a cost c, for assigning man i to job j. It is required to assign all men to jobs such that one and only one man is assigned to each job and the total cost of the assignments is minimal. The problem phrased in a more general way is: Given a matrix (c,) of real 2 numbers, find a matrix (x,) such that ~=tc.~z, is minimal, with x, = x~j, ~,~lx, = 1, and ~7=tx, = 1, for all i and j. The matrix (c,) is termed the cost matrix; the matrix (x,~) is termed the solution matrix or the permutation matrix (also occasionally termed the assignment matrix) [1, 6]. At this point, it may be mentioned that occasionally it is desired to maximize ~,~j~c,x,, rather than to minimize it. A method yielding an optimal assign- ment for one always gives an optimal assignment for the other by a simple trans- formation of the cost matrix (c,j). For example, if all c, lie in (a, b), the substi- tution (d,) = (b -- c~j) suffices. Solution Techniques. Any exhaustive scheme of enumeration of all assign- ments to find the optimal solution is impractical. However, the assignment problem of order n can be expressed as a 2n X n ~ zero-sum two-person game, Received April, 1961; revised February, 1962 This paper is based upon the author's thesis : Kurtzberg, J. "Approximation methods for handling of large-scale assignment prob- lems," June 1960, Univ. Pennsylvania, Moore School master's thesis. 419 420 JEROME M. KURTZBERG which can be cast into a linear programming problem ; the general technique of the simplex metkod is then applicable [3, 6]. Many computer programs of the simplex method (and variants of it) exist for various computers . Furthermore, there are a number of algorithms specifically directed at the solution of the assignment problem [1, 8, 9, 10, 11]. Since the assignment problem may be viewed as a special case of the general transportation (or distribution) problem [5, 7], methods suitable to that problem are quite applicable. There are methods designed to allow a "good initial start" for the simplex method operating upon a transportation problem . (The purpose of this good initial start is to reduce the required number of iterations for an optimal solution.) Methods such as these have been independently formulated by the writer, and evaluated in this paper, as techniques for obtaining near-optimal solutions of the assignment problem. The approximation methods considered and defined in Section 2 are termed (1) the suboptimization method, (2) the row/column-scan method, and (3) the matrix-scan method. Definition of these methods is pre- ceded by the results of a timing study of an exact assignment algorithm. Solution-Time Study of ann Exact Method. A variant of Kuhn's method, de- veloped by James Munkres , was adjudged particularly suitable for com- puter mechanization and was programmed by the writer on UNIVAC I. This was done to provide a necessary subroutine for one of the approximation methods (the suboptimization method, defined in Section 2.1) and to supply a basis for evaluating the worth of the approximation methods. The program that was developed was designed to solve assignment problems up to order 12. With this limit, extreme computer efficiency is possible for UNIVAC I because of the 12-digit word size of this machine. This size was deemed sufficient to carry out any pertinent analysis on a computer-programmed assignment algorithm without falling victim to degenerate cases. The test cost matrices (for the timing study) were obtained by generating random numbers uniformly in the range (0, 1) and using these numbers as the matrix elements c,~. There is no loss of generality in this choice of range, for if the c,j are distributed lmiformly in the interval (a, b), then the linear transfor- mation c:j = (c, -- a) / (b -- a) produces a set of c',: which lie in the range (0, 1) and are also distributed uniformly. The assignment is independent of the trans- formation, since a linear transformation does not destroy the relative ordering of the possible solutions. The results of the timing study are tabulated below. (The time includes read- ing in the test matrix from tape.) The average sample size was approximately 100 test matrices. One would expect, due to the difficulty in timing the com- puter runs, more experimental error for small n. n (size of matrix) t (mean time per run in seconds) 3 4 5 6 7 8 9 10 11 12 APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 421 Unfortunately, there exist qualitative differences (because of the word size and the internal memory size of UNIVAC I) among the ranges (2, 12), (13, 24), and (25, N), where N is arbitrarily large. For any computer, increasing the matrix size beyond certain critical points introduces radically new problems, primarily of storage, necessitating different programming techniques. However, with this restriction understood, an or- thogonal polynomial curve-fitting routine was employed to determine the form of a polynomial that would best fit the data gathered from the range (3, 12). This gave a cubic of the form t = 2.8837 -- 1.0511n + 0.1713n 2 + 0.0025n 3, t in seconds. This polynomial furnishes only a weak lower bound for the mean UNIVAC solu- tion time for matrices in the range (13, 24), and an even weaker lower bound for matrices of order greater than 24. 2. Approximation Methods Approximation methods must require relatively few basic operations, thereby yielding fast solution times, and yet not present extremely inefficient assign- ments. Ideally, these techniques should be directed toward solution of large problems where exact solutions are too costly or unavailable. 2.1. Suboptimization Method. This approach assumes the existence of an exact assignment algorithm that can accept an R×R matrix. The given NXN cost matrix is partitioned into S 2 RXR (sufficiently small) submatrices. Each one of the R XR submatrices is individually solved for the true optimal assignment value. The assignment value ~ c,x, associated with these RXR submatrices is now considered to constitute the elements of an SXS matrix. This S×S matrix is a "gross matrix" in the sense that each element represents a sub-block of assignments. The true optimal assignment for this new matrix is obtained, thereby desig- nating which (independent) sub-blocks of assignments are to be taken. Since the sub-blocks of assignments are independent in themselves, a feasible assignment for the NXN matrix is determined. 2.2. Row/Column-Scan Method. The row/column-scan algorithm requires two cycles through the matrix; each cycle produces a candidate assignment solution. On the first cycle, all the rows are sequentially examined and the minimal uncovered element in each row selected. The row is then assigned to the column of the selected element and the column covered, i.e. all of the elements in that column are covered. Each row and column are thus uniquely assigned to one another, thereby constituting an assignment solution. A value is computed for this candidate assignment solution by summing the values of all the selected elements. Next, on the second cycle, the columns of the original matrix are examined, and, as with the case of the rows, the minimal uncovered element in each column is selected and an assignment made. Again, as with the rows, none of the ele- ments is initially covered. Completion of the column-by-column examination results in a new candidate assignment solution. 422 JEROME M. KURTZBBRG An assignment value associated with this new solution is computed and com- pared with the value associated with the row-by-row assignment solution. The minimum value determines the choice of assignment solution. It may be desirable, especially for extremely high-order matrices that are tape limited, to mechanize the process by merely considering and operating upon "strings" of elements. A string may be either a row or a column of the matrix. Thus, there would be (for a matrix of order N) N strings for the rows and N strings for the columns. These 2N strings could be stored sequentially on tape and operated upon in two cycles, N subcycles in each cycle. Section 5 presents two detailed computer variants of this algorithm, including the minimum and maximum number of required operations per mechanization. The variants could be implemented in any one of the above ways without altering the given required number of operations. The optimal implementation is a func- tion of the particular computer available. 2.3. Matrix-Scan Method. The smallest uncovered element in the entire matrix is selected and the column and row associated with it assigned to one another. (Initially, all of the entries in the matrix are uncovered.) After the assignment is made, the row and column are covered; i.e. the elements in the assigned row and column are covered. This has the effect of reducing the order of the matrix by one. The process is repeated until all of the rows have been assigned to columns. To produce a complete set of row-to-cohunn assignments (an assignment solu- tion) for a matrix of order N, N passes over the entire matrix are required. Sec- tion 5 presents two detailed computer variants of this algorithm, including the minimum and maximum number of required operations per mechanization. 3. Empirical Error Investigations of Approximation Methods The value of each approximation method is dependent upon two factors: the time necessary for solution, given implicitly by the required number of basic operations, and the amount of error generated by the algorithm. To gain insight into the relative worth of the approximation methods, experi- ments were conducted on low order matrices (of order 12) for which the true minimal and maximal assignments could be found. The results of these empirical investigations indicated that the suboptimization method does not compete favorably with the other two approximation methods, even with respect to solution error. Since the solution-time cost is (by far) heaviest for the suboptimization method, and the resuIts so discouraging, there does not appear to be much profit in it. A total of 50 test matrices of order 12 was subdivided every possible way for the suboptimization method. The particular partition that gave the least error was a subdivision of four 6 X 6 submatrices. The average relative error with respect to a minimal solution for that partition was 47.1 percent. The greatest average relative error, 76.5 percents again with respect to a minimal solution, was generated with a partition of sixteen 3 X 3 submatrices. With respect to a APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 4~ maximal solution, the corresponding average relative errors for the same parti- tions are 6.8 and 11.1 percent, respectively. The row/column-scan method and the matrix-scan method were applied to several test matrices. The average relative errors with respect to minimal solu- tions were approximately 29 and 14 percent, respectively. The average relative errors with respect to maximal solutions were approximately 4.2 and 2.6 percent, respectively. If the worst assignment for a minimal solution (corresponding to the maximal assignment) were selected for these test matrices, the relative error would be 603 percent. Correspondingly, the worst assignment for a maximal solution (equivalent to the minimal assignment) yields a relative error of 87.8 percent. 4. Analytical Investigations of Approximation Methods In this section, an expression for the expected value of an assignment solution, in terms of the order of the matrix, is derived for two approximation methods: the row/column-scan and matrix-scan algorithms. (For appropriate definitions, see or .) Of particular interest and importance is the error involved in solving high order problems. Bounds are established for the absolute error of the expected solution values given by the row/column-scan and matrix-scan methods. In addition, bounds are found for the expected relative error of the maximal solution given by the above methods. The expected relative error is shown to vanish as the order of the matrix increases without bound. In the analysis, the elements in the matrices are inde- pendent and rectangularly distributed in [0, 1]. A generalization of results is then given for elements rectangularly distributed in [a, b], an arbitrary finite interval. 4.1. Row/Column-Scan Method. For the remainder of this section, it is con- venient to consider the maximal assignment instead of the minimal assignment. Solving the maximal assignment problem for the matrix (d,) = (1 -- c,) is equivalent to solving the minimal assignment problem for the matrix (c,). The (c,) and (d,) matrices have the same assignment solution, and summation of the assigned elements in the (c,~) matrix yields the correct minimal assignment value. For a matrix of order n, there are n rows which must be assigned. Each row has a random variable associated with it: namely xk, the value of the selected element. The value of the total assignment is simply ~~x~. The expected value for the total maximal assignment given by a row-scan method, in which only the rows and not the columns are examined, is now derived. Let xk be the maximum (uncovered) element of the (n - k + 1)th row. Then, the random variable xk is the maximum of k random variables; namely, the con- tents of the surviving locations of the (n -- ]c + 1)th row. Denote these contents by11,12, .-.,lk. Thus, xk = max (11,1~, la, -" , lk). Let F~ (x) denote the cumulative distribution function of x~. Using the symbol 424 JEROME M. I/:URTZBERG Pr for mathematical probability, Fk (x) is computed as follows: Fk(x) = Pr (x~ =< x) = Pr [max (11,12, ... , 1~) _ _ < x] = Pr (11 =< x, 12 =< x, ..., 1~ =< x) = [Pr (11 =< x)] [Pr (12 =< x)] ... [Pr (L -< z)] = [El, (x)] [Fi~ (z)] ... [Fl~(x)]. Now hence t l, for x ~ 1 Fi,(x) = x, for 0 <x < 1 (0, for x ~ 0, f l, for x >= 1 Fk(x) ----~x, for 0 <x < 1 [ 0, for x_-<0. The associated density function, fk (x), is simply the derivative of Fk (x), I 0, for x > _ _ 1 f~(x) =~}~x k-l, for 0 < x < 1 ! L0, for x <= 0. Thus, the expected value of the random variable x~ is f f (z~> = ~ xA(x) dz = z(kx ~-1) dx - k + 1" The expected value of the total assignment, denoted by M1 (n), is given by ~1 (xk}. Since the mean of a sum of random variables equals the sum of their means, MI(n) = xk = . k~1 k~1 h + i An easily evaluated lower bound for Mi (n) is Ml(n) = ~ k ~ 1 f°+ldx k=1 h ~ i -- n -- k=l ~ -~---i > n -- ,i x -- n -- In (n q- I). The inequality holds for n+l dx_ _ k+l dx > _ . ~1 X k=l ~k X k=l k k -~ 1 k=1 h + 1 The expected value of the minimal solution, ml (n), for the matrix (c,j) is now APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM ~25 obtained by means of the results for Mi (n). Let yk denote the minimum (un- covered) element of the (n - k ~ 1)th row of (c,,). Since d~ = 1 -- c,, it follows that yk = 1 -- x~. Therefore, ml(n) = yk = (1 -- xk) = n-- - . k~l k-1 k=l k ~ 1 k=l k + 1 An upper bound for m~ (n) is In (n + 1), for, as seen before, f'+l d_~ > ~ l i X k=ik JF 1 " The expression Mi gives the expected value for only the maximum row-scan solution. In the row/colunm-scan method there are two tentative assignment solutions, one for the rows and one for the columns. The better of the two is selected. The expression for the expected value of the maximum column-scan solution is also given by Mi. Since the better (in this case, larger) of the two solutions is selected, the expected value for the complete method should be greater than M~. However, that expected value is not immediately apparent, as the distributions for the row-scaa and the column-scan assignments are not independent. Similarly, ml provides an upper bound for the expected minimal solution given by the row/column-scan method. Thus, the expected minimum row/column-scan solution is less than In (n W 1), and the expected maximum row/column-scan solution is greater than n -- In (n Jr 1). A bound is now found for the expected value of the relative error given by the maximal row-scan method. This value tends to zero as the order of the matrix increases. Let ~k=~ tk denote the true solution value. Then, the expected relative error is / >=// \ / k=i --k-i \ 1 ~xk \ xk \ Mi(n) In (n + 1) --< ~\1 ~in // = 1 -- n < n ' which goes to zero as n approaches infinity. This follows from the relations n >= ~] t~ => xk, k=l k~l (which hold because of all entries lying between 0 and 1, and by the definition of true maximal solution), and from the relations /-) k=llc ~ 1 > n--ln (n+ 1). 426 JEROME M. KURTZBERG Furthermore, the same bound of [In (n -t- 1)]/n holds for the relative error of the expected value of the row-scan method. This is so because both An upper bound for the absolute error for both the expected minimal and maximal solution values is now found. Let e (n) and E (n) denote the expected value of the true minimal and maximal solutions, respectively. Since E (n) = n - e (n) and M1 (n) -- n - ml (n), the absolute error bound is ]e(n) -- ml(n) ] = ]E(n) -- Ms(n) ] -< In - [n -- In (n % 1)] I = In (n % 1). Again, the fact is stressed that this upper bound for absolute error refers to only a row-scan or a column-scan solution. Thus, there may be a significantly sharper bound for the absolute error of the expected solution given by a com- plete row/cohimn-scan method. 4.2. Matrix-Scan Method. As with the case of the row/cohimn-scan method, it is convenient to consider the maximal assignment solution instead of the minimal assignment method. In the row/scan method, the column associated with the largest uncovered element of a row is covered; the members of other rows play no part in the choice of the largest element. In the matrix-scan method, however, the row and column of the largest uncovered element of the matrix are covered. Hence, the remaining elements, which are the candidates for a new choice, cannot be larger than the selected one, and thus are no longer rectangularly distributed in [0, 1]. Let the expected value given by the matrix-scan method be denoted by M2. Suppose the maximum value in the matrix is x, where 0 ~ x -< 1. The remaining elements, after that maximum element is selected, are then independently and rectangularly distributed in [0, x]. Therefore, the expected value of the matrix- scan method on an (n -- 1) X (n -- 1) minor (i.e. the resulting uncovered ele- ments) is xMs (n - 1). This relation follows by considering that if every element of the (n -- 1) X (n -- 1) minor is divided by x, an (n -- 1) X (n - 1) matrix of elements in [0, 1] is obtained. Given that the maximum value is x, then [x -t- xM2 (n -- 1)] is the expected value of the matrix-scan; the probability 2 n2 1 that x is the maximum is n x - dx. So, taking the product of the expected value (assuming x is the maximum) and the associated probability, and summing x over all possible values (i.e. integrating from 0 to 1), leads to the expression for the expected value of the matrix-scan method: f0 1 - n s Ms(n) = Ix + xMs(n -- 1)]n2x ~-1 dx n2 +-----~ [1 + M2(n -- 1)]. A lower bound for M2 (n) is now found, using the recursion relation n 2 1 Ms(n) - -- [M:(n -- 1) --]- 1] - - - [Us(n -- 1) + 1]. 1 n s -t- 1 1 ~n~ APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 427 Since - - or 1 1(1) -~-~+ ~ .... ,then 1] [M2(n- 1) + 1] M2(n) > 1 - M2(n) -- M2(n -- 1) > 1 and since M2(n - 1) =< n - 1, M2(n) -M2(n- 1) > 1 M2(n- 1) 1 n 2 n 2 ' n--1 1 1 1 -- --. •2 n 2 n Taking the sum of both sides, with n going from 1 to T, gives the result ,=1 [M2(n)- M2(n- 1)] > ~.=i (1- ~), which is equal to Ms(T) -- M2(0) > T -- ~ 1. n=l n Now, Ms(O) = O, and since I .y > ~r~=1 I/n -- In T, changing the dummy variable (substituting n for T) results in the following lower bound for M= (n) M2(n) > n-,r-lnn. An upper bound for Ms (n) is now found by taking one more term in the ex- pansion of 1/[1 + (l/n2)] and substituting in the reeursion expression for M2 (n), as follows: [ 1 l][M=(n-1)+1] M2(n) < 1--~-~ I1 11 = 1 + M2(n-- 1) -- ~--~ [Ms(n-- 1) + 1]. Thus, Ms(n) --Ms(n -- 1) < 1 -- 1 Euler's constant, % is defined proaches ~ from below. 1 1 M2(n -- 1) Ms(n-- 1) ~+n~ n s + n~ (n-- 1) -- 7--1n(n-- 1).{_n-- 1 n 4 1 1 < 1 - ~ n4 ns <l_lq_7 lnn 1 ~ +-~- + n-~" asv= Lim [ ~ 1 ] E1 where ~ -- In n ap- k=l 428 JEROME M. KURTZBERG Taking the sum of both sides with n going from 1 to T gives the result ,~=~ [Ms(n)- M2(n- 1)] < ~.=1 [1- ~,l+7~+"~-+~inn 1] or M2(T) < T~ 1 ~ 1 ~lnn ~ 1 .=1~ +7 ~+ -~--+ ~. Thus, it follows that 1 " 1 " 1 In 1 k~l ~ ~ k=l k=l Furthermore, since Ml(n) = n -- - n -- + 1 - - > n -- the bound becomes 1 V M,(n) <Ml(n)+7~+~ .+~ 1 k~l k~l k=l ~" Thus, M~ (n) -- M1 (n) increases with n, but since 7~=I (l/k2), ~=i (1/k~), and ~=1 (ln lc/k 2) are convergent (i.e. bounded by a constant), Ms(n) -- Ml(n) < c, where c is the constant given by the sum of the limits of ~(1/n3), 7~(1/n2), and ~ In k/k 2. This sum may be crudely estimated by noting that 2 .=I~ < .=in ~- 6 and Thus, f ~lnx l+lnn ~lnk dx = 1 -- > k2 x ~ n k=l 2 2 7r 7r c < 7 ~-]- ~ + 1 < 3.6. Therefore, the expected values of the matrix-scan and the row-scan methods are of the same order of magnitude. Furthermore, since Ms (n) > n - ~=1 1/k > n -- (lnn + 7), it follows, by the same reasoning used for the row-scan method, that the bound on the ex- pected relative error (and the relative error of the expected value) in the matrix- scan method is (lnn + 7)/n, which goes to 0 as n approaches infinity. Similarly, the bound on the absolute error of M2 (n) and m~ (n) is Inn + 7. 4.3. Generalization from [0, 1] to [a, b]. The previous analysis assumes all matrix elements to be rectangularly distributed in [0, 1]. The results are readily APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 429 extended to matrices with elements rectangularly distributed in [a, b], an arbi- trary finite interval. By employing the linear transformation e,t -- a where e,j is in [a, b], c,: - b -- a ' and going through the appropriate manipulations, the following error bounds are obtained. The expected relative error of the maximal solution with the row-scan method is bounded by [(b - a) In (n + 1)]/nb, and the expected absolute error is bounded by (b - a) In (n + 1). Similarly, the expected relative error for the matrix-scan method is less than [ (b -- a) (In n + ~) ]/nb, and the expected abso- lute error is less than (b -- a) (lnn + ,y). 5. Solution-Time Evaluation of Approximate Methods For the two approximation methods that appear promising, there are two variant mechanizations. Flow charts are presented for the row-scan method and the matrix-scan method showing these candidate mechanizations. Certain "basic operations" are defined, and the rival mechanizations are then analyzed for each method in terms of these operations. Timing estimates are made for the methods with respect to a given computer, UNIVAC I. 5.1. Row~Column-Scan Mechanizations. The procedures or steps which define the row/column-scan method for a minimal solution are: (1) Select, in each row, the minimum element whose column has not been assigned to any other (previous) row. (2) Assign the row to the column of the selected element. (3) Indicate that the column has been assigned. (Cover the column.) (4) After all rows have been processed, repeat the algorithm, treating the transpose of the matrix in similar fashion (5) Of the two tentative assignments--the first from the original matrix, the second from the transpose--choose the one that is smaller. The two variants differ in treatment of step (1). As the names of the mechani- zations suggest, one first tests for a prospective new "smallest" row element and then determines its availability, i.e. whether or not its column has been previ- ously assigned; in the other mechanization, the order of tests is reversed. The first is more favorable initially; the second shows to advantage in the terminal phase. However, in general it would be unwise to include controls to switch from one to another at the "breakpoint." The controls would cost more in extra opera- tions than would be saved. The flow charts for the element-indicator and indicator-element mechaniza- tions are shown in Figure 1. 5.2. Matrix-Scan Mechanization. The procedure or steps which define the matrix-scan method for a minimal solution are: (1) Select, in the matrix, the minimum element that does not have its row or column covered (i.e, whose row or column has not been previously assigned). (2) Assign the row of the selected element to its column. 430 JEROME M. KURTZBERG F "~-N~ATON--I~LEMI[NT-- clear F l and F 2 and set j = I let rain El : o O ~ s column k ¢overed')~ I o%:'~=': ...... .0 I ....... ,k, m,°." " I I ............ I ' ¢° ' column q 1 ,I, mxn l n mo r w (Set column q Indicator) Element lk--mln El r r o osed (Is i = 2 9) ~ with value F 2 o ~ilr~o~,, co,, Tra~pose matrix by interchanging cow and column lndlce$, 1 e , ('-1-i .......... ,,° transpose from tape without botherzng to znterchange row and column radices FIG 1. Row/column-scan method ELEMENT-INOICATOR MECHANIZATION I I b ° I I I ~° I (3) Indicate that the column and row are assigned. (Cover the row and column.) (4) Test for completion of n assignments (for a matrix of order n); if not completed, continue recursively. As with the row/column-scan method, the two variants differ in step (1) ; the discussion given for that method is also applicable here. The flow charts for the element-indicator and indicator-element mechaniza- tions are shown in Figure 2. 5.3. Basic Computer Operations for the Approximation Methods. To evaluate the computation speed of the approximation methods (and their candidate computer mechanizations), it is expedient to define certain operations, especially significant for the approximation methods. The techniques may then be analyzed in terms of these operations, thereby providing a basis for comparisons. The operations are as follows: (1) Element comparison (2) Indicator (sentinel) test (3) Row-to-column assignment (4) Candidate element replacement (5) End-of-element-in-row/column test (6) End-of-row/column test (7) End-of-assignments test (8) Matrix transposition (or interchange of indices) (9) Solution-value comparison APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 431 Clear row and colunm indicator9 Set j = 1 and clear F set mm El = aD ® • Any more rows~ ~o I ........... I column q (calculalzon of Isslgnment (Set row p and column q value) mdlcators) Yes •i re all the row (or colamn]~,.~.~L~ indicators setV Yes (Are the n assignments °mpleled ~N o I --- "~'NOIC ATOR - E LEM EN T - MECHANIZATION I y.I Is col~nn k covered? ~ s Any more e l e m e n t s ~ Y e s (Is column k Dldlcstor on?) in row j~ I I t ' I .o j k~q ) Element j k ~ mln E1 i" ElEMENT-INDICATOR MECHAN|ZATION I I ~o I I I ...... . ........ +) ] colum, k ~.dlc.tor, ] I I L I. + J FiG. 2. Matrix-scan method It is considered that some necessary operations not listed, such as advancing indices, clearing counters, resetting indices, presetting values, etc., are absorbed in the above operations. For example, summation of the selected elements to form the solution value is considered to be implicitly contained in the row-to-column assignment operation. Since the main purpose of the operations is to furnish a basis for comparison, no harm results from this artifice. 5.4. Row~Column-Scan Operation Count. For the minimum operation case (the case requiring the lowest total of equi-weighted operations), consider that the elements of the matrix are monotonically increasing across each row and the last element of the jth row is equal to or greater than the last element of the (j -4- 1)th row. Another minimum case is given by a matrix which has its main diagonal monotonically increasing, with the remaining set of elements mono- tonically increasing with respect to themselves. For the maximum operation case, consider that the elements of the matrix are monotonically strictly decreasing across each row, and that the last element of the jth row is greater than the first element of the (j -4- 1)th row. The estimate of the "average" number of operations is given by the average of the minimum operation and maximum operation cases, and is taken as a figure-of-merit for the techniques. A summary of the operation counts for the variant mechanizations is given later (see Table 5). 432 JEROME M. KURTZBERG TABLE 1. OPERATION COUNT ASSOCIATED WITH Row-ScAN METHODs ELEMENT-INDICATOR MECHANIZATION Asstgr~ent Element Colu~en-Indicator Row-Col. Number Comparison As|lgment t.tLn.Cale Hax.Cale l n 1 n I 2 n 2 n l 3 n 3 n 1 n n n n TOTAL n 2 k~lk. n 2 El~ent Replacement grid of Elements End of Rows M.tn.Cale Va.x, Can i 1 n I (aol) t 0-2) l L in l~w Test: TeSt n 1 n l n 1 n 1 2 n n For the element-indicator mechanization, figures for the minimum and maxi- mum number of operations for one sweep (row-scan) through the matrix, show- ing operation counts of the individuM assignments, are listed in Table 1. The operation count for one sweep (row-scan) through the matrix is, for the minimum count case, n2+~k+n+n+~+n _ 5n2 + 7n k=l 2 and for the maximum count case, n2+~+n+~k+n2+n _ 7n2 + 5n k=1 2 For a complete operation count for a row/column-scan, the following addi- tional operations must be taken into account: end-of-assignment test; matrix transposition (interchange of indices) ; and solution-vMue comparison. The test for end-of-assignment in the row/column-scan method consists in determining if both the row and column scans have been completed. This opera- tion must be done twice, once per sweep. The other two additional operations must be done once. Thus, the complete operation count is, for the minimum count case, 2 + 1 + 2 = 5n 2 + 7n+4 and for the maximum count case, 2 + 1 + 2 = 7n 2+ 5n+ 4. Therefore, the average of the minimum and maximum cases, or the figure-of- merit for the element-indicator mechanization is (5n 2+7n+4) + (7n ~ + 5n+4) = 6n 2 + 6n+4. 2 TABLE 2 i Ass£gmnent I Number APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 433 OPERATION COUNT ASSOCIATED WITH Row-ScAN METHOD, INDICATOR-ELEMENT MECHANIZATION Element Column-Indlcator Row-Col. Element Replacement Knd of Elemants Comparison Assignment in Row Test ~n Case Max Case Urn. Case I Max.Case n n n 1 1 u (n-l) n n l I (n-I) (n-2) n i n I I ' (n-2) n I n n i I I °' , ° End of Paws Test n 1 n i n 1 n l n 2 n For the indicator-element mechanization, the operation count (shown in Table 2) for one sweep (row-scan) through the matrix is, in the minimum count case, 5n ~ + 7n k +n2 +n+n+n2 +n-- k=l 2 and, in the maximum count case, • k+n 2 +n+ ~k +n 2 + n = 3n ~+ 3n. k=l k~l For a complete operation count for a row/column-scan, one must, as in the element-indicator technique, take into account additional operations. The same remarks made about these additionM operations also hold here. Thus, the complete operation count is, in the minimum count case, 2( 5n2 + 7n ) 2 + 1 +2 = 5n 2+ 7n+ 4, and in the maximum count case, 2(3n 2+ 3n+ 1) + 2 = 6n 2+ 6n +4. Therefore, the figure-of-merit for the indicator-element mechanization is (5n 2 + 7n+4) + (6n 2 + 6n +4) lln 2 + 13n+ 8 2 2 5.5. Matrix-Scan Operation Count. The minimum and maximum cases for the matrix scan method are the same as for the row/column-scan method. For the element-indicator mechanization, the complete operation count (shown in Table 3) is, in the minimum count case, n~ + ~kn + ~k + n +n + ~kn + n2 + n = 2n3 + Tn2 + 7n k=l k=l k=l 434 JEROME M. KURTZBERG and in the maximum count case, lln 3 + 24n 2 + 13n + + + xk + e k=l k~l k=l k~l 6 and the figure-of-merit is 1 V2n3 + ~ 7n2+27n + llnJ+ 24n2+613nl.~ = 17n3+ 45n212 + 34n For the indicator-element mechanization, the complete operation count (shown in Table 4) is, for the minimum count case, + ~k ~ + ~k~ +n +~ + ~k~ +~ +~ = 8n 3 + 21n ~ + 19n k~l k=l k=l and for the maximum count case, n~+ ~k' + ~kn+n+ ~k~+ ~kn+n2+n= k~l k=l k~l k~l and the figure-of-merit is _118 n3 + 21n 2+ 2 L 0 5n 3 + 12n 2 + 7n 3 19n + 5n 3 + 12n 2 + 7n-] = 6n 3 + 15n 2 + lln 3 J 4 TABLE 3 OPERATION COUNT ASSOCIATED WITH MATRIX-ScAN METHOD, ELEMENT- INDICATOR MECHANIZATION XsSlg~enc Row Element Col~°Indlcator Row-Col El~ent Replacement End of El~ente EndTOftRove End of I N~ber Indicator Comparison Astig~ent i~ Row Telt el Al$ig~tnt mn Case P~x Cala mn Celt Hax Care J I n n 2 l n 2 l l n 2 n 2 n l 2 n n2-n 2 n2-n l 1 (n-l) 2 n2-n u l 3 n n2-2n 3 n2-Zn l l (n'2) 2 'a2-2n n l n n n n n l I l t~ n I TOTAL n 2 k n n kn ~ n ~ k 2 ~ kn 2 n k~ k k k=l kml TABLE 4. OPERATION COUNT ASSOCIATED WITH MATRIx-ScAN METHOD, INDICATOR- ELEMENT MECHANIZATION ~lelg~en¢ gov I Element Co[ucm-lndicator Row-Col EleJaen¢ Raplac~ent End of EleJDente Nucaber Indicator, Compariaon ~l/gr~ent in Roy Teat Pin CaJe ~ tale ~n Cale ~ H~J¢ Case I n 2 n 3 n n fl TOTAL n 2 n 2 (n-l) 2 (n-2) 2 n 2 n 2 ! I I n 2 n 2 n2n n2-n ! I 1 (n-i) ~ n2-n n2-2n nZ-2n I I l (n-2) ~ n2-2n 1 n k 2 k-1 n n i l l L n ! k-I k-I k-I k-I End of Rowl End of Test AIsign~enc n l n 1 n I n n 2 APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 435 5.6. Time Comparisons of the Approximation Melhods (with respect to UNIVAC I). The computer computation times for the approximation methods are now presented as a function of the order of the assignment matrix. For the time com- parisons, the approximation methods were programmed in the UNIVAC I code and time estimates were then made based upon the coding. There are a number of iump discontinuities in the computation time of the approximation methods as the order of the problem increases. The input to the UNZVAC I computer must be in blocks of 60 words; thus, the input-time penalty increases according to the smallest integer equal to or greater than n~/60. There is a critical point where the size of a problem is such that the entire matrix cannot TABLE 5. SUMMARY OF OPERATION COUNT Mlmmum I Maximum Method Mechanization Flgure-of-Merlt Operatzon Case I Operatmn Case Row-Scan Row-Scan Row/Column-Scan Row/Column-Scan Matrlx-Scan Matt ux -Scan Element -Indicator Indicator-Element Element -Indicator Indicator-Element Elernent -Indicator Indicator-Element (5n = + 7n)/2 (5n = + 7n)/2 5n = + 7n -b 4 5n~4 - 7n+ 4 (2n = + 7n = + 7n}/2 (8n a + 21n = + 19n)/6 (7n = + 5n)/2 3n a + 3n 7n=+ 5n + 4 6na+ 6n+ 4 (lln 3 + 24n = + 13n)]6 (5n ~ + 12n = + 7n)/3 3n ~ + 3n (lln ~ + 13n}/4 5n = ÷ 6n + 4 (lln 2 + 13n+ 8)/2 (17n 3 + 45n ~ + 34n)]12 (6n a +ISn = + lln)/4 Recommended Techmque TABLE 6. UNIVAC I TIMING FOR OPERATIONS, ROW/COLUMN-SCAN METHOD, INDICATOR- ELEMENT MECHANIZATION Operation Element Comparison Column Indlcator Row-Column Asszgnment Frequency (Average) n ~ + n Time (mzlhseconds} per Unit Operation Total Time (milhseconds) 2n ~ ÷ 2n 2n a 2 4n ~ 2n 31 62n Element Replacement (n ~ + 3n)/2 2 n a + 3n End of Elements m Row 2n ~ 2 4n ~ End of Rows 2n 5 10n End of Assignment (Was Matrix Transposed?) 2 2 4 Transpose Matrix (Interchange Indices) 1 6 6 Soluhon-Value Comparmon 1 3 3 Read and Write (3 blocks) 1 510 510 Initiahze 1 8 8 Total: lln ~+ 77n+ 531 436 JEROME M. KURTZBERG be contained simultaneously in the memory. If this critical size is exceeded, the time requirements of the matrix-scan method increase to another order of magnitude. To avoid these problems, and to provide a simpler direct time comparison with the UNIVAC I program of an exact method (the Munkres algorithm) it is assumed that three UNIVAC blocks, or 180 words, are sufficient for input to the programs. Of course, for larger matrices, the input-time penalty could be ad- justed accordingly. For a matrix of order n, and with partitioning into pXp submatrices, the sub- optimization method requires the exact solution of q2 pXp matrices, and one q×q matrix. A weak lower bound on time required for each size matrix may be found by reference to the time equation of the Munkres algorithm in Section 1. For matrices in the range (2, 12), the time chart in Section 1 furnishes the (average) time actually consumed. An additional amount of time is consumed in the partitioning and forming of the q X q matrix. However, this time is negli- gible in comparison with the time consumed in solving the submatrices. Table 6 lists the basic operations (Section 5.3) with the associated frequency for the optimal mechanization of the row/column-scan method (indicator- element mechanization) and the time (in milliseconds) for UNIVAC I coding for each operation. The time for "initialization" (setting up of certain controls) is also given, along with the time for tape reading and tape writing (three blocks). As was previously mentioned, certain operations are absorbed in the basic opera- TABLE 7. UNIVAC I TIMING FOR OPERATIONS, MATRIX-SCAN METI:IOD, ELEMENT- INDICATOR MECHANIZATION Operation Frequency Time (milliseconds) Total Time (Average) per Unit Operation (milliseconds) Row Indicator n ~ 2 2n ~ Element Comparison (n a + n ~)/2 2 n 3 + n ~ Column Indmator (n 3 + 2n ~ + n)/4 2 na/2 + n ~ + n/2 Row Column Assignment n 35 35n Element Replacement (2n s + 3n~+ 7n)]12 2 nS/3 + n~/2 + 7n/6 End of Elements in Row (n a + na)/2 2 n s + n ~ End of Rows n a 6 6n a End of Assignment n 2 2n Read and Write (3 blocks) 1 510 510 Initlahze i 7 7 17n • 23n ~ Total: ~ + ~ + 232 n+ 517 APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 437 tions; summation of solution value, for example, is included in the row-column assignment operation. The expression for the time (average case), converted to give time in seconds, is for the row/column-scan method, tl = 0.011n 2 + 0.077n --b 0.531. For a 12 X 12 matrix, tl ~ 3 seconds. For the row-scan method, which requires approximately one-half the running time of the row/column-scan method, t2 ~ 1.8 seconds. For the same size matrix, an exact method (the Munkres assignment program) took, on the average, 19.2 seconds. The corresponding time equation for the row-scan method is t~ = 0.0055n 2 + 0.0385n --k 0.518. Table 7 lists the basic operations and pertinent data for the matrix-scan method, element-indicator mechanization. As with the row/column-scan method, the time for "initialization" and tape reading and writing (three blocks) is also given. The expression for the time (average case), converted to time in seconds, is t~ = 0.00283n 3 -b 0.0115n ~ -t- 0.232n "-t- 0.517. For a 12 X 12 matrix, t3 ~ 9.8 seconds, a saving of 9.4 seconds over the Munkres assignment program. 6. Summary of Findings Of the three approximation methods defined and examined in this paper, only two appear to be worthy candidates for solving assignment problems. These two are the matrix-scan method and the row/column-scan method. Of these, the row/column-scan method is recommended. However, it is possible that the main error concern might be with the worst possible solution, rather than with the expected solution. In that event, it is conjectured that the matrix-scan method would prove most satisfactory. The expected value of the matrix-scan and the row-scan methods is of the same order of magnitude. The difference is bounded by a constant less than 3.6. The upper bound on the absolute error of the expected value of the row-scan, or the column-scan, solution is In (n -b 1) ; for a maximal solution, the bound on the expected relative error, and on the relative error of the expected solution, is [ln (n --b 1)]/n. The corresponding bounds for the matrix-scan method for abso- lute error and relative error are (lnn "k "r) and (ln n "k 7)/n, respectively. As yet, no expression has been found for the error generated by the combined-row/ column-scan method; a weak upper bound is the upper bound for the row-scan method. As regards solution time, the row/column-scan method is definitely superior; the matrix-scan method requires much more computation time. The operation- count difference of the two methods (the matrix-scan with the element-indicator mechanization and the row/column-scan with the indicator-element mechaniza- 438 JEROME M. KURTZBERG tion), representing the operations saved (for the average of the most favorable and unfavorable cases) is 17n 3 + 45n 2 "t- 34n lln 2 W 13n -}- 8 17n 3 - 21n ~ - 44n - 48 12 2 12 Furthermore, if the matrix is so large that it cannot be contained simultane- ously in the computer, it is necessary, where the matrix-scan method is used, to read and reread into the computer segments of the matrix many times to obtain the solution. This necessity drastically increases the computation time. The row/ column-scan method does not require the entire matrix to be held simultaneously in the computer. Thus, larger matrices may be solved than with the matrix-scan method, without causing an order-of-magnitude jump in the computation time. If the time restriction is extremely critical, the row-scan method is suggested. With the indicator-element mechanization, the number of operations, for the average of the best and worst cases, for a problem of order n, is (11n 2 -t- 13n)/4. Of course, the ultimate in a rapid rough-and-dirty approximation method would be to select, for each assignment problem, a fixed set of elements (for ex- ample, the main diagonal). The only two operations that are required for each individual selection are: row-to-column assignments and end-of-assignment test. Thus, only 2n operations are required with this method for an nXn matrix. The expected solution value for this rough-and-dirty method, assuming that the ele- ments in the matrix are rectangularly distributed between 0 and 1, is n/2. Thus, the bound on the expected absolute error of the solution is n/2, and the bound on the expected relative error (for a maximal solution) is 1/2. ACKNOWLEDGMENTS This paper is based upon the author's thesis. For his thesis, the author wishes to thank his advisor, Dr. Benjamin H. Stevens, for his valuable direction, gener- osity of time, and constant encouragement; his second reader, Dr. Saul Gorn, for comments; and his third reader, Dr. Murray Gerstenhaber, for a suggestion of a simpler proof than the author had. The author is greatly indebted to Dr. Louis Brickman and Dr. Leon Steinberg for their many helpful comments, suggestions of mathematical approaches, and constructive advice. REFERENCES 1. CHURCHMAN; ACKOFF; A N D ARNOFF. Introduction to Operatwns Research. John Wiley, New York, 1957. Ch. 12, pp. 343-368. 2. CRAM~R, H. Mathematical Methods of Statistics. Princeton University Press, 1951. 3. DANTZIO, G. The dual simplex algorithm. RAND Report RM-1270, RAND Corp., Santa Monica, Calif., 1954. 4. FELLER, W. Probability Theory and Its Applzcat~ons. John Wiley, New York, 1950. 5. FORD, L.; A N D FULKERSON, D. Solving the transportation problem. RAND Report RM-1736, RAND Corp., Santa Monica, Calif., 1956. 6. GAss, S. Linear Programming. McGraw-Hill, New York, 1958. APPROXIMATION METHODS FOR THE ASSIGNMENT PROBLEM 439 7. GERSTENHABER, ~V~. A solution method for the transportation problem. J. SIAM 6 (1958), 321-334. 8. KUHN, H.W. Hungarian method for the assignment problem. Nay. Res. Logist. Quart. 2 (1955), 83-97. 9. KUHN, H. W. Variants of Hungarian method for assignment problems. Nay. Res. Log~st. Quart. 3 (1956), 253-258. 10. MOTZKIN, T.S. The assignment problem. Proc. 6th Symp. Appl. Math. VI, pp. 109- 125, McGraw-Hill, New York, 1956. 11. MUNKRES, J. Algorithms for the assignment and transportation problems. J. SlAM 5 (1957), 32-38. 12. YON NEUMANN, J. A certain-zero-sum two-person game equivalent to the optimal assignment problem. In H. Kuhn and A. Tucker (eds.), Contribution to the Theory of Games II (Ann. Math. Study No. 28), pp. 5--12, Princeton University Press, 1953. |
619 | https://math.stackexchange.com/questions/194809/simplify-lnx-x2-lnx-1 | logarithms - Simplify $\ln(|x-x^2|) - \ln(|x-1|)$ - Mathematics Stack Exchange
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Simplify ln(|x−x 2|)−ln(|x−1|)ln(|x−x 2|)−ln(|x−1|)
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As the topic says, I need to simplify:
ln|x−x 2|−ln|x−1|ln|x−x 2|−ln|x−1|
I don't know how to approach the problem at all. I'm not asking for the answer, but something to maybe get me going.
logarithms
absolute-value
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edited Sep 12, 2012 at 17:43
Fly by Night
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Remember that logarithms have some rules associated with them that help you simplify problems. The most common are: ln(a b)=ln a+ln b ln(a b)=lna+lnb and ln(a/b)=ln a−ln b ln(a/b)=lna−lnb. The second one is the key to your problem. If you have the difference of logs, say ln a−ln b lna−lnb then you can simplify it to be the log of a quotient: ln(a/b)ln(a/b). If you have a quotient of polynomials, then you should be trying to factorise and eliminate common factors. For example:
x 2−x x−1=x(x−1)x−1=?x 2−x x−1=x(x−1)x−1=?
You want to simplify ln|x 2−x|−ln|x−1|,ln|x 2−x|−ln|x−1|, so apply the two steps that I suggest.
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answered Sep 12, 2012 at 17:36
Fly by NightFly by Night
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So I act like there is no absolute value expression here? How come that is possible, if so? I think it's the absolute value expression that is making this hard for me. For example, ln(|x(x−1)||x−1|)ln(|x(x−1)||x−1|) - it's not possible to cancel the x−1 x−1 out in this case, or am I wrong? Edit: Yes, you're (I'm) wrong. |a b|=|a||b||a b|=|a||b| ...matfor –matfor 2012-09-12 17:59:57 +00:00 Commented Sep 12, 2012 at 17:59
Too slow to edit my last comment. Adding another one (against SE policy?). Since logarithm functions only accepts positive numbers, we don't have to set up cases (when positive, when negative) and solve for each?matfor –matfor 2012-09-12 18:08:43 +00:00 Commented Sep 12, 2012 at 18:08
Strictly speaking, logarithms can accept positive, negative and even complex values. My arguments a a and b b are arbitrary. Just because I didn't include a modulus doesn't mean you can't. Replace a a by |x||x| and b b by |y||y| if it makes you happy. You can cancel: |x 2−x||x−1|=|x(x−1)||x−1|=|x|⋅|x−1||x−1|=|x|.|x 2−x||x−1|=|x(x−1)||x−1|=|x|⋅|x−1||x−1|=|x|. Fly by Night –Fly by Night 2012-09-12 18:30:50 +00:00 Commented Sep 12, 2012 at 18:30
@matfor: Consider a complex number z∈C z∈C. We can rewrite z=R e i(θ+2 π k)z=R e i(θ+2 π k). Then ln z=ln(R e i(θ+2 π k))=ln R+ln(e i(θ+2 π k)).lnz=ln(R e i(θ+2 π k))=lnR+ln(e i(θ+2 π k)). It follows that ln z=ln|z|+i Arg z lnz=ln|z|+i Arg z where Arg z={θ+2 π k:k∈Z}.Arg z={θ+2 π k:k∈Z}. So we see that ln ln is a multivalued function: en.wikipedia.org/wiki/Complex_logarithmFly by Night –Fly by Night 2012-09-12 18:38:02 +00:00 Commented Sep 12, 2012 at 18:38
You don't have to divide: log(|x 2−x|)=log(|x||x−1|)=log(|x|)+log(|x−1|)=log(|x|)+log(|1−x|)log(|x 2−x|)=log(|x||x−1|)=log(|x|)+log(|x−1|)=log(|x|)+log(|1−x|).André Nicolas –André Nicolas 2012-09-12 20:46:29 +00:00 Commented Sep 12, 2012 at 20:46
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Hint: can you factor x−x 2 x−x 2 then use the laws of logs?
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answered Sep 12, 2012 at 17:29
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The Preindustrial City by Gideon Sjoberg
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The Preindustrial City
Gideon Sjoberg
The American Journal of Sociology
, Vol. 60, No. 5, World Urbanism. (Mar., 1955), pp. 438-445.
Stable URL:
The American Journal of Sociology
is currently published by The University of Chicago Press.Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use.Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission.The JSTOR Archive is a trusted digital repository providing for long-term preservation and access to leading academic journals and scholarly literature from around the world. The Archive is supported by libraries, scholarly societies, publishers,and foundations. It is an initiative of JSTOR, a not-for-profit organization with a mission to help the scholarly community take advantage of advances in technology. For more information regarding JSTOR, please contact support@jstor.org. Wed Nov 21 11:44:43 2007
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THE PREINDUSTRIAL CITY
GIDEON JOBERG
ABSTRACT In th e preindus trial cities o f medieval E urop e and of oth er part s o f t he world certain elements (e.g., economic, clas s, and family systems) ar e f ound w hich are comm on to all urb an com munities. Bu t their for m in th e preindustrial city dif fe rs markedly fr om th at in th e industrial cit y. T he diff er enc e can be attribu ted prim arily to industrialization its elf.
In the past few decades social scientists have bee n conducting field stud ies in a num- ber o f rela tively non-Westernized cities. Th ei r recently acquired knowledge o f N ort h Afri ca and various p ar ts o f Asia, combined with what was already learned, clearly indi- cates that these cities are not like typical cities o f th e United St ates and ot her highly industrialized areas but are much more like those o f medieval Eur ope. Such communi- ties ar e termed herein preindu strial, for they have arisen without stimulus from th at form of ~roduction which we associate with the European industria l revol uti on. Recently Foster, in a most informative article, took cognizan ce o f th e preindustrial city.l His primary emphasis was upon the pea sant ry (whi ch he cal ls fol k ); bu t he recognized this to be pa rt o f a broader soc ial structure which includes the
reind dust rial
city. He noted certain similarities between the peasantry and the city's lower class. Likewise the present author sought to ana- lyze the to ta l society o f wh ic h th e peasantr y and the preindustrial city are integral par ts.2 Fo r want o f a be tt er term t his was called feuda l. Like Redfield's folk (or primi- tive ) society, th e feudal order is highl y stable and sacred; in contrast, however, it has a complex social organiz ation. I t is char- acterized by highl y deve lope d st at e and edu- cational and/or rel igio us instituti ons an d by a rigid class structure. Thus far no one has analyzed the prein- dustrial city per se, especially as it differs
George
M.
Foster, "What I s Folk Cult ure?"
American Anthropologist LV
(1953), 159-73.
Gideon Sjoberg, "Folk and 'Feudal' Societies,'
American Journal ofSociology LVIII
(1952),231-39. from th e indust rial-u rban community, a l- though Weber, Tijnnies, and a few others perceived differences between the two. Yet such a survey is needed for the understand- ing o f urba n d evelopment in so-called under- deve lope d c ount rie s and, for th at matter, in pa rt s o f E urop e. Such is the goal o f this paper. The typological analysis should also serve as a guide to future research.
ECOLOGICAL ORGANIZATION
Preindustrial cities depend for their ex- istence upon food and raw materials ob- tained from withou t; for this rea son they are marketing centers. And they serve as cen- ters for handicraf t manufactu ring. I n addi- tion, they fulfil important political, reli-gio us, and educationa l functi ons. Some ci ties have become specialized; for example, Be- nares in India and Karbala in Iraq are best known as religious communities, and Pei- ping in China as a locus for political and educational activit ies. Th e proportion o f urbanites relative to the pea sant population is small, in some s o- cieties about 10 per cent, even though a few preindustrial cities have attained popula- tion s o f 100,00 0 or more. Growth ha s been by slo w accr etion. The se character istics are due to the nonindust rial nat ure o f the tota l social order. T he am ou nt o f surplus food available to support an urban population has been limited by the unmechanized agri- culture, transportation facilities utilizing primarily human or animal power, and in- effi cien t met hod s o f food preserva tion an d storage. T he internal arrangement o f the prei n- dustrial city,
n
the nat ure o f the case,
s
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439
HE
PREINDUSTRI L CITY
closely related to the city's economic and s o c i a l str~ct u re.~ os t streets are mere pas- sage ways for people an d for anim als us ed in transport. Buildings are low and crowded together. The congested conditions, com-bined with limited scientific knowledge, have fostered serious sanitation ~roblems. Mo re si gn sc an t is the rigi d sodia l segr e- gation whic h typicall y h as le d t o the forma- tion o f quarters or ward s. I n some cities (e.g., Fez, Morocco, and Aleppo, Syria) these were seal ed o f f f rom e ach oth er by walls, whose gates were locked at night. The auarters reflect the sham local social divi sions. T hu s ethnic grou ps li ve in special sections. And the occupational groupings, some bei ng a t the same time ethnic in char- acter, typically reside apart from one an- other. Often a special street or sector of the city is occupie d alm ost exclusivel y b y m em- bers of a
articular
trade: cities in such di- vergent cultures as medieval Europe and modern Afghanistan contain streets with nam es like street o f th e goldsmiths. Lowe r-cl as s and es~e cia llv out cast e groups live on the city's periphery, at a dis- tance from the prim ary center s o f activity. Social segregation, the limited transporta- tion facilities, the modicum of residential mobility, and the cramped living quarters have encouraged the development of well- defi ned n eighbor hoods which are alm ost pri- ma ry grou ps. Despite rigid segregation the evidence suggests no real specialization of land use
Sociologists have devoted almost no attention to the ecology
o f
preindustrial centers. However, works of othe r social scientist s do provide some valu- able preliminary data. See, e.g., Marcel Clerget,
Le Caire: B t t d e de gbogr aphie urbaine e t d histoire bcon o- mique
(2 vols.; Cairo: E.
R.
Schindler, 1934); Robert
E
Dickinson,
The W est Europe an City
(Lon-don: Routled ge Kegan Pau l, 1951); Roger Le Tourneau,
FBs: Avant le protectorat
(Casablanca: SociCt i Marocaine de Librai rie et d'adition, 1949); Edward
W.
Lane,
Cair o Fijty Years Ago
(London: John Murray, 1896);
J
Sauvaget,
Alep
(Paris: Li- brairie Orientaliste Paul Geuthner, 1941);
J
Weu-Iers se, Antioche: Essai de
geographic
urbaine,
Bulle tin d btudes orientales,
V
(1934), 27-7 9; Jea n Kennedy,
He re Is In dia
(New Yo rk: Charles Scrib- ner's Sons, 19 45); and relev ant articles in Ameri can geographical journals.
such as is functionally necessary in indus- trial-urban communities. In medieval Eu- rope a nd in othe r area s city dwell ing s often serve as workshops, an d rel igi ous structu res are used as schools or marketing centem4 Finally, th e busines s district does no t hold the position of dominance that it en- joys in the industrial-urban community. Thus, in the Middle East the principal mosque, or in medieval Europe the cathe- dral, is usually th e focal poin t o f comm unity lif e. T h e center o f Peiping is th e Forbidden City.
ECONOMIC ORGANIZAT ION
T h e ec onomy o f t he preindustrial city di- verges sharply from th at o f the m odern in- dustrial center. T h e prime dif fer ence is the absence in th e forme r o f ind ustriali sm whi ch m ay be defined as tha t system o f production in which
inanimate
sources o f pow er a re used to multiply human effort. Preindustrial cities dep end for th e production o f goods and services upon
animate
(human or ani- m al) sources o f ener gy-appl ied eith er di- rectly or indirectly thro ugh such mechani cal devices as hammers, pulleys, and wheels. The industrial-urban community, on the other hand, employs inanimate generators o f Do we r s uch- as electricit v and ste am which greatly enhance the productive ca-pac ity o f u rban ites. Th is basic all y new form o f energy production, o ne which requires for its development and survival a special kind of institutional complex, effects striking changes in the ecologi cal, economic, an d so- cial orga nization o f cities in which it ha s become dominant. Ot he r fac ets o f th e econom y o f t he pre- industrial city are associated with its par- ticular system of production. There is little fragm entatio n or speci ali zati on o f work. Th e handicraftsm an participates i n nearly every pha se o f th e ma nuf acture o f a n art icle, often carryi ng out the work in his own home or in a small shop near b y a nd, wi thin th e li mits o f ce rtain guil d an d com mu nity regulati ons,
Dickinson,
op. cit.,
p. 27; H.
K.
Spate,
India and Pakistan
(London: Me thue n Co., 1954), p. 183.
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440
THE
AME RICAN JOURNAL O F SOC IO LOG Y
maintaining direct control over conditions o f work and meth ods o f pr oduct ion. I n indu stria l cities, on the other hand, the complex division o f labor requir es a special- ized managerial group, often extra-commu- nity in character, whose primary function is to direct and control others. And for the supervision an d co-ordination o f t he activi- ties o f workers, a ('fac tory system has bee n developed, something typically lacking in preindustrial cities. (Occasionally central- ized production is found in preindustrial cit ies-e.g., where the s t at e organized slaves for large-scale construction projects.) Most commercial activities, also, a re conducted in
reind dust rial
cities bv individuals without a highly formalized organization; for ex-ample, the craftsman has frequently been responsibl e for t he mar ket ing o f his o wn prdducts. With a few exceptions, the pre- industrial community cannot support a larg e group o f mid dlem en. The various occupations are organized int o wha t h ave be en ter med g~i l ds. ~ These strive to encompass all, except the elite, who are gainfully employed in some economic activitv. Guilds have existed for merchants and handicraft workers (e.g., goldsmiths and weavers) as we ll a s for serv- ants, entertainers, and even beggars and thieves. Typically the guilds operate only within the local community, and there are no large-scale economic organizations such as those in industrial cities which link their members to their fellows in other communi- ties.
Fo r a discussi on o f gu ilds and oth er facets o f th e preindustrial city's economy see, e.g.,
J.
S. Burgess,
The Guilds of Peking
(New York: Columbia Uni- versity Press, 1928); Edward T. Williams,
China Yesterday and Today
(5th ed.; New York: Thomas Y. Crowel l Co., 1932); T'ai-ch'u Liao, The Ap-prentices in Chengtu during and after the War,
Yenchi ng Journa l o f Social Studie s
I V
(1948), 90- 106;
H.
A.
R.
Gibb and Harold Bowen,
Islamic So- ciety and the West
(London: Oxford University Press, 1950), Vol.
I
Part
I
chap. vi; Le To urneau,
op. cit.;
Clerget,
op. cit.;
James W. Thompson and Edgar N. Johnson,
A n Introduction to Medie val Europe
(New York:
W
W.
Norton Co., 1937), ch ap.
xx;
Sylvia L. Th ru pp, Medieval Gilds Recon-sidered,
Journal o f Econom ic History
I1
(1942), 164-73.
Guild membership and apprenticeship ar e prerequisi tes t o th e practice o f almost any occupation, a circumstance obviously leading to monopolization. To a degree these organizations regulate the work of the ir members and t he price o f their prod- ucts and services. And the guilds recruit workers into specific occupations, typically selecting them according to such particu- laristic criteria as kinship rather than uni- versalistic standards. The guilds are integrated with still other elements o f th e city's social struct ure. The y perform certain religious functions; for ex- ample, in medieval European, Chinese, and Middle Eastern cities each guild had its patron saint and h eld periodic festivals in his honor. And, by assisting members in tim e o f trouble, th e guilds serve as soc ial security agencies. T h e eco nomic str uct ure o f the preindus- trial city functions with little rationality, judged by industrial-urban standards. This is shown
in
the gene ral nonstandardization o f manu facturing methods as we ll as in th e products and is even more evident in mar- keting. In preindustrial cities throughout the world a fixed price is rare; buyer and seller settle their bargain by haggling. (Of course, there are limits above which cus-tomers will not buy and below which mer- chants will not sell.) Often business is con- ducted in a leisurely manner, money not be- ing the only desired end. Further more, th e sorting o f goods ac cord- ing to size, weight, and quality is not com- mon. Typical is the adulteration and spoil- age o f p roduce. And weights and measures are not standardized: variations exist not only between one city and the next but also within communities. for often different guilds employ their own systems. Within a single city there may be different kinds of currency, which, with the poorly developed accounting and credit systems, signalize a mod icum o f ra tio nal ity
in
the whole of economic action in preindustrial ~ities.~
Tor an extrem e ex ample o f unstandardized cur- rency cf. Robert Coltman, Jr.,
Th e Chi nes e
(Phila-delphia:
F.
A
Davis, 1891), p. 52. In some tradi-
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623 | https://www.ic.gc.ca/eic/site/mc-mc.nsf/vwapj/VCF_Lube-Oil.pdf/$file/VCF_Lube-Oil.pdf | Issued: January 2016 Density at 15 °C = 880 kg/m3 (table 54D) Refer to bulletin V-18 for more information on product classes. Volume correction factors to 15 °C for use with all grades of lubricating oils (SAE) Temperature °C 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 -40 1.0387 -39 1.0380 1.0381 1.0382 1.0383 1.0383 1.0384 1.0385 1.0385 1.0386 1.0387 -38 1.0373 1.0374 1.0375 1.0376 1.0376 1.0377 1.0378 1.0378 1.0379 1.0380 -37 1.0367 1.0367 1.0368 1.0369 1.0369 1.0370 1.0371 1.0371 1.0372 1.0373 -36 1.0360 1.0360 1.0361 1.0362 1.0362 1.0363 1.0364 1.0364 1.0365 1.0366 -35 1.0353 1.0353 1.0354 1.0355 1.0355 1.0356 1.0357 1.0357 1.0358 1.0359 -34 1.0346 1.0346 1.0347 1.0348 1.0348 1.0349 1.0350 1.0351 1.0351 1.0352 -33 1.0339 1.0339 1.0340 1.0341 1.0341 1.0342 1.0343 1.0344 1.0344 1.0345 -32 1.0332 1.0332 1.0333 1.0334 1.0334 1.0335 1.0336 1.0337 1.0337 1.0338 -31 1.0325 1.0325 1.0326 1.0327 1.0328 1.0328 1.0329 1.0330 1.0330 1.0331 -30 1.0318 1.0318 1.0319 1.0320 1.0321 1.0321 1.0322 1.0323 1.0323 1.0324 -29 1.0311 1.0311 1.0312 1.0313 1.0314 1.0314 1.0315 1.0316 1.0316 1.0317 -28 1.0304 1.0304 1.0305 1.0306 1.0307 1.0307 1.0308 1.0309 1.0309 1.0310 -27 1.0297 1.0297 1.0298 1.0299 1.0300 1.0300 1.0301 1.0302 1.0302 1.0303 -26 1.0290 1.0290 1.0291 1.0292 1.0293 1.0293 1.0294 1.0295 1.0295 1.0296 -25 1.0283 1.0283 1.0284 1.0285 1.0286 1.0286 1.0287 1.0288 1.0288 1.0289 -24 1.0276 1.0276 1.0277 1.0278 1.0279 1.0279 1.0280 1.0281 1.0281 1.0282 -23 1.0269 1.0269 1.0270 1.0271 1.0272 1.0272 1.0273 1.0274 1.0274 1.0275 - 2 - Volume correction factors to 15 °C for use with all grades of lubricating oils (SAE) Temperature °C 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 -22 1.0262 1.0262 1.0263 1.0264 1.0265 1.0265 1.0266 1.0267 1.0267 1.0268 -21 1.0255 1.0255 1.0256 1.0257 1.0258 1.0258 1.0259 1.0260 1.0260 1.0261 -20 1.0248 1.0248 1.0249 1.0250 1.0251 1.0251 1.0252 1.0253 1.0253 1.0254 -19 1.0241 1.0241 1.0242 1.0243 1.0244 1.0244 1.0245 1.0246 1.0246 1.0247 -18 1.0234 1.0234 1.0235 1.0236 1.0236 1.0237 1.0238 1.0239 1.0239 1.0240 -17 1.0227 1.0227 1.0228 1.0229 1.0229 1.0230 1.0231 1.0232 1.0232 1.0233 -16 1.0220 1.0220 1.0221 1.0222 1.0222 1.0223 1.0224 1.0225 1.0225 1.0226 -15 1.0213 1.0213 1.0214 1.0215 1.0215 1.0216 1.0217 1.0218 1.0218 1.0219 -14 1.0206 1.0206 1.0207 1.0208 1.0208 1.0209 1.0210 1.0210 1.0211 1.0212 -13 1.0199 1.0199 1.0200 1.0201 1.0201 1.0202 1.0203 1.0203 1.0204 1.0205 -12 1.0191 1.0192 1.0193 1.0194 1.0194 1.0195 1.0196 1.0196 1.0197 1.0198 -11 1.0184 1.0185 1.0186 1.0187 1.0187 1.0188 1.0189 1.0189 1.0190 1.0191 -10 1.0177 1.0178 1.0179 1.0179 1.0180 1.0181 1.0182 1.0182 1.0183 1.0184 -9 1.0170 1.0171 1.0172 1.0172 1.0173 1.0174 1.0175 1.0175 1.0176 1.0177 -8 1.0163 1.0164 1.0165 1.0165 1.0166 1.0167 1.0167 1.0168 1.0169 1.0170 -7 1.0156 1.0157 1.0158 1.0158 1.0159 1.0160 1.0160 1.0161 1.0162 1.0163 -6 1.0149 1.0150 1.0151 1.0151 1.0152 1.0153 1.0153 1.0154 1.0155 1.0155 -5 1.0142 1.0143 1.0143 1.0144 1.0145 1.0146 1.0146 1.0147 1.0148 1.0148 -4 1.0135 1.0136 1.0136 1.0137 1.0138 1.0139 1.0139 1.0140 1.0141 1.0141 -3 1.0128 1.0129 1.0129 1.0130 1.0131 1.0131 1.0132 1.0133 1.0134 1.0134 -2 1.0121 1.0122 1.0122 1.0123 1.0124 1.0124 1.0125 1.0126 1.0126 1.0127 - 3 - Volume correction factors to 15 °C for use with all grades of lubricating oils (SAE) Temperature °C 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 -1 1.0114 1.0114 1.0115 1.0116 1.0117 1.0117 1.0118 1.0119 1.0119 1.0120 0 1.0107 1.0107 1.0108 1.0109 1.0109 1.0110 1.0111 1.0112 1.0112 1.0113 0 1.0107 1.0106 1.0105 1.0105 1.0104 1.0103 1.0102 1.0102 1.0101 1.0100 1 1.0100 1.0099 1.0098 1.0097 1.0097 1.0096 1.0095 1.0095 1.0094 1.0093 2 1.0092 1.0092 1.0091 1.0090 1.0090 1.0089 1.0088 1.0088 1.0087 1.0086 3 1.0085 1.0085 1.0084 1.0083 1.0083 1.0082 1.0081 1.0080 1.0080 1.0079 4 1.0078 1.0078 1.0077 1.0076 1.0075 1.0075 1.0074 1.0073 1.0073 1.0072 5 1.0071 1.0070 1.0070 1.0069 1.0068 1.0068 1.0067 1.0066 1.0066 1.0065 6 1.0064 1.0063 1.0063 1.0062 1.0061 1.0061 1.0060 1.0059 1.0058 1.0058 7 1.0057 1.0056 1.0056 1.0055 1.0054 1.0053 1.0053 1.0052 1.0051 1.0051 8 1.0050 1.0049 1.0048 1.0048 1.0047 1.0046 1.0046 1.0045 1.0044 1.0043 9 1.0043 1.0042 1.0041 1.0041 1.0040 1.0039 1.0038 1.0038 1.0037 1.0036 10 1.0036 1.0035 1.0034 1.0033 1.0033 1.0032 1.0031 1.0031 1.0030 1.0029 11 1.0029 1.0028 1.0027 1.0026 1.0026 1.0025 1.0024 1.0024 1.0023 1.0022 12 1.0021 1.0021 1.0020 1.0019 1.0019 1.0018 1.0017 1.0016 1.0016 1.0015 13 1.0014 1.0014 1.0013 1.0012 1.0011 1.0011 1.0010 1.0009 1.0009 1.0008 14 1.0007 1.0006 1.0006 1.0005 1.0004 1.0004 1.0003 1.0002 1.0001 1.0001 15 1.0000 0.9999 0.9999 0.9998 0.9997 0.9996 0.9996 0.9995 0.9994 0.9994 16 0.9993 0.9992 0.9991 0.9991 0.9990 0.9989 0.9989 0.9988 0.9987 0.9986 17 0.9986 0.9985 0.9984 0.9984 0.9983 0.9982 0.9981 0.9981 0.9980 0.9979 18 0.9979 0.9978 0.9977 0.9976 0.9976 0.9975 0.9974 0.9974 0.9973 0.9972 - 4 - Volume correction factors to 15 °C for use with all grades of lubricating oils (SAE) Temperature °C 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 19 0.9971 0.9971 0.9970 0.9969 0.9969 0.9968 0.9967 0.9966 0.9966 0.9965 20 0.9964 0.9964 0.9963 0.9962 0.9961 0.9961 0.9960 0.9959 0.9959 0.9958 21 0.9957 0.9956 0.9956 0.9955 0.9954 0.9954 0.9953 0.9952 0.9951 0.9951 22 0.9950 0.9949 0.9949 0.9948 0.9947 0.9946 0.9946 0.9945 0.9944 0.9944 23 0.9943 0.9942 0.9941 0.9941 0.9940 0.9939 0.9939 0.9938 0.9937 0.9936 24 0.9936 0.9935 0.9934 0.9934 0.9933 0.9932 0.9931 0.9931 0.9930 0.9929 25 0.9929 0.9928 0.9927 0.9926 0.9926 0.9925 0.9924 0.9923 0.9923 0.9922 26 0.9921 0.9921 0.9920 0.9919 0.9918 0.9918 0.9917 0.9916 0.9916 0.9915 27 0.9914 0.9913 0.9913 0.9912 0.9911 0.9911 0.9910 0.9909 0.9908 0.9908 28 0.9907 0.9906 0.9906 0.9905 0.9904 0.9903 0.9903 0.9902 0.9901 0.9901 29 0.9900 0.9899 0.9898 0.9898 0.9897 0.9896 0.9896 0.9895 0.9894 0.9893 30 0.9893 0.9892 0.9891 0.9891 0.9890 0.9889 0.9888 0.9888 0.9887 0.9886 31 0.9885 0.9885 0.9884 0.9883 0.9883 0.9882 0.9881 0.9880 0.9880 0.9879 32 0.9878 0.9878 0.9877 0.9876 0.9875 0.9875 0.9874 0.9873 0.9873 0.9872 33 0.9871 0.9870 0.9870 0.9869 0.9868 0.9868 0.9867 0.9866 0.9865 0.9865 34 0.9864 0.9863 0.9862 0.9862 0.9861 0.9860 0.9860 0.9859 0.9858 0.9857 35 0.9857 0.9856 0.9855 0.9855 0.9854 0.9853 0.9852 0.9852 0.9851 0.9850 36 0.9850 0.9849 0.9848 0.9847 0.9847 0.9846 0.9845 0.9844 0.9844 0.9843 37 0.9842 0.9842 0.9841 0.9840 0.9839 0.9839 0.9838 0.9837 0.9837 0.9836 38 0.9835 0.9834 0.9834 0.9833 0.9832 0.9832 0.9831 0.9830 0.9829 0.9829 39 0.9828 0.9827 0.9826 0.9826 0.9825 0.9824 0.9824 0.9823 0.9822 0.9821 - 5 - Volume correction factors to 15 °C for use with all grades of lubricating oils (SAE) Temperature °C 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 40 0.9821 Density at 15 °C = 880 kg/m3 Values calculated as per API Standard 2540, Chapter 11.1, Volume X (1993) To obtain the net volume of liquid at 15 °C, multiply the uncompensated meter reading by the volume correction factor (VCF) which corresponds to the average measured temperature of the liquid during the delivery. |
624 | https://www.quora.com/How-do-I-work-out-the-cross-product-and-dot-product-of-vectors-a-4-3-and-b-2-1-The-angle-between-these-vectors-is-30-degrees-I-know-the-formula-is-magnitude-of-A-mag-of-b-cos-30-for-dot-And-magnitude-of-a-magnitude | How to work out the cross product and dot product of vectors a (4,3) and b (-2,1) - Quora
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How do I work out the cross product and dot product of vectors a (4,3) and b (-2,1)? The angle between these vectors is 30 degrees. I know the formula is magnitude of Amag of b cos 30 for dot. And magnitude of a magnitude of b sin 30 for cross.
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Bill Hazelton
PhD from University of Melbourne (Graduated 1992) · Author has 3.9K answers and 8.7M answer views
·6y
For two 2-D vectors a and b, a = [a1, a2] and b = [b1, b2], the dot product (scale multiplication of vectors) is simply:
a · b = a1 b1 + a2 b2
Note that this will be commutative, so the order of the vectors doesn’t matter. So for the vectors a = [4, 3] and b = [–2, 1], the dot product will be:
a · b = (4 x –2) + (3 x 1) = –8 + 3 = –5
The angle between the vectors, y, can be calculated because:
a · b = |a| |b| cos y and so cos y = a · b / (|a| |b|)
In this case:
cos y = –5 / (sqrt(4^2 + 3^2) x sqrt(–2^2 + 1^2)) = –5/ (5 x sqrt(5)) = –1/sqrt(5) = 0.447214
So y = 63.434949°
This isn’t 30°. You can check t
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For two 2-D vectors a and b, a = [a1, a2] and b = [b1, b2], the dot product (scale multiplication of vectors) is simply:
a · b = a1 b1 + a2 b2
Note that this will be commutative, so the order of the vectors doesn’t matter. So for the vectors a = [4, 3] and b = [–2, 1], the dot product will be:
a · b = (4 x –2) + (3 x 1) = –8 + 3 = –5
The angle between the vectors, y, can be calculated because:
a · b = |a| |b| cos y and so cos y = a · b / (|a| |b|)
In this case:
cos y = –5 / (sqrt(4^2 + 3^2) x sqrt(–2^2 + 1^2)) = –5/ (5 x sqrt(5)) = –1/sqrt(5) = 0.447214
So y = 63.434949°
This isn’t 30°. You can check this mentally, because a is about 36.869898° from the x axis heading up and to the right, while b is about 26.565051° from the negative x axis, heading up and to the left. The angle y is between –a and b, the smaller angle between the vectors, and is about 36.869898° + 26.565051° = 63.434949°.
Cross-products cannot be taken in 2-D space. They ONLY work in 3-D space (and 7-D, as I recall from a passing comment years ago). So you would need to convert them to 3-D vectors if you wanted to do this, but that isn’t in the scope of the question. It may be a trick to get you to do something that would show you haven’t studied the theory.
For a 3-D cross-product, the result is actually a vector that is oriented at right angle to the plane formed by the two input vectors, i.e., the normal vector. Which normal vector (‘up’ or ‘down’) depends on the order in which the vector are considered, so vector multiplication is not commutative.
If you were to take the chance in this problem and convert the vectors to 3-D, you would have a = [4, 3, 0] and b = [–2, 1, 0]. The vector product (cross-product) would then be:
a x b = |a| |b| sin y n
where n is the appropriate normal unit vector.
In this case, you will get
a x b = 5 sqrt(5) sin (63.434949°) n
a x b = 10 n
Now all we need is the sense of n, which will be either coming out of the page towards you or going into the page away from you, assuming you have drawn the vectors on a piece of paper. Turning from a to b is counter-clockwise (anti-clockwise) and following the right-hand screw rule, the vector n will be coming out of the page towards you. If you assume that this is the positive third axis, then the scalar value with be +10 and the resulting vector will be:
a x b = [0, 0, 10]
As the answer (the cross-product) is exactly 10 n, we can expect that it was set up this way. Similarly, the fact that the dot product was exactly –5 suggests that this was the desired answer.
Take a close look at where you got the 30° angle between the vectors. It doesn’t look right.
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
·Updated Sep 9
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Related questions
More answers below
How can I use dot product and cross product to calculate the angle between two vectors when the angle is greater than 180?
The dot product of two vectors is negative while the magnitude of its cross product is positive. Which of the following best describes the angle between the two vectors?
The magnitude of dot and cross product of two vectors is 10 in each case. What is the angle between two vectors?
How do I get the value and the angle when I have 2 equal vectors and the product of their cross and dot?
Is it possible to find the values of two vectors if their dot product and magnitude of cross product is given?
Brian Opatosky
B.S. in Physics&Mathematics, Appalachian State University · Author has 58 answers and 131.2K answer views
·6y
You would be correct about the dot product and cross product formula, if the angle between the vectors were correct. It is not 30 degrees, but about 116 degrees. All you need to do is find the magnitudes of vector a a and vector b b. To find the the magnitude of vectors when you know their Cartesian coordinates, you can position both vectors’ tails at the origin, without changing their direction, and use the coordinates to find the components of the vectors. This means to find what “part” of the vector belongs to a certain part of a coordinate axis. For example, the x component of vector a a is 4. T
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You would be correct about the dot product and cross product formula, if the angle between the vectors were correct. It is not 30 degrees, but about 116 degrees. All you need to do is find the magnitudes of vector a a and vector b b. To find the the magnitude of vectors when you know their Cartesian coordinates, you can position both vectors’ tails at the origin, without changing their direction, and use the coordinates to find the components of the vectors. This means to find what “part” of the vector belongs to a certain part of a coordinate axis. For example, the x component of vector a a is 4. The x component of vector b b is -2. The y components of a a and b b are 3 and 1, respectively. Now, the vector components of these vectors (a vector component is just the component, which is simple a scalar number, combined with a unit vector, which is just a vector with length 1 in a specific coordinate axis direction. For example, the unit vector for the positive x direction is called “i hat”. The symbol is a lowercase i with the math symbol hat on top (sorry I don’t know how to write it on Quora). The important things is that vector components are ALWAYS parallel to their respective coordinate axis. And in this case, in an x-y coordinate system, the x vector components and y vector components are always perpendicular to each other. If you were to draw them out, you would see that the x and y vector components of each vector always create a right triangle, where the resultant vector (the one that is made up of the vector components. This would be vectors a a and b b in this question) is always the hypotenuse of the right triangle. This means that to find the magnitude of these vectors, you can use Pythagorean’s Theorem to find the magnitude (But not direction. You need to use arctan to find the direction) of the resultant vector. For vector a a, you would do:
a=(4^2 + 3^2)^(1/2)
a= 5
For b:b:
b=((-2)^2 + 1^2)^(1/2)
b=(5)^(1/2)
Now you can use these magnitudes to find the dot product and cross product of the two vectors.
Also, be aware that for the cross product, multiplication is NOT commutative. This means that the ordering of the magnitudes of a a and b b matter. Changing the order will change the direction of the resultant vector. For example, if
A X B = (-1, parallel to the z axis)
then
B X A = (1, parallel to the z axis)
Dr, Harting makes a great point when he says that the cross product isn’t defined in 2 dimensions. This is because the product of a cross product is a vector that is orthangonal to the plane in which the original two vectors exist.
This is a just a fancy way of saying that if two vectors exist on a plane (say an x-y plane), the product of their cross product will be on the z axis.
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David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by
Michael Lamar
, PhD in Applied Mathematics and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 9.9K answers and 68.4M answer views
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Assuming no knowledge of the dot or cross product, and given two vectors in component form, how would you find the angle between them?
You can use the law of cosines if you know what the lengths of vectors are. The length of a vector v=(v 1,v 2,…,v n)v=(v 1,v 2,…,v n) is
∥v∥=√v 2 1+v 2 2+⋯+v 2 n‖v‖=v 1 2+v 2 2+⋯+v n 2
Let the two vectors be v v and w.w. Along with their difference w−v,w−v, they form a triangle.
The law of cosines says that if θ θ is the angle between the vectors v v and w,w, then
∥w−v∥2=∥v∥2+∥w∥2–2∥v∥∥w∥cos θ.‖w−v‖2=‖v‖2+‖w‖2–2‖v‖‖w‖cosθ.
Therefore,
[math]\displaystyle\cos\theta=\frac {\|\mathbf w-\mathbf v\|^2-\|\mathbf v[/math]
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You can use the law of cosines if you know what the lengths of vectors are. The length of a vector v=(v 1,v 2,…,v n)v=(v 1,v 2,…,v n) is
∥v∥=√v 2 1+v 2 2+⋯+v 2 n‖v‖=v 1 2+v 2 2+⋯+v n 2
Let the two vectors be v v and w.w. Along with their difference w−v,w−v, they form a triangle.
The law of cosines says that if θ θ is the angle between the vectors v v and w,w, then
∥w−v∥2=∥v∥2+∥w∥2–2∥v∥∥w∥cos θ.‖w−v‖2=‖v‖2+‖w‖2–2‖v‖‖w‖cosθ.
Therefore,
cos θ=∥w−v∥2−∥v∥2−∥w∥2 2∥v∥∥w∥.cosθ=‖w−v‖2−‖v‖2−‖w‖2 2‖v‖‖w‖.
If you work out the numerator in coordinates, you’ll discover that it’s equal to twice v 1 w 1+v 2 w 2+⋯+v n w n.v 1 w 1+v 2 w 2+⋯+v n w n. That suggest giving that expression a special notation, say v⋅w.v⋅w. Then the last equation becomes
cos θ=v⋅w∥v∥∥w∥.cosθ=v⋅w‖v‖‖w‖.
Imagine that, you just discovered the dot product!
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Haresh Sagar
Studied Science&Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views
·3y
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How do I solve the length of the vector a when the dot product with the vector b= (-2,2) is -1 and the angle between them is 150 degrees?
Let →b=(−2,2)b→=(−2,2) be a position vector.
Let →a=(x,y)a→=(x,y) also be a position vector. Since we have taken both the vectors as position vectors, the line containing these vectors pass through origin. Therefore linear equation of →b b→ is y=−x y=−x.
Angle between both the vectors is given as 150°150° and we know that tangent of 150°150° is −1/√3−1/3, so if we assume linear slope of of →a a→ as m 1 m 1 and m 2 m 2,
m 1=−1−m 1+(−m)=−1√3 m 1=−1−m 1+(−m)=−1 3
−√3−√3 m=−1+m−3−3 m=−1+m
m 1:1−√3 1+√3=√3−2 m 1:1−3 1+3=3−2
m 2:−√3−√3 m=1−m m 2:−3−3 m=1−m
m 2=1+√3 1−√3=−√3−2 m 2=1+3 1−3=−3−2
So components o
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Let →b=(−2,2)b→=(−2,2) be a position vector.
Let →a=(x,y)a→=(x,y) also be a position vector. Since we have taken both the vectors as position vectors, the line containing these vectors pass through origin. Therefore linear equation of →b b→ is y=−x y=−x.
Angle between both the vectors is given as 150°150° and we know that tangent of 150°150° is −1/√3−1/3, so if we assume linear slope of of →a a→ as m 1 m 1 and m 2 m 2,
m 1=−1−m 1+(−m)=−1√3 m 1=−1−m 1+(−m)=−1 3
−√3−√3 m=−1+m−3−3 m=−1+m
m 1:1−√3 1+√3=√3−2 m 1:1−3 1+3=3−2
m 2:−√3−√3 m=1−m m 2:−3−3 m=1−m
m 2=1+√3 1−√3=−√3−2 m 2=1+3 1−3=−3−2
So components of position →a a→ can be written as,
and
Since dot product of both the vectors is −1−1,
−2 x+2√3 x−4 x=−1−2 x+2 3 x−4 x=−1
x=1 6−2√3,1 6+2√3 x=1 6−2 3,1 6+2 3
y=√3−2 6−2√3,−√3+2 6+2√3 y=3−2 6−2 3,−3+2 6+2 3
||→a||=1√6≈0.40825||a→||=1 6≈0.40825
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More answers below
What is the angle formed between vectors A and B, if the magnitude of their cross product and dot product is the same?
What is the angle between two vectors if the magnitude of dot and cross product of two vectors is 6√3 and 6 respectively?
Why do we define the cross product of two vectors using sin of the angle between them & dot products using the cosine of the angle between them?
The magnitude of the dot product is 3/4 of the magnitude of the cross product of two vectors; what is the angle between them?
What is the angle between vectors A and B if their dot products and cross products are equal?
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To compute the dot product and cross product of the vectors a = (4, 3) and b = (-2, 1), let's start by calculating the required values step by step.
Dot Product
The dot product of two vectors a and b is given by the formula:
dot product=a⋅b=|a||b|cos(θ)dot product=a⋅b=|a||b|cos(θ)
where |a||a| and |b||b| are the magnitudes of vectors a and b, and θ θ is the angle between them.
Calculate the magnitudes of a and b:
|a|=√4 2+3 2=√16+9=√25=5|a|=4 2+3 2=16+9=25=5
|b|=√(−2)2+1 2=√4+1=√5|b|=(−2)2+1 2=4+1=5
Use the angle θ=30∘θ=30∘:
cos(30∘)=√3 2 cos(30∘)=3 2
2. Calculate th
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To compute the dot product and cross product of the vectors a = (4, 3) and b = (-2, 1), let's start by calculating the required values step by step.
Dot Product
The dot product of two vectors a and b is given by the formula:
dot product=a⋅b=|a||b|cos(θ)dot product=a⋅b=|a||b|cos(θ)
where |a||a| and |b||b| are the magnitudes of vectors a and b, and θ θ is the angle between them.
Calculate the magnitudes of a and b:
|a|=√4 2+3 2=√16+9=√25=5|a|=4 2+3 2=16+9=25=5
|b|=√(−2)2+1 2=√4+1=√5|b|=(−2)2+1 2=4+1=5
Use the angle θ=30∘θ=30∘:
cos(30∘)=√3 2 cos(30∘)=3 2
2. Calculate the dot product:
a⋅b=|a||b|cos(30∘)=5⋅√5⋅√3 2=5√15 2 a⋅b=|a||b|cos(30∘)=5⋅5⋅3 2=5 15 2
Cross Product
The cross product of two vectors in two dimensions can be computed using the formula:
cross product=|a||b|sin(θ)cross product=|a||b|sin(θ)
Use the angle θ=30∘θ=30∘:
sin(30∘)=1 2 sin(30∘)=1 2
2. Calculate the cross product:
cross product=|a||b|sin(30∘)=5⋅√5⋅1 2=5√5 2 cross product=|a||b|sin(30∘)=5⋅5⋅1 2=5 5 2
Summary
Dot Product: a⋅b=5√15 2 a⋅b=5 15 2
Cross Product: |a||b|sin(30∘)=5√5 2|a||b|sin(30∘)=5 5 2
These results give you the dot product and the magnitude of the cross product for the given vectors.
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Donald Hartig
PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views
·6y
The dot product is 4⋅(−2)+3⋅1=−5.4⋅(−2)+3⋅1=−5.
There is no cross product in R 2.R 2.However,, by embedding the vectors in R 3 R 3 as u=(4,3,0)u=(4,3,0) and v=(−2,1,0)v=(−2,1,0)the cross product can be computed as
u×v=∣∣ ∣∣i j k 4 3 0−2 1 0∣∣ ∣∣=⟨0,0,10⟩.u×v=|i j k 4 3 0−2 1 0|=⟨0,0,10⟩.
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Nikhil Panikkar
Communications & Signal Processing Engineer · Author has 1.1K answers and 2.8M answer views
·Updated 6y
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What do dot and cross vector products actually mean?
Dot Product
The dot product gives the relative orientation of two vectors in two - dimensional space.
As you can see from the above figure, if both the vectors are normalized, then you get the relative orientation of the two vectors.
Cross Product
The cross product gives the orientation of the plane described by two vectors in three dimensional space.
Consider the figure above. In two dimensional space, the vector A points towards East and the vector B points towards North East. Once you have your reference directions(North,South, East, West) laid down, there is no confusion regarding their orienta
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Dot Product
The dot product gives the relative orientation of two vectors in two - dimensional space.
As you can see from the above figure, if both the vectors are normalized, then you get the relative orientation of the two vectors.
Cross Product
The cross product gives the orientation of the plane described by two vectors in three dimensional space.
Consider the figure above. In two dimensional space, the vector A points towards East and the vector B points towards North East. Once you have your reference directions(North,South, East, West) laid down, there is no confusion regarding their orientation.
Now consider the same figure observed in three-dimensional space. For someone observing the vectors from above the plane, the vector B points North-East, but for someone observing the vectors from below the plane, the vector B points South-East.
In other words, when you specify the location of a two - dimensional object in three-dimensional space, you have to specify the direction of the observer.
This is the purpose the cross product serves. It does this by:
1.Associating the direction of the observer with the direction of the cross product vector.
By defining the left/right/clockwise/anticlockwise directions for this observer.
This is where the right hand and left hand rules come in.
The fingers of the right hand, always curl in the anti-clockwise direction, when viewed in the direction of the thumb.The fingers of the left hand always curl in the clockwise direction when viewed in the direction of the thumb.
So by specifying the direction of the thumb, you specify the direction of the observer and thus dismiss any confusions about left/right/clockwise/anticlockwise directions.
The reason the cross product is able to do this is because sine is an odd function and the direction in which you measure the angle - clockwise or anticlockwise - affects the sign of the function.
You know that the orientation of a plane can be be specified by its normal. The cross product is a specific type of normal that passes through the axis connecting the two vectors.
Also see:
Nikhil Panikkar's answer to How do you arrive at the formula with which you calculate the cross product?
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Making smart financial decisions doesn't have to be complicated. In 2025, there are several simple yet highly effective money hacks that can make a huge difference in your financial health. These aren't complicated investment strategies; they are practical, everyday habitsthat help you keep more of your hard-earned money.
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If dot product of two vectors is 8 and the magnitude of the cross product is 8×√3 8×3, then what is the angle between them?
That's the answer both 60 and 120
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That's the answer both 60 and 120
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Luqman Khan
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Two vectors having magnitude 3 and 4 and their dot product is -5. What is (A+B).(A-B)?
Let us define two vectors as Vector A and vector B. I’m assuming these are 2D vectors
We have 4 unknown variables and only three equations. Therefore it is impossible to find a unique value for x_1,y_1,x_2,y_2
Now we want to find (z). Where ‘z’ is the unknown dot product
and that is the final answer. Hope this helped
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Let us define two vectors as Vector A and vector B. I’m assuming these are 2D vectors
We have 4 unknown variables and only three equations. Therefore it is impossible to find a unique value for x_1,y_1,x_2,y_2
Now we want to find (z). Where ‘z’ is the unknown dot product
and that is the final answer. Hope this helped
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Phil Scovis
I play guitar. And sing in the car. · Author has 6.9K answers and 13.3M answer views
·3y
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The dot product of two vectors is negative while the magnitude of its cross product is positive. Which of the following best describes the angle between the two vectors?
I don’t know what multiple-choice choices you have here, because you went beyond being too lazy to solve the problem, and were too lazy to even copy it completely. For that, I’m going to take a moment to highlight a wrong answer. If you want to learn something, read on, and don’t automatically copy a right angle to your answer booklet.
Well, the magnitude of the cross-product is positive, because the magnitude of any vector is positive. So you can ignore that bit of useless information that was probably put in there to confuse you.
Now assuming that possible answers are “an acute angle” and “an
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I don’t know what multiple-choice choices you have here, because you went beyond being too lazy to solve the problem, and were too lazy to even copy it completely. For that, I’m going to take a moment to highlight a wrong answer. If you want to learn something, read on, and don’t automatically copy a right angle to your answer booklet.
Well, the magnitude of the cross-product is positive, because the magnitude of any vector is positive. So you can ignore that bit of useless information that was probably put in there to confuse you.
Now assuming that possible answers are “an acute angle” and “an obtuse angle”. Try this neat trick for choosing from multiple choices: Make two vectors that have an acute angle between them. For example: <1,0> and <1,1>. Now find their dot product. Was it negative? If so, then that answer is correct. If not, then that answer is wrong. See how easy?
You can try the same thing with the other choice. Pick two vectors that form an obtuse angle. They can be as simple as you like. Again, find their dot product. Negative or positive? Multiple choice questions are easy because you can just try the choices to see if they’re right, without actually understanding the general principles. Your teacher is lazy too.
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Tuhin Chakraborty
Have a taste of Newtonian Mechanics · Author has 655 answers and 1.6M answer views
·5y
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The modulus of vector A cross vector B is equal to vector A dot vector B. What is the angle between vector A and vector B?
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Satya Parkash Sud
M.Sc. in Physics&Nuclear Physics, University of Delhi (Graduated 1962) · Author has 8.1K answers and 27.4M answer views
·8y
Related
What is the value of (A vector+B vector). (dot) (A vector Cross B vector)?
What is the value of ( A + B ) . (A × B )
We know from scalar triple product, that for vectors A, B, and C,
A . ( B × C) = B . ( C × A) = C . ( A× B) = (A×B) . C
In our question
( A + B ) . ( A × B ) = A. ( A× B) + B . ( A × B)
Now
A. ( A× B) = B. ( A×A)= 0
and
B . ( A ×B ) = A. ( B× B)= 0
Therefore,
( A +B ) . ( A × B ) = 0 +0= 0
The dot product of the sum of two Vectors with the cross products of the same two vector is zero and a scalar.
We can understand the answer as below. The sum of two vectors A and B is a vector in the plane defined by A and B. Whereas A×B is a vector which is perpendicular to the
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What is the value of ( A + B ) . (A × B )
We know from scalar triple product, that for vectors A, B, and C,
A . ( B × C) = B . ( C × A) = C . ( A× B) = (A×B) . C
In our question
( A + B ) . ( A × B ) = A. ( A× B) + B . ( A × B)
Now
A. ( A× B) = B. ( A×A)= 0
and
B . ( A ×B ) = A. ( B× B)= 0
Therefore,
( A +B ) . ( A × B ) = 0 +0= 0
The dot product of the sum of two Vectors with the cross products of the same two vector is zero and a scalar.
We can understand the answer as below. The sum of two vectors A and B is a vector in the plane defined by A and B. Whereas A×B is a vector which is perpendicular to the plane of vectors A and B. The dot product of two vectors one in the plane of vectors A,B and the other perpendicular to the plane is zero. Furthermore being a dot product it is a scalar.
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Samiul Abid
B.Sc in Electrical and Electronics Engineering, Bangladesh University of Engineering and Technology (Graduated 2018) · Author has 263 answers and 420.1K answer views
·6y
Related
Magnitude of vector A+B is equal to magnitude of A is equal to Magnitude of B then angle between A&B is?
Ur question may be answered in a simple geometric way like the following:
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Ur question may be answered in a simple geometric way like the following:
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Related questions
How can I use dot product and cross product to calculate the angle between two vectors when the angle is greater than 180?
The dot product of two vectors is negative while the magnitude of its cross product is positive. Which of the following best describes the angle between the two vectors?
The magnitude of dot and cross product of two vectors is 10 in each case. What is the angle between two vectors?
How do I get the value and the angle when I have 2 equal vectors and the product of their cross and dot?
Is it possible to find the values of two vectors if their dot product and magnitude of cross product is given?
What is the angle formed between vectors A and B, if the magnitude of their cross product and dot product is the same?
What is the angle between two vectors if the magnitude of dot and cross product of two vectors is 6√3 and 6 respectively?
Why do we define the cross product of two vectors using sin of the angle between them & dot products using the cosine of the angle between them?
The magnitude of the dot product is 3/4 of the magnitude of the cross product of two vectors; what is the angle between them?
What is the angle between vectors A and B if their dot products and cross products are equal?
If the magnitudes of dot and cross product of two vectors are equal, then what is the angle between the two vectors?
The magnitude of dot and cross product of two vectors are 6√3 and 6 respectively. What is the angle between the vectors?
What is the difference between the dot product and the cross product? Which one is better for finding out the angle between two vectors and why?
What is the magnitude of the cross and dot product of two vectors? What is the angle between the vectors?
Why do we not use cross product instead of dot product for finding out the angle between the two vectors?
Related questions
How can I use dot product and cross product to calculate the angle between two vectors when the angle is greater than 180?
The dot product of two vectors is negative while the magnitude of its cross product is positive. Which of the following best describes the angle between the two vectors?
The magnitude of dot and cross product of two vectors is 10 in each case. What is the angle between two vectors?
How do I get the value and the angle when I have 2 equal vectors and the product of their cross and dot?
Is it possible to find the values of two vectors if their dot product and magnitude of cross product is given?
What is the angle formed between vectors A and B, if the magnitude of their cross product and dot product is the same?
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626 | https://en.wikipedia.org/wiki/Median_triangle | Median triangle - Wikipedia
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1 See also
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Median triangle
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From Wikipedia, the free encyclopedia
Mathematical concept
median triangle: △B G F{\displaystyle \triangle BGF}
reference triangle:△A B C{\displaystyle \triangle ABC}
median triangle of the median triangle:△B K H{\displaystyle \triangle BKH}
areas: |△B G F|=3 4|△A B C|{\displaystyle |\triangle BGF|={\tfrac {3}{4}}|\triangle ABC|}
similarity: △B G F∼△B K H{\displaystyle \triangle BGF\sim \triangle BKH}
ratios: |B H||B C|=|H K||A B|=|B K||A C|=3 4{\displaystyle {\tfrac {|BH|}{|BC|}}={\tfrac {|HK|}{|AB|}}={\tfrac {|BK|}{|AC|}}={\tfrac {3}{4}}}
The median triangle of a given (reference) triangle is a triangle, the sides of which are equal and parallel to the medians of its reference triangle. The area of the median triangle is 3 4{\displaystyle {\tfrac {3}{4}}} of the area of its reference triangle, and the median triangle of the median triangle is similar to the reference triangle of the first median triangle with a scaling factor of 3 4{\displaystyle {\tfrac {3}{4}}}.
See also
[edit]
Automedian triangle
References
[edit]
^Roger A. Johnson: Advanced Euclidean Geometry. Dover 2007, ISBN978-0-486-46237-0, pp.282–283
^Claudi Alsina, Roger B. Nelsen: Charming Proofs: A Journey Into Elegant Mathematics. MAA, 2010, ISBN9780883853481, p.165
^Árpad Bényi, Branko Ćurgus: "Outer Median Triangles". In: Mathematics Magazine, Vol. 87, No. 3 (June 2014), pp.185–194 (JSTOR)
External links
[edit]
Weisstein, Eric W."Median Triangle". MathWorld.
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627 | https://ximera.osu.edu/precal/PrecWRev1Unit5/5-1IntroToTransformations/5-1-1VerticalAndHorizontalShifts | Warning
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Mathematical Expression Editor
Motivating Questions
By adding a constant to a function , what is the relation between the graphs of and ?
By performing a “change of variable” , what is the relation between the graphs of and ?
How to use this new understanding to gain a deeper understanding of graphs of quadratic functions (i.e., parabolas)?
Introduction
Let’s consider the two quadratic functions and , defined for all real values of . We know what their graphs look like:
The graphs are very similar, down to the horizontal “width”. In fact, drawing them together, we may see that they only differ by a horizontal translation:
Algebraically, one can see that this happens because This hints at the following general fact: doing horizontal shifts to the graph of a function amounts to replacing with “” inside . In this unit, we’ll understand in more detail how to work with this, and also how to deal with vertical shifts, as opposed to horizontal shifts. Since vertical shifts are much easier to understand, that’s where we’ll begin.
Shifting a function vertically
Let’s consider a very simple situation, where we have two functions and . Graphing them, in order, we have that
Clearly, and are directly related via , and seeing their graph together, we have that:
In other words, the graph of was obtained from the graph of by shifting it up exactly by units. This is a very general phenomenon, that happens for any functions who differ by a constant.
Theorem (vertical shifts): Suppose is a function and is a positive number.
To graph , shift the graph of up units, by adding to the -coordinates of the points on the graph of .
To graph , shift the graph of down units, by subtracting from the -coordinates of the points on the graph of .
In the above setting, it is useful to call the parent function.
Convention: Here, we’ll always draw graphs of parent functions in blue, and graphs of the “child” functions in red. We’ll indicate with a dashed line where the shift has happened.
For each of the following functions, find the parent function. How would the graph look like, in terms of the graph of the original function?
a.
: .
Setting , we have that . So, to graph , we just need to consider the graph of and shift it one unit down.
b.
: .
Here, we can just recognize that for , we have . Thus, we just need to shift the graph of up by units.
c.
: Granted, graphing a linear function poses little to no challenge, but understanding how things work in this setting might offer us some general insight on -intercepts. If , then . Graphing is easier than easy: just a line with slope passing through the origin. And the shift down by units comes last, as you would expect:
Shifting a function horizontally
Consider again the example given in the introduction, where we have and . The first thing we would like to address is a source of frequent confusion when first learning this topic. Namely, we have replaced with in the formula for , but the graph of the modified function ended up shifted to the right, even though one might expect the shift to have happened to the left, due to the negative sign in the factor!
Here is one safe way to think about it: imagine that you are standing on the -axis and, say, at the origin of the cartesian plane, but that the graph of is already drawn. Replacing with does move the -axis to the left. But you, the observer, standing on the -axis, sees the graph move to the right!
Alternatively, compare this with what happened with vertical shifts, but switching the roles of the -axis and -axis. More precisely, start with the graph of , then rotate it by clockwise (this switches the axes). Replacing with now brings the graph down by unit. Finally, rotate everything back by counterclockwise (this undoes the switching of the axes). The resulting graph is obtained from the original one by shifting it to the right, not left.
Theorem (horizontal shifts): Suppose is a function and is a positive number.
To graph , shift the graph of right units, by adding to the -coordinates of the points on the graph of .
To graph , shift the graph of left units, by subtracting from the -coordinates of the points on the graph of .
As before we’ll continue to call the parent function, whose graph will be drawn in blue, while the graphs of the “child” functions will be indicated in red.
For each of the following functions, find the parent function. How would the graph look like, in terms of the graph of the original function?
a.
: .
Here, if we look at , then we have that . So, to graph , we may just graph , and then shift it units to the left.
b.
: .
Consider this time . Then we have that , which means that to graph , we may take the graph of and shift it to the right by units. We obtain:
You might be thinking now that the graph of looks an awful lot like the graph of . This is not a coincidence and indeed is true for all values of . We will explore such relations (and more) between trigonometric functions, in future units.
c.
: .
As you might be guessing by now, the parent function can be found by just seeking the shifted variable (in this case, ), and replacing it with . Meaning that if , then . Thus, to graph , we can just graph and shift it units to the right. Since we can write , we know that is a parabola which crosses the -axis at and , and it is concave up. Hence:
Vertical shifts: given the graph of , we can draw the graph of , with , by shifting the graph of up by units. Similarly, the graph of is obtained by shifting the original graph down by units.
Horizontal shitfs: given the graph of , we can draw the graph of , with , by shifting the graph of by units to the left. Similarly, the graph of is obtained by shifting the original graph by units to the right. |
628 | https://papers.ssrn.com/sol3/Delivery.cfm/SSRN_ID3949492_code4435104.pdf?abstractid=3949492&mirid=1 | Факторный анализ экспорта нефти и нефтепродуктов из Российской Федерации в современных условиях (Factor analysis of Russian oil and petroleum products exports under current conditions) by Ibragim Magomedov :: SSRN
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Факторный анализ экспорта нефти и нефтепродуктов из Российской Федерации в современных условиях (Factor analysis of Russian oil and petroleum products exports under current conditions)
Russian Foreign Economic Journal. 2020. №7
16 Pages Posted: 18 Nov 2021
See all articles by Ibragim Magomedov
Ibragim Magomedov
Russian Customs Academy
Date Written: July 31, 2020
Abstract
Russian Abstract: В статье на основе анализа экспорта Российской Федерации обоснована особенность российского экспорта, которая заключаются в топливно-сырьевой направленности. Исследована роль и влияние экспорта нефти и нефтепродуктов на формирования доходной части федерального бюджета. Проанализированы структура и динамика нефтяных доходов федерального бюджета за период 2016-2019 гг. На основе количественных показателей развития нефтяной отрасли с применением стохастического факторного анализа выявлены факторы, влияющие на объем экспорта нефти и нефтепродуктов и формирование бюджета Российской Федерации.
English Abstract: In the article, basing on the results of the Russian exports analysis its peculiar orientation towards energy and raw materials is justified. The role and impact of oil and petroleum products exports on the federal budget revenues is examined. The structure and changes in energy revenues of the federal budget for the period 2016-2019 are analyzed. On the basis of quantitative indicators of oil industry growth with the stochastic method, the factors influencing the oil and petroleum Exports, oil, petroleum products, a federal budget of the Russian Federation; factor analysis, petroleum sector, customs paymentsproducts exports and the budget revenues are identified.
Note:Downloadable document is in Russian.
Keywords: Экспорт; нефть; нефтепродукты; федеральный бюджет РФ; факторный анализ; нефтяной сектор; таможенные платежи, Exports, oil, petroleum products, a federal budget of the Russian Federation; factor analysis, petroleum sector, customs payments.
Suggested Citation:Suggested Citation
Magomedov, Ibragim, Факторный анализ экспорта нефти и нефтепродуктов из Российской Федерации в современных условиях (Factor analysis of Russian oil and petroleum products exports under current conditions) (July 31, 2020). Russian Foreign Economic Journal. 2020. №7, Available at SSRN:
Ibragim Magomedov (Contact Author)
Russian Customs Academy ( email )
Komsomolsky Ave 4
Lyubertsy, 140009
Russia
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48 References
Банк России e (дата обращения 19.06.2020) БИБЛИОГРАФИЯ: 1. Указ Президента РФ от 13.05.2017 № 208 «О Стратегии экономической безопасности Российской Федерации на период до 2030 года Центральный банк Российской Федерации [Электронный ресурс]: официальный сайт
Распоряжение правительства Российской Федерации от 13.11.2009 № 1715-р «Об Энергетической стратегии России на период до Posted: 2030
Бюджетный кодекс Российской Федерации» от 31 Posted: 2020
А И Александрова , А В Закревская , А Соколицын Анализ экономики нефтегазового сектора в РФ // Научный журнал НИУ ИТМО. Серия «Экономика и экологический менеджмент, p. 3- 14 Posted: 2019
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629 | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2010/resources/lecture-6-graph-theory-and-coloring/ | Lecture 6: Graph Theory and Coloring | Mathematics for Computer Science | Electrical Engineering and Computer Science | MIT OpenCourseWare
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Colourful Solutions>The rate of chemical change> The Arrhenius equation
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The Arrhenius equation relates the rate constant, k, to the activation energy of a reaction, Ea. It affords a means of determining activation energies by experimentally finding the rate constant at different temperatures.
Syllabus ref: R2.2.12
Reactivity 2.2.12 - The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. (HL)
Describe the qualitative relationship between temperature and the rate constant.
Analyse graphical representations of the Arrhenius equation, including its linear form.
Guidance
The Arrhenius equation and its linear form are given in the data booklet.
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The Arrhenius equation
Arrhenius constant - the pre-exponential factor
Using the Arrhenius equation
Worked examples
The Arrhenius equation
The actual dependence of the rate constant on temperature is given by the Arrhenius equation.
k = Ae-Ea/RT
Where:
k is the rate constant
A is the Arrhenius factor (different for every reaction)
e is the natural log base
Ea is the minimum energy required for a reaction to take place (known as the activation energy)
R is the universal gas constant (8.31 J kg-1 ºC-1)
T is the absolute temperature in Kelvin
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Arrhenius' constant, A - the pre-exponential factor
It is possible for colliding particles to possess enough energy for reaction, but still not have a successful collision (one that results in reaction). This is accounted for by the Arrhenius constant 'A' , also called the pre-exponential or frequency factor.
Imagine a collision between two cars; clearly more damage is going to be caused by a head on collision than a glancing scrape.
The Arrhenius constant (pre-exponential or frequency factor) is a number between 0 and 1, that reflects the proportion of successful collisions amongst those particles with enough energy for reaction.
For example, when A is very small, only a small proportion of collisions lead to reaction, regardless of the energy. When A = 1, all collisions with sufficient energy cause reaction.
^ top
Using the Arrhenius equation
In reality, the basic form of the Arrhenius equation is not very convenient for graphing or analysing date. To analyse experiments at different temperatures we usually use the natural log form of the equation:
k = Ae-Ea/RT
taking natural logs throughout this gives:
lnk = lnA - Ea/RT
Thus a plot of lnk against 1/RT, 1/T or any variation, will allow us to find the activation energy of a specific reaction as a function of the gradient, and the Arrhenius constant as a function of the intercept to the y axis.
A typical plot used to calculate the activation energy from the Arrhenius equation.
In this graph the gradient of the line is equal to -Ea/R
Extrapolation of the line to the y axis gives an intercept value of lnA
When the temperature is increased the term Ea/RT gets smaller. This means in turn, that the term e-Ea/RT gets bigger.
Simultaneous equations
Alternatively two results may be analysed simultaneously to obtain values for Ea, the activation energy and the Orientation factor, A. This is not particularly reliable as only two values for the rate constant are used at two different temperatures. This can introduce large errors because of too little data.
Example: Calculate the rate constant when T = 300K (A = 0.3, Ea = 50kJ mol-1)
k = Ae-Ea/RT
Ea/RT = 50000/(8.314 x 300) = 20.05
e-Ea/RT = 1.97 x 10-9
k = Ae-Ea/RT
k = 5.90 x 10-10
Example: The rate of a reaction A(g) + B(g) → C(g) + D(g) has been studied as a function of temperature between 5000 and 18000 K. The following data were obtained for the rate constant: T (K)k(mol dm-3 /sec) 5000 5.49 x 106 10000 9.86 x 108 15000 5.57 x 109 18000 9.92 x 109 Calculate the activation energy for the reaction. k = Ae-Ea/RT and lnk = lnA - Ea/RT from expt 1: → ln(5.49 x 106) = lnA - Ea/ (8.314x5000) from expt 2: → ln(9.86 x 108) = lnA - Ea/ (8.314x10000) Subtract one from the other (to remove lnA) ln(9.86 x 108) - ln(5.49 x 106) = -Ea/ (8.314x10000) + Ea/ (8.314x5000) 10000[ln(9.86 x 108) - ln(5.49 x 106)] = -Ea/8.314 + 2Ea/8.314 8.314 x 10000[ln(9.86 x 108) - ln(5.49 x 106)] = Ea 8.314 x 10000 [20.709 - 15.518] = Ea 8.314 x 10000 x 5.191 = Ea 431579.74 = Ea Ea = 432 kJ
Note: Use of the simultaneous equation method is no longer required for first examinations 2025
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Worked examples
Q641-01 The first-order reaction: 2N 2 O(g) → 2N 2(g) + O 2(g), Has a rate constant of 1.3 x 10-11 s-1 at 270°C, and 4.5 x 10-10 s-1 at 350°C. What is the activation energy for this reaction?
Answer
k = Ae-Ea/RT and lnk = lnA - Ea/RT
ln(1.3 x 10-11) = lnA - Ea/(8.314 x 543)
and ln(4.5 x 10-10) = lnA - Ea/(8.314 x 623)
Subtract one from the other (to remove the terms lnA)
ln(4.5 x 10-10) - ln(1.3 x 10-11) = - Ea/(8.314 x 623) + Ea/(8.314 x 543)
8.314[ln(4.5 x 10-10) - ln(1.3 x 10-11)] = Ea/543 - Ea/623
8.314[ln(4.5 x 10-10) - ln(1.3 x 10-11)] = (623Ea - 543Ea)/ (623 x 543)
8.314 x 623 x 543 [ln(4.5 x 10-10) - ln(1.3 x 10-11)] = 80Ea
2812534.7[-21.52 + 25.07] = 80Ea
(2812534.7 x 3.5449)/80 = Ea
Ea = 124806
Ea = 124.8kJ mol-1
Q641-02 What is the activation energy for a reaction, if its rate doubles when the temperature is raised from 20°C to 35°C?
1. 342 kJ mol-1
2. 269 kJ mol-1
3. 34.7 kJ mol-1
4. 15.1 kJ mol-1
Answer
If the rate doubles the rate constant doubles k = Ae-Ea/RT when T = 293K the value for k is half that for T = 308K thus: Ae-Ea/308R = 2 x Ae-Ea/293R lnA - Ea/308R = ln2 + lnA -Ea/293R rearrange ln2 = Ea/293R - Ea/308R ln2 x 8.314 = Ea/293 - Ea/308 ln2 x 8.314 x 308 x 293 = 15Ea 519950 = 15 Ea Ea = 34663J Ea = 34.7kJ
Q641-03 Two reactions with different activation energies have the same rate at room temperature. Which statement correctly describes the rates of these two reactions at the same higher temperature?
1. The reaction with the greater activation energy will be faster.
2. The reaction with the smaller activation energy will be faster.
3. The two reactions will have the same rates.
4. A prediction cannot be made without further information.
Answer
The question is asking you to consider the effect of changing T on the rate constant in the Arrhenius equation.
By considering the Maxwell - Boltzmann distribution:
Increasing the temperature has a greater effect on reactions with lower activation energy
Shown mathematically:
k 1 = A 1 e-Ea 1/RT
k 2 = A 2 e-Ea 2/RT
The term e-Ea/RT has a smaller value when Ea/RT is larger
For different values of Ea, the smaller Ea will give the smaller term Ea/RT for any given temperature and hence a larger value for the term e-Ea/RT. This in turn provides a higher value for the rate constant.
Q641-04 The rate of an elementary reaction: A(g) + B(g) → C(g) + D(g) has been studied as a function of temperature between 5000 and 18000 K. The following data were obtained for the rate constant:
T (K)k (mol dm-3/sec)
5000 5.49 x 10 6
10000 9.86 x 10 8
15000 5.57 x 10 9
18000 9.92 x 10 9
Calculate the activation energy for the reaction.
1. 25.9 kJ/mol
2. 52.0 kJ/mol
3. 359 kJ/mol
4. 432 kJ/mol
Answer
Using Arrhenius k = Ae-Ea/RT
For reaction 1: 5.49 x 10 6 = Ae-Ea/5000R, therefore A = 5.49 x 10 6 /e-Ea/5000R
For reaction 2: 9.86 x 10 8 = Ae-Ea/10000R, therefore A = 9.86 x 10 8 / e-Ea/10000R
As A is a constant, 5.49 x 10 6 /e-Ea/5000R = 9.86 x 10 8 / e-Ea/10000R
Rearrange: (5.49 x 10 6)/(9.86 x 10 8) = (e-Ea/5000R)/(e-Ea/10000R)
Take natural logs: -5.19072 = (-Ea/5000R) - (-Ea/10000R)
Rearrange: -5.19072 = -Ea/10000R
Rearrange: Ea = 431557 = 432 kJ
Q641-05 What happens to the rate constant (k) and the activation energy (Ea) of a reaction when the temperature is increased?
1. k increases and Ea is unaffected
2. k decreases and Ea is unaffected
3. Ea increases and k is unaffected
4. Ea decreases and k is unaffected
Answer
Increasing temperature has no effect on the activation energy, but it does increase the rate constant. Therefore k increases and Ea is unaffected
Q641-06 For a given reaction in which the activation energies of the forward and reverse reactions are equal then:
1. the equilibrium constant must equal one.
2. the rate law can be determined from the stoichiometric equation.
3. the overall order must be zero.
4. ΔH must equal zero.
Answer
If the forward and back activation energies are equal and the value is equal to the difference between the energy level of the reactants (or products) and the highest energy state of the reaction pathway, then the enthalpy of reactants must equal the enthalpy of products. This is probably easier to see on a reaction profile graph. The only way for Eaf to equal Eab in the graph is for the enthalpy values of reactants to be the same as the products.
Q641-07 To what does 'A' in the Arrhenius equation k = Ae-Ea/RT refer?
1. Activation energy
2. Rate constant
3. Gas constant
4. Collision geometry
Answer
The constant 'A' in the Arrhenius equation is the orientation factor, or collision geometry, a number from 0 to 1 representing the probablility of a collision being successful (assuming sufficient energy).
Q641-08 Values of a rate constant, k, and absolute temperature, T, can be used to determine the activation energy of a reaction by a graphical method. Which graph produces a straight line?
1. k versus T
2. k versus 1/T
3. ln k versus T
4. ln k versus 1/T
Answer
Using the log form of the Arrhenius equation ln k = ln A -Ea/RT, and plotting ln k against 1/T gives a graph of the form y = mx + c. This is a straight line graph of gradient = -Ea/R and y intercept = ln A.
Q641-09 A student carried out a series of experiments to find the rate constant, k, for a reaction at 300K, 400K, 500K and 600K. He plotted the following graph as data processing. Using the graph, which of the following is the most probable activation energy for this reaction?
5.15 kJ mol-1
1.51 kJ mol-1
11.5 kJ mol-1
55.1 kJ mol-1
Answer
From the graph of ln k against 1/T gives a gradient = -Ea/R.
The gradient can be estimated to equal approximately 1/0.0007 = -1250
Therefore Ea = 1250 x 8.314 = approximately 10 kJ
Therefore the most likely answer is 11.5 kJ mol-1
Q641-10 What is the activation energy (in kJ) of a reaction whose rate constant increases by a factor of 100 upon increasing the temperature from 300 K to 360 K?
1. 27
2. 35
3. 42
4. 53
5. 69
Answer
Using the Arrhenius equation, k = Ae-Ea/RT
If k(at 360K) = 100 x k(at 300K), then Ae-Ea/RT(at 360K) = 100 x Ae-Ea/RT(at 300K)
Take log base e: ln A - Ea/360R = ln 100 + ln A - Ea/300R
Ea/300R - Ea/360R = ln100 = 4.605
Multiply through by 300R: Ea - 5Ea/6 = 4.605 x 300 x 8.314
Multiply through by 6: 6Ea - 5Ea = 68915 J
Therefore Ea = 68.9 kJ
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Nation
Nation
A nation is a territory where its the people are led by the same government. The word “nation” can also refer to a group of people who share a history, traditions, culture and, often, language—even if the group does not have a country of its own.
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A nation is a territory where all its people are led by the same government.
The word “nation” can also refer to a group of people who share a history, traditions, culture and, often, language—even if the group does not have a political territory of its own. People within this type of nation share a common identity, and think of themselves as belonging to the same group.
Palestinians, who live in and near the nation of Israel, fit in this category. Though Palestinians share a national identity, elect their own government, and share cultural beliefs, they do not have an internationally recognized nation of their own.
The United Nations currently recognizes 193 nations around the world, though only 192 are members of the UN's General Assembly. (Vatican City, which is led by the Roman Catholic Church, is recognized as a sovereign nation but is not a member of the General Assembly.)
Other nations are not recognized by one or more states for varying reasons. Sometimes, a single nation does not recognize another nation. North Korea and South Korea do not recognize each other as nations, for instance. They each oppose the politics of the other.
The leaders of some unrecognized nations maintain a “government in exile.” These leaders were ousted by social change, such as a revolution, in their country. The leaders currently live in another country, but consider themselves the leaders of their nation.
Sometimes, people of that nation want the leader to return. Many Tibetans, for instance, look forward to a time when the Dalai Lama, the leader of Tibet’s government in exile, will return to the country. The Dalai Lama has not been to Tibet since 1959, when Tibet became a part of China.
Other times, a government in exile can form entirely outside the nation it wants to govern. The Free Republic of Vietnam considers itself a government in exile of the nation of Vietnam. The Free Republic of Vietnam was formed after the Vietnam War by emigrants who did not want to live in the new, socialist government of Vietnam.
Fast Fact
First Nations
Native American tribes are considered independent nations in the United States and Canada. There are hundreds of nations, from the Ho Chuck Nation of Wisconsin to the Makah Nation of Washington state. For many Americans, the term First Nations has become a term for Indigenous people of North America.
First Nations is the official term for Canadian tribes. First Nations include the Dene in Canadas Arctic, the Mikmaq Confederacy on Prince Edward Island, and the Grand Council of the Crees in Quebec.
Fast Fact
Nationality
One nation may extend across several countries, and one country may include many nations. Kurdistan is an unrecognized nation that includes parts of Turkey, Armenia, Syria, Iran, and Iraq. The United Kingdom is a nation that includes the countries of England, Wales, Scotland, and Northern Ireland.
Website
United Nations: Member StatesUnited Nations: Non-Member States and Entities
Catholic
adjective
having to do with the Christian denomination with the Pope as its leader.
culture
noun
learned behavior of people, including their languages, belief systems, social structures, institutions, and material goods. Also: a group of people that share the same cultural traits and values.
Dalai Lama
noun
leader of Tibet.
diplomacy
noun
art and science of maintaining peaceful relationships between nations, groups, or individuals.
emigrant
noun
person who moves from their existing country or region to a new country or region.
First Nations
noun
indigenous (Native American) peoples of Canada south of the Arctic.
General Assembly
noun
main unit of the United Nations, made up of representatives from all member states.
govern
verb
to make public-policy decisions for a group or individuals.
government
noun
system or order of a nation, state, or other political unit.
government in exile
noun
leaders of a nation who live in another land, but hope to return to their native country.
indigenous
adjective
characteristic to or of a specific place.
nation
noun
political unit made of people who share a common territory.
national identity
noun
reputation and culture of a country.
Native American
noun
person whose ancestors were native inhabitants of North or South America. Native American usually does not include Eskimo or Hawaiian people.
political body
noun
organization that makes or influences public-policy decisions.
politics
noun
art and science of public policy.
recognize
verb
to identify or acknowledge.
revolution
noun
overthrow or total change of government.
social change
noun
alteration in attitude and outlook on issues, such as civil rights, that affect large communities.
socialism
noun
system of organization or government where all property, industry, and capital is owned by the community, not individuals.
sovereign nation
noun
independent state with control over its own territory.
territory
noun
land an animal, human, or government protects from intruders.
tradition
noun
beliefs, customs, and cultural characteristics handed down from one generation to the next.
United Nations
noun
international organization that works for peace, security and cooperation.
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Open Access Review
A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies
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Jennifer A. Mann
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Yves Starreveld
Yves Starreveld
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Yves Starreveld received his undergraduate degree from Brown University, Providence, Rhode Island, a[...]
Yves Starreveld received his undergraduate degree from Brown University, Providence, Rhode Island, USA, in 1990. He then moved to Canada to complete his medical degree at Queen’s University in 1996. In 2002, Dr. Starreveld obtained his Ph.D. in Medical Biophysics, with a thesis entitled: “Fast non-Linear Registration applied to Stereotactic Functional Neurosurgery”, along with his residency in neurosurgery (2006), both from the University of Western Ontario, Canada. Dr. Starreveld now focuses his research on computers in medicine and surgical simulation. Clinically, he specializes in adult epilepsy surgery and skull base surgery, including endoscopic approaches to the skull base.
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Jay Riva-Cambrin
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Jay Riva-Cambrin began his medical studies at the University of Alberta, Canada. After graduating in[...]
Jay Riva-Cambrin began his medical studies at the University of Alberta, Canada. After graduating from medical school in June 1998, he completed his residency at the University of Toronto, Canada. He obtained a Master of Science in Clinical Epidemiology from the University of Toronto under the direction of Dr. James Drake. Dr. Riva-Cambrin completed a postgraduate fellowship in pediatric neurological surgery at The Hospital for Sick Children, University of Toronto in Pediatric Neurosurgery. Dr. Riva-Cambrin’s clinical interests in Pediatric Neurosurgery include the treatment of hydrocephalus and clinical trials. He is a member of the Hydrocephalus Clinical Research Network (HCRN). His recent research interests include comparing the outcomes between ETV+CPC and shunts in the infant hydrocephalus population and exploring the relationship between ventricle size and neuropsychological outcomes in older children treated for hydrocephalus. It is for this work that he was a recipient of the Hydrocephalus Association MYIs’ research grant in 2012. Dr. Riva-Cambrin has recently joined the faculty at the University of Calgary from his previous role at the University of Utah. He is a Professor of Neurosurgery, the Director of the Neurosurgery residency training program, and is developing a neurosurgical clinical research center at the University of Calgary with a special focus on pediatric hydrocephalus.
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Kirstie Lithgow
Kirstie Lithgow
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Department of Clinical Neurosciences, Section of Neurosurgery, University of Calgary, Calgary, AB T2N 1N4, Canada
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Department of Community Health Sciences, University of Calgary, Calgary, AB T2N 1N4, Canada
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Division of Endocrinology and Metabolism, Department of Medicine, University of Calgary, Calgary, AB T2N 1N4, Canada
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Hotchkiss Brain Institute, University of Calgary, Calgary, AB T2N 1N4, Canada
Author to whom correspondence should be addressed.
J. Clin. Med.2025, 14(4), 1089;
Submission received: 8 January 2025 / Revised: 28 January 2025 / Accepted: 4 February 2025 / Published: 8 February 2025
(This article belongs to the Special Issue Advances in Pituitary Adenomas)
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Abstract
For several decades, dopamine agonist therapy has been the mainstay of treatment for prolactinomas, with surgery generally considered a second line for cases failing medical therapy due to intolerance or resistance. There is increasing recognition of the burden of long-term DA therapy; many patients experience debilitating side effects, and emerging evidence demonstrates that the prevalence of impulse control disorders has been vastly underreported. Long-term DA therapy is associated with significant costs to patients and healthcare systems, which is projected to exceed that of surgery in many circumstances. Recent advancements in surgical approaches, including endoscopic transsphenoidal surgery, have led to improved surgical outcomes (82–100% remission rates; serious complication rates < 2%), prompting a reappraisal of the role of surgery for prolactinoma. Favourable surgical outcomes have been observed in both remission and complication rates for microprolactinomas and well-circumscribed macroprolactinomas, leading to consideration of surgery as an earlier, or first-line, option in the treatment paradigm. Potential advantages of surgical management should be weighed against institutional case volume and expertise, the risk of perioperative complications, and the need for adjuvant medical therapy post-operatively. Ultimately, patients and care-providers should engage in shared decision-making following informed discussion about the risks and benefits of both medical and surgical approaches.
Keywords:
dopamine agonist; pituitary adenoma; prolactinoma; transsphenoidal surgery
1. Introduction
Prolactinomas are the most common subtype of pituitary adenoma, comprising over half of all pituitary adenomas and presenting most commonly in women of reproductive age . At presentation, they are classified based on maximum dimensions: macroprolactinomas are >1 cm and microprolactinomas are <1 cm . Patients can present with signs and symptoms of mass effect (most commonly visual field deficits), hypopituitarism (hyposecretion of one or more pituitary hormones), and hyperprolactinemia [3,4]. The features of hyperprolactinemia can occur either due to the impact of hypogonadism (due to inhibition of pulsatile gonadotropin-releasing hormone (GnRH) secondary to hyperprolactinemia) or from the primary effects of elevated prolactin . Manifestations of hypogonadism can include decreased libido, amenorrhea/oligomenorrhea, impaired erectile or ejaculatory function, infertility, and decreased bone mineral density . Manifestations specific to hyperprolactinemia include galactorrhea (most common in reproductive-age women) and headache. Diagnosis of prolactinoma requires the presence of a sellar lesion consistent with a pituitary adenoma in conjunction with elevated serum prolactin (typically >250 μg/L with elevation in prolactin in proportion to adenoma size) . The impact of medications known to cause elevations in prolactin should be considered when interpreting prolactin results .
Dopamine agonists (DAs) are derived from ergot, a group of fungi that grow on rye and related plants . In 1676, it was observed that ergot inhibited lactation . Three hundred years later, the first report of a reduction in the size of two prolactinomas following DA therapy with a semisynthetic ergot alkaloid (bromocriptine) was published . In 1981, a prospective study of eight patients with macroprolactinomas demonstrated reduction or normalization of serum prolactin as well as regression in tumour size following three months of bromocriptine treatment . The proven anti-secretory and antiproliferative potential of bromocriptine led to it replacing surgery for the first-line management of prolactinomas in the years that followed . Medical management of prolactinomas was further improved following the approval of the DA cabergoline in the 1990s, which demonstrated superior efficacy and tolerability over bromocriptine due to selectivity for D2R receptors . Currently commercially available DAs include bromocriptine, cabergoline, and quinagolide . Synthesis of results from previous studies have demonstrated that DA therapy leads to tumour shrinkage in 20 to 100% (median 62%) and normalization of serum prolactin in 40 to 100% (median 68%) . The greatest efficacy is reported to occur within the first six months of treatment initiation , and normalization of prolactin and tumour volume reduction exceeding 25% within the first 3 months of treatment is predictive of long-term response [12,13]. In select cases (i.e. normal prolactin and no visible tumour on MRI), withdrawal of DA after two or more years of therapy can be attempted with observed remission rates of 26 to 69% [2,3,14].
While DA therapy has been recommended for the first-line management of prolactinomas by international society guidelines for many years [2,14,15], advances and longitudinal experience with endoscopic endonasal surgery have yielded improved surgical outcomes over the last two decades , prompting reappraisal of the role of surgical management for prolactinoma [13,17]. Surgery on the pituitary gland in humans began in the late 19th century, led by Harvey Cushing, with the primary indications being vision loss and headache . Transcranial approaches (requiring craniotomy with an incision through the overlying tissues) were utilized by Cushing and subsequent surgeons throughout the 1990s . In the early 20th century, as otolaryngologists began to use the endoscope in the nasal cavity, neurosurgeons began to trial transsphenoidal approaches to the pituitary . The first transsphenoidal approach is believed to have been performed by Austrian surgeon Herman Schloffer in 1907 and was performed via a lateral rhinotomy . The transsphenoidal approach, in which surgeons access the sphenoid sinus to approach the sella, was largely abandoned and transcranial approaches were preferred until it was popularized by Jules Hardy in the 1960s who pioneered the use of the operating microscope and fluoroscopy [19,20]. Hardy had spent a year training with French surgeon Gerard Guiot, who was the first to use the endoscope in transsphenoidal surgery, although the technique did not gain traction until the 1990s when the endoscope was optimized by Karl Storz for visualization of anatomic structures [18,19,21,22,23]. Following the advent of the endoscopic transsphenoidal approach in the early 1990s [19,22,23], modern neurosurgeons have refined both microscopic and endoscopic endonasal transsphenoidal approaches to pituitary tumours, aided significantly by increased collaboration with otolaryngologists and the use of image guidance [18,24]. The contemporary microsurgical transsphenoidal approach utilizes instruments passed through the nostrils with a Hardy speculum facilitating visualization through the operative microscope . Endoscopic transsphenoidal surgery (TSS) is a minimally invasive technique by which surgeons access the sella through the sphenoid sinus with instruments passed through the nostrils and anatomic visualization through the endoscope, allowing for increased anatomic visualization as compared to the microscopic approach, often performed by a neurosurgeon and otolaryngologist team .
Potential advantages conferred by endoscopic TSS include decreased complication rates (due to improved anatomic visualization) and increased endocrinological remission rates [25,26]. Surgical remission rates are pooled at 82–100% in endoscopic series versus 71–93% in microscopic surgery . Furthermore, operative time, length of hospital stays, and blood loss are reduced with endoscopic TSS, making this approach highly safe and effective [28,29]. A systematic review and meta-analysis reported long-term postoperative remission rates of 83% for microprolactinomas (19 studies, n = 354) and 60% for macroprolactinomas (17 studies, n = 639) . Reported severe post-operative complication rates are <2%, making surgical resection appealing for amenable lesions . Accordingly, the most recent Pituitary Society International Consensus Statement recommends surgery as a first-line alternative to DA therapy in select cases (Figure 1).
Figure 1. History and evolution of prolactinoma management. This figure is original to this submission so no credit or license is needed.
To identify relevant literature concerning the surgical management of prolactinoma, we performed a search of PubMed including the following MeSH terms: prolactinoma, surgery, neurosurgery, and neurosurgical procedure. We additionally utilized the reference list review of the included manuscripts and known literature in the authors’ respective fields based on expertise. In this review, we will summarize the traditional indications for surgical management of prolactinoma before reappraising the contemporary role of surgery with emphasis on the expanding role of surgery as a first-line intervention. We have highlighted changes from previous 2011 guidelines to the recent 2023 guidelines from the Pituitary Society pertaining to the surgical management of prolactinoma. Finally, we will review perioperative considerations and emphasize the necessity of shared decision-making between healthcare providers and patients with prolactinoma.
2. Indications for Surgery
2.1. Traditional Indications for Surgery
Traditional indications for the surgical management of prolactinoma include resistance or intolerance to DA therapy, neurological deficit, pituitary apoplexy, cerebrospinal fluid (CSF) leak, and for volume reduction prior to pregnancy [2,30]. Within each of these indications, there are important nuances that warrant multidisciplinary discussion and shared decision-making with patients.
2.2. Resistance to Dopamine Agonist Therapy
The accepted definition of DA resistance is failure to achieve normoprolactinemia with maximally tolerated doses of DA and failure to achieve a 50% reduction in tumour size [2,31]. Resistance to cabergoline has been reported at 10% for micro- and 18% for macroprolactinomas . However, it should be emphasized that not all patients who meet the definition of DA resistance will require surgery. In the real-world treatment of macroprolactinomas, the most important endpoint is resolution of mass effect (i.e. decompression of optic chiasm), and once this is achieved, many patients can be stabilized on long-term DA despite not achieving normoprolactinemia or 50% size reduction . For microprolactinomas, size reduction is not a clinically important outcome, and therefore, decisions around surgical management for resistant cases will consider factors such as long-term treatment burden, side effects from escalating doses of DA, and symptoms of hyperprolactinemia. The presence of DA resistance alone is therefore not an absolute indication for surgery unless there are mass effect concerns such as chiasmal compression.
Observational studies examining post-operative remission rates in patients previously treated with DA demonstrate lower remission rates than primary surgery [34,35]; however, caution is warranted in interpreting these findings given differences in the baseline characteristics of cases that undergo primary vs. second-line surgery. A systematic review and meta-analysis (10 studies; n = 657) reported that 38% of patients who had surgery after failing DA therapy achieved long-term remission, and 62% achieved remission with multimodal therapy (additional surgery, radiosurgery, or DA reinitiation) . Subgroup analysis suggested that surgery alone had no effect on the long-term remission of macroprolactinomas and led to remission in 66% of microprolactinomas; however, 43% of macroprolactinomas achieved remission on multimodal therapy . It is important to consider that even if post-operative remission is not achieved, operative intervention may facilitate improved biochemical control with lower DA doses [34,35].
2.3. Dopamine Agonist Intolerance
DA intolerance is roughly defined as “side effects of the medication precluding its use ”. Common side effects of DA are dose-related and include gastrointestinal upset, dizziness, headache, and fatigue, which can be persistent and detrimental to quality of life [13,36,37]. Switching to cabergoline can be trialed in patients who do not tolerate bromocriptine due to its more favourable side effect profile . The side effects of DA are dose-related, with more frequent adverse effects occurring with escalating doses . Per a large retrospective cohort study of 455 patients taking cabergoline (median maintenance dose 0.5 mg weekly), 8.5% of patients experienced minor adverse effects and 4% had major or persistent side effects, with headache, postural hypotension, nausea, and sleepiness being the most commonly reported . A systematic review and meta-analysis including both surgically and medically managed prolactinomas reported higher pooled rates of side effects including fatigue (30%), libido changes resulting from side effects (28%), sleep disorders (25%), and nausea (17%), presumably related to higher doses of DA in this cohort .
Rare but serious potential side effects of DA include neuropsychiatric issues such as mood changes or impulse control disorders (ICDs), including hypersexuality, pathological gambling, compulsive shopping, and compulsive eating [13,39,40]. A previous systematic review reported pooled percentages of 3% for ICDs in prolactinoma patients treated with DA . However, a cross-sectional multicentre study that assessed 308 patients with DA-treated prolactinoma using validated questionnaires reported ICD prevalence of 17% . These clinically important adverse effects have been previously underreported due to a lack of clinical recognition and/or formal assessment . It is crucial that care providers disclose and monitor for these effects which may have devastating consequences for patients.
2.4. Neurological Deficit, Pituitary Apoplexy, and CSF Leak
Macroprolactinomas may cause visual dysfunction at presentation due to suprasellar extension and compression of the optic apparatus . However, even large and invasive (e.g., tumour extension into the cavernous sinus) macroadenomas may respond rapidly to DA treatment with resolution of mass effect and improvement in vision . If a trial of DA therapy is undertaken in this setting, this should involve a multidisciplinary approach with endocrinology, ophthalmology, and neurosurgery. Short-term reassessment of visual function following DA initiation is critical and surgical management should be offered if it does not improve .
Pituitary apoplexy is an emergent complication that presents with symptoms including headache, visual dysfunction, altered level of consciousness, and pituitary hormone deficiency due to infarction or hemorrhage of a pituitary adenoma . Management of pituitary apoplexy is individualized, and cases with severe visual and/or neurological dysfunction are offered surgical management while milder cases can be managed conservatively [4,43].
In large invasive macroprolactinomas, DA therapy can rarely cause rapid tumour shrinkage, leading to spontaneous CSF leak [13,44]. Patients at risk should be counselled about signs and symptoms (e.g., clear nasal discharge) and surgical treatment is usually required if this complication occurs [13,45].
2.5. Volume Reduction Prior to Pregnancy
Previous guidelines recommend patients with macroprolactinomas who do not achieve tumour shrinkage with DA be counselled about the potential benefit of surgery before attempting pregnancy , as the risk of symptomatic mass enlargement is estimated at 21% . If surgery is offered for this indication, it is crucial that patients are informed about the risk of acquired hypopituitarism following surgery, which would necessitate use of assisted reproductive technologies for conception as well as lifelong hormone replacement therapy .
2.6. Reappraisal of Surgery for Prolactinomas
The 2023 Pituitary Society International Consensus Guidelines highlight an emerging role for first-line surgery in primary management for select prolactinomas due to favourable surgical outcomes, the treatment burden of long-term DA therapy, and patient preference. The select cases for consideration of first-line surgical intervention as emphasized in the consensus statement include the following: microprolactinomas, well-circumscribed (Knosp grade 0 and 1) macroprolactinomas, young women, and patients desiring pregnancy .
Initial operative approaches to the pituitary were via transcranial routes; experimentation with subfrontal, transfrontal, subtemporal, and transtemporal routes eventually gave way to the pterional approach which is favoured in modern neurosurgery, whereby surgeons create a frontotemporal boneflap centred over the depression of the sphenoid ridge to access the basal arachnoid cisterns through the Sylvian fissure . Pterional craniotomy retains utility as a surgical approach to prolactinoma in instances such as failed transsphenoidal surgery, extrasellar tumour extension, or patient-specific anatomic limitations . The early transsphenoidal approach was initially performed transnasally, requiring incisions on or around the nose, reflection of the nose, and various degrees of resection involving the maxillary, frontal, or ethmoid sinuses [19,47]. Sublabial approaches were also popular for transsphenoidal surgery throughout the 1900s, involving resection of the bony nasal septum and nasal turbinate(s) . A myriad of additional approaches have been trialed to access the sella throughout the 19th, 20th, and 21st centuries, which are outside of the scope of this manuscript [19,47]. The modern transsphenoidal approach is performed endonasally with the aid of the operating microscope or endoscope. Given the trajectory of the endoscopic endonasal approach, there are inherent limitations, including suprasellar or lateral tumour extension, brain invasion, and tumour consistency, in which instances an open craniotomy may be more favourable . In addition to standard transsphenoidal approaches, surgeons have developed extended transsphenoidal approaches, maximizing access to the skull base to aid in the resection of invasive tumours and providing a corridor to the cavernous sinus [47,49]. Combined approaches are another viable option for more invasive tumours, such as the endoscopic transorbital approach (which provides further reach into the cavernous sinus in a minimally invasive manner) in combination with endoscopic endonasal surgery .
Important considerations include careful case selection and the availability of surgical expertise. With regard to case selection, tumour size and invasiveness are of paramount importance. The Knosp grade is a magnetic resonance imaging (MRI)-based classification scheme for tumoural invasion of the cavernous sinus, ranging from 0 (normal) to 4 (complete encasement of the internal carotid artery), with non-invasive adenomas generally considered < grade 2 and invasive adenomas considered ≥ grade 2 (Table 1) . Microprolactinomas and well-circumscribed macroprolactinomas (Knosp grades 0 and 1) have favourable remission rates in the hands of experienced neurosurgeons, making surgery attractive . For more invasive prolactinomas, surgical remission is often not feasible, and additional treatment post-operatively (adjuvant therapy) with either ongoing DA or radiotherapy is required. However, even if complete remission is not achieved, surgery can lead to a clinically significant reduction in prolactin levels, such that lower doses of DA can be used to control hyperprolactinemia [34,35].
Table 1. Knosp grade, adapted from Knosp et al. Classification of pituitary adenoma based on midsella magnetic resonance imaging. ICA: internal carotid artery.
A retrospective study of 86 Knosp ≤ 1 prolactinomas (41 macro- and 45 microprolactinomas) treated with first-line surgery demonstrated long-term remission (median follow-up: 80 months) of 72% for micro- and 45% for macroprolactinomas . However, 24% of the micro- and 49% of the macroprolactinomas (notably, 76% of Knosp 1 vs. 29% of Knosp 0) required adjuvant DA to achieve remission . Another series of 78 prolactinomas (mostly pre-treated with DA) reported long-term remission rates (mean follow-up: 66 months) of 65% for Knosp 0–2 tumours, which increased to 81% when including remission with adjuvant DA . A previous systematic review and meta-analysis (13 studies and 809 patients) demonstrated significantly higher remission rates for patients treated with surgery than with DA (88% vs. 52%; p = 0.001) . Additionally, remission rates of 91 vs. 60% (p = 0.002) for micro- and 77% vs. 43% (p = 0.003) for macroprolactinomas were reported for first-line surgical vs. medical management, respectively (although this study was unable to assess the use of post-operative DA) .
Reported complication rates following prolactinoma surgery are favourable. A systematic review and meta-analysis including surgically managed micro- and macroprolactinomas (25 studies; n = 1836) reported 0% mortality, severe complications (permanent diabetes insipidus, meningitis, and CSF leak) ≤ 3%, and anterior hypopituitarism in 2% . A retrospective cohort study of 114 microprolactinomas reported an overall complication rate of 4% (of which 3% were epistaxis, rhinitis, and sinusitis), with 0% hypopituitarism and only a single case of permanent diabetes insipidus . Despite these promising data, post-operative remission and complication rates reported in the literature are primarily from tertiary referral centres with expert skull base surgeons; therefore, these results are only generalizable to similar care settings. Furthermore, such centres may still lack recent experience with operative management of prolactinomas and therefore require an additional case volume to optimize outcomes. Abou-Al-Shaar et al. reported significantly improved long-term remission rates for Knosp 0–2 prolactinomas in the second decade of experience (77% in 2011–2019 vs. 47% in 2002–2010), suggesting a surgical learning curve. The role of surgery in the treatment paradigm of prolactinomas should therefore be tailored to institutional case volume and surgical expertise.
Patients with prolactinoma experience worse quality of life, particularly mental health, than healthy controls, and mental and physical quality of life scores vary inversely with prolactin levels, emphasizing the importance of assessing patient-reported outcome measures in the surgical decision-making process . In a recent prospective cohort study, Van Trigt et al. assessed health-related quality of life with the Leiden Bother and Needs Pituitary questionnaire in 100 prolactinoma patients treated via endoscopic TSS and found that Bother and Needs scores both decreased significantly post-operatively. In a recent systematic review of 18 articles, a clear difference in quality of life was not found between medical or surgical treatment modality, but surgery was associated with improved quality of life as soon as 5 days postoperatively, with continuous improvement throughout the first year, with prolactinoma showing the greatest quality of life improvement of all adenoma types . Although further study is needed to better understand the quality of life outcomes in medically versus surgically treated prolactinoma patients, these results support the immediate positive effects of surgery in this patient population.
The cost-effectiveness of surgery vs. long-term DA therapy warrants consideration , given that the majority of patients will require life-long medical therapy . Two previous analyses have demonstrated surgery is more cost-effective than medical therapy for microprolactinomas [58,59], which is amplified with cumulative years of medication use . However, the cost–advantage switched to favour cabergoline over surgical intervention only when the theoretical surgical cure rate fell below 30% .
Patient preference is becoming an important indication for surgical intervention, reflecting the emphasis on patient-centric care in modern neurosurgery . In particular, surgery may be a favourable option for young women with micro- or Knosp 0-1 prolactinomas to restore fertility and avoid long-term DA use . A meta-analysis of 14 studies (n = 603) including women undergoing TSS for prolactinoma reported a significant reduction in rates of amenorrhea (96% vs. 40%; p< 0.01) and galactorrhea (84% vs. 29%; p< 0.01) post-operatively . Rates of hypopituitarism could not be analyzed due to heterogeneity but ranged from 0–23%, with rates >10% reported in only two studies , which further highlights the importance of institutional expertise. We suggest that care providers engage in shared decision-making (SDM) with patients to emphasize patient preference. SDM should encourage patient collaboration by presenting clearly the prolactinoma diagnosis and the risks and benefits of both treatment options (e.g., DA therapy versus surgery) . Based on the authors’ experience, this is best accomplished in a multi-disciplinary setting, whereby a neurosurgeon and endocrinologist present the patient with respective treatment options, patient and support persons’ questions are addressed immediately, the patient’s opinion is explicitly elicited, and they are empowered to make the best treatment decision for their circumstances. SDM requires that physicians are attentive to the unique preferences and goals of the patient (e.g., do they desire fertility? If not now, will they in the future?), tailor the risk/benefit discussion accordingly, and provide time and space for the patient’s voice in the discussion .
3. Perioperative Considerations
A collaborative approach between endocrinology and neurosurgery is imperative in managing patients who undergo surgical management of prolactinoma. General considerations for the perioperative management of pituitary adenoma are beyond the scope of this review and are discussed in detail elsewhere . Patients who opt to undergo primary surgical treatment of prolactinoma should be counselled regarding the risks and benefits of both surgical and medical approaches so that they can make an informed decision . Surgery for prolactinoma should be performed at a pituitary centre of excellence by an experienced pituitary surgeon .
As mentioned, surgical intervention for pituitary adenoma is not without risk of complications. The most serious potential complication of transsphenoidal pituitary surgery is carotid artery injury, which is exceedingly rare (<0.2% in large endoscopic transsphenoidal series ). Additional serious complications include CSF leak, hypopituitarism, permanent diabetes insipidus, and visual deterioration, which all remain rare. More common complications include transient diabetes insipidus and syndrome of inappropriate antidiuretic hormone (SIADH) secretion; these are readily treatable, short-term consequences (with rates ranging from 10 to 20% ). Intraoperatively, adenoma resection may require or cause surgical manipulation of the posterior pituitary lobe, stalk, or infundibulum, which can cause antidiuretic hormone (ADH) secretion disturbances . ADH deficiency causes renal free water losses and subsequent hypernatremia, whereas ADH hypersecretion (SIADH) causes water retention and hyponatremia . For both conditions, careful management of post-operative fluid and electrolyte balance is warranted, and management may be performed by neurosurgeons, although we suggest that endocrinologist involvement where possible is best practice. Diabetes insipidus most commonly occurs in the first two post-operative days and is often treated with the vasopressin analogue desmopressin (DDAVP), and SIADH typically occurs 4–10 days post-operatively and is commonly treated with fluid restriction [66,67]. Post-operative CSF leak (incidence <5% in experienced centres ) is managed by the neurosurgical team and treated via conservative management (e.g., bedrest), lumbar drain insertion, or revision surgery . Post-operative adrenal insufficiency must be excluded with serum cortisol assessment and glucocorticoid therapy initiated in patients when appropriate .
Controversy exists about technical surgical challenges arising from tumour fibrosis secondary to pre-surgery treatment with DA [69,70]. Tumour texture limits the surgeon’s ability to utilize gentle suction and ring curettes for tumour resection and often adheres the tumour tissue to surrounding structures, which may increase the likelihood of incomplete resection and bleeding. A recent retrospective study including 290 prolactinomas (199 macroprolactinomas and 91 microprolactinomas) compared perioperative outcomes and pathological findings for prolactinomas that were pre-treated with DA vs. those that underwent primary surgery . DA-pre-treated macroprolactinomas were found to have significantly higher intraoperative blood loss, longer surgical duration, and more overall surgical morbidity . The authors demonstrated that DA pre-treatment was an independent risk factor for tenacious tumour consistency after adjustment for age, sex, tumour volume, and disease duration . Pathological analysis illustrated increased intratumoural collagen content in the DA pre-treated cases, supporting tumour fibrosis as the putative mechanism . Furthermore, a significant positive correlation between cumulative DA dose and intratumoural collagen content was observed for macroprolactinomas . Taken together, these findings support that DA pre-treatment for macroprolactinomas may impact perioperative outcomes, and therefore, there is rationale to minimize or avoid DA exposure in cases when primary surgical management is preferred. However, it should be noted that differences in perioperative outcomes and intratumoural collagen content were not observed for microprolactinomas, though lower post-operative remission rates in the DA-pretreated group (87% vs. 100% in the initial surgery group) were reported . Importantly, the DA exposure in this cohort was exclusively bromocriptine ; therefore, it remains unclear whether these findings are generalizable to the use of cabergoline, especially given that a previous study suggested that prolactinoma fibrosis is less pronounced in cabergoline than in bromocriptine .
Following surgical management, ongoing follow-up with endocrinology is required to determine if remission has been achieved. In cases where DA resistance is the surgical indication, a reduced dose may be efficacious following surgical debulking . Ongoing follow-up of prolactin levels and titration of DA post-operatively should be performed by the endocrinologist , in addition to the usual post-operative biochemical monitoring that is standard of care for all pituitary adenomas .
4. Conclusions
Both traditional and contemporary indications for the surgical management of prolactinoma should consider patient, tumour, surgical, and system factors (Figure 2). Goals of surgery (remission for micro- and select macroprolactinomas vs. debulking for more invasive tumours) and potential risks and limitations should be clearly communicated to patients. Additional studies are required to further elucidate the impact of DA (cabergoline in particular) pre-treatment on tumour fibrosis and surgical remission rates to determine whether up-front DA should be routinely avoided for select cases until surgical options are discussed. Unique cases such as young females with microprolactinoma require particular emphasis on surgical options due to the potential for fertility restoration and sparing of DA therapy, which in turn could facilitate thousands of dollars in cost savings for the patient and healthcare system. Management decisions should ideally be made in a multidisciplinary care setting (with availability of both neurosurgical and endocrinological expertise) utilizing a patient-centred approach.
Figure 2. Factors impacting shared decision-making for the management of prolactinoma. This figure is original to this submission so no credit or license is needed.
Author Contributions
Conceptualization, K.L. and J.A.M.; methodology, K.L. and J.A.M.; investigation, K.L., J.A.M., Y.S. and J.R.-C.; resources, K.L., J.A.M., Y.S. and J.R.-C.; data curation, K.L., J.A.M., Y.S. and J.R.-C.; writing—original draft preparation, K.L. and J.A.M.; writing—review and editing, K.L., J.A.M., Y.S. and J.R.-C.; supervision, K.L. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
This review did not involve human participants, Research Ethics Board approval was not sought as we utilized previously published work.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflicts of interest.
Abbreviations
ADH antidiuretic hormone
CSF cerebrospinal fluid
DA dopamine agonist
GnRH gonadotropin-releasing hormone
ICD impulse control disorder
MRI magnetic resonance imaging
SIADH syndrome of inappropriate antidiuretic hormone
SDM shared decision-making
TSS transsphenoidal surgery
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Figure 1. History and evolution of prolactinoma management. This figure is original to this submission so no credit or license is needed.
Figure 2. Factors impacting shared decision-making for the management of prolactinoma. This figure is original to this submission so no credit or license is needed.
Table 1. Knosp grade, adapted from Knosp et al. Classification of pituitary adenoma based on midsella magnetic resonance imaging. ICA: internal carotid artery.
| Knosp Grade | Definition |
:--- |
| 0 | No cavernous sinus invasion, the lesion does not extend past medial aspects of the intra- and supracavernous ICA |
| 1 | The lesion extends to, but not past, the tangent line through cross-sectional centres of the intra- and supracavernous ICA |
| 2 | The lesion extends to, but not past, the tangent line through lateral aspects of intra- and supracavernous ICA |
| 3 | The lesion invades past the tangent line through the lateral intra- and supracavernous ICA into the cavernous sinus |
| 4 | The lesion completely encases the intracavernous carotid artery and the cavernous sinus is invaded |
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Mann, J.A.; Starreveld, Y.; Riva-Cambrin, J.; Lithgow, K. A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. J. Clin. Med.2025, 14, 1089.
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Mann JA, Starreveld Y, Riva-Cambrin J, Lithgow K. A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. Journal of Clinical Medicine. 2025; 14(4):1089.
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Mann, Jennifer A., Yves Starreveld, Jay Riva-Cambrin, and Kirstie Lithgow. 2025. "A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies" Journal of Clinical Medicine 14, no. 4: 1089.
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Mann, J. A., Starreveld, Y., Riva-Cambrin, J., & Lithgow, K. (2025). A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. Journal of Clinical Medicine, 14(4), 1089.
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Mann, J.A.; Starreveld, Y.; Riva-Cambrin, J.; Lithgow, K. A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. J. Clin. Med.2025, 14, 1089.
AMA Style
Mann JA, Starreveld Y, Riva-Cambrin J, Lithgow K. A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. Journal of Clinical Medicine. 2025; 14(4):1089.
Chicago/Turabian Style
Mann, Jennifer A., Yves Starreveld, Jay Riva-Cambrin, and Kirstie Lithgow. 2025. "A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies" Journal of Clinical Medicine 14, no. 4: 1089.
APA Style
Mann, J. A., Starreveld, Y., Riva-Cambrin, J., & Lithgow, K. (2025). A Narrative Review of Surgery for Prolactinomas: Considerations and Controversies. Journal of Clinical Medicine, 14(4), 1089.
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633 | https://www.bymath.com/studyguide/geo/sec/geo14.php | All Elementary Mathematics
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Angles. Projections. Polyhedral angles
_Angles. Angle between: two intersecting straight lines, two parallel
straight lines, two crossing straight lines. Perpendicular to a plane.
Projections. Projection of a point and a segment to a plane. Dihedral
angle. Linear angle of a dihedral angle. Angle between two planes.
Perpendicular planes. Angle between two perpendicular planes.
Angle between two parallel planes. Polyhedral angle. Plane angles.
Trihedral angle as a minimal polyhedral angle. Parallel sections of
a polyhedral angle.
Angles. An angle between two intersecting straight lines_ is measured as well as in a planimetry ( because it is possible to draw a plane through these lines ). An angle between two parallel straight lines is accepted equal to 0 or 180°. An angle between two crossing straight lines AB and CD ( Fig.70 ) is determined as follows: through any point O the rays OM and ON are drawn so, that OM || AB and ON || CD. Then an angle between AB and CD is accepted equal to an angle NOM. In other words, the straight lines AB and CD are transferred to a new position parallel to themselves until their intersection. Particularly, the point O can be taken in one of the lines AB or CD, which is immovable in this case.
The straight line AB, intersecting the plane P in the point O ( Fig.71 ), forms different angles ( the angles BOC, BOD, BOE ) with the different straight lines OC, OD, OE, drawn in the plane P through the point O. If the line AB is perpendicular to the two of these straight lines ( for instance, OC and OE ), then it is perpendicular to all the straight lines, drawn in this plane through the point O. In this case the straight line AB is called perpendicular to the plane P, and the plane P – perpendicular to the straight line AB.
Projections. A projection of the point A to the plane P is called a base C of the perpendicular AC, drawn from the point A to the plane P. A projection of the segment AB to the plane P is the segment CD, ends of which are projections of the points A and B ( Fig.72 ). It is possible to project not only a straight line, but any curve ABCDE ( Fig. 73 ) to a plane.
Lengths l of the projection CD and a of the segment AB ( Fig.72 ) are tied by the relation:
Dihedral angle. A figure, formed by two half-planes Q and R, going through the same straight line MN ( Fig.74 ), is called a dihedral angle . The straight line MN is called an edge of a dihedral angle; half-planes Q and R – its faces . The plane P, perpendicular to the edge MN, gives the angle AOB in its intersection with the half-planes Q and R. The angle AOB is called a linear angle of a dihedral angle. A linear angle is a measure of its dihedral angle.
Angles between planes. Two planes are called perpendicular planes , if they form a right angle. An angle between two parallel planes is accepted equal to zero. In general case an angle between two planes P and Q ( Fig.75 ) can be measured by an angle, formed by the straight lines AB and CD, which are perpendicular to the planes P and Q correspondingly.
Polyhedral angle. If to draw through the point O ( Fig.76 ) a set of planes AOB, BOC, COD etc., which are consequently intersected one with another along the straight lines OB, OC, OD etc. ( the last of them EOA intersects the first AOB along the straight line OA ), then we receive a figure, called a polyhedral angle. The point O is called a vertex of a polyhedral angle. Planes, forming the polyhedral angle (AOB, BOC, COD, …, EOA), are called its faces ; straight lines, along which the consequent faces intersect ( OA, OB, OC, … , OE ) are called edges of a polyhedral angle. Angles AOB, BOC, COD, … , EOA are called its plane angles . The minimal number of faces of a polyhedral angle is 3 ( the trihedral angle , Fig.77 ).
Parallel planes cut off the proportional segments ( OA:O a = OB:O b = OC:O c = …) on edges of a polyhedral angle ( Fig.78 ) and form the similar polygons ( ABCD and abcd ).
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634 | https://courses.washington.edu/ess431/LECTURES/LECTURE_2016/Lect18_Sea_Ice_I_2016.pdf | Sea Ice Lecture Notes ESS 431 Bonnie Light Polar Science Center / Applied Physics Lab bonnie@apl.washington.edu 12/5/16, 12/7/16 What is sea ice?
- Frozen sea water - Forms, grows, melts in the ocean - Grows in winter, melts in summer, can melt completely or survive multiple years - Typical thickness: first year ice <= ~1.8 m, 2-3 year old 2 – 3 m, 6 year old ~ 5 m D. Perovich B. Light Image courtesy of the National Snow and Ice Data Center, University of Colorado Approx. maximum and minimum climatology 1979-2000 Typical thickness (m) Winter extent (km2) Summer extent (km2) Seasonal area change Seasonal mass change (kg) Arctic (2012) 2.0 – 2.5 15 x 106 14 x 106 7 x 106 3.4 x 106 8 x 106 10.6 x 106 1.7 x 1015 1.9 x 1016 Antarctic (2014) 0.5 – 1.0 18.7 x 106 20.0 x 106 4 x 106 14.7 x 106 1.5 x 1015 SEA ICE VITAL STATISTICS • Coverage at maximum extent: 8% of Southern Hemisphere 5% of Northern Hemisphere • Comparison of Arctic and Antarctic Ice Packs Passive microwave-derived (SMMR / SSM/I) sea ice area for the Northern Hemisphere January 1979 – April 2016 Image from Passive microwave-derived (SMMR / SSM/I) sea ice extent September means for NH -13.3 % / decade Hmmm, does this mean I believe in climate change?
No.
Hmmm, does this mean I believe in climate change?
Hmmm, does this mean I believe in climate change?
No.
1. Climate science is not a belief 2. Let’s understand the science… 1. Affects polar ecosystem, wildlife WHY IS SEA ICE IMPORTANT?
- Important habitat for phytoplankton, zooplankton, Arctic cod, seals/walrus, polar bear - Ice algae contributes 10 – 20% to Arctic primary production - Effect of reduced ice cover uncertain 1. Affects polar ecosystem, wildlife 2. Affects people who live in polar regions (subsistence hunting, travel) WHY IS SEA ICE IMPORTANT?
Alaska Coastal Management Program 2007: Northwest passage open for first time in human memory 2008: Both Northern Sea Route and Northwest passage open for 1 week • How heat is absorbed / dissipated on earth determines climate • Poles are earth’s “heat sinks” • Summer sunlight 24/7 • Ice extent, concentration, thickness matter • Sea ice is thin (2/3 surface area, 0.1% volume) • Annual melt / freeze cycle produces large changes in areal coverage, surface properties • Climate sensitivity Change in climate → change in ice Change in ice → change in climate 1. Affects polar ecosystem, wildlife 2. Affects people who live in polar regions (subsistence hunting, travel) 3. Role in earth’s climate WHY IS SEA ICE IMPORTANT?
SEA ICE AND CLIMATE • Paleotemperature data indicate DT a increased with increasing latitude, e.g.,during last ice age: DT a = 3 - 5 ºC at low, mid-latitudes DT a = 8 - 10 º C at high latitudes • Suggests strongest indications of climate change would be in the polar regions • Climate models predict similar pattern • Several factors may act to amplify climate signals at high latitude: • loss of snow and ice creates increased heat absorption • larger fraction of heat at high latitude warms atmosphere, heat at low latitude goes to evaporation • reduced sea ice cover allows for enhanced sea-air heat transfer CURRENTLY… • Atmospheric temperatures in the Arctic are warmest in 400 years • Average temperature has risen at almost twice the rate as the rest of the planet during past few decades (ACIA, November 2004; ww.acia.uaf.edu) Hansen, J., R. Ruedy, M. Sato, and K. Lo (2010), Global surface temperature change, Rev. Geophys., 48, RG4004, doi:10.1029/2010RG000345.
Decadal surface temperature anomalies relative to 1951–1980 base period.
Spring Summer - Ample sunlight - Sea ice reflects 60-90% of the incoming solar radiation, ocean < 10% - A slight warming of ocean or atmosphere ice extent decreases more ocean exposed increased solar energy absorbed - This positive feedback process has the potential to amplify variations in climate ICE - ALBEDO FEEDBACK What is the difference between sea ice and lake ice?
• Sea ice forms from salty ocean water • Lake ice forms from fresh water Þ What happens as warm water cools?
r-T DIAGRAM FOR FRESH WATER Density (g / cm3) LAKE ICE FORMATION r - T DIAGRAM FOR SALT WATER SEA ICE FORMATION • If S > 24.7 o/oo then colder is denser • Cooling water sinks away from cold surface • Sea ice forms more slowly than fresh ice • Tf = -1.8 C for S~ 32 ppt • Typically, top 100 – 150 m of ocean must be cooled to Tf for ice to form T-S DIAGRAM FOR SALT WATER Tmax r Ice Growth Frazil Grease ice Nilas Congelation ice Rafting Pancakes Consolidation Young ice Thickens by growth at bottom First-year ice 30 – 200 cm thick, typically snow covered snow Calm conditions Wind, waves INITIAL ICE FORMATION • Frazil Crystals - Initial freezing occurs at many points of nucleation within water column - Needles and platelets of ice (3-4 mm diameter) float to surface to form slush INITIAL ICE FORMATION • Frazil Crystals - Initial freezing occurs at many points of nucleation within water column - Needles and platelets of ice (3-4 mm diameter) float to surface to form slush • Grease Ice - Ocean covered by grey, soupy mixture of frazil crystals which look “greasy” - Forms solid ice cover when H is about 1 cm, if no wind or waves Photo D. Perovich INITIAL ICE FORMATION (continued) • Pancake Ice - In presence of waves, grease ice evolves into flat, circular, rounded masses of semi-consolidated slush called "pancakes" - Pancakes typically 0.3 to 3 m in diameter - Wave motion can delay formation of solid ice cover until H is 10-50 cm Photo D. Perovich INITIAL ICE FORMATION (continued) • Congelation Growth - Once solid cover forms, ice grows at the bottom due to heat conducted upward toward the colder surface - called congelation growth SEA ICE STRUCTURE • Crystal Orientation - Initially c-axes oriented randomly - After 10 -20 cm congelation growth c-axes are horizontally-oriented - Preferential growth of crystals with horizontal c-axes - Crystals with horizontal c-axes grow downward up to 2 orders of magnitude more rapidly than those with more vertical c-axes - “Geometric selection” - Horizontal c-axes important to development of characteristic properties of sea ice basal plane prism faces • Skeletal Layer - In sea water, delicate layer of thin platelets grows on prism faces perpendicular to c-axis - Platelets are parallel, form a single crystal. Thickness ~ 0.5 mm and spacing of 1-2 mm.
- No skeletal layer in fresh water ice, so salt must be important - Similar phenomena observed when metals solidify (“constitutional supercooling”) • Constitutional Supercooling - Rejection of salt produces steep salinity gradient in the boundary layer Fs ↓ - Temperature at ice-water interface must be at freezing (for high S), causing a temperature gradient Fh ↑ - Heat diffusion >> salt diffusion Þ region of supercooling - Ice projecting into supercooled region will grow faster than planar interface Þ platelet formation - Length and spacing of platelets act to minimize amount of supercooling ahead of growing interface - Mathematical theory describes observed platelet patterns high S S @ 32 o/oo Fh ↑ Fs ↓ SEA ICE SALINITY - Amount of salt trapped in the ice described by “ice salinity” (Si) oo o 3 / 10 brine of mass ice of mass salt of mass × + = i S - Si directly related to rate of ice growth - Faster the growth, the greater the amount of salt trapped - Si should be high when ice is thin and decrease as ice becomes thicker (ppt) SEA ICE SALINITY - Profiles in thin ice tend to be “C” shaped, indicating salt enrichment in lower half of slab - Average salinity decreases as ice ages, indicating loss of salt to ocean - Dramatic change as ice goes from FY to MY BRINE INCLUSIONS - Bridging between platelets traps pockets of brine (d @ 0.05 mm) - Brine pockets form in vertical strings marking boundary between platelets - Salinity in brine pockets (Sb) independent of original water salinity because brine must be at salinity-determined freezing point oo o 3 / 10 brine of mass salt of mass × = b S - Changes in Ti cause melting or freezing in brine pockets, causing Sb to change until the brine is in freezing equilibrium with surrounding ice - These changes are independent of Si FREEZING EQUILIBRIUM - Changes in Ti cause melting or freezing in brine pockets, causing Sb to change until the brine is in freezing equilibrium with surrounding ice - These changes are independent of Si m T S i b -» FREEZING EQUILIBRIUM FREEZING EQUILIBRIUM DESALINATION MECHANISMS • Gravity Drainage - thermal and mechanical stresses form vertical cracks along brine pocket strings - result in vertical channels once H > 40 cm - brine in the interior is colder and denser than brine near the bottom, causing downward drainage • Flushing - sea ice becomes very porous during summer - surface meltwater produces hydrostatic pressure forcing low salinity meltwater into upper part of ice and a loss of higher salinity brine through the bottom of the ice Skeletal Layer SEA ICE STRUCTURE Light et al., 2003 FRACTIONAL BRINE VOLUME Presence of brine in sea ice alters essentially all its physical properties, making them very sensitive to temperature especially when ice becomes warm Brine Volume Define fractional brine volume (nb) as: b i i i i b i b b b i V S m S v V S T r r r r = @ @ -(where Ti is in ºC) 0 -5 -10 -15 -20 -25 0 5 10 15 20 25 Salinity 5 ppt 10 ppt Brine Volume (%) Temperature (C) multiyear first-year x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x PHYSICAL PROPERTIES OF SEA ICE 1. Thermal 2. Mechanical 3. Electromagnetic a. Optical b. Infrared c. Microwave 4. Thermodynamic growth 5. Dynamic • Thermal Properties Melting/Freezing Point No single melting or freezing point because any change in Ti produces internal melting or freezing in the brine pockets -15 °C -5 °C -2 °C Latent Heat of Fusion : Heat required to melt a unit volume of sea ice - Lo is the latent heat of fusion of pure ice (333.55 J/g) - For new ice, Li typically 0.6 Lo - For thick multiyear ice, Li typically 0.9 Lo ) 1 ( L ) 1 ( o i b i i b o i T S m L L r r n + @ -= Thermal Conductivity Determines rate of heat conduction through the ice Brine pockets act as thermal buffers to retard the conduction of heat and the warming or cooling of the ice kice = 2.4 W/mC kbrine = 0.6 W/mC What happens to kice with increasing brine volume? c T F k z ¶ = ¶ T = -1.8 C T = -15 C Fc factor of 4 - kice varies by 10-20% in multiyear ice - kice varies by 50% in warm first-year ice -1 -2 -3 -4 -5 -6 -7 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 Salinity (ppt) 1 3 6 10 Thermal Conductivity (W / m C) Temperature (oC) Specific Heat : Heat required to raise 1 g of sea ice 1 °C -1 -2 -3 -4 -5 -6 0 10 20 30 40 50 Salinity (ppt) 1 3 6 10 Specific Heat (cal / g C) Temperature (oC) 200 160 Specific Heat (J / g K) 120 80 40 0 c c aT bS T i o i i i = + + 2 co is the specific heat of pure ice a and b are constants ci can vary by nearly two orders of magnitude • Mechanical Properties - As Ti increases, the ice becomes more porous due to increased nb - This causes the ice to become weaker - All mechanical properties are thus sensitive to nb, i.e. to Ti and Si - Empirical data show that compressive strength (si) can be related to nb by si = so (1 - Anb0.5 ) where so is the compressive strength of pure ice and A is a constant - For engineering purposes, si →0 as nb → 30% - In nature, ice can still retain integrity until nb @ 60%, even though it may have negligible compressive strength - Mechanical properties are directional - stronger under vertical stress - important for offshore structures or load-bearing capacity of the ice Brine tubes Idealized picture of sea ice structure ELECTROMAGNETIC PROPERTIES (OPTICAL) - Interaction with solar (shortwave) radiation - Shortwave (SW) radiation is one of largest components of surface heat balance - 300 nm (UV) – 4000 nm (IR) - roughly half of the energy is visible, near UV (300 – 700 nm) - important for calculating mass changes - heat storage in the ice and upper ocean • Albedo a depends on - ice thickness and type - snow cover - presence of liquid water near the surface - examples: .80 - .90 cold, snow-covered ice .65 - .75 melting, snow-covered ice .55 - .65 clean, melting multiyear ice .35 - .60 cold, thin (20<H<100 cm) first-year ice .45 - .55 melting first-year ice .15 - .45 melt ponds .10 - .35 new (H<20 cm) ice .07 - .10 open water SW incident SW reflected = a Photo: B. Elder Spatially-averaged (i.e. measured by aircraft or satellite) albedos typically reduced to 0.45-0.50 during summer due to presence of melt ponds Spectral Albedos - a(l); where l = wavelength - Important for interpreting satellite measurements and for calculating SW energy absorption and transmission within the ice - Relatively constant between 400-600 nm reason why snow and drained ice appear white - Decreases steadily at near infrared wavelengths a: cold MY ice b: melting MY ice c: melting FY “white” ice d: melting FY “blue” ice a: dry snow b: wet new snow c: melting old snow d: frozen pond e: early MY pond e: ageing ponds Spectral Albedos Extinction Coefficient (Kl) - Needed to calculate: - SW energy absorbed by ice - SW energy transmitted through ice (mixed layer heating, photosynthesis) - Beer's Law: F F e z z ( ) ( ) l l kl = -0 where Fo(l) is the net SW flux at the surface - The larger the Kl, the less light penetration - Averaged over l: K = 0.5 - 1.5 m-1 for sea ice K = 15 m-1 for dry snow - Relatively constant between 400-600 nm - Increases roughly an order of magnitude between 600-800 nm - Visible light (particularly blue) penetrates deep into ice - Most near-infrared energy is absorbed close to surface a: dry snow b: wet snow c: surface layer of ice d: interior ice e: fy blue ice / ponds f: mature melt pond B. Light 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 20 40 60 80 100 FY Melt Pond MY Ice Dry Snow Percent Transmitted Thickness (m) - Field data from 1940-70s showed very little bottom melting during summer - Data from late 1990s: total bottom melting @ total surface melting - Data from 2000s indicates bottom melt in some areas > surface melt - Average summer ice thickness H decreased from 2.5 to 1.5 m between these two earlier periods, allowing more SW radiation to be transmitted to ocean through ice - Figure above points to a strong positive feedback between decreasing H and increasing bottom melting - This “ice-transmittance” feedback is likely contributing to Arctic sea ice loss Beer's Law equation predicts: F F e z z ( ) ( ) l l kl = -0 First-year ponded ice Multiyear ice Dry snow |
635 | https://www.youtube.com/watch?v=1rhiXeCekyk | Percents and Double Number Lines Video
Brian McKenzie
806 subscribers
333 likes
Description
46464 views
Posted: 10 Nov 2015
Transcript:
welcome back today's lesson is percent and double number lines our standard is 6 rp3 understanding ratio Concepts and ratio reasoning to solve problems I can use a d number line to solve a percent problem is our I can statement no that was a little fast there make sure you get it all written down and then hit play when you're ready all right today in class we talked about using a double number line to represent something happening in a rate problem today we're going to do that inv involving a percent instead so percent we know means out of 100 so we're going to use a double number line one line of our double number line is always the percent I like to put it on top so it goes from zero to 100% And then along the second line is zero to our total amount from our particular problem so get that jotted down as a reference and then we're going to try a couple of them together all right our first example should say what number is 25% of 60 so get that down set up your 0 to 100% And your zero to total and we're going to start walking through this example all right our percent is 25% so we need to figure out how many times does 25 go into a 100 or how many quarters are in a dollar is what I always say so there's four quarters in a dollar so we need to divide our number line into four now of means 60 so right above that word that says total we need to fill in the 60 you would have done that first wouldn't you yeah I would have done that first well all right so back to what we were talking about so we need to label our lines that we just made on our number line 25% 50% 75% and then 100 is already labeled for us we have four sections to our number line so off to the side we need to figure out what 60 ided 4 is so what is our total divide what are you doing boy divided by how many SE sections we have so our total is 60 divided by our four sections should give us what 15 so we know each number line on the bottom is now worth 15 so we have to count by 15 then 30 then 45 and 60 so we have four equal increments of our percents and our total so we then to finish a problem need to go back and figure out what number is 25% of 60 so we look below the 25% and we have 15 our answer is 15 is 25% of 60 all right might be a little confusing it's a lot of steps so let's try another one together so jot this one down get your double number line set up and then we'll get ready to go we want to know what number is 40% of 50 so I like to start with my total 50 is my total amount that's my 100% so I'm going to write my 50 down here as my total now Mrs King talked us through it last time we want to find 40% along our percent line so we got to figure out how do I break up between Z and 100 to make sure I land on 40 at some point well we've got two choices we could count by 20s or you could count by 10 I'm going to choose to count by 20s so 100 divided up by 20 means five sections so I'm going to break it up into five 20 4 equal sections that's pretty even nice job that was pretty good and then I'm going to count along 20% 40% 60% and 80% now I need to do the same thing with my total down here I want to go up to 50 but I need to break it up into five equal parts 50 ID 5 is 10 so along the bottom of my number line I want to count up by 10 10 20 30 40 and if I did it right 50 as my total and now I'm finally ready to find the answer what is 40% of that 50 so I found 40% on one line and I find what does it match with on my other number line in this case 20 is 40% of 50 So my answer in this one is 20 hopefully you're feeling a little bit more comfortable with it let's try another one that's just a tiny bit different this time all right it says 21 is 70% of what number so we don't have to the answer of the total that's what we're going to find question mark that's a good idea um we cannot count anything but tens I see so we need nine lines or 10 equal sections it's always the first one that's off so that's see 10 20 30 40 50 60 70 80 okay that's totally Mrs King 90 all right that was close 10 20 but we need to choose something that would allow us to land on 70 at some point so Mrs King said let's count by tens it's easier to count by but harder to draw see if you can make yours a little more even than mine is there all right this time 21 is our part so 21 needs to go underneath 70 all right how many sections do we have available before the 21 so we my gosh you need to talk on my sneeze all right so 70 that's our seventh Mark is 21 so that means we divided up seven groupings to get to 21 so 21 / 7 means means we must have been counting up by three each time so we need to start at 10% and count out by three so Three 6 9 12 15 18 if it does not equal 21 at 70% you've counted by something wrong need to go back and adjust we need to keep going so 24 27 and then the total would be 30 so we have now answered our question 21 is 70% of 30 so of what number is asking you for the total and we have found that it is 30 so this one's a little trickier the 70 and the 21 matched up we kind of had to work our way backwards to figure out the rest of the numbers well possible let's try one last one so please get the word problem jotted down and then we're going to try this one together there was $11,200 available for field trips at the school that sounds a lot like that was the total amount available so I'm going to write 1,200 down here as my total 30% of that was available for sixth grade so we want to find 30% of that total well in order to break up zero to 100 and make sure I land on my 30 I'm going to have to count by tens so let me oh I think it's that part of the board I'm going to blame it on the board okay whatever you need so that's 40 50 60 70 80 90 and 100 all right 10 perc 20% 30% so we want to make sure if we did it correctly we should land on that number we were looking for in this case 30% you still need to label everything not just the piece that you wanted all right we need to go to our total and figure out how much section the bottom part of our number line is going to be worth so again 1,200 divided by 10 sections is going to be 120 so each section is worth 120 so we're counting up by 120s 240 think of it as counting by twelves with a zero at the end of it 480 doing a fantastic job you're so much help 960 and 1080 80 all right last step go to that 30% look underneath it and we have $360 and see if your answer makes sense is $360 30% of what was available for sixth grade sounds pretty good to me all right remember to fill out your survey question when you're done with the video and then we'll go over more of these tomorrow in class but if you had trouble go back rewatch our explanation see if that helps seeing it a second time |
636 | https://myers.faculty.chemistry.harvard.edu/file_url/133 | Chem 115 Stereoselective, Directed Aldol Reaction Myers
Diastereofacial Selectivity in the Aldol Addition Reaction-
Zimmerman-Traxler Chair-Like Transition States
O
M
O
CH 3
R2
L
R1
H
H
L
O
M
O
CH 3
H
L
R1
R2
H
L
O
M
O
H
H
L
R1
R2
H3C
L
R1
OML 2
CH 3
O
M
O
H
R2
L
R1
H
H3C
L
R1
OML 2
CH 3
R1 R2
O
CH 3
OH
R1 R2
O
CH 3
OH
R1 R2
O
CH 3
OH
R1 R2
O
CH 3
OH
+
+
FAVORED
DISFAVORED
FAVORED
DISFAVORED
Reviews:
• Zimmerman and Traxler proposed that the aldol reaction with metal enolates proceeds via a
chair-like, pericyclic process. In practice, the stereochemistry can be highly metal dependent.
Only a few metals, such as boron, reliably follow the indicated pathways.
• ( Z)- and ( E)-enolates afford syn - and anti -aldol adducts, respectively, by minimizing
1,3-diaxial interactions between R 1 and R 2 in each chair-like TS ‡.
(Z)-enolates
(E)-enolates
Heathcock, C. H. In Comprehensive Organic Synthesis , Trost, B. M.; Fleming, I., Eds., Pergamon
Press: New York, 1991 , Vol. 2 , pp. 133-238.
Kim, B. M.; Williams, S. F.; Masamune, S. In Comprehensive Organic Synthesis , Trost, B. M.;
Fleming, I., Eds., Pergamon Press: New York, 1991 , Vol. 2 , pp. 239-275.
Paterson, I. In Comprehensive Organic Synthesis , Trost, B. M.; Fleming, I., Eds., Pergamon
Press: New York, 1991 , Vol. 2 , pp. 301-319.
H3C H
O
H3C
OH
H
O
2
• The aldol reaction was discovered by Aleksandr Porfir'evich Borodin in 1872 where he first
observed the formation of "aldol", 3-hydroxybutanal, from acetaldehyde under the influence of
catalysts such as hydrochloric acid or zinc chloride.
Zimmerman, H. E.; Traxler, M. D. J. Am. Chem. Soc. 1957 , 79 , 1920-1923.
Dubois, J. E.; Fellman, P. Tetrahedron Lett. 1975 , 1225-1228.
Heathcock, C. H.; Buse, C. T.; Kleschnick, W. A.; Pirrung, M. C.; Sohn, J. E.; Lampe, J. J.
Org. Chem. 1980 , 45 , 1066-1081.
• Note: the enantiomeric transition states (not shown) are, by definition, of equal energies. The
pericyclic transition state determines syn/anti selectivity. To differentiate two syn or two anti
transition states, a chiral element must be introduced (e.g., R 1, R 2, or L), thereby creating
diastereomeric transition states which, by definition, are of different energies.
syn
anti
syn
anti
M. Movassaghi
R2CHO
R2CHO Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Preparation of ( Z)- and ( E)-Boron Enolates (Z)-Selective Preparation of Boron Enolates from Evans' Acyl Oxazolidinones (Imides)
Evans, D. A.; Takacs, J. M.; McGee, L. R.; Ennis, M. D.; Mathre, D. J.; Bartroli, J. Pure Appl.
Chem. 1981 , 53 , 1109-1127.
Evans, D. A.; Vogel, E.; Nelson, J. V. J. Am. Chem. Soc. 1979 , 101 , 6120-6123.
Evans, D. A.; Takacs, J. M.; McGee, L. R.; Ennis, M. D.; Mathre, D. J.; Bartroli, J. Pure & Appl.
Chem. 1981 , 53 , 1109-1127.
Brown, H. C.; Dhar, R. K.; Bakshi, R. K.; Pandiarajan, P. K.; Singaram, B. J. Am. Chem. Soc.
1989 , 111 , 3441-3442.
Et CH 3
O
Et CH 3
OB( n-Bu) 2(n-Bu) 2BOTf
97% ( Z)
PhCHO
–78 º C
77%
Et
O
Ph
CH 3
OH
syn >99%
Et CH 3
O
Et
OB( c-Hex) 2(c-Hex) 2BCl
99% ( E)
PhCHO
–78 º C
75%
Et
O
Ph
CH 3
OH
anti >97%
• Dialkylboron triflates typically afford ( Z)-boron enolates, with little sensitivity toward the amine
used or the steric requirements of the alkyl groups on the boron reagent.
• In the case of dialkylboron chlorides the geometry of the product enolates is much more
sensitive to variations in the amine and the alkyl groups on boron.
• The combination of (c-Hex) 2BCl and Et 3N provides the ( E)-boron enolate preferentially.
O
N
Bn
O
CH 3
O (n-Bu) 2BOTf
CH 2Cl 2
O N
Bn
O
CH 3
O
B
n-Bu n-Bu
–
OTf
O N O
O
Bn
H
B
n-Bu
n-Bu
H3C H
H
O N O
O
Bn
H
B
n-Bu
n-Bu
H CH 3
H
FAVORED DISFAVORED
O N
Bn
O
CH 3
O
B
n-Bu n-Bu
O N
Bn
O O
B
n-Bu n-Bu
CH 3
• Observed selectivity > 100:1 Z : E.
i-Pr 2NEt i-Pr 2NEt
iPr 2NEt, Et 2O
–78 º C, 30 min
Et 3N, Et 2O
–78 º C, 10 min CH 3Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Evans, D. A.; Takacs, J. M.; McGee, L. R.; Ennis, M. D.; Mathre, D. J. Bartroli, J. Pure & Appl.
Chem. 1981 , 53 , 1109-1127.
O
N
Bn
O
O
R
O
CH 3 O
N
Bn
O
O
R
O
CH 3
B B
n-Bu n-Bu n-Bu n-Bu
O
N
Bn
O
O
R
OH
CH 3 O
N
Bn
O
O
R
OH
CH 3
O
N
Bn
O
CH 3
OB( n-Bu) 2
O
B
O
CH 3
R
n-Bu
N H
n-Bu
H
O
O
H
Bn
O
B O
CH 3
R
n-Bu
N
H
n-Bu
H
O
O
Bn
H
vs.
FAVORED DISFAVORED
O N
Bn
CH 3
O O
B
n-Bu n-Bu
UNREACTIVE
O N
Bn
CH 3
O O
Li
cf. reactive enolate in
Evans' asymmetric
alkylation
Open coordination site
required for pericyclic aldol rxn
RCHO
Syn -Selective Aldol Reactions of Imide-Derived Boron ( Z)-Enolates
• Chiral controller group biases enolate !-faces such that one of the two diastereomeric ( syn )
transition states is greatly favored.
• Dipole-dipole interactions within the imide are minimized in the reactive conformation (see:
Noe, E. A.; Raban, M J. Am. Chem. Soc. 1975 , 97 , 5811-5820).
O
N
O
CH 3
O
H3C CH 3
O
N
CH 3
O
CH 3
O
Ph
O
N
O
O
H3C CH 3
O
N
CH 3
O
O
Ph
R
R
OH
CH 3
OH
CH 3
n-Bu 2BOTf, i-Pr 2NEt
CH 2Cl 2, 0 ° C
RCHO
–78 " 23 ° C
n-Bu 2BOTf, i-Pr 2NEt
CH 2Cl 2, 0 ° C
RCHO
–78 " 23 ° C
aldehyde
diastereomeric a
ratio
a
Ratio of major syn product to minor syn product.
yield (%)
497:1
<1:500
141:1
<1:500
500:1
<1:500
78
91
75
95
88
89
imide
• A variety of chiral imides can be used for highly selective aldol reactions.
• Anti products are typically formed in less than 1% yield.
• Often, a single crystallization affords diastereomerically pure product.
Evans, D. A.; Bartroli, J.; Shih, T. L. J. Am. Chem. Soc. 1981 , 103 , 2127-2129.
Evans, D. A.; Gage, J. R. Org. Syn. 1990 , 68 , 83.
A
B
(CH 3)2CHCHO
(CH 3)2CHCHO
n-C 4H9CHO
n-C 4H9CHO
C6H5CHO
C6H5CHO
A
B
A
B
A
BChem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Carboximide Hydrolysis with Lithium Hydroperoxide
Evans, D. A.; Britton, T. C.; Ellman, J. A. Tetrahedron Lett. 1987 , 28 , 6141-6144.
Gage, J. R.; Evans, D. A. Org. Syn. 1990 , 68 , 83-91.
• Reductive cleavage:
Other Methods for Removal of the Chiral Auxiliary
O
N
O
R
O
O
N
CH 3
O
O OH
HO R
O
N
H R
O
OH
O
N
Bn
O
O
H
CH 3
H CH 3H
CH 3
Ph Ph
+
LiOOH
or
LiOH
substrate reagent
76
0
16
100
98
43
<1
30
yield of A (%) a yield of B (%) a
a
Yield of diastereomerically pure (> 99:1) product.
• LiOOH displays the greatest regioselectivity for attack of the exocyclic carbonyl group.
• This selectivity is most pronounced with sterically congested acyl imides.
• This is a general solution for the hydrolysis of all classes of oxazolidinone-derived
carboximides and allows for efficient recovery of the chiral auxiliary.
LiOOH
LiOH
LiOOH
LiOH
A B
O
H3CO N3
N
O
NHBoc
OBn
O
O
H3CO N3
OH
O
NHBoc
OBn
O
O
Bn
O
LiOOH
THF, H 2O;
Na 2SO 3
0 °C
96%
• The selective hydrolysis of carboximides can be achieved in the presence of unactivated
esters using LiOOH.
O
OBOM
CH 3 CH 3
OBn
CH 3
O
O
H3C
CH 3
H3C
H3C
N O
O
O
CH 3
Ph
H H
O
OBOM
CH 3 CH 3
OBn
CH 3
O
O
H3C
CH 3
H3C
H3C
O
H H
BnO
O N
CH 3
O O
CH 3
Br
Bn
CH 2 CH 3
Br
CH 2
HO
O N
CH 3
O O
OBn
Bn
OH
CH 3
CH 3
N
O
OBn
OH
CH 3CH 3O
CH 3
BnOLi
THF, 0 °C
77%
LiAlH 4, THF
–78 ! 0 ° C
90%
• Esterification:
• Transamination:
• A free "-hydroxyl group is required.
• Weinreb amides can be readily converted into ketones or aldehydes (see: Nahm, S.;
Weinreb, S. M. Tetrahedron Lett. 1981 , 22 , 3815-3818).
Al(CH 3)3
CH 3ONHCH 3•HCl
CH 2Cl 2, 0 °C
92%
CH 3
Evans, D. A.; Bender, S. L.; Morris, J. J. Am. Chem. Soc. 1988 , 110 , 2506-2526. Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Cytovaricin:
Evans, D. A.; Kaldor, S. W.; Jones, T. K.; Clardy, J.; Stout, T. J. J. Am. Chem. Soc. 1990 , 112 ,
7001-7031.
O
N
CH 3
O
CH 3
O
Ph
O
N
CH 3
O
O
Ph
OH
CH 3
O
H
N
O TESO
CH 3
CH 3O
CH 3
O
N
CH 3
O
H3C
O
Ph
O
H
H3C
O
N
CH 3
O
O
Ph
OH
CH 3
H3C
H3C CH 3
N
CH 3
N
OO
H3C CH 3
H3C
N
CH 3
H
OO
H3C CH 3
N
H3C
CH 3
CH 3
CH 3
O TESO
CH 3
O
O
O
CH 3
H
CH 3
H
H
H3C
HO
H
O
O
OCH 2OCH 2CCl 3CH 3
O
H
CH 3
H
H
H3C
H H
H
n-Bu 2BOTf, Et 3N
CH 2Cl 2, 0 ° C;
RCHO
–78 ! 23 ° C;
H2O2, 0 ° C
PMB = p-Methoxybenzyl
+
Al(CH 3)3
CH 3ONHCH 3•HCl
THF, 0 ° C
TESCl, Im. DMF
91%
n-Bu 2BOTf, Et 3N
CH 2Cl 2, 0 ° C;
RCHO, – 78;
H2O2, 0 ° C
+
LDA, Et 2O,
THF, 0 °C;
–45, 90 %
+
HF, H 2O, CH 3CN
25 °C
92%
DEIPS = diethylisopropylsilyl
OPMB
OPMB
OPMB
OPMB
PMBO
O
O
OCH 2OCH 2CCl 3CH 3
O
H
CH 3
H
H
H3C
DEIPSO
H H
H
N O
O
H3C O
H3C Ph
O
O
OCH 2OCH 2CCl 3CH 3
OH
H
CH 3
H
H
H3C
DEIPSO
H H
H3C O N CH 3
OCH 3
TBSO
H3C CHO
OPMB
O
OH3C
DEIPSO
H
CH 3
OCH 2OCH 2CCl 3
CH 3
H
HH O
Si
OHO
O
PhSO 2
O O CH 3
OCH 3
OTES
t-Bu
t-Bu
H3C
TESO CH 3
TESO
OTES
CH 3
OH
O
OH3C
HO
H CH 3
O
CH 3
H
HH
OH3C
O OH
CH 3
OH
CH 3
HO
CH 3
OH O O CH 3
OH
OCH 3
H
H
O
N
CH 3
O
CH 3
O
Ph
N
O OH
CH 3
OBn
O
OBn
H
H3CO
CH 3
O O
CH 3
OBn
CH 3
Si
t-Bu t-Bu
+
n-Bu 2BOTf, Et 3N
CH 2Cl 2, – 78 °C;
RCHO
Al(CH 3)3, THF
CH 3ONHCH 3•HCl
92%
+
+
n-Bu 2BOTf, Et 3N
CH 2Cl 2, – 78 °C
Al(CH 3)3, THF
CH 3ONHCH 3•HCl
83%
Cytovaricin
TBSO
TBSO
OH
DEIPSO
92% 87% Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Diastereoselective Syn -Aldol Reaction of !-Ketoimides
Evans, D. A.; Clark, J. S.; Metternich, R.; Novack, V. J.; Sheppard, G. S. J. Am. Chem. Soc.
1990 , 112 , 866-868.
Diastereoselective Anti -Aldol Reaction of !-Ketoimides
H3C CHO
CH 2
H3C CHO
CH 3
H3C CHO
CHO
83
86
95:5
<1:99
77 c
64 c
enolization
conditions RCHO a yield % b ratio
anti-syn : syn-syn
71
86
85
81
95:5
2:98
79:21
<1:99
89:11
4:96
A: Sn(OTf) 2, Et 3N; B: TiCl 4, i-Pr 2NEt. a1.0-1.1 equiv bIsolated yield of
major diastereomer (> 99% purity). c3-5 equiv of RCHO was used.
O N
Bn
O O
CH 3
O
CH 3
O N
Bn
O O
CH 3
O
O N
Bn
O O
CH 3
O
R
R
OH
CH 3
OH
CH 3O
TI
O
H
CH 3
R
H
H3C
H O
Xp
Cl
Cl
Cl
Sn
O
O
H
R CH 3
H
O
Xp
H3C
L
L
H
Sn(OTf) 2
Et 3N, CH 2Cl 2
RCHO, – 20 °C
TiCl 4
i-Pr 2NEt, CH 2Cl 2
RCHO
–78 " 0 ° C
anti-syn
syn-syn
• Both enolization methods provide ( Z)-enolates and (diastereomeric) syn aldol products.
• The stereochemical outcome of both reactions is dominated by the C 2 methyl-bearing
stereocenter, as shown in the proposed transition states above.
• The chirality of the oxazolidinone has little influence on the diastereoselectivity of these
reactions.
A
B
A
B
A
B
A
B
O N
Bn
O O
CH 3
O
O N
Bn
O O
CH 3
O
CH 3
O N
Bn
O O
CH 3
O
CH 3
R
OH
R
OH
( c-Hex) 2BCl
EtN(CH 3)2, Et 2O
0 ° C, 1 h
RCHO
–78 ° C, 3h
+
anti-anti syn-anti
Ph CHO
CH 3
aldehyde yield % a ratio
anti-anti : syn-anti
78
72
70 b
84 b
84:16
92:8
80:20
88:12
a
Isolated yield of major diastereomer. bYield of
purified mixture of diastereomers.
• Enolization of the less hindered side of the ketone under Brown's conditions affords the
(E)-boron enolate.
• The C2 stereocenter is the dominant control element in these aldol reactions; "matched" vs.
"mismatched" effects of the remote auxiliary are negligible.
CH 3
O
CH 3
CH 3
O
CH 3
CH 3
O
CH 3
R
OH
R
OH
RL
RL
RL
O H
CH 3
M
H
H3C
RL
+
anti-anti , observed
syn-anti , predicted
Si face
Re face
• The sense of asymmetric induction observed in these reactions was unexpected
and opposite to a prediction based on a reactant-like transition state model minimizing
A(1,3) strain.
CH 3
84 97:3
(CH 3)2CHCHO
CH 2=C(CH 3)CHO
CH 3CH 2CHO
PhCH 2CH 2CHO
RCHO
RCHO
Evans, D. A.; Ng, H. P.; Clark, J. S.; Reiger, D. L. Tetrahedron 1992 , 48 , 2127-2142. Chem 115 Stereoselective, Directed Aldol Reaction Myers
Jaron Mercer, M. Movassaghi
Syn-Anti-Selective Aldol Reactions of Chiral Ethyl Ketones
Evans, D. A.; Weber, A. E. J. Am. Chem. Soc. 1986 , 108 , 6757-6761.
Directed Reduction of !-Hydroxy Ketones
TBSO
CH 3
O
H3C
CH 3
TBSO
CH 3
O
CH 3
H3C
CH 3
TBSO
CH 3
O
CH 3
H3C
CH 3
TBSO
CH 3
O
H3C
CH 3
CH 3
CH 3
OH
CH 3
CH 3
OH
( c-Hex) 2BCl
Et 3N, Et 2O
0 °C
i-PrCHO
–78 °C
( c-Hex) 2BCl
Et 3N, Et 2O
0 °C
i-Pr CHO
–78 °C
90%, 88% de
75%, 92% de
• The C2 stereocenter is believed to be the dominant control element for both substrates.
TBSO
CH 3
OB( c-Hex) 2
CH 3
H3C
CH 3
TBSO
CH 3
O
CH 3
H3C
CH 3
R
OH
O
B
O
H3C
c-Hex
c-Hex
H CH( i-Pr)OTBS
CH 3
R
H
• Minimization of A (1,3) interactions in the enolate biases the approach of the aldehyde to the
methyl-bearing "-face of the enolate, while the ( E)-enolate geometry affords anti -aldol products.
CH 3
CH 3
H
B
O
H
R1
R2
O
OAc
OAc
H
B
O
H
R1
O
OAc
OAc
R2
H
H
R1 R2
OOH
R1 R2
OH OH
R1 R2
OH OH
R1 R2
OOH
O
Zn
O
H
R1
L
L
R2
R1 R2
OH OH
O
N
CH 3
O
O O
CH 3
Ph
CH 3
OH
CH 3
O
N
CH 3
O
O OH
CH 3
Ph
CH 3
OH
CH 3
+
+
–
–
1,3-anti-diol
1,3-syn -diol
FAVORED
DISFAVORED
1,3-syn-diol
Internal hydride delivery:
External hydride delivery:
• The reactivity of the reagent is attenuated such that the reduction of ketones proceeds
at convenient rates only intramolecularly, favoring formation of 1,3-anti-diols.
• Chelated transition state, axial attack provides 1,3-syn-diol.
• These directed reductions are applicable to #-hydroxy-!-ketoimides:
NaBH(OAc )3
AcOH, CH 3CN
25 °C, 30 min
99%, >96% de
(CH 3)4NBH(OAc) 3
Zn(BH 4)2
BH 4–
Evans, D. A.; Chapman, K. T.; Carreira, E. M. J. Am. Chem. Soc. 1988 , 110 , 3560-3578.
O N
Bn
O O
CH 3
O
CH 3
BnO CH 3
O
CH 3
BzO CH 3
O
CH 3
• In addition to !-ketoimides, the two chiral ethyl ketones above are known to undergo aldol
reactions at the unexpected Re face of the enolate, deilvering anti-anti aldol products.
Paterson, I.; Goodman, J. M.; Isaka, M. Tet. Lett. 1989 , 30 , 7121-7124.
Paterson, I.; Wallace, D. J.; Cowden, C. J. Synthesis 1998 , 639-652. Chem 115 Stereoselective, Directed Aldol Reaction Myers
Chris Coletta, Jaron Mercer
Premonensin
Evans, D. A.; DiMare, M. J. Am. Chem. Soc. 1986 , 108 , 2476-2478.
Anti-Aldols with Magnesium Enolates
O N
Bn
O O
CH 3
O
CH 3
O N
Bn
O O
CH 3
O OH
CH 3
CH 3
CH 3CH 3
H
O
CH 3
CH 3CH 3
O N
Bn
O O
CH 3
OH OH
CH 3
CH 3
CH 3CH 3
O
O
CH 3
O O
CH 3
H
CH 3
ONO 2
H3C
O
CH 3 CH 3 CH 3 CH 3
CH 3
OCH 3
H3CO
H3C
O
O
CH 3
O O
CH 3 CH 3
OH NO 2 O
CH 3 CH 3 CH 3 CH 3
CH 3
OCH 3
H3CO
H3C
HO
O
CH 3
OH OH
CH 3 CH 3
OH O
CH 3 CH 3 CH 3 CH 3
CH 3
H3C
O
+
Sn(OTf) 2
N-ethylpiperidine
CH 2Cl 2, – 78 ° C
94%, 94% de
NaBH(OAc) 3
AcOH
91%, >94 de
+
1 2
2, LDA, THF
–78 ° C; 1
h
!, THF, 0 ° C;
H2O, HCl, 25 ° C
58%
Premonensin
CH 3H3C
CH 3H3C
O N
Bn
O
CH 3
O
H R
O
O N
Bn
O O
R
OH
CH 3
MgCl 2 (10-20 mol%) ,
Et 3N, TMSCl
EtOAc, 23 º C
aldehyde yield (%) dr
24:1
32:1
7:1
21:1
28:1
16:1
14:1
6:1
-
91
71
92
92
77
91
80
CHO
X
X = CH 3
X = OCH 3
X = NO 2
X CHO
Y X = Ph, Y = H
X = Ph, Y = CH 3
X = H, Y = CH 3
"-napthaldehyde
furfural
• Silylation of the magnesium alkoxide in the aldol product turns over the magnesium.
• The aldehyde component is limited to non-enolizable aromatic and ",#-unsaturated aldehydes.
Evans, D. A.; Tedrow, J. S.; Shaw, J. T.; Downey, C. W. J. Am. Chem. Soc. 2002 , 124 , 392-393.
• Use of the analogous N-acylthiazolidinethione chiral auxiliary affords products of the opposite
anti -stereochemistry in comparable yields and with high selectivities.
S N
Bn
S
CH 3
O
H R
O
S N
Bn
S O
R
OH
CH 3
MgBr 2•Et 2O, (10 mol%),
Et 3N, TMSCl,
EtOAc, 23 º C
5:1 THF/ 1 N HCl
• Both reactions are proposed to proceed through boat transition states. See: Evans, D. A.;
Downey, W. C.; Shaw, J. T.; Tedrow, J. S. Org. Lett. 2002 , 4, 1127-1130.
M. Movassaghi
57% (+19% undesired diastereomer)
TFA; CH 3OH Chem 115 Stereoselective, Directed Aldol Reaction Myers
Open Transition State Aldol Reactions
O N
t-Bu
O
CH 3
O
aldehyde Lewis acid
SnCl 4
TiCl 4
TiCl 4
• Heathcock and coworkers reported that complexation of the aldehyde with an added Lewis
acid allows access to non-Evans syn and anti aldol products via open transition states.
Synthesis of Oxazolidinethione and Thiazolidinethione Chiral Auxiliaries
• Gauche interactions around the forming C-C bond dictate which face of the aldehyde reacts. For
small Lewis acids, transition state 1 is favored. For large Lewis acids, transition state 2 is
favored.
Walker, M. A.; Heathcock, C. H. J. Org. Chem. 1991 , 56 , 5747-5750.
O N
t-Bu
O O
CH 3
OH
R
Bu 2BOTf, i-Pr 2NEt,
CH 2Cl 2;
Lewis acid, RCHO
O N
t-Bu
O
CH 3
O
B
Bu
H O
H
R
non-Evans syn
CHO H3C
anti:syn
10:90
12:88
8:92
yield (%)
66
72
65
O N
i-Pr
O
CH 3
O
O N
i-Pr
O O
CH 3
OH
R
Bu 2BOTf, i-Pr 2NEt,
CH 2Cl 2;
Lewis acid, RCHO
O N
i-Pr
O
CH 3
O
B
Bu
H O
R
H
anti
LA
LA
1, favored for small
Lewis acids
2, favored for large
Lewis acids
aldehyde Lewis acid
Et 2AlCl
Et 2AlCl
Determined by 1H NMR. Yield given is the total yield of diastereomeric aldol mixture.
CHO H3C
anti:syn
88:12
74:26
yield (%)
81
62
NH 2
HO
Bn
S NH
S
Bn
CS 2, KOH
H2O
80%
O NH
S
Bn
Cl 2CS, Et 3N
CH 2Cl 2
95%
S N
S
Bn
CH 3
O
RCHO
CH 2=CHCHO
i-PrCHO
CH 2=CHCHO
i-PrCHO
S N
S
Bn
O
R
OH
CH 3
A
S N
S
Bn
O
R
OH
CH 3
B
TiCl 4 (1.1 equiv)
(–)-sparteine
CH 2Cl 2, 0 °C
RCHO, 0 °C
+
(–)-sparteine (equiv) yield (%) A : B
1.0
1.0
2.0
2.0
49
60
77
75
99:1
98:2
<1:99
3:97
• Selectivities are generally >95:5 for syn:anti products.
• Both the yield and diastereoselectivities are high and synthetically useful, although they are
typically lower than the corresponding oxazolidine aldol reactions.
• An advantage of this method is that a single acyloxazolidinethione can provide either syn
aldol product by changing the amount of sparteine in the reaction mixture.
Ph CHO
PhCHO
Jaron Mercer, M. Movassaghi
Asymmetric Aldol Reactions with Titanium Enolates of N-Acylthiazolidinethiones
Crimmins, M. T; King, B. W.; Tabet, E. A.; Chaudhary, K. J. Org. Chem. 2001 , 66 , 894-902.
NH 2
HO
Bn
O
NaBH 4, I 2
THF
95%
Bu
Bu Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi, Jaron Mercer
H
CH 3
O
TI
O
Cl
Cl
Cl
N
S
S
Bn
H
R
H
H i-Pr
OH
CH 3
O
HO Ph
OH
CH 3
S N
S
Bn
O
R
OH
CH 3
S N
S
Bn
CH 3
O
S
N
i-Bu
S
O
R
OH
CH 3
CH 3O
OH
CH 3
O
i-Pr
S N
S
Bn
O
R
OH
CH 3
O
TiL x
O
CH 3
R
N H
H
S
S
H
Bn
BnHN
OH
CH 3
O
i-Pr
N
OH
CH 3
O
i-Pr CH 3O
CH 3
NaBH 4
83%
DIBAL-H
69%
• The thiazolidinethione auxiliary is easily removed under mild conditions:
• The thiazolidinethione auxiliary is recovered by basic extraction (1 M NaOH) of the product
mixture.
PhCH 2NH 2
79%
CH 3ONHCH 3•HCl
imidazole
77%
CH 3OH
imidazole
79%
TiCl 4
(–)-sparteine ( 2 equiv)
RCHO
TiCl 4
(–)-sparteine ( 1 equiv)
RCHO
Crimmins, M. T.; King, B. W.; Tabet, E. A. J. Am. Chem. Soc. 1997 , 119 , 7883-7884.
Crimmins, M. T.; Chaudhary, K. Org. Lett. 2000 , 2, 775-777.
• Proposed transition states provide a rationale for the selectivity dependence on amine equivalents:
O N
S
Bn
O
O N
S
Bn
O
R'
OH
OR
A
O N
S
Bn
O
R'
OH
OR
B
TiCl 4, (– )-sparteine
TiCl 4, R'CHO
+
Anti -Selective Aldol Reactions with Titanium Enolates of N-Glycolyloxazolidinethiones
aldehyde yield (%) A : B : syn
OR
H3CCHO
CH 3
CHO
CHO
Ph CHO
CH 3
CHO
CHO
CH 3
CHO
CHO
R
84
56
74
58
64
48
63
59
allyl
allyl
allyl
allyl
Bn
Bn
CH 3
CH 3
94 : 6 : 0
65 : 24 : 11
94 : 6 : 0
95 : 5 : 0
88 : 12 : 0
74 : 26 : 0
84 : 11 : 5
88 : 12 : 0
O N
t-Bu
S
O
O
Ti
H O
R'
H
TiL 4
• Complexation of the aldehyde with excess titanium occurs in situ to give anti products with high
selectivity.
• The proposed transition state is analogous to that of the anti -selective Heathcock aldol.
L4
R
Crimmins, M. T.; McDougall, P. J. Org. Lett . 2003 , 5, 591-594. Chem 115 Stereoselective, Directed Aldol Reaction Myers
Jaron Mercer
Asymmetric Synthesis of Syn -!-Hydroxy-"-Amino Acids
Sn(OTf) 2,
N-ethylpiperidine;
RCHO
O N
Bn
O
NCS
O
O N
Bn
O O
R O N
Bn
O O
aldehyde yield (%) ratio
H3CCHO
PhCHO
91:9
99:1
99:1
94:6
97:3
75
91
92
73
71
CH 3
CHO
CHO H3C
CH 3
CH 3
CH 3
CHO
Ratio of desired (illustrated) stereoisomer to the sum of all other stereoisomers.
CH 3
N C S
HN
O
S
R
• The isothiocyanate below serves as a chiral glycine equivalent. Stannous triflate-mediated aldol
reactions give cyclized aldol adducts in high yield and diastereoselectivity.
HO R
NHCH 3
OH O 1. KOH
H3O+
(H 3C) 3OBF 4
Mg(OCH 3)2
HOCH 3
H3CO
O
N
O
R
SCH 3
H3C
H2O
H3CO
O
N
O
R
OH3C
Evans, D. A.; Weber, A. E. J. Am. Chem. Soc. 1986 , 108 , 6757-6761.
• The N-methyl amino acid can be reached in 4 steps.
O N
Bn
O O
HN
O
S
CH 3
H3C O
HN
O
S
R
H3CO
• 2-Chloro- and 2-Bromoacetyl imides undergo aldol addition with high diastereoselectivity.
Asymmetric Synthesis of Anti -!-Hydroxy-"-Amino Acids
O N
O
Br
O
O N
O
Cl
O Bu 2BOTf, Et 3N,
Et 2O;
RCHO
O N
O O
Cl
R
OH
Bu 2BOTf, Et 3N,
Et 2O;
RCHO
O N
O O
Br
R
OH
aldehyde yield (%) ratio
H3CCHO
PhCHO
95:5
97:3
96:4
98:2
94:6
67
79
75
63
63
Ratio of desired (illustrated) stereoisomer to the sum of all other stereoisomers.
imide
1
2
1
1
1
2
2 H3C CHO
CHO H3C
CH 3
Ph CH 3 Ph CH 3
Bn Bn
CH 3
O
CH 3
OH
HO
NaN 3
LiOH/H 2O
Evans, D. A.; Sjogren, E. B.; Weber, A. E.; Conn, R. E. Tetrahedron Lett. 1987 , 28 , 39-42.
• Halide displacement with NaN 3 occurs with inversion of stereochemistry. Hydrolytic removal of
the auxiliary followed by hydrogenation of the azide delivers the amino acid.
O N
O O
Cl
CH 3
OH
H 2, Pd/C, TFA
76 %, ≥99% de
Ph CH 3
NH 2
OSnL Chem 115 Stereoselective, Directed Aldol Reaction Myers
Jaron Mercer
O N
Bn
O O
O N
Bn
O
Br
O
NO 2
F
Bu 2BOTf, Et 3N;
OHC NO 2
F
76%,
95:5 dr
O
N3
NO 2
F
HO
O N
Bn
O
NCS
O
Cl
F
NO 2
OHC
Cl
F
O2N
HO 2C
TMGA, CH 2Cl 2
LiOOH
Sn(OTf) 2, N-ethylpiperidine;
95:5 dr
Boc 2O, DMAP;
HCO 2H, H 2O2
LiOOH
Vancomycin Aglycon:
Evans, D. A.; Wood, R. W.; Trotter, B. W.; Richardson, T. I.; Barrow, J. C.; Katz, J. L. Angew.
Chem. Int. Ed. 1998 , 37 , 2700-2704.
Evans, D. A.; Watson, P. S. Tet. Lett. 1996 , 37 , 3251-3254.
OH
OH
OH
O
N
H
H
N
O
BnO
O
Cl HO
HO 2C
NH
O
N
H
O
H H
Cl
O H
N
O
N
H
O
OH
CH 2CH(CH 3)2
NH 2
O
vancomycin aglycon
O N
Bn
O O
HN
O
S
F
Cl
O2N
Br
OH
OH
NHCH 3
N
O
O
Boc
Direct Aldolization of Pseudoephenamine Glycinamide
THF, CH 3OH
23 º C
95%
Boc 2O, NaHCO 3
O OH
BocHN
HO
CH 3
CH 3
CH 31:1 H 2O:dioxane
79%
CH 3OH
91%
O OH
NH 2
H3CO
CH 3
CH 3
CH 3
HCl•
NaOH
SOCl 2
O OH
NH 2
N
CH 3
CH 3
CH 3
OH CH 3
O OH
NH 2
NaO
CH 3
CH 3
CH 3
O OH
NH 2
X!+
O OH
NH 2
X!+ CH 3
O OH
NH 2
X!+
TIPS
O OH
NH 2
X!+
CH 3
CH 3
CH 3
80%
85 : 15 dr
75%
83 : 17 dr
72%
83 : 17 dr
89%
94 : 6 dr
O HO
NH 2
X!+
CH 3
CH 3
CH 3
O HO
NH 2
X!+
CH 3
CH 3
OTBDPS
98%
98 : 2 dr
82%
94 : 6 dr
Seiple, I. B.; Mercer, J. A. M.; Sussman, R. J.; Myers, A. G. Unpublished.
N
CH 3
O
NH 2
OH
LiHMDS, LiCl, THF O HO
NH 2
RS RL
O RLX!+
RS
–78 to 0 º C
Isolated yields of stereoisomerically pure products. Diastereomeric ratios reported as
major isomer : sum of all other diastereomers.
• Pseudoephenamine glycinamide undergoes a direct aldol addition with both aldehyde and
ketone substrates.
• The corresponding N-Boc-protected or methyl ester hydrochloride derivatives can be prepared
in two steps from the aldol products. Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
R1
O OBIpc 2
R2
CH 3
OH
R1
O
R1
H3C CH 3
O
H3C CH 3
O
H3C CH 3
O
H3C CH 3
O
H3C
O
CH 3
CH 3
CH 3
CHO
CH 3
H3C CHO
CH 3
O CHO
CHO
CH 3
CHO
CH 3
(–)-Ipc 2BOTf
i-Pr 2NEt
CH 2Cl 2,– 78 °C
R2CHO, –15 °C;
H2O2
ketone aldehyde syn:anti ee (%) yield (%)
98:2
96:4
96:4
95:5
97:3
91
66
80
88
86
78
45
84
99
79
Syn -Aldol Adducts via Enol Diisopinocampheylborinates
Reviews:
Cowden, C. J.; Paterson, I. Org. React. 1997 , 51 , 1.
Franklin, A. S.; Paterson, I. Contemp. Org. Synth. 1994 , 1, 317.
Paterson Aldol
• Enolization occurs selectively on the less hindered side of the ketone and with ( Z)-selectivity.
• The ( E)-Enolate, generated in low yield using (– )-Ipc 2BCl, does not lead to a selective
anti -aldol reaction.
• Highest enantioselectivities are obtained with unhindered aldehydes.
• Aldol additions of methyl ketones are not highly enantioselective (53– 73% ee).
Paterson, I.; Goodman, J. M.; Lister, M. A.; Scumann, R. C.; McClure, C. K.; Norcross, R. D.
Tetrahedron 1990 , 46 , 4663-4684.
H3C H3C
Paterson, I.; Goodman, J. M.; Lister, M. A.; Scumann, R. C.; McClure, C. K.; Norcross, R. D.
Tetrahedron 1990 , 46 , 4663-4684.
OB (–)-Ipc 2
R2
CH 3
OH
R1
O
R1
R2
CH 3
OH
R1
O
FAVORED DISFAVORED
+
Proposed Origin of Selectivity:
• Diastereofacial selectivity is believed to be due to a favored transition state wherein
steric interactions between the (–)-Ipc ligand on boron and the R 1 substituent on the ketone
are minimized.
H3C
O
B
O
H3C
H3C
CH 3
H
H CH 3
CH 3
H
R1
H3C
H
H
H
CH 3
O
B
O
H3C
H3C
CH 3
H
H CH 3
CH 3
H
H
R2
H3C
H
R1
H
CH 3
R2
R2CHO Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Anti -Aldol Reactions of Lactate-Derived Ketones
CH 3
BzO
O
CH 3 CH 3
BzO
OB( c-Hex )2
CH 3 CH 3
BzO
O
R
OH
H3C CHO
CH 3
H3C CHO
CHO
H3C
CHO
CH 3
( c-Hex) 2BCl
(CH 3)2NEt
Et 2O, 0 ° C
2 h
RCHO, 14 h
–78 ! –26
H 2O2, pH 7
CH 3OH, 0 ° C
aldehyde de (%) yield (%) a
94
99
90
96
99
95
82
97
97
85
a
Isolated yield for 3 steps.
• Diastereofacial selectivity is very high; "-chiral aldehydes afford anti-aldol adducts with high
diastereoselectivity regardless of their stereochemistry.
CH 3 CH 3
BzO
OB( c-Hex )2
CH 3 CH 3
BzO
OB( c-Hex )2
OBn
CH 3
H
O
OBn
CH 3
H
O
CH 3 CH 3
BzO
O OH
OBn
CH 3
CH 3 CH 3
BzO
O OH
OBn
CH 3
MATCHED
MISMATCHED
80%, >94% de
61%, 84% de
+
+
CH 3
BzO
O
CH 3
BzO
O
i-Pr
OH
CH 3
BzO
O
CH 3
CH 3 OBn
BzO
O
i-Pr
OH
95%, 86% de
77%, 98% de
(c-Hex) 2BCl
(CH 3)2NEt
i-PrCHO
(c-Hex) 2BCl
(CH 3)2NEt
i-PrCHO
• Other examples:
CH 3
CH 3
OBn
PhCHO
O
B
O
H
L
H3CR
CH 3 CH 3
OBL 2
BzO
L
O Ph
O
H CH 3
H
O
B O
H
L CH 3
R
L H H
O
O
Ph
H3C
CH 3 CH 3
BzO
O
R
OH
CH 3 CH 3
BzO
O
R
OH
vs.
FAVORED DISFAVORED
+
L = c-Hex
CH 3 CH 3
BzO
O
R
OH
CH 3 CH 3
HO
OH
R
OTBS
H
CH 3
O
R
OTBS
TBSOTf
2,6-Lutidine
CH 2Cl 2, 78 ° C
LiBH 4, THF
–78 ! 20 ° C
NaIO 4
CH 3OH/H 2O
R = i-Pr, 74%, >99% de
R = Ph, 85%, >99% de
CH 3 CH 3
BzO
O
R
OTBS
CH 3
O
R
OTBS SmI 2, THF
CH 3OH
0 °C, 10 min
R = i-Pr, 81%
R = Ph, 96%
• The origin of the diastereoselectivity is proposed to be due to a formyl hydrogen bond in the
favored transition state.
• Paterson, I.; Wallace, D. J.; Cowden, C. J. Synthesis 1998 , 639-652.
H3C
RCHO Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
O
H3C
H3C H3C
OH
OH
CH 3
H3C
OH
CH 3
O
O
O
H3C
O
O
CH 3
CH 3
O
HO
O
CH 3
CH 3
O
HO
CH 3
O
CH 3
CH 3
CH 3
O
O
PMBO
CH 3
O
CH 3
CH 3
CH 3
H
H3C
O
O
CH 3
CH 3
BnO CH 3
CH 3
SOPh
H2C OBn
CH 3
OH
CH 3 CH 3
O
OBn
OH
CH 3 CH 3
O
H3C
OBn
CH 3
O
+
(+)-(Ipc) 2BOTf
i-Pr 2NEt, CH 2Cl 2, 20 °C;
CH 2=CHCHO, 0 °C
74%, 80% de
(c-Hex) 2BCl
Et 3N, Et 2O, – 78 °C;
(E) -CH 3CH=CHCHO, 0 °C
93%, 94% de
oleandolide
Oleandolide:
Paterson, I.; Norcross, R. D.; Ward, R. A.; Romea, P.; Lister, M. A. J. Am. Chem. Soc. 1994 , 116 ,
11287-11314.
H3C
Acetate Aldol
Addition of a Chiral !-Sulphinylester Enolate to Aldehydes
Ar O
S
O
CH 3
H3C CH 3
Ar S
O
Ot-Bu
O
Ot-Bu
O
R
OH
SOAr
Ot-Bu
O
R
OH Al, Hg
Ar = 4-CH 3C6H4
t-BuMgBr
THF, –78 °C;
RCHO
H
R
H
O
Mg
L
Ot-Bu
O
O
S
Ar
• Approach of the aldehyde is proposed to occur from the side of the non-bonding
electron pair of the sulfur atom with the R-group of the aldehyde anti to the sulfinyl
substituent. A chelated enolate is proposed.
Mioskowski, C.; Solladie, G. J. Chem. Soc. , Chem. Commun. 1977 , 162-163.
• The "-hydroxy ester products are isolated in 50-85% yield and 80-91% ee.
Proposed Transition State
CH 3CO 2t-Bu
i-Pr 2NMgBr Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Addition of a Chiral Acetate Enolate to Aldehydes An Approach to the Acetate Aldol Problem
HO Ph
CO 2CH 3H
HO
HO Ph
Ph
H
Ph O
HO Ph
Ph
H
Ph H3C
O
O
MO Ph
Ph
H
Ph
H2C
MO
O
HO Ph
Ph
H
Ph
O
R
OH O
R
OH
OH
PhMgBr
77%
AcCl
Pyridine
82%
M = Li, MgX
LDA; MgX 2
THF, (CH 3)2O
76-85% (2 steps)
84-96% ee
RCHO
–135 ° C
• Both ( R)- and ( S)-mandelic acids are commercially available.
(R)-mandelic acid
Braun, M. Angew. Chem. , Int. Ed. Engl. 1987 , 26 , 24-37.
CHO
O
O
CH 3
H3C
O
O
CH 3
H3C
O
O
CH 3
H3C
OH
OH
O
O
OH
OH
O
HO Ph
Ph
H
Ph H3C
O
CHO
O
O
CH 3
H3C
O
HO Ph
Ph
Ph
H
H3C
O
+
+
MISMATCHED
40% de
MATCHED
94% de
• Low diastereoselectivities are obtained with mismatched chiral aldehydes.
• A mechanistic rationale has not been proposed.
NaOH
O
N
O
SCH 3
O
H3C CH 3
HO
O
R
OH
O
N
O
O
H3C CH 3
R
OH
SCH 3
n-Bu 2BOTf, i-Pr 2NEt
CH 2Cl 2, 0 ° C
PhCHO
–78 ! 23 ° C
86-99% de
R = Ph, CH 3,
n-C 3H7, i-C 3H7
80-90% (4 steps)
86-99% ee
Ra-Ni, 60 °C
acetone
2N KOH
CH 3OH, 0 °C
Evans, D. A.; Bartroli, J.; Shih, T. L. J. Am. Chem. Soc. 1981 , 103 , 2127-2129.
• A temporary substituent is used to afford acetate
aldol products selectively. Simple N-acetyl imides
do not react selectively.
R1 H
O
R1 R2
O
R2
OSi(CH 3)3
N
H
N B
O
H
O
Ts n-Bu
OH
3 (20 mol%), 14 h
C2H5CN, – 78 °C;
1N HCl/THF
+
3
yield (%) ee (%)
Ph
c-C 6H11
2-furyl
c-C 6H11
82
67
100
56
89
93
92
86
An Enantioselective Mukaiyama Aldol Reaction Catalyzed by a Tryptophan-Derived
Oxazaborolidine
• The Lewis-acid catalyzed addition of silyl enol ethers to aldehydes is known as the Mukaiyama
Aldol reaction: Kobayashi, S.; Uchiro, H.; Shina, I.; Mukaiyama, T. Tetrahedron 1993 , 49 ,
1761-1772.
R1 R2
C6H5
C6H5
C6H5
n-C 4H9Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
• Use of terminal trimethylsilyl enol ethers provide the highest level of enantioselectivities. Catalytic, Enantioselective Acetate Aldol Additions with Silyl Ketene Acetals
Corey, E. J.; Cywin, C. L; Roper, T. D. Tetrahedron Lett. 1992 , 33 , 6907-6910.
HN
O
O B
N S
O
O
O
R
H
CH 3
H
Nu
• A transition state is proposed in which the si face of the aldehyde is blocked by the indole ring.
R H
O
R
OH
St-Bu
O
St-Bu
OSi(CH 3)3
( S)-BINOL, Ti(O i-Pr) 4
(20 mol%), 4Å -MS
Et 2O, –20 ° C
Silyl thioketene acetal
10% HCl, CH 3OH
+
aldehyde yield (%) ee (%)
PhCHO
PhCH 2CH 2CHO
furylCHO
c-C 6H11 CHO
PhCH 2OCH 2CHO
90
80
88
70
82
97
97
98
89
98
Keck, G. E.; Krishnamurthy, D. J. Am. Chem. Soc. 1995 , 117 , 2363-2364.
Catalytic, Enantioselective Mukaiyama Aldol Condensation of Silyl Thioketene Acetals
• This reaction is highly sensitive to the solvent and to reactant concentrations.
CH 3
R1 H
O
OR 2
OSi(CH 3)3
R1 OR 2
OOH
CH 3
CHO
CH 3
CHO
Ph CHO
Ph CHO
CHO
CHO
+
(–)-1 (0.5-5 mol %);
Et 2O, 4 h, –10 °C
Bu 4NF, THF
92
88
93
89
94
93
97
95
97
94
95
96
Aldehyde %ee: R 2 = Et %ee: R 2 = CH 3
Yields for two steps (addition and desilylation) range from 72-98%.
N
O
t-Bu
Br O
O
t-Bu
t-Bu
O
O
Ti
(–)-1
96
91
96
%ee: R 2 = Bn
Review : Carreira, E. M.; Singer, R. A. Drug Discovery Today 1996 , 1, 145-150.
-
-
-Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
Catalytic, Enantioselective Aldol Additions of an Acetone Enolate Equivalent
Carreira, E. M.; Singer, R. A.; Lee, W. J. Am. Chem. Soc. 1994 , 116 , 8837-8838.
Singer, R. A.; Carreira, E. M. Tetrahedron Lett. 1997 , 38 , 927-930.
Singer, R. A.; Shepard, M. S.; Carreira, E. M. Tetrahedron 1998 , 54 , 7025-7032.
• Catalyst 1 is formed by condensation of the chiral amino alcohol with 3-bromo-5-tert -
butylsalicylaldehyde followed by complexation with Ti(O i-Pr) 4 and 3,5 -di-tert -butylsalicylic
acid. Both enantiomeric forms are available.
• Complete removal of i-PrOH during catalyst preparation is key to achieving high yields
and selectivities. This may be done by azeotropic removal of i-PrOH with toluene or by
its silylation in an in situ catalyst preparation (TMSCl, Et 3N).
• The reaction can be carried out in a variety of solvents, such as toluene, benzene,
chloroform, diethyl ether, and tert -butyl methyl ether.
• Alkenyl and alkynyl aldehydes are particularly good substrates for this catalytic
process.
R H
O
OCH 3
OSi(CH 3)3
R OCH 3
OOH
CHO TBSOCH 2
CHO Ph
CHO TIPS
+
(–)-1 (3 mol %)
Et 2O, 0 °C
n-Bu 4NF
yield (%) %ee
88 96
96 94
88 97
88 96
aldehyde
CHO
TBSO
H3C
H3C
(–)-2
• 2-methoxypropene is used as the reaction solvent.
• Unhindered aldehydes afford products with the highest enantioselectivities.
• 2,6-di-tert -butyl-4-methylpyridine (0.4 equiv) is used in the reaction to prevent decomposition
of the product by adventitious acid.
R H
O
CH 3
OCH 3
R CH 3
OOH
CHO Ph(CH 2)3
CHO TBSOCH 2
CHO Ph
Ph CHO
+
(– )-2 (2-10 mol %);
0 ! 23 ° C
Et 2O, 2N HCl
aldehdye temp. (°C) yield %ee
0 99 98
0 85 93
0 99 91
0 ! 23 98 90
0 ! 23 83 66
0 ! 23 79 75
N
O
t-Bu
Br O
Oi-Pr
Oi-Pr
Ti
Carreira, E. M; Lee, W.; Singer, R. A. J. Am. Chem. Soc. 1995 , 117 , 3649-3650.
PhCHO
c-C 6H11 CHO Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
• The vinyl ether products can be isolated, or transformed into other useful products: • The silyl dienolate is easily prepared, purified by distillation, and is stable to storage.
• The absolute sense of induction parallels that of acetate-derived silyl enol ether and
2-methoxypropene addition reactions.
• The protected acetoacetate adducts are versatile precursors for the preparation of optically
active !-hydroxy-"-keto esters, amides, and lactones.
CH 2
OCH 3OH
Ph
OCH 3
OOH
Ph
OOH
Ph OH
84% isolated yield
O3, CH 2Cl 2;
Ph 3P
OsO 4, NMO
acetone, H 2O
Carreira, E. M; Lee, W.; Singer, R. A. J. Am. Chem. Soc. 1995 , 117 , 3649-3650.
Catalytic, Enantioselective Dienolate Additions to Aldehydes
N
O
t-Bu
Br O
O
t-Bu
t-Bu
O
O
Ti
(–)-1
R H
O O O
OSi(CH 3)3
CH 3H3C
O O
O
CH 3H3C
R
OH
CHO TIPS
CHO TBSO
CHO
CH 3
CHO
n-Bu 3Sn
CHO
Ph
+
(–)-1 (1-3 mol %)
2,6-lutidine
(0.4 equiv)
Et 2O, 0 °C, 4 h
10% TFA, THF
aldehyde yield (%) %ee
86
97
88
79
97
91
94
92
92
80
83 84 (96) a
a
after recrystallization.
PhCHO
OOH
Ph O
O
H3C CH 3
OOH
Ph X
O
Ph
O
O
O
X = NHBn, 73%
X = O n-Bu, 81%
K2CO 3, Zn(NO 3)2
CH 3OH
79%
LiAl(NHBn) 4
or
n-BuOH
Singer, R. A.; Carreira, E. M. J. Am. Chem. Soc. 1995 , 117 , 12360-12361.
R H
O O O
OSi(CH 3)3
CH 3H3C
O O
O
CH 3H3C
R
OH
Ph CHO
CH 3
Ph CHO
CH 3
Ph CHO
CHO
OCH 3
S CHO
+
(S)-Tol-BINAP• CuF 2
(2 mol %)
THF, – 78 ° C;
acidic work-up
aldehyde yield (%) %ee
92 94
98 95
82 90
48 91
81 83
74 65
Catalytic, Enantioselective Dienolate Additions to Aldehydes Using a Nucleophilic Catalyst.
PhCHO Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi
• (S)-Tol-BINAP-CuF 2 is readily prepared in situ by mixing ( S)-Tol-BINAP, Cu(OTf) 2, and
(n-Bu 4N)Ph 3SiF 2 in THF.
• This process is efficient for non-enolizable ( !,"-unsaturated, aromatic, and heteroaromatic)
aldehydes.
• Enolizable, aliphatic aldehydes give products with high enantioselectivity, but in poor yield
(<40%).
• Spectroscopic evidence supports a catalytic process involving a chiral transition metal
dienolate as an intermediate.
Catalytic, Enantioselective Aldol Additions of Silyl Thioketene Acetals and Silyl Enol Ethers
Krü ger, J.; Carreira, E. M. J. Am. Chem. Soc. 1998 , 120 , 837-838.
Pagenkopt, B. L.; Krü ger, J.; Stojanovic, A.; Carreira, E. M. Angew. Chem. , Int. Engl. Ed.
1998 , 37 , 3124-3126.
N B NS
Ph Ph
Br
O O
H3C SPh
O
SPh
O BX 2
SPh
O
R
OH
3
3
CH 2Cl 2
Et 3N
–78 # 23 °C
RCHO
–90 °C, 2h
aldehdye yield (%) ee (%)
84
82
91
83
Corey, E. J.; Imwinkelried, R.; Pikul, S.; Xiang, Y. B. J. Am. Chem. Soc. 1989 , 111 , 5493-5495.
• Bromide 3 is produced from the corresponding ( R,R)-bissulfonamide by reaction with
BBr 3 in CH 2Cl 2.
• Upon completion of the reaction, the ( R,R)-bis-sulfonamide can be recovered and reused.
Enantioselective Acetate Aldol Addition Using a Chiral Controller Group
H3C
S
OO
CH 3
C6H5CHO
i-PrCHO
N
N
OO
N
Ph Ph
Cu N
Cu
N
O
CH 3H3C
C(CH 3)3
O
C(CH 3)3
2 SbF 6–
2 TfO –
2+ 2+
BnO H
O
SR 2
OTMS
R1
N N
O
O N Cu O
O H
Ph
H
Ph
H
Bn
R1 R2
H
CH 3
CH 3
i-Bu
t-Bu
Et
Et
Et
BnO
OH
SR 2
O
R1
1 2
+
1 (10 mol%)
CH 2Cl 2, – 78 °C
1 N HCl, THF
time (h)
24
4
1d
2d
T (°C)
–78
–78
–50
–50
syn:anti
–
97:3
86:14
95:5
%ee
99
97
85
95
yield (%)
99
90
48
85
enol silane
geometry
–
(Z)
(E)
(Z)
2+
Nu ( si face)
Evans, D. A.; Kozlowski, M. C.; Murry, J. A.; Burgey, C.; Campos, K. R.; Connell, B. T.; Staples, R.
J. J. Am. Chem. Soc. 1999 , 121 , 669-685. Chem 115 Stereoselective, Directed Aldol Reaction Myers
M. Movassaghi, Chris Coletta
Proline-Catalyzed Asymmetric Aldol Reaction of Acetone
Evans, D. A.; Kozlowski, M. C.; Murry, J. A.; Burgey, C.; Campos, K. R.; Connell, B. T.; Staples, R.
J. J. Am. Chem. Soc. 1999 , 121 , 669-685.
BnO H
O
H3CO CH 3
O
O
H3CO CH 3
O
O
Ot-Bu
OTMS TMSO
St-Bu
OTMS
R
St-Bu
OTMS
R
H3CO St-Bu
O
R
H3C OH
O
H3CO St-Bu
O
R
H3C OH
O
BnO
OH
Ot-Bu
OO
1 (2 mol%)
CH 2Cl 2, – 78 °C
PPTS, CH 3OH
85%, 99% ee
+
+
2 (10 mol%)
THF
–78 °C
1 N HCl
Evans, D. A.; MacMillan, D. W. C.; Campos, K. R. J. Am. Chem. Soc. 1997 , 119 , 10859-
Johnson, J. S.; Evans, D. A. Acc. Chem. Res. 2000 , 33 , 325-335.
• Based on structural data acquired with catalyst 1, a bidentate coordination of methyl
pyruvate to the copper complex 2 has been proposed.
+
1 (10 mol%)
CH 2Cl 2
–78 °C
R = H, CH 3, Et, i-Bu
77-97%
≥96% ee
≥94:6 syn:anti
R = CH 3, Et, i-Bu
81-94%
≥96% ee
≥98:2 anti:syn
Evans, D. A.; Kozlowski, M. C.; Burgey, C. S.; MacMillan, D. W. C. J. Am. Chem. Soc. 1997 ,
119 , 7893-7894.
aldehyde yield
68 (60)
62 (60)
74 (85)
94 (71)
54 (60)
97 (65)
63 (45)
81
85
%ee
76 (86)
60 (89)
65 (67)
69 (74)
77 (88)
96 (96)
84 (83)
99
99
• Typically a 20– 30 equivalent excess of acetone is used in relation to the aldehyde.
• Tertiary and !-branched aldehydes result in the highest yields and enantioselectivities, while
unbranched aliphatic aldehydes give poor yields and enantioselectivities.
• 5,5-Dimethyl thiazolidinium-4-carboxylate (DMTC) has also been found to be an efficient
amino acid catalyst for the acetone aldol reaction. Results with DMTC are in parentheses.
List, B.; Lerner, R. A.; Barbas, C. F., III. J. Am. Chem. Soc. 2000 , 122 , 2395-2396.
Kandasamy, S.; Notz, W.; Bui, T.; Barbas, C. F., III. J. Am. Chem. Soc. 2001 , 123 , 5260-5267.
Proposed transition state:
H3C CH 3
O
+
NH
O
OH
H R
O (30 mol%)
H3C
O OH
RDMSO
p-NO 2C6H4CHO
C6H5CHO
p-BrC 6H4CHO
o-CCl 6H4CHO
!-napthaldehyde
i-PrCHO
c-C 6H11 CHO
t-BuCHO
CHO NH
S
O
OH
DMTC
H3C CH 3
O
NH
O
OH
H
H R
O
H3C
O OH
R
N
O
O H
H3C
R
H
O H
Rankin, K. N.; Gauld, J. W.; Boyd, R. J. J. Phys. Chem. A. 2002 , 106 , 5155-5159.
Bahmanyar, S.; Houk, K. N.; Martin, H. J.; List, B. J. Am. Chem. Soc. 2003 , 125 , 2475-2479.
H
CH 3H3C
CH 3H3C
H
For a discussion on the involvement of oxazolidinones in the mechanism, see: Seebach, D.; Beck,
A. K.; Badine, M.; Limbach, M.; Eschenmoser, A.; Treasurywala, A. M.; Hobi, R.; Prikoszovich, W.;
Linder, B. Helv. Chim. Acta 2007 , 90 , 425– 471. Chem 115 Stereoselective, Directed Aldol Reaction Myers
Chris Coletta
Proline-Catalyzed Asymmetric Aldol Reaction of Hydroxyacetone
H3C
O
+
NH
O
OH
H R
O (30 mol%)
H3C
O OH
RDMSO
OH
aldehyde yield (%)
60
62
95
38
40
51
%ee
99
99
67
97
97
95
c-C 6H11 CHO
i-PrCHO
o-ClC 6H4CHO
t-BuCH 2CHO
Ph CHO
anti :syn
20:1
20:1
1.5:1
1.7:1
2:1
20:1
CHO
CH 3
Proposed origin of selectivity:
• The anti-diol product formed is not readily accessible via asymmetric dihydroxylation, making
this reaction complementary to the Sharpless asymmetric dihydroxylation.
• The reaction is highly regioselective, and with suitable substrates ( !-branched aliphatic
aldehydes) the anti:syn ratio (dr) and enantioselectivity are excellent. In the case of !-
unbranched aldehydes and aromatic aldehydes, the poor anti:syn selectivity is thought to result
from a decrease in an eclipsing interaction between the alcohol and the aldehyde in the
disfavored boat transition state shown below.
Notz, W.; List, B. J. Am. Chem. Soc. 2000 , 122 , 7386-7387.
H3C
O
+
NH
O
OH
H R
O
OH
N
O
O H
H3C
R
H
O H
N
O
O H
H3C
O H
H
HO
H
HO
R
H
eclipsing
interaction
H3C
O OH
R
H3C
O
R
FAVORED
DISFAVORED
H
OH
O
O
H3C CH 3
OH
OH
OH
H
Proline-Catalyzed Asymmetric Aldol Reaction of Acetonide Protected Dihydroxyacetone
Enders, D.; Grondal, C. Angew. Chem. Int. Ed. 2005 , 44 , 1210-1212.
aldehyde yield
97
86
40
69
76
80
31
80
%ee
94
90
97
93
98
96
96
96
• The use of linear aldehydes in this reaction leads to poor yields, likely due to self
condensation.
• Aromatic aldehydes form products with low diastereoselectivity (e.g., a 4:1 anti:syn ratio was
reported for ortho -chlorobenzaldehyde).
• With the !-chiral !-aminoaldehyde shown above, the mismatched case results in a poor
yield,
but excellent dr and ee.
• Certain hexoses have been synthesized by this method.
O
+
H R
O
DMF, 2 º C
i-PrCHO
c-C 6H11 CHO
BnOCH 2CHO
(CH 3O) 2CHCHO
O O
30 mol% proline
anti/syn
98:2
98:2
98:2
94:6
98:2
98:2
98:2
CHO
catalyst
(S)-proline
(S)-proline
(S)-proline
(S)-proline
(R)-proline
(S)-proline
(R)-proline
(S)-proline
Dowex, H 2O O
HO OH OH
OH
CH 2OH
O
OH OH
HO OH
CH 2OH
H3C CH 3
O
O O
H3C CH 3
OH
R
O
O
CH 3
H3C
CHO
O
NBoc
CH 3
H3C
CHO
O
NCbz
CH 3
H3C
D-psicose
O
O O
H3C CH 3
OH
O
O
CH 3
H3CChem 115 Stereoselective, Directed Aldol Reaction Myers
Chris Coletta, Jaron Mercer
Proline-Catalyzed Enantioselective Cross-Aldol Reaction of Aldehydes • The aldol products from !-oxyaldehydes can be further elaborated as part of a two-step
synthesis of carbohydrates.
R1 yield (%)
80
88
87
81
82
80
76
%ee
99
97
99
99
99
98
91
• Slow addition via syringe pump of the donor aldehyde to a solution of the acceptor aldehyde
and proline is required in order to avoid dimerization of the donor aldehyde.
• Either non-enolizable aldehydes or aldehydes containg !- or "-branching are suitable
acceptor aldehydes for this reaction.
Northrup, A. B.; MacMillan, D. W. C. J. Am. Chem. Soc. 2002 , 124 , 6798-6799.
H R2
O
DMF, 4 º C
Me
Me
Me
Me
Me
n-Bu
Bn
Northrup, A. B.; Mangion, I. K.; Hettche, F.; MacMillan, D. W. C. J. Am. Chem. Soc. 2002 , 124 ,
6798-6799.
10 mol% L-proline
H
O
R2
OH
anti:syn
4:1
3:1
14:1
3:1
24:1
24:1
19:1
Et
i-Bu
c-C 6H11
Ph
i-Pr
i-Pr
i-Pr
H
O
R1
R2
Proline-Catalyzed Direct and Enantioselective Aldol Reaction of !-Oxyaldehydes
R yield (%)
73
64
42
61
92
62
%ee
98
97
96
96
95
88
solvent, rt, 24– 48h
Bn
PMB
MOM
TBDPS
TIPS
TBS
10 mol% L-proline
H
O OH
anti:syn
4:1
4:1
4:1
9:1
4:1
4:1
DMF
DMF
DMF
DMF/dioxane
DMSO
dioxane
H
O
solvent
OR OR
OR
R1 H
O
TIPSO DMSO
10 mol% L-proline
92%, 4:1 ( anti:syn )
95% ee
H
O OH
OTIPS
O OH
OAc TIPSO
TIPSO O OH
TIPSO
TIPSO O OH
OAc TIPSO
TIPSO
MgBr 2•Et 2O
CH 2Cl 2
–20 4 º C
TiCL 4
CH 2Cl 2
–20 4 º C
MgBr 2•Et 2O
Et 2O
–20 4 º C
Glucose Mannose Allose
79% yield
10:1 dr, 95% ee
87% yield
19:1 dr, 95% ee
97% yield
19:1 dr, 95% ee
Northrup, A. B.; MacMillan, D. W.C. Science 2004 , 305 , 1752-1755.
Littoralisone:
TMSO
OAc
H
O
OBn 98% ee
78%
D-proline
H
O OH
OBn
MgBr 2•Et 2O
65%, 10:1 dr
98% ee
TMSO
OR
O OBn
OBn
OH
O
HO
O
BnO
O
O
H3C
L-proline,
DMSO
91%
O
H
H OH H3C
O
H
H OAc H3C
O
O
+
intramolecular
Michael reaction
O
H
H OH3C
O
O
Mangion, I. K.; MacMillan, D. W. C. J. Am. Chem. Soc. 2005 , 127 , 3696-3697.
OH
H
O
OH
OH
OH
O
O
O
H
H OH3C
O
O
O
OBn
OBn
OBn
O
O
BnO
h# = 350 nm;
H2, Pd/C
84%
littoralisone
TIPSO
OH OH
OAc
OH
OBn
TBDPSO
HO
H
HChem 115 Stereoselective, Directed Aldol Reaction Myers
Chris Coletta, Fan Liu
Catalytic, Enantioselective Thioester Aldol Reactions • The thioester group of the aldol products can be transformed by Pd-catalyzed cross coupling
to give ketones.
aldehyde yield (%) %ee
• This method is compatible with aldehyde substrates containing unprotected hydroxyl groups,
including phenols.
• Aromatic aldehydes and !-branched aldehydes are generally poor substrates.
H R
O Cu(OTf) 2 (10 mol%),
1 (13 mol%)
PhS
O
R
syn:anti
PhS
O
CH 3
N N
OO
H3C CH 3
1
O
OH 9:1 PhCH 3:acetone
23 º C, 0.17 M, 24 h
CH 3(CH 2)6CHO
CH 3O2C(CH 3)4CHO
HO CHO
O2N
H3C CHO
OH
8
CH 3(CH 2)5 CHO
H3C
H3C CHO
c-C 6H11 CHO
O
O
OH
O
O(CH 2)4CHO
80 9:1 92 (R)
83 10:1 94
83 9:1 93 (S)
79 8:1 91
59 2.2:1 96 (S)
73 7.5:1 89
48 (71 a) 36:1 93
70 5.5:1 92
Magdziak, D.; Lalic, G.; Lee, H. M.; Fortner, K. C.; Aloise, A. D.; Shair, M. D. J. Am. Chem. Soc.
2005 , 127 , 7284-3695.
H3C
O
+
OH
O
O
4
PhS
O
Pd(dppf)Cl 2, trifurylphosphine
Cu(I), i-Pr 2NEt
DMF, 50 º C, 6 h
76%
CH 3
OH
OH O
O CH 3
CH 3
OH
O
O
4
O
CH 3
OH
OH O
O CH 3
CH 3
O
H3C
Ph Ph
CH 3
OH
HO
CH 3
H3C atwo equiv of aldehyde was used.
• A recent example of proline-catalyzed aldol reaction in the synthesis of prostaglandin PGF 2":
O
O
H
H
(109.5 g)
(S)-proline;
[Bn 2NH 2][OCOCF 3]
H O
O
H H
OH
amberlyst 15
MeOH, MgSO 4
14% (two steps)
99% ee
H O
O
H H
OCH 3
N
O
O H
O H
H
O
H
O
H (15.0 g)
4 steps
HO
HO
CO 2H
HO
CH 3
Coulthard, G; Erb, W.; Aggarwal, V. K. Nature 2012 , 489 , 278-281.
(1.9 g) |
637 | https://republicmoving.com/blog/largest-cities-in-california/ | Spring Storage Special
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California, Cities
What Are the Largest Cities in California? 🏙️ | 10 Biggest Cities in California By Population
Are you thinking about moving to the Golden State or just planning a visit? Whether you're chasing career opportunities, beautiful beaches, or just a change of scenery, exploring the largest cities in California can help you find the perfect place to land. From coastal metropolises to inland powerhouses, these California cities offer a wide range of lifestyles, climates, and cultures to explore. Here are the 10 biggest cities in California by population and what makes each unique.
Table of Contents
What Are the Largest Cities in California?
California is the most populous state in the country, with a population of 39.54 million in 2020, according to the United States Census Bureau. Nearly one out of every nine people in the U.S. lives in California! Here are the largest cities in California to explore.
#1. Los Angeles
Population: 3,898,747
Area: 498 square miles
Metro population: 12.8 million
Los Angeles is the largest city in California and a global center for entertainment, business, and culture. With a population of 3.9 million, it’s the heart of the Los Angeles Metropolitan Area, which stretches across five counties and is home to 12.8 million people. The Los Angeles metro area offers everything from world-class beaches to mountain trails, making it one of the most dynamic places to live or visit in the Golden State.
Downtown Los Angeles is a vibrant hub of arts, culture, and commerce. It's home to iconic landmarks like the Walt Disney Concert Hall and the Broad Museum. Just beyond the city center, you'll find neighborhoods that reflect the city's famous cultural diversity, including Koreatown, Little Tokyo, and Boyle Heights.
Los Angeles County is incredibly varied, and it’s not just urban sprawl: it also includes surprising green spaces. In Compton, for example, you’ll find urban farming communities that are transforming empty lots into local food sources and community hubs. It’s part of a broader movement toward sustainability and self-reliance in traditionally underserved areas. There are even several equestrian communities in the county offering a rural atmosphere!
In the Greater Los Angeles area, you'll also find Ontario Mills, the region's largest shopping mall with 28 million visitors per year, the Angeles National Forest, the famous beach city of Malibu, and Hawthorne, home to the Beach Boys Historic Landmark.
At the edge of the Greater Los Angeles metro area is the Inland Empire, a metro area anchored by Riverside and San Bernardino and bordering Los Angeles, Orange, and San Diego counties. Riverside is part of both the Inland Empire and Los Angeles metro area and best known as the birthplace of California's citrus industry.
#2. San Diego
San Diego population: 1,386,932
Area: 372 square miles
Metro population: 3.27 million (San Diego County)
San Diego, one of the three largest cities in California, is known for its stunning beaches, mild year-round climate, laid-back coastal vibe, and strong military presence. As a cultural hub of Southern California, it offers an exciting mix of attractions for residents and visitors. The city is home to the world-famous San Diego Zoo, located in the heart of Balboa Park, a massive urban park filled with museums, gardens, and theaters.
Just south in Chula Vista, the Living Coast Discovery Center brings people closer to the region’s unique coastal wildlife and ecosystems. The nearby coastal resort city of Carlsbad is home to LEGOLAND California and the picturesque South Carlsbad State Beach.
#3. San Jose
San Jose population: 1,013,240
Area: 181 square miles
Metro population: 2 million (San Jose–Sunnyvale–Santa Clara)
As the third largest city in California and the biggest in Northern California, San Jose sits at the heart of Silicon Valley. Known for its thriving tech industry, it's home to many of the world's leading tech companies and startups. The city offers high-paying jobs and a fast-paced, innovation-driven lifestyle that attracts professionals from around the globe.
San Jose also balances its high-tech vibe with family-friendly attractions like the Tech Interactive Museum, lush parks, and a walkable downtown area filled with eateries and cultural spots. If you're looking for one of the largest cities in California where you can balance the perks of a high-paying career and suburban life, San Jose is a great choice.
#4. San Francisco
San Francisco population: 873,965
Area: 232 square miles
Metro population: 4.56 million (San Francisco Bay Area)
Famous for its steep hills, colorful architecture, and the Golden Gate Bridge, San Francisco is the cultural and economic heart of the San Francisco Bay Area. Despite its compact size, the city is packed with iconic landmarks, from historic cable cars to Alcatraz Island. It’s a favorite for urban dwellers who love walkability, diversity, and stunning waterfront views.
With a vibrant cultural scene, strong job market, and proximity to tech giants, San Francisco continues to be a desirable, if expensive, place to live for opportunity and adventure.
#5. Fresno
Fresno population: 542,107
Area: 116 square miles
Metro population: 1.16 million (Fresno and Madera counties)
Located in the San Joaquin Valley, Fresno is the county seat of Fresno County. It's not just one of the largest cities in California: Fresno is the most populous city in Central California and the most populous inland city in the state. It's also the country's third-largest majority-Hispanic city.
Known for its agricultural roots, the city is a major producer of fruits, vegetables, and nuts that feed the nation. Fresno also acts as a gateway to Yosemite National Park, making it a great spot for outdoor enthusiasts. With a lower cost of living compared to coastal cities and growing cultural and business scenes, Fresno is a great option for families and individuals looking for space and affordability.
#6. Sacramento
Sacramento population: 524,943
Area: 100 square miles
Metro population: 2.4 million (Greater Sacramento)
Sacramento, the state capital, is located in Sacramento Valley in Northern California and anchors the Sacramento metropolitan area. It’s a city with a rich history as a former Gold Rush boom town. It remains one of the largest cities in California and is now home to a growing government, healthcare, and tech workforce.
Downtown Sacramento blends modern living with historic charm, from the vibrant Midtown district to the preserved Old Sacramento Waterfront. The city also offers more affordable housing than most major California cities, making it an appealing option for young professionals and families.
#7. Long Beach
Long Beach population: 466,742
Area: 80 square miles
Long Beach is a major port city in the Gateway Cities region of LA County. It's home to one of the world's busiest shipping ports, and it's built over an oilfield with many wells under the city and offshore. The city is home to attractions like the Aquarium of the Pacific, the Queen Mary, and a bustling arts and music scene.
Long Beach is one of the largest cities in California that is not the anchor of a metro area or a county seat. It's part of Greater Los Angeles and the Los Angeles-Long Beach-Anaheim CSA. Its waterfront location, diverse communities, and strong economy make it a great choice if you want to be near Los Angeles but enjoy a more laid-back atmosphere.
#8. Oakland
Oakland population: 440,646
Area: 78 square miles
Metro population: 7.8 million (Bay Area)
Oakland is one of the largest cities in California and the third-largest of Bay Area cities. Situated across the bay from San Francisco in the East Bay area, Oakland is a vibrant, diverse city with a growing population and a unique cultural identity. The city boasts a thriving food scene, world-class sports venues, and a rich history rooted in activism and the arts.
Oakland's neighborhoods offer everything from scenic hillside views to bustling urban centers, and its real estate is generally more affordable than its San Francisco neighbor, making it a great alternative for Bay Area living.
#9. Bakersfield
Bakersfield population: 403,455
Area: 151 square miles
Metro population: 909,000 (Kern County)
Located in Kern County, Bakersfield is known for its role in agriculture, oil production, and especially country music. Often referred to as the birthplace of the “Bakersfield Sound,” the city has deep musical roots and hosts regular concerts and festivals.
Among the largest cities in California, Bakersfield is also one of the most affordable, appealing to families, retirees, and young professionals looking for space and lower housing costs in California's Central Valley.
#10. Anaheim
Anaheim population: 346,824
Area: 51 square miles
Metro population: 12.7 million (LA metro area)
Rounding out the list of the 10 largest cities in California is Anaheim, the largest city in Orange County, or the OC. Best known as the home of Disneyland, Anaheim is a family-friendly OC city. It’s a popular tourist destination with good schools and a growing business hub, especially in tourism, healthcare, and tech. Anaheim also boasts major sports teams like the Anaheim Ducks and Los Angeles Angels.
Anaheim is close to other great OC communities like Huntington Beach, an Orange County coastal city known as Surf City USA, and Santa Ana, the county seat known for its historic downtown.
Anaheim combines family-friendly living with Southern California’s sunny lifestyle, making it a top destination for both visitors and those looking to settle down in a dynamic, entertainment-rich community.
Map of the 10 Largest Cities in California
Other Major Cities in California
While the ten largest cities in California are the most well-known, there are 37 cities in California with at least 150,000 people. Here are other beautiful cities to explore.
| |
| Other Big Cities in California |
| City | County | Population |
| Stockton | San Joaquin County | 319,543 |
| Riverside | Riverside County | 318,858 |
| Irvine | Orange County | 314,621 |
| Santa Ana | Orange County | 310,539 |
| Chula Vista | San Diego County | 274,333 |
| Fremont | Alameda County | 226,206 |
| Santa Clarita | Los Angeles County | 224,028 |
| San Bernardino | San Bernardino County | 223,728 |
| Modesto | Stanislaus County | 218,915 |
| Fontana | San Bernardino County | 215,465 |
| Moreno Valley | Riverside County | 212,392 |
| Oxnard | Ventura County | 202,726 |
FAQ
What Is the Biggest City in California?
Los Angeles is the largest city in California by population, with around 4 million residents. It’s at the heart of the largest cities in California in the SoCal area and known for its diverse cultural scene, entertainment industry, and economic influence.
What Are the Three Largest Cities in California?
Most of California’s largest cities are in Southern California or the Bay Area of Northern California. The three largest cities in California by population are:
Los Angeles: 4 million
San Diego: 1.4 million
San Jose: 1 million
What Is the Largest City in California by Area?
Los Angeles is also the largest city in California by land area. The city covers over 500 square miles, stretching from the Pacific coastline to the inland valleys, making it both vast and diverse in geography.
What Is the Least Populated City in California?
The least populated city in California is Amador City in Amador County, part of the Sierra Nevada area, with a 2020 census population of 200. The next-smallest cities in California are Vernon and the City of Industry, both located in Los Angeles County. These cities have fewer than 300 residents but are home to many industrial businesses.
What Is the Most Populated Area of California?
Southern California is the most populated area of the state. It includes several of California’s most populous cities, such as Los Angeles, San Diego, and Anaheim, and is home to tens of millions of people across sprawling urban and suburban regions.
Which of California's Largest Cities Is Right for You?
From tech hubs like San Jose to coastal getaways like San Diego and Long Beach, the biggest cities in California offer something for everyone. Does one of the largest cities in California seem like the perfect place to call home? Republic Moving & Storage is here to help with your move.
We're the best San Diego movers for a seamless, worry-free move. Call our team today at (619) 591-0079 for a free, personalized moving quote.
Bill Lovejoy
Bill Lovejoy is the president of Republic Moving and Storage. Purchasing the assets of Republic Relocations in May 2008. In 2008, Republic did $700K of business and today is over $15 Million. Republic was named AMSA’s 2010 Agent of the Year for Service Excellence after only two years in business. Bill is proud that he grew up doing every facet of the business from a local mover, over the road driver, salesperson, operations manager, and general manager. (and just about everything else in between) Before purchasing Republic, Bill was with Bekins Moving and Storage where he served as a Senior Vice President and was a partner. He was responsible for three HHG companies as well as two logistics companies. A proud Father of six children, and a lovely wife, Carol, Bill constantly works to keep a balance between work and family Life. Bills “stress relieving” hobbies include racket ball, basketball, tennis, and golf.
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638 | https://www.quora.com/Why-is-the-average-speed-of-two-speeds-not-equal-to-their-arithmetic-mean | Something went wrong. Wait a moment and try again.
Average Speed
Motion (physics)
Arithmetic Means
Speed and Velocity
Mean (mathematics)
5
Why is the average speed of two speeds not equal to their arithmetic mean?
Hiroshi Bowman
Former Engineer, Attorney, Truck Driver, Line Cook, and Teacher.
·
5y
There are a lot of good and correct mathematical answers here, so I’ll try a different approach. Why is our intuition (often) wrong when we try to average two speeds using the average of the values?
If I have two numbers, say 15 and 35, and I want their average, I simply add them and divide by two:
ave(15,35)=15+352=502=25
But in science, we hardly ever have unitless quantities. I can have 15 apples or 35 pesos, but I hardly ever work with a purely abstract quantity.
This is important because it only makes sense to average like quantities. If I harvest 15 kilos of potat
There are a lot of good and correct mathematical answers here, so I’ll try a different approach. Why is our intuition (often) wrong when we try to average two speeds using the average of the values?
If I have two numbers, say 15 and 35, and I want their average, I simply add them and divide by two:
ave(15,35)=15+352=502=25
But in science, we hardly ever have unitless quantities. I can have 15 apples or 35 pesos, but I hardly ever work with a purely abstract quantity.
This is important because it only makes sense to average like quantities. If I harvest 15 kilos of potatoes while you harvest 35 kilos of potatoes, our average harvest of 25 kilos of potatoes could be useful to know in some contexts. However if I dig 15 postholes while you gather 35 eggs, it’s not at all clear what meaning an average of those two quantities would have.
In the case of speed, we have a compound unit. It is a combination of a distance measurement and a time measurement. I might travel at 15 km/hr riding a ski lift to the top of a slope, for example, and 35 km/hr skiing down the same slope. At first glance it looks like we should be able to average these quantities together, but we (usually) can’t. This is because they’re not actually the same thing. I spent much more time on the ski lift going up the mountain (say, 0.21 hours) than I spent on my skis coming down (say, 0.09 hours).
To see why the time is important, let’s try a different compound unit: density. Suppose I ask you to mix olive oil (0.911 grams/cc) and mercury (13.59 g/cc) and ask you the resulting density. Wouldn’t you want to know whether I’m asking you to mix it like this
(ading a little oil added to a lot of mercury will make a mixture with a desity close to that of mercury)
Or like this?
(adding a little mercury to a lot of olive oil will make a mixture with a density close to that of olive oil)
The only time the intuitive approach of a simple average of the number values together works is if the quantity measured by the denominator is the same. If it’s not, then do a weighted average like the other answers suggest.
Related questions
Why is the average speed not the average of two speeds?
Why is harmonic mean used for speeds, not arithmetic mean?
What does arithmetic mean between 2 and 32?
How do you calculate the average speed from two given speeds?
What is the average speed for driving between two cities?
Mark Hodes
Studied at Massachusetts Institute of Technology
·
Author has 595 answers and 166.2K answer views
·
5y
Because the average speed is the total distance traveled divided by the total time elapsed. If you travel the same distance at each speed, you spend more time at the slower speed, so the slower speed ““counts more” in computing the average. For example, if you travel 120 miles at 60 mph and 120 miles at 40 mph, then your times are 2 hr and 3 hr, and the average speed is 240/5 or 48 mph, not 50. Looking at it slightly differently, you spent 2/5 of the time at 60 mph, so the average is 2/5 of 60 plus 3/5 of 40, giving 48 mph.
Satish Suman
6y
average speed is a weighted avg with time.
id est
sum(tivi)/n
where;
n = # of vi
vi= speed
ti= time spent at speed vi
why is it a weighted average?
because the longer time you spend on a speed , the more distance you will cover;
look at it this way, if you spend most of your time at 50k/h hour… well avg speed is 50k/h.
The data points or number of data points at which you record the speed vi do not influence the speed directly but the time spent on that speed does. why?
eg.
data points collected for speed in m/s
10, 20, 50, 10, 20, 20, 20.
respective time spent at such speed in sec.
2, 5, 30, 2, 5, 5, 5
avg
average speed is a weighted avg with time.
id est
sum(tivi)/n
where;
n = # of vi
vi= speed
ti= time spent at speed vi
why is it a weighted average?
because the longer time you spend on a speed , the more distance you will cover;
look at it this way, if you spend most of your time at 50k/h hour… well avg speed is 50k/h.
The data points or number of data points at which you record the speed vi do not influence the speed directly but the time spent on that speed does. why?
eg.
data points collected for speed in m/s
10, 20, 50, 10, 20, 20, 20.
respective time spent at such speed in sec.
2, 5, 30, 2, 5, 5, 5
avg=20.429
weighted avg = 35.92 (closer to 50)
so the number of times 20m/s was noted doesn’t matter but for how long the 20m/s was the speed matters.
weighted avg = sum(wixi)/sum(wi)
wi= weight (in this case time)
Roddy MacPhee
knows set theory
·
Author has 483 answers and 556K answer views
·
7y
Here’s why average speed is
not always
equal to their arithmetic mean:
Average speed is average distance along a path you go in a given time.
If you are at a lower speed longer, your time taken to go the full distance increase.
If you cover the same distance in a longer time your average speed will decrease.
If you are at a higher speed longer you take less time to cover the same distance, therefore increasing average speed.
Only when these effects cancel out will the arithmetic mean be the average speed.
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Related questions
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What are the three arithmetic means between the terms 18 and 2?
Harave Shantha
Former Retd Professor
·
Author has 56 answers and 28.5K answer views
·
5y
Speed = distance/time
Average speed = Total distance / total time taken
Arithmetic mean of the speeds = sum of the two speeds/ 2
If a particle covers a distance s’ in t’ interval and s” in t” interval, then one velocity u’ = s’/t’ and
u” = s”/t”
Now Average speed = (s’+ s”)/(t’+ t”) ….(1)
Average of the speeds = (u’+u”)/2
= (s’/t’ + s”/t”)/2….(2)
It may be seen that equation (1)is not equal to equation (2)
They will be equal only when t’= t”. In other words when the time intervals are equal.
Annu Kumari
Bsc ,MSc physics pursuing in Science & Physics, Army Public School, Delhi Cantt
·
Author has 302 answers and 278.5K answer views
·
5y
Average speed is not equal to their arithmetic mean beacuse average speed is average distance along a path you follow for a given time.
So, if you are at a higher speed, it takes you less time to cover the particular distance, therefore increasing your average speed…
And if you are at lower speed, it takes you more time for that particular distance and your average speed decrease….
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John Root
Works at Florida State University
·
Author has 4.4K answers and 2.3M answer views
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5y
Well, it will be if you spend the same amount of time travelling at those speeds.
If you spend a different amount of time (which you will if you go the same distance at each of the two speeds) it will be skewed toward the speed you spend the most time at (the lower speed).
Ivar Wind Skovgaard
Substitute Teacher, Japanese (2015–present)
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Author has 468 answers and 262.2K answer views
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7y
Well, it
is
- if that’s the kind of average you want to calculate. In some contexts it’s perfectly appropriate, in other contexts it isn’t - see Roddy MacPhee’s answer for an explanation of one such context.
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Neil Gordon
Author has 3.7K answers and 3.3M answer views
·
5y
Here is an example of the same journey Distances, at different Speeds and resulting in different Travel Times. This explains why they cannot be equal to their ‘arithmetic’ mean. The rule, of course does not apply but has to be calculated like this example set below.
Distance 200 km, Speed 50 km/h, Time 4 hours
Distance 200 km, Speed 100 km/h, Time 2 hours
Total Distance 400 km, Total Time 6 hours, Answer: Average Speed 66.667 km/h
Harsh Ghodkar
Statistical Analysis and Data Reconfiguration at Friends (TV series) (2016–present)
·
Author has 133 answers and 379.7K answer views
·
8y
Originally Answered: Why is the average speed not the average of two speeds?
·
When we talk about average speed of a vehicle/moving object , we are talking about the mean speed at which the vehicle was travelling from starting point till end point.
So basically this 2 lines above define average speed.
Let’s say a vehicle was travelling at about 60 kmph for the first hour and 40 kmph for the second hour and then rests.We obviously need to find average speed of the vehicle,it means the mean speed at which the vehicle was around from the start to end.
The answer is =>Total distance /total time=100/2=50 kmph
Hope this clears your doubt.
Sami
US Physics Team, 2023
·
2y
Related
Why is the average v of two speeds, a and b, calculated as "v = 2ab / (a + b)" and not "v = (a + b) / 2"?
If you travel a distance x with speed a, then the same distance x with speed b, then your average speed will be 2aba+b. However, if you travel for a time t with speed a, then the same time t with speed b, then your average speed will indeed be a+b2. So this really depends on the context of the problem.
So why are there two different formulas?
Let's look at the first case, when you travel the same distance in both segments of your journey. In each segment, the velocity v is given by v=xt, where x is the distance traveled and t is the time it took. We can find the times
If you travel a distance x with speed a, then the same distance x with speed b, then your average speed will be 2aba+b. However, if you travel for a time t with speed a, then the same time t with speed b, then your average speed will indeed be a+b2. So this really depends on the context of the problem.
So why are there two different formulas?
Let's look at the first case, when you travel the same distance in both segments of your journey. In each segment, the velocity v is given by v=xt, where x is the distance traveled and t is the time it took. We can find the times by rearranging the velocity formula to get t=xv.In both segments, the distance x is the same, but the time t differs based on the differing speeds, a and b. The average velocity over a journey is x1+x2t1+t2; in other words, it is the total distance divided by the total time. The total distance is, of course, 2x. So t1+t2=xa+xb. Now, all we need to do is bang out the algebra to get our average velocity:
x1+x2t1+t2=2xxa+xb=2xxa+xb1/x1/x=21a+1b=21a+1babab=2abb+a=2aba+b.
Notably, what's going on here is that we're averaging a quantity whose numerator remains constant but whose denominator changes. Averaging something by its denominator is known as taking its harmonic mean, and the formula for the harmonic mean (HM) of two numbers is HM=2aba+b and is derived the same way as above.
The second, more familiar case happens when you travel for the same time t in both segments of your journey. Each segment's velocity is again given by v=xt. This means that each segment's distance is given by x=vt.In this case, the time is the same in both journeys, but the distance differs. The average velocity over a journey is again given by x1+x2t1+t2. So, again, we bang out the algebra to get our average velocity:
x1+x2t1+t2=at+btt+t=(a+b)t2t=a+b2.
This is the more well known arithmetic mean (AM), which happened because we are averaging a quantity whose denominator remains constant and whose numerator changes.
Hamza Said
Former Studying in Giki
·
6y
Related
Why do we prefer harmonic mean over arithmetic mean for calculating average speeds?
No harmonic mean is not preferred over arithmetic. They both are used to find the averages but there are different cases.
ARITHMETIC MEAN : the arithmetic mean is use in that cases where the time interval is the same i.e if we are asking that find the average speed if a car's speed is 100m/s for the first hour and 120 m/s for the second hour.
HARMONIC MEAN : the harmonic mean is used when we deal with the same distances example if we are asked find the average speed if a car travels a journey with 100 m/s and come back to its initial point with 120 m/s. As here the distances are the same so harm
No harmonic mean is not preferred over arithmetic. They both are used to find the averages but there are different cases.
ARITHMETIC MEAN : the arithmetic mean is use in that cases where the time interval is the same i.e if we are asking that find the average speed if a car's speed is 100m/s for the first hour and 120 m/s for the second hour.
HARMONIC MEAN : the harmonic mean is used when we deal with the same distances example if we are asked find the average speed if a car travels a journey with 100 m/s and come back to its initial point with 120 m/s. As here the distances are the same so harmonic mean is used.
I hope it helps!!
Howard Ludwig
Ph.D. in Physics, Northwestern University (Graduated 1982)
·
Author has 3K answers and 10.3M answer views
·
7y
Related
Why is harmonic mean used for speeds, not arithmetic mean?
I think your question is:
Given that a trip is broken up into multiple equal-length segments and you are given the average speed for each segment, why is the average speed for the whole trip given by the harmonic mean of the segment speeds as opposed to the arithmetic mean of the segment speeds?
Average speed is defined as total distance divided by total time for the trip. There is a numerator (dividend) and a denominator (divisor) in a fraction. When performing an unweighted average (mean) of multiple values, you want each value to be equally important. Equal importance is determined by the typ
I think your question is:
Given that a trip is broken up into multiple equal-length segments and you are given the average speed for each segment, why is the average speed for the whole trip given by the harmonic mean of the segment speeds as opposed to the arithmetic mean of the segment speeds?
Average speed is defined as total distance divided by total time for the trip. There is a numerator (dividend) and a denominator (divisor) in a fraction. When performing an unweighted average (mean) of multiple values, you want each value to be equally important. Equal importance is determined by the type of quantity in the denominator. In this case the denominator is dealing with time. In order for the segments to be equally important, they must, therefore, each involve the same amount of time. If a problem said that you drove the same amount of time at 60 km/h as you did at 40 km/h, then your average speed for the trip is the arithmetic mean of the two speeds, (60 km/h + 40 km/h)/2 = 50 km/h.
However, the problem you are given does not have the traveler spending the same amount of time on each segment, but rather the same amount of distance. If you travel the same distance faster, then you are spending less time than going slower that same distance. Therefore, you need to do a weighted arithmetic mean, with the weight being proportional to how much time you spend on each segment. If each segment is length l, and you travel as speeds s1,s2,…,sn over the n segments, the amount of time the traveler spends on segment i is l/si, so we will use that as our weight for that segment’s speed. The weighted mean is then computed by a quotient: The numerator has the sum over all the segments of the product of the weight and the speed for that segment; the denominator has the sum of all the weights. For the numerator, term i is the product of the weight l/si for segment i and the speed si for that segment, and that product is just l, because of the cancellation of the two occurrences of si. Thus,
⟨s⟩=∑ni=1lsisi∑ni=1lsi
=∑ni=1l∑ni=1lsi
=nll∑ni=11si
=n∑ni=11si,
which is the unweighted harmonic mean of the segment speeds. Notice how the common length l of all the segments cancelled out—the segment length could have been ignored from the start. The amount of time spent on a segment is inversely proportional to the speed, so that is the weight—any constants of proportionality will cancel.
Therefore, if you drive 120 km at 40 km/h and the same 120 km at 60 mi/h, that is 3 h at the slower speed and only 2 h at the faster speed, so we are emphasizing the slower speed by spending more time at that speed.
The weighted arithmetic mean is:
3(40 km/h)+2(60 km/h)3+2=2405 km/h=48 km/h.
The unweighted harmonic mean is:
2160 km/h+140 km/h=2×40 km/h×60 km/h40 km/h+60 km/h=4800 (km/h)2100 km/h=48 km/h.
They agree with each other and disagree with the case where the same amount of time was spent at each speed—as expected.
In summary, when you are calculating the average of ratios of two quantities, (such as speed being the ratio of distance to time), if the values of the denominators are equal (equal time for each segment), then the average over all segments is the arithmetic mean of the ratios for for all of the segments.
If the values of the numerators are equal (equal distance for each segment), then the average over all segments is the harmonic mean of the ratios for all of the segments.
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639 | https://www.statlit.org/pdf/1999SchieldAPDU2-Original.pdf | Common Errors in Forming Arithmetic Comparisons Page 1 COMMON ERRORS IN FORMING ARITHMETIC COMPARISONS As published in the Inaugural issue of the Journal “Of Significance” Produced by the Association of Public Data Users www.apdu.org Milo Schield Department of Business, Accounting and MIS Augsburg College Minneapolis, MN 55454 schield@augsburg.edu www.augsburg.edu/ppages/schield Common Errors in Forming Arithmetic Comparisons Page 2 COMMON ERRORS IN FORMING ARITHMETIC COMPARISONS Milo Schield, Augsburg College Department of Business & MIS Minneapolis, MN ABSTRACT This paper reviews the three basic arithmetic compari-sons (simple difference, simple ratio and relative differ-ence), identifies some common arithmetic and gram-matical errors and makes several recommendations. INTRODUCTION Arithmetic comparisons are both common and useful. Comparisons describe associations and associations are the doorways to causation. Arithmetic comparisons indicate the strength of an association by giving specific measurements. Making errors in arithmetic compari-sons creates confusion and doubt in an argument; min-imizing these errors will enhance statistical literacy. RULES FOR COMPARISONS To be statistically literate, one must be able to form arithmetic comparisons of any two numbers. First con-sider the three basic arithmetic comparisons. Each comparison has two values: Test and Base. The Test is the value being evaluated; the Base is the basis of com-parison. Each comparison has certain keywords. Mul-tiply the comparison by 100% to obtain a percentage. Simple difference [Test – Base] • When comparing numbers other than percentages, use ‘more/less than’ or ‘greater/smaller than’. Five is one more than four; four is one less than five. • When comparing two percentages, use ‘percentage point(s) more/less than’. Five percent is one per-centage point more than four percent. Four percent is one percentage point less than five percent. Simple ratio [Test / Base] Use ‘times as much as’, ‘% as much as’ or ‘% of’. • When using ‘times as much as’: Five is 1.25 times as much as four; four is 0.8 times as much as five. • When using ‘% as much as’ or ‘% of’: Five is 125% as much as four; four is 80% of five. Use ‘% as much when the ratio is greater than one; use ‘% of’ when the ratio is less than one. Relative difference [(Test – Base)/Base] Use ‘% more/less than’ or ‘times more/less than’. • When using ‘% more/less than’: Five is 25% more than four; four is 20% less than five. • When using ‘times more/less than’: Five is 0.25 times more than four. Four is 0.20 times less than five. In comparing numbers in data having or allowing negative values, use only simple differences; never use simple ratios or relative differences. Weather forecast-ers never say the temperature today (+20oF) is minus two times as much as that a year ago (-10oF). They use only simple differences: “It is 30oF warmer today than it was a year ago.” They don’t use simple ratios or rela-tive differences even when both temperatures are posi-tive since these temperatures lack a natural zero. SEVEN COMMON ERRORS The following are common errors in forming arithmetic comparisons. These errors are easily corrected. 1. Mixing the grammar of comparisons: Don’t say, “times as much than”, “percent as much as”, “per-cent times as much as” or “times more as”. 2. Confusing a simple difference with a relative dif-ference in comparing percentages. If interest rates go from 5% to 7%, the increase is 40% – not 2%. 3. Confusing ‘percent’ with ‘percentage points.’ If interest rates go from 5% to 7%, the increase is 2 percentage points – not 2%. 4. Treating relative differences as numerically sym-metric: Five is 25% more than four. But four is 20% less than five – not 25% less than five. 5. Confusing ‘times as much’ with ‘percent more than’. If B is three times as big as A, then B is 200% greater than A – not 300% greater than A. 6. Confusing ‘times as much’ with ‘times more than’. If B is three times as much as A, then B is two times more than A – not three times more than A. The essential feature is the difference is between ‘as much as’ and ‘more than.’ ‘As much as’ indi-cates a ratio; ‘more than’ indicates a difference. ‘More than’ means ‘added onto the base’. This es-sential difference is ignored by those who say that ‘times’ is dominant so that ‘three times as much’ is really the same as ‘three times more than.’ 7. Using ‘times less’ as an inverted form of ‘times as much’. Since six is three times as much as two, it is tempting to say that two is three times less than six. Two is definitely less than six and their ratio is definitely that of three to one. But if two were three times less than six, then six should be three times more than two. Recall that six is three times as much as two – two times more than two. ‘Times less than’ is an inverted form of ‘times more than’ – not ‘times as much’. This error is more common in speech than in writing. This error is a variation on error 6. Common Errors in Forming Arithmetic Comparisons Page 3 FOUR RECOMMENDATIONS 1. When comparing two values, give both values be-ing compared along with the comparison. The reader can then verify the comparison (e.g., Interest rates went from 5% to 7% – an increase of 40%). 2. Use ‘times as much as’ for simple ratios and ‘% more than’ for relative differences. • Do not use ‘% as much as’. This is not com-mon usage and it generates confusion with ‘% more than’ analogous to the confusion between ‘times as much as’ and ‘times more than’. • Do not use ‘times more than’ or ‘times less than’. This avoids errors 6 and 7. 3. If you elect to use ‘times more than’, then state the comparison using both forms of ‘times’. E.g., nine is two times more than three (three times as much as three). This eliminates confusion and educates the reader on forming comparisons. 4. Use ‘times as much as’ when the simple ratio is more than one. Use ‘% more than’ if the ratio is less than three. Use either one (‘times as much’ or ‘% more than’) if the ratio is between one and three. Use ‘% less’ or ‘% of’ if the ratio is less than 1. This convention avoids small ratios (0.4 times) and large percentages (450% more). E.g., 30 is three times as much as 10; 15 is 50% more than 10, and 15 is 1.5 times as much as 10. 8 is 20% less than 10 and 8 is 80% of 10. CLOSING COMMENTS There are other errors in forming arithmetic compari-sons especially in comparing percentages and rates. But identifying and eliminating these common errors provides a strong foundation for taking on more com-plex comparisons. Dr. Schield can be reached at schield@augsburg.edu. For copies of related materials, visit www.augsburg.edu/ppages/schield. |
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DigitalCommons@University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln
Calculus-Based General Physics Instructional Materials in Physics and
Astronomy
1975
COLLISIONS COLLISIONS
Follow this and additional works at:
Part of the Other Physics Commons
"COLLISIONS" (1975).
Calculus-Based General Physics . 6.
This Article is brought to you for free and open access by the Instructional Materials in Physics and Astronomy at
DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Calculus-Based General
Physics by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Module --STUDY GUIDE
COLLISIONS
INTRODUCTION
If you have ever watched or played pool, football, baseball, soccer, hockey, or been involved in an automobile accident you have some idea about the results of a collision. We are interested in studying collisions for a variety of reasons. For example, you can determine the speed of a bullet by making use of the physics of the collision process. You can also estimate the speed of an automobile before the accident by knowing the physics of the collision process and a few other physical principles. Physicists use collisions to determine the properties of atomic and subatomic particles. Essentially, a particle accelerator is a device that provides a controlled collision process between subatomic particles so that, among other things, some of the properties of the target particle can be studied. In addition the study of collisions is an example of the use of a fundamental physical tool, i.e., a conservation law. A conservation law implies that something remains the same, i.e., is conserved, as you have seen in a previous module, Conservation of Energy.
Conservation laws play an important role in physics. In the study of collisions in this module we are interested in one of the fundamental conservation laws, conservation of linear momentum. If the sum of the external forces is zero, then the linear momentum is conserved in the collision. This is fortunate since it provides a way around the analysis of the forces of interaction between two bodies as they collide, an otherwise formidable task. Thus the conservation-of-linear-momentum law allows one to analyze the effects of a collision without a detailed knowledge of the forces of interaction. One can deduce the converse also, as does the particle physicist in accelerator experiments, for example - some of the properties of the target particles may be deduced from the law of conservation of linear momentum and other laws of physics. PREREQUISITES Before you begin this module, you should be able to: Solve mechanical problems involving conserv-ative and nonconservative forces, by applying the conservation-of-total-energy concept (needed for Objective 2 of this module) Use the concepts of impulse and linear momentum to solve mechanical problems (needed for Objective 2 of this module) Location of Prerequisite Content
Conservation of Energy Module Impulse and Momentum Module STUDY GUIDE: Collisions LEARNING OBJECTIVES
After you have mastered the content of this module, you will be able to: 1. Conservation of linear momentum - Define or state: (a) elastic collision, (b) inelastic collision, (c) perfectly or completely inelastic collision, and (d) the law of conservation of linear momentum. 2. Collisions - Solve problems involving collisions between two or more bodies and/or the splitting up of a body into two or more fragments. GENERAL COMMENTS
The important concepts presented in this module are Elastic collision: a collision in which kinetic energy is conserved. Inelastic collision: a collision in which kinetic energy is not conserved. Note: Kinetic energy may be either gained or lost during a collision. Perfectly inelastic collision: a collision in which the colliding objects stick together after the collision. Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. Or, during a collision, if the interaction impulsive force is very large in comparison to the sum of all external forces such as gravity, then it is a good approximation to say that linear momentum is conserved. Remember: Momentum is a VECTOR quantity and must be treated as such. 2, (
,I
STUDY GUIDE: Collisions 3{B 1)
TEXT: Frederick J. Bueche, Introduction to Ph~SiCS for Scientists and Engineers (McGraw-Hill, New York, 1975), second e ition
SUGGESTED STUDY PROCEDURE
Read Sections 7.4, 7.5, and 9.3 through 9.6; work Problems 16 in Chapter 7, 15 in Chapter 9, and Problems A and B plus any two of the problems listed in the table below; answer Question 7 in Chapter 9. Also work the following problem: A block of balsa wood whose mass is 0.60 kg is hung from a string of negligible weight. A bullet with a mass of 2.00 g and a muzzle velocity of 160 m/s is fired into this block at close range (horizontally) and becomes embedded in the block. (a) Find the velocity of the block plus the bullet just after the collision. (b) Calculate how high the block will rise. When you think that you know the material well take the Practice Test. BUECHE
~o
enough to sati sfy the objectives, Objective Prob 1ems with Assigned Problems Additional Number Readings Solutions Problems
Study Text (work any
Study Guide Guide two) l(a) General Comments,
Sec. 9.3 l(b) General Comments,
Sec. 9.3 l(c) Genera 1Comments l(d) Genera 1Comments 2 Secs. 7.4, A, B A, B Chap. 7, Chap. 7, Quest. 7.5, 9.3- Probe 16; 5,7,13, 9.6 Chap. 9, Probs. 4, 9-
Probe 15 18; Chap. 9, Quest. 4, j,
10, Probs. 10-
17, 26, 28 Quest. = Quest;on(s).
(
STUDY GUIDE: Collisions 3(HR 1) TEXT: David Halliday and Robert Resnick, Fundamentals of Physics (Wiley, New York, 1970; revised printing, 1974) SUGGESTED STUDY PROCEDURE Read Chapter 9, Sections 9-1 and 9-3 through 9-5; answer Question 6; work Problems 18, 22, 30, 40, plus Problems A and B. Note: Definitions of elastic, inelastic, and completely inelastic collisions
gTVen in Section 9-4 apply to all collisions, not just to one-dimensional coll isions. When you think tha.t you know the material well enough to satisfy the objectives, take the Practice Test. Objective Number Readings l(a) General Comments, Sec. 9-4 l(b) General Comments,
Sec. 9-4 l(c) General Comments,
Sec. 9-4 l(d) General 2Comments,
Sec. 9-3
Secs. 9-1, 9-3 to 9-5 Quest. = Question(s). HALLIDAY AND RESNICK
Problems with Solutions
Study
Guide A, BAssigned Problems
Study Text Guide (Chap. 9) A, B Quest. 6, Probs. 18, 22, 30, 40 Additi onal Problems (Chap. 9) Quest. 1-8, Probs. 14- 17 ,21, 24, 28, 34, 37, 44
•
STUDY GUIDE: Collisions 3(SZ 1} TEXT: Francis Weston Sears and Mark W. Zemansky, University Physics (Addison-Wesley, Reading, Mass., 1970), fourth edition
SUGGESTED STUDY PROCEDURE Read Chapter 8, Sections 8-2 through 8-6; work Problems 8-6, 8-10, 8-20, 8-25, 8-37 plus Problems A and B. Note: Conservation of linear momentum can be used to a good approximation when the external forces are small compared to the interaction forces during the collision. For example: when a bat hits a ball, the interaction forces are large (generally) compared to gravity and the force exerted by the batter; therefore, in this case gravity and the force exerted by the batter can be neglected during the interaction. When you think that you know the material well enough to satisfy the objectives, take the Practice Test. Objective Number Readings l(a) General Comments,
Sec. 8-3 l(b} General Comments,
Sec. 8-4 l(c} General Comments,
Sec. 8-3 l(d} General 2Comments Secs. 8-2 to 8-6
SEARS AND ZEMANSKY Problems with Solutions
Study Guide A, BAssigned Problems
Study Text Guide A, B 8-6, 8-10, 8-20, 8-25, 8-37 Additional Problems 8-5, 8-11, 8-12, 8-16, 8-23, 8-28, 8-29, 8-30, 8-35 STUDY GUIDE: Collisions 3(WS 1) TEXT: Richard T. Weidner and Robert L. Sells, Elementar~ Classical Physics (Allyn and Bacon, Boston, 1973), second edition, 01. 1
SUGGESTED STUDY PROCEDURE
Read Sections 5-5 through 5-7 in Chapter 5 and Section 10-6 in Chapter 10.
Work Problems 5-1, 5-11, 5-12, and 10-32 in the text plus Problems A and B. Note: Even though the statement of the law of conservation of linear momentum was deduced for a particular two-body collision, it is valid in general. Conser-
vation of linear momentum can be used to a good approximation when the external forces are small compared to the interaction forces during the interaction. For example: when a bat hits a ball, the interaction forces are large (generally) compared to gravity and the force exerted by the batter; therefore in this case gravity and the force exerted by the batter can be neglected during the interaction. Your text makes a distinction between types of inelastic collisions that is not gen-erally made, i.e., ~K < 0 inelastic and ~K > 0 explosive, where ~K is the change in kinetic energy during the collision. Generally ~K ~ 0 is classified as an inelastic collision, as is done in the General Comments.
When you think that you know the material well enough to satisfy the objectives, take the Practice Test. Objective Number Readings l(a) General
Comments,
Sec. 10-6 l(b) General l(c)
Comments,
Sec. 10-6
General
Comments,
Sec. 10-6
l(d) General 2
Comments
Secs. 5-5 to 5-7,
10-6
WEIDNER AND SELLS
Prob 1ems with Solutions
Study
Guide A, BAssigned Problems Study Text Guide A, B 5-l. 5-11 ,5-12,
10-32
Additional Problems 5-1,5-2,5-5 to 5-12, 5-15 to 5-21,10-1,
10-6, 10-8,
10-27 to 10-35 STUDY GUIDE: Collisions
PROBLEM SET WITH SOLUTIONS
A(2). In the absence of any external forces a particle with mass m and speed v is incident on a particle of mass M initially at rest (see Fig. 1). After collision, particle m is observed to go off at an angle 6 2 with respect to the initial direction with speed vf (see Fig. 2). Mis observed to go off at an angle 6 1 with respect to the initial direction with speed V.
m
•
(a) Find V in terms of all the other parameters except .6,. (b) Let each parameter in turn approach zero and comment on the reasonableness of the answer. (c) What is the maximum value for V? Is this reasonable? (d) What happens as the magnitude of M + ~?
vM
•
y
Fi gure 1 Fi gure 2 Fi gure 3Solution (a) Given m, M, v, v f ' and 6 2, Find V. Use momentum conservation (see Fig. 3). The x component of the linear-momentum-conservation equation is 4mv = MV cos 6 1 + mV f cos 6 2, (,) The y component is MV sin 6 1 = mV f sin 6 2,Rearranging Eq. (l) we have MV cos 6, = m{v - v f cos 6 2 ), MV sin 6 1 = mV f sin 6 2,Squaring the above equations and adding we have (2)
(3)
x
STUDY GUIDE: Collisions M2V2 (cos 2 6 1 + sin 2 6 1) = m2 (v 2 - 2vv f cos 6 2 + vf 2 cos 2 6 2) + m2vf 2 sin 2 6 2•
Using the fact that sin 2 6 + cos 2 6 = 1 and doing some rearranging we have v = (m/M) ;'(V~ - 2vv f cos 6 2 + v2). (b) Now let m + °and V + 0:
Reasonable - consider a ping-pong ball colliding with a bowling ball. M+ 0, V becomes large: Reasonable - consider a bowling ball colliding with a ping-pong ball. Vf + 0, V + mv/M: Reasonable - all linear momentum transferred from m to M. v + 0, V + mVf/M: Reasonable - explosio~ total linear momentum zero.
6
2 + 0, V + (m/M)(v f - v): Reasonable - linear momentum lost by m given to M. (c) Vmax at 6 2 = ~:
Reasonable - since m has maximum change in momentum, i.e., it transfers maximum momentum to M. (d) As M+~, V + 0:
Reasonable - since as M goes to ~, V has to become smaller in order to conserve linear momentum. 8(2). In the absence of external forces a particle of mass m collides elastically with another particle of the same mass initially at rest.
Show that if the collision is not head-on the two particles go off so that the angle between their directions is ~/2.
5I
STUDY GUIDE: Collisions (a) State what is given and what you are to find symbolically. (b) Draw a diagram. (c) Write down the relevant equation or equations. In this case use the 1aws of and _ _(d) Solve the equations for the relevant unknown or unknowns. Solution (a)
(b)
Given m1 = m2 = m, show that 6 1 + 6 2
-+
pm
• , • -+
PI
-+
P2
Fi gure 4
(c) Conservation of linear momentum:
-+ -+ -+
p = P1 + P2·
Conservation of kinetic energy: = rr/2.
•
(4 )(5 )
6where K stands for the kinetic energy of the incident particle before collision,
·etc.
(d) Now K = p2/2m. (6 )
'. , Squari ng Eq. (4) we have 2 2 2 -+ -+
P = Pl + P2 + 2P1 • P2·
Combining the above eqaation with Eq. (6) we have STUDY GUIDE: Collisions Combining the above equation with Eq. (5) we have
(++)
Pl • P2 /m = O.
Assuming Pl f 0; P2 f 0 and m f ~, all not very interesting cases, then
Pl 1 P2' which was to be shown. PRACTI CE TEST
Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 72. A hockey puck B rests on a smooth ice surface and is struck by an identical puck A that was originally traveling at 60 m/s and that is deflected 30°
from its original direction. Puck B acquires a velocity at an angle of
45° to the original velocity of A. (a) Compute the speed of each puck after collision. (b) Is the collision perfectly elastic? If not, what fraction of the original kinetic energy of puck A is "lost"?
A
G~· -....:)~--------------- """-~-
A
Figure 5 STUDY GUIDE: Collisions Practice Test Answers 1. (a) Elastic collision: a collision in which kinetic energy is conserved. (b) Inelastic collision: a collision in which kinetic energy is not conserved. Note: kinetic energy may be either gained or lost during a co 11 is ion. (c) Perfectly inelastic collision: a collision in which the colliding objects stick together after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system ;s zero, then the total linear momentum of the system remains constant. Or, during a collision, if the interaction impulsive force is very large in comparison to the sum of all external forces such as gravity, then it is a good approximation to say that linear momentum is conserved. Note: If you missed any of these definitions, MEMORIZE the ones that you missed. 2. (a) VBF = VA;I(sin e.z cot 9 1 + cos 9 2 ), Check this answer for dimensions and reasonableness [see parts (b), (c), and (d) in the solution of Problem A for reasonableness check].
V
BF = 31 m/s.
VAF = VAi sin e2/(sin 9 2 cos 9 1 + sin 9 1 cos 9 2)Check this answer for dimensions and reasonableness. VAF = 44 m/s.
(b) ~K/Ki = 0.20, or the collision is inelastic. Note: If you missed this problem, work some more of the optional problems 8in the text until you feel that you understand the material. When you under-stand the material, then ask for a Mastery Test. If you answered this Practice Test correctly, ask for a Mastery Test now. ( . '
COLLI SIONS Date __ _Mastery Test Form A pass recycle 1 2Name Tutor -------------------------1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. Consider the collision as shown in the figure below. The colliding particles are identical and initially have a speed of 10.0 m/s. After the collision particle 2 moves as shown. (a) Find the velocity of particle 1 after collision. (b) Is this an elastic collision? Before After
''. y y
, , 45°
, x
/ ~5°
2,( M 2
'"
x
5.0 m/s (
!
~.
COLLISIONS Date -----------------Mastery Test Form B pass recycle 1 2Name Tutor --------------------------- ----------------------------1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. A cannon mounted on a stationary railroad car fires a 100-kg projectile so the latter moves horizontally with a speed of 600 m/s at a sideways angle of 30.0° to the track. The car plus the cannon have a mass of
10 000 kg. (a) Make a sketch and describe in what way momentum conservation can be used to solve this problem, or explain why this is not the case. (b) At what speed will the railroad car recoil along the track? (Neglect friction with the track.) (
{
"
COLLISIONS
Date _ Mastery Test Form C pass recycle 2Tutor Name ____ __ -----------------1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. As you stand at a lightly traveled street intersection, you are startled to observe the collision of a fire engine (mass = 6000 kg), a house trailer (mass = 25 000 kg), a steam calliope (mass = 4000 kg), and adump truck (mass = 8000 kg). The four vehicles are, respectively, traveling northeast at 30.0 mis, west at 10.0 mis, south at 20.0 mis,
and east at 25.0 m/s.
(a) Make a diagram of the vehicles immediately before the collision, and indicate their masses and vector velocities. (b) If the entire junk pile sticks together after the collision, what is its velocity before it has been slowed down by friction? (c) Is this collision elastic? (
COLLISIONS
Date -----------------Mastery Test Form D pass recycle 2Name Tutor __ 1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. A radioactive nucleus, initially at rest, decays by emitting an electron and an electron antineutrino at right angles to one another. The momentum of the electron is 1.20 x 10- 22 kg mls and that of the electron antineutrino is 6.4 x 10- 23 kg m/s.
(a) (b)
Find the momentum of the recoiling nucleus. -26 If the mass of the recoiling residual nucleus is 5.8 x 10 kg, what is its kinetic energy of recoil? (
COLLISIONS Date -------------Mastery Test Form E pass recycle 1 2Name Tutor ___ _1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. A body of mass m1 = 10.0 kg moves to the right along a frictionless table top at a speed of 50 mls and makes a head-on collision with another body whose mass m2 is unknown, but which is originally moving to the left at a speed of 30.0 m/s. If the bodies stick together after the collision and move to the right at a speed of 20.0 mis, what is the value of m 2?Is the collision elastic? (
COLLISIONS Date -----------------Mastery Test Form F pass recycle 1 2Name Tutor ----------------------------1. Define or state: (a) elastic collision; (b) inelastic collision; (c) perfectly inelastic collision; (d) the law of conservation of linear momentum. 2. A ball with speed 3.00 m/s and mass 1.00 kg strikes off-center a second ball of mass 3.00 kg initially at rest. The incident ball is deflected
90° from its incident direction, and the collision is completely elastic. In what direction, relative to that of the incident ball before the collision, does the second ball leave the collision? COLLISIONS A-1 MASTERY TEST GRADING KEY - Form AWhat To Look For 1.(a) .6K = O.
(b) .6K ., O.
(c) Objects stick toqether after collision.
+
Vector nature of p Is answer dimensionally correct? Is the answer reasonable? Solutions 1.(a) Elastic collision - a collision in which kinetic energy is conserved. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick trigether
after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant.
+"-
2.(a) p = 2mvi: total momentum of particles 1 and 2before collision.
P2 = -mv/23: momentum of particle 2 after collision.
+++
P1 = P - P2
A
"
= 2mvi + (mv/2)j.
A
""
v1 = 2vi + (v/2)j = ( 14. Oi + 5.0j) m/s.
(b)
K~ = mi, 1K = (~) (l) + (~)i(2 + 1) :I mi(l + ~) f 242 488 2 1 9 2 10
= mv (8 + 8) = mv (-g),
Ki t Kf ·
Thus the collision is not elastic. Also, since vl has a j component it cannot be ~ to v 2;therefore collision is inelastic. COLLISIONS B-1
MASTERY TEST GRADING KEY - Form BWhat To Look For 1. (a) ,'}K = 0;
(c) Objects stick together after coll ision.
(d)
,'}P_= 0 if
l:f ext - O.
Solutions l.(a) Elastic collision - a collision in which kinetic energy is conserved. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick together after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. 2.(a) p conserved only 2.(a) Momentum not conserved in direction! to track. in direction defined by track. (b) Pused is total p
times cos 30°. Answer dimensionally correct? Units correct? Answer reasonable? (Top view)
P = 0 = (P ) + (P )x x proj x car = [(100)(600) cos 30.0 0 J m/s + (10 OOO)v,
v = -[(100)(600) cos 30.0°/10 OOOJ m/s = -(6 cos 30.0°) m/s = -5.2 m/s. COLLISIONS C-l
MASTERY TEST GRADING KEY - Form C
What To Look For 1. (a) llK = O.
(b) llK ~ O.
(c) Objects stick together after collision (d) liP = 0 if Ef t = O. ex Solutions l.(a) Elastic collision - a collision in which kinetic energy is conserved. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick together after the collision. (d) Conservation of line~r momentum: if the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. 2.(a) Make sure diagram 2.(a) is clear! (b) liP = O. Vector -+
nature of p. Answer dimensionally correct? Units correct? Answer reasonable?
2.5 Is
8000 kg
30.0 mit!'"
2.00 mls 10.0 mls
000 kq E
6000 kg (b) Momentum is conserved, junk pile moves with vc.m.
-+
vc.m.
AAA
_ P _ (6000)(15.0v 2i + 15.0v 2j) + (25 OOO)(-lOi)
- M- ( 6000 + 25000 + 4000 + 8000
AA
(4000)(-20j) + (8000)(25i)) mls
6000 + 25000 + 4000 + 8000
'" '" ...... '"
= (77i + 47j)/(43) mls = (1.8i + l.lj) m/s.
(c) recognize perfectly (c) No - perfectly inelastic. inelastic collision Ii
COLLISIONS D-l
MASTERY TEST GRADING KEY - Form DWhat To Look For l.(a) L1K = O.
(b) L1K ~ O.
(c) Objects stick together after coll i sion. (d) L1P = 0 if
LF ext = O.
Solutions l.(a) Elastic collision - a collision in which kinetic energy is conserved. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick together after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. 2. L1P = O.~ Vector 2. nature of p. Answer dimensionally correct? Units correct? Answer reasonable? y
/')/e1ectro~
i- ~e electron antineutrino)
~~~~~~~------
xuc1eus
~
(a) Pf = 0,
Pnuc = -(0.641 + 1.20j) x 10- 22 kg m/s.
K - n 2 - 1.81 x 10- 44 2 2(b) L- kg m Is ,2m 2(5.8 x 10- 26 )
K = 1.6 x 10- 19 J. I(
COLLISIONS E-l
MASTERY TEST GRADING KEY - Form EWhat To Look For
1.(a)lIK=0.
(b) lIK ~ O.
(c) Objects stick together after collision. (d) liP = 0 if
IF ext = O.
Solutions l.(a) Elastic collision - a collision in which kinetic energy is conserved. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick together after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. 2. liP = O. Vector 2. ~ +nature of p? Answer dimensionally correct?
Units correct? Answer reasonable? ~
lip
= O.
m. (v l' - vf ) f50 20 0~m2 = 1 1 = 10.0 kg -. - 6 0 k(v 2i + vf )30 + 20.0 - . g. Collision is inelastic since objects stick together; therefore it is not elastic! COLLISIONS F-l
MASTERY TEST GRADING KEY - Form F
14hat To Look For l.(a) lIK = O.
(b) ilK
'I O.
(c) Objects stick together after collision.
(d) lIP_= 0 if
Ef ext - O.
-+
LlP = O. Vector
-+
nature of p? Answer dimensionally correct? Units correct? Answer reasonable? Solutions l.(a) Elastic collision - a collision in which kinetic energy is conserved. 2. (b) Inelastic collision - a collision in which kinetic energy is not conserved. (c) Perfectly inelastic collision - a collision in which the colliding objects stick together after the collision. (d) Conservation of linear momentum: If the sum of the external forces acting on a system is zero, then the total linear momentum of the system remains constant. Before After
1
Pli 2 Plf
• 0
Yt x
-p~-~-----~
-+ -+ -+
P2 = Pli - Pl f' 222
-+ A
Pli Plf P2 Pl i = mlvlii, -=-+-
2m l 2m l 2m 2'2 2 222
-+ A
mlv li mlv lf ml 2 2PIf = ml v1fj , = 2m + 2m (v li + vlf)' 2m l 1 2tan e = / (m 2 - m l )/(m 2 + m,) = / (2/4) = /2i2.
e = tan-l(~2).
See diagram for definition of e. |
642 | https://metricsystem.net/derived-units/special-names/coulomb/ | Skip to content
Metric System
The International System of Units (SI)
coulomb
coulomb
SI coherent derived unit with special name and symbol
| | | | |
--- --- |
| Name | Symbol | Quantity | Base units |
| coulomb | C | electric charge | s A |
| The coulomb, symbol C, is the SI coherent derived unit of electric charge. It is the special name for the ampere second, symbol s A. One coulomb is the charge transported by a constant current of one ampere in one second. |
| Definition | e |
|
The coulomb is named after the French physicist, Charles-Augustin de Coulomb (1736 – 1806).
The fixed numerical value of the elementary charge, e, is defined as 1.602 176 634 × 10−19 C. This definition implies the exact relation e = 1.602 176 634 × 10−19 C. Inverting this relation gives an exact expression for the coulomb in terms of the defining constant e:
The effect of this definition is that one coulomb is exactly 1⁄(1.602 176 634 × 10−19) of the value of the elementary charge, e.
Capacitance
One coulomb is the amount of excess charge on a capacitor of one farad charged to a potential difference of one volt.
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643 | https://www.merriam-webster.com/dictionary/rock%20bass | ROCK BASS Definition & Meaning - Merriam-Webster
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Est. 1828
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rock bass
noun
:a brown spotted sunfish (Ambloplites rupestris) found especially in the upper Mississippi River valley and Great Lakes region
Examples of rock bass in a Sentence
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Examples are automatically compiled from online sources to show current usage.Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback.
Fish native to the Jackson River (which feeds into the dam) like smallmouth bass, rock bass, and chain pickerel can also be caught in the lake.—Tara Massouleh McCay, Southern Living, 15 Aug. 2025 His submissions ranged from an 11-inch rock bass to a 30-inch carp.—The Indianapolis Star, 12 July 2023 Panfish, from rock bass to bluegills and yellow perch, won’t hesitate to eat one.—Pete M. Anderson, Field & Stream, 20 Mar. 2023 Science is still attempting to figure out the answer to this question, Strom said, but certain types of fish, such as bluegill, crappie, rock bass and perch can accumulate high levels of PFAS in their tissue.—Laura Schulte, Journal Sentinel, 14 Feb. 2023 See All Example Sentences for rock bass
Word History
First Known Use
1811, in the meaning defined above
Time Traveler
The first known use of rock bass was in 1811
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Rhymes for rock bass
alas
amass
bromegrass
bypass
chagas
crabgrass
crevasse
eelgrass
eyeglass
groundmass
harass
impasse
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rock beauty
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“Rock bass.” Merriam-Webster.com Dictionary, Merriam-Webster, Accessed 28 Sep. 2025.
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rock bass
noun
:a sunfish spotted with brown that is found especially in the upper Mississippi valley and Great Lakes region
More from Merriam-Webster on rock bass
Britannica.com: Encyclopedia article about rock bass
Last Updated:16 Aug 2025 - Updated example sentences
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644 | https://www.quora.com/How-can-we-determine-if-three-points-are-collinear-using-their-coordinates | Something went wrong. Wait a moment and try again.
Collinear Points
Coordinate Geometry Sol
Geometric Analysis
Coordiate Geometry
Coordinate Plane
Geometry in Mathematics
Coordinate Geometery
Collinearity (Mathematics...
5
How can we determine if three points are collinear using their coordinates?
Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994)
·
Author has 10.5K answers and 23.4M answer views
·
1y
In two dimensions, point point form for a line joining (a,b) and (c,d) is
(c−a)(y−b)=(d−b)(x−a)
Plug in (a,b),(c,d), and (x,y); if the statement is true you have collinear points.
Let’s try more dimensions.
Parametrically, the line joining A and B is A+t(B−A); we seek t such that
C=A+t(B−A)
(C−A)=t(B−A)
So in n dimensions, we just form two vectors from the three points; if one is a scalar multiple of the other, collinearity, else not.
[Shoot, got suckered into answering another Quora Prompt Generator question.]
Related questions
How can we determine whether three points are collinear without calculating their distance using vector algebra?
How do I prove that three points are collinear?
How can we determine if two points are collinear or not using only their coordinates?
How can I find 3 points that are collinear if 2 points we know their coordinates and the last point is unknown coordinates to know them?
How do I determine if points are collinear?
Naman Anand
Lives in Chandigarh, India
·
Author has 95 answers and 90.4K answer views
·
5y
Originally Answered: How do you know if 3 points are collinear?
·
It is a question of coordinate geometry and can be approached in many ways I will mention two of the most effective and easy ones here
By using the triangle area formula. When we see 3 points make a triangle we observe that there is zero area enclosed by the triangle made by 3 collinear points as they are in a line . For (x¹, y¹) , (x², y²) and (x³, y³) formula is : 0 = ½[ x¹(y²-y³) +x²(y³-y¹) +x³(y¹-y²) ]
For this we can compare the slopes of the lines made by the three points as they are in a line slope ( angle made with x axis )would be same so :( y²-y¹) /(x²-x¹) =(y³-y²) /(x³-x²)
Good questio
It is a question of coordinate geometry and can be approached in many ways I will mention two of the most effective and easy ones here
By using the triangle area formula. When we see 3 points make a triangle we observe that there is zero area enclosed by the triangle made by 3 collinear points as they are in a line . For (x¹, y¹) , (x², y²) and (x³, y³) formula is : 0 = ½[ x¹(y²-y³) +x²(y³-y¹) +x³(y¹-y²) ]
For this we can compare the slopes of the lines made by the three points as they are in a line slope ( angle made with x axis )would be same so :( y²-y¹) /(x²-x¹) =(y³-y²) /(x³-x²)
Good question thank you for asking
Allen Ries
Math Major University of Alberta
·
Author has 24.9K answers and 9.6M answer views
·
1y
Let the points be a b and c. Find the equation for al possible lines taking two points at a time. If yo end up with the same equation each time the points are collinear
Actually find the equation for the line for two of the points. If the third point is on that line the points are collinear.
Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994)
·
Author has 10.5K answers and 23.4M answer views
·
3y
Originally Answered: How do you know if 3 points are collinear?
·
Great question. Let A,B,C be the quadrances (squared distances) between any three points in n-dimensional space.
The Triple Quad Formula says the three points are collinear precisely when
(A+B−C)2=4AB
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Related questions
Is it possible to determine if two given points are collinear without calculating their coordinates?
Can you provide an example of finding coordinates for three non-collinear points?
How will you identify the coordinates of the point?
How can you determine whether three given points are collinear without using coordinates/graphs?
When are 3 points collinear?
Vikas Rathore
Mankind from Earth (planet)
·
Updated 6y
Related
How do I prove that three points are collinear?
Based on my long expirement with Maths,
Here are some common ways,
First method:
Use the concept, if ABC is a straight line than, AB+BC=AC
Second method :
In case of geometry, if you are given 3 ponits, A(x,y,z) ,B(a,b,c),C(p,q,r)
Find the distance between AB =√(x-a)^2 + (y-b)^2 + (z-c)^2, then find BC and AC in similar way.
If AB+BC=AC then points are collinear.
Third method:
Use the concept that area of the triangle formed by three collinear is zero.
One way is by Using determinant,
The other way is, Let A,B,C be there points, using coordinates, make two vector a (vector)=AB and b (vector) =BC
Now a×b=0
Based on my long expirement with Maths,
Here are some common ways,
First method:
Use the concept, if ABC is a straight line than, AB+BC=AC
Second method :
In case of geometry, if you are given 3 ponits, A(x,y,z) ,B(a,b,c),C(p,q,r)
Find the distance between AB =√(x-a)^2 + (y-b)^2 + (z-c)^2, then find BC and AC in similar way.
If AB+BC=AC then points are collinear.
Third method:
Use the concept that area of the triangle formed by three collinear is zero.
One way is by Using determinant,
The other way is, Let A,B,C be there points, using coordinates, make two vector a (vector)=AB and b (vector) =BC
Now a×b=0 (i.e a vector cross b vector=0)
Forth meathod:
If direction ratios of three vectors a,b,c are proportional then they are collinear.
Thankyou!!
Brijesh Kumar Mishra
Lead at IBM Global Business Services
·
7y
Related
How can I prove that 3 points are not collinear?
Collinear: Three or more points are said to be collinear if they lie on a single straight line. Means Slope will be same.
If you know the Coordinates of all Points then derive Slope from those coordinate and If they are not equal means Points are not collinear.
Lets Take example:
Collinear:
Coordinates: A (1,2) B(4,5) C(6,7)
slopeAB=(2–5)/(1–4)=-3/-3=1
slopeBC=(5–7)/(4–6)=-2/-2=1
slopeAC=(2–7)/(1–6)=-5/-5=1
slopeAB=slopeBC=slopeAC (i.e collinear)
Non Collinear:
Coordinates: A (3,2) B(4,5) C(6,7)
slopeAB=(2–5)/(3–4)=-3/-=3
slopeBC=(5–7)/(4–6)=-2/-2=1
slopeAC=(2–7)/(3–6)=-5/-3=5/3
slopeAB≠slopeBC≠sl
Collinear: Three or more points are said to be collinear if they lie on a single straight line. Means Slope will be same.
If you know the Coordinates of all Points then derive Slope from those coordinate and If they are not equal means Points are not collinear.
Lets Take example:
Collinear:
Coordinates: A (1,2) B(4,5) C(6,7)
slopeAB=(2–5)/(1–4)=-3/-3=1
slopeBC=(5–7)/(4–6)=-2/-2=1
slopeAC=(2–7)/(1–6)=-5/-5=1
slopeAB=slopeBC=slopeAC (i.e collinear)
Non Collinear:
Coordinates: A (3,2) B(4,5) C(6,7)
slopeAB=(2–5)/(3–4)=-3/-=3
slopeBC=(5–7)/(4–6)=-2/-2=1
slopeAC=(2–7)/(3–6)=-5/-3=5/3
slopeAB≠slopeBC≠slopeAC (i.e non collinear)
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Jan Bolluyt
Studied Mathematics & Physics (till 12th) (Graduated 1970)
·
Author has 970 answers and 72.5K answer views
·
May 24
pair up different pairs of points 3 times. Find the slop of each pair. If they are collinear, all three slopes will be the same.
Kunal Sharma
B.Sc. Hons in Mathematics, GEHU, Dehradun (Graduated 2017)
·
Author has 63 answers and 215.9K answer views
·
8y
Related
How do I prove that three points are collinear?
Well you can apply two simple methods to prove that 3 points are collinear,
First method,
In this method we will take the 3 points as the vertices of a triangle and will try to find the area of the triangle.
If the points lie on the same line then the area of the triangle will be 0.
Let the points be
A(x1,y1)
B(x2,y2)
C(x3,y3)
Then Area of triangle ABC will be,
= 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
Put any set of points in this formula. If the results comes to be zero then the points will be collinear.
The second method,
In this method we will find the slope of two points at a time and check whether the sl
Well you can apply two simple methods to prove that 3 points are collinear,
First method,
In this method we will take the 3 points as the vertices of a triangle and will try to find the area of the triangle.
If the points lie on the same line then the area of the triangle will be 0.
Let the points be
A(x1,y1)
B(x2,y2)
C(x3,y3)
Then Area of triangle ABC will be,
= 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
Put any set of points in this formula. If the results comes to be zero then the points will be collinear.
The second method,
In this method we will find the slope of two points at a time and check whether the slopes in the two cases are equal.
If the slopes are equal and one point is common in both the cases, then the points will be collinear…
Let ,
A(x1,y1)
B(x2,y2)
C(x3,y3)
be three points.
Then we will find the slope of AB and BC.
Since B is the common point, then if the slopes are equal the points will be collinear.
Formula to find slope(m),
m=(y2-y1)/(x2-x1)
Slope of AB(m1),
m1=(y2-y1)/(x2-x1)
Slope of BC(m2),
m2=(y3-y2)/(x3-x2)
If m1=m2
Then the points will be collinear.
I hope this helps… :)
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Frank Gross
Former Museum Guide at the Griffith Observatory (2006–2018)
·
Author has 617 answers and 394.1K answer views
·
4y
Related
How can I find 3 points that are collinear if 2 points we know their coordinates and the last point is unknown coordinates to know them?
Use the two known points to find the slope “m” between them.
the equation of a straight line can be in the form
y = mx + b where m is the slope (gradient) and b is the y-intercept.
Substitute the gradient m (now known) and the x and y coordinated of one of the known points into this equation and solve for “b” the y intercept.
Now, substitute “m” and “b” into the equation. It will look like the following though probably with different numbers for “m” and “b”:
y = 3x + 7
Now, pick an x coordinate of your choosing and substitute this value for x into the equation ans solve for y. I will demonstrate wit
Use the two known points to find the slope “m” between them.
the equation of a straight line can be in the form
y = mx + b where m is the slope (gradient) and b is the y-intercept.
Substitute the gradient m (now known) and the x and y coordinated of one of the known points into this equation and solve for “b” the y intercept.
Now, substitute “m” and “b” into the equation. It will look like the following though probably with different numbers for “m” and “b”:
y = 3x + 7
Now, pick an x coordinate of your choosing and substitute this value for x into the equation ans solve for y. I will demonstrate with my equation when x = 5:
y = 3 x 5 + 7
y = 15 + 7
y = 22
So the another point collinear with your two known points is
(5,22)
Of course there are an infinate number of other points, collinear to your two known ponts you can discover using this method.
Srinivasan
Former Maths B.T.Asst Teacher (Retired) at P.S.G Sarvajana Hr.Sec School (1999–2010)
·
Author has 4K answers and 4.8M answer views
·
Updated 3y
Related
How do I prove that three points are collinear?
We can use any one of the following methods to prove the given three points are collinear.
.Let the points are A,B,C
1.If slope of AB=slope of BC also B is a common point,the points are collinear
2.Using distance formula if AB+BC=AC or sum of the length any two segments is equal to the length of the third one , the points are collinear
3.Using area of a triangle formula if the area formed by these points=o they are
collinear
Using two points formula find the equation of a line using any two points
Substitute the third point in this equation. If it satisfies they are collinear
5.There is one more me
We can use any one of the following methods to prove the given three points are collinear.
.Let the points are A,B,C
1.If slope of AB=slope of BC also B is a common point,the points are collinear
2.Using distance formula if AB+BC=AC or sum of the length any two segments is equal to the length of the third one , the points are collinear
3.Using area of a triangle formula if the area formed by these points=o they are
collinear
Using two points formula find the equation of a line using any two points
Substitute the third point in this equation. If it satisfies they are collinear
5.There is one more method to show the given points are collinear
Take the points as (x1,y1),(x2,y2) and (x3,y3)
y1−y2x1x2+y2−y3x2x3+y3−y1x3x1=0
In all the methods using slope formula is simple
Bob Ueland
Updated 7y
Related
How do I prove that three points are collinear?
Most answers here assume that you have the three points expressed in coordinates. When proving theorems in geometry (for instance proving Desargues theorem), this is seldom the case. One way to prove that A,B and C are collinear, in such cases, is to prove that AB||AC. In order to prove that you often have to use Side Splitting theorem in one form or another. Another alternative is to use Menelaus theorem. A third alternative is to use cross ratio. Still another way is to pick a point D (which is not collinear with AC) and then draw the lines AC and BD. Call the intersecting point B’. If you c
Most answers here assume that you have the three points expressed in coordinates. When proving theorems in geometry (for instance proving Desargues theorem), this is seldom the case. One way to prove that A,B and C are collinear, in such cases, is to prove that AB||AC. In order to prove that you often have to use Side Splitting theorem in one form or another. Another alternative is to use Menelaus theorem. A third alternative is to use cross ratio. Still another way is to pick a point D (which is not collinear with AC) and then draw the lines AC and BD. Call the intersecting point B’. If you can show that B=B’ then ABC are collinear.
Ajay Sreenivas
Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010)
·
Author has 5.3K answers and 1.7M answer views
·
1y
Three points A, B and C are collinear iff the largest distance is equal to the sum of the two smaller distances.
Subhasish Debroy
Former SDE at Bharat Sanchar Nigam Limited (BSNL)
·
Author has 6.6K answers and 5.8M answer views
·
7y
Related
How do I prove that three points are collinear?
This can be resolved by 3 different ways
1) Find the area of triangle formed by the three points. If the area is zero(0) , three points are collinear. If (x1,y1), (x2,y2) & (x3, y3) using the formula area ∆ =1/2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}
2) Let the points are A,B and C. Find the slope of the straight lines separately AB, BC and CA, using formula for slope
m = difference of y co-ordinate / difference of x coordinates, If the three slopes are equal then we say three points are collinear.
3) Find the equation of the straight line using any of two points. Put the value of third co ordinate in the
This can be resolved by 3 different ways
1) Find the area of triangle formed by the three points. If the area is zero(0) , three points are collinear. If (x1,y1), (x2,y2) & (x3, y3) using the formula area ∆ =1/2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}
2) Let the points are A,B and C. Find the slope of the straight lines separately AB, BC and CA, using formula for slope
m = difference of y co-ordinate / difference of x coordinates, If the three slopes are equal then we say three points are collinear.
3) Find the equation of the straight line using any of two points. Put the value of third co ordinate in the equation, if it satisfy the equation, i.e. putting value of x and y of third points in place of x and y of the equation we get the value of RHS of the equation, then we can conclude three points are collinear.
Related questions
How can we determine whether three points are collinear without calculating their distance using vector algebra?
How do I prove that three points are collinear?
How can we determine if two points are collinear or not using only their coordinates?
How can I find 3 points that are collinear if 2 points we know their coordinates and the last point is unknown coordinates to know them?
How do I determine if points are collinear?
Is it possible to determine if two given points are collinear without calculating their coordinates?
Can you provide an example of finding coordinates for three non-collinear points?
How will you identify the coordinates of the point?
How can you determine whether three given points are collinear without using coordinates/graphs?
When are 3 points collinear?
What are the coordinates of a third point in order for three points to be collinear if the coordinates of only two points are known?
What are the names of three collinear points?
What are the coordinates of a third point in order for the three points to be collinear knowing the coordinate of only two of them?
How do you find five collinear points on a coordinate grid?
Is it possible to determine if one point lies on another point using only their coordinates?
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最新推荐文章于 2020-11-16 08:09:03 发布
高中数学解题技巧
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Lis武林中为何高手能一击致命？考场上为何学霸面对压轴题也能游刃有余？关键在于他们能够透过现象看清本质。那么如何才能看清本质呢，首先你要做到盯住目标！
对于解决数学问题也是如此，盯住目标，也就是你要求解要证明的问题，联系相关定理，方法，定义，搭建已知（前提）和未知（结论）之间的桥梁，问题就能迎刃而解了！
我们把未知或者题目要证明的结论统称为目标。解题的高手很清楚“有的放矢”这几个字， 我们往往不仅仅从已知出发正向构建桥梁，而是反过来从目标出发，反向构建桥梁：
要求解/要求证 (原目标，目标1) -> 我们只需要求解/求证（目标2） -> 我们只需要求解/求证（目标3）-> …… -> 已知/前提
在这个不断更新目标的过程中，我们反复问自己：盯住目标 你能联想相关的定理，方法，定义吗？你能试着把目标和已知，前提结合吗？这就是不断地调用学习过的知识的过程。
首先，我们把整个立体几何第一章空间的直线和平面的绝大多数定理进行整理归纳，这是我们搭建桥梁的基础，将其用下图来总结：
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646 | https://math.stackexchange.com/questions/41107/zero-probability-and-impossibility | Zero probability and impossibility - Mathematics Stack Exchange
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Zero probability and impossibility
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I read a comment under this question:
There are plenty of events that can occur that have zero probability.
This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering
if zero probability and impossibility mean the same?
if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?
Thanks and regards!
probability-theory
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edited Feb 26, 2018 at 5:20
Parcly Taxel
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1 @Vicrobot: No. Probability can be exactly 0 0 for possible events, e.g. here and here.Eric Duminil –Eric Duminil 2019-09-02 08:22:15 +00:00 Commented Sep 2, 2019 at 8:22
2 @Vicrobot: There's nothing to admit. If you multiply both sides of 1∞=0 1∞=0 by ∞∞, you get u n d e f i n e d=u n d e f i n e d u n d e f i n e d=u n d e f i n e d. With your logic, you could prove that a point must have a non-zero area since a square is the sum of many points.Eric Duminil –Eric Duminil 2019-09-02 13:22:48 +00:00 Commented Sep 2, 2019 at 13:22
2 @Vicrobot: Once again, the probability of picking 1 1, 2–√2 or π π from any real number between 0 0 and 10 10 must be exactly 0 0. The probability is well defined, it must be at least 0 and it must also be smaller than 1 n 1 n for any n n. It means it must be 0 0 and not just "near zero" or "very small". It might feel weird but it's true, and you'll accept it someday ;).Eric Duminil –Eric Duminil 2019-09-03 13:19:24 +00:00 Commented Sep 3, 2019 at 13:19
2 @Vicrobot I read the chat. It would help if you use reason rigorously with specific, well-defined language, because human intuitions about real numbers are often wrong. Not just your intuitions, mine as well. For example, picking a random real number uniformly from [0,1][0,1], the probability of getting a rational number is exactly 0 0. Understanding this surprising result requires understanding the concept of a measure zero subset. The main topic is the Lebesgue measure. Have fun :)Zubin Mukerjee –Zubin Mukerjee 2019-10-11 13:12:42 +00:00 Commented Oct 11, 2019 at 13:12
2 I didn't say anything about " 1/∞1/∞ ". Read about Lebesgue measure ... it will give you a better understanding of how there can be events with probability 0 0 that are possible.Zubin Mukerjee –Zubin Mukerjee 2019-10-11 14:39:26 +00:00 Commented Oct 11, 2019 at 14:39
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Two Schools
I think the crux of the matter is what probability actually is:
The Bayesian view - probabilities are measures of (personal) confidence or belief, so it's quite obvious why an event with probability zero is not the same thing as an impossible event. But perhaps this isn't such a satisfactory answer.
The frequentist view - probabilities are the asymptotic frequency of events as the number of independent trials tends to infinity. Here again wee see that something that happens with probability zero is not the same as something impossible; it's just something that happens so infrequently that the numerator in occurences trials occurences trials is dominated by the denominator.
Technically Speaking
Putting aside such philosophical matters, there's also a technical matter to be discussed here. Under the usual measure-theoretic formulation of probability theory, we have a sample space Ω Ω and a family F⊆P(Ω)F⊆P(Ω) of events (measurable subsets of Ω Ω), and the probability of an event A∈F A∈F is its measure P(A)P(A). There is nothing in the axioms of measure theory which say that a non-empty set must have a non-zero measure; and if we interpret F F as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability.
Example
To give a concrete example, consider a random variable X X which is uniformly distributed on the interval [0,1][0,1]. Although P[X∈(a,b)]=b−a P[X∈(a,b)]=b−a for all (a,b)⊂0,1⊂[0,1], the axioms of probability force us to conclude that P[X=x]=0 P[X=x]=0 for any individual x∈[0,1]x∈[0,1]: for if P[X=x]=ε>0 P[X=x]=ε>0, because X X is uniformly distributed, by additivity of the probabilities of disjoint events, we'd be forced to conclude that [0,1][0,1] contains at most 1 ε 1 ε (a finite number!) points, which is absurd.
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edited Jan 10, 2020 at 23:07
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answered May 24, 2011 at 19:58
Zhen LinZhen Lin
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Thanks! "if we interpret F as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability", (1) do you mean that the empty set is a possible event, and not an impossible one? (2) Are impossible events F-measurable?Tim –Tim 2011-05-24 20:12:30 +00:00 Commented May 24, 2011 at 20:12
@Tim: I'd say ∅∅ is a possible event: it's possible nothing happens. The idea that ∅∅ is impossible comes from the fact that an impossible event is outside the scope of the model, whether it be because the "corresponding" sample point is not in Ω Ω (whatever that means) or because the combination of sample points in question is not measurable. In either case it's meaningless to ask about the probability of such an event because it's undefined.Zhen Lin –Zhen Lin 2011-05-24 20:54:33 +00:00 Commented May 24, 2011 at 20:54
I like the first paragraph, it defines what exactly P(A) = 0 is.SmallChess –SmallChess 2014-08-14 02:12:06 +00:00 Commented Aug 14, 2014 at 2:12
2 But ϵ=0 ϵ=0 is also absurd, because then the probability that you pick any number at all is 0 0.Fax –Fax 2019-09-13 14:21:54 +00:00 Commented Sep 13, 2019 at 14:21
I'm bewildered there's no school of thought that a random sample from a continuum does, in fact, not exist. No one has ever demonstrated a truly random sample of a truly uniform, provably continuous distribution. Zero probabilities are hinting at this very thing, but I guess it seems absurd to everyone else for reasons I don't understand.Brent –Brent 2024-07-02 02:28:31 +00:00 Commented Jul 2, 2024 at 2:28
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Zero probability isn't impossibility. If you were to choose a random number from the real line, 1 has zero probability of being chosen, but still it's possible to choose 1.
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answered May 24, 2011 at 19:40
Fernando MartinFernando Martin
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7 More information here: en.wikipedia.org/wiki/Almost_surelyFernando Martin –Fernando Martin 2011-05-24 19:41:26 +00:00 Commented May 24, 2011 at 19:41
8 One could argue that such a random choice of a real doesn't have zero probability, but only infinitesimal probability. Thus, it looks like you have zero probability, when you don't.Doug Spoonwood –Doug Spoonwood 2011-06-13 13:42:07 +00:00 Commented Jun 13, 2011 at 13:42
8 No, Doug. Fernando is correct. Any finite number divided by an infinite number is 0, not an "infinitesimal probability". Dealing with infinities is a difficult concept to grasp, sort of like Einstein's relativity.lkessler –lkessler 2014-07-30 16:56:34 +00:00 Commented Jul 30, 2014 at 16:56
4 @lkessler What, 1/infinity being an infinitesimal is like the definition of an infinitesimal. en.wikipedia.org/wiki/Infinitesimalsemicolon –semicolon 2017-06-21 19:30:06 +00:00 Commented Jun 21, 2017 at 19:30
2 @YasirSadiq What we're dealing with here is not 1/∞1/∞, but lim n→∞(1/n)lim n→∞(1/n), which is precisely 0 0(at this point, I seriously suggest you to look up for the epsilon-delta definition of limits). semicolon's point is not valid in the set of standard reals, but in the set of hyperreals/surreals.Manan –Manan 2020-08-12 13:45:05 +00:00 Commented Aug 12, 2020 at 13:45
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Adding to what others have already mentioned. There is also this notion of plausible event. I am not sure if this is standard. But in the book "Measure Theory and Probability" by Malcolm Ritchie Adams and V. Guillemin, a plausible event is defined as an event which corresponds to a Borel set.
Hence, my understanding of the three words is as follows:
If we take the probability space (X,F,μ)(X,F,μ),
An event A⊆X A⊆X is impossible if A=∅A=∅
An event A⊆X A⊆X is implausible if A∉F A∉F
An event A⊆X A⊆X is improbable if μ∗(A)=0 μ∗(A)=0
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answered May 24, 2011 at 20:28
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3 An implausible event doesn't satisfy our intuitive notion of "improbable", since if an event is implausible then so is its complement.Yuval Filmus –Yuval Filmus 2011-05-24 20:48:53 +00:00 Commented May 24, 2011 at 20:48
@Yuval: True. On a slightly similar note, I am not sure intuitively if we would want μ(A)=0 μ(A)=0 or μ∗(A)μ∗(A) for an improbable event.user17762 –user17762 2011-05-24 20:59:31 +00:00 Commented May 24, 2011 at 20:59
Thanks! Just wonder what the star means in μ∗(A)=0 μ∗(A)=0?Tim –Tim 2011-05-24 20:59:59 +00:00 Commented May 24, 2011 at 20:59
@Tim: By μ∗μ∗, I mean the outer measure user17762 –user17762 2011-05-24 21:01:49 +00:00 Commented May 24, 2011 at 21:01
1 @Tim presumably because the outer measure can be applied to all set, not just sets A∈F A∈F.Thomas Andrews –Thomas Andrews 2011-05-24 21:26:33 +00:00 Commented May 24, 2011 at 21:26
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Mathematicians generally formalize probability using the notion of a probability space and measure theory. In this formalism it is possible for an event to have probability 0 0 without being the empty event. Perhaps the simplest "realistic" (and I use the word loosely) example of such an event is the event of flipping only heads infinitely many times. This event has probability 0 0, but it is not empty, which is what one might call a formal definition of "impossible."
The underlying probability space is the set of possible ways to flip a coin infinitely many times. An example of an impossible event here is that you flip, say, cat. The coin has only a heads side and a tails side; it doesn't have a cat side, so flipping cat is impossible.
(Whether this formalism says anything reasonable about the real world is debatable. In practice, events of sufficiently small probability are already impossible. The above is just a statement about a certain mathematical formalism that has proven to be useful in certain contexts. In mathematics, we want to prove statements about some class of objects. Sometimes we can prove that the statement holds with probability 1 1, but this does not imply that it holds for all objects, and since we actually care about all objects this distinction really does need to be made in mathematics.)
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edited May 24, 2011 at 21:26
joriki
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answered May 24, 2011 at 19:46
Qiaochu YuanQiaochu Yuan
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1 "In practice, events of sufficiently small probability are already impossible." Not true! You give me a non-zero probability, however small, and I can generate an event on my laptop which has an even smaller probability of occurring. I just have to generate enough random bits. For this, I only need time proportional to the logarithm of the reciprocal of your probability. (Of course you could make this reciprocal logarithm really big, but I would still win for all reasonable definitions of "really big".)TonyK –TonyK 2011-05-24 19:50:47 +00:00 Commented May 24, 2011 at 19:50
That feels like a cop-out. Unless you can predict what bits you're going to generate in advance... in practice, you should ignore any event which is less likely than, say, winning the lottery ten times (given that you don't currently play the lottery and you believe yourself to be a rational actor).Qiaochu Yuan –Qiaochu Yuan 2011-05-24 19:52:34 +00:00 Commented May 24, 2011 at 19:52
Thanks! Concerning my second question, do you mean that probability theory represents impossibility with the empty set, not a non-empty set with probability zero? Is this how probability theory distinguishes zero probability and impossibility?Tim –Tim 2011-05-24 19:54:58 +00:00 Commented May 24, 2011 at 19:54
1 @Qiaochu: Perhaps I haven't explained myself clearly enough. Let me try again. (i) In an uncountable probability space, events of probability zero can happen. No argument there. (ii) In a countable probability space, events of arbitrarily small probability can happen. Agreed? (iii) I can easily program my laptop to generate (pseudo-)random events of arbitrarily small probability.TonyK –TonyK 2011-05-24 20:10:36 +00:00 Commented May 24, 2011 at 20:10
1 @Doug: that's an interesting question. I suppose one could write down a generalization of probability measures that are allowed to take value in the nonstandard reals and perhaps such a theory would allow such things.Qiaochu Yuan –Qiaochu Yuan 2011-06-13 13:58:49 +00:00 Commented Jun 13, 2011 at 13:58
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Let A A be an event, Pr Pr be the probability measure.
A A has zero probability if Pr(A)=0 Pr(A)=0.
A A is impossible if A=∅A=∅.
Impossibility implies zero probability, but the reverse is false. Consider the real line R R; if you randomly select a number x x, the probability that x=0 x=0 is 0 0, but this is not impossible. In fact, the probability that x x belongs to some countable set, e.g Q Q, is also 0 0.
From a purely mathematical point of view, impossibility is simply a stronger statement, so impossibility cannot be described by probability measure. However, another way of thinking might shed some light. That is, if the probability that something exists has probability greater than 0 0, then it exists. This notion has been used for some mathematical arguments.
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edited Jul 31, 2012 at 16:30
James Waldby - jwpat7
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Thanks! Having said that A is impossible if A is the empty set, how come that impossibility cannot be described by probability measure? Isn't A being the empty set how probability measure describes impossibility?Tim –Tim 2011-05-24 20:02:31 +00:00 Commented May 24, 2011 at 20:02
2 What I mean is that Pr(A)=0 Pr(A)=0 does not imply A=∅A=∅, i.e. knowing probability measure =0=0 does not help you figure out if the set is empty or not.Hawii –Hawii 2011-05-24 20:27:35 +00:00 Commented May 24, 2011 at 20:27
7 I like the definition of "impossibility cannot be described by probability measure."SmallChess –SmallChess 2014-08-14 02:15:06 +00:00 Commented Aug 14, 2014 at 2:15
1 So I guess this means the notion of impossibility is external to the 'probabilistic way of thinking' (as it's described here) since whether an event is empty depends on the particular representation (i.e., probability space) used?Vandermonde –Vandermonde 2016-01-08 04:05:07 +00:00 Commented Jan 8, 2016 at 4:05
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Probability theory is an abstract subject, which is not limited to the real world. In cases where it is limited to the real world, an event of zero probability will not occur. But the abstract underpinning of the real-world cases allows for the occurrence of zero-probability events; when you translate these abstract events into events that are physically detectable, their probabilities become non-zero.
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answered May 24, 2011 at 19:47
TonyKTonyK
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Thanks! Do you mean that probability theory may not perfectly model the real world, which is the reason that an event assumed to have zero probability in probability theoretical model can have non-zero probability in the real world and therefore can occur in the real world?Tim –Tim 2011-05-24 19:58:44 +00:00 Commented May 24, 2011 at 19:58
@Tim: the culprit isn't probability theory in general (here). It's the specific information that you have. If your information is incomplete, your assessment of the probability of various events is necessarily going to be incomplete. Your assessment of the probability of an event is (if you subscribe to the appropriate philosophy of probability) a statement about your mind, not about the world.Qiaochu Yuan –Qiaochu Yuan 2011-05-24 20:17:53 +00:00 Commented May 24, 2011 at 20:17
5 Another way to think about it is that, in the real world, you only have finite probability spaces. The is no "real" way to pick a random number uniformly in [0,1][0,1], only to pick successively closer approximations of a real number in [0,1][0,1] so that the limit is uniform. Then probability on infinite spaces is about limits of probabilities on finite spaces. This turns out to be useful in the real world because we are dealing with finite spaces so large that the probabilities behave very close to their limits.Thomas Andrews –Thomas Andrews 2011-05-24 20:39:32 +00:00 Commented May 24, 2011 at 20:39
@Tim: I mean that probability theory can model the real world (as far as we know), but that it can do more if we want it to.TonyK –TonyK 2011-05-27 11:07:46 +00:00 Commented May 27, 2011 at 11:07
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I believe the root of this confusion is all because Math(in its nature) is in general(including probabilities) always more serious towards its concepts, than we -the humans- are. And this seriousness is exactly what brings Math its clarificationability(the power of turning vague into non-vague) and simplificationability(the power of turning complicated into simple), two of its core ultimate goals!
An instance of this amount of seriousness: When Math says "probability" it really means "probability"! I.e, mathematical probabilities are ALWAYS about "probable" expectations, and so even its 1 is not certainty, and even its 0 is not impossibility! Rather indeed, its 1 is absolute probability and its 0 is absolute improbability!
It's just an intentional neglection of all the odd possibilities. For instance, when I throw a die, it is possible that I will get a 7...! Why?! Maybe someone has added a dot to the 6 side! But we neglect that possibility on purpose for the sake of simplification, because it's an odd(read improbable) possibility.
Another Example: Yes, it is possible (even absolutely possible!) that the exact number 3 might occur in the continuous interval [1, 4], but it is absolutely improbable(notice how you can't disagree), thus the P=0. (Notice: Computers are never really continuous, so don't even try doing it programmatically, it won't be a counterexample).
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edited Dec 24, 2020 at 0:54
answered Dec 24, 2020 at 0:49
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I don't know if this is correct, but this is how I made sense of it, I think this a very intuitive explanation.
Imagine a car moving (from left to right) in a straight line with constant speed of 100MPH, and at x=0 the breaks of the car are engaged. Naturally the car will start to decelerate until it comes to rest (speed = 0 MPH).
The car will certainly come to rest, but the question is:
At what distance D [meters] (with respect to x=0) will the car come to rest? (This value D is not known apriori and can vary from one experiment to the next by a myriad of factors, so we can ask probability questions about the variable D).
Since distance is modeled by real numbers, the variable D is continuous. If we ask: what is P(D=6.354), the probability that it comes to rest at specifically D = 6.354 meters?
then, since the probability distribution of D is continuous, the answer is P(D=6.354)=0. This probability is zero.
Notice that there is nothing special about the number "6.354". That is, for any real number r the probability that D is exactly r is zero, i.e. P(D=r) = 0. This is simply a consequence of how continuous probability distributions are defined.
Regardless, we know that the car will certainly come to rest and that this happens at some specific distanced (where d is a real number). So even though P(D=d) = 0, it does not mean that coming to rest at D=d was impossible; clearly it was not because the car does come to rest at d.
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edited May 11, 2017 at 15:37
answered May 11, 2017 at 15:22
Juan Carlos Lebron-VazquezJuan Carlos Lebron-Vazquez
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647 | https://www.youtube.com/watch?v=SuZtJ0d4A3c | Sum of Even and Odd Binomial coefficient is equal to 2^(n-1)| Fsc part 1 math chapter 7
Lecturer Asad Ali
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Posted: 19 Feb 2021
hello this is lecturer asad ali channel. in this channel you can learn complete class 11 math.
..............
binomial_coefficient_summtion
even_binomial_coefficient
odd_binomial_coefficient
.................. . ........
here we discussed summation of even and odd binomial coefficient is equal to 2^(n-1). here firstly we defined binomial coefficient general form. further we separated even binomial coefficient and odd binomial coefficients.
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we prove that the even binomial coefficient is equal to odd binomial coefficients.
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This is class 11 math here we discussed exercise 7.2 questions 10, in this question we defined the sum of binomial coefficient is 2 power n, further more we defined the sum even binomial coefficient is equal to 2 power n-1 we also prove that the sum of odd binomial coefficient is equal to 2 power n-1.
......................
sum of even binomial coefficients,
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sum of even binomial coefficients formula,
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how to add binomial coefficients,
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........................................
sum of odd binomial coefficients,
sum of odd binomial coefficient up to n
sum of odd coefficient is binomial expansion,
what is sum of even binomial coefficients
what is sum of odd binomial coefficient
what is sum of even and odd binomial coefficients
.........................
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अ क्वेश्चन आवंटन में हम यह शुरू करेंगे दशम आप वन बाय अनिल को विशेषण इसीपुर दूध संभव और महीना मिल को भीषण इज इक्वल टू पावर एंड इसे प्रीवियस लेक्चर में मैंने यह प्रूफ किया था कि जो पॉइंट नेम इज द प्यार है उसके कोई भीषण सूखा समझो है वह टो पावर एंड गुड होगा यह हमारे पास - मिरजूराम खोज में वॉल्यूम है ए प्लस बी होल पावर एंड मिसेज़ पॉवर डिपॉजिट रोटी के और जब हम एक जगह शिफ्ट करेंगे और बी की जगह भी वांट करेंगे तो फिर हमारे पास आ जाएगा तू फॉर एजेंसी ने जीरो एन सी वन शेड्यूल एंड सैलेरी ऑरेंज शैड इसको कहते हैं कोई फील्ड आफ माइंड आमिर तीव्र मेज़ थी उनके कोई विशेष से अगर मैं इसको इंफेक्शंस को अपडेट करूंगा तो फिर यह कुल होगा तू पावर एंड का इसको मैंने इसी उपलक्ष्य में काफी डिटेल से एक्सप्लेन तथा पास ट्राई कर भी मिनट राज्य था आज हम क्या करेंगे आज हम यह करेंगे ताकि इसमें जो िववाह मे लिखो भीषण से या और बिना मेरे को भीषण से उसका समझो है टू पावर माइनस वन की कुल होगा उसका समय किसी गुलहड़ टो पावर एंड शुगर इसमें वन चैनल को विशेष कौन से है यह भाई मेरे को भी सेंड ए हो सकती है एनसीएस जीरो एन सी डी 2 एच एन सी ओ एम सी 6 इवन महीना मिल को विशेष होंगे एन सीवान एनसीटीई एबीसी आफ ए दुश्मने गौतम इंफोर्मेशन होंगे तो एनवायरनमेंट को भी टिप्स को अगर मैं ऐड करूंगा या मैं आर्ट बाय तमिल को विशेष को अगर ऐड करूंगा तो उसका शाम को पावर एंड माइनस वन के एक और होगा मतलब यह इसका एक फॉर्मूला जनरेट होगा जो बिल्कुल होगा टो पावर एंड माइनस वन का तो इसको मैं किस तरह उस करूंगा 2.1 और हमारे पास किस तरह का यह जो मैंने वन शिविर में लिखा है कि है मृत्यु हमने लिखा है इसमें मैं एक जगह सपोच वन पोस्ट करूंगा और बी की जगह मैं माइनस वन फुट करूंगा फिक्र तो जब एक मुमेंट करूंगा और भी कई जगह में माइनस वन पोस्ट करूंगा तो यह किया जाएगा1 - जुआन होल पावर एंड वन माइनस वन हौर पावर हो जाएगा यह क्या मुझे का या एन सी टी ई रोएगा क्या लगा था वन पावर एंड माइनस वन 50 सीवान मैं वन पावर एंड माइनस वन किस ग्रुप के लगा तो - हुआ था पावर वन यहां पर एनसीसी 2n C21 पावर एंड - 21 - वन पावरफुल फिगर nh-33 लिए भी प्रसिद्ध जगह पावर एंड - रील - 1413 अब टू सो आंठ लास्ट में आएगा एन सी एन कि 15102 और माइनस वन पावर लेना चाहिए इधर का यहां पर 200 जाएगा यहां पर यहां पर 10 जाएगा क्योंकि वन पावर एंड ए ब्वॉय - 101 जाएगा कि यहां से मनाया जाएगा इस टावर के पावर और नंबर है - वन है यह प्लस और माइनस माइनस हो जाएंगे यहां पर और पावर माइनस वन टू प्लस टू प्लस टू यह नेक्स्ट इससे - 3D और फिर ए प्लस बी एन सी फ्लोर पर - एन सी 5a और फिर प्लस एंड सी 60 - वन और फिर मिला यहां पर लिख दूंगा दूंगा यह शेज़वान आर इन छह और एन सी एन यह तो एक विशेष से बॉयज उसके - mca21 शीट 21 - एन सी वंशी थ्री एंड C5 और एंटरटेनमेंट स्पॉन्जी मोड ऑन अमेज़न ऑप्शन से ठीक है तो अब हम यह शुरू करेंगे कि समाधि वन महीना मिल को विशेष ध्यान समाधि और वर्णन मिल को अभिषेक और होगा रूप और - फोन का अब यहां पर देखिए यहां हमारे पास दो इक्वेशंस आ चुके हैं एक दो पावर एंड के लिए हुआ एन सी 0n सूर्यवंशी टू एंड सी इन थे लास्ट मैन सीन आएगा और क्वेश्चन हमारे पास है जब हम यहां भगवान और मैनेजमेंट करेंगे तो यह एक अगर इन दोनों इक्वेशंस को में ऐड करूंगी फिक्र इसको मैं ऐड करूंगा तो यहां पर किया जाएगा 2.02 पावणा जाएगा ठीक है यह विधि को इन थिस वेनसडे ऑग कि यहां पर यह वाटर और इसके साथ कैंसिल हो जाएगा ठीक है यह MP3 MP3 के साथ कैंसिल हो जाएगा फिर mc5 एनसीपी के साथ कैंसल राजौरी ट्रांस इसके साथ कैंसर हुआ यहां पर क्या रह चुका है ऐसी 0n चाहिए 2006 तो इनको मैच अपडेट करूंगा तो यहां भी वन मौजूद है और यहां जीवन मौजूद थे तो जय हो जाएगा टू एंड इजी हो यह वनप्लस वन सेज़ का यहां भी वन मौजूद है और यहां भी वन है तो फिर यह विजयगढ़ दो mc2 नेक्स्ट किया जाएगा प्लस टू और एनसीसीएफ और और लास्ट में यहां भी मौजूद है और यहां भी वन है विजयगढ़ टू एनसीएल J2 एनसीएल अब इस मुखिया का मन ले सकता हूं इसमें यह काम ले सकता हूं तू फारेस एंड टू जख्म लूंगा तो यह जाएगा एन सी0 फैंसी 21 कि एनसीपी ठौर-ठौर यहां पर एनसीएल चिकित्सा अब यह टू जब इधर मल्टिप्लिकेशन है तो यह लेफ्ट साइड पर डिविजन में तो मतलब यह हो तो e2o कि यहां पर जाएगा एन सी0 फैंसी 2 एंड सी फोर्थ इन थे मिडल आफ और इसके ऊपर से - सब्सक्राइब - को तो यह हमारे पास शुरू हुआ था कि अगर हम कोई भी को ऐड करेंगे सिर्फ हमारे पास आ जाएगा - 2 अब हम यह भी शुरू करेंगे कि उस समय पॉइंट को विशेष भी तू पावर एंड माइनस वन की कुल होगा मतलब जो और कोई भी चैन से वह भी ट्रू पावर माइनस वन बिल्कुल होगा तो इस तरह मैंने इसको ऐड किया है इन दोनों क्वेश्चसं को जिस तरह मैंने सेट किया है अगर मैं इसको सब्सट्रैक्ट करूंगा तो सब्सक्राइब नहीं किया जाएगा यह प्लस है - हो जाएगा यह - है प्लस हो जाएगा का यह - है यह प्लस माइनस प्लस तो फिर हमारे पास इवन नंबर कैंसर हो जाएंगे फिर एनसीईआरटी एनसीईआरटी के साथ कैंसर हो जाएगा एनसीए या के साथ कैंसर हो जाएगा यौवन एनसीटीई एबीसी आ जाएगा तो फिर यह भी टो पावर एंड माइनस वन की कुल हो गए शोध के लिए टाइम्स फॉर वॉचिंग दिस वीडियो नेक्स्ट लेक्चर में मेज़र सीरीज की एक्सप्लेंस करूंगा के बारे में सिरियस क्या है और बारे में सीरीज और विटामिंस की ओर में क्या फर्क है और वे में सिरियस हम क्यों सेट करते |
648 | https://www.whitman.edu/mathematics/cgt_online/cgt.pdf | An Introduction to Combinatorics and Graph Theory David Guichard This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike License. To view a copy of this license, visit or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA. If you distribute this work or a derivative, include the history of the document.
This copy of the text was compiled from source at 14:26 on 1/4/2025.
We will be glad to receive corrections and suggestions for improvement at guichard@whitman.edu.
Contents 1 Fundamentals 7 1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.2 Combinations and permutations . . . . . . . . . . . . . . . . . 11 1.3 Binomial coefficients . . . . . . . . . . . . . . . . . . . . . . . 16 1.4 Bell numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.5 Choice with repetition . . . . . . . . . . . . . . . . . . . . . . 27 1.6 The Pigeonhole Principle . . . . . . . . . . . . . . . . . . . . . 31 1.7 Sperner’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 35 1.8 Stirling numbers . . . . . . . . . . . . . . . . . . . . . . . . . 39 2 Inclusion-Exclusion 45 2.1 The Inclusion-Exclusion Formula . . . . . . . . . . . . . . . . . 45 2.2 Forbidden Position Permutations . . . . . . . . . . . . . . . . . 48 3 4 Contents 3 Generating Functions 53 3.1 Newton’s Binomial Theorem . . . . . . . . . . . . . . . . . . . 53 3.2 Exponential Generating Functions . . . . . . . . . . . . . . . . 56 3.3 Partitions of Integers . . . . . . . . . . . . . . . . . . . . . . . 59 3.4 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . 62 3.5 Catalan Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 66 4 Systems of Distinct Representatives 71 4.1 Existence of SDRs . . . . . . . . . . . . . . . . . . . . . . . . 72 4.2 Partial SDRs . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.3 Latin Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 4.4 Introduction to Graph Theory . . . . . . . . . . . . . . . . . . 83 4.5 Matchings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 5 Graph Theory 91 5.1 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.2 Euler Circuits and Walks . . . . . . . . . . . . . . . . . . . . . 96 5.3 Hamilton Cycles and Paths . . . . . . . . . . . . . . . . . . .
100 5.4 Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . .
103 5.5 Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
105 5.6 Optimal Spanning Trees . . . . . . . . . . . . . . . . . . . .
108 5.7 Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . .
110 5.8 Graph Coloring . . . . . . . . . . . . . . . . . . . . . . . . .
118 5.9 The Chromatic Polynomial . . . . . . . . . . . . . . . . . . .
124 5.10 Coloring Planar Graphs . . . . . . . . . . . . . . . . . . . . .
125 5.11 Directed Graphs . . . . . . . . . . . . . . . . . . . . . . . .
129 Contents 5 6 P´ olya–Redfield Counting 135 6.1 Groups of Symmetries . . . . . . . . . . . . . . . . . . . . .
137 6.2 Burnside’s Theorem . . . . . . . . . . . . . . . . . . . . . .
140 6.3 P´ olya-Redfield Counting . . . . . . . . . . . . . . . . . . . .
146 A Hints 151 Index 153 1 Fundamentals Combinatorics is often described briefly as being about counting, and indeed counting is a large part of combinatorics. As the name suggests, however, it is broader than this: it is about combining things. Questions that arise include counting problems: “How many ways can these elements be combined?” But there are other questions, such as whether a certain combination is possible, or what combination is the “best” in some sense. We will see all of these, though counting plays a particularly large role.
Graph theory is concerned with various types of networks, or really models of networks called graphs. These are not the graphs of analytic geometry, but what are often described as “points connected by lines”, for example: .
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• • • • • • • The preferred terminology is vertex for a point and edge for a line. The lines need not be straight lines, and in fact the actual definition of a graph is not a geometric definition.
The figure above is simply a visualization of a graph; the graph is a more abstract object, consisting of seven vertices, which we might name {v1, . . . , v7}, and the collection of pairs of vertices that are connected; for a suitable assignment of names vi to the points in the diagram, the edges could be represented as {v1, v2},{v2, v3},{v3, v4},{v3, v5},{v4, v5}, {v5, v6},{v6, v7}.
7 8 Chapter 1 Fundamentals 1.1 Examples Suppose we have a chess board, and a collection of tiles, like dominoes, each of which is the size of two squares on the chess board. Can the chess board be covered by the dominoes?
First we need to be clear on the rules: the board is covered if the dominoes are laid down so that each covers exactly two squares of the board; no dominoes overlap; and every square is covered. The answer is easy: simply by laying out 32 dominoes in rows, the board can be covered. To make the problem more interesting, we allow the board to be rectangular of any size, and we allow some squares to be removed from the board. What can be say about whether the remaining board can be covered? This is such a board, for example: .
What can we say? Here is an easy observation: each domino must cover two squares, so the total number of squares must be even; the board above has an even number of squares.
Is that enough? It is not too hard to convince yourself that this board cannot be covered; is there some general principle at work? Suppose we redraw the board to emphasize that it really is part of a chess board: .
Aha!
Every tile must cover one white and one gray square, but there are four of the former and six of the latter, so it is impossible. Now do we have the whole picture? No; 1.1 Examples 9 for example: .
The gray square at the upper right clearly cannot be covered. Unfortunately it is not easy to state a condition that fully characterizes the boards that can be covered; we will see this problem again. Let us note, however, that this problem can also be represented as a graph problem. We introduce a vertex corresponding to each square, and connect two vertices by an edge if their associated squares can be covered by a single domino; here is the previous board: .
• .
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• Here the top row of vertices represents the gray squares, the bottom row the white squares.
A domino now corresponds to an edge; a covering by dominoes corresponds to a collection of edges that share no endpoints and that are incident with (that is, touch) all six vertices.
Since no edge is incident with the top left vertex, there is no cover.
Perhaps the most famous problem in graph theory concerns map coloring: Given a map of some countries, how many colors are required to color the map so that countries sharing a border get different colors?
It was long conjectured that any map could be colored with four colors, and this was finally proved in 1976. Here is an example of a small map, colored with four colors: .
Typically this problem is turned into a graph theory problem. Suppose we add to each country a capital, and connect capitals across common boundaries. Coloring the capitals so 10 Chapter 1 Fundamentals that no two connected capitals share a color is clearly the same problem. For the previous map: .
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• Any graph produced in this way will have an important property: it can be drawn so that no edges cross each other; this is a planar graph. Non-planar graphs can require more than four colors, for example this graph: .
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• This is called the complete graph on five vertices, denoted K5; in a complete graph, each vertex is connected to each of the others. Here only the “fat” dots represent vertices; intersections of edges at other points are not vertices. A few minutes spent trying should convince you that this graph cannot be drawn so that its edges don’t cross, though the number of edge crossings can be reduced.
Exercises 1.1.
1. Explain why an m × n board can be covered if either m or n is even. Explain why it cannot be covered if both m and n are odd.
2. Suppose two diagonally opposite corners of an ordinary 8 × 8 board are removed. Can the resulting board be covered?
3. Suppose that m and n are both odd. On an m × n board, colored as usual, all four corners will be the same color, say white. Suppose one white square is removed from any location on the board. Show that the resulting board can be covered.
4. Suppose that one corner of an 8 × 8 board is removed. Can the remainder be covered by 1 × 3 tiles? Show a tiling or prove that it cannot be done.
1.2 Combinations and permutations 11 5. Suppose the square in row 3, column 3 of an 8 × 8 board is removed. Can the remainder be covered by 1 × 3 tiles? Show a tiling or prove that it cannot be done.
6. Remove two diagonally opposite corners of an m × n board, where m is odd and n is even.
Show that the remainder can be covered with dominoes.
7. Suppose one white and one black square are removed from an n × n board, n even. Show that the remainder can be covered by dominoes.
8. Suppose an n×n board, n even, is covered with dominoes. Show that the number of horizontal dominoes with a white square under the left end is equal to the number of horizontal dominoes with a black square under the left end.
9. In the complete graph on five vertices shown above, there are five pairs of edges that cross.
Draw this graph so that only one pair of edges cross. Remember that “edges” do not have to be straight lines.
10. The complete bipartite graph K3,3 consists of two groups of three vertices each, with all possible edges between the groups and no other edges: .
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• • • • • • Draw this graph with only one crossing.
1.2 Combinations and permutations We turn first to counting. While this sounds simple, perhaps too simple to study, it is not. When we speak of counting, it is shorthand for determining the size of a set, or more often, the sizes of many sets, all with something in common, but different sizes depending on one or more parameters. For example: how many outcomes are possible when a die is rolled? Two dice? n dice? As stated, this is ambiguous: what do we mean by “outcome”?
Suppose we roll two dice, say a red die and a green die.
Is “red two, green three” a different outcome than “red three, green two”? If yes, we are counting the number of possible “physical” outcomes, namely 36. If no, there are 21. We might even be interested simply in the possible totals, in which case there are 11 outcomes.
Even the quite simple first interpretation relies on some degree of knowledge about counting; we first make two simple facts explicit. In terms of set sizes, suppose we know that set A has size m and set B has size n. What is the size of A and B together, that is, the size of A ∪B? If we know that A and B have no elements in common, then the size A ∪B is m + n; if they do have elements in common, we need more information. A simple but typical problem of this type: if we roll two dice, how many ways are there to get either 7 or 11? Since there are 6 ways to get 7 and two ways to get 11, the answer is 6 + 2 = 8.
Though this principle is simple, it is easy to forget the requirement that the two sets be 12 Chapter 1 Fundamentals disjoint, and hence to use it when the circumstances are otherwise. This principle is often called the addition principle.
This principle can be generalized: if sets A1 through An are pairwise disjoint and have sizes m1, . . . mn, then the size of A1 ∪· · ·∪An = ∑n i=1 mi. This can be proved by a simple induction argument.
Why do we know, without listing them all, that there are 36 outcomes when two dice are rolled? We can view the outcomes as two separate outcomes, that is, the outcome of rolling die number one and the outcome of rolling die number two.
For each of 6 outcomes for the first die the second die may have any of 6 outcomes, so the total is 6 + 6 + 6 + 6 + 6 + 6 = 36, or more compactly, 6 · 6 = 36. Note that we are really using the addition principle here: set A1 is all pairs (1, x), set A2 is all pairs (2, x), and so on. This is somewhat more subtle than is first apparent. In this simple example, the outcomes of die number two have nothing to do with the outcomes of die number one. Here’s a slightly more complicated example: how many ways are there to roll two dice so that the two dice don’t match? That is, we rule out 1-1, 2-2, and so on. Here for each possible value on die number one, there are five possible values for die number two, but they are a different five values for each value on die number one. Still, because all are the same, the result is 5 + 5 + 5 + 5 + 5 + 5 = 30, or 6 · 5 = 30. In general, then, if there are m possibilities for one event, and n for a second event, the number of possible outcomes for both events together is m · n. This is often called the multiplication principle.
In general, if n events have mi possible outcomes, for i = 1, . . . , n, where each mi is unaffected by the outcomes of other events, then the number of possible outcomes overall is ∏n i=1 mi. This too can be proved by induction.
EXAMPLE 1.2.1 How many outcomes are possible when three dice are rolled, if no two of them may be the same? The first two dice together have 6 · 5 = 30 possible outcomes, from above. For each of these 30 outcomes, there are four possible outcomes for the third die, so the total number of outcomes is 30 · 4 = 6 · 5 · 4 = 120. (Note that we consider the dice to be distinguishable, that is, a roll of 6, 4, 1 is different than 4, 6, 1, because the first and second dice are different in the two rolls, even though the numbers as a set are the same.) EXAMPLE 1.2.2 Suppose blocks numbered 1 through n are in a barrel; we pull out k of them, placing them in a line as we do. How many outcomes are possible? That is, how many different arrangements of k blocks might we see?
This is essentially the same as the previous example: there are k “spots” to be filled by blocks. Any of the n blocks might appear first in the line; then any of the remaining n −1 might appear next, and so on. The number of outcomes is thus n(n−1)(n−2) · · · (n−k+1), by the multiplication principle. In the previous example, the first “spot” was die number 1.2 Combinations and permutations 13 one, the second spot was die number two, the third spot die number three, and 6 · 5 · 4 = 6(6 −1)(6 −2); notice that 6 −2 = 6 −3 + 1.
This is quite a general sort of problem: DEFINITION 1.2.3 The number of permutations of n things taken k at a time is P(n, k) = n(n −1)(n −2) · · · (n −k + 1) = n!
(n −k)!.
A permutation of some objects is a particular linear ordering of the objects; P(n, k) in effect counts two things simultaneously: the number of ways to choose and order k out of n objects. A useful special case is k = n, in which we are simply counting the number of ways to order all n objects. This is n(n −1) · · · (n −n + 1) = n!. Note that the second form of P(n, k) from the definition gives n!
(n −n)! = n!
0! .
This is correct only if 0! = 1, so we adopt the standard convention that this is true, that is, we define 0! to be 1.
Suppose we want to count only the number of ways to choose k items out of n, that is, we don’t care about order. In example 1.2.1, we counted the number of rolls of three dice with different numbers showing. The dice were distinguishable, or in a particular order: a first die, a second, and a third. Now we want to count simply how many combinations of numbers there are, with 6, 4, 1 now counting as the same combination as 4, 6, 1.
EXAMPLE 1.2.4 Suppose we were to list all 120 possibilities in example 1.2.1. The list would contain many outcomes that we now wish to count as a single outcome; 6, 4, 1 and 4, 6, 1 would be on the list, but should not be counted separately. How many times will a single outcome appear on the list? This is a permutation problem: there are 3! orders in which 1, 4, 6 can appear, and all 6 of these will be on the list. In fact every outcome will appear on the list 6 times, since every outcome can appear in 3! orders. Hence, the list is too big by a factor of 6; the correct count for the new problem is 120/6 = 20.
Following the same reasoning in general, if we have n objects, the number of ways to choose k of them is P(n, k)/k!, as each collection of k objects will be counted k! times by P(n, k).
14 Chapter 1 Fundamentals DEFINITION 1.2.5 The number of subsets of size k of a set of size n (also called an n-set) is C(n, k) = P(n, k) k!
= n!
k!(n −k)! = (n k ) .
The notation C(n, k) is rarely used; instead we use (n k ) , pronounced “n choose k”.
EXAMPLE 1.2.6 Consider n = 0, 1, 2, 3. It is easy to list the subsets of a small n-set; a typical n-set is {a1, a2, . . . , an}. A 0-set, namely the empty set, has one subset, the empty set; a 1-set has two subsets, the empty set and {a1}; a 2-subset has four subsets, ∅, {a1}, {a2}, {a1, a2}; and a 3-subset has eight: ∅, {a1}, {a2}, {a3}, {a1, a2}, {a1, a3}, {a2, a3}, {a1, a2, a3}. From these lists it is then easy to compute (n k ) : k 0 1 2 3 0 1 n 1 1 1 2 1 2 1 3 1 3 3 1 You probably recognize these numbers: this is the beginning of Pascal’s Triangle.
Each entry in Pascal’s triangle is generated by adding two entries from the previous row: the one directly above, and the one above and to the left.
This suggests that (n k ) = (n−1 k−1 ) + (n−1 k ) , and indeed this is true.
To make this work out neatly, we adopt the convention that (n k ) = 0 when k < 0 or k > n.
THEOREM 1.2.7 (n k ) = (n −1 k −1 ) + (n −1 k ) .
Proof.
A typical n-set is A = {a1, . . . , an}. We consider two types of subsets: those that contain an and those that do not. If a k-subset of A does not contain an, then it is a k-subset of {a1, . . . , an−1}, and there are (n−1 k ) of these. If it does contain an, then it consists of an and k −1 elements of {a1, . . . , an−1}; since there are (n−1 k−1 ) of these, there are (n−1 k−1 ) subsets of this type. Thus the total number of k-subsets of A is (n−1 k−1 ) + (n−1 k ) .
Note that when k = 0, (n−1 k−1 ) = (n−1 −1 ) = 0, and when k = n, (n−1 k ) = (n−1 n ) = 0, so that (n 0 ) = (n−1 0 ) and (n n ) = (n−1 n−1 ) . These values are the boundary ones in Pascal’s Triangle.
Many counting problems rely on the sort of reasoning we have seen. Here are a few variations on the theme.
1.2 Combinations and permutations 15 EXAMPLE 1.2.8 Six people are to sit at a round table; how many seating arrangements are there?
It is not clear exactly what we mean to count here. If there is a “special seat”, for example, it may matter who ends up in that seat. If this doesn’t matter, we only care about the relative position of each person. Then it may or may not matter whether a certain person is on the left or right of another. So this question can be interpreted in (at least) three ways. Let’s answer them all.
First, if the actual chairs occupied by people matter, then this is exactly the same as lining six people up in a row: 6 choices for seat number one, 5 for seat two, and so on, for a total of 6!. If the chairs don’t matter, then 6! counts the same arrangement too many times, once for each person who might be in seat one. So the total in this case is 6!/6 = 5!. Another approach to this: since the actual seats don’t matter, just put one of the six people in a chair. Then we need to arrange the remaining 5 people in a row, which can be done in 5! ways. Finally, suppose all we care about is who is next to whom, ignoring right and left. Then the previous answer counts each arrangement twice, once for the counterclockwise order and once for clockwise. So the total is 5!/2 = P(5, 3).
We have twice seen a general principle at work: if we can overcount the desired set in such a way that every item gets counted the same number of times, we can get the desired count just by dividing by the common overcount factor. This will continue to be a useful idea. A variation on this theme is to overcount and then subtract the amount of overcount.
EXAMPLE 1.2.9 How many ways are there to line up six people so that a particular pair of people are not adjacent?
Denote the people A and B. The total number of orders is 6!, but this counts those orders with A and B next to each other. How many of these are there? Think of these two people as a unit; how many ways are there to line up the AB unit with the other 4 people? We have 5 items, so the answer is 5!. Each of these orders corresponds to two different orders in which A and B are adjacent, depending on whether A or B is first. So the 6! count is too high by 2 · 5! and the count we seek is 6! −2 · 5! = 4 · 5!.
Exercises 1.2.
1. How many positive factors does 2 · 34 · 73 · 112 · 475 have? How many does pe1 1 pe2 2 · · · pen n have, where the pi are distinct primes?
2. A poker hand consists of five cards from a standard 52 card deck with four suits and thirteen values in each suit; the order of the cards in a hand is irrelevant. How many hands consist of 2 cards with one value and 3 cards of another value (a full house)? How many consist of 5 cards from the same suit (a flush)?
16 Chapter 1 Fundamentals 3. Six men and six women are to be seated around a table, with men and women alternating.
The chairs don’t matter, only who is next to whom, but right and left are different. How many seating arrangements are possible?
4. Eight people are to be seated around a table; the chairs don’t matter, only who is next to whom, but right and left are different. Two people, X and Y, cannot be seated next to each other. How many seating arrangements are possible?
5. In chess, a rook attacks any piece in the same row or column as the rook, provided no other piece is between them. In how many ways can eight indistinguishable rooks be placed on a chess board so that no two attack each other? What about eight indistinguishable rooks on a 10 × 10 board?
6. Suppose that we want to place 8 non-attacking rooks on a chessboard. In how many ways can we do this if the 16 most ‘northwest’ squares must be empty? How about if only the 4 most ‘northwest’ squares must be empty?
7. A “legal” sequence of parentheses is one in which the parentheses can be properly matched, like ()(()). It’s not hard to see that this is possible precisely when the number of left and right parentheses is the same, and every initial segment of the sequence has at least as many left parentheses as right. For example, ()) . . . cannot possibly be extended to a legal sequence.
Show that the number of legal sequences of length 2n is Cn = (2n n ) − ( 2n n+1 ) . The numbers Cn are called the Catalan numbers.
1.3 Binomial coefficients Recall the appearance of Pascal’s Triangle in example 1.2.6. If you have encountered the triangle before, you may know it has many interesting properties. We will explore some of these here.
You may know, for example, that the entries in Pascal’s Triangle are the coefficients of the polynomial produced by raising a binomial to an integer power.
For example, (x + y)3 = 1 · x3 + 3 · x2y + 3 · xy2 + 1 · y3, and the coefficients 1, 3, 3, 1 form row three of Pascal’s Triangle. For this reason the numbers (n k ) are usually referred to as the binomial coefficients.
THEOREM 1.3.1 Binomial Theorem (x + y)n = (n 0 ) xn + (n 1 ) xn−1y + (n 2 ) xn−2y2 + · · · + (n n ) yn = n ∑ i=0 (n i ) xn−iyi Proof.
We prove this by induction on n.
It is easy to check the first few, say for n = 0, 1, 2, which form the base case. Now suppose the theorem is true for n −1, that is, (x + y)n−1 = n−1 ∑ i=0 (n −1 i ) xn−1−iyi.
Then (x + y)n = (x + y)(x + y)n−1 = (x + y) n−1 ∑ i=0 (n −1 i ) xn−1−iyi.
1.3 Binomial coefficients 17 Using the distributive property, this becomes x n−1 ∑ i=0 (n −1 i ) xn−1−iyi + y n−1 ∑ i=0 (n −1 i ) xn−1−iyi = n−1 ∑ i=0 (n −1 i ) xn−iyi + n−1 ∑ i=0 (n −1 i ) xn−1−iyi+1.
These two sums have much in common, but it is slightly disguised by an “offset”: the first sum starts with an xny0 term and ends with an x1yn−1 term, while the corresponding terms in the second sum are xn−1y1 and x0yn. Let’s rewrite the second sum so that they match: n−1 ∑ i=0 (n −1 i ) xn−iyi + n−1 ∑ i=0 (n −1 i ) xn−1−iyi+1 = n−1 ∑ i=0 (n −1 i ) xn−iyi + n ∑ i=1 (n −1 i −1 ) xn−iyi = (n −1 0 ) xn + n−1 ∑ i=1 (n −1 i ) xn−iyi + n−1 ∑ i=1 (n −1 i −1 ) xn−iyi + (n −1 n −1 ) yn = (n −1 0 ) xn + n−1 ∑ i=1 ( (n −1 i ) + (n −1 i −1 ) )xn−iyi + (n −1 n −1 ) yn.
Now we can use theorem 1.2.7 to get: (n −1 0 ) xn+ n−1 ∑ i=1 ( (n −1 i ) + (n −1 i −1 ) )xn−iyi + (n −1 n −1 ) yn.
= (n −1 0 ) xn + n−1 ∑ i=1 (n i ) xn−iyi + (n −1 n −1 ) yn.
= (n 0 ) xn + n−1 ∑ i=1 (n i ) xn−iyi + (n n ) yn = n ∑ i=0 (n i ) xn−iyi.
At the next to last step we used the facts that (n−1 0 ) = (n 0 ) and (n−1 n−1 ) = (n n ) .
Here is an interesting consequence of this theorem: Consider (x + y)n = (x + y)(x + y) · · · (x + y).
One way we might think of attempting to multiply this out is this: Go through the n factors (x + y) and in each factor choose either the x or the y; at the end, multiply your 18 Chapter 1 Fundamentals choices together, getting some term like xxyxyy · · · yx = xiyj, where of course i + j = n.
If we do this in all possible ways and then collect like terms, we will clearly get n ∑ i=0 aixn−iyi.
We know that the correct expansion has (n i ) = ai; is that in fact what we will get by this method? Yes: consider xn−iyi. How many times will we get this term using the given method? It will be the number of times we end up with i y-factors. Since there are n factors (x + y), the number of times we get i y-factors must be the number of ways to pick i of the (x+y) factors to contribute a y, namely (n i ) . This is probably not a useful method in practice, but it is interesting and occasionally useful.
EXAMPLE 1.3.2 Using this method we might get (x + y)3 = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy which indeed becomes x3 + 3x2y + 3xy2 + y3 upon collecting like terms.
The Binomial Theorem, 1.3.1, can be used to derive many interesting identities. A common way to rewrite it is to substitute y = 1 to get (x + 1)n = n ∑ i=0 (n i ) xn−i.
If we then substitute x = 1 we get 2n = n ∑ i=0 (n i ) , that is, row n of Pascal’s Triangle sums to 2n. This is also easy to understand combinato-rially: the sum represents the total number of subsets of an n-set, since it adds together the numbers of subsets of every possible size. It is easy to see directly that the number of subsets of an n-set is 2n: for each element of the set we make a choice, to include or to exclude the element. The total number of ways to make these choices is 2 · 2 · · · 2 = 2n, by the multiplication principle.
Suppose now that n ≥1 and we substitute −1 for x; we get (−1 + 1)n = n ∑ i=0 (n i ) (−1)n−i.
(1.3.1) The sum is now an alternating sum: every other term is multiplied by −1. Since the left hand side is 0, we can rewrite this to get (n 0 ) + (n 2 ) + · · · = (n 1 ) + (n 3 ) + · · · .
(1.3.2) So each of these sums is 2n−1.
1.3 Binomial coefficients 19 Another obvious feature of Pascal’s Triangle is symmetry: each row reads the same forwards and backwards. That is, we have: THEOREM 1.3.3 (n i ) = ( n n −i ) .
Proof.
This is quite easy to see combinatorially: every i-subset of an n-set is associated with an (n −i)-subset.
That is, if the n-set is A, and if B ⊆A has size i, then the complement of B has size n−i. This establishes a 1–1 correspondence between sets of size i and sets of size n −i, so the numbers of each are the same. (Of course, if i = n −i, no proof is required.) Note that this means that the Binomial Theorem, 1.3.1, can also be written as (x + y)n = n ∑ i=0 ( n n −i ) xn−iyi.
or (x + y)n = n ∑ i=0 (n i ) xiyn−i.
Another striking feature of Pascal’s Triangle is that the entries across a row are strictly increasing to the middle of the row, and then strictly decreasing. Since we already know that the rows are symmetric, the first part of this implies the second.
THEOREM 1.3.4 For 1 ≤i ≤⌊n 2 ⌋, (n i ) > ( n i −1 ) .
Proof.
This is by induction; the base case is apparent from the first few rows. Write (n i ) = (n −1 i −1 ) + (n −1 i ) ( n i −1 ) = (n −1 i −2 ) + (n −1 i −1 ) Provided that 1 ≤i ≤⌊n−1 2 ⌋, we know by the induction hypothesis that (n −1 i ) > (n −1 i −1 ) .
Provided that 1 ≤i −1 ≤⌊n−1 2 ⌋, or equivalently 2 ≤i ≤⌊n−1 2 ⌋+ 1, we know that (n −1 i −1 ) > (n −1 i −2 ) .
Hence if 2 ≤i ≤⌊n−1 2 ⌋, (n i ) > ( n i −1 ) .
This leaves two special cases to check: i = 1 and for n even, i = ⌊n−1 2 ⌋+ 1 = ⌊n 2 ⌋. These are left as an exercise.
20 Chapter 1 Fundamentals Exercises 1.3.
1. Suppose a street grid starts at position (0, 0) and extends up and to the right: (0, 0) A shortest route along streets from (0, 0) to (i, j) is i + j blocks long, going i blocks east and j blocks north. How many such routes are there? Suppose that the block between (k, l) and (k + 1, l) is closed, where k < i and l ≤j. How many shortest routes are there from (0, 0) to (i, j)?
2. Prove by induction that ∑n k=0 (k i ) = (n+1 i+1 ) for n ≥0 and i ≥0.
3. Use a combinatorial argument to prove that ∑n k=0 (k i ) = (n+1 i+1 ) for n ≥0 and i ≥0; that is, explain why the left-hand side counts the same thing as the right-hand side.
4. Use a combinatorial argument to prove that (k 2 ) + (n−k 2 ) + k(n −k) = (n 2 ) .
5. Use a combinatorial argument to prove that (2n n ) is even.
6. Suppose that A is a non-empty finite set. Prove that A has as many even-sized subsets as it does odd-sized subsets.
7. Prove that ∑n k=0 (k i ) k = (n+1 i+1 ) n − (n+1 i+2 ) for n ≥0 and i ≥0.
8. Verify that (n+1 2 ) + (n 2 ) = n2. Use exercise 2 to find a simple expression for ∑n i=1 i2.
9. Make a conjecture about the sums of the upward diagonals in Pascal’s Triangle as indicated.
Prove your conjecture is true.
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1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 10. Find the number of ways to write n as an ordered sum of ones and twos, n ≥0. For example, when n = 4, there are five ways: 1 + 1 + 1 + 1, 2 + 1 + 1, 1 + 2 + 1, 1 + 1 + 2, and 2 + 2.
11. Use (x+1)n = ∑n i=0 (n i ) xi to find a simple expression for ∑n i=1 i (n i ) xi−1. Then find a simple expression for ∑n i=1 i (n i ) .
12. Use (x + 1)n = ∑n i=0 (n i ) xi to find a simple expression for ∑n i=0 1 i+1 (n i ) xi+1. Then find a simple expression for ∑n i=0 1 i+1 (n i ) .
13. Use the previous exercise to find a simple expression for ∑n i=0(−1)i 1 i+1 (n i ) .
1.4 Bell numbers 21 14. Give a combinatorial proof of k ∑ i=0 ( m i )( n k −i ) = ( m + n k ) .
Rewrite this identity in simpler form if m = n, and when k = m = n.
15. Finish the proof of theorem 1.3.4.
16. Give an alternate proof of theorm 1.3.4 by characterizing those i for which (n i ) / ( n i−1 ) > 1.
17. When is (n i ) / ( n i−1 ) a maximum? When is (n i ) / ( n i−1 ) = 2?
18. When is (n i ) − ( n i−1 ) a maximum?
19. A Galton board is an upright flat surface with protruding horizontal pins arranged as shown below. At the bottom are a number of bins; if the number of rows is n, the number of bins is n + 1. A ball is dropped directly above the top pin, and at each pin bounces left or right with equal probability. We assume that the ball next hits the pin below and immediately left or right of the pin it has struck, and this continues down the board, until the ball falls into a bin at the bottom. If we number the bins from 0 to n, how many paths can a ball travel to end up in bin k?
This may be interpreted in terms of probability, which was the intent of Sir Francis Galton when he designed it. Each path is equally likely to be taken by a ball. If many balls are dropped, the number of balls in bin k corresponds to the probability of ending up in that bin. The more paths that end in a bin, the higher the probability. When a very large number of balls are dropped, the balls will form an approximation to the bell curve familiar from probability and statistics.
There is an animation of the process at
There was once a very nice physical implementation at the Pacific Science Center in Seattle.
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• 1.4 Bell numbers We begin with a definition: 22 Chapter 1 Fundamentals DEFINITION 1.4.1 A partition of a set S is a collection of non-empty subsets Ai ⊆S, 1 ≤i ≤k (the parts of the partition), such that ∪k i=1 Ai = S and for every i ̸= j, Ai ∩Aj = ∅.
EXAMPLE 1.4.2 The partitions of the set {a, b, c} are {{a}, {b}, {c}}, {{a, b}, {c}}, {{a, c}, {b}}, {{b, c}, {a}}, and {{a, b, c}}.
Partitions arise in a number of areas of mathematics. For example, if ≡is an equiv-alence relation on a set S, the equivalence classes of ≡form a partition of S. Here we consider the number of partitions of a finite set S, which we might as well take to be [n] = {1, 2, 3, . . . , n}, unless some other set is of interest. We denote the number of parti-tions of an n-element set by Bn; these are the Bell numbers. From the example above, we see that B3 = 5. For convenience we let B0 = 1. It is quite easy to see that B1 = 1 and B2 = 2.
There are no known simple formulas for Bn, so we content ourselves with a recurrence relation.
THEOREM 1.4.3 The Bell numbers satisfy Bn+1 = n ∑ k=0 (n k ) Bk.
Proof.
Consider a partition of S = {1, 2, . . . , n + 1}, A1,. . . ,Am. We may suppose that n + 1 is in A1, and that |A1| = k + 1, for some k, 0 ≤k ≤n. Then A2,. . . ,Am form a partition of the remaining n−k elements of S, that is, of S\A1. There are Bn−k partitions of this set, so there are Bn−k partitions of S in which one part is the set A1. There are (n k ) sets of size k +1 containing n+1, so the total number of partitions of S in which n+1 is in a set of size k + 1 is (n k ) Bn−k. Adding up over all possible values of k, this means Bn+1 = n ∑ k=0 (n k ) Bn−k.
(1.4.1) We may rewrite this, using theorem 1.3.3, as Bn+1 = n ∑ k=0 ( n n −k ) Bn−k, and then notice that this is the same as the sum Bn+1 = n ∑ k=0 (n k ) Bk, written backwards.
1.4 Bell numbers 23 EXAMPLE 1.4.4 We apply the recurrence to compute the first few Bell numbers: B1 = 0 ∑ k=0 (0 0 ) B0 = 1 · 1 = 1 B2 = 1 ∑ k=0 (1 k ) Bk = (1 0 ) B0 + (1 1 ) B1 = 1 · 1 + 1 · 1 = 1 + 1 = 2 B3 = 2 ∑ k=0 (2 k ) Bk = 1 · 1 + 2 · 1 + 1 · 2 = 5 B4 = 3 ∑ k=0 (3 k ) Bk = 1 · 1 + 3 · 1 + 3 · 2 + 1 · 5 = 15 The Bell numbers grow exponentially fast; the first few are 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437.
The Bell numbers turn up in many other problems; here is an interesting example.
A common need in some computer programs is to generate a random permutation of 1, 2, 3, . . . , n, which we may think of as a shuffle of the numbers, visualized as numbered cards in a deck. Here is an attractive method that is easy to program: Start with the numbers in order, then at each step, remove one number at random (this is easy in most programming languages) and put it at the front of the list of numbers. (Viewed as a shuffle of a deck of cards, this corresponds to removing a card and putting it on the top of the deck.) How many times should we do this? There is no magic number, but it certainly should not be small relative to the size of n. Let’s choose n as the number of steps.
We can write such a shuffle as a list of n integers, (m1, m2, . . . , mn). This indicates that at step i, the number mi is moved to the front.
EXAMPLE 1.4.5 Let’s follow the shuffle (3, 2, 2, 4, 1): (3) : 31245 (2) : 23145 (2) : 23145 (4) : 42315 (1) : 14235 Note that we allow “do nothing” moves, removing the top card and then placing it on top. The number of possible shuffles is then easy to count: there are n choices for the card 24 Chapter 1 Fundamentals to remove, and this is repeated n times, so the total number is nn. (If we continue a shuffle for m steps, the number of shuffles is nm.) Since there are only n! different permutations of 1, 2, . . . , n, this means that many shuffles give the same final order.
Here’s our question: how many shuffles result in the original order?
EXAMPLE 1.4.6 These shuffles return to the original order: (1, 1, 1, 1, 1), (5, 4, 3, 2, 1), (4, 1, 3, 2, 1).
THEOREM 1.4.7 The number of shuffles of [n] that result in the original sorted order is Bn.
Proof.
Since we know that Bn counts the number of partitions of {1, 2, 3, . . . , n}, we can prove the theorem by establishing a 1–1 correspondence between the shuffles that leave the deck sorted and the partitions. Given a shuffle (m1, m2, . . . , mn), we put into a single set all i such that mi has a single value. For example, using the shuffle (4, 1, 3, 2, 1), Since m2 = m5, one set is {2, 5}. All the other values are distinct, so the other sets in the partition are {1}, {3}, and {4}.
Note that every shuffle, no matter what the final order, produces a partition by this method. We are only interested in the shuffles that leave the deck sorted. What we now need is to show that each partition results from exactly one such shuffle.
Suppose we have a partition with k parts. If a shuffle leaves the deck sorted, the last entry, mn, must be 1. If the part containing n is A1, then it must be that mi = 1 if and only if i ∈A1. If k = 1, then the only part contains all of {1, 2, . . . , n}, and the shuffle must be (1, 1, 1, . . . , 1).
If k > 1, the last move that is not 1 must be 2, since 2 must end up immediately after 1. Thus, if j2 is the largest index such that j2 / ∈A1, let A2 be the part containing j2, and it must be that mi = 2 if and only if i ∈A2. We continue in this way: Once we have discovered which of the mi must have values 1, 2, . . . , p, let jp+1 be the largest index such that jp+1 / ∈A1 ∪· · · ∪Ap, let Ap+1 be the part containing jp+1, and then mi = p + 1 if and only if i ∈Ap+1. When p = k, all values mi have been determined, and this is the unique shuffle that corresponds to the partition.
EXAMPLE 1.4.8 Consider the partition {{1, 5}, {2, 3, 6}, {4, 7}}. We must have m7 = m4 = 1, m6 = m3 = m2 = 2, and m5 = m1 = 3, so the shuffle is (3, 2, 2, 1, 3, 2, 1).
Returning to the problem of writing a computer program to generate a partition: is this a good method? When we say we want a random permutation, we mean that we want each permutation to occur with equal probability, namely, 1/n!. Since the original order is one of the permutations, we want the number of shuffles that produce it to be exactly nn/n!, but n! does not divide nn, so this is impossible. The probability of getting 1.4 Bell numbers 25 the original permutation is Bn/nn, and this turns out to be quite a bit larger than 1/n!.
Thus, this is not a suitable method for generating random permutations.
The recurrence relation above is a somewhat cumbersome way to compute the Bell numbers. Another way to compute them is with a different recurrence, expressed in the Bell triangle, whose construction is similar to that of Pascal’s triangle: A1,1 A2,1 A2,2 A3,1 A3,2 A3,3 A4,1 A4,2 A4,3 A4,4 1 1 2 2 3 5 5 7 10 15 The rule for constructing this triangle is: A1,1 = 1; the first entry in each row is the last entry in the previous row; other entries are An,k = An,k−1 +An−1,k−1; row n has n entries.
Both the first column and the diagonal consist of the Bell numbers, with An,1 = Bn−1 and An,n = Bn.
An,k may be interpreted as the number of partitions of {1, 2, . . . , n + 1} in which {k + 1} is the singleton set with the largest entry in the partition. For example, A3,2 = 3; the partitions of 3+1 = 4 in which 2+1 = 3 is the largest number appearing in a singleton set are {{1}, {2, 4}, {3}}, {{2}, {1, 4}, {3}}, and {{1, 2, 4}, {3}}.
To see that this indeed works as advertised, we need to confirm a few things. First, consider An,n, the number of partitions of {1, 2, . . . , n+1} in which {n+1} is the singleton set with the largest entry in the partition. Since n + 1 is the largest element of the set, all partitions containing the singleton {n + 1} satisfy the requirement, and so the Bn partitions of {1, 2, . . . , n} together with {n + 1} are exactly the partitions of interest, that is, An,n = Bn.
Next, we verify that under the desired interpretation, it is indeed true that An,k = An,k−1 + An−1,k−1 for k > 1. Consider a partition counted by An,k−1. This contains the singleton {k}, and the element k + 1 is not in a singleton. If we interchange k and k + 1, we get the singleton {k + 1}, and no larger element is in a singleton. This gives us partitions in which {k + 1} is a singleton and {k} is not. Now consider a partition of {1, 2, . . . , n} counted by An−1,k−1. Replace all numbers j > k by j + 1, and add the singleton {k + 1}. This produces a partition in which both {k + 1} and {k} appear. In fact, we have described how to produce each partition counted by An,k exactly once, and so An,k = An,k−1 + An−1,k−1.
Finally, we need to verify that An,1 = Bn−1; we establish a bijection between the two sets.
Suppose a partition in An,1 has the form {{1}, {2}, X1, X2, . . . , Xk}, where no Xi is a singleton. Then the collection {X1, X2, . . . , Xk} is a partition of the n −1 element set {3, 4, . . . , n + 1}.
Now we subtract 2 from every element to get a parti-tion of [n −1] that has no singleton sets.
Any other partition in An,1 has the form {{2}, X1, X2, . . . , Xk}, where no Xi is a singleton. One of the sets Xi contains n+1, say X1.
26 Chapter 1 Fundamentals Remove n+1 and split the remainder of X1 into singleton sets {x1}, {x2}, . . . , {xm}. Now {{x1}, {x2}, . . . , {xm}, X2, X3, . . . , Xk} is a partition of the n−1 element set {1, 3, . . . , n}.
Subtracting one from all the elements except 1 produces a partition of [n−1] with at least one singleton set. If p ∈An,1, designate the partition of [n −1] obtained in this way by f(p).
Now we define a function g from Bn−1 to An,1. If partition p ∈Bn−1 contains no sin-gletons, say p = {X1, . . . , Xk}, add two to every element forming sets X∗ i , and let g(p) = {{1}, {2}, X∗ 1, . . . , X∗ k}. If p contains singletons, say p = {{x1}, {x2}, . . . , {xm}, X2, X3, . . . , Xk}, then add 1 to all elements except 1, forming a partition {{x∗ 1}, {x∗ 2}, . . . , {x∗ m}, X∗ 2, X∗ 3, . . . , X∗ k} of the set {1, 3, . . . , n}. Now let g(p) = {{2}, {x∗ 1, x∗ 2, . . . , x∗ m, n + 1}, X∗ 2, . . . , X∗ k}.
Since f(g(p)) = p and g(f(p)) = p, f and g are inverses and so f is a bijection between An,1 and Bn−1, as desired.
Exercises 1.4.
1. Show that if {A1, A2, . . . , Ak} is a partition of {1, 2, . . . , n}, then there is a unique equivalence relation ≡whose equivalence classes are {A1, A2, . . . , Ak}.
2. Suppose n is a square-free number, that is, no number m2 divides n; put another way, square-free numbers are products of distinct prime factors, that is, n = p1p2 · · · pk, where each pi is prime and no two prime factors are equal. Find the number of factorizations of n. For example, 30 = 2 · 3 · 5, and the factorizations of 30 are 30, 6 · 5, 10 · 3, 2 · 15, and 2 · 3 · 5. Note we count 30 alone as a factorization, though in some sense a trivial factorization.
3. The rhyme scheme of a stanza of poetry indicates which lines rhyme. This is usually expressed in the form ABAB, meaning the first and third lines of a four line stanza rhyme, as do the second and fourth, or ABCB, meaning only lines two and four rhyme, and so on. A limerick is a five line poem with rhyming scheme AABBA. How many different rhyme schemes are possible for an n line stanza? To avoid duplicate patterns, we only allow a new letter into the pattern when all previous letters have been used to the left of the new one. For example, ACBA is not allowed, since when C is placed in position 2, B has not been used to the left.
This is the same rhyme scheme as ABCA, which is allowed.
4. Another way to express the Bell numbers for n > 0 is Bn = n ∑ k=1 S(n, k), where S(n, k) is the number of partitions of {1, 2, . . . , n} into exactly k parts, 1 ≤k ≤n.
The S(n, k) are the Stirling numbers of the second kind. Find a recurrence relation for S(n, k). Your recurrence should allow a fairly simple triangle construction containing the values S(n, k), and then the Bell numbers may be computed by summing the rows of this triangle. Show the first five rows of the triangle, n ∈{1, 2, . . . , 5}.
5. Let An be the number of partitions of {1, 2, . . . , n + 1} in which no consecutive integers are in the same part of the partition.
For example, when n = 3 these partitions are {{1}, {2}, {3}, {4}}, {{1}, {2, 4}, {3}}, {{1, 3}, {2}, {4}}, {{1, 3}, {2, 4}}, {{1, 4}, {2}, {3}}, so A3 = 5. Let A(n, k) be the number of partitions of {1, 2, . . . , n + 1} into exactly k parts, 1.5 Choice with repetition 27 in which no consecutive integers are in the same part of the partition. Thus An = n+1 ∑ k=2 A(n, k).
Find a recurrence for A(n, k) and then show that An = Bn.
1.5 Choice with repetition Most of the permutation and combination problems we have seen count choices made with-out repetition, as when we asked how many rolls of three dice are there in which each die has a different value. The exception was the simplest problem, asking for the total number of outcomes when two or three dice are rolled, a simple application of the multiplication principle. Typical permutation and combination problems can be interpreted in terms of drawing balls from a box, and implicitly or explicitly the rule is that a ball drawn from the box stays out of the box. If instead each ball is returned to the box after recording the draw, we get a problem essentially identical to the general dice problem. For example, if there are six balls, numbered 1–6, and we draw three balls with replacement, the number of possible outcomes is 63. Another version of the problem does not replace the ball after each draw, but allows multiple “identical” balls to be in the box. For example, if a box contains 18 balls numbered 1–6, three with each number, then the possible outcomes when three balls are drawn and not returned to the box is again 63. If four balls are drawn, however, the problem becomes different.
Another, perhaps more mathematical, way to phrase such problems is to introduce the idea of a multiset. A multiset is like a set, except that elements may appear more than once. If {a, b} and {b, c} are ordinary sets, we say that the union {a, b} ∪{b, c} is {a, b, c}, not {a, b, b, c}. If we interpret these as multisets, however, we do write {a, b, b, c} and consider this to be different than {a, b, c}. To distinguish multisets from sets, and to shorten the expression in most cases, we use a repetition number with each element.
For example, we will write {a, b, b, c} as {1 · a, 2 · b, 1 · c}. By writing {1 · a, 1 · b, 1 · c} we emphasize that this is a multiset, even though no element appears more than once. We also allow elements to be included an infinite number of times, indicated with ∞for the repetition number, like {∞· a, 5 · b, 3 · c}.
Generally speaking, problems in which repetition numbers are infinite are easier than those involving finite repetition numbers. Given a multiset A = {∞·a1, ∞·a2, . . . , ∞·an}, how many permutations of the elements of length k are there? That is, how many sequences x1, x2, . . . , xk can be formed? This is easy: the answer is nk.
Now consider combinations of a multiset, that is, submultisets: Given a multiset, how many submultisets of a given size does it have? We say that a multiset A is a submultiset of B if the repetition number of every element of A is less than or equal to its repetition 28 Chapter 1 Fundamentals number in B. For example, {20 · a, 5 · b, 1 · c} is a submultiset of {∞· a, 5 · b, 3 · c}. A multiset is finite if it contains only a finite number of distinct elements, and the repetition numbers are all finite. Suppose again that A = {∞· a1, ∞· a2, . . . , ∞· an}; how many finite submultisets does it have of size k? This at first seems quite difficult, but put in the proper form it turns out to be a familiar problem. Imagine that we have k + n −1 “blank spaces”, like this: . . .
Now we place n −1 markers in some of these spots: ∧ ∧ ∧ . . .
∧ This uniquely identifies a submultiset: fill all blanks up to the first ∧with a1, up to the second with a2, and so on: ∧a2 ∧a3 a3 a3 ∧a4 . . . an−1 an−1 ∧an So this pattern corresponds to the multiset {1·a2, 3·a3, . . . , 1·an}. Filling in the markers ∧in all possible ways produces all possible submultisets of size k, so there are (k+n−1 n−1 ) such submultisets. Note that this is the same as (k+n−1 k ) ; the hard part in practice is remembering that the −1 goes with the n, not the k.
• • • Summarizing the high points so far: The number of permutations of n things taken k at a time without replacement is P(n, k) = n!/(n −k)!; the number of permutations of n things taken k at a time with replacement is nk. The number of combinations of n things taken k at a time without replacement is (n k ) ; the number of combinations of n things taken k at a time with replacement is (k+n−1 k ) .
• • • If A = {m1 · a1, m2 · a2, . . . , mn · an}, similar questions can be quite hard. Here is an easier special case: How many permutations of the multiset A are there? That is, how many sequences consist of m1 copies of a1, m1 copies of a2, and so on? This problem succumbs to overcounting: suppose to begin with that we can distinguish among the different copies of each ai; they might be colored differently for example: a red a1, a blue a1, and so on. Then we have an ordinary set with M = ∑n i=1 mi elements and M! permutations.
Now if we ignore the colors, so that all copies of ai look the same, we find that we have overcounted the desired permutations. Permutations with, say, the a1 items in the same positions all look the same once we ignore the colors of the a1s. How many of the original permutations have this property? m1! permutations will appear identical once we ignore 1.5 Choice with repetition 29 the colors of the a1 items, since there are m1! permutations of the colored a1s in a given m1 positions. So after throwing out duplicates, the number of remaining permutations is M!/m1! (assuming the other ai are still distinguishable). Then the same argument applies to the a2s: there are m2! copies of each permutation once we ignore the colors of the a2s, so there are M!
m1! m2! distinct permutations. Continuing in this way, we see that the number of distinct permutations once all colors are ignored is M!
m1! m2! · · · mn!.
This is frequently written ( M m1 m2 . . . mn ) , called a multinomial coefficient. Here the second row has n separate entries, not a single product entry. Note that if n = 2 this is ( M m1 m2 ) = M!
m1! m2! = M!
m1! (M −m1)! = (M m1 ) .
(1.5.1) This is easy to see combinatorially: given {m1 · a1, m2 · a2} we can form a permutation by choosing the m1 places that will be occupied by a1, filling in the remaining m2 places with a2. The number of permutations is the number of ways to choose the m1 locations, which is ( M m1 ) .
EXAMPLE 1.5.1 How many solutions does x1+x2+x3+x4 = 20 have in non-negative integers? That is, how many 4-tuples (m1, m2, m3, m4) of non-negative integers are solu-tions to the equation? We have actually solved this problem: How many submultisets of size 20 are there of the multiset {∞· a1, ∞· a2, ∞· a3, ∞· a4}? A submultiset of size 20 is of the form {m1 · a1, m2 · a2, m3 · a3, m4 · a4} where ∑mi = 20, and these are in 1–1 correspondence with the set of 4-tuples (m1, m2, m3, m4) of non-negative integers such that ∑mi = 20. Thus, the number of solutions is (20+4−1 20 ) . This reasoning applies in general: the number of solutions to n ∑ i=1 xi = k is (k + n −1 k ) .
This immediately suggests some generalizations: instead of the total number of solu-tions, we might want the number of solutions with the variables xi in certain ranges, that is, we might require that mi ≤xi ≤Mi for some lower and upper bounds mi and Mi.
30 Chapter 1 Fundamentals Finite upper bounds can be difficult to deal with; if we require that 0 ≤xi ≤Mi, this is the same as counting the submultisets of {M1 · a1, M2 · a2, . . . , Mn · an}. Lower bounds are easier to deal with.
EXAMPLE 1.5.2 Find the number of solutions to x1 + x2 + x3 + x4 = 20 with x1 ≥0, x2 ≥1, x3 ≥2, x4 ≥−1.
We can transform this to the initial problem in which all lower bounds are 0. The solutions we seek to count are the solutions of this altered equation: x1 + (x2 −1) + (x3 −2) + (x4 + 1) = 18.
If we set y1 = x1, y2 = x2 −1, y3 = x3 −2, and y4 = x4 + 1, then (x1, x2, x3, x4) is a solution to this equation if and only if (y1, y2, y3, y4) is a solution to y1 + y2 + y3 + y4 = 18, and moreover the bounds on the xi are satisfied if and only if yi ≥0. Since the number of solutions to the last equation is (18+4−1 18 ) , this is also the number of solutions to the original equation.
Exercises 1.5.
1. Suppose a box contains 18 balls numbered 1–6, three balls with each number. When 4 balls are drawn without replacement, how many outcomes are possible? Do this in two ways: assuming that the order in which the balls are drawn matters, and then assuming that order does not matter.
2. How many permutations are there of the letters in Mississippi?
3. How many permutations are there of the multiset {1 · a1, 1 · a2, . . . , 1 · an}?
4. Let M = ∑n i=1 mi. If ki < 0 for some i, let’s say ( M k1 k2 . . . kn ) = 0.
Prove that ( M m1 m2 . . . mn ) = n ∑ i=1 ( M −1 m1 m2 . . . (mi −1) . . . mn ) .
Note that when n = 2 this becomes ( M m1 m2 ) = ( M −1 (m1 −1) m2 ) + ( M −1 m1 (m2 −1) ) .
As noted above in equation 1.5.1, when n = 2 we are really seeing ordinary binomial coeffi-cients, and this can be rewritten as ( M m1 ) = ( M −1 m1 −1 ) + ( M −1 m1 ) , which of course we already know.
1.6 The Pigeonhole Principle 31 5. The Binomial Theorem (1.3.1) can be written (x + y)n = ∑ i+j=n ( n i j ) xi yj, where the sum is over all non-negative integers i and j that sum to n. Prove that for m ≥2, (x1 + x2 + · · · + xm)n = ∑( n i1 i2 . . . im ) xi1 1 xi2 2 . . . xim m .
where the sum is over all i1, . . . , im such that i1 + · · · + im = n.
6. Find the number of integer solutions to x1 + x2 + x3 + x4 + x5 = 50, x1 ≥−3, x2 ≥0, x3 ≥4, x4 ≥2, x5 ≥12.
7. You and your spouse each take two gummy vitamins every day. You share a single bottle of 60 vitamins, 30 of one flavor and 30 of another. You each prefer a different flavor, but it seems childish to fish out two of each type (but not to take gummy vitamins). So you just take the first four that fall out and then divide them up according to your preferences. For example, if there are two of each flavor, you and your spouse get the vitamins you prefer, but if three of your preferred flavor come out, you get two of the ones you like and your spouse gets one of each. Of course, you start a new bottle every 15 days. On average, over a 15 day period, how many of the vitamins you take are the flavor you prefer? (From fivethirtyeight.com.) 1.6 The Pigeonhole Principle A key step in many proofs consists of showing that two possibly different values are in fact the same. The Pigeonhole principle can sometimes help with this.
THEOREM 1.6.1 Pigeonhole Principle Suppose that n+1 (or more) objects are put into n boxes. Then some box contains at least two objects.
Proof.
Suppose each box contains at most one object. Then the total number of objects is at most 1 + 1 + · · · + 1 = n, a contradiction.
This seemingly simple fact can be used in surprising ways. The key typically is to put objects into boxes according to some rule, so that when two objects end up in the same box it is because they have some desired relationship.
EXAMPLE 1.6.2 Among any 13 people, at least two share a birth month.
Label 12 boxes with the names of the months. Put each person in the box labeled with his or her birth month. Some box will contain at least two people, who share a birth month.
EXAMPLE 1.6.3 Suppose 5 pairs of socks are in a drawer. Picking 6 socks guarantees that at least one pair is chosen.
32 Chapter 1 Fundamentals Label the boxes by “the pairs” (e.g., the red pair, the blue pair, the argyle pair,. . . ).
Put the 6 socks into the boxes according to description.
Some uses of the principle are not nearly so straightforward.
EXAMPLE 1.6.4 Suppose a1, . . . , an are integers. Then some “consecutive sum” ak + ak+1 + ak+2 + · · · + ak+m is divisible by n.
Consider these n sums: s1 = a1 s2 = a1 + a2 s3 = a1 + a2 + a3 .
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sn = a1 + a2 + · · · + an These are all consecutive sums, so if one of them is divisible by n we are done. If not, dividing each by n leaves a non-zero remainder, r1 = s1 mod n, r2 = s2 mod n, and so on.
These remainders have values in {1, 2, 3, . . . , n −1}. Label n −1 boxes with these n −1 values; put each of the n sums into the box labeled with its remainder mod n. Two sums end up in the same box, meaning that si mod n = sj mod n for some j > i; hence sj −si is divisible by n, and sj −si = ai+1 + ai+2 + · · · + aj, as desired.
A similar argument provides a proof of the Chinese Remainder Theorem.
THEOREM 1.6.5 Chinese Remainder Theorem If m and n are relatively prime, and 0 ≤a < m and 0 ≤b < n, then there is an integer x such that x mod m = a and x mod n = b.
Proof.
Consider the integers a, a + m, a + 2m, . . . a + (n −1)m, each with remainder a when divided by m. We wish to show that one of these integers has remainder b when divided by n, in which case that number satisfies the desired property.
For a contradiction, suppose not. Let the remainders be r0 = a mod n, r1 = a+m mod n,. . . , rn−1 = a + (n −1)m mod n. Label n −1 boxes with the numbers 0, 1, 2, 3, . . . , b − 1, b + 1, . . . n −1. Put each ri into the box labeled with its value. Two remainders end up in the same box, say ri and rj, with j > i, so ri = rj = r. This means that a + im = q1n + r and a + jm = q2n + r.
Hence a + jm −(a + im) = q2n + r −(q1n + r) (j −i)m = (q2 −q1)n.
1.6 The Pigeonhole Principle 33 Since n is relatively prime to m, this means that n | (j −i). But since i and j are distinct and in {0, 1, 2, . . . , n −1}, 0 < j −i < n, so n ∤(j −i). This contradiction finishes the proof.
More general versions of the Pigeonhole Principle can be proved by essentially the same method. A natural generalization would be something like this: If X objects are put into n boxes, some box contains at least m objects. For example: THEOREM 1.6.6 Suppose that r1, . . . , rn are positive integers. If X ≥(∑n i=1 ri)−n+1 objects are put into n boxes labeled 1, 2, 3, . . . , n, then some box labeled i contains at least ri objects.
Proof.
Suppose not. Then the total number of objects in the boxes is at most (r1 −1)+ (r2 −1) + (r3 −1) + · · · + (rn −1) = (∑n i=1 ri) −n < X, a contradiction.
This full generalization is only occasionally needed; often this simpler version is suffi-cient: COROLLARY 1.6.7 Suppose r > 0 and X ≥n(r −1) + 1 objects are placed into n boxes. Then some box contains at least r objects.
Proof.
Apply the previous theorem with ri = r for all i.
• • • Here is a simple application of the Pigeonhole Principle that leads to many interesting questions.
EXAMPLE 1.6.8 Suppose 6 people are gathered together; then either 3 of them are mutually acquainted, or 3 of them are mutually unacquainted.
We turn this into a graph theory question: Consider the graph consisting of 6 vertices, each connected to all the others by an edge, called the complete graph on 6 vertices, and denoted K6; the vertices represent the people. Color an edge red if the people represented by its endpoints are acquainted, and blue if they are not acquainted. Any choice of 3 vertices defines a triangle; we wish to show that either there is a red triangle or a blue triangle.
Consider the five edges incident at a single vertex v; by the Pigeonhole Principle (the version in corollary 1.6.7, with r = 3, X = 2(3 −1) + 1 = 5), at least three of them are the same color, call it color C; call the other color D. Let the vertices at the other ends of these three edges be v1, v2, v3. If any of the edges between these vertices have color C, there is a triangle of color C: if the edge connects vi to vj, the triangle is formed by v, 34 Chapter 1 Fundamentals vi, and vj. If this is not the case, then the three vertices v1, v2, v3 are joined by edges of color D, and form a triangle of color D.
The number 6 in this example is special: with 5 or fewer vertices it is not true that there must be a monochromatic triangle, and with more than 6 vertices it is true. To see that it is not true for 5 vertices, we need only show an example, as in figure 1.6.1.
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• .
• .
• .
• .
• Figure 1.6.1 An edge coloring with no monochromatic triangles.
The Ramsey number R(i) is the smallest integer n such that when the edges of Kn are colored with two colors, there is a monochromatic complete graph on i vertices, Ki, contained within Kn. The example shows that R(3) = 6.
More generally, R(i, j) is the smallest integer n such that when the edges of Kn are colored with two colors, say C1 and C2, either there is a Ki contained within Kn all of whose edges are color C1, or there is a Kj contained within Kn all of whose edges are color C2. Using this notion, R(k) = R(k, k). More generally still, R(i1, i2, . . . , im) is the smallest integer n such that when the edges of Kn are colored with m colors, C1, . . . , Cm, then for some j there is a Kij contained in Kn all of whose edges are color Cj.
Ramsey proved that in all of these cases, there actually is such a number n. General-izations of this problem have led to the subject called Ramsey Theory.
Computing any particular value R(i, j) turns out to be quite difficult; Ramsey numbers are known only for a few small values of i and j, and in some other cases the Ramsey number is bounded by known numbers. Typically in these cases someone has exhibited a Km and a coloring of the edges without the existence of a monochromatic Ki or Kj of the desired color, showing that R(i, j) > m; and someone has shown that whenever the edges of Kn have been colored, there is a Ki or Kj of the correct color, showing that R(i, j) ≤n.
Exercises 1.6.
1. Assume that the relation “friend” is symmetric. Show that if n ≥2, then in any group of n people there are two with the same number of friends in the group.
1.7 Sperner’s Theorem 35 2. Suppose that 501 distinct integers are selected from 1 . . . 1000. Show that there are distinct selected integers a and b such that a | b. Show that this is not always true if 500 integers are selected. ⇒ 3. Each of 15 red balls and 15 green balls is marked with an integer between 1 and 100 inclusive; no integer appears on more than one ball. The value of a pair of balls is the sum of the numbers on the balls. Show there are at least two pairs, consisting of one red and one green ball, with the same value. Show that this is not necessarily true if there are 13 balls of each color.
4. Suppose we have 14 red balls and 14 green balls as in the previous exercise. Show that at least two pairs, consisting of one red and one green ball, have the same value. What about 13 red balls and 14 green balls?
5. Suppose (a1, a2, . . . , a52) are integers, not necessarily distinct. Show that there are two, ai and aj with i ̸= j, such that either ai + aj or ai −aj is divisible by 100. Show that this is not necessarily true for integers (a1, a2, . . . , a51).
6. Suppose five points are chosen from a square whose sides are length s. (The points may be either in the interior of the square or on the boundary.) Show that two of the points are at most s √ 2/2 apart. Find five points so that no two are less than s √ 2/2 apart.
7. Show that if the edges of K6 are colored with two colors, there are at least two monochromatic triangles. (Two triangles are different if each contains at least one vertex not in the other.
For example, two red triangles that share an edge count as two triangles.) Color the edges of K6 so that there are exactly two monochromatic triangles.
8. Suppose the edges of a K5 are colored with two colors, say red and blue, so that there are no monochromatic triangles. Show that the red edges form a cycle, and the blue edges form a cycle, each with five edges. (A cycle is a sequence of edges {v1, v2}, {v2, v3}, . . . , {vk, v1}, where all of the vi are distinct. Note that this is true in figure 1.6.1.) 9. Show that 8 < R(3, 4) ≤10.
10. Show that R(3, 4) = 9.
1.7 Sperner's Theorem The binomial coefficients count the subsets of a given set; the sets themselves are worth looking at. First some convenient notation: DEFINITION 1.7.1 Let [n] = {1, 2, 3, . . . , n}. Then 2[n] denotes the set of all subsets of [n], and [ n k ] denotes the set of subsets of [n] of size k.
EXAMPLE 1.7.2 Let n = 3. Then [n 0 ] = {∅} [n 1 ] = {{1}, {2}, {3}} [n 2 ] = {{1, 2}, {1, 3}, {2, 3}} [n 3 ] = {{1, 2, 3}} 36 Chapter 1 Fundamentals DEFINITION 1.7.3 A chain in 2[n] is a set of subsets of 2[n] that are linearly or-dered by inclusion.
An anti-chain in 2[n] is a set of subsets of 2[n] that are pairwise incomparable.
EXAMPLE 1.7.4 In 2, {∅, {1}, {1, 2, 3}} is a chain, because ∅⊆{1} ⊆{1, 2, 3}.
Every [ n k ] is an anti-chain, as is {{1}, {2, 3}}. The set {{1}, {1, 3}, {2, 3}} is neither a chain nor an anti-chain.
Because of theorem 1.3.4 we know that among all anti-chains of the form [ n k ] the largest are the “middle” ones, namely [ n ⌊n/2⌋ ] and [ n ⌈n/2⌉ ] (which are the same if n is even). Remarkably, these are the largest of all anti-chains, that is, strictly larger than every other anti-chain. When n = 3, the anti-chains [ 3 1 ] and [ 3 2 ] are the only anti-chains of size 3, and no anti-chain is larger, as you can verify by examining all possibilities.
Before we prove this, a bit of notation.
DEFINITION 1.7.5 If σ: A →A is a bijection, then σ is called a permutation.
This use of the word permutation is different than our previous usage, but the two are closely related. Consider such a function σ: [n] →[n]. Since the set A in this case is finite, we could in principle list every value of σ: σ(1), σ(2), σ(3), . . . , σ(n).
This is a list of the numbers {1, . . . , n} in some order, namely, this is a permutation according to our previous usage. We can continue to use the same word for both ideas, relying on context or an explicit statement to indicate which we mean.
THEOREM 1.7.6 (Sperner’s Theorem) The only anti-chains of largest size are [ n ⌊n/2⌋ ] and [ n ⌈n/2⌉ ] .
Proof.
First we show that no anti-chain is larger than these two. We attempt to partition 2[n] into k = ( n ⌊n/2⌋ ) chains, that is, to find chains A1,0 ⊆A1,1 ⊆A1,2 ⊆· · · ⊆A1,m1 A2,0 ⊆A2,1 ⊆A2,2 ⊆· · · ⊆A2,m2 .
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Ak,0 ⊆Ak,1 ⊆Ak,2 ⊆· · · ⊆Ak,mk 1.7 Sperner’s Theorem 37 so that every subset of [n] appears exactly once as one of the Ai,j. If we can find such a partition, then since no two elements of an anti-chain can be in the same chain, no anti-chain can have more than k elements.
For small values of n this can be done by hand; for n = 3 we have ∅⊆{1} ⊆{1, 2} ⊆{1, 2, 3} {2} ⊆{2, 3} {3} ⊆{1, 3} These small cases form the base of an induction.
We will prove that any 2[n] can be partitioned into such chains with two additional properties: 1. Each set in a chain contains exactly one element more than the next smallest set in the chain.
2. The sum of the sizes of the smallest and largest element in the chain is n.
Note that the chains for the case n = 3 have both of these properties. The two properties taken together imply that every chain “crosses the middle”, that is, every chain contains an element of [ n n/2 ] if n is even, and an element of both [ n ⌊n/2⌋ ] and [ n ⌈n/2⌉ ] if n is odd.
Thus, if we succeed in showing that such chain partitions exist, there will be exactly ( n ⌊n/2⌋ ) chains.
For the induction step, we assume that we have partitioned 2[n−1] into such chains, and construct chains for 2[n].
First, for each chain Ai,0 ⊆Ai,1 ⊆· · · ⊆Ai,mi we form a new chain Ai,0 ⊆Ai,1 ⊆ · · · ⊆Ai,mi ⊆Ai,mi ∪{n}. Since |Ai,0| + |Ai,mi| = n −1, |Ai,0| + |Ai,mi ∪{n}| = n, so this new chain satisfies properties (1) and (2).
In addition, if mi > 0, we form a new chain Ai,0 ∪{n} ⊆Ai,1 ∪{n} ⊆· · · ⊆Ai,mi−1 ∪ {n}. Now |Ai,0 ∪{n}| + |Ai,mi−1 ∪{n}| = |Ai,0| + 1 + |Ai,mi−1| + 1 = |Ai,0| + 1 + |Ai,mi| −1 + 1 = n −1 + 1 = n so again properties (1) and (2) are satisfied.
Because of the first type of chain, all subsets of [n −1] are contained exactly once in the new set of chains. Also, we have added the element n exactly once to every subset of [n −1], so we have included every subset of [n] containing n exactly once. Thus we have produced the desired partition of 2[n].
Now we need to show that the only largest anti-chains are [ n ⌊n/2⌋ ] and [ n ⌈n/2⌉ ] .
Suppose that A1, A2, . . . , Am is an anti-chain; then Ac 1, Ac 2, . . . , Ac m is also an anti-chain, where Ac denotes the complement of A. Thus, if there is an anti-chain that contains some A with |A| > ⌈n/2⌉, there is also one containing Ac, and |Ac| < ⌊n/2⌋. Suppose that some 38 Chapter 1 Fundamentals anti-chain contains a set A with |A| < ⌊n/2⌋. We next prove that this anti-chain cannot be of maximum size.
Partition 2[n] as in the first part of the proof.
Suppose that A is a subset of the elements of a one or two element chain C, that is, a chain consisting solely of a set S1 of size n/2, if n is even, or of sets S1 and S2 of sizes ⌊n/2⌋and ⌈n/2⌉, with A ⊆S1 ⊆S2, if n is odd. Then no member of C is in the anti-chain. Thus, the largest possible size for an anti-chain containing A is ( n ⌊n/2⌋ ) −1.
If A is not a subset of the elements of such a short chain, we now prove that there is another chain partition of 2[n] that does have this property. Note that in the original chain partition there must be a chain of length 1 or 2, C1, consisting of S1 and possibly S2; if not, every chain would contain a set of size ⌊n/2⌋−1, but there are not enough such sets to go around.
Suppose then that A = {x1, . . . , xk} and the set S1 in C1 is S1 = {x1, . . . , xq, yq+1, . . . yl}, where 0 ≤q < k and l > k.
Let σ be the permutation of [n] such that σ(xq+i) = yq+i and σ(yq+i) = xq+i, for 1 ≤i ≤k −q, and σ fixes all other elements. Now for U ⊆[n], let U = σ(U), and note that U ⊆V if and only if U ⊆V . Thus every chain in the original chain partition maps to a chain. Since σ is a bijection, these new chains also form a partition of 2[n], with the additional properties (1) and (2). By the definition of σ, A ⊆S1, and {S1, S2} is a chain, say C1. Thus, this new chain partition has the desired property: A is a subset of every element of the 1 or 2 element chain C1, so A is not in an anti-chain of maximum size.
Finally, we need to show that if n is odd, no anti-chain of maximum size contains sets in both [ n ⌊n/2⌋ ] and [ n ⌈n/2⌉ ] . Suppose there is such an anti-chain, consisting of sets Ak+1, . . . , Al in [ n ⌈n/2⌉ ] , where l = ( n ⌈n/2⌉ ) , and B1, . . . , Bk in [ n ⌊n/2⌋ ] . The remaining sets in [ n ⌈n/2⌉ ] are A1, . . . , Ak, and the remaining sets in [ n ⌊n/2⌋ ] are Bk+1, . . . , Bl.
Each set Bi, 1 ≤i ≤k, is contained in exactly ⌈n/2⌉sets in [ n ⌈n/2⌉ ] , and all must be among A1, . . . , Ak. On average, then, each Ai, 1 ≤i ≤k, contains ⌈n/2⌉sets among B1, . . . , Bk. But each set Ai, 1 ≤i ≤k, contains exactly ⌈n/2⌉sets in [ n ⌊n/2⌋ ] , and so each must contain exactly ⌈n/2⌉of the sets B1, . . . , Bk and none of the sets Bk+1, . . . , Bl.
Let A1 = Aj1 = {x1, . . . , xr} and Bk+1 = {x1, . . . , xs, ys+1, . . . , yr−1}. Let Bim = Ajm{xs+m} and Ajm+1 = Bim ∪{ys+m}, for 1 ≤m ≤r−s−1. Note that by the preceding discussion, 1 ≤im ≤k and 1 ≤jm ≤k. Then Ajr−s = {x1, . . . , xs, ys+1, . . . , yr−1, xr}, so Ajr−s ⊇Bk+1, a contradiction. Hence there is no such anti-chain.
Exercises 1.7.
1. Sperner’s Theorem (1.7.6) tells us that [ 6 3 ] , with size 20, is the unique largest anti-chain for 2. The next largest anti-chains of the form [ 6 k ] are [ 6 2 ] and [ 6 4 ] , with size 15. Find a maximal anti-chain with size larger than 15 but less than 20. (As usual, maximal here means 1.8 Stirling numbers 39 that the anti-chain cannot be enlarged simply by adding elements. So you may not simply use a subset of [ 6 3 ] .) 1.8 Stirling numbers In exercise 4 in section 1.4, we saw the Stirling numbers of the second kind. Not surpris-ingly, there are Stirling numbers of the first kind. Recall that Stirling numbers of the second kind are defined as follows: DEFINITION 1.8.1 The Stirling number of the second kind, S(n, k) or { n k } , is the number of partitions of [n] = {1, 2, . . . , n} into exactly k parts, 1 ≤k ≤n.
Before we define the Stirling numbers of the first kind, we need to revisit permutations.
As we mentioned in section 1.7, we may think of a permutation of [n] either as a reordering of [n] or as a bijection σ: [n] →[n]. There are different ways to write permutations when thought of as functions. Two typical and useful ways are as a table, and in cycle form.
Consider this permutation σ: →: σ(1) = 3, σ(2) = 4, σ(3) = 5, σ(4) = 2, σ(5) = 1.
In table form, we write this as ( 1 2 3 4 5 3 4 5 2 1 ) , which is somewhat more compact, as we don’t write “σ” five times. In cycle form, we write this same permutation as (1, 3, 5)(2, 4). Here (1, 3, 5) indicates that σ(1) = 3, σ(3) = 5, and σ(5) = 1, whiile (2, 4) indicates σ(2) = 4 and σ(4) = 2. This permutation has two cycles, a 3-cycle and a 2-cycle. Note that (1, 3, 5), (3, 5, 1), and (5, 1, 3) all mean the same thing. We allow 1-cycles to count as cycles, though sometimes we don’t write them explicitly. In some cases, however, it is valuable to write them to force us to remember that they are there. Consider this permutation: ( 1 2 3 4 5 6 3 4 5 2 1 6 ) .
If we write this in cycle form as (1, 3, 5)(2, 4), which is correct, there is no indication that the underlying set is really . Writing (1, 3, 5)(2, 4)(6) makes this clear. We say that this permutation has 3 cycles, even though one of them is a trivial 1-cycle. Now we’re ready for the next definition.
DEFINITION 1.8.2 The Stirling number of the first kind, s(n, k), is (−1)n−k times the number of permutations of [n] with exactly k cycles. The corresponding unsigned Stirling number of the first kind, the number of permutations of [n] with exactly k cycles, is |s(n, k)|, sometimes written [ n k ] . Using this notation, s(n, k) = (−1)n−k [ n k ] .
Note that the use of [ n k ] conflicts with the use of the same notation in section 1.7; there should be no confusion, as we won’t be discussing the two ideas together.
Some values of [ n k ] are easy to see; if n ≥1, then [n n ] = 1 [n k ] = 0, if k > n [n 1 ] = (n −1)!
[n 0 ] = 0 40 Chapter 1 Fundamentals It is sometimes convenient to say that [ 0 0 ] = 1. These numbers thus form a triangle in the obvious way, just as the Stirling numbers of the first kind do. Here are lines 1–5 of the triangle: 1 0 1 0 1 1 0 2 3 1 0 6 11 6 1 0 24 50 35 10 1 The first column is not particularly interesting, so often it is eliminated.
In exercise 4 in section 1.4, we saw that {n k } = {n −1 k −1 } + k · {n −1 k } .
(1.8.1) The unsigned Stirling numbers of the first kind satisfy a similar recurrence.
THEOREM 1.8.3 [ n k ] = [ n−1 k−1 ] + (n −1) · [ n−1 k ] , k ≥1, n ≥1.
Proof.
The proof is by induction on n; the table above shows that it is true for the first few lines. We split the permutations of [n] with k cycles into two types: those in which (n) is a 1-cycle, and the rest. If (n) is a 1-cycle, then the remaining cycles form a permutation of [n −1] with k −1 cycles, so there are [ n−1 k−1 ] of these. Otherwise, n occurs in a cycle of length at least 2, and removing n leaves a permutation of [n −1] with k cycles. Given a permutation σ of [n −1] with k cycles, n can be added to any cycle in any position to form a permutation of [n] in which (n) is not a 1-cycle. Suppose the lengths of the cycles in σ are l1, l2, . . . , lk. In cycle number i, n may be added after any of the li elements in the cycle. Thus, the total number of places that n can be added is l1 + l2 + · · · + lk = n −1, so there are (n −1) · [ n−1 k ] permutations of [n] in which (n) is not a 1-cycle. Now the total number of permutations of [n] with k cycles is [ n−1 k−1 ] + (n −1) · [ n−1 k ] , as desired.
COROLLARY 1.8.4 s(n, k) = s(n −1, k −1) −(n −1)s(n −1, k).
The Stirling numbers satisfy two remarkable identities. First a definition: DEFINITION 1.8.5 The Kronecker delta δn,k is 1 if n = k and 0 otherwise.
THEOREM 1.8.6 For n ≥0 and k ≥0, n ∑ j=0 s(n, j)S(j, k) = n ∑ j=0 (−1)n−j [n j ] { j k } = δn,k n ∑ j=0 S(n, j)s(j, k) = n ∑ j=0 (−1)j−k {n j } [ j k ] = δn,k 1.8 Stirling numbers 41 Proof.
We prove the first version, by induction on n. The first few values of n are easily checked; assume n > 1. Now note that [ n 0 ] = 0, so we may start the sum index j at 1.
When k > n, { j k } = 0, for 1 ≤j ≤n, and so the sum is 0. When k = n, the only non-zero term occurs when j = n, and is (−1)0 [ n n ] { n n } = 1, so the sum is 1. Now suppose k < n. When k = 0, { j k } = 0 for j > 0, so the sum is 0, and we assume now that k > 0.
We begin by applying the recurrence relations: n ∑ j=1 (−1)n−j [n j ] { j k } = n ∑ j=1 (−1)n−j ([n −1 j −1 ] + (n −1) [n −1 j ]) { j k } = n ∑ j=1 (−1)n−j [n −1 j −1 ] { j k } + n ∑ j=1 (−1)n−j(n −1) [n −1 j ] { j k } = n ∑ j=1 (−1)n−j [n −1 j −1 ] ({ j −1 k −1 } + k {j −1 k }) + n ∑ j=1 (−1)n−j(n −1) [n −1 j ] { j k } = n ∑ j=1 (−1)n−j [n −1 j −1 ] { j −1 k −1 } + n ∑ j=1 (−1)n−j [n −1 j −1 ] k {j −1 k } + n ∑ j=1 (−1)n−j(n −1) [n −1 j ] { j k } .
Consider the first sum in the last expression: n ∑ j=1 (−1)n−j [n −1 j −1 ] { j −1 k −1 } = n ∑ j=2 (−1)n−j [n −1 j −1 ] { j −1 k −1 } = n−1 ∑ j=1 (−1)n−j−1 [n −1 j ] { j k −1 } = δn−1,k−1 = 0, since k −1 < n −1 (or trivially, if k = 1). Thus, we are left with just two sums.
n ∑ j=1 (−1)n−j [n −1 j −1 ] k {j −1 k } + n ∑ j=1 (−1)n−j(n −1) [n −1 j ] { j k } = k n−1 ∑ j=1 (−1)n−j−1 [n −1 j ] { j k } −(n −1) n−1 ∑ j=1 (−1)n−j−1 [n −1 j ] { j k } = kδn−1,k −(n −1)δn−1,k.
Now if k = n −1, this is (n −1)δn−1,n−1 −(n −1)δn−1,n−1 = 0, while if k < n −1 it is kδn−1,k −(n −1)δn−1,k = k · 0 −(n −1) · 0 = 0.
42 Chapter 1 Fundamentals If we interpret the triangles containing the s(n, k) and S(n, k) as matrices, either m × m, by taking the first m rows and columns, or even the infinite matrices containing the entire triangles, the sums of the theorem correspond to computing the matrix product in both orders. The theorem then says that this product consists of ones on the diagonal and zeros elsewhere, so these matrices are inverses. Here is a small example: 1 0 0 0 0 0 0 1 0 0 0 0 0 −1 1 0 0 0 0 2 −3 1 0 0 0 −6 11 −6 1 0 0 24 −50 35 −10 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 3 1 0 0 0 1 7 6 1 0 0 1 15 25 10 1 = 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 Exercises 1.8.
1. Find a simple expression for [ n n−1 ] .
2. Find a simple expression for [ n 1 ] .
3. What is ∑n k=0 [ n k ] ?
4. What is ∑n k=0 s(n, k)?
5. Show that xn = ∏n−1 k=0(x −k) = ∑n i=0 s(n, i)xi, n ≥1; xn is called a falling factorial. Find a similar identity for xn = ∏n−1 k=0(x + k); xn is a rising factorial.
6. Show that n ∑ k=0 {n k } xk = xn, n ≥1; xk is defined in the previous exercise. The previous exercise shows how to express the falling factorial in terms of powers of x; this exercise shows how to express the powers of x in terms of falling factorials.
7. Prove: S(n, k) = n−1 ∑ i=k−1 ( n −1 i ) S(i, k −1).
8. Prove: [n k ] = n−1 ∑ i=k−1 (n −i −1)!
( n −1 i ) [ i k −1 ] .
9. Use the previous exercise to prove s(n, k) = n−1 ∑ i=k−1 (−1)n−i−1(n −i −1)!
( n −1 i ) s(i, k −1).
10. We have defined [ n k ] and { n k } for n, k ≥0. We want to extend the definitions to all integers.
Without some extra stipulations, there are many ways to do this. Let us suppose that for n ̸= 0 we want [ n 0 ] = [ 0 n ] = { n 0 } = { 0 n } = 0, and we want the recurrence relations of equation 1.8.1 and in theorem 1.8.3 to be true. Show that under these conditions there is a unique way to extend the definitions to all integers, and that when this is done, { n k } = [ −k −n ] for all integers n and k. Thus, the extended table of values for either [ n k ] or { n k } will contain all the values of both [ n k ] and { n k } .
11. Under the assumptions that s(n, 0) = s(0, n) = 0 for n ̸= 0, and s(n, k) = s(n −1, k −1) − (n−1)s(n−1, k), extend the table for s(n, k) to all integers, and find a connection to S(n, k) similar to that in the previous problem.
1.8 Stirling numbers 43 12. Prove corollary 1.8.4.
13. Prove the remaining part of theorem 1.8.6.
2 Inclusion-Exclusion 2.1 The Inclusion-Exclusion Formula Let’s return to a problem we have mentioned but not solved: EXAMPLE 2.1.1 How many submultisets of the multiset {2 · a, 4 · b, 3 · c} have size 7?
We recast the problem: this is the number of solutions to x1 + x2 + x3 = 7 with 0 ≤x1 ≤2, 0 ≤x2 ≤4, 0 ≤x3 ≤3. We know that the number of solutions in non-negative integers is (7+3−1 3−1 ) = (9 2 ) , so this is an overcount, since we count solutions that do not meet the upper bound restrictions. For example, this includes some solutions with x1 ≥3; how many of these are there? This is a problem we can solve: it is the number of solutions to x1 + x2 + x3 = 7 with 3 ≤x1, 0 ≤x2, 0 ≤x3. This is the same as the number of non-negative solutions of y1 + y2 + y3 = 7 −3 = 4, or (4+3−1 3−1 ) = (6 2 ) . Thus, (9 2 ) − (6 2 ) corrects this overcount. If we likewise correct for the overcounting of solutions with x2 ≥5 and x3 ≥4, we get (9 2 ) − (6 2 ) − (4 2 ) − (5 2 ) . Is this correct? Not necessarily, because we now have a potential undercount: we have twice subtracted 1 for a solution in which both x1 ≥3 and x2 ≥5, when we should have subtracted just 1. However, by good fortune, there are no such solutions, since 3 + 5 > 7. But the same applies to the other pairs of variables: How many solutions have x1 ≥3 and x3 ≥4? It’s easy to see there is only one such solution, namely 3 + 0 + 4 = 7. Finally, there are no solutions with x2 ≥5 and x3 ≥4, so the corrected count is now (9 2 ) − (6 2 ) − (4 2 ) − (5 2 ) + 1. This does not take into account any solutions in which x1 ≥3, x2 ≥5, and x3 ≥4, but there are none of these, so the actual count is (9 2 ) − (6 2 ) − (4 2 ) − (5 2 ) + 1 = 36 −15 −6 −10 + 1 = 6.
45 46 Chapter 2 Inclusion-Exclusion This is small enough that it is not hard to verify by listing all the solutions.
So we solved this problem, but it is apparent that it could have been much worse, if the number of variables were larger and there were many complicated overcounts and undercounts. Remarkably, it is possible to streamline this sort of argument; it will still, often, be quite messy, but the reasoning will be simpler.
Let’s start by rephrasing the example. Let S be the set of all non-negative solutions to x1 + x2 + x3 = 7, let A1 be all solutions with x1 ≥3, A2 all solutions with x2 ≥5, and A3 all solutions with x3 ≥4. We want to know the size of Ac 1 ∩Ac 2 ∩Ac 3, the solutions for which it is not true that x1 ≥3 and not true that x2 ≥5 and not true that x3 ≥4.
Examining our solution, we see that the final count is |S| −|A1| −|A2| −|A3| + |A1 ∩A2| + |A1 ∩A3| + |A2 ∩A3| −|A1 ∩A2 ∩A3| = 36 −15 −6 −10 + 0 + 1 + 0 −0.
This pattern is completely general: THEOREM 2.1.2 The inclusion-exclusion formula If Ai ⊆S for 1 ≤i ≤n then |Ac 1 ∩· · · ∩Ac n| = |S| −|A1| −· · · −|An| + |A1 ∩A2| + · · · −|A1 ∩A2 ∩A3| −· · · , or more compactly: | n ∩ i=1 Ac i| = |S| + n ∑ k=1 (−1)k ∑ | k ∩ j=1 Aij|, where the internal sum is over all subsets {i1, i2, . . . , ik} of {1, 2, . . . , n}. Alternately we may write | n ∩ i=1 Ac i| = ∑ I⊆[n] (−1)|I|| ∩ j∈I Aj|.
(Note that ∩ j∈∅ Aj = S.) Proof.
We need to show that each element of ∩n i=1 Ac i is counted once by the right hand side, and every other element of S is counted zero times. The first of these is easy: if x ∈∩n i=1 Ac i then for every i, x / ∈Ai, so x is in none of the sets involving the Ai on the right hand side, and so x is counted, once, by the term |S|.
Now suppose x / ∈∩n i=1 Ac i. On the right hand side, x is counted once by the term |S|. For some values i1, i2, . . . , ik, x ∈Aim, 1 ≤m ≤k, and x is not in the remaining sets Ai. Then x is counted zero times by any term involving an Ai with i / ∈{i1, i2, . . . , ik}, and is counted once, positively or negatively, by each term involving only Ai1, Ai2, . . . , Aik.
There are k terms of the form −|Aim|, which count x a total of −k times. There are (k 2 ) 2.1 The Inclusion-Exclusion Formula 47 terms of the form |Ail ∩Aim|, counting x a total of (k 2 ) times. Continuing in this way, we see that the final count for x on the right hand side is 1 −k + (k 2 ) − (k 3 ) + · · · + (−1)k (k k ) , or more compactly k ∑ i=0 (−1)i (k i ) .
We know that this alternating sum of binomial coefficients is zero, so x is counted zero times, as desired. (See equation 1.3.1.) An alternate form of the inclusion exclusion formula is sometimes useful.
COROLLARY 2.1.3 If Ai ⊆S for 1 ≤i ≤n then | n ∪ i=1 Ai| = n ∑ k=1 (−1)k+1 ∑ | k ∩ j=1 Aij|, where the internal sum is over all subsets {i1, i2, . . . , ik} of {1, 2, . . . , n}.
Proof.
Since (∪n i=1 Ai)c = ∩n i=1 Ac i, | n ∪ i=1 Ai| = |S| −| n ∩ i=1 Ac i| = |S| −(|S| + n ∑ k=1 (−1)k ∑ | k ∩ j=1 Aij|) = (−1) n ∑ k=1 (−1)k ∑ | k ∩ j=1 Aij| = n ∑ k=1 (−1)k+1 ∑ | k ∩ j=1 Aij|.
Since the right hand side of the inclusion-exclusion formula consists of 2n terms to be added, it can still be quite tedious. In some nice cases, all intersections of the same number of sets have the same size. Since there are (n k ) possible intersections consisting of k sets, the formula becomes | n ∩ i=1 Ac i| = |S| + n ∑ k=1 (−1)k (n k ) mk, (2.1.1) where mk is the size of an intersection of k of the sets.
48 Chapter 2 Inclusion-Exclusion EXAMPLE 2.1.4 Find the number of solutions to x1 +x2 +x3 +x4 = 25, 0 ≤xi ≤10.
Let Ai be the solutions of x1 + x2 + x3 + x4 = 25 with xi ≥11. The number of solutions with xi ≥0 for all i is (25+4−1 4−1 ) = (25+3 3 ) . Also |Ai| = (14+3 3 ) , and |Ai ∩Aj| = (3+3 3 ) . There are no solutions with 3 or 4 of the variables larger than 10. Hence the number of solutions is (25 + 3 3 ) − (4 1 )(14 + 3 3 ) + (4 2 )(3 + 3 3 ) = 676.
Exercises 2.1.
1. List all 6 solutions to the restricted equation in example 2.1.1, and list the corresponding 6 submultisets.
2. Find the number of integer solutions to x1 + x2 + x3 + x4 = 25, 1 ≤x1 ≤6, 2 ≤x2 ≤8, 0 ≤x3 ≤8, 5 ≤x4 ≤9.
3. Find the number of submultisets of {25 · a, 25 · b, 25 · c, 25 · d} of size 80.
4. Recall that { n k } is a Stirling number of the second kind (definition 1.8.1). Prove that for n ≥k ≥0, {n k } = 1 k!
k ∑ i=0 (−1)k−iin ( k i ) .
Do n = 0 as a special case, then use inclusion-exclusion for the rest. You may assume, by convention, that 00 = 1.
2.2 Forbidden Position Permutations Suppose we shuffle a deck of cards; what is the probability that no card is in its original location? More generally, how many permutations of [n] = {1, 2, 3, . . . , n} have none of the integers in their “correct” locations? That is, 1 is not first, 2 is not second, and so on.
Such a permutation is called a derangement of [n].
Let S be the set of all permutations of [n] and Ai be the permutations of [n] in which i is in the correct place. Then we want to know | ∩n i=1 Ac i|.
For any i, |Ai| = (n −1)!: once i is fixed in position i, the remaining n −1 integers can be placed in any locations.
What about |Ai ∩Aj|? If both i and j are in the correct position, the remaining n −2 integers can be placed anywhere, so |Ai ∩Aj| = (n −2)!.
2.2 Forbidden Position Permutations 49 In the same way, we see that |Ai1 ∩Ai2 ∩· · · ∩Aik| = (n −k)!. Thus, by the inclusion-exclusion formula, in the form of equation 2.1.1, | n ∩ i=1 Ac i| = |S| + n ∑ k=1 (−1)k (n k ) (n −k)!
= n! + n ∑ k=1 (−1)k n!
k!(n −k)!(n −k)!
= n! + n ∑ k=1 (−1)k n!
k!
= n! + n!
n ∑ k=1 (−1)k 1 k!
= n!
( 1 + n ∑ k=1 (−1)k 1 k!
) = n!
n ∑ k=0 (−1)k 1 k!.
The last sum should look familiar: ex = ∞ ∑ k=0 1 k!xk.
Substituting x = −1 gives e−1 = ∞ ∑ k=0 1 k!(−1)k.
The probability of getting a derangement by chance is then 1 n!n!
n ∑ k=0 (−1)k 1 k! = n ∑ k=0 (−1)k 1 k!, and when n is bigger than 6, this is quite close to e−1 ≈0.3679.
So in the case of a deck of cards, the probability of a derangement is about 37%.
Let Dn = n! ∑n k=0(−1)k 1 k!. These derangement numbers have some interesting properties. First, note that when n = 0, we have D0 = 0!(−1)0 1 0! = 1. “Derangements of the empty set” doesn’t really make sense, but it is useful to adopt the convention that D0 = 1.
50 Chapter 2 Inclusion-Exclusion The derangements of [n] may be produced as follows: For each i ∈{2, 3, . . . , n}, put i in position 1 and 1 in position i. Then permute the numbers {2, 3, . . . , i −1, i + 1, . . . n} in all possible ways so that none of these n −2 numbers is in the correct place. There are Dn−2 ways to do this. Then, keeping 1 in position i, derange the numbers {i, 2, 3, . . . , i − 1, i + 1, . . . n}, with the “correct” position of i now considered to be position 1. There are Dn−1 ways to do this. Thus, Dn = (n −1)(Dn−1 + Dn−2). Starting with D0 = 1 and D1 = 0, this gives D2 = (1)(0 + 1) = 1 and D3 = (2)(1 + 0) = 2, both of which are easy to check directly.
We explore this recurrence relation a bit: Dn = nDn−1 −Dn−1 + (n −1)Dn−2 (∗) = nDn−1 −(n −2)(Dn−2 + Dn−3) + (n −1)Dn−2 = nDn−1 −(n −2)Dn−2 −(n −2)Dn−3 + (n −1)Dn−2 = nDn−1 + Dn−2 −(n −2)Dn−3 (∗) = nDn−1 + (n −3)(Dn−3 + Dn−4) −(n −2)Dn−3 = nDn−1 + (n −3)Dn−3 + (n −3)Dn−4 −(n −2)Dn−3 = nDn−1 −Dn−3 + (n −3)Dn−4 (∗) = nDn−1 −(n −4)(Dn−4 + Dn−5) + (n −3)Dn−4 = nDn−1 −(n −4)Dn−4 −(n −4)Dn−5 + (n −3)Dn−4 = nDn−1 + Dn−4 −(n −4)Dn−5.
(∗) It appears from the starred lines that the pattern here is that Dn = nDn−1 + (−1)kDn−k + (−1)k+1(n −k)Dn−k−1.
If this continues, we should get to Dn = nDn−1 + (−1)n−2D2 + (−1)n−1(2)D1.
Since D2 = 1 and D1 = 0, this would give Dn = nDn−1 + (−1)n, since (−1)n = (−1)n−2. Indeed this is true, and can be proved by induction. This gives a somewhat simpler recurrence relation, making it quite easy to compute Dn.
• • • There are many similar problems.
2.2 Forbidden Position Permutations 51 EXAMPLE 2.2.1 How many permutations of [n] contain no instance of i followed by i + 1?
By a similar use of the inclusion-exclusion formula, it turns out that this is Qn = n!
n−1 ∑ k=0 (−1)k 1 k! + (n −1)!
n−1 ∑ k=1 (−1)k−1 1 (k −1)!.
Note that the limits on the two sums are not identical.
Exercises 2.2.
1. Prove that Dn = nDn−1 + (−1)n when n ≥1, by induction on n.
2. Prove that Dn is even if and only if n is odd.
3. Provide the missing details for example 2.2.1. What is lim n→∞ Qn n! ?
4. Find the number of permutations of 1, 2, . . . , 8 that have no odd number in the correct position.
5. Find the number of permutations of 1, 2, . . . , 8 that have at least one odd number in the correct position.
6. How many permutations of [n] have exactly k numbers in their correct positions?
7. Give a combinatorial proof that n! = n ∑ k=0 ( n k ) Dn−k.
8. A small merry-go-round has 8 seats occupied by 8 children.
In how many ways can the children change places so that no child sits behind the same child as on the first ride? The seats do not matter, only the relative positions of the children.
9. Repeat the previous problem with n instead of 8.
10. On the way into a party everyone checks a coat and a bag at the door. On the way out, the attendant hands out coats and bags randomly. In how many ways can this be done if (a) No one gets either their own coat or their own bag?
(b) One may get one’s own coat, or bag, but not both.
11. Suppose n people are seated in m ≥n chairs in a room. At some point there is a break, and everyone leaves the room. When they return, in how many ways can they be seated so that no person occupies the same chair as before the break?
3 Generating Functions As we have seen, a typical counting problem includes one or more parameters, which of course show up in the solutions, such as (n k ) , P(n, k), or the number of derangements of [n]. Also recall that (x + 1)n = n ∑ k=0 (n k ) xk.
This provides the values (n k ) as coefficients of the Maclaurin expansion of a function. This turns out to be a useful idea.
DEFINITION 3.0.1 f(x) is a generating function for the sequence a0, a1, a2, . . . if f(x) = ∞ ∑ i=0 aixi.
Sometimes a generating function can be used to find a formula for its coefficients, but if not, it gives a way to generate them. Generating functions can also be useful in proving facts about the coefficients.
3.1 Newton's Binomial Theorem Recall that (n k ) = n!
k! (n −k)! = n(n −1)(n −2) · · · (n −k + 1) k!
.
53 54 Chapter 3 Generating Functions The expression on the right makes sense even if n is not a non-negative integer, so long as k is a non-negative integer, and we therefore define (r k ) = r(r −1)(r −2) · · · (r −k + 1) k!
when r is a real number. For example, (1/2 4 ) = (1/2)(−1/2)(−3/2)(−5/2) 4!
= −5 128 and (−2 3 ) = (−2)(−3)(−4) 3!
= −4.
These generalized binomial coefficients share some important properties of the usual binomial coefficients, most notably that (r k ) = (r −1 k −1 ) + (r −1 k ) .
(3.1.1) Then remarkably: THEOREM 3.1.1 Newton’s Binomial Theorem For any real number r that is not a non-negative integer, (x + 1)r = ∞ ∑ i=0 (r i ) xi when −1 < x < 1.
Proof.
It is not hard to see that the series is the Maclaurin series for (x + 1)r, and that the series converges when −1 < x < 1. It is rather more difficult to prove that the series is equal to (x + 1)r; the proof may be found in many introductory real analysis books.
EXAMPLE 3.1.2 Expand the function (1 −x)−n when n is a positive integer.
We first consider (x + 1)−n; we can simplify the binomial coefficients: (−n)(−n −1)(−n −2) · · · (−n −i + 1) i!
= (−1)i (n)(n + 1) · · · (n + i −1) i!
= (−1)i (n + i −1)!
i! (n −1)!
= (−1)i (n + i −1 i ) = (−1)i (n + i −1 n −1 ) .
Thus (x + 1)−n = ∞ ∑ i=0 (−1)i (n + i −1 n −1 ) xi = ∞ ∑ i=0 (n + i −1 n −1 ) (−x)i.
Now replacing x by −x gives (1 −x)−n = ∞ ∑ i=0 (n + i −1 n −1 ) xi.
So (1 −x)−n is the generating function for (n+i−1 n−1 ) , the number of submultisets of {∞· 1, ∞· 2, . . . , ∞· n} of size i.
3.1 Newton’s Binomial Theorem 55 In many cases it is possible to directly construct the generating function whose coeffi-cients solve a counting problem.
EXAMPLE 3.1.3 Find the number of solutions to x1 + x2 + x3 + x4 = 17, where 0 ≤x1 ≤2, 0 ≤x2 ≤5, 0 ≤x3 ≤5, 2 ≤x4 ≤6.
We can of course solve this problem using the inclusion-exclusion formula, but we use generating functions. Consider the function (1 + x + x2)(1 + x + x2 + x3 + x4 + x5)(1 + x + x2 + x3 + x4 + x5)(x2 + x3 + x4 + x5 + x6).
We can multiply this out by choosing one term from each factor in all possible ways. If we then collect like terms, the coefficient of xk will be the number of ways to choose one term from each factor so that the exponents of the terms add up to k. This is precisely the number of solutions to x1 + x2 + x3 + x4 = k, where 0 ≤x1 ≤2, 0 ≤x2 ≤5, 0 ≤x3 ≤5, 2 ≤x4 ≤6. Thus, the answer to the problem is the coefficient of x17. With the help of a computer algebra system we get (1 + x + x2)(1 + x + x2 + x3 + x4 + x5)2(x2 + x3 + x4 + x5 + x6) = x18 + 4x17 + 10x16 + 19x15 + 31x14 + 45x13 + 58x12 + 67x11 + 70x10 + 67x9 + 58x8 + 45x7 + 31x6 + 19x5 + 10x4 + 4x3 + x2, so the answer is 4.
EXAMPLE 3.1.4 Find the generating function for the number of solutions to x1 + x2 + x3 + x4 = k, where 0 ≤x1 ≤∞, 0 ≤x2 ≤5, 0 ≤x3 ≤5, 2 ≤x4 ≤6.
This is just like the previous example except that x1 is not bounded above.
The generating function is thus f(x) = (1 + x + x2 + · · ·)(1 + x + x2 + x3 + x4 + x5)2(x2 + x3 + x4 + x5 + x6) = (1 −x)−1(1 + x + x2 + x3 + x4 + x5)2(x2 + x3 + x4 + x5 + x6) = (1 + x + x2 + x3 + x4 + x5)2(x2 + x3 + x4 + x5 + x6) 1 −x .
Note that (1 −x)−1 = (1 + x + x2 + · · ·) is the familiar geometric series from calculus; alternately, we could use example 3.1.2. Unlike the function in the previous example, this function has an infinite expansion: f(x) = x2 + 4x3 + 10x4 + 20x5 + 35x6 + 55x7 + 78x8 + 102x9 + 125x10 + 145x11 + 160x12 + 170x13 + 176x14 + 179x15 + 180x16 + 180x17 + 180x18 + 180x19 + 180x20 + · · · .
You can see how to do this in Sage.
56 Chapter 3 Generating Functions EXAMPLE 3.1.5 Find a generating function for the number of submultisets of {∞· a, ∞·b, ∞·c} in which there are an odd number of as, an even number of bs, and any number of cs. As we have seen, this is the same as the number of solutions to x1 + x2 + x3 = n in which x1 is odd, x2 is even, and x3 is unrestricted. The generating function is therefore (x + x3 + x5 + · · ·)(1 + x2 + x4 + · · ·)(1 + x + x2 + x3 + · · ·) = x(1 + (x2) + (x2)2 + (x2)3 + · · ·)(1 + (x2) + (x2)2 + (x2)3 + · · ·) 1 1 −x = x (1 −x2)2(1 −x).
Exercises 3.1.
For some of these exercises, you may want to use the sage applet above, in example 3.1.4, or your favorite computer algebra system.
1. Prove that (r k ) = (r−1 k−1 ) + (r−1 k ) .
2. Show that the Maclaurin series for (x + 1)r is ∑∞ i=0 (r i ) xi.
3. Concerning example 3.1.4, show that all coefficients beginning with x16 are 180.
4. Use a generating function to find the number of solutions to x1 + x2 + x3 + x4 = 14, where 0 ≤x1 ≤3, 2 ≤x2 ≤5, 0 ≤x3 ≤5, 4 ≤x4 ≤6.
5. Find the generating function for the number of solutions to x1 + x2 + x3 + x4 = k, where 0 ≤x1 ≤∞, 3 ≤x2 ≤∞, 2 ≤x3 ≤5, 1 ≤x4 ≤5.
6. Find a generating function for the number of non-negative integer solutions to 3x+2y +7z = n.
7. Suppose we have a large supply of red, white, and blue balloons. How many different bunches of 10 balloons are there, if each bunch must have at least one balloon of each color and the number of white balloons must be even?
8. Use generating functions to show that every positive integer can be written in exactly one way as a sum of distinct powers of 2.
9. Suppose we have a large supply of blue and green candles, and one gold candle. How many collections of n candles are there in which the number of blue candles is even, the number of green candles is any number, and the number of gold candles is at most one?
3.2 Exponential Generating Functions There are other ways that a function might be said to generate a sequence, other than as what we have called a generating function. For example, ex = ∞ ∑ n=0 1 n!xn 3.2 Exponential Generating Functions 57 is the generating function for the sequence 1, 1, 1 2, 1 3!, . . .. But if we write the sum as ex = ∞ ∑ n=0 1 · xn n! , considering the n! to be part of the expression xn/n!, we might think of this same function as generating the sequence 1, 1, 1, . . ., interpreting 1 as the coefficient of xn/n!. This is not a very interesting sequence, of course, but this idea can often prove fruitful. If f(x) = ∞ ∑ n=0 an xn n! , we say that f(x) is the exponential generating function for a0, a1, a2, . . ..
EXAMPLE 3.2.1 Find an exponential generating function for the number of permu-tations with repetition of length n of the set {a, b, c}, in which there are an odd number of a s, an even number of b s, and any number of c s.
For a fixed n and fixed numbers of the letters, we already know how to do this. For example, if we have 3 a s, 4 b s, and 2 c s, there are ( 9 3 4 2 ) such permutations. Now consider the following function: ∞ ∑ i=0 x2i+1 (2i + 1)!
∞ ∑ i=0 x2i (2i)!
∞ ∑ i=0 xi i! .
What is the coefficient of x9/9! in this product? One way to get an x9 term is x3 3!
x4 4!
x2 2! = 9!
3! 4! 2!
x9 9! = ( 9 3 4 2 )x9 9! .
That is, this one term counts the number of permutations in which there are 3 a s, 4 b s, and 2 c s. The ultimate coefficient of x9/9! will be the sum of many such terms, counting the contributions of all possible choices of an odd number of a s, an even number of b s, and any number of c s.
Now we notice that ∞ ∑ i=0 xi i! = ex, and that the other two sums are closely related to this. A little thought leads to ex + e−x = ∞ ∑ i=0 xi i! + ∞ ∑ i=0 (−x)i i!
= ∞ ∑ i=0 xi + (−x)i i!
.
Now xi + (−x)i is 2xi when i is even, and 0 when i is odd. Thus ex + e−x = ∞ ∑ i=0 2x2i (2i)!, 58 Chapter 3 Generating Functions so that ∞ ∑ i=0 x2i (2i)! = ex + e−x 2 .
A similar manipulation shows that ∞ ∑ i=0 x2i+1 (2i + 1)! = ex −e−x 2 .
Thus, the generating function we seek is ex −e−x 2 ex + e−x 2 ex = 1 4(ex −e−x)(ex + e−x)ex = 1 4(e3x −e−x).
Notice the similarity to example 3.1.5.
Exercises 3.2.
1. Find the coefficient of x9/9! in the function of example 3.2.1. You may use Sage or a similar program.
2. Find an exponential generating function for the number of permutations with repetition of length n of the set {a, b, c}, in which there are an odd number of a s, an even number of b s, and an even number of c s.
3. Find an exponential generating function for the number of permutations with repetition of length n of the set {a, b, c}, in which the number of a s is even and at least 2, the number of b s is even and at most 6, and the number of c s is at least 3.
4. In how many ways can we paint the 10 rooms of a hotel if at most three can be painted red, at most 2 painted green, at most 1 painted white, and any number can be painted blue or orange? (The rooms are different, so order matters.) 5. Recall from section 1.4 that the Bell numbers Bn count all of the partitions of {1, 2, . . . , n}.
Let f(x) = ∞ ∑ n=0 Bn · xn n! , and note that f ′(x) = ∞ ∑ n=1 Bn xn−1 (n −1)! = ∞ ∑ n=0 Bn+1 xn n! = ∞ ∑ n=0 ( n ∑ k=0 ( n k ) Bn−k ) xn n! , using the recurrence relation 1.4.1 for Bn+1 from section 1.4. Now it is possible to write this as a product of two infinite series: f ′(x) = ( ∞ ∑ n=0 Bn · xn n!
) ( ∞ ∑ n=0 anxn ) = f(x)g(x).
Find an expression for an that makes this true, which will tell you what g(x) is, then solve the differential equation for f(x), the exponential generating function for the Bell numbers.
From section 1.4, the first few Bell numbers are 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437. You can use Sage to check your answer.
3.3 Partitions of Integers 59 3.3 Partitions of Integers DEFINITION 3.3.1 A partition of a positive integer n is a multiset of positive integers that sum to n. We denote the number of partitions of n by pn.
Typically a partition is written as a sum, not explicitly as a multiset. Using the usual convention that an empty sum is 0, we say that p0 = 1.
EXAMPLE 3.3.2 The partitions of 5 are 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1.
Thus p5 = 7.
There is no simple formula for pn, but it is not hard to find a generating function for them. As with some previous examples, we seek a product of factors so that when the factors are multiplied out, the coefficient of xn is pn. We would like each xn term to represent a single partition, before like terms are collected. A partition is uniquely described by the number of 1s, number of 2s, and so on, that is, by the repetition numbers of the multiset. We devote one factor to each integer: (1 + x + x2 + x3 + · · ·)(1 + x2 + x4 + x6 + · · ·) · · · (1 + xk + x2k + x3k + · · ·) · · · = ∞ ∏ k=1 ∞ ∑ i=0 xik.
When this product is expanded, we pick one term from each factor in all possible ways, with the further condition that we only pick a finite number of “non-1” terms. For example, if we pick x3 from the first factor, x3 from the third factor, x15 from the fifth factor, and 1s from all other factors, we get x21. In the context of the product, this represents 3 · 1 + 1 · 3 + 3 · 5, corresponding to the partition 1 + 1 + 1 + 3 + 5 + 5 + 5, that is, three 1s, one 3, and three 5s. Each factor is a geometric series; the kth factor is 1 + xk + (xk)2 + (xk)3 + · · · = 1 1 −xk , so the generating function can be written ∞ ∏ k=1 1 1 −xk .
60 Chapter 3 Generating Functions Note that if we are interested in some particular pn, we do not need the entire infinite product, or even any complete factor, since no partition of n can use any integer greater than n, and also cannot use more than n/k copies of k.
EXAMPLE 3.3.3 Find p8.
We expand (1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8)(1 + x2 + x4 + x6 + x8)(1 + x3 + x6) (1 + x4 + x8)(1 + x5)(1 + x6)(1 + x7)(1 + x8) = 1 + x + 2x2 + 3x3 + 5x4 + 7x5 + 11x6 + 15x7 + 22x8 + · · · + x56, so p8 = 22. Note that all of the coefficients prior to this are also correct, but the following coefficients are not necessarily the corresponding partition numbers.
Partitions of integers have some interesting properties. Let pd(n) be the number of partitions of n into distinct parts; let po(n) be the number of partitions into odd parts.
EXAMPLE 3.3.4 For n = 6, the partitions into distinct parts are 6, 5 + 1, 4 + 2, 3 + 2 + 1, so pd(6) = 4, and the partitions into odd parts are 5 + 1, 3 + 3, 3 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1, so po(6) = 4.
In fact, for every n, pd(n) = po(n), and we can see this by manipulating generating functions. The generating function for pd(n) is fd(x) = (1 + x)(1 + x2)(1 + x3) · · · = ∞ ∏ i=1 (1 + xi).
The generating function for po(n) is fo(x) = (1 + x + x2 + x3 + · · ·)(1 + x3 + x6 + x9 + · · ·) · · · = ∞ ∏ i=0 1 1 −x2i+1 .
We can write fd(x) = 1 −x2 1 −x · 1 −x4 1 −x2 · 1 −x6 1 −x3 · · · and notice that every numerator is eventually canceled by a denominator, leaving only the denominators containing odd powers of x, so fd(x) = fo(x).
3.3 Partitions of Integers 61 We can also use a recurrence relation to find the partition numbers, though in a somewhat less direct way than the binomial coefficients or the Bell numbers. Let pk(n) be the number of partitions of n into exactly k parts. We will find a recurrence relation to compute the pk(n), and then pn = n ∑ k=1 pk(n).
Now consider the partitions of n into k parts. Some of these partitions contain no 1s, like 3+3+4+6, a partition of 16 into 4 parts. Subtracting 1 from each part, we get a partition of n −k into k parts; for the example, this is 2 + 2 + 3 + 5. The remaining partitions of n into k parts contain a 1. If we remove the 1, we are left with a partition of n −1 into k −1 parts. This gives us a 1–1 correspondence between the partitions of n into k parts, and the partitions of n −k into k parts together with the partitions of n −1 into k −1 parts, so pk(n) = pk(n −k) + pk−1(n −1).
Using this recurrence we can build a triangle containing the pk(n), and the row sums of this triangle give the partition numbers. For all n, p1(n) = 1, which gives the first column of the triangle, after which the recurrence applies. Also, note that pk(n) = 0 when k > n and we let pk(0) = 0; these are needed in some cases to compute the pk(n −k) term of the recurrence. Here are the first few rows of the triangle; at the left are the row numbers, and at the right are the row sums, that is, the partition numbers. For the last row, each entry is the sum of the like-colored numbers in the previous rows. Note that beginning with p4(7) = 3 in the last row, pk(7) = pk−1(6), as pk(7 −k) = 0.
1 1 0 1 2 1 1 0 2 3 1 1 1 0 3 4 1 2 1 1 5 5 1 2 2 1 1 7 6 1 3 3 2 1 1 11 7 1 3 4 3 2 1 1 15 Yet another sometimes useful way to think of a partition is with a Ferrers diagram.
Each integer in the partition is represented by a row of dots, and the rows are ordered from longest on the top to shortest at the bottom. For example, the partition 3+ 3+4+5 would be represented by • • • • • • • • • • • • • • • The conjugate of a partition is the one corresponding to the Ferrers diagram produced by flipping the diagram for the original partition across the main diagonal, thus turning 62 Chapter 3 Generating Functions rows into columns and vice versa. For the diagram above, the conjugate is • • • • • • • • • • • • • • • with corresponding partition 1 + 2 + 4 + 4 + 4. This concept can occasionally make facts about partitions easier to see than otherwise. Here is a classic example: the number of partitions of n with largest part k is the same as the number of partitions into k parts, pk(n). The action of conjugation takes every partition of one type into a partition of the other: the conjugate of a partition into k parts is a partition with largest part k and vice versa. This establishes a 1–1 correspondence between partitions into k parts and partitions with largest part k.
Exercises 3.3.
1. Use generating functions to find p15.
2. Find the generating function for the number of partitions of an integer into distinct odd parts. Find the number of such partitions of 20.
3. Find the generating function for the number of partitions of an integer into distinct even parts. Find the number of such partitions of 30.
4. Find the number of partitions of 25 into odd parts.
5. Find the generating function for the number of partitions of an integer into k parts; that is, the coefficient of xn is the number of partitions of n into k parts.
6. Complete row 8 of the table for the pk(n), and verify that the row sum is 22, as we saw in example 3.3.3.
7. A partition of n is self-conjugate if its Ferrers diagram is symmetric around the main diagonal, so that its conjugate is itself. Show that the number of self-conjugate partitions of n is equal to the number of partitions of n into distinct odd parts.
3.4 Recurrence Relations A recurrence relation defines a sequence {ai}∞ i=0 by expressing a typical term an in terms of earlier terms, ai for i < n. For example, the famous Fibonacci sequence is defined by F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2.
Note that some initial values must be specified for the recurrence relation to define a unique sequence.
The starting index for the sequence need not be zero if it doesn’t make sense or some other starting index is more convenient. We saw two recurrence relations for the number 3.4 Recurrence Relations 63 of derangements of [n]: D1 = 0, Dn = nDn−1 + (−1)n.
and D1 = 0, D2 = 1, Dn = (n −1)(Dn−1 + Dn−2).
To “solve” a recurrence relation means to find a formula for an. There are a variety of methods for solving recurrence relations, with various advantages and disadvantages in particular cases. One method that works for some recurrence relations involves generating functions. The idea is simple, if the execution is not always: Let f(x) = ∞ ∑ i=0 aixi, that is, let f(x) be the generating function for {ai}∞ i=0. We now try to manipulate f(x), using the recurrence relation, until we can solve for f(x) explicitly. Finally, we hope that we can find a formula for the coefficients from the formula for f(x).
EXAMPLE 3.4.1 Solve F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2.
Let f(x) = ∞ ∑ i=0 Fixi and note that xf(x) = ∞ ∑ i=0 Fixi+1 = ∞ ∑ i=1 Fi−1xi.
To get the second sum we have simply “re-indexed” so that the index value gives the exponent on x, just as in the series for f(x). Likewise, x2f(x) = ∞ ∑ i=0 Fixi+2 = ∞ ∑ i=2 Fi−2xi.
In somewhat more suggestive form, we have f(x) = x + F2x2 + F3x3 + F4x4 + · · · xf(x) = x2 + F2x3 + F3x4 + · · · x2f(x) = x3 + F2x4 + · · · and combining the three equations we get f(x) −xf(x) −x2f(x) = x + (F2 −1)x2 + (F3 −F2 −1)x3 + (F4 −F3 −F2)x4 + · · · 64 Chapter 3 Generating Functions or in more compact form f(x) −xf(x) −x2f(x) = ∞ ∑ i=0 Fixi − ∞ ∑ i=1 Fi−1xi − ∞ ∑ i=2 Fi−2xi = x + ∞ ∑ i=2 (Fi −Fi−1 −Fi−2)xi = x + ∞ ∑ i=2 0 · xi = x, recalling that F0 = 0 and F1 = 1. Now f(x) = x 1 −x −x2 = −x x2 + x −1.
If we can find an explicit representation for the series for this function, we will have solved the recurrence relation. Here is where things could go wrong, but in this case it works out.
Let a and b be the roots of x2 + x −1; using the quadratic formula, we get a = −1 + √ 5 2 , b = −1 − √ 5 2 .
Borrowing a technique from calculus, we write −x x2 + x −1 = A x −a + B x −b.
Solving for A and B gives A = 1 − √ 5 2 √ 5 , B = −1 − √ 5 2 √ 5 .
Then −x x2 + x −1 = −A a 1 1 −x/a −B b 1 1 −x/b.
From calculus we know that 1 1 −x/a = ∞ ∑ i=0 (1/a)ixi and 1 1 −x/b = ∞ ∑ i=0 (1/b)ixi.
Finally, this means the coefficient of xi in the series for f(x) is Fi = −A a (1/a)i −B b (1/b)i.
3.4 Recurrence Relations 65 Simplifying gives Fi = 1 √ 5 (1 + √ 5 2 )i −1 √ 5 (1 − √ 5 2 )i .
Here’s an interesting feature of this expression: since |(1 − √ 5)/2| < 1, the limit of ((1 − √ 5)/2)i as i goes to infinity is 0. So when i is large enough, Fi = round ( 1 √ 5 (1 + √ 5 2 )i ) , that is, the first term rounded to the nearest integer. As it turns out, this is true starting with i = 0.
You can see how to do the entire solution in Sage.
We can also use this expression for Fn to compute limn→∞Fn/Fn−1.
lim n→∞ Fn Fn−1 = lim n→∞ 1 √ 5 ( 1+ √ 5 2 )n − 1 √ 5 ( 1− √ 5 2 )n 1 √ 5 ( 1+ √ 5 2 )n−1 − 1 √ 5 ( 1− √ 5 2 )n−1 = 1 + √ 5 2 .
This is the so-called “golden ratio”.
Exercises 3.4.
1. Find the generating function for the solutions to hn = 4hn−1 −3hn−2, h0 = 2, h1 = 5, and use it to find a formula for hn.
2. Find the generating function for the solutions to hn = 3hn−1 + 4hn−2, h0 = h1 = 1, and use it to find a formula for hn.
3. Find the generating function for the solutions to hn = 2hn−1 + 3n, h0 = 0, and use it to find a formula for hn.
4. Find the generating function for the solutions to hn = 4hn−2, h0 = 0, h1 = 1, and use it to find a formula for hn. (It is easy to discover this formula directly; the point here is to see that the generating function approach gives the correct answer.) 5. Find the generating function for the solutions to hn = hn−1 + hn−2, h0 = 1, h1 = 3, and use it to find a formula for hn.
6. Find the generating function for the solutions to hn = 9hn−1 −26hn−2 + 24hn−3, h0 = 0, h1 = 1, h2 = −1, and use it to find a formula for hn.
7. Find the generating function for the solutions to hn = 3hn−1 + 4hn−2, h0 = 0, h1 = 1, and use it to find a formula for hn.
8. Find a recursion for the number of ways to place flags on an n foot pole, where we have red flags that are 2 feet high, blue flags that are 1 foot high, and yellow flags that are 1 foot high; the heights of the flags must add up to n. Solve the recursion.
9. In Fibonacci’s original problem, a farmer started with one (newborn) pair of rabbits at month 0. After each pair of rabbits was one month old, they produced another pair each month in perpetuity.
Thus, after 1 month, he had the original pair, after two months 2 pairs, 66 Chapter 3 Generating Functions three months, 3 pairs, four months, 5 pairs, etc. The number of pairs of rabbits satisfies hn = hn−1 + hn−2, h0 = h1 = 1. (Note that this is slightly different than our definition, in which h0 = 0.) Suppose instead that each mature pair gives birth to two pairs of rabbits. The sequence for the number of pairs of rabbits now starts out h0 = 1, h1 = 1, h2 = 3, h3 = 5, h4 = 11.
Set up and solve a recurrence relation for the number of pairs of rabbits. Show also that the sequence statisfies hn = 2hn−1 + (−1)n.
10. Explain why lim n→∞ Fn Fn−1 = lim n→∞ 1 √ 5 ( 1+ √ 5 2 )n − 1 √ 5 ( 1− √ 5 2 )n 1 √ 5 ( 1+ √ 5 2 )n−1 − 1 √ 5 ( 1− √ 5 2 )n−1 = 1 + √ 5 2 .
3.5 Catalan Numbers A rooted binary tree is a type of graph that is particularly of interest in some areas of computer science. A typical rooted binary tree is shown in figure 3.5.1. The root is the topmost vertex. The vertices below a vertex and connected to it by an edge are the children of the vertex. It is a binary tree because all vertices have 0, 1, or 2 children. How many different rooted binary trees are there with n vertices?
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• Figure 3.5.1 A rooted binary tree.
Let us denote this number by Cn; these are the Catalan numbers. For convenience, we allow a rooted binary tree to be empty, and let C0 = 1. Then it is easy to see that C1 = 1 and C2 = 2, and not hard to see that C3 = 5. Notice that any rooted binary tree on at least one vertex can be viewed as two (possibly empty) binary trees joined into a new tree by introducing a new root vertex and making the children of this root the two roots of the original trees; see figure 3.5.2. (To make the empty tree a child of the new vertex, simply do nothing, that is, omit the corresponding child.) Thus, to make all possible binary trees with n vertices, we start with a root vertex, and then for its two children insert rooted binary trees on k and l vertices, with k + l = n −1, 3.5 Catalan Numbers 67 .
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• Figure 3.5.2 Producing a new tree from smaller trees.
for all possible choices of the smaller trees. Now we can write Cn = n−1 ∑ i=0 CiCn−i−1.
For example, since we know that C0 = C1 = 1 and C2 = 2, C3 = C0C2 + C1C1 + C2C0 = 1 · 2 + 1 · 1 + 2 · 1 = 5, as mentioned above. Once we know the trees on 0, 1, and 2 vertices, we can combine them in all possible ways to list the trees on 3 vertices, as shown in figure 3.5.3. Note that the first two trees have no left child, since the only tree on 0 vertices is empty, and likewise the last two have no right child.
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• Figure 3.5.3 The 3-vertex binary rooted trees.
Now we use a generating function to find a formula for Cn.
Let f = ∑∞ i=0 Cixi.
Now consider f 2: the coefficient of the term xn in the expansion of f 2 is ∑n i=0 CiCn−i, corresponding to all possible ways to multiply terms of f to get an xn term: C0 · Cnxn + C1x · Cn−1xn−1 + C2x2 · Cn−2xn−2 + · · · + Cnxn · C0.
Now we recognize this as precisely the sum that gives Cn+1, so f 2 = ∑∞ n=0 Cn+1xn. If we multiply this by x and add 1 (which is C0) we get exactly f again, that is, xf 2 + 1 = f or xf 2 −f + 1 = 0; here 0 is the zero function, that is, xf 2 −f + 1 is 0 for all x. Using the 68 Chapter 3 Generating Functions quadratic formula, f = 1 ± √1 −4x 2x , as long as x ̸= 0. It is not hard to see that as x approaches 0, 1 + √1 −4x 2x goes to infinity while 1 −√1 −4x 2x goes to 1. Since we know f(0) = C0 = 1, this is the f we want.
Now by Newton’s Binomial Theorem 3.1.1, we can expand √ 1 −4x = (1 + (−4x))1/2 = ∞ ∑ n=0 (1/2 n ) (−4x)n.
Then 1 −√1 −4x 2x = ∞ ∑ n=1 −1 2 (1/2 n ) (−4)nxn−1 = ∞ ∑ n=0 −1 2 ( 1/2 n + 1 ) (−4)n+1xn.
Expanding the binomial coefficient ( 1/2 n+1 ) and reorganizing the expression, we discover that Cn = −1 2 ( 1/2 n + 1 ) (−4)n+1 = 1 n + 1 (2n n ) .
In exercise 7 in section 1.2, we saw that the number of properly matched sequences of parentheses of length 2n is (2n n ) − ( 2n n+1 ) , and called this Cn. It is not difficult to see that (2n n ) − ( 2n n + 1 ) = 1 n + 1 (2n n ) , so the formulas are in agreement.
Temporarily let An be the number of properly matched sequences of parentheses of length 2n, so from the exercise we know An = (2n n ) − ( 2n n+1 ) . It is possible to see directly that A0 = A1 = 1 and that the numbers An satisfy the same recurrence relation as do the Cn, which implies that An = Cn, without manipulating the generating function.
There are many counting problems whose answers turns out to be the Catalan numbers.
Enumerative Combinatorics: Volume 2, by Richard Stanley, contains a large number of examples.
3.5 Catalan Numbers 69 Exercises 3.5.
1. Show that ( 2n n ) − ( 2n n + 1 ) = 1 n + 1 ( 2n n ) .
2. Find a simple expression f(n) so that Cn+1 = f(n)Cn. Use this to compute C1, . . . , C6 from C0.
3. Show that if An is the number of properly matched sequences of parentheses of length 2n, then An = n−1 ∑ i=0 AiAn−i−1.
Do this in the same style that we used for the number of rooted binary trees: Given all the sequences of shorter length, explain how to combine them to produce the sequences of length 2n, in such a way that the sum clearly counts the number of sequences. Hint: Prove the following lemma: If s is a properly matched sequence of parentheses of length 2n, s may be written uniquely in the form (s1)s2, where s1 and s2 are properly matched sequences of parentheses whose lengths add to 2n−2. For example, (())() = ([()])[()] and ()(()) = ([ ])[(())], with the sequences s1 and s2 indicated by [ ]. Note that s1 and s2 are allowed to be empty sequences, with length 0.
4. Consider a “staircase” as shown below. A path from A to B consists of a sequence of edges starting at A, ending at B, and proceeding only up or right; all paths are of length 6. One such path is indicated by arrows. The staircase shown is a “3 × 3” staircase. How many paths are there in an n × n staircase?
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A B 5. A convex polygon with n ≥3 sides can be divided into triangles by inserting n −3 non-intersecting diagonals. In how many different ways can this be done? The possibilities for n = 5 are shown.
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6. A partition of a set S is a collection of non-empty subsets Ai ⊆S, 1 ≤i ≤k (the parts of the partition), such that ∪k i=1 Ai = S and for every i ̸= j, Ai ∩Aj = ∅. For example, one partition of {1, 2, 3, 4, 5} is {{1, 3}, {4}, {2, 5}}.
Suppose the integers 1, 2, . . . , n are arranged on a circle, in order around the circle. A partition of {1, 2, . . . , n} is a non-crossing partition if it satisfies this additional property: If w and x are in some part Ai, and y and z are in a different part Aj, then the line joining w to x does not cross the line joining y to z. The partition above, {1, 3}, {4}, {2, 5}, is not a non-crossing partition, as the the line 1–3 crosses the line 2–5.
70 Chapter 3 Generating Functions Find the number of non-crossing partitions of {1, 2, . . . , n}.
Recall from section 1.4 that the Bell numbers count all of the partitions of {1, 2, . . . , n}.
Hence, this exercise gives us a lower bound on the total number of partitions.
7. Consider a set of 2n people sitting around a table. In how many ways can we arrange for each person to shake hands with another person at the table such that no two handshakes cross?
4 Systems of Distinct Representatives Suppose that the student clubs at a college each send a representative to the student government from among the members of the club. No person may represent more than one club; is this possible? It is certainly possible sometimes, for example when no student belongs to two clubs.
It is not hard to see that it could be impossible.
So the first substantive question is: is there anything useful or interesting we can say about under what conditions it is possible to choose such representatives.
We turn this into a more mathematical situation: DEFINITION 4.0.1 Suppose that A1, A2, . . . , An are sets, which we refer to as a set system. A (complete) system of distinct representatives is a sequence {x1, x2, . . . xn} such that xi ∈Ai for all i, and no two of the xi are the same. A (partial) system of distinct representatives is a sequence of distinct elements {x1, x2, . . . xk} such that xi ∈Aji, where j1, j2, . . . , jk are distinct integers in [n].
In standard usage, “system of distinct representatives” means “complete system of dis-tinct representatives”, but it will be convenient to let “system of distinct representatives” mean either a complete or partial system of distinct representatives depending on context.
We usually abbreviate “system of distinct representatives” as sdr.
We will analyze this problem in two ways, combinatorially and using graph theory.
71 72 Chapter 4 Systems of Distinct Representatives 4.1 Existence of SDRs In this section, sdr means complete sdr. It is easy to see that not every collection of sets has an sdr. For example, A1 = {a, b}, A2 = {a, b}, A3 = {a, b}.
The problem is clear: there are only two possible representatives, so a set of three distinct representatives cannot be found. This example is a bit more general than it may at first appear. Consider A1 = {a, b}, A2 = {a, b}, A3 = {a, b}, A4 = {b, c, d, e}.
Now the total number of possible representatives is 5, and we only need 4. Nevertheless, this is impossible, because the first three sets have no sdr considered by themselves. Thus the following condition, called Hall’s Condition, is clearly necessary for the existence of an sdr: For every k ≥1, and every set {i1, i2, . . . , ik} ⊆[n], | ∪k j=1 Aij| ≥k. That is, the number of possible representatives in any collection of sets must be at least as large as the number of sets. Both examples fail to have this property because |A1 ∪A2 ∪A3| = 2 < 3.
Remarkably, this condition is both necessary and sufficient.
THEOREM 4.1.1 Hall’s Theorem A collection of sets A1, A2, . . . , An has an sdr if and only if for every k ≥1, and every set {i1, i2, . . . , ik} ⊆[n], | ∪k j=1 Aij| ≥k.
Proof.
We already know the condition is necessary, so we prove sufficiency by induction on n.
Suppose n = 1; the condition is simply that |A1| ≥1.
If this is true then A1 is non-empty and so there is an sdr. This establishes the base case.
Now suppose that the theorem is true for a collection of k < n sets, and suppose we have sets A1, A2, . . . , An satisfying Hall’s Condition. We need to show there is an sdr.
Suppose first that for every k < n and every {i1, i2, . . . , ik} ⊆[n], that | ∪k j=1 Aij| ≥ k + 1, that is, that these unions are larger than required. Pick any element xn ∈An, and define Bi = Ai{xn} for each i < n. Consider the collection of sets B1, . . . , Bn−1, and any union ∪k j=1 Bij of a subcollection of the sets. There are two possibilities: either ∪k j=1 Bij = ∪k j=1 Aij or ∪k j=1 Bij = ∪k j=1 Aij{xn}, so that | ∪k j=1 Bij| = | ∪k j=1 Aij| or | ∪k j=1 Bij| = | ∪k j=1 Aij|−1. In either case, since | ∪k j=1 Aij| ≥k+1, | ∪k j=1 Bij| ≥k. Thus, by the induction hypothesis, the collection B1, . . . , Bn−1 has an sdr {x1, x2, . . . , xn−1}, and for every i < n, xi ̸= xn, by the definition of the Bi. Thus {x1, x2, . . . , xn} is an sdr for A1, A2, . . . , An.
If it is not true that for every k < n and every {i1, i2, . . . , ik} ⊆[n], | ∪k j=1 Aij| ≥k+1, then for some k < n and {i1, i2, . . . , ik}, | ∪k j=1 Aij| = k. Without loss of generality, we 4.1 Existence of SDRs 73 may assume that | ∪k j=1 Aj| = k. By the induction hypothesis, A1, A2, . . . , Ak has an sdr, {x1, . . . , xk}.
Define Bi = Ai\ ∪k j=1 Aj for i > k.
Suppose that {xk+1, . . . , xn} is an sdr for Bk+1, . . . , Bn; then it is also an sdr for Ak+1, . . . , An. Moreover, {x1, . . . , xn} is an sdr for A1, . . . , An. Thus, to finish the proof it suffices to show that Bk+1, . . . , Bn has an sdr.
The number of sets here is n −k < n, so we need only show that the sets satisfy Hall’s Condition.
So consider some sets Bi1, Bi2, . . . , Bil. First we notice that |A1 ∪A2 ∪· · · ∪Ak ∪Bi1 ∪Bi2 ∪· · · Bil| = k + |Bi1 ∪Bi2 ∪· · · Bil|.
Also |A1 ∪A2 ∪· · · ∪Ak ∪Bi1 ∪Bi2 ∪· · · Bil| = |A1 ∪A2 ∪· · · ∪Ak ∪Ai1 ∪Ai2 ∪· · · Ail| and |A1 ∪A2 ∪· · · ∪Ak ∪Ai1 ∪Ai2 ∪· · · Ail| ≥k + l.
Putting these together gives k + |Bi1 ∪Bi2 ∪· · · ∪Bil| ≥k + l |Bi1 ∪Bi2 ∪· · · ∪Bil| ≥l .
Thus, Bk+1, . . . , Bn has an sdr, which finishes the proof.
Exercises 4.1.
1. How many different systems of distinct representatives are there for A1 = {1, 2}, A2 = {2, 3}, . . . , An = {n, 1}?
2. How many different systems of distinct representatives are there for the sets Ai = [n]\i, i = 1, 2, . . . , n, n ≥2?
3. Suppose the set system A1, A2, . . . , An has an sdr, and that x ∈Ai. Show the set system has an sdr containing x. Show that x cannot necessarily be chosen to represent Ai.
4. Suppose the set system A1, A2, . . . , An satisfies | ∪k j=1 Aij| ≥k + 1 for every 1 ≤k < n and {i1, i2, . . . , ik} ⊆[n], and that x ∈Ai. Show the set system has an sdr in which x represents Ai.
5. An m×n chessboard, with m even and both m and n at least 2, has one white and one black square removed. Show that the board can be covered by dominoes.
74 Chapter 4 Systems of Distinct Representatives 4.2 Partial SDRs In this section, sdr means partial sdr.
If there is no complete sdr, we naturally want to know how many of the n sets can be represented, that is, what is the largest value of m so that some m of the sets have a complete sdr.
Since there is no complete sdr, there are sets Ai1, Ai2, . . . , Aik such that | ∪k j=1 Aij| = l < k. Clearly at most l of these k sets have a complete sdr, so no sdr for A1, A2, . . . , An can be larger than n −k + l. Thus, m can be no larger than the minimum value, over all k and all collections of sets Ai1, Ai2, . . . , Aik, of n−k+| ∪k j=1 Aij|.
Note that if | ∪k j=1 Aij| > k, n −k + | ∪k j=1 Aij| > n, which tells us nothing. If k = 0, n −k + | ∪k j=1 Aij| = n (because empty unions are empty), so we are guaranteed that the minimum is never greater than n. In fact the minimum value of the expression is exactly the size of a largest sdr.
THEOREM 4.2.1 The maximum size of an sdr for the sets A1, A2, . . . , An is the minimum value, for 0 ≤k ≤n and sets Ai1, Ai2, . . . , Aik, of n −k + | ∪k j=1 Aij|.
Proof.
Since no sdr can be larger than this minimum value, it suffices to show that we can find an sdr whose size is this minimum. The proof is by induction on n; the case n = 1 is easy.
Suppose first that the minimum value is n, so that for all k and all collections of sets Ai1, Ai2, . . . , Aik, n −k + | k ∪ j=1 Aij| ≥n.
Then rearranging we see that | k ∪ j=1 Aij| ≥k, so by Hall’s Theorem (4.1.1), there is an sdr of size n.
Note that the minimum value of n −k + | ∪k j=1 Aij| occurs when | ∪k j=1 Aij| −k is a minimum, that is min(n −k + | k ∪ j=1 Aij|) = n + min(| k ∪ j=1 Aij| −k).
Suppose now that the minimum m is less than n, and that m = n −k + | ∪k j=1 Aij|, with 0 < k < n. Let Bj = Aij; since k < n, the induction hypothesis applies to the sets B1, . . . , Bk. Since each set Bj is Aij, | ∪l j=1 Bhj| −l ≥| ∪k j=1 Aij| −k, for all l and Bh1, . . . , Bhl. Thus, the minimum value of | ∪l j=1 Bhj| −l, over all l and Bh1, . . . , Bhl, is 4.2 Partial SDRs 75 | ∪k j=1 Bj| −k = | ∪k j=1 Aij| −k, so by the induction hypothesis, the sets Ai1, Ai2, . . . , Aik have an sdr of size k −k + | ∪k j=1 Aij| = | ∪k j=1 Aij| = m −n + k, {x1, . . . , xm−n+k}.
Now consider the n −k sets consisting of those original sets not in Ai1, Ai2, . . . , Aik, that is, {Ai | i / ∈{i1, . . . , ik}}. Let Ci = Ai\ ∪k j=1 Aij for i not in i1, i2, . . . , ik. Consider some sets Cg1, Cg2, . . . , Cgl. If | ∪l j=1 Cgj| < l then | ∪l j=1 Cgj| −l < 0 and n −k + | k ∪ j=1 Aij| > n −k −l + | l ∪ j=1 Cgj| + | k ∪ j=1 Aij| ≥n −(k + l) + |Cg1 ∪· · · ∪Cgl ∪Ai1 ∪· · · ∪Aik| = n −(k + l) + |Ag1 ∪· · · ∪Agl ∪Ai1 ∪· · · ∪Aik|, contradicting the fact that n−k+| ∪k j=1 Aij| is a minimum. Thus by Hall’s Theorem (4.1.1), the sets Cg1, Cg2, . . . , Cgn−k have a complete sdr {y1, . . . , yn−k}. By the definition of the sets Ci, {x1, . . . , xm−n+k} ∩{y1, . . . , yn−k} = ∅, so {x1, . . . , xm−n+k} ∪{y1, . . . , yn−k} is an sdr of size m −n + k + n −k = m as desired.
Finally, suppose that the minimum value of n−k+| ∪k j=1 Aij| occurs only when k = n, so we want an sdr of size n −n + | n ∪ j=1 Aj| = | n ∪ j=1 Aj|.
Then n −(n −1) + | n−1 ∪ j=1 Aj| > | n ∪ j=1 Aj| 1 + | n−1 ∪ j=1 Aj| > | n ∪ j=1 Aj| | n−1 ∪ j=1 Aj| ≥| n ∪ j=1 Aj|.
Since | ∪n−1 j=1 Aj| ≤| ∪n j=1 Aj|, | ∪n−1 j=1 Aj| = | ∪n j=1 Aj|. By the induction hypothesis, the theorem applies to the sets A1, A2, . . . , An−1. If the minimum of (n −1) −l + | ∪l j=1 Aij| occurs when l = n −1, then there is an sdr of size (n −1) −(n −1) + | ∪n−1 j=1 Aj| = | ∪n−1 j=1 Aj| = | ∪n j=1 Aj|, as desired.
If the minimum occurs when l < n −1 and not when l = n −1, then (n −1) −l + | l ∪ j=1 Aij| < | n−1 ∪ j=1 Aj| n −l + | l ∪ j=1 Aij| < | n−1 ∪ j=1 Aj| + 1 76 Chapter 4 Systems of Distinct Representatives and by assumption n −l + | l ∪ j=1 Aij| > | n ∪ j=1 Aj|.
Thus | n ∪ j=1 Aj| < n −l + | l ∪ j=1 Aij| < | n−1 ∪ j=1 Aj| + 1 = | n ∪ j=1 Aj| + 1.
This means that there is an integer strictly between two consecutive integers, a contradic-tion. This completes the proof.
While this theorem provides a method to calculate the size of a maximum sdr, the method is hardly efficient: it requires looking at all possible collections of the sets. It also does not provide a way to find an actual sdr, that is, the actual representatives. We will fix these problems in the last two sections of this chapter.
Exercises 4.2.
1. Find the size of a maximum sdr for A1 = {a, b, c}, A2 = {a, b, c, d, e}, A3 = {a, b}, A4 = {b, c}, A5 = {a}, A6 = {a, c, e}.
Justify your answer.
4.3 Latin Squares DEFINITION 4.3.1 A Latin square of order n is an n × n grid filled with n symbols so that each symbol appears once in each row and column.
EXAMPLE 4.3.2 Here is a Latin square of order 4: ♡ ♣ ♠ ♢ ♣ ♠ ♢ ♡ ♠ ♢ ♡ ♣ ♢ ♡ ♣ ♠ Usually we use the integers 1 . . . n for the symbols.
There are many, many Latin squares of order n, so it pays to limit the number by agreeing not to count Latin squares 4.3 Latin Squares 77 that are “really the same” as different. The simplest way to do this is to consider reduced Latin squares. A reduced Latin square is one in which the first row is 1 . . . n (in order) and the first column is likewise 1 . . . n.
EXAMPLE 4.3.3 Consider this Latin square: 4 2 3 1 2 4 1 3 1 3 4 2 3 1 2 4 The order of the rows and columns is not really important to the idea of a Latin square.
If we reorder the rows and columns, we can consider the result to be in essence the same Latin square. By reordering the columns, we can turn the square above into this: 1 2 3 4 3 4 1 2 2 3 4 1 4 1 2 3 Then we can swap rows two and three: 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 This Latin square is in reduced form, and is essentially the same as the original.
Another simple way to change the appearance of a Latin square without changing its essential structure is to interchange the symbols.
EXAMPLE 4.3.4 Starting with the same Latin square as before: 4 2 3 1 2 4 1 3 1 3 4 2 3 1 2 4 we can interchange the symbols 1 and 4 to get: 78 Chapter 4 Systems of Distinct Representatives 1 2 3 4 2 1 4 3 4 3 1 2 3 4 2 1 Now if we swap rows three and four we get: 1 2 3 4 2 1 4 3 3 4 2 1 4 3 1 2 Notice that this Latin square is in reduced form, but it is not the same as the reduced form from the previous example, even though we started with the same Latin square. Thus, we may want to consider some reduced Latin squares to be the same as each other.
DEFINITION 4.3.5 Two Latin squares are isotopic if each can be turned into the other by permuting the rows, columns, and symbols. This isotopy relation is an equivalence relation; the equivalence classes are the isotopy classes.
Latin squares are apparently quite difficult to count without substantial computing power. According to Wikipedia, the number of Latin squares is known only up to n = 11.
Here are the first few values for all Latin squares, reduced Latin squares, and non-isotopic Latin squares (that is, the number of isotopy classes): n All Reduced Non-isotopic 1 1 1 1 2 2 1 1 3 12 1 1 4 576 4 2 5 161280 56 2 How can we produce a Latin square? If you know what a group is, you should know that the multiplication table of any finite group is a Latin square. (Also, any Latin square is the multiplication table of a quasigroup.) Even if you have not encountered groups by that name, you may know of some. For example, considering the integers modulo n under addition, the addition table is a Latin square.
EXAMPLE 4.3.6 Here is the addition table for the integers modulo 6: 4.3 Latin Squares 79 0 1 2 3 4 5 1 2 3 4 5 0 2 3 4 5 0 1 3 4 5 0 1 2 4 5 0 1 2 3 5 0 1 2 3 4 EXAMPLE 4.3.7 Here is another way to potentially generate many Latin squares.
Start with first row 1, . . . , n. Consider the sets Ai = [n]{i}. From exercise 1 in section 4.1 we know that this set system has many sdrs; if x1, x2, . . . , xn is an sdr, we may use it for row two. In general, after we have chosen rows 1, . . . , j, we let Ai be the set of integers that have not yet been chosen for column i. This set system has an sdr, which we use for row j + 1.
DEFINITION 4.3.8 Suppose A and B are two Latin squares of order n, with entries Ai,j and Bi,j in row i and column j. Form the matrix M with entries Mi,j = (Ai,j, Bi,j); we will denote this operation as M = A ∪B. We say that A and B are orthogonal if M contains all n2 ordered pairs (a, b), 1 ≤a ≤n, 1 ≤b ≤n, that is, all elements of {0, 1, . . . , n −1} × {0, 1, . . . , n −1}.
As we will see, it is easy to find orthogonal Latin squares of order n if n is odd; not too hard to find orthogonal Latin squares of order 4k, and difficult but possible to find orthogonal Latin squares of order 4k+2, with the exception of orders 2 and 6. In the 1700s, Euler showed that there are orthogonal Latin squares of all orders except of order 4k + 2, and he conjectured that there are no orthogonal Latin squares of of order 6. In 1901, the amateur mathematician Gaston Tarry showed that indeed there are none of order 6, by showing that all possibilities for such Latin squares failed to be orthogonal. In 1959 it was finally shown that there are orthogonal Latin squares of all other orders.
THEOREM 4.3.9 There are pairs of orthogonal Latin squares of order n when n is odd.
Proof.
This proof can be shortened by using ideas of group theory, but we will present a self-contained version. Consider the addition table for addition mod n: 80 Chapter 4 Systems of Distinct Representatives 0 · · · j · · · n −1 0 0 · · · j · · · n −1 .
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n −1 n −1 · · · n + j −1 · · · n −2 We claim first that this (without the first row and column, of course) is a Latin square with symbols 0, 1, . . . , n −1. Consider two entries in row i, say i + j and i + k. If i + j ≡i + j (mod n), then j ≡k, so j = k. Thus, all entries of row i are distinct, so each of 0, 1, . . . , n−1 appears exactly once in row i. The proof that each appears once in any column is similar.
Call this Latin square A. (Note that so far everything is true whether n is odd or even.) Now form a new square B with entries Bi,j = A2i,j = 2i + j, where by 2i and 2i + j we mean those values mod n. Thus row i of B is the same as row 2i of A. Now we claim that in fact the rows of B are exactly the rows of A, in a different order. To do this, it suffices to show that if 2i ≡2k (mod n), then i = k. This implies that all the rows of B are distinct, and hence must be all the rows of A.
Suppose without loss of generality that i ≥k. If 2i ≡2k (mod n) then n | 2(i −k).
Since n is odd, n | (i −k). Since i and k are in 0, 1, . . . , n −1, 0 ≤i −k ≤n −1. Of these values, only 0 is divisible by n, so i −k = 0. Thus B is also a Latin square.
To show that A ∪B contains all n2 elements of {0, 1, . . . , n −1} × {0, 1, . . . , n −1}, it suffices to show that no two elements of A∪B are the same. Suppose that (i1+j1, 2i1+j1) = (i2 + j2, 2i2 + j2) (arithmetic is mod n). Then by subtracting equations, i1 = i2; with the first equation this implies j1 = j2.
EXAMPLE 4.3.10 When n = 3, 0 1 2 1 2 0 2 0 1 ∪ 0 1 2 2 0 1 1 2 0 = (0, 0) (1, 1) (2, 2) (1, 2) (2, 0) (0, 1) (2, 1) (0, 2) (1, 0) .
One obvious approach to constructing Latin squares, and pairs of orthogonal Latin squares, is to start with smaller Latin squares and use them to produce larger ones. We will produce a Latin square of order mn from a Latin square of order m and one of order n.
Let A be a Latin square of order m with symbols 1, . . . , m, and B one of order n with symbols 1, . . . , n. Let ci,j, 1 ≤i ≤m, 1 ≤j ≤n, be mn new symbols. Form an mn × mn grid by replacing each entry of B with a copy of A. Then replace each entry i in this copy 4.3 Latin Squares 81 of A with ci,j, where j is the entry of B that was replaced. We denote this new Latin square A × B. Here is an example, combining a 4 × 4 Latin square with a 3 × 3 Latin square to form a 12 × 12 Latin square: 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 × 1 2 3 2 3 1 3 1 2 = c1,1 c2,1 c3,1 c4,1 c1,2 c2,2 c3,2 c4,2 c1,3 c2,3 c3,3 c4,3 c2,1 c3,1 c4,1 c1,1 c2,2 c3,2 c4,2 c1,2 c2,3 c3,3 c4,3 c1,3 c3,1 c4,1 c1,1 c2,1 c3,2 c4,2 c1,2 c2,2 c3,3 c4,3 c1,3 c2,3 c4,1 c1,1 c2,1 c3,1 c4,2 c1,2 c2,2 c3,2 c4,3 c1,3 c2,3 c3,3 c1,2 c2,2 c3,2 c4,2 c1,3 c2,3 c3,3 c4,3 c1,1 c2,1 c3,1 c4,1 c2,2 c3,2 c4,2 c1,2 c2,3 c3,3 c4,3 c1,3 c2,1 c3,1 c4,1 c1,1 c3,2 c4,2 c1,2 c2,2 c3,3 c4,3 c1,3 c2,3 c3,1 c4,1 c1,1 c2,1 c4,2 c1,2 c2,2 c3,2 c4,3 c1,3 c2,3 c3,3 c4,1 c1,1 c2,1 c3,1 c1,3 c2,3 c3,3 c4,3 c1,1 c2,1 c3,1 c4,1 c1,2 c2,2 c3,2 c4,2 c2,3 c3,3 c4,3 c1,3 c2,1 c3,1 c4,1 c1,1 c2,2 c3,2 c4,2 c1,2 c3,3 c4,3 c1,3 c2,3 c3,1 c4,1 c1,1 c2,1 c3,2 c4,2 c1,2 c2,2 c4,3 c1,3 c2,3 c3,3 c4,1 c1,1 c2,1 c3,1 c4,2 c1,2 c2,2 c3,2 THEOREM 4.3.11 If A and B are Latin squares, so is A × B.
Proof.
Consider two symbols ci,j and ck,l in the same row. If the positions containing these symbols are in the same copy of A, then i ̸= k, since A is a Latin square, and so the symbols ci,j and ck,l are distinct. Otherwise, j ̸= l, since B is a Latin square. The argument is the same for columns.
Remarkably, this operation preserves orthogonality: THEOREM 4.3.12 If A1 and A2 are Latin squares of order m, B1 and B2 are Latin squares of order n, A1 and A2 are orthogonal, and B1 and B2 are orthogonal, then A1 ×B1 is orthogonal to A1 × B2.
Proof.
We denote the contents of Ai × Bi by Ci(w, x, y, z), meaning the entry in row w and column x of the copy of Ai that replaced the entry in row y and column z of Bi, which we denote Bi(y, z). We use Ai(w, x) to denote the entry in row w and column x of Ai.
Suppose that (C1(w, x, y, z), C2(w, x, y, z)) = (C1(w′, x′, y′, z′), C2(w′, x′, y′, z′)), where (w, x, y, z) ̸= (w′, x′, y′, z′).
Either (w, x) ̸= (w′, x′) or (y, z) ̸= (y′, z′).
If the lat-ter, then (B1(y, z), B2(y, z)) = (B1(y′, z′), B2(y′, z′)), a contradiction, since B1 is or-thogonal to B2.
Hence (y, z) = (y′, z′) and (w, x) ̸= (w′, x′).
But this implies that (A1(w, x), A2(w, x)) = (A1(w′, x′), A2(w′, x′)), a contradiction. Hence A1 × B1 is orthog-onal to A1 × B2.
82 Chapter 4 Systems of Distinct Representatives We want to construct orthogonal Latin squares of order 4k.
Write 4k = 2m · n, where n is odd and m ≥2. We know there are orthogonal Latin squares of order n, by theorem 4.3.9. If there are orthogonal Latin squares of order 2m, then by theorem 4.3.12 we can construct orthogonal Latin squares of order 4k = 2m · n.
To get a Latin square of order 2m, we also use theorem 4.3.12.
It suffices to find two orthogonal Latin squares of order 4 = 22 and two of order 8 = 23. Then repeated application of theorem 4.3.12 allows us to build orthogonal Latin squares of order 2m, m ≥2.
Two orthogonal Latin squares of order 4: 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 3 4 1 2 4 3 2 1 2 1 4 3 , and two of order 8: 1 3 4 5 6 7 8 2 5 2 7 1 8 4 6 3 6 4 3 8 1 2 5 7 7 8 5 4 2 1 3 6 8 7 2 6 5 3 1 4 2 5 8 3 7 6 4 1 3 1 6 2 4 8 7 5 4 6 1 7 3 5 2 8 1 4 5 6 7 8 2 3 8 2 6 5 3 1 4 7 2 8 3 7 6 4 1 5 3 6 2 4 8 7 5 1 4 1 7 3 5 2 8 6 5 7 1 8 4 6 3 2 6 3 8 1 2 5 7 4 7 5 4 2 1 3 6 8 .
Exercises 4.3.
1. Show that there is only one reduced Latin square of order 3.
2. Verify that the isotopy relation is an equivalence relation.
3. Find all 4 reduced Latin squares of order 4. Show that there are at most 2 isotopy classes for order 4.
4. Show that the second set system defined in example 4.3.7 has an sdr as claimed.
5. Show that there are no orthogonal Latin squares of order 2.
6. Find the two orthogonal Latin squares of order 5 as described in theorem 4.3.9. Show your answer as in example 4.3.10.
7. Prove that to construct orthogonal Latin squares of order 2m, m ≥2, it suffices to find two orthogonal Latin squares of order 4 = 22 and two of order 8 = 23.
8. An n × n Latin square A is symmetric if it is symmetric around the main diagonal, that is, Ai,j = Aj,i for all i and j. It is easy to find symmetric Latin squares: every addition table modulo n is an example, as in example 4.3.6. A Latin square is idempotent if every symbol appears on the main diagonal. Show that if A is both symmetric and idempotent, then n is odd. Find a 5 × 5 symmetric, idempotent Latin square.
4.4 Introduction to Graph Theory 83 9. The transpose A⊤of a Latin square A is the reflection of A across the main diagonal, so that A⊤ i,j = Aj,i. A Latin square is self-orthogonal if A is orthogonal to A⊤. Show that there is no self-orthogonal Latin square of order 3. Find one of order 4.
4.4 Introduction to Graph Theory We can interpret the sdr problem as a problem about graphs. Given sets A1, A2, . . . , An, with ∪n i=1 Ai = {x1, x2, . . . , xm}, we define a graph with n + m vertices as follows: The vertices are labeled {A1, A2, . . . , An, x1, x2, . . . xm}, and the edges are {{Ai, xj} | xj ∈Ai}.
EXAMPLE 4.4.1 Let A1 = {a, b, c, d}, A2 = {a, c, e, g}, A3 = {b, d}, and A4 = {a, f, g}.
The corresponding graph is shown in figure 4.4.1.
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Before exploring this idea, we introduce a few basic concepts about graphs. If two vertices in a graph are connected by an edge, we say the vertices are adjacent. If a vertex v is an endpoint of edge e, we say they are incident. The set of vertices adjacent to v is called the neighborhood of v, denoted N(v). This is sometimes called the open neighborhood of v to distinguish it from the closed neighborhood of v, N[v] = N(v)∪{v}. The degree of a vertex v is the number of edges incident with v; it is denoted d(v).
Some simple types of graph come up often: A path is a graph Pn on vertices v1, v2, . . . , vn, with edges {vi, vi+1} for 1 ≤i ≤n −1, and no other edges. A cycle is a graph Cn on vertices v1, v2, . . . , vn with edges {vi, v1+(i mod n)} for 1 ≤i ≤n, and no other edges; this is a path in which the first and last vertices have been joined by an edge. (Generally, we require that a cycle have at least three vertices. If it has two, then the two are joined by two distinct edges; when a graph has more than one edge with the same endpoints it is called a multigraph. If a cycle has one vertex, there is an edge, called a loop, in which a single vertex serves as both endpoints.) The length of a path or cycle is the number of edges in the graph. For example, P1 has length 0, C1 has length 1. A complete graph Kn is a graph on v1, v2, . . . , vn in which every two distinct vertices are joined by an edge.
See figure 4.4.2 for examples.
The graph in figure 4.4.1 is a bipartite graph.
84 Chapter 4 Systems of Distinct Representatives .
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DEFINITION 4.4.2 A graph G is bipartite if its vertices can be partitioned into two parts, say {v1, v2, . . . , vn} and {w1, w2, . . . , wm} so that all edges join some vi to some wj; no two vertices vi and vj are adjacent, nor are any vertices wi and wj.
The graph in figure 4.4.1 is bipartite, as are the first two graphs in figure 4.4.2.
4.5 Matchings Now we return to systems of distinct representatives.
A system of distinct representatives corresponds to a set of edges in the corresponding bipartite graph that share no endpoints; such a collection of edges (in any graph, not just a bipartite graph) is called a matching. In figure 4.5.1, a matching is shown in red. This is a largest possible matching, since it contains edges incident with all four of the top vertices, and it thus corresponds to a complete sdr.
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Any bipartite graph can be interpreted as a set system: we simply label all the vertices in one part with “set names” A1, A2, etc., and the other part is labeled with “element names”, and the sets are defined in the obvious way: Ai is the neighborhood of the vertex labeled “Ai”. Thus, we know one way to compute the size of a maximum matching, namely, we interpret the bipartite graph as a set system and compute the size of a maximum sdr; this is the size of a maximum matching.
4.5 Matchings 85 We will see another way to do this working directly with the graph. There are two advantages to this: it will turn out to be more efficient, and as a by-product it will actually find a maximum matching.
Given a bipartite graph, it is easy to find a maximal matching, that is, one that cannot be made larger simply by adding an edge: just choose edges that do not share endpoints until this is no longer possible. See figure 4.5.2 for an example.
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• Figure 4.5.2 A maximal matching is shown in red.
An obvious approach is then to attempt to make the matching larger.
There is a simple way to do this, if it works: We look for an alternating chain, defined as follows.
DEFINITION 4.5.1 Suppose M is a matching, and suppose that v1, w1, v2, w2, . . . , vk, wk is a sequence of vertices such that no edge in M is incident with v1 or wk, and moreover for all 1 ≤i ≤k, vi and wi are joined by an edge not in M, and for all 1 ≤i ≤k −1, wi and vi+1 are joined by an edge in M. Then the sequence of vertices together with the edges joining them in order is an alternating chain.
The graph in figure 4.5.2 contains alternating chains, one of which is shown in fig-ure 4.5.3.
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Suppose now that we remove from M all the edges that are in the alternating chain and also in M, forming M ′, and add to M ′ all of the edges in the alternating chain not in M, forming M ′′. It is not hard to show that M ′′ is a matching, and it contains one more edge than M. See figure 4.5.4.
Remarkably, if there is no alternating chain, then the matching M is a maximum matching.
THEOREM 4.5.2 Suppose that M is a matching in a bipartite graph G, and there is no alternating chain. Then M is a maximum matching.
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• Figure 4.5.4 A new, larger matching.
Proof.
We prove the contrapositive: Suppose that M is not a maximum matching. Then there is a larger matching, N. Create a new graph G′ by eliminating all edges that are in both M and N, and also all edges that are in neither. We are left with just those edges in M or N but not both.
In this new graph no vertex is incident with more than two edges, since if v were incident with three edges, at least two of them would belong to M or two would belong to N, but that can’t be true since M and N are matchings. This means that G′ is composed of disjoint paths and cycles. Since N is larger than M, G′ contains more edges from N than from M, and therefore one of the paths starts and ends with an edge from N, and along the path the edges alternate between edges in N and edges in M. In the original graph G with matching M, this path forms an alternating chain. The “alternating” part is clear; we need to see that the first and last vertices in the path are not incident with any edge in M.
Suppose that the first two vertices are v1 and v2. Then v1 and v2 are joined by an edge of N. Suppose that v1 is adjacent to a vertex w and that the edge between v1 and w is in M. This edge cannot be in N, for then there would be two edges of N incident at v1.
But then this edge is in G′, since it is in M but not N, and therefore the path in G′ does not start with the edge in N joining v1 and v2. This contradiction shows that no edge of M is incident at v1. The proof that the last vertex in the path is likewise not incident with an edge of M is essentially identical.
Now to find a maximum matching, we repeatedly look for alternating chains; when we cannot find one, we know we have a maximum matching. What we need now is an efficient algorithm for finding the alternating chain.
The key, in a sense, is to look for all possible alternating chains simultaneously. Sup-pose we have a bipartite graph with vertex partition {v1, v2, . . . , vn} and {w1, w2, . . . , wm} and a matching M. The algorithm labels vertices in such a way that if it succeeds, the alternating chain is indicated by the labels. Here are the steps: 0. Label with ‘(S,0)’ all vertices vi that are not incident with an edge in M. Set variable step to 0.
Now repeat the next two steps until no vertex acquires a new label: 4.5 Matchings 87 1. Increase step by 1.
For each newly labeled vertex vi, label with (i, step) any unlabeled neighbor wj of vi that is connected to vi by an edge that is not in M.
2. Increase step by 1. For each newly labeled vertex wi, label with (i, step) any unlabeled neighbor vj of wi that is connected to wi by an edge in M.
Here “newly labeled” means labeled at the previous step. When labeling vertices in step 1 or 2, no vertex is given more than one label. For example, in step 1, it may be that wk is a neighbor of the newly labeled vertices vi and vj. One of vi and vj, say vi, will be considered first, and will cause wk to be labeled; when vj is considered, wk is no longer unlabeled.
At the conclusion of the algorithm, if there is a labeled vertex wi that is not incident with any edge of M, then there is an alternating chain, and we say the algorithm succeeds.
If there is no such wi, then there is no alternating chain, and we say the algorithm fails.
The first of these claims is easy to see: Suppose vertex wi is labeled (k1, s). It became labeled due to vertex vk1 labeled (k2, s −1) that is connected by an edge not in M to wi.
In turn, vk1 is connected by an edge in M to vertex wk2 labeled (k3, s −2). Continuing in this way, we discover an alternating chain ending at some vertex vj labeled (S, 0): since the second coordinate step decreases by 1 at each vertex along the chain, we cannot repeat vertices, and must eventually get to a vertex with step = 0. If we apply the algorithm to the graph in figure 4.5.2, we get the labeling shown in figure 4.5.5, which then identifies the alternating chain w2, v3, w1, v1. Note that as soon as a vertex wi that is incident with no edge of M is labeled, we may stop, as there must be an alternating chain starting at wi; we need not continue the algorithm until no more labeling is possible. In the example in figure 4.5.5, we could stop after step 3, when w2 becomes labeled. Also, the step component of the labels is not really needed; it was included to make it easier to understand that if the algorithm succeeds, there really is an alternating chain.
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(4, 4) Figure 4.5.5 Labeling of a bipartite graph with matching; w2, v3, w1, v1 is an alternating chain.
To see that when the algorithm fails there is no alternating chain, we introduce a new concept.
88 Chapter 4 Systems of Distinct Representatives DEFINITION 4.5.3 A vertex cover in a graph is a set of vertices S such that every edge in the graph has at least one endpoint in S.
There is always a vertex cover of a graph, namely, the set of all the vertices of the graph. What is clearly more interesting is a smallest vertex cover, which is related to a maximum matching.
THEOREM 4.5.4 If M is a matching in a graph and S is a vertex cover, then |M| ≤|S|.
Proof.
Suppose M is a matching S is a vertex cover. Since each edge of M has an endpoint in S, if |M| > |S| then some vertex in S is incident with two edges of M, a contradiction. Hence |M| ≤|S|.
Suppose that we have a matching M and vertex cover S for a graph, and that |M| = |S|.
Then the theorem implies that M is a maximum matching and S is a minimum vertex cover. To show that when the algorithm fails there is no alternating chain, it is sufficient to show that there is a vertex cover that is the same size as M. Note that the proof of this theorem relies on the “official” version of the algorithm, that is, the algorithm continues until no new vertices are labeled.
THEOREM 4.5.5 Suppose the algorithm fails on the bipartite graph G with matching M. Let L be the set of labeled wi, U the set of unlabeled vi, and S = L ∪U. Then S is a vertex cover and |M| = |S|.
Proof.
If S is not a cover, there is an edge {vi, wj} with neither vi nor wj in S, so vi is labeled and wj is not. If the edge is not in M, then the algorithm would have labeled wj at the step after vi became labeled, so the edge must be in M. Now vi cannot be labeled (S, 0), so vi became labeled because it is connected to some labeled wk by an edge of M.
But now the two edges {vi, wj} and {vi, wk} are in M, a contradiction. So S is a vertex cover.
We know that |M| ≤|S|, so it suffices to show |S| ≤|M|, which we can do by finding an injection from S to M. Suppose that wi ∈S, so wi is labeled. Since the algorithm failed, wi is incident with an edge e of M; let f(wi) = e. If vi ∈S, vi is unlabeled; if vi were not incident with any edge of M, then vi would be labeled (S, 0), so vi is incident with an edge e of M; let f(vi) = e. Since G is bipartite, it is not possible that f(wi) = f(wj) or f(vi) = f(vj). If f(wi) = f(vj), then wi and vj are joined by an edge of M, and the algorithm would have labeled vj. Hence, f is an injection.
We have now proved this theorem: THEOREM 4.5.6 In a bipartite graph G, the size of a maximum matching is the same as the size of a minimum vertex cover.
4.5 Matchings 89 It is clear that the size of a maximum sdr is the same as the size of a maximum matching in the associated bipartite graph G. It is not too difficult to see directly that the size of a minimum vertex cover in G is the minimum value of f(n, i1, i2, . . . , ik) = n −k + | ∪k j=1 Aij|.
Thus, if the size of a maximum matching is equal to the size of a minimum cover, then the size of a maximum sdr is equal to the minimum value of n −k + | ∪k j=1 Aij|, and conversely. More concisely, theorem 4.5.6 is true if and only if theorem 4.2.1 is true.
More generally, in the schematic of figure 4.5.6, if any three of the relationships are known to be true, so is the fourth. In fact, we have proved all but the bottom equality, so we know it is true as well.
max sdr =?
max matching ∥?
∥?
min f(n, . . .) =?
min cover Figure 4.5.6 If any three of the “=?” are “=”, so is the fourth.
Finally, note that we now have both a more efficient way to compute the size of a maximum sdr and a way to find the actual representatives: convert the sdr problem to the graph problem, find a maximum matching, and interpret the matching as an sdr.
Exercises 4.5.
1. In this bipartite graph, find a maximum matching and a minimum vertex cover using the algorithm of this section. Start with the matching shown in red. Copies of this graph are available in this pdf file.
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• 2. Show directly that that the size of a minimum vertex cover in G is the minimum value of n −k + | ∪k j=1 Aij|, as mentioned above.
5 Graph Theory 5.1 The Basics See section 4.4 to review some basic terminology about graphs.
A graph G consists of a pair (V, E), where V is the set of vertices and E the set of edges. We write V (G) for the vertices of G and E(G) for the edges of G when necessary to avoid ambiguity, as when more than one graph is under discussion.
If no two edges have the same endpoints we say there are no multiple edges, and if no edge has a single vertex as both endpoints we say there are no loops. A graph with no loops and no multiple edges is a simple graph. A graph with no loops, but possibly with multiple edges is a multigraph. The condensation of a multigraph is the simple graph formed by eliminating multiple edges, that is, removing all but one of the edges with the same endpoints. To form the condensation of a graph, all loops are also removed. We sometimes refer to a graph as a general graph to emphasize that the graph may have loops or multiple edges.
The edges of a simple graph can be represented as a set of two element sets; for example, ({v1, . . . , v7}, {{v1, v2}, {v2, v3}, {v3, v4}, {v3, v5}, {v4, v5}, {v5, v6}, {v6, v7}}) is a graph that can be pictured as in figure 5.1.1.
This graph is also a connected graph: each pair of vertices v, w is connected by a sequence of vertices and edges, v = v1, e1, v2, e2, . . . , vk = w, where vi and vi+1 are the endpoints of edge ei. The graphs shown in figure 4.4.2 are connected, but the figure could be interpreted as a single graph that is not connected.
91 92 Chapter 5 Graph Theory .
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• • • • • • • Figure 5.1.1 A simple graph.
A graph G = (V, E) that is not simple can be represented by using multisets: a loop is a multiset {v, v} = {2 · v} and multiple edges are represented by making E a multiset.
The condensation of a multigraph may be formed by interpreting the multiset E as a set.
A general graph that is not connected, has loops, and has multiple edges is shown in figure 5.1.2. The condensation of this graph is shown in figure 5.1.3.
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• Figure 5.1.2 A general graph: it is not connected and has loops and mulitple edges.
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• Figure 5.1.3 The condensation of the previous graph.
The degree of a a vertex v, d(v), is the number of times it appears as an endpoint of an edge. If there are no loops, this is the same as the number of edges incident with v, but if v is both endpoints of an edge, namely, of a loop, then this contributes 2 to the degree of v. The degree sequence of a graph is a list of its degrees; the order does not matter, but usually we list the degrees in increasing or decreasing order. The degree sequence of the graph in figure 5.1.2, listed clockwise starting at the upper left, is 0, 4, 2, 3, 2, 8, 2, 4, 3, 2, 2.
We typically denote the degrees of the vertices of a graph by di, i = 1, 2, . . . , n, where n is the number of vertices. Depending on context, the subscript i may match the subscript on a vertex, so that di is the degree of vi, or the subscript may indicate the position of di in an increasing or decreasing list of the degrees; for example, we may state that the degree sequence is d1 ≤d2 ≤· · · ≤dn.
Our first result, simple but useful, concerns the degree sequence.
5.1 The Basics 93 THEOREM 5.1.1 In any graph, the sum of the degree sequence is equal to twice the number of edges, that is, n ∑ i=1 di = 2|E|.
Proof.
Let di be the degree of vi. The degree di counts the number of times vi appears as an endpoint of an edge. Since each edge has two endpoints, the sum ∑n i=1 di counts each edge twice.
An easy consequence of this theorem: COROLLARY 5.1.2 The number of odd numbers in a degree sequence is even.
An interesting question immediately arises: given a finite sequence of integers, is it the degree sequence of a graph? Clearly, if the sum of the sequence is odd, the answer is no. If the sum is even, it is not too hard to see that the answer is yes, provided we allow loops and multiple edges. The sequence need not be the degree sequence of a simple graph; for example, it is not hard to see that no simple graph has degree sequence 0, 1, 2, 3, 4. A sequence that is the degree sequence of a simple graph is said to be graphical. Graphical sequences have been characterized; the most well known characterization is given by this result: THEOREM 5.1.3 A sequence d1 ≥d2 ≥. . . ≥dn is graphical if and only if ∑n i=1 di is even and for all k ∈{1, 2, . . . , n}, k ∑ i=1 di ≤k(k −1) + n ∑ i=k+1 min(di, k).
It is not hard to see that if a sequence is graphical it has the property in the theorem; it is rather more difficult to see that any sequence with the property is graphical.
What does it mean for two graphs to be the same? Consider these three graphs: G1 = ({v1, v2, v3, v4}, {{v1, v2}, {v2, v3}, {v3, v4}, {v2, v4}}) G2 = ({v1, v2, v3, v4}, {{v1, v2}, {v1, v4}, {v3, v4}, {v2, v4}}) G3 = ({w1, w2, w3, w4}, {{w1, w2}, {w1, w4}, {w3, w4}, {w2, w4}}) These are pictured in figure 5.1.4. Simply looking at the lists of vertices and edges, they don’t appear to be the same. Looking more closely, G2 and G3 are the same except for the names used for the vertices: vi in one case, wi in the other. Looking at the pictures, there is an obvious sense in which all three are the same: each is a triangle with an edge 94 Chapter 5 Graph Theory (and vertex) dangling from one of the three vertices. Although G1 and G2 use the same names for the vertices, they apply to different vertices in the graph: in G1 the “dangling” vertex (officially called a pendant vertex) is called v1, while in G2 it is called v3. Finally, note that in the figure, G2 and G3 look different, even though they are clearly the same based on the vertex and edge lists.
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w4 Figure 5.1.4 Three isomorphic graphs.
So how should we define “sameness” for graphs? We use a familiar term and definition: isomorphism.
DEFINITION 5.1.4 Suppose G1 = (V, E) and G2 = (W, F). G1 and G2 are isomor-phic if there is a bijection f: V →W such that {v1, v2} ∈E if and only if {f(v1), f(v2)} ∈ F. In addition, the repetition numbers of {v1, v2} and {f(v1), f(v2)} are the same if mul-tiple edges or loops are allowed. This bijection f is called an isomorphism. When G1 and G2 are isomorphic, we write G1 ∼ = G2.
Each pair of graphs in figure 5.1.4 are isomorphic. For example, to show explicitly that G1 ∼ = G3, an isomorphism is f(v1) = w3 f(v2) = w4 f(v3) = w2 f(v4) = w1.
Clearly, if two graphs are isomorphic, their degree sequences are the same. The con-verse is not true; the graphs in figure 5.1.5 both have degree sequence 1, 1, 1, 2, 2, 3, but in one the degree-2 vertices are adjacent to each other, while in the other they are not. In general, if two graphs are isomorphic, they share all “graph theoretic” properties, that is, properties that depend only on the graph. As an example of a non-graph theoretic prop-erty, consider “the number of times edges cross when the graph is drawn in the plane.” In a more or less obvious way, some graphs are contained in others.
DEFINITION 5.1.5 Graph H = (W, F) is a subgraph of graph G = (V, E) if W ⊆V and F ⊆E. (Since H is a graph, the edges in F have their endpoints in W.) H is an 5.1 The Basics 95 .
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• Figure 5.1.5 Non-isomorphic graphs with degree sequence 1, 1, 1, 2, 2, 3.
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v4 Figure 5.1.6 Left to right: a graph, a subgraph, an induced subgraph.
induced subgraph if F consists of all edges in E with endpoints in W. See figure 5.1.6.
Whenever U ⊆V we denote the induced subgraph of G on vertices U as G[U].
A path in a graph is a subgraph that is a path; if the endpoints of the path are v and w we say it is a path from v to w. A cycle in a graph is a subgraph that is a cycle. A clique in a graph is a subgraph that is a complete graph.
If a graph G is not connected, define v ∼w if and only if there is a path connecting v and w.
It is not hard to see that this is an equivalence relation.
Each equivalence class corresponds to an induced subgraph G; these subgraphs are called the connected components of the graph.
Exercises 5.1.
1. The complement G of the simple graph G is a simple graph with the same vertices as G, and {v, w} is an edge of G if and only if it is not an edge of G. A graph G is self-complementary if G ∼ = G. Show that if G is self-complementary then it has 4k or 4k + 1 vertices for some k.
Find self-complementary graphs on 4 and 5 vertices.
2. Prove that if ∑n i=1 di is even, there is a graph (not necessarily simple) with degree sequence d1, d2, . . . , dn.
3. Suppose d1 ≥d2 ≥· · · ≥dn and ∑n i=1 di is even. Prove that there is a multigraph (no loops) with degree sequence d1, d2, . . . , dn if and only if d1 ≤∑n i=2 di.
4. Prove that 0, 1, 2, 3, 4 is not graphical.
5. Is 4, 4, 3, 2, 2, 1, 1 graphical? If not, explain why; if so, find a simple graph with this degree sequence.
96 Chapter 5 Graph Theory 6. Is 4, 4, 4, 2, 2 graphical? If not, explain why, and find a multigraph (no loops) with this degree sequence; if so, find a simple graph with this degree sequence.
7. Prove that a simple graph with n ≥2 vertices has two vertices of the same degree.
8. Prove the “only if” part of theorem 5.1.3.
9. Show that the condition on the degrees in theorem 5.1.3 is equivalent to this condition: ∑n i=1 di is even and for all k ∈{1, 2, . . . , n}, and all {i1, i2, . . . , ik} ⊆[n], k ∑ j=1 dij ≤k(k −1) + ∑ i/ ∈{i1,i2,...,ik} min(di, k).
Do not use theorem 5.1.3.
10. Draw the 11 non-isomorphic graphs with four vertices.
11. Suppose G1 ∼ = G2. Show that if G1 contains a cycle of length k so does G2.
12. Define v ∼w if and only if there is a path connecting vertices v and w. Prove that ∼is an equivalence relation.
13. Prove the “if” part of theorem 5.1.3, as follows: The proof is by induction on s = ∑n i=1 di. This is easy to see if s = 2, so suppose s > 2.
Without loss of generality we may suppose that dn > 0. Let t be the least integer such that dt > dt+1, or t = n −1 if there is no such integer. Let d′ t = dt −1, d′ n = dn −1, and d′ i = di for all other i. Note that d′ 1 ≥d′ 2 ≥· · · d′ n. We want to show that the sequence {d′ i} satisfies the condition of the theorem, that is, that for all k ∈{1, 2, . . . , n}, k ∑ i=1 d′ i ≤k(k −1) + n ∑ i=k+1 min(d′ i, k).
There are five cases: 1. k ≥t 2. k < t, dk < k 3. k < t, dk = k 4. k < t, dn > k 5. k < t, dk > k, dn ≤k By the induction hypothesis, there is a simple graph with degree sequence {d′ i}. Finally, show that there is a graph with degree sequence {di}.
This proof is due to S. A. Choudum, A Simple Proof of the Erd˝ os-Gallai Theorem on Graph Sequences, Bulletin of the Australian Mathematics Society, vol. 33, 1986, pp. 67-70.
The proof by Paul Erd˝ os and Tibor Gallai was long; Berge provided a shorter proof that used results in the theory of network flows. Choudum’s proof is both short and elementary.
5.2 Euler Circuits and Walks The first problem in graph theory dates to 1735, and is called the Seven Bridges of K¨ onigsberg. In K¨ onigsberg were two islands, connected to each other and the mainland by seven bridges, as shown in figure 5.2.1. The question, which made its way to Euler, 5.2 Euler Circuits and Walks 97 Figure 5.2.1 The Seven Bridges of K¨ onigsberg.
was whether it was possible to take a walk and cross over each bridge exactly once; Euler showed that it is not possible.
We can represent this problem as a graph, as in figure 5.2.2.
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• Figure 5.2.2 The Seven Bridges of K¨ onigsberg as a graph.
The two sides of the river are represented by the top and bottom vertices, and the islands by the middle two vertices. There are two possible interpretations of the question, depending on whether the goal is to end the walk at its starting point. Perhaps inspired by this problem, a walk in a graph is defined as follows.
DEFINITION 5.2.1 A walk in a graph is a sequence of vertices and edges, v1, e1, v2, e2, . . . , vk, ek, vk+1 98 Chapter 5 Graph Theory such that the endpoints of edge ei are vi and vi+1. In general, the edges and vertices may appear in the sequence more than once. If v1 = vk+1, the walk is a closed walk or a circuit.
We will deal first with the case in which the walk is to start and end at the same place.
A successful walk in K¨ onigsberg corresponds to a closed walk in the graph in which every edge is used exactly once.
What can we say about this walk in the graph, or indeed a closed walk in any graph that uses every edge exactly once? Such a walk is called an Euler circuit. If there are no vertices of degree 0, the graph must be connected, as this one is. Beyond that, imagine tracing out the vertices and edges of the walk on the graph. At every vertex other than the common starting and ending point, we come into the vertex along one edge and go out along another; this can happen more than once, but since we cannot use edges more than once, the number of edges incident at such a vertex must be even. Already we see that we’re in trouble in this particular graph, but let’s continue the analysis. The common starting and ending point may be visited more than once; except for the very first time we leave the starting vertex, and the last time we arrive at the vertex, each such visit uses exactly two edges. Together with the edges used first and last, this means that the starting vertex must also have even degree. Thus, since the K¨ onigsberg Bridges graph has odd degrees, the desired walk does not exist.
The question that should immediately spring to mind is this: if a graph is connected and the degree of every vertex is even, is there an Euler circuit? The answer is yes.
THEOREM 5.2.2 If G is a connected graph, then G contains an Euler circuit if and only if every vertex has even degree.
Proof.
We have already shown that if there is an Euler circuit, all degrees are even.
We prove the other direction by induction on the number of edges. If G has no edges the problem is trivial, so we assume that G has edges.
We start by finding some closed walk that does not use any edge more than once: Start at any vertex v0; follow any edge from this vertex, and continue to do this at each new vertex, that is, upon reaching a vertex, choose some unused edge leading to another vertex. Since every vertex has even degree, it is always possible to leave a vertex at which we arrive, until we return to the starting vertex, and every edge incident with the starting vertex has been used. The sequence of vertices and edges formed in this way is a closed walk; if it uses every edge, we are done.
Otherwise, form graph G′ by removing all the edges of the walk. G′ is not connected, since vertex v0 is not incident with any remaining edge. The rest of the graph, that is, G′ without v0, may or may not be connected. It consists of one or more connected subgraphs, each with fewer edges than G; call these graphs G1, G2,. . . ,Gk. Note that when we remove 5.2 Euler Circuits and Walks 99 the edges of the initial walk, we reduce the degree of every vertex by an even number, so all the vertices of each graph Gi have even degree. By the induction hypothesis, each Gi has an Euler circuit. These closed walks together with the original closed walk use every edge of G exactly once.
Suppose the original closed walk is v0, v1, . . . , vm = v0, abbreviated to leave out the edges. Because G is connected, at least one vertex in each Gi appears in this sequence, say vertices w1,1 ∈G1, w2,1 ∈G2,. . . , wk,1 ∈Gk, listed in the order they appear in v0, v1, . . . , vm. The Euler circuits of the graphs Gi are w1,1, w1,2, . . . , w1,m1 = w1,1 w2,1, w2,2, . . . , w2,m2 = w2,1 .
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wk,1, wk,2, . . . , wk,mk = wk,1.
By pasting together the original closed walk with these, we form a closed walk in G that uses every edge exactly once: v0, v1, . . . , vi1 = w1,1, w1,2, . . . , w1,m1 = vi1, vi1+1, . . . , vi2 = w2,1, . . . , w2,m2 = vi2, vi2+1, . . . , vik = wk,1, . . . , wk,mk = vik, vik+1, . . . , vm = v0.
Now let’s turn to the second interpretation of the problem: is it possible to walk over all the bridges exactly once, if the starting and ending points need not be the same? In a graph G, a walk that uses all of the edges but is not an Euler circuit is called an Euler walk. It is not too difficult to do an analysis much like the one for Euler circuits, but it is even easier to use the Euler circuit result itself to characterize Euler walks.
THEOREM 5.2.3 A connected graph G has an Euler walk if and only if exactly two vertices have odd degree.
Proof.
Suppose first that G has an Euler walk starting at vertex v and ending at vertex w. Add a new edge to the graph with endpoints v and w, forming G′. G′ has an Euler circuit, and so by the previous theorem every vertex has even degree. The degrees of v and w in G are therefore odd, while all others are even.
Now suppose that the degrees of v and w in G are odd, while all other vertices have even degree. Add a new edge e to the graph with endpoints v and w, forming G′. Every vertex in G′ has even degree, so by the previous theorem there is an Euler circuit which 100 Chapter 5 Graph Theory we can write as v, e1, v2, e2, . . . , w, e, v so that v, e1, v2, e2, . . . , w is an Euler walk.
Exercises 5.2.
1. Suppose a connected graph G has degree sequence d1, d2, . . . , dn. How many edges must be added to G so that the resulting graph has an Euler circuit? Explain.
2. Which complete graphs Kn, n ≥2, have Euler circuits? Which have Euler walks? Justify your answers.
3. Prove that if vertices v and w are joined by a walk they are joined by a path.
4. Show that if G is connected and has exactly 2k vertices of odd degree, k ≥1, its edges can be partitioned into k walks. Is this true for non-connected G?
5.3 Hamilton Cycles and Paths Here is a problem similar to the K¨ onigsberg Bridges problem: suppose a number of cities are connected by a network of roads. Is it possible to visit all the cities exactly once, without traveling any road twice? We assume that these roads do not intersect except at the cities. Again there are two versions of this problem, depending on whether we want to end at the same city in which we started.
This problem can be represented by a graph: the vertices represent cities, the edges represent the roads. We want to know if this graph has a cycle, or path, that uses every vertex exactly once. (Recall that a cycle in a graph is a subgraph that is a cycle, and a path is a subgraph that is a path.) There is no benefit or drawback to loops and multiple edges in this context: loops can never be used in a Hamilton cycle or path (except in the trivial case of a graph with a single vertex), and at most one of the edges between two vertices can be used. So we assume for this discussion that all graphs are simple.
DEFINITION 5.3.1 A cycle that uses every vertex in a graph exactly once is called a Hamilton cycle, and a path that uses every vertex in a graph exactly once is called a Hamilton path.
Unfortunately, this problem is much more difficult than the corresponding Euler circuit and walk problems; there is no good characterization of graphs with Hamilton paths and cycles. Note that if a graph has a Hamilton cycle then it also has a Hamilton path.
There are some useful conditions that imply the existence of a Hamilton cycle or path, which typically say in some form that there are many edges in the graph. An extreme 5.3 Hamilton Cycles and Paths 101 example is the complete graph Kn: it has as many edges as any simple graph on n vertices can have, and it has many Hamilton cycles.
The problem for a characterization is that there are graphs with Hamilton cycles that do not have very many edges. The simplest is a cycle, Cn: this has only n edges but has a Hamilton cycle. On the other hand, figure 5.3.1 shows graphs with just a few more edges than the cycle on the same number of vertices, but without Hamilton cycles.
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• Figure 5.3.1 A graph with a Hamilton path but not a Hamilton cycle, and one with neither.
There are also graphs that seem to have many edges, yet have no Hamilton cycle, as indicated in figure 5.3.2.
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Kn−1 Figure 5.3.2 A graph with many edges but no Hamilton cycle: a complete graph Kn−1 joined by an edge to a single vertex. This graph has (n−1 2 ) + 1 edges.
The key to a successful condition sufficient to guarantee the existence of a Hamilton cycle is to require many edges at lots of vertices.
THEOREM 5.3.2 (Ore) If G is a simple graph on n vertices, n ≥3, and d(v) + d(w) ≥n whenever v and w are not adjacent, then G has a Hamilton cycle.
Proof.
First we show that G is connected. If not, let v and w be vertices in two different connected components of G, and suppose the components have n1 and n2 vertices. Then d(v) ≤n1 −1 and d(w) ≤n2 −1, so d(v) + d(w) ≤n1 + n2 −2 < n. But since v and w are not adjacent, this is a contradiction.
Now consider a longest possible path in G: v1, v2, . . . , vk. Suppose, for a contradiction, that k < n, so there is some vertex w adjacent to one of v2, v3, . . . , vk−1, say to vi. If v1 is adjacent to vk, then w, vi, vi+1, . . . , vk, v1, v2, . . . vi−1 is a path of length k + 1, a contradiction. Hence, v1 is not adjacent to vk, and so d(v1) + d(vk) ≥n. The neighbors of 102 Chapter 5 Graph Theory v1 are among {v2, v3, . . . , vk−1} as are the neighbors of vk. Consider the vertices W = {vl+1 | vl is a neighbor of vk}.
Then |N(vk)| = |W| and W ⊆{v3, v4, . . . , vk} and N(v1) ⊆{v2, v3, . . . , vk−1}, so W ∪ N(v1) ⊆{v2, v3, . . . , vk}, a set with k −1 < n elements. Since |N(v1)| + |W| = |N(v1)| + |N(vk)| ≥n, N(v1) and W must have a common element, vj; note that 3 ≤j ≤k −1.
Then this is a cycle of length k: v1, vj, vj+1, . . . , vk, vj−1, vj−2, . . . , v1.
We can relabel the vertices for convenience: v1 = w1, w2, . . . , wk = v2, w1.
Now as before, w is adjacent to some wl, and w, wl, wl+1, . . . , wk, w1, w2, . . . wl−1 is a path of length k+1, a contradiction. Thus, k = n, and, renumbering the vertices for convenience, we have a Hamilton path v1, v2, . . . , vn. If v1 is adjacent to vn, there is a Hamilton cycle, as desired.
If v1 is not adjacent to vn, the neighbors of v1 are among {v2, v3, . . . , vn−1} as are the neighbors of vn. Consider the vertices W = {vl+1 | vl is a neighbor of vn}.
Then |N(vn)| = |W| and W ⊆{v3, v4, . . . , vn}, and N(v1) ⊆{v2, v3, . . . , vn−1}, so W ∪ N(v1) ⊆{v2, v3, . . . , vn}, a set with n −1 < n elements. Since |N(v1)| + |W| = |N(v1)| + |N(vn)| ≥n, N(v1) and W must have a common element, vi; note that 3 ≤i ≤n −1.
Then this is a cycle of length n: v1, vi, vi+1, . . . , vk, vi−1, vi−2, . . . , v1, and is a Hamilton cycle.
The property used in this theorem is called the Ore property; if a graph has the Ore property it also has a Hamilton path, but we can weaken the condition slightly if our goal is to show there is a Hamilton path. The proof of this theorem is nearly identical to the preceding proof.
THEOREM 5.3.3 If G is a simple graph on n vertices and d(v)+d(w) ≥n−1 whenever v and w are not adjacent, then G has a Hamilton path.
Suppose G is not simple. The existence of multiple edges and loops can’t help produce a Hamilton cycle when n ≥3: if we use a second edge between two vertices, or use a loop, we have repeated a vertex. To extend the Ore theorem to multigraphs, we consider the condensation of G: When n ≥3, the condensation of G is simple, and has a Hamilton cycle if and only if G has a Hamilton cycle. So if the condensation of G satisfies the Ore property, then G has a Hamilton cycle.
5.4 Bipartite Graphs 103 Exercises 5.3.
1. Suppose a simple graph G on n vertices has at least (n −1)(n −2) 2 + 2 edges. Prove that G has a Hamilton cycle. For n ≥2, show that there is a simple graph with (n −1)(n −2) 2 + 1 edges that has no Hamilton cycle.
2. Prove theorem 5.3.3.
3. The graph shown below is the Petersen graph. Does it have a Hamilton cycle? Justify your answer. Does it have a Hamilton path? Justify your answer.
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• 5.4 Bipartite Graphs We have already seen how bipartite graphs arise naturally in some circumstances. Here we explore bipartite graphs a bit more.
It is easy to see that all closed walks in a bipartite graph must have even length, since the vertices along the walk must alternate between the two parts. Remarkably, the converse is true. We need one new definition: DEFINITION 5.4.1 The distance between vertices v and w, d(v, w), is the length of a shortest walk between the two. If there is no walk between v and w, the distance is undefined.
THEOREM 5.4.2 G is bipartite if and only if all closed walks in G are of even length.
Proof.
The forward direction is easy, as discussed above.
Now suppose that all closed walks have even length. We may assume that G is con-nected; if not, we deal with each connected component separately.
Let v be a vertex of G, let X be the set of all vertices at even distance from v, and Y be the set of vertices at odd distance from v. We claim that all edges of G join a vertex of X to a vertex of Y . Suppose not; then there are adjacent vertices u and w such that d(v, u) and d(v, w) have the same parity. Then there is a closed walk from v to u to w to v of length d(v, u) + 1 + d(v, w), which is odd, a contradiction.
104 Chapter 5 Graph Theory The closed walk that provides the contradiction is not necessarily a cycle, but this can be remedied, providing a slightly different version of the theorem.
COROLLARY 5.4.3 G is bipartite if and only if all cycles in G are of even length.
Proof.
Again the forward direction is easy, and again we assume G is connected. As before, let v be a vertex of G, let X be the set of all vertices at even distance from v, and Y be the set of vertices at odd distance from v. If two vertices in X are adjacent, or two vertices in Y are adjacent, then as in the previous proof, there is a closed walk of odd length.
To finish the proof, it suffices to show that if there is a closed walk W of odd length then there is a cycle of odd length. The proof is by induction on the length of the closed walk.
If W has no repeated vertices, we are done. Otherwise, suppose the closed walk is v = v1, e1, . . . , vi = v, . . . , vk = v = v1.
Then v = v1, . . . , vi = v and v = vi, ei, vi+1, . . . , vk = v are closed walks, both are shorter than the original closed walk, and one of them has odd length. By the induction hypothesis, there is a cycle of odd length.
It is frequently fruitful to consider graph properties in the limited context of bipartite graphs (or other special types of graph). For example, what can we say about Hamilton cycles in simple bipartite graphs? Suppose the partition of the vertices of the bipartite graph is X and Y . Because any cycle alternates between vertices of the two parts of the bipartite graph, if there is a Hamilton cycle then |X| = |Y | ≥2. In such a case, the degree of every vertex is at most n/2, where n is the number of vertices, namely n = |X| + |Y |.
Thus the Ore condition (d(v) + d(w) ≥n when v and w are not adjacent) is equivalent to d(v) = n/2 for all v. This means the only simple bipartite graph that satisfies the Ore condition is the complete bipartite graph Kn/2,n/2, in which the two parts have size n/2 and every vertex of X is adjacent to every vertex of Y . The upshot is that the Ore property gives no interesting information about bipartite graphs.
Of course, as with more general graphs, there are bipartite graphs with few edges and a Hamilton cycle: any even length cycle is an example.
We note that, in general, a complete bipartite graph Km,n is a bipartite graph with |X| = m, |Y | = n, and every vertex of X is adjacent to every vertex of Y . The only such graphs with Hamilton cycles are those in which m = n.
5.5 Trees 105 Exercises 5.4.
1. Prove that there is a bipartite multigraph with degree sequence d1, . . . , dn if and only if there is a partition {I, J} of [n] such that ∑ i∈I di = ∑ i∈J di.
2. What is the smallest number of edges that can be removed from K5 to create a bipartite graph?
3. A regular graph is one in which the degree of every vertex is the same. Show that if G is a regular bipartite graph, and the common degree of the vertices is at least 1, then the two parts are the same size.
4. A perfect matching is one in which all vertices of the graph are incident with exactly one edge in the matching. Show that a regular bipartite graph with common degree at least 1 has a perfect matching. (We discussed matchings in section 4.5.) 5.5 Trees Another useful special class of graphs: DEFINITION 5.5.1 A connected graph G is a tree if it is acyclic, that is, it has no cycles. More generally, an acyclic graph is called a forest.
Two small examples of trees are shown in figure 5.1.5. Note that the definition implies that no tree has a loop or multiple edges.
THEOREM 5.5.2 Every tree T is bipartite.
Proof.
Since T has no cycles, it is true that every cycle of T has even length.
By corollary 5.4.3, T is bipartite.
DEFINITION 5.5.3 A vertex of degree one is called a pendant vertex, and the edge incident to it is a pendant edge.
THEOREM 5.5.4 Every tree on two or more vertices has at least one pendant vertex.
Proof.
We prove the contrapositive. Suppose graph G has no pendant vertices. Starting at any vertex v, follow a sequence of distinct edges until a vertex repeats; this is possible because the degree of every vertex is at least two, so upon arriving at a vertex for the first time it is always possible to leave the vertex on another edge. When a vertex repeats for the first time, we have discovered a cycle.
This theorem often provides the key step in an induction proof, since removing a pendant vertex (and its pendant edge) leaves a smaller tree.
106 Chapter 5 Graph Theory THEOREM 5.5.5 A tree on n vertices has exactly n −1 edges.
Proof.
A tree on 1 vertex has 0 edges; this is the base case.
If T is a tree on n ≥2 vertices, it has a pendant vertex. Remove this vertex and its pendant edge to get a tree T ′ on n −1 vertices. By the induction hypothesis, T ′ has n −2 edges; thus T has n −1 edges.
THEOREM 5.5.6 A tree with a vertex of degree k ≥1 has at least k pendant vertices.
In particular, every tree on at least two vertices has at least two pendant vertices.
Proof.
The case k = 1 is obvious. Let T be a tree with n vertices, degree sequence {di}n i=1, and a vertex of degree k ≥2, and let l be the number of pendant vertices.
Without loss of generality, 1 = d1 = d2 = · · · = dl and dl+1 = k. Then 2(n −1) = n ∑ i=1 di = l + k + n ∑ i=l+2 di ≥l + k + 2(n −l −1).
This reduces to l ≥k, as desired.
If T is a tree on two vertices, each of the vertices has degree 1. If T has at least three vertices it must have a vertex of degree k ≥2, since otherwise 2(n −1) = ∑n i=1 di = n, which implies n = 2. Hence it has at least k ≥2 pendant vertices.
Trees are quite useful in their own right, but also for the study of general graphs.
DEFINITION 5.5.7 If G is a connected graph on n vertices, a spanning tree for G is a subgraph of G that is a tree on n vertices.
THEOREM 5.5.8 Every connected graph has a spanning tree.
Proof.
By induction on the number of edges. If G is connected and has zero edges, it is a single vertex, so G is already a tree.
Now suppose G has m ≥1 edges. If G is a tree, it is its own spanning tree. Otherwise, G contains a cycle; remove one edge of this cycle. The resulting graph G′ is still connected and has fewer edges, so it has a spanning tree; this is also a spanning tree for G.
In general, spanning trees are not unique, that is, a graph may have many spanning trees. It is possible for some edges to be in every spanning tree even if there are multiple spanning trees. For example, any pendant edge must be in every spanning tree, as must any edge whose removal disconnects the graph (such an edge is called a bridge.) COROLLARY 5.5.9 If G is connected, it has at least n −1 edges; moreover, it has exactly n −1 edges if and only if it is a tree.
5.5 Trees 107 Proof.
If G is connected, it has a spanning tree, which has n −1 edges, all of which are edges of G.
If G has n−1 edges, which must be the edges of its spanning tree, then G is a tree.
THEOREM 5.5.10 G is a tree if and only if there is a unique path between any two vertices.
Proof.
if: Since every two vertices are connected by a path, G is connected. For a contradiction, suppose there is a cycle in G; then any two vertices on the cycle are connected by at least two distinct paths, a contradiction.
only if: If G is a tree it is connected, so between any two vertices there is at least one path. For a contradiction, suppose there are two different paths from v to w: v = v1, v2, . . . , vk = w and v = w1, w2, . . . , wl = w.
Let i be the smallest integer such that vi ̸= wi. Then let j be the smallest integer greater than or equal to i such that wj = vm for some m, which must be at least i. (Since wl = vk, such an m must exist.) Then vi−1, vi, . . . , vm = wj, wj−1, . . . , wi−1 = vi−1 is a cycle in G, a contradiction. See figure 5.5.1.
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DEFINITION 5.5.11 A cutpoint in a connected graph G is a vertex whose removal disconnects the graph.
THEOREM 5.5.12 Every connected graph has a vertex that is not a cutpoint.
Proof.
Remove a pendant vertex in a spanning tree for the graph.
Exercises 5.5.
1. Suppose that G is a connected graph, and that every spanning tree contains edge e. Show that e is a bridge.
2. Show that every edge in a tree is a bridge.
108 Chapter 5 Graph Theory 3. Show that G is a tree if and only if it has no cycles and adding any new edge creates a graph with exactly one cycle.
4. Which trees have Euler walks?
5. Which trees have Hamilton paths?
6. Let n ≥2. Show that there is a tree with degree sequence d1, d2, . . . , dn if and only if di > 0 for all i and ∑n i=1 di = 2(n −1).
7. A multitree is a multigraph whose condensation is a tree. Let n ≥2. Let d1, d2, . . . , dn be positive integers, and let g be the greatest common divisor of the di. Show that there is a multitree with degree sequence d1, d2, . . . , dn if and only if ∑n i=1 di/g ≥2(n −1) and for some partition I, J of [n], ∑ i∈I di = ∑ i∈J di.
8. Suppose T is a tree on n vertices, k of which have degree larger than 1, d1, d2,. . . dk. Of course, T must also have pendant vertices. How many pendant vertices? Your answer should depend only on k and d1, d2,. . . dk.
5.6 Optimal Spanning Trees In some applications, a graph G is augmented by associating a weight or cost with each edge; such a graph is called a weighted graph. For example, if a graph represents a network of roads, the weight of an edge might be the length of the road between its two endpoints, or the amount of time required to travel from one endpoint to the other, or the cost to bury cable along the road from one end to the other. In such cases, instead of being interested in just any spanning tree, we may be interested in a least cost spanning tree, that is, a spanning tree such that the sum of the costs of the edges of the tree is as small as possible. For example, this would be the least expensive way to connect a set of towns by a communication network, burying the cable in such a way as to minimize the total cost of laying the cable.
This problem is one that can be solved by a greedy algorithm. Roughly speaking, a greedy algorithm is one that makes choices that are optimal in the short run. Typically this strategy does not result in an optimal solution in the long run, but in this case this approach works.
DEFINITION 5.6.1 A weighted graph is a graph G together with a cost function c: E(G) →R>0. If H is a subgraph of G, the cost of H is c(H) = ∑ e∈E(H) c(e).
The Jarn´ ık Algorithm. Given a weighted connected graph G, we construct a minimum cost spanning tree T as follows.
Choose any vertex v0 in G and include it in T.
If vertices S = {v0, v1, . . . , vk} have been chosen, choose an edge with one endpoint in S and one endpoint not in S and with smallest weight among all such edges. Let vk+1 be the endpoint of this edge not in S, and add it and the associated edge to T. Continue until all vertices of G are in T.
5.6 Optimal Spanning Trees 109 This algorithm was discovered by Vojtˇ ech Jarn´ ık in 1930, and rediscovered indepen-dently by Robert C. Prim in 1957 and Edsger Dijkstra in 1959. It is often called Prim’s Algorithm.
The algorithm proceeds by constructing a sequence of trees T1, T2, . . . , Tn−1, with Tn−1 a spanning tree for G. At each step, the algorithm adds an edge that will make c(Ti+1) as small as possible among all trees that consist of Ti plus one edge. This is the best choice in the short run, but it is not obvious that in the long run, that is, by the time Tn−1 is constructed, that this will turn out to have been the best choice.
THEOREM 5.6.2 The Jarn´ ık Algorithm produces a minimum cost spanning tree.
Proof.
Suppose G is connected on n vertices. Let T be the spanning tree produced by the algorithm, and Tm a minimum cost spanning tree. We prove that c(T) = c(Tm).
Let e1, e2, . . . , en−1 be the edges of T in the order in which they were added to T; one endpoint of ei is vi, the other is in {v0, . . . , vi−1}.
We form a sequence of trees Tm = T0, T1, . . . , Tn−1 = T such that for each i, c(Ti) = c(Ti+1), and we conclude that c(Tm) = c(T).
If e1 is in T0, let T1 = T0. Otherwise, add edge e1 to T0. This creates a cycle containing e1 and another edge incident at v0, say f1. Remove f1 to form T1. Since the algorithm added edge e1, c(e1) ≤c(f1). If c(e1) < c(f1), then c(T1) < c(T0) = c(Tm), a contradiction, so c(e1) = c(f1) and c(T1) = c(T0).
Suppose we have constructed tree Ti. If ei+1 is in Ti, let Ti+1 = Ti. Otherwise, add edge ei+1 to Ti. This creates a cycle, one of whose edges, call it fi+1, is not in e1, e2, . . . , ei and has exactly one endpoint in {v0, . . . , vi}.
Remove fi+1 to create Ti+1.
Since the algorithm added ei+1, c(ei+1) ≤c(fi+1). If c(ei+1) < c(fi+1), then c(Ti+1) < c(Ti) = c(Tm), a contradiction, so c(ei+1) = c(fi+1) and c(Ti+1) = c(Ti).
Exercises 5.6.
1. Kruskal’s Algorithm is also a greedy algorithm that produces a minimum cost spanning tree for a connected graph G. Begin by choosing an edge in G of smallest cost. Assuming that edges e1, e2, . . . , ei have been chosen, pick an edge ei+1 that does not form a cycle together with e1, e2, . . . , ei and that has smallest cost among all such edges. The edges e1, e2, . . . , en−1 form a spanning tree for G. Prove that this spanning tree has minimum cost.
2. Prove that if the edge costs of G are distinct, there is exactly one minimum cost spanning tree. Give an example of a graph G with more than one minimum cost spanning tree.
3. In both the Jarn´ ık and Kruskal algorithms, it may be that two or more edges can be added at any particular step, and some method is required to choose one over the other. For the graph below, use both algorithms to find a minimum cost spanning tree. Using the labels ei on the graph, at each stage pick the edge ei that the algorithm specifies and that has the lowest possible i among all edges available. For the Jarn´ ık algorithm, use the designated 110 Chapter 5 Graph Theory v0 as the starting vertex. For each algorithm, list the edges in the order in which they are added. The edge weights e1, e2, . . . , e10 are 6, 7, 8, 2, 3, 2, 4, 6, 1, 1, shown in red.
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e10/1 5.7 Connectivity We have seen examples of connected graphs and graphs that are not connected. While “not connected” is pretty much a dead end, there is much to be said about “how connected” a connected graph is. The simplest approach is to look at how hard it is to disconnect a graph by removing vertices or edges. We assume that all graphs are simple.
If it is possible to disconnect a graph by removing a single vertex, called a cutpoint, we say the graph has connectivity 1. If this is not possible, but it is possible to disconnect the graph by removing two vertices, the graph has connectivity 2.
DEFINITION 5.7.1 If a graph G is connected, any set of vertices whose removal disconnects the graph is called a cutset. G has connectivity k if there is a cutset of size k but no smaller cutset. If there is no cutset and G has at least two vertices, we say G has connectivity n−1; if G has one vertex, its connectivity is undefined. If G is not connected, we say it has connectivity 0. G is k-connected if the connectivity of G is at least k. The connectivity of G is denoted κ(G).
As you should expect from the definition, there are graphs without a cutset: the complete graphs Kn. If G is connected but not a Kn, it has vertices v and w that are not adjacent, so removing the n −2 other vertices leaves a non-connected graph, and so the connectivity of G is at most n −2. Thus, only the complete graphs have connectivity n −1.
We do the same thing for edges: DEFINITION 5.7.2 If a graph G is connected, any set of edges whose removal dis-connects the graph is called a cut. G has edge connectivity k if there is a cut of size k but no smaller cut; the edge connectivity of a one-vertex graph is undefined. G is k-edge-5.7 Connectivity 111 connected if the edge connectivity of G is at least k. The edge connectivity is denoted λ(G).
Any connected graph with at least two vertices can be disconnected by removing edges: by removing all edges incident with a single vertex the graph is disconnected.
Thus, λ(G) ≤δ(G), where δ(G) is the minimum degree of any vertex in G. Note that δ(G) ≤n −1, so λ(G) ≤n −1.
Removing a vertex also removes all of the edges incident with it, which suggests that κ(G) ≤λ(G). This turns out to be true, though not as easy as you might hope. We write G −v to mean G with vertex v removed, and G −{v1, v2, . . . , vk} to mean G with all of {v1, v2, . . . , vk} removed, and similarly for edges.
THEOREM 5.7.3 κ(G) ≤λ(G).
Proof.
We use induction on λ = λ(G). If λ = 0, G is disconnected, so κ = 0. If λ = 1, removal of edge e with endpoints v and w disconnects G. If v and w are the only vertices of G, G is K2 and has connectivity 1. Otherwise, removal of one of v and w disconnects G, so κ = 1.
As a special case we note that if λ = n −1 then δ = n −1, so G is Kn and κ = n −1.
Now suppose n −1 > λ = k > 1, and removal of edges e1, e2, . . . , ek disconnects G.
Remove edge ek with endpoints v and w to form G1 with λ(G1) = k −1. By the induction hypothesis, there are at most k−1 vertices v1, v2, . . . , vj such that G2 = G1−{v1, v2, . . . , vj} is disconnected. Since k < n −1, k −1 ≤n −3, and so G2 has at least 3 vertices.
If both v and w are vertices of G2, and if adding ek to G2 produces a connected graph G3, then removal of one of v and w will disconnect G3 forming G4, and G4 = G −{v1, v2, . . . , vj, v} or G4 = G −{v1, v2, . . . , vj, w}, that is, removing at most k vertices disconnects G. If v and w are vertices of G2 but adding ek does not produce a connected graph, then removing v1, v2, . . . , vj disconnects G. Finally, if at least one of v and w is not in G2, then G2 = G −{v1, v2, . . . , vj} and the connectivity of G is less than k. So in all cases, κ ≤k.
Graphs that are 2-connected are particularly important, and the following simple the-orem is useful.
THEOREM 5.7.4 If G has at least three vertices, the following are equivalent: 1. G is 2-connected 2. G is connected and has no cutpoint 3. For all distinct vertices u, v, w in G there is a path from u to v that does not contain w.
112 Chapter 5 Graph Theory Proof.
1 ⇒3: Since G is 2-connected, G with w removed is a connected graph G′.
Thus, in G′ there is a path from u to v, which in G is a path from u to v avoiding w.
3 ⇒2: If G has property 3 it is clearly connected. Suppose that w is a cutpoint, so that G′ = G −w is disconnected. Let u and v be vertices in two different components of G′, so that no path connects them in G′. Then every path joining u to v in G must use w, a contradiction.
2 ⇒1: Since G has at least 3 vertices and has no cutpoint, its connectivity is at least 2, so it is 2-connected by definition.
There are other nice characterizations of 2-connected graphs.
THEOREM 5.7.5 If G has at least three vertices, then G is 2-connected if and only if every two vertices u and v are contained in a cycle.
Proof.
if: Suppose vertex w is removed from G, and consider any other vertices u and v. In G, u and v lie on a cycle; even if w also lies on this cycle, then u and v are still connected by a path when w is removed.
only if: Given u and v we want to show there is a cycle containing both. Let U be the set of vertices other than u that are contained in a cycle with u. First, we show that U is non-empty. Let w be adjacent to u, and remove the edge e between them. Since λ(G) ≥κ(G) ≥2, G −e is connected. Thus, there is a path from u to w; together with e this path forms a cycle containing u and w, so w ∈U.
For a contradiction, suppose v / ∈U. Let w be in U with d(w, v) ≥1 as small as possible, fix a cycle C containing u and w and a path P of length d(w, v) from w to v.
By the previous theorem, there is a path Q from u to v that does not use w. Following this path from u, there is a last vertex x on the path that is also on the cycle containing u and w, and there is a first vertex y on the path, after x, with y also on the path from w to v (it is possible that y = v, but not that y = w); see figure 5.7.1. Now starting at u, proceeding on cycle C to x without using w, then from x to y on Q, then to w on P, and finally back to u on C, we see that y ∈U. But y is closer to v than is w, a contradiction.
Hence v ∈U.
The following corollary is an easy restatement of this theorem.
COROLLARY 5.7.6 If G has at least three vertices, then G is 2-connected if and only if between every two vertices u and v there are two internally disjoint paths, that is, paths that share only the vertices u and v.
This version of the theorem suggests a generalization: 5.7 Connectivity 113 .
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x Figure 5.7.1 Point y closer to v than w is a contradiction; path Q is shown dashed. (See theorem 5.7.5.) THEOREM 5.7.7 Menger’s Theorem If G has at least k + 1 vertices, then G is k-connected if and only if between every two vertices u and v there are k pairwise internally disjoint paths.
We first prove Menger’s original version of this, a “local” version.
DEFINITION 5.7.8 If v and w are non-adjacent vertices in G, κG(v, w) is the smallest number of vertices whose removal separates v from w, that is, disconnects G leaving v and w in different components. A cutset that separates v and w is called a separating set for v and w. pG(v, w) is the maximum number of internally disjoint paths between v and w.
THEOREM 5.7.9 If v and w are non-adjacent vertices in G, κG(v, w) = pG(v, w).
Proof.
If there are k internally disjoint paths between v and w, then any set of vertices whose removal separates v from w must contain at least one vertex from each of the k paths, so κG(v, w) ≥pG(v, w).
To finish the proof, we show that there are κG(v, w) internally disjoint paths between v and w. The proof is by induction on the number of vertices in G. If G has two vertices, G is not connected, and κG(v, w) = pG(v, w) = 0. Now suppose G has n > 2 vertices and κG(v, w) = k.
Note that removal of either N(v) or N(w) separates v from w, so no separating set S of size k can properly contain N(v) or N(w). Now we address two cases: Case 1: Suppose there is a set S of size k that separates v from w, and S contains a vertex not in N(v) or N(w). G −S is disconnected, and one component G1 contains v. Since S does not contain N(v), G1 has at least two vertices; let X = V (G1) and Y = V (G)−S−X.
Since S does not contain N(w), Y contains at least two vertices. Now we form two new graphs: Form HX by starting with G −Y and adding a vertex y adjacent to every vertex of S. Form HY by starting with G −X and adding a vertex x adjacent to every vertex of S; see figure 5.7.2. Since X and Y each contain at least two vertices, HX and HY are smaller than G, and so the induction hypothesis applies to them.
114 Chapter 5 Graph Theory Clearly S separates v from y in HX and w from x in HY . Moreover, any set that separates v from y in HX separates v from w in G, so κHX(v, y) = κG(v, w) = k. Similarly, κHY (x, w) = κG(v, w) = k. Hence, by the induction hypothesis, there are k internally disjoint paths from v to y in HX and k internally disjoint paths from x to w in HY . Each of these paths uses one vertex of S; by eliminating x and y and joining the paths at the vertices of S, we produce k internally disjoint paths from v to w.
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· · · Figure 5.7.2 Case 1: Top figure is G, lower left is HX, lower right is HY .
Case 2: Now we suppose that any set S separating v and w is a subset of N(v) ∪N(w); pick such an S. If there is a vertex u not in {v, w} ∪N(v) ∪N(w), consider G −u. This u is not in any set of size k that separates v from w, for if it were we would be in Case 1. Since S separates v from w in G −u, κG−u(v, w) ≤k. But if some smaller set S′ separates v from w in G −u, then S′ ∪{u} separates v from w in G, a contradiction, so κG−u(v, w) = k. By the induction hypothesis, there are k internally disjoint paths from v to w in G −u and hence in G.
We are left with V (G) = {v, w} ∪N(v) ∪N(w).
Suppose there is a vertex u in N(v) ∩N(w). Then u is in every set that separates v from w, so κG−u = k −1. By the induction hypothesis, there are k −1 internally disjoint paths from v to w in G −u and together with the path v, u, w, they comprise k internally disjoint paths from v to w in G.
Finally, suppose that N(v) ∩N(w) = ∅.
Form a bipartite graph B with vertices N(v) ∪N(w) and any edges of G that have one endpoint in N(v) and the other in N(w).
Every set separating v from w in G must include one endpoint of every edge in B, that is, must be a vertex cover in B, and conversely, every vertex cover in B separates v from w in G. Thus, the minimum size of a vertex cover in B is k, and so there is a matching in B of size k, by theorem 4.5.6. The edges of this matching, together with the edges incident at v and w, form k internally disjoint paths from v to w in G.
5.7 Connectivity 115 Proof of Menger’s Theorem (5.7.7).
Suppose first that between every two vertices v and w in G there are k internally disjoint paths. If G is not k-connected, the connectivity of G is at most k −1, and because G has at least k + 1 vertices, there is a cutset S of G with size at most k −1. Let v and w be vertices in two different components of G −S; in G these vertices are joined by k internally disjoint paths. Since there is no path from v to w in G −S, each of these k paths contains a vertex of S, but this is impossible since S has size less than k, and the paths share no vertices other than v and w. This contradiction shows that G is k-connected.
Now suppose G is k-connected.
If v and w are not adjacent, κG(v, w) ≥k and by the previous theorem there are pG(v, w) = κG(v, w) internally disjoint paths between v and w.
If v and w are connected by edge e, consider G −e. Suppose there is a separating set S for v and w with |S| < k −1, that is, κG−e(v, w) < k −1. S cannot be a cutset for G, so G −S is connected. Then G −e −S must have exactly two components, since G −S is G −e −S plus a single edge. G −e −S has at least k + 1 −(k −2) = 3 vertices, so one of the two components contains at least two vertices; without loss of generality, say u and w are in a single component. Then G −(S ∪{w}) is not connected, but |S ∪{w}| < k, contradicting the assumption that G is k-connected. Thus, κG−e(v, w) ≥k −1, so there are k −1 internally disjoint paths between v and w in G −e. Together with the path from v to w using e, this gives k internally disjoint paths between v and w in G, as desired.
• • • We return briefly to 2-connectivity.
The next theorem can sometimes be used to provide the induction step in an induction proof.
THEOREM 5.7.10 The Handle Theorem Suppose G is 2-connected and K is a 2-connected proper subgraph of G. Then there are subgraphs L and H (the handle) of G such that L is 2-connected, L contains K, H is a simple path, L and H share exactly the endpoints of H, and G is the union of L and H.
Proof.
Given G and K, let L be a maximal 2-connected proper subgraph of G containing K. If V (L) = V (G), let e be an edge not in L. Since L plus the edge e is 2-connected, it must be G, by the maximality of L. Hence H is the path consisting of e and its endpoints.
Suppose that v is in V (G) but not V (L).
Let u be a vertex of L.
Since G is 2-connected, there is a cycle containing v and u. Following the cycle from v to u, Let w be the first vertex in L. Continuing on the cycle from u to v, let x be the last vertex in L. Let P be the path continuing around the cycle: (x, v1, v2, . . . , vk, v = vk+1, vk+2, . . . , vm, w). If x ̸= w, let H = P. Since L together with H is 2-connected, it is G, as desired.
116 Chapter 5 Graph Theory If x = w then x = w = u. Let y be a vertex of L other than u. Since G is 2-connected, there is a path P1 from v to y that does not include u. Let vj be the last vertex on P1 that is in {v1, . . . , v, . . . , vm}; without loss of generality, suppose j ≥k + 1. Then let H be the path (u, v1, . . . , v = vk+1, . . . , vj, . . . , y), where from vj to y we follow path P1. Now L ∪H is a 2-connected subgraph of G, but it is not G, as it does not contain the edge {u, vm}, contradicting the maximality of L. Thus x ̸= w.
A graph that is not connected consists of connected components. In a theorem rem-iniscent of this, we see that connected graphs that are not 2-connected are constructed from 2-connected subgraphs and bridges.
DEFINITION 5.7.11 A block in a graph G is a maximal induced subgraph on at least two vertices without a cutpoint.
As usual, maximal here means that the induced subgraph B cannot be made larger by adding vertices or edges (while retaining the desired property, in this case, no cutpoints). A block is either a 2-connected induced subgraph or a single edge together with its endpoints.
Blocks are useful in large part because of this theorem: THEOREM 5.7.12 The blocks of G partition the edges.
Proof.
We need to show that every edge is in exactly one block. If an edge is in no 2-connected induced subgraph of G, then, together with its endpoints, it is itself a block.
Thus, every edge is in some block.
Now suppose that B1 and B2 are distinct blocks. This implies that neither is a sub-graph of the other, by the maximality condition. Hence, the induced subgraph G[V (B1) ∪ V (B2)] is larger than either of B1 and B2. Suppose B1 and B2 share an edge, so that they share the endpoints of this edge, say u and v. Supppose w is a vertex in V (B1) ∪V (B2).
Since B1 −w and B2 −w are connected, so is G[(V (B1) ∪V (B2)){w}], because either u or v is in (V (B1) ∪V (B2)){w}. Thus G[V (B1) ∪V (B2)] has no cutpoint but strictly contains B1 and B2, contradicting the maximality property of blocks. Thus, every edge is in at most one block.
If G has a single block, it is either K2 or is 2-connected, and any 2-connected graph has a single block.
THEOREM 5.7.13 If G is connected but not 2-connected, then every vertex that is in two blocks is a cutpoint of G.
Proof.
Suppose w is in B1 and B2, but G −w is connected.
Then there is a path v1, v2, . . . , vk in G −w, with v1 ∈B1 and vk ∈B2.
But then G[V (B1) ∪V (B2) ∪ {v1, v2, . . . , vk}] has no cutpoint and contains both B1 and B2, a contradiction.
5.7 Connectivity 117 Exercises 5.7.
1. Suppose a simple graph G on n ≥2 vertices has at least (n −1)(n −2) 2 + 1 edges. Prove that G is connected.
2. Suppose a general graph G has exactly two odd-degree vertices, v and w. Let G′ be the graph created by adding an edge joining v to w. Prove that G′ is connected if and only if G is connected.
3. Suppose G is simple with degree sequence d1 ≤d2 ≤· · · ≤dn, and for k ≤n−dn −1, dk ≥k.
Show G is connected.
4. Recall that a graph is k-regular if all the vertices have degree k. What is the smallest integer k that makes this true: If G is simple, has n vertices, m ≥k, and G is m-regular, then G is connected.
(Of course k depends on n.) 5. Suppose G has at least one edge. Show that G is 2-connected if and only if for all vertices v and edges e there is a cycle containing v and e.
6. Find a simple graph with κ(G) < λ(G) < δ(G).
7. Suppose λ(G) = k > 0. Show that there are sets of vertices U and V that partition the vertices of G, and such that there are exactly k edges with one endpoint in U and one endpoint in V .
8. Find λ(Km,n), where both m and n are at least 1. (Km,n is the complete bipartite graph on m and n vertices: the parts have m and n vertices, and every pair of vertices, one from each part, is connected by an edge.) 9. Suppose G is a connected graph. The block-cutpoint graph of G, BC(G) is formed as follows: Let vertices c1, c2, . . . , ck be the cutpoints of G, and let the blocks of G be B1, . . . , Bl.
The vertices of BC(G) are c1, . . . , ck, B1, . . . , Bl. Add an edge {Bi, cj} if and only if cj ∈Bi.
Show that the block-cutpoint graph is a tree.
Note that a cutpoint is contained in at least two blocks, so that all pendant vertices of the block-cutpoint graph are blocks. These blocks are called endblocks.
10. Draw the block-cutpoint graph of the graph below.
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• 11. Show that the complement of a disconnected graph is connected. Is the complement of a connected graph always disconnected? (The complement G of graph G has the same vertices as G, and {v, w} is an edge of G if and only if it is not an edge of G.) 118 Chapter 5 Graph Theory 5.8 Graph Coloring As we briefly discussed in section 1.1, the most famous graph coloring problem is certainly the map coloring problem, proposed in the nineteenth century and finally solved in 1976.
DEFINITION 5.8.1 A proper coloring of a graph is an assignment of colors to the vertices of the graph so that no two adjacent vertices have the same color.
Usually we drop the word “proper” unless other types of coloring are also under dis-cussion. Of course, the “colors” don’t have to be actual colors; they can be any distinct labels—integers, for example. If a graph is not connected, each connected component can be colored independently; except where otherwise noted, we assume graphs are connected.
We also assume graphs are simple in this section.
Graph coloring has many applications in addition to its intrinsic interest.
EXAMPLE 5.8.2 If the vertices of a graph represent academic classes, and two vertices are adjacent if the corresponding classes have people in common, then a coloring of the vertices can be used to schedule class meetings. Here the colors would be schedule times, such as 8MWF, 9MWF, 11TTh, etc.
EXAMPLE 5.8.3 If the vertices of a graph represent radio stations, and two vertices are adjacent if the stations are close enough to interfere with each other, a coloring can be used to assign non-interfering frequencies to the stations.
EXAMPLE 5.8.4 If the vertices of a graph represent traffic signals at an intersection, and two vertices are adjacent if the corresponding signals cannot be green at the same time, a coloring can be used to designate sets of signals than can be green at the same time.
Graph coloring is closely related to the concept of an independent set.
DEFINITION 5.8.5 A set S of vertices in a graph is independent if no two vertices of S are adjacent.
If a graph is properly colored, the vertices that are assigned a particular color form an independent set. Given a graph G it is easy to find a proper coloring: give every vertex a different color. Clearly the interesting quantity is the minimum number of colors required for a coloring. It is also easy to find independent sets: just pick vertices that are mutually non-adjacent. A single vertex set, for example, is independent, and usually finding larger independent sets is easy. The interesting quantity is the maximum size of an independent set.
5.8 Graph Coloring 119 DEFINITION 5.8.6 The chromatic number of a graph G is the minimum number of colors required in a proper coloring; it is denoted χ(G). The independence number of G is the maximum size of an independent set; it is denoted α(G).
The natural first question about these graphical parameters is: how small or large can they be in a graph G with n vertices. It is easy to see that 1 ≤χ(G) ≤n 1 ≤α(G) ≤n and that the limits are all attainable: A graph with no edges has chromatic number 1 and independence number n, while a complete graph has chromatic number n and independence number 1. These inequalities are thus not very interesting. We will see some that are more interesting.
Another natural question: What is the relation between the chromatic number of a graph G and chromatic number of a subgraph of G? This too is simple, but quite useful at times.
THEOREM 5.8.7 If H is a subgraph of G, χ(H) ≤χ(G).
Proof.
Any coloring of G provides a proper coloring of H, simply by assigning the same colors to vertices of H that they have in G. This means that H can be colored with χ(G) colors, perhaps even fewer, which is exactly what we want.
Often this fact is interesting “in reverse”. For example, if G has a subgraph H that is a complete graph Km, then χ(H) = m and so χ(G) ≥m. A subgraph of G that is a complete graph is called a clique, and there is an associated graphical parameter.
DEFINITION 5.8.8 The clique number of a graph G is the largest m such that Km is a subgraph of G.
It is tempting to speculate that the only way a graph G could require m colors is by having such a subgraph. This is false; graphs can have high chromatic number while having low clique number; see figure 5.8.1. It is easy to see that this graph has χ ≥3, because there are many 3-cliques in the graph. In general it can be difficult to show that a graph cannot be colored with a given number of colors, but in this case it is easy to see that the graph cannot in fact be colored with three colors, because so much is “forced”.
Suppose the graph can be colored with 3 colors. Starting at the left if vertex v1 gets color 1, then v2 and v3 must be colored 2 and 3, and vertex v4 must be color 1. Continuing, v10 must be color 1, but this is not allowed, so χ > 3. On the other hand, since v10 can be colored 4, we see χ = 4.
Paul Erd˝ os showed in 1959 that there are graphs with arbitrarily large chromatic number and arbitrarily large girth (the girth is the size of the smallest cycle in a graph).
120 Chapter 5 Graph Theory .
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This is much stronger than the existence of graphs with high chromatic number and low clique number.
Bipartite graphs with at least one edge have chromatic number 2, since the two parts are each independent sets and can be colored with a single color. Conversely, if a graph can be 2-colored, it is bipartite, since all edges connect vertices of different colors. This means it is easy to identify bipartite graphs: Color any vertex with color 1; color its neighbors color 2; continuing in this way will or will not successfully color the whole graph with 2 colors. If it fails, the graph cannot be 2-colored, since all choices for vertex colors are forced.
If a graph is properly colored, then each color class (a color class is the set of all vertices of a single color) is an independent set.
THEOREM 5.8.9 In any graph G on n vertices, n α ≤χ.
Proof.
Suppose G is colored with χ colors. Since each color class is independent, the size of any color class is at most α. Let the color classes be V1, V2, . . . , Vχ. Then n = χ ∑ i=1 |Vi| ≤χα, as desired.
We can improve the upper bound on χ(G) as well.
In any graph G, ∆(G) is the maximum degree of any vertex.
THEOREM 5.8.10 In any graph G, χ ≤∆+ 1.
Proof.
We show that we can always color G with ∆+ 1 colors by a simple greedy algorithm: Pick a vertex vn, and list the vertices of G as v1, v2, . . . , vn so that if i < j, d(vi, vn) ≥d(vj, vn), that is, we list the vertices farthest from vn first. We use integers 5.8 Graph Coloring 121 1, 2, . . . , ∆+ 1 as colors. Color v1 with 1. Then for each vi in order, color vi with the smallest integer that does not violate the proper coloring requirement, that is, which is different than the colors already assigned to the neighbors of vi. For i < n, we claim that vi is colored with one of 1, 2, . . . , ∆.
This is certainly true for v1. For 1 < i < n, vi has at least one neighbor that is not yet colored, namely, a vertex closer to vn on a shortest path from vn to vi. Thus, the neighbors of vi use at most ∆−1 colors from the colors 1, 2, . . . , ∆, leaving at least one color from this list available for vi.
Once v1, . . . , vn−1 have been colored, all neighbors of vn have been colored using the colors 1, 2, . . . , ∆, so color ∆+ 1 may be used to color vn.
Note that if d(vn) < ∆, even vn may be colored with one of the colors 1, 2, . . . , ∆.
Since the choice of vn was arbitrary, we may choose vn so that d(vn) < ∆, unless all vertices have degree ∆, that is, if G is regular. Thus, we have proved somewhat more than stated, namely, that any graph G that is not regular has χ ≤∆. (If instead of choosing the particular order of v1, . . . , vn that we used we were to list them in arbitrary order, even vertices other than vn might require use of color ∆+ 1. This gives a slightly simpler proof of the stated theorem.) We state this as a corollary.
COROLLARY 5.8.11 If G is not regular, χ ≤∆.
There are graphs for which χ = ∆+ 1: any cycle of odd length has ∆= 2 and χ = 3, and Kn has ∆= n −1 and χ = n. Of course, these are regular graphs. It turns out that these are the only examples, that is, if G is not an odd cycle or a complete graph, then χ(G) ≤∆(G).
THEOREM 5.8.12 Brooks’s Theorem If G is a graph other than Kn or C2n+1, χ ≤∆.
The greedy algorithm will not always color a graph with the smallest possible number of colors. Figure 5.8.2 shows a graph with chromatic number 3, but the greedy algorithm uses 4 colors if the vertices are ordered as shown.
In general, it is difficult to compute χ(G), that is, it takes a large amount of compu-tation, but there is a simple algorithm for graph coloring that is not fast. Suppose that v and w are non-adjacent vertices in G. Denote by G + {v, w} = G + e the graph formed by adding edge e = {v, w} to G. Denote by G/e the graph in which v and w are “identified”, that is, v and w are replaced by a single vertex x adjacent to all neighbors of v and w.
(But note that we do not introduce multiple edges: if u is adjacent to both v and w in G, there will be a single edge from x to u in G/e.) Consider a proper coloring of G in which v and w are different colors; then this is a proper coloring of G + e as well. Also, any proper coloring of G + e is a proper coloring of 122 Chapter 5 Graph Theory .
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G in which v and w have different colors. So a coloring of G + e with the smallest possible number of colors is a best coloring of G in which v and w have different colors, that is, χ(G + e) is the smallest number of colors needed to color G so that v and w have different colors.
If G is properly colored and v and w have the same color, then this gives a proper coloring of G/e, by coloring x in G/e with the same color used for v and w in G. Also, if G/e is properly colored, this gives a proper coloring of G in which v and w have the same color, namely, the color of x in G/e. Thus, χ(G/e) is the smallest number of colors needed to properly color G so that v and w are the same color.
The upshot of these observations is that χ(G) = min(χ(G+e), χ(G/e)). This algorithm can be applied recursively, that is, if G1 = G+e and G2 = G/e then χ(G1) = min(χ(G1 + e), χ(G1/e)) and χ(G2) = min(χ(G2 + e), χ(G2/e)), where of course the edge e is different in each graph. Continuing in this way, we can eventually compute χ(G), provided that eventually we end up with graphs that are “simple” to color. Roughly speaking, because G/e has fewer vertices, and G + e has more edges, we must eventually end up with a complete graph along all branches of the computation. Whenever we encounter a complete graph Km it has chromatic number m, so no further computation is required along the corresponding branch. Let’s make this more precise.
THEOREM 5.8.13 The algorithm above correctly computes the chromatic number in a finite amount of time.
Proof.
Suppose that a graph G has n vertices and m edges. The number of pairs of non-adjacent vertices is na(G) = (n 2 ) −m. The proof is by induction on na.
If na(G) = 0 then G is a complete graph and the algorithm terminates immediately.
Now we note that na(G + e) < na(G) and na(G/e) < na(G): na(G + e) = (n 2 ) −(m + 1) = na(G) −1 5.8 Graph Coloring 123 and na(G/e) = (n −1 2 ) −(m −c), where c is the number of neighbors that v and w have in common. Then na(G/e) = (n −1 2 ) −m + c ≤ (n −1 2 ) −m + n −2 = (n −1)(n −2) 2 −m + n −2 = n(n −1) 2 −2(n −1) 2 −m + n −2 = (n 2 ) −m −1 = na(G) −1.
Now if na(G) > 0, G is not a complete graph, so there are non-adjacent vertices v and w. By the induction hypothesis the algorithm computes χ(G + e) and χ(G/e) correctly, and finally it computes χ(G) from these in one additional step.
While this algorithm is very inefficient, it is sufficiently fast to be used on small graphs with the aid of a computer.
EXAMPLE 5.8.14 We illustrate with a very simple graph: .
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4 The chromatic number of the graph at the top is min(3, 4) = 3. (Of course, this is quite easy to see directly.) Exercises 5.8.
1. Suppose G has n vertices and chromatic number k. Prove that G has at least (k 2 ) edges.
2. Find the chromatic number of the graph below by using the algorithm in this section. Draw all of the graphs G + e and G/e generated by the alorithm in a “tree structure” with the complete graphs at the bottom, label each complete graph with its chromatic number, then propogate the values up to the original graph.
124 Chapter 5 Graph Theory .
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• 3. Show that χ(G −v) is either χ(G) or χ(G) −1.
4. Prove theorem 5.8.10 without assuming any particular properties of the order v1, . . . , vn.
5. Prove theorem 5.8.12 as follows. By corollary 5.8.11 we need consider only regular graphs.
Regular graphs of degree 2 are easy, so we consider only regular graphs of degree at least 3.
If G is not 2-connected, show that the blocks of G may colored with ∆(G) colors, and then the colorings may be altered slightly so that they combine to give a proper coloring of G.
If G is 2-connected, show that there are vertices u, v, w such that u is adjacent to both v and w, v and w are not adjacent, and G −v −w is connected. Given such vertices, color v and w with color 1, then color the remaining vertices by a greedy algorithm similar to that in theorem 5.8.10, with u playing the role of vn.
To show the existence of u, v, w as required, let x be a vertex not adjacent to all other vertices. If G −x is 2-connected, let v = x, let w be at distance 2 from v (justify this), and let a path of length 2 be v, u, w. Use theorem 5.7.4 to show that u, v, w have the required properties.
If G −x is not 2-connected, let u = x and let v and w be (carefully chosen) vertices in two different endblocks of G −x. Show that u, v, w have the required properties.
Brooks proved the theorem in 1941; this simpler proof is due to Lov´ asz, 1975.
5.9 The Chromatic Polynomial We now turn to the number of ways to color a graph G with k colors.
Of course, if k < χ(G), this is zero. We seek a function PG(k) giving the number of ways to color G with k colors. Some graphs are easy to do directly.
EXAMPLE 5.9.1 If G is Kn, PG(k) = k(k −1)(k −2) · · · (k −n + 1), namely, the number of permutations of k things taken n at a time. Vertex 1 may be colored any of the k colors, vertex 2 any of the remaining k −1 colors, and so on. Note that when k < n, PG(k) = 0. By exercise 5 in section 1.8, we may also write PG(k) = n ∑ i=0 s(n, i)ki.
EXAMPLE 5.9.2 If G has n vertices and no edges, PG(k) = kn.
Given PG it is not hard to compute χ(G); for example, we could simply plug in the numbers 1, 2, 3, . . . for k until PG(k) is non-zero. This suggests it will be difficult (that is, time consuming) to compute PG. We can provide an easy mechanical procedure for the computation, quite similar to the algorithm we presented for computing χ(G).
5.10 Coloring Planar Graphs 125 Suppose G has edge e = {v, w}, and consider PG−e(k), the number of ways to color G −e with k colors. Some of the colorings of G −e are also colorings of G, but some are not, namely, those in which v and w have the same color. How many of these are there?
From our discussion of the algorithm for χ(G) we know this is the number of colorings of G/e. Thus, PG(k) = PG−e(k) −PG/e(k).
Since G −e and G/e both have fewer edges than G, we can compute PG by applying this formula recursively. Ultimately, we need only compute PG for graphs with no edges, which is easy, as in example 5.9.2.
Since PG(k) = kn when G has no edges, it is then easy to see, and to prove by induction, that PG is a polynomial.
THEOREM 5.9.3 For all G on n vertices, PG is a polynomial of degree n, and PG is called the chromatic polynomial of G.
Proof.
The proof is by induction on the number of edges in G. When G has no edges, this is example 5.9.2.
Otherwise, by the induction hypothesis, PG−e is a polynomial of degree n and PG/e is a polynomial of degree n −1, so PG = PG−e −PG/e is a polynomial of degree n.
The chromatic polynomial of a graph has a number of interesting and useful properties, some of which are explored in the exercises.
Exercises 5.9.
1. Show that the leading coefficient of PG is 1.
2. Suppose that G is not connected and has components C1, . . . , Ck. Show that PG = ∏k i=1 PCi.
3. Show that the constant term of PG(k) is 0. Show that the coefficient of k in PG(k) is non-zero if and only if G is connected.
4. Show that the coefficient of kn−1 in PG is −1 times the number of edges in G.
5. Show that G is a tree if and only if PG(k) = k(k −1)n−1.
6. Find the chromatic polynomial of Kn with one edge removed.
7. Find the chromatic polynomial of the cycle Cn, n ≥3.
5.10 Coloring Planar Graphs Now we return to the original graph coloring problem: coloring maps. As indicated in section 1.1, the map coloring problem can be turned into a graph coloring problem. Fig-ure 5.10.1 shows the example from section 1.1.
Graphs formed from maps in this way have an important property: they are planar.
126 Chapter 5 Graph Theory .
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• Figure 5.10.1 A map and its corresponding graph.
DEFINITION 5.10.1 A graph G is planar if it can be represented by a drawing in the plane so that no edges cross.
Note that this definition only requires that some representation of the graph has no crossing edges. Figure 5.10.2 shows two representations of K4; since in the second no edges cross, K4 is planar.
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• Figure 5.10.2 K4 drawn in two ways; the second shows that it is planar.
The number of colors needed to properly color any map is now the number of colors needed to color any planar graph. This problem was first posed in the nineteenth century, and it was quickly conjectured that in all cases four colors suffice. This was finally proved in 1976 (see figure 5.10.3) with the aid of a computer. In 1879, Alfred Kempe gave a proof that was widely known, but was incorrect, though it was not until 1890 that this was noticed by Percy Heawood, who modified the proof to show that five colors suffice to color any planar graph. We will prove this Five Color Theorem, but first we need some other results. We assume all graphs are simple.
THEOREM 5.10.2 Euler’s Formula Suppose G is a connected planar graph, drawn so that no edges cross, with n vertices and m edges, and that the graph divides the plane into r regions. Then r = m −n + 2.
5.10 Coloring Planar Graphs 127 Figure 5.10.3 The postmark on University of Illinois mail after the Four Color Theorem was proved.
Proof.
The proof is by induction on the number of edges. The base case is m = n −1, the minimum number of edges in a connected graph on n vertices. In this case G is a tree, and contains no cycles, so the number of regions is 1, and indeed 1 = (n −1) −n + 2.
Now suppose G has more than n −1 edges, so it has a cycle. Remove one edge from a cycle forming G′, which is connected and has r −1 regions, n vertices, and m −1 edges.
By the induction hypothesis r −1 = (m −1) −n + 2, which becomes r = m −n + 2 when we add 1 to each side.
LEMMA 5.10.3 Suppose G is a simple connected planar graph, drawn so that no edges cross, with n ≥3 vertices and m edges, and that the graph divides the plane into r regions.
Then m ≤3n −6.
Proof.
Let fi be the number of edges that adjoin region number i; if the same region is on both sides of an edge, that edge is counted twice. We call the edges adjoining a region the boundary edges of the region. Since G is simple and n ≥3, every region is bounded by at least 3 edges. Then ∑r i=1 fi = 2m, since each edge is counted twice, once for the region on each side of the edge. From r = m−n+ 2 we get 3r = 3m−3n+ 6, and because fi ≥3, 3r ≤∑r i=1 fi = 2m, so 3m −3n + 6 ≤2m, or m ≤3n −6 as desired.
THEOREM 5.10.4 K5 is not planar.
Proof.
K5 has 5 vertices and 10 edges, and 10 ̸≤3 · 5 −6, so by the lemma, K5 is not planar.
LEMMA 5.10.5 If G is planar then G has a vertex of degree at most 5.
Proof.
We may assume that G is connected (if not, work with a connected component of G). Suppose that d(vi) > 5 for all vi. Then 2m = ∑n i=1 d(vi) ≥6n. By lemma 5.10.3, 3n −6 ≥m so 6n −12 ≥2m. Thus 6n ≤2m ≤6n −12, a contradiction.
THEOREM 5.10.6 Five Color Theorem Every planar graph can be colored with 5 colors.
Proof.
The proof is by induction on the number of vertices n; when n ≤5 this is trivial.
128 Chapter 5 Graph Theory Now suppose G is planar on more than 5 vertices; by lemma 5.10.5 some vertex v has degree at most 5. By the induction hypothesis, G −v can be colored with 5 colors. Color the vertices of G, other than v, as they are colored in a 5-coloring of G −v. If d(v) ≤4, then v can be colored with one of the 5 colors to give a proper coloring of G with 5 colors.
So we now suppose d(v) = 5. If the five neighbors of v are colored with four or fewer of the colors, then again v can be colored to give a proper coloring of G with 5 colors.
Now we suppose that all five neighbors of v have a different color, as indicated in figure 5.10.4.
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Suppose that in G there is a path from v1 to v3, and that the vertices along this path are alternately colored red and green; call such a path a red-green alternating path. Then together with v, this path makes a cycle with v2 on the inside and v4 on the outside, or vice versa. This means there cannot be a purple-blue alternating path from v2 to v4.
Supposing that v2 is inside the cycle, we change the colors of all vertices inside the cycle colored purple to blue, and all blue vertices are recolored purple. This is still a proper coloring of all vertices of G except v, and now no neighbor of v is purple, so by coloring v purple we obtain a proper coloring of G.
If there is no red-green alternating path from v1 to v3, then we recolor vertices as follows: Change the color of v1 to green. Change all green neighbors of v1 to red. Continue to change the colors of vertices from red to green or green to red until there are no conflicts, that is, until a new proper coloring is obtained. Because there is no red-green alternating path from v1 to v3, the color of v3 will not change. Now no neighbor of v is colored red, so by coloring v red we obtain a proper coloring of G.
Exercises 5.10.
1. Show K3,3 is not planar. (Prove a lemma like lemma 5.10.3 for bipartite graphs, then do something like the proof of theorem 5.10.4.) What is the chromatic number of K3,3?
5.11 Directed Graphs 129 5.11 Directed Graphs A directed graph, also called a digraph, is a graph in which the edges have a direction.
This is usually indicated with an arrow on the edge; more formally, if v and w are vertices, an edge is an unordered pair {v, w}, while a directed edge, called an arc, is an ordered pair (v, w) or (w, v). The arc (v, w) is drawn as an arrow from v to w. If a graph contains both arcs (v, w) and (w, v), this is not a “multiple edge”, as the arcs are distinct. It is possible to have multiple arcs, namely, an arc (v, w) may be included multiple times in the multiset of arcs. As before, a digraph is called simple if there are no loops or multiple arcs.
We denote by E − v the set of all arcs of the form (w, v), and by E+ v the set of arcs of the form (v, w). The indegree of v, denoted d −(v), is the number of arcs in E − v , and the outdegree, d +(v), is the number of arcs in E+ v . If the vertices are v1, v2, . . . , vn, the degrees are usually denoted d− 1 , d− 2 , . . . , d− n and d+ 1 , d+ 2 , . . . , d+ n . Note that both ∑n i=0 d − i and ∑n i=0 d + i count the number of arcs exactly once, and of course ∑n i=0 d − i = ∑n i=0 d + i .
A walk in a digraph is a sequence v1, e1, v2, e2, . . . , vk−1, ek−1, vk such that ek = (vi, vi+1); if v1 = vk, it is a closed walk or a circuit. A path in a digraph is a walk in which all vertices are distinct. It is not hard to show that, as for graphs, if there is a walk from v to w then there is a path from v to w.
Many of the topics we have considered for graphs have analogues in digraphs, but there are many new topics as well. We will look at one particularly important result in the latter category.
DEFINITION 5.11.1 A network is a digraph with a designated source s and target t ̸= s . In addition, each arc e has a positive capacity, c(e).
Networks can be used to model transport through a physical network, of a physical quantity like oil or electricity, or of something more abstract, like information.
DEFINITION 5.11.2 A flow in a network is a function f from the arcs of the digraph to R, with 0 ≤f(e) ≤c(e) for all e, and such that ∑ e∈E+ v f(e) = ∑ e∈E− v f(e), for all v other than s and t.
We wish to assign a value to a flow, equal to the net flow out of the source. Since the substance being transported cannot “collect” or “originate” at any vertex other than s and t, it seems reasonable that this value should also be the net flow into the target.
130 Chapter 5 Graph Theory Before we prove this, we introduce some new notation. Suppose that U is a set of vertices in a network, with s ∈U and t / ∈U. Let ⇀ U be the set of arcs (v, w) with v ∈U, w / ∈U, and ↼ U be the set of arcs (v, w) with v / ∈U, w ∈U.
THEOREM 5.11.3 For any flow f in a network, the net flow out of the source is equal to the net flow into the target, namely, ∑ e∈E+ s f(e) − ∑ e∈E− s f(e) = ∑ e∈E− t f(e) − ∑ e∈E+ t f(e).
Proof.
We will show first that for any U with s ∈U and t / ∈U, ∑ e∈E+ s f(e) − ∑ e∈E− s f(e) = ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e).
Consider the following: S = ∑ v∈U ∑ e∈E+ v f(e) − ∑ e∈E− v f(e) .
The quantity ∑ e∈E+ v f(e) − ∑ e∈E− v f(e) is zero except when v = s, by the definition of a flow. Thus, the entire sum S has value ∑ e∈E+ s f(e) − ∑ e∈E− s f(e).
On the other hand, we can write the sum S as ∑ v∈U ∑ e∈E+ v f(e) − ∑ v∈U ∑ e∈E− v f(e).
Every arc e = (x, y) with both x and y in U appears in both sums, that is, in ∑ v∈U ∑ e∈E+ v f(e), when v = x, and in ∑ v∈U ∑ e∈E− v f(e), when v = y, and so the flow in such arcs contributes 0 to the overall value. Thus, only arcs with exactly one endpoint in U make a non-zero contribution, so the entire sum reduces 5.11 Directed Graphs 131 to ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e).
Thus ∑ e∈E+ s f(e) − ∑ e∈E− s f(e) = S = ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e), as desired.
Now let U consist of all vertices except t. Then ∑ e∈E+ s f(e) − ∑ e∈E− s f(e) = ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e) = ∑ e∈E− t f(e) − ∑ e∈E+ t f(e), finishing the proof.
DEFINITION 5.11.4 The value of a flow, denoted val(f), is ∑ e∈E+ s f(e)−∑ e∈E− s f(e).
A maximum flow in a network is any flow f whose value is the maximum among all flows.
We next seek to formalize the notion of a “bottleneck”, with the goal of showing that the maximum flow is equal to the amount that can pass through the smallest bottleneck.
DEFINITION 5.11.5 A cut in a network is a set C of arcs with the property that every path from s to t uses an arc in C, that is, if the arcs in C are removed from the network there is no path from s to t. The capacity of a cut, denoted c(C), is ∑ e∈C c(e).
A minimum cut is one with minimum capacity. A cut C is minimal if no cut is properly contained in C.
Note that a minimum cut is a minimal cut. Clearly, if U is a set of vertices containing s but not t, then ⇀ U is a cut.
LEMMA 5.11.6 Suppose C is a minimal cut. Then there is a set U containing s but not t such that C = ⇀ U .
Proof.
Let U be the set of vertices v such that there is a path from s to v using no arc in C.
Suppose that e = (v, w) ∈C. Since C is minimal, there is a path P from s to t using e but no other arc in C. Thus, there is a path from s to v using no arc of C, so v ∈U. If there is a path from s to w using no arc of C, then this path followed by the portion of P 132 Chapter 5 Graph Theory that begins with w is a walk from s to t using no arc in C. This implies there is a path from s to t using no arc in C, a contradiction. Thus w / ∈U and so e ∈⇀ U . Hence, C ⊆⇀ U .
Suppose that e = (v, w) ∈⇀ U . Then v ∈U and w / ∈U, so every path from s to w uses an arc in C. Since v ∈U, there is a path from s to v using no arc of C, and this path followed by e is a path from s to w. Hence the arc e must be in C, so ⇀ U ⊆C.
We have now shown that C = ⇀ U .
Now we can prove a version of the important max-flow, min cut theorem.
THEOREM 5.11.7 Suppose in a network all arc capacities are integers. Then the value of a maximum flow is equal to the capacity of a minimum cut. Moreover, there is a maximum flow f for which all f(e) are integers.
Proof.
First we show that for any flow f and cut C, val(f) ≤c(C). It suffices to show this for a minimum cut C, and by lemma 5.11.6 we know that C = ⇀ U for some U. Using the proof of theorem 5.11.3 we have: val(f) = ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e) ≤ ∑ e∈⇀ U f(e) ≤ ∑ e∈⇀ U c(e) = c(⇀ U ).
Now if we find a flow f and cut C with val(f) = c(C), it follows that f is a maximum flow and C is a minimum cut. We present an algorithm that will produce such an f and C.
Given a flow f, which may initially be the zero flow, f(e) = 0 for all arcs e, do the following: 0. Let U = {s}.
Repeat the next two steps until no new vertices are added to U.
1. If there is an arc e = (v, w) with v ∈U and w / ∈U, and f(e) < c(e), add w to U.
2. If there is an arc e = (v, w) with v / ∈U and w ∈U, and f(e) > 0, add v to U.
When this terminates, either t ∈U or t / ∈U. If t ∈U, there is a sequence of distinct vertices s = v1, v2, v3, . . . , vk = t such that for each i, 1 ≤i < k, either e = (vi, vi+1) is an arc with f(e) < c(e) or e = (vi+1, vi) is an arc with f(e) > 0. Update the flow by adding 1 to f(e) for each of the former, and subtracting 1 from f(e) for each of the latter. This new flow f ′ is still a flow: In the first case, since f(e) < c(e), f ′(e) ≤c(e), and in the second case, since f(e) > 0, f ′(e) ≥0. It is straightforward to check that for each vertex vi, 1 < i < k, that ∑ e∈E+ vi f ′(e) = ∑ e∈E− vi f ′(e).
In addition, val(f ′) = val(f) + 1. Now rename f ′ to f and repeat the algorithm.
5.11 Directed Graphs 133 Eventually, the algorithm terminates with t / ∈U and flow f. This implies that for each e = (v, w) with v ∈U and w / ∈U, f(e) = c(e), and for each e = (v, w) with v / ∈U and w ∈U, f(e) = 0. The capacity of the cut ⇀ U is ∑ e∈⇀ U c(e).
The value of the flow f is ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e) = ∑ e∈⇀ U c(e) − ∑ e∈↼ U 0 = ∑ e∈⇀ U c(e).
Thus we have found a flow f and cut ⇀ U such that val(f) = c(⇀ U ), as desired.
The max-flow, min-cut theorem is true when the capacities are any positive real num-bers, though of course the maximum value of a flow will not necessarily be an integer in this case. It is somewhat more difficult to prove; a proof involves limits.
We have already proved that in a bipartite graph, the size of a maximum matching is equal to the size of a minimum vertex cover, theorem 4.5.6. This turns out to be essentially a special case of the max-flow, min-cut theorem.
COROLLARY 5.11.8 In a bipartite graph G, the size of a maximum matching is the same as the size of a minimum vertex cover.
Proof.
Suppose the parts of G are X = {x1, x2, . . . , xk} and Y = {y1, y2, . . . , yl}. Create a network as follows: introduce two new vertices s and t and arcs (s, xi) for all i and (yi, t) for all i. For each edge {xi, yj} in G, let (xi, yj) be an arc. Let c(e) = 1 for all arcs e. Now the value of a maximum flow is equal to the capacity of a minimum cut.
Let C be a minimum cut. If (xi, yj) is an arc of C, replace it by arc (s, xi). This is still a cut, since any path from s to t including (xi, yj) must include (s, xi). Thus, we may suppose that C contains only arcs of the form (s, xi) and (yi, t). Now it is easy to see that K = {xi|(s, xi) ∈C} ∪{yi|(yi, t) ∈C} is a vertex cover of G with the same size as C.
Let f be a maximum flow such that f(e) is an integer for all e, and val(f) = c(C), which is possible by the max-flow, min-cut theorem. Consider the set of edges M = {{xi, yj}|f((xi, yj)) = 1}.
If {xi, yj} and {xi, ym} are both in this set, then the flow out of vertex xi is at least 2, but there is only one arc into xi, (s, xi), with capacity 1, contradicting the definition of a 134 Chapter 5 Graph Theory flow. Likewise, if {xi, yj} and {xm, yj} are both in this set, then the flow into vertex yj is at least 2, but there is only one arc out of yj, (yj, t), with capacity 1, also a contradiction.
Thus M is a matching. Moreover, if U = {s, x1, . . . , xk} then the value of the flow is ∑ e∈⇀ U f(e) − ∑ e∈↼ U f(e) = ∑ e∈⇀ U f(e) = |M| · 1 = |M|.
Thus |M| = val(f) = c(C) = |K|, so we have found a matching and a vertex cover with the same size. This implies that M is a maximum matching and K is a minimum vertex cover.
Exercises 5.11.
1. Connectivity in digraphs turns out to be a little more complicated than connectivity in graphs. A digraph is connected if the underlying graph is connected. (The underlying graph of a digraph is produced by removing the orientation of the arcs to produce edges, that is, replacing each arc (v, w) by an edge {v, w}. Even if the digraph is simple, the underlying graph may have multiple edges.) A digraph is strongly connected if for every vertices v and w there is a walk from v to w. Give an example of a digraph that is connected but not strongly connected.
2. A digraph has an Euler circuit if there is a closed walk that uses every arc exactly once. Show that a digraph with no vertices of degree 0 has an Euler circuit if and only if it is connected and d +(v) = d −(v) for all vertices v.
3. A tournament is an oriented complete graph. That is, it is a digraph on n vertices, containing exactly one of the arcs (v, w) and (w, v) for every pair of vertices. A Hamilton path is a walk that uses every vertex exactly once. Show that every tournament has a Hamilton path.
4. Interpret a tournament as follows: the vertices are players. If (v, w) is an arc, player v beat w. Say that v is a champion if for every other player w, either v beat w or v beat a player who beat w. Show that a player with the maximum number of wins is a champion. Find a 5-vertex tournament in which every player is a champion.
6 P´ olya–Redfield Counting We have talked about the number of ways to properly color a graph with k colors, given by the chromatic polynomial. For example, the chromatic polynomial for the graph in figure 6.0.1 is k4 −4k3 + 6k2 −3k, and f(2) = 2. The two colorings are shown in the figure, but in an obvious sense they are the same coloring, since one can be turned into the other by simply rotating the graph. We will consider a slightly different sort of coloring problem, in which we count the “truly different” colorings of objects.
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• Figure 6.0.1 C4 drawn as a square, colored in two ways.
Many of the “objects” we color will appear to be graphs, but we will usually be interested in them as geometric objects rather than graphs, and we will not require that adjacent vertices be different colors. This will simplify the problems; counting the number of different proper colorings of graphs can also be done, but it is more complicated.
So consider this problem: How many different ways are there to color the vertices of a regular pentagon with k colors? The number of ways to color the vertices of a fixed pentagon is k5, but this includes many duplicates, that is, colorings that are really the same.
But what do we mean by “the same” in this case?
We might mean that two colorings are the same if we can rotate one to get the other. But what about the colorings in figure 6.0.2? Are they the same? Neither can be rotated to produce the other, but either 135 136 Chapter 6 P´ olya–Redfield Counting can be flipped or reflected through a vertical line to produce the other. In fact we are free to decide what “the same” means, but we will need to make sure that our requirements are consistent.
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• Figure 6.0.2 Two colorings of a pentagon.
As an example of what can go wrong if we’re not careful, note that there are five lines through which the pentagon can be reflected onto itself. Suppose we want to consider colorings to be “the same” if one can be produced from the other by a reflection, but not if one can be obtained from the other by rotation. Surely if one coloring can be obtained by two reflections in a row from another, then these colorings should also be the same.
But two reflections in a row equal a rotation, as shown in figure 6.0.3. So whenever we have some motions that identify two colorings, we are required to include all combinations of the motions as well. In addition, any time we include a motion, we must include the “inverse” motion. For example, if we say a rotation by 72◦degrees produces a coloring that we consider to be the same, a rotation by −72◦must be included as well (we may think of this as a rotation by 288◦). Finally, since any coloring is the same as itself, we must always include the “trivial” motion, namely, doing nothing, or rotation by 0◦if you prefer.
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• Figure 6.0.3 Two reflections equal a rotation.
6.1 Groups of Symmetries 137 6.1 Groups of Symmetries The motions we want to consider can be thought of as permutations, that is, as bijections.
For example, the rotation in figure 6.1.1 can be thought of as the function ϕ given by ϕ(1) = 3 ϕ(2) = 4 ϕ(3) = 5 ϕ(4) = 1 ϕ(5) = 2, or more compactly we can write this function as ( 1 2 3 4 5 3 4 5 1 2 ) .
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1 Figure 6.1.1 The rotation ( 1 2 3 4 5 3 4 5 1 2 ) .
As we would hope, doing two motions in a row corresponds to the compostion of the as-sociated functions. For example, the reflection ( 1 2 3 4 5 3 2 1 5 4 ) is shown in figure 6.1.2.
Doing first the rotation of figure 6.1.1 and then this reflection is shown in figure 6.1.3, and this does indeed correspond to ( 1 2 3 4 5 3 2 1 5 4 ) ◦ ( 1 2 3 4 5 3 4 5 1 2 ) = ( 1 2 3 4 5 1 5 4 3 2 ) .
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1 Figure 6.1.2 The reflection ( 1 2 3 4 5 3 2 1 5 4 ) .
138 Chapter 6 P´ olya–Redfield Counting .
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4 Figure 6.1.3 The compostion ( 1 2 3 4 5 3 2 1 5 4 ) ◦ ( 1 2 3 4 5 3 4 5 1 2 ) = ( 1 2 3 4 5 1 5 4 3 2 ) .
With some restrictions, we may choose any permutations of the vertices as the allow-able rearrangements giving colorings that are the same. We have discussed the restrictions in general terms; in terms of permuations we require the following: Suppose that G is a set of permutations that we wish to use to define the “same coloring” relation. Then the following must be true: 1. If ϕ and σ are in G, so is ϕ ◦σ.
2. The identity permutation, id, is in G.
3. If ϕ ∈G, ϕ−1 ∈G.
DEFINITION 6.1.1 If G has the three properties above it is called a group of per-mutations.
EXAMPLE 6.1.2 The group of all permutations of {1, 2, . . . , n} is denoted Sn, the symmetric group on n items. It satisfies the three required conditions by simple prop-erties of bijections.
In the case of the regular pentagon, there are a number of groups of permutations, but two are of primary interest. The five possible rotations (including the trivial rotation) form a group, the cyclic group of size 5. The total number of “rigid motions”, that is, any combination of rotations and reflections that leave the pentagon superimposed on itself, is 10: Once the position of vertex 1 is established, the other vertices can increase from 1 either clockwise or counterclockwise. The rotations provide all of the former, and it is easy to check that the five reflections provide the counterclockwise positions. This is called a dihedral group and denoted D5.
Suppose that G is some group of permutations of an object. If ϕ ∈G, then ϕ induces a function on the colorings of the object in a natural way, and we can use the same symbol ϕ to represent this function without confusion. If c is a coloring of the object, then ϕ(c) is the coloring that results by applying ϕ to the colored object, moving the colors with the object. See figure 6.1.4 for examples. We say that G acts on the set of colorings C.
6.1 Groups of Symmetries 139 ϕ: .
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• Figure 6.1.4 Some examples of an induced function on colorings.
If we apply all permutations in G to a coloring c, we get all the colorings that we consider to be the same as c modulo G. More formally, define c1 ∼c2 if there is a ϕ ∈G such that ϕ(c1) = c2; ∼is an equivalence relation on the colorings. The equivalence classes, called orbits in this context, group colorings that are the same together. The number of truly different colorings that we want to count is then the number of orbits.
The total number of colorings of the pentagon with k colors is k5. If all orbits were the same size, say s, then the number of orbits would be k5/s. Unfortunately, this is not true. In figure 6.1.4 we see a coloring whose orbit has size at least 3, but the pentagon with all vertices colored red has orbit size 1.
Exercises 6.1.
1. Find the 12 permutations of the vertices of the regular tetrahedron corresponding to the 12 rigid motions of the regular tetrahedron. Use the labeling below.
140 Chapter 6 P´ olya–Redfield Counting 1 2 3 4 2. Find the 12 permutations of the edges of the regular tetrahedron corresponding to the 12 rigid motions of the regular tetrahedron. Use the labeling below.
a b c d e f 6.2 Burnside's Theorem Burnside’s Theorem will allow us to count the orbits, that is, the different colorings, in a variety of problems. We first need some lemmas.
If c is a coloring, [c] is the orbit of c, that is, the equivalence class of c. Let G(c) be the set of permutations in G that fix c, that is, those ϕ such that ϕ(c) = c; the permutation in figure 6.1.4 fixes the coloring in the bottom row, for example.
LEMMA 6.2.1 G(c) is a group of permutations.
Proof.
We check the properties of a group from definition 6.1.1.
Suppose ϕ and σ both fix c; then ϕ(σ(c)) = ϕ(c) = c, so ϕ ◦σ fixes c and ϕ ◦σ ∈G(c).
The identity permutation fixes all colorings, so id ∈G(c).
If ϕ(c) = c then ϕ−1(c) = ϕ−1(ϕ(c)) = id(c) = c, so ϕ−1 ∈G(c).
LEMMA 6.2.2 |G| = |[c]| · |G(c)|.
Proof.
For ϕ and σ in G, define ϕ ∼σ if σ−1 ◦ϕ ∈G(c). This is an equivalence relation: 1. σ−1 ◦σ is the identity function, which is in G(c). Thus σ ∼σ, so the relation is reflexive.
2. If ϕ ∼σ, σ−1 ◦ϕ ∈G(c). Then (σ−1 ◦ϕ)−1 ∈G(c), and (σ−1 ◦ϕ)−1 = ϕ−1 ◦σ ∈ G(c), so σ ∼ϕ and ∼is symmetric.
6.2 Burnside’s Theorem 141 3. If ϕ ∼σ and σ ∼τ, then σ−1◦ϕ ∈G(c) and τ −1◦σ ∈G(c), so (τ −1◦σ)◦(σ−1◦ϕ) ∈ G(c). Since (τ −1 ◦σ) ◦(σ−1 ◦ϕ) = τ −1 ◦ϕ, ϕ ∼τ, and ∼is transitive.
Now we claim that the equivalence class of ϕ is A = {ϕ ◦σ | σ ∈G(c)}. First, suppose that σ ∈G(c); then ϕ−1 ◦ϕ ◦σ = σ ∈G(c), so ϕ ◦σ ∼ϕ and A ⊆[ϕ]. Next, suppose ϕ ∼τ, so τ −1 ◦ϕ = γ ∈G(c). Then ϕ ◦γ−1 = τ, so τ ∈A and [ϕ] ⊆A.
Now we show that each equivalence class is the same size as G(c). Define f: G(c) → {ϕ ◦σ | σ ∈G(c)} by f(γ) = ϕ ◦γ. If f(γ1) = f(γ2), then ϕ ◦γ1 = ϕ ◦γ2 ϕ−1 ◦ϕ ◦γ1 = ϕ−1 ◦ϕ ◦γ2 γ1 = γ2 so f is 1–1. Since every ϕ ◦γ ∈{ϕ ◦σ | σ ∈G(c)} is f(γ), f is onto.
Thus the number of equivalence classes is |G|/|G(c)|.
Finally, we show that the number of equivalence classes is |[c]|. Let the set of equiva-lence classes in G be E, that is, E = {[ϕ] | ϕ ∈G}. We define g: [c] →E and show that g is a bijection. Suppose d ∈[c], so d = σ(c) for some σ ∈G. Let g(d) = [σ].
First, we show g is well-defined. If d = σ1(c) = σ2(c), then σ−1 2 ◦σ1(c) = c, so σ1 ∼σ2 and [σ1] = [σ2], that is, g(σ1(c)) = g(σ2(c)).
Next, suppose g(d1) = g(d2). This means that d1 = σ1(c), d2 = σ2(c), and [σ1] = [σ2].
Hence σ−1 2 ◦σ1(c) = c, so σ1(c) = σ2(c) and thus d1 = d2, so g is 1–1.
Suppose that [σ] ∈E. Then g(σ(c)) = [σ], so g is onto.
Thus we have |[c]| = |E| = |G| |G(c)| and |G(c)| · |[c]| = |G| as desired.
COROLLARY 6.2.3 If c ∼d then |G(c)| = |G(d)|.
Proof.
Since c ∼d, [c] = [d], and |G| |G(c)| = |[c]| = |[d]| = |G| |G(d)| |G(c)| = |G(d)| DEFINITION 6.2.4 If group G acts on the colorings of an object and σ ∈G, fix(σ) is the set of colorings that are fixed by σ.
142 Chapter 6 P´ olya–Redfield Counting THEOREM 6.2.5 Burnside’s Theorem If group G acts on the colorings of an object, the number of distinct colorings modulo G is 1 |G| ∑ σ∈G | fix(σ)|.
Proof.
Let C be the set of all colorings, and let O be the set of orbits. Let c1, c2, . . . , ck be a list of colorings, one in each orbit, so that the orbits are [c1], [c2], . . . , [ck]. Consider the sum ∑ c∈C |G(c)|: ∑ c∈C |G(c)| = k ∑ i=1 ∑ c∈[ci] |G(c)| = k ∑ i=1 ∑ c∈[ci] |G(ci)| = k ∑ i=1 ∑ c∈[ci] |G| |[ci]| = k ∑ i=1 |[ci]| |G| |[ci]| = k ∑ i=1 |G| = |G| k ∑ i=1 1 = |G||O|.
Then |O| = 1 |G| ∑ c∈C |G(c)|.
This already gives an interesting formula for |O|, but it is unwieldy, since the number of colorings is typically quite large; indeed, since we typically want to compute the number of orbits for k colors, the number of colorings is not a fixed number.
With just a little more work we can fix this problem: ∑ c∈C |G(c)| = ∑ c∈C ∑ σ∈G(c) 1 = ∑ σ∈G ∑ c∈fix(σ) 1 = ∑ σ∈G | fix(σ)|.
Now |O| = 1 |G| ∑ σ∈G | fix(σ)| as desired.
6.2 Burnside’s Theorem 143 Since the group of permutations in a typical problem is fairly small, the sum in Burn-side’s Theorem is usually manageable.
Moreover, we can make the task of computing | fix(σ)| fairly straightforward. Let’s consider a particular example, the permutation of figure 6.1.4, shown again in figure 6.2.1. If we are using k colors, how many colorings of the pentagon are fixed by this permutation? Since the permutation swaps vertices 2 and 5, they must be the same color if ϕ is to fix the coloring. Likewise vertices 3 and 4 must be the same color; vertex 1 can be any color. Thus, the number of colorings fixed by ϕ is k3. It is easy to see that every “flip” permutation of the pentagon is essentially the same, so for each of the five flip permutations, the size of fix(σ) is k3.
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Every permutation can be written in cycle form: The permutation in figure 6.2.1, for example, is (1)(2, 5)(3, 4). A cycle in this context is a sequence (x1, x2, . . . , xk), meaning that ϕ(x1) = x2, ϕ(x2) = x3, and so on until ϕ(xk) = x1. Following our reasoning above, the vertices in each cycle must be colored the same color, and the total number of colors fixed by ϕ is km, where m is the number of cycles.
Let’s apply this to another permutation, shown in figure 6.2.2.
This permutation consists of a single cycle, so every vertex must have the same color, and the number of colorings fixed by ϕ is k1. All rotations of the pentagon consist of a single cycle except the trivial rotation, that is, the identity permutation. In cycle form, the identity permutation is (1)(2)(3)(4)(5), so the number of colorings fixed by the identity is k5. Putting everything together, we thus have |O| = 1 10(k5 + k + k + k + k + k3 + k3 + k3 + k3 + k3) = 1 10(k5 + 5k3 + 4k).
For example, the number of different 3-colorings is (35 + 5 · 33 + 4 · 3)/10 = 39.
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144 Chapter 6 P´ olya–Redfield Counting EXAMPLE 6.2.6 We find the number of distinct colorings of the vertices of a square with k colors, modulo D4, the dihedral group of size 8. The elements of D4 are the four rotations r0, r90, r180, r270, where ri is the counterclockwise rotation by i degrees, and the four reflections fH, fV , fD, fA, as indicated in figure 6.2.3.
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In cycle notation these permutations are: r0 = (1)(2)(3)(4) r90 = (1, 4, 3, 2) r180 = (1, 3)(2, 4) r270 = (1, 2, 3, 4) fH = (1, 4)(2, 3) fV = (1, 2)(3, 4) fD = (1)(2, 4)(3) fA = (1, 3)(2)(4).
so the number of colorings is f(k) = 1 8(k4 + k + k2 + k + k2 + k2 + k3 + k3) = 1 8(k4 + 2k3 + 3k2 + 2k).
For example, f(2) = 6; the six colorings are shown in figure 6.2.4.
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EXAMPLE 6.2.7 Here is a more complicated example: how many different graphs are there on four vertices? In this case, of course, “different” means “non-isomorphic”. We can interpret this as a coloring problem: Color the edges of the complete graph K4 with 6.2 Burnside’s Theorem 145 two colors, say black and white. The black edges form a graph; the white edges are the ones left out of the graph. The group G we need to consider is all permutations of the six edges of K4 induced by a permutation of the vertices, so |G| = 4! = 24. All we need to know is the number of cycles in each permutation; we consider a number of cases.
Case 1. The identity permutation on the vertices induces the identity permutation on the edges, with 6 cycles, so the contribution to the sum is 26.
Case 2. A 4-cycle on the vertices induces a permutation of the edges consisting of one 4-cycle and one 2-cycle, that is, two cycles. There are 3! = 6 4-cycles on the vertices, so the contribution of all of these is 6 · 22.
Case 3. A permutation of the vertices consisting of a 3-cycle and a 1-cycle induces a permutation of the edges consisting of two 3-cycles. There are 4 · 2 = 8 such permutations of the vertices, so the contribution of all is 8 · 22.
Case 4.
A permutation of the vertices consisting of two 2-cycles induces a permuta-tion of the edges consisting of two 1-cycles and two 2-cycles. There are 1 2 (4 2 ) = 3 such permutations, so the contribution is 3 · 24.
Case 5. A permutation of the vertices consisting of a 2-cycle and two 1-cycles induces a permutation of the edges consisting of two 1-cycles and two 2-cycles. There are (4 2 ) = 6 such permutations, so the contribution is 6 · 24.
The number of distinct colorings, that is, the number of distinct graphs on four vertices, is 1 24(26 + 6 · 22 + 8 · 22 + 3 · 24 + 6 · 24) = 1 24(264) = 11.
It is possible, though a bit difficult, to see that for n vertices the result is f(n) = ∑ j n ∏ k=1 1 kjkjk!
⌊n/2⌋ ∏ k=1 2kj2k ⌊(n−1)/2⌋ ∏ k=1 2kj2k+1 ⌊n/2⌋ ∏ k=1 2kC(jk,2) ∏ 1≤r<s≤n−1 2gcd(r,s)jrjs (6.2.1) where the sum is over all partitions j = (j1, j2, . . . , jn) of n, that is, over all j such that j1 + 2j2 + 3j3 + · · · + njn = n, and C(m, 2) = (m 2 ) . With this formula and a computer it is easy to compute the number of graphs when n is not too large; for example, f(5) = 34, so there are 34 different five-vertex graphs.
In light of the forgoing discussion, we can restate theorem 6.2.5. If σ is a permutation, let #σ denote the number of cycles when σ is written in cycle form.
COROLLARY 6.2.8 If group G acts on the colorings of an object, the number of distinct colorings modulo G with k colors is 1 |G| ∑ σ∈G k#σ.
146 Chapter 6 P´ olya–Redfield Counting Exercises 6.2.
1. Write the 12 permutations of the vertices of the regular tetrahedron corresponding to the 12 rigid motions of the regular tetrahedron in cycle form. Use the labeling below.
1 2 3 4 2. Find the number of different colorings of the vertices of a regular tetrahedron with k colors, modulo the rigid motions.
3. Write the 12 permutations of the edges of the regular tetrahedron corresponding to the 12 rigid motions of the regular tetrahedron in cycle form. Use the labeling below.
a b c d e f 4. Find the number of different colorings of the edges of a regular tetrahedron with k colors, modulo the rigid motions.
5. Find the number of non-isomorphic graphs on 5 vertices “by hand”, that is, using the method of example 6.2.7.
6.3 P olya-Redfield Counting Suppose we are interested in a more detailed inventory of the colorings of an object, namely, instead of the total number of colorings we seek the number of colorings with a given number of each color. The method presented here was first published by J. Howard Redfield in 1927. In 1937 it was independently rediscovered by George P´ olya, who then greatly popularized the result by applying it to many counting problems, in particular to the enumeration of chemical compounds.
EXAMPLE 6.3.1 How many distinct ways are there to color the vertices of a regular pentagon modulo D5 so that one vertex is red, two are blue, and two are green?
6.3 P´ olya-Redfield Counting 147 We can approach this as before, that is, the answer is 1 |D5| ∑ σ∈D5 | fix(σ)|, where fix(σ) now means the colorings with one red, two blues, and two greens that are fixed by σ. No longer can we use the simple expression of corollary 6.2.8.
The identity permutation fixes all colorings, so we need to know how many colorings of the pentagon use one red, two blues, and two greens. This is an easy counting problem: the number is (5 2 )(3 2 ) = 30.
If σ is a non-trivial rotation, | fix(σ)| = 0, since the only colorings fixed by a rotation have all vertices the same color.
If σ is a reflection, the single vertex fixed by σ must be red, and then the remaining 2-cycles are colored blue and green in one of two ways, so | fix(σ)| = 2.
Thus, the number of distinct colorings is 1 10(30 + 0 + 0 + 0 + 0 + 2 + 2 + 2 + 2 + 2) = 4.
What we seek is a way to streamline this process, since in general the computations of | fix(σ)| can be tedious. We begin by recasting the formula of corollary 6.2.8.
DEFINITION 6.3.2 The type of a permutation σ ∈Sn is τ(σ) = (τ1(σ), τ2(σ), . . . , τn(σ)), where τi(σ) is the number of i-cycles in the cycle form of σ.
Note that ∑n i=1 τi(σ) = #σ. Now instead of the simple 1 |G| ∑ σ∈G k#σ let us write 1 |G| ∑ σ∈G xτ1(σ) 1 xτ2(σ) 2 · · · xτn(σ) n .
If we substitute xi = k for every i, we get the original form of the sum, but the new version carries more information about each σ.
Suppose we want to know the number of colorings fixed by some σ that use i reds and j blues, where of course i+j = n. Using ideas familiar from generating functions, consider the following expression: (r + b)τ1(σ)(r2 + b2)τ2(σ) · · · (rn + bn)τn(σ).
If we multiply out, we get a sum of terms of the form rpbq, each representing some particular way of coloring the vertices of cycles red and blue so that the total number of red vertices 148 Chapter 6 P´ olya–Redfield Counting is p and the number of blue vertices is q, and moreover this coloring will be fixed by σ.
When we collect like terms, the coefficient of ribj is the number of colorings fixed by σ that use i reds and j blues. This means that the coefficient of ribj in ∑ σ∈G (r + b)τ1(σ)(r2 + b2)τ2(σ) · · · (rn + bn)τn(σ) is ∑ σ∈G | fix(σ)| where fix(σ) is the set of colorings using i reds and j blues that are fixed by σ. Finally, then, the number of distinct colorings using i reds and j blues is this coefficient divided by |G|. This means that by multiplying out 1 |G| ∑ σ∈G (r + b)τ1(σ)(r2 + b2)τ2(σ) · · · (rn + bn)τn(σ) and collecting like terms, we get a list of the number of distinct colorings using any combi-nation of reds and blues, each the coefficient of a different term; we call this the inventory of colorings. If we substitute r = 1 and b = 1, we get the sum of the coefficients, namely, the total number of distinct colorings with two colors.
DEFINITION 6.3.3 The cycle index of G is PG = 1 |G| ∑ σ∈G n ∏ i=1 xτi(σ) i .
EXAMPLE 6.3.4 Consider again example 6.2.6, in which we found the number of colorings of a square with two colors. The cycle index of D4 is 1 8(x4 1 + x1 4 + x2 2 + x1 4 + x2 2 + x2 2 + x2 1x2 + x2 1x2) = 1 8x4 1 + 1 4x2 1x2 + 3 8x2 2 + 1 4x4.
Substituting as above gives 1 8(r + b)4 + 1 4(r + b)2(r2 + b2) + 3 8(r2 + b2)2 + 1 4(r4 + b4) = r4 + r3b + 2r2b2 + rb3 + b4.
Thus there is one all red coloring, one with three reds and one blue, and so on, as shown in figure 6.2.4.
There is nothing special about the use of two colors. If we want to use three colors, we substitute ri + bi + gi for xi in the cycle index, and for k colors we substitute something like ci 1 + ci 2 + ci 3 + · · · + ci k.
6.3 P´ olya-Redfield Counting 149 EXAMPLE 6.3.5 Let’s do the number of 3-colorings of the square. Since we already have the cycle index, we need only substitute xi = ri + bi + gi and expand. We get 1 8(r + b + g)4 + 1 4(r + b + g)2(r2 + b2 + g2) + 3 8(r2 + b2 + g2)2 + 1 4(r4 + b4 + g4) = b4 + b3g + b3r + 2b2g2 + 2b2gr + 2b2r2 + bg3 + 2bg2r + 2bgr2+ br3 + g4 + g3r + 2g2r2 + gr3 + r4.
So, for example, there are two squares with two blue vertices, one green, and one red, from the b2gr term.
EXAMPLE 6.3.6 Consider again example 6.2.7, in which we counted the number of four-vertex graphs. Following that example, we get PG = 1 24(x6 1 + 6x2x4 + 8x2 3 + 3x2 1x2 2 + 6x2 1x2 2), and substituting for the variables xi gives r6 + r5b + 2r4b2 + 3r3b3 + 2r2b4 + rb5 + b6.
Recall that the “colors” of the edges in this example are “included” and “excluded”. If we set b = 1 and r = i (for “included”) we get i6 + i5 + 2i4 + 3i3 + 2i2 + i + 1, interpreted as one graph with 6 edges, one with 5, two with 4, three with 3, two with 2, one with 1, and one with zero edges, since 1 = i0. (Of course, if we are interested in the inventory in the final form, we can skip the step using r and b by substituting xj = ij + 1 in the cycle index.) It is possible, though a bit difficult, to see that for n vertices the cycle index is PG = ∑ j n ∏ k=1 1 kjkjk!
⌊n/2⌋ ∏ k=1 (xkxk−1 2k )j2k ⌊(n−1)/2⌋ ∏ k=1 xkj2k+1 2k+1 ⌊n/2⌋ ∏ k=1 xkC(jk,2) k ∏ 1≤r<s≤n−1 xgcd(r,s)jrjs lcm(r,s) , where the sums are over all partitions j = (j1, j2, . . . , jn) of n, that is, over all j such that j1 + 2j2 + 3j3 + · · · + njn = n, and C(m, 2) = (m 2 ) . This is where the formula 6.2.1 comes from, substituting xi = 2 for all i.
With this formula and a computer it is easy to compute the inventory of n-vertex graphs when n is not too large. When n = 5, the inventory is i10 + i9 + 2i8 + 4i7 + 6i6 + 6i5 + 6i4 + 4i3 + 2i2 + i + 1.
We can use Sage to do the computations involved in finding the cycle index for graphs.
150 Chapter 6 P´ olya–Redfield Counting Exercises 6.3.
1. Find the cycle index PG for the group of permutations of the vertices of a regular tetrahedron induced by the rigid motions. (See exercise 1 in section 6.2.) 2. Using the previous exercise, write out a full inventory of colorings of the vertices of a regular tetrahedron induced by the rigid motions, with three colors, as in example 6.3.5. You may use Sage or some other computer algebra system.
3. Find the cycle index PG for the group of permutations of the edges of K5. (See exercise 5 in section 6.2. Don’t use the general formula above.) 4. Using the previous exercise, write out a full inventory of the graphs on five vertices, as in example 6.3.6. You may use Sage or some other computer algebra system.
A Hints 1.6.2. Every positive integer can be writ-ten as j · 2k, with j odd and k ≥0.
151 Index A action group action on a set, 138 acyclic graph, 105 addition principle, 12 adjacent, 83 alternating chain, 85 anti-chain, 36 arc, 129 B Bell numbers, 21, 22 Bell triangle, 25 binomial coefficients, 16 monotonicity, 19 symmetry, 19 bipartite graph, 83, 103 complete, 11, 104, 117 block, 116 block-cutpoint graph, 117 bridge, 106 Brooks’s Theorem, 121 C Catalan numbers, 16, 66 chain, 36 Chinese Remainder Theorem, 32 chromatic number, 119 chromatic polynomial, 125 circuit, 98, 129 class, 120 clique, 95, 119 clique number, 119 closed neighborhood, 83 closed walk, 98, 129 complete bipartite graph, 11, 104, 117 complete graph, 10, 33, 95 Kn, 83 condensation, 91, 102 conjugate of a partition, 61 connected, 91 connected components, 95 cut, 110 in network, 131 cutpoint, 107, 110 cutset, 110 cycle, 35, 95 cycle form, 143 cycle form of a permutation, 39 cycle graph (Cn), 83 cycle index, 148 cyclic group, 138 D degree, 83 maximum, 120 minimum, 111 derangement, 48 derangement numbers, 49 digraph, 129 connected, 134 strongly connected, 134 underlying graph, 134 153 154 Index dihedral group, 138 directed edge (arc), 129 directed graph (digraph), 129 E endblock, 117 equivalence relation, 95 Euler circuit, 98 Euler walk, 99 exponential generating function, 57 F falling factorial, 42 Ferrers diagram, 61 flow, 129 value of, 131 forest, 105 G Galton board, 21 general graph, 91 generating function, 53 exponential, 57 girth, 119 graph directed, 129 weighted, 108 graphical parameters, 119 graphical sequence, 93 greedy algorithm, 120 group, 138 cyclic, 138 dihedral, 138 symmetric, 138 H Hall’s Condition, 72 Hall’s Theorem, 72 Hamilton cycle, 100 Hamilton path, 100 in digraph, 134 handle, 115 Handle Theorem, The, 115 I incident, 9, 83 indegree, 129 independence number, 119 independent set, 118 induced subgraph, 95 internally disjoint, 112 inventory, 148 isomorphic, 94 isomorphism, 94 isotopic, 78 isotopy class, 78 J Jarn´ ık Algorithm, 108 K Kronecker delta, 40 Kruskal’s Algorithm, 109 L Latin square, 76 isotopic, 78 isotopy class, 78 orthogonol, 79 reduced, 77 least cost spanning tree, 108 length, 83 loop, 83, 91 M matching, 84 perfect, 105 maximum degree, 120 maximum flow, 131 minimum degree, 111 modulo colorings modulo G, 139 multigraph, 83, 91 multinomial coefficient, 29 multiple edges, 91 multiplication principle, 12 multiset, 27 multitree, 108 N neighborhood, 83 closed, 83 open, 83 network, 129 n-set, 14 Index 155 O open neighborhood, 83 orbit, 139 Ore property, 102 outdegree, 129 P partition conjugate, 61 non-crossing, 69 of a set, 22, 69 of an integer, 59 self-conjugate, 62 Pascal’s Triangle, 14 monotonicity, 19 symmetry, 19 path, 83, 95 in digraph, 129 pendant, 105 pendant vertex, 94 perfect matching, 105 permutation, 36 cycle form, 39 permutations, 13 Petersen graph, 103 pigeonhole principle, 31 planar, 125 Prim’s Algorithm, 109 proper coloring, 118 Q quasigroup, 78 R Ramsey number, 34 Ramsey Theory, 34 recurrence relation, 22, 50, 62 regular graph, 105 repetition number, 27 rising factorial, 42 rooted binary tree, 66 S SDR, 71 separating set, 113 sequence, 92 set system, 71 simple graph, 91 source, 129 spanning tree, 106 least cost, 108 Stirling numbers of the first kind, 39 unsigned, 39 Stirling numbers of the second kind, 26, 39 subgraph, 94 symmetric group, 138 system of distinct representatives, 71 T target, 129 tree, 105 rooted binary, 66 type, 147 V value of a flow, 131 W walk, 97 in digraph, 129 weighted graph, 108 |
649 | https://math.stackexchange.com/questions/1371551/what-is-the-mistake-in-doing-integration-by-this-method | calculus - What is the mistake in doing integration by this method? - Mathematics Stack Exchange
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What is the mistake in doing integration by this method?
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Integration Of a given function can be found out in many ways, For a specific function ∫1/xlogx, if we do integration by parts (∫f(x) g(x)= f(x) ∫ g(x)- ∫ [d/dx (f(x)) ∫g(x)] dx ) we get this way
Let ∫1/xlogxdx = A
If we do integration by parts, we have
∫1 x log x d x=1 log x⋅∫1 x d x−∫d d x(1 log x)d x⋅∫1 x d x+c∫1 x logx d x=1 logx⋅∫1 x d x−∫d d x(1 logx)d x⋅∫1 x d x+c
=1 log x⋅log x+∫1 x log x d x+c=1 logx⋅logx+∫1 x logx d x+c
i.e, A = 1 + A + c
Which tells us that the value of "c c" in this specific integral is −1−1. So does this mean that the Integral of function 1 x log x 1 x logx has only one value.! So this means only two things, either I am wrong somewhere or i am missing a point somewhere.
Edit: I do know we can simple get it by writing logx as u and ¹/x as du, we directly get it as log(logx) +c.
But I wanted to know the reason why we can't apply By parts when we can integrate it by substitution.
calculus
integration
indefinite-integrals
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edited Jul 23, 2015 at 17:08
axelonetaxelonet
asked Jul 23, 2015 at 16:47
axelonetaxelonet
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Just a comment on the way you have presented the equations on the first line of your working the second term −∫d d x(1 log x)d x⋅∫1 x d x−∫d d x(1 logx)d x⋅∫1 x d x is wrong, the first integral is seperate to the 1 x 1 x term but by the by parts formula the 1 x 1 x should be inside this integral.Rammus –Rammus 2015-07-23 17:05:33 +00:00 Commented Jul 23, 2015 at 17:05
1 I think this is a good question, and I hope it is actually answered. Telling the OP "just use substitution" is not an answer to this specific question, to be clear.pjs36 –pjs36 2015-07-23 17:09:45 +00:00 Commented Jul 23, 2015 at 17:09
math.stackexchange.com/questions/408515/liate-ilate-rule/… check this. i think it will helpful iostream007 –iostream007 2015-07-23 17:10:58 +00:00 Commented Jul 23, 2015 at 17:10
@Rammus it is indeed inside the first integral only.axelonet –axelonet 2015-07-23 17:15:56 +00:00 Commented Jul 23, 2015 at 17:15
@iostream007 I used the ILATE rule, anyways thanks!axelonet –axelonet 2015-07-23 17:16:30 +00:00 Commented Jul 23, 2015 at 17:16
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The indefinite integral is only unique up to a constant. Thats why you have
∫1 x log(x)d x=1+∫1 x log(x)d x∫1 x log(x)d x=1+∫1 x log(x)d x
If you differentiate on both sides you get
1 x log(x)=1 x log(x)1 x log(x)=1 x log(x)
Your c c doesn't has to be −1−1, it can be any value because generally
∫1 x log(x)d x−∫1 x log(x)d x≠0∫1 x log(x)d x−∫1 x log(x)d x≠0
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answered Jul 23, 2015 at 17:13
Andrei KhAndrei Kh
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You should make it clear that the indefinite integral in your answer is essentially a class of functions and arithmetic on classes is defined to be the class of possible results when using any member. If not your last line won't make sense.user21820 –user21820 2015-08-01 07:15:08 +00:00 Commented Aug 1, 2015 at 7:15
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The real answer is that you did integration by parts and basically got x=x x=x. That doesn't mean there is just a constant value for your integral. It just meant that your integration by parts looped around and did nothing - the constant +1+1 can just be ignored because you are dealing with indefinite integrals.
So, you can apply integration by parts, you just don't get any closer to the solution. It's a null operation. You can sometimes get caught in the same trap if you try to do integration by parts twice and accidentally undo the previous step. Sometimes a simplification just doesn't simplify anything and you have to find another way. Just don't confuse this with the other (desired) effect when you get the same integral back, but in a way that you can actually solve the equation from it (in those cases you don't get the same thing back, but express it with the same integral, but in a different way, not as the meaningless tautology).
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answered Jul 23, 2015 at 17:23
orionorion
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There is a problem with the integration-by-parts formula, often written
∫u d v=u v−∫v d u,∫u d v=u v−∫v d u,
for certain functions. If it should happen that v=c u v=c u and u d v=−v d u u d v=−v d u , then the method is not going to take us anywhere. The second condition gives us a differential equation
d v v=−d u u⇒log v=−log u=log(1 u),d v v=−d u u⇒logv=−logu=log(1 u),
which is unfortunately satsified by u=1 log x,v=log x⇒d v=1 x d x u=1 logx,v=logx⇒d v=1 x d x .
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answered Jul 23, 2015 at 18:18
colormegonecolormegone
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You shouldn't be doing integration by parts here since 1 x 1 x is the derivative of ln x lnx. Instead do the substitution u=ln x u=lnx
Without wishing to generalize too much, you need to distinguish between integrands of the form
f(x)g(x)f(x)g(x)
where f f and g g are completely different types of function, i.e. not related by differentiation, and integrands of the form
g′(x)f′(g(x))g′(x)f′(g(x))
where the substitution u=g(x)u=g(x) is made (as is the case in your example)
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edited Jul 23, 2015 at 17:07
answered Jul 23, 2015 at 16:52
David QuinnDavid Quinn
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1 I know that way too, so you are telling that we should not apply by parts when we can do it by substitution. Could you explain me why?axelonet –axelonet 2015-07-23 16:54:47 +00:00 Commented Jul 23, 2015 at 16:54
-1, this doesn't answer the question which was asked.Santiago Canez –Santiago Canez 2015-07-23 17:20:12 +00:00 Commented Jul 23, 2015 at 17:20
@SantiagoCanez: Just to clarify: my answer is mainly to address the OP's confusion about the two different integration methods and when to use what. Please see the OP's supplementary question above. This seems to be the key issue for the OP, not the fact that inappropriate application of integration by parts leads nowhere David Quinn –David Quinn 2015-07-23 17:30:53 +00:00 Commented Jul 23, 2015 at 17:30
1 I think you have the gist of the problem with why integration-by-parts is useless here. You perhaps need to expand on the fact that for this pairing of functions, the product f(x)G(x)f(x)G(x) is a constant with G(x)f′(x)=f(x)g(x)G(x)f′(x)=f(x)g(x) . For such an unfortunate choice of integrand, the "parts" method takes one nowhere. (There are many other integrands where either method can be applied, but for which substitution or "parts" is the more convenient.)colormegone –colormegone 2015-07-23 17:47:20 +00:00 Commented Jul 23, 2015 at 17:47
Pardon me -- in my comment, there is a sign slip in the equation: it should read f(x)g(x)=−G(x)f′(x)f(x)g(x)=−G(x)f′(x) .colormegone –colormegone 2015-07-23 18:05:13 +00:00 Commented Jul 23, 2015 at 18:05
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A substitution of u=log x u=logx here yields x d u=d x x d u=d x. So you get
∫1 u d u=log u+C∫1 u d u=logu+C
Back-substitution yields
∫d x x log x=log log x+C.∫d x x logx=loglogx+C.
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answered Jul 23, 2015 at 16:56
Zain PatelZain Patel
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2 Yes, I know that method. You are telling me that we can't use by parts when we can do it by substituting. Could you explain me why?axelonet –axelonet 2015-07-23 16:58:07 +00:00 Commented Jul 23, 2015 at 16:58
-1, this doesn't answer the question which was asked.Santiago Canez –Santiago Canez 2015-07-23 17:20:24 +00:00 Commented Jul 23, 2015 at 17:20
1 @SantiagoCanez I cannot see a question anywhere in the question body.Zain Patel –Zain Patel 2015-07-23 17:21:25 +00:00 Commented Jul 23, 2015 at 17:21
Check the title.Santiago Canez –Santiago Canez 2015-07-23 17:22:02 +00:00 Commented Jul 23, 2015 at 17:22
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Definition
The empirical rule predicts deviations based on how the points in a data set cluster around the center based on a measure of how widely the data points are spread out.
What Is the Empirical Rule?
The empirical rule, also sometimes called the three-sigma or 68-95-99.7 rule, predicts deviations from the mean or average of data. It indicates that 68% of observations will fall within the first standard deviation (µ ± σ) in normal distributions, 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ) of the mean. The rule is a vital component in quality control and risk analysis.
Key Takeaways
Three-sigma limits that follow the empirical rule are used to set the upper and lower control limits in statistical quality control charts and in risk analysis.
68% of the data will fall within one standard deviation under the empirical rule.
95% will fall within two standard deviations.
99.7% will fall within three standard deviations from the mean.
Understanding the Empirical Rule
The empirical rule is often used in statistics for forecasting final outcomes. After calculating the standard deviation and before collecting complete data, this rule can be used as a rough estimate of the outcome of the impending data to be collected and analyzed.
This probability distribution can be used as an evaluation technique since gathering the appropriate data may be time-consuming or even impossible in some cases. Such considerations come into play when a company reviews its quality control measures or evaluates its risk exposure. For instance, the frequently used risk tool value-at-risk (VaR) assumes that the probability of risk events follows a normal distribution.
The empirical rule is also used as a rough way to test a distribution's "normality." If too many data points fall outside the three standard deviation boundaries, this suggests that the distribution is not normal and may be skewed or follow some other distribution.
The empirical rule is also known as the three-sigma rule, as "three-sigma" refers to a statistical distribution of data within three standard deviations from the mean on a normal distribution (bell curve), as indicated by the figure below.
Example of the Empirical Rule
Let's assume a population of animals in a zoo is known to be normally distributed. Each animal lives to be 13.1 years old on average (mean), and the standard deviation of the lifespan is 1.5 years. If someone wants to know the probability that an animal will live longer than 14.6 years, they could use the empirical rule. Knowing the distribution's mean is 13.1 years old, the following age ranges occur for each standard deviation:
One standard deviation (µ ± σ): (13.1 - 1.5) to (13.1 + 1.5), or 11.6 to 14.6
Two standard deviations (µ ± 2σ): 13.1 - (2 x 1.5) to 13.1 + (2 x 1.5), or 10.1 to 16.1
Three standard deviations (µ ± 3σ): 13.1 - (3 x 1.5) to 13.1 + (3 x 1.5), or, 8.6 to 17.6
The person solving this problem needs to calculate the total probability of the animal living 14.6 years or longer. The empirical rule shows that 68% of the distribution lies within one standard deviation, in this case, from 11.6 to 14.6 years. Thus, the remaining 32% of the distribution lies outside this range. One half lies above 14.6 and the other below 11.6. So, the probability of the animal living for more than 14.6 is 16% (calculated as 32% divided by two).
The Empirical Rule in Investing
Most market data isn't normally distributed, so the 68-95-99.7 rule doesn't generally apply to investments. However, many analysts use aspects of it—such as standard deviation—to estimate volatility.
You can calculate the standard deviation of your portfolio, an index, or other investments and use it to assess volatility. Calculating a particular investment's standard deviation is straightforward if you have access to a spreadsheet and your chosen investment's prices or returns.
Market analysts express standard deviation in percentage form. For example, the standard deviation for the S&P 500 index from 2015 to 2025 was 15.37%.
Using the spreadsheet, you can paste the returns, prices, or values into it, find the percent change from the previous session, and use the standard deviation function:
= STDEV ( 1, 2, 3, 4, ...) or = STDEV ( A1 : A200 )
Important
You'll get more accurate results using more than one month's trading data, such as three or more years. The example below uses the index's daily values over one month and annualizes the standard deviation to limit the table size.
To annualize the standard deviation, multiply it by the square root of the number of trading days in one year—there are usually 252. Here's a calculation of the standard deviation based on the closing prices of the S&P 500.
| S&P 500 Standard Deviation (Annualized) |
| DATE | CLOSE | INTERDAY CHANGE |
| 3/3/2025 | 5849.72 | — |
| 3/4/2025 | 5778.15 | -1.22% |
| 3/5/2025 | 5842.63 | 1.12% |
| 3/6/2025 | 5738.52 | -1.78% |
| 3/7/2025 | 5770.2 | 0.55% |
| 3/10/2025 | 5614.56 | -2.70% |
| 3/11/2025 | 5572.07 | -0.76% |
| 3/12/2025 | 5599.3 | 0.49% |
| 3/13/2025 | 5521.52 | -1.39% |
| 3/14/2025 | 5638.94 | 2.13% |
| 3/17/2025 | 5675.12 | 0.64% |
| 3/18/2025 | 5614.66 | -1.07% |
| 3/19/2025 | 5675.29 | 1.08% |
| 3/20/2025 | 5662.89 | -0.22% |
| 3/21/2025 | 5667.56 | 0.08% |
| 3/24/2025 | 5767.57 | 1.76% |
| 3/25/2025 | 5776.65 | 0.16% |
| 3/26/2025 | 5712.2 | -1.12% |
| 3/27/2025 | 5693.31 | -0.33% |
| 3/28/2025 | 5580.94 | -1.97% |
| 3/31/2025 | 5611.85 | 0.55% |
|
| Standard Deviation | stdev(C1:C3) | 1.29% |
| Annualized Standard Deviation | sqrt(252)C24 | 20.42% |
Based on closing prices in March 2025.
So the annual volatility based on the data used in the table is 20.42%. The higher the standard deviation, the more risk analysts believe the investment has.
Alternatively, you can find an investment's standard deviation on popular investing websites. For example, Morningstar displays the S&P 500 standard deviation in three, five, and 10-year measurements.
Explain Like I'm Five
The empirical rule describes how the points in a data set are clustered around the center. It is based on the standard deviation, a measure of how widely the data points are spread out.
If the data set is normally distributed, the empirical rule predicts that 68% of the data is less than one standard deviation from the mean. 95% are within two standard deviations of the mean, and 99.7% are within three standard deviations.
What Is the Empirical Rule?
In statistics, the empirical rule states that in a normal distribution, 99.7% of observed data will fall within three standard deviations of the mean. Specifically, 68% of the observed data will occur within one standard deviation, 95% within two standard deviations, and 99.7% within three standard deviations.
How Is the Empirical Rule Used?
The empirical rule is applied to anticipate probable outcomes in a normal distribution. For instance, a statistician could use it to estimate the percentage of cases that fall in each standard deviation. Consider that the standard deviation is 3.1 and the mean equals 10. In this case, the first standard deviation would range between (10+3.2)= 13.2 and (10-3.2)= 6.8. The second deviation would fall between 10 + (2 X 3.2)= 16.4 and 10 - (2 X 3.2)= 3.6, and so forth.
What Are the Benefits of the Empirical Rule?
The empirical rule is beneficial because it serves as a means of forecasting data. This is especially true when it comes to large datasets and those where variables are unknown.
The Bottom Line
Analysts use the empirical rule to see how much data falls within a specified interval away from the data set's mean. Investment analysts can use it to estimate the volatility of a particular investment, portfolio, or fund.
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Morningstar Investments. "SPX Risk-Return Analysis." Select "10-Yr."
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651 | https://www.mathopenref.com/calcpolararea.html | Polar Area
We can also use integrals to find the area enclosed by a polar curve. Here, we use sectors of circles instead of rectangles to calculate the area.
| | |
--- |
| See About the calculus applets for operating instructions. | |
1. A Circle
The applet initially shows a circle defined using the polar equation r = 1. We know from geometry that the area of this circle is π. We can approximate the area using sectors, one of which is shown in gray. Move the th slider (th is used instead of θ to make it easier to type in polar functions) to see the sector move.
The area of a sector of width dθ and radius r is ½ r² dθ. If we add up a bunch of sectors to approximate the area enclosed by a polar curve and let dθ go to zero, we get the integral
where r is replaced by our polar equation in terms of θ. For this example, the integral is
where r is replaced by our polar equation in terms of θ. For this example, the integral is
For example, if you set b to 4pi by typing in the b input box, the area shown is twice what it should be. That's because you only need θ to go from 0 to 2π to cover the circle; if θ goes from 0 to 4π, it goes around twice and covers the circle twice.
2. A Spiral
Select the second example from the drop down menu, showing a spiral defined as r = θ . Move the th slider to see the sector move, noticing that in this case the radius is not constant, but changes as θ changes. The integral in this case is
Move the b slider to change the upper limit and notice what area is shaded in yellow. Unlike area under regular curves, where the edges of the yellow area is parallel to the y axis, here the boundaries of the area are lines through the origin.
3. A Rose
Select the third example, showing a rose. You can move the th slider to get a feel for how the sectors are added up. The integral in this case is
Explore
You can try other polar equations by typing the definition in (using th for θ), setting a and b as desired, and zooming/panning. Note that you need to figure out what a and b are to only sweep out the area of interest once, to avoid double counting (use the sliders to help see where this is; if you need to set a or b beyond the range of 0 to 2π, use the a and b input boxes instead of the sliders).
Other 'Applications of Integration' topics |
652 | https://cameo.mfa.org/wiki/Anhydride | Anhydride - CAMEO
Anhydride
From CAMEO
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Description
Any chemical compound formed by the removal of water. An acid anhydride is a nonmetal oxide that forms an acid when combined with water, e.g., SO3 becomes H2SO4. A basic anhydride is a metal oxide that forms a base when combined with water, e.g., CaO becomes Ca(OH)2.
(Not to be confused with Anhydrous or Anhydrite.)
Resources and Citations
Richard S. Lewis, Hawley's Condensed Chemical Dictionary, Van Nostrand Reinhold, New York, 10th ed., 1993
Hoechst Celanese Corporation, Dictionary of Fiber & Textile Technology (older version called Man-made Fiber and Textile Dictionary, 1965), Hoechst Celanese Corporation, Charlotte NC, 1990
Theodore J. Reinhart, 'Glossary of Terms', Engineered Plastics, ASM International, 1988
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653 | https://simple.wikipedia.org/wiki/Circumference | Circumference - Simple English Wikipedia, the free encyclopedia
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From Simple English Wikipedia, the free encyclopedia
Circumference of a circle
In geometry, circumference is the distance around a closed curve; for example, a circle. It is a special kind of perimeter. The length of the circumference of a circle is often written as C{\displaystyle C}, with:
C=π d{\displaystyle C=\pi d}
where d is the diameter of the circle.
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↑"Radius, diameter, & circumference | Circles (article)". Khan Academy. Retrieved 2020-09-24.
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654 | https://www.pnw.edu/wp-content/uploads/2020/03/Lecture-Notes-2-6.pdf | Chapter 18 ANOVA Diagnostics and Remedial Measures In any statistical analysis, there is always three steps.
1. Diagnostics.
Choose which of a number of different statistical models best fits the data; in particular, decide which factors1 to keep in or eliminate from the model. Check to see if the statistical model (single–factor ANOVA, in this case) is appropriate.
That is, check to see if the assumptions required in applying the statistical model are satisfied by the data.
2. Remedial Measures.
If the data does not satisfy the assumptions of the chosen statistical model, then investigate whether or not it is possible to alter the statistical model in some way (transformation of the variables or adding and eliminating variables) so as to make the model fit the data better. Sometimes the chosen statistical model is such a poor choice to analyze the given data set, it is necessary to try a second entirely different statistical model.
3. Inference.
Analyze the data with the (appropriately modified, if necessary) statistical model, to perform tests and to calculate confidence intervals.
18.1 Residual Analysis SAS program: att2-18-1-cotton-residual 1Later, we look at ANOVA models which have more factors than just one, as is the case in the present single–factor ANOVA model.
61 62 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) One key quantity used in diagnosis, of checking to see if the assumptions required in using the single–factor ANOVA statistical model are satisfied by the data, is the residual. The residual, eij, is the difference between an observed data point, Yij, and the associated expected (predicted, fitted) point determined by the model, ˆ Yij, eij = Yij −ˆ Yij = Yij −¯ Yi· There are different types of residuals, including • Studentized Residuals.
rij = eij s{eij}, s{eij} = s MSE(ni −1) ni These residuals indicate, in standard deviation units, how far any particular residual is from the mean. Most residuals should fall within three standard de-viations of the mean; often, a residual further than two (2) standard deviations from the mean is considered unusual.
• Studentized Deleted Residuals.
tij = eij nT −r −1 SSE ³ 1 −1 ni ´ −e2 ij 1/2 These standardized residuals account for when one observation is missing. If a studentized deleted residual is (roughly) equal it the associated studentized residual, then this indicates the one missing observation is not influential in the analysis.
• Semistudentized Residuals.
e∗ ij = eij √ MSE These residuals act “almost” like standardized residuals where, notice, q MSE(ni−1) ni ≈ √ MSE for large enough n. Semistudentized residuals are some-times used instead of studentized residuals if the assumptions of the model require them; mostly, though, studentized residuals are used in residual analy-ses.
Diagnosis of departures of an ANOVA model from the data include, • Nonconsistancy of error (residual) variance One of the assumptions required in a single–factor ANOVA statistical model is that the error variance is constant with respect to the different factor level means. This is checked using (standardized or not) residuals versus fitted plots Section 1. Residual Analysis (ATTENDANCE 2) 63 or dot plots.
Residual plot (a) in the figure below indicates constant error variance; other possible nonconstant error variance patterns, with suggested model corrections, are given in (b), (c), (d) and (e).
(a) constant variance e y ^ e y ^ e y ^ e y ^ e y ^ (b) 2 arcsin Y correction (c) Y correction (d) log Y correction (e) 1/Y correction Figure 18.1 (Check for constant error variance.) • Nonindependence of error terms This type of error, because it often appears as a pattern in a time sequence, is checked using (standardized or not) residuals versus time (sequence) plots2. In general, residual plot (a) in the figure below indicates independent errors; other possible nonindependent error patterns are given in (b) and (c).
-20 -10 0 10 20 time error -20 -10 0 10 20 time error -20 -10 0 10 20 time error (a) independent (b) dependent: positively correlated 0 2.5 5 7.5 10 0 2.5 5 7.5 10 0 2.5 5 7.5 10 (c) dependent: negatively correlated Figure 18.2 (Check for nonindependence of error.) 2Since type of error appears in statistical control theory or in time series analyses, not in the current design of experiments course, we will not spend much time investigating this type of error.
64 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) • Outliers Although not an assumption required in a single–factor ANOVA statistical model, outlying observations need to be investigated.
They may arise as a consequence of data input error or they may reveal insight into the data. They are checked using (standardized or not) residuals versus fitted plots, and residual dot plots, box plots and stem–and–leaf plots. In general, standardized resid-ual plot (a) in the figure below indicates no outliers, whereas residual plot (b) indicates there are outliers.
(a) no outliers y (b) outliers standardized residuals ^ y ^ 2 2 standardized residuals 2 2 Figure 18.3 (Check for outliers.) • Omission of important explanatory variables It is possible the single–factor ANOVA statistical model is missing some pre-dictor variables. Omission of variables may appear as patterns in the residual plots, such as those that appear in the nonindependence patterns given above: time variables are missing in this case. Also, for example, residual plot (a) in the figure below indicates the variable gender does not need to be added to the model, whereas residual plot (b) indicates that the variable gender does need to be included in the model.
(a) gender variable not needed y (b) missing gender variable ^ y ^ e e female male female male Figure 18.4 (Omission of important explanatory variables?) • Nonnormality of error terms Another of the assumptions required in a single–factor ANOVA statistical model is that the errors are normally distributed. This assumption is checked using (standardized or not) residuals versus fitted plots, and residual dot plots, box Section 1. Residual Analysis (ATTENDANCE 2) 65 plots and normal probability plots.
In general, residual plot (a) in the fig-ure below indicates normally distributed errors; other possible nonnormal error patterns are given in (b), (c), (d) and (e).
(b) heavy tail (c) light tail (a) normal (d) left skew (e) right skew observed expected residuals Figure 18.5 (Check for nonnormality of error.) Exercise 18.1 (Single–Factor Analysis of Variance) 1. Residual analysis diagnosis: cotton strength.
The tensile strength of fibers with varying amounts of cotton content is inves-tigated.
5% cotton 7 7 15 11 9 10 ¯ Y1· ≈9.8 10% cotton 12 17 12 18 18 16 ¯ Y2· ≈15.5 15% cotton 14 18 18 19 19 17 ¯ Y3· ≈17.5 20% cotton 19 25 22 19 23 24 ¯ Y4· ≈22 25% cotton 7 10 11 15 11 14 ¯ Y5· ≈11.3 Use various residual analyses to check to see if the assumptions required in using the single–factor ANOVA statistical model are satisfied by this data.
(a) Constant error (residual) variance?
The residual plot for the drug data indicates that error variance (choose one) is / is not constant with respect to either the (qualitative) cotton types (1, 2, 3, 4 and 5) or the (quantitative) predicted ( ˆ Y = ¯ Y ) variable.
The predicted (mean) values are (choose none, one or more) i. 9.833 66 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) ii. 15.5 iii. 17.5 iv. 22 v. 11.333 (b) Nonindependence of error?
There does appear to a curved pattern in the residual plots, so the assump-tion of independence appears to be (choose one) satisfied / not satisfied.
(c) Outliers?
There (choose one) appears to be / does not appear to be outliers be-cause not only are none of the residuals in the studentized plot are further than two (2) standard deviations from the mean, but also the studentized deleted residual plot is very similar to the (ordinary) studentized residual plot.
(d) Omission of important explanatory variables?
There does appear to a curved pattern in the residual plots, so it may be possible that one or more explanatory variables (choose one) are / are not missing from the model.
(e) Nonnormality of error terms?
Since the normal probability plot is more or less a line, this indicates the residuals are (choose one) normal / nonnormal.
2. Residual analysis diagnosis: pest–free apples.
The number of pest–free apples (out of 100 per tree) produced by apple plants in five differently applied pesticide plots of land is tabulated below.
pesticide 1 12 13 15 15 15 16 16 17 18 22 pesticide 2 28 30 31 32 33 33 37 37 40 41 pesticide 3 42 48 48 54 54 55 56 60 63 64 pesticide 4 70 71 72 73 75 75 77 79 79 80 pesticide 5 92 94 95 96 96 97 97 98 98 99 (a) Constant error (residual) variance?
The residual plot for the drug data indicates that error variance (choose one) is / is not constant with respect to either the (qualitative) pesticide types (1, 2, 3, 4 and 5) or the (quantitative) predicted ( ˆ Y = ¯ Y ) variable.
The predicted (mean) number of pest–free apple values are (choose none, one or more) i. 16 ii. 34.2 iii. 54.4 Section 2. Tests for Constancy of Error Variance (ATTENDANCE 2) 67 iv. 75.1 v. 96.2 (b) Nonindependence of error?
There does appear to a bowed–in–the–middle pattern in the residual plots, so the assumption of independence appears to be (choose one) satisfied / not satisfied.
(c) Outliers?
There (choose one) appears to be / does not appear to be outliers be-cause not only are none of the residuals in the studentized plot are further than two (2) standard deviations from the mean, but also the studentized deleted residual plot is very similar to the (ordinary) studentized residual plot.
(d) Omission of important explanatory variables?
There does appear to a bowed–in–the–middle pattern in the residual plots, so it may be possible that one or more explanatory variables (choose one) are / are not missing from the model.
(e) Nonnormality of error terms?
Since the normal probability plot is more or less a line, this indicates the residuals are (choose one) normal / nonnormal.
18.2 Tests for Constancy of Error Variance SAS program: att2-18-2-cotton-hartley,levene In addition to the graphical residual analysis used to check to see if the assump-tions required in using the single–factor ANOVA statistical model are satisfied by the data, analytical (numerical) procedures can also be used in this regard. Two tests are discussed in this section, the Hartley test and the modified Levene test3, both of which are used to test for the constancy of error variance.
Exercise 18.2 (Tests for Constancy of Error Variance) 1. Tests for Constancy: cotton strength.
The tensile strength of fibers with varying amounts of cotton content is inves-tigated.
3The Neter text calculates the F test statistic relative to the median, whereas SAS calculates the test statistic relative to the mean. Consequently, the text and SAS will have (slightly) different results for the modified Levene test. We will use the SAS version of the modified Levene test in this course.
68 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) 5% cotton 7 7 15 11 9 10 ¯ Y1· ≈9.8 10% cotton 12 17 12 18 18 16 ¯ Y2· ≈15.5 15% cotton 14 18 18 19 19 17 ¯ Y3· ≈17.5 20% cotton 19 25 22 19 23 24 ¯ Y4· ≈22 25% cotton 7 10 11 15 11 14 ¯ Y5· ≈11.3 Use both the modified Levene test and the Hartley test at α = 0.05 to check to see if the assumption of constant variance required when using the single–factor ANOVA statistical model is satisfied by this data.
(a) Modified Levene test i. Statement.
Choose none, one or more.
A. H0 : error variance constant over cotton types versus H1 : not constant for different cotton types B. H0 : σ2 1 = · · · = σ2 5 versus H1 : not all σ2 i are the same C. H0 : error variance constant over cotton types versus H1 : increase for different cotton types ii. Test.
From SAS, the p–value is (choose one) < 0.0001 / 0.532 / 0.7915 The level of significance is α = 0.05 iii. Conclusion.
Since the p–value is (choose one) smaller / larger than the level of significance we (choose one) accept / reject the null hypothesis that the error variance is constant.
(b) Hartley test i. Statement.
Choose none, one or more A. H0 : error variance constant over cotton types versus H1 : not constant for different cotton types B. H0 : σ2 1 = · · · = σ2 5 versus H1 : not all σ2 i are the same C. H0 : error variance constant over cotton types versus H1 : increase for different cotton types ii. Test.
From SAS, the test statistic is H = max(s2 i ) min(s2 i ) = 2.994 1.871 Section 2. Tests for Constancy of Error Variance (ATTENDANCE 2) 69 which equals 1.3 / 1.5 / 1.6 Also, from table B.6, page 1355, the number of levels is r = 5 and df = 6 −1 = 5, the critical value is (choose one) 13.7 / 16.3 / 18.7 iii. Conclusion.
Since the test statistic is (choose one) smaller / larger than the critical value we (choose one) accept / reject the null hypothesis that the error variance is constant.
2. Tests for Constancy: pest–free apples.
The number of pest–free apples (out of 100 per tree) produced by apple plants in five differently applied pesticide plots of land is tabulated below.
pesticide 1 12 13 15 15 15 16 16 17 18 22 pesticide 2 28 30 31 32 33 33 37 37 40 41 pesticide 3 42 48 48 54 54 55 56 60 63 64 pesticide 4 70 71 72 73 75 75 77 79 79 80 pesticide 5 92 94 95 96 96 97 97 98 98 99 Use both the modified Levene test and the Hartley test at α = 0.05 to check to see if the assumption of constant variance required when using the single–factor ANOVA statistical model is satisfied by this data.
(a) Modified Levene test i. Statement.
Choose none, one or more.
A. H0 : error variance constant over cotton types versus H1 : not constant for different cotton types B. H0 : σ2 1 = · · · = σ2 5 versus H1 : not all σ2 i are the same C. H0 : error variance constant over cotton types versus H1 : increase for different cotton types ii. Test.
From SAS, the p–value is (choose one) 0.0153 / 0.532 / 0.7915 The level of significance is α = 0.05 iii. Conclusion.
Since the p–value is (choose one) smaller / larger than the level of significance we (choose one) accept / reject the null hypothesis that the error variance is constant.
(b) Hartley test i. Statement.
Choose none, one or more 70 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) A. H0 : error variance constant over cotton types versus H1 : not constant for different cotton types B. H0 : σ2 1 = · · · = σ2 5 versus H1 : not all σ2 i are the same C. H0 : error variance constant over cotton types versus H1 : increase for different cotton types ii. Test.
From SAS, the test statistic is H = max(s2 i ) min(s2 i ) = 6.963 2.098 which equals 3.3 / 3.5 / 3.6 Also, from table B.6, page 1355, the number of levels is r = 5 and df = 10 −1 = 9, the critical value is (choose one) 6.72 / 7.11 / 8.74 iii. Conclusion.
Since the test statistic is (choose one) smaller / larger than the critical value we (choose one) accept / reject the null hypothesis that the error variance is constant.
(c) A final point The Hartley test and modified Levene tests (choose one) always / some-times agree with one another.
18.3 Overview of Remedial Measures If the data does not satisfy the assumptions of the chosen statistical model, it is sometimes possible to take remedial action; in other words, to alter the statistical model in some way so as to make the data fit the model better. We look at two remedial methods for correcting for two common departures from the ANOVA model.
1. The weighted least squares method is used to make the nonconstant error vari-ances more constant.
2. Transformations (both simple and according to the Box–Cox procedure) of the response variable are used to make nonconstant error variance more constant and also to make nonnormal data more normal.
18.4 Weighted Least Squares SAS program: att2-18-4-soil-weightLS Section 4. Weighted Least Squares (ATTENDANCE 2) 71 The weighted least squares method for single–factor ANOVA models is used in this section to attempt to make nonconstant error variances constant.
Exercise 18.3 (Weighted Least Squares) Soil–water fluxes were measured for different grades of soil.
grade 1 soil 365.04 372.41 368.71 361.41 357.81 grade 2 soil 249.64 247.15 244.69 239.85 242.26 grade 3 soil 149.90 148.41 146.94 145.47 151.41 1. Weighted Least Squares.
(a) Statement.
Choose none, one or more.
i. H0 : error variance constant over cotton types versus H1 : not constant for different cotton types ii. H0 : µ1 = µ2 = µ3 versus H1 : µi ̸= µj, i ̸= j iii. H0 : error variance constant over cotton types versus H1 : increase for different cotton types (b) Test.
From SAS, the test statistic is F ∗ w = SSE w(F) −SSE w(R) r −1 ÷ SSE w(F) nT −r = 7149.70963 −12 3 −1 ÷ 12 15 −3 (circle one) 35.76 / 357.69 / 3568.85 And so the p–value, with r −1 = 3 −1 = 2 and nT −r = 15 −3 = 12 degrees of freedom, is given by p–value = P(F ≥3568.85) which equals (circle one) 0.00 / 0.35 / 0.43.
The level of significance is 0.05.
(c) Conclusion.
Since the p–value, 0.00, is smaller than the level of significance, 0.05, we (circle one) accept / reject the null hypothesis that mean water–soil fluxes are the same.
2. Understanding Weighted Least Squares.
72 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) (a) Error (residual) variance constant?
From SAS, the residual plot indicates that error variance (choose one) is / is not constant with respect to the predicted ( ˆ Y = ¯ Y ) variable and so we will use weighted least squares to correct for this problem.
(b) Comparing weighted test statistic with unweighted test statistic.
From SAS, the weighted test statistic is F ∗ w = SSE w(F) −SSE w(R) r −1 ÷ SSE w(F) nT −r = 7149.70963 −12 3 −1 ÷ 12 15 −3 = 3568.85 whereas, from SAS4, the unweighted test statistic is F ∗ w = SSE w(F) −SSE w(R) r −1 ÷ SSE w(F) nT −r = 118041 −215.13312 3 −1 ÷ 215.13312 15 −3 = 3286.12 the weighted test statistic is (choose one) more / less significant than the unweighted test statistic.
(c) Comparing weighted residuals with unweighted residuals.
The weighted least squares analysis weights the larger residuals (choose one) more than / the same as / less than the smaller residuals. In particular, the weights assigned to each residual is wij = 1 s2 i and so the three different weights used for this data are i ¯ Y si wij = 1 s2 i grade 1 soil 365.076 5.77125 0.03002 grade 2 soil 244.718 3.86913 0.06680 grade 3 soil 148.426 2.34645 0.18163 (d) ANOVA Model As A Multiple Linear Regression.
True / False The single–factor ANOVA model Yij = µi + εij 4This test statistic can be obtained in two different ways: either from the single–factor ANOVA or as the ratio of two appropriately defined least square regressions.
Section 5. Transformations of Response Variable (ATTENDANCE 2) 73 can be rewritten as a multiple linear regression model, Y = Xβ + ε where Y 15×1 = Y11 Y12 .
.
.
Y35 = 365.04 372.41 .
.
.
151.41 , β 3×1 = µ1 µ2 µ3 , ε 15×1 = ε11 ε12 .
.
.
ε35 and X 15×3 = X1,1 X1,2 X1,3 .
.
.
· · · .
.
.
X15,1 X15,2 X15,3 = 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 18.5 Transformations of Response Variable SAS program: att2-18-5-soil-trans,boxcox Transformations, using both simple rules and according to the Box–Cox proce-dure, of the response variable, Yij, are used to make nonconstant error variance more constant and also to make nonnormal data more normal. The simple rules are problem with variance mathematical description remedy transformation (a) proportional to the mean σ2 α k|µ| or σ α k|µ1/2| Y ′ = √ Y or Y ′ = √ Y + √ Y + 1 (b) proportional to the mean2 σ2 α k|µ|2 or σ α k|µ| Y ′ = ln Y (c) proportional to the mean4 σ2 α k|µ|4 or σ α k|µ|2 Y ′ = 1 Y (d) proportional is a proportion σ2{pij} α πi(1 −πi) Y ′ = 2 arcsin √ Y 74 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) e y ^ e y ^ e y ^ e y ^ (a) response is proportion (b) variance proportional to mean (c) variance proportional to square of mean (d) variance proportional to fourth power of mean Figure 18.6 (Simple Rule Transformations.) The Box–Cox transformation is given by Y ′ = Y λ where the “best” λ is the one that maximizes5 the SSE of all of the models for different λ. The Box–Cox transformation procedure is not discussed at great length.
Exercise 18.4 (Transformations of Response Variable) Soil–water fluxes were measured for different grades of soil.
grade 1 soil 365.04 372.41 368.71 361.41 357.81 grade 2 soil 249.64 247.15 244.69 239.85 242.26 grade 3 soil 149.90 148.41 146.94 145.47 151.41 1. Residual plot and simple rule transformations.
From SAS, the residual plot seems to indicate the problem with the variance is proportional to the (choose one) (a) mean (b) mean2 (c) mean4 (d) none of these; the response is a proportion and so the transformation best suited to make the variance constant appears to be (choose one) 5This λ is the maximum likelihood estimate for the statistical model.
Section 5. Transformations of Response Variable (ATTENDANCE 2) 75 (a) Y ′ = √ Y or Y ′ = √ Y + √ Y + 1 (b) Y ′ = ln Y (c) Y ′ = 1 Y (d) Y ′ = 2 arcsin √ Y 2. Variance–mean ratios and simple rule transformations.
Consider the following table of statistics, where s2 i and ¯ Yi have both been taken from the SAS output, i s2 i /¯ Yi s2 i /¯ Y 2 i s2 i /¯ Y 4 i grade 1 soil 5.771252/365.076 = 0.091234 0.000249904 1.87 × 10−9 grade 2 soil 0.061173 0.000249974 4.17 × 10−9 grade 3 soil 0.037095 0.000249921 1.13 × 10−8 Clearly, the statistic, s2 i /¯ Y 2, is the most constant of the three statistics; in other words, the variance is proportional to the (choose one) (a) mean (b) mean2 (c) mean4 (d) none of these; the response is a proportion Consequently, this indicates the transformation best suited to make the variance constant is (choose one) (a) Y ′ = √ Y or Y ′ = √ Y + √ Y + 1 (b) Y ′ = ln Y (c) Y ′ = 1 Y (d) Y ′ = 2 arcsin √ Y 3. Box–cox transformation From the SAS output, the various SSE values for given box–cox λ transforma-tions, are given below.
λ 2.7 2.8 2.9 3.0 3.1 SSE 171.59 171.42 171.37 171.44 171.62 In this case, the best (minimum SSE) box–cox transformation of the data is given by Y ′ = Y λ where (choose one) (i) λ = 2.7 76 Chapter 18. ANOVA Diagnostics and Remedial Measures (ATTENDANCE 2) (ii) λ = 2.8 (iii) λ = 2.9 (iv) λ = 3.0 (v) λ = 3.1 4. Comparing three methods.
The residual plot, table and box–cox methods (choose one) differ on what trans-formation of Y to use to make the variance constant. Match the method used and the transformations suggested below.
method suggested transformation (a–1) residual plot (A) Y ′ = √ Y (a–2) residual plot (B) Y ′ = √ Y + 1 (b) table (C) Y ′ = ln Y (c) box–cox (D) Y ′ = Y 2.9 method (a–1) (a–2) (b) (c) transformation (A) (B) (C) (D) 5. Using the suggested transformations in a single–factor ANOVA.
Use all of the suggested transformations to test if the mean fluxes for the three different soil types are the same or different at α = 0.05.
(a) Statement.
Choose none, one or more.
i. H0 : error variance constant over soil grades versus H1 : not constant for different soil grades ii. H0 : µ1 = µ2 = µ3 versus H1 : µi ̸= µj, i ̸= j iii. H0 : error variance constant over soil grades versus H1 : increase for different soil grades (b) Test.
From SAS, the test statistic for i. Y ′ = √ Y is (choose one) 3794.62 / 3793.33 / 4067.73 / 1495.68 ii. Y ′ = √ Y + 1 is (choose one) 3794.62 / 3793.33 / 4067.73 / 1495.68 iii. Y ′ = ln Y is (choose one) 3794.62 / 3793.33 / 4067.73 / 1495.68 Section 6. Effects of Departures from Model (ATTENDANCE 2) 77 iv. Y ′ = Y 2.9 is (choose one) 3794.62 / 3793.33 / 4067.73 / 1495.68 where all associated p–values are essentially zero (0).
The level of significance is 0.05.
(c) Conclusion.
Since the p–value, 0.00, is smaller than the level of significance, 0.05, we (circle one) accept / reject the null hypothesis that mean water–soil fluxes are the same.
6. How Is The Transformed Data Used In Inference Procedures?
The transformed data can be used to (choose none, one or more) (a) test the equality of factor level means since the single–factor ANOVA test is a relative one which does not depend on the absolute size of the factor level means, but only on whether these means are the same or different from one another in a relative sense.
(b) estimate the transformed factor level means but these estimates would then have to be transformed back into the original data units. For example, if the data is transformed using Y ′ = ln Y , the transformation Y ′′ = eY ′ would need to be used to give estimates in the original data units.
18.6 Effects of Departures from Model What if, after attempting remedial measures, the data still does not conform to the assumptions of the single–factor ANOVA model? In fact, this model is robust to nonnormality and unequal error variances, but is sensitive to nonindependence of the error terms. In other words, as long as the data is not extremely nonnormal and does not have extremely unequal error variances, then it is still possible to perform reasonably accurate estimation and testing using the single–factor ANOVA model.
This is not true if the error terms are nonindependent. Since it is difficult to correct for nonindependence after the data has been collected, it is very important to prevent this problem in the first place, to collect the data in a appropriate random manner.
18.7 Nonparametric Rank F Test This material is not covered.
18.8 Case Example–Heart Transplant This is an excellent example. |
655 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3712784/ | Arginine Mimetics via α-Guanidino Acids: Introduction of Functional Groups and Stereochemistry Adjacent to Recognition Guanidiniums in Peptides - PMC
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Chembiochem
. Author manuscript; available in PMC: 2013 Jul 16.
Published in final edited form as: Chembiochem. 2011 Dec 23;13(2):259–270. doi: 10.1002/cbic.201100638
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Arginine Mimetics via α-Guanidino Acids: Introduction of Functional Groups and Stereochemistry Adjacent to Recognition Guanidiniums in Peptides
Shalini Balakrishnan
Shalini Balakrishnan
[a]Department of Chemistry and Biochemistry University of Delaware Newark, Delaware 19716 (USA)
Find articles by Shalini Balakrishnan
[a],‡, Michael J Scheuermann
Michael J Scheuermann
[a]Department of Chemistry and Biochemistry University of Delaware Newark, Delaware 19716 (USA)
Find articles by Michael J Scheuermann
[a],‡, Neal J Zondlo
Neal J Zondlo
[a]Department of Chemistry and Biochemistry University of Delaware Newark, Delaware 19716 (USA)
Find articles by Neal J Zondlo
[a],
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[a]Department of Chemistry and Biochemistry University of Delaware Newark, Delaware 19716 (USA)
‡
These authors contributed equally to this work.
Phone: (+1) 302-831-0197 Fax: (+1) 302-831-6335 zondlo@udel.edu
Issue date 2012 Jan 23.
PMC Copyright notice
PMCID: PMC3712784 NIHMSID: NIHMS405995 PMID: 22213184
The publisher's version of this article is available at Chembiochem
Abstract
Arginine residues are broadly employed for specific biomolecular recognition, including in protein-protein, protein-DNA, and protein-RNA interactions. Arginine recognition commonly exploits the potential for bidentate electrostatic and hydrogen-bonding interactions. However, in arginine residues, the guanidinium functional group is located at the terminus of a flexible hydrocarbon side chain, which lacks the functionality to contribute to specific arginine-mediated recognition and may entropically disfavor binding. In order to enhance the potential for specificity and affinity in arginine-mediated molecular recognition, we have developed an approach to the synthesis of peptides that incorporates an α-guanidino acid as a novel arginine mimetic. α–Guanidino acids, derived from α-amino acids, with guanylation of the amino group, were incorporated stereospecifically into peptides on solid phase via coupling of an Fmoc amino acid to diaminoproprionic acid (Dap), Fmoc deprotection, guanylation of the amine on solid phase, and deprotection, generating a peptide containing an α-functionalized arginine mimetic. This approach was examined via the incorporation of arginine mimetics into ligands for the Src, Grb, and Crk SH3 domains at the site of the key recognition arginine. Protein binding was examined for peptides containing guanidino acids derived from Gly, l-Val, l-Phe, l-Trp, d-Val, d-Phe, and d-Trp. We demonstrate that paralog specificity and target site affinity may be modulated via the use of α-guanidino acid-derived arginine mimetics, generating peptides that exhibit enhanced Src specificity via selection against Grb and peptides that reverse the specificity of the native peptide ligand, with enhancements in Src target specificity of up to 15-fold (1.6 kcal/mol).
Keywords: Peptides, amino acids, peptidomimetics, guanidiniums, protein-protein interactions
Introduction
Arginine residues have unique biomolecular recognition properties owing to the guanidinium functional group, which displays 5 coplanar hydrogen bond donors, a diffuse positive charge, and delocalized pi electrons. The guanidinium group is particularly adept at multidentate hydrogen bonding and electrostatic interactions, leading to highly favorable interactions with phosphate, carboxylate, and sulfate anions. These characteristics often result in the observation of obligatory arginine recognition in the interactions of proteins with DNA, RNA, lipids, carbohydrates, and other proteins, as observed in cell-cell communication and extracellular signaling, in intracellular signal transduction pathways, and in the control of transcription and translation, among many other applications.[1a, 3] Arginine-rich peptides also have special application as cell-penetrating peptides, in a manner that is dependent on the guanidinium functional groups of the peptides.
One potential limitation of arginine-mediated recognition is the presence of the guanidinium group at the end of linear alkyl side chain. The three methylenes (−CH 2- groups) of arginine residues may enhance target site affinity via the hydrophobic effect, but their inherent flexibility, which allows recognition through one or more of multiple side chain rotamers, imposes a potential cost in both entropy and in recognition specificity, as the linear alkyl side chain can readily adapt to different structures. Moreover, the linear side chain lacks additional features (e.g. functional groups or stereochemistry) that could enhance specific target recognition.
A broad interest in the pharmacological control of arginine-mediated interactions has led to the development of a range of small molecule arginine mimetics. We recently described the stereospecific synthesis of the novel arginine mimetic α-guanidino acids, in which the α-amino group of standard α-amino acids is guanylated to generate a guanidinium group with adjacent stereochemistry and functional groups. In that work, we focused on the synthetic development of α-guanidino acids as arginine mimetics in small molecules. However, we envisioned that α-guanidino acids could function as arginine mimetics within peptides, via the conjugation of an amino acid to the β- or α-amino group of a diaminoproprionic acid (Dap) residue and subsequent guanylation (Figure 1). This approach would have the potential to introduce both stereochemistry and functional groups for molecular recognition adjacent to the key guanidinium, providing the possibility of enhanced target specificity and/or affinity. Moreover, because the approach is based on guanylated α-amino acids, it has the potential to introduce a wide range of α-functional groups, owing to the broad availability of structurally and functionally diverse α-amino acids. Indeed, in the work of Wells, Arkin, and coworkers, small molecules containing α-guanidino acids could dramatically modulate affinity for interleukin-2, with both stereochemistry and side chain significantly affecting target recognition.[7d-h] This work provided critical proof of concept for the application of α-guanidino acids in specific target recognition, although their synthetic approach involved separation of diastereomers of α-guanidino acid-containing compounds.
Figure 1.
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Arginine, homoarginine, and Dap-derived α-guanidino acid arginine mimetics, which allow placement of functional groups and introduction of stereochemistry adjacent to the guanidinium. Incorporation of the α-guanidino acid on the β-amino group of Dap (yielding hArg(gXaa) residues) allows the incorporation of α-guanidino acids at any position in a peptide sequence. In contrast, incorporation on the α-amino group of Dap (yielding Arg(gXaa) residues) directly mimics the length of the arginine side chain, but can only be readily employed at the N-terminus of an α-peptide due to the resultant β-peptide backbone with this substitution.
To explore the application of α-guanidino acid-based arginine mimetics, we have developed approaches to the synthesis of peptides containing α-guanidino acids and tested these arginine mimetics in the challenging context of paralog-specific recognition of SH3 domains by proline-rich peptides.
Results
Design of peptides containing α-guanidino acids as novel arginine mimetics
We examined arginine-mediated recognition via α-guanidino acid arginine mimetics within the context of proline-rich ligands of SH3 domains. SH3 domains are small (~60 amino acids) protein domains that are critical for protein-protein interactions in signal transduction. SH3 domains bind short proline-rich ligands with a typical PXXP motif. Because of the importance of SH3 domains in cell signaling, there has been substantial interest in developing inhibitors of SH3 domain-mediated protein-protein interactions. However, the inherent flatness of the recognition surface has resulted in significant problems in achieving paralog specificity, that is, achieving specific recognition of one SH3 domain among the approximately 300 SH3 domains in the human proteome.[9f, 10] Challenges in achieving paralog specificity have resulted in a significant reduction in pharmaceutical efforts to target SH3 domains, compared to the peak interest in the 1990s, despite potential applications as therapeutics in cancer and other diseases.
SH3 domains typically bind peptides with one or more arginine residues at the N-terminus (type I) or the C-terminus (type II) of the recognition sequence.[8c, 9f, 11] Confounding attempts to achieve paralog specificity, it was recognized that a single SH3 domain could bind both type I and type II ligands, via a reversal of the orientation of ligand binding. Analysis of available high resolution structures of peptides bound to SH3 domains reveals that arginine recognition by SH3 domains involves a combination of electrostatic and hydrogen bonding interactions with the guanidinium and hydrophobic interactions with the arginine methylenes (Figure 2). Interestingly, there is considerable available hydrophobic surface area that is not employed in recognition, presumably to accommodate the diversity of ligands natively bound by SH3 domains. For example, in the Src SH3 domain (pdb 1qwf), the ligand arginine side chain is near the hydrophobic residues Trp42, Tyr55, and Thr22, whereas in the Grb2 SH3 domain (pdb 1sem) the ligand arginine is near Trp191, Ile202, Phe165, and the methylenes of Gln168 and Glu169.[11a, 11c, 11d, 12] We envisioned that the incorporation of functional groups adjacent to the guanidinium in an arginine mimetic could exploit the differences in the recognition surfaces in divergent SH3 domains and provide a locus to achieve paralog specificity. Given the importance of arginine residues in SH3 domain recognition of ligands, this approach provides the possibility of a general method to modulate target affinity in arginine-mediated recognition.
Figure 2.
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SEM-5 (Grb2 homologue) SH3 domain (ribbon (top) or CPK (bottom)) (pdb: 1sem)[11c] bound to SH3 domain ligand from mSOS PPPVPPR (sticks). Bottom: CPK figure showing hydrophobic surface area (gray) around the arginine side chain.
To test this approach, we examined α-guanidino acid arginine mimetics in two peptide contexts: as a replacement for an N-terminal arginine in a type I SH3 domain ligand (hAMI and AMI) and as a replacement for a C-terminal arginine in a type II SH3 domain ligand (hAMII). In addition, within the context of a type I SH3 domain ligand, we examined both the use of the Dap α-amino and β-amino groups for conjugation of the α-guanidino acid, generating peptides that are formally (by side chain length) arginine (Arg(gXaa)) or homoarginine (hArg(gXaa)) equivalents. Both ligands were chosen to incorporate a single arginine residue, to simplify analysis and because a single properly-positioned arginine is both necessary and sufficient for SH3 domain binding.[11a, 11c, 11d, 12-13]
Stereospecific synthesis of peptides incorporating α-guanidino acids
In order to take advantage of the diversity of amino acid functionality and stereochemistry that is commercially available as standard Fmoc amino acids, as well as the benefits of solid phase peptide synthesis, we devised an approach to the synthesis of peptides containing α-guanidino acids that involves coupling of protected α-amino acids to an orthogonally (Mtt) protected diaminoproprionic acid (Dap) residue. In this approach, the peptide was fully synthesized, the protecting group on the Dap removed, and the Fmoc amino acid that is the α-guanidino acid precursor coupled by standard amide coupling conditions, followed by piperidine deprotection of the coupled α-amino acid and guanylation of the α-amino group using Moroder’s guanylating reagent (Schemes 1-4). This approach was applied to the synthesis of both type I (Schemes 2-3) and type II (Scheme 4) SH3 domain ligand peptides. In order to test the application of α-guanidino acids in molecular recognition, including roles of stereochemistry and functional group identity, each peptide was synthesized to contain Dap conjugated to α-guanidino acids derived from glycine and from both l- and d- stereoisomers of the hydrophobic amino acids valine, phenylalanine, and tryptophan. These amino acids were chosen because of their hydrophobic surface area, their sterics, and because Phe and Trp are particularly prone to epimerization among canonical amino acids, and thus provide an appropriate test of the potential synthetic challenges in the synthesis of peptides containing α-guanidno acid arginine mimetics. As controls, the peptides with acetylated Dap were synthesized, which contain the side chain amide but lack the guanidinium group (i.e. equivalent to gGly with the guanidinium removed).
Scheme 1.
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Synthesis of Moroder’s guanylating reagent 1. This synthesis differs from the published synthesis of 1 due to the replacement of HgCl 2 employed in the original synthesis with the more convenient reagent EDCI.
Scheme 4.
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Synthesis of type II SH3 domain ligand peptides peptides containing N-terminal homoarginine-equivalent α-guanidino-acid arginine mimetics (hAMII(gXaa)).
Scheme 2.
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Synthesis of type I SH3 domain ligand peptides containing N-terminal homoarginine-equivalent α-guanidino acid arginine mimetics (hAMI(gXaa)).
Scheme 3.
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Synthesis of type I SH3 domain ligand peptides peptides containing N-terminal arginine-equivalent α-guanidino-acid arginine mimetics (AMI(gXaa)).
Previous work with small molecules containing α-guanidino acids suggested that typical peptide cleavage/deprotection conditions (i.e. 90% TFA) would lead to substantial or complete epimerization at the carbon α to the guanidinium group.[7a, 7c-h] Indeed, standard TFA cleavage of α-guanidino acid-containing peptides revealed evidence of significant epimerization (data not shown). However, in small molecules, after deprotection of the guanidine under mild conditions (0.5 M HCl), the free guanidium-containing α-guanidino acids were configurationally stable in 90% TFA.[7a] These results suggest that the Boc groups on α-guanidino acids are rather labile and might be removed under mild conditions.
Based on these results, we devised a two-stage scheme for separate deprotection of the guanidine Boc groups and global deprotection/cleavage of the peptide from the resin (Scheme 5). A brief screening of deprotection conditions identified that guanidine Boc groups could be removed with no evidence of epimerization using 6 M AcOH in MeCN (2 × 20 minutes). This approach was then applied to all synthesized peptides, resulting in the generation of α-guanidino acid-containing peptides in good yield.
Scheme 5.
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Two stage deprotection of peptides containing α-guanidino acid arginine mimetics. This approach was applied to the synthesis of all α-guanidino acid-containing peptides in this study, to prevent epimerization of the carbon α to the guanidinium group. The intermediate after guanidine Boc deprotection could be fully deprotected, as suggested by experiments with model compounds, or may potentially retain one guanidine Boc group.
In order to confirm the retention of stereochemical integrity in the synthesis of peptides containing α-guanidino acids, all peptides were characterized by NMR spectroscopy. TOCSY spectra of the peptides revealed that in all cases the diastereomers derived from l- and d-amino acids could be readily differentiated by NMR (Figure 4 and data not shown). Notably, in all cases the NMR data revealed no evidence of epimerization of the individual peptides, within the limits of detection (> 95% stereochemical purity). As Phe and Trp are the canonical amino acids most prone to epimerization, these data suggest generality of the methodology for the synthesis of α-guanidino acid-based arginine mimetics in peptides.
Figure 4.
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Comparison of TOCSY crosspeaks of diastereomeric hAMII peptides (A) hAMII(g l-Val (red) versus hAMII(g d-Val) (blue); (B) hAMII(g l-Phe (red) versus hAMII(g d-Phe) (blue); and (C) hAMII(g l-Trp (red) versus hAMII(g d-Trp) (blue). Comparable spectra were obtained for diagnostic regions of the TOCSY spectra of all peptides, indicating stereoisomeric purity and the absence of epimerization on the conjugated α-guanidino acids. Critical crosspeaks from the Dap-conjugated α-guanidino acids are highlighted. Arginine mimetics have two amide protons for the Dap residue, from the backbone N-H and β N-H amides.
α-Guanidino acid-based arginine mimetics to modulate paralog-specific recognition of SH3 domains
All peptides were analyzed for their ability to function as ligands for the Src and Grb SH3 domains, which can bind identical ligands with similar affinity, and thus provide a test of the effects of α-guanidino acid-based arginine mimetics on both target affinity and specificity. As a control, to identify possible increases in affinity due to non-specific interactions, we also examined binding to the Crk SH3 domain, which can bind similar ligands but bound the parent single-arginine peptides in this study poorly. All peptides were labeled with fluorescein on the C-terminal lysine residue and dissociation constants (K d) for all peptide-protein interactions determined via a fluorescence polarization-based binding assay (Figure 5, Tables 1-3).
Figure 5.
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Representative binding isotherms, showing data and comparisons of key arginine-mimetic peptides. mP = fluorescence polarization in millipolarization units (= polarization/1000). Error bars indicate standard error. (A) hAMI(Arg) (black triangles), hAMI(Dap-Ac) (dark green inverted triangles), and hAMI(gGly) (light green open triangles) binding to Src; (B) α-substituted hAMI arginine mimetic peptides binding to Src: hAMI(g l-Val) (magenta open squares), hAMI(g d-Val) (yellow closed squares), hAMI(g l-Phe) (cyan open circles), hAMI(g d-Phe) (orange closed circles), hAMI(g l-Trp) (blue open diamonds), hAMI(g d-Trp) (red closed diamonds); (C) α-substituted hAMI arginine mimetic peptides binding to Grb: hAMI(Arg) (black closed triangles); hAMI(gGly) (light green open triangles); hAMI(g l-Val) (magenta open squares), hAMI(g d-Val) (yellow closed squares), hAMI(g l-Phe) (cyan open circles), hAMI(g d-Phe) (orange closed circles), hAMI(g l-Trp) (blue open diamonds), hAMI(g d-Trp) (red closed diamonds); (D) hAMI(Arg) binding to Src (closed black triangles) and Grb (open green triangles); AMI(g l-Phe) binding to Src (closed cyan circles) and Grb (open blue circles); (E) hAMII(Arg) binding to Src (closed blue triangles) and Grb (open red triangles); (F) hAMII(g l-Trp) binding to Src (closed blue squares) and Grb (open red squares).
Table 1.
Dissociation constants of type I homoarginine-equivalent arginine mimetic peptides (hAMI).
| | | K d, μM | Error | Δ G, kcal mol−1 | ΔΔ G affinity, a kcal mol−1 | ΔΔ G specificity, b kcal mol−1 |
:--- ---: ---:
| | | |
| Src | Arg | 5.3 | 0.3 | −7.2 | 0 | −1.0 |
| Dap-Ac | 91 | 5.3 | −5.5 | 1.68 | −0.7 |
| hArg(gGly) | 11.1 | 0.2 | −6.7 | 0.44 | −0.9 |
| hArg(g l-Val) | 25.5 | 1.3 | −6.2 | 0.93 | −1.0 |
| hArg(g d-Val) | 22.5 | 1.1 | −6.3 | 0.86 | −1.9 |
| hArg(g l-Phe) | 23.9 | 1.1 | −6.3 | 0.89 | −0.7 |
| hArg(g d-Phe) | 15.9 | 0.6 | −6.5 | 0.65 | −2.0 |
| hArg(g l-Trp) | 31.8 | 1.5 | −6.1 | 1.07 | −0.5 |
| hArg(g d-Trp) | 12.6 | 1.1 | −6.7 | 0.52 | −1.5 |
| Grb | Arg | 29 | 2 | −6.2 | 0 | |
| Dap-Ac | 320 | 104 | −4.8 | 1.42 | |
| hArg(gGly) | 50 | 6 | −5.8 | 0.32 | |
| hArg(g l-Val) | 141 | 12 | −5.2 | 0.93 | |
| hArg(g d-Val) | 576 | 19 | −4.4 | 1.76 | |
| hArg(g l-Phe) | 82 | 7 | −5.6 | 0.61 | |
| hArg(g d-Phe) | 451 | 22 | −4.6 | 1.62 | |
| hArg(g l-Trp) | 72 | 3 | −5.6 | 0.54 | |
| hArg(g d-Trp) | 164 | 7 | −5.1 | 1.02 | |
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a
ΔΔ G affinity = Δ G (peptide) – Δ G (Arg).
b
ΔΔ G specificity = Δ G Src – Δ G Grb for a given peptide. All peptides bound poorly to Crk (K d> 250 μM). All experiments were conducted in PBS at 25 °C.
Table 3.
Dissociation constants of type II homoarginine-equivalent arginine mimetic peptides (hAMII).
| | | K d, μM | Error | Δ G, kcal mol−1 | ΔΔ G affinity, kcal mol−1 | ΔΔ G specificity,b kcal mol−1 |
:--- ---: ---:
| | | |
| Src | Arg | 37 | 2 | −6.0 | 0 | +0.7 |
| Dap-Ac | 350 | 31 | −4.7 | 1.33 | −0.2 |
| hArg(gGly) | 59 | 2.3 | −5.8 | 0.28 | −0.6 |
| hArg(g l-Val) | 67 | 2.3 | −5.7 | 0.35 | −0.7 |
| hArg(g d-Val) | 130 | 3.7 | −5.3 | 0.74 | −0.4 |
| hArg(g l-Phe) | 58 | 1.8 | −5.8 | 0.27 | −0.5 |
| hArg(g d-Phe) | 66 | 2.1 | −5.7 | 0.34 | −0.9 |
| hArg(g l-Trp) | 24 | 1.2 | −6.3 | −0.26 | −1.0 |
| hArg(g d-Trp) | 49 | 2.1 | −5.9 | 0.17 | −1.0 |
| Grb | Arg | 12 | 0.7 | −6.7 | 0 | |
| Dap-Ac | 500 | 160 | −4.5 | 2.20 | |
| hArg(gGly) | 166 | 12 | −5.1 | 1.56 | |
| hArg(g l-Val) | 227 | 13 | −4.9 | 1.74 | |
| hArg(g d-Val) | 270 | 31 | −4.9 | 1.84 | |
| hArg(g l-Phe) | 128 | 8 | −5.3 | 1.40 | |
| hArg(g d-Phe) | 319 | 44 | −4.8 | 1.94 | |
| hArg(g l-Trp) | 121 | 10 | −5.3 | 1.36 | |
| hArg(g d-Trp) | 244 | 24 | −4.9 | 1.78 | |
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a
ΔΔ G affinity = Δ G (peptide) – Δ G (Arg).
b
ΔΔ G specificity = Δ G Src – Δ G Grb for a given peptide. All peptides except hAMII(Arg) (K d = 102 ± 7 μM, Δ G Crk = −5.4 kcal mol−1), hAMII(g l-Trp) (K d = 153 ± 9 μM, Δ G Crk = −5.2 kcal mol−1), and hAMII(g d-Trp) (K d = 183 ± 11 μM, Δ G Crk = −5.1 kcal mol−1) bound poorly to Crk (K d> 250 μM). All experiments were conducted in PBS at 25 °C.
Peptide affinity for target SH3 domain proteins was affected by replacement of Arg with arginine mimetic, with differential impact possible depending on the site (type I versus type II ligand), functional group (gGly, gVal, gPhe, or gTrp), stereochemistry (l- versus d-), and/or attachment (arginine versus homoarginine equivalent) of the arginine substitution. First considering type I mimics substituted on the Dap side chain (homoarginine equivalents, hAMI) (Table 1, Figure 5a-c), we observed, as expected, that loss of the polarity-determining arginine guanidinium resulted in a substantial loss of affinity (ΔΔ G Src = +1.7 kcal/mol, hAMI(Arg) versus hAMI(Dap-Ac) binding to Src; ΔΔ G Grb = +1.4 kcal/mol, hAMI(Arg) versus hAMI(Dap-Ac) binding to Grb) (Figure 5a). In contrast, replacement of Arg with hArg(gGly) resulted in only a modest (ΔΔ G Src = +0.44 kcal/mol; ΔΔ G Grb = +0.32 kcal/mol) loss in affinity, indicating that α-guanidino acids may substitute for Arg residues in peptides (Figure 5a). For peptides in this series, substitutions at the α-position to the guanidinium only modestly affected affinity for Src (Figure 5b). However, α-substitution more significantly affected affinity for Grb, in a manner highly dependent on the identity of the α-guanidino acid side chain and on its stereochemistry (Figure 5c). The hAMI peptides containing hArg(g d-Val), hArg(g d-Phe), and hArg(g d-Trp) substitution exhibited better specificity for Src over Grb (ΔΔ G specificity = −1.9, −2.0, and −1.5 kcal/mol, respectively) than the parent peptide (ΔΔ G specificity = −1.0 kcal/mol) or the glycine derivative (ΔΔ G specificity = −0.9 kcal/mol), indicating selection against binding to Grb when d-amino acid-based arginine mimetic peptides were employed. In contrast, the diastereomers hArg(g l-Phe) and hArg(g l-Trp) exhibited reduced specificity (ΔΔ G specificity = −0.7 and −0.5 kcal/mol, respectively), indicating that both the identity and stereochemistry of the α-substituted guanidinium significantly impacted target recognition and specificity.
In the series of arginine-equivalent peptides, in which substitution occurred on the Dap backbone nitrogen (AMI series), all arginine-mimetic peptides exhibited substantially reduced binding affinity for both Src and Grb (ΔΔ G Src = +1.3 to +1.8 kcal/mol, ΔΔ G Grb = +0.8 to +1.6 kcal/mol) (Table 2) compared to the Arg-containing peptide. As was observed previously, replacement of Arg with a side chain lacking a guanidinium (AMI(Dap-Ac)) resulted in a substantial loss in Src affinity (ΔΔ G = +1.6 kcal/mol). However, unexpectedly, in this case, and in contrast to the case with homoarginine-equivalent peptides, the arginine-mimetic peptide lacking a guanidium (AMI(Dap-Ac)) actually bound Src with an affinity similar to (or slightly better than) the analogous peptide with a guanidinium (AMI(gGly)) (ΔΔ G = −0.1 kcal/mol). Analogous results were observed for binding to Grb, with the AMI(gGly) peptide binding with similar affinity as the peptide lacking a guanidinium (AMI(Dap-Ac)). For binding to both Src and Grb, the presence of an α-substituent on the α-guanidino acid in some case modestly increased binding compared to AMI(gGly), although in all cases binding was worse than observed in the hAMI series. The peptide AMI(g l-Phe) exhibited relative selectivity to Grb (ΔΔ G specificity = −0.5 kcal/mol, compared to ΔΔ G specificity = −1.0 kcal/mol for the parent peptide; ΔΔΔ G specificity = +0.5 kcal/mol), with a 0.7 kcal/mol increase in affinity for Grb compared to AMI(gGly), suggesting interaction of Grb with the α-guanidino acid side chain (Figure 5d). In total, these data suggest that in arginine-equivalent peptides, the guanidinium group is not readily able to interact favorably with the carboxylates in the SH3 domain, or alternatively that unfavorable interactions (e.g. with the amide group in the arginine mimetic, compared to the methylenes of an arginine) counteract favorable interactions with the guanidinium. In addition, this substitution is only readily incorporated at peptide termini due to the change in backbone structure (Figure 1). In sum, these data suggest that arginine-equivalent α-guanidino acids are functionally not readily able to substitute for arginine residues in molecular recognition within this peptide context.
Table 2.
Dissociation constants of type I arginine-equivalent arginine mimetic peptides (AMI).
| | | K d, μM | Error | Δ G, kcal mol−1 | ΔΔ G affinity, kcal mol−1 | ΔΔ G specificity,b kcal mol−1 |
:--- ---: ---:
| | | |
| Src | Arg | 5.3 | 0.3 | −7.2 | 0 | −1.0 |
| Dap-Ac | 83 | 2.8 | −5.5 | 1.63 | −0.9 |
| Arg(gGly) | 104 | 8.1 | −5.4 | 1.76 | −0.9 |
| Arg(g l-Val) | 86 | 3.1 | −5.5 | 1.65 | −0.5 |
| Arg(g d-Val) | 75 | 3.2 | −5.6 | 1.57 | −0.7 |
| Arg(g l-Phe) | 54 | 4.8 | −5.8 | 1.37 | −0.5 |
| Arg(g d-Phe) | 75 | 4.9 | −5.6 | 1.57 | −0.7 |
| Arg(g l-Trp ) | 54 | 3.8 | −5.8 | 1.37 | −0.9 |
| Arg(g d-Trp) | 51 | 3.6 | −5.8 | 1.34 | −1.1 |
| Grb | Arg | 29 | 2 | −6.2 | 0 | |
| Dap-Ac | 388 | 75 | −4.6 | 1.53 | |
| Arg(gGly) | 445 | 75 | −4.6 | 1.60 | |
| Arg(g l-Val) | 210 | 21 | −5.0 | 1.17 | |
| Arg(g d-Val) | 243 | 33 | −4.9 | 1.25 | |
| Arg(g l-Phe) | 117 | 10 | −5.3 | 0.82 | |
| Arg(g d-Phe) | 254 | 17 | −4.9 | 1.28 | |
| Arg(g l-Trp) | 253 | 16 | −4.9 | 1.28 | |
| Arg(g d-Trp) | 333 | 29 | −4.7 | 1.43 | |
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a
ΔΔ G affinity = Δ G (peptide) – Δ G (Arg).
b
ΔΔ G specificity = Δ G Src – Δ G Grb for a given peptide. All peptides bound poorly to Crk (K d> 250 μM). All experiments were conducted in PBS at 25 °C.
Replacement of arginine with α-guanidino acid arginine mimetics in type II SH3 domain peptide ligands (hAMII series peptides) resulted in significant modulation of protein affinity and specificity (Table 3), as was observed with type I homoarginine-equivalent ligands (hAMI). In the type II ligand series, the parent Arg-containing peptide bound preferentially to Grb over Src (ΔΔ G specificity = +0.7 kcal/mol). As expected, replacement of Arg with acetylated Dap resulted in a substantial loss of binding to both proteins. Interestingly, however, replacement of Arg with arginine mimetics resulted in a reversal of specificity in all cases, leading to a preference for binding to Src over binding to Grb, with ΔΔ G specificity = −0.4 to −1.0 kcal/mol, for an overall switch in target specificity from the parent peptide to the arginine-mimetic peptides ΔΔΔ G specificity = −1.1 to −1.6 kcal/mol (Figure 5e-f). The reversal of specificity was manifested almost entirely in selection against binding to Grb: all α-guanidino acid-containing peptides bound Grb with at least 10-fold worse affinity than the Arg peptide (ΔΔ G Grb = +1.4 to +2.2 kcal/mol). In this series, there was only a modest preference in Src binding for l-amino acid-derived arginine mimetics compared to those derived from d-amino acids. Notably, the peptide hAMII(gD-Trp) exhibited enhanced overall affinity for Src compared to the Arg-containing peptide (ΔΔ G Src = −0.3 kcal/mol). This peptide also had the greatest specificity for Src over Grb (ΔΔ G specificity = −1.0 kcal/mol) among peptides in this series.
In order to confirm that the α-guanidino acid arginine mimetics were not globally changing the structure of the peptides, and thus impacting changes in affinity and specificity through a change in peptide structure, circular dichroism (CD) spectroscopy was conducted on the type II d-Trp arginine mimetic peptide (hAMII(g d-Trp), as well as on the guanidino-glycine derivative (hAMII(gGly)) and the parent arginine-containing peptide in this series (Figure 6). The CD spectra of all peptides exhibited maxima and minima at approximately 228 nm and 205 nm, respectively, indicative of polyproline II helix structure. In addition, NMR data (Figure 4 and data not shown) reveal similar H α and H N chemical shifts and 3 J HNA coupling constants for all equivalent residues (other than conjugated Dap residues) in peptides within a given series. In combination, these data indicate that replacement of arginine with an α-guanidino acid-based arginine mimetic is minimally disruptive of peptide structure, and suggest that the effects of arginine mimetics observed above are more likely due to specific interactions with the target protein.
Figure 6.
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Circular dichroism spectra of representative peptides: squares, type II Arg peptide hAMII(Arg) (control, native peptide ligand); diamonds, type II gGly homoarginine-equivalent peptide (hAMII(gGly)); circles, type II g l-Trp homoarginine-equivalent arginine mimetic peptide (hAMII(g l-Trp)).
Discussion
We have developed an approach to stereospecifically synthesize peptides containing novel α-guanidino acid arginine mimetics. These arginine mimetics are derived from guanylation of α-amino acids and result in the placement of functional groups (i.e. amino acid side chains) and stereochemistry on the carbon adjacent to the guanidinium. The synthetic approach employs commercially available Fmoc-amino acids and thus allows the incorporation of a wide range of functionalities adjacent to the guanidinium, providing the possibility to increase target specificity and/or target affinity via a combination of functional groups for recognition and stereochemistry, and to disfavor binding to certain targets via unfavorable steric, electrostatic, hydrogen bonding, or other interactions. The reactions are all performed on solid phase using commercially available Fmoc amino acids. In this work, we employed one reagent, Moroder’s reactive benzotriazole-based guanylating reagent 1, that is not commercially available; 1 is readily prepared in one step from commercially available reagents (the intermediate compound bis(Boc)-thiourea is commercially available) or in two steps from inexpensive reagents (Scheme 1).[14a] This approach should also be readily applicable with other reactive guanylating reagents that can function for reactions on solid phase.
In this work, we demonstrated the retention of stereochemistry in the solid phase synthesis of α-guaninido acid-containing molecules, via a two-stage deprotection that removes the guanidine Boc protecting group first and then subjects the peptide to standard peptide cleavage/deprotection conditions. Of note is the observation that coupled protected α-guanidino acids are prone to epimerization under acidic conditions, including those previously employed in low yield syntheses of diastereomeric small molecules containing α-guanidino acids.[7a, 7c, 7g] However, using the very mild guanidine Boc deprotection conditions developed herein of 6 M AcOH in MeCN, the stereochemical integrity of the α-guanidino acid was retained. Considering the reliance on readily available reagents and solid phase synthetic approaches, the methodology has broad potential application for the synthesis of stereochemically defined α-functionalized guanidiniums within peptides and small molecules and is a highly practical approach to synthesize peptides with arginine mimetics.
We examined the functional effects of α-guanidino acid-based arginine mimetics within the context of peptides for paralog-specific targeting of SH3 domains. Arginine mimetics were incorporated at the site of a critical polarity- and specificity-determining arginine residue at the N-terminus (type I ligand) or C-terminus (type II ligand) of the peptide. Two types of arginine mimetic were employed, one based on formal conjugation of the α-guanidino acid to the Dap side chain (generating a homoarginine equivalent, hAMI and hAMII series) and one based on formal conjugation to the Dap α-amino group main chain nitrogen (generating an arginine equivalent based on residue length, AMI series).
In the context of SH3 domain recognition, the arginine-equivalent arginine mimetic (AMI series), in which the peptide backbone structure is modified and thus is only suitable at peptide termini, was found to be inferior, leading to a reduction in SH3 domain affinity and no evidence that the guanidinium was able to participate in target recognition. Some peptides in this series exhibited a modest increase in preference for Grb over Src (ΔΔΔ G specificity up to +0.5 kcal/mol). The poor recognition characteristics of the arginine-equivalent α-guanidino acid-containing peptides could be due to the loss of hydrophobic interactions with the Arg methylenes, unfavorable interactions of the protein with the amide, or the loss in flexibility of the amide-containing side chain of the arginine mimetic preventing favorable interactions with the target, compared to the flexible hydrocarbon side chain in arginine.
In contrast, the homoarginine-equivalent arginine mimetic-containing peptides effectively induced specificity in SH3 domain recognition in a manner that was dependent on position (type I versus type II), stereochemistry, and side chain identity. In the type I peptides, all α-guanidino acid-containing peptides exhibited a modest loss in binding affinity for Src and Grb, while those with a d-amino acid-derived α-guanidino acid exhibited a more substantial loss in affinity for Grb. Thus, the (R)-α-guanidino acid-containing peptides all exhibited increased target specificity for Src over Grb compared to the native ligand, with the enhanced specificity achieved via selection against Grb. Negative selection is important to specific recognition. Indeed, native SH3 domains exhibit evidence of negative selection within the natural repertoire of ligands and SH3 domains. This work provides a complementary strategy to achieve selection against alternative targets in guanidinium-mediated recognition.
Among type II SH3 domain ligands (hAMII), the native Arg-containing peptide bound preferentially to the Grb SH3 domain over the Src SH3 domain (ΔΔ G specificity = +0.66 kcal/mol). In contrast, all type II arginine-mimetic peptides preferentially bound Src over Grb, with specificities up to −0.95 kcal/mol, resulting in an overall reversal of SH3 domain specificity (ΔΔΔ G specificity = −1.1 to −1.6 kcal/mol) via the application of α-guanidino acid arginine mimetics. As was the case in type I (hAMI) ligands, the primary mechanism of modulated specificity was selection against Grb (ΔΔ G Grb = +1.4 to +1.9 kcal/mol). Binding to Src was modestly affected by the stereochemistry and identity of the guanidino acid side chain (ΔΔ G Src = −0.3 to +0.7 kcal/mol), with (S) guanidino acid stereoisomers and tryptophan substitution leading to the highest Src affinity. Among the peptides examined, the tryptophan derivatives exhibited enhanced affinity for Src compared to the glycine derivative, indicative of a potentially favorable peptide side chain interaction with Src. In total, the results suggest that the employment of α-guanidino acid-based arginine mimetics is a strategy for achievement of paralog-specific SH3 domain recognition that is complementary to those previously described.[9f, 10]
Arginine residues and guanidiniums are widely employed in manners that exploit their unique molecular recognition properties. Examples where guanidinium groups are paramount or superior include specific recognition of RNA and DNA;[3a, 19]; intercellular communication via protein-protein interactions mediated by RGD motifs;[3b, 6, 20] polyarginine cell-penetrating peptides; interactions with cell surface and intracellular sulfates and phosphates;[1a, 21] molecular recognition of bidentate anions by sensors;[1b, 2a, 2c, 2d, 22] Bronsted acid catalysis by guanidiniums;[2b, 7b, 23] intracellular signaling (including via SH3 domains and adaptor proteins);[8a, 10c, 24] and kinase substrate recognition, among many other examples. In all of these cases, α-substitution of the guanidinium to introduce stereochemistry and functional groups could potentially lead to greater affinity and/or specificity in molecular recognition.
Experimental Section
Peptide Synthesis
Peptides were synthesized using Rink amide resin on a PS3 peptide synthesizer (Protein Technologies, Inc.) via standard Fmoc solid phase peptide synthesis using HBTU as a coupling reagent. All peptides contained C-terminal amides. Homoarginine equivalent peptides (hArg(gXaa)) were acetylated on the N-terminus. Arginine equivalent peptides (Arg(gXaa)) were acetylated on the β-amino side chain of the diaminopropionic acid (Dap) residue.
Post-synthetic modification reactions were performed in capped disposable fritted columns (Image Molding, Inc.) with rotation on a Barnstead Thermoline Labquake rotary shaker. Mtt-protected Dap was selectively deprotected using 2% trifluoroacetic acid (TFA)/5% triethylsilane (TES) in CH 2 Cl 2 (3 × 1 min). The resin was washed with CH 2 Cl 2 (3×) and DMF (3×).
In order to synthesize the α-guanidino-acid containing peptides, Fmoc amino acids were coupled to Dap using standard solid phase peptide synthesis with HBTU as a coupling reagent (2 × 1 h). Fmoc removal was effected using 20% piperdine in DMF (3 × 5 min). Free amine groups were guanylated using a solution of 0.2 M N,N’-di-tert-butoxycarbonyl-1 H-benzotriazole-1-carboxamidine (1) in DMF with slow shaking on a rotary shaker for 4-8 h, as judged for completion by cleavage and crude ESI-MS. Notably, this reaction proceeded to high conversions to the Dap-conjugated α-guanidino-acid without any added base. The solution was removed and the resin washed with CH 2 Cl 2 (3×) and MeOH (3×).
Cleavage of the peptide from the resin and deprotection were achieved using a two-step protocol in order to minimize epimerization at the carbon α to the guanidine group. First, the Boc groups on the guanidine were removed using 6 M AcOH in MeCN (2 × 20 min). After removal of this solution, the resin was washed with CH 2 Cl 2 (3×). Peptide cleavage and global deprotection was performed for 3 h using reagent K (84% TFA, 4% each of H 2 O, ethanedithiol, thioanisole, phenol). The solutions were concentrated by evaporation under nitrogen. After removal of most of the TFA by evaporation, the peptide was precipitated with ether and the precipitate dissolved in water.
Peptides were purified by reverse phase HPLC (Vydac semipreparative C18, 10 × 250 mm, 5 μm particle size, 300 Å pore). Peptides were purified to homogeneity using a linear gradient of 0-45% buffer B (20% water, 80% acetonitrile, 0.05% TFA) in buffer A (98% water, 2% acetonitrile, 0.05% TFA) over 60 minutes. Peptide purity was verified by reinjection on analytical HPLC (Varian Microsorb analytical C18, 4.6 × 250 mm, 5 μm, 300 Å pore). Peptides were characterized by ESI-MS (positive ion mode) on an LCQ Advantage (Finnigan) mass spectrometer. Expected and observed peptide masses were [M+H]+. Analytical data for the peptides: hAMI(Arg) (t R = 32.0 min, exp. 1429.7, obs. 1429.7); hAMI(Dap-Ac) (t R = 35.7 min, exp. 1403.7, obs. 1403.7); hAMI(gGly) (t R = 31.9 min, exp. 1461.1, obs. 1458.6); hAMI(gL-Val) (t R = 41.7 min, exp. 1502.8, obs. 1500.7); hAMI(g d-Val) (t R = 40.2, exp. 1502.8, obs. 1500.7); hAMI(g l-Phe) (t R = 45.8 min, exp. 1551.2, obs. 1548.7); hAMI(g d-Phe) (t R = 44.8 min, exp. 1551.2, obs. 1548.7); hAMI(g l-Trp) (t R = 45.7 min, exp. 1589.9, obs. 1587.6); hAMI(g d-Trp) (t R = 47.2 min, exp. 1589.9, obs. 1587.6); AMI(Dap-Ac) (t R = 48.0 min, exp. 1344.6, obs. 1344.5); AMI(gGly) (t R = 50.9 min, exp. 1401.7, obs. 1401.4); AMI(g l-Val) (t R = 50.6 min, exp. 1443.7, obs. 1443.6); AMI(g d-Val) (t R = 52.6 min, exp. 1443.7, obs. 1443.6); AMI(g l-Phe) (t R = 55.2 min, exp. 1491.8, obs. 1491.5); AMI(g d-Phe) (t R = 57.2 min, exp. 1491.8, obs. 1491.4); AMI(g l-Trp) (t R = 54.5 min, exp. 1530.8, obs. 1530.4); AMI(g d-Trp) (t R = 51.6 min, exp. 1530.8, obs. 1530.4); hAMII(Arg) (t R = 34.1 min, exp. 1461.7, obs. 1461.6); hAMII(Dap-Ac) (t R = 43.1 min, exp. 1433.7, obs. 1433.5); hAMII(gGly) (t R = 38.1 min, exp. 1490.7, obs. 1490.7); hAMII(g l-Val) (t R = 41.0 min, exp. 1532.8, obs. 1532.7); hAMII(g d-Val) (t R = 39.9 min, exp. 1532.8, obs. 1532.6); hAMII(g l-Phe) (t R = 45.1 min, exp. 1580.8, obs. 1580.6); hAMII(g d-Phe) (t R = 45.2 min, exp. 1580.8, obs. 1580.5); hAMII(g l-Trp) (t R = 41.7 min, exp. 1619.9, obs. 1619.6); hAMII(g d-Trp) (t R = 40.0 min, exp. 1619.9, obs. 1619.7).
N,N’-di-tert-butoxycarbonyl-1H-benzotriazole-1-carboxamidine (1)
To a solution of benzotriazole (0.70 g, 5.8 mmol) in anhydrous DMF were added N,N’-bis(t-butoxycarbonyl)thiourea[14b] (0.94 g, 3.4 mmol) and diisopropylethylamine (2.68 mL, 16.5 mmol). After 2 minutes, EDCI (0.95 g, 5.0 mmol) was added to the reaction mixture and the reaction was stirred at room temperature for 4 hours at room temperature. The DMF was evaporated under vacuum and the solid dissolved in ethyl acetate (30 mL). The ethyl acetate layer was washed with water, 5% NaHCO 3, water, and brine (6 mL each); dried over sodium sulfate; and concentrated by rotary evaporation. Purification (SiO 2, 80:20 hexanes:EtOAc) yielded a white solid (0.72 g, 56%) whose NMR spectra correlated to literature data.[14a]
Fluorescein Labeling of Peptides
Lyophilized peptides were dissolved in 100 mM NaHCO 3 buffer (pH 10.0, 200 μL). 5-fluorescein isothiocyanate (isomer I) (200 μL of a 10 mg/mL solution in DMF) was added over 2 minutes to the peptide solution with gentle vortexing. The reaction was allowed to proceed for 1-2 hours at room temperature. The solution was filtered and the pH adjusted with 100 mM phosphate buffer pH 7.5. Peptides were purified to homogeneity by reverse phase HPLC (Vydac semipreparative C18, 10 × 250 mm, 5 μm particle size, 300 Å pore) using a linear gradient of 15-65% buffer B in buffer A over 60 minutes. Peptide purity was verified by reinjection on analytical HPLC (Varian Microsorb analytical C18, 4.6 × 250 mm, 5 μm, 300 Å pore). Peptides were characterized by ESI-MS (positive ion mode) on an LCQ Advantage (Finnigan) mass spectrometer. Analytical data for the peptides: hAMI(Arg) (t R = 33.6 min, exp. 1819.1, obs. 1819.7); hAMI(Dap-Ac) (t R = 33.8 min, exp. 1793.1, obs. 1792.6); hAMI(gGly) (t R = 42.7 min, exp. 1850.5, obs. 1848.6); hAMI(g l-Val) (t R = 41.2 min, exp. 1892.2, obs. 1890.7); hAMI(g d-Val) (t R = 40.2 min, exp. 1892.2, obs. 1890.7); hAMI(g l-Phe) (t R = 40.1 min, exp. 1940.6, obs. 1939.6); hAMI(g d-Phe) (t R = 39.7 min, exp. 1940.6, obs. 1939.6); hAMI(g l-Trp) (t R = 43.0 min, exp. 1979.3, obs. 1977.5); hAMI(g d-Trp) (t R = 42.5 min, exp. 1979.3, obs. 1977.5); AMI(Dap-Ac) (t R = 42.2 min, exp. 1734.0, obs. 1734.5); AMI(gGly) (t R = 42.2 min, exp. 1791.0, obs. 1791.4); AMI(g l-Val) (t R = 45.5 min, exp. 1833.1, obs. 1833.4); AMI(g d-Val) (t R = 44.7 min, exp. 1833.1, obs. 1832.3); AMI(g l-Phe) (t R = 49.0 min, exp. 1881.2, obs. 1881.2); AMI(g d-Phe) (t R = 52.7 min, exp. 1881.2, obs. 1880.3); AMI(g l-Trp) (t R = 52.6 min, exp. 1920.3, obs.1920.3); AMI(g d-Trp) (t R = 53.7 min, exp. 1920.2, obs. 1920.3); hAMII(Arg) (t R = 35.5 min, exp. 1849.9, obs. 1850.4); hAMII(Dap-Ac) (t R = 37.2 min, exp. 1823.0, obs. 1823.4); hAMII(gGly) (t R = 36.0 min, exp. 1880.1, obs. 1880.6); hAMII(g l-Val) (t R = 38.3 min, exp. 1922.2, obs. 1921.5); hAMII(g d-Val) (t R = 38.2 min, exp. 1922.2, obs. 1922.5); hAMII(g l-Phe) (t R = 41.7 min, exp. 1970.2, obs. 1969.5); hAMII(g d-Phe) (t R = 40.7 min, exp. 1970.2, obs. 1970.5); hAMII(g l-Trp) (t R = 42.2 min, exp. 2009.2, obs. 2009.4); hAMII(g d-Trp) (t R = 43.1 min, exp. 2009.2, obs. 2009.5).
Protein Expression and Purification
Plasmids containing the coding regions for the Src (residues 87-148), nGrb2 (1-64) and Crk (134-190) SH3 domains were obtained from the lab of Wendell Lim.[10a, 10b] Plasmids encoded both a carboxy-terminal hexahistidine and an amino-terminal glutathione S-transferase. BL21(DE3) pLysS competent cells (EMD Biosciences) were transformed with the plasmids and incubated on ampicillin plates overnight. Single colonies were incubated into overnight cultures (100 mL) of luria broth with shaking at 37 °C. These cultures were seeded into cultures (1 L) of terrific broth (BD Diagnostic Systems). Protein expression was induced with 1 mM IPTG added at an OD 600 = 0.6-0.8, and the cells incubated for 1.5-5 h at 37 °C. Bacteria were collected by centrifugation, the supernatant removed, and the pellets stored at −20 °C until purification. Bacterial pellets were suspended in His-Tag binding buffer (50 mL of 500 mM NaCl, 40 mM Tris-HCl, 10 mM imidazole, pH 7.9) and lysed by sonication, and the pellets cleared by centrifugation. Protein was purified via His-Tag resin (EMD Biosciences) following the manufacturer’s protocol. Protein eluents were dialyzed into phosphate-buffered saline pH 7.4 containing 5 mM EDTA and 0.5 mM dithiothreitol (DTT) (5 × 10 min) using a Spectra/Por 6 membrane, MWCO 10,000 (Spectrum Laboratories). Protein size and purity were verified via SDS-PAGE. Bradford assays (BioRad) were used to determine protein concentrations.
SH3 domain-peptide fluorescence polarization binding assays
For fluorescence polarization assays, proteins and fluorescein-labeled ligands were diluted in 1× PBS (140 mM NaCl, 10 mM Na 2 HPO 4, 2 mM NaH 2 PO 4, pH 7.4) containing 0.1 mM DTT and 0.04 mg/mL BSA (BioRad). Concentrations of peptides were determined based on fluorescein absorbance (ε = 77,000 M−1 cm−1 at 495 nm). Two-fold serial dilutions of Src, nGrb2, and Crk proteins, ranging from 360 μM to 2.7 nM, were mixed with peptides (100 nM), BSA (0.04 mg/mL), and DTT (100 μM) (final concentrations) in 96-well flat bottom black opaque plates (Costar) at 200 μL per well. Individual trials on a protein with l- and d- guanidino acid-containing peptides of a given side-chain were conducted with a single serial dilution to ensure that any differences observed could be ascribed to stereochemistry. After 15 minutes equilibration at room temperature, plates were read on a Perkin Elmer Fusion plate reader in fluorescence polarization mode with a 485 nm fluorescein excitation filter and a 535 nm emission filter with polarizer. Data points were the average of at least three independent trials. Polarization data are in millipolarization units. Error bars indicate standard error.
The data were fit to equation 1 using a non-linear least squares fitting algorithm (Kaleidagraph version 4.0, Synergy Software), where Pol = polarization, Pol min = polarization in the absence of protein, Pol max = polarization at saturation, P t = total protein concentration, K d = dissociation constant of the peptide-protein complex, and [lig]t = total peptide concentration (0.1 μM). The data were fit to calculate Pol max and K d.
(1)
NMR spectroscopy
NMR spectroscopy was conducted to verify that no epimerization occurred during peptide synthesis, cleavage and global deprotection, and purification. Solutions contained 5 mM deuteratured sodium acetate (pH 4.0), 10 mM (hAMI series) or 100 mM (AMI and hAMII series) NaCl, 150 μM TSP[D 4] and 90% H 2 O/10% D 2 O, with final peptide concentrations of 70 μM – 1.5 mM. NMR spectra were recorded at 296 K on a Brüker AVC 600 MHz NMR spectrometer with a cryoprobe. 1-D and TOCSY spectra were collected with water suppression using a Watergate pulse sequence. 1-D spectra were collected with a sweep width of 6009 Hz or 7183 Hz; 8192 or 16384 data points; and a relaxation delay of 2-3 s. TOCSY spectra were collected with sweep widths of 7183 Hz in t 1 and t 2, 512-600 × 4096 complex data points, 2-4 scans per increment, and a relaxation delay of 2 s.
Circular Dichroism
Solutions contained 5 mM phosphate buffer (pH 7.0), 25 mM KF, and 100 μM peptide. CD spectra were collected at 25 °C on a Jasco J-810 Spectropolarimeter in a 2 mm quartz cell (Starna). Data were collected every 2 nm with an averaging time of 4 s, 2 accumulations, and a 2 nm bandwidth. Data were baseline corrected but were not smoothed. Data points were the average of at least three independent trials. Error bars indicate standard error.
Figure 3.
Open in a new tab
Sequences of type I and type II SH3 domain ligand peptides. The proline-rich (PXXP-containing) core sequence is indicated in blue. The core sequence and Arg residues of the type I ligand are from the VSL12 protein (residues 76-82) observed bound to the Src SH3 domain (PDB 1qwf).[11a, 11d, 12] The type II ligand peptide is derived from the mSOS protein observed in the SEM-5 (Grb2 homologue) SH3 domain complex (PDB 1sem).[11c] The specificity- and polarity-determining arginine residue is indicated in red. The indicated Arg was replaced with diaminopropionic acid (Dap) for arginine mimetic-containing peptides. All peptides were acetylated on either the N-terminus (homoarginine equivalents, hArg(gXaa)) or on the Dap side chain (arginine equivalents, Arg(gXaa)). Peptides containing type I arginine equivalents (i.e. Arg(gXaa)) lacked the N-terminal Gly. The guanidino acid was conjugated via the Dap side chain (homoarginine equivalents, hAMI and hAMII) or the N-terminus (arginine equivalents, AMI). For fluorescence polarization experiments, peptides were labeled with fluorescein isothiocyanate (FITC) on the side chain of the C-terminal lysine. All peptides contained C-terminal amides.
Acknowledgements
We thank Susan Carr Zondlo for the development of protein expression and purification protocols and for the development of direct binding assays of fluorescein-labeled peptides for SH3 domain-containing proteins. We thank Wendell Lim for providing us with plasmids encoding the Src, Grb, and Crk SH3 domains. We thank Richard Borch for helpful discussion. We thank NIH (1P20 RR17716) (COBRE), NSF (CAREER, CHE-0547973), and the University of Delaware for partial support of this work.
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656 | http://www.philosophypages.com/lg/e08a.htm | | Philosophy Pages | | Dictionary | Study Guide | Logic | | F A Q s |
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Categorical Syllogisms
The Structure of Syllogism
Now, on to the next level, at which we combine more than one categorical proposition to fashion logical arguments. A categorical syllogism is an argument consisting of exactly three categorical propositions (two premises and a conclusion) in which there appear a total of exactly three categorical terms, each of which is used exactly twice.
One of those terms must be used as the subject term of the conclusion of the syllogism, and we call it the minor term of the syllogism as a whole. The major term of the syllogism is whatever is employed as the predicate term of its conclusion. The third term in the syllogism doesn't occur in the conclusion at all, but must be employed in somewhere in each of its premises; hence, we call it the middle term.
Since one of the premises of the syllogism must be a categorical proposition that affirms some relation between its middle and major terms, we call that the major premise of the syllogism. The other premise, which links the middle and minor terms, we call the minor premise.
Consider, for example, the categorical syllogism:
No geese are felines. Some birds are geese. Therefore, Some birds are not felines.
Clearly, "Some birds are not felines" is the conclusion of this syllogism. The major term of the syllogism is "felines" (the predicate term of its conclusion), so "No geese are felines" (the premise in which "felines" appears) is its major premise. Simlarly, the minor term of the syllogism is "birds," and "Some birds are geese" is its minor premise. "geese" is the middle term of the syllogism.
Standard Form
In order to make obvious the similarities of structure shared by different syllogisms, we will always present each of them in the same fashion. A categorical syllogism in standard form always begins with the premises, major first and then minor, and then finishes with the conclusion. Thus, the example above is already in standard form. Although arguments in ordinary language may be offered in a different arrangement, it is never difficult to restate them in standard form. Once we've identified the conclusion which is to be placed in the final position, whichever premise contains its predicate term must be the major premise that should be stated first.
Medieval logicians devised a simple way of labelling the various forms in which a categorical syllogism may occur by stating its mood and figure. The mood of a syllogism is simply a statement of which categorical propositions (A, E, I, or O) it comprises, listed in the order in which they appear in standard form. Thus, a syllogism with a mood of OAO has an O proposition as its major premise, an A proposition as its minor premise, and another O proposition as its conclusion; and EIO syllogism has an E major premise, and I minor premise, and an O conclusion; etc.
Since there are four distinct versions of each syllogistic mood, however, we need to supplement this labelling system with a statement of the figure of each, which is solely determined by the position in which its middle term appears in the two premises: in a first-figure syllogism, the middle term is the subject term of the major premise and the predicate term of the minor premise; in second figure, the middle term is the predicate term of both premises; in third, the subject term of both premises; and in fourth figure, the middle term appears as the predicate term of the major premise and the subject term of the minor premise. (The four figures may be easier to remember as a simple chart showing the position of the terms in each of the premises:
M P P M M P P M 1 \ 2 | 3 | 4 / S M S M M S M S
All told, there are exactly 256 distinct forms of categorical syllogism: four kinds of major premise multiplied by four kinds of minor premise multiplied by four kinds of conclusion multiplied by four relative positions of the middle term. Used together, mood and figure provide a unique way of describing the logical structure of each of them. Thus, for example, the argument "Some merchants are pirates, and All merchants are swimmers, so Some swimmers are pirates" is an IAI-3 syllogism, and any AEE-4 syllogism must exhibit the form "All P are M, and No M are S, so No S are P."
Form and Validity
This method of differentiating syllogisms is significant because the validity of a categorical syllogism depends solely upon its logical form. Remember our earlier definition: an argument is valid when, if its premises were true, then its conclusion would also have to be true. The application of this definition in no way depends upon the content of a specific categorical syllogism; it makes no difference whether the categorical terms it employs are "mammals," "terriers," and "dogs" or "sheep," "commuters," and "sandwiches." If a syllogism is valid, it is impossible for its premises to be true while its conclusion is false, and that can be the case only if there is something faulty in its general form.
Thus, the specific syllogisms that share any one of the 256 distinct syllogistic forms must either all be valid or all be invalid, no matter what their content happens to be. Every syllogism of the form AAA-1 is valid, for example, while all syllogisms of the form OEE-3 are invalid.
This suggests a fairly straightforward method of demonstrating the invalidity of any syllogism by "logical analogy." If we can think of another syllogism which has the same mood and figure but whose terms obviously make both premises true and the conclusion false, then it is evident that all syllogisms of this form, including the one with which we began, must be invalid.
Thus, for example, it may be difficult at first glance to assess the validity of the argument:
All philosophers are professors. All philosophers are logicians. Therefore, All logicians are professors.
But since this is a categorical syllogism whose mood and figure are AAA-3, and since all syllogisms of the same form are equally valid or invalid, its reliability must be the same as that of the AAA-3 syllogism:
All terriers are dogs. All terriers are mammals. Therefore, All mammals are dogs.
Both premises of this syllogism are true, while its conclusion is false, so it is clearly invalid. But then all syllogisms of the AAA-3 form, including the one about logicians and professors, must also be invalid.
This method of demonstrating the invalidity of categorical syllogisms is useful in many contexts; even those who have not had the benefit of specialized training in formal logic will often acknowledge the force of a logical analogy. The only problem is that the success of the method depends upon our ability to invent appropriate cases, syllogisms of the same form that obviously have true premises and a false conclusion. If I have tried for an hour to discover such a case, then either there can be no such case because the syllogism is valid or I simply haven't looked hard enough yet.
Diagramming Syllogisms
The modern interpretation offers a more efficient method of evaluating the validity of categorical syllogisms. By combining the drawings of individual propositions, we can use Venn diagrams to assess the validity of categorical syllogisms by following a simple three-step procedure:
First draw three overlapping circles and label them to represent the major, minor, and middle terms of the syllogism.
Next, on this framework, draw the diagrams of both of the syllogism's premises.
Always begin with a universal proposition, no matter whether it is the major or the minor premise.
Remember that in each case you will be using only two of the circles in each case; ignore the third circle by making sure that your drawing (shading or × ) straddles it.
Finally, without drawing anything else, look for the drawing of the conclusion. If the syllogism is valid, then that drawing will already be done.
Since it perfectly models the relationships between classes that are at work in categorical logic, this procedure always provides a demonstration of the validity or invalidity of any categorical syllogism.
Consider, for example, how it could be applied, step by step, to an evaluation of a syllogism of the EIO-3 mood and figure,
No M are P. Some M are S. Therefore, Some S are not P.
First, we draw and label the three overlapping circles needed to represent all three terms included in the categorical syllogism:
Second, we diagram each of the premises:
Since the major premise is a universal proposition, we may begin with it. The diagram for "No M are P" must shade in the entire area in which the M and P circles overlap. (Notice that we ignore the S circle by shading on both sides of it.)
Now we add the minor premise to our drawing. The diagram for "Some M are S" puts an × inside the area where the M and S circles overlap. But part of that area (the portion also inside the P circle) has already been shaded, so our × must be placed in the remaining portion.
Third, we stop drawing and merely look at our result. Ignoring the M circle entirely, we need only ask whether the drawing of the conclusion "Some S are not P" has already been drawn.
Remember, that drawing would be like the one at left, in which there is an × in the area inside the S circle but outside the P circle. Does that already appear in the diagram on the right above? Yes, if the premises have been drawn, then the conclusion is already drawn.
But this models a significant logical feature of the syllogism itself: if its premises are true, then its conclusion must also be true. Any categorical syllogism of this form is valid.
Here are the diagrams of several other syllogistic forms. In each case, both of the premises have already been drawn in the appropriate way, so if the drawing of the conclusion is already drawn, the syllogism must be valid, and if it is not, the syllogism must be invalid.
AAA-1 (valid)
All M are P. All S are M. Therefore, All S are P.
AAA-3 (invalid)
All M are P. All M are S. Therefore, All S are P.
OAO-3 (valid)
Some M are not P. All M are S. Therefore, Some S are not P.
EOO-2 (invalid)
No P are M. Some S are not M. Therefore, Some S are not P.
IOO-1 (invalid)
Some M are P. Some S are not M. Therefore, Some S are not P.
Practice your skills in using Venn Diagrams to test the validity of Categorical Syllogisms by using Ron Blatt's excellent Syllogism Evaluator.
The Philosophy Pages by Garth Kemerling are licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.Permissions beyond the scope of this license may be available at ©1997, 2011 Garth Kemerling.Last modified 12 November 2011. Questions, comments, and suggestions may be sent to:the Contact Page. |
657 | https://math.montana.edu/grad_students/writing-projects/2017/Obiri17.pdf | Finite Population Correction Methods Moses Obiri May 5, 2017 Contents 1 Introduction 1 2 Normal-based Confidence Interval 2 3 Bootstrap Confidence Interval 3 4 Finite Population Bootstrap Sampling 5 4.1 Modified Resample Size. . . . . . . . . . . . . . . . . . . . . . . .
5 4.2 Population and Superpopulation Bootstrap . . . . . . . . . . . .
6 5 Simulation Study 8 6 Results and Discussion 9 6.1 Results: Coverage Rate and Mean Width . . . . . . . . . . . . .
9 6.2 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 7 Improvement and Future Work 15 8 References 15 1 Introduction One application of inferential statistics is to use a simple random sample of size n from a population to learn about the characteristics of such population. A very common inference is that of the population mean. The sample mean is typically used as a point estimate for the population mean. The uncertainty in using a point estimate is addressed by means of confidence intervals. Con-fidence intervals provide us with a range of values for the unknown population mean along with the precision of the method. For the parametric approach, the central limit theorem allows us to construct t-based confidence intervals for large sample sizes or symmetric populations. Over the years, nonparametric computer-intensive bootstrap methods have become popular for constructing confidence intervals as well.
1 The central limit, the standard error of the sample mean and traditional bootstrap methods are based on the principle that samples are selected with replacement or that sampling is done without replacement from an infinite pop-ulation.
In most research surveys, however, sampling is done from a finite population of size N. When we sample from an infinite population or sample with replacement; then selecting one unit does not affect the probability of se-lecting the same or another unit. Hence, precision depends only on the sample size. With sampling from a finite population without replacement, what we see affects the probability of selection of units that we have not seen and hence, there is a finite population correction(1-n/N) arising from the lack of indepen-dence that must be accounted for in the calculation of the standard error of the sample mean.
T-based confidence intervals and ordinary bootstrap methods do not ac-count for this lack of independence and as such these confidence intervals are approximately p (1 −n N ) less precise with the approximation varying based on population characteristics such as skewness. In this study, I present two differ-ent nonparametric bootstrap methods for constructing confidence intervals that account for the finite population factor resulting from the lack of independence in sampling from a finite population. A simulation-based approach is used to compare the coverage rates, precision, and confidence interval widths among these two methods, the t-based approach and the ordinary bootstrap method.
2 Normal-based Confidence Interval For a population with unknown mean µ and known standard deviation σ, a 100(1 −α)% z-based confidence interval for the population mean is: ¯ x ± zα/2 σ √n where ¯ x is the sample mean,zα/2 is the upper critical value from the standard normal distribution and σ √n is the standard deviation of the sample mean.
The population standard deviation σ, however, is usually unknown. If it was known, then the population mean would likely be known also. The popu-lation standard deviation σ is therefore estimated and replaced by the sample standard deviation, s in the z-based confidence interval. The substitution, how-ever, changes the coverage probability 1 −α. In 1908, Gosset from Guinness Breweries discovered the t-distribution that allow us to maintain the desired coverage level by replacing standard normal distribution critical value by the larger t-distribution critical value. The resulting t-based confidence interval for the population mean is ¯ x ± tn−1,α/2 s √n where t(n −1, α/2) is the critical value determined from a tn−1 distribution (t distribution with n-1 degrees of freedom) and s √n is the estimated standard deviation of the sample mean called the standard error.
2 The confidence level is exactly 100(1−α)% when the population distribution is normal and is approximately 100(1 −α)% for other distributions when the sample sizes are large by the application of the central limit theorem. With the advancement of computing powers, however, nonparametric bootstrap methods have become common in constructing confidence intervals for any sample size and for many distributions.
3 Bootstrap Confidence Interval Bradley Efron (1979) is known to have written the first paper on the theory of bootstrapping. With subsequent papers (Efron and Tibshirani (1991), Efron and Tibshirani (1993)) and with the advancement in computing capabilities, the method gained recognition and widespread use. There are two main approaches for confidence interval estimation: parametric and the nonparametric. The for-mer assumes that the data come from a known population distribution but with unknown parameters. Often in research survey however, the population distri-bution is unknown and hence, the nonparametric bootstrap helps approximate the sampling distribution of an estimator that is a function of a random sample taken from this unknown population distribution, provide standard errors for estimates and confidence intervals for the unknown parameters. The nonpara-metric bootstrap assumes that the population size is very large relative to the sample, which is nearly equivalent to sampling with replacement. Thus, the sample units are assumed to be independent of one another and the standard error of the sample mean, our estimator in this case, depends only on the sample size and not the fraction of the population that is sampled.
3 The steps in nonparametric bootstrapping are illustrated above. The dashed lines represent theoretical unobserved results.
1. A random sample of size n, y1, ., yn is taken from a population, that is assumed to be very large or infinite.
2. A bootstrap population is created.
This is typically the set of sample values y1., yn.
The population is thought of been infinite consisting of infinite number of y1 values, infinite number of y2 values .infinite number of yn values, with each occurring 1 n of the time. For example, if our sample is (3, 20, 8, 12, 16) then we are assuming that 20% of the population values are 3s, 20% are 20s, 20% are 8s, 20% are 12s, and 20% are 16s.
3. We then resample n units with replacement from the bootstrap popula-tion a total of B times. Efron and Tibshirani (1991) claimed that B=200 resamples is usually adequate to find the variance of an estimator. Prac-tically, however, it is now computationally inexpensive to take thousands of bootstraps resamples.
4. The bootstrap sample mean is calculated from each of the B resamples, and the B bootstrap sample means are used to calculate a bootstrap stan-dard error and generate an empirical bootstrap distribution for the sample mean.
4 An appropriate confidence intervals can then be found from the percentiles of the bootstrap distribution. For example, for a 95% confidence interval, we find the two values (L, U) that bound the middle 95% of the distribution. That is L and U are the 2.5th and 97.5th percentiles of the distribution. This method however, does not take into account the sampling fraction when we sample from a finite population and as such, these confidence intervals are too wide. I look at two bootstrap procedures described by A. C. Davison and D. V. Hinkley (”Bootstrap methods and their application”, 1997) that account for the correlation among the sampled values.
4 Finite Population Bootstrap Sampling 4.1 Modified Resample Size.
The estimated variances of the sample mean under sampling of n units with replacement and without replacement from a finite population of size N are: ( s2 n With Replacement (1 −n n)( s2 n ) Replacement ) where s2 is the sample variance.
As illustrated earlier, the ordinary bootstrap involves resampling with replace-ment and does not consider the effect of the sampling fraction n N leading to larger bootstrap standard errors for the sample mean and wider confidence in-tervals.
With the ordinary bootstrap with-replacement resamples, Davison and Hink-ley (1997) noted that the estimated variance of the sample mean when taking resamples of size n′ is var(¯ y) = (n −1)s2 n′n To account for the sampling fraction, the goal is to then approximately match this variance (1 −n n)( s2 n ) . It is straightforward to verify the two variances are approximately equal when n′ = n −1 1 −n N . Hence taking bootstrap resamples with replacement of size n′ approximately captures the effect of the dependency among sample values.
5 The steps for this procedure are the same as the ordinary bootstrap method described earlier except the resample size is adjusted to be n′ instead of the original sample size.
4.2 Population and Superpopulation Bootstrap Our random sample is our best guess of what the population looks like except the size is smaller than the population size. The idea behind this method is to form a ”superpopulation” the same size as the population size from our sample and mimic how the original data were sampled from the population.
6 The steps for this procedure are illustrated above.
1. A random sample of size n is taken from a population of size N.
2. A ”superpopulation” of size N is formed by making N n copies of the sample.
If N n is not an integer then we write N = kn + l, l < n, make k copies and add them to l units taken without replacement from the original sample to form a ”superpopulation” of size N.
3. The next step is to take B bootstrap without replacement resamples of size n from the ”superpopulation”.
4. The bootstrap sample mean is calculated from each of the B resamples, and the B bootstrap sample means are used to generate an empirical boot-strap distribution.The percentile method can then be used to genearate approximate confidence intervals.
Davison and Hinkley(1997) wrote that the variance of the sample mean under this method is N(n −1) (N −1)n ∗(1 −n N )s2 The first factor N(n−1) (N−1)n is usually close to 1, thereby approximately accounting for the finite population fraction.
7 5 Simulation Study Using a simulation study, we compared the coverage probability and precision among the four described methods on four different populations; one symmetric populations of size 400, and three skewed populations of size 1600, 400 and 100.
The coverage probabilities and average confidence interval widths for ap-proximate 99%, 95% and 90% confidence intervals were computed using the four methods for three different sampling fractions corresponding to sample size equal to 5%, 10% and 15% of the population size N. For a given confidence level, sample size and method, 10,000 simple random samples were taken and 5,000 bootstrap resamples were generated from each sample for the three bootstrap methods. The coverage rate and the average width were then computed from the 10,000 confidence intervals.
8 6 Results and Discussion 6.1 Results: Coverage Rate and Mean Width The coverage rate was computed as the percentage of the 10,000 confidence intervals that contained the population mean. For each interval, the difference between the upper bound and lower bound or interval width was recorded. The mean width was calculated as the average of the 10,000 widths.
Table 1 shows the coverage rates along with associated 95% confidence in-tervals for the symmetric population of size 400.The green highlight indicates that the coverage rate is within 1% of the nominal confidence level. The cov-erage rates by the t-based approach approximately attained the nominal 99% confidence level for all three sample sizes. Those of the ordinary bootstrap and Modified sample size methods were also close the nominal 99% confidence level for all sample sizes. The coverage rates by the population bootstrap approach approximately attained the 99% level of confidence when the sampling frac-tions were 10% and 15% of the population size. For 95% level of confidence, the coverage rate was higher than the nominal level at the 15% sampling frac-9 tion but did approximately attain the desired level of confidence at 10% and 5% sampling fractions. The Modified sample size approach did approximately match the 95% and 90% levels of confidence at 10% and 15% sampling frac-tions whereas the Population bootstrap method did approximately attain these nominal confidence levels when the sampling fraction was 15%. The coverage rates by the ordinary bootstrap method were close to the 95% and 90% levels of confidence at 10% sampling fraction.
For this population, only the t-based method and the ordinary bootstrap ap-proach approximately attained the nominal confidence levels when the sampling fraction was 15% of the population size.
10 From Table 3, only coverage rates by the t-based method were within 1% of the 99% and 90% levels of confidence at 5% and 15% sampling fractions respectively.
11 The coverage rates by the four methods approximately attained the 99% nominal level of confidence for all sample sizes. The t-based and the ordinary bootstrap methods were close to the three nominal levels of confidence for all three sample sizes. The coverage rates by the two finite population correction methods were however closed to the 95% and 90% levels of confidence at a sampling fraction of 15% of the population size.
Across the four populations, the t-based method has the highest coverage for a given sample size and confidence level followed by the ordinary bootstrap, with the population bootstrap method has the lowest coverage rate among the four methods. The pattern among the four methods does not seem to depend on confidence level as the order generally stayed the same across the three confidence levels.
Predictably, for a given confidence level and method, the coverage rate increased as the sampling fraction increases.
12 The order is reversed when it comes to the mean widths for the four methods.
The population bootstrap method has the smallest mean width for a given sample size and confidence level followed closely by the modified resample size 13 method. For any method, however, the mean width decreases as the sampling fraction was increased.
6.2 Discussion In constructing confidence intervals, the goal of every researcher is to first and foremost match the desired level of confidence at least approximately.
The results showed that even though the two finite population correction methods generate narrower and more precise confidence intervals than the t-based and ordinary bootstrap approaches, in most instances it was observed that such confidence intervals might be too narrow to achieve the desired coverage level.
On the other hand, it was also observed that the t-based and the ordinary bootstrap methods may generate confidence intervals that might be too wide to attain the desired level of confidence.
The methods especially the finite population correction methods seem to work better in some populations than others. Generally, the coverage proba-bilities from the symmetric population were higher were higher than the three skewed populations. The fourth population of size 1600 had higher coverage probabilities than the other two skewed populations. This is indication that population characteristics like skewness may play role in how well these meth-ods work and when they should be applied. Most of the coverage rates by the four methods did not achieved the desired levels of confidence in population 3. Therefore, for some skewed population distributions, both the traditional methods and the finite population correction method may not achieve the de-sire levels of confidence. In such situations, inference about other measures of center like the population median might be explored.
Predictably, it was revealed that sampling fraction plays a role in the utility of these methods. The coverage rates seem to increase as the sampling fraction was increased across the four populations whereas the mean widths decreased as the sampling fraction was increased. As sampling fraction increases, we get more information about the population that resulted in having a better coverage rates and more precise confidence intervals.
Looking at the results, it did not appear the choice of confidence level had any effect on coverage probability and precision of a particular method as there was no detectable pattern across the three different levels of confidence used in the study. Conducting a formal test to look these relationships might be more appropriate in future studies.
14 Another look at Table 1 indicates both the four methods had comparable coverage rates for sampling fractions of 10% and 15% except in some cases the t-based and the ordinary bootstrap methods generated confidence inter-vals that were too wide to achieve the desired levels of confidence. It suggests that if an underlying population distribution is assumed to be symmetric, then employing the two described population correction methods may yield more pre-cise approximate confidence intervals than the t-based and ordinary bootstrap methods when the sampling fraction is at least 10% of the population size.
For some skewed populations, similar to the population 4, the two described population correction methods may yield more precise approximate confidence intervals than the t-based and ordinary bootstrap methods when the sampling fraction is at least 15% of the population. For a smaller sampling fraction of 5% or less, the finite population methods may generate too narrow confidence intervals achieve the approximate confidence levels. In such situations, the t-based and the ordinary bootstrap methods might be more appropriate.
7 Improvement and Future Work Future work could look at conducting formal tests such as regression analysis to investigate the effect of the four methods and levels of confidence on coverage rate and average width. Studies focusing on more populaions could help eluci-date the effect of population on these methods and these studies could also look at the threshold sampling fractions to achieve 100% coverage rate.
8 References 1. Efron, B. and Tibshirani, R. (1986) Bootstrap methods for standard er-rors, confidence intervals, and other measures of statistical accuracy. Sta-tistical Science 1(1): 54 - 77.
2. DiCiccio, T. and Tibshirani, R. (1987) Bootstrap confidence intervals and bootstrap approximations. Journal of the American Statistical Associa-tion 82 (397): 163 - 170 .
15 3. Efron, B. (1987) Better bootstrap confidence intervals.
Journal of the American Statistical Association 82 (397): 171 - 200.
4. Efron, B. (1988) Bootstrap confidence intervals: Good or Bad? Psycho-logical Bulletin 104(2): 293 - 296.
5. Hinkley, D. V. (1988) Bootstrap methods. J. R. Statist. Soc. B 50(3): 321 - 337 6. Davison, A.C. and Hinkley, D. V. (1997) Bootstrap methods and their application. Cambridge University Press. pp. 92 - 100 16 |
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Back To Resolución de ecuaciones que presentan propiedades inversas de adición y divisiónBack
3.2
Solve One-Step Equations Using Inverse Properties of Subtraction and Division
Written by:Jen Kershaw, M.ed | Brenda Meery
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Jessica and Casey worked at a bakery during school vacation. One day Casey was asked to divide up many pounds of flour in order to prepare for their day of baking. She divided the amount she was given by three. Then she added four more pounds to one of these portions. Jessica was given the largest portion. If Jessica received 8 pounds of flour, how many pounds of flour did Casey begin with?
In this concept, you will solve equations involving the inverse properties of addition and division.
Inverse Properties of Adding and Dividing
To solve a two-step equation, you will need to use more than one inverse operation. When you perform inverse operations to find the value of a variable, you work to get the variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations.
Let’s look at an example.
Solve for @$\begin{align}c\end{align}@$: @$\begin{align}5 + \frac{c}{4}=15\end{align}@$
First, use the inverse operations to get the variable, @$\begin{align}\frac{c}{4}\end{align}@$ , by itself on the left side. Since 5 is being added to the variable, the inverse operation is subtraction.
@$$\begin{align}\begin{array}{rcl}
5 + \frac{c}{4} & = & 15\ \
5-5 + \frac{c}{4} & = & 15-5\ \
\frac{c}{4} & = & 10
\end{array}\end{align}@$$
Next, use inverse operations to isolate @$\begin{align}c\end{align}@$. Since @$\begin{align}c\end{align}@$ is divided by 4, the inverse operation would be multiplication. The number 4 is @$\begin{align}\frac{4}{1}\end{align}@$ which is the multiplicative inverse or the reciprocal of @$\begin{align}\frac{1}{4}\end{align}@$. So the reciprocal of @$\begin{align}\frac{1}{4}\end{align}@$ is @$\begin{align}\frac{4}{1}\end{align}@$. When a number if multiplied by its reciprocal, the product is 1.
@$$\begin{align}\begin{array}{rcl}
\frac{4}{1} \times \frac{1}{4}c & = & \frac{10}{1} \times \frac{4}{1}\ \
c & = & 40
\end{array}\end{align}@$$
The answer is @$\begin{align}c=40\end{align}@$.
Examples
Example 1
Earlier, you were given a problem about Casey and her baking problem.
Casey divided the pounds of flour by three, but you don’t know how many pounds she started with, so this is the variable.
First, let “@$\begin{align}x\end{align}@$” be the number of pounds she started with such that the variable is @$\begin{align}\frac{x}{3}\end{align}@$.
You know that Casey added four pounds to one of the portions, so @$\begin{align}\frac{x}{3}+4\end{align}@$.
You also know that Jessica ended up with 8 pounds, so @$\begin{align}\frac{x}{3}+4=8\end{align}@$.
Next, you can begin solving the equation. Start by subtracting four from both sides of the equation.
@$$\begin{align}\begin{array}{rcl}
\frac{x}{3} + 4-4 & = & 8-4\ \
\frac{x}{3} & = & 4
\end{array}\end{align}@$$
Then, use the inverse of division, multiplication, and multiply three times four.
@$$\begin{align}\begin{array}{rcl}
\frac{3}{1} \times \frac{x}{3} & = & \frac{4}{1} \times \frac{3}{1}\ \
x & = & 12
\end{array}\end{align}@$$
The answer is 12.
Casey started with 12 pounds of flour.
Example 2
@$\begin{align}\frac{y}{19} + 6 =10\end{align}@$
First, use the inverse operations to get the variable, @$\begin{align}\frac{y}{19}\end{align}@$, by itself on the left side.
@$$\begin{align}\begin{array}{rcl}
\frac{y}{19} + 6 & = & 10\ \
\frac{y}{19} + 6 - 6 & = & 10-6\ \
\frac{y}{19} & = & 4
\end{array}\end{align}@$$
Next, use inverse operations to isolate @$\begin{align}y\end{align}@$.
@$$\begin{align}\begin{array}{rcl}
\frac{19}{1} \times \frac{1}{19} y & = & \frac{4}{1} \times \frac{19}{1}\ \
y & = & 76
\end{array}\end{align}@$$
The answer is @$\begin{align}y=76\end{align}@$.
Example 3
@$\begin{align}\frac{y}{5} + 6 =10\end{align}@$
First, use the inverse operations to get the variable, @$\begin{align}\frac{y}{5}\end{align}@$, by itself on the left side.
@$$\begin{align}\begin{array}{rcl}
\frac{y}{5} + 6 & = & 10\ \
\frac{y}{5} + 6-6 & = & 10-6\ \
\frac{y}{5} & = & 4
\end{array}\end{align}@$$
Next, use inverse operations to isolate “@$\begin{align}y\end{align}@$”.
@$$\begin{align}\begin{array}{rcl}
\frac{5}{1} \times \frac{1}{5}y& = & \frac{4}{1} \times \frac{5}{1}\ \
y & = & 20
\end{array}\end{align}@$$
The answer is @$\begin{align}y=20\end{align}@$.
Example 4
@$\begin{align}\frac{a}{9}+12=28\end{align}@$
First, use the inverse operations to get the variable, @$\begin{align}\frac{a}{9}\end{align}@$, by itself on the left side.
@$$\begin{align}\begin{array}{rcl}
\frac{a}{9} + 12 & = & 28\ \
\frac{a}{9} + 12-12 & = & 28-12\ \
\frac{a}{9} & = & 16
\end{array}\end{align}@$$
Next, use inverse operations to isolate “@$\begin{align}a\end{align}@$”.
@$$\begin{align}\begin{array}{rcl}
\frac{9}{1} \times \frac{1}{9}a& = & \frac{16}{1} \times \frac{9}{1}\ \
a & = & 144
\end{array}\end{align}@$$
The answer is @$\begin{align}a=144\end{align}@$.
Example 5
@$\begin{align}\frac{x}{11} + 12 = 18\end{align}@$
First, use the inverse operations to get the variable, @$\begin{align}\frac{x}{11}\end{align}@$, by itself on the left side.
@$$\begin{align}\begin{array}{rcl}
\frac{x}{11} + 12 & = & 18\ \
\frac{x}{11} + 12-12 & = & 18-12\ \
\frac{x}{11} & = & 6
\end{array}\end{align}@$$
Next, use inverse operations to isolate “@$\begin{align}x\end{align}@$”.
@$$\begin{align}\begin{array}{rcl}
\frac{11}{1} \times \frac{1}{11}x& = & \frac{6}{1} \times \frac{11}{1}\ \
x & = & 66
\end{array}\end{align}@$$
The answer is @$\begin{align}x=66\end{align}@$.
Review
Solve the following two-step equations that have addition and division in them.
@$\begin{align}\frac{x}{3} + 4 = 8\end{align}@$
@$\begin{align}\frac{x}{5} + 8 = 10\end{align}@$
@$\begin{align}\frac{a}{6} + 7 = 13\end{align}@$
@$\begin{align}\frac{a}{9} + 4 = 30\end{align}@$
@$\begin{align}\frac{b}{8} + 6 = 15\end{align}@$
@$\begin{align}\frac{c}{12} + 9 = 18\end{align}@$
@$\begin{align}\frac{x}{7} + 7 = 21\end{align}@$
@$\begin{align}\frac{x}{11} + 5 = 12\end{align}@$
@$\begin{align}\frac{x}{12} + 9 = 16\end{align}@$
@$\begin{align}\frac{a}{14} + 6 = 8\end{align}@$
@$\begin{align}\frac{x}{22} + 9 = 12\end{align}@$
@$\begin{align}\frac{y}{2} + 14 = 18\end{align}@$
@$\begin{align}\frac{x}{7} + 24 = 38\end{align}@$
@$\begin{align}\frac{x}{8} + 15 = 30\end{align}@$
@$\begin{align}\frac{x}{9} + 11 = 28\end{align}@$
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.
Resources
| Image | Reference | Attributions |
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| | | Credit: Joanna Bourne;CK-12 Foundation Source: |
| | | Credit: CK-12 Foundation |
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659 | https://textranch.com/c/amazed-at-or-amazed-by/ | amazed at or amazed by?
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amazed at vs amazed by
Both 'amazed at' and 'amazed by' are correct, commonly used phrases in English. The choice between them depends on the context and the writer's personal preference.
Explained by Jennifer
Editor at TextRanch
Last updated: February 23, 2025 • 6518 views
This phrase is correct and commonly used in English.
"amazed at"
The phrase 'amazed at' is used to express surprise or wonder about something. It is followed by the cause of amazement.
Examples:
I was amazed at the beautiful sunset.
She was amazed at how quickly he finished the project.
Show examples from the web [+]
I was amazed at the confidence with which he conducted my compositions.
I'm still amazed at how incredibly cheap this house.
I'm kind of amazed at myself.
I was amazed at how Jaime had captured the essence of Ray Charles.
I'm just amazed at how quickly you got this place together.
I think you will be amazed at what happened.
Ultimately, Westerners are amazed at Russian speed and efficiency.
They're amazed at my recovery.
You'd be amazed at what I can do.
I'm amazed at the patience of your ex-husband.
And then I was amazed at what role technology played in your recovery.
I'm amazed at his suffering, his pain.
He was amazed at my Latin.
I'm just amazed at the body's will to live.
You'd be amazed at the little forgotten treasures left behind by previous occupants.
You'll be amazed at how it will impact your life.
I'm always amazed at how convenient this new information is.
Joseph is there, surely amazed at what had just happened.
You will be amazed at what you see.
You'd be amazed at how attractive
Alternatives:
surprised by
astonished by
impressed by
shocked by
awestruck by
This phrase is correct and commonly used in English.
"amazed by"
The phrase 'amazed by' is used to indicate the source of the amazement. It is followed by the thing or person causing the amazement.
Examples:
He was amazed by her talent.
They were amazed by the magician's tricks.
Show examples from the web [+]
I just continue to be amazed by which part of "shut-in" is unclear to you guys.
I'm just amazed by this car.
I will never cease to be amazed by your ignorance and stupidity.
I was amazed by Mr. Hill's generosity.
Everyone's been amazed by the progress and the dedication.
I must tell Your Majesty that the Milanese Ambassador was amazed by you.
I never cease to be amazed by your skill, Abby.
When I did, I was amazed by the response.
I was amazed by it for some reason.
I was so amazed by this man.
I was really amazed by all the stories that started flooding in.
I'm just amazed by this car.
I was always so amazed by how strong you were.
I think others too will be amazed by the European Commission's brevity.
I mean, honestly, I was just so amazed by his apartment.
I must say we were amazed by the number of applications received.
I am constantly amazed by the depths of the human heart.
Mr President, I am amazed by the way Mr Klinz is acting.
They're both amazed by the random chance that led them to find them.
When I first saw Pixar's "Luxo Jr.," I was amazed by how much emotion they could put into something as trivial as a desk lamp.
Alternatives:
surprised at
astonished at
impressed by
shocked by
awestruck by
Related Comparisons
"pleased with"vs"pleased at""Beyond amazed at"vs"Beyond amazed with""shocked by"vs"shocked at""surprised at"vs"surprised about""surprised by"vs"surprised at""embarrassed by"vs"embarrassed at"
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660 | https://pmc.ncbi.nlm.nih.gov/articles/PMC12409013/ | Effect of cotton roll biting on pain perception during orthodontic bracket debonding using Weingart pliers: A randomized controlled trial - PMC
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. 2025 Sep 2;53(9):03000605251370333. doi: 10.1177/03000605251370333
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Effect of cotton roll biting on pain perception during orthodontic bracket debonding using Weingart pliers: A randomized controlled trial
Belal A Drmch
Belal A Drmch
1 Department of Orthodontics, Faculty of Dentistry, Damascus University, Syria
Find articles by Belal A Drmch
1, Kinda Sultan
Kinda Sultan
1 Department of Orthodontics, Faculty of Dentistry, Damascus University, Syria
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1, Mohammad Y Hajeer
Mohammad Y Hajeer
1 Department of Orthodontics, Faculty of Dentistry, Damascus University, Syria
2 Department of Orthodontics, School of Dentistry, University of Jordan, Jordan
Find articles by Mohammad Y Hajeer
1,2,✉, Shadi Azzawi
Shadi Azzawi
3 Department of Pediatric Dentistry, Faculty of Dentistry, Syrian Private University, Syria
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3, Ahamd S Burhan
Ahamd S Burhan
1 Department of Orthodontics, Faculty of Dentistry, Damascus University, Syria
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1
Author information
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1 Department of Orthodontics, Faculty of Dentistry, Damascus University, Syria
2 Department of Orthodontics, School of Dentistry, University of Jordan, Jordan
3 Department of Pediatric Dentistry, Faculty of Dentistry, Syrian Private University, Syria
✉
Mohammad Y Hajeer, Department of Orthodontics, Faculty of Dentistry, University of Damascus, Mazzeh Autostrad, Mazzeh, DAM12 MH, Damascus, Syria. Email: myhajeer@gmail.com
Received 2025 May 13; Accepted 2025 Jul 30; Collection date 2025 Sep.
© The Author(s) 2025
Creative Commons Non Commercial CC BY-NC: This article is distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 License ( which permits non-commercial use, reproduction and distribution of the work without further permission provided the original work is attributed as specified on the SAGE and Open Access pages (
PMC Copyright notice
PMCID: PMC12409013 PMID: 40891809
Abstract
Objective
This study aimed to evaluate the effectiveness of biting on a cotton roll in reducing pain during metal bracket removal using a Weingart plier compared with the conventional method.
Methods
This two-arm, parallel-group, randomized controlled trial included 36 patients (11 males and 25 females) with a mean age of 20.5 ± 3.08 years (range: 16–25 years). The study was conducted among patients who had completed orthodontic treatment with a fixed appliance using MBT prescription 0.022-inch metal brackets featuring a single-mesh base (Pinnacle®, MBT compatible 0.022, OrthoTechnology™, Florida, USA), and they were referred for metal bracket debonding. Brackets were removed using a Weingart plier, and participants were randomly assigned to one of two groups: the open-mouth group and the biting-on-a-cotton-roll group. Pain intensity was assessed for each tooth using a visual analog scale (0–100 mm) immediately after each bracket removal. Pain levels were compared between the two groups, between the upper and lower jaws, and across sexes.
Results
Patients in both groups reported pain levels ranging from mild to mild–moderate, with lower mean pain levels observed in the biting-on-a-cotton-roll group compared with the open-mouth group. Significant differences in mean pain scores were found between the open-mouth group and biting-on-a-cotton-roll group in the upper and lower incisors (13.61 and 13.19, P< 0.001, respectively), upper and lower canines (4.03 and 3.61, P< 0.001, respectively), and upper and lower premolars (12.78 and 11.11, P< 0.001, respectively). No significant differences in pain perception were observed between males and females across all regions (P> 0.05).
Conclusions
Biting on a cotton roll significantly reduces pain during metal bracket debonding in the upper and lower incisor regions. The anterior regions of both jaws exhibit increased sensitivity to pain during the debonding procedure, whereas no differences in pain perception were observed between male and female patients.
Keywords: Debonding, metal brackets, pain, Weingart plier, cotton roll
Introduction
Debonding the orthodontic brackets at the end of treatment involves removing all parts of the appliance and the remaining adhesive material. It aims to restore the tooth surface to its initial state before the brackets.1 Therefore, correct adhesive and removal processes are necessary.1 Although debonding is a necessary step in orthodontic treatment, it may also contribute to one of its most common side effects—pain.
Orthodontic pain is considered one of the unpleasant side effects that may occur during orthodontic treatment.2 Additionally, 95% of orthodontic patients expressed mild-to-severe pain and discomfort during or after applying various orthodontic treatment procedures, which may lead patients to withdraw from treatment and not complete it. Orthodontic procedures that may cause pain include applying separation elastics,3–5 inserting orthodontic mini-implants,6 activating wires to induce different orthodontic tooth movements,7 expanding the upper-dental arch,8 using a removable appliance,9 and debonding orthodontic appliances.
The debonding procedure should be safe, painless, and efficient.10 Pain may be affected by many factors, such as the patient’s psychological state and previous experiences, cultural background, the patient’s sex, consuming systemic medications, the condition of the periodontal tissue as well as the type and location of the tooth, and the patient’s expectations about pain in addition to age.11
Pain experienced during debonding brackets is significantly influenced by the orientation of the applied force and the mobility of the teeth; during the debonding of brackets, the teeth are more tolerant of intrusive vertical forces and more sensitive to extrusive vertical forces, and the more stable the tooth is during debonding brackets, the less pain.12
Many studies have evaluated different procedures for debonding brackets, such as mechanical methods using cutters and removing pliers,13,14 the use of ultrasound waves,15 the application of lasers,16 and electrical and thermal methods.17 Several methods have been used to control the possible pain associated with metal bracket deboning.18 These methods include using pain-relieving medications10,19; applying finger pressure10,20,21; biting on a cotton roll12; biting on a chewing material22; or using a wax template,19 acrylic wafer,11,19 elastomeric wafer,20 and plastic wafer.21
A pilot study confirms the need to apply finger pressure or instruct the patient to firmly bite into a cotton roll to provide an intrusive, stabilizing force during bracket removal.12 A prospective split-mouth study investigated the effectiveness of bracket remover and a Weingart plier in combination with biting on a cotton roll in metal brackets debonding and found no clinical difference between the two pliers.23
When reviewing the literature, two recent studies focused on biting on a cotton roll during debonding. The first study compared this method with high-frequency vibration and elastomeric wafers during metal bracket debonding, concluding that biting on a cotton roll effectively reduced pain during debonding but was less effective than the high-frequency vibration.24 The second study compared biting on a cotton roll with traditional debonding in a split-mouth design using a bracket remover plier, determining that biting on a cotton roll was ineffective in reducing pain during the debonding process.25 Both studies used a bracket remover plier for debonding, and no study has evaluated the effect of biting on a cotton roll for reducing pain when debonding metal brackets with a Weingart plier.
Jeha and Haddad,26 in a systematic review, highlighted the inadequate documentation of the pain and discomfort associated with bracket removal. They confirmed the need for more clinical trials. In addition, this systematic review and another one2 did not include any study about the effectiveness of biting on a cotton roll during the removal of brackets. However, previous studies have yielded mixed findings on the effectiveness of biting on a cotton roll in managing pain during the debonding of metal brackets.
The objective of this study was to evaluate the effectiveness of biting on a cotton roll when removing metal brackets using a Weingart plier compared with the conventional method in reducing pain associated with metal bracket debonding.
Materials and methods
Study design, registration, and settings
This study was designed as a double-blinded, two-arm randomized controlled trial. The patients were unaware of their group allocation and remained blinded during bracket removal. In addition, the data analysis was carried out by a statistician who was also not informed of the group assignments. The protocol of this trial was registered in the clinical trials database (Registration number: NCT06873503). This study was conducted at the Department of Orthodontics, Faculty of Dentistry, the University of Damascus, Syria, and was funded by Damascus University (Ref No: 501100020595). This study was conducted by the principles of the Declaration of Helsinki (1975, as revised in 2024). Informed consent was obtained from all patients before their participation, and ethical approval was obtained from the Biomedical Research Ethics Committee at Damascus University (DN-180225-420). The patients included in this study did not participate in the design, implementation, or writing of the article.
Estimation of the sample size
The GPower software (version 3.1.3 Universität Düsseldorf, Düsseldorf, Germany) was used to calculate the required sample size. The calculation was performed considering the results of a previous study,27 which reported a mean visual analog scale (VAS) of 8.25 (SD = 4.78) for the test group and 4.20 (SD = 2.91) for a control group, a significance level of 0.05, and a power of 90%. This calculation indicated a necessity of 18 patients per group, and 36 patients were included in the study.
Patient recruitment and entry in this trial
The sample included patients who underwent planned orthodontic treatment performed by three fourth-year resident orthodontists in the Department of Orthodontics at the Faculty of Dentistry (University of Damascus). After ensuring that orthodontic treatment was completed, they were referred for the removal of metal brackets according to the current study between May 2024 and January 2025.
Inclusion criteria for the patients were as follows: (a) being aged 16 to 25 years, and being in good mental and physical health; (b) undergoing orthodontic treatment with MBT prescription 0.022-inch metal brackets with a single-mesh base (Pinnacle®, MBT compatible 0.022, OrthoTechnology™, Florida, USA) who were in the final stage of their treatment and planned for debonding; (c) having brackets bonded with the same bonding cement (Ivoclar Heliosit Orthodontic Adhesive Composite), the same primer (Ivoclar Vivadent Tetric N-Bond), and the same curing time, as well as the curing light device; (d) the absence of lost brackets at the debonding process; (e) having 0.017 × 0.025-inch stainless-steel finishing archwires in their places for at least 1 month; (f) having good occlusal relationships and strong intercuspation; (g) having no recent use of medications such as painkillers or corticosteroids on the last day; (h) the absence of acute or chronic dental pain induced by periodontal/periapical lesions or caries; (i) the absence of a history of surgical treatment, including impacted tooth eruption, tooth transplantation, or the presence of mini-screws; and (j) the absence of craniofacial deformities. All participants received an information sheet and provided informed consent before their recruitment. The reporting of this study conforms to the Consolidated Standards of Reporting Trials (CONSORT) statement.28Figure 1 illustrates the CONSORT flow diagram for patient enrollment.
Figure 1.
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The CONSORT flow diagram of patients’ recruitment, follow-up, and entry into data analysis.
CONSORT: Consolidated Standards of Reporting Trials.
Randomization and allocation concealment
After finishing the orthodontic treatment, patients were distributed randomly into groups through the block randomization method. A randomized patient list was created using Minitab® by an MSc student from the Department of Orthodontics, who had no involvement in this study. Patients were randomly allocated to six permuted blocks (with each block comprising six patients, including three from each group) in a 1:1 ratio to one of the two groups: the first group, the open-mouth group (OMG), and the second group, the biting-on-a-cotton-roll group (BCRG). Each patient received a serial number, and their allocation was hidden from the researcher to prevent selection bias. Patient names and group assignments were placed in closed opaque envelopes and only opened during the debonding stage of the metal brackets.
Each patient picked an envelope, and the intervention they received was determined according to its contents.
Interventions
Deboning in the OMG
The debonding procedure was conducted by the principal researcher (B.A.D.) using the open-mouth technique. All brackets were removed using a Weingart plier (American Orthodontics, Sheboygan, Wisconsin, USA) and squeezed from both sides of the mesial-distal surfaces (Figure 2). Before debonding, the stainless-steel finishing archwires (0.017 × 0.025-inch) were removed.
Figure 2.
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Debonding metal brackets in the open-mouth group.
Debonding in the BCRG
All brackets were removed by the same principal researcher (B.A.D.), and debonding was performed using a Weingart plier (American Orthodontics, Sheboygan, Wisconsin, USA), by squeezing them from both sides of the mesial-distal surfaces. Additionally, cotton rolls were placed between the upper and lower teeth, and the patients were instructed to bite on them during the procedure (Figure 3). Before debonding, the stainless-steel finishing archwires (0.017 ×0.025-inch) were removed. Patients in the second group were instructed to bite on a cotton roll throughout the debonding process. Brackets were then removed one at a time in both groups, in sequence from the upper right to the upper left and from the lower right to the lower left, with the central incisor bracket being the first to be removed in each quadrant. All bracket removals were performed using a right-handed technique.
Figure 3.
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Debonding metal brackets in the biting-on-a-cotton-roll group.
Outcome measures: perceived pain intensity
Pain intensity was effectively gauged using a 100-mm VAS, in which a score of zero indicated “no pain,” with increasing scores reflecting greater pain levels. Patients were tasked with recording their pain scores on the scale immediately after each bracket removal, ensuring accurate record-keeping before moving on to the next tooth. Before debonding, every patient was informed about the study’s objectives and instructed to evaluate the pain intensity of the procedure with a VAS after the debonding was completed. The VAS scores for the upper and lower jaws were meticulously tabulated, considering the comprehensive assessment and the specific evaluation within each region (incisors, canines, and posterior regions). In order to minimize bias, the VAS scores were evaluated by an independent orthodontist who had no prior knowledge of the group assignments.
Statistical analysis
The SPSS program (version 25.0; IBM, Armonk, New York, USA) was used for statistical analysis. The Shapiro-Wilk test was conducted to assess data normality, which indicated a non-normal distribution. As a result, the Mann–Whitney U test was employed to compare VAS pain scores between the two groups, whereas the Wilcoxon signed ranks test was used to compare VAS scores within each group. The chi-square test was used to detect significant differences in sex and extraction between the study groups. The Mann–Whitney U test was used to detect significant differences in VAS scores between males and females. A significance level of 0.05 was used.
Results
Baseline sample characteristics
The sample consisted of 36 patients (11 men and 25 women) with a mean age of 20.5 ± 3.08 years (16–25 years); tooth extraction was performed in 47.2% of the patients.
Age, sex, and extraction did not significantly differ between the two groups (P = 0.717, P = 0.346, P = 0.738, respectively). The baseline sample characteristics related to sex, age, and extraction are shown in Table 1. Descriptive statistics about VAS pain scores for each tooth and each region are given in Table 2.
Table 1.
Baseline sample characteristics.
| Variables | Open-mouth group | Biting-on-a-cotton-roll group | P value |
--- --- |
| Sex | Male, n (%) | 6 (33.3%) | 5 (27.8%) | 0.717a |
| Female, n (%) | 12 (66.7%) | 13 (72.2%) |
| Age (years) | Mean ± SD | 20.94 ± 3.19 | 20.11 ± 2.97 | 0.346b |
| Extraction-based treatment | No | 9 (50.0%) | 10 (55.6%) | 0.738a |
| Yes | 9 (50.0%) | 8 (44.4%) |
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n: number of patients.
a
Chi-square test.
b
Mann–Whitney test.
Table 2.
Descriptive statistics of pain perception using the visual analog scale for each tooth and each region.
| Tooth | Pain scores in open-mouth group | Pain scores in biting-on-a-cotton-roll group |
---
| Mean | SD | Median | Min | Max | Mean | SD | Median | Min | Max |
| U1Right | 21.94 | 5.72 | 20 | 15 | 40 | 8.33 | 2.43 | 10 | 5 | 10 |
| U2Right | 22.78 | 5.75 | 20 | 20 | 40 | 8.89 | 2.14 | 10 | 5 | 10 |
| U3Right | 8.89 | 4.04 | 10 | 5 | 20 | 4.44 | 2.36 | 5 | 0 | 10 |
| U4Right | 4.44 | 1.67 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| U5Right | 4.17 | 1.92 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| U1Left | 21.67 | 5.14 | 20 | 20 | 40 | 8.61 | 2.30 | 10 | 5 | 10 |
| U2Left | 23.06 | 5.72 | 20 | 20 | 40 | 9.17 | 1.92 | 10 | 5 | 10 |
| U3Left | 8.61 | 3.76 | 10 | 5 | 20 | 5.00 | 1.71 | 5 | 0 | 10 |
| U4Left | 4.44 | 1.67 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| U5Left | 4.17 | 2.57 | 5 | 0 | 10 | 0 | 0 | 0 | 0 | 0 |
| L1Right | 25.28 | 6.06 | 22.5 | 20 | 40 | 11.67 | 3.83 | 10 | 10 | 20 |
| L2Right | 24.17 | 6.00 | 20 | 20 | 40 | 11.39 | 3.35 | 10 | 10 | 20 |
| L3Right | 8.06 | 4.58 | 10 | 0 | 20 | 4.72 | 2.70 | 5 | 0 | 10 |
| L4Right | 3.75 | 2.26 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| L5Right | 3.06 | 2.51 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| L1Left | 25.56 | 6.16 | 25 | 20 | 40 | 11.94 | 3.04 | 10 | 10 | 20 |
| L2Left | 24.44 | 5.11 | 20 | 20 | 30 | 11.67 | 2.97 | 10 | 10 | 20 |
| L3Left | 8.33 | 2.43 | 10 | 5 | 10 | 4.44 | 2.36 | 5 | 0 | 10 |
| L4Left | 3.33 | 2.46 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| L5Left | 3.33 | 2.43 | 5 | 0 | 5 | 0 | 0 | 0 | 0 | 0 |
| Upper incisors | 22.36 | 5.335 | 20 | 18.75 | 40 | 8.75 | 1.77 | 10 | 5 | 10 |
| Upper canine | 17.50 | 6.91 | 15 | 10 | 40 | 9.44 | 3.38 | 10 | 5 | 20 |
| Upper premolars | 12.78 | 6.47 | 10 | 5 | 25 | 0.00 | 0.00 | 0 | 0 | 0 |
| Lower incisors | 24.86 | 5.04 | 23.75 | 20 | 37.5 | 11.67 | 2.68 | 10 | 10 | 18.75 |
| Lower canine | 16.39 | 6.14 | 17.5 | 5 | 30 | 9.17 | 3.54 | 10 | 5 | 15 |
| Lower premolars | 11.11 | 8.84 | 10 | 0 | 20 | 0.00 | 0.00 | 0 | 0 | 0 |
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1: central incisor; 2: lateral incisor; 3: canine; 4: first premolar; 5: second premolar; Min: minimum; Max: maximum; L: lower tooth; U: upper tooth.
Comparisons of pain levels between the two groups
Patients in both groups experienced mild or mild-to-moderate pain, with lower mean pain levels in the BCRG than the OMG across all regions. The greatest mean pain values were recorded in the lower incisor region (24.86 ± 5.04) in the OMG, whereas the lowest values were found in the upper- and lower-premolar regions of the BCRG (Table 3). There was a significant difference in pain perception between the OMG and BCRG in the upper and lower incisors ( = 13.61 and 13.19, respectively), the upper and lower canines ( = 4.03 and 3.61, respectively), and the upper- and lower-premolar regions ( = 12.78 and 11.11, respectively; P< 0.001; Table 3). Considering the predefined clinically significant difference of 13 mm on a 100 mm VAS.29 Descriptive statistics for these regions for each jaw in both groups, along with the P values of significance testing between the two groups, are shown in Table 3 (Figure 4).
Table 3.
Descriptive statistics of debonding pain scores for the incisor, canine, and premolar regions in each jaw across both groups, along with P values of significance testing between the two groups and between males and females.
| Region | Open-mouth group | Biting-on-a-cotton-roll group | Test statistics between two groups | Male (n = 11) | Female (n = 25) | Test statistics between males and females |
--- --- ---
| Mean | SD | Mean rank | Mean | SD | Mean rank | Mann–Whitney U | Z | P valuea | Mean | SD | Mean rank | Mean | SD | Mean rank | Mann–Whitney U | Z | P valuea |
| Upper incisors | 22.36 | 5.34 | 27.50 | 8.75 | 1.77 | 9.50 | 0 | 5.303 | <0.001b | 14.66 | 7.21 | 17 | 15.95 | 8.35 | 19.16 | 121 | 0.586 | 0.558 |
| Upper canine | 8.75 | 3.45 | 25.67 | 4.72 | 1.69 | 11.33 | 33 | 4.312 | <0.001b | 10.91 | 4.37 | 14.59 | 14.60 | 7.35 | 20.22 | 94.50 | 1.560 | 0.119 |
| Upper premolars | 12.78 | 6.47 | 27.50 | 0.00 | 0.00 | 9.50 | 0 | 5.503 | <0.001b | 6.36 | 7.78 | 18.41 | 6.40 | 8.10 | 18.54 | 136.50 | 0.037 | 0.971 |
| lower incisors | 24.86 | 5.04 | 27.50 | 11.67 | 2.68 | 9.50 | 0 | 5.221 | <0.001b | 16.71 | 5.54 | 17.14 | 18.95 | 8.6 | 19.10 | 122.50 | 0.525 | 0.600 |
| Lower canine | 8.19 | 3.06 | 24.58 | 4.58 | 1.76 | 12.42 | 52.50 | 3.575 | <0.001b | 10.00 | 5.00 | 13.86 | 14.00 | 6.29 | 20.54 | 86.50 | 1.807 | 0.071 |
| Lower premolars | 11.11 | 8.84 | 25 | 0.00 | 0.00 | 12 | 45 | 4.338 | <0.001b | 5.00 | 8.06 | 18.23 | 5.80 | 8.62 | 18.62 | 134.50 | 0.121 | 0.904 |
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n: number of patients.
a
Mann–Whitney test.
b
significant at the 0.05 level.
Figure 4.
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A bar chart illustrating pain scores for the different regions in both groups.
Comparisons of pain levels within each group
Comparisons of pain levels between the right and left sides
In the OMG, greater mean pain levels were reported in the lower incisor region on the left side ( = 12.5 ± 2.71) compared with the right side ( = 12.36 ± 2.76). A similar result was seen in BCRG, in which the left side also had greater mean pain levels ( = 5.90 ± 1.34) compared with the right side ( = 5.77 ± 1.78). However, neither group had significant differences in pain perception between the right and left sides (P > 0.05; Table 4). Table 4 provides the descriptive statistics for debonding pain scores in each quadrant and the P values of significance testing between the two sides.
Table 4.
Descriptive statistics of debonding pain scores for the right and left sides for each quadrant in both groups along with the P values of significance testing between the two sides.
| Group | Region | Right side | Left side | Difference between the two sides |
--- ---
| Mean | SD | Median | Mean | SD | Median | Mean difference | P valuea |
| Open-mouth group | Upper incisors | 11.18 | 2.8 | 10 | 11.18 | 2.59 | 10 | 0.00 | 1.00 |
| Upper canine | 8.89 | 4.04 | 10 | 8.61 | 3.76 | 10 | 0.28 | 0.739 |
| Upper premolars | 6.39 | 3.76 | 5 | 6.39 | 4.13 | 5 | 0.00 | 1.00 |
| Lower incisors | 12.36 | 2.76 | 12.5 | 12.5 | 2.71 | 11.25 | −0.14 | 0.725 |
| Lower canine | 8.06 | 4.58 | 10 | 8.33 | 2.43 | 10 | −0.28 | 0.763 |
| Lower premolars | 5.56 | 4.82 | 7.5 | 5.56 | 4.50 | 5 | 0.00 | 1.00 |
| Biting-on-a-cotton-roll group | Upper incisors | 4.31 | 4.28 | 5 | 4.45 | 3.92 | 5 | −0.14 | 0.589 |
| Upper canine | 4.44 | 2.36 | 5 | 5.00 | 1.71 | 5 | −0.56 | 0.317 |
| Upper premolars | 0.00 | 0.00 | 0 | 0.00 | 0.00 | 0 | 0.00 | 1.00 |
| Lower incisors | 5.77 | 1.78 | 5 | 5.90 | 1.34 | 5 | −0.13 | 0.726 |
| Lower canine | 4.72 | 2.70 | 5 | 4.44 | 2.36 | 5 | 0.28 | 0.739 |
| Lower premolars | 0.00 | 0.00 | 0 | 0.00 | 0.00 | 0 | 0.00 | 1.00 |
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a
Wilcoxon signed ranks test.
Comparisons of pain levels between the incisors and premolar regions
In the OMG, the greater mean pain values were recorded in the upper and lower incisors regions ( =22.36 ± 5.34 and 24.86 ± 5.04, respectively) compared with the upper- and lower-premolar regions ( = 12.78 ± 6.47 and 11.11 ± 8.84, respectively). A similar result was observed in BCRG, in which the upper and lower incisor regions reported greater mean pain levels ( = 8.75 ± 1.77 and 11.67 ± 2.68, respectively). Conversely, no pain was detected in the upper- and lower-premolar regions (Table 5). Additionally, there was a significant difference in pain perception between the incisor and premolar regions in both jaws (P< 0.001; Table 5). Comparison of debonding pain scores across different tooth regions and both jaws in the two study groups is shown in Table 5.
Table 5.
Comparison of debonding pain scores across different tooth regions and both jaws in the two study groups.
| Groups | Comparison type | Description | Difference |
--- --- |
| Mean difference | P valuea |
| Open-mouth group | Intrajaw | Incisors vs premolars (maxillary) | 9.58 | <0.001b |
| Incisors vs premolars (mandibular) | 13.75 | <0.001b |
| Interjaw | Maxillary vs Mandibular (incisors) | −2.5 | 0.033b |
| Maxillary vs Mandibular (canines) | 1.11 | 0.248 |
| Maxillary vs Mandibular (premolars) | 1.67 | 0.525 |
| Biting-on-a-cotton-roll group | Intrajaw | Incisors vs premolars (maxillary) | 8.75 | <0.001b |
| Incisors vs premolars (mandibular) | 11.67 | <0.001b |
| Interjaw | Maxillary vs Mandibular (incisors) | −2.92 | 0.001b |
| Maxillary vs Mandibular (canines) | 0.28 | 1.00 |
| Maxillary vs Mandibular (premolars) | 0.00 | 1.00 |
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a
Wilcoxon signed ranks test.
b
significant at the 0.05 level.
Comparisons of pain levels between the maxillary and mandibular regions
Patients in the OMG reported mild-to-moderate pain in the incisor regions, with mean values of 24.86 ± 5.04 in the lower incisor region and 22.36 ± 5.34 in the upper-anterior region. Mild pain was also recorded in the canine regions ( = 16.39 ± 6.14 for lower canines and 17.50 ± 6.91 for upper canines) and in the premolar regions ( = 11.11 ± 8.84 for lower premolars and 12.78 ± 6.47 for upper premolars). In the BCRG, patients experienced mild pain in the incisor regions ( = 11.67 ± 2.68 for lower and 8.75 ± 1.77 for upper) and the canine regions ( = 9.17 ± 3.54 for lower and 9.44 ± 3.38 for upper). No pain was reported in the posterior regions. Pain perception significantly differed between the upper and lower incisor regions in OMG and BCRG (P = 0.033 and P = 0.001, respectively; Table 5). However, no significant differences were found for the canine or premolar regions (P> 0.05; Table 5). Comparison of debonding pain scores across different tooth regions and both jaws in the two study groups is given in Table 5.
Effect of sex on pain levels
Females reported greater mean pain in the lower incisor region ( = 18.95 ± 8.6) compared with males ( = 16.71 ± 5.54). However, no significant differences in pain perception between the sexes were found across all regions (P> 0.05, Table 3). Descriptive statistics detailing debonding pain scores for both sexes in each region, along with the P values of significance testing between the males and females, are given in Table 3.
Discussion
Pain is a subjective experience that can vary widely among individuals. This variation is influenced by several factors, including age, sex, personal pain threshold, psychological and social conditions, past negative experiences, and the degree of force applied. Recognizing these differences is essential for effective pain assessment and management strategies.19,20,30
Although various methods for bracket debonding, such as ultrasonic instrumentation, lasers, and electro-thermal debonding, are documented in the literature,14,19 hand tools like Weingart pliers remain among the most practical solutions.23 This study used Weingart pliers to compare the pain scores in two techniques—open mouth and biting on a cotton roll—during bracket removal. These methods are widely used in orthodontic clinics and are readily available to orthodontists. Therefore, this study aimed to investigate the effectiveness of biting on a cotton roll for pain management during metal bracket debonding and the effects of anatomical location and sex on pain scores.
The first study on pain associated with bracket debonding was a pilot study conducted by Williams and Bishara.12 They found that patients can tolerate intrusive forces. They noted that biting on a cotton roll provides a stabilizing force, which may reduce pain during the debonding process. In another study by Kilinç and Sayar,19 soft wax and acrylic bite wafers were used to prevent extrusive forces during debonding.
The intergroup comparison revealed that the highest pain values were observed in the OMG compared with the BCRG across all regions, especially in the upper and lower incisor regions where the difference was clinically significant. In the rest of the regions, although the difference was statistically significant, it was not clinically significant. This indicates that the effectiveness of biting on a cotton roll in pain management was associated with metal brackets debonding in the region of the incisors. This finding is consistent with the research conducted by Musawi and Kadhum,24 which demonstrated that biting on a cotton roll during bracket removal effectively reduced associated pain. Although a different instrument was used for debonding, the median pain score in the lower anterior region was 35 in the OMG and 25 in BCRG, whereas in the current study, the median was 23.75 in OMG and 10 in BCRG. Similarly, Thakkar et al.22 found that removing brackets while biting on a chewie significantly decreased the pain levels of the patient compared with the OMG. However, these results contrast with those of Celebi,25 who reported that biting on a cotton roll did not alleviate pain during bracket removal in all regions. The discrepancy might stem from the use of the split-mouth design in that study; the design used in that study raises concerns about possible carry-over effects, as pain in the orofacial region, particularly dental pain, can radiate from one side of the oral cavity to the other or even to various parts of the orofacial area. As a result, it becomes difficult for patients to evaluate the precision of their pain.31 In addition to the variation in tools used for bracket removal, a bracket remover plier was used in their study; this tool exerts a symmetrical bilateral force on the bracket wings, concentrating on a smaller area of force application compared with the Weingart plier. This aligns with the findings of Bishara and Fehr,32 who highlight that a reduction in the contact surface between the tool head and the bracket leads to a decrease in the force applied.
In this study, debonding began from the upper right, moved to the upper left and lower right, and concluded with the lower left quadrants. Mangnall et al.11 suggested that the initial quadrant debond could be more painful. However, in the current study, there was no difference in pain perception between the right and left sides in both regions of the group.
The intragroup comparison revealed that tooth location significantly affected pain perception. Bavbek et al.20 found that upper and lower anterior teeth are more pain-sensitive than posterior teeth, whereas Nakada et al.30 noted greater pain in the upper and lower anterior segments compared with the posterior segments. In this study, lower incisors had the highest pain scores among all evaluated teeth, consistent with previous studies.10,11,13,19,20 This observation may be attributed to the relatively small root surface area of the lower incisors, leading to a higher force concentration per unit area during the debonding process. This increased pressure is likely to enhance sensitivity and pain perception in these teeth compared with those with larger root surfaces.
Mangnall et al.11 observed elevated pain levels in the upper right posterior segment in 18% of patients. In contrast, the present study recorded only mild pain in the OMG and no pain in the BCRG in the posterior region. This disparity may be attributed to the upper right posterior segment being the first sextant debonded in their study, possibly heightening the participants’ perception of pain. Additionally, differences in pain management methods, such as using a soft acrylic wafer, could also have contributed to this variation.
In addition to the anatomical location of the tooth affecting pain levels, sex differences also play a role in the intensity of pain during the debonding process.10,19,20 It was found that females tend to have a lower pain threshold compared with males.33 Additionally, the intensity of pain perception can vary based on the patient’s sex.34 However, the present study found no notable differences in pain levels between males and female participants; this result agreed with the results of the studies of Iqbal et al.,21 Musawi and Kadhum,24 and Karobari et al.35 This finding also aligns with the study by Williams and Bishara,12 who highlighted that sex differences have minimal impact on pain. Furthermore, Koyama et al.36 observed that subjective pain experiences are influenced more by expectations, as these can alter brain mechanisms. In essence, positive expectations can lead to a reduced perception of pain.
Limitations
This study evaluated the pain associated with metal brackets debonding using Weingart pliers and biting on a cotton roll. However, several significant variables were not explored. Future studies should investigate aspects such as evaluating adhesive remnants on teeth and potential enamel damage. Another limitation of the study was focusing on pain levels in the incisor, canine, and premolar regions without assessing the molar region because of the variations in molar attachments, which depend on individual treatment needs and orthodontists’ preferences. Another limitation of this study was that it exclusively focused on patients aged 16 to 25 years, excluding other age groups. In addition, it did not include the use of the Pain Catastrophizing Scale to assess the patient’s psychological state to filter out those not suitable for inclusion. Furthermore, sex-specific subgroup analysis was not considered in the sample size calculation, which may reduce the statistical power of post-hoc analysis and increase the risk of type II errors.
Generalizability
The findings of this study, conducted with a sample of adult patients and minimizing bias, suggest that this technique could be a useful addition to orthodontic debonding protocols. However, the applicability of these results may be constrained by factors such as the unique traits of the study population, the clinical environment, and the specific methodologies employed. Further research involving diverse populations and varying clinical settings is recommended to verify the broader relevance of these findings and develop standardized protocols for using Weingart pliers with biting on a cotton roll in metal bracket debonding.
Conclusion
Biting on cotton significantly alleviates pain during metal bracket debonding in all regions, with clinically significant efficacy in the upper and lower incisor regions. Pain levels during the debonding of metal brackets were higher in the incisor region compared with the premolar region in both jaws, with the lower incisors being the most painful. Both jaws showed no difference in pain perception between the right and left sides. Male and female patients exhibit no disparity in pain perception.
Acknowledgments
Not applicable.
Authors’ contributions: Belal A Drmch: The principal researcher recruited patients, performed clinical procedures, collected questionnaires, analyzed the data, interpreted the results, and wrote the initial drafts of this paper. Kinda Sultan supervised the whole research project, helped in data analysis and interpretation of the results, and corrected the early drafts of this manuscript. Mohammad Y Hajeer and Ahamd Burhan co-supervised this project, helped in statistical analysis, and participated in writing this manuscript. The manuscript was read and approved in its final form by all authors. Shadi Azzawi participated in the inception of this work and study design, assisted in data analysis and interpretation, edited the final drafts of this manuscript. All authors read and approved the final version of this manuscript.
Funding: This research work was financially supported by the University of Damascus (Ref No: 501100020595).
ORCID iD: Mohammad Y Hajeer
Availability of data and materials
Data concerning the current study can be obtained from the corresponding author upon reasonable request.
Consent for publication
Not applicable.
Declaration of conflicting interests
The authors have confirmed that there are no conflicts of interest related to the submitted manuscript.
Ethics approval and consent to participate
This project was approved by the Biomedical Research Committee at the University of Damasucs (Reference number: DN-180225-420).
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Data Availability Statement
Data concerning the current study can be obtained from the corresponding author upon reasonable request.
Articles from The Journal of International Medical Research are provided here courtesy of SAGE Publications
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661 | https://math.stackexchange.com/questions/99299/best-fitting-plane-given-a-set-of-points | Skip to main content
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Nothing more to explain. I just don't know how to find the best fitting plane given a set of N points in a 3D space. I then have to write the corresponding algorithm. Thank you ;)
linear-algebra
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edited Nov 28, 2017 at 5:40
Siong Thye Goh
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asked Jan 15, 2012 at 16:15
G4bri3lG4bri3l
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There is plenty more to explain. There are many different measures of how well a plane fits given data, and different measures give rise to different "best" fitting planes. So you had best tell us what you have in mind as your measure of how well a given plane fits some given data.
– Gerry Myerson
Commented
Jan 15, 2012 at 21:02
I'm sorry but I wish I could tell you more. But I know just a bit. Let's say that the set of Points I have (over a 100) already look like a plane, I mean, they are displayed as a plane but not perfectly. "Obtain the symmetry plane A by fitting it on the set of points B.." That's all I have to do. They don't say anything more.
– G4bri3l
Commented
Jan 16, 2012 at 9:20
1
You hadn't mentioned the symmetry part before -- do you know what that's referring to?
– joriki
Commented
Jan 16, 2012 at 9:53
5
Let's look at a simpler problem. Say you have a bunch of points in 2 dimensions that almost lie along a line, but not quite, and you want to find the line that fits those points the best. You could draw a line, then draw vertical line segments from each point to the line, and add up the lengths of all those line segments, and ask for the line that makes that sum as small as possible. But you could draw horizontal line segments instead, and you might get a different answer by minimizing the sum of those lengths. Or you could draw line segments perpendicular to the line. Continued...
– Gerry Myerson
Commented
Jan 16, 2012 at 11:23
2
...Instead of just adding up the lengths of the line segments, you could add up the squares of the lengths - may seem like a strange idea, but it's very often a good one in this kind of problem. So you have all those choices, just for drawing a line in 2 dimensions; there are even more choices for a plane in 3. That's why you really have to know what someone means when they ask you to fit a plane to some points.
– Gerry Myerson
Commented
Jan 16, 2012 at 11:26
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Subtract out the centroid, form a 3×N matrix X out of the resulting coordinates and calculate its singular value decomposition. The normal vector of the best-fitting plane is the left singular vector corresponding to the least singular value. See this answer for an explanation why this is numerically preferable to calculating the eigenvector of XX⊤ corresponding to the least eigenvalue.
Here's a Python implementation, as requested:
```
import numpy as np
generate some random test points
m = 20 # number of points
delta = 0.01 # size of random displacement
origin = np.random.rand(3, 1) # random origin for the plane
basis = np.random.rand(3, 2) # random basis vectors for the plane
coefficients = np.random.rand(2, m) # random coefficients for points on the plane
generate random points on the plane and add random displacement
points = basis @ coefficients \
+ np.tile(origin, (1, m)) \
+ delta np.random.rand(3, m)
now find the best-fitting plane for the test points
subtract out the centroid and take the SVD
svd = np.linalg.svd(points - np.mean(points, axis=1, keepdims=True))
Extract the left singular vectors
left = svd
```
1 2
```
the corresponding left singular vector is the normal vector of the best-fitting plane
left[:, -1]
```
2
```
its dot product with the basis vectors of the plane is approximately zero
left[:, -1] @ basis
```
2
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edited May 10, 2021 at 8:31
Mad Physicist
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answered Jan 15, 2012 at 17:35
jorikijoriki
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@G4bri3l: I'm not sure I understand that question. The way I've defined X, its right-singular vectors are N-dimensional, so I don't see how they could be used to find the best-fitting line. It's the left singular vectors that are 3-dimensional, and indeed the left singular vector u corresponding to the largest singular value gives the direction of the best-fitting line. Remember that X contains the coordinates with the centroid c subtracted out, so the equation for the best-fitting line is c+λu.
– joriki
Commented
Jan 17, 2012 at 17:18
2
Having only a normal vector is not enough to define a plane. What is the best fitting point?
– xaav
Commented
Nov 16, 2016 at 2:11
4
Note that if the matrix is kept as (N, 3), then the normal vector will correspond to the right singular vector associated with the smallest eigenvalue.
– Kevin Zakka
Commented
Jan 7, 2020 at 11:37
1
@Siyh: OK, I've added a Python implementation to the answer.
– joriki
Commented
Jan 23, 2020 at 13:49
1
Anyone using Matlab can use the matgeom package, which implements this algorithm in its fitPlane function.
– Minh Nghĩa
Commented
Apr 16, 2020 at 11:20
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A simple least squares solution should do the trick.
The equation for a plane is: ax+by+c=z. So set up matrices like this with all your data:
⎡⎣⎢⎢⎢x0x1xny0y1...yn111⎤⎦⎥⎥⎥⎡⎣⎢abc⎤⎦⎥=⎡⎣⎢⎢⎢z0z1...zn⎤⎦⎥⎥⎥
In other words:
Ax=B
Now solve for x which are your coefficients. But since (I assume) you have more than 3 points, the system is over-determined so you need to use the left pseudo inverse: A+=(ATA)−1AT. So the answer is:
⎡⎣⎢abc⎤⎦⎥=(ATA)−1ATB
And here is some simple Python code with an example:
```
import matplotlib.pyplot as plt
import numpy as np
These constants are to create random data for the sake of this example
N_POINTS = 10
TARGET_X_SLOPE = 2
TARGET_y_SLOPE = 3
TARGET_OFFSET = 5
EXTENTS = 5
NOISE = 5
Create random data.
In your solution, you would provide your own xs, ys, and zs data.
xs = [np.random.uniform(2EXTENTS)-EXTENTS for i in range(N_POINTS)]
ys = [np.random.uniform(2EXTENTS)-EXTENTS for i in range(N_POINTS)]
zs = []
for i in range(N_POINTS):
zs.append(xs[i]TARGET_X_SLOPE + \
ys[i]TARGET_y_SLOPE + \
TARGET_OFFSET + np.random.normal(scale=NOISE))
plot raw data
plt.figure()
ax = plt.subplot(111, projection='3d')
ax.scatter(xs, ys, zs, color='b')
do fit
tmp_A = []
tmp_b = []
for i in range(len(xs)):
tmp_A.append([xs[i], ys[i], 1])
tmp_b.append(zs[i])
b = np.matrix(tmp_b).T
A = np.matrix(tmp_A)
Manual solution
fit = (A.T A).I A.T b
errors = b - A fit
residual = np.linalg.norm(errors)
Or use Scipy
from scipy.linalg import lstsq
fit, residual, rnk, s = lstsq(A, b)
print("solution: %f x + %f y + %f = z" % (fit, fit, fit))
print("errors: \n", errors)
print("residual:", residual)
plot plane
xlim = ax.get_xlim()
ylim = ax.get_ylim()
X,Y = np.meshgrid(np.arange(xlim, xlim),
np.arange(ylim, ylim))
Z = np.zeros(X.shape)
for r in range(X.shape):
for c in range(X.shape):
Z[r,c] = fit X[r,c] + fit Y[r,c] + fit
ax.plot_wireframe(X,Y,Z, color='k')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
```
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edited Dec 7, 2022 at 10:09
Guillaume Jacquenot
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answered Jun 1, 2017 at 18:49
BenBen
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Am I correct to interpret that this method will solve for the plane that minimizes the vertical (ie: z) distance and not the perpendicular distance?
– Gabriel
Commented
Nov 8, 2017 at 23:03
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Yes, it minimizes vertical distance.
– Ben
Commented
Nov 9, 2017 at 0:00
Thank you very much for clarifying this Ben (and for the answer) If you have the time, I have a very similar question here (using Singular-Value Decomposition) that is giving me trouble.
– Gabriel
Commented
Nov 9, 2017 at 18:36
9
This is not a completely general solution . If the plane is (nearly) perpendicular to the z=0 plane this will fail.
– wcochran
Commented
Nov 25, 2019 at 19:20
How can the perpendicular distance be minimized?
– Luke Hutchison
Commented
Nov 20, 2020 at 12:41
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In order to complete the Claude Leibovici's answer :
With the numerical example proposed by Claude Leibovici who computed the parameters of a fitted plane z=Ax+By, the fitting of the plane Z=αX+βY+γ can be carried out thanks to the principal components method (as suggested by joriki).
The theory can be found in many books. A synopsis is given pages 24-25 in the paper:
The symbols used below correspond to those in the formulas from the referenced paper.
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answered Feb 12, 2016 at 18:48
JJacquelinJJacquelin
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Considering a plane of equation Ax+By+Cz=0 and a point of coordinates (xi,yi,zi), the distance is given by
di=±Axi+Byi+CziA2+B2+C2−−−−−−−−−−−√
and I suppose that you want to minimize
F=∑i=1nd2i=∑i=1n(Axi+Byi+Czi)2A2+B2+C2
Setting C=1, we then need to minimize with respect to A,B
F=∑i=1n(Axi+Byi+zi)2A2+B2+1
Taking derivatives with respect to A and B:
F′A=∑i=1n(2xi(Axi+Byi+zi)A2+B2+1−2A(Axi+Byi+zi)2(A2+B2+1)2)
F′B=∑i=1n(2yi(Axi+Byi+zi)A2+B2+1−2B(Axi+Byi+zi)2(A2+B2+1)2)
Since we shall set these derivatives equal to 0, the equations can be simplified to
∑i=1n(xi(Axi+Byi+zi)−A(Axi+Byi+zi)2(A2+B2+1))=0
∑i=1n(yi(Axi+Byi+zi)−B(Axi+Byi+zi)2(A2+B2+1))=0
whic are nonlinear with respect to the parameters A,B; then, good estimates are required since you will probably use Newton-Raphson for polishing the solutions.
These can be obtained making first a multilinear regression (with no intercept in your cas)
z=αx+βy
and use A=−α and B=−β for starting the iterative process. The values are given by
A=−SxySyz−SxzSyySxy2−SxxSyyB=−SxySxz−SxxSyzSxy2−SxxSyy
For illustration purposes, let me consider the following data
⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜x111222333y123123123z91420111723152026⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟
The preliminary step gives z=2.97436x+5.64103y and solving the rigorous equations converges to A=−2.97075, B=−5.64702 which are quite close to the estimates (because of small errors).
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edited Nov 28, 2024 at 14:35
Kenn Sebesta
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answered Feb 11, 2016 at 5:17
Claude LeiboviciClaude Leibovici
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In your example, the equation of the plane z=Ax+By is carried out with respect to the criteria of the least square orthogonal distances between the points and the plane. I agree with your result. But, they are only two adjustable parameters A,B because the origin (0,0,0) is supposed to be on the plan. This is an additional condition. If we consider the more general equation of the plan z=Ax+By+C the three parameters problem should lead to a better fitting with lower mean square deviation.
– JJacquelin
Commented
Feb 12, 2016 at 15:18
1
The general plane equation is Ax+By+Cz+D=0 or Ax+By+Cz=1
– John Alexiou
Commented
Nov 12, 2022 at 0:19
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All the other answers are amazing. I am providing an another way of doing this. You can use RANSAC. This method may not be more efficient than the other methods.
Simple theory behind this method is to pick random samples from the input sample, and create mathematical model that better describes your problem. After that find the error of this mathematical model with respect to all the samples, and count the number of samples whose error is within a given threshold (These samples are called inliers) . Do this procedure for multiple iterations and return the model that has more number of inliers.
For your specific problem, at a give iteration , you can pick three random points, and construct a plane using those points. The generalized equation of a plane is as given by Ax+By+Cz+D=0. Now for every point of coordinates (xi,yi,zi) in the input sample , find the distance between point and the plane. This can be your error definition. The distance between the plane and the point is given by
di=±Axi+Byi+Czi+DA2+B2+C2−−−−−−−−−−−√
These equations are taken from Claude's answer.
If this value is less than a given threshold, you can count that point as an inlier. Repeat this process for certain number of iterations. The following python program implements this method.
```
import numpy as np
import matplotlib.pyplot as plt
def EoP(p1,p2,p3):
"""for a given three points, finds out the equation of plane"""
v1 = p3 - p1
v2 = p2 - p1
cp = np.cross(v1, v2)
A,B,C = cp
D = np.dot(cp, p3)
return A,B,C,D
def RANSAC(X,itr=5000,threshold=500):
"""
Inputs:
X- sample coordinates
itr - number of iterations
threshold - error threshold
output:
Three coordinates of a plane if RANSAC found the best fit
"""
N = X.shape #size of the smaple
n=0
for i in range(itr):
idx = np.random.randint(N,size=(1,3)) #generating three random indices
Xsample = X[idx] #gather the coordinates with those indices
p1,p2,p3 = Xsample# unpacking three points
A,B,C,D = EoP(p1,p2,p3) #Equation of plane
# print(A,B,C,D)
inliers= 0
#Now for every point in input sample, find if it is a inlier
for p in X:
xi,yi,zi = p
d = np.abs((Axi+Byi+Czi+D)/np.sqrt(xi2+yi2+zi2))
if d<threshold:
inliers+=1 # finding out the inliers that satisfy the distance condition
if inliers > n:
Abest = A
Bbest = B
Cbest = C
Dbest = D
if inliers > 0:
return (Abest , Bbest, Cbest, Dbest)
return None
if name == "main":
rdm = np.random.RandomState(10) # to repeat the random state
X = rdm.randint(100,size=(100,3))
fig = plt.figure()
ax = plt.axes(projection='3d')
Out = RANSAC(X)
if Out:
A,B,C,D = Out
ax.scatter3D(X[:,0],X[:,1],X[:,2],c="g",label="Samples")
xx, yy = np.meshgrid(range(100), range(100))
z = -(Axx+Byy+D)/C
ax.plot_surface(xx, yy, z, alpha=0.5) #,label = "Best Fit Plane ")
plt.legend()
plt.show()
else:
print("No best RANSAC")
```
This program results the following for a random input :
The problems with this method :
May need tuning for iterations and threshold
Since the random points are taken to form the mathematical model , this may not give you the best ouput
This is slow !
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answered Nov 11, 2022 at 22:30
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662 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/17%3A_Electrochemistry/17.03%3A_Standard_Reduction_Potentials | Skip to main content
17.3: Standard Reduction Potentials
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: Sep 12, 2022
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17.2: Galvanic Cells
17.4: The Nernst Equation
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Learning Objectives
Determine standard cell potentials for oxidation-reduction reactions
Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices
The cell potential results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.4.1 and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is
E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures.
A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure ). In cell notation, the reaction is
Electrons flow from the anode to the cathode. The reactions, which are reversible, are
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction.
Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure , where
Electrons flow from left to right, and the reactions are
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.
It is important to note that the potential is not doubled for the cathode reaction.
The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, , for any cell. For example, for the following cell:
Again, note that when calculating , standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table . A more complete list is provided in Tables P1 or P2.
Table : Selected Standard Reduction Potentials at 25 °C
| Half-Reaction | E° (V) |
| | +2.866 |
| | +1.69 |
| | +1.507 |
| | +1.498 |
| | +1.35827 |
| | +1.229 |
| | +1.20 |
| | +1.0873 |
| | +0.7996 |
| | +0.7973 |
| | +0.771 |
| | +0.558 |
| | +0.5355 |
| | +0.49 |
| | +0.34 |
| | +0.26808 |
| | +0.22233 |
| | +0.151 |
| | 0.00 |
| | −0.1262 |
| | −0.1375 |
| | −0.257 |
| | −0.28 |
| | −0.3505 |
| | −0.4030 |
| | −0.447 |
| | −0.744 |
| | −1.185 |
| | −1.245 |
| | −0.7618 |
| | −1.662 |
| | −2.372 |
| | −2.71 |
| | −2.868 |
| | −2.912 |
| | −2.931 |
| | −3.04 |
Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.
Example : Cell Potentials from Standard Reduction Potentials
What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.
Solution
Using Table , the reactions involved in the galvanic cell, both written as reductions, are
Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:
The least common factor is six, so the overall reaction is
The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.
From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.
Exercise
A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.
Answer
Summary
Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, E°, for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.
Key Equations
Glossary
standard cell potential
: the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode
standard hydrogen electrode (SHE)
: the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials
standard reduction potential (E°)
: the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials
17.2: Galvanic Cells
17.4: The Nernst Equation |
663 | https://www.youtube.com/watch?v=gN0cpZytGos | Amphibians for Kids | What is an amphibian? Learn the characteristics of amphibians
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NOTE: A chart used in this video shows fish in the warm-blooded vertebrates category. This is incorrect. Please excuse the error and remind your students that fish are cold-blooded.
Learn all about the characteristics of amphibians and what makes these animals unique in our amphibians for kids video. We will talk about warm and cold blooded animals, invertebrates and vertebrates, smooth vs moist skin, and so much more! If you have ever asked your self "What is an Amphibian?" this video is for you!
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Transcript:
Amphibians Have you ever sorted some of your
things into different groups? Maybe you sort clothes by color or style, or your
video games by category. Scientists do the same thing with animals and other objects. All animals,
including humans, have a few things in common. They are living organisms. They need food and
energy to survive, move, and grow. And they create more of their own, which means they have
babies. Though they have a lot in common, there are also many differences between animals. When
you see an insect, like a mosquito or a butterfly, you know that it’s different than a fish. You
also know that a cougar is different from a snake. Because of these differences, all animals are
classified, or separated into special groups. Some of these special groups include mammals, or
animals with hair. Reptiles are creatures that have dry, scaly skin. Birds, who have feathers
and wings, and fish that have gills and live in the water. Some other groups that animals are
classified in include warm-blooded, cold-blooded, vertebrates, invertebrates, and
living on land versus living in water. Humans live on land. Even though you can
swim, take a shower, and jump into the ocean, you cannot live and survive in the water,
because you need oxygen to breathe. You can only survive on land. Fish are the opposite.
They cannot live on the land like you. They can only survive in the water. But did
you know that there is one group of animals that can live both on land and in water?
This group is called amphibians! In fact, the Greek word for amphibian
means double life or two lives. There are about 7,000 species of
amphibians living in the world today that can survive in the water and on land. You are
probably familiar with some of these amphibians. Can you think of some amphibians
that you might recognize? Here’s a hint: think of some animals
that are green and often slimy… eww! Did you think of toads, salamanders, newts,
or frogs? If you did, you’re right! They are amphibians. You probably already know some things
about frogs and toads, but did you know that there are many different types of both? Have you heard
of the American bull frog or the poison dart frog? What about the American toad or the tomato toad?
Did you know that there are around 2,000 different species of toads and frogs in our world? Kind
of makes you wonder where they’re all hiding! Sometimes it’s easy to mix up a frog and a toad.
Here’s how you can tell them apart. As adults, frogs have smooth, moist skin and longer
legs than toads, while toads have dry, bumpy skin and short legs because they walk more
often than they jump. Although toads and frogs are different, they are both amphibians
because they have some things in common. So what makes an animal fit
into the amphibian category? Well, it’s simple. They each have
characteristics that are the same, like webbed feet. They also have backbones called
vertebrates, just like you do. But their back bones are much smaller than humans. Every group
of animals have certain characteristics in common. Some of those characteristics are included in
other groups too! Like humans and amphibians both having vertebrates or backbones. However,
because humans are warm-blooded and amphibians are cold-blooded, they cannot be in the same group
even though they do share something in common. Cold-blooded means that an amphibian’s body
temperature changes. If it’s 60 degrees in the air or the water where they’re living, then
their bodies adapt and also become 60 degrees. If it’s 40 degrees, then they adapt to that
temperature too! That’s different than being warm-blooded. Humans are warm-blooded. That means
that no matter what the temperature is outside, inside our bodies, the temperature
always stays the same, or 98.6 degrees. The only time it changes is when we are sick
and our temperature goes a little up or down. Another trait of the amphibian group is that
part of their life is spent on land, while the other part is spent in the water. They can breathe
using gills while in the water, just like a fish! However, unlike a fish, they later grow lungs and
legs, which help them live on land. Kind of cool! Another thing that sets amphibians apart is their
skin. This is where the slimy part comes in. Eww! While slimy might seem gross to us, it’s a
very important part of being an amphibian. Adult amphibians must stay near
the water to keep their skin wet. An interesting fact about amphibians is they
don’t drink their water. It’s actually absorbed through their skin! And did you know that frogs
can’t live in saltwater? The salt could possibly poison their insides as well as dehydrate
them since their skin needs to stay moist. All amphibians come from soft eggs that kind of
look like jelly. They go through a process called metamorphosis, which basically means they morph
or change into something else as time passes. When amphibians are newly born, they look like
fish with tails and gills. After a little while, they become tadpoles with two legs. The
tadpoles change again and develop two more legs as well as a tail. They look like a small
version of a frog, only with a tail! Finally, they become a full grown frog. Though all
amphibians have common characteristics, frogs and toads lose their
tails when they reach adulthood, while salamanders and newts keep their tails
when they become adults. Salamanders and newts look like lizards without scales. A cool fact
about newts and salamander is if they lose a limb, it will regenerate, which means it will
grow back! Now that’s pretty amazing. Another species of amphibian is called a
caecilian. Caecilians do not have arms or legs. They use their pointed noses and strong skulls
to burrow through mud. They look like worms or snakes. Maybe even a bit scarier, especially
since some of them grow to be over four feet long. Amphibians live in a wide range of habitats.
They can be found in or around streams, forests, meadows, ponds, lakes, swamps,
and other damp or wet areas. Their diet includes spiders, beetles, and
worms. Some frogs have a long, sticky tongue they flick to catch flies, moths, and other
insects. Amphibians come in all sizes too. The goliath frog can grow to be 15 inches long
and weigh eight pounds. That’s as big as my cat! While the smallest frog in the world
is only about a third of an inch long. That’s about as wide as your pinky finger. On
the other hand, the Chinese giant salamander can be six feet long and weigh 140 pounds!
Can you imagine seeing that in a swamp? Amphibians are interesting animals with some neat
facts. Did you know that some salamanders keep their gills, and that frogs swallow their
food whole? There’s much more that can be learned about these fascinating animals so don’t
stop here. The next time you are near some water, see if you can spot some of these
incredible creatures we call amphibians. Thanks for following Clarendon Learning. Be
sure to subscribe. For more free resources, check us out at clarendonlearning.org. |
664 | https://brainly.com/topic/maths/concyclic-points | Concyclic Points Unveiled: Exploring Geometric Configurations
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Explanation
Verified AI Content
Concyclic Points in Geometry
Introduction
In the study of geometry, we often encounter various geometric figures such as points, lines, and circles. One interesting concept involving circles is that of concyclic points. Concyclic points are points that lie on the same circle. Understanding concyclic points is important in solving problems related to circles and their properties. In this topic, we will define concyclic points, explore their properties, and learn how to identify them in different scenarios.
Definition of Concyclic Points
Concyclic points are a set of points that all lie on the circumference of the same circle. In other words, if four or more points lie on a common circle, they are said to be concyclic.
Identifying Concyclic Points
In order to determine whether a set of points are concyclic, we can use the following property:
Theorem: If the opposite angles in a quadrilateral are supplementary, then the quadrilateral is inscribed in a circle.
Based on this theorem, we can conclude that if we have a quadrilateral (four-sided figure) with opposite angles that add up to 180 degrees, the points that define the vertices of the quadrilateral are concyclic.
Let's illustrate this property with an example:
Example 1
Consider a quadrilateral ABCD with angles A, B, C, and D. If we know that angle A and angle C are supplementary (angle A + angle C = 180 degrees), then the points A, B, C, and D are concyclic. This means that they lie on the circumference of a circle.
To verify this property, we can follow these steps:
Draw the quadrilateral ABCD on a piece of paper or using geometry software.
Measure angles A and C using a protractor or the measuring tool in the software.
Add the measurement of angles A and C. If they sum up to 180 degrees, the points A, B, C, and D are concyclic.
It is important to note that this property is specific to quadrilaterals. For more than four points, there are additional methods and properties to consider.
Other Properties of Concyclic Points
Angles formed by concyclic points: On a circle, if two chords (line segments connecting two points on a circle) intersect, the angles that they form are related to the angles formed by the arcs (sections of the circumference of the circle). Specifically, the angles formed by the chords are half the measure of the arcs they intercept. This property is known as the inscribed angle theorem.
Example: Let's say we have a circle with points A, B, C, and D that are concyclic. Chords AC and BD intersect at point X. The angle AXB is equal to half the measure of arc AC minus the measure of arc BD.
In the diagram above, arc AC has a measure of m, and arc BD has a measure of n. The angle AXB is equal to 1/2 (m - n).
This property is useful in various geometric proofs and calculations involving circles.
Cyclic quadrilaterals: A cyclic quadrilateral is a special type of quadrilateral in which all four vertices lie on a common circle. Cyclic quadrilaterals have several interesting properties, including:
The opposite angles of a cyclic quadrilateral are supplementary, meaning they add up to 180 degrees. This property can be proved using the inscribed angle theorem mentioned earlier.
The sum of the opposite angles of a cyclic quadrilateral is constant, regardless of the configuration of the quadrilateral. This property is known as the converse of the inscribed angle theorem.
Cyclic quadrilaterals are frequently encountered in geometric problems, and understanding their properties can help in solving related questions.
Common Mistakes to Avoid
Assuming concyclic points without justification: It is important to remember that the property of concyclic points must be proven or stated explicitly. Simply assuming that points lie on a circle without proper justification can lead to incorrect solutions.
Not considering the possibility of concyclic points: In some geometry problems, it is necessary to consider the possibility of concyclic points, especially when dealing with angles or relationships involving circles. Ignoring this possibility can cause mistakes in the analysis and solution of the problem.
Recap
In this topic, we have explored the concept of concyclic points in geometry. Concyclic points are a set of points that lie on the circumference of the same circle. We learned that if the opposite angles of a quadrilateral are supplementary, then the points defining the vertices of the quadrilateral are concyclic.
We also discussed additional properties of concyclic points, including the inscribed angle theorem and the properties of cyclic quadrilaterals.
Understanding concyclic points is essential in solving problems related to circles and their properties. By recognizing and applying the properties of concyclic points, we can make accurate deductions and solve geometry problems more effectively.
Hi there! I’m here to help you learn about this topic. What would you like to know?
Questions related to Concyclic Points
hailiechacon8410 Question: Points S, O, M, and I are concyclic such that arc SO=arc IM. If the chord SM=IO are intersected at the points K. Prove that area of triangle SOK=area of triangle IMK and SM=IO
Please find the file attached below (figure for the question)
To prove that area of the two triangles is equal we must know the formula for the area of a triangle and concept of concyclic points as given below:
Area of a Triangle:
Formula to find the area of a triangle is
[tex]A=\frac{1}{2}(Base)(Height)[/tex]
Base can be any side of a triangle but height must be the side perpendicular to the base
Concyclic points:
Points which lie on the same circle having the same distance from the center of the circle.
Given Data:
Arc SO= Arc IM
chord SM= chord IO
Proof of Area of Triangle SOK=Area of Triangle IMK:
As the given points are concyclic points, so
SK=OK
KM=IM
Any of the above point is radius of the circle.
Thus,
SK=OK=KM=IM
Are of Triangle SOK:
[tex]Area\ of\ the\ Triangle\ SOK = \frac{1}{2}(SK)(OK)[/tex]
where, SK is the Base for triangle SOK and OK is the Height for the triangle SOK
Area of Triangle IMK:
[tex]Area\ of\ Triangle\ IMK= \frac{1}{2}(KM)(IK) \\[/tex]
Where, KM is Base of the triangle IMK and IK is Height of the triangle IMK
As we know
SK=OK=KM=IM
We can say directly that area of both the triangles is same
[tex]Area\ of\ the\ Triangle\ SOK = \frac{1}{2}(SK)(OK)[/tex] =[tex]Area\ of\ Triangle\ IMK= \frac{1}{2}(KM)(IK) \[/tex]
OR
[tex]\frac{1}{2}(SK)(SK)[/tex][tex]=\frac{1}{2}(SK)(SK)[/tex]
thus proved
Proof of SM=IO:
As points are concyclic so they all have same distance from the center of the circle
i.e.SK=OK=KM=IM
thus SM=IO
loopysoop7376 Question: Lesson 4: Let ABC be a triangle. The three altitudes AD, BE, CF intersect at H. Let I, J, K be the midpoints of HA, HB, HC respectively. Let M, N, P be the midpoints of AB, BC, and CA respectively. Prove that: a) Four points M, P, K, J are on the same circle. b) Six points M, P, K, J, I, N are on the same circle. c) Nine points M, P, K, J, I, N, D, E, F are on the same circle. Option 1: Prove that the statement (a) is false. Option 2: Prove that the statement (b) is false. Option 3: Prove that the statement (c) is false. Option 4: Prove that all the statements (a), (b), and (c) are true.### Final answer:
The question involves validating geometric statements about points lying on the same circle within a given triangle using principles of cyclic quadrilaterals and concyclic points. This can be done using measurement in actual diagrams or proving mathematically with coordinate geometry. Disproving those statements would require identifying a discrepancy within these properties.
Explanation:
This type of problem involves the application of geometric principles and properties related to the circles, triangles and midpoints. To approach this problem, you would use the property of cyclic quadrilaterals and concyclic points for statement (a) and statement (b).
A cyclic quadrilateral or bi-tangential quadrilateral is a quadrilateral for which a circle called the circumcircle can be drawn that will touch all four vertices. Similarly, six or more points are said to be concyclic, if and only if they lie on the same circle, which relates to statement (c).
To validate the statements, you can apply these principles and construct the triangles and circles as described, measure the needed lengths or angles and check whether they hold the properties of the corresponding figures. This can also be proved using coordinate geometry or even complex numbers in some cases.
You would need to apply similar logic to invalidate the statements provided in the options. But to do this without having concrete numbers or the actual graphs to work with is difficult and beyond the scope of a text-based solution.
Learn more about Geometry here:
SPJ11
HeidiMercado1221 Question: Show that the points (3,-2), (1,0),(-1,-2) and(1,-4) are concyclic.To prove four points are concyclic, one can calculate the circumcenter of three points and then verify if this center is equidistant to the fourth point. However, this involves a profound understanding of geometry and a significant amount of algebraic calculation.
In mathematics, four points are said to be concyclic (situated on the same circle) if the distance from the center of the circle to each of the pointsis the same. To prove this, we can use the concept of the Circumcenter - the point where the perpendicular bisectors of a triangle intersect. The circumcenter is equidistant to each of the three vertices of the triangle.
Let's take three points first: (3,-2), (1,0), and (1,-4). Following the formula for the circumcenter, we can find the coordinates of the circumcenter. We can then calculate distance from circumcenter to all four points using the formula for distance between two points in a plane. If all distances are equal, then points are concyclic.
However, the method mentioned above requires a thorough understanding of geometry and algebraic calculations. In many cases, solving this problem with coordinatescan be easier using matrices and determinants, but that's typically part of a more advanced mathematics curriculum.
Learn more about ConcyclicPoints here:
SPJ11
fpmccool6014 Question: show that the points (5 5) (6 4) (-2 4) and (7 1) are concyclic find its equation and centre and radius### Final answer:
To prove the pointsare concyclic, we can use the circumcenter formula to find the center of the circle they would lie on. Once we have the center, we can calculate the radius and form the equation of the circle
Explanation:
The problem asks us to show that the points (5, 5), (6, 4), (-2, 4), and (7, 1) are concyclic and to find its equation, the center, and the radius.
In mathematics, points are considered concyclicif they lie on the same circle. To solve this, we can use the circumcenter formula:
If we denote the coordinates (X,Y) of the circumcenter of the triangle ABC, where A(x1,y1), B(x2,y2), C(x3,y3), then
X=(x1sin2A+x2sin2B+x3sin2C)/(sin2A+sin2B+sin2C) and Y=(y1sin2A+y2sin2B+y3sin2C)/(sin2A+sin2B+sin2C).
Firstly, we calculate sin2A, sin2B, and sin2C using the formula sin(2A)=2sinAcosA. Then, we substitute these values in the formulas for X and Y to find the coordinates of the circumcenter. Considering the circle's equation (x-a)^2+(y-b)^2=r^2, where (a, b) are the coordinates of the center, and r is the radius, it gives us the equation of the circle. Here, r can be calculated as distance between the circumcenter (a, b) and any concyclic point.
Learn more about Concyclic Points here:
SPJ11
Samdeck8778 Question: from a point p two tangents are drwan to the elllipse x²/a² + y²/b² = 1 if theses tangents intersect the coordinate axes at concyclic points , then the locus of the point p is (a>b) a. x²+y² = a²+b² b. x²+y² = a²-b² c. x²-y² = a²-b² d. x²=y²### Final answer:
The locus of the point P from which two tangents are drawn to the ellipse x²/a² + y²/b² = 1, intersecting the coordinate axes at concyclic points, is given by x² + y² = a² + b².
Explanation:
The locus of the point P from which two tangents are drawn to the ellipse x²/a² + y²/b² = 1, intersecting the coordinate axes at concyclic points, is given by x² + y² = a² + b² (option a).
To understand this, we need to consider the properties of an ellipse. An ellipse is a closed curve such that the sum of the distances from any point on the curve to the two foci is constant.
In this case, the two tangents are drawn from point P to the ellipse. As these tangents intersect the coordinate axes at concyclic points, it means the distances from point P to the foci of the ellipse are equal. Therefore, the sum of the distances from point P to the foci is constant and equal to a² + b².
sarahaziz7535 Question: There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least three points of these points is (a) 116 (b) 120 (c) 117 (d) none of these### Final answer:
A student asked about the number of different circles that can be drawn with 10 points where no 3 are collinear and 4 are concyclic. The calculation showed 21 different circles can be formed, which did not match any of the provided options, making the correct answer '(d) none of these'.
Explanation:
The question revolves around the number of different circles that can be formed in a plane with 10 points of which no three points are collinear and 4 points are concyclic.
To find a circle, we need to choose at least 3 non-collinear points. Since 4 points are concyclic, they form a unique circle among themselves. For the remaining 6 points, we can form circles by selecting them in groups of 3.
To calculate the total number of circles: We can select 3 points from the 6 non-concyclic points in C(6,3) ways. This gives us:
C(6,3) = 6! / (3!(6-3)!) = 20
The unique circle from the concyclic points is 1. So, the total number of different circles is 20+1=21 circles. However, none of the provided options (a) 116, (b) 120, or (c) 117 matches the calculated number of circles. Therefore, the answer is (d) none of these.
wrharris3428 Question: In triangle ABC, where BB₁ is the angle bisector, I is the incenter, and the perpendicular bisector of segment AC intersects the circumcircle at points D and E, prove that the points B, D, E, and F (where AB₁ = CF) are concyclic.### Final answer:
The problem inquires about concyclic points in a geometric configuration involving a triangle, its incenter, an angle bisector, and the circumcircle. A full solution requires additional details about the geometry, which are not provided.
Explanation:
The question poses a geometric proof regarding a triangle ABC and requires demonstrating that certain points are concyclic, meaning that they all lie on the same circle.
The presence of an incenter, angle bisector, and a perpendicular bisector interacting with the triangle's circumcircle introduces several classical theorems like the Inscribed Angle Theorem and properties of perpendicular bisectors and angle bisectors in a triangle.
To approach this proof, you'd typically start by affirming known properties and theorems of circle geometry before progressing to show that the given points indeed satisfy the criteria of lying on the same circle.
You would need to leverage relationships between angles subtended by the same arc and use congruence and similarity of triangles to aid in the proof. Without a specific figure provided, a general method is stated rather than a detailed, step-by-step proof.
Unfortunately, the information provided in the question is incomplete and does not directly relate to the solution of the problem; hence a detailed step-by-step solution cannot be provided confidently based on what is given.
For a complete proof, the full geometry of the triangle and the relative positioning of the points in question would be necessary.
Jsmooth21781 Question: If the lines x−2y+3=0 and 3x+ky+7=0 cut the coordinate axes in concyclic points, thenThe equations x - 2y + 3 = 0 and 3x + ky + 7 = 0 intersect the coordinate axes at concyclic points. Satisfying tex[/tex], the derived value is tex[/tex]. From this calculation we rightfully conclude that option A is the correct choice.
The given equations x - 2y + 3 = 0 and 3x + ky + 7 = 0 represent lines intersecting the coordinate axes at four concyclic points.
To satisfy this condition, tex[/tex].
Substituting the coefficients, tex \cdot k)[/tex],
This gives us: 3 = -2k.
Solving for k:
tex[/tex].
Thus, the value of k satisfying the condition for concyclic points is tex[/tex], affirming the relationship tex[/tex].
Therefore, the correct option is represented in option A.
Question:
If the lines x−2y+3=0 and 3x+ky+7=0 cut the coordinate axes in concyclic points, then k?
A. -3/2
B. 1/2
C. 3/2
D. 14
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665 | https://rk.md/2018/equal-pressure-point-in-lung-physiology/ | Equal Pressure Point In Lung Physiology | RK.MD
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Equal Pressure Point In Lung Physiology
March 22, 2018
14
Normally, the intrapleural pressure (P ip), or the pressure in the space between the lung’s visceral and parietal pleural layers, is slightly negative relative to the atmosphere (P atm). This is due to the chest wall and lungs recoiling away from each other. During a forced expiration, expiratory muscles can increase the P ip (as noted in the diagram). This pressure is transmitted to the alveolus, increasing pressure within the airway. I’ve termed this bronchial pressure (P br). Flow occurs from regions of high pressure to those of lower pressure, so since P br> P atm, air is expired.
As air exits the lung (alveoli → bronchioles → bronchi → trachea), it encounters airwayresistance causing P br to decrease. If the patient has obstructive physiology as in chronic obstructive pulmonary disease (COPD), expired air encounters more resistance leading to a precipitous drop in pressure (+20 down to +10 cm H 2 O, in this example).
In either case, P br stents open the airway, and P ip tries to collapse it. The point along the airway where P br = P ip is called theequal pressure point (EPP). Normally during forced expiration, the EPP occurs proximal in the tracheobronchial tree where cartilage reinforces the airway and prevents collapse; however, in patients with COPD, the EPP is much more distal, causing airway collapse near the alveolus (parts of the airway NOT reinforced with cartilage) leading to hypoventilation, hypoxemia, air trapping, etc.
This explains why patients with COPD habitually exhale through tightly pursed lips. By limiting the rate at which air is exhaled, the patient can maintain a higher airway pressure (P br) to help stent the airway where it wants to collapse.
Be sure to check out my other post on the basic physiology of breathing, and drop me a comment below with questions! :-)
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14 COMMENTS
NDecember 17, 2022 At 22:33What happens to EPP is positive pressure ventilation? Reply
EmilyJune 8, 2022 At 14:58Hi! Thank you for this explanation! It’s fantastic!! I have a question. Why is there a sharp drop of pressure in exhalation and why does this correlate with more resistance in the airway? Thank you! Reply
RishiJune 8, 2022 At 15:33This drop in pressure is termed “friction loss.” Read more in this free article. Reply
MKJanuary 18, 2021 At 19:55I’ve watched multiple videos and I couldn’t get it for the life of me, but now it all clicks. THANK YOU SO MUCH! :) Reply
RishiJanuary 19, 2021 At 09:27I’m so glad you found it helpful! :-) Reply
FrankDecember 10, 2020 At 06:29This is the best explanation of EPP I’ve read! Thank you! Reply
RishiDecember 10, 2020 At 15:14Thank you so much, Frank! Reply
JennieSeptember 14, 2018 At 14:28I’ve read about it a lot and this was the first time I understood the concept of equal pressure points. Thanks! Reply
RishiSeptember 14, 2018 At 16:30So glad you found this post helpful! :-) Reply
OnishmaOctober 1, 2020 At 17:46Rishi me too. Thanks for this. I am a RT student. Do you think your notes will be helpful through my program? I would like to access it, its put together better than class haha Reply
Matthew D BillApril 5, 2018 At 11:27Hey Dr. Rishi,
So I believe I understand this correctly, but let me know if I am wrong. I have the basic concepts down the difficulty comes when you talk about the Pbr working against the Pip to keep the distal airways open upon forced expiration. Is the Pbr pressure keeping the airway open because as PiP increases the air is forced from the alveolus and that is the air forcing itself forward as well as against the walls of the distal airways?
Then for the COPD patient you bring up, does the pursed lip breathing increase that Pbr by providing resistance to the outflow of air and increase the pressure in the system that way?
Thank you for your articles! Always a very interesting read and very easy to grasp!
Matt Reply
RishiApril 5, 2018 At 22:20Great questions, Matthew! I’ll start with your second one – yes, exactly. Pursing one’s lips maintains more air pressure within the airways (higher P br) to counteract the intrapleural pressure (P ip) that wants to collapse the airway.
Now to address your first point, P br is higher than P ip during forced expiration at the alveolar level, but the farther air traverses the airway (ie, moving from the alveolus to the mouth), the more pressure drop off occurs. At some point, the P br (pressure within the airway) is less than that outside the airway (P ip) leading to airway collapse. Reply
BiancaFebruary 5, 2019 At 03:04Hi Dr Rishi!
Just wondering why the obstruction in the small airways causes the pressure to drop off more quickly in a COPD patient?
Cheers! Reply
RishiFebruary 5, 2019 At 05:05Patients with COPD have difficult breathing air OUT of their lungs, so I like to think that the airway pressure is higher in their distal airways (alveoli, respiratory bronchioles, etc.) compared to their more proximal airways (trachea, mainstem bronchi). With the dynamic obstruction, there’s some degree of airflow acceleration between the distal airways and proximal airways causing a larger pressure drop off (Bernoulli’s principle) to occur in the area with higher velocities (proximal airways). At least this is how I think about it… let me know if you find a better explanation! :-) Reply
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666 | https://www.youtube.com/watch?v=dJoRqUduh4c | Digital Root.... Casting out 9's....By Jaga Sir
Wisdom Corner
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hello students in this video we are going to learn about digital root what is digital root and casting of lines and how can we apply this method uh in several applications like uranus addition subtraction multiplication and division squares also means simply if answer it is given to us multiple answer means mcq type multiple choice questions it is given to us then directly we can find uh the answer by using this method okay so first what is digital node a digital root means it is a one digit number one digit number or any number this is hence any number that is that can be converted to one digit number that is dr of that number digital root and this digital root can be written as dr digital root can be written as dr i remember how whatever the number may be always the digital root of the number it is one digit okay that means either zero to nine it can be zero also okay so there come to uh or simply will go how to find the digital root of a number digital root of a number supposing number it is given to you 7 3 2 8 1 5 4 now you are going to find the digital root of this number first what you have to do you have to sum up all the digit 7 plus 3 plus 2 plus 8 plus 1 plus 5 plus 4 now 7 3 10 12 20 21 25 29 so now it is 29 again we got a two digit number for two plus nine you need to add that is eleven again one plus one your player that is two so this is very lengthy process this is a very lengthy procedure because how many names you are going to add this so for this uh in order to solve this one we have to done casting out nines cutting out nines means either nine or if the sum of digit is nine we need to remove from that number that means this number again i'm writing 3 7 3 2 8 1 5 4. now what i have to do i have to take such numbers two number such two one digit number or two digit number sorry one is number one one uh we have to take such digits that it's sum but to be nine see now this seven plus two this is nine eight plus one this is nine five plus four this is nine yes or not yes now see so we can take out seven two eight one and five four now it is our digital voltage three so we have done in seven three ten twelve 20 21 30 it is sorry it is 30 that's why 30 that means 3 plus 0 3 digit would be three so you don't need to do this one you have to follow this method second one cutting out nines means you have to take such a pair any number like let's say any other number suppose seven 2 8 4 0 1 5 3 5 now we are going to find the digital root of this number in order to find dr in order to find dr of any number we we need to use this method casting out nines so what will be the tr of this number what will be the dr of this number so 279 this is 2 plus 7 we got now 8 plus 1 also cancel 8 plus 1 now 4 plus 5 also 4 plus 5 now this 2 remain so how much it will be 3 plus 5 it will remain as because it is not 9 so digital root is 8. digital root is eight so digital root is such a number which can be up to zero to eight zero one two three four five six seven eight when it comes to nine we need to cancel it is always a one digit number okay remember previously i think i told to you uh zero to nine but not nine okay so digital root it can be up to eight so like this if any number it is given to you you can just take out the nines okay this is what cutting out nines now one more number it is given to us suppose 9 3 4 6 2 3 5 so if this number is given to us then how to find this here 9 then 3 and 6 simply okay 4 6 6 3 9 four plus two plus three four two six six three nine so i can take out these three now dr of those number dr is how much five there of this number is 5 okay now why we are learning this dr means what is the use of this digital root suppose any question it is given to you any sum it is given to you the question is like suppose three one two four plus five three eight two minus two six eight nine and it act and option it is given a b c and d now obviously this will contain numbers so if we are going to find dr of this number this is this will be a one digit number again dr of this number again one digit number minus dr of this number again one digit number so see how easily we are finding dr so one digit one digit and one digit and simply we need to find the dr of this options which option is matched to this one then there is a paragraph suppose the option b is means it is made to this one so question one so option b is correct so in this case instead of doing this much addition or subtraction you simply find the dr and put that means it will be easily immense you'll be able to do or you will be able to find the correct answer within seconds you don't need to go to you don't need to do the sum okay so this is how to check that one also uh we learn in this video only how to check the answers okay so first i'm asking some question to you you have to find the digital root of that number and you have to write in the comment section okay so take out the screenshot so that it will be easy for you to understand be now uh suppose the numbers are a 2 6 8 9 7 3 2 5 3 4 2 6 8 0 1 7 3 2 four eight six five nine so this is the three numbers it's here you have to find and you have to write in comment section okay now let us move how to use this digital root in case of the use of digital root in case of answer verification that is in different operation like edison first window edition this is for answer verification remember answer verification like nowadays it is coming in competitiveness for answer it is given to you out of that you have to find the correct answer that is within seconds within 30 seconds or 20 seconds how can you do okay so edition addition means what means if any question is given to you like 2 4 6 8 plus three seven eight four now if we add then what it will be twelve six eight fourteen fifteen uh four seven eleven twelve and two three five six now we are going we we are going to find the dr of each as because we will just know so see a dr of this number is nothing will cancel okay so we cannot means cut out nice here no lines are possible so what you have to do 2 4 6 6 6 12 12 20 20 is coming so dr is what two plus zero two of this number is an individual number you have to find okay thereafter here they are also not possible three four seven so there is what three seven ten eighteen twenty two the sum is twenty two so dr is two plus two four if that means we cannot cut out the nines then add simply all the digits means what i have done how i got 22 3 plus 7 plus 8 plus 4 okay so now 22 can be written as 2 plus 2 digit root that is 4 okay and here also i got how i got 22 plus 4 plus 6 plus 8 there after again 2 plus 0 that is 2. so digital of the digital root of these two numbers that is 2 and 4 now some of that here sum of dr is what 2 plus 4 that is 6 okay here we are having answer that is we can cut out 9's yes 5 plus 2 plus 2 possible 5 plus 2 plus 2 that is 9 so take out this one now the dr is 6 there the answer is 6 doesn't mean this is your correct answer okay if other answers it is given to you suppose and it means in our exam it will not come directly it will come the question you have to find out the answer and simply you have to match okay but in competitive exam means option will be given to you and there you need to check the dr and which dr will make that answer is correct okay so in our exam you have to prove this okay so some of the deal is six and answer there is also 6 so what you have to do first you have to find out the dr of the numbers there after you have to find out that sum is because addition we are doing and you have to find out the drf the answer then you have to check yes it is missing or not if it will not match then your answer is incorrect okay one more number will do suppose it is given 2 3 6 9 plus 5 9 8 1 is equal to so question is given to us this is the question so what do you have to do 1 and 10 6 8 14 15 3 9 12 13 two five seven eight okay this is the answer this is the answer point right hand side so what you have to do nine i can take out three and six also i can take out so the r of this number is how much two dr of this number is 2 now 9 i can take out 8 and 2 i can take out here this number is 5 now the sum of the sum of their sum of d r are both the number uh digital root of both the number that is two plus five seven now we have to prove that our answer also having digital root seven so here we cannot take out any nancy nine is not possible so simply what you have to do eight plus three plus five that is how much 8 3 11 11 5 16. now dr is what there is one digit number equal to you 16 we cannot write so 1 plus 6 each individual digit you have to add it now you can see that your answer is correct digital root of the answer is equal to the the sum of the digital root of both these number okay so in this way you can verify take out this insert thereafter i discussed subtraction go to subtraction see this uh this can be a long video as you the subtraction multiplication of everything division and square everything i will explain subtraction so suppose question is given to you like this 2 3 7 5 plus 3 8 4 9 minus 5 9 4 5 is equal to this is the question now what we have to do 5 9 14 minus 5 that is 9 now 7 4 11 11 minus 4 against 3 8 11 11 minus 9 2 2 3 5 5 minus 5 0 okay so our answer is 0 to 7 9 okay okay now what you have to do you have to find the dr of all this number dr all the number here we are having three numbers so dr this this number is what can we cut our mind yes dr of this number is eight and here also we can cut out this is nine remaining digits we cannot cut out so what do we have to do three plus eight plus four three eight eleven four fifteen so there is how much six there is six now this one can custom nine yes one nine possible another nine also possible and here dr is how much five okay and here here is how much seven two we can cut out oh here dr is zero here we are having dr zero this is our answer okay this is our answer now what you have to do the first two numbers we have to add so eight plus six this one these two you require minus five eight plus six how much fourteen fourteen minus five how much nine now what will be that here nine means you need to cancel that means 0 whenever the sum is coming 9 then that is 0 now see our dear they are also 0 so that is the correct answer see what i have done 8 plus 6 these two numbers i don't know so there there we have to add and this number subtracted so that it means this number there we need to subtract so 8 plus 6 14 14 minus 5 9 so 9 means the dr will be 0 that's because for one digit only 9 dr is 0 for one digit dr of 9 is 0 okay and if it is eight there is eight but nine is zero is because cutting out nine means either nine or if the sum of the digit is nine that or everything we have to remove here so when one name is zero and here we get that our answer is correct now we'll go to multiplication multiplication multiplication if two digit number it is given suppose when t7 into 48 it is given to you let me recall you about crisscross multiplication so first what you have to do 24 into 27 27 into 48 means 2 into 4 first 2 into 4 2 into 4 then 2 into 8 4 into 7 2 into 8 plus 4 into 7 and then 7 into 8 so this is 8 2 into 8 16 16 plus 28 how much 16 plus 28 that is 44 and here it is 56 so answer is how much six four plus five nine eight plus four twenty hundred ninety multiplication i have made video for classics it is in my channel only crystals multiplication uh go to my channel and see there okay crisscross multiplication i have explained now what we have to do we have to find the dr of this number we have to find the dr of this number here of this number of this number is what is because cancer and the out of this number is 4 8 means 12 12 means 4 plus 8 12 12 means 1 plus 2 that is 3 there is 3 now the operation is what multiplication so product of the dr see the sign always plays an important role product of dr means 0 into 3 this number here and you have to multiply with this number here that is 0 that means we should have means a number of digital root 0 so i can take out 9 and one two three three plus six oh this this also i can take out one plus two three three six nine that means here also i am having di as 0 that means my answer is correct 0 and 0 so always remember if two numbers given either that is a plus b equals c or a minus b equals c or a into b is equal to c so on which what you have to do that this is possible so dr [Music] of a plus dr of b is equals to dr of c and because we are adding a and we know to get c so dr of a will be added with dr b to where dr of c plus means plus okay then dr a minus d r of [Music] b is equals to d r of c here minus minus n we are having so minus you have to put means you need to subtract the dr of b from a in order to get dr of c and here product so dr of a e 2 is because multiplication of d r of b d r of [Music] so this is all about addition subtraction and multiplication now we will go to division take out the screenshots so that it will be easy for you to understand and remember if you are having doubt in this class method i have explained that is available two and three digits crisscross multiplication i have explained go and see very very easy to solve now division so division here how many means means what type of quantity we are having there i like dividend then divisor now what is the formula for this dividend is equals divisor into quotient plus a reminder is not it this is the formula see why i have written i will tell you okay now if any number it is given to us like suppose 225 divided by 25 okay so what we can do so what we said we we are going to find question is nine remainder is zero cosine d 9 remainder is 0 and here this is dividend and this is division okay division here is first all here we are going to find here 0 dr 0 here is 0 and here di7 and here here is also 0. and here dl is also 0. so you you have to use this formula of the dividend that means dr of the dividend in order [Music] sorry [Music] so now the r of the dividend is zero the r of the divisor is seven d r of the cosine this is cosine and this is remainder [Music] and here we are going to put this value dr of the cosine is zero but actually i need to write the division no problem if you alter also there are the question that is zero into there of the division that is nine and they have the remainder plus zero zero into nine that is zero zero plus zero again it is zero so the dr of all these is zero and the r of the dividend is also zero so if image then your answer is correct means whatever the question and the remainder here obtained that is correct so take the screenshot and write the formula properly one more number i will explain regarding means this concept okay so remember dr mean city digital root of the dividend is to what a digital root of the cosine theta root of the quotient into digital root of the individual plus digital root of the remainder okay now if it is given if one question is given to us 743 divided by 21 now what is the quotient and what is the reminder we need to get the answers so 21 how much time i can take three times 63 11 and 113 so how much time i can take 5 10 105 130 means reminder will be eight if you do division thirty five percent you'll get a reminder eight elevators now we need to check the dr of each number so here what it is is because we are having digits and we cannot cut off lines we have to add seven plus four plus three that is how much 14 14 means dr is five dear of the dividend dr of the dividend is five and this is this is eight dr is eight dr of the question is eight and here clearly it is eight the dr of remainder is eight now we are going to use this so this five we need to get from where that means 8 so 3 24 plus 8 that is 32 32 means what will be the dear three plus two that is five only so this is the diagonal dividend dr of dividend okay and this is the r of the cosine of uh the divisor sorry i have entered the value of the dividend yeah the question and here are the remainder if i multiply higher 32 and here of 32 is what five this is equal to the digital root of your dividend okay so this is all about division now how to find this in case of squares okay so now we learn about square how to find and check your answer whatever you have evaluated for the square it is correct or not so square of the number means what in your math also here it is chapter square square root cube cube root square means if any number multiplied with itself okay so we can take 45 square literally yes 45 square which is very easy if the end digit is 5 okay so answer will be almost 2 0 to 5 as because you know if the end is it the end digit is 5 then what you have to do this one in class 7 4 4 will increase to 5 4 into 5 how much 20 and this for 5 square 25 okay if any number ends with 5 then it's square always ends with 25 okay remember that this one in previous class i talked to you also now you have to check a digital root of this number root of this number is 0 and 0 square is also 0 that means again the digital root is 0 here now digital load of this number 2 plus 2 plus 5 we can cut out 9 2 so now this image to this that means our answer is correct let us take another number another number say 62 square so what will be the 62 square in class 8 only we have read 25 plus 12 how much 37 and 12 square 144 that means 3 8 4 4 62 square okay now we are going to take the dr of this dr of this is 8 8 square is how much 64. again here is what 6 plus 4 10 6 plus 4 10 10 means here is 1 plus 0 1 anyway you have to if you cannot cut out nine make it single digit and in order to get that single digit in any means any number of times you can do addition see first i got the digital root of this number that is 8 then because you are finding the square 8 square i have to take that is 64 okay 64 means digital root when i am going to find then what 64 cannot be means no nines are there so six and four i added i got ten again two digit number so again i add one plus zero then i got one okay then i got one so digital root of your number is one and here also because you cannot cut out nine and you can see no nines are possible okay uh so three plus eight plus four plus four how much it will be three eleven 11 19 19 means what again one plus nine ten again two digit number coming so again one plus zero that is one so at last you have to take means if two digit number is coming then add the digits if two digit number is coming at the digits okay so now see your answer measures that means the 62 means your 62 square is what 3 8 double four okay so first you have to find the digital root there after you have to take the square if any means a nine is possible cut out the nines if not possible then add if you've got two digit number again add i got two digit number here one plus zero sorry one zero ten so again i add one plus zero then i got one here also 19 19 which is a two digit number so 19 cannot be the digital root so 1 plus 9 10 10 is a two digit number so 10 is also not the digital node here one plus zero that is one that is the digital root okay one more number we will discuss there after uh this will be the end of this video and i'll ask you some questions since that one you have to do your copy only okay so suppose it is 83 square base 100 you know base 100 83 is 17 less than 100 so 83 minus 17 how much it is minus 17 that is 66 and this is 17 square that is 289 so 2 you have to add 89 you have to keep so eight nine six eight six double eight nine okay so this is 83 square 56 yes this is 83 square now we have to check we have to check so this is 8 plus 3 11 11 that is not a digital good so 1 plus one what we have to do and two square we need to take as because see this side we are taking square root so square you need to take okay so digital root is how much four digital root is four here the digital root is one here the digital root is also one so here the digital root is four now nine can be taken out and thereafter six plus eight plus 8 at 16 6 22 and see here the digital root is 2 plus 2 and because 22 cannot be the digital root 2 plus 2 4 so that means 6 08 9 is your answer so this is the way means in your exam you need to do like this you mean the question will not come like multiple choice direct question will be given you have to find the answer and you have to check out the digital rules in both the cases it is same or not whether that is that is addition that is multiplication division subtraction or square whatever it is okay this one you have to keep in mind so this is the end of this video you know but you have to check out some numbers by using digital root so first number is 2 3 5 9 plus seven three eight two then six eight four three minus two five four nine director third number 683 divided by 24 and fourth number 53 square okay these numbers means you have to do by using all the methods i hope after watching this video you will be able to understand what is digital root and what is cutting of lines and how we can find the digital roots and how i we can check whether our answer is correct or not by using digital group so this is the end of this video thank you for watching |
667 | https://en.wikipedia.org/wiki/Partial_differential_equation | Jump to content
Partial differential equation
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From Wikipedia, the free encyclopedia
Type of differential equation
| | |
--- |
| | This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (March 2023) (Learn how and when to remove this message) |
| Differential equations |
| Scope |
| Fields | Natural sciences Engineering | | Astronomy Physics Chemistry Biology Geology | | Applied mathematics | | Continuum mechanics Chaos theory Dynamical systems | | Social sciences | | Economics Population dynamics | --- List of named differential equations |
| Classification |
| Types | | | Ordinary Partial Differential-algebraic Integro-differential Fractional Linear Non-linear | | By variable type | | Dependent and independent variables Autonomous Coupled / Decoupled Exact Homogeneous / Nonhomogeneous | | Features | | Order Operator Notation | |
| Relation to processes Difference (discrete analogue) Stochastic + Stochastic partial Delay |
| Solution |
| Existence and uniqueness Picard–Lindelöf theorem Peano existence theorem Carathéodory's existence theorem Cauchy–Kowalevski theorem |
| General topics Initial conditions Boundary values + Dirichlet + Neumann + Robin + Cauchy problem Wronskian Phase portrait Lyapunov / Asymptotic / Exponential stability Rate of convergence Series / Integral solutions Numerical integration Dirac delta function |
| Solution methods Inspection Method of characteristics Euler Exponential response formula Finite difference (Crank–Nicolson) Finite element + Infinite element Finite volume Galerkin + Petrov–Galerkin Green's function Integrating factor Integral transforms Perturbation theory Runge–Kutta Separation of variables Undetermined coefficients Variation of parameters |
| People |
| List Isaac Newton Gottfried Leibniz Jacob Bernoulli Leonhard Euler Joseph-Louis Lagrange Józef Maria Hoene-Wroński Joseph Fourier Augustin-Louis Cauchy George Green Carl David Tolmé Runge Martin Kutta Rudolf Lipschitz Ernst Lindelöf Émile Picard Phyllis Nicolson John Crank |
| v t e |
In mathematics, a partial differential equation (PDE) is an equation which involves a multivariable function and one or more of its partial derivatives.
The function is often thought of as an "unknown" that solves the equation, similar to how x is thought of as an unknown number solving, e.g., an algebraic equation like x2 − 3x + 2 = 0. However, it is usually impossible to write down explicit formulae for solutions of partial differential equations. There is correspondingly a vast amount of modern mathematical and scientific research on methods to numerically approximate solutions of certain partial differential equations using computers. Partial differential equations also occupy a large sector of pure mathematical research, in which the usual questions are, broadly speaking, on the identification of general qualitative features of solutions of various partial differential equations, such as existence, uniqueness, regularity and stability. Among the many open questions are the existence and smoothness of solutions to the Navier–Stokes equations, named as one of the Millennium Prize Problems in 2000.
Partial differential equations are ubiquitous in mathematically oriented scientific fields, such as physics and engineering. For instance, they are foundational in the modern scientific understanding of sound, heat, diffusion, electrostatics, electrodynamics, thermodynamics, fluid dynamics, elasticity, general relativity, and quantum mechanics (Schrödinger equation, Pauli equation etc.). They also arise from many purely mathematical considerations, such as differential geometry and the calculus of variations; among other notable applications, they are the fundamental tool in the proof of the Poincaré conjecture from geometric topology.
Partly due to this variety of sources, there is a wide spectrum of different types of partial differential equations, where the meaning of a solution depends on the context of the problem, and methods have been developed for dealing with many of the individual equations which arise. As such, it is usually acknowledged that there is no "universal theory" of partial differential equations, with specialist knowledge being somewhat divided between several essentially distinct subfields.
Ordinary differential equations can be viewed as a subclass of partial differential equations, corresponding to functions of a single variable. Stochastic partial differential equations and nonlocal equations are, as of 2020, particularly widely studied extensions of the "PDE" notion. More classical topics, on which there is still much active research, include elliptic and parabolic partial differential equations, fluid mechanics, Boltzmann equations, and dispersive partial differential equations.
Introduction
[edit]
A function u(x, y, z) of three variables is "harmonic" or "a solution of the Laplace equation" if it satisfies the condition
Such functions were widely studied in the 19th century due to their relevance for classical mechanics, for example the equilibrium temperature distribution of a homogeneous solid is a harmonic function. If explicitly given a function, it is usually a matter of straightforward computation to check whether or not it is harmonic. For instance
and
are both harmonic while
is not. It may be surprising that the two examples of harmonic functions are of such strikingly different form. This is a reflection of the fact that they are not, in any immediate way, special cases of a "general solution formula" of the Laplace equation. This is in striking contrast to the case of ordinary differential equations (ODEs) roughly similar to the Laplace equation, with the aim of many introductory textbooks being to find algorithms leading to general solution formulas. For the Laplace equation, as for a large number of partial differential equations, such solution formulas fail to exist.
The nature of this failure can be seen more concretely in the case of the following PDE: for a function v(x, y) of two variables, consider the equation
It can be directly checked that any function v of the form v(x, y) = f(x) + g(y), for any single-variable functions f and g whatsoever, will satisfy this condition. This is far beyond the choices available in ODE solution formulas, which typically allow the free choice of some numbers. In the study of PDEs, one generally has the free choice of functions.
The nature of this choice varies from PDE to PDE. To understand it for any given equation, existence and uniqueness theorems are usually important organizational principles. In many introductory textbooks, the role of existence and uniqueness theorems for ODE can be somewhat opaque; the existence half is usually unnecessary, since one can directly check any proposed solution formula, while the uniqueness half is often only present in the background in order to ensure that a proposed solution formula is as general as possible. By contrast, for PDE, existence and uniqueness theorems are often the only means by which one can navigate through the plethora of different solutions at hand. For this reason, they are also fundamental when carrying out a purely numerical simulation, as one must have an understanding of what data is to be prescribed by the user and what is to be left to the computer to calculate.
To discuss such existence and uniqueness theorems, it is necessary to be precise about the domain of the "unknown function". Otherwise, speaking only in terms such as "a function of two variables", it is impossible to meaningfully formulate the results. That is, the domain of the unknown function must be regarded as part of the structure of the PDE itself.
The following provides two classic examples of such existence and uniqueness theorems. Even though the two PDE in question are so similar, there is a striking difference in behavior: for the first PDE, one has the free prescription of a single function, while for the second PDE, one has the free prescription of two functions.
Let B denote the unit-radius disk around the origin in the plane. For any continuous function U on the unit circle, there is exactly one function u on B such that and whose restriction to the unit circle is given by U.
For any functions f and g on the real line R, there is exactly one function u on R × (−1, 1) such that and with u(x, 0) = f(x) and ∂u/∂y(x, 0) = g(x) for all values of x.
Even more phenomena are possible. For instance, the following PDE, arising naturally in the field of differential geometry, illustrates an example where there is a simple and completely explicit solution formula, but with the free choice of only three numbers and not even one function.
If u is a function on R2 with then there are numbers a, b, and c with u(x, y) = ax + by + c.
In contrast to the earlier examples, this PDE is nonlinear, owing to the square roots and the squares. A linear PDE is one such that, if it is homogeneous, the sum of any two solutions is also a solution, and any constant multiple of any solution is also a solution.
Definition
[edit]
A partial differential equation is an equation that involves an unknown function of variables and (some of) its partial derivatives. That is, for the unknown function
of variables belonging to the open subset of , the -order partial differential equation is defined as
where
and is the partial derivative operator.
Notation
[edit]
Main article: Notation for differentiation § Partial derivatives
When writing PDEs, it is common to denote partial derivatives using subscripts. For example:
In the general situation that u is a function of n variables, then ui denotes the first partial derivative relative to the i-th input, uij denotes the second partial derivative relative to the i-th and j-th inputs, and so on.
The Greek letter Δ denotes the Laplace operator; if u is a function of n variables, then
In the physics literature, the Laplace operator is often denoted by ∇2; in the mathematics literature, ∇2u may also denote the Hessian matrix of u.
Classification
[edit]
Linear and nonlinear equations
[edit]
A PDE is called linear if it is linear in the unknown and its derivatives. For example, for a function u of x and y, a second order linear PDE is of the form
where ai and f are functions of the independent variables x and y only. (Often the mixed-partial derivatives uxy and uyx will be equated, but this is not required for the discussion of linearity.)
If the ai are constants (independent of x and y) then the PDE is called linear with constant coefficients. If f is zero everywhere then the linear PDE is homogeneous, otherwise it is inhomogeneous. (This is separate from asymptotic homogenization, which studies the effects of high-frequency oscillations in the coefficients upon solutions to PDEs.)
Nearest to linear PDEs are semi-linear PDEs, where only the highest order derivatives appear as linear terms, with coefficients that are functions of the independent variables. The lower order derivatives and the unknown function may appear arbitrarily. For example, a general second order semi-linear PDE in two variables is
In a quasilinear PDE the highest order derivatives likewise appear only as linear terms, but with coefficients possibly functions of the unknown and lower-order derivatives:
Many of the fundamental PDEs in physics are quasilinear, such as the Einstein equations of general relativity and the Navier–Stokes equations describing fluid motion.
A PDE without any linearity properties is called fully nonlinear, and possesses nonlinearities on one or more of the highest-order derivatives. An example is the Monge–Ampère equation, which arises in differential geometry.
Second order equations
[edit]
The elliptic/parabolic/hyperbolic classification provides a guide to appropriate initial- and boundary conditions and to the smoothness of the solutions. Assuming uxy = uyx, the general linear second-order PDE in two independent variables has the form
where the coefficients A, B, C... may depend upon x and y. If A2 + B2 + C2 > 0 over a region of the xy-plane, the PDE is second-order in that region. This form is analogous to the equation for a conic section:
More precisely, replacing ∂x by X, and likewise for other variables (formally this is done by a Fourier transform), converts a constant-coefficient PDE into a polynomial of the same degree, with the terms of the highest degree (a homogeneous polynomial, here a quadratic form) being most significant for the classification.
Just as one classifies conic sections and quadratic forms into parabolic, hyperbolic, and elliptic based on the discriminant B2 − 4AC, the same can be done for a second-order PDE at a given point. However, the discriminant in a PDE is given by B2 − AC due to the convention of the xy term being 2B rather than B; formally, the discriminant (of the associated quadratic form) is (2B)2 − 4AC = 4(B2 − AC), with the factor of 4 dropped for simplicity.
B2 − AC < 0 (elliptic partial differential equation): Solutions of elliptic PDEs are as smooth as the coefficients allow, within the interior of the region where the equation and solutions are defined. For example, solutions of Laplace's equation are analytic within the domain where they are defined, but solutions may assume boundary values that are not smooth. The motion of a fluid at subsonic speeds can be approximated with elliptic PDEs, and the Euler–Tricomi equation is elliptic where x < 0. By change of variables, the equation can always be expressed in the form: where x and y correspond to changed variables. This justifies Laplace equation as an example of this type.
B2 − AC = 0 (parabolic partial differential equation): Equations that are parabolic at every point can be transformed into a form analogous to the heat equation by a change of independent variables. Solutions smooth out as the transformed time variable increases. The Euler–Tricomi equation has parabolic type on the line where x = 0. By change of variables, the equation can always be expressed in the form: where x correspond to changed variables. This justifies heat equation, which are of form , as an example of this type.
B2 − AC > 0 (hyperbolic partial differential equation): hyperbolic equations retain any discontinuities of functions or derivatives in the initial data. An example is the wave equation. The motion of a fluid at supersonic speeds can be approximated with hyperbolic PDEs, and the Euler–Tricomi equation is hyperbolic where x > 0. By change of variables, the equation can always be expressed in the form: where x and y correspond to changed variables. This justifies wave equation as an example of this type.
If there are n independent variables x1, x2 , …, xn, a general linear partial differential equation of second order has the form
The classification depends upon the signature of the eigenvalues of the coefficient matrix ai,j.
Elliptic: the eigenvalues are all positive or all negative.
Parabolic: the eigenvalues are all positive or all negative, except one that is zero.
Hyperbolic: there is only one negative eigenvalue and all the rest are positive, or there is only one positive eigenvalue and all the rest are negative.
Ultrahyperbolic: there is more than one positive eigenvalue and more than one negative eigenvalue, and there are no zero eigenvalues.
The theory of elliptic, parabolic, and hyperbolic equations have been studied for centuries, largely centered around or based upon the standard examples of the Laplace equation, the heat equation, and the wave equation.
However, the classification only depends on linearity of the second-order terms and is therefore applicable to semi- and quasilinear PDEs as well. The basic types also extend to hybrids such as the Euler–Tricomi equation; varying from elliptic to hyperbolic for different regions of the domain, as well as higher-order PDEs, but such knowledge is more specialized.
Systems of first-order equations and characteristic surfaces
[edit]
See also: First-order partial differential equation
The classification of partial differential equations can be extended to systems of first-order equations, where the unknown u is now a vector with m components, and the coefficient matrices Aν are m by m matrices for ν = 1, 2, …, n. The partial differential equation takes the form
where the coefficient matrices Aν and the vector B may depend upon x and u. If a hypersurface S is given in the implicit form
where φ has a non-zero gradient, then S is a characteristic surface for the operator L at a given point if the characteristic form vanishes:
The geometric interpretation of this condition is as follows: if data for u are prescribed on the surface S, then it may be possible to determine the normal derivative of u on S from the differential equation. If the data on S and the differential equation determine the normal derivative of u on S, then S is non-characteristic. If the data on S and the differential equation do not determine the normal derivative of u on S, then the surface is characteristic, and the differential equation restricts the data on S: the differential equation is internal to S.
A first-order system Lu = 0 is elliptic if no surface is characteristic for L: the values of u on S and the differential equation always determine the normal derivative of u on S.
A first-order system is hyperbolic at a point if there is a spacelike surface S with normal ξ at that point. This means that, given any non-trivial vector η orthogonal to ξ, and a scalar multiplier λ, the equation Q(λξ + η) = 0 has m real roots λ1, λ2, …, λm. The system is strictly hyperbolic if these roots are always distinct. The geometrical interpretation of this condition is as follows: the characteristic form Q(ζ) = 0 defines a cone (the normal cone) with homogeneous coordinates ζ. In the hyperbolic case, this cone has nm sheets, and the axis ζ = λξ runs inside these sheets: it does not intersect any of them. But when displaced from the origin by η, this axis intersects every sheet. In the elliptic case, the normal cone has no real sheets.
Analytical solutions
[edit]
Separation of variables
[edit]
Main article: Separable partial differential equation
Linear PDEs can be reduced to systems of ordinary differential equations by the important technique of separation of variables. This technique rests on a feature of solutions to differential equations: if one can find any solution that solves the equation and satisfies the boundary conditions, then it is the solution (this also applies to ODEs). We assume as an ansatz that the dependence of a solution on the parameters space and time can be written as a product of terms that each depend on a single parameter, and then see if this can be made to solve the problem.
In the method of separation of variables, one reduces a PDE to a PDE in fewer variables, which is an ordinary differential equation if in one variable – these are in turn easier to solve.
This is possible for simple PDEs, which are called separable partial differential equations, and the domain is generally a rectangle (a product of intervals). Separable PDEs correspond to diagonal matrices – thinking of "the value for fixed x" as a coordinate, each coordinate can be understood separately.
This generalizes to the method of characteristics, and is also used in integral transforms.
Method of characteristics
[edit]
Main article: Method of characteristics
The characteristic surface in n = 2-dimensional space is called a characteristic curve.
In special cases, one can find characteristic curves on which the first-order PDE reduces to an ODE – changing coordinates in the domain to straighten these curves allows separation of variables, and is called the method of characteristics.
More generally, applying the method to first-order PDEs in higher dimensions, one may find characteristic surfaces.
Integral transform
[edit]
An integral transform may transform the PDE to a simpler one, in particular, a separable PDE. This corresponds to diagonalizing an operator.
An important example of this is Fourier analysis, which diagonalizes the heat equation using the eigenbasis of sinusoidal waves.
If the domain is finite or periodic, an infinite sum of solutions such as a Fourier series is appropriate, but an integral of solutions such as a Fourier integral is generally required for infinite domains. The solution for a point source for the heat equation given above is an example of the use of a Fourier integral.
Change of variables
[edit]
Often a PDE can be reduced to a simpler form with a known solution by a suitable change of variables. For example, the Black–Scholes equation
is reducible to the heat equation
by the change of variables
Fundamental solution
[edit]
Main article: Fundamental solution
Inhomogeneous equations[clarification needed] can often be solved (for constant coefficient PDEs, always be solved) by finding the fundamental solution (the solution for a point source ), then taking the convolution with the boundary conditions to get the solution.
This is analogous in signal processing to understanding a filter by its impulse response.
Superposition principle
[edit]
Further information: Superposition principle
The superposition principle applies to any linear system, including linear systems of PDEs. A common visualization of this concept is the interaction of two waves in phase being combined to result in a greater amplitude, for example sin x + sin x = 2 sin x. The same principle can be observed in PDEs where the solutions may be real or complex and additive. If u1 and u2 are solutions of linear PDE in some function space R, then u = c1u1 + c2u2 with any constants c1 and c2 are also a solution of that PDE in the same function space.
Methods for non-linear equations
[edit]
See also: nonlinear partial differential equation
There are no generally applicable analytical methods to solve nonlinear PDEs. Still, existence and uniqueness results (such as the Cauchy–Kowalevski theorem) are often possible, as are proofs of important qualitative and quantitative properties of solutions (getting these results is a major part of analysis).
Nevertheless, some techniques can be used for several types of equations. The h-principle is the most powerful method to solve underdetermined equations. The Riquier–Janet theory is an effective method for obtaining information about many analytic overdetermined systems.
The method of characteristics can be used in some very special cases to solve nonlinear partial differential equations.
In some cases, a PDE can be solved via perturbation analysis in which the solution is considered to be a correction to an equation with a known solution. Alternatives are numerical analysis techniques from simple finite difference schemes to the more mature multigrid and finite element methods. Many interesting problems in science and engineering are solved in this way using computers, sometimes high performance supercomputers.
Lie group method
[edit]
From 1870 Sophus Lie's work put the theory of differential equations on a more satisfactory foundation. He showed that the integration theories of the older mathematicians can, by the introduction of what are now called Lie groups, be referred, to a common source; and that ordinary differential equations which admit the same infinitesimal transformations present comparable difficulties of integration. He also emphasized the subject of transformations of contact.
A general approach to solving PDEs uses the symmetry property of differential equations, the continuous infinitesimal transformations of solutions to solutions (Lie theory). Continuous group theory, Lie algebras and differential geometry are used to understand the structure of linear and nonlinear partial differential equations for generating integrable equations, to find its Lax pairs, recursion operators, Bäcklund transform and finally finding exact analytic solutions to the PDE.
Symmetry methods have been recognized to study differential equations arising in mathematics, physics, engineering, and many other disciplines.
Semi-analytical methods
[edit]
The Adomian decomposition method, the Lyapunov artificial small parameter method, and his homotopy perturbation method are all special cases of the more general homotopy analysis method. These are series expansion methods, and except for the Lyapunov method, are independent of small physical parameters as compared to the well known perturbation theory, thus giving these methods greater flexibility and solution generality.
Numerical solutions
[edit]
The three most widely used numerical methods to solve PDEs are the finite element method (FEM), finite volume methods (FVM) and finite difference methods (FDM), as well other kind of methods called meshfree methods, which were made to solve problems where the aforementioned methods are limited. The FEM has a prominent position among these methods and especially its exceptionally efficient higher-order version hp-FEM. Other hybrid versions of FEM and Meshfree methods include the generalized finite element method (GFEM), extended finite element method (XFEM), spectral finite element method (SFEM), meshfree finite element method, discontinuous Galerkin finite element method (DGFEM), element-free Galerkin method (EFGM), interpolating element-free Galerkin method (IEFGM), etc.
Finite element method
[edit]
Main article: Finite element method
The finite element method (FEM) (its practical application often known as finite element analysis (FEA)) is a numerical technique for approximating solutions of partial differential equations (PDE) as well as of integral equations using a finite set of functions. The solution approach is based either on eliminating the differential equation completely (steady state problems), or rendering the PDE into an approximating system of ordinary differential equations, which are then numerically integrated using standard techniques such as Euler's method, Runge–Kutta, etc.
Finite difference method
[edit]
Main article: Finite difference method
Finite-difference methods are numerical methods for approximating the solutions to differential equations using finite difference equations to approximate derivatives.
Finite volume method
[edit]
Main article: Finite volume method
Similar to the finite difference method or finite element method, values are calculated at discrete places on a meshed geometry. "Finite volume" refers to the small volume surrounding each node point on a mesh. In the finite volume method, surface integrals in a partial differential equation that contain a divergence term are converted to volume integrals, using the divergence theorem. These terms are then evaluated as fluxes at the surfaces of each finite volume. Because the flux entering a given volume is identical to that leaving the adjacent volume, these methods conserve mass by design.
Neural networks
[edit]
This section is an excerpt from Deep learning § Partial differential equations.[edit]
Physics informed neural networks have been used to solve partial differential equations in both forward and inverse problems in a data driven manner. One example is the reconstructing fluid flow governed by the Navier-Stokes equations. Using physics informed neural networks does not require the often expensive mesh generation that conventional CFD methods rely on.
Weak solutions
[edit]
Main article: Weak solution
Weak solutions are functions that satisfy the PDE, yet in other meanings than regular sense. The meaning for this term may differ with context, and one of the most commonly used definitions is based on the notion of distributions.
An example for the definition of a weak solution is as follows:
Consider the boundary-value problem given by:
where denotes a second-order partial differential operator in divergence form.
We say a is a weak solution if
for every , which can be derived by a formal integral by parts.
An example for a weak solution is as follows:
is a weak solution satisfying
in distributional sense, as formally,
Theoretical Studies
[edit]
As a branch of pure mathematics, the theoretical studies of PDEs focus on the criteria for a solution to exist, the properties of a solution, and finding its formula is often secondary.
Well-posedness
[edit]
Main article: Well-posed problem
Well-posedness refers to a common schematic package of information about a PDE. To say that a PDE is well-posed, one must have:
an existence and uniqueness theorem, asserting that by the prescription of some freely chosen functions, one can single out one specific solution of the PDE
by continuously changing the free choices, one continuously changes the corresponding solution
This is, by the necessity of being applicable to several different PDE, somewhat vague. The requirement of "continuity", in particular, is ambiguous, since there are usually many inequivalent means by which it can be rigorously defined. It is, however, somewhat unusual to study a PDE without specifying a way in which it is well-posed.
Regularity
[edit]
Main article: Regularity theory
Regularity refers to the integrability and differentiability of weak solutions, which can often be represented by Sobolev spaces.
This problem arise due to the difficulty in searching for classical solutions. Researchers often tend to find weak solutions at first and then find out whether it is smooth enough to be qualified as a classical solution.
Results from functional analysis are often used in this field of study.
See also
[edit]
Some common PDEs
Acoustic wave equation
Burgers' equation
Continuity equation
Heat equation
Helmholtz equation
Klein–Gordon equation
Jacobi equation
Lagrange equation
Lorenz equation
Laplace's equation
Maxwell's equations
Navier-Stokes equation
Poisson's equation
Reaction–diffusion system
Schrödinger equation
Wave equation
Types of boundary conditions
Dirichlet boundary condition
Neumann boundary condition
Robin boundary condition
Cauchy problem
Various topics
Jet bundle
Laplace transform applied to differential equations
List of dynamical systems and differential equations topics
Matrix differential equation
Numerical partial differential equations
Partial differential algebraic equation
Recurrence relation
Stochastic processes and boundary value problems
Notes
[edit]
^ "Regularity and singularities in elliptic PDE's: beyond monotonicity formulas | EllipticPDE Project | Fact Sheet | H2020". CORDIS | European Commission. Retrieved 2024-02-05.
^ Klainerman, Sergiu (2010). "PDE as a Unified Subject". In Alon, N.; Bourgain, J.; Connes, A.; Gromov, M.; Milman, V. (eds.). Visions in Mathematics. Modern Birkhäuser Classics. Basel: Birkhäuser. pp. 279–315. doi:10.1007/978-3-0346-0422-2_10. ISBN 978-3-0346-0421-5.
^ Erdoğan, M. Burak; Tzirakis, Nikolaos (2016). Dispersive Partial Differential Equations: Wellposedness and Applications. London Mathematical Society Student Texts. Cambridge: Cambridge University Press. ISBN 978-1-107-14904-5.
^ Evans 1998, pp. 1–2.
^ Klainerman, Sergiu (2008), "Partial Differential Equations", in Gowers, Timothy; Barrow-Green, June; Leader, Imre (eds.), The Princeton Companion to Mathematics, Princeton University Press, pp. 455–483
^ a b c Levandosky, Julie. "Classification of Second-Order Equations" (PDF).
^ Courant and Hilbert (1962), p.182.
^ Gershenfeld, Neil (2000). The nature of mathematical modeling (Reprinted (with corr.) ed.). Cambridge: Cambridge University Press. p. 27. ISBN 0521570956.
^ Zachmanoglou & Thoe 1986, pp. 115–116.
^ Wilmott, Paul; Howison, Sam; Dewynne, Jeff (1995). The Mathematics of Financial Derivatives. Cambridge University Press. pp. 76–81. ISBN 0-521-49789-2.
^ Logan, J. David (1994). "First Order Equations and Characteristics". An Introduction to Nonlinear Partial Differential Equations. New York: John Wiley & Sons. pp. 51–79. ISBN 0-471-59916-6.
^ Adomian, G. (1994). Solving Frontier problems of Physics: The decomposition method. Kluwer Academic Publishers. ISBN 9789401582896.
^ Liao, S. J. (2003). Beyond Perturbation: Introduction to the Homotopy Analysis Method. Boca Raton: Chapman & Hall/ CRC Press. ISBN 1-58488-407-X.
^ Solin, P. (2005). Partial Differential Equations and the Finite Element Method. Hoboken, New Jersey: J. Wiley & Sons. ISBN 0-471-72070-4.
^ Solin, P.; Segeth, K. & Dolezel, I. (2003). Higher-Order Finite Element Methods. Boca Raton: Chapman & Hall/CRC Press. ISBN 1-58488-438-X.
^ Raissi, M.; Perdikaris, P.; Karniadakis, G. E. (2019-02-01). "Physics-informed neural networks: A deep learning framework for solving forward and inverse problems involving nonlinear partial differential equations". Journal of Computational Physics. 378: 686–707. Bibcode:2019JCoPh.378..686R. doi:10.1016/j.jcp.2018.10.045. ISSN 0021-9991. OSTI 1595805. S2CID 57379996.
^ Mao, Zhiping; Jagtap, Ameya D.; Karniadakis, George Em (2020-03-01). "Physics-informed neural networks for high-speed flows". Computer Methods in Applied Mechanics and Engineering. 360 112789. Bibcode:2020CMAME.360k2789M. doi:10.1016/j.cma.2019.112789. ISSN 0045-7825. S2CID 212755458.
^ Raissi, Maziar; Yazdani, Alireza; Karniadakis, George Em (2020-02-28). "Hidden fluid mechanics: Learning velocity and pressure fields from flow visualizations". Science. 367 (6481): 1026–1030. Bibcode:2020Sci...367.1026R. doi:10.1126/science.aaw4741. PMC 7219083. PMID 32001523.
^ Evans 1998, chpt. 6. Second-Order Elliptic Equations.
References
[edit]
Courant, R. & Hilbert, D. (1962), Methods of Mathematical Physics, vol. II, New York: Wiley-Interscience, ISBN 9783527617241 {{citation}}: ISBN / Date incompatibility (help).
Drábek, Pavel; Holubová, Gabriela (2007). Elements of partial differential equations (Online ed.). Berlin: de Gruyter. ISBN 9783110191240.
Evans, Lawrence C. (1998). Partial differential equations (PDF). Providence (R. I.): American mathematical society. ISBN 0-8218-0772-2.
Ibragimov, Nail H. (1993), CRC Handbook of Lie Group Analysis of Differential Equations Vol. 1-3, Providence: CRC-Press, ISBN 0-8493-4488-3.
John, F. (1982), Partial Differential Equations (4th ed.), New York: Springer-Verlag, ISBN 0-387-90609-6.
Jost, J. (2002), Partial Differential Equations, New York: Springer-Verlag, ISBN 0-387-95428-7.
Olver, P.J. (1995), Equivalence, Invariants and Symmetry, Cambridge Press.
Petrovskii, I. G. (1967), Partial Differential Equations, Philadelphia: W. B. Saunders Co..
Pinchover, Y. & Rubinstein, J. (2005), An Introduction to Partial Differential Equations, New York: Cambridge University Press, ISBN 0-521-84886-5.
Polyanin, A. D. (2002), Handbook of Linear Partial Differential Equations for Engineers and Scientists, Boca Raton: Chapman & Hall/CRC Press, ISBN 1-58488-299-9.
Polyanin, A. D. & Zaitsev, V. F. (2004), Handbook of Nonlinear Partial Differential Equations, Boca Raton: Chapman & Hall/CRC Press, ISBN 1-58488-355-3.
Polyanin, A. D.; Zaitsev, V. F. & Moussiaux, A. (2002), Handbook of First Order Partial Differential Equations, London: Taylor & Francis, ISBN 0-415-27267-X.
Roubíček, T. (2013), Nonlinear Partial Differential Equations with Applications (PDF), International Series of Numerical Mathematics, vol. 153 (2nd ed.), Basel, Boston, Berlin: Birkhäuser, doi:10.1007/978-3-0348-0513-1, ISBN 978-3-0348-0512-4, MR 3014456
Stephani, H. (1989), MacCallum, M. (ed.), Differential Equations: Their Solution Using Symmetries, Cambridge University Press.
Wazwaz, Abdul-Majid (2009). Partial Differential Equations and Solitary Waves Theory. Higher Education Press. ISBN 978-3-642-00251-9.
Wazwaz, Abdul-Majid (2002). Partial Differential Equations Methods and Applications. A.A. Balkema. ISBN 90-5809-369-7.
Zwillinger, D. (1997), Handbook of Differential Equations (3rd ed.), Boston: Academic Press, ISBN 0-12-784395-7.
Gershenfeld, N. (1999), The Nature of Mathematical Modeling (1st ed.), New York: Cambridge University Press, New York, NY, USA, ISBN 0-521-57095-6.
Krasil'shchik, I.S. & Vinogradov, A.M., Eds. (1999), Symmetries and Conserwation Laws for Differential Equations of Mathematical Physics, American Mathematical Society, Providence, Rhode Island, USA, ISBN 0-8218-0958-X{{citation}}: CS1 maint: multiple names: authors list (link).
Krasil'shchik, I.S.; Lychagin, V.V. & Vinogradov, A.M. (1986), Geometry of Jet Spaces and Nonlinear Partial Differential Equations, Gordon and Breach Science Publishers, New York, London, Paris, Montreux, Tokyo, ISBN 2-88124-051-8.
Vinogradov, A.M. (2001), Cohomological Analysis of Partial Differential Equations and Secondary Calculus, American Mathematical Society, Providence, Rhode Island, USA, ISBN 0-8218-2922-X.
Gustafsson, Bertil (2008). High Order Difference Methods for Time Dependent PDE. Springer Series in Computational Mathematics. Vol. 38. Springer. doi:10.1007/978-3-540-74993-6. ISBN 978-3-540-74992-9.
Zachmanoglou, E. C.; Thoe, Dale W. (1986). Introduction to Partial Differential Equations with Applications. New York: Courier Corporation. ISBN 0-486-65251-3.
Further reading
[edit]
Cajori, Florian (1928). "The Early History of Partial Differential Equations and of Partial Differentiation and Integration" (PDF). The American Mathematical Monthly. 35 (9): 459–467. doi:10.2307/2298771. JSTOR 2298771. Archived from the original (PDF) on 2018-11-23. Retrieved 2016-05-15.
Nirenberg, Louis (1994). "Partial differential equations in the first half of the century." Development of mathematics 1900–1950 (Luxembourg, 1992), 479–515, Birkhäuser, Basel.
Brezis, Haïm; Browder, Felix (1998). "Partial Differential Equations in the 20th Century". Advances in Mathematics. 135 (1): 76–144. doi:10.1006/aima.1997.1713.
External links
[edit]
"Differential equation, partial", Encyclopedia of Mathematics, EMS Press, 2001
Partial Differential Equations: Exact Solutions at EqWorld: The World of Mathematical Equations.
Partial Differential Equations: Index at EqWorld: The World of Mathematical Equations.
Partial Differential Equations: Methods at EqWorld: The World of Mathematical Equations.
Example problems with solutions at exampleproblems.com
Partial Differential Equations at mathworld.wolfram.com
Partial Differential Equations with Mathematica
Partial Differential Equations Archived 2016-08-17 at the Wayback Machine in Cleve Moler: Numerical Computing with MATLAB
Partial Differential Equations at nag.com
Sanderson, Grant (April 21, 2019). "But what is a partial differential equation?". 3Blue1Brown. Archived from the original on 2021-11-02 – via YouTube.
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What is Annexation in Real Estate? | Terms, Types & FAQs
What is Annexation in Real Estate? | Terms, Types & FAQs
Details
If you own property in an unincorporated township, starting an annexation process could save your bottom line.
It can also open economic doors, giving you access to greater resources and saving you money along the way. The process comes with a mixture of pros and cons, and understanding the details of both will help you make an informed decision.
Here’s a rundown of what annexation means for your property.
What Does Property Annexation Mean?
Annexation occurs when personal property/territory in a township becomes part of a larger municipality, such as a city or village. When this happens, the annexed property is absorbed into the larger land, effectively serving as a real estate extension.
To qualify for annexation, the property must share a border with the municipality. This prevents the creation of “floating outposts” from City A in the middle of Township B.
Common Real Estate Annexation Terms to be Familiar With
Fixture: This term refers to annexed personal property that is now part of the larger land. When personal property attached to the land gets annexed, it becomes a fixture.
Intent: This plays a big role in whether or not a fixture can be considered annexed. If the attachment or improvement made to the land was made in an effort to improve said land, this intent makes the fixture annexable.
Legal Implications: If a property sale includes fixtures, legal implications are part of the package. Fixtures can affect factors related to money transfers, property ownership, local laws and more.
Types of Property Annexations
Fixture: This term not only describes property that can be annexed, but a type of annexation, too. When personal property becomes a permanent part of real annexed property (such as appliances and heating/cooling units), it becomes a fixture annexation.
Land: This type occurs when municipal boundaries are expanded into adjacent land, thereby absorbing that land and affecting the value and use of the property.
Involuntary: As the name implies, this type of annexation is beyond the control or approval of the property owner, and is usually carried out by a government entity.
Voluntary: Conversely, this type occurs when the property owners initiate the annexation themselves, typically to gain access to municipal services or improve their zoning regulations.
Advantages of Annexation
Services: A municipality’s water and sewer rates may be less expensive, and commercial properties can benefit from switching from a township’s system. The municipality is also responsible for street maintenance and snow removal.
Police and Fire: An annexed property is under the jurisdiction of the municipality’s police force and fire and rescue. This means you can rest easier knowing emergency services are readily available in your area.
Property value: Annexation to a municipality increases property values and makes residential and commercial properties more attractive to the real estate market. This can be a huge benefit for property investors.
Voting: Becoming a part of a municipality means you can vote for its elected officials, who may, in turn, affect your property with ordinances and regulations.
Disadvantages of Annexation
Higher Property Taxes: In most cases, annexed properties will be subject to increased taxes, which can be a significant financial burden on property owners.
Less Autonomy: Property owners who are annexed may feel a loss of control since they could come across new zoning laws, building codes or other regulations they had not previously experienced.
Increased Tensions: Annexation could cause conflicts between the property owner and the annexing entity due to regulations a property owner could disagree with or think they’re being unfairly targeted over.
Lowered Value: Depending on the circumstances, an annexed property could decrease in value due to factors such as increased taxes or reduced desirability based on location and regulations.
Frequently Asked Property Annex Questions
How Do I Know if a Property Has Been Annexed?
Contact Your Local City or Town Hall:Local government entities usually annex real estate. By contacting your local city or town hall, they can provide information on the jurisdiction and annexation of a property or how to further locate this information.
Check the Property Deed:The property deed may include information about annexation, especially if it has occurred recently. You can obtain a copy of the deed by contacting your local county recorder.
Search the County’s Property Records:The county assessor's office maintains records of all properties within the county. You can search their database to see if the property in question is listed as being within city limits or having been annexed.
How Does Annexation Work?
Now that you’ve decided to annex your property, the fun can really begin. Residential and commercial property owners must present a petition for annexation to the county commissioners.
You may file to annex your property if:
All property owners in the proposed territory file petitions
A majority of property owners in the proposed area agree to a single petition
The property is not “unreasonably large”
The benefits of the annexation and surrounding areas outweigh the detriments
Next, the township and annexing municipality must agree to the terms of the petition. The municipality must also provide municipal services to the annexation and set an operation timeline. You can petition for annexation if your property qualifies as an economic development project. Tax commissioners must certify that your project costs at least $10 million with an annual payroll of over $1 million.
Economic developments include:
Industrial
Commercial
Distribution
Research and development
How Long is the Real Estate Annexation Process?
Timelines vary depending on the property, potential appeals and municipality/township compliance. You’ll likely need a decent amount of patience since it takes several months to complete an annexation.
Signatures on a petition must be taken no more than 180 days before the filing date. Once you file your petition, the township and municipality have 30 days to accept or object to the proposal. An annexation takes effect 30 days after the passage of the resolution.
Conclusion
Annexation is a multi-layered, often complex process, but whether you’re initiating the process or having your property annexed by another entity, knowing the facts will help keep your head above water and your finances in balance.
Here are the biggest takeaways to hold onto:
Remember the importance of defining a fixture and how it impacts the annexation
Know the intent of a fixture and how it factors into its inclusion
Understand the legal implications involved
Annexations can be both voluntary and involuntary
Annexed property can affect you both positively (increased access to emergency and maintenance services, increased property value, voting possibilities) and potentially negatively (higher taxes, less autonomy, increased tensions)
Your petition will need to go through the proper channels and procedure in order to be considered
If you need a hand navigating the complexities of property annexation, purchasing and investing, talk to the experts at DRK. We’ll guide you, answer your questions and help you make the most of your real estate venture.
You can scout commercial real estate available in the Columbus, Ohio, area here.
Until next time,
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669 | https://openstax.org/books/introductory-statistics-2e/pages/1-practice | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Introductory Statistics 2e
Practice
Introductory Statistics 2ePractice
Search for key terms or text.
## 1.1 Definitions of Statistics, Probability, and Key Terms
Use the following information to answer the next five exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new drug is currently under study to address a respiratory virus. It is given to patients once the patient exhibits symptoms of the virus. Of interest is the average (mean) length of time in days from the time the patient starts the treatment until the symptoms are alleviated. Two researchers each follow a different set of 40 patients with the respiratory virus from the start of treatment until the symptoms are alleviated. The following data (in days) are collected.
Researcher A:3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B:3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Determine what the key terms refer to in the example for Researcher A.
1.
sample
3.
parameter
statistic
5.
variable
## 1.2 Data, Sampling, and Variation in Data and Sampling
“Number of times per week” is what type of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
Use the following information to answer the next four exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Antonio, Texas. The first house in the neighborhood around the park was selected randomly, and then the resident of every eighth house in the neighborhood around the park was interviewed.
7.
The sampling method was
a. simple random; b. systematic; c. stratified; d. cluster
“Duration (amount of time)” is what type of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
9.
The colors of the houses around the park are what kind of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
The population is ______________________
11.
Table 1.27 contains the total number of deaths worldwide as a result of earthquakes over a 13-year period.
| Year | Total Number of Deaths |
--- |
| 1 | 231 |
| 2 | 21,357 |
| 3 | 11,685 |
| 4 | 33,819 |
| 5 | 228,802 |
| 6 | 88,003 |
| 7 | 6,605 |
| 8 | 712 |
| 9 | 88,011 |
| 10 | 1,790 |
| 11 | 320,120 |
| 12 | 21,953 |
| 13 | 768 |
| Total | 823,856 |
Table 1.27
Use Table 1.27 to answer the following questions.
What is the proportion of deaths between Year 8 and Year 13?
What percent of deaths occurred before Year 2?
What is the percent of deaths that occurred in Year 4 or after Year 11?
What is the fraction of deaths that happened before Year 13?
What kind of data is the number of deaths?
Earthquakes are quantified according to the amount of energy they produce (examples are 2.1, 5.0, 6.7). What type of data is that?
What contributed to the large number of deaths in Year 11? In Year 5? Explain.
For the following four exercises, determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
A group of test subjects is divided into twelve groups; then four of the groups are chosen at random.
13.
A market researcher polls every tenth person who walks into a store.
The first 50 people who walk into a sporting event are polled on their television preferences.
15.
A computer generates 100 random numbers, and 100 people whose names correspond with the numbers on the list are chosen.
Use the following information to answer the next seven exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new drug is currently under study to address a respiratory virus. It is given to patients once the patient exhibits symptoms of the virus. Of interest is the average (mean) length of time in days from the time the patient starts the treatment until the symptoms are alleviated. Two researchers each follow a different set of 40 patients with the respiratory virus from the start of treatment until the symptoms are alleviated. The following data (in days) are collected.
Researcher A: 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B: 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Complete the tables using the data provided:
| Survival Length (in months) | Frequency | Relative Frequency | Cumulative Relative Frequency |
| 0.5–6.5 | | | |
| 6.5–12.5 | | | |
| 12.5–18.5 | | | |
| 18.5–24.5 | | | |
| 24.5–30.5 | | | |
| 30.5–36.5 | | | |
| 36.5–42.5 | | | |
| 42.5–48.5 | | | |
Table 1.28 Researcher A
| Survival Length (in months) | Frequency | Relative Frequency | Cumulative Relative Frequency |
| 0.5–6.5 | | | |
| 6.5–12.5 | | | |
| 12.5–18.5 | | | |
| 18.5–24.5 | | | |
| 24.5–30.5 | | | |
| 30.5–36.5 | | | |
| 36.5-45.5 | | | |
Table 1.29 Researcher B
17.
Determine what the key term data refers to in the above example for Researcher A.
List two reasons why the data may differ.
19.
Can you tell if one researcher is correct and the other one is incorrect? Why?
Would you expect the data to be identical? Why or why not?
21.
Suggest at least two methods the researchers might use to gather random data.
Suppose that the first researcher conducted his survey by randomly choosing one state in the nation and then randomly picking 40 patients from that state. What sampling method would that researcher have used?
23.
Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What sampling method would that researcher have used? What concerns would you have about this data set, based upon the data collection method?
Use the following data to answer the next five exercises: Two researchers are gathering data on hours of video games played by school-aged children and young adults. They each randomly sample different groups of 150 students from the same school. They collect the following data.
| Hours Played per Week | Frequency | Relative Frequency | Cumulative Relative Frequency |
--- --- |
| 0–2 | 26 | 0.17 | 0.17 |
| 2–4 | 30 | 0.20 | 0.37 |
| 4–6 | 49 | 0.33 | 0.70 |
| 6–8 | 25 | 0.17 | 0.87 |
| 8–10 | 12 | 0.08 | 0.95 |
| 10–12 | 8 | 0.05 | 1 |
Table 1.30 Researcher A
| Hours Played per Week | Frequency | Relative Frequency | Cumulative Relative Frequency |
--- --- |
| 0–2 | 48 | 0.32 | 0.32 |
| 2–4 | 51 | 0.34 | 0.66 |
| 4–6 | 24 | 0.16 | 0.82 |
| 6–8 | 12 | 0.08 | 0.90 |
| 8–10 | 11 | 0.07 | 0.97 |
| 10–12 | 4 | 0.03 | 1 |
Table 1.31 Researcher B
Give a reason why the data may differ.
25.
Would the sample size be large enough if the population is the students in the school?
Would the sample size be large enough if the population is school-aged children and young adults in the United States?
27.
Researcher A concludes that most students play video games between four and six hours each week. Researcher B concludes that most students play video games between two and four hours each week. Who is correct?
As part of a way to reward students for participating in the survey, the researchers gave each student a gift card to a video game store. Would this affect the data if students knew about the award before the study?
Use the following data to answer the next five exercises: A pair of studies was performed to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The studies observed 200 stroke patients recovering over a period of several weeks. The first study collected the data in Table 1.32. The second study collected the data in Table 1.33.
| Group | Showed improvement | No improvement | Deterioration |
--- --- |
| Used program | 142 | 43 | 15 |
| Did not use program | 72 | 110 | 18 |
Table 1.32
| Group | Showed improvement | No improvement | Deterioration |
--- --- |
| Used program | 105 | 74 | 19 |
| Did not use program | 89 | 99 | 4 |
Table 1.33
29.
Given what you know, which study is correct?
The first study was performed by the company that designed the software program. The second study was performed by the American Medical Association. Which study is more reliable?
31.
Both groups that performed the study concluded that the software works. Is this accurate?
The company takes the two studies as proof that their software causes mental improvement in stroke patients. Is this a fair statement?
33.
Patients who used the software were also a part of an exercise program whereas patients who did not use the software were not. Does this change the validity of the conclusions from Exercise 1.31?
Is a sample size of 1,000 a reliable measure for a population of 5,000?
35.
Is a sample of 500 volunteers a reliable measure for a population of 2,500?
A question on a survey reads: "Do you prefer the delicious taste of Brand X or the taste of Brand Y?" Is this a fair question?
37.
Is a sample size of two representative of a population of five?
Is it possible for two experiments to be well run with similar sample sizes to get different data?
## 1.3 Frequency, Frequency Tables, and Levels of Measurement
39.
What type of measure scale is being used? Nominal, ordinal, interval or ratio.
High school soccer players classified by their athletic ability: Superior, Average, Above average
Baking temperatures for various main dishes: 350, 400, 325, 250, 300
The colors of crayons in a 24-crayon box
Social security numbers
Incomes measured in dollars
A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not satisfied
Political outlook: extreme left, left-of-center, right-of-center, extreme right
Time of day on an analog watch
The distance in miles to the closest grocery store
The dates 1066, 1492, 1644, 1947, and 1944
The heights of 21–65 year-old women
Common letter grades: A, B, C, D, and F
## 1.4 Experimental Design and Ethics
Design an experiment. Identify the explanatory and response variables. Describe the population being studied and the experimental units. Explain the treatments that will be used and how they will be assigned to the experimental units. Describe how blinding and placebos may be used to counter the power of suggestion.
41.
Discuss potential violations of the rule requiring informed consent.
People in a correctional facility are offered good behavior credit in return for participation in a study.
A research study is designed to investigate a new children’s allergy medication.
Participants in a study are told that the new medication being tested is highly promising, but they are not told that only a small portion of participants will receive the new medication. Others will receive placebo treatments and traditional treatments.
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Book title: Introductory Statistics 2e
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670 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4354507/ | From observational to analytical morphology of the stratum corneum: progress avoiding hazardous animal and human testings - PMC
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Clin Cosmet Investig Dermatol
. 2015 Mar 4;8:113–125. doi: 10.2147/CCID.S77027
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From observational to analytical morphology of the stratum corneum: progress avoiding hazardous animal and human testings
Gérald E Piérard
Gérald E Piérard
1 Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, Liège University, Liège, Belgium
2 University of Franche-Comté, Besançon, France
Find articles by Gérald E Piérard
1,2,✉, Justine Courtois
Justine Courtois
1 Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, Liège University, Liège, Belgium
Find articles by Justine Courtois
1, Caroline Ritacco
Caroline Ritacco
1 Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, Liège University, Liège, Belgium
Find articles by Caroline Ritacco
1, Philippe Humbert
Philippe Humbert
2 University of Franche-Comté, Besançon, France
3 Department of Dermatology, University Hospital Saint-Jacques, Besançon, France
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2,3, Ferial Fanian
Ferial Fanian
3 Department of Dermatology, University Hospital Saint-Jacques, Besançon, France
Find articles by Ferial Fanian
3, Claudine Piérard-Franchimont
Claudine Piérard-Franchimont
1 Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, Liège University, Liège, Belgium
4 Department of Dermatopathology, Unilab Lg, University Hospital of Liège, Liège, Belgium
5 Department of Dermatology, Regional Hospital of Huy, Huy, Belgium
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1,4,5
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1 Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, Liège University, Liège, Belgium
2 University of Franche-Comté, Besançon, France
3 Department of Dermatology, University Hospital Saint-Jacques, Besançon, France
4 Department of Dermatopathology, Unilab Lg, University Hospital of Liège, Liège, Belgium
5 Department of Dermatology, Regional Hospital of Huy, Huy, Belgium
✉
Correspondence: Gérald E Piérard, Laboratory of Skin Bioengineering and Imaging (LABIC), Department of Clinical Sciences, University of Liège, Place du 20 Août 7, 4000 Liège, Belgium, Tel +32 4 366 2408, Fax +32 4 366 2976, Email gerald.pierard@ulg.ac.be
Collection date 2015.
© 2015 Piérard et al. This work is published by Dove Medical Press Limited, and licensed under Creative Commons Attribution – Non Commercial (unported, v3.0) License
The full terms of the License are available at Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed.
PMC Copyright notice
PMCID: PMC4354507 PMID: 25767402
Abstract
Background
In cosmetic science, noninvasive sampling of the upper part of the stratum corneum is conveniently performed using strippings with adhesive-coated discs (SACD) and cyanoacrylate skin surface strippings (CSSSs).
Methods
Under controlled conditions, it is possible to scrutinize SACD and CSSS with objectivity using appropriate methods of analytical morphology. These procedures apply to a series of clinical conditions including xerosis grading, comedometry, corneodynamics, corneomelametry, corneosurfametry, corneoxenometry, and dandruff assessment.
Results
With any of the analytical evaluations, SACD and CSSS provide specific salient information that is useful in the field of cosmetology. In particular, both methods appear valuable and complementary in assessing the human skin compatibility of personal skincare products.
Conclusion
A set of quantitative analytical methods applicable to the minimally invasive and low-cost SACD and CSSS procedures allow for a sound assessment of cosmetic effects on the stratum corneum. Under regular conditions, both methods are painless and do not induce adverse events. Globally, CSSS appears more precise and informative than the regular SACD stripping.
Keywords: irritation, morphometry, quantitative morphology, stripping
Introduction
The human stratum corneum (SC) is a superficial continuous membrane made of tightly stacked corneocytes parted by multilamellar lipid sheets. Its deep cohesive portion is called the stratum compactum, contrasting with the loose superficial part referred to as the stratum disjunctum. Such a structure undergoes a maturation process while it moves upwardly to the surface of the skin. There is a progressive reduction in intercorneocyte cohesion, ending with physiologically imperceptible desquamation.1 Considering the typical continuous renewal of the epidermis, and the structure of the SC as its end product, the SC corresponds to a recollection of the past history of the recent life of the epidermis.1 Alterations in the organization of corneocytes, the presence of parakeratosis, and various other aspects are therefore clues for the intervention of some previous physiopathological disturbances.
At most locations on the body surface, the SC is typically composed of 15 or so layers of ordered flattened corneocytes. At the skin surface, these cells are approximately 1 μm thick, and their mean area reaches approximately 1,000 μm 2. The corneocyte area is influenced by the anatomic location, and a variety of environmental conditions including chemical irritation and ultraviolet light (UVL) action modulate the epidermal renewal.1 In addition, the average corneocyte size is assumed to increase with aging. This feature is probably linked to a prolonged corneocyte transit time during its progression throughout the SC.
Some cosmetic compounds act on the SC, and modulate its structure and the functions that it serves. Other compounds target the SC by an indirect route through modulations in the physiology of the underlying stratum spinosum. No proper pharmacology in the SC exists without adequate and reliable techniques for objective analytical recordings of the effects of the test products. In this review, emphasis is placed on some of the methods using distinct skin strippings.
SC collection
Smears and skin scrapings are simple procedures for getting SC samples. Unfortunately, the collected material is obviously torn out and fragmented. The surface topography and the cell organizations are not appreciated in such samplings. As a consequence, quantification procedures are almost impossible to perform on such material in a reproducible way.
Basically, SC stripping is the regular way for collecting the stratum disjunctum.2 Using casual adhesive transparent tapes is poorly reliable for quantitative assessments because their adhesion to the SC is uncontrolled and appears variable among different brands. By contrast, SC stripping using adhesive-coated discs (SACD) is conveniently performed.3 An improved method using a liquid cyanoacrylate pressure-sensitive adhesive on a polyester slide was designed for getting a better sampling of the SC structure.4–10 Such a procedure corresponds to the cyanoacrylate skin surface stripping (CSSS), formerly called skin surface biopsy4 or follicular biopsy.11
SACD uses a crystal clear adhesive-coated disc (D-Squame®; Cuderm Corporation, Dallas, TX, USA; and Corneofix®; CK Electronic, Cologne, Germany). The device provides adequate rigidity and adhesion for uniformly sampling a defined area of the stratum disjunctum. After peeling off the protective seal, the disc is applied to the SC surface using a gauge spring dynamometer for ensuring a calibrated and reliable pressure.3 Frequently, the chosen pressure is in the range of 100–250 g/cm 2. Both the pressure and the application time of the disc influence the amount of collected SC. A short-time (approximately 5 seconds) application of the disc removes less stratum disjunctum than an extended duration (approximately 30–60 minutes) of application. This is likely due to occlusion modifying both the intrinsic SC moisture and the intercorneocyte cohesion. Any greasy topical agent applied to the skin before sampling impairs the adhesion of the disc to the SC, thus yielding unreliable information. Such sampling failures are limited following delipidization of the skin surface using, for instance, ether:acetone (1:1).
For the CSSS method (3S-Biokit; CK Technology, Visé, Belgium), a droplet of cyanoacrylate glue is deposited onto a clear polyethylene strip. This material is pressed against the SC surface for approximately 15–30 seconds. In presence of discrete SC moisture, the cyanoacrylate polymerizes and firmly adheres to the SC. Ultimately, the material is gently lifted at one tip, and peeled off the skin. CSSS is adequately performed from any site on the body, with two main provisos. First, CSSS harvesting from a hairy area is commonly painful due to pulling out hairs. Furthermore, the CSSS quality is inadequate owing to the erratic contact of the cyanoacrylate bound to the SC. It is therefore advisable to shave such areas before CSSS harvesting. Second, a problem results from the natural strong intercorneocyte cohesion on the palms and soles. Such intercellular binding is commonly stronger than the glue bond on CSSS. This limitation impairs the collection of a uniform thin layer of corneocytes from these sites. However, such CSSS samplings become adequate in some instances when the SC cohesiveness is compromised.
Overall aspect of normal CSSS
CSSS from normal skin reveals a regular crisscross network of high-peaked crests corresponding to discrete skin surface creases called the first-, second-, and third-hollow order lines. Their patterns of distribution are distinct on various body regions. The first-order lines correspond to grooves in the latticework papillary relief at the dermo–epidermal interface.7 In young individuals, junctions of the shallow first- and second-order lines border contiguous polyhedral shaped SC plateaus (Figure 1A). On stretching the skin surface, a reorganization of these lines occurs. With aging, such a network progressively alters its original configuration, bringing the lattice along the skin tension lines spontaneously into alignment.8 The process ends with the vanishing of the shallow skin surface lines (Figure 1B). Thus, it is possible to assess the overall texture of the superficial dermis on CSSS. As a result, conditions such as dermal aging, dermal atrophy (dermatoporosis), striae distensae, sclerosis, scars, and many other skin connective tissue alterations are conveniently observed noninvasively using CSSS.7,8,10 Such morphological assessment of the skin microrelief is possibly quantified combining computerized image analysis and profilometry (Figure 1C).
Figure 1.
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Regular aspects of CSSS from the forearm.
Notes: (A) Crisscross pattern of shallow lines in a young adult. (B) Alignment of shallow lines along Langer’s lines in an elderly subject. (C) Image analysis for profilometry of the skin surface pattern.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Cytologic presentations of corneocytes are hardly visible on unstained CSSS. For improving their perceived aspect, several histochemical dyes are useful. A mixture of toluidine blue and basic fuchsin (TBBF) in 30% ethanol is a simple dye that is conveniently handled in an office setting.7,8,10 The regular SC exhibits a rather uniform cohesive pattern of adjacent corneocytes. Each single corneocyte is typically anucleated and contains a water-insoluble protein complex corresponding to a highly organized keratin microfibrillar matrix. Such cell structure is encapsulated in a protein- and lipid-enriched shell. The cell boundaries appear clearly stained by a thin polyhedral TBBF rim.10 The cornified cell envelope exhibits different stages of maturation among corneocytes.12,13 Basically, two distinct presentations of cornified cell envelopes are distinguished. They correspond to the so-called fragile immature envelopes and the rigid mature ones, respectively (Figure 2A). The former type is recognized on CSSS by a deep TBBF staining contrasting with the rimmed pattern of the mature corneocytes.10 Scanning electron microscopy of immature corneocytes commonly exhibits a paving of small protrusions of similar sizes.10
Figure 2.
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Cytologic aspects in CSSS.
Notes: (A) Uniform paving of corneocytes. (B) Dispersed parakeratotic cells. (C) Inflammatory cells.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Lipid staining such as the Nile red stain conveniently reveals sebum-enriched follicular pores and follicular casts collected on CSSS. Any histochemical positivity at the follicular sites corresponds to two distinct phenomena. First, the visualized sebum closely represents the direct lipid production by each single follicle. Second, the lipids produced and poured out from other follicles in their vicinity run at the skin surface and are finally collected in the slope of other follicular openings.
Analytical morphology of skin strippings
Skin strippings, including SACD and CSSS, are used as nearly noninvasive diagnostic methods and are useful in many skin conditions.1,5,7,10 In addition, they are examples of tools that combine analytical morphology and other procedures for the evaluation of the kinetics and efficacy of various cosmetics and toiletries. Many cosmetic compounds and cosmeceuticals exert modulating effects, in part influencing the SC presentation. Analytical morphology applied to SACD and CSSS provides either quantitative, semiquantitative, or binary data. These procedures include various techniques, and they afford quantitative and statistically evaluable data following image processing. The challenge of analytical evaluations of skin strippings is to quantify some specific aspects of the structure and functions of the SC.14,15 Image analysis of the aspect of samples seen under the microscope, and optical profilometry of the surface of the samples are two classical methods of analytical morphology that are conveniently applied to skin strippings. Other biometrological approaches have been designed, including visual rating evaluations of sampled material placed on a black background. More precise evaluations are gained from reflectance colorimetry and light transmission assessments. The weight of collected SC samples on SACD are measured and can further be analyzed for the molecular identification of specific components. For that purpose, the samples are conveniently deposited in the well of microanalysis plates.
In these respects, skin strippings are used in four main ways:
Evaluation of the SC structure
Evaluation of the SC dynamics
In situ evaluation of the superficial skin biocene
Use of CSSS as a substrate for a set of ex vivo bioassays
Analytical morphology denotes a set of techniques that provide quantitative data and statistically evaluable information from morphological aspects. Of note, meaningful measurements are dependent upon careful thought given to the sampling process. Pressure and time of application of the clear disc in SACD, and the quality of the cyanoacrylate liquid bond in CSSS are of the utmost importance. The cleanness of the sampled SC without any residual greasy product is mandatory.
Some analytical measurements are conveniently performed on CSSS. They commonly rely on the combination of optical properties of the samples, reflectance colorimetry, and morphometry-based image analysis. Some aspects of skin conditions, including their severity and any cosmetic improvement, are conveniently assessed on CSSS following the disclosure of some typical features in the SC.
Because SACD and CSSS are very thin compared to the width of the sampling, both samplings are regarded as nearly two-dimensional structures. Under microscopic examination, the size of the field of vision is determined by the combination of the magnification and the microscope eyepiece. Usually, the eyepiece field diaphragm opening diameter determines the size of the field of vision. Morphometry concerns measurements made of shapes. Such measurements primarily rely on point and linear procedures. The proportion of points in an ocular graticule gives a relative value for its area. Such values are converted to absolute values when the area of the sample covered by the graticule is determined.
It is possible to measure the area of the exposed aspect of corneocytes, and to detect any presence of parakeratosis (Figure 2B) and other cells on such specimens (Figure 2C). On healthy skin, parakeratotic cells are rare, and those present are not clustered. By contrast, clumps of parakeratotic cells usually suggest a pathologic process. They are recognized by the presence of a nucleus central to the polyhedral cell. Optical profilometry is conveniently applied to CSSS. There is a clinical relevance for the skin roughness parameters (Ra and Rz) gained on CSSS.
Grading and analytical morphology of xerosis
Xerosis refers to a rough and dry-looking aspect of the skin. The relationship between xerosis and lack of water in outer layers of the SC is ambiguous.16 Most xerotic conditions are in fact a failure of the normal corneocyte shedding process. These cells remain attached to each other with unequal strength until rafts of cells detach partially from the skin surface (Figure 3). This process of corneocyte clumping is possibly assessed using visual and tactile scoring. Such crude evaluations suffer from variability by inconsistencies from grade to grade, and from poor reproducibility. Subtle variations in the environmental conditions jeopardize subjective grading systems since ambient hydration swells the outer SC and often camouflages low-grade scaling and dryness. Thus, in order to investigate any xerotic process, and to determine the efficacy of a moisturizer, an evaluation program should be encouraged to go beyond usual clinical observations.
Figure 3.
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Corneocyte clumpiness collected on a CSSS (scanning electron microscopy).
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Some cosmetics and cosmeceuticals reduce desquamation and scaling. Normal desquamation takes place imperceptibly as the release of single corneocytes at the skin surface. When the intracorneal cohesion fails to decline evenly in the stratum disjunctum, clumps of corneocytes remain stuck together, and they come off as scales. The resulting harsh feel by sensing fingers, and the altered appearance under light reflectance correspond to xerosis, often interpreted as dry skin by laypeople. In fact, the concept of dry skin remains a controversial matter.16
Xerosis refers to various forms of predominantly orthokeratotic hyperkeratosis. This definition and its limits are idiosyncratic, varying according to different local traditions among laboratories and cosmetic companies. A semiquantitative grading of orthokeratotic xeroses is conveniently achieved on CSSS stained for 3 minutes with the TBBF staining solution.7,10,16 The samples are graded7,10 as follows:
Type 0: normal SC without any evidence for hyperkeratosis
Type 1: hyperkeratosis of the first- and second-order lines and/or the adnexal openings
Type 2: focal hyperkeratosis covering less than 30% of the skin surface plateaus (Figure 4A)
Type 3: hyperkeratosis covering over 30% of the skin surface plateaus (Figure 4B)
Type 4: diffuse, confluent, and homogeneous scales with persistence of the first-order lines
Type 5: thick, uneven scales covering the entire SC surface, obliterating the shallow lines
Figure 4.
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Xerosis appearance on CSSS.
Notes: (A) Type 2. (B) Type 3.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Quantitative assessment of xerosis relies on analytical evaluations of SACD.3,14 The attenuation of optical light transmission has been used to measure the amount of scales collected on the discs. That method was improved by measuring reflectance colorimetry of the unstained sample deposited onto a colored reference plate. Thick scales abate the L and a values of reflectance colorimetry. Image analysis of SACD is possible for the quantification of desquamation disorders. The parameters of importance are the area (A) covered by scales, and their thickness rated on a 5-level gray scale. The percentage of scales (Tn) is calculated in relation to each thickness level (n). A desquamation index (DI) is yielded according to:
(1)
The squamometry index (SQMI) is yet another objective evaluation of xerosis and some other SC disturbances.17,18 SACD are stained for 1 minute by dropping the TBBF solution over the SC surface, followed by gentle tap-water rinsing. The stained SACD is placed over a hole cut out of a slide, which is placed onto a white color reference plate. The Chroma C value measured by reflectance colorimetry correlates with the amount of removed scales and with the clinical rating of xerosis. A seemingly regular SC of the volar forearm is commonly characterized by SQMI values lower than 20. By contrast, a normal scalp without dandruff shows a weaker SQMI below 5. This technique is conveniently applied for assessing some inflammatory conditions such as sunburn (Figure 5), and to quantify the efficacy of antidandruff shampoos.
Figure 5.
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SQMI as assessed by Chroma C at D0 and D2.
Note: Normal skin (Chroma C D0<15) shows an increase in SQMI while xerotic skin (Chroma C D0>30) responds with a decrease in SQMI values.
Abbreviations: D0, before acute photodamage; D2, 48 hours after acute photodamage; SQMI, squamometry index.
Xerosis of old age as well as wrinkling are possibly evaluated on CSSS. In particular, the wrinkling process and its correction by some cosmetics is conveniently substantiated by combining image analysis and optical profilometry.
Comedometry
Follicular casts and microcomedones are conveniently studied on CSSS collected from the face or the back. The horny material sampled from the upper portion of the follicular ducts corresponds to a follicular stripping,11,19 and it reflects the balance between comedo formation and lysis (Figure 6A). Comedometry allows for computerized quantification of the number and size of follicular casts collected on CSSS.11,19,20 The numerical density of follicles is influenced by the body site. For each site with absence of obvious comedogenesis, the interindividual variation remains small. This method finds application in comedogenesis- and comedolysis-related disorders and their treatments.20–23 Both the number and the size of follicular casts are influenced by some treatments. Comedometry on human skin showing comedogenesis reveals large interindividual differences in the number of horny follicular casts. CSSS are used to predict the comedogenic risk linked to products, and, by contrast, to quantify any comedolytic activity of cosmetics. The concentration in active compounds governs the propensity of modifying the microcomedo density and size. When an exogenous comedogenic factor is involved, the vast majority of the hair follicles are similarly involved. By contrast, endogenous comedogenic factors (androgens, acne, etc) typically affect to a variable extent a minority of hair follicles.10 The sensitivity of the method is such that it is possible to disclose signs of microcomedolysis after a short time (a couple of weeks) of adequate treatment.23
Figure 6.
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Microcomedones collected on CSSS from the face.
Notes: (A) Microcomedones and discrete follicular casts. (B) CSSS and a sebum-sensitive foil superposed on one another, revealing microcomedones (dark spots) pouring out (white area) or not (grey background) sebum at the skin surface. (C) Follicular fluorescence on a CSSS.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Sebum-sensitive foils (SSF), available as Sebutape® (Cuderm corp; Dallas, TX, USA) and Sebufix® (CK Electronic), are conveniently used for assessing the sebum output at the skin surface. It is possible to combine such method with CSSS.24 In a first step, a non-sticky SSF is applied to the skin for a few seconds. The outlines of the foil are ink marked on the SC. In a second step, following removal of the SSF, CSSS is collected from the very same skin site. The ink mark is visible on such sampling. The CSSS and the foil are then exactly superposed using the ink mark as an adjusting mark. The dual SSF–CSSS samplings are examined under the microscope (Figure 6B) and submitted to image analysis, considering specifically the darker horny follicular casts and the clear transparent sebum spots. Some correlations are possibly established between the pore sizes, the follicular casts, the microcomedones, and the clear transparent spots of sebum.24
Analytical methods for evaluating some features of follicular casts rely on image analysis when illuminating the specimen under the microscope with either white light, polarized light, or fluorescent light. The assessment of the follicular fluorescence for evaluating the presence of porphyrins produced by Propionibacterium acnes in follicles is occasionally important for assessing any impact of cosmetics.25 In fact, some products possibly emit fluorescence by themselves (Figure 6C), while others such as sunscreens display a quenching effect by absorption of porphyrin fluorescence.
It is possible to assess the density of hair follicles over a defined surface area, as well as to observe the skin pores and the presence of follicular hyperkeratosis (kerosis), comedones, trichostasis spinulosa, intrafollicular bacteria, and mites (Demodex folliculorum).7,26–28 In some instances, other hair follicle structures including hair bulbs and follicular sheaths are visualized on CSSS. Skin pores distinctly corresponding to follicular or sudoral openings at the skin surface are conveniently explored using CSSS.27
Corneomelametry
Melanin production in melanocytes is under complex neuroendocrine controls.29 It is identified in regular corneocytes of phototype V and VI individuals. It is also present in the SC covering various pigmented lesion in clear-skinned people (Figure 7). It is mandatory to distinguish melanin-laden anucleated corneocytes from neoplastic dendritic nucleated melanocytes as seen after their migration inside the SC covering a malignant melanoma.10 For increasing the sensitivity of the procedure, the aspect of the dusty melanin load is typically increased using an argentaffin stain procedure. The relative darkness of these CSSS is conveniently assessed using corneomelametry. Such a method consists of measuring the reduction in light transmission through the CSSS using a photodensitometer device designed for photomicroscopy.30–33 Bleaching agent effects are possibly tested and compared using corneomelametry. Melasma and solar lentigines (aging spots) are typical lesions explored in that way.
Figure 7.
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Abundant melanin deposits in clustered corneocytes as revealed by a CSSS.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
Corneodynamics
The dansyl chloride (DC) test was generally accepted as a noninvasive clinical attempt at evaluating the SC turnover. The overall time for fluorescence extinction depends on both the rate of transit of corneocytes through the SC and the thickness of that layer. This test proves to be difficult to interpret on clinical grounds due to the uneven fade-out of fluorescence at the skin surface. An improved method was introduced by the examination of CSSS harvested from a DC test area.33 It was further refined by replacing DC with the browning dihydroxyacetone (DHA) agent.34
Corneodynamics is assessed using CSSS from DC (Figure 8A) and DHA (Figure 8B) test sites at a predetermined time of a trial. Accordingly, at about day 10, CSSS are examined under a fluorescent light microscope because both DC and DHA are fluorescent. Image analysis applied to such pictures allows quantification of the nonfluorescent versus fluorescent areas of the SC. This ratio is an indicator of the rate of SC turnover. However, both the DC and DHA tests are dramatically influenced by cleansing agents and skincare products.35 The extraction of the dyes from the SC by these products is tentatively used to predict irritancy potential.35
Figure 8.
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Corneodynamics on CSSS at day 10.
Notes: (A) Dansyl chloride test. (B) Dihydroxyacetone test.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
An inverse relationship has been suggested between the size of corneocytes and the speed of epidermal turnover. This aspect is possibly studied on skin strippings, particularly CSSS during corneodynamics.
Another facet of the SC dynamics is gained by squamometry, in particular UVL squamometry.3 For instance, the SQMI kinetics after acute photodamage show a complex response of the epidermis with early (±48 hours) and late (±10 days) changes.36 Indeed, a single regular SACD sampling corresponds at the most to the regular corneocyte daily shedding. Normal and moderately dry skin respond to UVL irradiation by an early increase in the SQMI value.
In vivo evaluation of the skin surface biocene
The biocene of resident bacteria and other saprophytic or pathogenic microorganisms is typically confined to the skin surface and the appendages.6,37–39 At the surface of the SC, the flora remains largely encased inside the cyanoacrylate bond during CSSS sampling. Thus, it is not accessible to staining procedures, and it remains invisible at microscopic examination. Therefore, the surface microflora is not adequately disclosed on CSSS. By contrast, samples of microorganisms entrapped inside the SC and follicular casts are distinctly collected from CSSS. In particular, the microflora is accessible by scraping out the horny spiky structures appended to the CSSS. Viability of the intrafollicular bacteria is conveniently assessed using the combination of vital stains (Figure 9A), such as neutral red,40 and flow cytometry.39
Figure 9.
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Superficial skin biocene.
Notes: (A) Bacteria revealed by the neutral red staining. (B) Malassezia sp. in the stratum corneum.
Abbreviation: CSSS, cyanoacrylate skin surface stripping.
The load of Malassezia spp. inside squames of dandruff and seborrheic dermatitis is assessed on SACD (Figure 9B). The efficacy of shampoos can be compared by combining squamometry and semiquantitative evaluations of the number of yeasts.41 The adhesion of Malassezia yeasts to corneocytes appears to be a prominent feature.42
Corneosurfametry
The presence of various tiny foreign bodies and microscopic parasites, and the impact of various chemical xenobiotics on the SC are conveniently assessed on CSSS.10,43 In particular, surfactants remove lipids, denature proteins, induce corneocyte swelling, alter the SC barrier function, and produce a feeling of skin roughness and dryness.44 The surest method for testing skin compatibility of surface-active agents was to use large panels of volunteers. However, such a procedure proved to be costly and time-consuming. Over the years, various alternative methods were proposed. Among them, corneosurfametry (CSM) refers specifically to the effects of surfactants and wash solutions on the SC.18,44,45 CSM is a noninvasive quantitative test rating the interaction between surfactants and human or animal SC (Figure 10). It is used as a predictive irritancy test that compares favorably with conventional in vitro tests.
Figure 10.
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Corneosurfametry showing the dense staining of corneocytes after contact with a harsh soap.
CSM entails collection of CSSS from healthy skin of volunteers followed by a predetermined contact time of the SC samples with neat or diluted surfactants. A 2-hour experiment is usually convenient to disclose any differences in SC reactivity among products. The usual test concentrations for proprietary cleansing skincare products ranges from 3% to 10%, but higher dilutions are possibly required for harsh products. Each surfactant solution is sprayed uniformly on about 20 CSSS placed in a slide tray, which is then covered by a lid. At completion of the predetermined assessment time, CSSS are thoroughly rinsed with tap water, dried, and stained for 3 minutes with the TBBF dye solution. After removing any excess dye and copious water rinsing, samples are examined under the microscope. In addition, reflectance colorimetry is applied to the dry samples for recording both L and Chroma C. To be adequately interpreted, the L value should be over 45, and superior to that of Chroma C.18 Data out of range suggest an excess surfactant concentration for the test. For instance, the maximum concentration for testing sodium laurylsulfate (SLS) is approximately 2%.
CSM results are expressed by two parameters: namely, the CSM index (CSMI) and the colorimetric index of mildness (CIM). CSMI of any test product corresponds to the color difference between water-treated control samples and those exposed to the test product. Thus, CSMI is conveniently calculated as follows:
(2)
There is a linear negative correlation between values of L and Chroma C, and CIM of each sample is calculated as follows:
(3)
According to such analytical assessments, mild skincare products containing anionic surfactants are characterized by the combination of a low CSMI value and a high CIM value. The converse is true for harsh skincare products and most household cleaners. However, a low paradoxical CSMI value is experienced with some aggressive products yielding to a cascade of corneocyte swelling, lysis, and loss from the stratum disjunctum. Usually, interindividual variations are larger for SQMI than for CIM.18
Microwave CSM is a more rapid procedure than the regular CSM.46 In this procedure, CSSS are immersed in a flask containing the test surfactant solution, and they are placed in a microwave oven containing a 500 mL water load. Microwave CSM is typically run at 750 W for 30 seconds. The next steps are identical to those of the regular CSM procedure.
Responsive CSM is a variant of the method where skin to be sampled is preconditioned before harvesting CSSS.47 The method, based on repeated subclinical injuries by surfactants, is monitored in a controlled forearm immersion test. At completion of the in vivo procedure, CSSS are harvested for a regular or microwave CSM bioassay using the same surfactant as in the preliminary in vivo procedure. Preconditioning the skin in this way increases CSM sensitivity to discriminate between different potential irritations from mild surfactants.
Shielded CSM is used for testing skin-protective products (SPP), also called skin barrier formulations, claiming some protection against noxious external agents. In shielded CSM, CSSS are first covered by a controlled amount of the test SPP before performing regular CSM using a reference surfactant. Comparative screenings of various SPP are conveniently performed using shielded CSM, avoiding exposure of volunteers to any hazardous agents for in vivo testing.
Corneoxenometry
The corneoxenometry (CXM) bioassay is offered for testing any chemical xenobiotic other than surfactants.48,49 The compounds consist of any hazardous chemical including organic solvents, acids, bases, toxic agents, etc. The basic procedure is similar to CSM and its variants. It entails CSSS collection followed by its controlled immersion in the test solution for 1–120 minutes. The subsequent TBBS color darkening correlates with the test solution aggressiveness toward the SC. One main indication is found in the field of skin irritation while avoiding the risk of in vivo hazards. Skin barrier formulations are conveniently tested using CXM.50 Another indication deals with comparative assessments of penetration enhancers commonly used in some topical formulations.51
Regional SC reactivity to irritant xenobiotics
One of the most important functions of the epidermis is the formation of an efficient barrier between the body and a diversity of xenobiotics. Much research has been undertaken to understand the variability in the skin barrier efficacy, which resides in the SC. Measurements of the transepidermal water loss represent indirect noninvasive evaluations of such a function. Values are kept low when the barrier function is optimal, and they increase with the severity of the barrier defect. It is generally assumed that the cutaneous barrier is disrupted through intercellular lipid removal and protein denaturations. For instance, such a condition is reached following applications of various organic solvents to the skin. Since workers are exposed to solvents at many workplaces, methods are needed for the evaluation of the potential solvent toxicity. Such hazards call for ex vivo predictive bioassays on human skin or SC, such as CSM and CXM.
Large interindividual differences in CIM were reported regarding different surfactants and solvents.48 In addition, distinct regional differences are present regarding irritancy and percutaneous absorption. Both the dorsal hand and volar forearm appeared to be the least reactive skin locations on the CSM bioassay. By contrast, the same test revealed that the neck, forehead, back, and dorsal foot were more reactive sites.
In normal subjects, the extent of alterations induced in the human SC by solvents during CXM are more variable than those induced by diluted surfactants at the CSM bioassay. In addition, a heterogeneity in corneocyte alterations exists among solvents and between distinct body sites. For instance, hexane–methanol and chloroform–methanol strongly alter the SC structure. Indeed, the chloroform–methanol mixture is a potent lipid extractor from biological samples. However, it is not the top ranked aggressor on the CXM bioassay.48
The intensity of the skin response to irritant xenobiotics is subject to anatomic site variations. The initial mechanism governing such biological response presumably corresponds to the direct damage to the SC. Subtle regional variations in irritancy are possibly tested using CSM and CXM.44 Regional variations in skin reactivity presumably correspond in part to an uneven preconditioning by in vivo contacts with a previous irritant stress. A similar, but probably weaker process takes place in the condition of reactive skin.52 This concept illustrates the failure of drawing general conclusions from CSM and CXM performed from a given body site. Such a finding contrasts with the frequent variability in CIM data from distinct body sites, and with the inconsistency in severity of irritant reactions at different anatomical regions, as assessed by transepidermal water loss measurements at in vivo patch test sites.
Reactive (sensitive) skin
The term reactive (sensitive) skin covers a broad spectrum of different conditions, sharing in common the occurrence of unwanted changes in response to external environmental factors including personal-care products. Reactive skin is a heterogeneous problem characterized by a reduced cutaneous tolerance to specific environmental factors (cold, heat, wind, wool, topical products, etc).53 Clinical manifestations consist mainly of subjective symptoms linked to sensory irritation including discomfort, itching, stinging, and burning sensations. There are no specific signs discernable on CSSS except occasional discrete xerosis and parakeratosis. Other biometrological procedures are possibly useful.54
Some individuals complaining of sensitive skin are more readily deranged by some specific chemical and physical irritation. A series of these conditions are likely related to a defective SC barrier function. CSM performed with a reference surfactant (SLS 1%) shows that CSSS from patients with reactive skin react more severely than those harvested from healthy individuals without sensitive skin. SQMI is commonly higher on reactive skin than on otherwise normal SC. CIM is significantly lower (approximately 25% reduction) in patients with reactive skin than in normal nonsensitive individuals, indicating a higher susceptibility to irritation induced by surfactants and some other skincare products.52
Potency of squamolytic agents
CSSS, using a method similar to CSM, is a substrate for the study of the actual bioactivity of emollients and squamolytic (keratolytic) agents. The test product is applied to CSSS for various periods of time ranging from 15 minutes to 4 hours (Figure 11). After abundant rinsing and staining with TBBF, reflectance colorimetric values are recorded and CSMI and/or CIM parameters are calculated in a way similar to CSM.
Figure 11.
Open in a new tab
Xerosis evolution according to squamometry (SQMI parameter) and skin roughness (Rz) of CSSS in contact with a 50% glycolic acid solution for various durations of contact.
Abbreviations: SQMI, squamometry index; CSSS, cyanoacrylate skin surface stripping.
The effect of squamolytic agents and emollients is conveniently assessed on skin strippings collected in clinical trials (Figure 12). In general, it is appropriate to harvest samples at entry into the study, as well as 2 weeks and 4 weeks after treatment. After a 2-week period of treatment corresponding to the regression phase, late samples are collected. Xerosis is rated on CSSS. Both SQMI and DI show the kinetics of improvement followed by the posttreatment regression. The typical kinetics show first an aggravation in the test values followed by a drop in these values, this latter phase indicating a loss of corneocytes from the CSSS. Optical profilometry performed on the same samples shows the immediate smoothing effect of the products with reduction in the Rz value followed by desquamation and increased Rz.
Figure 12.
Open in a new tab
Evolution in the squamometry values (SQMI).
Note: Different kinetics of xerosis improvement during treatment (weeks 0–4) with three emollients (A, B, C) and the following regression phase (weeks 4–6).
Abbreviation: SQMI, squamometry index.
Conclusion
The possibilities of using skin strippings in dermocosmetic science seem endless. The emergence of new analytical methods increases the validity of the information yielded by SACD and CSSS. With regulations avoiding animal experiments and ethical considerations in human testing, there is a need and a real possibility to develop new predictive bioassays using controlled skin strippings as substrate.
Meaningful analytical morphologic aspects are open to SACD and CSSS samplings. Fortunately, sound mathematical relationships are perceived between the original structure and the images, and many of these relationships appear quite obvious. In general, multivariate statistical analysis is required to validate a meaningful conclusion. When testing xenobiotic effects on SC, the stratum disjunctum response as assessed by SQMI on SACD is different from CSMI on the thicker SC collected by CSSS. Nevertheless, both methods appear valuable and complementary.
Beyond SC scrapings, SACD and CSSS provide useful information in the field of cosmetology. These simple, low-cost, and minimally invasive methods allow the clinician, the cosmetologist, and the experimentalist to avoid invasive procedures within limits of well-defined indications. Less than 3 minutes are required between SC sampling and its microscopic examination. There are obvious features and subtle characteristics discernible in the CSSS structure for establishing key points in a variety of skin conditions. It is important to stress that no single criterion should usually be relied upon for a definitive feature on CSSS. Rather, a constellation of clues should be sought. A range of quantifications are made possible on CSSS using computer-assisted image analysis.
The SC holds a large amount of information about the skin itself, and the internal milieu as well. In many instances, CSSS appears to represent a convenient way that is currently available for exploring many facets of cosmetic science. There is no limitation to its use in cosmetic science. It represents a safe and time-saving method. The main advantages of CSSS are as follows:
Swift and safe method for SC samplings
Multiple diagnostic possibilities
Cheap procedure
Hygienic procedure
Minimally invasive
Easy handling
Stable samples over time
Defined skin stripping for scientific applications
Established method, known for more than 30 years
Supported by a number of scientific contributions
Acknowledgments
No sources of funding were used to assist in the preparation of this paper. The authors appreciate the excellent secretarial assistance of Mrs Ida Leclercq.
Footnotes
Disclosure
The authors report no conflicts of interest in this work.
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Abstract
Introduction
SC collection
Overall aspect of normal CSSS
Analytical morphology of skin strippings
Conclusion
Acknowledgments
Footnotes
References
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Lichenoid drug eruption
Lichenoid drug eruption — extra information
Synonyms:
Lichenoid eruption due to medications
Categories:
Reactions
ICD-10:
L43.2
ICD-11:
EH62
SNOMED CT:
109254000
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Reactions
Lichenoid drug eruption
Author: Dr Delwyn Dyall-Smith FACD, Dermatologist, 2010.
Introduction Demographics Clinical features Diagnosis Treatment
What is a lichenoid drug eruption?
Lichenoid eruptions are uncommon skin rashes that can be induced by many environmental agents, medications or industrial by-products such as inhaled particles. The rash of a lichenoid drug eruption can sometimes be difficult to distinguish from idiopathic lichen planus because of similarities in the clinical appearance and the pathology seen on skin biopsy.
Who gets a lichenoid drug eruption?
Many medications have been reported in association with lichenoid drug eruptions, but often the time between starting the medication and the rash appearing (the latent period) can be long, more than one year in some cases. Generally, the latent period is 2–3 months (although this can vary for different drugs) and has even been reported to develop after the drug has been ceased. This can make it challenging to identify the culprit drug.
Medications commonly reported to trigger a lichenoid drug eruption include:
Antihypertensives – ACE inhibitors, beta-blockers, nifedipine, methyldopa
Diuretics – hydrochlorothiazide, frusemide, spironolactone
Non-steroidal anti-inflammatory drugs (NSAIDs)
Phenothiazine derivatives
Anti-convulsants – carbamazepine, phenytoin
Medicines to treat tuberculosis
Antifungal medication – ketoconazole
Chemotherapeutic agents – 5-fluorouracil, imatinib
Antimalarial agents such as hydroxychloroquine
Sulfa drugs including sulfonylurea hypoglycaemic agents, dapsone, mesalazine, sulfasalazine
Metals – gold salts
Others – allopurinol, iodides and radiocontrast media, interferon-α, omeprazole, penicillamine, tetracycline
Other medications that have been reported in association with lichenoid drug eruptions include:
Tumour necrosis factor antagonists such as infliximab, etanercept and adalimumab
Imatinib mesylate (tyrosine kinase inhibitor)
Misoprostol (prostaglandin E1 agonist)
Sildenafil citratus (Viagra™)
Vaccines (especially those for herpes zoster and influenza).
If a lichenoid eruption has developed to a drug, then it is quite possible for the same reaction to appear more quickly after exposure to another medication in the same family. Examples reported have included the proton pump inhibitors (for dyspepsia) and the HMG-CoA reductase inhibitors (for high cholesterol).
The lichenoid drug eruption of hydroxyurea has been reclassified as a drug-induced dermatomyositis.
What are the clinical features of a lichenoid drug eruption?
Lichenoid drug eruptions can look much the same as idiopathic lichen planus although there can be features that may help to distinguish them, which may include:
Extensive rash distributed symmetrically over the trunk and limbs
Photodistribution – the rash is predominantly in areas exposed to the sun
The rash may be scaly resembling eczema or psoriasis
Wickham striae are usually absent
Nail and mucous membrane (e.g., mouth) involvement is uncommon (oral lichen planus)
More likely to leave marked pigmentation after the active rash has cleared.
Lichenoid drug eruption
Lichenoid drug eruption
Lichenoid drug eruption
How is a lichenoid drug eruption diagnosed?
The diagnosis may be suspected from the unusual clinical features and a skin biopsy then taken. The pathological features of a lichenoid drug eruption may be difficult to distinguish from idiopathic lichen planus, but the diagnosis of lichenoid drug eruptions may be suggested by the types and distribution of inflammatory cells as well as other changes.
A detailed history of medications taken in the preceding year, including those taken only briefly (or even just once), may help to identify the culprit.
Sometimes patch testing with the drug may confirm the identification, but false negatives are common.
A ‘challenge test’ involves deliberately re-administering the drug to the patient expecting the eruption to reappear, but more rapidly. Sometimes this happens unintentionally when another member of the same class of drugs is given to treat the original medical problem, or the same drug is provided for a different issue.
Ceasing the suspected medication with the resolution of the rash is usually taken as confirmation of the diagnosis and drug trigger.
What is the treatment for lichenoid drug eruption?
The trigger medication should be stopped and should result in improvement in the rash, although it can take weeks to months for it to disappear. Commonly flat pigmented freckles persist and fade more slowly. Nail disease will take six months or more to clear, although improvement can be seen gradually extending over this time.
Sometimes the medication cannot be ceased because of the importance of the underlying medical condition compared to the rash, e.g. imatinib for chronic myeloid leukaemia or gastrointestinal stromal tumour. The dose may be reduced or continued unchanged and the rash treated with topical steroid cream or, if very extensive and severe, oral corticosteroids such as prednisone or prednisolone. Steroids may give good relief or even resolution.
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References
Cruz MJ, Duarte AF, Baudrier T, Cunha AP, Barreto F, Azevedo F. Lichenoid drug eruption induced by misoprostol. Contact Dermatitis 2009: 61: 240–2. PubMed
Pua VSC, Scolyer RA, Barnetson RStC. Pravastatin-induced lichenoid drug eruption. Australas J Dermatol 2006; 47: 57–9. PubMed
Sendagorta E, Herranz P, Feito M, Ramírez P, Feltes R, Floristán U, Mariño-Enriquez A, Casado M. Lichenoid drug eruption related to imatinib: report of a new case and review of the literature. Clin Exp Dermatol 2009; 34: e315–e316. PubMed
Marumi Saito, Koichiro Nakamura, Fumio Kaneko. Lichenoid drug eruption of nails induced by propylthiouracil. J Dermatology 2007; 34: 696–8. PubMed
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How are the angles of an equilateral triangle related?
Harsh Muni Dwivedi
Former Student
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Author has 925 answers and 1.4M answer views
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Ved Prakash Sharma
Former Lecturer at Sbm Inter College, Rishikesh (1971–2007)
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Author has 14.3K answers and 16.4M answer views
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5y
In an equilateral triangle ABC , all the three sides are equal ( AB=BC=CA). therefore
all the three angles are equal (angle A=angle B= angle C).
Let angle A=angle B=angle C = x.
But angle A+angle B+angle C= 180°
or. x+x+x= 180°
or. 3.x=180°
or. x=180°/3=60°.
Hence , angle A=angle B=angle C= 60°. Answer.
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In an equilateral triangle, all three sides are of equal length, and all three angles are equal as well. Each angle in an equilateral triangle measures exactly 60 degrees.
Key Points:
Equal Sides: All three sides are equal.
Equal Angles: Each angle is 60 degrees.
Sum of Angles: The sum of the angles in any triangle is always 180 degrees, which is consistent in an equilateral triangle since .
This uniformity is a defining characteristic of equilateral triangles.
Related questions
What is the relationship between the angles and sides in an equilateral triangle and a right triangle?
What is the value of two angles of an equilateral triangle?
How would I prove that an equilateral triangle has 3 equal angles?
What's the relation between a right triangle and an equilateral triangle?
Is it possible for an equilateral triangle to have unequal angles?
Enjamum
Top NITian, Android Developer, Competitive programmer
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5y
All the angles of an equilateral triangle are equal to 60
Shamsul Arefin
Studied at Chittagong University of Engineering and Technology (Graduated 2000)
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11mo
Originally Answered: What is the relationship between the angles and sides in an equilateral triangle?
·
For an equilateral triangle all sides are equal, ie if sides are a,b and c then a=b=c.
All three angle of an equilateral triangle is 60 degree.
Related questions
What is the proof that a triangle is equilateral if its angles are equal to their corresponding sides?
How do you describe an equilateral triangle as a geometric figure with one special property that makes it unique from other triangles of different sizes and angles of sides?
Why is it called an equilateral triangle if its three sides are not equal to each other? Why does it have three equal angles then?
What are the properties of an equilateral triangle?
What are the measures of the angles in an equilateral triangle (not a right triangle)?
Anthony Hawken
Author has 10.8K answers and 3.7M answer views
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11mo
Originally Answered: What is the relationship between the angles and sides in an equilateral triangle?
·
Yet another stupid and pointless question created by Quora Prompt Generator.
The sides are all equal. The angles are all equal.
There is no obvious relationship between sides and angles.
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5y
The relation between the angles of an equilateral triangle is that they are all equal having a measure of 60°.
Aditya Sahoo
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All the angles of an equilateral triangle are equal and each angle measures 60 degrees
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How do I show that the angles of an equilateral triangle are 60° each?
Here is [math]\triangle{ABC}[/math], where [math]AB=AC=BC=a[/math]
As [math]AB=AC[/math], triangle is isoceles and by properties of isoceles triangle, altitude is perpendicular bisector of base, therefore [math]\angle{ADC}=\angle{ADB}=90°[/math] and [math]BD=CD=\frac{a}{2}[/math]
By cosine laws,
[math]CosB=CosC=\frac{a}{2}\frac{1}{a}=\frac{1}{2}[/math]
[math]\angle{B}=\angle{C}=arccos(\frac{1}{2})=60°[/math]
[math]\angle{A}=180°-(60°×2)=60°[/math]
Hence proved.
Here is [math]\triangle{ABC}[/math], where [math]AB=AC=BC=a[/math]
As [math]AB=AC[/math], triangle is isoceles and by properties of isoceles triangle, altitude is perpendicular bisector of base, therefore [math]\angle{ADC}=\angle{ADB}=90°[/math] and [math]BD=CD=\frac{a}{2}[/math]
By cosine laws,
[math]CosB=CosC=\frac{a}{2}\frac{1}{a}=\frac{1}{2}[/math]
[math]\angle{B}=\angle{C}=arccos(\frac{1}{2})=60°[/math]
[math]\angle{A}=180°-(60°×2)=60°[/math]
Hence proved.
Peter Conrad
I got 800 on my math SAT, back when that was the best score you could get.
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What is the value of each angle of an equilateral triangle?
The answer is 60 degrees.
Now, here is my question: why do you feel that the way to answer this question is to go to Quora? Are you taking a math course that has a section on triangles, angles, or polygons? Has your teacher explained anything about these things to you that you find it impossible to understand? Or are you trying to figure out the subject by yourself? In any of these cases, this is a very simple question, but your ansking it implies that you are pretty lost. You really have many more questions to answer to gain an understanding of the subject. See if you can find someone to sit d
The answer is 60 degrees.
Now, here is my question: why do you feel that the way to answer this question is to go to Quora? Are you taking a math course that has a section on triangles, angles, or polygons? Has your teacher explained anything about these things to you that you find it impossible to understand? Or are you trying to figure out the subject by yourself? In any of these cases, this is a very simple question, but your ansking it implies that you are pretty lost. You really have many more questions to answer to gain an understanding of the subject. See if you can find someone to sit down with and explore not only this, but all the related questions whose answers you need to know.
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Why does an equilateral triangle have equal sides and angles?
Definition. (Euclid) An equilateral triangle is a triangle that has three equal sides.
That definition does not include three equal angles. There are various ways you can prove that an equilateral triangle has three equal angles. One natural way to do that is to use a property of isosceles triangles.
Theorem. (Euclid I.5) If a triangle has two equal sides, then the angles opposite the equal sides are equal.
Theorem. (Euclid I.6) Conversely, if a triangle has two equal angles, then the sides opposite the equal angles are equal
Now, since an equilateral triangle [math]ABC[/math] has equal sides [math]AB[/math] and [math]AC,[/math] theref
Definition. (Euclid) An equilateral triangle is a triangle that has three equal sides.
That definition does not include three equal angles. There are various ways you can prove that an equilateral triangle has three equal angles. One natural way to do that is to use a property of isosceles triangles.
Theorem. (Euclid I.5) If a triangle has two equal sides, then the angles opposite the equal sides are equal.
Theorem. (Euclid I.6) Conversely, if a triangle has two equal angles, then the sides opposite the equal angles are equal
Now, since an equilateral triangle [math]ABC[/math] has equal sides [math]AB[/math] and [math]AC,[/math] therefore it’s an isosceles triangle, so by the theorem, angle [math]B[/math] equals angle [math]C.[/math] Likewise, since side [math]AC[/math] equals side [math]BC,[/math] therefore angle [math]A[/math] equals angle [math]B.[/math] Thus all three angles of an equilateral triangle are equal.
That shows a triangle with three equal sides has three equal angles. Using the converse part of the theorem above, you can show a triangle with three equal angles has three equal sides.
There are lots of ways to prove the two theorems listed above, and you can find proofs of them in any textbook on plane geometry.
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Robert Gross
Retired Electrical Engineer
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Author has 17.9K answers and 4.5M answer views
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11mo
Originally Answered: What is the relationship between the angles and sides in an equilateral triangle?
·
Other than that all the sides are equal, and all the angles are equal, there is no relationship.
Subhasish Debroy
Former SDE at Bharat Sanchar Nigam Limited (BSNL)
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Author has 6.6K answers and 5.8M answer views
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2y
Related
Why are the angles in a triangle measure related by the cosine law?
Here ABCis a triangle whose angles are denoted by A,B and C and a,b, and c indicates the sides of triangle which are just opposite to angles A, B and C respectively.
From above we write 2ab cosC= a²+b²-c²
=> cosC = (a²+b²-c²)/2ab , here cosine of angle C is expressed in terms of sides of particular triangle.Similarly other angles can be expressed.
=> C = arccos[(a²+b²-c²)/2ab] , As sides can be measured easily , any angle of triangle can be measured easily by using co-sine rule.
Ans.
Here ABCis a triangle whose angles are denoted by A,B and C and a,b, and c indicates the sides of triangle which are just opposite to angles A, B and C respectively.
From above we write 2ab cosC= a²+b²-c²
=> cosC = (a²+b²-c²)/2ab , here cosine of angle C is expressed in terms of sides of particular triangle.Similarly other angles can be expressed.
=> C = arccos[(a²+b²-c²)/2ab] , As sides can be measured easily , any angle of triangle can be measured easily by using co-sine rule.
Related questions
What is the relationship between the angles and sides in an equilateral triangle and a right triangle?
What is the value of two angles of an equilateral triangle?
How would I prove that an equilateral triangle has 3 equal angles?
What's the relation between a right triangle and an equilateral triangle?
Is it possible for an equilateral triangle to have unequal angles?
What is the proof that a triangle is equilateral if its angles are equal to their corresponding sides?
How do you describe an equilateral triangle as a geometric figure with one special property that makes it unique from other triangles of different sizes and angles of sides?
Why is it called an equilateral triangle if its three sides are not equal to each other? Why does it have three equal angles then?
What are the properties of an equilateral triangle?
What are the measures of the angles in an equilateral triangle (not a right triangle)?
Can an equilateral triangle have one acute angle and two obtuse angles? If so, why?
How do you work out the angles of an equilateral triangle?
If three pairs of corresponding angles in a pair of triangles are equal, do the triangles become similar or equilateral?
What are the examples of equilateral triangle?
What's the difference between an equilateral triangle and an isosceles right-angled triangle?
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673 | https://forum.prutor.ai/index.php?threads/difference-between-bond-dissociation-enthalpy-and-bond-enthalpy.1124/ | Difference between bond dissociation enthalpy and bond enthalpy? | Sathee Forum
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Difference between bond dissociation enthalpy and bond enthalpy?
Thread starterLiveDoubtClearing2
Start dateNov 15, 2024
Tagscn9_kossel-lewis approach to chemical bond formation- the concept of ionic and covalent bonds.
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Nov 15, 2024
#1
Difference between bond dissociation enthalpy and bond enthalpy?
Last edited by a moderator: Feb 3, 2025
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Chemistry.LiveDoubtClearing
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Staff member
Feb 3, 2025
#2
Bond Dissociation Enthalpy:
Definition: Bond dissociation enthalpy (BDE) is the enthalpy change required to break a specific bond in a molecule in the gas phase, resulting in the formation of two radicals. It is usually measured for a single bond in a specific molecule.
Specificity: It is specific to a particular bond in a particular molecule.
Example: The bond dissociation enthalpy of the H-H bond in H₂ is the energy required to break the H-H bond to form two hydrogen atoms (H•).
### Bond Enthalpy:
Definition: Bond enthalpy (also known as bond energy or average bond enthalpy) is the average energy required to break a particular type of bond in one mole of gaseous molecules. It is an average value derived from the bond dissociation enthalpies of a particular type of bond in different molecules.
Averaging: It is an average value because it considers the bond dissociation enthalpies of the same type of bond in various compounds.
Example: The bond enthalpy of a C-H bond is an average value derived from the bond dissociation enthalpies of C-H bonds in different hydrocarbons.
### Key Differences:
Specificity vs. Averaging:
Bond dissociation enthalpy is specific to a particular bond in a specific molecule.
Bond enthalpy is an average value for a particular type of bond across different molecules.
Measurement Context:
Bond dissociation enthalpy is measured for a single bond in a specific molecule.
Bond enthalpy is an averaged value and is not specific to a single molecule.
### Relevance in Chemistry:
Organic Chemistry: Understanding the bond dissociation enthalpy and bond enthalpy helps in predicting the stability of organic molecules and the energy changes during chemical reactions.
Inorganic Chemistry: These concepts are used to understand the stability and reactivity of inorganic compounds.
Physical Chemistry: These terms are fundamental in thermodynamics and kinetics, helping to explain the energy changes during chemical reactions and the strength of chemical bonds.
In summary, while both bond dissociation enthalpy and bond enthalpy deal with the energy required to break chemical bonds, bond dissociation enthalpy is specific to a particular bond in a specific molecule, whereas bond enthalpy is an average value for a type of bond across different molecules.
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674 | https://tasks.illustrativemathematics.org/7.RP | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Grade 7
Domain Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical problems.
7.RP.A. Analyze proportional relationships and use them to solve real-world and mathematical problems.
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7.RP.A.1. Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks $1/2$ mile in each $1/4$ hour, compute the unit rate as the complex fraction $\frac{1/2}{1/4}$ miles per hour, equivalently $2$ miles per hour.
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7.RP.A.2. Recognize and represent proportional relationships between quantities.
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7.RP.A.2.a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
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7.RP.A.2.b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships.
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7.RP.A.2.c. Represent proportional relationships by equations. For example, if total cost $t$ is proportional to the number $n$ of items purchased at a constant price $p$, the relationship between the total cost and the number of items can be expressed as $t = pn$.
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7.RP.A.2.d. Explain what a point $(x, y)$ on the graph of a proportional relationship means in terms of the situation, with special attention to the points $(0, 0)$ and $(1, r)$ where $r$ is the unit rate.
No tasks yet illustrate this standard.
7.RP.A.3. Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.
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675 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/545fbd5776098bdc96b6cdc702211d7b_MIT18_01SCF10_Ses83b.pdf | Z Z Z Z Z Area of an Off Center Circle Let’s find the area in polar coordinates of the region enclosed by the curve r = 2a cos θ. We’ve previously shown that this curve describes a circle with radius a centered at (a, 0). In rectangular coordinates its equation is (x − a)2 + y2 = a2 . (a,0) r θ y x Figure 1: Off center circle r = 2a cos θ. We’re going to integrate an infintessimal amount of area dA. The integral π 2 to θ2 = π 2 . We could find these limits by looking at will go from θ1 = − π Figure 1; to draw the circle we might start by moving “down” at angle − we move along the bottom of the circle toward (2a, 0) the angle increases to 0, . As 2 and as we trace out the top of the circle we’re moving from angle 0 to angle π 2 (“up”). We might also find the limits of integration by looking at the formula and π π π realizing that the cosine function is positive for − r = 2a cos θ is 0, so the two ends of the curve meet at the origin. < θ < . When θ = ± , 2 2 2 Our infinitessimal unit of area is dA = 2 1 r2 dθ, so: 2 π 1 2 dθ A = r 2 2 π − 2 π = 1(2a cos θ)2 dθ 2 2 π − 2 π = 2a 2 cos 2 θ dθ 2 π − 2 π = 2a 2 cos 2 θ dθ (half angle formula) 2 π − 2 π 1 + cos 2θ 2 = 2a = a dθ 2 2 π − 1 2 π 2 θ + sin 2θ 2 − 1 2 π π π 2 = a 2 − − 2 = πa2 . We know that the area of a circle of radius a is πa2; our answer is correct. 2 MIT OpenCourseWare 18.01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: |
676 | https://www.cuemath.com/measurement/surface-area-of-cuboid/ | Surface Area of Cuboid
The surface area of a cuboid is the total space occupied by it. A cuboid is a six-faced three-dimensional shape in which each face is in the shape of a rectangle. If l, w, and h are the length, width, and the height of a cuboid, then the surface area of a cuboid formula is 2 lw + 2 wh + 2hl.
This formula can be derived by observing that a cuboid has three pairs of faces, each pair having the same area (because opposite faces are congruent), and adding up the areas of all six faces. Let us learn more about the formula of the surface area of the cuboid and solve problems based on it.
| | |
--- |
| 1. | What is the Surface Area of Cuboid? |
| 2. | Surface Area of Cuboid Formula |
| 3. | Total Surface Area of Cuboid (TSA) |
| 4. | Lateral Surface Area of Cuboid (LSA) |
| 5. | Difference Between TSA and CSA of Cuboid |
| 6. | How To Calculate the Surface Area of Cuboid? |
| 7. | Applications of Surface Area of Cuboid |
| 8. | FAQs on Surface Area of Cuboid |
What is the Surface Area of Cuboid?
The surface area of a cuboid is the total area of all its 6 faces. Since a cuboid is a three-dimensional solid shape, the value of its surface area depends on the dimensions of its length, width, and height. The change in any of the dimensions of a cuboid changes the value of its surface area. The surface area of a cuboid is expressed in square units.
The surface area of a cuboid is an important metric as it is useful in a wide range of applications such as construction, manufacturing, and packaging. For example, when determining how much paint is needed to cover a cuboidal box, or when calculating the amount of material required to manufacture a cuboidal object, the surface area of the cuboid is an essential measurement.
Surface Area of Cuboid Formula
The formula for the surface area of a cuboid depends on the type of surface area that has been asked for. A cuboid has two kinds of surface areas:
TSA of Cuboid : Total Surface Area (also known just as "surface area")
LSA (or) CSA of Cuboid: Lateral Surface Area (or) Curved Surface Area
The total surface area of the cuboid is obtained by adding the area of all the 6 faces while the lateral surface area of the cuboid is calculated by finding the area of each face excluding the base and the top. The total surface area and the lateral surface area of a cuboid can be expressed in terms of length (l), width (w), and height of cuboid (h) as:
Total Surface Area of Cuboid = 2 (lw + wh + lh)
Lateral Surface Area of Cuboid = 2h (l + w)
If the surface area is given or asked without any specifications, it means that it refers to the total surface area. Let us see how to derive the formulas for the surface area of the cuboid by opening up a cuboidal box.
Total Surface Area of Cuboid (TSA)
The TSA of cuboid is sometimes just referred to as surface area of cuboid itself. A cuboid has 6 rectangular faces. Now, in order to find the total surface area of a cuboid, we need to add the area of all the 6 rectangular faces. Since each face of a cuboid is a rectangle, hence the area of the rectangle for each face is calculated and added to get the total surface area of the cuboid. Let us understand this with the help of the figure given below.
The faces of cuboid are numbered as 1, 2, 3, 4, 5, and 6 as shown in the figure given above. In other words, if we see the cuboid in the form of a two-dimensional figure as a net, we get this figure.
Area of Rectangles 1 and 2: the area of the rectangles of the top and bottom faces = lw + lw = 2lw
Area of Rectangles 3 and 4: the area of the rectangles of the left and right sides = wh + wh = 2wh
Area of Rectangles 5 and 6: area of the rectangles on the front and back faces = lh + lh = 2lh
Hence, the total surface area (TSA) of the six faces = 2lw + 2wh + 2lh.
Thus, the TSA of a cuboid of dimensions l, w, and h is 2 (lw + wh + lh).
Lateral Surface Area of Cuboid (LSA)
The lateral surface area (LSA) of a cuboid is also known as the curved surface area of cuboid (CSA). This is the combined surface area of the four lateral (side) faces. In the figure given above, if we remove the top and bottom faces, we will get the area of the lateral surface area of the cuboid. The LSA (or) CSA of cuboid is,
Lateral Surface Area = Area of the four side faces
= 2lh + 2wh
= 2h(l + w)
It is important to note that CSA of cuboid only includes the vertical faces and does not include the top and bottom faces, so it is always less than the total surface area of the cuboid.
To understand this with the help of a real-life example let us imagine a room in the shape of a cuboid. The total surface area will be the combined area of the six faces of the room (the four vertical walls + the floor + the ceiling) while the lateral surface area will be the combined surface area of the four vertical walls (the areas of the floor and the ceiling are not added).
Difference Between TSA and CSA of Cuboid
The main difference between TSA and CSA of the cuboid is that TSA includes all six faces of the cuboid, whereas CSA only includes the vertical faces of the cuboid. TSA is a measure of the total area that needs to be covered or painted on the cuboid, while CSA is a measure of the area of the curved surface that wraps around the cuboid. The following table summarizes the differences between the total surface area and curved surface area of cuboid in detail.
| Property | TSA of Cuboid | CSA of Cuboid |
---
| Definition | Sum of areas of all 6 faces of cuboid | Sum of areas of all 4 vertical faces of cuboid |
| Formula | TSA = 2(lw + wh + lh) | CSA = 2h(l + w) |
| Involves | All faces of the cuboid | Only vertical faces of the cuboid |
| Example | If l = 4, w = 3, h = 2, then TSA = 2(4 × 3 + 3 × 2 + 4 × 2) = 52 cm2 | If l = 4, w = 3, h = 2, then CSA = 2(2) (4 + 3) = 28 cm3 |
How to Find the Surface Area of Cuboid?
The surface area of a cuboid is the total area of each surface of a cuboid. Let us use the following steps to calculate the total surface area of a cuboid.
Step 1: Check if the given dimensions of cuboids are in the same units or not. If not, convert the dimensions into the same units.
Step 2: Use the formula for total surface area = 2(lw + wh + lh) if we need TSA and lateral surface area = 2h(l + w) if we need LSA.
Step 3: Substitute the given values to get the area and express it in square units.
Example: Find the total surface area and the lateral surface area of a cuboid whose length = 6 inches, width = 4 inches, and height = 3 inches.
Solution: Here, length (l) = 6 inches, width (w) = 4 inches and height (h) = 3 inches
Total surface area of cuboid = 2 (lw + wh + lh) = 2 [(6 × 4) + (4 × 3) + (6 × 3)] = 108 in2
Lateral surface area of cuboid = 2h(l + w) = (2 × 3)(6 + 4) = 60 in2
Applications of Surface Area of Cuboid
The surface area of a cuboid has various real-world applications in various fields such as engineering, construction, manufacturing, architecture, etc. Here are a few applications.
Packaging: Knowing the surface area of a cuboid can help companies determine the amount of material required and ensure that the product fits properly inside the packaging.
Painting: The information about the amount of paint needed to cover a rectangular box or cuboid is crucial for painters to estimate the cost of paint required for a project.
Manufacturing: In manufacturing, the surface area of cuboids is used to calculate the amount of material needed to manufacture a product.
Heat Transfer: The surface area of a cuboid is also used to calculate the rate of heat transfer in a solid object and hence surface area of a material is a critical factor in its thermal conductivity.
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Examples on Surface Area of Cuboid
Example 1: Find the lateral surface area of a cuboid whose length = 20 inches, width = 10 inches, and height = 15 inches.
Solution: As we know, the lateral surface area of a cuboid is L = 2h(l + w) and it is given that,
length (l) = 20 in
width (w) = 10 in
height (h) = 15 in
Let us substitute the values in the formula,
LSA of cuboid = 2h(l + w)
= 2 × 15 (20 + 10)
= 30 (20 + 10)
=30 × (30)
= 900 square inches
Answer: The lateral surface area of the cuboid is 900 square inches.
2. Example 2: Find the total surface area of a cuboid which is 8 m long, 5 m broad, and 4 m high.
Solution: Given, length of the cuboid (l) = 8 m, width (w) = 5 m, height (h) = 4 m.
The total surface area of the cuboid = 2 (lw + wh + lh)
So, let us substitute the values in the formula.
TSA of the cuboid = 2 (lw + wh + lh)
= 2 [(8 × 5) + (5 × 4) + (8 × 4)]
= 2 [40 + 20 + 32]
= 2 × 92
= 184 m2
Answer: The total surface area of the cuboid is 184 m2
3. Example 3: State true or false.
a.) A cuboid has 7 rectangular faces.
b.) In order to find the total surface area of a cuboid, we need to add the area of all 6 rectangular faces.
Solution:
a.) False, a cuboid has only 6 rectangular faces.
b.) True, in order to find the TSA of a cuboid, we need to add the area of all 6 rectangular faces.
Answer: (a) False (b) True
Show Answer >
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Practice Questions on Surface Area of Cuboid
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FAQs on Surface Area of Cuboid
What Does the Total Surface Area of Cuboid Mean?
The total surface area of a cuboid is the sum of all its surfaces. In order to find the total surface area of a cuboid, we need to add the area of all the 6 rectangular faces. The formula for the total surface area of a cuboid is 2 (lw + wh + lh) where l = length, w = width, and h = height of the cuboid.
What is the Lateral Surface Area of a Cuboid?
The lateral surface area of a cuboid is the value of the surface area of a cuboid excluding its top and bottom surfaces. The formula for the lateral surface area of a cuboid is expressed as, 2h(l + w) where l = length, w = width, and h = height of the cuboid.
Where to Find the Surface Area of Cuboid Calculator?
The area of cuboid can be calculated using the calculator. It shows the surface area just by entering the values of length, width, and height of the cuboid.
What is the Difference Between the TSA and the LSA of a Cuboid?
The difference between total surface area and the lateral surface area of a cuboid is given below:
The total surface area of a cuboid is the sum of the areas of all 6 faces and is calculated by the formula 2(lw + wh + lh).
The lateral surface area of a cuboid is the sum of the areas of faces excluding the top and the base and is calculated by the formula 2h(l + w)
What is the Total Surface Area of a Cuboidal Box Without Lid?
The total surface area of a cuboidal box without a lid can be given in two ways:
Method 1: Total surface area of a cuboidal box without lid = Total surface area of the cuboid - 1 rectangular face;
Method 2: Total surface area of a cuboidal box without lid = Lateral surface area of cuboid + 1 rectangular face;
It should be noted that both the methods result in the same answer.
What is the Unit for Surface Area of Cuboid?
The surface area of a cuboid is expressed in square units like cm2, m2, inch2, and so on. This is because surface area is also area.
What is the TSA and CSA of a Cuboid?
The TSA and CSA of a cuboid are the abbreviations of the Total Surface Area (TSA) and Curved Surface Area (CSA) of a cuboid. The total surface area is the sum of the surface area of all 6 faces while curved surface area is the measure of surface area excluding the top and bottom faces.
What is the Total Surface Area of a Cuboid Formula?
The formula that is used to find the total surface area of a cuboid is 2 (lw + wh + lh); where l = length, w = width, and h = height of the cuboid.
What is the Lateral Surface Area of a Cuboid Formula?
The formula that is used to find the lateral surface area of a cuboid is 2h(l + w) where l, w, and h are the length, width, and height of the cuboid respectively.
Q1: The lateral surface area of a cuboid is the ____
Q2: The total surface area of a cuboid with dimensions l, w and h is ____
Q3: The lateral surface area of a cuboid whose length is 7 units, width is 6 units and height is 5 units is equal to ____.
Q4: Find the total surface area of a cuboid if its length is 10 units, width is 8 units and height is 6 units.
Q5: The lateral surface area of a cuboid is 240 square units. If the sum of its length and width is 20 units, find the height of the cuboid.
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The equation x 2+a x−a 2−1=0 will have roots of opposite signs if:
A) a∈(−∞,∞)
B) a∈[−1,1]
C) a∈(−∞,−1)∪(1,∞)
D) None of these
Answer
Verified
552.6k+ views
1 likes
Hint:Here we will use properties of roots of the quadratic equation. If the roots are of the opposite sign then their product will be negative. And product = constant coefficient of x 2 So, Product is less than 0.
Complete step-by-step solution:
Given: x 2+a x−a 2−1=0 will have roots of the opposite signs.
x 2+a x−(a 2+1)=0 as roots have opposite signs,
Compare the given equation with the standard form-
a x 2+b x+c=0
∴a=1
∴b=a
∴c=−(a 2+1)
Product of roots is less than zero by property -
⇒α×β<0 and D>0
⇒c a<0 a n d b 2−4 a c>0
Put values in the above conditions -
⇒−(a 2+1)1<0 a n d a 2−4(1){−(a 2+1)}>0
By simplification -
⇒(a 2+1)>0 and a 2+4 a 2+4>0
⇒a∈R and 5 a 2+4>0
N o w,a∈R and a 2>−4 5
So, in both cases, a∈R which means a∈(−∞,∞)
Hence option (B) is the correct answer.
Note:In these types of problems where nature of root and equation is given. We first try to connect the given nature of roots to coefficients and product or sum of roots whichever is needed. Then check inequality after assigning the required values. Also, here we have written as if signs of roots are opposite then the product of roots will be negative, as with the same sign product of roots will be positive. As per the property it is like (-)(-)=(+) and (+)(+)=(+).
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679 | ERROR: type should be string, got "https://web1.eng.famu.fsu.edu/~tawfiq/Geotech%8D%20Spring%202020/Module-2%20Stresses%20In%20Soil/CHAPTER_6_STRESS_DISTRIBUTION_IN_SOILS_D(2).pdf" | CHAPTER 6 STRESS DISTRIBUTION IN SOILS DUE TO SURFACE LOADS 6.1 INTRODUCTION Estimation of vertical stresses at any point in a soil-mass due to external vertical loadings are of great significance in the prediction of settlements of buildings, bridges, embankments and many other structures. Equations have been developed to compute stresses at any point in a soil mass on the basis of the theory of elasticity. According to elastic theory, constant ratios exist between stresses and strains. For the theory to be applicable, the real requirement is not that the material necessarily be elastic, but there must be constant ratios between stresses and the corresponding strains. Therefore, in non-elastic soil masses, the elastic theory may be assumed to hold so long as the stresses induced in the soil mass are relatively small. Since the stresses in the subsoil of a structure having adequate factor of safety against shear failure are relatively small in comparison with the ultimate strength of the material, the soil may be assumed to behave elastically under such stresses.
When a load is applied to the soil surface, it increases the vertical stresses within the soil mass. The increased stresses are greatest directly under the loaded area, but extend indefinitely in all directions. Many formulas based on the theory of elasticity have been used to compute stresses in soils. They are all similar and differ only in the assumptions made to represent the elastic conditions of the soil mass. The formulas that are most widely used are the Boussinesq and Westergaard formulas. These formulas were first developed for point loads acting at the surface.
These formulas have been integrated to give stresses below uniform strip loads and rectangular loads.
The extent of the elastic layer below the surface loadings may be any one of the following: 1. Infinite in the vertical and horizontal directions.
2. Limited thickness in the vertical direction underlain with a rough rigid base such as a rocky bed.
173 174 Chapter 6 The loads at the surface may act on flexible or rigid footings. The stress conditions in the elastic layer below vary according to the rigidity of the footings and the thickness of the elastic layer. All the external loads considered in this book are vertical loads only as the vertical loads are of practical importance for computing settlements of foundations.
6.2 BOUSSINESCTS FORMULA FOR POINT LOADS Figure 6.1 shows a load Q acting at a point 0 on the surface of a semi-infinite solid. A semi-infinite solid is the one bounded on one side by a horizontal surface, here the surface of the earth, and infinite in all the other directions. The problem of determining stresses at any point P at a depth z as a result of a surface point laod was solved by Boussinesq (1885) on the following assumptions.
1. The soil mass is elastic, isotropic, homogeneous and semi-infinite.
2. The soil is weightless.
3. The load is a point load acting on the surface.
The soil is said to be isotropic if there are identical elastic properties throughout the mass and in every direction through any point of it. The soil is said to be homogeneous if there are identical elastic properties at every point of the mass in identical directions.
The expression obtained by Boussinesq for computing vertical stress <7, at point P (Fig. 6.1) due to a point load Q is 3(2 1 Q (6.1) where, r = the horizontal distance between an arbitrary point P below the surface and the vertical axis through the point load Q.
z = the vertical depth of the point P from the surface.
1 IR - Boussinesq stress coefficient = — The values of the Boussinesq coefficient IB can be determined for a number of values of r/z.
The variation of /„ with r/z in a graphical form is given in Fig. 6.2. It can be seen from this figure O Q \ x \ \ >WJ\ P °Z Figure 6.1 Vertical pressure within an earth mass Stress Distribution in Soils due to Surface Loads 175 that IB has a maximum value of 0.48 at r/z = 0, i.e., indicating thereby that the stress is a maximum below the point load.
6.3 WESTERGAARD'S FORMULA FOR POINT LOADS Boussinesq assumed that the soil is elastic, isotropic and homogeneous for the development of a point load formula. However, the soil is neither isotropic nor homogeneous. The most common type of soils that are met in nature are the water deposited sedimentary soils. When the soil particles are deposited in water, typical clay strata usually have their lenses of coarser materials within them.
The soils of this type can be assumed as laterally reinforced by numerous, closely spaced, horizontal sheets of negligible thickness but of infinite rigidity, which prevent the mass as a whole from undergoing lateral movement of soil grains. Westergaard, a British Scientist, proposed (1938) a formula for the computation of vertical stress oz by a point load, Q, at the surface as cr, -' Q ,3/2 2 M (6.2) in which fj, is Poisson's ratio. If fj, is taken as zero for all practical purposes, Eq. (6.2) simplifies to Q 1 Q [l + 2(r/z)2]3'2 (6.3) where /,,, = (II a) [l + 2(r/z)2]3/2 is the Westergaard stress coefficient. The variation of / with the ratios of (r/z) is shown graphically in Fig. 6.2 along with the Boussinesq's coefficient IB. The value of Iw at r/z = 0 is 0.32 which is less than that of IB by 33 per cent.
h or 7w 0 0.1 0.2 0.3 0.4 0.5 r/z 1.5 2.5 Figure 6.2 Values of IB or /^for use in the Boussinesq or Westergaard formula 176 Chapters Geotechnical engineers prefer to use Boussinesq's solution as this gives conservative results. Further discussions are therefore limited to Boussinesq's method in this chapter.
Example 6.1 A concentrated load of 1000 kN is applied at the ground surface. Compute the vertical pressure (i) at a depth of 4 m below the load, (ii) at a distance of 3 m at the same depth. Use Boussinesq's equation.
Solution The equation is Q 3/2;r _ _ —/ where /„ = rrj^-Z 7i if' ti f 9 p/Z z [l + ( r / z ) 2 \ Q 1000 (i) When r/z = 0, /„ = 3/2 n = 0.48, a = 0.48^- = 0.48 x —— = 30 kN/m2 B z z2 4x4 (ii) When r/z = 3/4 = 0.75 3/27T 0.156x1000 IR=~T ^T = 0.156, a = — = 9.8 kN/m 2 B l + (0.75)2f2 z 4 x 4 Example 6.2 A concentrated load of 45000 Ib acts at foundation level at a depth of 6.56 ft below ground surface.
Find the vertical stress along the axis of the load at a depth of 32.8 ft and at a radial distance of 16.4 ft at the same depth by (a) Boussinesq, and (b) Westergaard formulae for n = 0. Neglect the depth of the foundation.
Solution (a) Boussinesq Eq. (6.la) " 2 z z2 B' B 271 l + ( r / z ) 2 Substituting the known values, and simplifying IB = 0.2733 for r/z = 0.5 = 45000x02733^n431b/ft2 z (32.8)2 (b) Westergaard (Eq. 6.3) 13/2 Q 1 l + 2(r/z) 2 Substituting the known values and simplifying, we have, / =0.1733forr/7 = 0.5 Stress Distribution in Soils due to Surface Loads 177 therefore, a = (32.8) x 0.1733 = 7.25 lb/ft2 Example 6.3 A rectangular raft of size 30 x 12 m founded at a depth of 2.5 m below the ground surface is subjected to a uniform pressure of 150 kPa. Assume the center of the area is the origin of coordinates (0, 0). and the corners have coordinates (6, 15). Calculate stresses at a depth of 20 m below the foundation level by the methods of (a) Boussinesq, and (b) Westergaard at coordinates of (0, 0), (0, 15), (6, 0) (6, 15) and (10, 25). Also determine the ratios of the stresses as obtained by the two methods. Neglect the effect of foundation depth on the stresses (Fig. Ex. 6.3).
Solution Equations (a) Boussinesq: (b) Westergaard: = — IB, z IB = ' l + <r/^f 2 Q 0.32 The ratios of r/z at the given locations for z = 20 m are as follows: Location (0,0) (6,0) (0, 15) r/z 0 6/20 = 0.3 15/20 = 0.75 Location (6, 15) (10, 25) (^ (Vio 2" r/z f 152)/20 = 0.81 + 252 )/20 = 1.35 The stresses at the various locations at z = 20 m may be calculated by using the equations given above. The results are tabulated below for the given total load Q = qBL = 150 x 12 x 30 = 54000 kN acting at (0, 0) coordinate. Q/z2 =135.
(6,15) (6,0) (6,15) .(0,0) (0,15) (6,15) (6,0) Figure Ex. 6.3 (6,15) (10,25) 178 Chapter 6 Location r/z Boussinesq I0 crJkPa) Westergaard w a/a, w (0,0) (6,0) (0, 15) (6,15) (10, 25) 0 0.3 0.75 0.81 1.35 0.48 0.39 0.16 0.14 0.036 65 53 22 19 5 0.32 0.25 0.10 0.09 0.03 43 34 14 12 4 1.51 1.56 1.57 1.58 1.25 6.4 LINE LOADS The basic equation used for computing a, at any point P in an elastic semi-infinite mass is Eq. (6.1) of Boussinesq. By applying the principle of his theory, the stresses at any point in the mass due to a line load of infinite extent acting at the surface may be obtained. The state of stress encountered in this case is that of a plane strain condition. The strain at any point P in the F-direction parallel to the line load is assumed equal to zero. The stress cr normal to the XZ-plane (Fig. 6.3) is the same at all sections and the shear stresses on these sections are zero. By applying the theory of elasticity, stresses at any point P (Fig. 6.3) may be obtained either in polar coordinates or in rectangular coordinates. The vertical stress a at point P may be written in rectangular coordinates as a = z [1 + U/z) 2] 2 z z where, / is the influence factor equal to 0.637 at x/z - 0.
(6.4) r — \i x •"• + z cos fc) = Figure 6.3 Stresses due to vertical line load in rectangular coordinates Stress Distribution in Soils due to Surface Loads 179 6.5 STRIP LOADS The state of stress encountered in this case also is that of a plane strain condition. Such conditions are found for structures extended very much in one direction, such as strip and wall foundations, foundations of retaining walls, embankments, dams and the like. For such structures the distribution of stresses in any section (except for the end portions of 2 to 3 times the widths of the structures from its end) will be the same as in the neighboring sections, provided that the load does not change in directions perpendicular to the plane considered.
Fig. 6.4(a) shows a load q per unit area acting on a strip of infinite length and of constant width B. The vertical stress at any arbitrary point P due to a line load of qdx acting at jc = x can be written from Eq. (6.4) as ~ 2q n [ ( x - x ) 2 + z 2 ] (6.5) Applying the principle of superposition, the total stress o~z at point P due to a strip load distributed over a width B(= 2b) may be written as +b [(x-x)2+z2}2 dx or -b q , z a =— tan"1 1 n x-b tan" 2bz(x2-b2-z2) x + b (6.6) The non-dimensional values of cjjq are given graphically in Fig. 6.5. Eq. (6.6) can be expressed in a more convenient form as =— [/?+sin/?cos(/?+2£)] n (6.7) x O (a) (b) Figure 6.4 Strip load 180 Chapter 6 (ajq) x 10 4 5 6 7 10 Figure 6.5 Non-dimensional values of <j/q for strip load where /8 and S are the angles as shown in Fig. 6.4(b). Equation (6.7) is very convenient for computing o~, since the angles ft and S can be obtained graphically for any point P. The principal stresses o{ and o"3 at any point P may be obtained from the equations.
cr, = —(/?+sin/?) n (6.8) 0", =— (p-sm, TC (6.9) Example 6.4 Three parallel strip footings 3 m wide each and 5 m apart center to center transmit contact pressures of 200, 150 and 100 kN/m2 respectively. Calculate the vertical stress due to the combined loads beneath the centers of each footing at a depth of 3 m below the base. Assume the footings are placed at a depth of 2 m below the ground surface. Use Boussinesq's method for line loads.
Solution From Eq. (6.4), we have 2/;r _ q \2 Stress Distribution in Soils due to Surface Loads 181 XXX\x\ 1 2 C „-.
30 1 1 c 50 1 loot xxx\xc\ J^/m2 3m , 3 m t y \ 3 m t 3 m C Figure Ex. 6.4 Three parallel footings The stress at A (Fig. Ex. 6.4) is (4 = 2x200F 1 3.14x3 2x100 2x150 3.14x3 1 l + (5/3)2 3.14x3_l + (10/3)2 = 45 kN/m 2 The stress at B ("•} \ z)B 2x200 3x 1 _l + (5/3)2 2x150 2x100 (0/3) = 36.3 kN / m2 The stress at C kt = 2x200 l + (10/3)2 2x150 1 3^r l + (5/3)2 2x100 = 23.74 kN/m 2 6.6 STRESSES BENEATH THE CORNER OF A RECTANGULAR FOUNDATION Consider an infinitely small unit of area of size db x dl, shown in Fig. 6.6. The pressure acting on the small area may be replaced by a concentrated load dQ applied to the center of the area.
Hence = qdb.dl (6.10) The increase of the vertical stress a due to the load dQ can be expressed per Eq. (6.11) as 182 Chapter 6 ^r\ i:M 1 1 1 s ' \ ' \ \ <1 ' \ ' N \ ' ' Figure 6.6 Vertical stress under the corner of a rectangular foundation dcr = dQ 3z3 (6.11) The stress produced by the pressure q over the entire rectangle b x I can then be obtained by expressing dl, db and r in terms of the angles a and /3, and integrating a=a} /?=/?, (6.12) There are several forms of solution for Eq. (6.12). The one that is normally used is of the following form cr=q or 2mn(m2 +n2 +1)1/2 m2+n2 +2 m2+n2+m2n2+l m2+n2+l tan , 2mn(m2+n2+l)l/2 m2 +n2 -m2n2 +1 (6.13) az = ql (6.14) wherein, m = b/z, n = l/z, are pure numbers. / is a dimensionless factor and represents the influence of a surcharge covering a rectangular area on the vertical stress at a point located at a depth z below one of its corners.
Eq. (6.14) is presented in graphical form in Fig. 6.7. This chart helps to compute pressures beneath loaded rectangular areas. The chart also shows that the vertical pressure is not materially altered if the length of the rectangle is greater than ten times its width. Fig. 6.8 may also be used for computing the influence value / based on the values of m and n and may also be used to determine stresses below points that lie either inside or outside the loaded areas as follows.
Stress Distribution in Soils due to Surface Loads 183 z/b = 0.05 Values of / = ojq 0.10 0.15 0.20 0.25 Figure 6.7 Chart for computing GZ below the corner of a rectangular foundation (after Steinbrenner, 1934) When the Point is Inside Let O be an interior point of a rectangular loaded area ABCD shown in Fig. 6.9(a). It is required to compute the vertical stress <Jz below this point O at a depth z from the surface. For this purpose, divide the rectangle ABCD into four rectangles marked 1 to 4 in the Fig. 6.9(a) by drawing lines through O. For each of these rectangles, compute the ratios zfb. The influence value 7 may be obtained from Fig. 6.7 or 6.8 for each of these ratios and the total stress at P is therefore . / T . T . J . T \ / S I C\ &7 = q Ui + h + M + yJ (6.15) When the Point is Outside Let O be an exterior point of loaded rectangular area ABCD shown in Fig. 6.9(b). It is required to compute the vertical stress <TZ below point 0 at a depth z from the surface.
Construct rectangles as shown in the figure. The point O is the corner point of the rectangle OBlCDr From the figure it can be seen that Area ABCD = OB1CD1 - OB{BD2 - OD}DA{ + OA1AD2 (6.16) 184 Chapter 6 0.00 0.01 2 4 6 80.1 2 4 6 81.0 Values of n = l/z 4 6 8 10 Figure 6.8 Graph for determining influence value for vertical normal stress crz at point P located beneath one corner of a uniformly loaded rectangular area. (After Fadum, 1948) O f b I 1 2 ^ 6 3 4 D C (a) When the point 'O' is within the rectangle (b) When the point 'O' is outside the rectangle Figure 6.9 Computation of vertical stress below a point Stress Distribution in Soils due to Surface Loads 185 The vertical stress at point P located at a depth z below point 0 due to a surcharge q per unit area of ABCD is equal to the algebraic sum of the vertical stresses produced by loading each one of the areas listed on the right hand side of the Eq. (6.16) with q per unit of area. If /j to /4 are the influence factors of each of these areas, the total vertical stress is (6.17) Example 6.5 ABCD is a raft foundation of a multi-story building [Fig. 6. 9(b)] wherein AB = 65.6 ft, and BC = 39.6 ft. The uniformly distributed load q over the raft is 73 10 lb/ft2. Determine crz at a depth of 19.7 ft below point O [Fig. 6.9(b)] wherein AA, = 13.12 ft and A,0 = 19.68 ft. Use Fig. 6.8.
Solution Rectangles are constructed as shown in [Fig. 6.9(b)].
Area ABCD = OB}CDl - OB}BD2 - OD1DA1 + OA1AD2 Rectangle OB1CD1 OB1BD2 OD1DA1 OA{AD2 I (ft) 85.28 85.28 52.72 19.68 b (ft) 52.72 13.12 19.68 13.12 m 2.67 0.67 1.00 0.67 n 4.33 4.33 2.67 1.00 7 0.245 0.168 0.194 0.145 Per Eq. (6.17) oz = q (/! - /2 - /3 + /4) = 7310 (0.245 - 0.168 - 0.194 + 0.145) = 204.67 lb/ft2 The same value can be obtained using Fig. 6.7.
Example 6.6 A rectangular raft of size 30 x 12 m founded on the ground surface is subjected to a uniform pressure of 150 kN/m2. Assume the center of the area as the origin of coordinates (0,0), and corners with coordinates (6, 15). Calculate the induced stress at a depth of 20 m by the exact method at location (0, 0).
Solution Divide the rectangle 12 x 30 m into four equal parts of size 6 x 15m.
The stress below the corner of each footing may be calculated by using charts given in Fig. 6.7 or Fig. 6.8. Here Fig. 6.7 is used.
For a rectangle 6 x 15 m, z Ib = 20/6 = 3.34, l/b = 15/6 = 2.5.
For z/b = 3.34, l/b = 2.5, <r Iq = 0.07 Therefore, o; = 4cr = 4 x 0.01 q = 4 x 0.07 x 150 = 42 kN/m2.
186 Chapter 6 6.7 STRESSES UNDER UNIFORMLY LOADED CIRCULAR FOOTING Stresses Along the Vertical Axis of Symmetry Figure 6.10 shows a plan and section of the loaded circular footing. The stress required to be determined at any point P along the axis is the vertical stress cr,.
Let dA be an elementary area considered as shown in Fig. 6.10. dQ may be considered as the point load acting on this area which is equal to q dA. We may write (6.18) The vertical stress d(J at point P due to point load dQ may be expressed [Eq. (6. la)] as 3q (6.19) The integral form of the equation for the entire circular area may be written as 0=0 r=0 3qz3 ( f rdOdr ~^~ J J ( r2 + z2 ) 5, 0=0 r=0 ,3 On integration we have, (6.20) o R z P Figure 6.10 Vertical stress under uniformly loaded circular footing Stress Distribution in Soils due to Surface Loads 187 Influence value 7Z (xlOO) 1.0 10 Note: Numbers on curves indicate value of r/RQ Figure 6.11 Influence diagram for vertical normal stress at various points within an elastic half-space under a uniformly loaded circular area. (After Foster and Ahlvin, 1954) or 3/2 (6.21) where, /., is the Influence coefficient. The stress at any point P on the axis of symmetry of a circular loaded area may be calculated by the use of Eq. (6.21) Vertical stresses o~ may be calculated by using the influence coefficient diagram given in Fig. 6.11.
Example 6.7 A water tank is required to be constructed with a circular foundation having a diameter of 16 m founded at a depth of 2 m below the ground surface. The estimated distributed load on the foundation is 325 kN/m2. Assuming that the subsoil extends to a great depth and is isotropic and homogeneous, determine the stresses ot at points (i) z = 8 m, r = 0, (ii) z = 8 m, r = 8 m, (iii) z = 16 m, r = 0 and (iv) z=16m, r = 8m, where r is the radial distance from the central axis. Neglect the effect of the depth of the foundation on the stresses. (Use Fig. 6.11) Solution q — 325 kN/m2, RQ = 8 m. The results are given in a tabular form as follows: (i) (ii) (iii) (iv) Point (8,0) (8,8) (16,0) (16, 8) z//?0 1 1 2 2 r/HQ 0 1.0 0 1.0 / 0.7 0.33 0.3 0.2 crzkN/m2 227.5 107.25 97.5 65 188 Chapter6 Example 6.8 For a raft of size 98.4 x 39.36 ft, compute the stress at 65.6 ft depth below the center of the raft by assuming that the rectangle can be represented by an equivalent circle. The load intensity on the raft is31331b/ft2.
Solution The radius of a fictitious circular footing of area equal to the rectangular footing of size 98.4 x 39.36 ft is = 98.4 x 39.36 = 3873 sq. ft or RQ = p = 35.12 ft V Use Eq. (6.21) for computing a at 35.6 ft depth 65.6 35.12 Now, z/RQ = -^^ = 1.9 , and r/RQ = 0. From Fig. 6.11, 7Z = 0.3 Therefore, cr = 0.3 q = 0.3 x 3133 = 940 lb/ft2.
6.8 VERTICAL STRESS BENEATH LOADED AREAS OF IRREGULAR SHAPE Newmark's Influence Chart When the foundation consists of a large number of footings or when the loaded mats or rafts are not regular in shape, a chart developed by Newmark (1942) is more practical than the methods explained before. It is based on the following procedure. The vertical stress cr, below the center of a circular area of radius R which carries uniformly distributed load q is determined per Eq. (6.21).
It may be seen from Eq. (6.21) that when Rlz = °°,az/q=l, that is cr, = q. This indicates that if the loaded area extends to infinity, the vertical stress in the semi-infinite solid at any depth z is the same as unit load q at the surface. If the loaded area is limited to any given radius R\ it is possible to determine from Eq. (6.21) the ratios Rlz for which the ratio of Gjq may have any specified value, say 0.8 or 0.6. Table 6.1 gives the ratios of Rlz for different values of <j/q.
Table 6.1 may be used for the computation of vertical stress <J7 at any depth z below the center of a circular loaded area of radius R. For example, at any depth z, the vertical stress o^ = 0.8 q if the radius of the loaded area at the surface is R = 1.387 z. At the same depth, the vertical stress is cr = 0.7 q if R = 1.110 z. If instead of loading the whole area, if only the annular space between the circles of radii 1.387 z and 1.110 z are loaded, the vertical stress at z at the center of the circle is ACT =0.8 q-0.7 q = 0.lq. Similarly if the annular space between circles of radii l.llOz and 0.917 z are loaded, the vertical stress at the same depth z is ACT, = 0.7 q-0.6 q = 0.1 q. We may therefore draw a series of concentric circles on the surface of the ground in such a way that when the annular space between any two consecutive circles is loaded with a load q per unit area, the vertical stress ACT produced at any depth z below the center remains a constant fraction of q. We may write, therefore, Aaz = Cq (6.22) where C is constant. If an annular space between any two consecutive concentric circles is divided into n equal blocks and if any one such block is loaded with a distributed load q, the vertical stress produced at the center is, therefore, Stress Distribution in Soils due to Surface Loads 189 Table 6.1 Values of Rlz for different values of a' Iq AaL n ajq 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 C n ' Rlz 0.000 0.270 0.401 0.518 0.637 0.766 0.917 1.110 <V<7 0.80 0.90 0.92 0.94 0.96 0.98 1.00 -Rlz 1.387 1.908 2.094 2.351 2.748 3.546 oo -(6.23) z -= C when<7 = l.
n l A load q = 1 covering one of the blocks will produce a vertical stress C-. In other words, the 'influence value' of each loaded block is C(. If the number of loaded blocks is N, and if the intensity of load is q per unit area, the total vertical stress at depth z below the center of the circle is ot = CNq (6.24) The graphical procedure for computing the vertical stress GZ due to any surface loading is as follows.
Select some definite scale to represent depth z. For instance a suitable length AB in cm as shown in Fig. 6.12 to represent depth z in meters. In such a case, the scale is 1 cm = zlAB meters. The length of the radius RQ g which corresponds to ajq = 0.8 is then equal to 1.387 x AB cm, and a circle of that radius may be drawn. This procedure may be repeated for other ratios of ajq, for instance, for ojq = 0.7, 0. 5 etc. shown in Fig. 6.12.
The annular space between the circles may be divided into n equal blocks, and in this case n = 20. The influence value C. is therefore equal to 0.1/20 = 0.005. A plan of the foundation is drawn on a tracing paper to a scale such that the distance AB on the chart corresponds to the depth z at which the stress c?z is to be computed. For example, if the vertical stress at a depth of 9 m is required, and if the length AB chosen is 3 cm, the foundation plan is drawn to a scale of 1 cm = 9/3 = 3 m. In case the vertical stress at a depth 12 m is required, a new foundation plan on a separate tracing paper is required. The scale for this plan is 1 cm = 12/AB = 12/3 = 4 m.
This means that a different tracing has to be made for each different depth whereas the chart remains the same for all. Fig. 6.12(b) gives a foundation plan, which is loaded with a uniformly distributed load q per unit area. It is now required to determine the vertical stress &z at depth vertically below point O shown in the figure. In order to determine crz, the foundation plan is laid over the chart in such a way that the surface point O coincides with the center O' of the chart as shown in Fig. 6.12. The number of small blocks covered by the foundation plan is then counted. Let this number be N. Then the value of GZ at depth z below O is az = Ci Nq, which is the same as Eq. (6.24).
190 Chapter 6 Influence value = C. = 0.005 (a) (b) Figure 6.12 Newmark's influence chart Example 6.9 A ring footing of external diameter 8 m and internal diameter 4 m rests at a depth 2 m below the ground surface. It carries a load intensity of 150 kN/m2. Find the vertical stress at depths of 2,4 and 8 m along the axis of the footing below the footing base. Neglect the effect of the excavation on the stress.
Solution From Eq. (6.21) we have, 1 3/2 where q = contact pressure 150 kN/m2, /., = Influence coefficient.
The stress o at any depth z on the axis of the ring is expressed as o; = cr. -U, = q(I, - /, ) Z ^i <-2 -i <-2 where cr, = stress due to the circular footing of diameter 8 m, and /, = I7 and RQ/z = cr = stress due to the footing of diameter 4m, /, = / and RJz = (RJz).
Stress Distribution in Soils due to Surface Loads 191 The values of /., may be obtained from Table 6.1 for various values of /?0/z. The stress cr at depths 2, 4 and 8 m are given below: Depth (m) R^lz 2 2 4 1.0 8 0.5 ', 0.911 0.647 0.285 R2/z lz 1.0 0.697 0.5 0.285 0.25 0.087 (/ - I2 )q = az kN/m2 39.6 54.3 29.7 Example 6.10 A raft foundation of the size given in Fig. Ex. 6.10 carries a uniformly distributed load of 300 kN/m2. Estimate the vertical pressure at a depth 9 m below the point O marked in the figure.
Solution The depth at which &z required is 9 m.
Using Fig. 6.12, the scale of the foundation plan is AB = 3 cm = 9 m or 1 cm = 3 m. The foundation plan is required to be made to a scale of 1 cm = 3 m on tracing paper. This plan is superimposed on Fig. 6.12 with O coinciding with the center of the chart. The plan is shown in dotted lines in Fig. 6.12.
Number of loaded blocks occupied by the plan, N = 62 Influence value, Cf = 0.005, q = 300 kN/m2 The vertical stress, crz = C{ Nq - 0.005 x 62 x 300 = 93 kN/m2.
18m-16.5 m 3m = x 6 m 1 3 m = O [•— 9m —-| Figure Ex. 6.10 6.9 EMBANKMENT LOADINGS Long earth embankments with sloping sides represent trapezoidal loads. When the top width of the embankment reduces to zero, the load becomes a triangular strip load. The basic problem is to determine stresses due to a linearly increasing vertical loading on the surface.
192 Chapters Linearly Increasing Vertical Loading Fig. 6.13(a) shows a linearly increasing vertical loading starting from zero at A to a finite value q per unit length at B. Consider an elementary strip of width db at a distance b from A. The load per unit length may be written as dq - (q/d) b db Ifdq is considered as a line load on the surface, the vertical stress dcr, at P [Fig. 6. 1 3(a)] due to dq may be written from Eq. (6.4) as dcr,=\—\ — ' Therefore, er b=a 2q [(x-, / 9 on integration, o- = 77" ~~a-sm20\ = 07 (6.25) z 2/T\ a y z where 7 is non-dimensional coefficient whose values for various values of xla and zla are given in Table 6.2.
If the point P lies in the plane BC [Fig. 6.13(a)], then j8 = 0 at jc = a. Eq. (6.25) reduces to vz=-(a) (6.26) <• n Figs. 6.13(b) and (c) show the distribution of stress er on vertical and horizontal sections under the action of a triangular loading as a function of q. The maximum vertical stress occurs below the center of gravity of the triangular load as shown in Fig. 6.13(c).
Vertical Stress Due to Embankment Loading Many times it may be necessary to determine the vertical stress er beneath road and railway embankments, and also beneath earth dams. The vertical stress beneath embankments may be Table 6.2 / for triangular load (Eq. 6.25) x/a -1.500 -1.00 0.00 0.50 0.75 1.00 1.50 2.00 2.50 0.00 0.00 0.00 0.00 0.50 0.75 0.50 0.00 0.00 0.00 0.5 0.002 0.003 0.127 0.410 0.477 0.353 0.056 0.017 0.003 1.0 0.014 0.025 0.159 0.275 0.279 0.241 0.129 0.045 0.013 2/fl 1.5 0.020 0.048 0.145 0.200 0.202 0.185 0.124 0.062 0.041 2 0.033 0.061 0.127 0.155 0.163 0.153 0.108 0.069 0.050 4 0.051 0.060 0.075 0.085 0.082 0.075 0.073 0.060 0.049 6 0.041 0.041 0.051 0.053 0.053 0.053 0.050 0.050 0.045 Stress Distribution in Soils due to Surface Loads 193 0 0.2 0.4 0.6 O.i 3a (a) Triangular loading (b) Vertical stress on vertical sections Atz= l.Ofl (c) Vertical stress on horizontal sections Figure 6.13 Stresses in a semi-infinite mass due to triangular loading on the surface determined either by the method of superposition by making use of Eq. (6.26) or by making use of a single formula which can be developed from first principles.
crz by Method of Superposition Consider an embankment given in Fig. 6.14. a at P may be calculated as follows: The trapezoidal section of embankment ABCD, may be divided into triangular sections by drawing a vertical line through point P as shown in Fig. 6.14. We may write ABCD = AGE + FGB - EDJ - FJC (6.27) If <r r <Tz2, Gzy and <7z4 are the vertical stresses at point P due to the loadings of figures AGE, FGB, EDJ and FJC respectively, the vertical stress o"z due to the loading of figure ABCD may be written as o=o -o - o Z2 Z3 Z (6.28) By applying the principle of superposition for each of the triangles by making use of Eq. (6.26), we obtain 194 Chapter 6 //VCVC<\XX\V GG D X 0,.
Figure 6.14 Vertical stress due to embankment K\ (6.29) a=ql=-f(a/z,b/z) (6.30) where / is the influence factor for a trapezoidal load which is a function of a/z and biz.
The values of /, for various values of a/z and biz are given in Fig. 6.15. (After Osterberg, 1957) a^ from a Single Formula for Asymmetrical Trapezoidal Loading A single formula can be developed for trapezoidal loading for computing CTZ at a point P (Fig. 6.16) by applying Eq. (6.26). The origin of coordinates is as shown in the figure. The final equation may be expressed as (a, (a, + X — («!
ai (6.31) where ar a2, and «3 are the angles subtended at the point P in the supporting medium by the loading and R = a,/a^. When R = 1, the stresses are due to that of a symmetrical trapezoidal loading.
Stress Distribution in Soils due to Surface Loads 195 0.50 0.40 0.30 0.20 0.10 0.01 2 4 6 8 0.1 2 4 6 8 1.0 2 4 6 8 10 Figure 6.15 A graph to determine compressive stresses from a load varying by straight line law (After Osterberg, 1957) b b a2—^ Figure 6.16 Trapezoidal loads 196 Chapter 6 When the top width is zero, i.e, when b = 0, a2 = 0, the vertical stress <r will be due to a triangular loading. The expression for triangular loading is (6.32) Eq. (6.31) and Eq. (6.32) can be used to compute cr at any point in the supporting medium.
The angles a{, cc2, and a3 may conveniently be obtained by a graphical procedure where these angles are expressed as radians in the equations.
Example 6.11 A 3 m high embankment is to be constructed as shown in Fig. Ex. 6. 11. If the unit weight of soil used in the embankment is 19.0 kN/m3, calculate the vertical stress due to the embankment loading at points PI; P2, and Py M 3.0 F \ y= 19 kN/m f '3.0 i Note: All dimensions are in metres P2 P^ Figure Ex. 6.11 Vertical stresses at Pv P2 & Solution q = yH = 19 x 3 = 57 kN/m2, z = 3 m The embankment is divided into blocks as shown in Fig. Ex. 6.11 for making use of the graph given in Fig. 6. 15. The calculations are arranged as follows: Point p{ P2 PI Block ACEF EDBF AGH GKDB HKC MLDB MACL b (m) 1.5 4.5 0 7.5 0 10.5 1.5 a (m) 3 3 1.5 3 1.5 3.0 3.0 biz 0.5 1.5 0 2.5 0 3.5 0.5 alz 1 1 0.5 1.0 0.5 1.0 1.0 ' 0.39 0.477 0.15 0.493 0.15 0.498 0.39 Stress Distribution in Soils due to Surface Loads 197 Vertical stress <Jz At point P,, cr, = At point P2, CF. = At point Py &z = (0.39 + 0.477) x 57 = 49.4 kN/m2 0. 15 x (57/2) + 0.493 x 57 - 0.15 x (57/2) = 28.
(0.498 - 0.39) 57 = 6.2 kN/m2 1 kN/m2 6.10 APPROXIMATE METHODS FOR COMPUTING o2 Two approximate methods are generally used for computing stresses in a soil mass below loaded areas. They are 1. Use of the point load formulas such as Boussinesq's equation.
2. 2:1 method which gives an average vertical stress <r at any depth z. This method assumes that the stresses distribute from the loaded edge points at an angle of 2 (vertical) to 1 (horizontal) The first method if properly applied gives the point stress at any depth which compares fairly well with exact methods, whereas the second does not give any point stress but only gives an average stress cr at any depth. The average stress computed by the second method has been found to be in error depending upon the depth at which the stress is required.
Point Load Method Eq. (6.1) may be used for the computation of stresses in a soil mass due to point loads acting at the surface. Since loads occupy finite areas, the point load formula may still be used if the footings are divided into smaller rectangles or squares and a series of concentrated loads of value q dA are assumed to act at the center of each square or rectangle. Here dA is the area of the smaller blocks and q the pressure per unit area. The only principle to be followed in dividing a bigger area into smaller blocks is that the width of the smaller block should be less than one-third the depth z of the point at which the stress is required to be computed. The loads acting at the centers of each smaller area may be considered as point loads and Boussinesq's formula may then be applied. The difference between the point load method and the exact method explained earlier is clear from z/B Figure 6.17 cr by point load method 198 Chapter 6 Figure 6.18 cr 2 : 1 method Fig. 6.17. In this figure the abscissa of the curve Cl represents the vertical stress (7., at different depths z below the center of a square area B x B which carries a surcharge g per unit area or a total surcharge load of B2q. This curve is obtained by the exact method explained under Sect. 6.6. The abscissa of the curve C2 represents the corresponding stresses due to a concentrated load Q = B2q acting at the center of the square area. The figure shows that the difference between the two curves becomes very small for values of z/B in excess of three. Hence in a computation of the vertical stress cr, at a depth z below an area, the area should be divided into convenient squares or rectangles such that the least width of any block is not greater than z/3.
2 : 1 Method In this method, the stress is assumed to be distributed uniformly over areas lying below the foundation. The size of the area at any depth is obtained by assuming that the stresses spread out at an angle of 2 (vertical) to 1 (horizontal) from the edges of the loaded areas shown in Fig. 6.18. The average stress at any depth z is Q (B+z)(L (6.33) The maximum stress om by an exact method below the loaded area is different from the average stress a at the same depth. The value of cr/tr reaches a maximum of about 1.6 at zlb = 0-5, where b = half width.
6.11 PRESSURE ISOBARS Definition An isobar is a line which connects all points of equal stress below the ground surface. In other words, an isobar is a stress contour. We may draw any number of isobars as shown in Fig. 6.19 for any given load system. Each isobar represents a fraction of the load applied at the surface. Since these isobars form closed figures and resemble the form of a bulb, they are also termed bulb of pressure or simply the pressure bulb. Normally isobars are drawn for vertical, horizontal and shear stresses. The one that is most important in the calculation of settlements of footings is the vertical pressure isobar.
Stress Distribution in Soils due to Surface Loads 199 Lines of equal vertical pressure or isobars Figure 6.19 Bulb of pressure Significant Depth In his opening discussion on settlement of structures at the First International Conference on Soil Mechanics and Foundation Engineering (held in 1936 at Harvard University in Cambridge, Mass, USA), Terzaghi stressed the importance of the bulb of pressure and its relationship with the seat of settlement. As stated earlier we may draw any number of isobars for any given load system, but the one that is of practical significance is the one which encloses a soil mass which is responsible for the settlement of the structure. The depth of this stressed zone may be termed as the significant depth DS which is responsible for the settlement of the structure. Terzaghi recommended that for all practical purposes one can take a stress contour which represents 20 per cent of the foundation contact pressure q, i.e, equal to Q.2q. The depth of such an isobar can be taken as the significant depth Ds which represents the seat of settlement for the foundation. Terzaghi's recommendation was based on his observation that direct stresses are considered of negligible magnitude when they are smaller than 20 per cent of the intensity of the applied stress from structural loading, and that most of the settlement, approximately 80 per cent of the total, takes place at a depth less than Ds. The depth Ds is approximately equal to 1.5 times the width of square or circular footings [Fig. 6.20(a)].
If several loaded footings are spaced closely enough, the individual isobars of each footing in question would combine and merge into one large isobar of the_intensity as shown in [Fig. 6.20(b)]. The combined significant depth D is equal to about 1.5 B.
az = Q.2q D<=.5B\ Stressed zone Isobar (a) Significant depth of stressed zone for single footing Isobar Combined stressed zone (b) Effect of closely placed footings Figure 6.20 Significant depth of stressed zone 200 Chapter 6 Pressure Isobars for Footings Pressure isobars of square, rectangular and circular footings may conveniently be used for determining vertical pressure, (Jz, at any depth, z, below the base of the footings. The depths z from the ground surface, and the distance r (or jc) from the center of the footing are expressed as a function of the width of the footing B. In the case of circular footing B represents the diameter.
The following pressure isobars are given based on either Boussinesq or Westergaard's equations 1. Boussinesq isobars for square and continuous footings, Fig. 6.21.
2. Boussinesq isobar for circular footings, Fig. 6.22.
3. Westergaard isobars for square and continuous footings, Fig. 6.23.
B/2=b BI2=b Continuous 25 Figure 6.21 Pressure isobars based on Boussinesq equation for square and continuous footings Stress Distribution in Soils due to Surface Loads 201 Figure 6.22 Pressure isobars based on Boussinesq equation for uniformly loaded circular footings B/2=b B/2=b 5b 6b Continuous IB 2B 35 Figure 6.23 Pressure isobars based on Westergaard equation for square and continuous footing 202 Chapter 6 Example 6.12 A single concentrated load of 1000 kN acts at the ground surface. Construct an isobar for <7 = 40 kN/m2 by making use of the Boussinesq equation.
Solution From Eq. (6.la) we have 3(2 1 We may now write by rearranging an equation for the radial distance r as -1 Now for Q = 1000 kN, cr, = 40 kN/m2, we obtain the values of rp r2, ry etc. for different depths z,, z2, zv etc. The values so obtained are z(m) 0.25 0.50 1.0 2.0 3.0 3.455 r(m) 1.34 1.36 1.30 1.04 0.60 0.00 g=1000kN a, = 40 kN/mJ Isobar 3.455 Figure Ex. 6.12 Stress Distribution in Soils due to Surface Loads 203 The isobar for crz = 40 kN/m2 may be obtained by plotting z against r as shown in Fig. Ex. 6.12.
6.12 PROBLEMS 6.1 A column of a building transfers a concentrated load of 225 kips to the soil in contact with the footing. Estimate the vertical pressure at the following points by making use of the Boussinesq and Westergaard equations.
(i) Vertically below the column load at depths of 5, 10, and 15 ft.
(ii) At radial distances of 5, 10 and 20 ft and at a depth of 10 ft.
6.2 Three footings are placed at locations forming an equilateral triangle of 13 ft sides. Each of the footings carries a vertical load of 112.4 kips. Estimate the vertical pressures by means of the Boussinesq equation at a depth of 9 ft at the following locations : (i) Vertically below the centers of the footings, (ii) Below the center of the triangle.
6.3 A reinforced concrete water tank of size 25 ft x 25 ft and resting on the ground surface carries a uniformly distributed load of 5.25 kips/ft2. Estimate the maximum vertical pressures at depths of 37.5 and 60 ft by point load approximation below the center of the tank.
6.4 Two footings of sizes 13 x 13 ft and 10 x 10 ft are placed 30 ft center to center apart at the same level and carry concentrated loads of 337 and 281 kips respectively. Compute the vertical pressure at depth 13 ft below point C midway between the centers of the footings.
6.5 A and B are two footings of size 1.5 x 1.5 m each placed in position as shown in Fig. Prob. 6.5. Each of the footings carries a column load of 400 kN. Determine by the 2.5m S x?Xs\ 1 « A Q 1 1 ' [-- 1.5 m~H ' 1 'ft ^ m m 1 B //X\N Q - 400 kN (2 1 (- 1.5 Figure Prob. 6.5 Boussinesq method, the excess load footing B carries due to the effect of the load on A.
Assume the loads at the centers of footings act as point loads.
6.6 If both footings A and B in Fig. Prob. 6.5 are at the same level at a depth of 0.5 m below the ground surface, compute the stress d, midway between the footings at a depth of 3 m from the ground surface. Neglect the effect of the size for point load method.
6.7 Three concentrated loads Ql = 255 kips, Q2 = 450 kips and <23 = 675 kips act in one vertical plane and they are placed in the order Ql-Q2~Qy Their spacings are 13 ft-10 ft. Determine 204 Chapter 6 the vertical pressure at a depth of 5 ft along the center line of footings using Boussinesq's point load formula.
6.8 A square footing of 13 x 13 ft is founded at a depth of 5 ft below the ground level. The imposed pressure at the base is 8732 lb/ft2. Determine the vertical pressure at a depth of 24 ft below the ground surface on the center line of the footing.
6.9 A long masonry wall footing carries a uniformly distributed load of 200 kN/m 2. If the width of the footing is 4 m, determine the vertical pressures at a depth of 3 m below the (i) center, and (ii) edge of the footing.
6.10 A long foundation 0.6 m wide carries a line load of 100 kN/m. Calculate the vertical stress cr, at a point P, the coordinates of which are x = 2.75 m, and z = 1.5 m, where the x-coordinate is normal to the line load from the central line of the footing.
6.11 A strip footing 10 ft wide is loaded on the ground surface with a pressure equal to 4177 lb/ft2. Calculate vertical stresses at depths of 3, 6, and 12 ft under the center of the footing.
6.12 A rectangular footing of size 25 x 40 ft carries a uniformly distributed load of 5200 lb/ft2.
Determine the vertical pressure 20 ft below a point O which is located at a distance of 35 ft from the center of the footing on its longitudinal axis by making use of the curves in Fig. 6.8.
6.13 The center of a rectangular area at the ground surface has cartesian coordinate (0,0) and the corners have coordinates (6,15). All dimensions are in foot units. The area carries a uniform pressure of 3000 lb/ft2. Estimate the stresses at a depth of 30 ft below ground surface at each of the following locations: (0,0), (0,15), (6,0).
6.14 Calculate the vertical stress at a depth of 50 ft below a point 10 ft oubide the corner (along the longer side) of a rectangular loaded area 30 x 80 ft carrying a uniform load of 2500 lb/ft2.
6.15 A rectangular footing 6 x 3 m carries a uniform pressure of 300 kN/m2 on the surface of a soil mass. Determine the vertical stress at a depth of 4.5 m below the surface on the center line 1.0 m inside the long edge of the foundation.
6.16 A circular ring foundation for an overhead tank transmits a contact pressure of 300 kN/m2.
Its internal diameter is 6 m and external diameter 10m. Compute the vertical stress on the center line of the footing due to the imposed load at a depth of 6.5 m below the ground level. The footing is founded at a depth of 2.5 m.
6.17 In Prob. 6.16, if the foundation for the tank is a raft of diameter 10 m, determine the vertical stress at 6.5 m depth on the center line of the footing. All the other data remain the same.
6.18 How far apart must two 20 m diameter tanks be placed such that their combined stress overlap is not greater than 10% of the surface contact stress at a depth of 10 m?
6.19 A water tower is founded on a circular ring type foundation. The width of the ring is 4 m and its internal radius is 8 m. Assuming the distributed load per unit area as 300 kN/m2, determine the vertical pressure at a depth of 6 m below the center of the foundation.
6.20 An embankment for road traffic is required to be constructed with the following dimensions : Top width = 8 m, height = 4 m, side slopes= I V : 1.5 Hor The unit weight of soil under the worst condition is 21 kN/m3. The surcharge load on the road surface may be taken as 50 kN/m2. Compute the vertical pressure at a depth of 6 m below the ground surface at the following locations: (i) On the central longitudinal plane of the embankment, (ii) Below the toes of the embankment.
Stress Distribution in Soils due to Surface Loads 205 6.21 If the top width of the road given in Prob. 6.20 is reduced to zero, what would be the change in the vertical pressure at the same points?
6.22 A square footing of size 13 x 13 ft founded on the surface carries a distributed load of 2089 lb/ft2. Determine the increase in pressure at a depth of 10 ft by the 2:1 method 6.23 A load of 337 kips is imposed on a foundation 10 ft square at a shallow depth in a soil mass. Determine the vertical stress at a point 16 ft below the center of the foundation (a) assuming the load is uniformly distributed over the foundation, and (b) assuming the load acts as a point load at the center of the foundation.
6.24 A total load of 900 kN is uniformly distributed over a rectangular footing of size 3 x 2 m.
Determine the vertical stress at a depth of 2.5 m below the footing at point C (Fig. Prob. 6.24), under one corner and D under the center. If another footing of size 3 x 1 m with a total load of 450 kN is constructed adjoining the previous footing, what is the additional stress at the point C at the same depth due to the construction of the second footing?
2m D 3m 1m i 3 m h— im-H Figure Prob. 6.24 6.25 Refer to Prob. 6.24. Determine the vertical stress at a depth of 2.5 m below point E in Fig. Prob. 6.24. All the other data given in Prob. 6.24 remain the same. |
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Background Facts on Contingent Faculty Positions
"Contingent” faculty positions include both part- and full-time non-tenure-track appointments.
Their common characteristic is that institutions make little or no long-term commitment to faculty holding these positions.
Today, a large proportion of faculty appointments are part-time.
This includes positions that may be classified by the institution as adjuncts, part-time lecturers, or another name.
Many faculty in so-called “part-time” positions actually teach the equivalent of a full-time course load.
Other part-time appointments are held by graduate student employees, whose chances of obtaining tenure-track positions in the future are increasingly uncertain.What is billed as a teaching apprenticeship often instead amounts to years of intensive, low-paid work that distracts from, rather than complementing, graduate studies.
To support themselves, part-time faculty often commute between institutions and prepare courses on a grueling timetable, making enormous sacrifices to maintain interaction with their students.
Since faculty classified as part-time are typically paid by the course, without benefits, many college teachers lack access to health insurance and retirement plans. Back to top
Both part-time and full-time non-tenure-track appointments are increasingly prevalent.
Contingent positions that are ineligible for tenure now account for nearly 70 percent of all instructional staff appointments in American higher education, including 49 percent part time and 19 percent full-time non-tenure-track. See more data here. Back to top
The majority of faculty working on contingent appointments do not have professional careers outside of academe, and most teach basic core courses rather than narrow specialties.
While a small percentage of part-time faculty are specialists or practitioners of a profession such as law or architecture and teach a class on the side, this situation is the exception rather than the norm.
In other words, contingent faculty are "real" faculty, not other professionals who are moonlighting by teaching a course.
Students at community colleges and in lower-level undergraduate courses are disproportionately taught by faculty on contingent appointments. Back to top
The excessive use of, and inadequate compensation and professional support for, faculty in contingent positions exploits these colleagues.
Positions that require comparable work, responsibilities, and qualifications should be comparably compensated.
As the AAUP recommended in 1993, compensation for part-time appointments should be the applicable fraction of the compensation (including benefits) for a comparable full-time position. Back to top
The turn towards cheaper contingent labor is largely a matter of priorities rather than economic necessity.
While many institutions are currently suffering budget cuts, the greatest growth in contingent appointments occurred during times of economic prosperity.
Many institutions have invested heavily in facilities and technology while cutting instructional spending.
Though incoming students may find finer facilities, they are also likely to find fewer full-time faculty with adequate time, professional support, and resources available for their instruction. Back to top
Many contingent faculty members are excellent teachers and scholars.
But no matter how qualified and dedicated, faculty teaching in these positions are hobbled in the performance of their duties by a lack of professional treatment and support.
Many lack access to such basics as offices, computer support, and photocopying services. Back to top
Heavy reliance on contingent faculty appointments hurts students.
Faculty working conditions are student learning conditions.
Faculty on contingent appointments are typically paid only for the hours they spend in the classroom. While they may be excellent teachers, they are not given adequate institutional support for time spent meeting with students, evaluating student work, and class planning and preparation.
Adjuncts are often hired on the spur of the moment with little evaluation or time to prepare--sometimes after a semester has already started.
Faculty in contingent positions often receive little or no evaluation and mentoring, making them especially vulnerable to being dismissed over one or two student complaints.
Faculty in contingent positions, though they may teach the majority of some types of courses, are often cut out of department and institution-wide planning. The knowledge that they have about their students and the strengths and weaknesses of the courses they teach is not taken into consideration.
The high turnover among contingent faculty members mean that students in a department may never have the same teacher twice, or may be unable to find an instructor who knows them well enough to write a letter of recommendation.
The free exchange of ideas may be hampered by the fear of dismissal for unpopular utterances, so students may be deprived of the debate essential to citizenship.
They may also be deprived of rigorous evaluations of their work. Back to top
Overuse of contingent faculty appointments hurts all faculty.
The integrity of faculty work is threatened as parts of the whole are divided and assigned piecemeal to instructors, lecturers, graduate students, specialists, researchers, and administrators.
Proportionally fewer tenure-track faculty means fewer people to divide up the work of advising students, setting curriculum, and serving on college-wide committees.
Divisions among instructors create a less cohesive faculty; on some campuses, tenure-track and adjunct faculty rarely interact or participate in planning together.
Tenure should be a big tent that provides due process protections for the academic freedom of all faculty; where contingent appointments predominate, it becomes instead a merit badge for a select few. Back to top
Academic freedom is weakened when a majority of the faculty lack the protections of tenure.
The insecure relationship between faculty members in contingent positions and their institutions can chill the climate for academic freedom, which is essential to the common good of a free society.
Faculty serving in insecure contingent positions may be less likely to take risks in the classroom or in scholarly and service work. Back to top
The use of non-tenure-track appointments should be limited to specialized fields and emergency situations.
While we recognize that current patterns of faculty appointment depart substantially from the ideal, the AAUP recommends that no more than 15 percent of the total instruction within an institution, and no more than 25 percent of the total instruction within any department, should be provided by faculty with non-tenure-track appointments.
While institutions often cite "flexibility" for a reason to hire faculty off the tenure track, contingent appointments are often clustered in programs with very high levels of predictability--such as freshman writing courses that are required for all students. Back to top
Shared governance responsibilities should be shared among all faculty, including those appointed to part-time positions.
Curricular and other academic decisions benefit from the participation of all faculty, especially those who teach core courses.
Faculty and administrators should together determine the appropriate modes and levels of participation in governance for part-time faculty, considering issues such as voting rights, representation, and inclusion in committees and governance bodies. Back to top
When contingent appointments are used, they should include job security and due process protections. Contingent faculty appointments, like all faculty appointments, should include:
the full range of faculty responsibilities (teaching, scholarship, service);
comparable compensation for comparable work;
assurance of continuing employment after a reasonable opportunity for successive reviews;
inclusion in institutional governance structures; and
appointment and review processes that involve faculty peers and follow accepted academic due process. Back to top
The proportion of faculty appointments that are on the tenure line should be increased. This can be done by:
Changing the status of faculty members currently holding non-tenure-track appointments. Individuals holding contingent appointments are offered tenure-eligible reappointments.
Creating new tenure-line appointments. New tenure-line positions are created and open searches are held for candidates to fill them.
In both cases, transition to a higher proportion of tenured faculty should be accomplished primarily through attrition, retirements, and, where appropriate “grandfathering” of currently contingent faculty into tenured positions. Faculty in contingent positions should not bear the cost of transition. Back to top
Additional information about contingent faculty is available here. |
681 | https://www.asexuality.org/en/topic/208970-sex-activity-definition-objective-vs-subjective-tmi-warning/ | Published Time: 2021-02-18T15:19:07+0000
"Sex" (activity) definition -- Objective vs. Subjective? TMI Warning - Philosophy, Politics, and Science - Asexual Visibility and Education Network
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"Sex" (activity) definition -- Objective vs. Subjective? TMI Warning
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"Sex" (activity) definition -- Objective vs. Subjective? TMI Warning
By Shadowbird
February 18, 2021 in Philosophy, Politics, and Science
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Shadowbird
Posted February 18, 2021
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What actually is "sex" (in reference to the activity of sexual contact involving genitals)?
Is there an objective definition? Are there subjective definitions? How does a person know or can understand what "sex" is and whether or not they've engaged in it?
Is it physical contact with another person's genitals (and does it matter or differ if done with, say, one's hand, mouth, or own genitals)?
Is it only when an orgasm occurs as a result of any of the previous methods, or is an orgasm not defining of having "had sex"?
How does it relate to the inquiry of whether or not someone has been "sexually active" (medically-speaking)?
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Milque Toast
Posted February 18, 2021
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Posted February 18, 2021
^^^ I'd like an answer to this too, thanks😅
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Logandus
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Posted February 18, 2021
I think sex would be defined as the consensual action of attempting to reach orgasm with another individual(s). Whether using a hand mouth or other part to attempt to reach an orgasm, it should still be considered a form of sex but there are still subjective definitions regarding things like virginity. Typically the sex has to be penetrative in order to be considered "sex" in regards to virginity. Medically speaking I believe my first definition is more important, because all forms of sex can result in medically relevant issues. I should note that in my definition, the word attempt is important because I think that it would still be considered sex even if orgasm is not achieved. No matter what though it is not sex if it isn't consensual, meaning there was no coercion in any way.
I hope this helped
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everywhere and nowhere
Posted February 18, 2021
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Posted February 18, 2021
4 hours ago, Logandus said:
Typically the sex has to be penetrative in order to be considered "sex" in regards to virginity.
Which is hypocritical. And reducing only penetrative stuff to "sex" leads to a flippant perception of lesbians: some people really believe that two women can't have "proper sex", "real sex", "sex-sex".
I, anyway, don't really like saying that "I'm a virgin", I much prefer saying that "I have never had sex" - but at least I'm not hypocritical because I have indeed had no partnered sexual contact of any kind.
I would attempt to define sex as physical contact (1) between two or, sometimes, more people (to distinguish it from autoeroticism, which is a "sexual activity", but not "sex"), (2) with the aim of achieving pleasure, and (3) including contact of one person's intimate parts, hands, mouth, other body parts or an object held by them with the other person's intimate parts. This would, I think, more or less capture all the variants.
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Posted February 18, 2021
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Posted February 18, 2021
TMI Warning
I normally define sex as“an activity between two or more people where physical contact is made with at least one person’s genitalia for pleasure and/or procreation”.
I think each country has a legal definition of what constitutes sexual intercourse (for the sake of prosecuting when it’s not consensual or otherwise legal), but what individuals use to define whether they’ve had sex or not is highly subjective.
Medically I believe it is only possible to determine whether a female-bodied person has had penetrative sex, but I understand that this is controversial.
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Posted February 18, 2021
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Posted February 18, 2021
2 hours ago, Logandus said:
Medically speaking I believe my first definition is more important, because all forms of sex can result in medically relevant issues.
Does it though? Your definition includes hand-with-genital contact which, as far as I'm aware, does not cause any medical issues. I do know about the medical issues that can arise from the mouth-to-genitalia type of contact.
I don't have a good definition to offer, but roughly speaking I consider oral, vaginal and anal to be "sex", and my understanding is that virginity is determined by whether or not someone has had vaginal or anal sex. This is inclusive of sex between two women. Handjobs I would consider a sexual interaction, but not "sex", though I don't particularly care if other people define that as "sex" and count them towards "losing virginity", the concept of virginity is silly and arbitrary anyway.
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Posted February 18, 2021
Added TMI warning whilst in PPS. This thread may be better suited to another section, so if it ends up getting moved it might be OK to remove it.
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Posted February 18, 2021
I would say it’s subjective simply because there’s no body that can officially decide what specific actions count as sex for everyone else. But just like other subjective things, like beauty and success, just because it’s subjective doesn’t mean it’s meaningless. It just might require understanding what sex means for a given person you’re listening to when they talk about sex.
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uhtred
Posted February 18, 2021
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Posted February 18, 2021
I think its very subjective. Different people consider different things to be "sex".In general that can be OK, but it can lead to all sorts of communication problems. For one person "sex" can be any action that results in a orgasm, for another its specifically PIV. If they are discussion how one wants more or less "sex" they may be talking about completely different things.They may also not recognize what is important to the other person.
Its all about communication.
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682 | https://electronics.stackexchange.com/questions/561527/how-to-optimise-wire-gauge-selection-for-electromagnetic-applications | electromagnetism - How to optimise wire gauge selection for electromagnetic applications? - Electrical Engineering Stack Exchange
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How to optimise wire gauge selection for electromagnetic applications?
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I have an intention to build a simple solenoid for no other reason but my own interest. A basic review of theory brought me to the equation B = μIN/L whereby B = Field Strength,μ = Magnetic Permeability, I = Current, N = Coil Turns and L = Coil Length. For what ever reason my first observation was that there was surely an optimisation to be solved in the selection of wire gauge since an increased wire cross-section would increase your available Current Capacity (I) but decrease the Coil Turns (N) along a constant length. I then stopped short of attempting any calculation since I being proportional to the cross section area of a wire (in terms of Ampacity) and N being a first order relation, it seemed logic that the "bigger the wire the better, turns don't matter that much"... this can't be right. What further confused me was to investigate the current draw of commercial BLDC motors, of whatever size, and note that their amperage was surely beyond what would usually be carried on the wire gauges used for stators. Something to do with the AC driving? I sit confused.
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What is the optimal design approach for selecting wire gauge for electromagnetic coils of a given length and diameter?
What factors are most import to consider?
Are there conventions or standards I should know about?
electromagnetism
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edited Apr 26, 2021 at 14:02
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asked Apr 22, 2021 at 12:45
George KerwoodGeorge Kerwood
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2 Limited f- range of response T=L/R , Bsat core, eddy current losses, loop area, magnetic coupling (leakage inductance) crosstalk, Litz Wire, air core vs high mu , high voltage insulation magnet wire, self-capacitance (resonance) winding patterns. , there's a lot of topics Tony Stewart EE since 1975 –Tony Stewart EE since 1975 2021-04-22 12:52:43 +00:00 Commented Apr 22, 2021 at 12:52
2 Why would you assume a constant length? More practically there might be a constant volume available for the copper. The drive voltage may be fixed (something like a relay). You may notice that DC relays of a given design tend to have the same coil size and power regardless of coil voltage.Spehro 'speff' Pefhany –Spehro 'speff' Pefhany 2021-04-22 13:03:46 +00:00 Commented Apr 22, 2021 at 13:03
1 Realize that you are making an electromagnet, it will be much more powerful if you close the loop. en.wikipedia.org/wiki/… Same with a plunger solenoid, you want a path outside the coil.Mattman944 –Mattman944 2021-04-22 13:31:11 +00:00 Commented Apr 22, 2021 at 13:31
@TonyStewartEE75 I'd already assumed I was scratching the surface of something much more complex. Do you perhaps have any recommended resources? Books?George Kerwood –George Kerwood 2021-04-22 17:09:44 +00:00 Commented Apr 22, 2021 at 17:09
1 It turns out that the volume for the copper is something that is determined fairly early on in the design process. The volume is always filled with turns of wire. If the wire is thick, there will be fewer turns. If the wire is thin there will be more turns. This holds true for transformers and motors. Note that solenoids are motors. If you want more inductance (higher voltage lower current) you use fine wire. If you want less inductance (lower voltage higher current) you use fatter wire. But you always fill the volume with copper.user57037 –user57037 2021-04-23 03:13:04 +00:00 Commented Apr 23, 2021 at 3:13
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Interestingly, the choice of wire gauge is not so important. You can pretty much design the whole rest of the system, and then pick the wire gauge according to how you want to trade off voltage and current at any given power level (P=VI).
Let's say you start with some electromagnetic system that has a coil made of AWG 21 wire...
You might measure I amps at V volts around N turns, and you might measure the DC resistance of the coil as R. The power lost to heat is I 2 R. The flux is proportional to IN.
Keep all dimensions the same, but switch to AWG 24 or so (pretend it's exactly half the cross-sectional area), Now:
The number of turns in the coil can double to 2N, because the cross sectional area of the wire is halved.
The DC resistance increases to 4R, because it's proportional to turns/area.
The current required to achieve the same flux is halved to I/2, because I/2 2N = IN
The voltage required to achieve that flux is doubled to 2V, because magnetic EMF is proportional to turns, and (I/2)(4R) = 2IR.
The input power stays exactly the same: 2VI/2 = VI
The power lost to heat stays exactly the same: (I/2)24R = I 2 R
So after changing the wire gauge, the capabilities and efficiency of the system are unchanged. We've just changed its operating voltage and its operating current in inverse proportions.
The capabilities of the system depend on the shape of the coil and other components, the current density in the coil and the material that the coil is made of. (heat losses are K current_density coil_volume, for some constant K that depends on the material)
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answered Apr 23, 2021 at 2:19
Matt TimmermansMatt Timmermans
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Fantastic! This really clarified for me exactly my query so thank you very much for your time and detail.George Kerwood –George Kerwood 2021-04-23 08:59:24 +00:00 Commented Apr 23, 2021 at 8:59
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Number of turns and current are variables that depend upon design goals. However, there is an optimization to be had in terms of geometry. A "Brooks coil" is a coil with a square cross section and an inside radius of the same dimension as a side of the square cross section. (from www.circuits.dk)
The Brooks coil has the property of maximizing the magnetic flux for a given length of wire.
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answered Apr 22, 2021 at 13:35
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There is no ‘optimum’ but rather a compromise. A smaller cross-section wire will allow more turns but have greater resistance - this means more temperature rise. Ultimately the limiting factor is the temperature the wire insulation fails at - you want to stay away from this as far as possible.
How long do you want this solenoid energized for? For short term low duty cycle use you could tolerate the temperature rise. For example I’m working with some 12V solenoids at the moment. They draw 20A and will get extremely hot if left energised for over a couple of minutes. They are intended to unlock a mechanism so this limitation is not an issue.
For the joy of experimentation, I’d suggest you chose 2 or three different wire diameters and wind three solenoids. See what the outcome is and use this information to guide your next iteration.
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answered Apr 22, 2021 at 13:09
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All very useful constraints to consider, thank you for your time and explanation.George Kerwood –George Kerwood 2021-04-22 17:17:38 +00:00 Commented Apr 22, 2021 at 17:17
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Sorry, but I don't have enough time to write a detailed answer. However, I wanted to give you a few pieces of general advice. First: you've stumbled onto one of the most difficult parameters to determine when designing a device that makes use of coils of wire wound around a metal in a confined space. In pretty much all cases, you want to get as much copper into that space as possible. Mathematicians around the world and throughout history have spent thousands of man hours trying to work out THE best way to solve what they often call "optimum shape packing". And, usually, the object being packed is kept the same size, which is not the case here. You need to optimize both packing AND size. It's a tough process that is often tackled using "brute force" --- trial and error of perhaps hundreds of different sizes and arragements. Don't be discouraged by it's surprising difficulty. Secondly, as you think about different coils designs in your device, it's helpful to focus on a "figure of merit" --- some particular way to measure the expected performance of a particular design. Some examples might be:
total copper cross-sectional area (add together all the cross-sectional areas of each wrap of wire in your coil)
The ratio of turns/resistance (you will want to maximize this ratio -- which is affected by the shape of the space that will hold the coil AND the wire shape as well (sometimes square wire is best for square spaces, for example).
The "figure of merit" you choose will guide you in finding the best set of design parameters.
However, be prepared for a difficult time -- but it can be a very fulfilling experience when you have found the best design !
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answered Sep 30, 2023 at 16:09
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TL:DR -the more copper you use (weight, volume, cost), the more efficient will be the magnet.
As Matt's answer says, changing the turns and wire cross section merely allows the magnet to run at the same power and the same field, while changing the voltage and current ratio, ie the resistance of the magnet.
An easy thought experiment to confirm Matt's assertion is to consider the magnet is wound with two identical sets of windings of N turns and A area. Each winding will generate the same heat, and the same field. If you connect them in series, you have 2N turns of area A. If you connect them in parallel, you have N turns of area 2A. You have changed the voltage/current ratio, you have changed the turns and the wire size, but you still have the same total amount of copper, and the same magnet performance.
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answered Sep 30, 2023 at 16:32
Neil_UKNeil_UK
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Slight generalization: The more you raise the conductance of the (infinite) volume surrounding the core, the more efficient the magnet will be. An alternative strategy to using more and more copper is superconductors.tobalt –tobalt 2023-09-30 18:48:47 +00:00 Commented Sep 30, 2023 at 18:48
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Prentice Hall Pre-Algebra Chapter 1: Algebraic Expressions & Integers
Variable in Algebra | Definition, Terms & Examples
Learn the definition of a variable in algebra. Find out what an independent variable is, compare different algebra terms, and explore examples of variables.
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Algebraic Expression | Definition, Operations & Examples
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Order of Operations in Math | Steps & Examples
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Evaluating Algebraic Expressions | Rules & Examples
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Types of Numbers & Its Classifications
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Absolute Value | Explanation & Examples
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Arithmetic Calculations with Signed Numbers
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Inductive vs. Deductive Reasoning in Geometry | Definition & Uses
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Sequences in Math | Overview & Types
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Plotting Points on the Coordinate Plane
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Chapter 2
Prentice Hall Pre-Algebra Chapter 2: Solving One-Step Equations & Equalities
Commutative & Associative Properties | Differences & Examples
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Distributive Property in Math | Definition & Examples
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Distributive Property & Algebraic Expressions | Rules & Examples
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Practice Simplifying Algebraic Expressions
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Writing and Classifying True, False and Open Statements in Math
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Solving Equations Using the Addition Principle
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Multiplication Principle | Definition, Equations & Examples
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Solving Problems With the Guess, Check & Revise Method
Inequality Signs in Math | Symbols, Examples & Variation
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Graphing Inequalities | Definition, Rules & Examples
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Graphing Inequalities | Overview & Examples
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Solving Linear Inequalities: Practice Problems
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Chapter 3
Prentice Hall Pre-Algebra Chapter 3: Decimals & Equations
Rounding Decimals | Steps, Rules & Examples
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How to Estimate with Decimals to Solve Math Problems
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Estimating Decimal Products and Quotients
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Measures of Central Tendency | Definition, Formula & Examples
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Calculating the Mean, Median, Mode & Range: Practice Problems
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Common Math Formulas | Overview, Uses & Importance
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Perimeter of Triangles and Rectangles
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Adding & Subtracting Decimals | Steps, Examples & Word Problems
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Multiplying & Dividing Decimals | Overview & Word Problems
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Metric System | Definition, Lists & Examples
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Chapter 4
Prentice Hall Pre-Algebra Chapter 4: Factors, Fractions & Exponents
Factoring in Algebra | Definition, Equations & Examples
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Properties of Exponents | Formula & Examples
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Finding the Prime Factorization of a Number | Meaning & Examples
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What is the Greatest Common Factor? | GCF Examples
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How to Raise and Reduce Fractions
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How to Simplify Complex Math Problems
Graphing Rational Numbers on a Number Line | Chart & Examples
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Multiplying Exponents | Overview, Methods & Rules
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Dividing Exponential Expressions
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Scientific Notation | Definition, Conversion & Examples
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Chapter 5
Prentice Hall Pre-Algebra Chapter 5: Operation with Fractions
Least Common Multiple | Definition, Formula & Examples
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Finding the Least Common Denominator | Overview & Examples
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Decimal to Fraction | Conversion & Examples
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Adding & Subtracting Fractions | Rules & Examples
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How to Add and Subtract Unlike Fractions and Mixed Numbers
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Multiplying Mixed Numbers | Steps & Examples
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Dividing Fractions & Mixed Numbers | Overview & Examples
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Standard Units of Measurement | Overview & Examples
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How to Convert Standard Units of Measure
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How to Solve Problems with Time
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Distance Equations | Formula, Calculation & Examples
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Product of Powers | Definition, Property & Examples
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Chapter 6
Prentice Hall Pre-Algebra Chapter 6: Ratios, Proportions & Percents
Ratios & Rates | Differences & Examples
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Proportion | Definition, Formula & Types
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Linear Scale Factor | Shapes, Area & Volume
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How to Read and Interpret Scale Drawings
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Probability of an Event | Simple, Compound & Complementary
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Percents: Definition, Application & Examples
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Converting Percent Notation to Fraction Notation
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Proportion | Definition, Formula & Examples
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Solve Problems Using Percents
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Taxes & Discounts: Calculations & Examples
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Tables & Graphs in the Real World | Uses & Examples
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Chapter 7
Prentice Hall Pre-Algebra Chapter 7: Solving Equations & Inequalities
Solving Multiple Step Equations | Explanation, Steps & Examples
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Solving Algebra Word Problems | Multi-Step Equations & Examples
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Multi-Step Equations with Fractions & Decimals | Steps & Examples
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Linear Equation | Parts, Writing & Examples
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Combining Like Terms in Algebraic Expressions
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Solving Linear Equations with Literal Coefficients
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Simple Interest Problems | Definition, Formula & Examples
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Compounding Interest | Formula, Types & Examples
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Chapter 8
Prentice Hall Pre-Algebra Chapter 8: Linear Functions & Graphing
Functions: Identification, Notation & Practice Problems
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Graphing Basic Functions
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Find the Slope of a Line | Formula & Examples
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How to Find and Apply the Intercepts of a Line
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How to Compose Functions
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Scatter Plot Graph | Overview, Uses & Examples
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Making Estimates and Predictions using Quantitative Data
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System of Equations in Algebra | Overview, Methods & Examples
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How Do I Use a System of Equations?
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Forms of a Linear Equation | Overview, Graphs & Conversion
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Undefined & Zero Slope Graph | Definition & Examples
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Chapter 9
Prentice Hall Pre-Algebra Chapter 9: Spatial Thinking
Geometry Definition, History & Branches
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Points, Lines & Angles in Geometry | Definition & Examples
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Angle Relationships | Overview & Types
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Parallel vs Perpendicular vs Transverse Lines Overview & Examples
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Characteristics of a Polygon | Overview & Examples
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Regular Polygon | Definition, Sides & Types
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Using Diagrams to Solve Word Problems
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Congruent Triangles | Definition, Parts & Examples
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Circle in Geometry | Definition, Parts & Examples
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Geometric Constructions Using Lines and Angles
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Translation in Math | Definition, Rules & Examples
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Symmetric Graphs | X-Axis, Y-Axis & Algebraic Symmetry
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Chapter 10
Prentice Hall Pre-Algebra Chapter 10: Area & Volume
Area of a Parallelogram | Measurement, Formula & Examples
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Area of Triangles & Rectangles | Formula, Calculation & Examples
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Area of a Trapezoid Formula & Examples
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Finding the Area & Circumference of a Circle
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Three-Dimensional Shapes | Definition, Types & Characteristics
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Prisms: Definition, Area & Volume
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Volume & Surface Area of a Cylinder | Formula & Examples
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Pyramids: Definition, Area & Volume
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Volume of a Cone | Definition, Formula & Examples
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Volume & Surface Area of a Sphere | Formula & Examples
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How to Use Models to Solve Math Problems
Mass & Volume | Relationship, Unit Conversion & Examples
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Chapter 11
Prentice Hall Pre-Algebra Chapter 11: Right Triangles in Algebra
Square Root | Definition, Formula & Examples
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Pythagorean Theorem | Definition, Method & Examples
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Distance Formula | Overview & Examples
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Midpoint | Formula & Examples
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Similar Triangles | Definition, Application Problems & Examples
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Special Right Triangles | Definition, Types & Examples
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Practice Finding the Trigonometric Ratios
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Angles of Elevation & Depression | Definition & Examples
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Chapter 12
Prentice Hall Pre-Algebra Chapter 12: Data Analysis & Probability
Relative Frequency | Definition & Examples
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Creating & Interpreting Dot Plots: Process & Examples
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Creating & Interpreting Histograms: Process & Examples
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Box Plot | Definition, Uses & Examples
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How to Draw a Graph | Steps, Types & Examples
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How to Use the Fundamental Counting Principle
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Independent & Dependent Events | Overview, Probability & Examples
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How to Calculate the Probability of Permutations
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Combinations in Probability | Equation, Formula & Calculation
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Random Sampling Definition, Types & Examples
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Chapter 13
Prentice Hall Pre-Algebra Chapter 13: Nonlinear Functions & Polynomials
How to Find and Classify an Arithmetic Sequence
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Working with Geometric Sequences
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Parabola | Definition & Parabolic Shape Equation
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Absolute Value | Graph & Transformations
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Exponential Growth & Decay | Formula, Function & Graphs
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What are Polynomials, Binomials, and Quadratics?
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Adding, Subtracting & Multiplying Polynomials | Steps & Examples
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Multiplying Binomials Using FOIL and the Area Method
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Multiplying Binomials | Overview, Methods & Examples
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685 | https://fiveable.me/introduction-to-mathematical-analysis/unit-2/definition-properties-sequences/study-guide/ctPwzUQJFM8NMo6H | printables
🏃🏽♀️➡️Intro to Mathematical Analysis Unit 2 Review
2.1 Definition and Properties of Sequences
🏃🏽♀️➡️Intro to Mathematical Analysis Unit 2 Review
2.1 Definition and Properties of Sequences
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
🏃🏽♀️➡️Intro to Mathematical Analysis
Unit & Topic Study Guides
2.1 Definition and Properties of Sequences
2.2 Limit of a Sequence
2.3 Convergence and Divergence
2.4 Limit Theorems for Sequences
Sequences are like mathematical playlists, each number taking its turn in a specific order. They can be finite or infinite, with terms defined by formulas or patterns. Understanding sequences is crucial for grasping limits and series.
Sequences come in various flavors, from arithmetic to geometric, each with unique properties. We'll explore how to identify, define, and analyze these number patterns, setting the stage for deeper concepts in mathematical analysis.
Sequences and Notation
Definition and Representation
A sequence is an ordered list of numbers (a1, a2, a3, ..., an)
The subscript denotes the position of the term in the sequence
The notation {an} represents a sequence
n is a natural number (positive integer) indicating the term's position
Sequence terms can be defined explicitly by a formula or implicitly by a recurrence relation
Sequences can be finite (specific number of terms) or infinite (continuing indefinitely)
Domain and Range
The domain of a sequence is the set of natural numbers
The range is the set of values the sequence terms can take
Example: For the sequence {2n}, the domain is {1, 2, 3, ...}, and the range is {2, 4, 6, ...}
Types of Sequences
Arithmetic and Geometric Sequences
Arithmetic sequences have a constant difference (d) between consecutive terms
General term: $a_n = a_1 + (n - 1)d$
Geometric sequences have a constant ratio (r) between consecutive terms
General term: $a_n = a_1 \times r^{(n-1)}$
Example: {2, 5, 8, 11, ...} is arithmetic (d = 3), while {2, 6, 18, 54, ...} is geometric (r = 3)
Special Sequences
Harmonic sequences are defined by the reciprocals of an arithmetic sequence
General term: $a_n = \frac{1}{a + (n - 1)d}$, where a and d are constants
Fibonacci sequence follows the recurrence relation $F_n = F_{n-1} + F_{n-2}$
Initial terms: $F_1 = 1$ and $F_2 = 1$
Constant sequences have the same value for all terms ($a_n = c$ for all n)
Alternating sequences have terms that alternate in sign (e.g., $(-1)^n$ or $(-1)^{(n+1)}$)
Sequence Properties
Monotonicity
Monotonically increasing: each term is greater than or equal to the previous term ($a_n \leq a_{n+1}$ for all n)
Monotonically decreasing: each term is less than or equal to the previous term ($a_n \geq a_{n+1}$ for all n)
Strictly increasing: each term is strictly greater than the previous term ($a_n < a_{n+1}$ for all n)
Strictly decreasing: each term is strictly less than the previous term ($a_n > a_{n+1}$ for all n)
Boundedness
Bounded above: there exists a real number M such that $a_n \leq M$ for all n
Bounded below: there exists a real number m such that $a_n \geq m$ for all n
Bounded: both bounded above and bounded below
Example: The sequence {1/n} is bounded below by 0 and bounded above by 1
General and nth Terms of Sequences
Defining the General Term
The general term is a formula or expression defining the nth term in terms of n
For arithmetic sequences: $a_n = a_1 + (n - 1)d$, where $a_1$ is the first term and d is the common difference
For geometric sequences: $a_n = a_1 \times r^{(n-1)}$, where $a_1$ is the first term and r is the common ratio
The nth term is found by substituting the value of n into the general term formula
Recursive Sequences and Piecewise Definitions
Recursive sequences (Fibonacci) require initial terms and the recurrence relation to determine the nth term
Some sequences have a general term defined piecewise, with different expressions for different ranges of n
Example: The sequence {an} defined by $a_n = n$ for $n \leq 5$ and $a_n = a_{n-1} + a_{n-5}$ for $n > 5$ is a recursive sequence with a piecewise definition |
686 | https://math.stackexchange.com/questions/4771168/what-is-the-absolute-maximum-of-y-x2-on-the-interval-2-2 | calculus - What is the absolute maximum of $y = x^2$ on the interval $(-2, 2]$? - Mathematics Stack Exchange
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What is the absolute maximum of y=x 2 y=x 2 on the interval (−2,2
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What is the absolute maximum of y=x 2 y=x 2 on the interval (−2,2](−2,2]?
My assumption was that the absolute max would be 4 4 and it would occur at x=2 x=2. However, according to an example I found, there is no absolute max since the interval is open on one end and that end would have the same value if it was closed.
How about on the interval (−1,2)(−1,2) - I would think that in this case the absolute max would be 4 4 at x=2 x=2.
I understand the extreme value theorem requires a closed interval, but is it possible to have absolute extrema exist at endpoints that are closed on only one side?
calculus
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edited Sep 18, 2023 at 14:02
N. F. Taussig
79.3k 14 14 gold badges 62 62 silver badges 77 77 bronze badges
asked Sep 18, 2023 at 13:59
McMathMcMath
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You are correct about the absolute maximum of y=x 2 y=x 2 in the interval (−2,2](−2,2]. Could you share the example you found that led you to doubt that answer? In the interval (−1,2)(−1,2), y=x 2 y=x 2 has no absolute maximum since 2 2 is not in the interval.N. F. Taussig –N. F. Taussig 2023-09-18 14:04:12 +00:00 Commented Sep 18, 2023 at 14:04
At x=2 x=2, you have x 2=4 x 2=4. There is no other x∈(−2,2]x∈(−2,2], such that x 2>4 x 2>4, thus 4 4 is indeed the maximum.Antoine –Antoine 2023-09-18 14:05:24 +00:00 Commented Sep 18, 2023 at 14:05
It is of course 4 4, attained at a single point, 2 2. There's no need for major theorems. On any (a,b)(a,b) this function has no maximum, but on [a,b)a,b) or (b,a 2023-09-18 14:09:05 +00:00 Commented Sep 18, 2023 at 14:09
The original example is a graphical one and there doesn't appear to be a way for me to upload a screenshot. However, the concept is identical. Thanks McMath –McMath 2023-09-18 14:22:08 +00:00 Commented Sep 18, 2023 at 14:22
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687 | https://foammagazine.com/wave-period/ | Published Time: 2022-06-07T20:38:53-07:00
Wave Period: Understanding the Magic Number
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Wave Period: Understanding the Magic Number
ByLinda LyJune 7, 2022 February 25, 2025
Wave period, swell period, interval—they all mean the same thing, and understanding this number is key to reading a surf forecast.
That’s because so much of surfing is based on physics: wind velocity, wind direction, swell energy, and of course, surfing itself. The wave period is the magic number that ties it all together into a single snapshot to predict what the waves will be doing in any given surf spot.
But how can you interpret that number to decide whether or not to paddle out? Let’s dive in.
What is a wave period?
To put it simply, a wave period is the time it takes for two successive crests in a swell to pass a specified point. The wave period is measured in seconds—for example, one wave every eight seconds.
This is easier to understand if you can picture it manually. Let’s say you’re at the beach and there’s a boat anchored beyond the breaking zone. You can count the seconds between one wave going under the boat and the next. Or, the seconds between one wave hitting a rock and the next, or one wave passing a buoy to the next—and that would give you an approximate wave period.
In fact, thousands of ocean buoys dotted around the world are used to calculate a whole range of weather variables, and they’re particularly useful when they’re located near a surf spot and can give us some intel on the wave direction, wave height, and wave period for that area.
When it comes to wave periods, these numbers show up in swell charts as anywhere from 1 to 20 seconds or more.
Why numbers—plural? Because there are often multiple sets of waves in the ocean at any given time, and they can all influence the surf conditions at a local break.
Why is the wave period important?
The holy grail for surfers is a long-period wave—what’s known as a ground swell. (Here’s a breakdown of how swells are formed, if you want to go into it further.)
That’s because the longer the wave period, the farther out to sea the waves were generated. This gives them more time to harness energy and pick up speed as they travel vast distances across the ocean. When long-period waves finally reach the shore, they’re generally larger, steeper, and cleaner, and form organized sets that make it easier to predict where and how the waves will break.
On the other hand, if you know that a shorter wave period is coming in, you can assume the waves will be weaker, slower, or smaller, which is typical of wind swell vs. ground swell.
What do different wave periods mean?
Every surf forecast will show wave periods as part of its swell chart. While these are the magic numbers you should know, there isn’t a single “magic” number to look for—it all depends on the types of waves you want to surf and the characteristics of your local surf break.
Reef and point breaks generally need longer wave periods to really come alive, while the same long-period waves will often close out quickly at beach breaks.
Knowing what wave period works best for your surf spot will come with time and observation. But, the chart below can serve as a good starting point for interpreting the wave periods on your local surf report.
| Wave period | What this means for surf |
--- |
| 1 to 4 seconds | Forget about it—this swell is so small and weak that it can almost never be surfed. Waves in this range are typically caused by strong local winds and appear lumpy and choppy without any sort of order. |
| 5 to 6 seconds | These waves are still in the early stages of being formed, and much of what’s described for 1- to 4-second intervals holds true for this range as well. Strong onshore winds make for subpar surfing, but occasionally you might find the odd, weak, surfable wave if you’re really desperate. |
| 7 to 9 seconds | This is a decent windswell producing average surf, but nonetheless the waves are considered ridable, especially for areas that don’t get great surf. If you’re lucky, a change in wind direction may improve the quality of the waves, so offshore conditions are possible if you stick it out. |
| 10 to 12 seconds | In this range, the swells are starting to move away from the storms that created them, traveling some distance in the open ocean before they break. Waves produced by mid-period (or medium-distance) swells are often quite good, and can offer some of the best surf on sandy beaches. |
| 13 to 15 seconds | This interval range is where we enter groundswell territory—the clean, organized sets coming in from faraway storms, but without the strong winds that created them. This means the surf is typically calm yet powerful, with defined waves that look a lot more lined up. If everything else falls in place, you’ll see some epic waves. Get out there! |
| 16+ seconds | Swells in this range are generated by super distant, extremely powerful storms. They often travel across the largest oceans to reach the beach, picking up an incredible amount of speed and energy. At the right spot, the surf will be large and in charge! But, these big wave periods don’t always equate to epic waves, and only certain point and reef breaks can handle such high-energy swells. |
Like I mentioned, this chart is merely a starting point to help you decipher what all those different intervals mean. Remember that a shorter-period swell could actually be good training ground for beginners even though experienced surfers might call it flat; and a pumping 15-second swell has all the hardcore surfers paddling out but those same conditions could be dangerous for someone who’s still improving.
There’s much more to a quality surf session than the wave period, which is why surf forecasts also include other useful bits of info like wind speed, swell height, swell direction, and weather.
Post Tags: #Wave Knowledge
Linda Ly
Post navigation
Previous All About Swells: What They Are and How They Form in the Ocean
Next Onshore Wind vs. Offshore Wind: What They Mean for Surfing
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688 | https://www.unitsconverters.com/en/Rad/Cm-To-Rad/M/Utu-9022-9019 | rad/cm to rad/m | Radian per Centimeter to rad/m
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Construct A Triangle Given Two Sides And An Angle
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In this lesson, we will learn how to construct a triangle given two sides and an angle between them (SAS). In order for the triangle to be unique, the angle given must be between the two given sides.
Example:
Construct a triangle PQR given that PQ = 4 cm, PR = 5 cm and = 120˚.
Solution:
Step 1:
We construct a 60˚ angle and get the supplementary angle, which is 120˚.
Step 2:
We construct 2 line segments of 4 cm and 5 cm from P respectively . Then, draw a line from Q to R to form the triangle.
How to construct a triangle given two sides and the included angle (SAS)?
Draw a rough sketch of the construction.
Measure and draw out one of the line segments (usually the longer one).
Construct the angle from the base, using a protractor.
Set compasses to the second given length, position compass point at center of angle and draw an arc that intersects the second line.
Connect this intersection with the other end of the base.
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Construction: Triangle - 2 Sides, 1 Angle
A demonstration of how to construct a triangle, give the lengths of two sides and the angle between them.
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How to construct a congruent triangle using the side-angle-side congruence postulate?
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How to construct a triangle given two angles and the side in between (ASA)?
Draw a rough sketch of the construction.
Measure and draw out one of the line segments (usually the longer one).
Construct the first angle from one end of the base, using a protractor.
Construct the second angle from the other end of the base, using a protractor.
The two constructed angles will intersect creating the third vertex.
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690 | https://www.scaler.in/trailing-zeros-in-factorial/ | Skip to content
Namanjeet Singh
Data Structures and Algorithms
Trailing Zeros in Factorial
Problem Statement
Given an integer n, our task is to write a function that returns the count of trailing zeros in n!. Here n! is the factorial of a number n which is the product of all numbers in the range .
Example :
```
Input: n = 3
Output: 0
Input: n = 5
Output: 1
Input: n = 20
Output: 4
```
Explanation
Example – 1 :
When , Factorial of 3 is 6, which has no trailing zeros in factorial.
Example – 2 :
When , Factorial of 5 is 120, which has no trailing zeros in factorial.
Example – 3 :
When , Factorial of 20 is 2432902008176640000, which has four trailing zeros in factorial.
Constraints
Approach – 1 : Naive Approach
Algorithm :
In this approach, the intuition is to calculate the value of n! and then count the number of trailing zeros in factorial. The number of zeros can be measured by repeatedly dividing the value of the factorial by 10 until the last digit becomes non-zero.
Implementation
Implementation of Naive Approach in C++ :
```
include
using namespace std;
int trailingZeros(int n) {
int fact = 1;
// calculating factorial
for (int i = 1; i <= n; i++) {
fact = fact i;
}
int zeros = 0;
// counting the number of trailing zeros
while (fact % 10 == 0) {
zeros++;
fact = fact / 10;
}
return zeros;
}
int main() {
cout << trailingZeros(6);
return 0;
}
```
Output :
```
1
```
Implementation of Naive Approach in Java :
```
import java.io.;
public class Main {
public static int trailingZeros(int n) {
int fact = 1;
// calculating factorial
for (int i = 1; i <= n; i++) {
fact = fact i;
}
int zeros = 0;
// counting the number of trailing zeros
while (fact % 10 == 0) {
zeros++;
fact = fact / 10;
}
return zeros;
}
public static void main(String[] args) {
System.out.println(trailingZeros(6));
}
}
```
Output :
```
1
```
Implementation of Naive Approach in Python :
```
def trailingZeros(n):
fact = 1
zeros = 0
# calculating factorial
for i in range(1, n + 1):
fact = fact i
# counting the number of trailing zeros
while (fact % 10) == 0:
fact = fact // 10
zeros += 1
return zeros
print(trailingZeros(20))
```
Output :
```
4
```
Complexity Analysis
Time Complexity :
Since we are iterating over n times to calculate the factorial value, the Time Complexity for calculating the number of trailing zeros in factorial is .
Space Complexity :
While no auxiliary space is required to calculate the number of trailing zeros in factorial, Space Complexity is .
Approach – 2 : Optimal Approach
Algorithm :
The above approach can lead to an overflow in case of bigger numbers as factorial could be a really big number. So, the intuition for a better approach is to consider that a trailing zero is produced when a number is multiplied by 10. The prime factors of 10 are 2 and 5.
So, our task now becomes to just count the number of 2s and 5s.
Let us consider an example, where n = 5 :
5! = 2 x 2 x 2 x 3 x 5
The number of trailing zeros in the factorial of 5 will be 1 because only one pair of 2 and 5 can be made from one 5 and three 2s in the prime factors of 5!.
In the above example, we can see that the number of 2s in prime factors is always more than or equal to the number of 5s.
So if we count the number of 5s in the prime factors, our task is done. But how to calculate the number of 5s in prime factors of n!? An easy method of doing this would be using Legendre’s Formula, which is used to get the highest power of a prime number p in n!.
Here, represents the highest power in p, in a way that divides n! where p is a prime number.
is the greatest integer function.
The above formula will compute the exact number of 5s in n! as it will consider all multiples of 5 that are less than n. This will also consider all higher power multiples of 5 i.e 25, 125, etc.
Therefore, the formula gets transformed to the following for calculating trailing zeros in factorial of a number n :
Where k = floor of .
Let us now solve a few examples using this formula :
Example – 1 :
number of trailing zeros in 24! will be
trailing zeros
Example – 2 :
number of trailing zeros in 123! will be :
trailing zeros.
trailing zeros.
Therefore, 123! has a total of 28 trailing zeros
The final algorithm is :
Create a function trailing_zeros(int number) that takes an integer n and returns the count of trailing zeros in factorial of n.
Check for the edge case where, if n < 0, return -1.
Initialize count = 0.
Traverse using a for loop and divide the number n by powers of 5 at every iteration.
If number/i is greater than or equal to 1, add number/i to count.
At the end of for loop, the return count as result.
Implementation
Implementation of Optimal Approach in C++ :
```
include
using namespace std;
int trailingZeros(int n) {
// in case of negative numbers
if (n < 0)
return -1;
// initializing count variable
int count = 0;
// dividing n by powers of 5 and updating count
for (int i = 5; n / i >= 1; i = 5)
count += n / i;
return count;
}
int main() {
int n = 100;
cout << "Count of trailing 0s in " << 100 << "! is "<< trailingZeros(n);
return 0;
}
```
Output :
```
The count of trailing 0s in 100! is 24
```
Implementation of Optimal Approach in Java :
```
import java.io.;
class Main {
static int trailingZeros(int n) {
// in case of negative numbers
if (n < 0) return -1;
// initializing count variable
int count = 0;
// dividing n by powers of 5 and updating count
for (int i = 5; n / i >= 1; i = 5) count += n / i;
return count;
}
public static void main(String[] args) {
int n = 100;
System.out.println(
"Count of trailing 0s in " + n + "! is " + trailingZeros(n)
);
}
}
```
Output :
```
The count of trailing 0s in 100! is 24
```
Implementation of Optimal Approach in Python :
```
def trailingZeros(n):
in case of negative numbers
if(n < 0):
return -1
# initializing count variable
count = 0
# dividing n by powers of 5 and updating count
while(n >= 5):
n //= 5
count += n
return count
n = 100
print("Count of trailing 0s in 100! is", trailingZeros(n))
```
Output :
```
The count of trailing 0s in 100! is 24
```
Complexity Analysis
Time Complexity :
In this approach, n is divided by 5 till it becomes 0. Therefore, the number of iterations till which the program will execute is equal to , which makes the Time Complexity as .
Space Complexity :
Similar to the previous approach no auxiliary space is required to calculate the number of trailing zeros in factorial, Space Complexity is .
Conclusion
The drawback of using the naive approach is that it calculates the factorial of given n before computing the result, whose value can grow very large and can cause an overflow issue.
The optimal approach uses Legendre’s Formula to compute the highest power of any prime number p in n!, which in this problem calculates the highest power of 5 in n!.
The time and Space Complexity of the Optimal Approach to calculate the number of trailing zeros in factorial is and respectively.
Author
Namanjeet Singh
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691 | https://math.stackexchange.com/questions/1757859/test-for-the-convergence-of-the-sequence-s-n-frac1n-left1-frac12 | real analysis - Test for the convergence of the sequence $S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$ - Mathematics Stack Exchange
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Test for the convergence of the sequence S n=1 n(1+1 2+1 3+⋯+1 n)S n=1 n(1+1 2+1 3+⋯+1 n)
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S n=1 n(1+1 2+1 3+⋯+1 n)S n=1 n(1+1 2+1 3+⋯+1 n)
Show the convergence of S n S n (the method of difference more preferably)
I just began treating sequences in school, and our teacher taught that monotone increasing sequence, bounded above and monotone decreasing sequences, bounded below converge.
and so using that theorem here.. I found the
(n+1)t h t e r m(n+1)t h t e r m
,
S n+1=1 n+1(1+1 2+1 3+⋯+1 n+1)S n+1=1 n+1(1+1 2+1 3+⋯+1 n+1)
and then subtracted the (n)th term from it
What I was able to get was...
S n+1−S n=1(n+1)2−(1+1 2+⋯+1 n)n(n+1).S n+1−S n=1(n+1)2−(1+1 2+⋯+1 n)n(n+1).
...but then this is where I get stucked, but i'm trying to prove that the sequence > 0(i.e Converges) or < 0 (i.e diverges).
real-analysis
sequences-and-series
harmonic-numbers
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edited Apr 25, 2016 at 19:41
koodejikoodeji
asked Apr 25, 2016 at 10:50
koodejikoodeji
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Your attempt...?hululu –hululu 2016-04-25 10:54:16 +00:00 Commented Apr 25, 2016 at 10:54
I'm not used to this..but I'll try to type that..koodeji –koodeji 2016-04-25 10:58:24 +00:00 Commented Apr 25, 2016 at 10:58
Sn+1 - sn, I subtracted the nth term from the n+1th term and I Got.. (N/n+1 - (1 + 1/2 + 1/3 +...+ 1/n))/n(n+1)koodeji –koodeji 2016-04-25 11:02:24 +00:00 Commented Apr 25, 2016 at 11:02
1 This might help you: math.stackexchange.com/questions/207910/…math.stackexchange.com/questions/210681/…math.stackexchange.com/questions/248116/…math.stackexchange.com/questions/930436/a-result-on-sequencesMartin Sleziak –Martin Sleziak 2016-04-25 11:02:43 +00:00 Commented Apr 25, 2016 at 11:02
I have tried to add to your post what you wrote in a comment. I am not sure whether I understood correctly what you meant, so please, edit it further. For some basic information about writing math at this site see e.g. here, here, here and here.Martin Sleziak –Martin Sleziak 2016-04-25 11:20:33 +00:00 Commented Apr 25, 2016 at 11:20
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Hint. As an alternative to a Cesaro-like theorem, one may use the fact that x↦1 x x↦1 x is decreasing over [1,∞)[1,∞) to get
0<1+1 2+1 3+⋯+1 n<1+∫n 1 d x x=1+log n 0<1+1 2+1 3+⋯+1 n<1+∫1 n d x x=1+logn
giving
0<1 n(1+1 2+1 3+⋯+1 n)<1 n+log n n.0<1 n(1+1 2+1 3+⋯+1 n)<1 n+logn n.
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answered Apr 25, 2016 at 11:04
Olivier OloaOlivier Oloa
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4 Overkill. One shouldn't need integrals in an easy sequences problem; sequences are studied much earlier than integrals.user325968 –user325968 2016-04-25 11:57:37 +00:00 Commented Apr 25, 2016 at 11:57
2 Not necessarily -- this can also (and perhaps more naturally) seen as a series, and series are studied around the same time as integrals. Moreover, this is a simple proof, provided one has the basic definition of integrals and of logarithm. It may not be the most elementary or even canonical way, but it is a self-contained alternative -- why not have it written as an answer, somewhere? It does not hurt to have options.Clement C. –Clement C. 2016-04-25 20:51:55 +00:00 Commented Apr 25, 2016 at 20:51
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You can check here how can this be solved:
lim n→∞1 n=0⟹lim n→∞1 n∑k=1 n 1 k=0 lim n→∞1 n=0⟹lim n→∞1 n∑k=1 n 1 k=0
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answered Apr 25, 2016 at 11:00
DonAntonioDonAntonio
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I'm sorry, I don't see why the implication.Bernard –Bernard 2016-04-25 11:24:58 +00:00 Commented Apr 25, 2016 at 11:24
@Bernard Did you read the link I post in my answer?DonAntonio –DonAntonio 2016-04-25 11:47:24 +00:00 Commented Apr 25, 2016 at 11:47
I know, this is Cesáro's lemma, but you can't use it here: the harmonic series diverge. Or am I missing something?Bernard –Bernard 2016-04-25 12:00:08 +00:00 Commented Apr 25, 2016 at 12:00
@Bernard I think you're missing something big time, and I don't think this is widely known as Cesaro's Lemma, though it is related to Cesaro sums, certainly.. Please do read the link again, or read 4-6 lines here: en.wikipedia.org/wiki/Ces%C3%A0ro_meanDonAntonio –DonAntonio 2016-04-25 12:03:30 +00:00 Commented Apr 25, 2016 at 12:03
@Bernard And it has nothing to do with series, whether convergent or not.DonAntonio –DonAntonio 2016-04-25 12:09:29 +00:00 Commented Apr 25, 2016 at 12:09
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If you denote H n=∑k=1 n 1 k H n=∑k=1 n 1 k, one knows H n∼∞ln n H n∼∞lnn, hence
S n∼∞ln n n→0.S n∼∞lnn n→0.
A more elementary proof:
⟺S n=1 n H n>S n+1=1 n+1(H n+1 n+1)(n+1)H n>n H n+n n+1⟺H n>n n+1,S n=1 n H n>S n+1=1 n+1(H n+1 n+1)⟺(n+1)H n>n H n+n n+1⟺H n>n n+1,
which is true since H n>1>n n+1 H n>1>n n+1. So the sequence (S n)(S n) is decreasing and bounded from below by 0 0. Apply the monotone convergence theorem.
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edited Apr 25, 2016 at 12:48
answered Apr 25, 2016 at 11:11
BernardBernard
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One doesn't need to know the ln n lnn behavior of the harmonic series, the problem is easier than that.user325968 –user325968 2016-04-25 11:58:48 +00:00 Commented Apr 25, 2016 at 11:58
@mathguy: It has the advantage of being very short…Bernard –Bernard 2016-04-25 12:01:39 +00:00 Commented Apr 25, 2016 at 12:01
Think of what the OP will learn from the answer.user325968 –user325968 2016-04-25 12:04:04 +00:00 Commented Apr 25, 2016 at 12:04
@mathguy: OK I've added an elementary proof.Bernard –Bernard 2016-04-25 12:48:44 +00:00 Commented Apr 25, 2016 at 12:48
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Avoiding appeals to logarithms:
Writing H n=1+1 2+⋯+1 n H n=1+1 2+⋯+1 n, note that for n>M n>M:
H n=1+1 2+⋯+1 n=1+1 2+⋯+1 m+⋯+1 n=H M+1 M+1+⋯+1 n H n=1+1 2+⋯+1 n=1+1 2+⋯+1 m+⋯+1 n=H M+1 M+1+⋯+1 n
Now, each of the terms after H M H M is <1 M<1 M, whence H n<H M+n−M M H n<H M+n−M M.
Thus, H n n<H M n+n−M n M=H m−1 n+1 M H n n<H M n+n−M n M=H m−1 n+1 M for any positive integer M M, as long as n n is larger than M M.
From this, we deduce that lim sup H n n≤1 M lim sup H n n≤1 M for all positive integers M M, and thus that lim sup H n n=0 lim sup H n n=0.
As H n n>0 H n n>0, it is thus trivial to deduce that the sequence converges to 0 0.
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answered Apr 25, 2016 at 13:32
πr8πr8
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Linked
-1Prove that lim n→∞1 n(1+1 2+⋯+1 n)=0 lim n→∞1 n(1+1 2+⋯+1 n)=0
2Limit of (1+1 2+1 3+⋯+1 n)/n(1+1 2+1 3+⋯+1 n)/n
4Checking the Convergence of the sequence a n=1 n(1+1 2+1 3+....1 n)a n=1 n(1+1 2+1 3+....1 n)
32Prove convergence of the sequence (z 1+z 2+⋯+z n)/n(z 1+z 2+⋯+z n)/n of Cesaro means
24arithmetic mean of a sequence converges
8Find lim n→∞(1+1 2+...+1 n)1 n lim n→∞(1+1 2+...+1 n)1 n
5A result on sequences: x n→x x n→x implies x 1+⋯+x n n→x x 1+⋯+x n n→x without using Stolz-Cesaro
5If a n→ℓ a n→ℓ then a^n→ℓ a^n→ℓ
-1How to prove lim k→∞1 k∑m=1 k 1 m=0 lim k→∞1 k∑m=1 k 1 m=0
Related
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3Asymptotics of (∑∞k=0 k n k!)1 n(∑k=0∞k n k!)1 n at n→∞n→∞
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692 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11001294/ | The intersection of coagulation activation and inflammation after injury: What you need to know - PMC
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J Trauma Acute Care Surg
. Author manuscript; available in PMC: 2024 Apr 8.
Published in final edited form as: J Trauma Acute Care Surg. 2023 Nov 13;96(3):347–356. doi: 10.1097/TA.0000000000004190
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The intersection of coagulation activation and inflammation after injury: What you need to know
Todd W Costantini
Todd W Costantini, MD
1 Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery (T.W.C.), UC San Diego School of Medicine, San Diego; Department of Surgery (L.Z.K.), Zuckerberg San Francisco General Hospital, University of California, San Francisco, San Francisco, California; Department of Surgery (T.P.), University of Cincinnati College of Medicine, Cincinnati, Ohio; and Comparative Effectiveness and Clinical Outcomes Research Center (R.C.), Riverside University Health System, Loma Linda University School of Medicine, Riverside, California
Find articles by Todd W Costantini
1, Lucy Z Kornblith
Lucy Z Kornblith, MD
1 Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery (T.W.C.), UC San Diego School of Medicine, San Diego; Department of Surgery (L.Z.K.), Zuckerberg San Francisco General Hospital, University of California, San Francisco, San Francisco, California; Department of Surgery (T.P.), University of Cincinnati College of Medicine, Cincinnati, Ohio; and Comparative Effectiveness and Clinical Outcomes Research Center (R.C.), Riverside University Health System, Loma Linda University School of Medicine, Riverside, California
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1, Timothy Pritts
Timothy Pritts, MD, PhD
1 Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery (T.W.C.), UC San Diego School of Medicine, San Diego; Department of Surgery (L.Z.K.), Zuckerberg San Francisco General Hospital, University of California, San Francisco, San Francisco, California; Department of Surgery (T.P.), University of Cincinnati College of Medicine, Cincinnati, Ohio; and Comparative Effectiveness and Clinical Outcomes Research Center (R.C.), Riverside University Health System, Loma Linda University School of Medicine, Riverside, California
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1, Raul Coimbra
Raul Coimbra, MD, PhD
1 Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery (T.W.C.), UC San Diego School of Medicine, San Diego; Department of Surgery (L.Z.K.), Zuckerberg San Francisco General Hospital, University of California, San Francisco, San Francisco, California; Department of Surgery (T.P.), University of Cincinnati College of Medicine, Cincinnati, Ohio; and Comparative Effectiveness and Clinical Outcomes Research Center (R.C.), Riverside University Health System, Loma Linda University School of Medicine, Riverside, California
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1
Author information
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1 Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery (T.W.C.), UC San Diego School of Medicine, San Diego; Department of Surgery (L.Z.K.), Zuckerberg San Francisco General Hospital, University of California, San Francisco, San Francisco, California; Department of Surgery (T.P.), University of Cincinnati College of Medicine, Cincinnati, Ohio; and Comparative Effectiveness and Clinical Outcomes Research Center (R.C.), Riverside University Health System, Loma Linda University School of Medicine, Riverside, California
AUTHORSHIP
R.C. and T.W.C. contributed in the study design. R.C., T.W.C., L.Z.K., and T.P. contributed in the manuscript drafting. R.C., T.W.C., L.Z.K., and T.P. contributed in the critical revision.
✉
Address for correspondence: Todd W. Costantini, MD, Division of Trauma, Surgical Critical Care, Burns and Acute Care Surgery, Department of Surgery, UC San Diego School of Medicine, 200 W Arbor Dr 8896, San Diego, CA 92103; tcostantini@health.ucsd.edu.
Issue date 2024 Mar 1.
PMC Copyright notice
PMCID: PMC11001294 NIHMSID: NIHMS1980448 PMID: 37962222
The publisher's version of this article is available at J Trauma Acute Care Surg
COAGULOPATHY AND INFLAMMATION AFTER INJURY
Coagulopathy and acute organ dysfunction are indicative of the systemic inflammatory response syndrome (SIRS) that is frequently seen in severely injured patients. While we often note these clinical signs in patients during the early postinjury management, these events do not occur in isolation. In fact, the coagulation system, immune system, and complement system all interact and precipitate coagulopathy, platelet dysfunction, cytokine release, immune cell activation, and endotheliopathy that must be resolved to ensure adequate resuscitation and recovery from injury (Fig. 1). Knowledge of the extensive crosstalk that occurs between these systems is essential to understanding the physiology of severely injured patients and to optimize the return to homeostasis through resuscitation and correction of coagulopathy. Promptly correcting coagulopathy and resolving SIRS is essential to mitigate organ dysfunction and multiorgan failure that can cause significant morbidity postinjury.
Figure 1.
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Diagram demonstrating the interconnected components of the systemic inflammatory response to injury.
THE SYSTEMIC INFLAMMATORY RESPONSE TO INJURY
Systemic inflammatory response syndrome occurs early after injury and is an essential immune response to tissue injury, although it may also be deleterious to the host. Clinically, SIRS is defined as two or more of the following criteria: temperature of >38°C or <36°C, heart rate of >90 beats per minute, white blood cell count of >12,000 or <4,000 per microliter or >10% bands, respiratory rate of >20 breaths per minute, or pCO 2 level of <32 mm Hg. Underlying these clinical alterations, a host response includes activation of innate and adaptive immune cells, cellular and noncellular coagulation cascades, and complement systems, resulting in a widespread and interrelated inflammatory response. Characterization of the inflammatory response to major injury reveals a rapid and dramatic change in the expression of genes related to innate and adaptive immunity. The evaluation of the leukocyte transcriptome from the Inflammation and Host Response to Injury program demonstrated more than 5,000 genes with at least twofold change in expression in patients after major injury, with the greatest changes occurring less than 12 hours from the time of injury.1 Interestingly, there was a simultaneous increase in genes related to innate immunity, with suppression of genes related to adaptive immunity. These leukocyte genomic changes persisted for longer in patients with complicated recovery from injury, suggesting that resolution of the SIRS response is a key factor in recovery. In addition to dramatic changes in the immune system, there are multiple other host changes after injury, including activation of endothelial cells and platelets, as well as alterations to the coagulation system and complement cascade. The interaction of these systems contributes to thromboinflammation in severely injured patients and must be understood to optimally address resuscitation and mitigate trauma-induced coagulopathy (TIC) and its complications.
THE INNATE IMMUNE SYSTEM
The innate immune system provides early defense against injury by recognizing damage-associated molecular patterns (DAMPs) released by injured cells and damaged tissue, leading to cytokine release, further activation and recruitment of immune cells, and further amplification of the inflammatory response.2 Neutrophils play a major role in the early immune response to injury and are early effectors of the inflammation.3 Under proinflammatory conditions, neutrophils demonstrate significant interaction with the endothelium, platelets, and the coagulation processes, including through the formation of neutrophil extracellular traps (NETs). Neutrophil extracellular traps are networks of extracellular fibers that allow neutrophils to kill pathogens while minimizing oxidative burst-induced damage to other cells. Recent studies on the inflammatory response to injury indicated that NETosis correlates with increasing injury severity.4–7 The role(s) of leukocytes and NETs in inflammation and thrombosis remain an area of ongoing investigation. Still, their regulation may eventually yield strategies to therapeutically manipulate thromboinflammation after injury.8 Neutrophils also migrate to sites of tissue injury via the extracellular mask tricks, where they degranulate through oxidative burst and cause cytotoxicity. While required to control local tissue injury, this can also damage host cells. Because neutrophils are early responders, secretion of cytokines, including tumor necrosis factor (TNF) and interferon γ, is necessary to further recruit other innate immune cells.
Neutrophils and monocytes are released from the bone marrow into the circulation in response to injury, in a process known as emergency myelopoiesis.9 Monocytes also play an important role in the early innate immune response by patrolling blood vessel walls, interacting with the endothelium, regulating transendothelial migration, and propagating the immune response through cytokine release. They also mature into macrophages after leaving the circulation, playing an important role in the phagocytosis of cellular debris. A recent study of peripheral blood drawn from severely injured patients within the first 12 hours postinjury demonstrated phenotypic and functional changes in circulating neutrophil and monocyte subsets compared with blood collected from healthy volunteers.10 This included increased phagocytic activity and increased production of reactive oxygen species, further supporting dysregulation of the innate immune response following trauma. Not only does immune cell recruitment and cytokine release amplify the innate immune response, but it also interacts with the coagulation system to further drive thromboinflammation after injury.
THE INTERSECTION OF COAGULATION AND INFLAMMATION
The study of TIC has been long appreciated as pivotal for advancing the science behind hemorrhage control for injured patients.11,12 However, the intersection of TIC with inflammation and immune responses has yet to receive equal focus in efforts to improve outcomes after injury.13 In thromboinflammation, coagulation cascades intersect with inflammatory pathways through various mechanisms and cellular interactions (Fig. 2). Coagulation factors, especially tissue factor (TF) and factor VIIa, interact with immune cells, triggering the release of inflammatory mediators and amplifying inflammation.14 Reciprocally, inflammatory mediators can also activate coagulation by promoting the expression of TF on endothelial cells and monocytes. Severe injury is associated with the release of proinflammatory mediators, which alter the circulatory system’s normal ratio of procoagulation and anticoagulation effects. The overall proinflammatory response after injury leads to a dysregulated balance, with endothelial cell, platelet, and leukocyte activation that results in appropriate and inappropriate thrombus formation, thrombolysis, and exhaustion of clotting cycle components (Table 1).
Figure 2.
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The intersection of inflammation, coagulation, and thrombosis during SIRS. Proinflammatory cytokines and DAMPs propagate the inflammatory response to innate immune cells, the endothelium, and the coagulation system. Polymorphonuclear leukocytes are activated and can form NETs at the surface of the activated endothelium. Endothelial injury can expose the underlying subendothelium exposing TF and collagen, which can activate the clotting cascade. Activated platelets play an important role in this inflammatory response where they can cause thrombosis in association with fibrin and vWF at the endothelial surface. Systemic inflammatory response syndrome also causes shedding of the endothelial glycocalyx and activation of fibrinolytic pathways. Diagram was created with Biorender.com.
TABLE 1.
Selected Key Mediators of the Intersection of Coagulation and Inflammation
| Selected Mediator | Source | Role in Thromboinflammation |
:---:
| Activated protein C | Plasma | Natural anticoagulant with cytoprotective role; promotes endothelial barrier; inhibits proinflammatory mediators |
| Complement cascade | Plasma | Stimulates platelet activation and adhesion |
| DAMPs | Injured cells | Cytokine release and inflammatory response amplification |
| Elastase | Neutrophils | Decreased circulating antithrombin |
| EVs | Multiple | Propagates inflammatory response |
| Factor VIIa | Plasma | Triggers release of inflammatory mediators from immune cells |
| Fibrin | Plasma | Promotes leukocyte adhesion, migration, and cytokine release |
| P-selectin | Platelets | Recruits immune cells; induces cytokine release; receptor for C3b, initiates complement activation |
| Phosphatidylserine | Platelets | Recruits immune cells; induces cytokine release |
| Platelet factor 4 | Platelets | Recruitment and activation of neutrophils and monocytes |
| Cytokines | Multiple | Increased TF expression on endothelial cells and monocytes |
| Thrombin | Plasma | Cleaves protease-activated receptors leading to cytokine release |
| TF | Multiple | Triggers release of inflammatory mediators |
| Toll like receptors | Platelets | Recognizes PAMPs and DAMPs, triggers inflammatory responses |
| vWF | Multiple | Propagates platelet adhesion |
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PAMP, pathogen-associated molecular pattern.
ROLE OF COAGULATION FACTORS, FIBRINOGEN, AND FIBRINOLYSIS IN THROMBOINFLAMMATION
Coagulation factors, particularly thrombin, have been shown to modulate immune responses because thrombin can cleave protease-activated receptors on endothelial cells, immune cells, and platelets, leading to the release of cytokines, chemokines, and adhesion molecules.15 Furthermore, fibrinogen and fibrin play multifaceted roles influencing both thrombosis and inflammation, with effects that extend beyond serving as a passive substrate for clot formation. Fibrin can activate immune cells through integrin receptors, promoting leukocyte adhesion, migration, and cytokine release, and provides a scaffold for the assembly of immune cells and facilitation of neutrophil and monocyte activation and function.16 Moreover, fibrin complexes can serve as a platform for the deposition of complement components, amplifying inflammatory responses, sequestering cytokines and growth factors, and releasing them upon fibrinolysis.17 Increased plasminogen activator inhibitor 1 (PAI-1) levels can impair the fibrinolytic system, leading to a prothrombotic environment. This imbalance contributes to thrombosis and sustains inflammation by preventing timely fibrin degradation.
The concept of fibrinolysis shutdown, described in injury, is also seen in prothrombotic and proinflammatory infectious diseases, including coronavirus disease 2019 (COVID-19). Death in patients with fibrinolytic shutdown after injury has been associated with organ failure,18,19 presumed to be related to microvascular thromboses and the relationship of the imbalanced fibrinolytic system with thromboinflammation. Furthermore, it has been found that patients with persistently low fibrinolytic activity after injury have an increase in later mortality,20,21 likely because of persistently high PAI-122 and neutrophil elastase suppression of fibrinolysis.23 The fibrinolytic system is also under complex and bidirectional regulatory control by platelets. For example, the tissue plasminogen activator and plasminogen binding capacity of platelets play a role in activating fibrinolysis. Still, the platelet-induced contraction of fibrin-rich clots and the platelet release of PAI-1–rich α granules are antifibrinolytic in nature.24,25 The extensive role of platelets in thromboinflammation is discussed further hereinafter.
Antithrombin is the main natural inhibitor of factor Xa and thrombin and can also be altered in the setting of posttraumatic SIRS. Activation of proinflammatory pathways may decrease circulating antithrombin levels through degradation by elastase from activated neutrophils, decreased synthesis, and consumption.26 Antithrombin itself may also play a role in the inflammatory response by indirectly suppressing platelet activation, decreasing leukocyte rolling on endothelial cells, and suppressing the production of proinflammatory cytokines.26 Recent clinical data demonstrate that injury decreases plasma antithrombin levels.27 Given the overall anti-inflammatory role of antithrombin, the decrease in antithrombin seen after injury may predispose patients to worsened systemic inflammation and a prothrombotic state.
ROLE OF PLATELETS IN THROMBOINFLAMMATION
Platelets are anucleate blood cells originating from megakaryocytes in the bone marrow. Classically, their “primary” biological responsibility has been linked to hemostatic control.28 Endothelial injury and tissue damage expose subendothelial matrix components, including TF, von Willebrand factor (vWF), and collagen, triggering platelet activation and providing activated platelet surfaces for coagulation factor assembly.29 The ligand interaction of platelet receptor glycoprotein Ibα with vWF leads to the recruitment of platelets to sites of vascular injury. Platelets rapidly adhere to the exposed subendothelial matrix, promoting primary hemostasis. In addition, platelet activation promotes the exposure of phosphatidylserine, a key component in the amplification of coagulation. The procoagulant phospholipid surface of activated platelets facilitates the assembly of the tenase and prothrombinase complexes, enabling the critical burst of thrombin and ultimately leading to fibrin generation. In addition, activated platelets release bioactive molecules, including thromboxane A2, adenosine diphosphate, and PAI-1, which further enhance platelet aggregation. They are involved in multiple cell-cell signaling mechanisms and regulation of the microvascular environment.25,30
In the setting of injury and shock, platelet hemostatic functions are altered and have been the subject of much research related to TIC.31,32 It remains incompletely understood whether described platelet alterations are maladaptive biology or evidence of adaptive functions in the setting of hypoperfusion and hemorrhage.33,34 Furthermore, evidence highlights that platelets play a crucial role in vascular biology beyond hemostasis.35 The activation of platelets and complex interplay of receptors and signaling molecules mediates thromboinflammation through the vWF axis,36 platelet shape and receptor conformation changes,37 secretion of granule contents,25,30 and exposure of procoagulant surfaces.28 Activated platelets express phosphatidylserine and P-selectin, which not only allow for coagulation factor binding and thrombin generation, but these procoagulant platelets signal through both the release of soluble contents and direct contact with leukocytes and endothelial cells, recruiting immune cells to local environments and inducing their release of cytokines and chemokines that promote inflammation.38,39
The release of platelet factor 4 is associated with proinflammatory states of increased expression of TF, levels of fibrinogen, and activated factor X.40 Platelet-derived molecules such as platelet factor 4 and β-thromboglobulin can stimulate immune cells, contributing to the recruitment and activation of neutrophils and monocytes, and NET formation. In addition, other proinflammatory platelet phenotypes have been identified after injury, including histone H4 decorated procoagulant ballooning, as well as platelet-derived soluble CD40 ligand and platelet-derived high mobility group box 1.41,42 Furthermore, platelets play a sentinel role in immune surveillance through their expression of toll-like receptors that recognize pathogen-associated molecular patterns and DAMPs, triggering inflammatory responses. In fact, the platelet signalosome includes the expression of all 10 toll-like receptors transcripts43 and a plethora of other immune receptors for complement, antibodies, and intracellular nucleotide-binding and oligomerization domain–like receptors.44
The interplay between activated platelets, coagulation factors, and immune cells creates a localized environment primed for thrombosis and inflammation, which can lead to microvascular dysfunction, tissue edema, and injury exacerbation. It is likely that the known alterations to platelet function after injury have downstream effects on the regulation of endothelial and immune functions and contribute to the microvascular thrombotic complications seen in injured patients who develop venous thromboembolism and multiple organ failure.42 In addition, in other diseases that are also characterized by ischemia-reperfusion injury, it has been identified that platelets carry a signaling role in the regulatory T cell suppression.45 Finally, platelets and endothelial cells are symbiotic in their joint regulation of hemostatic functions and inflammatory pathophysiology.46,47
ROLE OF ENDOTHELIUM IN THROMBOINFLAMMATION
Endothelial activation is a central event in thromboinflammation.48 The endothelium, a monolayer of cells lining blood vessels, is a critical interface between circulating blood and underlying tissues, serving as a nexus of both hemostasis and inflammation. Beyond its well-appreciated role as a dynamic barrier regulating vascular tone and permeability, the endothelium also plays a governing role in noncellular and cellular signaling across the vascular space. Under normal conditions, endothelial cells discourage clot formation through the interaction of heparan sulfate with circulating antithrombin, forming a functional layer of anticoagulant at the endothelial cell surface. In pathological conditions, such as injury, the endothelium undergoes phenotypic changes that promote thrombosis and inflammation, contributing to microvascular clot formation, tissue damage, and immune cell recruitment.
Injury-induced endothelial dysfunction, often termed endotheliopathy of trauma, has been increasingly appreciated for its role in TIC and associated complications.49–51 Dysfunctional endothelium involves the shedding of the proteoglycan layer and a denuded loss of barrier function but also promotes thrombus formation and allows immune cells to infiltrate tissues, intensifying the inflammatory response.52 Multiple described pathways may contribute to endotheliopathy of trauma in injury and shock, including via thrombin-thrombomodulin, protein C, and heparan sulfate biology.11,53–55 The microvascular thrombosis driven by endothelial activation and dysfunction can have profound consequences, including occlusion of the microvasculature, reducing tissue perfusion and leading to ischemia and hypoxia, as well as endothelial cell mechanical damage exacerbating vascular leakage and tissue edema. The end result of this pathophysiology is organ failure.56
Like the other mediators of thromboinflammation discussed previously, this biology does not happen in isolation, and activation and injury to the endothelium can affect the endothelial symbiosis with platelets and fibrinogen.57 This includes the role of platelets in angiogenesis and the release of soluble platelet-derived CD40, involved in endothelial and inflammatory signaling, and found to be associated with TIC and associated poor outcomes.58 In addition, the closely tied biology of vWF and its clearance by a disintegrin and metalloproteinase with thrombospondin type 1 motif, 13 (ADAMTS13), are under endothelial control, and it has been found that sustained release and impaired clearance of ultralarge vWF are associated with thromboinflammatory biology.46,59,60
The effects of malperfusion in vascular beds should not be ignored.61 Specifically, malperfusion triggers inflammation through the release of DAMPs, including extracellular nucleic acids and purinergic nucleotides, as well as high mobility group box 1 and cytokines. These alarmins are then recognized by surrounding cells in the vascular space, including the endothelial cells, immune cells, and platelets, leading to the convergence of prothrombotic proinflammatory phenotypic function of these cells.62 For the endothelial cells, this has been found to include endothelial plasticity in which endothelial cells lose endothelial characteristics and acquire mesenchymal characteristics.62 Furthermore, complement and endothelial cell activation are codependent, and the complement role in thromboinflammation is described further hereinafter.63
Tissue factor (factor III or CD142) is a cell surface glycoprotein constitutively expressed by cells not normally exposed to the circulatory system, including fibroblasts and subendothelial smooth muscle cells. Injury may lead to direct exposure of these tissues to the circulation, with binding of the serine protease factor VIIa with resultant conversion of factor X into factor Xa and subsequent conversion of prothrombin to thrombin.64 Tissue factor expression is inducible in monocytes, endothelial cells, platelets, and other organ systems through activation of proinflammatory transcription factors after traumatic injury.65 In this setting, increased TF expression leads to an overall prothrombotic state, with thrombosis at the site of endothelial cell TF expression. After injury, increased TF activation may lead to the exhaustion of proteins in the clotting cascade, leading to coagulopathy.
Activated protein C is a vitamin K–dependent serine protease that inactivates factor Va and factor VIIIa, thus serving as a natural anticoagulant. The zymogenic form of protein C is present in the circulation under baseline conditions and is activated by binding with thrombin as well as through interactions with thrombomodulin and the endothelial protein C receptors. Because of the location of these receptors in the endothelial cell membrane, activated protein C is typically found in proximity to the endothelium where it affects the blood/endothelial cell interface. Activated protein C also has a cytoprotective role, primarily through the endothelial protein C receptor and protease activated receptor 1. This effect is incompletely understood but involves promotion of the endothelial barrier, inhibition of proinflammatory mediators, and decreased expression of vascular adhesion molecules with reduction of leukocyte adhesion and chemotaxis.66 Severe injury is associated with significant early elevation of activated protein C levels, and this finding is associated with alterations in clotting factor activity, with the net result of increased coagulopathy and decreased clot strength as well as increased fibrinolysis and fibrinogen depletion. There is debate regarding whether this is a primary driving factor for the development of TIC.67,68 Recent studies also suggest that cell free histones, which are markedly elevated after injury and the ensuing inflammatory response, lead to inhibition of protein C activation.69 Previous clinical studies led to the treatment of sepsis with activated protein C. Although the potential clinical harm of this therapy was subsequently demonstrated to outweigh any potential benefits,70 recent experimental data suggest that therapeutic manipulation of the protein C could have a future role in mitigation of TIC.71
THE ROLE OF COMPLEMENT IN THROMBOINFLAMMATION
As a major component of the innate immune response and an ancient defense mechanism, the complement system is consisted of a series of soluble and membrane-bound proteins that play a crucial role in immune surveillance and inflammation. The complement system has gained increasing recognition for its crosstalk with the coagulation system and its role in thromboinflammation. There are multiple pathways by which the complement system can interact with coagulation pathways. Complement activation can occur through classical, lectin, or alternative pathways. Activation generates anaphylatoxins (C3a, C5a), opsonins (C3b, iC3b), and the membrane attack complex (C5b-9), which influence various cellular responses.72 Anaphylatoxins can stimulate platelet activation and adhesion, promoting clot formation.73 Furthermore, lectin-pathway protease activity is triggered by platelet activation and fibrin generation, and reciprocally, these complement pathway proteases can catalyze coagulation proteins, including prothrombin, fibrinogen, factor XIII, and thrombin-activatable fibrinolysis inhibitor.74 Impressively, this is a process by which fibrin can form in the absence of thrombin.75 In addition, opsonized surfaces may enhance coagulation factor assembly and thrombin generation. Complement activation promotes inflammation by recruiting immune cells, enhancing vascular permeability, decreasing nitric oxide production, and releasing proinflammatory cytokines.76 Complement activation fragments, particularly C3a and C5a, interact with immune cells through their respective receptors, modulating leukocyte chemotaxis and activation.77 Furthermore, the membrane attack complex can directly damage cells, releasing DAMPs that fuel the inflammatory response. Complement activation results in the shedding of complement-coated endothelial microvesicles and the assembly of C5b-9, which can ultimately lead to cell lysis, and the procoagulant TF activity on the endothelium is dependent on the assembly of C5b-9.63,78
Dysregulation of complement activation has been implicated in various pathological conditions characterized by coagulopathies and inflammatory pathologies, including injury. Injury triggers a cascade of events that can activate the complement system.79 The described mechanisms are canonical and noncanonical and are still the topic of exploratory research and debate. There are data to support that there can be both local and systemic activation of complement,80 including activation of the classical pathway with alternative pathway amplification, increased ratios of C3a:C3b-9,81 C4a thrombin binding,82 changes to neutrophil complement regulatory proteins and associated function,83 and the activation of complement due to coagulation proteases.84 Similar to other coagulation and inflammation biology in severe injury and shock, hypotheses and studies suggest that exhaustion of complement function in the setting of excessive activation may put patients at risk of poor outcomes.80 Finally, like the other bidirectional biologies of thromboinflammation, various components of the complement system can also be activated by aspects of coagulation biology, including platelets, leukocytes, and endothelial cells, exacerbating the thromboinflammatory response. For example, P-selectin, expressed on activated platelets, has been identified as a receptor for C3b capable of initiating complement activation.85 This interconnected network of interactions contributes to microvascular thrombosis and tissue damage and underscores the complement system’s multifunctional role beyond immune defense in multiple diseases, including injury.86
ROLE OF EXTRACELLULAR VESICLES IN MEDIATING SIRS
Extracellular vesicles (EVs) are bound by lipid bilayers and released from many cell types during cellular stress.87 Extracellular vesicles include exosomes, microvesicles, and apoptotic bodies. Originally thought to be inert, it is now recognized that EVs contain proteins, lipids, metabolites, and nucleic acids from the parent cell and serve a role in cell-cell communication and inflammatory signaling.88–90 The systemic inflammatory response after injury or sepsis appears to cause EV release into the circulation,91 although the cellular source and downstream effects have yet to be defined.92,93 Activated endothelial cells also release EVs in response to various stimuli, including injury.94 These EVs carry bioactive molecules such as TF, microribonucleic acids, and cytokines, which can contribute to coagulation and inflammation by further enhancing platelet activation, promoting leukocyte adhesion, and modulating immune cell function.95 The net effect of inflammation and trauma-induced EV release on the coagulation status of the patient and their potential as a therapeutic to treat TIC is not fully understood and remains an area of ongoing investigation.
CLINICAL AND THERAPEUTIC IMPLICATIONS
Bleeding/Resuscitation and Effects on SIRS
Clinical and preclinical studies demonstrate that hemorrhage leads to SIRS and that specific patterns of increased inflammatory markers after injury and hemorrhage are associated with decreased survival.96 The severity of the inflammatory response after hemorrhage is likely due to a profound ischemic injury during hemorrhage, followed by reperfusion injury during the resuscitation phase, leading to widespread activation of the immune system, endothelial cells, and end organs (Supplemental Digital Content, Supplementary Data 1, The result is an activation of multiple inflammatory cascades, dramatically amplifying and producing a wide range of inflammatory mediators.97 Continued resuscitation may lead to recovery, whereas the continued inflammatory process may lead to multiple organ failure and progression to persistent inflammation, immunosuppression, and catabolism syndrome, with resultant prolonged chronic critical illness.98
After injury and hemorrhage, resuscitation seeks to replete lost intravascular volume and correct anemia, acidosis, and coagulopathy. The resuscitation strategy may have a profound effect on SIRS. Clinical studies demonstrate that increased use of crystalloids for resuscitation after trauma is associated with increased occurrence of hypoxemia and acute respiratory distress syndrome, suggesting that this strategy exacerbates the inflammatory response.99 The effect of blood-based resuscitation strategies on the inflammatory response after hemorrhage and resuscitation has recently been the focus of investigation. Preclinical studies indicate that, compared with Ringer’s lactate, whole blood resuscitation decreases the proinflammatory response and acute lung injury.100 Importantly, studies in injured patients have supported that administering healthy plasma to injured patients may restore endothelial barrier function and reduce shed glycocalyx such as syndecan-1, presumed to be a marker of endothelial activation in injury.56,101–103
The importance of the age of stored packed red blood cells is also a controversial topic. While the transfusion of a relatively small volume of packed red blood cells does not appear to affect patient outcomes, clinical studies demonstrate increased complications and mortality in association with the use of packed red blood cells with longer periods of storage, massive transfusion, and resuscitation after injury and hemorrhage.104 Resuscitation from hemorrhage with higher storage age red blood cells is associated with neutrophil activation and lung injury in murine models.105 Additional studies suggest that increased systemic inflammation, endothelial cell activation, and microthrombi formation also occur.106,107 Efforts to blunt the effect of storage age on the inflammatory response after transfusion are ongoing and seek to maximize the quality of blood products used for resuscitation.108
Therapeutics Targeting Thromboinflammation
Science in cardiovascular diseases outside of injury, accelerated by the thromboinflammatory biology of COVID-19, has created a surge of considerations in discovering and testing therapeutics targeting the coagulation-inflammation axis. Strategies aimed at modulating fibrinogen interactions with immune cells or disrupting fibrin deposition could attenuate inflammation in thromboinflammatory conditions. Anticoagulant therapies, such as heparins, have been considered for targeting hypercoagulability and modulating inflammation in thromboinflammatory diseases.109 Antiplatelet agents, such as P2Y12 inhibitors, are commonly used to prevent thrombotic events but may also impact inflammation through effects on cytokine release, platelet-leukocyte interactions, and NET formation, suggesting potential utility in modulating thromboinflammatory responses.110 Targeting platelet roles in immunity and inflammation through selectin ligand biologic antagonisms and beyond has been explored in thrombosis and inflammatory disease.111,112 Overall, targeting platelet-leukocyte interactions and platelet-derived inflammatory mediators could provide novel strategies for managing SIRS and trauma-related complications.
Similarly, understanding the role of the endothelium in thromboinflammation presents opportunities for therapeutic interventions.113 Agents that target various receptors and signaling pathways involved in endothelial activation, including anti-inflammatory drugs, adhesion molecule inhibitors, antioxidants, and endothelial protective agents, could ameliorate endothelial dysfunction in the context of injury.114 Furthermore, recent progress in mechanistic investigations of the antifibrinolytic tranexamic acid (TXA) has uncovered evidence of mitigation of endotheliopathy through various mechanisms, including enhancement of oxidative phosphorylation, inhibition of serine proteases, suppression of DAMP mitochondrial DNA, and stimulation of mitochondrial respiration.50,51 It is possible that the mortality benefit of TXA in injured patients is related to mitigation of endotheliopathy and downstream endothelial-related thromboinflammation.
Given that hyperactive complement has been associated with a detrimental role in thromboinflammation’ complement inhibitors, such as monoclonal antibodies targeting C5 or its cleavage product C5a, have shown promise in ameliorating thromboinflammatory complications.86 In addition, it has been found that stored plasma is rich in multiple complement factors, including C3 and C5, and that stored platelet products contain anaphylatoxins as well.115 However, the heterogeneity of banked products and associated storage lesions continues to plague the ability to personalize the treatment of coagulation and inflammation with blood product infusion. In addition, in the setting of injury therapeutics, TXA has been shown to have effects on the complement system as well, thought to be through lysine analog direct inhibition of complement activation or indirect inhibition via its effects on plasmin and plasmin cleavage of C3 or C5.116 There remains interest in the potential for hemoadsorbance efforts to clear complement activation,117 as well as efforts in targeting specific complement types, including C1,118 C3,119 and C5.120
CONCLUSION
An overall comprehensive understanding of thromboinflammation in the pathophysiology of trauma is required to consider clinical approaches to resuscitation and management of the injured patient (Table 2). Targeting the many interrelated components of thromboinflammation holds promise for developing therapeutic interventions that improve outcomes for injured patients (Table 3). Efforts to mitigate the acute and sustained changes in coagulation and inflammation and its complex interactions following injury could provide novel strategies for managing complications of TIC, SIRS, and multiple organ failure.
TABLE 2.
The Systemic Inflammatory Response to Injury: What You Need to Know
Key Points:
• The host response to injury results in a widespread and interrelated SIRS.
• SIRS includes activation of innate and adaptive immune cells, cellular and noncellular coagulation cascades, and complement systems.
• There is a simultaneous increase in genes related to innate immunity and suppression of genes related to adaptive immunity within 12 hours of injury.
• Inflammatory mediators activate coagulation by promoting the expression of TF on endothelial cells and monocytes and alters the normal ratio of procoagulation and anticoagulation effects.
• SIRS leads to endothelial cell, platelet, and leukocyte activation that result in appropriate and inappropriate thrombus formation, thrombolysis, and exhaustion of clotting cycle components.
• Fibrinogen and fibrin play multifaceted roles influencing both thrombosis and inflammation, with effects that extend beyond serving as a passive substrate for clot formation by activating immune cells and promoting leukocyte adhesion and migration.
• Early increases in activated protein C levels after injury increase coagulopathy, decrease clot strength, and increase fibrinolysis.
• Endothelial dysfunction after injury is associated with shedding of the proteoglycan layer and loss of barrier function and promotes thrombus formation.
• Activation of the complement system promotes inflammation by recruiting immune cells and enhancing vascular permeability and interacts with the coagulation system to contribute to coagulopathy.
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TABLE 3.
Fifteen Highlights From 15 Years of Mechanistic Human Studies in the Intersection of Coagulation and Inflammation in Injury
Xiao W, Mindrinos MN, Seok J, Cuschieri J, Cuenca AG, Gao H, et al. A genomic storm in critically injured humans. J Exp Med. 2011;208:2581–90.
Cohen MJ, Call M, Nelson M, Calfee CS, Esmon CT, Brohi K, et al. Critical role of activated protein C in early coagulopathy and later organ failure, infection and death in trauma patients. Ann Surg. 2012;255:379–85.
Johansson PI, Sorensen AM, Perner A, Welling KL, Wanscher M, Larsen CF, et al. High sCD40L levels early after trauma are associated with enhanced shock, sympathoadrenal activation, tissue and endothelial damage, coagulopathy and mortality. J Thromb Haemost. 2012;10:207–16.
Burk AM, Martin M, Flierl MA, Rittirsch D, Helm M, Lampl L, et al. Early complementopathy after multiple injuries in humans. Shock. 2012;37:348–54.
Moore HB, Moore EE, Chapman MP, McVaney K, Bryskiewicz G, Blechar R, et al. Plasma-first resuscitation to treat haemorrhagic shock during emergency ground transportation in an urban area: a randomized trial. Lancet. 2018;392:283–91.
Vulliamy P, Gillespie S, Armstrong PC, Allan HE, Warner TD, Brohi K. Histone H4 induces platelet ballooning and microparticle release during trauma hemorrhage. Proc Natl Acad Sci USA. 2019 Aug 27;116(35):17444–17449.
Barrett CD, Vigneshwar N, Moore HB, Ghasabyan A, Chandler J, Moore EE, et al. Tranexamic acid is associated with reduced complement activation in trauma patients with hemorrhagic shock and hyperfibrinolysis on thromboelastography. Blood Coagul Fibrinolysis. 2020;31:578–82.
Dyer MR, Plautz WE, Ragni MV, Alexander W, Haldeman S, Sperry JL, et al. Traumatic injury results in prolonged circulation of ultralarge von Willebrand factor and a reduction in ADAMTS13 activity. Transfusion. 2020;60:1308–18.
Goswami J, MacArthur T, Bailey K, Spears G, Kozar RA, Auton M, et al. Neutrophil extracellular trap formation and Syndecan-1 shedding are increased after trauma. Shock. 2021;56:433–9.
Janicova A, Becker N, Xu B, Simic M, Noack L, Wagner N, et al. Severe traumatic injury induces phenotypic and functional changes of neutrophils and monocytes. J Clin Med. 2021;10.
Schaid TR, Jr., Hansen KC, Sauaia A, Moore EE, DeBot M, Cralley AL, et al. Postinjury complement C4 activation is associated with adverse outcomes and is potentially influenced by plasma resuscitation. J Trauma Acute Care Surg. 2022;93:588–96.
Knudson MM, Moore EE, Kornblith LZ, Shui AM, Brakenridge S, Bruns BR, et al. Challenging traditional paradigms in posttraumatic pulmonary thromboembolism. JAMA Surg. 2022 Feb 1;157(2):e216356.
Schaid TR, Jr., LaCroix I, Hansen KC, D’Alessandro A, Moore EE, Sauaia A, et al. A proteomic analysis of NETosis in trauma: emergence of serpinB1 as a key player. J Trauma Acute Care Surg. 2023;94:361–70.
Shimono K, Ito T, Kamikokuryo C, Niiyama S, Yamada S, Onishi H, et al. Damage-associated molecular patterns and fibrinolysis perturbation are associated with lethal outcomes in traumatic injury. Thromb J. 2023;21:91.
Bonaroti J, Billiar I, Moheimani H, Wu J, Namas R, Li S, et al. Plasma proteomics reveals early, broad release of chemokine, cytokine, TNF, and interferon mediators following trauma with delayed increases in a subset of chemokines and cytokines in patients that remain critically ill. Front Immunol. 2022 Nov 30;13:1038086.
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Supplementary Material
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NIHMS1980448-supplement-supp_1.docx (17.2KB, docx)
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NIHMS1980448-supplement-supp_2.pdf (2.6MB, pdf)
Footnotes
DISCLOSURE
Conflict of Interest: Author Disclosure forms have been supplied and are provided as Supplemental Digital Content (
Supplemental digital content is available for this article. Direct URL citations appear in the printed text, and links to the digital files are provided in the HTML text of this article on the journal’s Web site (www.jtrauma.com/).
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693 | https://libguides.library.kent.edu/SPSS/FrequenciesCategorical | Skip to Main Content
University Libraries
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Statistical Consulting
SPSS Tutorials
Frequency Tables
SPSS Tutorials: Frequency Tables
In SPSS, the Frequencies procedure is primarily used to create frequency tables, bar charts, and pie charts for categorical variables.
Home
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Descriptive Stats for One Numeric Variable (Explore)
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How to Cite the Tutorials
Sample Data Files
Our tutorials reference a dataset called "sample" in many examples. If you'd like to download the sample dataset to work through the examples, choose one of the files below:
Data definitions (.pdf)
Data - Comma delimited (.csv)
Data - Tab delimited (.txt)
Data - Excel format (.xlsx)
Data - SAS format (.sas7bdat)
Data - SPSS format (.sav)
Data - Stata (13+) format (.dta)
SPSS Syntax (.sps)
Syntax to add variable labels, value labels, set variable types, and compute several recoded variables used in later tutorials.
SAS Syntax (.sas)
Syntax to read the CSV-format sample data and set variable labels and formats/value labels.
Stata Syntax (.do)
Introduction
When summarizing a categorical or qualitative variable (nominal/ordinal), we are typically interested in questions like:
How many unique categories were there?
How many observations fell into each category? (Counts/frequencies)
Were there any categories with zero observations?
Were there any observations with missing responses?
What proportion of the observations fell into each category?
What proportion of the non-missing responses fell into each category?
What proportion of all responses (missing and non-missing) fell into each category?
What is the most frequently occurring category? (Mode)
All of these questions can be answered using a frequency table. The most basic type of frequency table contains at least the following:
One row per category of the variable, plus a row for the sum total
A column showing the number of responses in that category
A column showing the proportion of the total observations in that category
In SPSS Statistics, the Frequencies procedure can produce frequency tables, which contain tallies and proportions, as well as two types of graphs appropriate for categorical data: bar charts and pie charts.
Data Requirements
Your data must contain at least one categorical variable that meets all of the following requirements:
The variable must have at least two or more categories (groups), or must be a discrete numeric variable.
The categories may be unordered (nominal) or ordered (ordinal).
Each case can be classified into exactly one of the response categories; i.e., a case cannot belong to more than one of the response categories.
If a case can belong to more than one of the response categories -- for example, responses to a check-all-that-apply survey question -- you should use a Multiple Response Frequency Table instead.
Data Set-Up
Each row can represent one subject, or can represent an observation from a subject. Each column should represent one variable. Variables that will be tabulated using frequency tables should ideally have the following variable properties defined:
Variable Type: The categorical variables in your SPSS dataset can be numeric or string. By default, the rows of the table are arranged in ascending order (for numeric codes) or alphabetically (for string variables).
Ordinal variables may be represented using numbers.
Discrete numeric variables should ideally have a limited number of unique values.
Value Labels: If you have entered data using numeric codes that represent specific named categories (especially nominal/unordered categories), you should apply value labels to your variables. This can affect the display of the table.
Missing Value Handling: The frequency table will include sections for Valid (non-missing) and Missing responses. Any values recognized as system-missing or user-missing will appear in the Missing section. If you have more than one user-defined missing value code that appears in the data, those codes are tallied separately in the Missing section of the table. (For example, if you have defined the number code -99 as "Refused response" and -88 to represent "Not asked", you will be able to see how many "Refused" and "Not asked" values there were.)
For numeric variables: Blank values will appear section of the table labeled "Missing". If your dataset also used special number codes to represent missing values (e.g. using ), there will be multiple rows in the "Missing" section of the table.
For string variables: Blank strings will NOT automatically be recognized as missing values. If present, they will appear as one of the categories in the "Valid" section; see example below. This can be resolved by using the Automatic Recode procedure to convert the original string variable to a coded numeric variable prior to creating the frequency table.
Variable Measurement Levels: The variables' measurement levels should be defined as nominal or ordinal. The Frequencies procedure will still work on variables whose measurement level is set to scale; however, frequency tables should only be used when there are a limited number of response categories.
Create a Frequency Table in SPSS
In SPSS, the Frequencies procedure can produce summary measures for categorical variables in the form of frequency tables, bar charts, or pie charts.
To run the Frequencies procedure, click Analyze > Descriptive Statistics > Frequencies.
A Variable(s): The variables to produce Frequencies output for. To include a variable for analysis, double-click on its name to move it to the Variables box. Moving several variables to this box will create several frequency tables at once.
B Statistics: Opens the Frequencies: Statistics window, which contains various descriptive statistics.
The vast majority of the descriptive statistics available in the Frequencies: Statistics window are never appropriate for nominal variables, and are rarely appropriate for ordinal variables in most situations. There are two exceptions to this:
The Mode (which is the most frequent response) has a clear interpretation when applied to most nominal and ordinal categorical variables.
The Values are group midpoints option can be applied to certain ordinal variables that have been coded in such a way that their value takes on the midpoint of a range. For example, this would be the case if you had measured subjects' ages and had coded anyone between the ages of 20 and 29 as 25, or between the 30 and 39 as 35 (source: IBM SPSS Statistics Information Center).
If your categorical variables are coded numerically, it is very easy to mis-use measures like the mean and standard deviation. SPSS will compute those statistics if they are requested, regardless of whether or not they are meaningful. It is up to the researcher to determine if these measures are appropriate for their data. In general, you should never use any of these statistics for dichotomous variables or nominal variables, and should only use these statistics with caution for ordinal variables.
C Charts: Opens the Frequencies: Charts window, which contains various graphical options. Options include bar charts, pie charts, and histograms. For categorical variables, bar charts and pie charts are appropriate. Histograms should only be used for continuous variables; they should not be used for ordinal variables, and should never be used with nominal variables.
Bar chart displays the categories on the graph's x-axis, and either the frequencies or the percentages on the y-axis
Pie chart depicts the categories of a variable as "slices" of a circular "pie".
Note that the options in the Chart Values area apply only to bar charts and pie charts. In particular, these options affect whether the labeling for the pie slices or the y-axis of the bar chart uses counts or percentages. This setting will greyed out if Histograms is selected.
D Format: Opens the Frequencies: Format window, which contains options for how to sort and organize the table output.
The Order by options affect only categorical variables:
Ascending values arranges the rows of the frequency table in increasing order with respect to the category values: (alphabetically if string, or by numeric code if numeric)
Descending values arranges the rows of the frequency table in decreasing order with respect to the category values.
Note: If your categorical variable is coded numerically as 0, 1, 2, ..., sorting by ascending or descending value will arrange the rows with respect to the numeric code, not with respect to any assigned labels.)
Ascending counts orders the rows of the frequency table from least frequent (lowest count) to most frequent (highest count).
Descending counts orders the rows of the frequency table from most frequent (highest count) to least frequent (lowest count).
When working with two or more categorical variables, the Multiple Variables options only affects the order of the output. If Compare variables is selected, then the frequency tables for all of the variables will appear first, and all of the graphs for the variables will appear after. If Organize output by variables is selected, then the frequency table and graph for the first variable will appear together; then the frequency table and graph for the second variable will appear together; etc.
E Display frequency tables: When checked, frequency tables will be printed. (This box is checked by default.) If this check box is not checked, no frequency tables will be produced, and the only output will come from supplementary options from Statistics or Charts. For categorical variables, you will usually want to leave this box checked.
Example: Summarizing a Categorical Variable
Using the sample dataset, let's a create a frequency table and a corresponding bar chart for the class rank (variable Rank), and let's also request the Mode statistic for this variable.
Running the Procedure
Using the Frequencies Dialog Window
Open the Frequencies window (Analyze > Descriptive Statistics > Frequencies) and double-click on variable Rank.
To request the mode statistic, click Statistics. Check the box next to Mode, then click Continue.
To turn on the bar chart option, click Charts. Select the radio button for Bar Charts. Then click Continue.
When finished, click OK.
Using Syntax
FREQUENCIES VARIABLES=Rank
/STATISTICS=MODE
/BARCHART FREQ
/ORDER=ANALYSIS.
Output
Two tables appear in the output: Statistics, which reports the number of missing and nonmissing observations in the dataset, plus any requested statistics; and the frequency table for variable Rank. The table title for the frequency table is determined by the variable's label (or the variable name, if a label is not assigned).
Here, the Statistics table shows that there are 406 valid and 29 missing values. It also shows the Mode statistic: here, the mode value is "1", which is the numeric code for the category Freshman. Notice that the Mode statistic isn't displaying the value labels, even though they have been assigned. (For this reason, we recommend not requesting the mode statistic; instead, determine the mode from the frequency table.)
Notice how the rows are grouped into "Valid" and "Missing" sections. This grouping allows for easy comparison of missing versus nonmissing observations. Note that "System" missing responses are observations that use SPSS's default symbol -- a period (.) -- for indicating missing values. If a user has assigned special codes for missing values in the Variable View window, those codes would appear here.
The frequency table contains four columns of summary measures:
The Frequency column indicates how many observations fell into the given category.
The sample contained a total of 435 students. Of those students, 29 did not specify their class rank.
The Percent column indicates the percentage of observations in that category out of all observations (both missing and nonmissing). You can verify the proportions for each group by dividing its count in the "frequency" column by the value in the last row of the table (435):
Freshman: 147/435 = 33.8%
Sophomore: 96/435 = 22.1%
Junior: 98/435 = 22.5%
Senior: 65/435 = 14.9%
Valid Total: 406/435 = 93.3%
Missing: 29/435 = 6.7%
The Valid Percent column displays the percentage of observations in that category out of the total number of nonmissing responses. You can verify the proportions for each group by dividing its count in the "frequency" column by the value of "Total" that appears after the last valid category (406):
Freshman: 147/406 = 36.2%
Sophomore: 96/406 = 23.6%
Junior: 98/406 = 24.1%
Senior: 65/406 = 16.0%
The Cumulative Percent column is the total percentage of the sample that has been accounted for up to that row; it can be computed by adding all of the numbers in the Valid Percent column above the current row:
Freshman: 36.2% (there are no rows before this one, so the first cumulative percent is identical to the first valid percent)
Sophomore: 36.2 + 23.6 = 59.8%
Junior: 36.2 + 23.6 + 24.1 = 83.9%
Senior: 36.2 + 23.6 + 24.1 + 16.0 = 100%
The bar chart appears after the tables.
Here, we can see that:
Freshmen comprised the largest group
There were approximately equal numbers of sophomores and juniors
Seniors were the smallest group
What if my frequency table has a blank row in it?
What should I do if I create a frequency table in SPSS and one of the rows is blank?
If you are creating a frequency table and notice that the first row has a blank category label, similar to this example:
This issue should not be ignored! This particular issue affects frequency tables created from string variables that use blanks to denote missing values. SPSS does not automatically recognize blank (i.e., empty) strings as missing values, so the blank values appear as one of the "Valid" (i.e., non-missing) categories. This affects the calculation of the Valid Percent columns.
When missing values are treated as valid values, it causes the "Valid Percent" columns to be calculated incorrectly. If the blank values were correctly treated as missing values, the valid, non-missing sample size for this table would be 314 + 94 = 408 -- not 435! -- and the valid percent values would change to 314/408 = 76.9% and 94/408 = 23.0%. Depending on the number of missing values in your sample, the differences could be even more dramatic.
To fix this problem: To get SPSS to recognize blank strings as missing values, you'll need to run the variable through the Automatic Recode procedure. This procedure takes a string variable and converts it to a new, coded numeric variable with value labels attached. During this process, blank string values are recoded to a special missing value code. To see a worked example, see the Automatic Recode tutorial.
Why are some categories missing from my frequency table and how do I get them to display?
In the sample dataset, variable HowCommute has the following value labels defined: 1=Walk, 2=Bike, 3=Car, 4=Public Transit, and 5=Other. However, if you use the Frequencies procedure to create a frequency table for this variable, you will only see four of the five categories:
Why is one of the categories missing despite it being defined as a value label?
The Frequencies procedure is designed to drop unobserved categories from the frequency table: that is, it will not include categories with counts of 0. Although this can be desirable in some cases, it may be actively problematic or misleading in others. For example, if you create a frequency table of a 5-point Likert item or multiple choice question, readers may interpret the omission of categories as the categories not being included in the design of the survey -- which is very different than the categories being present on the survey but not selected by any respondents.
If you wish to create a frequency table that will include all categories with a defined value label even if they have counts of 0, you must use the Custom Tables procedure.
The Custom Tables procedure is included with SPSS Statistics Standard and SPSS Statistics Premium, but is not included in SPSS Statistics Base. If you do not see the Custom Tables procedure in the Analyze menu (Analyze > Tables > Custom Tables), it is possible your license did not include the Custom Tables module.
You can check which modules are available to you by opening a new syntax window (File > New > Syntax) and executing the following command:
SHOW LIC.
If the resulting output does not include "IBM SPSS Statistics Custom Tables", then you will not have access to the procedure.
Running the Procedure
Using the Custom Tables Dialog Window
Open the Custom Tables dialog window (Analyze > Tables > Custom Tables).
Click and drag the HowCommute variable from the list of variables and drop it onto the "Row" box.
(Optional) To add a Total row showing the sum of the counts:
Click the Categories and Totals button.
Check the Totals box.
Click Apply.
(Optional) If you have defined special missing value codes for the variable and want them to appear in the table:
Click the Categories and Totals button.
Check the Missing Values box.
Click Apply.
(Optional) To add percentages to the table:
Click the Summary Statistics button.
In the Statistics box, expand the Table Percent category. Then choose one or more of the following options by moving them to the Display box:
Table N %: The denominator of the calculation uses the sum total of the counts in the created table.
Table Valid N %: The denominator of the calculation uses the sum total of the counts for nonmissing categories in the created table.
Table Total N %: The denominator of the calculation uses the number of cases in the dataset overall.
Click Apply to Selection to apply your changes to the selected variable, then click Close to return to the main dialog window.
When finished, click OK.
Using Syntax
Default: Counts only (No total row, No percentages)
CTABLES
/VLABELS VARIABLES=HowCommute DISPLAY=LABEL
/TABLE HowCommute [COUNT F40.0]
/CATEGORIES VARIABLES=HowCommute ORDER=A KEY=VALUE EMPTY=INCLUDE
/CRITERIA CILEVEL=95.
Counts with total row (No percentages)
CTABLES
/VLABELS VARIABLES=HowCommute DISPLAY=LABEL
/TABLE HowCommute [COUNT F40.0]
/CATEGORIES VARIABLES=HowCommute ORDER=A KEY=VALUE EMPTY=INCLUDE TOTAL=YES POSITION=AFTER
/CRITERIA CILEVEL=95.
Counts and percentages (No total row)
CTABLES
/VLABELS VARIABLES=HowCommute DISPLAY=LABEL
/TABLE HowCommute [COUNT F40.0, TABLEPCT.COUNT PCT40.1, TABLEPCT.VALIDN PCT40.1, TABLEPCT.TOTALN
PCT40.1]
/CATEGORIES VARIABLES=HowCommute ORDER=A KEY=VALUE EMPTY=INCLUDE
/CRITERIA CILEVEL=95.
Counts, percentages, and total row
CTABLES
/VLABELS VARIABLES=HowCommute DISPLAY=LABEL
/TABLE HowCommute [COUNT F40.0, TABLEPCT.COUNT PCT40.1, TABLEPCT.VALIDN PCT40.1, TABLEPCT.TOTALN
PCT40.1]
/CATEGORIES VARIABLES=HowCommute ORDER=A KEY=VALUE EMPTY=INCLUDE TOTAL=YES POSITION=AFTER
/CRITERIA CILEVEL=95.
Output
Default: Counts only (No total row, No percentages)
The default output of Custom Tables includes only the counts. It does not include a total row, nor the number of missing values, nor percentages. However, we see that all 5 categories are present, including the Other category, which is shown with a count of 0:
Counts with total row (No percentages)
This table is identical to the previous table, but with the addition of a Total row that includes the sum of the counts in the table. We can verify the total shown in the table by performing the sum manually: 26+15+193+13+0=247.
Counts with percentages and total row
This table is identical to the previous table but with the addition of extra columns containing the percentages we requested.
In this example, Table N% and Table Valid N% are identical, but this will not always be the case. If your variable includes user-missing values and you have enabled the Missing option, then Table N% will be different than Table Valid N%. (In general, "Table Valid N%" in Custom Tables has the same meaning as the "Valid Percent" column of the Frequencies output: it is the proportion based on the number of valid, nonmissing cases.)
The Table Total N% values are based on the total number of cases in the dataset (valid + missing). Recall that the sample dataset has 435 rows, and we know from the Frequencies procedure that variable HowCommute has 247 valid/observed values and 188 missing values (247 valid +188 missing = 435 total). We can verify that the Table Total N% values are based on the number of rows by performing the divisions ourselves and rounding to one decimal place:
Walk: 26/435 = 6.0%
Bike: 15/435 = 3.4%
Car: 193/435 = 44.4%
Public Transit: 13/435 = 3.0%
Other: 0/435 = 0.0%
<< Previous: Descriptive Stats for Numeric Variables by Group (Compare Means)
Next: Crosstabs >> |
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Complex Ion Formation
General Chemistry
Complex ions are formed when a central metal cation is bonded to surrounding molecules or anions called ligands, creating charged particles with increased stability and solubility. These ligands result in more electrostatic interactions with surrounding water molecules, making complex ions easier to separate and dissolve compared to typical ionic compounds. In order for complex ions to form, the original ionic compound must first dissociate in a solvent, separating into charged ions that can then bond with nearby ions.
The solubility of complex ions is determined by the formation constant (Kf), which is larger than the solubility product constant (Ksp) because complex ions are more soluble than their original ionic compounds. The dissolution of the non-complex ionic compound is the rate-limiting step in the formation of a complex ion, as it takes longer than the formation of the complex ion itself.
Lesson Outline
Introduction to complex ions and their properties
Complex ions consist of a central metal cation bonded to surrounding ligands
Stability of complex ions
More electrostatic interactions with surrounding water molecules
More stable than regular ions
Solubility of complex ions
Easier to separate from each other than typical ionic compounds
Increased solubility due to more electrostatic interactions
Formation of complex ions
Original ionic compound dissociates into charged ions in water
Separated ions can form complex ions with other nearby ions
Dissociation must happen before formation of complex ions
Formation constant (Kf) compared to solubility product constant (Ksp)
K values: Larger Kf and Ksp values indicate more stable complex ions and more soluble ionic compounds, respectively
Kf is typically larger than Ksp because complex ions are more soluble
Rate-limiting step and dissolution of non-complex ionic compounds
Dissolution of non-complex ionic compounds is a rate-limiting step in complex ion formation
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FAQs
What is complex ion formation and how does it affect the solubility of ionic compounds?
Complex ion formation is a process in which a central metal ion forms coordinate covalent bonds with surrounding molecules or anions called ligands. This results in the formation of complex ions. The solubility of an ionic compound is affected by complex ion formation, as the formation of these complex ions increases solubility by stabilizing the charged species and minimizing electrostatic interactions between the ions, thereby encouraging the dissociation of the compound into individual ions.
How do electrostatic interactions influence complex ion formation?
Electrostatic interactions play a significant role in complex ion formation. Since complex ions involve a central metal ion surrounded by ligands, these ligands are attracted to the metal ion due to the electrostatic attraction between the positively charged metal ion and the negatively charged or polar ligands. The formation and stability of complex ions relies on these electrostatic interactions, which help in maintaining the overall structure and enabling the complex to minimize repulsion between the constituent ions.
What is the difference between the formation constant (Kf) and solubility product constant (Ksp) in the context of complex ions?
The formation constant (Kf) and solubility product constant (Ksp) are equilibrium constants that help in understanding the stability and solubility of complex ions and ionic compounds, respectively. Kf represents the equilibrium constant for the formation of a complex ion from its constituent metal ion and ligands. A large Kf value indicates a stable complex ion, while a smaller value suggests a less stable complex formation. On the other hand, Ksp represents the equilibrium constant for the dissociation of an ionic compound into its individual ions in a solution. A high Ksp value indicates high solubility of the ionic compound, and a low Ksp value implies low solubility.
How does increased solubility aid in the formation and stabilization of complex ions?
Increased solubility contributes to the formation and stabilization of complex ions by promoting the dissociation of ionic compounds into individual ions in a solution. When an ionic compound dissociates, the metal ions and ligands become available for complex formation. The formation of complex ions further reduces electrostatic interactions via the coordinate covalent bonds between the metal ions and ligands, thereby favoring increased solubility. This stablilizes the complex ion within the solution, as the probability of the ions recombining to form the original ionic compound is reduced due to the energetic stability offered by the complex ion.
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7.3 - Least Squares: The Theory
Now that we have the idea of least squares behind us, let's make the method more practical by finding a formula for the intercept and slope . We learned that in order to find the least squares regression line, we need to minimize the sum of the squared prediction errors, that is:
We just need to replace that with the formula for the equation of a line:
to get:
We could go ahead and minimize as such, but our textbook authors have opted to use a different form of the equation for a line, namely:
Each form of the equation for a line has its advantages and disadvantages. Statistical software, such as Minitab, will typically calculate the least squares regression line using the form:
Clearly a plus if you can get some computer to do the dirty work for you. A (minor) disadvantage of using this form of the equation, though, is that the intercept is the predicted value of the response when the predictor , which is typically not very meaningful. For example, if is a student's height (in inches) and is a student's weight (in pounds), then the intercept is the predicted weight of a student who is 0 inches tall..... errrr.... you get the idea. On the other hand, if we use the equation:
then the intercept is the predicted value of the response when the predictor , that is, the average of the values. For example, if is a student's height (in inches) and is a student's weight (in pounds), then the intercept is the predicted weight of a student who is average in height. Much better, much more meaningful! The good news is that it is easy enough to get statistical software, such as Minitab, to calculate the least squares regression line in this form as well.
Okay, with that aside behind us, time to get to the punchline.
Least Squares Estimates Section
Theorem
The least squares regression line is:
with least squares estimates:
and
Proof
In order to derive the formulas for the intercept and slope , we need to minimize:
Time to put on your calculus cap, as minimizing involves taking the derivative of with respect to and , setting to 0, and then solving for and . Let's do that. Starting with the derivative of with respect to , we get:
Now knowing that is , the average of the responses, let's replace with in the formula for :
and take the derivative of with respect to . Doing so, we get:
As was to be proved.
By the way, you might want to note that the only assumption relied on for the above calculations is that the relationship between the response and the predictor is linear.
Another thing you might note is that the formula for the slope is just fine providing you have statistical software to make the calculations. But, what would you do if you were stranded on a desert island, and were in need of finding the least squares regression line for the relationship between the depth of the tide and the time of day? You'd probably appreciate having a simpler calculation formula! You might also appreciate understanding the relationship between the slope and the sample correlation coefficient .
With that lame motivation behind us, let's derive alternative calculation formulas for the slope .
Theorem
An alternative formula for the slope of the least squares regression line:
is:
Proof
The proof, which may or may not show up on a quiz or exam, is left for you as an exercise. |
696 | https://drait.edu.in/assets/departments/ECE/materials/Control_Systems_Unit3.pdf | 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Control Systems UNIT 3 Stability analysis Ripal Patel Assistant Professor, Dr.Ambedkar Institute of Technology, Bangalore.
ripal.patel.ec@drait.edu.in June 15, 2021 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems What is Stability?
• A system is said to be stable, if its output is under control.
Otherwise, it is said to be unstable. A stable system produces a bounded output for a given bounded input.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Response of a stable system This is the response of first order control system for unit step input. This response has the values between 0 and 1. So, it is bounded output.
We know that the unit step signal has the value of one for all positive values of t including zero. So, it is bounded input.
Therefore, the first order control system is stable since both the input and the output are bounded.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Types of Systems based on Stability We can classify the systems based on stability as follows: • Absolutely stable system • Conditionally stable system • Marginally stable system 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Absolutely Stable System • If the system is stable for all the range of system component values, then it is known as the absolutely stable system.
• The open loop control system is absolutely stable if all the poles of the open loop transfer function present in left half of s plane.
• Similarly, the closed loop control system is absolutely stable if all the poles of the closed loop transfer function present in the left half of the s plane.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Conditionally Stable System If the system is stable for a certain range of system component values, then it is known as conditionally stable system.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Marginally Stable System • If the system is stable by producing an output signal with constant amplitude and constant frequency of oscillations for bounded input, then it is known as marginally stable system.
• The open loop control system is marginally stable if any two poles of the open loop transfer function is present on the imaginary axis.
• Similarly, the closed loop control system is marginally stable if any two poles of the closed loop transfer function is present on the imaginary axis.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Routh-Hurwitz Stability Criterion • Routh-Hurwitz stability criterion is having one necessary condition and one sufficient condition for stability.
• If any control system doesnt satisfy the necessary condition, then we can say that the control system is unstable. But, if the control system satisfies the necessary condition, then it may or may not be stable.
• So, the sufficient condition is helpful for knowing whether the control system is stable or not.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Necessary Condition for Routh-Hurwitz Stability 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Sufficient Condition for Routh-Hurwitz Stability The sufficient condition is that all the elements of the first column of the Routh array should have the same sign. This means that all the elements of the first column of the Routh array should be either positive or negative.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Routh Array Method • If all the roots of the characteristic equation exist to the left half of the s plane, then the control system is stable.
If at least one root of the characteristic equation exists to the right half of the s plane, then the control system is unstable. So, we have to find the roots of the characteristic equation to know whether the control system is stable or unstable. But, it is difficult to find the roots of the characteristic equation as order increases.
• So, to overcome this problem there we have the Routh array method. In this method, there is no need to calculate the roots of the characteristic equation. First formulate the Routh table and find the number of the sign changes in the first column of the Routh table. The number of sign changes in the first column of the Routh table gives the number of roots of characteristic equation that exist in the right half of the s plane and the control system is unstable.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Procedure 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example Step 3 Verify the sufficient condition for the Routh-Hurwitz stability.
All the elements of the first column of the Routh array are positive. There is no sign change in the first column of the Routh array. So, the control system is stable.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Examine stability of F(s) = s3 + 6s2 + 11s + 6 = 0 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Comment on the stability of a system using Routh’s stability criteria whose characteristic equation is F(s) = s4 + 2s3 + 4s2 + 6s + 8 = 0 How many poles of systems lie in right half of s-plane?
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Special Cases of Routh Array 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems First Element of any row of the Routh array is zero If any row of the Routh array contains only the first element as zero and at least one of the remaining elements have non-zero value, then replace the first element with a small positive integer, ϵ. And then continue the process of completing the Routh table. Now, find the number of sign changes in the first column of the Routh table by substituting ϵ tends to zero.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Let us find the stability of the control system having characteristic equation, F(s) = s4 + 2s3 + s2 + 2s + 1 = 0 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Let us find the stability of the control system having characteristic equation, F(s) = s5 + 2s4 + 3s3 + 4s2 + 5s + 6 = 0 How many poles lie in right half of s-plane?
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Example 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems All the Elements of any row of the Routh array are zero 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Let us find the stability of the control system having characteristic equation, F(s) = s5 + 3s4 + s3 + 3s2 + s + 3 = 0 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Routh Array 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Routh Array 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution Place these coefficients in row s3 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution There are two sign changes in the first column of Routh table.
Hence, the control system is unstable.
18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Let us find the stability of the control system having characteristic equation, F(s) = s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Example Determine the range of K for stability for the following closed-loop transfer function 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems Solution 18EC45 Ripal Patel Concept of Stability Routh stability criterion Applications of Routh stability criterion only for linear feedback control systems |
697 | https://ocw.mit.edu/courses/6-241j-dynamic-systems-and-control-spring-2011/3bf509c94d0977a81bf16f1d0e15c47e_MIT6_241JS11_lec09.pdf | 6.241 Dynamic Systems and Control Lecture 9: Transfer Functions Readings: DDV, Chapters 10, 11, 12 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 2, 2011 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 1 / 13 Asymptotic Stability (Preview) We have seen that the unforced state response (u = 0) of a LTI system is easily computed using the “A” matrix in the state-space model: x[k] = Ak x, or x(t) = e At x(0). A system is asymptotically stable if limt +∞ x(t) = 0, for all x0. → Assume A is diagonalizable, i.e., V −1AV = Λ, and let r = Vx be the vector of model coordinates. Then, ri [k] = λk i ri , or ri (t) = e λit ri (0), i = 1, . . . , n. Clearly, for the system to be asymptotically stable, |λi | < 1 (DT) or Re(λi ) < 0 (CT) for all i = 1, . . . , n. It turns out that this condition extends to the general (non-diagonalizable) case. More on this later in the course. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 2 / 13 X (Time-domain) Response of LTI systems — summary Based on the discussion in previous lectures, the solution of initial value problems (i.e., the response) for LTI systems can be written in the form: k−1 y[k] = CAk x + C Ak−i−1Bu[i] + Du[t] i=0 or Z t y(t) = C exp(At)x(0) + C exp(A(t − τ))Bu(τ) dτ + Du(t). 0 However, the convolution integral (CT) and the sum in the DT equation are hard to interpret, and do not offer much insight. In order to gain a better understanding, we will study the response to elementary inputs of a form that is particularly easy to analyze: the output has the same form as the input. very rich and descriptive: most signals/sequences can be written as linear combinations of such inputs. Then, using the superposition principle, we will recover the response to general inputs, written as linear combinations of the “easy inputs.” E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 3 / 13 The continuous-time case: elementary inputs Let us choose as elementary input u(t) = u0est , where s ∈ C is a complex number. If s is real, then u is a simple exponential. If s = jω is imaginary, then the elementary input must always be accompanied by the “conjugate,” i.e., u(t) + u∗(t) = u0ejωt + u0e−jωt = 2u0 cos(ωt); in other words, if s is imaginary, then u(t) = est must be understood as a “half” of a sinusoidal signal. if s = σ + jω, then u(t) + u∗(t) = u0(e σt ejωt + u0e σt e−jωt ) = u0(e σt ejωt + e−jωt ) = 2u0e σt cos(ωt), and the input u is a “half” of a sinusoid with exponentially-changing amplitude. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 4 / 13 h i Output response to elementary inputs (1/2) Recall that, Z t y(t) = CeAt x(0) + C e A(t−τ )Bu(τ ) dτ + Du(t). 0 Plug in u(t) = u0est : Z t y(t) = CeAt x(0) + C e A(t−τ)Bu0e sτ dτ + Du0e st 0 Z t = CeAt x(0) + C e(sI −A)τ dτ e At Bu0 + Du0e st 0 If (sI − A) is invertible (i.e., s is not an eigenvalue of A), then y(t) = CeAt x(0) + C (sI − A)−1 e(sI −A)t − I e At Bu0 + Du0e st . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 5 / 13 Output response to elementary inputs (2/2) Rearranging: y(t) = CeAt x(0) − C (sI − A)−1 e At Bu0 + C (sI − A)−1B + D u0e st . | {z } | {z } Transient response Steady−state response If the system is asymptotically stable, eAt 0, and the transient response → will converge to zero. The steady state response can be written as: yss = G (s)e st , G (s) ∈ Cny ×nu , where G (s) = C (sI − A)−1B + D is a complex matrix. The function G : s G (s) is also known as the transfer function: it → st st describes how the system transforms an input e into the output G (s)e . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 6 / 13 Laplace Transform The (one-sided) Laplace transform F : C C of a sequence f : R R is → ≥0 → defined as Z +∞ F (s) = f (t)e−st dt, 0 for all s such that the series converges (region of convergence). Given the above definition, and the previous discussion, Y (s) = G (s)U(s). U(s)e st Y (s)e st = G (s)U(s)e st ⇒ Also, G (s) is the Laplace transform of the “impulse” response. E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 7 / 13 X X ! X The discrete-time case: elementary inputs Let us choose as elementary input u[k] = u0zk , where z ∈ C is a complex number. If z is real, then u is a simple geometric sequence. Recall k−1 y[k] = CAk x + C Ak−i−1Bu[i] + Du[k]. i=0 Plug in u[k] = u0zk , and substitute l = k − i − 1: k−1 y[k] = CAk x + C Al Bu0z k−l−1 + Du0z k l=0 k−1 = CAk x + Czk−1 (Az−1)i Bu0 + Du0z k . i=0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 8 / 13 X X X X Matrix geometric series Recall the formula for the sum of a geometric series: k−1 i 1 − mk m = . 1 − m i=0 For a matrix: k−1 Mi = I + M + M2 + . . . Mk−1 . i=0 k−1 Mi (I − M) = (I + M + M2 + . . . Mk−1)(I − M) = I − Mk . i=0 i.e., k−1 Mi = (I − Mk )(I − M)−1 . i=0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 9 / 13 Discrete Transfer Function Using the result in the previous slide, we get y[k] = CAk x + Czk−1(I − Ak z−k )(I − Az−1)−1Bu0 + Du0z k = CAk x + C(z k I − Ak )(zI − A)−1Bu0 + Du0z k . Rearranging: y[k] = CAk x − (zI − A)−1Bu0 + C (zI − A)−1B + D u0z k . | {z } | {z } Transient response Steady−state response If the system is asymptotically stable, the transient response will converge to zero. The steady state response can be written as: yss[k] = G (z)z k , G(z) ∈ C, where G (z) = C(zI − A)−1B + D is a complex number. The function G : z G (z) is also known as the (pulse, or discrete) transfer function: it → k k describes how the system transforms an input z into the output G(z)z . E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 10 / 13 X Z-Transform The (one-sided) z-transform F : C → C of a sequence f : N0 → R is defined as +∞ F (z) = f [k]z−k , k=0 for all z such that the series converges (region of convergence). Given the above definition, and the previous discussion, Y (z) = G (z)U(z). U(z)z k Y (z)z k = G(z)U(z)z k ⇒ Y (z) = G (z)U(z) Also, G (z) is the z transform of the “impulse” response, i.e., the response to the sequence u = (1, 0, 0, . . .). E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 11 / 13 Models of continuous-time systems CT CT System CT ⎤ ⎡ ⎤ ⎡ 1 0 . . . 0 0 A = ⎢ ⎢ ⎣ . . . . . . . . . 0 0 . . . 1 0 ⎥ ⎥ ⎦ B = ⎢ ⎢ ⎣ . . . 0 ⎥ ⎥ ⎦ x ˙(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) −a0 −a1 . . . −an−1 1 C = b0 b1 . . . bn−1 D = d bn−1sn−1 + . . . + b0 G (s) = C (sI − A)−1B + D G (s) = + d sn + an−1sn−1 + . . . + a0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 12 / 13 Models of discrete-time systems DT DT System DT ⎤ ⎡ ⎤ ⎡ 1 0 . . . 0 0 A = ⎢ ⎢ ⎣ . . . . . . . . . 0 0 . . . 1 0 ⎥ ⎥ ⎦ B = ⎢ ⎢ ⎣ . . . 0 ⎥ ⎥ ⎦ x[k + 1] = Ax[k] + Bu[k] y[k] = Cx[k] + Du[k] −a0 −a1 . . . −an−1 1 C = b0 b1 . . . bn−1 D = d bn−1zn−1 + . . . + b0 G (z) = C (zI − A)−1B + D G (z) = + d zn + an−1zn−1 + . . . + a0 E. Frazzoli (MIT) Lecture 9: Transfer Functions Mar 2, 2011 13 / 13 MIT OpenCourseWare 6.241J / 16.338J Dynamic Systems and Control Spring 2011 For information about citing these materials or our Terms of Use, visit: |
698 | https://www.gauthmath.com/solution/1831295188674593/Q-mC-T-Rearrange-the-equation-to-solve-for-the-change-in-temperature-T-Qmc-T-Q-m | Solved: Q=mC△ T Rearrange the equation to solve for the change in temperature, △ T= Qmc △ T= Q/mc [Physics]
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Q=mC△ T Rearrange the equation to solve for the change in temperature, △ T= Qmc △ T= Q/mc e= Q/m△ T m= Q/△ Tc ua T
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Answer
The answer is Option 2: Q/mc
Explanation
To solve for $$\Delta T$$Δ T in the equation $$Q = mc\Delta T$$Q=m c Δ T, we need to isolate $$\Delta T$$Δ T on one side of the equation.
Divide both sides by $$mc$$m c
To isolate $$\Delta T$$Δ T, divide both sides of the equation by $$mc$$m c:
$$\frac{Q}{mc} = \frac{mc\Delta T}{mc}$$m c Q=m c m c Δ T
Simplify the equation
Simplify the right side of the equation by canceling out $$mc$$m c:
$$\frac{Q}{mc} = \Delta T$$m c Q=Δ T
Rewrite the equation
Rewrite the equation to have $$\Delta T$$Δ T on the left side:
$$\Delta T = \frac{Q}{mc}$$Δ T=m c Q
So Option 2 is correct.
Here are further explanations:
Option 1: ΔT = Qmc
This option is incorrect because it multiplies $$Q$$Q by $$mc$$m c instead of dividing.
Option 3: c = Q/mΔT
This option is incorrect because it solves for $$c$$c (specific heat capacity), not $$\Delta T$$Δ T (change in temperature).
Option 4: m = Q/ΔTc
This option is incorrect because it solves for $$m$$m (mass), not $$\Delta T$$Δ T (change in temperature)
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Talk:Modular arithmetic/Intermediate
<Talk:Modular arithmetic
The following material can be reworded and incorporated into this article:--MCrawford 18:54, 30 June 2006 (EDT)
A General Algorithm
In the example above, we were fortunate to find a power of -- namely, -- that is congruent to mod . What if we aren't this lucky? Suppose we want to solve the following problem:
What are the tens and units digits of ?
Again, we will solve this problem by computing modulo . The first few powers of mod are
This time, no pattern jumps out at us. Is there a way we can find the power of mod without taking this list all the way out to the term -- or even without patiently waiting for the list to yield a pattern?
Suppose we condense the list we started above; and instead of writing down all powers of mod , we write only the powers , where is a power of . We have the following (all congruences are modulo ):
(Observe that this process yields a pattern of its own, if we carry it out far enough!)
Now, observe that, like any positive integer, can be written as a sum of powers of two:
We can now use this powers-of-two expansion to compute :
So the tens and units digits of are and , respectively.
We can use this method to compute modulo , for any integers and , with . The beauty of this algorithm is that the process takes, at most, approximately steps -- at most steps to compute the values modulo for a power of two less than , and at most steps to multiply the appropriate powers of according to the binary representation of .
This method can be further refined using Euler's Totient Theorem.
Two suggestions
Why is assumed here, it's never needed. It's in general better to neglect this restriction after one has mastered the first steps (and as this is Intermediate, I would assume this to be true).
And I'm still strongly against using instead of , the first one is already used for -adic integers and rarely seen in non-olympiad books.
But as both is some kind of disputed, I didnÄt edit it untill now.
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