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Solving Minimum Coin Change
Try K.
8 min readSep 24, 2020
For many people that are preparing for coding interviews, the fear of dynamic programming (DP) questions is quite common. This is in part due to the type of problems that often requires DP to solve. These questions often concern combinatorics and repetitive calculations.
Now, if someone was learning about DP, it’s likely that they would come across the minimum coin change problem (or at least a variant of it). In this post, we’ll discuss the logic involved in solving this problem.
But before we can dive into the problem, let’s talk a bit about what dynamic programming is.
A Brief Explanation of DP
Here’s a Quora answer that I believe nails the philosophy of dynamic programming:
Simply put, DP isa method in which we store previously calculated values so that we can easily retrieve them again without having to recalculate.
Being able to store values allows us to quickly retrieve them and use as smaller sub-solutions to solve for even larger ones.
Now, I believe what makes DP problems so frightening is not so much the method of approach, but the nature of the problem and exactly how to apply DP.
In order to better understand it, let’s look at the minimum coin change problem.
Defining the Problem
The minimum coin change problem goes as follow:
Suppose you’re given an array of numbers that represent the values of each coin. Then you’re given an amount and asked to find the minimum number of coins that are needed to make that amount.
Assume the number of coins you have are infinite, so you don’t need to worry about how many coins are at your disposal.
Example:
Coins: [1, 2, 5]
Amount: 11
Answer: 3 coins (because 5 + 5 + 1 = 11)
Approaching the Problem
From a glance, this problem seems really daunting. An initial thought might be to determine which of the combinations of coins will have the minimum number of coins.
Using this thinking, we’d determine that 11 can be made up in the following ways:
1 + 1 + 1 + … + 1 = 11
1 + 1 + 1 + 1 + 1 (9 ones) + 2 = 11
…
5 + 5 + 1= 11
Now, this approach looks rather simple. But as you can see, I got tired of writing the various combination of coins. Just imagine if we have a larger array of coins and an even larger amount. This approach would be impractical.
As a matter of fact, it’s unnecessary and tedious to record all the combination of coins. But fortunately, there’s a much simpler and arguably more elegant solution.
DP Approach
Recall the Quora answer from earlier. Let’s start with:
1 + 1 + 1 +1 + … + 1 = 10
Now, the first time, it may take a while to add up all the 1s to get 10. But, if I was to ask you “How could you make 11?”, you would be able to tell me right away that all we have to do is: 10+ 1 = 11.
Remember, DP consists of storing previously calculated values to prevent repetitive calculations and allow for quick retrievals as well as using smaller values to solve for even larger ones.
But exactly how are we going to store our previous calculations?
Well, we could use an array. Suppose that we have this array of numbers, that will hold the minimum number of coins for each amount, starting from 0 to the amount, which in this case is 11:
Now the question we face is, what values do we initialize each index with? Well, since we’re dealing with minimums, values are often initialized to Infinity (∞). The logic is that at this moment, the minimum number of coins to make each amount is infinite:
But, there’s one important thing to note. Assuming that we’re only given positive value coins, we know that it is impossible to make an amount of 0 using any of the coins. So, we can say that there are 0 ways of making 0:
Okay, now let’s look at each coin separately to see how many of a specific coin can make each amount. Since our first coin is the 1-coin, we’re going to ask: “Using just 1-coins, how many coins does it take to make a value of amount: 1? 2? … 11?”
First, let’s ask: “From 0, how many 1-coins do I need to make 1?”
Well, that’s simply 1.
0 + 1 = 1
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Now, we ask: “Okay now, which is smaller? 1 or ∞ ?”
It’s 1.
Let’s visualize it:
What we’re basically doing is we’re looking at how many coins it takes to make amount 0 and then using that amount to make 1. Another way of thinking about it is if we imagined having a pile of coins that make up 0 and then adding another coin to make 1.
Then, we determine if the new pile uses up less coins than the current current. Since 1 < ∞, we replace ∞ with 1.
We repeat this process with 2, 3, … , 11:
Here, we’re saying that we already know that it takes one 1-coin to make the amount 1. In order to make amount 2, we just put a 1-coin on top of the pile that makes up amount 1 and determine if that new pile is less than ∞. If so, we replace ∞ with the new minimum number of coins.
Okay, before I go to the next step, you’re probably asking: “What about the other coins, 2 and 5?” We’re getting to them next. What I wanted to emphasize is the idea that we’re solving this problem one coin at a time. Using just one coin, we ask ourselves: “Using just this coin, what’s the minimum number of coins that we can use to get this specific amount?”
Now, let’s look at what happens when use both 1 and 2. We’ve already seen the ways in which using 1-coins will get us a certain amount. But now, we’re going to look at whether we can replace some of those 1-coins with 2-coins.
It’s clear that we can’t make the amounts 0 and 1 using a 2-coin. So those values stay the same. But what about the amount 2?
Since we can now use a 2-coin, can’t we just use that coin instead of using two 1-coins? As a matter of fact, isn’t this just the same as having an amount of 0 and then adding a 2-coin to make the amount 2?:
0 + 2 = 2 (using one 2-coin)
Now, that we can use 2-coins, we can ask how we can make 3 using both coins 1 and 2. Well, we already know how to make 3 using just 1-coins. But, what about 1-coins and 2-coins? Well, we can:
1 + 2 = 3
Or simply put, from the amount 1, we can put a 2-coin on top of it to make 3:
Now, this is where it gets interesting and I think you might start to see the pattern emerge. In order to create the amount 4 using 1-coins and 2-coins, we can see:
1 + 1 + 1 + 1 = 4 (only 1s)
1 + 1 + 2 = 4 (both 1 and 2)
2 + 2 = 4 (all 2s)
As we can see, that using only two 2-coins will give us the minimum number of coins it takes to make the amount 4. Or another way to put it, from the amount 2, we can add a 2-coin to make 4:
From amount 6, we can add a 2-coin. That will give us an amount of 8. Since the minimum number of coins needed to make 6 is 3 (2 + 2 + 2), the new minimum number of ways to make 8 is by putting a 2-coin on top of the amount 6, thus making it 4 coins.
Now that we’ve finished looking at the minimum number of coins needed to make each amount using 1-coins and 2-coins, let’s look at 5-coins:
It’s becoming clear that the first argument in the min() is array[current_amount — current_coin] + 1, while the second argument is just the array[current_amount].
And at last, we’ve exhausted all our coin options and see that at amount 11, there is a minimum value of 3 coins (5 + 5 + 1) that are required to make 11.
Now, let’s look at the code.
Code
The code is written as follow using JavaScript:
Analysis
From the code, we can see that we’re using two for-loops. Since we’re iterating over the entire minCoins (which has a length equal to amount) array every time we iterate over an element in the coins array, we say that the runtime of this algorithm is O(|coins|•|amount|) where |coins| is the length of the coins array and |amount| is the length of the minCoins array.
And since we’re using an array to keep track of our minimum coins for each amount, minCoins, the space complexity is O(|amount|).
Summary
Dynamic Programming (DP) is simply the method of storing previously calculated values so that we don’t have to recalculate them, which saves us time and allows us to use smaller sub-solutions to solve larger ones.
Look at one coin at a time and find out what is the minimum number of coins that are needed to make each amount from 0 to amount.
Runtime: O(|coins|•|amount|), where |coins| represent the length of the coins array and |amount| represents the length of minCoins array.
Space Complexity: O(|amount|) because we used an array to keep track of the minimum number of coins for each amount.
Dynamic Programming
Coin Change Problem
Algorithms
Data Structures
Software Development
## Written by Try K.
300 followers
·16 following
I just write what I've learned and what's on my mind. Welcome!
Responses (4)
Write a response
What are your thoughts?
Rakesh Venkatesh
Feb 3, 2021
```
I think the inner loop can be optimized as for (let i = coin; i <= amount; i++)
You dont need to start always from 0
```
6
Developerlite
Jun 10, 2022
```
Hi Guys
I made a small video to explain the need for DP approach to this problem. Please check this out :)
```
Sarah Smith
Apr 17, 2021
```
The number of ways to make up 0 is 1.
```
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9101 | https://artofproblemsolving.com/downloads/printable_post_collections/255106.pdf?srsltid=AfmBOoqrr7RC_FjVIpY8ZQORAMfyHMqxYsyLXh5wFo53Cnf7wXXVA4pH | AoPS Community 2016 China National Olympiad
China National Olympiad 2016
www.artofproblemsolving.com/community/c255106 by rkm0959, sqing, fattypiggy123
Day 1 December 16th
1 Let a1, a 2, · · · , a 31 ; b1, b 2, · · · , b 31 be positive integers such that a1 < a 2 < · · · < a 31 ≤ 2015 ,
b1 < b 2 < · · · < b 31 ≤ 2015 and a1 + a2 + · · · + a31 = b1 + b2 + · · · + b31 .
Find the maximum value of S = |a1 − b1| + |a2 − b2| + · · · + |a31 − b31 |.
2 In 4AEF , let B and D be on segments AE and AF respectively, and let ED and F B intersect at C. Define K, L, M, N on segments AB, BC, CD, DA such that AK KB = AD BC and its cyclic equivalents. Let the incircle of 4AEF touch AE, AF at S, T respectively; let the incircle of
4CEF touch CE, CF at U, V respectively. Prove that K, L, M, N concyclic implies S, T, U, V concyclic.
3 Let p be an odd prime and a1, a 2, ..., a p be integers. Prove that the following two conditions are equivalent: 1) There exists a polynomial P (x) with degree ≤ p−12 such that P (i) ≡ ai (mod p) for all 1 ≤
i ≤ p
2) For any natural d ≤ p−12 , p∑
i=1
(ai+d − ai)2 ≡ 0 (mod p)
where indices are taken (mod p)
Day 2 December 17th
4 Let n ≥ 2 be a positive integer and define k to be the number of primes ≤ n. Let A be a subset of S = {2, ..., n } such that |A| ≤ k and no two elements in A divide each other. Show that one can find a set B such that |B| = k, A ⊆ B ⊆ S and no two elements in B divide each other.
5 Let ABCD be a convex quadrilateral. Show that there exists a square A′B′C′D′ (Vertices maybe ordered clockwise or counter-clockwise) such that A 6 = A′, B 6 = B′, C 6 = C′, D 6 = D′
and AA ′, BB ′, CC ′, DD ′ are all concurrent.
6 Let G be a complete directed graph with 100 vertices such that for any two vertices x, y one can find a directed path from x to y.a) Show that for any such G, one can find a m such that for any two vertices x, y one can find a directed path of length m from x to y (Vertices can be repeated in the path)
©2019 AoPS Incorporated
1AoPS Community 2016 China National Olympiad
b) For any graph G with the properties above, define m(G) to be smallest possible m as defined in part a). Find the minimim value of m(G) over all such possible G’s.
©2019 AoPS Incorporated
2 |
9102 | https://jingyan.baidu.com/article/425e69e69f07fbbe15fc161f.html | 中位数怎么求-百度经验
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什么是中位数？所谓中位数就是一组数据从小到大排列中间的那个数字。但是有的时候一组数据是偶数的话就是中间两个数字相加除以2.知道了定义就很好求了？这篇经验将进行介绍。
一组数据是奇数
1 所谓奇数也就是单数了。还是具例子比较清楚。比如原始数据：10 20 60 80 70我要看重新排列一下。怎么排列呢？按照从小到大的顺序排列。排列完数据是重新排列：10 20 60 70 80。
2 排列完之后就要看这组数据有几个，我们可以数一下是5个。五个中间的自然是第三个数据位中位数。算法（5+1）/2=3.我们可以看一下这组数据的中位数是60. END
一组数据是偶数
1 所谓偶数也就是双数了。还是具例子比较清楚。比如原始数据：10 20 60 80 70 50我要看重新排列一下。怎么排列呢？按照从小到大的顺序排列。排列完数据是重新排列：10 20 50 60 70 80。
2 排列完之后就要看这组数据有几个，我们可以数一下是6个。算法（6+1）/2=3.5.
这个3.5的数字是那个呢？如果算的数字是类似这样的就要选择这组数据中的第三和第四两位。
3. 3 我们知道中位数字只有一个，这出现了两个怎么算呢？很简单两个数字相加除以2.本例（50+60）/2=55.这组数据的中位数也就是55. END
特殊数据
1 所谓特殊数据就是一个数字在这组数据中重复出现。比如原始数据10 20 20 30 20 20要把它重新排列：10 20 20 20 20 30
2 我们可以看到这个数据是是偶数。计算方法为（6+1）/2=3.5.
这个时候第三和第四两个数字都是20.那么这组数据的中位数就是20.计算方法相同。 END
经验内容仅供参考，如果您需解决具体问题(尤其法律、医学等领域)，建议您详细咨询相关领域专业人士。
作者声明： 本篇经验系本人依照真实经历原创，未经许可，谢绝转载。
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9103 | https://courses.lumenlearning.com/intermediatealgebra/chapter/read-variation/ | Variation
Learning Outcomes
Define direct variation and solve problems involving direct variation
Define inverse variation and solve problems involving inverse variation
Define joint variation and solve problems involving joint variation
So many cars, so many tires.
Direct Variation
Variation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation.
number of tires=4⋅number of cars
The number 4 tells you the rate at which cars and tires are related. You call the rate the constant of variation. It is a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of direct variation, where the number of tires varies directly with the number of cars.
You can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output, 4 is the constant, and the number of cars is the input. Let us enter those generic terms into the equation. You get y=kx. That is the formula for all direct variation equations.
number of tires=4⋅number of carsoutput=constant⋅input
Example
Solve for k, the constant of variation, in a direct variation problem where y=300 and x=10.
Show Solution
Write the formula for a direct variation relationship.
y=kx
Substitute known values into the equation.
300=k(10)
Solve for k by dividing both sides of the equation by 10.
30010=10k1030=k
In the video that follows, we present an example of solving a direct variation equation.
Inverse Variation
Another kind of variation is called inverse variation. In these equations, the output equals a constant divided by the input variable that is changing. In symbolic form, this is the equation y=kx.
One example of an inverse variation is the speed required to travel between two cities in a given amount of time.
Let us say you need to drive from Boston to Chicago which is about 1,000 miles. The more time you have, the slower you can go. If you want to get there in 20 hours, you need to go 50 miles per hour (assuming you do not stop driving!) because 1,00020=50. But if you can take 40 hours to get there, you only have to average 25 miles per hour since 1,00040=25.
The equation for figuring out how fast to travel from the amount of time you have is speed=milestime. This equation should remind you of the distance formula d=rt. If you solve d=rt for r, you get r=dt, or speed=milestime.
In the case of the Boston to Chicago trip, you can write s=1,000t. Notice that this is the same form as the inverse variation function formula y=kx.
Example
Solve for k, the constant of variation, in an inverse variation problem where x=5 and y=25.
Show Solution
Write the formula for an inverse variation relationship.
y=kx
Substitute known values into the equation.
25=k5
Solve for k by multiplying both sides of the equation by 5.
5⋅25=k5⋅5125=5k5125=k
In the next example, we will find the water temperature in the ocean at a depth of 500 meters. Water temperature is inversely proportional to depth in the ocean.
Water temperature in the ocean varies inversely with depth.
Example
The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of 1,000 meters, the water temperature is 5º Celsius. What is the water temperature at a depth of 500 meters?
Show Solution
You are told that this is an inverse relationship and that the water temperature (y) varies inversely with the depth of the water (x).
y=kxtemp=kdepth
Substitute known values into the equation.
5=k1,000
Solve for k.
1,000⋅5=k1,000⋅1,0005,000=1,000k1,0005,000=k
Now that k, the constant of variation is known, use that information to solve the problem: find the water temperature at 500 meters.
temp=kdepthtemp=5,000500temp=10
At 500 meters, the water temperature is 10º C.
In the video that follows, we present an example of inverse variation.
Joint Variation
A third type of variation is called joint variation. Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula A=lw, where l is the length of the rectangle and w is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle “varies jointly with the length and the width of the rectangle.”
The formula for the volume of a cylinder, V=πr2h, is another example of joint variation. The volume of the cylinder varies jointly with the square of the radius and the height of the cylinder. The constant of variation is π.
Example
The area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is 30 inches2 when the base is 10 inches and the height is 6 inches, find the variation constant and the area of a triangle whose base is 15 inches and height is 20 inches.
Show Solution
You are told that this is a joint variation relationship and that the area of a triangle (A) varies jointly with the lengths of the base (b) and height (h).
y=kxzArea=k(base)(height)
Substitute known values into the equation and solve for k.
30=k(10)(6)30=60k3060=60k6012=k
Now that k is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.
Area=k(base)(height)Area=(15)(20)(12)Area=3002Area=150square inches
The constant of variation, k, is 12, and the area of the triangle is 150 square inches.
Finding k to be 12 should not be surprising. You know that the area of a triangle is one-half base times height, A=12bh. The 12 in this formula is exactly the same 12 that you calculated in this example!
In the following video, we show an example of finding the constant of variation for a jointly varying relation.
Direct, Joint, and Inverse Variation
k is the constant of variation. In all cases, k≠0.
Direct variation: y=kx
Inverse variation: y=kx
Joint variation: y=kxz
Summary
Rational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship—as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship—as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.
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9104 | https://mathematica.stackexchange.com/questions/296482/how-to-verify-a-solution-of-an-ordinary-differential-equation | Skip to main content
How to verify a solution of an ordinary differential equation?
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Given an ordinary differential equation with initial conditions
```
eq = u a[u] + (16 + u^2 + 2 u a[u] (12 + u a[u] (6 + u a[u]))) a'[u] == 0
ic = a == -1/2
```
How can I analytically verify that
```
sol = a[u]^2 (2 + a[u] u)^2 == 1 + a[u] u
```
represents its implicit solution? Please, avoid numerical demonstrations. Assume also that DSolve cannot solve the given equation. I am explicitly interested in algebraic demonstration that would work in more complicated cases.
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edited Jan 16, 2024 at 19:36
user64494
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asked Jan 15, 2024 at 20:40
yarchikyarchik
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Did you try implicitly differentiating the solution to find a’[u] and substituting that into the equation?
– Ghoster
Commented
Jan 15, 2024 at 21:01
1
@Ghoster Clever! Why didn't you post an answer?
– Ulrich Neumann
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Jan 15, 2024 at 22:08
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You can't plugin in the implicit solution as is, need to first solve for a[u] from it. This gives 3 solutions. Then verify each one, one by one.
Maple can verify implicit solutions directly, without first solving for each solution. But Mathematica as far as I know, can't as of now.
```
eq = u a[u] + (16 + u^2 + 2 u a[u] (12 + u a[u] (6 + u a[u]))) a'[u] == 0
ic = a == -1/2
sol = a[u]^2 (2 + a[u] u)^2 == 1 + a[u] u
explicit=a[u]/.Solve[sol,a[u]]
trialSol = a -> Function[{u}, Evaluate[First@explicit]]
eq /. trialSol // FullSimplify
```
But this solution did not verify the IC
```
ic /. trialSol // FullSimplify
```
Gives
Do the same for the result of the solutions, like this
Verified in Maple that your solution verifies the ode and the IC, using odetest directly. So need to work more on figuring out why Mathematica does not verify the IC. Looks like a limit issue,
```
eq := ua(u)+(16 + u^2 + 2ua(u)(12 + ua(u)(6 + ua(u))))diff(a(u),u)=0;
ic := a(0) = -1/2;
sol := a(u)^2(2 + a(u)u)^2 = 1 + a(u)u;
odetest(sol,[eq,ic])
```
Gives [0,0] which means both the solution and IC was verified correct.
It will be useful if Mathematica had odetest function which is built-in to make it easier to verify the solutions of an ode.
Notice also that Mathematica DSolve has no option to return an implicit solution. It tries to always return an explicit solution. It will also be good to have such an option in future versions of DSolve
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edited Jan 15, 2024 at 22:55
answered Jan 15, 2024 at 21:05
NasserNasser
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Indeed, would be good if MA had this functionality. It seems to be a pure algebraic problem. I tried to handle it with Eliminate and PolynomialReduce, but somehow stuck. The problem with your solution is that it generates multiple branches and one needs to verify one by one. Out of curiosity, is Maple able to return this implicit solution?
– yarchik
Commented
Jan 15, 2024 at 22:33
and one needs to verify one by one. Yes. This is the limitation of using Function method to verify solution. This is the method that Mathematica help pages recommend to use to verify ode solution (added link above). Otherwise, you'd have to do it the hard way, i.e. plugin in the dependent variable and all its derivatives into the ode and simplify.
– Nasser
Commented
Jan 15, 2024 at 22:51
Out of curiosity, is Maple able to return this implicit solution? Maple returns 3 solutions. One of them implicit, It is written different, using ln but it looks like it could be simplified to the one you showed. I did not try. Need to raise both sides to exp and simplify. Here it is in plain text -1/2ln(4a(u)^2u^2+16ua(u)+u^2+16)+ln(2+ua(u))-ln(a(u))+PiI = 0
– Nasser
Commented
Jan 15, 2024 at 22:54
Thank you, very useful. I will wait a day or two before accepting your solution as I still have hope that someone will propose how to emulate Maple's odetest function for implicit polynomial solutions and how to test the initial conditions.
– yarchik
Commented
Jan 16, 2024 at 8:56
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9105 | https://math.cumt.edu.cn/_upload/article/files/17/d1/b3a73286480996cd092544a85f3a/306e83f7-607f-4784-bf4a-730f33e7c82b.pdf | 华师大数学分析(第五版)讲义 第四章 函数的连续性 第四章 函数的连续性 §1 连续性概念 连续函数是数学分析中着重讨论的一类函数． 从几何形象上粗略地说， 连续函数在坐标平面上的图象是一条连绵不断的曲线． 当然我 们不能满足于这种直观的认识， 而应给出函数连续性的精确定义， 并由此出发研究连续函数 的性质．本节中先定义函数在一点的连续性和在区间上的连续性． 一、函数在一点的连续性 中国矿业大学数学学院 1 定义1 设函数 在某U 内有定义．若 f 0 x 0 0 lim x x f x f x 则称 在点 连续． f 0 x 设函数 在某 内有定义．若 在点 无定义，或 在点 有定义而不连续， 则称点 为函数 的间断点或不连续点． f 0 U x f f 0 x f 0 x 0 x 若 为函数 的间断点，则必出现下列情形之一： 0 x f （1） 在点 无定义 f 0 x （2）极限 x f x x 0 lim 不存在； （3） 在点 有定义且极限 f 0 x x f x x 0 lim 存在，但 x f x x 0 lim 0 x f 例如：函数 1 sin , 0 0, 0 x x f x x x ,在点 0 x 连续 函数 sin , 0 1, 0 x x f x x x ,在点 0 x 连续 增量形式的定义 华师大数学分析(第五版)讲义 第四章 函数的连续性 记 ，称为自变量 0 x x x x (在点 )的增量．设 0 x 0 0 x f y ，相应的函数 (在 点 )的增量记为 y 0 x 0 0 0 0 y y x f x x f x f x f y 则 x f y 在点 连续 0 x 0 lim 0 y x 定义2 设函数 在某 f 0 U x 0 U x 内有定义．若 0 0 lim x x f x f x 0 0 lim x x f x f x , 则称 在点 右(左)连续． f 0 x 显然 函数 在点 连续 f 0 x f 在点 既是右连续又是左连续 0 x 例如： 在点 0 , 2 2 , 2 x x x x x f 0 x 右连续，但不左连续，从而它在 不连续。 0 x 例1 狄利克雷函数 1, 0, x Q D x x Q 没有连续点。 因为 0 x R ， 不存在（由归结原则） 0 lim x x D x 例2 函数 仅在点 x xD x f 0 x 连续． 显然 由 ， 0 lim ( ) 0 0 x f x f f 在点 0 x 连续. 另外， ， 0 0 x 0 lim x x f x 不存在。这是因为 取有理点列 0 n x x ， 0 0 n n n n f x x D x x x 取无理点列 0 n x x ， 0 0 n n n f x x D x 例3 黎曼函数 1 , 0 0 1 01 p x p q p q q q R x x 当 为正整数 为既约真分数 当 及 内无理数 ， ， ， ， （， ） 中国矿业大学数学学院 2 在 内任何无理点处都连续，任何有理点处都不连续． ） ， （1 0 华师大数学分析(第五版)讲义 第四章 函数的连续性 中国矿业大学数学学院 3 x 由上一章例题 0 0 [0,1], lim ( ) 0 x x x R 便知。 （间断点是可数无穷多） 二、间断点的分类 第一类间断点：左右极限都存在的间断点。 第二类间断点：非第一类的间断点。即左右极限至少有一个不存在。 第一类间断点又分为可去间断点与跳跃间断点 1．可去间断点 若 0 lim x x f x A ，而 在点 无定义，或有定义但 f 0 x 0 f x A ， 则称 为 的可去间断点． 0 x f 例如， x x f sgn ， 是 0 x f 的可去间断点．又如 x x x g sin ， 是 的可 去间断点． 0 x g 设 为函数 的可去间断点，且 0 x f 0 lim x x f x A ，定义 0 0 ( ), ˆ( ) , f x x x f x A x x 则对于函数ˆ f 在点 连续． 0 x 2．跳跃间断点 若函数 在点 的左、右极限都存在，但 f 0 x x f x f x x x x 0 0 lim lim ，则 称点 为函数 的跳跃间断点． 0 x f 例如， x x f ，当 n x (n 为整数)时有 1 lim n x n x ， n x n x lim 整数点都是函数 的跳跃间断点． x x f 第二类间断点的例子： 1 ( ) f x x ，当 时，不存在有限的极限（无穷间断点） 0 x 1 ( ) sin f x x 在点 处左、右极限都不存在（振荡间断点） 0 x 狄利克雷函数 ，其定义域 x D R 上每一点x 都是第二类间断点． 华师大数学分析(第五版)讲义 第四章 函数的连续性 三、区间上的连续函数 若函数 在区间 f I 上的每一点都连续，则称 为 f I 上的连续函数，记为 ( ) f C I 。对 于闭区间或半开半闭区间的端点，函数在这些点上连续是指左连续或右连续． 如函数 sin y x 是R 上连续函数， 又如 2 1 x y 在 ) 1 , 1 ( 每一点处都连续， 在 为左连续，在 为右连续，因而它在 1 x 1 x 1 , 1 上连续． 若函数 在区间 上仅有有限个第一类间断点，则称 在 f b a, f b a, 上分段连续． 例如，函数 x y 和 x x y ]在区间 3 , 3 上是分段连续的． 中国矿业大学数学学院 4 华师大数学分析(第五版)讲义 第四章 函数的连续性 §2 连续函数的性质 一、连续函数的局部性质 定理1(局部有界性) 若函数 在点 连续，则 在某 f 0 x f 0 x U 内有界． 定理2(局部保号性) 若函数 在点 连续，且 f 0 x 0 x f 0 (或 0 )，则对任何正数 (或 0 x f r 0 x f r )，存在某 0 x U ，使得对一切 x 0 x U 有 （或 r x f f x r ） 。 【注】 在具体应用局部保号性时， 常取 0 2 1 x f r ， 则当 0 x f 0 时， 存在某 使在其内有 0 x U x f 0 2 1 x f . 定理3(四则运算) 若函数 和 在点 连续，则 f g 0 x g f g f g f , , (这里 ) 也都在点 连续． 0 0 x g 0 x 定理4（复合函数的连续）若函数 在点 连续， g 0 x 0 u g x 0 ，f 在点 连续，则复 合函数 0 u ( )( ) [ ( )] f g x f g x 在点 连续． 0 x 【注】见上一章。 二、闭区间上连续函数的基本性质 定义1 设 为定义在数集 上的函数．若存在 f D D x 0 ，使得对一切 有 D x 0 0 f x f x f x f x ， 则称 在 上有最大(最小)值，并称 f D 0 x f 为 在 上的最大(最小)值，称 f D 0 x 为 在 上 的最大(最小)值点。 f D 定理4 (最值性定理) 若 , f C a b ，则 在 f b a, 上有最大值与最小值． 证 （用致密性定理证明）记 [ , ] sup ( ) x a b M f x ( ) M 中国矿业大学数学学院 5 华师大数学分析(第五版)讲义 第四章 函数的连续性 下面证明 [ , ] n x a b ，使得 lim ( ) n n f x M 如果 M ，根据上确界的定义，取 1,2, n ，则 [ , ] n x a b ，使得 1 ( ) n M f x M n 两边取极限，即得lim ( ) n n f x M 。 如果M ，即 ( ) f x 无上界，取 1,2, n ，则 [ , ] n x a b ，使得 ( ) n f x n 显然有 lim ( ) n n f x 又 [ , ] n x a b ， n x 有界，由致密性定理， n x 有收敛子列，不妨假设就是它本身。 于是设 0 lim n n x x ，显然 0 [ , ] x a b ，再由f 的连续性得 0 lim ( ) ( ) n n f x f x M 上式说明M 必为有限数，就是 ( ) f x 在[ , 上的最大值。 ] a b 同理可证， ( ) f x 在[ , 上有最小值。 ] a b 由最值定理立即得下面有界性定理。 定理5 (有界性定理) 若 , f C a b ，则 在 f b a, 上有界． 定理6(根的存在定理) 若 , f C a b ，且 a f 与 b f 异号(即 0 b f a f )，则至 少存在一点 ，使得 b a x , 0 0 0 x f ，即方程 0 x f 在 b a, 上至少有一个根． 这个定理的几何解释如下图所示：若点 a f a A , 与 b f b B , 分别在x 轴的两侧，则 连接 、 A B 的连续曲线 与 x f y x 轴至少有一个交点． 中国矿业大学数学学院 6 华师大数学分析(第五版)讲义 第四章 函数的连续性 证 不妨设 。记 0, 0 f a f b | ( ) 0, [ , ] E x f x x a b 显然E 且E 有界，记 0 sup [ , ] x E a b 。 首先证明 0 , x a b ，即 0 ( , ) x a b 。由 0, 0 f a f b 以及连续函数的局部保号性 知， 0 ，使得 0, [ , ) f x x a a 和 0, ( , ] f x x b b 这说明 0 , x a b ，即 0 ( , ) x a b 。. 其次再证明 0 0 x f 。倘若 0 0 f x ，不妨设 0 0 f x 。再由局部保号性， 0 ( , ) ( U x a, ) b ，使得 0 0 f x ， 0 ( , ) x U x 特别地 0 0 2 f x ，因此 0 2 x E ，这与 0 sup x E 相矛盾。 【注】如果 ，也称 0 0 0 x f x 是函数f 的一个零点。因此，根的存在定理也称零点 存在定理。 定理7 (介值性定理1) 设 , f C a b ，且 a f b f ．若为介于 与 之 间的任何实数 a f b f f a f b ( 或 a f b f )，则至少存在一点 ，使得 a, 0 b x .
0 x f 证 令( ) ( ) g x f x 。由根的存在定理立即得证。 中国矿业大学数学学院 7 华师大数学分析(第五版)讲义 第四章 函数的连续性 中国矿业大学数学学院 8 上 这个定理表明， 若f 在 连续， 又不妨设 b a, b f a f ， 在 上 中 则f a, 必能取得 区间 f 的一切值，即有 b b f a , b a f b f a f , , 其几何意义如下图所示． 定理8(介值性定理2) 若函数 在闭区间 f b a, 上连续， 在 f b a, 上的最大值为M ， 最小值为m ，则 在 上必能取得区间 f b a, , m M 中的一切值，即 M m b , a f , 。 证 由最值性定理，记 1 1 [ , ] [ , ] min ( ), max ( ) x a b x a b a f x b f x 1 不妨设 ，则对区间 用介值性定理1 即得证。 1 a b 1 1 [ , ] a b 例1（教材例3） 证明：若 ， 为正整数，则存在唯一正数 ，使得 0 r n 0 x 0 n x r 。 证 先证存在性．由于当 x 时有 ，故必存在正数 ，使得 n x a n a r ．因 在 上连续，并有 n x x f a , 0 a f r f 0 ，故由介值性定理1，至少存在一点 ，使得 a x , 0 0 x x 0 0 r n f ． 再证唯一性．设正数 使得 ，则有 1 x r x n 1 0 1 1 1 2 0 1 0 1 0 1 0 n n n n n x x x x x x x x ， 由于第二个括号内的数为正，所以只能 0 1 0 x x ，即 0 1 x x . 【注】上面 0 x 称为r 的 次正根（即算术根） ，记作 n n r x 0 )． 例2（教材例4） 设 在 上连续，满足 f b a, 华师大数学分析(第五版)讲义 第四章 函数的连续性 b a b a f , , 证明：存在 ，使得 b a x , 0 0 0 x x f . 证 条件意味着：对任何 有 b a x , b x f a ，特别有 a f a 以及 b b f ． 若 或 a f a b b f ，则取 a x 0 或b 即可． 现设 与 ．令 a f a b b f x x f x F 则 ， , 0 a a f a F 0 b b f b F .
0 x ．故由根的存在性定理，存在 ， 使得 ，即 （参见下图） 0 x b a, 0 0 x F x f 0 a b b a 例3（习题4.2：10）证明：任一实系数奇次方程至少有一个实根。 证 设实系数奇次方程为 2 1 2 2 1 2 1 0 ( ) 0 n n n n f x a x a x a x a 2 1 0 n a ， 不妨 。由 2 1 0 n a 2 1 2 0 1 2 1 2 2 ( ) ( ) n n n n n a a a f x x a x x x 1 知 ，故存在 lim ( ) , lim ( ) x x f x f x a b ， ( ) 0, ( ) 0 f a f b 因f 在[ , 上连续，于是由根的存在定理，存在 ] a b 0 ( , ) x a b ，使得 。 0 ( ) 0 f x 例4（习题4.2：19）设函数f 在 , a b 上连续， 1 2 , , , , n x x x a b 。证明：存在 中国矿业大学数学学院 9 华师大数学分析(第五版)讲义 第四章 函数的连续性 , a b ，使得 1 2 1 ( ) ( ) ( ) ( ) n f f x f x f x n 。 证 记 max ( ), min ( ) a x b a x b M f x m f x ，则 1 1 ( ) n i i m f x n M 由介值性定理2，存在 , a b ，使得 1 1 ( ) ( ) n i i f f x n 。 三、反函数的连续性 定理9 若函数 在 上严格单调并连续，则反函数 在其定义域 或 上连续． f b a, 1 f b f a f , a f b f , 证 不妨设 在 上严格增． f b a, 此时 的值域即反函数 的定义域为 f 1 f a f [ ， ．任取 b f a f y , 0 ，设 0 x 0 1 y f ，则 ．于是对任给的 ] b f b a x , 0 0 ，可在 b 0 x a, 内 的 两侧各取异于 的点 0 x 2 1, x x 2 x 0 x 1 x ，使它们与 的距离小于 0 x (见图)． 设与 对应的函数值分别为 ， ，由 的严格增性知 2 1, x 1 y 2 y f 2 0 1 y y y ，令 x 1 0 0 2 , min y y y y ， 则当 ; 0 y U 时，对应的 y f x 1 的值都落在 与 之间，故有 1 x 2 x y 0 0 1 1 x x y f y f ，所以 在点 连续，从而 在 1 f 0 y 1 f b f , a f 内连续． 类似地可证 在其定义区间的端点 1 f a f 与 b f 分别为右连续与左连续． 所以 在 上连续． 1 f b f a f , 【注】上面定理是按闭区间来叙述，由于连续是对点来定义的，显然可改为开区间或 无穷区间等。 中国矿业大学数学学院 10 华师大数学分析(第五版)讲义 第四章 函数的连续性 例如 由于 x y sin 在区间 2 , 2 上严格单调且连续，故其反函数 在 区间 上连续． y x arcsin 1 , 1 同理可得其它反三角函数也在相应的定义区间上连续．如 x y arccos 在 上连 续， 1 , 1 x y arctan 在 , 上连续等． 四、一致连续性 函数 在区间上连续，是指 在该区间上每一点都连续．本段中讨论的一致连续性概 念反映了函数在区间上更强的连续性． f f 引例 考察函数 1 ( ) f x x 。它在 上每一点 (0,1) 0 x 都连续，用定义写出为 0 (0,1), 0, 0, x 当 0 x x 时，有 0 ( ) ( ) f x f x 注意这里不仅与有关，而且与 0 x 有关。对同一个， 0 x 越靠近0 ，就越小。 作图说明（待补） 。 问题： 为定义在区间 f I 上的函数，对 0 ，能否找到公共的 0 ，对 0 x I ， 当 0 x x 时，有 0 ( ) ( ) f x f x 。 定义2 设 为定义在区间 f I 上的函数．若对任给的 0 ，存在 0 ，使得 对任何 , ，只要 x I x x x ，就有 x f x f ， 则称函数 在区间 f I 上一致连续． 直观地说， 在 f I 上一致连续意味着：不论两点x与x 在I 中处于什么位置，只要它 们的距离小于，就可使 x f x f ． 思考 如果叙述 在区间 f I 上不一致连续． 存在常数 0 0 ，对任何正数(不论多么小)，总存在两点 ，尽管 I x x , 中国矿业大学数学学院 11 华师大数学分析(第五版)讲义 第四章 函数的连续性 x x ，但 0 x f x f . 例5 证明 sin f x x 在 , 上一致连续． 证 sin sin 2cos sin 2 2 x x x x x x x x 例6（教材例9）证明函数 x y 1 在 1 , 0 内不一致连续． 证 取 1 0 ，对无论多么小的正数 2 1 ，只要取 x 与 2 x ，则虽有 2 x x ，但 0 1 1 1 1 x x 所以 x y 1 在 内不一致连续． 1 , 0 定理10（教材例10） ( ) f x 在I 上一致连续的充要条件为：对 , n n x x I ，若 ，则 lim 0 n n n x x lim ) 0 n n n ( ) f x ( f x 证 （必要性） 若 ( ) f x 在I 上一致连续， 则 0 ， 0 ， , x x I ， x x ， 有 x f x f 。 设 , n n x x I ，满足 lim 0 n n n x x ，于是对上述 0 ， 0 N ， ， n N n n x x ，由一致连续性，有 n n f x f x ，即 lim ( ) ( ) 0 n n n f x f x （充分性）设 , n n x x I ，若 lim 0 n n n x x ，则 lim ( ) ( ) 0 n n n f x f x 。现 证 ( ) f x 在I 上一致连续。 用反证法证明。若 ( ) f x 在I 上不一致连续，则 0 0, 0, , , x x I 满足 x x ，但有 0 ) f x ( ) ( f x 。 取 1 1 , , , n n n n n x x I x x n n ，使得 0 ( ) ( ) , 1,2, n n f x f x n 中国矿业大学数学学院 12 华师大数学分析(第五版)讲义 第四章 函数的连续性 于是 ，但 lim 0 n n n x x lim ( ) ( ) 0 n n n f x f x 。矛盾。 例7 证明 2 ( ) sin f x x 在[0, ) 上不一致连续。 证 取 , 2 n n x n x n ，则 / 2 0 / 2 n n x x n n 但sin sin 1 0 n n x x ，由定理10 得证。 0 1 2 3 4 5 6 7 8 -1 -0.5 0 0.5 1 例8 证明 2 ( ) f x x 在 上不一致连续。 ( , ) 证 取 1 , n n x n x n n ，由定理10 易证。 f 在区间I 上每一点都连续，并不能推出 在 f I 上一致连续．然而，对于定义在闭区 间上的函数来说， 由它在每一点都连续却可推出在区间上的一致连续性， 即有如下重要定理： 定理11 (一致连续性定理， Cantor 定理) 若函数 在闭区间 f b a, 上连续， 则 在 ] 上一致连续． f , [a b 证 反证。假设则 在[ , 上不一致连续，则 f ] a b 0 0 ， 0 ， 1 2 , [ , ] x x a b ， 满足 1 2 x x ，但 1 ( ) ( f x2 0 ) f x . 取 ) , 2 , 1 ( 1 n n ， 相应地满足上述条件的两点记为 , [ , n n ] x x a b ， 由致密性定理， 有子列 n x k n x 收敛，设 0 lim [ , ] k n k x x a b 。 由 中国矿业大学数学学院 13 华师大数学分析(第五版)讲义 第四章 函数的连续性 0 0 1 k k k k k k n n n n n n k x x x x x x x n x 得 0 lim k n k x x 但是由我们的做法 0 ( ) ( ) k k n n f x f x 再由 的连续，上式两边取极限，得 f 0 0 0 ( ) ( ) f x f x 0 ，矛盾。 例9 若 分别在 和[ , 上一致连续，则 在 f , ] a c c b f , a b 上也一致连续． 证 任给 0 ， 1 0 ， , , ] x x a c ，只要 1 x x ，就有 x f x f 2 0 ， , [ , x x c b ，只要 2 x x ，就有 x f x f f 在点c 为既左连续又右连续，所以 在点 连续．故对上述 f c 0 ，存在 0 3 ，当 3 x c 时，有 2 c f x f 令 3 2 1 , , min ，对任何 I x x , ， x x ，分别讨论以下两种情形： (i) 同时属于 或[ , ，则 x x , , ] a c c b x f x f 成立； （ii） 分属 x x , , ] a c 与[ , ，设 c b , ],, [ , x a c x c b ，则 3 x x x c c x ， 故 2 c f x f .同理 2 c f x f ，从而 x f x f 。 例10（第四章总练习习题：1）设函数f 在 上连续，且 ( , ) a b ( 0 f a ) 与 为有 限值。证明： ( 0 f b ) （1）f 在 上有界； ( , ) a b （2）若存在 ( , ) a b ，使得 ( ) max ( 0), ( 0) f f a f b ，则f 在 内能取 到最大值。 ( , ) a b （3）f 在 上一致连续。 ( , ) a b 证 (1) 定义 中国矿业大学数学学院 14 华师大数学分析(第五版)讲义 第四章 函数的连续性 ( 0), ( ) ( ), ( 0), f a x F x f x a x b a f b x b 则 在[ , 上连续，从而 在[ , 上有界，当然也在( , 内有界，即 ( ) F x ] a b ( ) F x ] a b ) a b f 在( , 内有界。 ) a b (2) 因为 在[ , 上连续，从而 在 ( ) F x ] a b ( ) F x [ , ] a b 内取得最大值 0 ( ) M F x 。 如 0 ( ) ( ) F x f ，即为f 的最大值点。则得证。 如 0 ( ) ( ) F x f ，则 0 ( ) ( ) ( ) F x F f max ( ), ( ) F a F b ， 说明 0 , x a b ， 的最大值在 上达到，即 ( ) F x ( , ) a b f 的最大值在 上达到。 ( , ) a b （3） 在[ , 上一致连续，从而 ( ) F x ] a b f 在( , 上一致连续。 ) a b 中国矿业大学数学学院 15 华师大数学分析(第五版)讲义 第四章 函数的连续性 §3 初等函数的连续性 我们已经证得下面函数在其定义域上都是连续的： y c sin , cos , tan , cot y x y x y x y x arcsin , arccos , arctan , arccot y x y x y x y x 定理1 指数函数 在R 上是连续的. ( 0, 1 x y a a a ) 证 先设 ．由第三章知 1 a 0 0 1 lim a a x x , 这表明 在 连续．现任取 x a 0 x 0 x R ． 0 0 0 0 ) ( x x x x x x x a a a a 令 ，则当 时有 ，从而有 0 x x t 0 x x 0 t 0 0 0 0 0 0 0 lim lim lim x t t x x x x x x x x x a a a a a a . 这就证明了 在任一点 连续． x a 0 x 当 时，令 1 0 a a b 1 ，则有 ，而 1 b x x x b b a ) 1 ( 可看作函数 与 u b x u 的复合，所以此时 亦在 上连续． x a R 推论1 对数函数 loga y x 在其定义域 ) , 0 ( 上连续． 推论2 幂函数 在其定义域 x y ) , 0 ( 上连续． 由 ln x y x e 便知。 例1 设 ， 0 ) ( lim 0 a x u x x b x v x x ) ( lim 0 ．证明 b x v x x a x u ) ( ) ( lim 0 . 证 补充定义 b x v a x u ) ( , ) ( 0 ， 则 在点 连续， 从而 在 连 ) ( ), ( x v x u 0 x ) ( ln ) ( x u x v 0 x 中国矿业大学数学学院 16 华师大数学分析(第五版)讲义 第四章 函数的连续性 续，所以 在 连续．由此得 b a b x u x v x v a e e x u ln ) ( ln ) ( ) ( ) ( x x x v x x x u ) ( 0 0 lim ) ( lim (1 ) ( 0) x x x 0 x x v e ) ( b a b x u a e ln ) ( ln . 定理2 一切基本初等函数都是其定义域上的连续函数． 定理3 任何初等函数都是在其定义区间上的连续函数． 例2 证明ln 中国矿业大学数学学院 17 1 1 0 0 0 ln(1 ) lim limln(1 ) ln lim(1 ) ln 1 x x x x x x x x e x ) 例3 证明 1 ( 0 x e x x 令 1, 0 0 x y e x y 0 0 im ln 1 lim l 1 (1 ) x x y e y x y 例4 证明(1 ) 1 ( 0, x x x 0) ) 令 ， (1 ) 1 y x ln(1 ) ln(1 x y ， 0 0 x y 0 0 (1 ) 1 ln(1 ) lim lim 1 ln(1 ) x x x x y x x y |
9106 | https://www.reddit.com/r/AskPhysics/comments/18qa30m/when_is_the_acceleration_due_to_gravity/?tl=es-419 | ¿Cuándo la aceleración debida a la gravedad es positiva/negativa en mecánica? : r/AskPhysics
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httpshassan
¿Cuándo la aceleración debida a la gravedad es positiva/negativa en mecánica?
O sea, mi profe dijo que la gravedad normalmente es -9.8 m/s² porque va hacia abajo, pero un libro que estoy leyendo dice que es positiva, y lo recalca un montón.
Estoy re confundida con esto, ¿siempre es positiva o negativa, o depende de la situación?
Lo que yo pensaba era que si tirás un objeto hacia abajo, es positiva, pero si lo tirás en cualquier ángulo por encima de la horizontal, es negativa. ¿Estoy en lo cierto?
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9107 | https://courses.lumenlearning.com/wm-biology1/chapter/prokaryotic-translation/ | Prokaryotic Translation
Learning Outcomes
Understand the basics of prokaryotic translation and how it differs from eukaryotic translation
As with mRNA synthesis, protein synthesis can be divided into three phases: initiation, elongation, and termination. The process of translation is similar in prokaryotes and eukaryotes. Here we’ll explore how translation occurs in E. coli, a representative prokaryote, and specify any differences between prokaryotic and eukaryotic translation.
Initiation
Protein synthesis begins with the formation of an initiation complex. In E. coli, this complex involves the small 30S ribosome, the mRNA template, three initiation factors (IFs; IF-1, IF-2, and IF-3), and a special initiator tRNA, called tRNAMetf.
In E. coli mRNA, a sequence upstream of the first AUG codon, called the Shine-Dalgarno sequence (AGGAGG), interacts with the rRNA molecules that compose the ribosome. This interaction anchors the 30S ribosomal subunit at the correct location on the mRNA template. Guanosine triphosphate (GTP), which is a purine nucleotide triphosphate, acts as an energy source during translation—both at the start of elongation and during the ribosome’s translocation. Binding of the mRNA to the 30S ribosome also requires IF-III.
The initiator tRNA then interacts with the start codon AUG (or rarely, GUG). This tRNA carries the amino acid methionine, which is formylated after its attachment to the tRNA. The formylation creates a “faux” peptide bond between the formyl carboxyl group and the amino group of the methionine. Binding of the fMet-tRNAMetf is mediated by the initiation factor IF-2. The fMet begins every polypeptide chain synthesized by E. coli, but it is usually removed after translation is complete. When an in-frame AUG is encountered during translation elongation, a non-formylated methionine is inserted by a regular Met-tRNAMet. After the formation of the initiation complex, the 30S ribosomal subunit is joined by the 50S subunit to form the translation complex. In eukaryotes, a similar initiation complex forms, comprising mRNA, the 40S small ribosomal subunit, eukaryotic IFs, and nucleoside triphosphates (GTP and ATP). The methionine on the charged initiator tRNA, called Met-tRNAi, is not formylated. However, Met-tRNAi is distinct from other Met-tRNAs in that it can bind IFs.
Instead of depositing at the Shine-Dalgarno sequence, the eukaryotic initiation complex recognizes the 7-methylguanosine cap at the 5′ end of the mRNA. A cap-binding protein (CBP) and several other IFs assist the movement of the ribosome to the 5′ cap. Once at the cap, the initiation complex tracks along the mRNA in the 5′ to 3′ direction, searching for the AUG start codon. Many eukaryotic mRNAs are translated from the first AUG, but this is not always the case. According to Kozak’s rules, the nucleotides around the AUG indicate whether it is the correct start codon. Kozak’s rules state that the following consensus sequence must appear around the AUG of vertebrate genes: 5′-gccRccAUGG-3′. The R (for purine) indicates a site that can be either A or G, but cannot be C or U. Essentially, the closer the sequence is to this consensus, the higher the efficiency of translation.
Once the appropriate AUG is identified, the other proteins and CBP dissociate, and the 60S subunit binds to the complex of Met-tRNAi, mRNA, and the 40S subunit. This step completes the initiation of translation in eukaryotes.
In eukaryotes, initiation complex formation is similar, with the following differences:
The initiator tRNA is a different specialized tRNA carrying methionine, called Met-tRNAi
Instead of binding to the mRNA at the Shine-Dalgarno sequence, the eukaryotic initiation complex recognizes the 5′ cap of the eukaryotic mRNA, then tracks along the mRNA in the 5′ to 3′ direction until the AUG start codon is recognized. At this point, the 60S subunit binds to the complex of Met-tRNAi, mRNA, and the 40S subunit.
Figure 1. Translation in bacteria begins with the formation of the initiation complex, which includes the small ribosomal subunit, the mRNA, the initiator tRNA carrying N-formyl-methionine, and initiation factors. Then the 50S subunit binds, forming an intact ribosome.
Elongation
Figure 2. Translation begins when an initiator tRNA anticodon recognizes a start codon on mRNA bound to a small ribosomal subunit. The large ribosomal subunit joins the small subunit, and a second tRNA is recruited. As the mRNA moves relative to the ribosome, successive tRNAs move through the ribosome and the polypeptide chain is formed. Entry of a release factor into the A site terminates translation and the components dissociate.
In prokaryotes and eukaryotes, the basics of elongation are the same, so we will review elongation from the perspective of E. coli. When the translation complex is formed, the tRNA binding region of the ribosome consists of three compartments. The A (aminoacyl) site binds incoming charged aminoacyl tRNAs. The P (peptidyl) site binds charged tRNAs carrying amino acids that have formed peptide bonds with the growing polypeptide chain but have not yet dissociated from their corresponding tRNA. The E (exit) site releases dissociated tRNAs so that they can be recharged with free amino acids. The initiating methionyl-tRNA, however, occupies the P site at the beginning of the elongation phase of translation in both prokaryotes and eukaryotes.
During translation elongation, the mRNA template provides tRNA binding specificity. As the ribosome moves along the mRNA, each mRNA codon comes into register, and specific binding with the corresponding charged tRNA anticodon is ensured. If mRNA were not present in the elongation complex, the ribosome would bind tRNAs nonspecifically and randomly.
Elongation proceeds with charged tRNAs sequentially entering and leaving the ribosome as each new amino acid is added to the polypeptide chain. Movement of a tRNA from A to P to E site is induced by conformational changes that advance the ribosome by three bases in the 3′ direction. The energy for each step along the ribosome is donated by elongation factors that hydrolyze GTP. GTP energy is required both for the binding of a new aminoacyl-tRNA to the A site and for its translocation to the P site after formation of the peptide bond. Peptide bonds form between the amino group of the amino acid attached to the A-site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the 50S ribosomal subunit. The energy for each peptide bond formation is derived from the high-energy bond linking each amino acid to its tRNA. After peptide bond formation, the A-site tRNA that now holds the growing peptide chain moves to the P site, and the P-site tRNA that is now empty moves to the E site and is expelled from the ribosome (Figure 2). Amazingly, the E. colitranslation apparatus takes only 0.05 seconds to add each amino acid, meaning that a 200-amino-acid protein can be translated in just 10 seconds.
Termination
Termination of translation occurs when a nonsense codon (UAA, UAG, or UGA) is encountered. Upon aligning with the A site, these nonsense codons are recognized by protein release factors that resemble tRNAs. The releasing factors in both prokaryotes and eukaryotes instruct peptidyl transferase to add a water molecule to the carboxyl end of the P-site amino acid. This reaction forces the P-site amino acid to detach from its tRNA, and the newly made protein is released. The small and large ribosomal subunits dissociate from the mRNA and from each other; they are recruited almost immediately into another translation initiation complex. After many ribosomes have completed translation, the mRNA is degraded so the nucleotides can be reused in another transcription reaction.
In summary, there are several key features that distinguish prokaryotic gene expression from that seen in eukaryotes. These are illustrated in Figure 3 and listed in Table 1.
Figure 3. (a) In prokaryotes, the processes of transcription and translation occur simultaneously in the cytoplasm, allowing for a rapid cellular response to an environmental cue. (b) In eukaryotes, transcription is localized to the nucleus and translation is localized to the cytoplasm, separating these processes and necessitating RNA processing for stability.
| Table 1. Comparison of Translation in Bacteria Versus Eukaryotes | | |
---
| Property | Bacteria | Eukaryotes |
| Ribosomes | 70S 30S (small subunit) with 16S rNA subunit 50S (large subunit) with 5S and 23S rRNA subunits | 80S 40S (small subunit) with 18S rRNA subunit 60S (large subunit) with 5S, 5.8S, and 28S rRNA subunits |
| Amino acid carried by initiator tRNA | fMet | Met |
| Shine-Dalgarno sequence in mRNA | Present | Absent |
| Simultaneous transcription and translation | Yes | No |
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9108 | https://www.wordreference.com/definition/corpulent | WordReference.com | Online Language Dictionaries
English Dictionary | corpulent
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| + See Also: + - corporator - corporeal - corporeity - corps - corps de ballet - corps diplomatique - Corps of Engineers - corpse - corpsman - corpulence - corpulent - corpus - corpus callosum - Corpus Christi - Corpus Christi Bay - corpus delicti - corpus juris - Corpus Juris Civilis - corpus luteum - corpus striatum - corpuscle + Recent searches: + View All | corpulent [links] UK:UK and possibly other pronunciationsUK and possibly other pronunciations/ˈkɔːrpjʊlənt/US:USA pronunciation: respellingUSA pronunciation: respelling(kôr′pyə lənt) ⓘ One or more forum threads is an exact match of your searched term in Spanish | in French | in Italian | English synonyms | English Usage | Conjugator | in context | images "images") WordReference Random House Unabridged Dictionary of American English © 2025 cor•pu•lent (kôr′pyə lənt),USA pronunciation adj. 1. large or bulky of body; portly; stout; fat. Latin corpulentus, equivalent. to corp(us) body + -ulentus -ulent Middle English 1350–1400 cor′pu•lent•ly, adv. Collins Concise English Dictionary © HarperCollins Publishers:: corpulent /ˈkɔːpjʊlənt/ adj 1. physically bulky; fat Etymology: 14th Century: from Latin corpulentus fleshyˈcorpulence, ˈcorpulency n 'corpulent' also found in these entries (note: many are not synonyms or translations): obese - portly - stout - ventricose - corpulence - fat - overstuffed - pursy - rotund - -ulent Synonyms: overweight, fleshy, beefy, fat, pudgy, more... 🗣️Forum discussions with the word(s) "corpulent" in the title: corpulent Visit the English Only Forum.Help WordReference: Ask in the forums yourself. Look up "corpulent" at Merriam-Webster Look up "corpulent" at dictionary.com Go to Preferences page and choose from different actions for taps or mouse clicks. In other languages: Spanish | French | Italian | Portuguese | Romanian | German | Dutch | Swedish | Russian | Polish | Czech | Greek | Turkish | Chinese | Japanese | Korean | Arabic Links: ⚙️Preferences | Abbreviations | Pron. Symbols | Privacy Policy | Terms of Service | Support WR | Forums | Suggestions | | | | Advertisements | | | | Advertisements | | | | Report an inappropriate ad. | | WordReference.com WORD OF THE DAY travel | rack GET THE DAILY EMAIL! | | | | Become a WordReference Supporter to view the site ad-free. | | Chrome users: Use search shortcuts for the fastest search of WordReference. | |
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9109 | https://testbook.com/question-answer/successive-discounts-of-a-and-b-are-equivalent-t--61c4257532eee36ad7d99a01 | [Solved] Successive discounts of a% and b% are equivalent to a single
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Profit and Loss
Discount and MP
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Successive discounts of a% and b% are equivalent to a single discount of:
This question was previously asked in
CG TET 2016 Paper 2 (Maths & Science)
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(a+b 2)%
(a + b)%
(a+b−a b 100)%
(a+b+a b 100)%
Answer (Detailed Solution Below)
Option 3 : (a+b−a b 100)%
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Detailed Solution
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Given:
Discount 1 = a%
Discount 2 = b%
Formula Used:
Successive Discount = Discount 1 + Discount 2 -D i s c o u n t 1×D i s c o u n t 2 100
Calculation:
Successive discount= a+b−a b 100
The correct option is 3 i.e.a+b−a b 100
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More Discount and MP Questions
Q1.A person purchases a two item from a shop. The item I is bought at Rs. 350. The item II is bought at Rs. 360. For both items, He marks up the price by 60% over the cost price. Difference between marked price of two item is y. Then, he offers a [y -6] % discount on the marked price of item I and [y + 4] %. Let the total profit earned from both items be Rs. x. Based on this information, determine which of the following statements is correct: I. The difference between the selling prices of the two items is Rs. [2y +10.2] II. The profit made on the item bought from the first shop is Rs. [x - 100] III. If selling price item II is Rs. 432 then discount amount is Rs. [x – y]
Q2.A shopkeeper offers two successive discounts on a certain item. The second discount is 20% of the first. If the overall discount on the item is p% of the marked price, and the selling price after applying the first discount is Rs y, the marked price of the item is Rs x. Which one of the following is correct in respect of the question and the statements given below? Statement 1: The values of p,y, and x are respectively given as 18, 9600, and 12000. Statement 2: The values of p,y, and x are respectively given as 22, 9500, and 12500.
Q3.Find a single discount (rounded up to two decimal places) equal to three consecutive discounts of 13%, 20% and 4%.
Q4.A seller marks their goods 28% above the cost price. On customer demand, they offer a 25% reduction on the bills. What percentage of profit or loss does the seller make?
Q5.Find a single discount (rounded up to two decimal places) equal to three consecutive discounts of 3%, 27% and 23%.
Q6.The marked price of a laptop is ₹750. It sells with two successive discounts of 4% and 25%. An additional 25% discount is given on cash payment. What is the selling price (in ₹) of the laptop on cash payment?
Q7.A dealer marks his goods at 5% above the cost price and allows a discount of 67% on the marked price. What is his gain or loss percentage (correct to two decimal places)?
Q8.Neeraj sells sarees for ₹3,944 each, after giving two successive discounts of 15% and 20% on its marked price. If he sells a saree giving no discount on its marked price, then he earns a profit of 45%. If he sells a saree for ₹4,496, then his profit percentage will be ______
Q9.Naman sells an item at Rs x, after giving two successive discounts each of 10% on its marked price. If he does not give any discount on the marked price, he earns a profit of 25%. The cost price of the item is ₹800. What is the value of x ?
Q10.An article is marked x% (0 < x < 40) above its cost price. It is sold by giving x 2% discount on its marked price. If there is a profit of 10 1 2% then what is the value of x ?
More Profit and Loss Questions
Q1.A and B start a business with Rs. 16000 and Rs. 20000. After x months B left the business. After one year total profit is Rs. 3300 and share of A is Rs. 1800. Find the value of x?
Q2.On a particular day, each of Alok, Sameer, and Amit sold three types of gadgets in their respective shops. Alok and Sameer sold an identical number of Type X gadgets while Sameer sold twice as many of Type Y gadgets as Alok. The ratio of the number of Type Z gadgets sold by Alok, Sameer, and Amit was 4:6:5, respectively. Each of the three sellers sold an identical number of Type X gadgets at different prices per unit. Assertion (A): It is possible that Alok sold each Type X gadget at a profit of Rs 5, each Type Y gadget at a profit of Rs 3, and each Type Z gadget at a loss of Rs 2, and made an overall profit of Rs 150; Sameer sold each Type X gadget at a profit of Rs 4, each Type Y gadget at a profit of Rs 2, and each Type Z gadget at a loss of Rs 1, and made an overall profit of Rs 200; and Amit sold each Type X gadget at a profit of Rs 3, each Type Y gadget at a loss of Rs 2, and each Type Z gadget at a profit of Rs 5, and made an overall profit of Rs 180. Reason (R): Framing and s
Q3.A Question is given below followed by two Statements I and II. Question: Arjun works in an Apple store. What profit did Arjun make by selling a laptop for Rs. 70500? Statement-I: Cost price of 10 laptops is equal to the selling price of 8 laptops. Statement-II: 25% profit is earned by selling each laptop. Which one of the following is correct in respect of the above Question and the Statements?
Q4.When an article is sold for Rs 600 loss incurred is 33.33% more than the profit earned on selling it at Rs 900. What is the selling price when it earned a profit of 16.66%?
Q5.A person purchases a two item from a shop. The item I is bought at Rs. 350. The item II is bought at Rs. 360. For both items, He marks up the price by 60% over the cost price. Difference between marked price of two item is y. Then, he offers a [y -6] % discount on the marked price of item I and [y + 4] %. Let the total profit earned from both items be Rs. x. Based on this information, determine which of the following statements is correct: I. The difference between the selling prices of the two items is Rs. [2y +10.2] II. The profit made on the item bought from the first shop is Rs. [x - 100] III. If selling price item II is Rs. 432 then discount amount is Rs. [x – y]
Q6.A shopkeeper offers two successive discounts on a certain item. The second discount is 20% of the first. If the overall discount on the item is p% of the marked price, and the selling price after applying the first discount is Rs y, the marked price of the item is Rs x. Which one of the following is correct in respect of the question and the statements given below? Statement 1: The values of p,y, and x are respectively given as 18, 9600, and 12000. Statement 2: The values of p,y, and x are respectively given as 22, 9500, and 12500.
Q7.On a particular day, each of Rahul, Priya, and Samir sold three types of bags from their respective shops. Rahul and Priya sold an identical number of bags of Type X, while Samir sold 2/3 times as many bags of Type X as Rahul and Priya together sold. The ratio of the numbers of bags of Type Y sold by Rahul, Priya, and Samir was 5:4:2 respectively, and Rahul and Samir sold an identical number of bags of Type Z each, while Priya sold 3/2 as many bags of Type Z as each of Rahul and Samir sold. The three sellers sold each of the types of bags at different prices per unit. Assertion (A): It is possible that Rahul sold each bag of Type X at a profit of Rs 3, each bag of Type Y at a loss of Rs 2, and each bag of Type Z at a profit of Rs 4, and made an overall profit of Rs 120; Priya sold each bag of Type X at a loss of Rs 5, each bag of Type Y at a profit of Rs 6, and each bag of Type Z at a loss of Rs 3, and made an overall profit of Rs 180; and Samir sold each bag of Type X at a profit of
Q8.Find a single discount (rounded up to two decimal places) equal to three consecutive discounts of 13%, 20% and 4%.
Q9.Pankaj, Meera, and Ashok invest ₹41,000, ₹39,000, and ₹84,000 respectively to start a business. At the end of the year, the total profit earned is ₹49,200. 29% of the total profit earned is given to charity and the rest is divided among them in the ratio of their investments. What will be the share of Ashok (in ₹)?
Q10.The selling price of 34 books is equal to the cost price of 17 books. Find the loss or gain percentage.
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Q7.On a particular day, each of Alok, Sameer, and Amit sold three types of gadgets in their respective shops. Alok and Sameer sold an identical number of Type X gadgets while Sameer sold twice as many of Type Y gadgets as Alok. The ratio of the number of Type Z gadgets sold by Alok, Sameer, and Amit was 4:6:5, respectively. Each of the three sellers sold an identical number of Type X gadgets at different prices per unit. Assertion (A): It is possible that Alok sold each Type X gadget at a profit of Rs 5, each Type Y gadget at a profit of Rs 3, and each Type Z gadget at a loss of Rs 2, and made an overall profit of Rs 150; Sameer sold each Type X gadget at a profit of Rs 4, each Type Y gadget at a profit of Rs 2, and each Type Z gadget at a loss of Rs 1, and made an overall profit of Rs 200; and Amit sold each Type X gadget at a profit of Rs 3, each Type Y gadget at a loss of Rs 2, and each Type Z gadget at a profit of Rs 5, and made an overall profit of Rs 180. Reason (R): Framing and s
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9110 | https://pubmed.ncbi.nlm.nih.gov/11057979/ | Transmission of the nocturnal periodic strain of Wuchereria bancrofti by Culex quinquefasciatus: establishing the potential for urban filariasis in Thailand - PubMed
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. 2000 Aug;125(1):207-12.
doi: 10.1017/s0950268899004355.
Transmission of the nocturnal periodic strain of Wuchereria bancrofti by Culex quinquefasciatus: establishing the potential for urban filariasis in Thailand
S Triteeraprapab1,K Kanjanopas,S Suwannadabba,S Sangprakarn,Y Poovorawan,A L Scott
Affiliations Expand
Affiliation
1 Department of Parasitology, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand.
PMID: 11057979
PMCID: PMC2869589
DOI: 10.1017/s0950268899004355
Item in Clipboard
Transmission of the nocturnal periodic strain of Wuchereria bancrofti by Culex quinquefasciatus: establishing the potential for urban filariasis in Thailand
S Triteeraprapab et al. Epidemiol Infect.2000 Aug.
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. 2000 Aug;125(1):207-12.
doi: 10.1017/s0950268899004355.
Authors
S Triteeraprapab1,K Kanjanopas,S Suwannadabba,S Sangprakarn,Y Poovorawan,A L Scott
Affiliation
1 Department of Parasitology, Faculty of Medicine, Chulalongkorn University, Bangkok, Thailand.
PMID: 11057979
PMCID: PMC2869589
DOI: 10.1017/s0950268899004355
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Abstract
Control programmes have reduced the prevalence of Bancroftian filariasis in Thailand to low levels. Recently, there has been an influx of more than one million Myanmar immigrants into urban centres of Thailand. The prevalence of patent Wuchereria bancrofti infection in these immigrants (2-5%) has prompted concern in the public health community that the potential now exists for a re-emergence of Bancroftian filariasis in Thailand. It is possible that an urban cycle of transmission could become established. The Myanmar immigrants are infected with the nocturnal periodic (urban) type W. bancrofti for which Culex quinquefasciatus serves as the main vector. The Thai strains of Cx. quinquefasciatus have never been reported to transmit Bancroftian filariasis. Our results of feeding experiments demonstrated that the Thai Cx. quinquefasciatus are permissive for the development of Myanmar W. bancrofti to infective third-stage larvae thus establishing the potential for establishing an urban cycle of transmission in Thailand. We also adapted the SspI repeat PCR assay for the identification of infective mosquitoes that was capable of detecting a single infective stage larvae in a pool of 100 mosquitoes.
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9111 | https://math.stackexchange.com/questions/2894484/rational-functions-reciprocal-of-linear-and-quadratic-functions-center-of-curv | differential geometry - Rational functions: reciprocal of linear and quadratic functions: center of curvature - Mathematics Stack Exchange
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Rational functions: reciprocal of linear and quadratic functions: center of curvature
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When one graphs rational functions in Pre-Calculus type course, one usually graphs functions that are reciprocals of linear functions and reciprocals of quadratics. One of the other properties that maybe asked is to find the invariant points. Sometimes the invariant points correspond to the center of curvature of the arms of the hyperbolas that are created from these functions.
Is there a way to determine the center of curvature for the hyperbola arms of these rational functions without using Calculus based methods. When i say center of curvature that means i am looking for the coordinates(x,y) of the center. To make this more clear I have added an image i generated:
As you can see, I put the quadratic as well to get intersection with the reciprocal which gives the invariant points. I thought I would try to add a line that connect the peak of the bottom quadratic like parabola with vertical asymptote, but it went lower than the center of curvature of the hyperbolic arm. (The line is in green color)
I just looked at the graph and decided to try to put a line thru the asymptote at x =2 to with a slope of 1, and it looks like its going thru the center of curvature of the hyperbolic arm. See attached image, the new line is in orange color:
Also wish to add further analogy to the actual conic section Hyperbola, in this the center is called the Vertex of the Hyperbola. See attached image:
I also want to note something I just noticed, that the line that goes thru the hyperbolic arms of this reciprocal quadratic, cuts the arms in a very symmetrical way such that the line acts like a mirror, cutting it symmetrically in half.
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edited Aug 26, 2018 at 4:51
PaluPalu
asked Aug 25, 2018 at 20:54
PaluPalu
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y=1/(x 2−4)y=1/(x 2−4) is not a hyperbola. What is your definition for center of curvature?dxiv –dxiv 2018-08-26 01:09:25 +00:00 Commented Aug 26, 2018 at 1:09
Yes, I know, I made that clear that its not a hyperbola. I said this rational has hyperbolic arms. It is well known from conic sections that these types of functions have a center/vertex like structure. I know that in differential geometry they would discuss the concept of curvature, which the greatest rate of change on a section of the curve. I had friends in university who took course on differential geometry, and tried to explain these things to me. I took other maths like complex analysis, I wish i had taken differential geometry.Palu –Palu 2018-08-26 04:22:43 +00:00 Commented Aug 26, 2018 at 4:22
I am trying to visualize the center of curvature thru the graphs, i believe its obvious. For a more rigorous definition, I need help from people on this forum, because if i knew the rigorous proper definition, I would know how to do this.Palu –Palu 2018-08-26 04:35:22 +00:00 Commented Aug 26, 2018 at 4:35
1 The hyperbola has an axis of symmetry, and its vertices have a clear-cut definition. Your curve has neither. The curvature is well-defined point-wise, but it's not clear what you mean by "center of curvature" for an entire curve here, or even if it's related. For a more rigorous definition, I need help from people on this forum But you first need to spell out what exactly it is that you are after.dxiv –dxiv 2018-08-26 05:23:18 +00:00 Commented Aug 26, 2018 at 5:23
1 You seem to want to identify the pt where one "arm" of your curve-of-interest ends and the other arm begins. The lack of symmetry in the arms makes this tricky, but one possible definition is "the point of maximal curvature". "Curavature", here, is formally defined, via Calculus, as the reciprocal of the radius of the "osculating circle", which best approximates the curve (very!) near the point in question. (BTW: the term "center of curvature" is reserved for the center of that circle; you'll want to choose something else for your notion.)Blue –Blue 2018-08-26 07:52:54 +00:00 Commented Aug 26, 2018 at 7:52
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9112 | https://math.stackexchange.com/questions/105231/show-that-there-is-no-affine-transformation-that-takes-a-circle-to-a-hyperbola-i | Stack Exchange Network
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Show that there is no affine transformation that takes a circle to a hyperbola in $\mathbb{R}^2$
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$\begingroup$
"Show that the standard circle (defined by $f(x,y) = x^2 + y^2 - 1$) is not equivalent to the standard hyperbola (defined by $g(x,y) = x^2 - y^2 - 1$). That is, show that there is no $[A,\overline{s}] \in \text{Aff}(\mathbb{R}^2)$ such that $[A,\overline{s}] \cdot f(x,y) = g(x,y)$. Check that there is such an $[A,\overline{s}]$ if we allow $A \in \text{GL}_2(\mathbb{C}).$"
I've reduced this to showing that there are no $a,b,c,d,s,t \in \mathbb{R}$ such that $$f(ax+as+by+bt,\: cx + cs + dy+dt) = g(x,y).$$ $$\Rightarrow(ax+as+by+bt)^2+(cx + cs + dy+dt)^2 - 1=x^2-y^2-1$$ How should I proceed? Expanding that expression probably isn't the best way to do it.
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edited May 29, 2017 at 7:02
Martin Sleziak
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asked Feb 3, 2012 at 6:09
WillWill
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$\endgroup$
8
$\begingroup$ Let $y\to\infty$. $\endgroup$
Jonas Meyer
– Jonas Meyer
2012-02-03 06:12:57 +00:00
Commented Feb 3, 2012 at 6:12
$\begingroup$ @Jonas Meyer I'm not following. Would you explain that further? $\endgroup$
Will
– Will
2012-02-03 06:17:32 +00:00
Commented Feb 3, 2012 at 6:17
$\begingroup$ Fix $x$. The limit of the right-hand side as $y\to\infty$ is $-\infty$, but the same is not true for the left-hand side. $\endgroup$
Jonas Meyer
– Jonas Meyer
2012-02-03 06:24:47 +00:00
Commented Feb 3, 2012 at 6:24
3
$\begingroup$ «It seems daunting»?! Have you tried? I hate when my students say that kind of things to me... :/ $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 06:29:24 +00:00
Commented Feb 3, 2012 at 6:29
1
$\begingroup$ @Mariano: Endurance and persistence seem to be rather uncommon traits with young ones these days, I'm told. $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2012-02-03 06:31:19 +00:00
Commented Feb 3, 2012 at 6:31
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4 Answers 4
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$\begingroup$
In $\mathbb R^2$, a circle is a bounded set and a hyperbola is not.
An invertible affine transformation of the plane maps bounded sets to bounded sets.
(One can replace «bounded» by «compact» or «connected»...)
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edited Feb 3, 2012 at 7:09
answered Feb 3, 2012 at 6:23
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
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$\endgroup$
5
$\begingroup$ An enlightening question is: why does this break over the complex numbers? $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 06:24:32 +00:00
Commented Feb 3, 2012 at 6:24
$\begingroup$ Yes, but I was hoping for a way that doesn't use that fact (we haven't proved it in class yet). My apologies for not making that clear in the question. $\endgroup$
Will
– Will
2012-02-03 06:46:16 +00:00
Commented Feb 3, 2012 at 6:46
$\begingroup$ What haven't you proved in class? In any case, my two bullet points are simply suggestions for you to actually prove them, so if you did not cover them in class it does not matter :D $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 06:51:35 +00:00
Commented Feb 3, 2012 at 6:51
$\begingroup$ We have not proved the second statement. My mistake for interpreting your two bullet points as an answer to the question. $\endgroup$
Will
– Will
2012-02-03 06:53:27 +00:00
Commented Feb 3, 2012 at 6:53
$\begingroup$ I also did not prove Sylvester's Law of Inertia! :P $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 07:02:30 +00:00
Commented Feb 3, 2012 at 7:02
Add a comment |
6
$\begingroup$
The other Mariano is of course cheating: this is algebraic geometry after all...
The gentlemanly way of doing this is to observe how the coefficients of the homogeneous part of top degree changes when you do an affine transformation, and then invoke Sylvester's Law of Inertia.
In more detail... Let $f\in \mathbb R[x,y]$ be a polynomial of degree two, which we can write uniquely as $$f(x,y) = v^tav+b^tv+c$$ with $v$ the vector $\left(\begin{smallmatrix}x\y\end{smallmatrix}\right)$, $a$ a symmetric $2\times 2$ matrix, $b\in\mathbb R^2$ and $c\in\mathbb R$.
If $[A,\bar s]$ is an affine transformation, then $$[A,\bar s]\cdot f=v^ta'v+b'^tv+c'$$ for some new $a'$, $b'$ and $c'$ which we can compute explicitly. Most interestingly, we have $$a'=A^taA.$$ Sylvester's Law of Inertia then implies that $a$ and $a'$ have the same number of positive eigenvalues, the same number of negative eigenvalues, and the same number of zero eigenvalues.
Now, the matrix $a$ corresponding to your $f$ is $\begin{pmatrix}1&0\0&1\end{pmatrix}$, while that of $g$ is $\begin{pmatrix}1&0\0&-1\end{pmatrix}$. Since they have different numbers of negative eigenvalues, there is no affine transformation mapping one to the other.
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edited Feb 3, 2012 at 6:49
J. M. ain't a mathematician
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answered Feb 3, 2012 at 6:38
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
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$\endgroup$
2
2
$\begingroup$ I had to double-take when you said "the other Mariano." $\endgroup$
anon
– anon
2012-02-03 06:56:43 +00:00
Commented Feb 3, 2012 at 6:56
2
$\begingroup$ @anon: there are way too many Marianos in this thread... ;) $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2012-02-03 07:22:10 +00:00
Commented Feb 3, 2012 at 7:22
Add a comment |
5
$\begingroup$
Why think when you can simply compute?
So $f(x,y)=x^2+y^2-1$. Consider the affine transformation given by $$\left{\begin{array}{l}x\leftarrow ax+by+s\y\leftarrow cx+dy+t\end{array}\right.$$ Applying it to $f$ we get $$f(ax+by+s, cx+dy+t)=(a^2+c^2) x^2+2(a b + c d)xy+(b^2 + d^2)y^2+\text{terms of lower degree}.$$ If this is to be equal to $g(x,y)=x^2-y^2-1$, then, in particular, looking at the coefficient of $y^2$ we see that we must have $$b^2+d^2=-1.$$ Of course, this is not going to work...
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answered Feb 3, 2012 at 7:00
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
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$\endgroup$
2
$\begingroup$ I would say that for this particular problem, expanding is the best way of doing it :D $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 07:06:09 +00:00
Commented Feb 3, 2012 at 7:06
$\begingroup$ Agreed. When looking at the expanded form I failed to notice the significance of the $(b^2 + d^2)y^2$ term. $\endgroup$
Will
– Will
2012-02-03 18:27:52 +00:00
Commented Feb 3, 2012 at 18:27
Add a comment |
2
$\begingroup$
Let us call $f_0$, $f_1$ and $f_2$ the parts of $f$ which are homogeneous of degree $0$, $1$ and $2$, respectively, so that in particular $f=f_2+f_1+f_0$, and similarly for $g$.
If $[A,s]$ is an invertible affine transformation and $[A,s]\cdot f=g$, then one checks easily that $[A,s]\cdot f_i=g_i$ for each $i\in{0,1,2}$. In particular, $$[A,s]\cdot(x^2+y^2)=x^2-y^2.$$ But two polynomials which are affinely equivalent are either both irreducible or both reducible, yet $x^2+y^2$ is irreducible and $x^2-y^2$ is not.
This argument is «geometric»: we are using the fact that a circle has no asymptotic directions while the hyperbola has two, and that the property of having asymtotic directions is preserved under affine equivalence.
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answered Feb 3, 2012 at 7:17
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
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$\endgroup$
1
$\begingroup$ This has the added benefit of almost telling us what to do in the complex case: since an affine equivalence must take asymptotic directions to asymptotic directions, the matrix of the equivalence is determined. $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2012-02-03 07:21:24 +00:00
Commented Feb 3, 2012 at 7:21
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Imaginary Eigenvalue
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Imaginary eigenvalues refer to the eigenvalues of a matrix that have a non-zero imaginary component, indicating that the matrix exhibits certain dynamic behaviors. In computational settings, determining the presence of imaginary eigenvalues in the associated matrix can be complex and is critical for iterative algorithms.
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Chapter
Water-Waves as a Spatial Dynamical System
2003, Handbook of Mathematical Fluid DynamicsFrèdèric Dias, Gèrard Iooss
4.2 The case of one layer with surface tension
Let us consider the system (1), (4). The eigenvalues of the linear operatorL μ (for c 1 = 0) are given by σ = ik, with k satisfying (12). and Figure 5 gives the position of the four closest eigenvalues to the imaginary axis. For the study of solutions of (1) near 0, there are three main “interesting” cases to be considered:
(i)
(b, λ) is near Δ+ = {λ = 1, b> 1/3}: in this case, L μ 0 has only a double 0 eigenvalue on the imaginary axis (0 2+ resonance),
(ii)
(b, λ) is near Δ− = {λ = 1, 0 <b< 1/3): in this case, L μ 0 has a double 0 eigenvalue, and a pair of simple imaginary eigenvalues ±iq on the imaginary axis (0 2+ (iq) resonance),
(iii)
(b, λ) is near Γ (left part); in this case L μ 0 has only a pair of double imaginary eigenvalues ±iq on the imaginary axis ((iq)2 resonance).
The system (1), (6) leads to cases (ii) and (iii) as well, and can be treated similarly. When (b, λ) is close to (1/3, 1) a specific study is needed, because at this point the eigenvalue 0 is quadruple (see for this case). There are other interesting cases, for instance when one has two pairs of resonating eigenvalues on the imaginary axis (the (iq)(2 iq) resonance is the most special because of the occurrence of heteroclinics between periodic solutions, see ). However, we shall not detail their study here, since it is always in the same spirit, and we restrict our presentation to the most typical cases.
4.2.1 Case (i): 0 2+ resonance.
This case was first solved by Amick and Kirchgässner . It is also studied in particular in the papers [74,101,64]. The method used in is the one we present here (however without fixing the parameter c 1 = 0). Here the center manifold is two-dimensional. Let us define by (A, B) the (real) coordinates (or “amplitudes”) associated with the choice of eigenvectors
ξ 0=(0,0,1)t,ξ 1=(−1,−(y+1),0)t.
Then, we need to know how the reversibility symmetry So acts on (A, B). There are two theoretical possibilities: (A, B) → (A, – B) or (– A, B). Here, as in all water-wave problems, the first case holds. This is the 0 2+ resonance. Then the normal form (see, for instance, ), truncated at leading orders, reads
(24){d A d x=B.d B d x=v A+a A 2+δ c 1,
where one can compute explicitly the coefficients (see, for instance. ) as functions of the parameters:
v=(b−1/3)−1(λ−1)+O(|c 1|+|λ−1|2),a=−3/2(b−1/3)−1[1+O(|λ−1|+|c 1|)],δ=(b−1/3)−1 b[1+O(|λ−1|+|c 1|)].
We notice the blow-up of the coefficients when b tends towards 1/3. due to the change of dimension of the central system at this point (it becomes 4-dimensional). Here, the two conjugate equilibria (both corresponding to a flat free surface) are denoted by A−<A+. They exist provided that c 1> –(λ – 1)2/(6 b). The equilibrium A− is hyperbolic while the equilibrium A+ is elliptic. The vector field (24) is integrable, and its phase portrait is given in Figure 10. For any fixed c 1, there is a one-parameter family of periodic solutions, and a solution homoclinic to the hyperbolic equilibrium. All these solutions disappear after the saddle–node bifurcation when c 1<−(λ−1)2/(6 b). For c 1 = 0 and λ > 1. the homoclinic solution of the truncated system is given by
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Fig. 10. Phase portrait of the 2D vector field (24) [case(i)], for c 1> -(λ – 1)2/(6 b).
(25)A(x)=λ−1 cosh╡2(v 1/2 x/2).
Because our system is two-dimensional and reversible, it is easy to show that these phase portraits fully persist for the complete system. We sum up these results in the following theorem:
Theorem 3
Assume that a two-dimensional reversible vector field has a fixed point at the origin for the parameter value 0. and assume that it has a 0 2+resonance for its linearized part. Then the phase portraits near 0 of the vector field, for fixed values of the parameter ((v. c 1) here) near 0, are genetically the same as for the normal form(24).
Corollary 4
The above theorem applies for describing travelling waves solutions of the water-wave Problem 2.1.2 (finite depth, large surface tension such that b> 1/3) for λ near 1.
We observe that A(x) >A−, hence the homoclinic solution corresponds to a “solitary wave” of depression (see Figure 11) for the problem (1), (4), whose principal part follows directly from (25).
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Fig. 11. Shape of the solitary wave of depression in the case of one layer with strong surface tension.
Remark
Fixing c 1 = 0 leads to an artificial distinction between the cases λ > 1 and λ < 1, since this is just a matter of choosing the suitable constant flow for the velocity scaling (the one which is an hyperbolic equilibrium).
4.2.2 Case (ii): 0 2+ (iq) resonance.
This case was treated in the spirit of this review in the work . Here the center manifold is four-dimensional. Let us denote by ±iq the pair of simple eigenvalues depending on b, such that q = (1 + bq 2) tanh q, and define by (A, B) the (real) amplitudes and C the complex one, corresponding to the oscillating mode. Then the reversibility symmetry S 0 acts on (A, B, C, C¯) as follows: (A, B, C, C¯) → (A, – B, C¯). This is a 0 2+ (iq) resonance. The normal form, truncated at quadratic order, reads
(26){d A d x=B,d B d x=v A+a A 2+c|C|2+δ c 1,d C d x=i C(q+v 1+d A),
where the (real) coefficients v, a, Δ are the same as for case (i), and c, v 1, d may also be explicitly computed in terms of the parameters (λ, b, c 1) (see where they are computed for c 1 = 0). We notice that a> 0 and we have
v 1=O(|λ−1|+|c 1|),d=O(1),c=(1/3−b)−1(1+sinh╡2 q q+h.o.t)>0
where (λ – 1, c 1) is close to 0. This system is indeed integrable, with the two first integrals
(27)H=|C|2,K=B 2−(2/3)a A 3−v A 2−2(c H+δ c 1)A.
Fixing c 1 = 0 to simplify the discussion, we see again on Figure 8 (after an obvious scaling) the various graphs of the functions f H.K(A)=(2/3)a A 3+ν A 2+2 c H A+K depending on (K, H), for v> 0 (left), and for v< 0 (right) (v has the sign opposite of λ – 1, since b< 1/3).
In this case, for the normal form vector field, the curves Γ h, and Γ e in the (K, H) plane correspond to families of periodic solutions, where the C component is not 0, except for H = 0, where this gives the conjugate constant flow (as above). Now, we have other types of periodic solutions and quasi-periodic solutions corresponding to the interior of the triangular region in (K, H) plane. The curves Γ h correspond to the existence of homoclinic solutions, one homoclinic to 0 for λ < I (v> 0, H = K = 0), and all others homoclinic to some periodic solution. Figure 12 gives in the (A, B) plane the phase portrait of all small bounded solutions (left side) for 0 <c H<v 2/4 a, c 1 = 0, and for H = 0 (right side). Notice that the homoclinic solution to A+ corresponds here to a generalized solitary wave for the problem (1), (4), tending at infinity towards a periodic wave. Note that A+ ~ –cH/v when |H| is very small, meaning that oscillations at infinity are then very small in this case. For H = 0 this corresponds to a solitary wave of elevation for the problem (1), (4). For λ > 1 we have analogous phase portraits where, for instance, for H = K = 0, we have a solution homoclinic to A+ ≠ 0. This limit equilibrium corresponds to the flow conjugate to 0, and might be chosen a priori as the origin (instead of the previous origin) if we change the scale c for velocities (see the discussion for case (i)). Then λ would become λ′ with the new scaling, and λ > 1 would become λ′ < 1.
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Fig. 12. Phase portraits in the (A, B) plane of the vector field (26) for c 1 = 0. v> 0 (λ < 1); left side:0 < cH< v 2/4 a, right side: H = 0.
The natural mathematical problem consists now in proving persistence results when considering the full system, not only reduced to its normal form. In summary, the persistence of periodic solutions of the normal form can in general be performed, through an adaptation of the Lyapunov–Schmidt technique [64,82]. The persistence of quasi-periodic solutions is much more delicate, and can only be performed in a subset of the 2-dimensional space of first integrals, where these solutions exist for the normal form. For a fixed value of the bifurcation parameter v, quasi-periodic solutions of the perturbed reversible vector field exist for (H, K) lying in a region which is locally the product of a line by a Cantor set (see ). The persistence of pairs of reversible solutions (invariant under the reversibility symmetry) homoclinic to periodic solutions, provided that they are not too small, is proved. for instance, in [109,64] (see Figure 13). For the normal form, there is a family of orbits homoclinic to a family of periodic solutions whose amplitude can be chosen arbitrarily small. Such a case (ii) has been investigated by many authors (see for instance [15.113, 82]). There are homoclinic solutions to oscillations at infinity whose size is smaller than any power of the bifurcation parameter, corresponding to the fact that we cannot avoid such oscillations when we consider the full untruncated system. The extremely delicate aspect of exponentially small and still existing oscillations was proved by Sun and Shen on the water-wave problem (1), (4), and is being thoroughly studied by Lombardi for a wide class of problems (including the water-wave problem) in . Moreover, despite the fact that a solution homoclinic to 0 exists for the normal form (26), this is not true in general for the full system (see ), even though one can compute an asymptotic expansion up to any order of such a homoclinic (non-existing) “solution”! (see for an extensive study of the phenomenon). The difficulty comes from the fact that there is only one unstable direction and one symmetric stable direction for the origin (v> 0). Indeed, the two-dimensional unstable manifold of a periodic orbit near 0 becomes one-dimensional when the amplitude of the oscillation vanishes. In fact, this two-dimensional unstable manifold (identical to the two-dimensional stable manifold for the normal form) intersects transversally the two-dimensional subspace invariant under the symmetry reversibility (B = 0, C real) (2 intersection points) for the 4-dimensional normal form vector field. For a large enough size of the periodic orbit for the perturbed vector field, its unstable manifold is shown to intersect transversally the plane (B = 0, C real) in two points, as for the normal form. This shows the persistence, for the full vector field, of two reversible solutions homoclinic to this closed orbit. Now, it results from that, as soon as the radius of the periodic orbit is smaller than a critical value, there is a loss of transversality for the perturbed vector field, and that the generic minimal size of the limiting oscillation is O(C(l)exp[–lq/v 1/2]), with l< π (l = π would be the optimal result here, but not yet obtained, see ). While the result on non-existence of solitary waves is generic here, there is a precise proof that there are no true solitary waves (of elevation here) near b = 1/3 (see ), and the result for b< 1/3 (not near 1/3) is not known, although a not completely rigorous analysis suggests that there are no such solutions (see [124,27]). We summarize these results in the following rough theorem:
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Fig. 13. Shape of the generalized solitary waves in case (ii).
Theorem 5
Assume that a 4-dimensional reversible vector field has a fixed point at 0 and has a 0 2+(iq) resonance for its linear part. Then, in a neighborhood of 0, near the critical value of the bifurcation parameter, the small periodic solutions of frequencies close to q are generically given by the normal form(26), and lie in a one-parameter family. There is in addition a two-parameter family of periodic and quasi-periodic solutions. Moreover, each hyperbolic periodic solution of frequency close to q has two reversible homoclinic connections, provided that the diameter of this periodic solution is larger than an exponentially small quantity.
More details may be found in .
Corollary 6
The above theorem applies for describing travelling waves of the water-wave Problem 2.1.2 for λ near 1 and b< 1/3 (small surface tension). It also applies to the problem of travelling waves 2.2 under an elastic plate(6)for λ near 1 (see the line Δ in Figure 6 and see).
4.2.3 Case (iii): (iq)2 resonance.
This case occurs for (b, λ) near the curve Γ of the parameter plane in Figure 5, and was first treated in . Here the center manifold is four-dimensional. Let us denote by ±iq the pair of double eigenvalues at criticality, and define by (A, B) the complex amplitudes corresponding respectively to the eigenmode and to the generalized eigenmode. This case is often denoted by “1:1 reversible resonance”. We can always assume that the reversibility symmetry S 0 acts as (A, B) → ( A¯, – B¯). The normal form (see, for instance, ) reads at any order (making c 1 = 0, which does not restrict the study, since λ is not close to 1):
(28)d A d x=i q A+B+i A P[v,|A|2,i/2(A B¯−A¯B)],d B d x=i q B+i B P[v,|A|2,i/2(A B¯−A¯B)]+A Q[v,|A|2,i/2(A B¯−A¯B)],
where P and Q are real polynomial of degree one in their arguments, for the cubic normal form. Let us define more precisely the coefficients of Q:
(29)Q(v,u,υ)=v+q 2 u+q 3 υ,
which means that for v> 0 the eigenvalues are at a distance √v from the imaginary axis, while for v< 0 they sit on the imaginary axis. Values v> 0 correspond to points in the plane (b, λ) above the curve Γ, and v is of the order of the distance to this curve (the precise expression of v in terms of the parameters is given in ). The vector field (28) is integrable, with the two following first integrals:
(30)K=i/2(A B¯−A¯B),H=|B|2−∫0|A|2 Q(v,u,K)d u.
It is then possible to describe all small bounded solutions of (28). Indeed, we obtain
(d|A|2 d x)2=f K,H(|A|2),
where
(31)f K,H(|A|2)=2 q 2|A|6+4(v+q 3 K)|A|4+4 H|A|2−4 K 2.
We show in Figure 14 various graphs for the functions f K,H depending on (K. H), for v> 0, q 2< 0 (right), and for v< 0, q 2> 0 (left), which are the most interesting cases. The change (q 2.H,u)↦(−q 2,−H,−u) leaves f H.K(u) unchanged. It results that the relevant graph (we need |A|2>0 of f K.H for v< 0, and q 2> 0 corresponds to the side H> 0 of the left part of Figure 14, while for v< 0, q 2< 0 we need to consider the side H< 0 of the left part of Figure 14. Notice that for q 2> 0, v> 0 there is no small bounded solution other than 0.
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Fig. 14. Different graphs of u↦f K,H(u) depending on the parameters H and K.
Looking at these graphs, where in particular any tangency to the u axis on the positive side corresponds to a periodic solution of frequency close to q, it is clear that we obtain for a fixed v two-parameter families of periodic and quasi-periodic solutions and, for q 2< 0, v> 0, a circle of solutions homoclinic to 0 with exponentially damped oscillations at infinity, while for q 2> 0, v< 0, we have a one-parameter family of circles of solutions homoclinic to periodic solutions (as in case (ii)) where the amplitude is minimum at x = 0.
The computation of the coefficients of the normal form (28) corresponding to the system (4) is performed by Dias and looss in ; it is shown that q 2< 0 holds all along Γ. For the ice problem (6), the present case (iii) occurs along the curve Γ of Figure 6. and the coefficient q 2 can have either sign (see ), depending on the water depth. The mathematical problem of persistence of the above solutions of the normal form system for the full vector field is done in an analogous way as for case (ii). This means in particular that we have a one-parameter family of pairs of reversible homoclinics to periodic orbits. For the homoclinic to 0, it is in fact simpler than case (ii). This is due to the fact that the unstable manifold of 0 (identical to the stable manifold, for the normal form) is two-dimensional, and intersects transversally in two points the plane of symmetric points (A real, B pure imaginary). It gives the persistence of two reversible homoclinic orbits, corresponding to two different “bright” solitary waves, with exponentially damped oscillations at infinity: one has a crest in the middle (elevation wave), and the other has a trough in the middle (depression wave) (see Figure 15 for the depression wave). Note that this type of solutions has been experimentally observed [86,126,107,115] at least when some forcing is present. The forcing can be an obstacle at the bottom, wind on the surface, a moving load on the surface in the case of ice experiments.
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Fig. 15. Depression wave for case (iii).
The complete proofs on persistence for periodic and homoclinics can be found in . and for quasi-periodic solutions, it is shown in (the method applies directly here with very slight modifications), that persistence holds true in a subset, locally looking like the product of a curve by a Cantor set, of the region of the (K, H) plane where these quasi-periodic solutions exist for the normal form. We sum up these results in a rough theorem:
Theorem 7
Assume that a 4-dimensional reversible vector field has a fixed point at the origin, and has a (iq)2 resonance for its linear part. Then, in a neighborhood of 0, and near the critical value of the bifurcation parameter (possibly only on one side of criticality), there is a one-parameter family of periodic solutions of frequency near q and a two-parameter family of other periodic and quasi-periodic solutions. Moreover, we have generically one of the two cases, depending on the sign of a certain nonlinear coefficient (q 2 in (29)):
Case 1
for bifurcation parameter values which lead to four non-purely imaginary eigenvalues for the linearized operator, and for q 2< 0, there are two reversible orbits homoclinic to 0;
Case 2
for bifurcation parameter values which lead to four purely imaginary eigenvalues for the linearized operator, and for q 2> 0, there are two one-parameter families of reversible orbits homoclinic to the “hyperbolic” periodic solutions of frequencies near q.
Corollary 8
The above theorem applies for describing travelling waves of the water-wave problem 2.1.2 for (b, λ) near the curve Γ of Figure 5 (case 1) (see). This theorem also applies to the problem of travelling waves under an elastic plate 2.2 for (D, λ) near the curve Γ of Figure 6 (cases 1 and 2) (see).
Remarks
For these results on homoclinics, it should be mentioned that the decay at infinity is exponential. There are degenerate cases (codimension two situations) where this is no longer true. For example, when the coefficient q 2 is close to 0 (see ), there exists in general (for v = 0) an homoclinic to 0, with a polynomial decay at infinity. This case may occur in examples having more parameters, such as with several superposed layers. It should be noticed that this phenomenon is in fact different from the similar property of polynomial decay that we shall meet for cases with an infinitely deep layer. Both phenomena are due to different causes.
For problems with several bounded superposed layers, with surface and (or) interfacial tension, there are always values of the parameters where cases (i), (ii), (iii) occur. They can be treated in the same way. More complex bifurcations may occur, for example in case (iii) when q 2 changes sign. Such a case is a codimension-2 singularity, and is partly studied in and completed in .
4.2.4 Bifurcation of plethora of solitary waves.
So far, we have discussed the solitary waves that one can obtain via the normal form technique. In fact, this technique provides only a small portion of the existing solitary waves, in particular in cases (ii) and (iii) above. What happens is that it is possible to combine several solitary wave solutions together and still obtain a solution of the problem, a so-called multibump solitary wave (see Figure 16 for the profile of such a wave – this type of profile has been obtained numerically in on the full water-wave equations). This can be done as soon as one is sufficiently far from the bifurcation curve. In practice, this distance can be exponentially small. The formation of multibump solutions has been studied in [26,23,25,24] (the first three works deal with a model differential equation and all of them use a Hamiltonian formulation). We do not intend to describe this process in detail here (it would make the paper much longer!). Rather, we follow the formal approach used in , and we concentrate on case (iii). Similar results occur for case (ii).
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Fig. 16. Multibump solitary wave resulting from the superposition of two depression waves near the (iq)2 resonance.
In the last subsection, we showed that at each order the (iq)2 resonance normal form admits two reversible homoclinic solutions provided certain coefficients have the correct sign. The corresponding solutions of the water-wave problem are modulated wavepackets whose envelopes are symmetric and decay exponentially to zero at infinity. In the middle, one wave has a central crest (elevation wave), while the other wave has a central trough (depression wave). As is well known, the normal form (28) yields the nonlinear Schrödinger (NLS) equation to leading order, and of course the NLS equation admits two symmetric envelope-soliton solutions. But one can also construct small-amplitude asymmetric solitary waves, by translating the crests of a symmetric solitary wave relative to its wave envelope. The problem is that such asymmetric waves do not persist when considering the full system. Exponentially small terms come into play! Shifting the carrier oscillations relative to the envelope leads to the appearance of growing oscillations of exponentially small amplitude on one side of the wave packet. However, due to nonlinearity, this growing tail evolves into a new wavepacket and it can be shown that, for certain values of the phase of the carrier oscillations, the whole disturbance terminates, resulting in a solitary wave with two wavepackets. Otherwise, a third wavepacket is generated and the process continues. The main result is that there exists a countable infinity of symmetric and asymmetric multibump solutions. But, unlike the solitary waves obtained in the previous subsection, each of these multibump solitary waves bifurcates at a certain finite amplitude.
When the parameters b and λ are close to the critical point (b, λ) = (1/3. 1). which corresponds to the occurrence of a quadruple eigenvalue 0, it can be shown, via center manifold reduction and a normal form argument (see ), that the problem essentially reduces to the fourth-order differential equation
(32)y x x x x+P y x x+y−y 2=0,
where y is directly related to the elevation of the free surface and x to the horizontal coordinate. When the parameter P is equal to 2, one is along the curve Γ in Figure 5. When the parameter P is equal to −2, one is along the dashed line in Figure 5. Equation (32) has been studied excessively in [23,26]. Using the fact that (32) is a Hamiltonian system, these authors proved the existence of an infinity of homoclinic orbits and the presence of spatial chaos. They showed rigorously that (32) admits a unique (up to translations) homoclinic orbit for P ≤ -2 (i.e., in-between the dashed curve and the half line Δ of Figure 5), while for P in (−2 + ε, 2) it has at least two small-amplitude homoclinic orbits. What happens at P = -2 is that the unique orbit can bifurcate into a countable infinity of multimodal homoclinic orbits. As P is increased towards 2 (i.e., one goes from the dashed curve towards the curve Γ in Figure 5), the domain of existence of each orbit reaches a limit (turning) point before the value P = 2, except for one orbit which can be followed all the way towards P = 2. This orbit is nothing else than the depression wave found earlier (near the half line Δ)! Some of these multimodal homoclinic orbits have been computed numerically for the full water-wave problem in . As said above it was found in that, close to each turning point of a branch of homoclinic, for P ~ 2 (i.e., next to the curve Γ in our context), there is a bifurcation into a branch of asymmetric homoclinic orbits.
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Handbook2003, Handbook of Mathematical Fluid DynamicsFrèdèric Dias, Gèrard Iooss
Chapter
Guided waves in aerospace composites via the semi-analytical finite element method
2022, Stress, Vibration, and Wave Analysis in Aerospace CompositesVictor Giurgiutiu
7.2.8 Finite element method eigenvalue problem
According to the variational principle, Eq. (7.53) must be true for any arbitrary variation δU; it follows that
(7.54)(K 1+i ξ K 2+ξ 2 K 3−ω 2 M)U=0
Eq. (7.54) is a homogeneous equation describing the displacement at nodal locations in the thickness direction z as the wave propagates along the longitudinal directionx with frequency ω and wavenumber ξ. Eq. (7.54) accepts nontrivial solutions only at ξ−ω combinations that make zero the system determinant.
For a fixed frequency ω, Eq. (7.54) can be seen as a generalized polynomial eigenvalue problem which is second order in ξ and has ω as a parameter. If M is the dimension of the vector U, then, at each frequency ω, Eq. (7.54) yields 2 M eigenvalues ξ and 2 MeigenvectorsU. However, the eigenvalues occur pairs: the real eigenvalues occur as positive/negative pairs, as appropriate for forward and backward propagation. For the imaginary and complex eigenvalues, complex-conjugate pairing also appears.
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Book 2022, Stress, Vibration, and Wave Analysis in Aerospace CompositesVictor Giurgiutiu
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Algebraic Riccati Equation
2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
10.3.1 No pure imaginary eigenvalues
Theorem 10.3
Let Θ⊂ℂ 2 n be an n-dimensional invariant subspace of H and let P 1,P 2∈ℂ n×n be two complex matrices such that Θ=Im[P 1 P 2].Then the assumption
(10.10)λ i+λ¯j≠0 f o r a l l i,j=1,…,2 n
where λ i,λ¯j are the eigenvalues of H, implies
1.
P 1P 2 _is Hermitian, that is,_
P 1P 2=P 2P 1
2.
if, in addition, P 1 is nonsingular, the matrix P=P 2 P 1−1 is Hermitian too, that is,
P=(P 2 P 1−1)=P
Remark 10.2
The condition(10.10)is equivalent to the restriction
(10.11)Re λ i≠0 f o r a l l i=1,…,2 n
which means that H has no eigenvalues on the imaginary axis.
Proof
Since Θ is an invariant subspace of H, then there exists a matrix Λ such that spectrums of the eigenvalues of Λ and H coincide, that is,
σ(Λ)=σ(H)
and (10.7) holds:
(10.12)H[P 1 P 2]=[P 1 P 2]Λ
Pre-multiplying this equation by [P 1 P 2] J, we get
[P 1 P 2]J H[P 1 P 2]=[P 1 P 2]J[P 1 P 2]Λ
By (10.4), it follows that J H is symmetric and, hence, is Hermitian (since H is real). So, we obtain that the left-hand side is Hermitian, and, as a result, the right-hand side is Hermitian too:
[P 1 P 2]J[P 1 P 2]Λ=Λ[P 1 P 2]J[P 1 P 2]=−Λ[P 1 P 2]J[P 1 P 2]
which implies
X Λ+ΛX=0 X:=(−P 1P 2+P 2P 1)
But this is a Lyapunov equation which has a unique solution X = 0 if λ i(Λ)+λ¯j(Λ)≠0. But, since the spectrum of eigenvalues of Λ and H coincides, we obtain the proof of the claim. Moreover, if P 1 is nonsingular, then for P=P 2 P 1−1 it follows that
P=(P 2 P 1−1)=(P 1−1)P 2=(P 1−1)(P 1P 2 P 1−1)=(P 1)−1 P 1P 2 P 1−1=P
Theorem is proven.
Theorem 10.4
Suppose a Hamiltonian matrix H(10.2)has no pure imaginary eigenvalues and X−(H)and X+(H)are n-dimensional invariant subspaces corresponding to eigenvalues λ i(H)(i=1,…,n)in Re s<0 and to λ i(H)(i=n+1,…,2 n)in Re s>0 respectively, that is, X−(H)has the basics
[υ 1…υ n]=[υ 1,1.υ n,1...υ 1,n.υ n,n υ 1,n+1.υ n,n+1...υ 1,2 n.υ n,2 n]=[P 1 P 2]P 1=[υ 1,1.υ n,1...υ 1,n.υ n,n],P 2=[υ 1,n+1.υ 1,n+1...υ 1,2 n.υ n,2 n]
Then P 1 is invertible, i.e. P 1−1 exists if and only if the pair (A,B) is stabilizable.
Proof
Sufficiency. Let the pair (A,B) be stabilizable. We want to show that P 1 is nonsingular. Contrariwise, suppose that there exists a vector x 0≠0 such that P 1 x 0=0. Then we have the following. First, notice that
(10.13)x 0P 2(B R−1 B T)P 2 x 0=‖R−1/2 B T P 2 x 0‖2=0
or, equivalently,
(10.14)R−1/2 B T P 2 x 0=0
Indeed, the pre-multiplication of (10.12) by [I 0] implies
(10.15)A P 1−(B R−1 B T)P 2=P 1 Λ
where Λ=d i a g(λ 1,…,λ n) is a diagonal matrix with elements from Re s<0. Then, premultiplying the last equality by x 0P 2, post-multiplying by x 0 and using the symmetricity of P 2P 1=P 1P 2 we get
x 0P 2[A P 1−(B R−1 B T)P 2]x 0=−x 0P 2(B R−1 B T)P 2 x 0=x 0P 2P 1 Λ x 0=x 0P 1P 2 Λ x 0=(P 1 x 0)P 2 Λ x 0=0
which implies (10.13). Pre-multiplying (10.12) by [I 0], we get
(10.16)−Q P 1−A T P 2=P 2 Λ
Post-multiplying (10.16) by x 0 we obtain
(−Q P 1−A T P 2)x 0=−A T P 2 x 0=P 0 Λ x 0=λ 0 P 2 x 0
where
λ 0=x 0Λ x 0‖x 0‖2
which implies
0=A T P 2 x 0+P 2 λ 0 x 0=(A T+λ 0 I)P 2 x 0
Taking into account that, by (10.13),
(B R−1 B T)P 2 x 0=0
it follows that
[(A T+λ 0 I)]⋮(B R−1 B T)P 2 x 0=0
Then, the stabilizability of (A,B) (see Criterion 1 of stabilizability) implies that P 2 x 0=0. So,
[P 1 P 2]x 0=0
and, since [P 1 P 2] forms the basis and, hence, has a full rank, we get x 0=0, which is a contradiction.
Necessity. Let P 1 be invertible. Hence, by (10.15)
A−(B R−1 B T)P 2 P 1−1=P 1 Λ P 1−1
Since the spectrum of the eigenvalues of P 1 Λ P 1−1 coincides with one of Λ, we may conclude that the matrix
A c l o se d:=A−(B R−1 B T)P 2 P 1−1
is stable and, hence, the pair (A,B R−1 B T) is stabilizable (in the corresponding definition K=P 2 P 1−1). It means that for all λ and x such that A x=λ x and Re λ≥0, in other words, for all unstable modes of A
(10.17)xB R−1 B T≠0
which implies
xB≠0
Indeed, by the contradiction, assuming that xB=0, we obtain xB R−1 B T=0 which violates (10.17).
Corollary 10.2
The stabilizability of the pair (A,B) implies that the matrix
(10.18)A c l o s e d:=A−(B R−1 B T)P 2 P 1−1
is stable (Hurwitz).
Proof
Post-multiplying (10.12) by P 1−1 we get
H[I P]=[I P 2]P 1 Λ P 1−1,P=P 2 P 1−1
which after pre-multiplication by [I 0] gives
[I 0]H[I P 2 P 1−1]=[I 0][A−(B R−1 B T)P−Q−A T P]=A−(B R−1 B T)P=A c l o s e d=[I 0][I P 2]P 1 Λ P 1−1=P 1 Λ P 1−1
But P 1 Λ P 1−1 is stable, and hence A c l o s e d is stable too.
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Book 2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
Chapter
FEEDBACK STABILIZATION, EIGENVALUE ASSIGNMENT, AND OPTIMAL CONTROL
2004, Numerical Methods for Linear Control SystemsBISWA NATH DATTA
Theorem 10.5.2.
Let (A, B) be stabilizable and (A, Q) be detectable. Then the Hamiltonian matrix H in(10.5.3)has n eigenvalues with negative real parts , no eigenvalues on the imaginary axis and n eigenvalues with positive real parts. In this case the CARE(10.5.2)has a unique stabilizing solution X. Furthermore, the closed-loop eigenvalues, that is, the eigenvalues of A – BK, are the stable eigenvalues of H.
A note on the solution of the CARE: It will be shown in Chapter 13 that the unique stabilizing solution to (10.5.2) can be obtained by constructing an invariant subspace associated with the stable eigenvalues of the Hamiltonian matrix H in (10.5.3). Specifically, if H does not have any imaginary eigenvalue and (X 1 X 2) is the matrix with columns composed of the eigenvectors corresponding to the stable eigenvalues of H, then, assuming that X 1 is nonsingular, the matrix X=X 2 X 1−1 is a unique stabilizing solution of the CARE. For details, see Chapter 13.
The MATLAB function care solves the CARE. The matrix S in CARE is assumed to be nonnegative definite.
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Book 2004, Numerical Methods for Linear Control SystemsBISWA NATH DATTA
Chapter
Water-Waves as a Spatial Dynamical System
2003, Handbook of Mathematical Fluid DynamicsFrèdèric Dias, Gèrard Iooss
5 The case of infinite depth
The aim of this section is to present typical results for two-dimensional travelling waves in fluid layers when one layer (the bottom one) is infinitely deep. We consider in more detail the systems (1), (9) and (1), (7∞). We observed in Section 3 that the spectrum of the linear operatorL μ contains the full real line, and that, on it, the only possible eigenvalue is 0. There are other eigenvalues ik in the complex plane, solutions of the dispersion relationΔ|μ.sgn╡(Re╡k)k|=0..
In the case of system (1), (9), we note an interesting situation where we have a pair of double eigenvalues ik = ±i/2 on the imaginary axis for μ 1 = 1/4 [see (15)]. This is again a 1 : 1 resonance, but the rest of the spectrum is the full real line, which indeed crosses the imaginary axis, so we cannot use a center manifold reduction into an ODE (no gap between the imaginary axis and the rest of the spectrum). Notice that the point 0 in the spectrum is “resonant” with the purely imaginary eigenvalues, which may lead to problems even for the existence of periodic solutions.
In the case of system (1), (7∞), we note that 0 is always an eigenvalue with eigenvector ξ 0 [see (16) and Section 4.3], and we have a pair of simple eigenvaluesik = ±i λ on the imaginary axis, and for (1 – ρ)λ ≤ 1 there are no other eigenvalues on the imaginary axis. For (1 – ρ)λ > 1 we have another pair ±ik 1 of simple eigenvalues appearing on the imaginary axis, emerging from the continuous spectrum, and this new pair does not meet the other pair. For this system, interesting cases are the strong resonances when k 1/λ = 1/2 or 1/3, and a particularly interesting one is when (1 – ρ)λ is near 1, when one pair of eigenvalues disappears (is melting) in the continuous spectrum.
5.1 Periodic waves
For periodic solutions, we follow the Lyapunov–Schmidt method, except that the presence of 0 in the spectrum gives some resonant terms. It appears that we can formulate all these problems in a way that there are no such resonant terms. As a result, there are us many periodic solutions (with period near 2π/k 0) as in the truncated normal form (see ).
In this section, we consider the case of system (1). (7∞) when there is a pair of simple imaginary eigenvalues ±ik 0 of the operator L μ, such that other pairs ±ik 1 which might be on this axis satisfy k 1 ≠ nk 0 for n = 1, 2, …. This condition is satisfied in our case for k 0 = λ, and is in general (for other similar problems) satisfied because there is only a finite number of eigenvalues on the imaginary axis, whose positions depend on the parameter set μ. The method developed below assumes that the basic pair ±ik 0 is not close to 0, because this would imply that other pairs of eigenvalues would be close to some multiples of this pair.
It results that in general the only point in the spectrum of L μ in resonance with our pair is 0. The “exotic” character of this point of the spectrum leads to a specific difficulty we are dealing with below, by adapting the classical Lyapunov–Schmidt method for periodic orbits.
Let s = (k 0 + γ)x, where Γ is close to 0, and k 0 + γ is the wave number of the periodic solution we are looking for. We denote by H p# (E) the space of (2π-periodic) functions of s, such that their derivatives up to order p are in L 2( R/2π Z), taking values in the Banach spaceE. For such a space we use the norm defined by
‖u‖H z p 2=∑n∈Z(1+n 2 p)‖u n‖E 2
which gives a Banach space structure. Let us define, in the space H#=H#1(H),, the linear operator T μ=k 0 d d x−L μ, with domain D#=H#2(H)∩H#1(D). The basic tool is the inversion of
(37)T μ U=V,
where V is given in H z, and where we look for U ∈ D z. Now expanding in Fourier seriesV and U we have for n ≠ 0, 1, −1
U n=(i n λ I−L μ)−1 V n,
which, with the resolvent estimate (17), insures that if we define
(38)U′=∑n∈Z(0,1,−1)U n e n i s∈D z,then‖U′‖D z≤C 1‖V‖H z holds.
It then remains to study the equations (i n k 0 I−L μ)U n=V n for n = 0, 1, −1. For n = 0 the compatibility condition on V 0 and the additional condition for being in the range of L μ are automatically satisfied by the 0th Fourier component of the nonlinear term, for reversible solutions (in particular the nonlinear term has 0 components on –∞ <y< 0). Then the 0th Fourier component U 0 of the solution U in (37) is uniquely determined in D. up to an arbitrary multiple of the eigenvector ξ′0. For n = 1, −1 this is a classical Fredholm alternative (one compatibility condition for V 1 and for V-1).
Let us now consider the system (1), (7∞0) rewritten as follows
(39)T μ U=G(μ,γ,U),
where we look for solutions in D, and
(40)G(μ,γ,U)=−γ d U d s+F(μ,U)−L μ U.
We observe that G(⋅,⋅,⋅):R 2×D#→H# is analytic in the neighborhood of 0, and is such that
G(μ,γ,0)=0,D(γ,U)G(μ,0,0)=0.
Now, let us define the symmetry S^ in L(H#)∩L(D#) by
(S^U)(s)=S U(−s).
It is then easy to check that
S^T μ=−T μ S^,S^G(μ,γ,U)=−G(μ,γ,S^U)
holds.
For solving (39), (40) we use a classical Lyapunov–Schmidt method. We are then able to prove the following (see proofs in ), where we denote by ζ μ the eigenvector of L μ belonging to ik 0:
Theorem 9
For (u 0, |A| lying in a neighborhood of 0, there is a family of periodic solutions of(1), (7∞), bifurcating from 0, which possess the following converging power series in u 0, A, Ā:
U(s)=u 0 ξ 0′+A e i s ζ μ+A¯e−i s ζ¯μ+∑n+p+q≥2 u 0 n A p A¯q e i(p−q)s U n p q∈D z,γ=∑n+r≥1 γ n r u 0 n|A|2 r∈R,s=[k 0+γ]x.
These solutions are reversible for A real, and we have S U n p q=U n p q=U¯n p q. Moreover, we can find a unique analytic function u 0=h(μ,|A|2)=O(|A|2)such that at the interface, the averages of the velocities in the two fluids are equal (we can as well fix a Bernoulli first integral).
For other similar problems, the proofs are analogous, the number of arbitrary constants n 0 depending on the dimension of the kernel of L μ.
For cases of strong resonances, the additional difficulty is the same as in finite-dimensional reversible systems. See, for instance, for the study of periodic solutions for the 1 : 2 resonance, and for the 1 : 1 resonance. Roughly speaking, there are as many periodic solutions as would be given by the analysis of the corresponding generic reversible bifurcation in finite dimension.
A new difficulty occurs when one considers solutions with very large periods, for example in the case when the basic frequency is given by a pair of eigenvalues of L μ close to 0. The difficulty is that 0 also belongs to the spectrum (since it is on the real line), and that any other pair of imaginary eigenvalues is also quasi-resonant with the basic pair. This problem needs further investigation.
5.2 Normal forms in infinite dimensions
Since we cannot reduce our infinite depth problems to finite-dimensional ODEs, we still would like to believe that eigenvalues near the imaginary axis are ruling the bounded solutions. This is a motivation for developing a theory of normal forms in separating the finite-dimensional critical space from the rest (the “hyperbolic” part of the spectrum, including 0). This leads to “partial normal forms”, where there are coupling terms, especially “bad” in the infinite-dimensional part of the system (see [35,65]). Indeed, there are some additional difficulties:
(i)
when 0 is an eigenvalue embedded in the essential spectrum, we need the explicit form of the resolvent operator near the real axis, to extract the corresponding eigenmode from the rest of space, by a suitable projection (see [58,66]),
(ii)
in space H the linear operator does not have an “easy” (even formal) adjoint. This adjoint and some of its eigenvectors are usually necessary for expressing projections on the critical finite-dimensional space. In our problems, we use again the explicit form of the resolvent operator near the (double) eigenvalues, to make the projections explicit (see ). Let us give below some details on these infinite-dimensional normal forms.
Consider our system under the form
(41)d U d x=F(μ,U),
where F(μ, 0) = 0 and L μ U denotes its linear part. Assume that for μ = 0, one has a spectral decomposition of the form H = E 0 ⊕ E h where E 0 is finite-dimensional, and is spanned by all eigenvectors and generalized eigenvectors belonging to eigenvalues of L 0 lying on the imaginary axis. Then E 0 ∪ D. The space E h is a complementary subspace invariant under L 0 which, in this subspace, has no eigenvalue on the imaginary axis, and is such that its spectrum contains the full real line (with no eigenvalue in 0, and close to 0), and is bounded by a double sector in C centered on the real axis. In all the water-wave problems, with an infinitely deep bottom layer, we also have the estimate (17) for the resolvent of L 0 on the imaginary axis, far from 0 (this indeed implies the sectorial bound of the spectrum of L μ, for μ near 0). The above spectral decomposition of H implies that the 0 eigenvalue which occurs, for instance, in problem (1), (7∞) is extracted from the continuous spectrum, by using a suitable projection for being incorporated into E 0 (this is done explicitly in and ). The result (see [65.35]) is that one can find a polynomial change of variables in H such that
U=V+W+Φ(μ,V,W),V∈E 0,W∈E h,
where Φ(μ,⋅,⋅) is a polynomial with values in D, of degree 1 in W and degree m in V, with regular coefficients in μ, such that (41) reads
(42)d V d x=L 0 V+N(μ,V)+R 0(μ,V,W)in E 0,
(43)d W d x=L h W+R h(μ,V,W)in H,
where L 0=L 0|E o,L h=L 0|E h, and N(μ, V) corresponds to the usual finite-dimensional polynomial (degree m) normal form, i.e., satisfies the following additional symmetry (see )
e L 0x N(μ,V)=N(μ,e L 0x V),∀x∈R,∀V∈E 0.
Now the remaining coupling terms satisfy the following estimates
‖R 0(μ,V,W)‖E 0=O(‖V‖m+1+‖V‖‖W‖D+‖W‖D 2),‖R h(μ,V,W)‖H=O[(‖V‖+‖W‖D)2+|μ|‖W‖D],
which means in particular that we have no linear term in W in (42), and no linear term in V in (43). Moreover, we observe that in general, there are quadratic terms in V in (43), which determine the size of W in this method. These quadratic terms are due to the resonance of the pure imaginary eigenvalues of L 0 with the point 0 in the continuous spectrum. Notice now that we simplify the system (42), (43) in using Bernoulli first integrals which allow to eliminate the coordinates belonging to the 0 eigenvalue. Indeed, this elimination is in general not singular with respect to μ, however it becomes singular in the case of problem (1), (7∞) when (1 – ρ)λ is close to 1. Section 5.4 dealing with a new reversible bifurcation describes below what happens in such a case. It then appears that the above belief that eigenvalues on the imaginary axis govern bounded solutions, becomes wrong at least in this case. In fact our system (42), (43) still contains all small bounded solutions of (41), but a scaling based on solutions corresponding to a truncated normal form in V implies a “slaving” for W which may not be satisfied for other types of solutions, such as the ones which occur in Section 5.4.
5.3 Results for the (iq)2 resonance with continuous spectrum
With the method we use now, we need to give a priori the type of solution we are looking for. This is a major difference with the case where a (center manifold) reduction to an ODE is possible.
For periodic waves, with periods close to the basic one, one may use a method similar to the one given for the nonresonant case, as it is done in . We may also use the formulation (42), (43), which is more transparent for such solutions, easily found on the reduced normal form in the variable V. The result is roughly that there are as many periodic solutions as in the finite depth case.
For solutions homoclinic to 0 (solitary waves), we first invert the infinite-dimensional part (43) in W, using Fourier transform. Indeed, the linearized Fourier transform uses the resolvent operator of L h. The fact that the resolvent operator is not analytic near 0 (0 is not a pole, since we eliminated it, but there is still a jump of the resolvent in crossing the real axis ) leads to the fact that this “hyperbolic part” of the solution cannot decay exponentially, but instead decays polynomially at infinity. The finite-dimensional part (42) where W is replaced by its expression in terms of V, is an integro-differential equation, because of the non-local term coming from W. The principal part of this equation comes from the usual normal form. In the case of 1 : 1 resonance (for instance, as in problem (1), (9) for μ 1 = 1/4, with double eigenvalues ik = ±i/2), the normal form is given by (28) and in case of problem (1), (9), q 2< 0 holds (see ). In other problems, with several layers, the 1 : 1 resonance may occur [except in problem (1), (7∞)] with q 2> 0 or < 0 or even cancels (see ), leading to solutions similar to the ones given by the finite depth case, except that the convergence at infinity towards 0 or to a periodic wave is no longer exponential. For problem (1), (9), and μ 1>∼ 1/4, a fixed point argument in a space of polynomially decaying functions leads to the existence of two reversible homoclinics like in the finite-dimensional case, except for the decay rate which is indeed proved to be in 1/x 2 in . See also [11 where this decay is checked numerically.
The principal part of the solution at finite distance still comes from the finite-dimensional truncated normal form, but it decays faster at infinity than the other part of the solution, which makes this tail part predominant at infinity. This is the main difference with the finite depth case, where the principal part coming from the normal form is valid for all values of x [see for the proofs related with problem (1), (9)s].
5.4 A new reversible bifurcation: pair of eigenvalues diving in the essential spectrum through the origin
It appears that the technique described above may miss important solutions. This occurs precisely for the problem (1), (7∞) when the parameter ε = 1 – (1 – ρ)λ is close to 0. In such a case the singularity of the resolvent operator (i q I−L μ)−1 is a little worse when ε → 0. Indeed the projection operator on the eigenvector belonging to the 0 eigenvalue becomes singular, having ε in its denominator! One may then suspect that this changes relative orders of magnitude for various components of the variable U. The dispersion relation (16) in this case shows a first factor giving two isolated eigenvalues ik = ±iλ, and for k real near 0, a second factor |k|[ε+p|k|+O(|k|3)] giving the 0 eigenvalue, and a pair of imaginary eigenvalues near 0 only for ε < 0. For ε > 0. the factor of |k| corresponds to the linear part of the dispersion relation for the Benjamin–Ono model (see [16.31,95]). Indeed, a suitable scaling here allows to find an asymptotic expansion in powers of ε of a formal solitary wave, with a 1/x 2 decay at infinity , with the same principal part as the solitary wave solution of the Benjamin–Ono equation. In addition, as shown above we also obtain a two parameter family of bifurcating periodic waves (their amplitude is one of the parameters) with periods close to 2π/λ. We can use the form of the family of periodic solutions (given in Theorem 1) to construct Ψ in the change of variables leading to the normal form (42), (43). It then appears that the manifold W = 0 contains all the family of periodic solutions, and that the W part of the system possesses an approximate homoclinic to 0, close to the Benjamin–Ono solitary wave. Finally, the formal solitary wave is not a solution of (1), (7∞), because of the additional frequency λ. It is shown in that there are two reversible solutions homoclinic to each of the above periodic wave, provided that their amplitude is not too small (exponentially small in ε as it is proved in ). The principal part at finite distance of all these “generalized solitary waves” is given by the approximate homoclinic described above (see Figure 17). One difficulty here is that it is not possible to “morally” reduce the problem to a finite-dimensional one, as in [82,84,109,113], because of the essential spectrum filling the real axis (especially near 0). Another difficulty, not appearing in previous work, is that the decay at infinity, towards the periodic solution (with a shift opposite at both infinities), is only polynomial, instead of exponential. This implies the use of refinement techniques for being able to use a fixed point technique in a good function space. See for a numerical computation of periodic as well as generalized solitary waves.
Notice that if we consider the problem of two superposed layers, the bottom one being, as here, infinitely deep, but the top layer being bounded by a rigid horizontal wall, we may formulate the problem as a dynamical system, in a simpler way than here, and the spectrum of the linearized operator is simplified, in the sense that we do not have the pair of eigenvalues ±i λ. on the imaginary axis. This avoids the problems above provoked by the additional frequency, and one finds a solution homoclinic to 0, close to the Benjamin–Ono solitary wave (see [3,112] for proofs, not relying on a dynamical system formulation).
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Handbook2003, Handbook of Mathematical Fluid DynamicsFrèdèric Dias, Gèrard Iooss
Chapter
NUMERICAL SOLUTIONS AND CONDITIONING OF ALGEBRAIC RICCATI EQUATIONS
2004, Numerical Methods for Linear Control SystemsBISWA NATH DATTA
Theorem 13.2.5
Let (A, B) be stabilizable and (A, Q) be detectable. Assume that Q ⩾ 0, S ⩾ 0. Then the Hamiltonian matrix:
H=(A−S−Q−A T)
associated with the CARE does not have a purely imaginary eigenvalue.
Proof. The proof is by contradiction.
Suppose that H has a purely imaginary eigenvalue j α, where α is a nonnegative real number, and let (r s) be the corresponding eigenvector.
Then,
(13.2.20)H(r s)=j α(r s),(r s)≠(0 0).
Multiplying both sides of (13.2.20) by (s, r) to the left, we obtain
sA r−rQ r−sS s−rA T s=j α(sr+rs)
or
(sA r−rA T s)−rQ r−sS s=j α(sr+rs).
Considering the real part of this equation, we get
−rQ r−sS s=0.
Since S ⩾ 0 and Q ⩾ 0, we conclude that
(13.2.21)S s=0
and
(13.2.22)Q r=0.
So, from (13.2.20), we have
(13.2.23)A r=j α r
and
(13.2.24)−A T s=j α s.
Thus, combining (13.2.23) and (13.2.22), we have (A−j α I Q)r=0. Since (A, Q) is detectable, we have r = 0. Similarly, using (13.2.24) and (13.2.21), one can show that s = 0. This gives us a contradiction that (r s) is an eigenvector. Thus, H cannot have a purly imaginary eigenvalue. ▪
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Book 2004, Numerical Methods for Linear Control SystemsBISWA NATH DATTA
Chapter
Algebraic Riccati Equation
2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
10.3.2 Unobservable modes
Theorem 10.5
Assuming that the pair (A,B) is stabilizable, the Hamiltonian matrix H(10.2)has no pure imaginary eigenvalues if and only if the pair (C,A), where Q=C T C,has no unobservable mode on the imaginary axis, that is, for all λ and x 1≠0 such that A x 1=λ x 1,λ=i ω,it follows that C x 1≠0
Proof
Suppose that λ=i ω is an eigenvalue and the corresponding eigenvector[x 1 x 2]≠0.
Then
H[x 1 x 2]=[A x 1−B R−1 B T x 2−C T C x 1−A T x 2]=i ω[x 1 x 2]=[i ω x 1 i ω x 2]
After rearranging, we have
(10.19)(A−i ω I)x 1=B R−1 B T x 2−(A T−i ω I)x 2=C T C x 1
which implies
(x 2,(A−i ω I)x 1)=(x 2,B R−1 B T x 2)=||R−1/2 B T x 2||−(x 1,(A T−i ω I)x 2)=−((A−i ω I)x 1,x 2)=(x 1,C T C x 1)=||C x 1||2
As a result, we get
‖R−1/2 B T x 2‖+‖C x 1‖2=0
and, hence,
B T x 2=0,C x 1=0
In view of this, from (10.19) it follows that
(A−i ω I)x 1=B R−1 B T x 2=0−(A T−i ω I)x 2=C T C x 1=0
Combining the four last equations we obtain
x 2[(A−i ω I)B]=0[(A−i ω I)C]x 1=0
The stabilizability of (A,B) provides the full range for the matrix [(A−i ω I)B] and implies that x 2=0. So, it is clear that i ω is an eigenvalue of H if and only if it is an unobservable mode of (C,A), that is, the corresponding x 1=0 too.
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Book 2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
Chapter
Orbit and Attitude Actuators
2018, Spacecraft Dynamics and ControlEnrico Canuto, ... Carlos Perez Montenegro
Exercise 21
Prove that the four eigenvalues of the LTI part of the state matrix of Eq. (9.138), with ζ r=0, are the positive (λ 1 and λ 3 for the leftmost equation) and negative (λ 2 and λ 4 for the conjugate equation) imaginary eigenvalues of Eq. (9.141), that is
(9.142)λ 1=j γ ω¯w−ω r 2+γ 2 ω¯w 2=j 2 π−f w 1,λ 2=-λ 1,f w 1≥0 λ 3=±j γ ω¯w+ω r 2+γ 2 ω¯w 2=j 2 π f w 2,λ 4=-λ 3,f w 2≥0 γ=J w s 2 J w t<1,ω r 2=k r J w t,f w 1=ω r 2+γ 2 ω¯w 2−γ ω¯w<f w 2.□
The vibration modes of the spin axis defined by Eq. (9.142) are known as whirl modes. The negative frequency, −f w 1,0<f w 10 is a different pair from λ 1,2=±j 2 π(−f w 1). The remark implies that a trigonometric signal a r tuned on ω w 1=2 π f w 1 will not induce any resonance, unlike a trigonometric signal tuned on the positive whirl frequency, namely on ω w 2=2 π f w 2. Resonance and nonresonance conditions can be checked from the following limits:
(9.143)lim s⇒j 2 π f|P r s|=lim s⇒j 2 π f|q r s a r s|,lim s⇒j 2 π f|P rs|=lim s⇒j 2 π f|q rs a rs|,
where f is any of the frequencies in {−f w 1,f w 2,f w 1}. An unbounded limit implies resonance. A bounded limit implies nonresonance.
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Book 2018, Spacecraft Dynamics and ControlEnrico Canuto, ... Carlos Perez Montenegro
Chapter
Algebraic Riccati Equation
2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
Theorem 10.5
Assuming that the pair (A,B) is stabilizable, the Hamiltonian matrix H(10.2)has no pure imaginary eigenvalues if and only if the pair (C,A), where Q=C T C,has no unobservable mode on the imaginary axis, that is, for all λ and x 1≠0 such that A x 1=λ x 1,λ=i ω,it follows that C x 1≠0
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Book 2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
Chapter
The Algebraic Eigenvalue Problem
2015, Numerical Linear Algebra with ApplicationsWilliam Ford
18.10 Computing Both Eigenvalues and Their Corresponding Eigenvectors
We have developed the implicit double-shift QR iteration for computing eigenvalues and the Hessenberg inverse iteration for computing eigenvectors. It is now time to develop a function, eigb, that computes both. The function applies the Francis iteration of degree two to compute the eigenvalues, followed by the use of the shifted inverse Hessenberg iteration to determine an eigenvector corresponding to each eigenvalue. Inverse iteration is economical because we do not have to accumulate transformations during the Francis iteration. Inverse iteration deals with H−σ I using O(n 2) flops, and normally one or two iterations will produce a suitable eigenvector [2, pp. 394-395]. If A has a multiple eigenvalue, σ, Hessenberg inverse iteration can result in vector entries NaN or Inf. The MATLAB implementation checks for this, perturbs σ slightly, and executes eigvechess again with the perturbed eigenvalue. It is hoped this will produce another eigenvector corresponding to σ. All the pieces are in place, so we will not state the formal algorithm that is implemented in the software distribution. Note that it uses the MATLAB features of variable input arguments and variable output to make its use more flexible.
The function can be called in the following ways:
•
[V D] = eigb(A,tol);
–
Returns a diagonal matrix D of eigenvalues and a matrix V whose columns are the corresponding normalized eigenvectors so that AV = VD. tol is the error tolerance for the computations.
If tol is not present, the default is 1.0 × 10−8.
•
E = eigb(A,tol);
–
Assigns E a column vector containing the eigenvalues. The default for tol is the same as for the previous calling sequence.
•
eigb(A,tol)
–
Returns a vector of eigenvalues. The default value of tol is as before.
Example 18.18
Generate a 75×75 random real matrix that most certainly will have imaginary eigenvalues and use eigb to compute its eigenvalues and eigenvectors. A function, checkeigb, in the software distribution computes the minimum and maximum values of ‖A v i−λ i v i‖2,1≤i≤n.
A = randn(75,75);
[V D] = eigb(A,1.0e-12);
[min max] = checkeigb(A,V,D)
min =
7.4402e-15
max =
4.3307e-12
E = diag(D);
E(1:6)
ans =
0.25436 +0i
0.62739 +0i
1.2352 +1.8955i
1.2352 -1.8955i
2.3356 +0.90619i
2.3356 -0.90619i
E(70:75)
ans =
6.8625 -1.1636i
2.7378 +7.5951i
2.7378 -7.5951i
7.2777 +5.6828i
7.2777 -5.6828i
9.4579 +0i
Show more
View chapterExplore book
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URL:
Book 2015, Numerical Linear Algebra with ApplicationsWilliam Ford
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Transfer Function
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9114 | https://www.ixl.com/math/grade-7/area-of-semicircles-and-quarter-circles | IXL | Area of semicircles and quarter circles | 7th grade math
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9115 | https://en.wikipedia.org/wiki/Horvitz%E2%80%93Thompson_estimator | Jump to content
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1 The method
2 Proof of Horvitz–Thompson unbiased estimation of the mean
3 Notes
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From Wikipedia, the free encyclopedia
Statistical estimation method
In statistics, the Horvitz–Thompson estimator, named after Daniel G. Horvitz and Donovan J. Thompson, is a method for estimating the total and mean of a pseudo-population in a stratified sample by applying inverse probability weighting to account for the difference in the sampling distribution between the collected data and the target population. The Horvitz–Thompson estimator is frequently applied in survey analyses and can be used to account for missing data, as well as many sources of unequal selection probabilities.
The method
[edit]
Formally, let be an independent sample from of distinct strata with an overall mean . Suppose further that is the inclusion probability that a randomly sampled individual in a superpopulation belongs to the th stratum. The Horvitz–Thompson estimator of the total is given by:: 51
and the Horvitz–Thompson estimate of the mean is given by:
In a Bayesian probabilistic framework is considered the proportion of individuals in a target population belonging to the th stratum. Hence, could be thought of as an estimate of the complete sample of persons within the th stratum. The Horvitz–Thompson estimator can also be expressed as the limit of a weighted bootstrap resampling estimate of the mean. It can also be viewed as a special case of multiple imputation approaches.
For post-stratified study designs, estimation of and are done in distinct steps. In such cases, computating the variance of is not straightforward. Resampling techniques such as the bootstrap or the jackknife can be applied to gain consistent estimates of the variance of the Horvitz–Thompson estimator. The "survey" package for R conducts analyses for post-stratified data using the Horvitz–Thompson estimator.
Proof of Horvitz–Thompson unbiased estimation of the mean
[edit]
For this proof it will be useful to represent the sample as a random subset of size . We can then define indicator random variables representing whether for each in whether it is present in the sample. Note that for any observation in the sample, the expectation is the definition of the inclusion probability: . [a]
Taking the expectation of the estimator we can prove it is unbiased as follows:
The Hansen–Hurwitz (1943) is known to be inferior to the Horvitz–Thompson (1952) strategy, associated with a number of Inclusion Probabilities Proportional to Size (IPPS) sampling procedures.
Notes
[edit]
^ Technically, the indexing scheme in the proof is different from the indexing in the description of the estimator. In the proof, is the th value in a global ordering out of strata. In the description, is the th value in the sample, out of . To unify these two, we could explicitly define a function mapping sample-indices to global indices.
References
[edit]
^ Horvitz, D. G.; Thompson, D. J. (1952) "A generalization of sampling without replacement from a finite universe", Journal of the American Statistical Association, 47, 663–685, . JSTOR 2280784
^ William G. Cochran (1977), Sampling Techniques, 3rd Edition, Wiley. ISBN 0-471-16240-X
^ Särndal, Carl-Erik; Swensson, Bengt; Wretman, Jan Hȧkan (1992). Model Assisted Survey Sampling. ISBN 9780387975283.
^ Roderick J.A. Little, Donald B. Rubin (2002) Statistical Analysis With Missing Data, 2nd ed., Wiley. ISBN 0-471-18386-5
^ Quatember, A. (2014). "The Finite Population Bootstrap - from the Maximum Likelihood to the Horvitz-Thompson Approach". Austrian Journal of Statistics. 43 (2): 93–102. doi:10.17713/ajs.v43i2.10.
^ "CRAN - Package survey". 19 July 2021.
^ PRABHU-AJGAONKAR, S. G. "Comparison of the Horvitz–Thompson Strategy with the Hansen–Hurwitz Strategy." Survey Methodology (1987): 221. (pdf)
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9116 | https://www.teacherspayteachers.com/Product/Combining-like-terms-Equivalent-expressions-Using-Algebra-Tile-Models-2436409 | Combining like terms - Equivalent expressions Using Algebra Tile Models
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9117 | https://zhuanlan.zhihu.com/p/700879134?utm_psn=1779840488620310529 | 博弈论（5）——海盗分金博弈 - 知乎
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博弈论（5）——海盗分金博弈
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博弈论（5）——海盗分金博弈
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喜欢胜过所有道理，原则抵不过我乐意。
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博弈论
盐选小说：海盗分金
以下是正文：
经济学上有个“海盗分金”模型：
5个海盗抢得100枚金币，他们按抽签的顺序依次提方案：首先由1号提出分配方案，然后5人表决，超过半数同意方案才被通过，否则他将被扔入大海喂鲨鱼，依此类推。
“海盗分金”其实是一个高度简化和抽象的模型，体现了博弈的思想。在“海盗分金”模型中，任何“分配者”想让自己的方案获得通过的关键是事先考虑清楚“挑战者”的分配方案是什么，并用最小的代价获取最大收益，拉拢“挑战者”分配方案中最不得意的人们。
假定“每个海盗都是绝顶聪明且很理智”，那么“第一个海盗提出怎样的分配方案才能够使自己的收益最大化？”
〖情形一〗允许自己投票，超过（不含）半数时通过，必须超过半数才能通过。
对于这种题目，我们通常采用倒推法：假设现在只剩下4号和5号，则5号一定投反对票让4号喂鲨鱼，以独吞全部金币。所以，4号必须支持3号才能保命。
3号知道这一点，就会提出“100，0，0”的分配方案，对4号、5号一毛不拔而将全部金币归为已有，因为他知道4号一无所获但还是会投赞成票，再加上自己一票，他的方案即可通过。
不过，2号推知3号的方案，就会提出“98，0，1，1”的方案，即放弃3号，而给予4号和5号各一枚金币。由于该方案对于4号和5号来说比在3号分配时更为有利，他们将支持他而不希望他出局而由3号来分配。这样，2号将拿走98枚金币。
同样，2号的方案也会被1号所洞悉，1号并将提出（97，0，1，2，0）或（97，0，1，0，2）的方案，即放弃2号，而给3号一枚金币，同时给4号（或5号）2枚金币。由于1号的这一方案对于3号和4号（或5号）来说，相比2号分配时更优，他们将投1号的赞成票，再加上1号自己的票，1号的方案可获通过，97枚金币可轻松落入囊中。这无疑是1号能够获取最大收益的方案了！
因此最终的答案是：1号强盗分给3号1枚金币，分给4号或5号强盗2枚，自己独得97枚。分配方案可写成（97，0，1，2，0）或（97，0，1，0，2）。
〖情形二〗允许自己投票，超过（含）半数时通过，不必超过半数才能通过。
我们依然采用倒推法：假设现在只剩下4号和5号，则4号只能给自己投一票，就可以独吞全部金币。所以，4号一定会提出“100，0”的分配方案。
3号知道这一点，就会提出“99，0，1”的分配方案，这样5号就一定会同意自己的分配方案，再加上自己一票，他的方案即可通过。
2号推知3号的方案，就会提出“99，0，1，0”的方案，由于该方案对于4号来说，比在3号分配时更为有利，他们将支持他而不希望他出局而由3号来分配。且由于2号此时只需要拉一票就行，所以2号只需要给4号或5号其中一个人分配一个金币即可，没必要都分配。这样，2号将拿走99枚金币。
同样，2号的方案也会被1号所洞悉，1号并将提出（98，0，1，0，1）的方案，即放弃2号和4号，给3号和5号各一枚金币。这样，1号的方案可获通过，98枚金币可轻松落入囊中。这无疑是1号能够获取最大收益的方案了！
因此最终的答案是：1号强盗分给3号1枚金币，分给4号或5号强盗1枚，自己独得98枚。分配方案可写成（98，0，1，0，1）。
〖举一反三〗
试分析“不允许自己投票，超过（不含）半数时通过，必须超过半数才能通过”时的情况。
核心思路：
先假设只剩下2个人，根据“分配者是否有同意权”，以及“半数是否算通过”，分析是否有必要给对方留一个，以及谁独吞大头。
人数依次递增，分析“需要拉几票”和“拉谁的票更划算”，依次写出分配策略。
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博弈论
编辑于 2025-09-17 10:19・山西
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默认
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叶零清秋
情形一下4号海盗在只剩两人时不是必死，他选择自己不拿就行了，所以三其实拿捏不了4，因为对于4而言，3拿全部和5拿全部没有区别。
08-21 · 陕西
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叶零清秋
假定自身预期收入不变情况下海盗更倾向于反对的话，随着人数的增多，最后一个海盗的预期收入会逐渐增加。
08-21 · 陕西
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y江
既然大家都如此聪明，为什么会想出这个规则来的呢？难道他们都去赌自己会抽中一号位置吗？
08-20 · 浙江
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年轻人
作者
这是假设的一个理想化模型而已，就是假设五个人已经被这个规则束缚了，至于是怎么束缚的，不需要关心。就像物理上的绝对光滑一样，你不需要考虑绝对光滑的平面是哪里来的。
08-21 · 山西
回复喜欢
不见销语
〖情形二〗允许自己投票，超过（含）半数时通过，不必超过半数才能通过。
我们依然采用倒推法：假设现在只剩下4号和5号，则4号只能给自己投一票就能让5号喂鲨鱼，以独吞全部金币。所以，5号惟有支持3号才能保命。
是否有瑕疵，设定没有无金币就需要喂鲨鱼这一项，5号的最低待遇只是分不到金币，忘记加上的话当我没说
06-14 · 广东
回复喜欢
王勇
五号无论如何不会喂鲨鱼的。
07-07 · 福建
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《博弈论》课程学习笔记（8）-信号传递博弈 ===================== Walde《博弈论》读后感 ======== 第一章：博弈论入门博弈论是指双方或者多方在竞争、合作、冲突等情况下，充分了解各方信息，并依此选择一种能为本方争取最大利益的最优决策的理论。博弈论就是在一定情况下，充分了解各方面… 爱喝长岛冰茶 《博弈论》（翟文明）读书笔记 ============== 君莫邪书籍推荐《博弈论》 ========= 冲着博弈论三个字加上冯诺依曼的名头点开的这本书，看书过程中一直在走神，有点不知所云的感觉，像是说了什么又像是什么也没说，也没见着让人觉得精彩绝伦的地方，可能是普通人的我无法理解… 知否智否
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<Linear Algebra and the C Language
In linear algebra, a circulant matrix is a square matrix in which all rows are composed of the same elements and each row is rotated one element to the right relative to the preceding row. It is a particular kind of Toeplitz matrix.... Wikipedia: Circulant matrix
Install and compile this file in your working directory.
/ ------------------------------------ /
/ Save as: c00f.c /
/ ------------------------------------ /
include "v_a.h"
/ ------------------------------------ /
define RC RC6
/ ------------------------------------ /
int main(void)
{
/ Toeplitz Matrix
V
U 1 5 6 7
2
3
4
/
double u[R1RC]={ 1,2,3,4,5,6};
double v[RCC1]={ 1,
6,
5,
4,
3,
2};
double V = ca_A_mR(v,i_mR(RC,C1));
double U = ca_A_mR(u,i_mR(R1,RC));
double A = i_mR(RC,RC);
clrscrn();
rToeplitz_mR(U,V,A);
printf(" In linear algebra, a circulant matrix is\n"
" a square matrix in which the coefficients\n"
" are moved from one row to the next by\n"
" circular permutation (right shift). \n\n\n");
printf(" A: Circulating matrix");
p_mR(A, S4,P0,C10);
stop();
f_mR(U);
f_mR(V);
f_mR(A);
return 0;
}
/ ------------------------------------ /
/ ------------------------------------ /
Screen output example:
In linear algebra, a circulant matrix is
a square matrix in which the coefficients
are moved from one row to the next by
circular permutation (right shift).
A: Circulating matrix
+1 +2 +3 +4 +5 +6
+6 +1 +2 +3 +4 +5
+5 +6 +1 +2 +3 +4
+4 +5 +6 +1 +2 +3
+3 +4 +5 +6 +1 +2
+2 +3 +4 +5 +6 +1
Press return to continue.
Retrieved from "
Category:
Book:Linear Algebra and the C Language
This page was last edited on 17 September 2025, at 21:54.
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9119 | https://www.semanticscholar.org/paper/35fc064eb12f704110ffcc1ab3391ebbef2cccc6 | [PDF] On the minimum diameter of plane integral point sets | Semantic Scholar
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Corpus ID: 12061851
On the minimum diameter of plane integral point sets
@article{Kurz2008OnTM,
title={On the minimum diameter of plane integral point sets},
author={Sascha Kurz and Alfred Wassermann},
journal={Ars Comb.},
year={2008},
volume={101},
pages={265-287},
url={
}
Sascha Kurz, A. Wassermann
Published in Ars Comb.8 April 2008
Mathematics
Ars Comb.
TLDR
It turns out that plane integral point sets with minimum diameter consist very likely of subsets with many collinear points, and a lower bound for d(2, n) is proved achieving the upper bound n2 log log n up to a constant.Expand
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### On diameter bounds for planar integral point sets in semi-general position
N. Avdeev
Mathematics
Чебышевский сборник
2021
A point set $M$ in the Euclidean plane is called a planar integral point set if all the distances between the elements of $M$ are integers, and $M$ is not situated on a straight line. A planar… Expand
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### Refinement of the upper bound of the radius of a circular integral points set
Batzorig UndrakhGanbileg Bat-Ochir
Mathematics
Gulf Journal of Mathematics
2022
Let n be a product of the prime numbers whose positive integer powers are of the form a2+Db2 where D> 4 is a square-free number and a, b are positive integers. For n≤ 3072, we obtained a refinement… Expand
1
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### Maximal integral point sets over ℤ2
A. AntonovSascha Kurz
Mathematics
International Journal of Computational…
2010
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This work describes an algorithm to prove or disprove the maximality of a given integral point set and considers such sets for a given cardinality and with minimum possible diameter.Expand
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### On the characteristic and diameter of planar integral point sets
N. AvdeevE. A. Lushina
Mathematics
2024
A point set $M$ in Euclidean plane is called an integral point set in semi-general position if all the distances between the elements of $M$ are integers, and $M$ does not contain collinear triples.… Expand
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### Maximal integral point sets in affine planes over finite fields
Michael KiermaierSascha Kurz
Mathematics
Discrete Mathematics
2009
16
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### On the number of points with pairwise integral distances on a circle
Ganbileg Bat-Ochir
Mathematics
Discrete Applied Mathematics
2019
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### On integer distance sets
Rachel GreenfeldMarina IliopoulouSarah Peluse
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2024
We develop a new approach to address some classical questions concerning the size and structure of integer distance sets. Our main result is that any integer distance set in the Euclidean plane is… Expand
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### Integral point sets over finite fields
Sascha Kurz
Mathematics
The Australasian Journal of Combinatorics
2009
TLDR
This work considers point sets in the affine plane Fq where each Euclidean distance of two points is an element of Fq and determines their maximal cardinality I(Fq, 2).Expand
16
[PDF]
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### On existence of integral point sets and their diameter bounds
N. Avdeev
Mathematics
The Australasian Journal of Combinatorics
2020
TLDR
The linear lower bound for diameter of planar integral point sets in Euclidean space is improved, taking into account some results related to the Point Packing in a Square problem.Expand
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### Maximal Circular Point Sets over Arbitrary Fields and an Application to Cryptography
Chris Busenhart
Computer Science, Mathematics
2024
TLDR
A possible application in cryptography of the rotation group similar to the Diffie-Hellman key exchange is described and the cardinality of maximal circular point sets which depends on the radius of the corresponding circle and the characteristic of the underlying field is determined.Expand
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### An upper bound for the minimum diameter of integral point sets
H. HarborthA. KemnitzM. Möller
Computer Science, Mathematics
Discrete & Computational Geometry
1993
TLDR
It is proved that the minimum diameter of such integral point sets has an upper bound of 2c logn log logn.Expand
32
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### On the characteristic of integral point sets in 𝔼m
Sascha Kurz
Mathematics
The Australasian Journal of Combinatorics
2006
TLDR
The definition of the characteristic of an integral triangle is generalised to integral simplices and it is proved that each simplex in an integral point set has the same characteristic.Expand
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### Note on Integral Distances
J. Solymosi
Mathematics
Discrete & Computational Geometry
2003
TLDR
It is proved that any integral set contains many collinear points or the minimum distance should be relatively large if |S| is large.Expand
35
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2 Excerpts
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### Enumeration of integral tetrahedra
Sascha Kurz
Mathematics
2007
We determine the numbers of integral tetrahedra with diameter d up to isomorphism for all d ≤ 1000 via computer enumeration. Therefore we give an algorithm that enumerates the integral tetrahedra… Expand
6
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### A note on Erdös-Diophantine graphs and Diophantine carpets
A. KohnertSascha Kurz
Mathematics
2005
A Diophantine figure is a set of points on the integer grid $\mathbb{Z}^{2}$ where all mutual Euclidean distances are integers. We also speak of Diophantine graphs. In this language a Diophantine… Expand
9
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1 Excerpt
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### There Are Integral Heptagons, no Three Points on a Line, no Four on a Circle
Tobias KreiselSascha Kurz
Mathematics
Discrete & Computational Geometry
2008
Abstract We give two configurations of seven points in the plane, no three points in a line, no four points on a circle with pairwise integral distances. This answers a famous question of Paul Erdős.
45
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### On the number of 8×8 latin squares
G. KolesovaC. LamL. Thiel
Mathematics
Journal of Combinatorial Theory
1990
40
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### Classification Algorithms for Codes and Designs
P. KaskiPatric R. J. stergrd
Mathematics
2005
A new starting-point and a new method are requisite, to insure a complete [classi?cation of the Steiner triple systems of order 15]. This method was furnished, and its tedious and di?cult execution… Expand
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### A fast algorithm for the maximum clique problem
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Discrete Applied Mathematics
2002
648
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### An Introduction to the Theory of Numbers
E. T.
Mathematics
Nature
1946
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Home›Forums›Chart Support›grid lines in between each datapoint in a bar chart
grid lines in between each datapoint in a bar chart
This topic has 1 reply, 2 voices, and was last updated 4 years, 6 months ago by Adithya Menon.
Viewing 2 posts - 1 through 2 (of 2 total)
March 11, 2021 at 4:50 pm#33215 Theo Hi,
I have long lines in a bar chart and I want to separate each data point with a line. Using gridlines doesn’t separate each datapoint and striplines requires exact numbers.
Thanks March 12, 2021 at 7:07 pm#33234 Adithya Menon @Theo,
It is possible to add a line between each dataPoint using stripLines. You can dynamically calculate the value between the dataPoints to find the value where stripline has to be added. Please take a look at the code snippet below,
```
function addStripLines() {
var dataPoints = chart.options.data.dataPoints;
for(var i = 1; i < dataPoints.length; i++) {
chart.axisX.addTo("stripLines", {
value: dataPoints[i].x - (dataPoints[i].x - dataPoints[i-1].x) / 2,
showOnTop: true,
color: "black",
thickness: 1,
labelPlacement: "outside"
}, false)
}
chart.render();
}
```
Kindly take a look at this JSFiddle for an example on creating Striplines inbetween Datapoints.
—
Adithya Menon
Team CanvasJS
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9121 | https://community.the-hospitalist.org/content/what-your-diagnosis-herpes-zoster | What Is Your Diagnosis? Herpes Zoster | MDedge
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What Is Your Diagnosis? Herpes Zoster
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What Is Your Diagnosis? Herpes Zoster
Author(s)
Mona Shahriari, MD
Adrienne Berke, MD
Michael Payette, MD, MBA
The Diagnosis: Herpes Zoster
Herpes zoster is a common infection that clinically presents with moderate to intense pain in the involved dermatome (1–3 days before the outbreak 1) followed by a unilateral dermatomal eruption. The most common sites of involvement include the trunk (dermatomes T3–L2) and upper face (dermatome V1).2 Herpes zoster characteristically begins as erythematous macules and papules that progress to vesicles and sometimes pustules, which subsequently crust over (7–10 days after the initial outbreak).1 Regional lymphadenopathy is present in most cases. Due to the classic presentation of herpes zoster, clinical diagnosis is the mainstay for most cases. Although it can present in any age group, herpes zoster is most commonly associated with advancing age.3
During the course of primary varicella infection, the herpes virus spreads from infected lesions to the contiguous endings of sensory nerves and travels to the dorsal root ganglion cells where it remains latent.1 It is hypothesized that cell-mediated immunity suppresses viral activity and maintains viral latency. A decline in varicella zoster virus–specific cell-mediated immunity can result in reactivation of the latent virus.1 The virus is then transported to the skin from the dorsal root ganglion via myelinated nerves, which terminate at the isthmus of hair follicles, and subsequent infection of the folliculosebaceous unit occurs.4
Laboratory tests that can assist in the diagnosis of herpes zoster, especially atypical cases, include Tzanck smear and viral culture of the vesicle fluid.1 When vesicles are not present, biopsy of the lesion followed by immunohistochemical staining and polymerase chain reaction assay can aid in the diagnosis. The differential diagnosis for our patient included pseudolymphoma, herpes simplex virus, lymphomatoid papulosis, sarcoidosis, trigeminal trophic syndrome, and Sweet syndrome.
Vesicles were not present in our patient, but the dermatomal nature of the eruption and the pain she experienced made the clinical scenario suspicious for herpes zoster. A 4-mm punch biopsy of a single folliculosebaceous unit revealed herpetic, cytopathic features including prominent keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (Figure 1). Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (Figure 2). Staining for herpes simplex virus types 1 and 2 was negative. Within days of starting valacyclovir 1000 mg (every 8 hours for 1 week), the patient’s symptoms resolved.
Figure 1. Keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (H&E, original magnification ×40).
Figure 2. Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (aminoethyl carbazole, original magnification ×40).
Atypical presentations of herpes zoster (eg, presentations that are not completely dermatomal) are becoming more common. Herpetic infections should always be in the differential diagnosis for cutaneous ulcerations. Misdiagnosis of herpes zoster due to atypical features is common and can delay prompt and adequate treatment.
References
Rockley PF, Tyring SK. Pathophysiology and clinical manifestations of varicella zoster virus infections . Int J Dermatol. 1994;33:227-232.
Chen TM, George S, Woodruff CA, et al. Clinical manifestations of varicella-zoster virus infection. Dermatol Clin. 2002;20:267-282.
Hope-Simpson RE. The nature of herpes zoster: a long-term study and a new hypothesis . Proc R Soc Med. 1965;58:9-20.
Walsh N, Boutilier R, Glasgow D, et al. Exclusive involvement of folliculosebaceous units by herpes: a reflection of early herpes zoster . Am J Dermatopathol. 2005;27:189-194.
Article PDF
media_586b252_CT096012364.PDF
Author and Disclosure Information
From the Department of Dermatology and Dermatopathology, University of Connecticut Health Center, Farmington.
The authors report no conflict of interest.
Correspondence: Mona Shahriari, MD, Department of Dermatology and Dermatopathology, University of Connecticut Health Center, 21 South Rd, Farmington, CT 06032 (mshahriari@resident.uchc.edu).
Issue
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herpes, herpes zoster, erythematous plaque, varicella, virus, infection
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Adrienne Berke, MD
Michael Payette, MD, MBA
Author(s)
Mona Shahriari, MD
Adrienne Berke, MD
Michael Payette, MD, MBA
Author and Disclosure Information
From the Department of Dermatology and Dermatopathology, University of Connecticut Health Center, Farmington.
The authors report no conflict of interest.
Correspondence: Mona Shahriari, MD, Department of Dermatology and Dermatopathology, University of Connecticut Health Center, 21 South Rd, Farmington, CT 06032 (mshahriari@resident.uchc.edu).
Author and Disclosure Information
From the Department of Dermatology and Dermatopathology, University of Connecticut Health Center, Farmington.
The authors report no conflict of interest.
Correspondence: Mona Shahriari, MD, Department of Dermatology and Dermatopathology, University of Connecticut Health Center, 21 South Rd, Farmington, CT 06032 (mshahriari@resident.uchc.edu).
Article PDF
media_586b252_CT096012364.PDF
Article PDF
media_586b252_CT096012364.PDF
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The Diagnosis: Herpes Zoster
Herpes zoster is a common infection that clinically presents with moderate to intense pain in the involved dermatome (1–3 days before the outbreak 1) followed by a unilateral dermatomal eruption. The most common sites of involvement include the trunk (dermatomes T3–L2) and upper face (dermatome V1).2 Herpes zoster characteristically begins as erythematous macules and papules that progress to vesicles and sometimes pustules, which subsequently crust over (7–10 days after the initial outbreak).1 Regional lymphadenopathy is present in most cases. Due to the classic presentation of herpes zoster, clinical diagnosis is the mainstay for most cases. Although it can present in any age group, herpes zoster is most commonly associated with advancing age.3
During the course of primary varicella infection, the herpes virus spreads from infected lesions to the contiguous endings of sensory nerves and travels to the dorsal root ganglion cells where it remains latent.1 It is hypothesized that cell-mediated immunity suppresses viral activity and maintains viral latency. A decline in varicella zoster virus–specific cell-mediated immunity can result in reactivation of the latent virus.1 The virus is then transported to the skin from the dorsal root ganglion via myelinated nerves, which terminate at the isthmus of hair follicles, and subsequent infection of the folliculosebaceous unit occurs.4
Laboratory tests that can assist in the diagnosis of herpes zoster, especially atypical cases, include Tzanck smear and viral culture of the vesicle fluid.1 When vesicles are not present, biopsy of the lesion followed by immunohistochemical staining and polymerase chain reaction assay can aid in the diagnosis. The differential diagnosis for our patient included pseudolymphoma, herpes simplex virus, lymphomatoid papulosis, sarcoidosis, trigeminal trophic syndrome, and Sweet syndrome.
Vesicles were not present in our patient, but the dermatomal nature of the eruption and the pain she experienced made the clinical scenario suspicious for herpes zoster. A 4-mm punch biopsy of a single folliculosebaceous unit revealed herpetic, cytopathic features including prominent keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (Figure 1). Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (Figure 2). Staining for herpes simplex virus types 1 and 2 was negative. Within days of starting valacyclovir 1000 mg (every 8 hours for 1 week), the patient’s symptoms resolved.
Figure 1. Keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (H&E, original magnification ×40).
Figure 2. Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (aminoethyl carbazole, original magnification ×40).
Atypical presentations of herpes zoster (eg, presentations that are not completely dermatomal) are becoming more common. Herpetic infections should always be in the differential diagnosis for cutaneous ulcerations. Misdiagnosis of herpes zoster due to atypical features is common and can delay prompt and adequate treatment.
The Diagnosis: Herpes Zoster
Herpes zoster is a common infection that clinically presents with moderate to intense pain in the involved dermatome (1–3 days before the outbreak 1) followed by a unilateral dermatomal eruption. The most common sites of involvement include the trunk (dermatomes T3–L2) and upper face (dermatome V1).2 Herpes zoster characteristically begins as erythematous macules and papules that progress to vesicles and sometimes pustules, which subsequently crust over (7–10 days after the initial outbreak).1 Regional lymphadenopathy is present in most cases. Due to the classic presentation of herpes zoster, clinical diagnosis is the mainstay for most cases. Although it can present in any age group, herpes zoster is most commonly associated with advancing age.3
During the course of primary varicella infection, the herpes virus spreads from infected lesions to the contiguous endings of sensory nerves and travels to the dorsal root ganglion cells where it remains latent.1 It is hypothesized that cell-mediated immunity suppresses viral activity and maintains viral latency. A decline in varicella zoster virus–specific cell-mediated immunity can result in reactivation of the latent virus.1 The virus is then transported to the skin from the dorsal root ganglion via myelinated nerves, which terminate at the isthmus of hair follicles, and subsequent infection of the folliculosebaceous unit occurs.4
Laboratory tests that can assist in the diagnosis of herpes zoster, especially atypical cases, include Tzanck smear and viral culture of the vesicle fluid.1 When vesicles are not present, biopsy of the lesion followed by immunohistochemical staining and polymerase chain reaction assay can aid in the diagnosis. The differential diagnosis for our patient included pseudolymphoma, herpes simplex virus, lymphomatoid papulosis, sarcoidosis, trigeminal trophic syndrome, and Sweet syndrome.
Vesicles were not present in our patient, but the dermatomal nature of the eruption and the pain she experienced made the clinical scenario suspicious for herpes zoster. A 4-mm punch biopsy of a single folliculosebaceous unit revealed herpetic, cytopathic features including prominent keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (Figure 1). Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (Figure 2). Staining for herpes simplex virus types 1 and 2 was negative. Within days of starting valacyclovir 1000 mg (every 8 hours for 1 week), the patient’s symptoms resolved.
Figure 1. Keratinocyte necrosis involving sebaceous, isthmic, and infundibular epithelium; ballooning of epithelial cells with steel gray nuclei; and multinucleation with nuclear molding (H&E, original magnification ×40).
Figure 2. Strong nuclear and cytoplasmic staining was seen in the affected keratinocytes under anti–varicella zoster virus immunohistochemical analysis (aminoethyl carbazole, original magnification ×40).
Atypical presentations of herpes zoster (eg, presentations that are not completely dermatomal) are becoming more common. Herpetic infections should always be in the differential diagnosis for cutaneous ulcerations. Misdiagnosis of herpes zoster due to atypical features is common and can delay prompt and adequate treatment.
References
Rockley PF, Tyring SK. Pathophysiology and clinical manifestations of varicella zoster virus infections . Int J Dermatol. 1994;33:227-232.
Chen TM, George S, Woodruff CA, et al. Clinical manifestations of varicella-zoster virus infection. Dermatol Clin. 2002;20:267-282.
Hope-Simpson RE. The nature of herpes zoster: a long-term study and a new hypothesis . Proc R Soc Med. 1965;58:9-20.
Walsh N, Boutilier R, Glasgow D, et al. Exclusive involvement of folliculosebaceous units by herpes: a reflection of early herpes zoster . Am J Dermatopathol. 2005;27:189-194.
References
Rockley PF, Tyring SK. Pathophysiology and clinical manifestations of varicella zoster virus infections . Int J Dermatol. 1994;33:227-232.
Chen TM, George S, Woodruff CA, et al. Clinical manifestations of varicella-zoster virus infection. Dermatol Clin. 2002;20:267-282.
Hope-Simpson RE. The nature of herpes zoster: a long-term study and a new hypothesis . Proc R Soc Med. 1965;58:9-20.
Walsh N, Boutilier R, Glasgow D, et al. Exclusive involvement of folliculosebaceous units by herpes: a reflection of early herpes zoster . Am J Dermatopathol. 2005;27:189-194.
Issue
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What Is Your Diagnosis? Herpes Zoster
Display Headline
What Is Your Diagnosis? Herpes Zoster
Legacy Keywords
herpes, herpes zoster, erythematous plaque, varicella, virus, infection
Legacy Keywords
herpes, herpes zoster, erythematous plaque, varicella, virus, infection
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A 32-year-old woman with no remarkable medical history presented with a progressively worsening erythematous and edematous plaque on the right cheek of 8 days’ duration. She had previously been treated by her primary care physician with cephalexin for 3 days and trimethoprim-sulfamethoxazole for 2 days, which resulted in “flattening” of the plaque, but the lesion did not resolve. She was referred to our dermatology clinic for further evaluation. She denied any trauma to the cheek or scratching of the lesion. On physical examination, a 2-cm pink, erythematous, edematous plaque with a central eschar was noted on the right cheek with crusting of the right nasal wall.
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9122 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3401535/ | Current Trends in the Management of Gastroesophageal Reflux Disease: A Review - PMC
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ISRN Gastroenterol
. 2012 Jul 11;2012:391631. doi: 10.5402/2012/391631
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Current Trends in the Management of Gastroesophageal Reflux Disease: A Review
Sylvester Chuks Nwokediuko
Sylvester Chuks Nwokediuko
1 Gastroenterology Unit, Department of Medicine, University of Nigeria Teaching Hospital Ituku/Ozalla, PMB, Enugu 01129, Nigeria
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1 Gastroenterology Unit, Department of Medicine, University of Nigeria Teaching Hospital Ituku/Ozalla, PMB, Enugu 01129, Nigeria
✉
Sylvester Chuks Nwokediuko: scnwokediuko@yahoo.com
Academic Editors: P. Correa and C.-T. Shun
Received 2012 Apr 28; Accepted 2012 May 28; Collection date 2012.
Copyright © 2012 Sylvester Chuks Nwokediuko.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC3401535 PMID: 22844607
Abstract
Gastroesophageal reflux disease (GERD) is a chronic disorder of the upper gastrointestinal tract with global distribution. The incidence is on the increase in different parts of the world. In the last 30 to 40 years, research findings have given rise to a more robust understanding of its pathophysiology, clinical presentation, and management. The current definition of GERD (The Montreal definition, 2006) is not only symptom-based and patient-driven, but also encompasses esophageal and extraesophageal manifestations of the disease. The implication is that the disease can be confidently diagnosed based on symptoms alone. Nonerosive reflux disease (NERD) remains the predominant form of GERD. Current thinking is that NERD and erosive reflux disease (ERD) are distinct phenotypes of GERD rather than the old concept which regarded them as components of a disease spectrum. Non erosive reflux disease is a very heterogeneous group with significant overlap with other functional gastrointestinal disorders. There is no gold standard for the diagnosis of GERD. Esophageal pH monitoring and intraluminal impedance monitoring have thrown some light on the heterogeneity of NERD. A substantial proportion of GERD patients continue to have symptoms despite optimal PPI therapy, and this has necessitated research into the development of new drugs. Several safety concerns have been raised about chronic use of proton pump inhibitors but these are yet to be substantiated in controlled studies. The debate about efficacy of long-term medical treatment compared to surgery continues, however, recent data indicate that modern surgical techniques and long-term PPI therapy have comparable efficacy. These and other issues are subjects of further research.
1. Introduction
Gastroesophageal reflux disease (GERD) is a common chronic disorder prevalent in many countries . Apart from the economic burden of the disease and its associated impact on quality of life [2–5], it is the most common predisposing factor for adenocarcinoma of the esophagus. As a consequence of the irritation caused by the reflux of acid and bile, adenocarcinoma may develop in these patients, representing the last of a sequence that starts with the development of GERD and progresses to metaplasia (Barrett's esophagus), low-grade dysplasia, high-grade dysplasia, and adenocarcinoma. Although there has been a decrease in the incidence of squamous cell cancers, the rate of esophageal adenocarcinoma has increased rapidly, and this has been traced to the advent of obesity epidemic, GERD and Barrett's esophagus [6, 7].
Over the years, several issues have emerged regarding the definition, classification, natural history and treatment of GERD, and complications associated with its treatment. This paper focuses on some of these evolving issues. Recent studies, limited to English language, were identified via PubMed searches (1990–2011) with the search terms GERD, NERD, prevalence, incidence, epidemiology, and management. Recent reviews on epidemiology and management were also examined for appropriate references.
2. Definition
Until recently, there were many definitions of GERD. The lack of a gold standard for diagnosis made it difficult to adopt a satisfactory definition. The first ever global consensus definition was published in 2006. According to that document, GERD is defined as “a condition which develops when the reflux of stomach contents causes troublesome symptoms and/or complications” . Based on this definition, GERD can be classified into 2 syndromes: esophageal and extraesophageal syndromes (Table 1). This definition recognizes that GERD can be diagnosed in primary care on the basis of symptoms alone without additional diagnostic testing. This approach is appropriate for most patients and does not use unnecessary resources. Symptoms reach a threshold where they constitute disease when they are troublesome to patients and affect their functioning during usual activities of living. This patient-centered approach to diagnosis includes asking patients how their symptoms affect their everyday lives.
Table 1.
The Montreal definition of GERD and its constituent syndromes .
| Esophageal syndromes |
| Syndromes with symptoms |
| (i) Typical reflux syndrome |
| (ii) Reflux chest pain |
| Syndromes with esophageal injury |
| (i) Reflux esophagitis |
| (ii) Reflux stricture |
| (iii) Barrett's esophagus |
| (iv) Esophageal adenocarcinoma |
| |
| Extraesophageal syndromes |
| |
| Established associations |
| (i) Reflux cough syndrome |
| (ii) Reflux laryngitis syndrome |
| (iii) Reflux asthma syndrome |
| (iv) Reflux dental erosion syndrome |
| Proposed associations |
| (i) Pharyngitis |
| (ii) Sinusitis |
| (iii) Idiopathic pulmonary fibrosis |
| (iv) Recurrent otitis media |
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Heartburn and regurgitation are the characteristic symptoms of GERD. Heartburn is defined as a burning sensation in the retrosternal area. Regurgitation is defined as the perception of flow of refluxed gastric contents into the mouth or hypopharynx. These symptoms are sufficiently descriptive to be diagnostic. Esophageal and extraesophageal symptoms and syndromes that form part of the framework of GERD also include chest pain, sleep disturbances, cough, hoarseness, asthma, and dental erosions (Table 1) .
3. Epidemiology
Gastroesophageal reflux disease is now the most common upper gastrointestinal disease in the western countries, with 10% to 20% of the population experiencing weekly symptoms [4, 8]. In Asia, the prevalence has been variously reported but is generally lower (2.3% by Wong et al. and 6.2% by Chen et al.) [9, 10]. Population-based survey studies indicate that the prevalence is rising . Possible explanations for this include aging population, the obesity epidemic (and associated changes in diet or physical activity), and changes in sleep pattern .
A limited number of studies have reported GERD and its complications to be rare in Africa . However, a recent study of Nigerian medical students showed a prevalence of 26.3% . Nonerosive reflux disease (NERD) accounts for over 60% of cases of GERD in Nigeria .
4. Classification
Gastroesophageal reflux disease is broadly classified into 2 groups on the basis of endoscopy findings: having esophageal mucosal damage (erosive esophagitis and Barrett's esophagus) and no mucosal damage (endoscopy-negative reflux disease or nonerosive reflux disease, NERD).
Traditionally, GERD had been approached as a spectrum disease, with NERD at the mild end and complicated GERD (stricture, Barrett's esophagus, or adenocarcinoma) at the other end of the spectrum. However, emerging evidence indicates that the vast majority of NERD and erosive esophagitis (ER) patients remain within their respective GERD groups throughout their lifetime [15, 16]. This new paradigm proposes that the genetic makeup of each individual subject exposed to similar environmental factors may ultimately determine the specific phenotypic presentation of GERD. In other words, GERD phenotypes once determined remain true to form [15, 16].
Nonerosive reflux disease (NERD) patients have been subclassified into 3 types on the basis of the results of 24-hour pH evaluation:
Type 1: Patients who demonstrate an abnormal acid exposure time in a manner similar to those with erosive esophagitis .
Type 2: Patients with a normal acid exposure time, but with symptoms and reflux events that are significantly correlated, suggesting acid hypersensitivity. This is also referred to as “the hypersensitive esophagus” [17–19].
Type 3: Patients with typical reflux symptoms, but normal pH studies, and no correlation between symptoms and acid exposure. Within this group are 2 subgroups; namely: those who respond to proton pump inhibitor therapy and those who do not respond. The latter subgroup represents functional heartburn (according to Rome III guideline) .
A combination of conventional esophageal pH monitoring and intraluminal impedance monitoring now offers the opportunity to detect acid and non-acid reflux and their association with symptoms . Using this technique, NERD patients with normal pH studies were found to have a positive symptom association for acid reflux in 15% but also a positive association for non-acid reflux in 12% of patients . These findings have led to the narrowing down of the proportion of patients who were otherwise labeled as presenting with functional heartburn, leading to the identification of a new subgroup of patients whose symptoms are due to reflux other than acid; a subgroup of patients with nonacid reflux disease (NARD) or weakly acidic reflux disease (WARD).
5. Risk Factors
There is a potential genetic component to the development of GERD and perhaps Barrett's esophagus . In the US, although the frequency of GERD symptoms does not differ between Caucasians and African Americans, the latter group have a persistently lower risk of esophagitis . In a study from Johannesburg, of the 216 consecutive Barrett's esophagus patients only 5% were black despite the ratio of Blacks to Whites in the city being 5 : 1 .
There is evidence to suggest that age and male sex are associated with a higher incidence of esophagitis [25–27]. Obese subjects are 2.5 times more likely to have GERD than those with normal body mass index (BMI) . Several other researchers have reported similar relationship between body mass and GERD [29, 30]. Alcohol consumption and the presence of a hiatus hernia are risk factors for GERD and esophagitis [25, 31]. The presence and size of a hiatal hernia are associated with a more incompetent LES, defective peristalsis, more severe mucosal damage, and increased acid exposure . A Japanese study identified cigarette smoking and alcohol as risk factors for GERD . In Nigeria, increased consumption of cola and coffee by medical students in order to stay awake to read for examinations was associated with an increased prevalence of GERD . One study showed that an initial diagnosis of either GERD or irritable bowel syndrome raised the risk of a subsequent diagnosis of the other three fold . Gastroesophageal reflux disease is frequently found in patients with connective tissue disease, especially scleroderma , as well as patients with chronic obstructive airway disease . In addition, a number of common drugs and hormonal products have been associated with GERD. These include anticholinergics, benzodiazepines, calcium channel blockers, dopamine, nicotine, nitrates, theophylline, estrogen, progesterone, glucagon, and some prostaglandins. Heartburn is a very common gastrointestinal manifestation of pregnancy.
6. Pathophysiology
Reflux is a normal physiologic occurrence and is produced most often by transient relaxation of the lower esophageal sphincter (LES). In patients with GERD, these transient relaxations occur more frequently than normal. The basal pressure of this sphincter is 10–45 mmHg. The crural diaphragm and gastric sling fibres provide structural support and contribute to LES pressure and competence. The ability of the LES to maintain a tone higher than structures proximal and distal is a result of spikes of calcium influx that are mediated by excitatory cholinergic neurons . Higher intracellular calcium levels are present in the resting LES compared with nonsphincteric esophageal muscle. Other defects of the LES that may contribute to GERD include a chronically hypotensive LES and the effects of a hiatal hernia.
Under normal situations, endogenous defense mechanisms either limit the amount of noxious material that is introduced into the esophagus or rapidly clear the material from the esophagus so that symptoms and esophageal mucosal irritation are minimized. Examples of such defense mechanisms include actions of the LES and normal esophageal motility. When the defense mechanisms are defective or become overwhelmed so that the esophagus is bathed in acid or bile-containing fluid for prolonged periods, GERD can be said to exist.
The esophagus, LES, and stomach can be likened to a simple plumbing circuit . The esophagus functions as an anterograde pump, the LES as a valve, and the stomach as a reservoir. The abnormalities that contribute to GERD can stem from any component of the system. A dysfunctional LES allows reflux of large amounts of gastric juice. Delayed gastric emptying can increase volume and pressure in the reservoir until the valve mechanism is overwhelmed, leading to GERD. Esophageal defense mechanisms include esophageal clearance and mucosal resistance. Esophageal clearance has a mechanical arm (esophageal peristalsis) and a chemical component (saliva), both of which limit the amount of time the esophagus is exposed to refluxed gastric juice.
Transient relaxation of the LES can be caused by foods (coffee, alcohol, chocolate, fatty and meals), medications (beta-blockers, nitrates, calcium channel blockers, anticholinergics), hormones (progesterone), and nicotine.
Regarding the effect of hiatal hernia, not all patients with hiatal hernias have symptomatic reflux. In the presence of a hiatal hernia, the LES may migrate proximally into the chest and lose its abdominal high-pressure zone (HPZ), or the length of the HPZ may decrease. The diaphragmatic hiatus may be widened by a large hernia, which impairs the ability of the crura to function as an external sphincter. Also the gastric contents may be trapped in the hernia sac and reflux proximally into the esophagus during relaxation of the LES. Reduction of the hernias and crural closure result in the restoration of an adequate intra-abdominal length of esophagus and recreating the HPZ.
7. Diagnosis
There is no gold standard for the diagnosis of GERD. Endoscopy is positive in only about 40% of cases . Furthermore, the evaluation of antireflux therapies is based on resolution of symptoms and this suffers greatly from subjectivity. The Society of American Gastrointestinal Endoscopic surgeons (SAGES) Practice Guidelines stipulates that the diagnosis of GERD can be confirmed if at least one of the following conditions exists: a mucosal break seen on endoscopy in a patient with typical symptoms, Barrett's esophagus on biopsy, a peptic stricture in the absence of malignancy, or positive pH-metry . This definition obviously excludes patients with NERD who are negative on pH-metry. Therefore, an objective diagnostic tool with acceptable sensitivity and specificity remains an unmet need for clinicians and researchers.
7.1. Clinical Diagnosis
Heartburn and regurgitation are characteristic symptoms of the typical reflux syndrome . The typical reflux syndrome can be diagnosed on the basis of characteristic symptoms without diagnostic testing , provided that alarm symptoms have been excluded. Alarm symptoms are symptoms which raise a strong suspicion of malignant disease or complication. They include vomiting, gastrointestinal bleeding, anemia, abdominal mass, unexplained weight loss, and progressive dysphagia.
Over the years, several symptom-based diagnostic questionnaires have been developed to help primary care physicians in making provisional categorization of patients presenting with upper abdominal complaints and in the selection of patients with reflux symptoms for empirical treatment. The original reflux disease questionnaire developed by Carlsson et al. and a modified version of it have proved to be useful in this respect.
7.2. Radiology
This has a low sensitivity and specificity for the diagnosis of erosive esophagitis. It has no place in the diagnosis of NERD.
7.3. Endoscopy
This has a high specificity but low sensitivity as over 60% of patients with GERD actually have NERD . In future, new imaging procedures are likely to shed more light on cases that were hitherto classified as NERD by standard white light endoscopy. Such emerging procedures include high-resolution magnification endoscopy, chromoendoscopy, narrow-band imaging, and confocal endomicroscopy [41–43].
7.4. Histology
Various histological lesions have been described in NERD. These include dilated intercellular spaces (DIS) , basal cell hyperplasia , papilla elongation , intraepithelial eosinophils , and neutrophils , with varying sensitivities and specificities. Zentilin et al. proposed a scoring system that takes multiple possible histologic abnormalities into account. Using a receiver operator characteristic curve analysis, a score of 2 was identified as optimal cut-off value for separating GERD patients from controls. A recent study of Nigerian patients with NERD showed a high degree of intraepithelial neutrophil infiltration of the esophageal mucosa; a finding that may be related to the relative rarity of Barrett's esophagus in Nigerians, and indeed black patients . Despite the diagnostic potential of histology, the widespread use of histopathology in clinical practice is hampered by the need for standardization of biopsy and microscopy techniques.
7.5. Proton Pump Inhibitor (PPI) Test
In this test, a short trial of PPI to determine if the patient is going to have symptom relief is carried out. Significant symptom improvement suggests GERD. False positive and false negative results can occur in this test. If the patient's history is typical for uncomplicated GERD, an initial PPI trial (including lifestyle modification) is appropriate . This is the position of the American Gastroenterological Association. The Asia-Pacific Consensus on the management of GERD also favors this approach .
7.6. Manometry
In patients with persistent reflux symptoms despite PPI therapy and normal findings on endoscopy a further evaluation with manometry is indicated to identify alternative diagnosis, such as motor esophageal abnormalities. Manometry helps to analyze the function and the peristaltic activity of the body of the esophagus and the lower esophageal sphincter (LES) prior to antireflux surgery. However manometry is not indicated for confirming a suspected diagnosis of GERD. It is mainly used to establish the diagnosis of dysphagia in cases in which a mechanical obstruction (e.g., stricture) cannot be found. It is also indicated for the preoperative assessment of candidates for antireflux surgery, to exclude achalasia or ineffective peristalsis . Moreover, manometry serves to localize the LES for subsequent pH monitoring for documentation of abnormal esophageal acid exposure.
7.7. Ambulatory pH Monitoring
Patients with NERD who do not respond to medications are best evaluated by ambulatory pH monitoring. The test should be performed-off therapy if the diagnosis is under question but should be performed-on therapy if one is trying to determine the adequacy of treatment. The wireless pH radiotelemetry capsule eliminates the need for the uncomfortable nasogastric tube and increases diagnostic yield by allowing for longer monitoring. Ambulatory esophageal pH monitoring is based upon the duration of time the intraesophageal pH is less than 4, with normal defined as less than 4% over a 24-hour period . Up to 50% of patients with NERD have a normal 24-hour pH monitoring study.
Esophageal impedance pH monitoring is a very promising technique. Multichannel intraluminal impedance monitoring with pH sensor (MII-pH) can detect all types of reflux (acidic, weakly acidic, and weakly alkaline). This test measures the resistance of electrical conductivity of the esophageal content, thus detecting any change of esophageal pH due to the presence of liquid or gas reflux [55, 56].
8. Treatment
The goals of treatment include relief of symptoms, healing of esophagitis, prevention of recurrence, and prevention of complications. The principles of treatment include lifestyle modifications and control of gastric acid secretion using drugs or surgical treatment with corrective antireflux surgery.
8.1. Lifestyle/Dietary Modifications
These are considered the first line of treatment. They include weight loss (for patients who are overweight); avoiding alcohol, chocolate, citrus juice, tomato-based products, peppermint, coffee, and onion. Other measures include avoiding large meals, decreasing fat intake, cessation of smoking, elevation of head of the bed, and avoiding recumbency for 3 hours postprandial . Although there are no randomized trials to test the efficacy of these measures, most gastroenterologists are of the opinion that it is reasonable to employ them. Pregnant women who have GERD should be offered lifestyle modification as first-line therapy.
8.2. Antacids/Alginates
These are effective in symptom relief [58–60] and should be taken after each meal and at bed time.
8.3. Acid Suppressive Therapy
Currently, acid suppressive therapy forms the mainstay of GERD treatment . Histamine 2 receptor antagonists (H2RAs) can decrease gastric acid secretion after a meal and are better than antacids . however, they are not efficacious in the healing of esophagitis and maintenance therapy with standard doses of H2RAs cannot prevent relapses . Today they are used for the treatment of milder forms of the disease and for on-demand therapy, especially for nocturnal symptoms .
Proton pump inhibitors (PPI) are the most potent type of acid suppressants. They are substituted benzimidazoles that irreversibly bind the H+K+ATPase, the final step in gastric acid secretion . Several trials and reviews have shown the superiority of PPIs over H2RAs in the treatment of reflux esophagitis [61, 62, 66, 67]. For patients with NERD, resolution of symptoms with PPIs is inferior to the response in erosive esophagitis as only 61% of patients experience resolution of heartburn, which is still better than 40% reported for H2RAs [68, 69].
Clinical experience shows that 20–30% of patients with GERD continue to have persistent reflux symptoms even while taking PPI daily and one quarter of patients report the use of additional over-the-counter therapies to aid in symptom control . Putative mechanisms for failure of PPI treatment include compliance, improper dosing time, weakly acidic reflux, duodenogastroesophageal reflux (DGER), delayed gastric emptying, esophageal hypersensitivity, eosinophilic esophagitis, nocturnal reflux, residual acid reflux, reduced PPI bioavailability, and psychological comorbidity [72, 73].
Prokinetic agents are somewhat effective but only in patients with mild symptoms; other patients usually require additional acid-suppressing medications such as PPIs. Metoclopramide is a commonly used member of this group. Domperidone has the advantage of less extrapyramidal effects. Long-term use of prokinetic agents may have serious, even potentially fatal complications and should be discouraged. Randomized controlled trials provide moderate-quality evidence that prokinetic drugs improve symptoms in patients with reflux esophagitis and low-quality evidence that they have impact on endoscopic healing .
8.4. Maintenance Therapy
Recurrence of esophagitis is substantially reduced in patients who receive daily PPI therapy . Maintenance therapy for GERD is recommended at the lowest effective dose. Evidence from randomized controlled trials demonstrate that subjects treated with an H2RA as maintenance are twice as likely to have recurrent esophagitis as those treated with a PPI. However, among patients with NERD, on-demand regimens may be effective .
8.5. Issues with Chronic PPI Therapy
Proton pump inhibitors are generally well tolerated but there are reports of minor side effects such as headache, diarrhea, and abdominal pain [75, 76]. In general, these occur in about 1–4% of patients and resolve when the treatment is discontinued. Over the short term, PPIs are safe.
The long-term safety of PPIs is not completely understood. Some safety issues have been raised, although most of these have been in epidemiologic, case-control studies. Epidemiologic data are useful in looking for associations, which of course, should not be confused with causality.
Proton pump inhibitors cause hypergastrinemia in response to acid suppression. Enterochromaffin-like cell (ECL), hyperplasia, and carcinoid tumors have been described in rats , raising a safety concern in humans. However, several studies in humans did not show similar lesions [78–81]. The associations of fractures of hip, wrist, forearm, and other sites appear weak and only slightly higher than the risks in control populations matched for age [82–85]. However, there is an urgent need for careful prospective studies of the effects of PPIs on bone metabolism and for epidemiological studies carefully designed to minimize confounding by various clinical variables. The risks of Clostridium difficile colitis, other enteric infections, small bowel bacterial overgrowth, and possibly spontaneous bacterial peritonitis also appear increased [86–88]. Impaired gastric secretion may adversely affect the absorption of various nutrients but their clinical impact is still ill-defined . Interaction of PPI with other drugs has assumed tremendous importance recently. Co-therapy with clopidogrel and low-dose PPI therapy is widely used to minimize the risk of serious gastrointestinal bleeding, particularly in high-risk patients, so a balancing of risks in the individual patient is appropriate. Although the FDA has recently promulgated some cautionary statements, these remain controversial .
The true importance of these concerns regarding the safety of long-term PPI use can only be estimated from prospective and where possible randomized studies designed solely to measure safety, with minimal confounding.
9. Newer Treatments
Acid-suppressive therapy currently forms the mainstay of treatment for GERD, and PPI is the drug of choice in this regard . However, a substantial proportion of patients diagnosed with GERD continue to experience symptoms despite PPI treatment [70, 91], and 22% of PPI users report taking additional over-the-counter (OTC) medicines to control their symptoms .
Transient lower esophageal sphincter relaxation (TLESR) is an important factor behind the occurrence of reflux, and preclinical studies have identified gamma aminobutyric acid (GABA) type B receptor (GABA B) agonists and metabotropic glutamate receptor 5 (mG1uR5) modulators as candidate drugs for modifying TLESR. Baclofen is an example of the former, while ADX10059 is an example of the latter. Both drugs reduce the incidence of TLESR but poor tolerability is the key issue with these drugs .
Potassium-competitive acid blockers (P-CAB) are a group of acid-suppressive drugs that inhibit gastric H+K+-ATPase (proton pump) reversibly rather than irreversibly. Whereas the PPIs covalently and irreversibly block the proton pump of the gastric parietal cell [93, 94], P-CABs exert their effect by reversible, potassium-competitive binding at, or near, the potassium-binding site on the proton pump . Unfortunately randomized, double-blind trials have not demonstrated any superiority of P-CABs over PPIs [95, 96]. However, there are two other molecules in the same group that are showing some promise [97, 98].
5-hydroxytryptamine type 4 (5-HT 4) receptor agonists increase gastric smooth muscle contractility. This receptor is a potential new target in GERD . Drugs in this class include cisapride, monsapride, and togaserod (which is also used in the treatment of constipation and irritable bowel syndrome). However, safety issues have limited their usefulness in contemporary clinical practice [100, 101]. ATI-7505 is a cisapride analogue that is currently undergoing trial .
Known modulators of visceral pain such as tricyclic antidepressants and selective serotonin reuptake inhibitors (SSRIs) may present an attractive option for GERD patients. A randomized double-blind trial assessing the efficacy of pain modulation with nortriptyline, a tricyclic antidepressant in patients with GERD who have failed to respond to standard-dose PPI therapy is currently on-going .
10. Surgery
To address the chronic and relapsing nature of GERD, two treatment options are available and these are long-term medication and surgery. The advantages and disadvantages of long-term medical treatment and surgery are shown in Table 2 . A multicentre study which compared optimized esomeprazole therapy and standard laparoscopic antireflux surgery (LARS) in patients with GERD demonstrated that both approaches are equally effective as most patients achieve and remain in remission at 5 years .
Table 2.
Potential advantages and disadvantages of medical therapy and antireflux surgery in the management of chronic gastroesophageal reflux disease .
| Medical | |
:---: |
| Advantages | |
| (i) Noninvasive | |
| (ii) Simple and easy to use | |
| (iii) Reproducible effect | |
| (iv) Very effective on symptoms and lesions of GERD | |
| (v) Excellent tolerance and safety profile of PPI | |
| (vi) Relatively cheap especially since the development of PPI generics | |
| Disadvantages | |
| (i) Does not correct underlying pathophysiological mechanisms | |
| (ii) Continuous maintenance therapy frequently required to control the disease | |
| (iii) Persistence of symptoms in at least 10% of patients | |
| (iv) Rare side effects and potential drug-drug interactions | |
| |
| Surgery | |
| |
| Advantages | |
| (i) The only treatment capable of physically controlling reflux | |
| (ii) Very effective (improved quality of heartburn control, reduction of regurgitation, better sleep pattern, increased activities and exercise, etc.) | |
| (iii) Avoids the need to take medication | |
| (iv) Psychological effects of not having chronic disease | |
| (v) Particular clinical groups of cystic fibrosis, lung transplant, and congenital hernia | |
| Disadvantages | |
| (i) Invasive | |
| (ii) Small risk of mortality | |
| (iii) Measurable postoperative mortality | |
| (iv) Recurrence is possible | |
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10.1. Indications for Surgery
Failed medical management (inadequate symptom control, severe regurgitation not controlled with acid suppression, or medication side effects).
Patients who opt for surgery despite successful medical management (due to quality of life considerations, life-long need for medication intake, expense of medication etc.)
Complications of GERD (Barrett's esophagus, peptic stricture) [106, 107].
Extraesophageal manifestations (asthma, hoarseness, cough, chest pain, and aspiration) [108–111]. The coexistence of Barrett's esophagus with reflux symptoms is considered by many as clear indication for antireflux surgery .
Over the past 50 years, surgery for GERD has evolved from an open to a laparoscopic procedure and recently to a new incisionless procedure called transoral incisionless fundoplication. The most common procedure is Nissen fundoplication, which can be open or laparoscopic. Fundoplication can involve a complete (360 degrees) or partial (varying degrees) wrap of the LES with a portion of the stomach, thereby increasing the LES pressure. In the era of open antireflux surgery, symptom response rates of 80–90% were reported [113, 114]. Even at that, many patients avoided it because of high morbidity. With the introduction of laparoscopic techniques, there has been an exponential growth in the number of antireflux operations. The advantages include fewer incisional hernias, shorter hospital stay, less pain, quicker return to work, and fewer defective wraps at follow-up endoscopy .
Complications of fundoplication include persistent dysphagia, inability to belch and vomit, epigastric fullness, bloating and postprandial pain, temporary swallowing discomfort, and sometimes intense flatus . Inability to belch, epigastric fullness, bloating, and flatus constitute the syndrome of “gas bloat”.
Endoluminal fundoplication is a new, modified version of open or laparoscopic fundoplication which accesses the stomach through the mouth, thereby eliminating the need for incisions.
11. Overlap of GERD with Other Gastrointestinal Disorders
Patients with NERD have other functional gastrointestinal symptoms, such as functional dyspepsia and irritable bowel syndrome (IBS), with a frequency higher than that observed in most studies of erosive reflux disease [117–119]. A common denominator may well be visceral hypersensitivity . The NERD patient group incorporates subgroups which differ significantly in terms of presentation, pathophysiology, and management. Patients with functional heartburn are more likely to have psychopathology, similar to functional dyspepsia patients . Abdominal symptoms appear to be independent predictors of severity of reflux symptoms in NERD patients when compared to control subjects who do not have such symptoms . The significance of this overlap is still a subject of serious research, and it does appear that more revisions await the classification of functional gastrointestinal disorders.
12. Conclusion
In conclusion, GERD is one aspect of gastroenterology that has undergone tremendous innovations in the last 30–40 years and is still an area of intensive research. There have been innovations in the definition, classification, diagnosis, clinical course, and management of GERD. Nonerosive reflux disease (NERD) is the variant of GERD that affects over 60% of patients with GERD and it is not only more heterogeneous than erosive esophagitis but has a different pathophysiology and response to standard medical therapy. Because GERD is a chronic, relapsing disease, patients have to be managed with either long-term medical treatment or surgery after a thorough analysis of the pros and cons of each modality. A number of issues remain unresolved about GERD and it is hoped that the next couple of years would come with more discoveries in this important disease.
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9123 | https://archive.nytimes.com/learning.blogs.nytimes.com/2012/02/09/its-all-an-allusion-identifying-allusions-in-literature-and-in-life/ | It's All an Allusion: Identifying Allusions, in Literature and in Life - The New York Times
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It’s All an Allusion: Identifying Allusions, in Literature and in Life
By Shannon Doyne, Katherine Schulten and Holly Ojalvo February 9, 2012 3:06 pm February 9, 2012 3:06 pm
Illustration by Shannon MayGo to related essay »
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Overview | What is an allusion? How often do you spot them, whether in your reading, in pop culture, in advertising or anywhere else? In this lesson, students read a Book Review essay about allusions in literature, take a quiz in which they identify allusions made in New York Times articles and headlines, then choose from a variety of activities to go deeper.
Materials | Computers with Internet access and printing capability.
Warm-Up | Ask students to define “allusion.” Check that they understand it as a “brief, usually indirect reference to another place, event” or to words spoken by or that depict a person or fictional character. Give a few common examples, like someone being described as a “Romeo,” an allusion to Shakespeare’s romantic but doomed tragic hero, or a person saying, “I never thought I’d move back to my hometown, but I guess deep down, I’m a Dorothy,” alluding to the “Wizard of Oz” character who learns “there’s no place like home.” You can also ask what is meant by calling a group of women “the real housewives of (name of your city or town)” and asking the source (wealthy, drama-prone women who have a similar look to those seen on the “Real Housewives” franchise). Ask, what would you expect if I called a certain boy an Edward? What about a Jacob? (main characters of the “Twilight” book and movie series). Have students name more allusions, explaining their meanings and sources.
Then, lead a discussion about the pros and cons of making allusions. Pros might include conveying much information in a single word or two or bonding over a shared interest in the source. Cons include allusions only making sense to those who know the source material or, in the case of pop culture phenomena, losing their meaning as time passes. To prove this, ask students to describe a “Jeannie Bueller,” or an “Eddie Haskell.” Then ask colleagues in their 40’s or 50’s the same question to share the answers: a jealous sister who has a wildly popular brother (from “Ferris Bueller’s Day Off”), and a sycophant who is overly polite to adults for his own gain (“Leave It to Beaver”).
End by turning the tables on yourself and your colleagues, having students ask you to explain allusions to things that most in their peer group will understand immediately but might not be so clear to those being asked. Or, have them brainstorm allusions today that most would understand (“Brangelina,” for example), but that may be impenetrable twenty-five years from now.
Related | In the essay “Grand Allusion,” Elizabeth D. Samet writes about the pleasures and perils of this literary device:
Allusion can feel like something of a parlor game even in the best of times. In the 1940s, in a discussion of T. S. Eliot’s densely allusive poem “The Waste Land,” the formalist critics William K. Wimsatt and Monroe C. Beardsley questioned prevailing assumptions about the value of allusion-hunting. Eschewing the role of literary detective, they rejected the notion that we “do not know what a poet means unless we have traced him in his reading.” “Eliot’s allusions work,” they argued in “The Intentional Fallacy,” “when we know them — and to a great extent even when we do not know them, through their suggestive power. . . . It would not much matter if Eliot invented his sources,” as Walter Scott and Coleridge had done. Wimsatt and Beardsley’s warning that identifying an allusion does not amount to the same thing as understanding its significance has renewed urgency in the current age of allusion-automation, for if the Web makes it that much easier for the allusion-hunter to bag his quarry, it does not necessarily tell him how to dress it.
Read the entire article with your class, using the questions below.
Questions | For discussion and reading comprehension:
Why do allusions that fail to convey their intended meaning or are not understood by the audience “leave us all exposed,” as Ms. Samet contends?
How is the “Vronsky’s horse” anecdote an example of this?
Why does the classroom have “its own special dangers” when it comes to allusions?
How, according to the essay, has the Internet affected people’s abilities to use and verify allusions?
How does your experiences with allusions support or challenge the author’s statement that “In trying to illuminate an allusion in class, I sometimes feel as if I’m opening one nesting doll after another until there’s nothing left at all.”
Related resources:
RELATED RESOURCES
From The Learning Network
Lesson: “Rounding Up the Usual Suspects: Interpreting Famous Quotations”
Lesson: “Remix, Reuse, Recombine: Holding a Seminar on Mash-Up Culture“
Lesson: “Making a Commitment: Memorizing and Reciting”
From NYTimes.com
After Deadline: “Allusions We Love Too Much”
Reading Room: “A Flood of Allusions”
On Language: Poetic Allusion Watch
Around the Web
Buzzle.com’s Allusion Examples
The Write Practice: “How to Use Allusion Like Taylor Swift”
Newspapers in Education’s “Literary Allusion vs. Pop Culture”
Activity | Begin by having students reread the article, finding the unexplained allusions it contains. Have them identify the sources, marking “K” for those they already knew and therefore recognized immediately, “I” for those they found via Internet searching, and “O” for other, meaning that they found the source in some other way, whether asking other people, looking in books, or even just guessing.
The allusions include:
“I was reminded that each unhappy allusion is unhappy in its own way.” –Leo Tolstoy, “Anna Karenina”
“Don’t make it sad, Cricket, I don’t feel that way.” — Ernest Hemingway, “To Have and Have Not” and the 1944 film adaptation
“Gimme a whiskey. . . . And don’t be stingy, baby.” — Eugene O’Neill’s play “Anna Christie”
“Bastard Normans, Norman bastards.” — William Shakespeare, “The Life of King Henry the Fifth”
“Scylla of the swindle to the Charybdis of condescension” — Greek mythology: Scylla is a sea creature who devours sailors and Charybdis is a whirlpool opposite Scylla’s cave
“… several keyboarding Natty Bumppos of my acquaintance” — James Fenimore Cooper’s The Leatherstocking Tales
The title of the essay “Grand Allusion” — a play on “Grand Illusion,” a 1937 French war film directed by Jean Renoir
Lead a discussion about the experience of identifying the above allusions. Compare the pride of knowing, or at least being being familiar with, one or more, versus the experience of searching for the phrase online. Ask: What is the difference between relying on Google versus your own memory of favorite books, movies or plays? Is this distinction relevant today for most people, do you think?
Next, ask students to discuss this idea from today’s essay:
Wimsatt and Beardsley’s warning that identifying an allusion does not amount to the same thing as understanding its significance has renewed urgency in the current age of allusion-automation, for if the Web makes it that much easier for the allusion-hunter to bag his quarry, it does not necessarily tell him how to dress it.
Ask for examples that illustrate the point. You might begin by asking students if they share a love of a certain sports team or musician with a parent or other older person, and the difference, if there is any, between, say, watching live as a famous game unfolded (or listening just as a seminal album was released) versus watching highlights online or downloading an entire music catalog in a single sitting.
Then use the Scylla and Charybdis allusion from the essay as an example. Ask: Is it enough to know it’s a reference to a Greek myth? Or to truly “get” the allusion, do you have to know the story it comes from and what it means?
Next, have students take the New York Times Literary Allusions Quiz, below, alone, in partners or in small groups.
New York Times Literary Allusions Quiz
Fill in the blanks in the lines below taken from New York Times articles or headlines. Each refers to an often-taught work of literature. For extra points, identify that work and give more context to explain its meaning and usage here. (For answers, scroll to the very bottom of this lesson.)
1. “I left Dickens World after a couple of days. As a literary experience, it had been pretty thin gruel. But like ___, I wanted more.”
2. “More Than Kin, and Less Than ___”
3. “In his previous books the journalist Ron Rosenbaum has tackled big topics — Hitler’s evil, Shakespeare’s genius — with acuity and irreverence, believing, correctly, that some things are too important to leave to the experts. He’s proud of his gonzo amateur status, so much so that you half suspect he has a scarlet ‘___’ tattooed across his chest, where Superman wore his ‘S.'”
4. “Without a Bang or a ___, The School Board Fades Away”
5. “… I’ve come to think something is ___ in the state of economics. The dismal science, as Thomas Carlyle called it, has been ravaged by the same virus that has corrupted the rest of our national discourse.”
6. “Qaddafi might have maneuvered himself into a gilded overseer’s role and gifted power to his bespectacled son Seif al-Islam el-Qaddafi, the nice, educated boy who lost it when he realized — The horror! The ___! — that he might have to give up all his toys”
7. “To paraphrase William Carlos Williams, so much ___ upon a Chicago bank.”
8. “Decentralizing the Internet So Big ___ Can’t Find You”
9. “Destruction, thy ___ is Bieber.”
10. “The Kushner Flap: Much Ado About ___”
11. “Call me, ___” (Please note: In the original, there is no comma; in the Times essay we are quoting here, there is. Read it to see why.)
12. “It was the best of times, it was the ___ of times. O.K., maybe not literally the worst, but definitely bad.”
Have students share their findings and check their answers, then discuss the meaning and source, including, again, whether they knew it or looked it up. Students might then look up the original and explain its context, then relate it to the allusion. This can be done independently or in groups.
Going Further | There are several options to extend this lesson:
Allusion Hunting Every edition of The New York Times is full of allusions. Challenge teams to find and list as many as they possibly can as they read articles that interest them. Then, have them read the After Deadline column “Allusions We Love Too Much” on the line between allusion and cliché. Which, if any, of the allusions they found do they believe have been overused enough to become clichés? Why?
Digital Remix Culture and Allusion Though this lesson has mostly dealt with literary allusions, students will likely bring up the dizzying ways in which artists across categories remix, reuse, update and mash-up material today, from musicians who “sample” bits of other music in their pieces to new versions of Jane Austen and Shakespeare to advertising that references or remakes cultural touchstones. Have them do a version of the Allusion Hunting activity above, but with the whole world as fair game: how many “allusions” can they spot in one day? (Use our lesson plan Remix, Reuse, Recombine: Holding a Seminar on Mash-Up Culture for more help.)
Build Your Own Allusion Ask students to write an original essay on a topic unrelated to allusions, planting as many allusions as they can. Or, they can retrofit an old essay, sewing in references. Have them read one another’s work, identifying and discussing the allusions and weighing in on whether each one “works” or seems tacked on for the sake of the assignment. Have them incorporate a response to this sentence from today’s article: “Unlike most tricks, the allusion triumphs only when people know precisely how it is done.”
Frank O’Hara’s World and Ours Have students read the Frank O’Hara poem “The Day Lady Died,” which is mentioned in the article. Have them identify the references and then address Ms. Samet’s description of her students’ responses to it. They can write a reaction essay or write a poem of their own in the style of O’Hara’s.
Standards | This lesson is correlated to McREL’s national standards (it can also be aligned to the new Common Core State Standards):
Art Connections
Understands connections among the various art forms and other disciplines.
Theater
Understands how informal and formal theater, film, television, and electronic media productions create and communicate meaning.
Understands the context in which theater, film, television and electronic media are performed today as well as in the past.
Historical Understanding
Understands and knows how to analyze chronological relationships and patterns.
Understands the historical perspective.
Language Arts
Uses the general skills and strategies of the writing process.
Uses the stylistic and rhetorical aspects of writing.
Uses grammatical and mechanical conventions in written compositions.
Gathers and uses information for research purposes.
Uses the general skills and strategies of the reading process.
Uses skills and strategies to read a variety of literary texts.
Uses skills and strategies to read a variety of informational texts.
Uses listening and speaking strategies for different purposes.
Uses viewing skills and strategies to understand and interpret visual media.
New York Times Literary Allusion Quiz answers:
1. Oliver Twist, alluding to the character in Charles Dickens’s novel of the same name (from a 2012 article, “The World of Charles Dickens, Complete With Pizza Hut”).
2. Kind, alluding to a line in Shakespeare’s “Hamlet” (The headline was for a 2011 Opinon piece about the News of the World scandal.)
3. “A,” alluding to Nathaniel Hawthorne’s “The Scarlet Letter” (from “Thinking the Unthinkable Again in a Nuclear Age,” a 2011 book review).
4. Whimper, alluding to T.S. Eliot’s “The Hollow Men” (from this 2002 article).
5. rotten, alluding to Shakespeare’s “Hamlet” again (from “The Politics of Economics in the Age of Shouting,” a 2011 Op-Ed).
6. horror, alluding to Joseph Conrad’s “Heart of Darkness” (from a 2011 Op-Ed, “The Price of Delusion”).
7. depends, alluding to Williams’s poem, “The Red Wheel Barrow” (from a 2007 Business article, “Chicago Is Major Battleground for ABN Amro Bid War”).
8. Brother, alluding to George Orwell’s “1984” (from a2011 New York Region article).
9. name, alluding to “Frailty, thy name is woman,” from, yep, Shakespeare’s “Hamlet” again (from a 2009 piece, “20-Year-Old Fogy Cedes Audience to 15-Year-Old”).
10. Nothing, alluding to the Shakespearean play of the same name (from a 2011 Opinion piece by Stanley Fish).
11. Ishmael, alluding to the first line of Herman Melville’s “Moby Dick” (the headline of a 2007 Book Review essay).
12. worst, alluding to the first line of Charles Dickens’s “Tale of Two Cities” (from “The Banks Are Not All Right,” a 2009 Op-Ed).
Comments are no longer being accepted.
ReginaFebruary 9, 2012·4:03 pm
It’s not “Call me,Ishmael” (as enticing as that might sound), but “Call me Ishmael”. At least in the novel I’m thinking of. If another novel begins “Call me, Ishmael,” please let me know!
Hi Regina–We’re quoting a Times essay there, in which there IS a comma. To make it clearer to students, however, I’ve just added a note on that question. “(Please note: In the original, there is no comma; in the Times essay we are quoting here, there is. Read it to see why.)” –Katherine
JimTongeFebruary 9, 2012·6:29 pm
I once heard a comment about an elusive allusion to an illusion by a Frenchman. His speech employed so much elision that it was difficult for me to understand him.
“Elisive, elusive, allusive and illusive.” No wonder some people are confused.
joanFebruary 10, 2012·3:14 pm
awesome lesson
Stephanie MeyerFebruary 10, 2012·11:57 pm
For the kids with less literary knowledge, you could do a lesson on allusions in popular music i.e. Simon & Garfunkel’s “Sound of Silence” & its allusion to John Donne’s essay “No Man Is an Island.” Of course there are also “allusions” to songs in other songs like the allusion to Stevie Wonder’s “Pastime Paradise” in Coolio’s “Gangsta’s Paradise.” There are also TONS of allusions in “Family Guy”–I once used the Seven Plagues episode to go with chapter 7 of Chinua Achebe’s Things Fall Apart (when the locusts come to Umuofia).
Stacey SimienFebruary 14, 2012·1:44 pm
Common Core lesson plan
MaddySeptember 1, 2012·6:07 pm
hey. My name is Maddy and i am a freshman. I am in Advanced freshman English, and i kinda procrastinated all summer so now i only have till this Tuesday to finish my book reports. It says that i need “major allusions” for the book “Pygmalion” and “Black like me” i found one for Pygmalion which is “She rose like Eliza Dolittle from the gutter to the heights of society” but it states that i have to identify it…..i am so lost and i googled how to identify allusions, and this was the first site linked to the page so i clicked on itand figured i should give it a try. at this point anything works
Thanks(:
Goetz KlugeNovember 23, 2013·4:58 am
Lewis Carroll’s “The Hunting of the Snark” (1876) is an example for a poem that lead to many discussion about allusions. Interestingly, Henry Holiday’s illustrations to the “Snark” didn’t receive the same attention. That is a mistake. //snrk.de
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9124 | https://designexptr.org/powchpt.html | Design and Analysis of Experiments and Observational Studies using R: A Volume in the Chapman & Hall/CRC Texts in Statistical Science Series
4 Power and Sample Size
4.1 Introduction
Consider a scenario where a continuous study outcome, with variance equal to one, is measured in two groups. The group means are (\mu_1) and (\mu_2). An investigator tests (H_0:\mu_1-\mu_2=0) versus (H_1:\mu_1 \ne\mu_2) with a sample of fifteen experimental units in each group.
The function my_ttest() simulates data to test this hypothesis with a t-test, where (\mu_1=0), (\mu_2=0.1), and (\sigma^2=1) in both groups.
my_ttest <- function(sampsize) {my_ttest<- function(sampsize){ set.seed(6)set.seed(6) x1 <- rnorm(sampsize, mean = 0, sd = 1) x1<- rnorm(sampsize = 0 = 1) x2 <- rnorm(sampsize, mean = 0.1, sd = 1) x2<- rnorm(sampsize =0.1 = 1) myttest <- t.test(x1, x2, var.equal = TRUE) myttest<-t.test(x1 x2 = TRUE) return(myttest$p.value) return(myttest $p.value) }}
The simulation below shows that the t-test failed to detect a significant difference even though a difference exists (i.e., (H_0) is false).
my_ttest(15) my_ttest(15)#> 0.4446615#> 0.4446615
In this case, we know (\mu_1 \ne \mu_2), so why does the the p-value indicate that there is no difference between the group means? Let’s change the sample size to 1,000 per group.
my_ttest(1000) my_ttest(1000)#> 0.004706338#> 0.004706338
Now, the p-value indicates that there is strong evidence that (H_0) is false. The test failed to detect a difference when the sample size was 15 due to low power, and in this case, low power is due to a small sample size in each group.
Power is important in many fields of study. Suppose that researchers would like to compare two versions of a web page to investigate whether one page leads to an increase in sales. A pharmaceutical company is comparing a novel treatment for cancer against the standard treatment to investigate whether patient mortality decreases when receiving the novel treatment. A psychologist is studying how physical expression influences psychological processes such as risk taking, so she plans to randomize subjects into two groups where one group was instructed to pose in a high-power (this has nothing to do with statistical power) position and the second group in a low-power position. In all these scenarios, the investigators can calculate how many experimental units are required so that the statistical test used will have a high probability of detecting a significant difference between the groups if indeed there is a difference.
In studies with large sample sizes in each group the power might be large, but the differences detected may be small and not practically important.
4.2 Statistical Hypotheses and the Number of Experimental Units
Suppose that experimental units are randomized to treatments A or B with equal probability. Let (\mu_A) and (\mu_B) be the mean responses in groups A and B. The null hypothesis is that there is no difference between A and B; the alternative claims there is a difference: (H_0:\mu_A=\mu_B \thinspace \text{ vs } \thinspace H_1:\mu_A \ne \mu_B)
The type I error rate is defined as:
[\alpha =P\left(\text{type I error}\right) =P_{H_0}\left(\text{Reject } H_0\right).] (P_{H_0}) means the probability calculated using the distribution induced by (H_0). For example, a testing of (\mu), the meanof a normal distribution, (H_0:\mu=\mu_0 \mbox{ vs. } H_1:\mu\ne \mu_0), at the 5% level would have (\text{p-value}=P_{H_0}\left(\text{Reject } H_0\right)=P\left(|t_{n-1}| > |t^{obs}|\right)), with (t^{obs}=\sqrt{n}(\bar x - \mu_0)/S_x).
The type II error rate is defined as:
[\beta=P\left(\text{type II error}\right) =P_{H_1}\left(\text{Accept }H_0\right).]
Power is defined as:
[ \begin{aligned} \text {power} &= 1-\beta \ &= 1-P_{H_1}\left(\text{Accept }H_0\right) = P_{H_1}\left(\text{Reject } H_0\right). \end{aligned}]
4.3 Power of the One-Sample z-test
Let (X_1,...,X_n) be a random sample from a (N(\mu,\sigma^2)) distribution. A test of the hypothesis
[H_0:\mu=\mu_0 \thinspace \text{ versus } \thinspace H_1:\mu \ne \mu_0]
will reject at level (\alpha) if and only if
[ \left|\frac{{\bar X} - \mu_0}{\sigma/{\sqrt{n}}} \right| \ge z_{1-\alpha/2},]
[ \left|{\bar X} -\mu_0 \right| \ge \frac{\sigma}{\sqrt{n}} z_{1-\alpha/2},]
where (z_{p}) is the (p^{th}) quantile of the (N(0,1)).
The power of the test at (\mu=\mu_1) (i.e., when (H_1:\mu=\mu_1)) is
[\begin{aligned} 1-\beta &= 1-P\left(\text{type II error}\right) \ &= P_{H_1}\left(\text{Reject } H_0\right) \ &= P_{H_1}\left(\left|{\bar X} -\mu_0 \right| \ge \frac{\sigma}{\sqrt{n}} z_{1-\alpha/2}\right) \ &= P_{H_1}\left({\bar X} -\mu_0 \ge \frac{\sigma}{\sqrt{n}} z_{1-\alpha/2}\right) + P_{H_1}\left({\bar X} -\mu_0 < \frac{-\sigma}{\sqrt{n}} z_{1-\alpha/2}\right). \ \end{aligned}]
Subtract the mean (\mu_1) and divide by (\sigma/\sqrt{n}) to obtain:
[\begin{aligned} 1-\beta = 1-\Phi\left( z_{1-\alpha/2}-\left(\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}\right) \right)+\Phi\left( -z_{1-\alpha/2}-\left(\frac{\mu_1-\mu_0}{\sigma/\sqrt{n}}\right) \right). \end{aligned} \tag{4.1}]
The power function of the one-sample z-test depends on (\alpha), (\mu_1), (\mu_0), (\sigma), and (n).
4.3.1 Computation Lab: Power of the One-Sample z-test
An R function to compute the power of the one-sample z-test (4.1) using qnorm() to calculate (z_{1-\alpha/2}) and pnorm() to calculate (\Phi(\cdot)) is
pow_ztest <- function(alpha, mu1, mu0, sigma, n) {pow_ztest<- function(alpha mu1 mu0 sigma n){ arg1 <- qnorm(1 - alpha / 2) - (mu1 - mu0) / (sigma / sqrt(n)) arg1<- qnorm(1 - alpha/ 2) -(mu1 - mu0)/(sigma/ sqrt(n)) arg2 <- -1 qnorm(1 - alpha / 2) - arg2<- - 1 qnorm(1 - alpha/ 2) - (mu1 - mu0) / (sigma / sqrt(n))(mu1 - mu0)/(sigma/ sqrt(n)) return(1 - pnorm(arg1) + pnorm(arg2)) return(1 - pnorm(arg1) + pnorm(arg2))}}
Example 4.1 Calculate the power of (H_0:\mu = 0 \thinspace \text{ vs. } \thinspace H_1:\mu = 0.2) with 30 experimental units per group, (\sigma = 0.2, \alpha = 0.05).
pow_ztest(alpha = 0.05, mu1 = 0.15, mu0 = 0, sigma = 0.2, n = 30) pow_ztest(=0.05 =0.15 = 0 =0.2 = 30)#> 0.9841413#> 0.9841413
A study with 30 experimental units will have a 0.98 probability of detecting a mean difference of 0.2, when (\sigma=0.2, \alpha=0.05).
4.4 Power of the One-Sample t-test
Let (X_1,...,X_n) be i.i.d. (N(\mu,\sigma^2)). A test of the hypothesis
[H_0:\mu=\mu_0 \thinspace \text{ versus } \thinspace H_1:\mu \ne \mu_0 ]
will reject at level (\alpha) if and only if
[ \left|\frac{{\bar X} - \mu_0}{S/{\sqrt{n}}} \right| \ge t_{n-1, 1-\alpha/2},]
where (t_{n-1, p}) is the (p^{th}) quantile of the (t_{n-1}).
It can be shown that
[\sqrt{n} \left[\frac{{\bar X}-\mu_0}{S}\right] = \frac{Z + \gamma}{\sqrt{V/(n-1)}},]
where,
[\begin{aligned} Z &= \frac{\sqrt{n}({\bar X}-\mu_1)}{\sigma}, \ \gamma &= \frac{\sqrt{n}(\mu_1-\mu_0)}{\sigma}, \text{ and}\ V &= \frac{(n-1)}{\sigma^2} S^2. \end{aligned}]
(Z \sim N(0,1)), (V \sim \chi^2_{n-1}), and (Z) is independent of (V).
If (\gamma = 0) then (\sqrt{n} \left[\frac{{\bar X}-\mu_0}{S}\right] \sim t_{n-1}). But, if (\gamma \ne 0), then (\sqrt{n} \left[\frac{{\bar X}-\mu_0}{S}\right] \sim t_{n-1, \gamma}), where (t_{n-1, \gamma}) is the non-central t-distribution with non-centrality parameter (\gamma). If (\gamma = 0), this is sometimes called the central t-distribution (see Figure 4.1).
Figure 4.1: Noncentral t-distribution
The power of the test at (\mu=\mu_1) is
[\begin{aligned} 1-\beta &= 1-P\left(\text{type II error}\right) \notag \ &= P_{H_1}\left(\text{Reject } H_0\right) \notag\ &= P_{H_1}\left(\text{Reject } H_0\right) \notag\ &= P_{H_1}\left(\left|\frac {{\bar X} -\mu_0} {\frac{S}{\sqrt{n}}} \right| \ge t_{n-1, 1-\alpha/2}\right) \notag \ &= P_{H_1}\left(\frac {{\bar X} -\mu_0} {\frac{S}{\sqrt{n}}} \ge t_{n-1, 1-\alpha/2}\right) + P_{H_1}\left(\frac {{\bar X} -\mu_0} {\frac{S}{\sqrt{n}}} < - t_{n-1, 1-\alpha/2}\right) \notag \ &= P(t_{n-1,\gamma}\ge t_{n-1,1-\alpha/2})+P(t_{n-1,\gamma}< -t_{n-1,1-\alpha/2}). \end{aligned} \tag{4.2}]
4.4.1 Computation Lab: Power of the One-Sample t-test
The following function calculates the power function using (4.2) for the one-sample t-test in R:
onesampttestpow <- function(alpha, n, mu0, mu1, sigma) {onesampttestpow<- function(alpha n mu0 mu1 sigma){ delta <- mu1 - mu0 delta<- mu1 - mu0 t.crit <- qt(1 - alpha / 2, n - 1)t.crit<- qt(1 - alpha/ 2 n - 1) t.gamma <- sqrt(n) (delta / sigma)t.gamma<- sqrt(n)(delta/ sigma) t.power <-t.power<- 1 - pt(t.crit, n - 1, ncp = t.gamma) + 1 - pt(t.crit n - 1 =t.gamma) + pt(-t.crit, n - 1, ncp = t.gamma) pt(-t.crit n - 1 =t.gamma) return(t.power) return(t.power)}}
Example 4.2 Calculate the power of (H_0:\mu = 0 \mbox{ vs. } H_1:\mu = 0.15) with (n = 10), (\sigma = 0.2), and (\alpha = 0.05) by calling the above function.
onesampttestpow(onesampttestpow( alpha = 0.05, =0.05 n = 10, = 10 mu0 = 0, = 0 mu1 = 0.15, =0.15 sigma = 0.2 =0.2))#> 0.5619533#> 0.5619533
This means that the test will reject (H_0) in 56.2% of studies testing this hypothesis (i.e., the study testing this hypothesis was replicated a large number of times).
stats::power.t.test() is part of the default R packages. Using this function on the previous example we get
power.t.test(power.t.test( n = 10, = 10 delta = 0.15, =0.15 sd = 0.2, =0.2 sig.level = 0.05, =0.05 type = "one.sample" ="one.sample"))#> #> #> One-sample t test power calculation #> One-sample t test power calculation #> #> #> n = 10#> n = 10#> delta = 0.15#> delta = 0.15#> sd = 0.2#> sd = 0.2#> sig.level = 0.05#> sig.level = 0.05#> power = 0.5619339#> power = 0.5619339#> alternative = two.sided#> alternative = two.sided
Exactly one of the parameters n, delta, power, sd, and sig.level must be passed as NULL, and that parameter is determined from the others. In this example, the function calculates power given the other parameters. If sample size is required for, say, 80% power then use
power.t.test(power.t.test( power = 0.8, =0.8 delta = 0.15, =0.15 sd = 0.2, =0.2 sig.level = 0.05, =0.05 type = "one.sample" ="one.sample"))#> #> #> One-sample t test power calculation #> One-sample t test power calculation #> #> #> n = 15.98026#> n = 15.98026#> delta = 0.15#> delta = 0.15#> sd = 0.2#> sd = 0.2#> sig.level = 0.05#> sig.level = 0.05#> power = 0.8#> power = 0.8#> alternative = two.sided#> alternative = two.sided
The calculation shows that a study testing (H_0:\mu=\mu_0 \mbox{ vs } H_1:\mu=\mu_1), where delta= (\mu_0-\mu_1) requires sixteen experimental units for 80% power.
4.5 Power of the Two-Sample t-test
The two-sample t-test, described in Section 3.9, is often used to test if the difference between two means is zero. This section develops the power of this test using the same notation as Section 3.9.
Recall from Section 3.9 that (T_n) is the two-sample t-statistic, where (T_n \sim t_{n_1+n_2-2}) under (H_0) and (t_{n_1+n_2-2,\gamma}) with non-centrality parameter (\gamma ={\theta}/{\sigma\sqrt{1/n_1+1/n_2}},) under (H_1).
(H_0) is rejected if (\left| T_n \right| \ge t_{n_1+n_2-2,1-\alpha/2}) where (t_{\lambda,p}) is the (pth) quantile of the central t-distribution with (\lambda) degrees of freedom.
The sample size can be determined by specifying the type I and type II error rates, the standard deviation, and the difference in treatment means that the study aims to detect.
In a derivation similar to the power of the one-sample t-test in Section 4.4, the power of the two-sample t-test is
[1-\beta = P(t_{n_1+n_2-2, \gamma} \ge t_{n_1+n_2-2,\gamma ,1-\alpha/2})+P(t_{n_1+n_2-2, \gamma} < -t_{n_1+n_2-2,\gamma , 1-\alpha/2}),]
where (t_{\lambda,\gamma ,p}) is the (pth) quantile of the non-central t-distribution with non-centrality parameter (\gamma) and (\lambda) degrees of freedom. The power is not a closed form expression so it’s not possible to derive a formula for the study sample size. Nevertheless, if the variance is assumed to be known, and if (n_1=n_2=n/2,) then the total sample size for a study at level (\alpha), power (1-\beta), that tests (H_0:\theta=\theta_0 \mbox{ vs. } H_1:\theta = \theta_1) is
[\begin{equation} n=\frac{4\sigma^2\left(z_{1-\alpha/2}+z_{1-\beta}\right)}{\theta_1^2}. \tag{4.3} \end{equation}]
4.5.1 Effect Size
In some studies, instead of specifying the difference in treatment means and standard deviation separately, the ratio
[\text{ES} = \frac{\mu_1-\mu_2}{\sigma}]
can be specified. This ratio is called the scaled effect size. Jacob Cohen36 suggests that effect sizes of 0.2, 0.5, and 0.8 correspond to small, medium, and large effects.
4.5.2 Sample Size—Known Variance and Equal Allocation
Consider a study where experimental units are randomized into two treatment groups and the investigator would like an equal number of experimental units in each group.
If the variance is known, then a test statistic for testing if the means of two populations are equal is
[ Z=\frac {{\bar Y}_1 - {\bar Y}_2}{\sigma \sqrt{(1/n_1+1/n_2)}} \sim N(0,1).]
This is known as the two-sample z-test.
The power at (\theta=\theta_1) is given by
[\begin{equation} 1-\beta = P\left( Z \ge z_{1-\alpha/2} - \frac{\theta_1}{\sigma \sqrt{1/n_1+1/n_2}} \right) \ + P\left( Z < -z_{1-\alpha/2} - \frac{\theta_1}{\sigma \sqrt{1/n_1+1/n_2}} \right). \tag{4.4} \end{equation}]
Ignoring terms smaller than (\alpha/2) and combining positive and negative (\theta) in (4.4)
[\begin{equation} \beta \approx \Phi\left( z_{1-\alpha/2} - \frac{\left|\theta_1\right|}{\sigma \sqrt{1/n_1+1/n_2}} \right). \tag{4.5} \end{equation}]
Applying (\Phi^{-1}) to (4.5) and using 2.1, it follows that
[\begin{equation} z_{1-\beta}+z_{1-\alpha/2} = \left( \frac{\left|\theta_1\right|}{\sigma \sqrt{1/n_1+1/n_2}} \right). \tag{4.6} \end{equation}]
If we assume that there will be an equal allocation of subjects to each group, then (n_1 = n_2 = n/2), and the total sample size is
[\begin{equation} n= \frac {4\sigma^2 \left(z_{1-\beta}+z_{1-\alpha/2}\right)^2}{\theta^2}, \tag{4.7} \end{equation}]
where (\alpha, \beta \in (0,1).)
4.5.3 Sample Size—Known Variance and Unequal Allocation
In many studies comparing two treatments, it is desirable to put more experimental units into the experimental group to learn more about this treatment. If the allocation of experimental units between the two groups is (r = n_1/n_2) then (n_1 = r\cdot n_2). Plugging this into (4.6) the sample size for (n_2) is
[\begin{equation} n_2=\frac {(1+1/r)\sigma^2 \left(z_{\beta}+z_{\alpha/2}\right)^2} {\theta^2}. \tag{4.8} \end{equation}]
4.5.4 Computation Lab: Power of the Two-Sample t-test
The following R function uses qt() to compute (t_{n,\gamma,\lambda}) and pt() to compute the (t_{n,\gamma}) CDF.
twosampttestpow <- function(alpha, n1, n2, mu1, mu2, sigma) {twosampttestpow<- function(alpha n1 n2 mu1 mu2 sigma){ delta <- mu1 - mu2 delta<- mu1 - mu2 t.crit <- qt(1 - alpha / 2, n1 + n2 - 2)t.crit<- qt(1 - alpha/ 2 n1 + n2 - 2) t.gamma <- delta / (sigma sqrt(1 / n1 + 1 / n2))t.gamma<- delta/(sigma sqrt(1/ n1 + 1/ n2)) t.power <-t.power<- (1 - pt(t.crit, n1 + n2 - 2, ncp = t.gamma) +(1 - pt(t.crit n1 + n2 - 2 =t.gamma) + pt(-t.crit, n1 + n2 - 2, ncp = t.gamma)) pt(-t.crit n1 + n2 - 2 =t.gamma)) return(t.power) return(t.power)}}
The power of a study to detect (\theta = 1) with (\sigma = 3,n_1 = n_2 = 50) is
twosampttestpow(twosampttestpow( alpha = .05, =.05 n1 = 50, = 50 n2 = 50, = 50 mu1 = 1, = 1 mu2 = 2, = 2 sigma = 3 = 3))#> 0.3785749#> 0.3785749
stats::power.t.test() can also be used and gives the same results.
power.t.test(power.t.test( n = 50, = 50 delta = 1, = 1 sd = 3, = 3 sig.level = 0.05 =0.05))#> #> #> Two-sample t test power calculation #> Two-sample t test power calculation #> #> #> n = 50#> n = 50#> delta = 1#> delta = 1#> sd = 3#> sd = 3#> sig.level = 0.05#> sig.level = 0.05#> power = 0.3784221#> power = 0.3784221#> alternative = two.sided#> alternative = two.sided#> #> #> NOTE: n is number in each group#> NOTE: n is number in each group
So the study would require 50 subjects per group to achieve 38% power to detect a difference of (\theta = 1) at the 5% significance level assuming (\sigma = 3). power.t.test() can also output the number of subjects required to achieve a certain power. Suppose the investigators want to know how many subjects per group would have to be enrolled in each group to achieve 80% power under the same conditions?
power.t.test(power.t.test( power = 0.8, =0.8 delta = 1, = 1 sd = 3, = 3 sig.level = 0.05 =0.05))#> #> #> Two-sample t test power calculation #> Two-sample t test power calculation #> #> #> n = 142.2466#> n = 142.2466#> delta = 1#> delta = 1#> sd = 3#> sd = 3#> sig.level = 0.05#> sig.level = 0.05#> power = 0.8#> power = 0.8#> alternative = two.sided#> alternative = two.sided#> #> #> NOTE: n is number in each group#> NOTE: n is number in each group
142 subjects would be required in each group to achieve 80% power.
Figure 4.2 shows power of the two-sample t-test as a function of (n) and (\theta) (left), and (n) and (\sigma) (right) with a horizontal line drawn at 0.8 power for reference.
f <- function(n, d, sd) f<- function(n d sd) power.t.test(power.t.test( n = n, = n delta = d, = d sd = sd, = sd type = "two.sample", ="two.sample" alternative = "two.sided", ="two.sided" sig.level = 0.05 =0.05 )$power) $ powerp1 <- p1<- ggplot() + ggplot() + xlim(0, 3.0) + xlim(03.0) + scale_y_continuous(breaks = seq(0, 1, by = 0.1)) + scale_y_continuous(= seq(0 1 =0.1)) + geom_function(fun = f, geom_function(= f args = list(n = 30, sd = 1.5), = list(= 30 =1.5) aes(linetype = "N = 30")) + aes(="N = 30")) + geom_function(fun = f, geom_function(= f args = list(n = 20, sd = 1.5), = list(= 20 =1.5) aes(linetype = "N = 20")) + aes(="N = 20")) + geom_function(fun = f, geom_function(= f args = list(n = 10, sd = 1.5), = list(= 10 =1.5) aes(linetype = "N = 10")) + aes(="N = 10")) + ylab("Power") + ylab("Power") + xlab(TeX("$\\theta$")) + xlab(TeX("$\\theta$")) + geom_hline(yintercept = 0.8) + geom_hline(=0.8) + guides(linetype = guide_legend(title = "Sample size")) guides(= guide_legend(= "Sample size"))p2 <- p2<- ggplot() + ggplot() + xlim(0.15, 3.0) + xlim(0.153.0) + scale_y_continuous(breaks = seq(0, 1, by = 0.1)) + scale_y_continuous(= seq(0 1 =0.1)) + geom_function(fun = f, geom_function(= f args = list(n = 30, d = 1), = list(= 30 = 1) aes(linetype = "N = 30")) + aes(="N = 30")) + geom_function(fun = f, geom_function(= f args = list(n = 20, d = 1), = list(= 20 = 1) aes(linetype = "N = 20")) + aes(="N = 20")) + geom_function(fun = f, geom_function(= f args = list(n = 10, d = 1), = list(= 10 = 1) aes(linetype = "N = 10")) + aes(="N = 10")) + ylab("Power") + ylab("Power") + xlab(TeX("$\\sigma$")) + xlab(TeX("$\\sigma$")) + geom_hline(yintercept = 0.8) + geom_hline(=0.8) + guides(linetype = guide_legend(title = "Sample size")) guides(= guide_legend(= "Sample size"))p1 + p2 p1 + p2
Figure 4.2: Power of Two-Sample t-test
Power as a function of effect size can be investigated by rearranging (4.4). In the R function below we assume (n_1=n_2=10, ]\alpha=0.05).
pow.t <- function(ES) {pow.t<- function(ES){ alpha <- 0.05 alpha<-0.05 nA <- 10 nA<- 10 nB <- 10 nB<- 10 t.crit <- qt(1 - alpha / 2, nA + nB - 2)t.crit<- qt(1 - alpha/ 2 nA + nB - 2) t.gamma <- ES / (sqrt(1 / nA + 1 / nB))t.gamma<- ES/(sqrt(1/ nA + 1/ nB)) t.power <- (1 - pt(t.crit, nA + nB - 2, ncp = t.gamma) +t.power<-(1 - pt(t.crit nA + nB - 2 =t.gamma) + pt(-t.crit, nA + nB - 2, ncp = t.gamma)) pt(-t.crit nA + nB - 2 =t.gamma)) t.powert.power}}
The code below was used to create Figure 4.3.
ggplot() + ggplot() + scale_x_continuous(limits = c(-2, 2), scale_x_continuous(= c(- 2 2) breaks = seq(-2, 2, by = 0.5)) + = seq(- 2 2 =0.5)) + scale_y_continuous(limits = c(0, 1), scale_y_continuous(= c(0 1) breaks = seq(0, 1, by = 0.1)) + = seq(0 1 =0.1)) + geom_function(fun = pow.t) + geom_function(=pow.t) + ylab("Power") + ylab("Power") + xlab("Effect Size") xlab("Effect Size")
Figure 4.3: Two-Sample t-test Power and Effect Size, N = 10
The R code below defines a function to compute the sample size in groups 1 and 2 for unequal allocation using (4.8).
size2z.uneq.test <- function(theta, alpha, beta, sigma, r)size2z.uneq.test<- function(theta alpha beta sigma r){{ zalpha <- qnorm(1 - alpha / 2) zalpha<- qnorm(1 - alpha/ 2) zbeta <- qnorm(1 - beta) zbeta<- qnorm(1 - beta) n2 <- (1 + 1 / r) (sigma (zalpha + zbeta) / theta) ^ 2 n2<-(1 + 1/ r)(sigma(zalpha + zbeta)/ theta) ^ 2 n1 <- r n2 n1<- r n2 c(n1, n2) c(n1 n2)}}
The number of patients in group 1 is computed using size2z.uneq.test.
```
sample size for theta =1, alpha = 0.05, # sample size for theta =1, alpha = 0.05, # beta = 0.1, sigma = 2, r = 2# beta = 0.1, sigma = 2, r = 2size2z.uneq.test(size2z.uneq.test( theta = 1, = 1 alpha = .05, =.05 beta = .1, =.1 sigma = 2, = 2 r = 2 = 2) # group 1 sample size (experimental group))# group 1 sample size (experimental group)#> 126.0891#> 126.0891
```
So, the number of patients in the experimental group is 126. The number in the control group is computed below.
size2z.uneq.test(size2z.uneq.test( theta = 1, = 1 alpha = .05, =.05 beta = .1, =.1 sigma = 2, = 2 r = 2 = 2) # group 2 sample size (control group))# group 2 sample size (control group)#> 63.04454#> 63.04454
The sample size required for 90% power to detect (\theta = 1) with (\sigma = 2) at the 5% level in a trial where two patients will be enrolled in the experimental arm for every patient enrolled in the control arm is 126 in the control group and 63 in the experimental group. The total sample size is 189.
The power of the two-sample z-test (4.5) can be studied as a function of the allocation ratio (r). The code chunk below produces Figure 4.4.
```
power of z test as a function of allocation ratio r,# power of z test as a function of allocation ratio r,# total sample size n, alpha, theta, and sigma# total sample size n, alpha, theta, and sigmapow_ztest <- function(r, n, alpha, theta, sigma) pow_ztest<- function(r n alpha theta sigma){{ n2 <- n / (r + 1) n2<- n/(r + 1) x <- qnorm(1 - alpha / 2) - abs(theta) / x<- qnorm(1 - alpha/ 2) - abs(theta)/ (sigma sqrt(1 / (r n2) + 1 / n2))(sigma sqrt(1/(r n2) + 1/ n2)) pow <- 1 - pnorm(x) pow<- 1 - pnorm(x) return(pow) return(pow)}}ggplot() + xlim(0.1, 5) + ggplot() + xlim(0.1 5) + geom_function(geom_function( fun = pow_ztest, = pow_ztest args = list(= list( n = 25, = 25 alpha = 0.05, =0.05 theta = 1, = 1 sigma = 1 = 1 ),) aes(linetype = "N = 25") aes(="N = 25") ) +) + geom_function(geom_function( fun = pow_ztest, = pow_ztest args = list(= list( n = 50, = 50 alpha = 0.05, =0.05 theta = 1, = 1 sigma = 1 = 1 ),) aes(linetype = "N = 50") aes(="N = 50") ) +) + ylab("Power") + xlab(TeX("$\r$")) + ylab("Power") + xlab(TeX("$\r$")) + guides(linetype = guide_legend(title = "Allocation ratio")) guides(= guide_legend(= "Allocation ratio"))
```
Figure 4.4: Power of z-test as a Function of Allocation Ratio and Sample Size
As the allocation ratio increases the power decreases assuming other parameters remain fixed (see Figure 4.4).
4.6 Power and Sample Size for Comparing Proportions
This section discusses power and sample size for studies where the primary study outcomes are dichotomous.
Let (p_1) denote the response rate in group 1, (p_2) the response rate in group 2, and let the difference be (\theta= p_1—p_2). A binary outcome for subject (i) in arm (k) is
[Y_{ik} = \left{ \begin{array}{ll} 1 & \mbox{with probability } p_k \ 0 & \mbox{with probability } 1-p_k, \end{array} \right.]
for (i = 1,...,n_k) and (k = 1,2). The sum of independent and identically distributed Bernoulli random variables has a binomial distribution,
[ \sum_{i = 1}^{n_k} Y_{ik} \sim Bin(n_k,p_k), \thinspace k = 1,2.]
The sample proportion for group (k) is
[{\hat p}_k= \frac{1}{n_k}\sum_{i = 1}^{n_k} Y_{ik}, \thinspace k = 1,2,]
and (E\left( {\hat p}_k\right)=p_k) and (Var\left({\hat p}_k \right)=p_k(1-p_k)/{n_k}).
Consider a study where the aim is to determine if there is a difference between the two groups. That is we want to test (H_0: \theta = 0) versus (H_1: \theta \ne 0.)
If (H_0) is true then the test statistic
[Z = \frac {{\hat p}_1-{\hat p}_2} {\sqrt{p_1(1-p_1)/n_1+p_2(1-p_2)/n_2}} \sim N(0,1).]
The test rejects at level (\alpha) if and only if
[\left|Z\right| \ge z_{1-\alpha/2}.]
Using the same argument as the case with continuous endpoints in Section 4.5.2 and ignoring terms smaller than (\alpha/2), we can solve for (\beta)
[\begin{equation} \beta \approx \Phi\left( z_{1-\alpha/2}- \frac{|\theta_1|}{\sqrt{p_1(1-p_1)/n_1+p_2(1-p_2)/n_2}}\right). \tag{4.9} \end{equation}]
A formula for sample size can be derived using (4.9). If (n_1 = r \cdot n_2) then
[n_2= \frac {\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{\theta^2} \left(p_1(1-p_1)/r+p_2(1-p_2) \right),] where (\alpha,\beta \in (0,1).)
4.6.1 Computation Lab: Power and Sample Size for Comparing Proportions
stats::power.prop.test() can be used to calculate sample size or power.
Example 4.3 The standard treatment for a disease has a response rate of 20%, and an experimental treatment is anticipated to have a response rate of 28%. The investigators want both groups to have an equal number of subjects. How many patients should be enrolled if the study will conduct a two-sided test at the 5% level with 80% power?
power.prop.test(p1 = 0.2, p2 = 0.28, power = 0.8)power.prop.test(=0.2 =0.28 =0.8)#> #> #> Two-sample comparison of proportions power calculation #> Two-sample comparison of proportions power calculation #> #> #> n = 446.2054#> n = 446.2054#> p1 = 0.2#> p1 = 0.2#> p2 = 0.28#> p2 = 0.28#> sig.level = 0.05#> sig.level = 0.05#> power = 0.8#> power = 0.8#> alternative = two.sided#> alternative = two.sided#> #> #> NOTE: n is number in each group#> NOTE: n is number in each group
This means that 446 patients should be enrolled in each group for a study to have a power of 0.8 to detect (p_1=0.2) and (p_2=0.28) at the (\alpha=0.05) level.
4.7 Calculating Power by Simulation
Statistical power is the probability that the test correctly rejects the null hypothesis. Consider a thought experiment: imagine being able to replicate a study with different random samples a large number of times, and each time the null hypothesis is tested. If the null hypothesis is false, it’s reasonable to expect a large proportion of these tests reject the null hypothesis. This thought experiment can be operationalized via simulation.
4.7.1 Algorithm for Simulating Power
Use an underlying model to generate random data with (a) specified sample sizes, (b) parameter values that are being tested via the hypothesis test, and (c) other (nuisance) parameters such as variances.
Run an estimation program (e.g., a two-sample t-test) on these randomly generated data.
Calculate the test statistic and p-value.
Repeat steps 1–3 many times, say, N, and save the p-values. The estimated power for a level alpha test is the proportion of observations (out of N) for which the p-value is less than alpha.
4.7.2 Computation Lab: Simulating Power of a Two-Sample t-test
If the test statistic and distribution of the test statistic are known, then the power of the test can be calculated via simulation.
Example 4.4 Consider a two-sample t-test with 30 subjects per group and the standard deviation known to be 1. What is the power of the test (H_0:\mu_1-\mu_2 = 0) versus (H_1:\mu_1-\mu_2 = 0.5), at the 5% significance level?
Power is the proportion of times that the test correctly rejects the null hypothesis in repeated testing of the same hypothesis based on different randomly drawn samples.
A single study is simulated below. Let’s assume that (n_1 = n_2 = 30), (\mu_1 = 3.5), (\mu_2 = 3), (\sigma = 1), and (\alpha = 0.05).
set.seed(2301)set.seed(2301)samp1 <- rnorm(30, mean = 3.5, sd = 1) samp1<- rnorm(30 =3.5 = 1)samp2 <- rnorm(30, mean = 3, sd = 1) samp2<- rnorm(30 = 3 = 1)twosampt <- t.test(samp1, samp2, var.equal = T) twosampt<-t.test(samp1 samp2 = T)twosampt$p.value twosampt $p.value#> 0.03604893#> 0.03604893
The null hypothesis would be rejected at the 5% level.
Suppose that 10 studies are simulated. What proportion of these 10 studies will reject the null hypothesis at the 5% level? To investigate how many times the two-sample t-test will reject at the 5% level, the replicate() function will be used to generate 10 studies and calculate the p-value in each study. It will still be assumed that (n_1 = n_2 = 30), (\mu_1 = 3.5), (\mu_2 = 3), (\sigma = 1), and (\alpha = 0.05).
set.seed(2301)set.seed(2301)tpval <- function() {tpval<- function(){ n <- 30 n<- 30 mu1 <- 3.5 mu1<-3.5 mu2 <- 3.0 mu2<-3.0 sigma <- 1 sigma<- 1 samp1 <- rnorm(n, mean = mu1, sd = sigma) samp1<- rnorm(n = mu1 = sigma) samp2 <- rnorm(n, mean = mu2, sd = sigma) samp2<- rnorm(n = mu2 = sigma) twosamp <- t.test(samp1, samp2, var.equal = T) twosamp<-t.test(samp1 samp2 = T) return(twosamp$p.value) return(twosamp $p.value)}}reps <- 10 reps<- 10pvals <- replicate(reps, tpval()) pvals<- replicate(reps tpval()) #power is the number of times the test rejects at the 5% level #power is the number of times the test rejects at the 5% levelsum(pvals <= 0.05) / reps sum(pvals<=0.05)/ reps#> 0.3#> 0.3
But, since we only simulated 10 studies the estimate of power will have a large standard error. So let’s try simulating 10,000 studies so that we can obtain a more precise estimate of power.
set.seed(2301)set.seed(2301)reps <- 10000 reps<- 10000pvals <- replicate(reps, tpval()) pvals<- replicate(reps tpval())sum(pvals <= 0.05) / reps sum(pvals<=0.05)/ reps#> 0.4881#> 0.4881
This is much closer to the theoretical power obtained from stats::power.t.test().
powt <- power.t.test(powt<-power.t.test( n = 30, = 30 delta = 0.5, =0.5 sd = 1, = 1 sig.level = 0.05 =0.05)) powt$power powt $ power#> 0.477841#> 0.477841
Example 4.5 Suppose that the standard treatment for a disease has a response rate of 20%, and an experimental treatment is anticipated to have a response rate of 28%. The researchers are considering enrolling 1,500 patients in the standard group and 500 patients in the experimental arm. What is the power of this study?
The number of subjects in the experimental arm that have a positive response to treatment will be an observation from a (Bin(1500,0.20)) and the number of subjects that have a positive response to the standard treatment will be an observation from a (Bin(500,0.28)). We can obtain simulated responses from these distributions using the rbinom() function.
set.seed(2301)set.seed(2301)rbinom(1, 500, 0.28) rbinom(1 5000.28)#> 132#> 132rbinom(1, 1500, 0.20) rbinom(1 15000.20)#> 324#> 324
In this simulated study, 132 of the 500 patients in the experimental arm had a positive response to the experimental treatment and 324 of the 1,500 patients in the control arm had a positive response to the standard treatment. The p-value for this simulated study can be obtained using prop.test().
set.seed(2301)set.seed(2301)samp1 <- rbinom(1, 500, 0.28) samp1<- rbinom(1 5000.28)samp2 <- rbinom(1, 1500, 0.20) samp2<- rbinom(1 15000.20)twosamp <- prop.test(twosamp<-prop.test( x = c(samp1, samp2), = c(samp1 samp2) n = c(500, 1500), = c(500 1500) correct = F = F))twosamp$p.value twosamp $p.value#> 0.0267226#> 0.0267226
In this study, the p-value is 0.03, which is less than 0.05 so there would be evidence that the new treatment is significantly better than the standard treatment. A power simulation repeats this process a large number of times. The replicate() command can be used for the repetition.
set.seed(2301)set.seed(2301)proppval <- function() {proppval<- function(){ n1 <- 500 n1<- 500 p1 <- 0.28 p1<-0.28 n2 <- 1500 n2<- 1500 p2 <- 0.20 p2<-0.20 samp1 <- rbinom(n = 1, size = n1, prob = p1) samp1<- rbinom(= 1 = n1 = p1) samp2 <- rbinom(n = 1, size = n2, prob = p2) samp2<- rbinom(= 1 = n2 = p2) twosamp <- prop.test(x = c(samp1, samp2), twosamp<-prop.test(= c(samp1 samp2) n = c(n1, n2), = c(n1 n2) correct = F) = F) return(twosamp$p.value) return(twosamp $p.value)}}reps <- 10000 reps<- 10000pvals <- replicate(reps, proppval()) pvals<- replicate(reps proppval())sum(pvals <= 0.05) / reps sum(pvals<=0.05)/ reps#> 0.9534#> 0.9534
The power of the study in this case is 0.95.
4.8 Exercises
Exercise 4.1 Consider the power function of one-sample z-test shown in Equation (4.1). What is the limit of the power function as (n \rightarrow \infty)? How about when (\mu_1\rightarrow \mu_0)? What do the results tell you about the power of the one-sample z-test?
Exercise 4.2 Show that
[\begin{aligned} & P_{H_1}\left(\frac {{\bar X} -\mu_0} {\frac{S}{\sqrt{n}}} \ge t_{n-1, 1-\alpha/2}\right) + P_{H_1}\left(\frac {{\bar X} -\mu_0} {\frac{S}{\sqrt{n}}} < - t_{n-1, 1-\alpha/2}\right) \notag \ =& P(t_{n-1,\gamma}\ge t_{n-1,1-\alpha/2})+P(t_{n-1,\gamma}< -t_{n-1,1-\alpha/2}). \end{aligned}]
The right-hand side is the final expression for the power of the t-test in Section 4.4.
Exercise 4.3 Derive equation (4.3) for the total sample of the two-sample t-test.
Exercise 4.4 Consider the derivation of Equation (4.7) for the sample size of a two-sample z-test with known variance and equal allocation.
Identify terms in Equation (4.4) that are smaller than (\alpha/2) and derive line (4.5) based on the information.
Show how Equation (4.6) is derived from Equation (4.5).
Exercise 4.5 Consider Figure 4.2 and the R code that generated the plots.
Modify the code so that sample size is on the x-axis, and three different lines show the relationships between sample size and power for a difference of (\theta=1,2,3). Fix the standard deviation to 1.5 for all lines.
Interpret the relationships between power, effect size (\theta), standard deviation (\sigma), and sample size (n), on Figure 4.2 and the figure from part a. For example, what happens to power as (\theta) decreases with (n) and (\sigma) fixed? How does (\theta) change as power increases with other parameters fixed?
Exercise 4.6 Consider the plot of power as a function of effect size for the two-sample t-test shown in Figure 4.3.
Create a plot to show power as a function of both effect size and sample size while keeping other parameters fixed.
The curve is shaped like an inverted bell curve, or a “bathtub” shape. Interpret the shape of the curve.
Exercise 4.7 Consider the plot of power as a function of the allocation ratio, (r), for the two-sample z-test shown in Figure 4.4 and the R code for generating plot.
Recall that n is the total sample size for the function pow_ztest(). What does n2 represent, and why is it n/(r+1)?
Does sample size imbalance between the two groups lead to an increase or decrease in power? Explain.
Modify the function pow_ztest() to return the value of n2 and plot n2 versus r. Describe the relationship between the two variables.
Exercise 4.8 The R function size2z.test() shown below implements the sample size formula for calculating the sample size for a test of (H_0:\theta = 0) versus (H_1: \theta \ne 0), where (\theta=\mu_1-\mu_2).
size2z.test <- function(theta, alpha, beta, sigma)size2z.test<- function(theta alpha beta sigma){{ zalpha <- qnorm(1-alpha/2) zalpha<- qnorm(1 - alpha/ 2) zbeta <- qnorm(1-beta) zbeta<- qnorm(1 - beta) (2sigma(zalpha+zbeta)/theta)^2(2 sigma(zalpha + zbeta)/ theta) ^ 2}}
What are the main assumptions behind this sample size formula?
Assuming all other parameters remain fixed, indicate whether the sample size increases or decreases in each of the following cases.
As (\sigma) decreases.
As (\alpha) decreases.
As (\theta) decreases.
Exercise 4.9 A statistician is designing a phase III clinical trial comparing a continuous outcome in two groups receiving experimental versus standard therapy with a total sample size of 168 patients. The team requires the study have 80% power at the 5% significance level to detect a difference of 1. Assume that the standard deviation of the outcome is 2. The design team would like to investigate whether it’s possible to have four times as many patients in the experimental group versus the control group without having to increase the total sample size.
What is the power if there are four times as many patients in the experimental group? What should the statistician recommend to the team in order for the study to have at least 80% power?
Exercise 4.10 Let (X_1, X_2, ..., X_n) be iid (N\left(\mu,\sigma^2\right)).
Show that the power function of the test (H_0:\mu = 0) versus (H_1:\mu > 0) at (\mu=1) is
where (z_{\frac{\alpha}{2}}) is the (100\left(1-\frac{\alpha}{2}\right)^{th}) percentile of the (N\left(0,1\right)).
2. Use R to calculate the power when (n = 10), (\alpha = 0.01), and (\sigma = 1).
3 Comparing Two Treatments
5 Comparing More Than Two Treatments |
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1.4.1 Independence
Let A be the event that it rains tomorrow, and suppose that P(A)=1 3. Also suppose that I toss a fair coin; let B be the event that it lands heads up. We have P(B)=1 2.
Now I ask you, what is P(A|B)? What is your guess? You probably guessed that P(A|B)=P(A)=1 3. You are right! The result of my coin toss does not have anything to do with tomorrow's weather. Thus, no matter if B happens or not, the probability of A should not change. This is an example of two independent events. Two events are independent if one does not convey any information about the other. Let us now provide a formal definition of independence.
Two events A and B are independent if and only if P(A∩B)=P(A)P(B).
Now, let's first reconcile this definition with what we mentioned earlier, P(A|B)=P(A). If two events are independent, then P(A∩B)=P(A)P(B), so
P(A|B)=P(A∩B)P(B)
=P(A)P(B)P(B)
=P(A).
Thus, if two events A and B are independent and P(B)≠0, then P(A|B)=P(A). To summarize, we can say "independence means we can multiply the probabilities of events to obtain the probability of their intersection", or equivalently, "independence means that conditional probability of one event given another is the same as the original (prior) probability".
Sometimes the independence of two events is quite clear because the two events seem not to have any physical interaction with each other (such as the two events discussed above). At other times, it is not as clear and we need to check if they satisfy the independence condition. Let's look at an example.
Example
I pick a random number from {1,2,3,⋯,10}, and call it N. Suppose that all outcomes are equally likely. Let A be the event that N is less than 7, and let B be the event that N is an even number. Are A and B independent?
Solution
We have A={1,2,3,4,5,6}, B={2,4,6,8,10}, and A∩B={2,4,6}. Then P(A)=0.6,P(B)=0.5,P(A∩B)=0.3 Therefore, P(A∩B)=P(A)P(B), so A and B are independent. This means that knowing that B has occurred does not change our belief about the probability of A. In this problem the two events are about the same random number, but they are still independent because they satisfy the definition.
The definition of independence can be extended to the case of three or more events.
Three events A, B, and C are independent if all of the following conditions hold P(A∩B)=P(A)P(B),P(A∩C)=P(A)P(C),P(B∩C)=P(B)P(C),P(A∩B∩C)=P(A)P(B)P(C).
Note that all four of the stated conditions must hold for three events to be independent. In particular, you can find situations in which three of them hold, but the fourth one does not. In general, for n events A 1,A 2,⋯,A n to be independent we must have P(A i∩A j)=P(A i)P(A j),for all distinct i,j∈{1,2,⋯,n};
P(A i∩A j∩A k)=P(A i)P(A j)P(A k),for all distinct i,j,k∈{1,2,⋯,n};......P(A 1∩A 2∩A 3⋯∩A n)=P(A 1)P(A 2)P(A 3)⋯P(A n).
This might look like a difficult definition, but we can usually argue that the events are independent in a much easier way. For example, we might be able to justify independence by looking at the way the random experiment is performed. A simple example of an independent event is when you toss a coin repeatedly. In such an experiment, the results of any subset of the coin tosses do not have any impact on the other ones.
Example
I toss a coin repeatedly until I observe the first tails at which point I stop. Let X be the total number of coin tosses. Find P(X=5).
Solution
Here, the outcome of the random experiment is a number X. The goal is to find P(A)=P(5). But what does X=5 mean? It means that the first 4 coin tosses result in heads and the fifth one results in tails. Thus the problem is to find the probability of the sequence H H H H T when tossing a coin five times. Note that H H H H T is a shorthand for the event "(The first coin toss results in heads) and (The second coin toss results in heads) and (The third coin toss results in heads) and (The fourth coin toss results in heads) and (The fifth coin toss results in tails)." Since all the coin tosses are independent, we can write
P(H H H H T)=P(H)P(H)P(H)P(H)P(T)
=1 2.1 2.1 2.1 2.1 2
=1 32.
Discussion: Some people find it more understandable if you look at the problem in the following way. I never stop tossing the coin. So the outcome of this experiment is always an infinite sequence of heads or tails. The value X (which we are interested in) is just a function of the beginning part of the sequence until you observe a tails. If you think about the problem this way, you should not worry about the stopping time. For this problem it might not make a big difference conceptually, but for some similar problems this way of thinking might be beneficial.
We have seen that two events A and B are independent if P(A∩B)=P(A)P(B). In the next two results, we examine what independence can tell us about other set operations such as complements and unions.
Lemma
If A and B are independent then
A and B c are independent,
A c and B are independent,
A c and B c are independent.
Proof
We prove the first one as the others can be concluded from the first one immediately. We have
P(A∩B c)=P(A−B)
=P(A)−P(A∩B)
=P(A)−P(A)P(B)since A and B are independent
=P(A)(1−P(B))
=P(A)P(B c).
Thus, A and B c are independent.
Sometimes we are interested in the probability of the union of several independent events A 1,A 2,⋯,A n. For independent events, we know how to find the probability of intersection easily, but not the union. It is helpful in these cases to use De Morgan's Law: A 1∪A 2∪⋯∪A n=(A c 1∩A c 2∩⋯∩A c n)c
Thus we can write
P(A 1∪A 2∪⋯∪A n)=1−P(A c 1∩A c 2∩⋯∩A c n)
=1−P(A c 1)P(A c 2)⋯P(A c n)(since the A i's are independent)
=1−(1−P(A 1))(1−P(A 2))⋯(1−P(A n)).
If A 1,A 2,⋯,A n are independent then P(A 1∪A 2∪⋯∪A n)=1−(1−P(A 1))(1−P(A 2))⋯(1−P(A n)).
Example
Suppose that the probability of being killed in a single flight is p c=1 4×10 6 based on available statistics. Assume that different flights are independent. If a businessman takes 20 flights per year, what is the probability that he is killed in a plane crash within the next 20 years? (Let's assume that he will not die because of another reason within the next 20 years.)
Solution
The total number of flights that he will take during the next 20 years is N=20×20=400. Let p s be the probability that he survives a given single flight. Then we have p s=1−p c. Since these flights are independent, the probability that he will survive all N=400 flights is P(Survive N flights)=p s×p s×⋯×p s=p N s=(1−p c)N. Let A be the event that the businessman is killed in a plane crash within the next 20 years. Then P(A)=1−(1−p c)N=9.9995×10−5≈1 10000.
Warning! One common mistake is to confuse independence and being disjoint. These are completely different concepts. When two events A and B are disjoint it means that if one of them occurs, the other one cannot occur, i.e., A∩B=∅. Thus, event A usually gives a lot of information about event B which means that they cannot be independent. Let's make it precise.
Lemma
Consider two events A and B, with P(A)≠0 and P(B)≠0. If A and B are disjoint, then they are not independent.
Proof
Since A and B are disjoint, we have P(A∩B)=0≠P(A)P(B).
Thus, A and B are not independent. ◻
Table 1.1 summarizes the two concepts of disjointness and independence.
Concept Meaning Formulas
Disjoint A and B cannot occur at the same time A∩B=∅,
P(A∪B)=P(A)+P(B)
Independent A does not give any information about B P(A|B)=P(A),P(B|A)=P(B)
P(A∩B)=P(A)P(B)
Table 1.1: Differences between disjointness and independence.
Example (A similar problem is given in )
Two basketball players play a game in which they alternately shoot a basketball at a hoop. The first one to make a basket wins the game. On each shot, Player 1 (the one who shoots first) has probability p 1 of success, while Player 2 has probability p 2 of success (assume 0<p 1,p 2<1). The shots are assumed to be independent.
Find P(W 1), the probability that Player 1 wins the game.
For what values of p 1 and p 2 is this a fair game, i.e., each player has a 50 percent chance of winning the game?
Solution
In this game, the event W 1 can happen in many different ways. We calculate the probability of each of these ways and then add them up to find the total probability of winning. In particular, Player 1 may win on her first shot, or her second shot, and so on. Define A i as the event that Player 1 wins on her i'th shot. What is the probability of A i? A i happens if Player 1 is unsuccessful at her first i−1 shots and successful at her i th shot, while Player 2 is unsuccessful at her first i−1 shots. Since different shots are independent, we obtain P(A 1)=p 1,P(A 2)=(1−p 1)(1−p 2)p 1,P(A 3)=(1−p 1)(1−p 2)(1−p 1)(1−p 2)p 1,⋯P(A k)=[(1−p 1)(1−p 2)]k−1 p 1,⋯ Note that A 1,A 2,A 3,⋯ are disjoint events, because if one of them occurs the other ones cannot occur. The event that Player 1 wins is the union of the A i's, and since the A i's are disjoint, we have
P(W 1)=P(A 1∪A 2∪A 3∪⋯)
=P(A 1)+P(A 2)+P(A 3)+⋯
=p 1+(1−p 1)(1−p 2)p 1+[(1−p 1)(1−p 2)]2 p 1+⋯
=p 1[1+(1−p 1)(1−p 2)+[(1−p 1)(1−p 2)]2+⋯].
Note that since 0<p 1,p 2<1, for x=(1−p 1)(1−p 2) we have 0<x<1. Thus, using the geometric sum formula (∑∞k=0 a x k=a 1 1−x for |x|<1), we obtain P(W 1)=p 1 1−(1−p 1)(1−p 2)=p 1 p 1+p 2−p 1 p 2. It is always a good idea to look at limit cases to check our answer. For example, if we plug in p 1=0,p 2≠0, we obtain P(W 1)=0, which is what we expect. Similarly, if we let p 2=0,p 1≠0, we obtain P(W 1)=1, which again makes sense.
Now, to make this a fair game (in the sense that P(W 1)=.5), we have P(W 1)=p 1 p 1+p 2−p 1 p 2=0.5 and we obtain p 1=p 2 1+p 2. Note that this means that p 1<p 2, which makes sense intuitively. Since Player 1 has the advantage of starting the game, she should have a smaller success rate so that the whole game is fair.
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Practical uncertainty:Useful Ideas in Decision-Making, Risk, Randomness, & AI
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0 Preface
1 Basic Concepts
1.0 Introduction
1.1 Introduction
1.1.0 What Is Probability?
1.1.1 Example
1.2 Review of Set Theory
1.2.0 Review
1.2.1 Venn Diagrams
1.2.2 Set Operations
1.2.3 Cardinality
1.2.4 Functions
1.2.5 Solved Problems
1.3 Random Experiments and Probabilities
1.3.1 Random Experiments
1.3.2 Probability
1.3.3 Finding Probabilities
1.3.4 Discrete Models
1.3.5 Continuous Models
1.3.6 Solved Problems
1.4 Conditional Probability
1.4.0 Conditional Probability
1.4.1 Independence
1.4.2 Law of Total Probability
1.4.3 Bayes' Rule
1.4.4 Conditional Independence
1.4.5 Solved Problems
1.5 Problems
1.5.0 End of Chapter Problems
2 Combinatorics: Counting Methods
2.1 Combinatorics
2.1.0 Finding Probabilities with Counting Methods
2.1.1 Ordered with Replacement
2.1.2 Ordered without Replacement
2.1.3 Unordered without Replacement
2.1.4 Unordered with Replacement
2.1.5 Solved Problems
2.2 Problems
2.2.0 End of Chapter Problems
3 Discrete Random Variables
3.1 Basic Concepts
3.1.1 Random Variables
3.1.2 Discrete Random Variables
3.1.3 Probability Mass Function
3.1.4 Independent Random Variables
3.1.5 Special Distributions
3.1.6 Solved Problems
3.2 More about Discrete Random Variables
3.2.1 Cumulative Distribution Function
3.2.2 Expectation
3.2.3 Functions of Random Variables
3.2.4 Variance
3.2.5 Solved Problems
3.3 Problems
3.3.0 End of Chapter Problems
4 Continuous and Mixed Random Variables
4.0 Introduction
4.1 Continuous Random Variables
4.1.0 Continuous Random Variables and their Distributions
4.1.1 Probability Density Function
4.1.2 Expected Value and Variance
4.1.3 Functions of Continuous Random Variables
4.1.4 Solved Problems
4.2 Special Distributions
4.2.1 Uniform Distribution
4.2.2 Exponential Distribution
4.2.3 Normal (Gaussian) Distribution
4.2.4 Gamma Distribution
4.2.5 Other Distributions
4.2.6 Solved Problems
4.3 Mixed Random Variables
4.3.1 Mixed Random Variables
4.3.2 Using the Delta Function
4.3.3 Solved Problems
4.4 Problems
4.4.0 End of Chapter Problems
5 Joint Distributions
5.1 Two Discrete Random Variables
5.1.0 Two Random Variables
5.1.1 Joint Probability Mass Function (PMF)
5.1.2 Joint Cumulative Distribution Function (CDF)
5.1.3 Conditioning and Independence
5.1.4 Functions of Two Random Variables
5.1.5 Conditional Expectation
5.1.6 Solved Problems
5.2 Two Continuous Random Variables
5.2.0 Two Continuous Random Variables
5.2.1 Joint Probability Density Function
5.2.2 Joint Cumulative Distribution Function
5.2.3 Conditioning and Independence
5.2.4 Functions of Two Continuous Random Variables
5.2.5 Solved Problems
5.3 More Topics
5.3.1 Covariance and Correlation
5.3.2 Bivariate Normal Distribution
5.3.3 Solved Problems
5.4 Problems
5.4.0 End of Chapter Problems
6 Multiple Random Variables
6.0 Introduction
6.1 Methods for More Than Two Random Variables
6.1.1 Joint Distributions and Independence
6.1.2 Sums of Random Variables
6.1.3 Moment Generating Functions
6.1.4 Characteristic Functions
6.1.5 Random Vectors
6.1.6 Solved Problems
6.2 Probability Bounds
6.2.0 Probability Bounds
6.2.1 Union Bound and Extension
6.2.2 Markov Chebyshev Inequalities
6.2.3 Chernoff Bounds
6.2.4 Cauchy Schwarz Inequality
6.2.5 Jensen's Inequality
6.2.6 Solved Problems
6.3 Problems
6.3.0 End of Chapter Problems
7 Limit Theorems and Convergence of Random Variables
7.0 Introduction
7.1 Limit Theorems
7.1.0 Limit Theorems
7.1.1 Law of Large Numbers
7.1.2 Central Limit Theorem (CLT)
7.1.3 Solved Problems
7.2 Convergence of Random Variables
7.2.0 Convergence of Random Variables
7.2.1 Convergence of Sequence of Numbers
7.2.2 Sequence of Random Variables
7.2.3 Different Types of Convergence for Sequences of Random Variables
7.2.4 Convergence in Distribution
7.2.5 Convergence in Probability
7.2.6 Convergence in Mean
7.2.7 Almost Sure Convergence
7.2.8 Solved Problems
7.3 Problems
7.3.0 End of Chapter Problems
8 Statistical Inference I: Classical Methods
8.1 Introduction
8.1.0 Introduction
8.1.1 Random Sampling
8.2 Point Estimation
8.2.0 Point Estimation
8.2.1 Evaluating Estimators
8.2.2 Point Estimators for Mean and Variance
8.2.3 Maximum Likelihood Estimation (MLE)
8.2.4 Asymptotic Properties of MLEs
8.2.5 Solved Problems
8.3 Interval Estimation (Confidence Intervals)
8.3.0 Interval Estimation (Confidence Intervals)
8.3.1 The general framework of Interval Estimation
8.3.2 Finding Interval Estimators
8.3.3 Confidence Intervals for Normal Samples
8.3.4 Solved Problems
8.4 Hypothesis Testing
8.4.1 Introduction
8.4.2 General Setting and Definitions
8.4.3 Hypothesis Testing for the Mean
8.4.4 P-Values
8.4.5 Likelihood Ratio Tests
8.4.6 Solved Problems
8.5 Linear Regression
8.5.0 Linear Regression
8.5.1 Simple Linear Regression Model
8.5.2 The First Method for Finding beta
8.5.3 The Method of Least Squares
8.5.4 Extensions and Issues
8.5.5 Solved Problems
8.6 Problems
8.6.0 End of Chapter Problems
9 Statistical Inference II: Bayesian Inference
9.1 Bayesian Inference
9.1.0 Bayesian Inference
9.1.1 Prior and Posterior
9.1.2 Maximum A Posteriori (MAP) Estimation
9.1.3 Comparison to ML Estimation
9.1.4 Conditional Expectation (MMSE)
9.1.5 Mean Squared Error (MSE)
9.1.6 Linear MMSE Estimation of Random Variables
9.1.7 Estimation for Random Vectors
9.1.8 Bayesian Hypothesis Testing
9.1.9 Bayesian Interval Estimation
9.1.10 Solved Problems
9.2 Problems
9.2.0 End of Chapter Problems
10 Introduction to Random Processes
10.1 Basic Concepts
10.1.0 Basic Concepts
10.1.1 PDFs and CDFs
10.1.2 Mean and Correlation Functions
10.1.3 Multiple Random Processes
10.1.4 Stationary Processes
10.1.5 Gaussian Random Processes
10.1.6 Solved Problems
10.2 Processing of Random Signals
10.2.0 Processing of Random Signals
10.2.1 Power Spectral Density
10.2.2 Linear Time-Invariant (LTI) Systems with Random Inputs
10.2.3 Power in a Frequency Band
10.2.4 White Noise
10.2.5 Solved Problems
10.3 Problems
10.3.0 End of Chapter Problems
11 Some Important Random Processes
11.1 Poisson Processes
11.1.0 Introduction
11.1.1 Counting Processes
11.1.2 Basic Concepts of the Poisson Process
11.1.3 Merging and Splitting Poisson Processes
11.1.4 Nonhomogeneous Poisson Processes
11.1.5 Solved Problems
11.2 Discrete-Time Markov Chains
11.2.1 Introduction
11.2.2 State Transition Matrix and Diagram
11.2.3 Probability Distributions
11.2.4 Classification of States
11.2.5 Using the Law of Total Probability with Recursion
11.2.6 Stationary and Limiting Distributions
11.2.7 Solved Problems
11.3 Continuous-Time Markov Chains
11.3.1 Introduction
11.3.2 Stationary and Limiting Distributions
11.3.3 The Generator Matrix
11.3.4 Solved Problems
11.4 Brownian Motion (Wiener Process)
11.4.0 Brownian Motion (Wiener Process)
11.4.1 Brownian Motion as the Limit of a Symmetric Random Walk
1.4.2 Definition and Some Properties
11.4.3 Solved Problems
11.5 Problems
11.5.0 End of Chapter Problems
12 Introduction to Simulation Using MATLAB
13 Introduction to Simulation Using R
14 Introduction to Simulation Using Python
15 Recursive Methods
Appendix
Some Important Distributions
Review of the Fourier Transform
Bibliography
Introduction to Probability by Hossein Pishro-Nik is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License |
9127 | https://arxiv.org/pdf/math/0208056 | arXiv:math/0208056v1 [math.NT] 7 Aug 2002
Curves Dy 2 = x3 − x of odd analytic rank
Noam D. Elkies 1
Department of Mathematics, Harvard University, Cambridge, MA 02138 USA
elkies@math.harvard.edu
Abstract. For nonzero rational D, which may be taken to be a square-free integer, let ED be the elliptic curve Dy 2 = x3 − x over Q arising in the “congruent number” problem. 1 It is known that the L-function of
ED has sign −1, and thus odd analytic rank ran (ED ), if and only if |D|
is congruent to 5, 6, or 7 mod 8. For such D, we expect by the conjec-ture of Birch and Swinnerton-Dyer that the arithmetic rank of each of these curves ED is odd, and therefore positive. We prove that ED has positive rank for each D such that |D| is in one of the above congruence classes mod 8 and also satisfies |D| < 10 6. Our proof is computational: we use the modular parametrization of E1 or E2 to construct a rational point PD on each ED from CM points on modular curves, and compute
PD to enough accuracy to usually distinguish it from any of the rational torsion points on ED . In the 1375 cases in which we cannot numerically distinguish PD from ( ED )tors , we surmise that PD is in fact a torsion point but that ED has rank 3, and prove that the rank is positive by searching for and finding a non-torsion rational point. We also report on the conjectural extension to |D| < 10 7 of the list of curves ED with odd
ran (ED ) > 1, which raises several new questions.
1 Introduction
1.1 Review: The curves ED and their arithmetic
For nonzero rational D let ED be the elliptic curve
ED : Dy 2 = x3 − x (1) over Q. Since ED and Ec2 D are isomorphic for any nonzero rational c, D , we may assume without loss of generality that D is a squarefree integer. The change of variable x ↔ − x shows that ED is also isomorphic with E−D; this may also be seen from the Weierstrass equation y2 = x3 − D2x for ED.
1
The problem is: for which D does ED have nontrivial rational points, or equivalently positive rank? Such D are called “congruent”, because they are precisely the num-bers that arise as the common difference (“congruum”) of a three-term arithmetic progression of rational squares, namely the squares of ( x2 − 2x − 1) /2y, ( x2 + 1) /2y,and ( x2 + 2 x − 1) /2y. See the Preface and Chapter XVI of [Di] for the early history of this problem, and [Kob] for a more modern treatment of the curves ED .2 Noam D. Elkies
The arithmetic of the curves ED has long attracted interest, both for its connection with the classical “congruent number” problem (see [Di, Ch.XVI];
|D| is a “congruent number” if and only if ED has positive rank) and, more recently, as a paradigmatic example and test case for results and constructions concerning elliptic curves in general (see for instance [Kob]). The curves ED have some special properties that make them more accessible than general elliptic curves over Q. They have complex multiplication and are quadratic twists of the curve E1. This led to the computation of the sign of the functional equation of the L-function L(ED/Q, s ): it depends on |D| mod 8, and equals +1 or −1according as |D| is in {1, 2, 3} or {5, 6, 7} mod 8. We shall be concerned with the case of sign −1. The conjecture of Birch and Swinnerton-Dyer (BSD) predicts that the (arith-metic) rank of any elliptic curve E over a number field K, defined as the Z-rank of its Mordell-Weil group E(K), should equal the order of vanishing at s = 1 of
L(E/K, s ), known as the “analytic rank” ran (E/K ). The BSD conjecture im-plies the “BSD parity conjecture”: the arithmetic rank is even or odd according as the functional equation of L(E/K, s ) has sign +1 or −1. It would follow that if the sign is −1 then E always has positive rank. In our context, where K = Q
and E = ED , this leads to the conjecture that ED has positive rank (and thus that |D| is a “congruent number”) if |D| is any 2 integer of the form 8 k +5, 8 k +6, or 8 k + 7.
1.2 New results and computations
We prove:
Theorem 1. Let D be an integer such that |D| is congruent to 5, 6, or 7 mod 8
and also satisfies |D| < 10 6. Then ED has positive rank over Q.
In our ANTS-1 paper [E1] we announced such a result for |D| < 2·10 5. Our main tool for proving Theorem 1 is the same: we use the modular parametrization of E1
or E2 to construct a rational point PD on each ED from CM points on modular curves, and usually compute PD to enough accuracy to distinguish it from any of the rational torsion points on ED. Faster computer hardware and new software were both needed to extend the computation to 10 6. The faster machine made it feasible to compute PD for more and larger D. Cremona’s program mwrank , not available when [E1] was written, found rational points on the curves ED on which we could neither distinguish PD from a torsion point nor find a rational nontor-sion point by direct search. This happened for 1375 values of |D| — less than 0 .5% of the total, but too many to list here a rational point on ED for each such D.These tables, and further computational data on the curves ED , can be found on the Web starting from .
2We have dropped the hypothesis that Dbe squarefree because c2D≡Dmod 8 for any odd integer c. Our integers Dare not divisible by 4, and therefore cannot be of the form c2Dfor any even c.Curves Dy 2=x3−xof odd analytic rank 3
Our computations also yield conjectural information on the rank of ED: the rank should equal 1 if and only if PD is nontorsion. In half the cases, those for which |D| or |D|/2 is of the form 8 k + 7, we obtain this connection from Kolyvagin’s theorem [Kol], which gives the “if” direction unconditionally, and the Gross-Zagier formula [GZ], which gives the “only if” direction under the BSD conjecture. Neither Kolyvagin nor Gross-Zagier has been proved to extend to the remaining cases, when |D| or |D|/2 is of the form 8 k + 5. But we expect that similar results do hold in these cases, and hence that ED has rank 1 if and only if PD is nontorsion also when |D| or |D|/2 is congruent to 5 mod 8. One piece of evidence in this direction is that whenever we found PD to be numerically indistinguishable from a torsion point, the Selmer groups for the 2-isogenies between ED and the curve Dy 2 = x3 + 4 x were large enough for
ED to have arithmetic rank at least 3. We extended the list of curves ED of conjectural rank ≥ 3 to |D| < 10 7 by imposing the 2-descent condition from the start and computing PD only for those D that pass this test. We find a total of 8740 values of |D|. The list not only provides new numerical data on the distribution of quadratic twists of rank > 1 with large |D|, but also suggests unexpected biases in the distribution that favor some congruence classes of |D|’s.
2 Proof of Theorem 1
Let D be a squarefree integer such that |D| is congruent to 5, 6, or 7 mod 8. Set
KD = Q(√−| D| ) if D is odd, and KD = Q(√−| D|/2 ) if D is even. Then KD
is an imaginary quadratic field in which the rational prime 2 splits if D = 8 k + 7 or D = 16 k + 14, ramifies if D = 8 k + 5, and is inert if D = 16 k + 6. A point
P ∈ ED(Q) is equivalent to a KD-rational point Q of E1 or E2 (according as
D is odd or even) whose complex conjugate Q equals −Q. If Q′ is any point of
E1 or E2 over KD then Q = Q′ − Q′ satisfies Q = −Q, and thus amounts to a point of ED over Q. To prove Theorem 1 for ED, it will be enough to find
QD ∈ E1(KD) or E2(KD) and show that the point PD ∈ ED(Q) corresponding to QD − QD is not in ( ED (Q)) tors = ED. We use the modular parametrizations of E1 and E2 by the modular curves X0(32) and X 0(64). These curves have “CM points” parametrizing cyclic isoge-nies of degree 32 or 64 between elliptic curves of complex multiplication by the same order in KD. If the prime 2 splits in KD, these points are defined over the class field of KD; otherwise they are defined over a ray class field. (In the former case, the CM points are often called “Heegner points”; in the latter, [Mo] applies the term “mock Heegner points”, though Birch points out that Heegner’s seminal paper [He] already used both kinds of points to construct rational points on ED , and the distinction between the two cases was a later development.) In either case, we obtain a point QD defined over KD by taking a suitable subset of these CM points, mapping them to E1 or E2 by the modular parametrization, and adding their images using the group law of the curve. See [Bi1,Bi2,Mo] for more details on these subsets. 4 Noam D. Elkies
Now the key computational point is that the size of each subset is propor-tional to the class number of KD, and thus to |D|1/2 when averaged over D.This is much smaller than the number of terms of the series needed to numeri-cally estimate L′(ED/Q, 1), which is on the order of D: as explained for instance in [BGZ], for a general elliptic curve E/ Q of conductor N (E) it takes N 1/2+ ǫ
terms to adequately estimate L′(E/ Q, 1), and N (ED) = 32 D2 or 64 D2 (accord-ing as D is odd or even) so N 1/2 is of order D. As explained in [E1], the numerical computation of each CM point as a point on the complex torus E1(C) or E2(C)to within say 10 −25 takes essentially constant time: find a representative τ in a fundamental domain for the upper half-plane mod Γ0(32) or Γ0(64), and sum enough terms of a power series for ∫ τ
∞
ϕ dq/q where ϕ is the modular form for
E1 or E2. Thus it takes time ∆3/2+ ǫ (and negligible space) to approximate QD
for each |D| < ∆ .3
We implemented this computation in gp and ran it for ∆ = 10 6. For all but 1375 of the 303979 squarefree values of |D| < 10 6 congruent to 5, 6, or 7 mod 8, we found that PD is at distance at least 10 −8 from the nearest 2-torsion point of ED, and is thus a rational point of infinite order. For each of the remaining D, the point PD is numerically indistinguishable (at distance 4 at most 10 −20 , usually much less) from a 2-torsion point. We believe that PD then actually is a torsion point, and thus that we must find a nontorsion rational point on ED in some other way. We did this as follows. We first searched for rational numbers x = r/s with |r|, |s| < 5 · 10 7 such that s4x = rs (r2 − s2) is
D times a square for |D| < 10 6. This is a reasonable search since we may assume that gcd( r, s ) = 1, require that one of the factors r, s, r +s, r −s of rs (r2 −s2) have squarefree part f < (4 · 10 6)1/4 and that another have squarefree part at most (4 · 10 6/f )1/3, and loop over those factors. 5 This took several hours and found points on all but 70 of our 1375 ED’s. The remaining curves were handled by Cremona’s mwrank program, which used a 2-descent on each curve (exploiting its full rational 2-torsion) to locate a rational point. This completed the proof of Theorem 1.
3 Curves ED of conjectural rank ≥ 3
It might seem surprising that we were able to find a rational point on each of the 1375 ED’s for which we could not use PD. Many curves ED, even with D
3This computation is particularly efficient in our setting, in which ϕis a CM form (so most of its coefficients vanish) and the normalizers of Γ0(32), Γ0(64) in SL 2(R) can be used to obtain an equivalent τwith imaginary part at least 1 /8 and √3/16 re-spectively. These efficiencies represent a considerable practical improvement, though they contribute negligible factors O(∆ǫ) to the asymptotic running time of the com-putation.
4Here, as in the preceding paragraph, the distance is measured on the complex torus representing E1(C) or E2(C).
5In fact we removed the factors of 4 by using the squarefree parts of ( r±s)/2 instead of r±swhen r≡smod 2. Curves Dy 2=x3−xof odd analytic rank 5
well below our upper limit of 10 6, have rank 1 but generator much too large to locate with repeated 2-descents (see for instance [E1]). The reason we could find nontorsion points on the curves ED with PD ∈ ED is that these are precisely the curves ED of odd sign that should have rank at least 3, which makes the minimal height of a non-torsion point much smaller than it can get in the rank-1 case. We explain these connections below, and then report on our computations that extend to 10 7 the list of |D| such that ran (ED ) is odd and conjecturally at least 3.
3.1 PD and the rank of ED
Consider first the cases D = 8 k + 7 and D = 16 k + 14. In these cases the prime 2, which is the only prime factor of the conductors of E1 and E2, is split in KD.Therefore the results of Gross-Zagier [GZ] and Kolyvagin [Kol] apply to PD . The former result gives the canonical height of PD as a positive multiple of L′(ED , 1). Therefore ran (ED) > 1 if and only if PD is torsion. The latter result shows that if
PD is nontorsion then in fact the arithmetic rank of ED also equals 1. Hence any
ED of rank 3 or more must be among those for which we could not distinguish
PD from a torsion point. The hypotheses of the theorems of Gross-Zagier and Kolyvagin are not sat-isfied in the remaining cases D = 8 k + 5 and D = 16 k + 6. However, numerical evidence suggests that both theorems generalize to these cases as well. For in-stance, when PD is numerically indistinguishable from a torsion point, ED seems to have rank 3. For small |D| we readily find three independent points; for all
|D| in the range of our search, ED and each of the curves Dy 2 = x3 + 4 x and
Dy 2 = x3 − 11 x ± 14 isogenous with ED has a 2-Selmer group large enough to accommodate three independent points. When PD is nontorsion but has small enough height to be recovered from its real approximation by continued frac-tions, we find that it is divisible by 2 if and only if the 2-Selmer group has rank at least 5, indicating that ED has either rank ≥ 3 or nontrivial X. (The for-mer possibility should not occur, and can often be excluded by 2-descent on one of the curves isogenous to ED .) Both of these observations are consistent with a generalized Gross-Zagier formula and the conjecture of Birch and Swinnerton-Dyer, and would be most unlikely to hold if the vanishing of PD had no relation with the arithmetic of ED. We thus expect that also in these cases ED should have rank > 1 if and only if PD is a torsion point.
3.2 Rank and minimal nonzero height
The conjecture of Birch and Swinnerton-Dyer also explains why curves ED of rank ≥ 3 have nontorsion points of height much smaller than is typical of curves
ED of rank 1. This conjecture relates the regulator of the Mordell-Weil group of ED with various invariants of the curve, including its real period and the leading coefficient L(r)(ED, 1) /r ! (where r = ran (ED )). Now the real period is proportional to |D|−1/2. The leading coefficient is ≪ | D|o(1) under the gener-alized Riemann hypothesis for L(Ed, s ), or even the weaker assumption of the 6 Noam D. Elkies
Lindel¨ of conjecture for this family of L-series (see for instance [IS, p.713]). One expects, and in practice finds, that it is also ≫ | D|−o(1) (otherwise L(Ed, s ) has zeros 1 + it for very small positive t). Thus we expect the regulator to grow as
|D|1/2+ o(1) , at least if X is small, which should be true for most |D|. Hence the minimal nonzero height would be at most |D|1/2r . When r = 1 this grows so fast that already for |D| < 10 4 there are many curves ED with generators much too large to be found by 2-descents. 6 But for r ≥ 3 the minimal nonzero height is at most |D|1/6+ o(1) , so |D| must grow much larger before a 2-descent search becomes infeasible.
Remark on curves curves ED of even sign: For such curves we readily determine whether ran (ED ) > 0 by using the Waldspurger-Tunnell formula [Tu] to compute
L(ED , 1). If L(ED , 1) 6 = 0 then ran (ED ) = 0 and ED also has arithmetic rank 0 by Kolyvagin (or even Coates-Wiles [CW] because ED has CM). If L(ED , 1) = 0 then
ran (ED ) ≥ 2, and we can prove that ED has positive arithmetic rank if we find a nontorsion point. We expect that the minimal height of such a point is |D|1/4+ o(1) . This grows slower than the |D|1/2+ o(1) estimate for rank 1, but fast enough that 2-descent searches fail for |D| much smaller than our bound of 10 6. Even in the odd-rank case that concerns us in this paper, it is the curves of rank 3 that make it hard to extend Theorem 1 much beyond ∆ = 10 6 : searching for points on those curves take time roughly exp ∆1/6, which eventually swamps the polynomial time ∆3/2+ ǫ required to find those curves.
3.3 Computing ED of conjectural rank ≥ 3 with |D| < 10 7
We extended to ∆ = 10 7 our search for PD numerically indistinguishable from torsion points. These are the curves that we expect to have rank at least 3. Since we do not expect to extend Theorem 1 to 10 7, we saved time by requiring that the Selmer groups for the isogenies between ED and Dy 2 = x3 + 4 x be large enough to together accommodate an arithmetic rank of 3. For very large ∆ this is a negligible saving because most D pass this test. But it saved a substantial factor in practice for ∆ = 10 7: the test eliminated all but 35% of choices of
|D| = 16 k + 14, all but 32 .1% of |D| = 16 k + 6, all but 21 .6% of |D| = 8 k + 5, and all but 16 .2% of |D| = 8 k + 7. We found a total of 8740 values of D for which
PD appears to be a torsion point. We expect that each PD is in fact torsion and that the corresponding ED all have rank at least 3. Some PD might conceivably be a nontorsion point very close to ED, but this seems quite unlikely; at any rate no PD came closer than 10 −8 but far enough to distinguish from ED . All the curves probably have rank exactly 3: the smallest |D| known for a curve ED
of rank 5 exceeds 4 · 10 9 [Ro]. At any rate none of our curves with |D| < 2 · 10 6
can have rank 5: we applied mwrank ’s descents-only mode to each of these ED
6
The generators can be obtained using the CM-point construction in time |D|O(1) ,but not |D|1/2+ o(1) because PD must be computed to high accuracy to recognize its coordinates as rational numbers from their real approximations. Note that in our computations we showed only that PD is nontorsion and did not attempt to determine it explicitly in ED (Q). Curves Dy 2 = x3 − x of odd analytic rank 7
and the isogenous curves, and in each case obtained an upper bound of 3 or 4 on the rank. Our curves ED and the isogenous curves include many examples of conjectural rank 3 and nontrivial X. There are striking disparities in the distribution of our 8740 values of |D|
among the allowed congruence classes. The odd classes 8 k + 5 and 8 k + 7 account for 2338 and 2392 curves ED of presumed rank 3. But even |D|’s are much more plentiful: there are 4010 of them, almost as many as in the two odd classes combined. This might be explained by the behavior of the 2-descent, which depends on the factorization of |D|, or the fact that we are twisting a different curve: E1 for odd D and E2 for even D. But the 4010 even D’s are themselves unequally distributed between the 16 k + 6 and 16 k + 14 cases, the former being significantly more numerous: 2225 as against 1785. (See Figure 1.) This disparity is much larger than would be predicted by the 2-descent test, which in the range
|D| < 10 7 favors 16 k + 16 but only by a factor of 1 .09 whereas 2225 exceeds 1785 by almost 25%. Note too that the 2-descent survival rates would predict a preponderance of |D| = 8 k + 7 over 8 k + 5, whereas the two counts are almost identical. Do these disparities persist as ∆ increases, and if so why? Naturally we would also like to understand the overall distribution of quadratic twists of rank ≥ 3, not only for the “congruent number” family but for an arbitrary initial curve in place of ED. We hope that the computational data reported here, and more fully at , might suggest reasonable ideas and conjectures in this direction.
Acknowledgments
Thanks to Peter Sarnak for the reference [IS], and to the referee for several suggestions that improved the paper. I am grateful to the Packard Foundation for partial financial support during the preparation of this work.
References
Bi1. Birch, B.J.: Elliptic curves and modular functions. Symp. Math. 4(1970), 27–37. Bi2. Birch, B.J.: Heegner points of elliptic curves. Symp. Math. 15 (1975), 441–445. BGZ. Buhler, J.P., Gross, B.H., Zagier, D.: On the conjecture of Birch and Swinnerton-Dyer for an elliptic curve of rank 3. Math. of Computation 44 (1985) #175, 473–481. CW. Coates, J., Wiles, A.: On the conjecture of Birch and Swinnerton-Dyer. Invent. Math. 39 (1977) #3, 223–251. Di. Dickson, L.E.: History of the Theory of Numbers, Vol. II: Diophantine Analysis.
New York: Stechert 1934. E1. Elkies, N.D.: Heegner point computations, Lecture Notes in Computer Science
877 (proceedings of ANTS-1, 5/94), 122–133. GZ. Gross, B.H., Zagier, D.: Heegner points and derivatives of L-series. Invent. Math.
84 (1986), 225–320. He. Heegner, K.: Diophantische Analysis und Modulfunktionen. Math. Z. 56 (1952), 227–253.
8 Noam D. Elkies IS. Iwaniec, H., Sarnak, P.: Perspectives on the analytic theory of L-functions.
Geom. Funct. Anal. 2000 , Special Volume (GAFA 2000, Tel Aviv 1999), Part II, 705–741. Kob. Koblitz, N.: Introduction to elliptic curves and modular forms. New York: Springer 1984. Kol. Kolyvagin, V.A.: Euler systems. Pages 435–483 of The Grothendieck Festscrhift
Vol. II, Birkh¨ auser: Boston 1990. Mo. Monsky, P.: Mock Heegner points and congruent numbers. Math. Z. 204 (1990) #1, 45–67. Ro. Rogers, N.F.: Rank Computations for the Congruent Number Elliptic Curves.
Experimental Mathematics 9 (2000) #4, 591–594. Si. Silverman, J.H.: The Arithmetic of Elliptic Curves . New York: Springer 1986. Tu. Tunnell, J.B.: A classical Diophantine problem and modular forms of weight 3/2, Invent. Math. 72 (1983) #2, 323–334.
f(N) := number of D<N of the form 16k+6 (upper curve) or 16k+14 (lower curve) such that the elliptic curve D y 2= x 3- x has presumed rank at least 3 f(N) = 2000 1000 5.10 6N=10 7
Figure 1. Twists with |D| ≡ 6 mod 16 seem to have rank 3 much more often than those with |D| ≡ 14 mod 16 |
9128 | https://www.doubtnut.com/qna/645254418 | Show that the length of the chord intercepted by the ellipse x2a2+y2b2=1 on the line y=mx+c is 2aba2m2+b2√(1+m2)(a2m2+b2−c2)
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Inequality
The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads.
Contents
1 Overview
2 Solving Inequalities
2.1 Linear Inequalities
2.2 Polynomial Inequalities
2.3 Rational Inequalities
3 Complete Inequalities
4 List of Theorems
4.1 Introductory
4.2 Advanced
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 Resources
6.1 Books
6.1.1 Intermediate
6.1.2 Olympiad
6.2 Articles
6.2.1 Olympiad
6.3 Classes
6.3.1 Olympiad
7 See also
Overview
Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: .
For two numbers and :
if is greater than , that is, is positive.
if is smaller than , that is, is negative.
if is greater than or equal to , that is, is nonnegative.
if is less than or equal to , that is, is nonpositive.
Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa.
Some properties of inequalities are:
If , then , where .
If , then , where .
If , then , where .
Solving Inequalities
In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable.
We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict.
Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means .
Linear Inequalities
Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign.
Polynomial Inequalities
The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots.
Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality.
Rational Inequalities
A more complex example is .
Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either.
A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$.
For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set.
One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by .
Complete Inequalities
A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications.
List of Theorems
Here are some of the more useful inequality theorems, as well as general inequality topics.
Introductory
Arithmetic Mean-Geometric Mean Inequality
Cauchy-Schwarz Inequality
Titu's Lemma
Chebyshev's Inequality
Geometric inequalities
Jensen's Inequality
Nesbitt's Inequality
Rearrangement Inequality
Power mean inequality
Triangle Inequality
Trivial inequality
Schur's Inequality
Advanced
Aczel's Inequality
Callebaut's Inequality
Carleman's Inequality
Hölder's inequality
Radon's Inequality
Homogenization
Isoperimetric inequalities
Maclaurin's Inequality
Muirhead's Inequality
Minkowski Inequality
Newton's Inequality
Ptolemy's Inequality
Can someone fix that Ptolemy's is in Advanced?
Problems
Introductory
Practice Problems on Alcumus
Inequalities (Prealgebra)
Solving Linear Inequalities (Algebra)
Quadratic Inequalities (Algebra)
Basic Rational Function Equations and Inequalities (Intermediate Algebra)
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3)
Intermediate
Practice Problems on Alcumus
Quadratic Inequalities (Algebra)
Advanced Rational Function Equations and Inequalities (Intermediate Algebra)
General Inequality Skills (Intermediate Algebra)
Advanced Inequalities (Intermediate Algebra)
Given that , and show that . (weblog_entry.php?t=172070 Source)
Olympiad
See also Category:Olympiad Inequality Problems
Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2)
Resources
Books
Intermediate
Introduction to Inequalities
Geometric Inequalities
Olympiad
Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess.
The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
Problem Solving Strategies by Arthur Engel contains significant material on inequalities.
Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya.
Articles
Olympiad
Inequalities by MIT Professor Kiran Kedlaya.
Inequalities by IMO gold medalist Thomas Mildorf.
Classes
Olympiad
The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities.
See also
Mathematics competitions
Math books
Retrieved from "
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Algebra
Inequalities
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Medical Jurisprudence and Related Laws in India
Swati GuptaArticles,Forensic4 years ago 404 ViewsShort URL
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Medical Jurisprudence and Related Laws in India
MEDICAL JURISPRUDENCE is the application of medical knowledge in the legal field for providing justice in both civil as well as criminal cases. It provides the basic legal guidelines which should be followed by a medical practitioner.
The Latin term ‘juris’ stands for ‘law’ and ‘prudentia’ stands for ‘knowledge’. Thereby, it is the domain which makes the use of the medically relevant facts and then integrates them with the legal system, providing assistance to the criminal justice system.
In order to convict a probable offender, it is utmost important to know the relevancy of the evidences. Thus, the need of using scientific principles was felt.
As this field grew, it gave immense power to the medical practitioner/doctor as they were now playing a very important role by having an expert opinion in the cases.
But with this power came huge responsibilities. The doctor patient relation, medical negligence, ethical conducts, professional misconducts are a few to name.
The field of medical jurisprudence is a very ancient field but with the advent of technology and the reforms being added in the legal system, this branch is always under development.
In order to completely understand true meaning and its importance one should understand how actually did this field came into action.
Fig. 1 – “This image is taken from Practical forensic medicine [electronic resource]: a police-surgeon’s emergency guide” by Medical Heritage Library, Inc. is licensed under CC BY-NC-SA 2.0
The HISTORY of medical jurisprudence dates back to 4000-3000 BC and the data recorded can be studied from the Materia Medica. Imhotep (around 2300 BC), the chief justice and the personal physician to the ruler of Egypt was considered as the first medicolegal expert.
Fig. 2 – “NYC – Brooklyn Museum – Statuette of Imhotep”bywallygis licensed underCC BY-NC-ND 2.0
While in India, the Charaka Samhita (around the 7 th century BC) contained rules related to the ethics, duties, privilege, etc. mandatory to be followed by a physician. Various other texts like Manusmriti, Sushruta Samhita, Yajnavalkya Smriti, etc. also played a major role in maintaining and regulating the medical practice.
An autopsy is considered as the most important tool in medicolegal practice. In the 18 th century Dr. Edward Bulkley was the first to perform a medico-legal autopsy in India.
With the introduction of this branch in India, it didn’t take much of a time for it to grow in all directions. In 1822, the state of Calcutta got the countries first medical school. This development further proceeded to the states of Madras and Bombay.
Fig. 3 – “This image is taken from Page 160 of Clinical manual for India: compiled for the use of the students of the Madras Medical College”byMedical Heritage Library, Inc.is licensed underCC BY-NC-SA 2.0
Over the last few decades, the scientific techniques have advanced by multiple folds. The tests performed and the results obtained are so precise that they can act as evidence, sufficient in itself to prove or disprove a convict.
This growth has also led to the enlargement and thus diversification of medical jurisprudence into numerous small branches.
To fit more accurately in the present time, the term medical jurisprudence has mostly been replaced by the new terminology ‘FORENSIC MEDICINE’.
Some of the major branches in FORENSIC MEDICINE are:
·forensic pathology,
·forensic odontology,
·forensic toxicology,
·forensic anthropology, etc.
Due to the indispensable efforts in this field, Dr. J.P. Modi is renowned as the Father of Indian Forensic Medicine.
Before getting into the details, it is very crucial to understand the true significance and seriousness of forensic medicine to different professions.
The major professions involved in legal cases are:
·medicine department; consisting of doctors, nurses, pharmacists
·legal department; consisting of lawyers, judges, police
For enhancing the knowledge of a personnel of medical background different courses in forensic medicine have been introduced in their curriculum. Along with providing legal
guidance this course focuses on training the doctors in the major advancements in the forensic medicine field, their responsibilities, rights, ethics, etc.
By providing scientific evidence and assisting the judiciary and legal system, the medical field helps in strengthening the ultimate of providing justice and thereby, benefitting all.
The legal practitioners i.e., the lawyers who may have to time and again encounter the medical issues in their practice are hardly aware about the medical language, knowledge, profession and the associated complexities of this profession.
The profession of a medical personnel is considered as being the next to God. But are doctors really God? This is a question that arises time and again in the medical fraternity.
William Osler wrote: “Medicine is a science of uncertainty and an art of probability. Absolute diagnoses are unsafe and are made at expense of the conscience”
The doctors are also humans and it is impossible for them to be unerring. When a simple mistake of a doctor leads to the loss of a life, he is answerable to the system.
In a hypothetical case, if a patient dies due to a hypersensitive allergic reaction from a certain drug, which is widely accepted by the majority of the population then is the death to be considered as the fault of the doctor. The patient’s family may file a report against the doctor for medical negligence. The lawyers working on the case on behalf of the doctor and the patient, should have ample knowledge for understanding the reasons which led to the sudden death. This is how medical jurisprudence will help in determining whether the doctor is supposed to be held guilty or not.
Many times, in a bizarre circumstance, the doctors keep aside their legal restrictions and prioritize the patient’s wellbeing and this often leads them as guilty in the court. To avoid and keep such actions under a check, hospitals have their own legal firms who timely monitor the work of the doctors.
While considering medical jurisprudence one cannot brush aside certain controversial areas like role of doctor for evidence collection from crime scene, mercy killing, organ transplantation, medical negligence, civil negligence, malpractice, therapeutic misadventure, etc.
From the world of fiction to reality, the cases of Anaesthetic and Operative Deaths are always under the limelight.
The causes of death due to anaesthetic agents can be found out via post-mortem examination. These causes may include; hypersensitivity, cardiac arrest, respiratory inadequacy, etc. The cardiac arrest and respiratory failure are the most common modes of death.
These cases often invite an investigation on the medical personals involved in the procedure. Legal department overlaps here with the medical department and helps in figuring out whether it was a medical negligence or a medical maloccurence.
Some of the most common cases under medical jurisprudence include:
·paternity testing
·injury and wounds
·death due to poisons
·cause and manner of death
·violent death, etc.
Fig. 4 – “Know your future for only 999$ with DNA test _DDC6259.JPG”byAbode of Chaosis licensed underCC BY 2.0
To understand the legal portion under medical jurisprudence, it is essential to understand structure under which the laws are classified in The Indian Judicial System. The following ACTS can be seen as the pillars of justice.
1.INDIAN PENAL CODE, 1860
2.INDIAN EVIDENCE ACT, 1872
3.CODE FOR CRIMINAL PROCEDURE, 1973
4.INDIAN MEDICAL COUNCIL ACT,1956
LAWS RELATED TO MEDICAL JURISPRUDENCE IN INDIA:
ØINDIAN PENAL CODE, 1860
·Section 44 of IPC: Definition of Injury. Any harm whatever illegally caused to any person in body, mind, reputation, or property.
·Section 319 IPC: Hurt. Hurt means bodily pain, disease, or infirmity caused to any person.
·Section 320 IPC: Grievous Injury Any of the following injuries is grievous:
Emasculation (Depriving a male of masculine vigor)
Permanent privation of sight of either eye
Permanent privation of the hearing of either ear
Privation of any member or joint (member means an organ or a limb being part of man capable of performing a distinct function)
Destruction or permanent impairing of powers of any member or joint
Permanent disfiguration of the head or face
Fracture or dislocation of bone or tooth
Any hurt which endangers life, or which causes the victim to be in severe bodily pain, or unable to follow his ordinary pursuits for a period of 20-day.
·Section 321 IPC: Defines “Voluntarily Causing Hurt”.
·Section 322 IPC: Defines “Voluntarily Causing Grievous Hurt”.
·Section 351 IPC: Defines Assault: Threat/attempt to apply force. Whoever makes any gesture or preparation intending or knowing, it to be likely that such gesture or preparation will cause any person present to apprehend that he who makes the gesture or preparation is about to use criminal force to that person, is said to commit an assault.
·Section 323 IPC: Describes Punishment for Voluntarily Causing Hurt. Shall be imprisonment which may extend for 1 year with or without fine which may be Rs. 1000.
·Section 324 IPC: Describes Punishment for Voluntarily Causing Hurt by dangerous weapon shall be imprisonment for up to 3 years with or without fine.
·Section 325 IPC: Describes Punishment for Voluntarily Causing Grievous Hurt. Shall be imprisonment which may extend for 7 years with or without fine.
·Section 326 IPC: Describes Punishment for Voluntarily Causing Grievous Hurt by dangerous weapon or means. Shall be imprisonment for life or for 10 years with or without fine.
·Section 328 IPC: Punishment of causing hurt using poison, etc., shall be imprisonment up to 10 years with or without fine.
ØINDIAN EVIDENCE ACT, 1872
·Section 45: Opinions of experts. When the Court has to form an opinion upon a point of foreign law or of science or art, or as to identity of handwriting, the opinions upon that point of persons especially skilled in such foreign law, science or art,are relevant facts. Such persons are called experts.
·Section 114A: In a prosecution for rape, where the question is whether sexual intercourse was without the consent of the woman, and she states in her evidence that she did not consent, the court shall presume that she did not consent.
ØCODE FOR CRIMINAL PROCEDURE, 1973
·Section 53 (i) CrPC: An accused may be examined by a medical practitioner at the request of a police officer using reasonably necessary force.
·Section 53 (ii) CrPC: Whenever the person of a female accused is to be examined, the examination shall be made only by or under the supervision of a female registered medical practitioner.
·Section 54 CrPC: An arrested person may be examined at hi s request by a medical practitioner to detect evidence in his favour.
·Section 174 CrPC: Police to enquire and report on suicide, etc.
·Section 176 CrPC: Inquiry by Magistrate into cause of death.
ØINDIAN MEDICAL COUNCIL ACT,1956
·Section 20A- Professional conduct: 1) T he Council may prescribe standards of professional conduct and etiquette and a code of ethics for medical practitioners.
2) Regulations made by the Council under sub-section
(1) may specify which violations thereof shall constitute infamous conduct in any professional respect, that is to say, professional misconduct, and such provision shall have effect notwithstanding anything contained in any law for the time being in force.
According to the Current Indian Scenario, there are certain drawbacks with respect to these laws.
Irrespective of having several indispensable benefits, the evidences procured via medical jurisprudence are still considered under the expert opinion rather than a primary evidence.
Under the Indian Evidence Act, 1872 only in certain cases the report of an autopsy procedure is considered as a documentary evidence.
Doctors are bounded by the doctor-patient confidentiality and this often places them in a dilemma of whether to share the information or not with the legal system.
If it is in the doctor’s knowledge that any of his patient has committed a crime (other than suicide), then he is bound to inform the legal authorities. In failing to do so, he may be punished.
Thus, clear briefing to the doctors on these pre-existing laws and amendment in certain conflicting laws is the need of the hour. A few suggestions would be;
o legal drafting of legislation which will consider the evidences of medical jurisprudence as a primary evidence.
o Other scientific test results, when performed under standard testing protocol, must also be considered as documentary evidences.
o If a patient tries to commit suicide the doctor must report to the concerned authorities as this may help in resolving the factors which may have led to such a huge step by the victim and also ensure the patients safety in the near future.
Medical jurisprudence plays a very crucial role in bridging this gap between the doctors and the lawyers. This field has contributed to the society by opening the new doors to those old pending cases, which were once considered to be stuck at a dead end.
The personnel working in the legal and medical professions must join hands and help each other in providing the field of medical jurisprudence the due credit that it deserves since a very long time.Hence, Medical jurisprudence can be seen as the kaleidoscope of new possibilities.
REFERENCES:
·BIBLIOGRAPHY Aggrawal, A. (n.d.). Textbook of Forensic Medicine and Toxicology . AVICHAL PUBLISHING COMPANY.
·Britannica, T. E. (August 18, 2011). Medical jurisprudence. Encyclopædia Britannica.
·CODE FOR CRIMINAL PROCEDURE. (1973).
·CREATIVECOMMONS. (n.d.). Retrieved from
·Dr. K.S. NARAYAN REDDY, D. O. (1973). THE ESSENTIALS OF FORENSIC MEDICINE AND TOXICOLOGY. Jaypee Brothers Medical Publishers (P) Ltd .
·DUHAN, R. (2016). FORENSIC MEDICINE AND INDIAN CRIMINAL LAWS: A STUDY OF RELEVANCY WITH LEGAL. Innovare Journal Of Medical Sciences.
·INDIAN EVIDENCE ACT. (1872).
·INDIAN KANOON. (n.d.). Retrieved from
·INDIAN MEDICAL COUNCIL ACT. (1956).
·INDIAN PENAL CODE. (1860).
·Rao, N. G. (2000). Textbook of Forensic Medicine and Toxicology. JAYPEE BROTHERS MEDICAL PUBLISHERS (P) LTD.
·Venkataramani, S. (n.d.). Legal Service India – Lawyers. Retrieved from
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9131 | https://northernontario.travel/sunset-country/online-map-find-out-which-fish-species-are-ontario-lakes | Skip to main content
Northern Ontario Travel
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Northern Ontario Travel • The Official Magazine
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Great Online Map to Find Out Which Fish Species Are in Ontario Lakes
Do you ever wonder which fish species are in your favourite lake? Well, with the MNR's Fish On-Line Interactive Map, it's easy to find out.
By Erin Rody
Erin Rody is a staff writer for Sunset Country.
October 2, 2024
The Ontario Ministry of Natural Resources (MNR) oversees the Fish ON-Line tool for anglers. We no longer need to find an in-the-know local fisherman or resort owner to gather information about the various fishing lakes in Northwest Ontario. The tool provides a lot of information that was hard to find. Now information about the different species of fish found in a particular lake, fish stocked in lakes, regulations for the lakes, and some information about the lake itself is available online. In an area like Ontario's Sunset Country, where there are thousands upon thousands of lakes, the information is undoubtedly beneficial.
For example, take Red Lake. If you click on the Lake Depth Contours in the left-hand column, you can see the contours of Red Lake. Click the fish icon, and you'll see the surface area of the lake is 8838 ha, the maximum depth is 25.6 m, the average depth is 10.1 m, and it's in the Fisheries Management Zone 4. Click on the Regulations tab, and you'll see specific Red Lake fishing regulations and lake facts. Under the Fish tab, you'll see that Red Lake has Walleye, Lake Trout, Northern Pike, Muskie, Smallmouth Bass, Yellow Perch, Whitefish, Sauger, Rock Bass, Rainbow Smelt, White Sucker and Burbot.
You can quickly link to a popular waterway or search by lake, fish species, stocked water body, and GPS Coordinates. Use the Get Directions Tool to find your way around quite easily.
I did a quick test of the search by species. I chose Brook trout and set the parameters to 300 km around Ignace, Ontario. It showed me 137 lakes with brook trout in them in that radius.
Search for lakes with brook trout within 300 km of Ignace, Ontario
Everything You Need to Know About Planning a Fishing Trip to Northwest Ontario
While the map is imperfect in that it doesn't show all the lakes and their information, the 13,000 lakes it does have are certainly a significant amount!
You can show the following on the map:
Water Body
Fish Sanctuary
Fishing Access Points
Lake Depth Contours (For the bigger lakes)
Fisheries Management Zones
ServiceOntario Licence Issuers (You can easily order fishing licenses online if you don't want to go to a Service Ontario office or you can get one from a Northwestern Ontario tourist outfitter that sells licenses.)
All in all, it is a pretty impressive fishing information tool. If you are planning a fishing trip to Ontario, check out the Fish ON-Line website for more information on the lake you'll be fishing on. If you are looking for a fishing resort or lodge to stay at in Sunset Country, visit these tourist outfitters in Northwest Ontario or visit this Northern Ontario list of lakes to find out which lodge is on the lake you want to fish.
### Find out more about the fishing lodges in Sunset Country
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About Erin Rody
I grew up on Black Sturgeon Lake in Northwestern Ontario. I am a staff writer for the Sunset Country Travel Association. Through my articles I hope to entice you to visit the wonderful region I call home. We are all about outdoor adventure; with 70,000 lakes and rivers and a whole lot of forests how can we not be? Whether you like to fish, hunt, canoe, kayak, boat or go camping, Sunset Country has something for you. Enjoy!
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9132 | https://www.vedantu.com/biology/differences-between-protostomes-and-deuterostomes | Biology
Protostomes vs Deuterostomes: Core Differences and Significance
Protostomes vs Deuterostomes: Core Differences and Significance
Comparing Embryonic Development: Protostomes and Deuterostomes Explained
In biology, embryonic development is classified into two kinds on the basis of the complex animals and can be given as deuterostomes and protostomes. One of the major differences seen in both the development types is the development of blastopore which is also the first opening of the animal’s embryo. The opening can either be the mouth as in the case of protostomes or it can be anus as in the case of deuterostomes. To know more about the Differences Between Protostomes and Deuterostomes here is a detailed guide via
Vedantu
that lets you know regarding the same.
Deuterostomes
The embryonic development of deuterostomes goes through radial cleavage. While the blastula is forming via cleavage of the embryo, the cell division process occurs in radial. In this group of animals, the initial cavity created by the blastopore matures as the anus of the organism. Also, the mouth of the concerned organism is created on the other side after the formation of the anus.
Based on the similarities of structure, the following clades of deuterostomes animals are available-
Echinodermata
Echinodermata is exclusively identified as marine animals. The adult echinoderms are recognisable by the radial symmetry.
Example:
Sea cucumbers, Starfish, Sand dollars, Sea lilies, etc.
Chordates
It includes both terrestrial and marine animals.
Example:
Frog, Tiger, Bat, Turtle, Snake, Jawless fish, etc.
Cephalochordata
They are small marine organisms with segmented bodies.
Example:
Lancelets.
Urochordata
They are also sea animals and known as Tunicata.
Example:
Sea squirts, Thaliacea, Ascidians, Larvacea, etc.
Vertebrata
It includes all animals that fall under subphylum vertebrates.
Example:
humans, Birds, Amphibians, Fish, etc.
Hemichordata
Hemichordata is a marine species and can be grouped as sister organisms of echinoderms.
Example:
Graptolithina, Acorn worm, Pterobranchia, etc.
Protostomes
include the lower invertebrate species in which the formation of the mouth happens before the creation of the anus during embryo development. These animals have determinate and spiral cleavage and through the dividing of mesoderm their coelom forms.
Some examples of protostomes are octopuses, squid, snail, centipedes, millipedes, spiders, butterflies, ants, bees, earthworms, clams, oysters, etc.
Thus, the fundamental difference between protostomes and deuterostomes animals is the conversion of blastopore into the mouth or anus.
Differences between Protostomes and Deuterostomes
| Deuterostomes | Protostomes |
| In deuterostomes, an anus develops from the blastopore. | In protostomes, a mouth develops from the blastopore. |
| Their coelom is developed from the longitudinal pouches of the archenteron. Hence, they are known as enterococcus. | Their coelom is developed by dividing the embryonic mesoderm. So, they are known as schizocoelomates. |
| The gastrointestinal tract is channelled into the embryo and develops the mouth. | The anus forms as the gastrointestinal tract are channelled into the embryo. |
| Their archenteron development occurs during the initial stage of embryo creation. | They don’t have archenteron development. |
| By nature, these animals are enterococcus. | By nature, protostomes can be priapulids. |
| Deuterostomes’ nervous system consists of hollow nerve fibres and gill slits. | Their nervous system consists of ventral and solid nerve cords. |
| These animals have evolved more, and their body compositions are complex. | These animals have evolved less, and their body compositions are simple. |
| Their cell ciliation process is based on a single cell. | Multiple cells are involved in their cell ciliation process. |
| Deuterostomes are divided into Echinodermata, Chordates, Urochordata, Cephalochordata, Vertebrata, and Hemichordata. | Protostomes include the remaining bilaterian species. |
| Deuterostomes include a significantly smaller number of species. | Most bilaterian phyla come under protostomes. |
| An anal opening forms at first. | A mouth forms at first. |
| These animals have indeterminate cleavage. | They have determinate cleavage. |
| The cleavage is radial. | The cleavage is a spiral structure. |
| Complex and higher organisms like humans, other terrestrial animals like tigers, monkeys, etc. are examples of deuterostomes. | Simpler and lower organisms like Arthropods, Flatworms, Annelids, etc. are examples of protostomes. |
Thus, the difference between protostomes and deuterostomes lies in the position of their organs. In deuterostomes, the blastopore transforms into an anus, and another cavity on the opposite side creates a mouth. However, in protostomes, the mouth is created from the blastopore.
To learn about the difference between protostomes and deuterostomes in-depth, go through the study materials available on Vedantu’s website. For quick learning, download our Vedantu app.
Cleavage Types are Seen in Both Deuterostomes and Protostomes
The two classes differ entirely in the types of cleavages that they undergo while the formation of blastopore takes place. The cleavage that is seen can be of two types:
Determinate cleavage is also called mosaic cleavage
indeterminate cleavage is called regulative cleavage.
Protostomes:
These undergo the determinate type of cleavage during the formation of the blastopore. During this type of cleavage, the developmental fate of cells is already determined during the early stages of the embryo.
Deuterostomes:
Deuterostomes on the other hand have an indeterminate type of blastomere cleavage. In this type, the fate of the cells is not set and will only be decided after each cell is formed. It is also seen that any of the cells present in the blastopore has an equal potency to become any of the cell types that the cell wants to be. This condition is specifically called the pluripotency of the cells where each cell can undergo a different function.
Characteristics of Protostomes:
Protostomes are also referred to as schizocoelomates as the coelom is developed by splitting the solid mass of embryonic mesoderm
In protostomes, the gut is tunnelled through the embryo and it reaches up to the anus
Protostomes are said to be multi-ciliated cells.
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FAQs on Protostomes vs Deuterostomes: Core Differences and Significance
What is the primary difference between protostomes and deuterostomes?
The primary difference between protostomes and deuterostomes lies in their embryonic development, specifically the fate of the blastopore. In protostomes (from Greek, meaning "first mouth"), the blastopore develops into the mouth. In deuterostomes (meaning "second mouth"), the blastopore develops into the anus, and the mouth forms later at the opposite end of the embryo.
Which animal phyla are examples of protostomes and deuterostomes?
These developmental patterns classify major animal groups.
Protostome examples include phyla like Annelida (earthworms, leeches), Mollusca (snails, clams, octopuses), and Arthropoda (insects, spiders, crustaceans).
Deuterostome examples include phyla like Echinodermata (starfish, sea urchins), Hemichordata (acorn worms), and Chordata (fishes, amphibians, reptiles, birds, and mammals, including humans).
How does embryonic cleavage differ between protostomes and deuterostomes?
The pattern of cell division (cleavage) in the early embryo is a key distinguishing feature:
Protostomes typically exhibit spiral cleavage, where the dividing cells are offset from the cells below them, creating a spiral pattern. This cleavage is also determinate, meaning the developmental fate of each cell is fixed very early.
Deuterostomes exhibit radial cleavage, where the dividing cells are aligned directly above one another. This cleavage is indeterminate, meaning that early embryonic cells are not fated and retain the potential to develop into a complete organism if separated.
What is the difference in how the coelom (body cavity) forms in protostomes and deuterostomes?
The formation of the true body cavity, or coelom, also differs significantly. In many protostomes, the coelom forms through a process called schizocoely, where a solid mass of mesodermal tissue splits apart to form the coelomic cavity. In deuterostomes, the coelom forms through enterocoely, where folds of the archenteron (the embryonic gut) grow outwards and pinch off to form the coelomic cavity.
Are humans protostomes or deuterostomes, and why?
Humans are deuterostomes. As members of the phylum Chordata, we exhibit the key developmental characteristics of this group. During our embryonic development, the blastopore forms the anus, our cleavage pattern is radial and indeterminate, and our coelom forms via enterocoely. This places us firmly within the deuterostome lineage alongside other vertebrates and echinoderms.
What is the biological significance of indeterminate cleavage in deuterostomes?
The indeterminate cleavage in deuterostomes is biologically significant because it allows for greater developmental flexibility. Since the fate of early cells is not fixed, each cell has the potential to form a complete embryo on its own. This is the biological basis for the natural occurrence of identical twins in humans and other deuterostomes, where a single embryo splits into two, and each part develops into a complete individual.
Despite their differences, what important biological features do protostomes and deuterostomes share?
Although their developmental paths diverge, both protostomes and deuterostomes share several advanced biological features. Both groups are bilaterally symmetrical, meaning they have a distinct left and right side. They are also triploblastic, possessing three primary germ layers (ectoderm, mesoderm, and endoderm) during development. Most members of both groups also possess a true coelom and a complete digestive tract with a separate mouth and anus.
If protostome development involves schizocoelous coelom formation, how are acoelomate animals like flatworms (Platyhelminthes) classified?
This is an excellent question that highlights evolutionary complexity. Flatworms (Phylum Platyhelminthes) are classified within the protostome lineage based on other developmental and molecular evidence, such as spiral cleavage. However, they are acoelomates, meaning they lack a true body cavity. It is believed that their acoelomate condition is a derived or possibly a primitive trait within the protostome group. Their ancestors may have had a coelom that was subsequently lost during evolution, or they may have diverged before the evolution of a true coelom was established in other protostomes.
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9133 | https://www.cuemath.com/questions/what-is-the-formula-for-calculating-the-length-of-the-diagonal-of-a-right-rectangular-prism/ | What is the formula for calculating the length of the diagonal of a right rectangular prism?
Solution:
The diagonal of a right rectangular prism is the line that connects opposite vertices of a prism.
Look at the figure of a right rectangular prism shown below.
The formula for the length of the diagonal of a right rectangular prism is: d = √(l2+w2+h2)
Here, d = length of the diagonal, l = length of the rectangular base of the prism, w = width of the rectangular base of the prism, and h = height of the prism.
Let us find the diagonal of a right rectangular prism with a length of 20 units, the width of 10 units, and the height of 5 units.
The formula for the length of the diagonal of a right rectangular prism = √(l2+ w2+ h2) units.
Substitute the value in the formula,
= √(202+ 102+ 52) units
= 22.91 units
So, the formula for the length of the diagonal of a right rectangular prism is √(l2+w2+h2), where l is the length, b is the breadth and h is the height of a right rectangular prism.
What is the formula for calculating the length of the diagonal of a right rectangular prism?
Summary:
The formula for calculating the length of the diagonal of a right rectangular prism is √(l2+ w2+ h2). where l is the length, w is the width and h is the height.
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9134 | https://www.opentextbookstore.com/arithmetic/arith4-2.pdf | 4.11 4.2 Proportions Learning Objective(s) 1 Determine whether a proportion is true or false. 2 Find an unknown in a proportion. 3 Solve application problems using proportions. 4 Solve application problems using similar triangles. Introduction A true proportion is an equation that states that two ratios are equal. If you know one ratio in a proportion, you can use that information to find values in the other equivalent ratio. Using proportions can help you solve problems such as increasing a recipe to feed a larger crowd of people, creating a design with certain consistent features, or enlarging or reducing an image to scale. For example, imagine you want to enlarge a 5-inch by 8-inch photograph to fit a wood frame that you purchased. If you want the shorter edge of the enlarged photo to measure 10 inches, how long does the photo have to be for the image to scale correctly? You can set up a proportion to determine the length of the enlarged photo. Determining Whether a Proportion Is True or False A proportion is usually written as two equivalent fractions. For example: = 12 inches 36 inches 1foot 3 feet Notice that the equation has a ratio on each side of the equal sign. Each ratio compares the same units, inches and feet, and the ratios are equivalent because the units are consistent, and 12 1 is equivalent to 36 3 . Proportions might also compare two ratios with the same units. For example, Juanita has two different-sized containers of lemonade mix. She wants to compare them. She could set up a proportion to compare the number of ounces in each container to the number of servings of lemonade that can be made from each container. = 40 ounces 10 servings 84 ounces 21servings Since the units for each ratio are the same, you can express the proportion without the units: 40 10 84 21 = Objective 1 4.12 When using this type of proportion, it is important that the numerators represent the same situation – in the example above, 40 ounces for 10 servings – and the denominators represent the same situation, 84 ounces for 21 servings. Juanita could also have set up the proportion to compare the ratios of the container sizes to the number of servings of each container. = 40 ounces 84 ounces 10 servings 21servings Sometimes you will need to figure out whether two ratios are, in fact, a true or false proportion. Below is an example that shows the steps of determining whether a proportion is true or false. Example Problem Is the proportion true or false? 100 miles 50 miles = 4 gallons 2 gallons miles The units are consistent across the numerators. gallons The units are consistent across the denominators. 100 4 25 4 4 1 ÷ = ÷ 50 2 25 2 2 1 ÷ = ÷ 25 25 1 1 = Write each ratio in simplest form. Since the simplified fractions are equivalent, the proportion is true. Answer The proportion is true. Identifying True Proportions To determine if a proportion compares equal ratios or not, you can follow these steps. 1. Check to make sure that the units in the individual ratios are consistent either vertically or horizontally. For example, = miles miles hour hour or = miles hour miles hour are valid setups for a proportion. 2. Express each ratio as a simplified fraction. 3. If the simplified fractions are the same, the proportion is true; if the fractions are different, the proportion is false. 4.13 Sometimes you need to create a proportion before determining whether it is true or not. An example is shown below. Example Problem One office has 3 printers for 18 computers. Another office has 20 printers for 105 computers. Is the ratio of printers to computers the same in these two offices? printers printers = computers computers Identify the relationship. 3 printers 20 printers = 18 computers 105 computers Write ratios that describe each situation, and set them equal to each other. printers Check that the units in the numerators match. computers Check that the units in the denominators match. ÷ = ÷ 3 3 1 18 3 6 ÷ = ÷ 20 5 4 105 5 21 1 4 6 21 ≠ Simplify each fraction and determine if they are equivalent. Since the simplified fractions are not equal (designated by the ≠ sign), the proportion is not true. Answer The ratio of printers to computers is not the same in these two offices. There is another way to determine whether a proportion is true or false. This method is called “finding the cross product” or “cross multiplying”. To cross multiply, you multiply the numerator of the first ratio in the proportion by the denominator of the other ratio. Then multiply the denominator of the first ratio by the numerator of the second ratio in the proportion. If these products are equal, the proportion is true; if these products are not equal, the proportion is not true. 4.14 This strategy for determining whether a proportion is true is called cross-multiplying because the pattern of the multiplication looks like an “x” or a criss-cross. Below is an example of finding a cross product, or cross multiplying. In this example, you multiply 3 • 10 = 30, and then multiply 5 • 6 = 30. Both products are equal, so the proportion is true. To see why this works, let’s start with a true proportion: = 4 5 8 10 . If we multiplied both sides by 10, we’d get = 4 5 10 10 8 10 . The right side of this equation would simplify to 5, leaving = 4 10 5 8 . Now if we multiplied both sides by 8, we’d get = 4 10 8 5 8 8 , and the left side would simplify to = 10 4 5 8. Notice this is the same equation we would get by cross-multiplying, so cross-multiplying is just a quick way to do these operations. Below is another example of determining if a proportion is true or false by using cross products. Example Problem Is the proportion true or false? 5 9 6 8 = Identify the cross product relationship. 5 • 8 = 40 6 • 9 = 54 Use cross products to determine if the proportion is true or false. 40 ≠ 54 Since the products are not equal, the proportion is false. Answer The proportion is false. Self Check A Is the proportion 3 24 5 40 = true or false? 4.15 Finding an Unknown Quantity in a Proportion If you know that the relationship between quantities is proportional, you can use proportions to find missing quantities. Below is an example. Example Problem Solve for the unknown quantity, n. 25 4 20 n = 20 • n = 4 • 25 Cross multiply. 20n = 100 5 20 100 n = 5 You are looking for a number that when you multiply it by 20 you get 100. You can find this value by dividing 100 by 20. Answer n = 5 Self Check B Solve for the unknown quantity, x. = 15 6 10 x Now back to the original example. Imagine you want to enlarge a 5-inch by 8-inch photograph to make the length 10 inches and keep the proportion of the width to length the same. You can set up a proportion to determine the width of the enlarged photo. 5 inches 8 inches 10 inches ? inches Objective 2 4.16 Example Problem Find the length of a photograph whose width is 10 inches and whose proportions are the same as a 5- inch by 8-inch photograph. width length Determine the relationship. Original photo: 5 inches wide 8 inches long Enlarged photo: 10 inches wide inches long n Write a ratio that compares the length to the width of each photograph. Use a letter to represent the quantity that is not known (the width of the enlarged photo). 5 10 8 n = Write a proportion that states that the two ratios are equal. 5 • n = 8 • 10 5n = 80 Cross multiply. You are looking for a number that when it is multiplied by 5 will give you 80. 5 80 5 5 n = 80 5 n = n = 16 Divide both sides by 5 to isolate the variable. 16 5 80 Answer The length of the enlarged photograph is 16 inches. Solving Application Problems Using Proportions Setting up and solving a proportion is a helpful strategy for solving a variety of proportional reasoning problems. In these problems, it is always important to determine what the unknown value is, and then identify a proportional relationship that you can use to solve for the unknown value. Below are some examples. Objective 3 4.17 Example Problem Among a species of tropical birds, 30 out of every 50 birds are female. If a certain bird sanctuary has a population of 1,150 of these birds, how many of them would you expect to be female? Let x = the number of female birds in the sanctuary. Determine the unknown item: the number of female birds in the sanctuary. Assign a letter to this unknown quantity. = 30 female birds female birds in sanctuary 50 birds 1 ,150 birds in sanctuary x Set up a proportion setting the ratios equal. 30 10 3 50 10 5 ÷ = ÷ 3 5 1 ,150 x = Simplify the ratio on the left to make the upcoming cross multiplication easier. 3 • 1,150 = 5 • x 3,450 = 5x Cross multiply. 690 5 3,450 x = 690 birds What number when multiplied by 5 gives a product of 3,450? You can find this value by dividing 3,450 by 5. Answer You would expect 690 birds in the sanctuary to be female. Example Problem It takes Sandra 1 hour to word process 4 pages. At this rate, how long will she take to complete 27 pages? 4 pages 27 pages = 1hour hours x Set up a proportion comparing the pages she types and the time it takes to type them. 4 • x = 1 • 27 4x = 27 Cross multiply. You are looking for a number that when it is multiplied by 4 will give you 27. 6.75 4 27.00 x = 6.75 hours You can find this value by dividing 27 by 4. Answer It will take Sandra 6.75 hours to complete 27 pages. 4.18 Self Check C A map uses a scale where 2 inches represents 5 miles. If the distance between two cities is shown on a map as 20 inches, how many miles apart are the two cities? Solving Application Problems Using Similar Triangles In the photograph problem from earlier, we created an enlargement of the picture, and both the width and height scaled proportionally. We would call the two rectangles similar. With triangles, we say two triangles are similar triangles if the ratios of the pairs of corresponding sides are equal sides. Consider the two triangles below. We see that side AB corresponds with side DE and so on, and we can see that each of the ratios of corresponding sides are equal: 6 18 4 12 3 9 = = , so these triangles are similar. If two triangles have the same angles, then they will also be similar. You can find the missing measurements in a triangle if you know some measurements of a similar triangle. Let’s look at an example. Objective 4 4.19 Example Problem ABC ∆ and XYZ ∆ are similar triangles. What is the length of side BC? BC AB YZ XY = In similar triangles, the ratios of corresponding sides are proportional. Set up a proportion of two ratios, one that includes the missing side. 6 2 1.5 n = Substitute in the known side lengths for the side names in the ratio. Let the unknown side length be n. 2• 6 1.5• 12 1.5 8 n n n = = = Solve for n using cross multiplication. Answer The missing length of side BC is 8 units. This process is fairly straightforward—but be careful that your ratios represent corresponding sides, recalling that corresponding sides are opposite corresponding angles. Applying knowledge of triangles, similarity, and congruence can be very useful for solving problems in real life. Just as you can solve for missing lengths of a triangle drawn on a page, you can use triangles to find unknown distances between locations or objects. 4.20 Let’s consider the example of two trees and their shadows. Suppose the sun is shining down on two trees, one that is 6 feet tall and the other whose height is unknown. By measuring the length of each shadow on the ground, you can use triangle similarity to find the unknown height of the second tree. First, let’s figure out where the triangles are in this situation! The trees themselves create one pair of corresponding sides. The shadows cast on the ground are another pair of corresponding sides. The third side of these imaginary similar triangles runs from the top of each tree to the tip of its shadow on the ground. This is the hypotenuse of the triangle. If you know that the trees and their shadows form similar triangles, you can set up a proportion to find the height of the tree. Example Problem When the sun is at a certain angle in the sky, a 6-foot tree will cast a 4-foot shadow. How tall is a tree that casts an 8-foot shadow? Tree 1 Shadow 1 Tree 2 Shadow 2 = The angle measurements are the same, so the triangles are similar triangles. Since they are similar triangles, you can use proportions to find the size of the missing side. Set up a proportion comparing the heights of the trees and the lengths of their shadows. 6 4 = 8 h Substitute in the known lengths. Call the missing tree height h. 4.21 6• 8 4 48 4 12 h h h = = = Solve for h using cross-multiplication. Answer The tree is 12 feet tall. Self Check D Find the unknown side. Summary A proportion is an equation comparing two ratios. If the ratios are equivalent, the proportion is true. If not, the proportion is false. Finding a cross product is another method for determining whether a proportion is true or false. Cross multiplying is also helpful for finding an unknown quantity in a proportional relationship. Setting up and solving proportions is a skill that is useful for solving a variety of problems. 4.2 Self Check Solutions Self Check A Is the proportion 3 24 5 40 = true or false? True Using cross products, you find that 3 • 40 = 120 and 5 • 24 = 120, so the cross products are equal and the proportion is true. 50 cm x cm 30 cm 20 cm 4.22 Self Check B Solve for the unknown quantity, x. = 15 6 10 x Cross-multiplying, you get the equation 6x = 150. Dividing, you find x = 25. Self Check C A map uses a scale where 2 inches represents 5 miles. If the distance between two cities is shown on a map as 20 inches, how many miles apart are the two cities? Setting up the proportion = 2 inches 20 inches 5 miles x , you find that x = 50 miles. Self Check D Find the unknown side. To see the similar triangles, it may be helpful to split apart the picture, as shown to the right above. Setting up the proportion = 20 cm cm 50 30 cm x cm , you find x = 12 cm. 50 cm x cm 30 cm 20 cm 50 cm x cm 30 cm 20 cm |
9135 | https://link.springer.com/book/10.1007/978-1-4612-4190-4 | Skip to main content
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Classical Descriptive Set Theory
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: Alexander S. Kechris
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Alexander S. Kechris
Alfred P. Sloan Laboratory of Mathematics and Physics Mathematics 253-37, California Institute of Technology, Pasadena, USA
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Graduate Texts in Mathematics (GTM, volume 156)
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Descriptive set theory has been one of the main areas of research in set theory for almost a century. This text attempts to present a largely balanced approach, which combines many elements of the different traditions of the subject. It includes a wide variety of examples, exercises (over 400), and applications, in order to illustrate the general concepts and results of the theory.
This text provides a first basic course in classical descriptive set theory and covers material with which mathematicians interested in the subject for its own sake or those that wish to use it in their field should be familiar. Over the years, researchers in diverse areas of mathematics, such as logic and set theory, analysis, topology, probability theory, etc., have brought to the subject of descriptive set theory their own intuitions, concepts, terminology and notation.
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Keywords
Baire space
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Table of contents (40 chapters)
Front Matter
Pages i-xviii
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2. ### Polish Spaces
Topological and Metric Spaces
Alexander S. Kechris
Pages 1-4
2. #### Trees
Alexander S. Kechris
Pages 5-12
3. #### Polish Spaces
Alexander S. Kechris
Pages 13-17
4. #### Compact Metrizable Spaces
Alexander S. Kechris
Pages 18-28
5. #### Locally Compact Spaces
Alexander S. Kechris
Pages 29-30
6. #### Perfect Polish Spaces
Alexander S. Kechris
Pages 31-34
7. #### Zero-dimensional Spaces
Alexander S. Kechris
Pages 35-40
8. #### Baire Category
Alexander S. Kechris
Pages 41-57
9. #### Polish Groups
Alexander S. Kechris
Pages 58-64
3. ### Borel Sets
Measurable Spaces and Functions
Alexander S. Kechris
Pages 65-67
2. #### Borel Sets and Functions
Alexander S. Kechris
Pages 68-72
3. #### Standard Borel Spaces
Alexander S. Kechris
Pages 73-81
4. #### Borel Sets as Clopen Sets
Alexander S. Kechris
Pages 82-84
5. #### Analytic Sets and the Separation Theorem
Alexander S. Kechris
Pages 85-88
6. #### Borel Injections and Isomorphisms
Alexander S. Kechris
Pages 89-93
7. #### Borel Sets and Baire Category
Alexander S. Kechris
Pages 94-102
8. #### Borel Sets and Measures
Alexander S. Kechris
Pages 103-119
9. #### Uniformization Theorems
Alexander S. Kechris
Pages 120-128
10. #### Partition Theorems
Alexander S. Kechris
Pages 129-136
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Authors and Affiliations
Alfred P. Sloan Laboratory of Mathematics and Physics Mathematics 253-37, California Institute of Technology, Pasadena, USA
Alexander S. Kechris
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Book Title: Classical Descriptive Set Theory
Authors: Alexander S. Kechris
Series Title:
Graduate Texts in Mathematics
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Publisher: Springer New York, NY
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Copyright Information: Springer-Verlag New York, Inc. 1995
Hardcover ISBN: 978-0-387-94374-9Published: 06 January 1995
Softcover ISBN: 978-1-4612-8692-9Published: 21 December 2011
eBook ISBN: 978-1-4612-4190-4Published: 06 December 2012
Series ISSN:
0072-5285
Series E-ISSN:
2197-5612
Edition Number: 1
Number of Pages: XVIII, 404
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Mathematical Logic and Foundations,
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9136 | https://web.ma.utexas.edu/users/m408m/Display12-1-4.shtml | | | |
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| M408M Learning Module Pages Main page Chapter 10: Parametric Equations and Polar Coordinates Chapter 12: Vectors and the Geometry of Space Learning module LM 12.1: 3-dimensional rectangular coordinates: Rectangular coordinates in $3$-space Rectangular coordinates in $3$-space p2 Terminology and notation Learning module LM 12.2: Vectors: Learning module LM 12.3: Dot products: Learning module LM 12.4: Cross products: Learning module LM 12.5: Equations of lines and planes: Learning module LM 12.6: Surfaces: Chapter 13: Vector Functions Chapter 14: Partial Derivatives Chapter 15: Multiple Integrals | Terminology and notationTerminology and Notation In the $x$-$y$ plane, the coordinate axes break the plane into four quadrants, and the first quadrant is the region where $x \ge 0$ and $y \ge 0$. In 3-dimensional space, the coordinate planes break space into eight regions, called octants. The first octant is the region where $x \ge 0$, $y \ge 0$ and $z \ge 0$. Unlike in the plane, there is no standard numbering for the other octants. We usually think of the $x$-$y$ plane as being horizontal, with the $x$ axis pointing East, the $y$ axis pointing North, and the $z$ axis pointing straight up. This is described by the right hand rule. If you hold your right hand so that the fingers are pointing in the positive $x$ direction, and if you can bend your fingers so that they point in the positive $y$ direction, then your outstretched thumb is pointing in the positive $z$ direction. (How would things be different if we applied the left hand rule instead? Try it and see.) Points are written as triples of numbers between parentheses, sometimes preceded by the name of the point. So $P(3,2,1)$ is a point named $P$ at $x=3$, $y=2$, $z=1$, while $(3,2,1)$ is an unnamed point at the same spot. Angle brackets are used to denote vectors, as explained in the next learning module. The displacement from $P(3,2,1)$ to $Q(5,6,7)$ is the vector ${\overrightarrow{PQ}} = \langle 2,4,6 \rangle$. | |
9137 | https://www.khanacademy.org/math/8th-engage-ny/engage-8th-module-1/8th-module-1-topic-a/v/products-and-exponents-raised-to-an-exponent-properties | Exponent properties with parentheses (video) | Khan Academy
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8th grade (Eureka Math/EngageNY)
Course: 8th grade (Eureka Math/EngageNY)>Unit 1
Lesson 1: Topic A: Exponential notation and properties of integer exponents
Exponent properties with products
Multiply powers
Exponent properties with parentheses
Powers of powers
Exponent properties with quotients
Divide powers
Powers of products & quotients (structured practice)
Powers of products & quotients
Exponent properties review
Negative exponents
Negative exponent intuition
Negative exponents
Negative exponents review
Multiply & divide powers (integer exponents)
Powers of products & quotients (integer exponents)
Properties of exponents challenge (integer exponents)
Powers of zero
Multiplying multiples of powers of 10
Multiplication and division with powers of ten
Approximating with powers of 10
Approximating with powers of 10
Math>
8th grade (Eureka Math/EngageNY)>
Module 1: Integer exponents and scientific notation>
Topic A: Exponential notation and properties of integer exponents
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Exponent properties with parentheses
CCSS.Math: 8.EE.A.1
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About About this video Transcript
Learn two exponent properties: (ab)^c = (a^c)(b^c) and (a^b)^c = a ^ (bc). See WHY they work and HOW to use them. In other words, multiplying two numbers, then raising the product to an exponent is the same as raising each number to that exponent and then multiplying. Raising a number to an exponent and then to another exponent equals raising the base to the product of the two exponents.Created by Sal Khan.
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Cybernetic Organism 12 years ago Posted 12 years ago. Direct link to Cybernetic Organism's post “a^3 a^3 = a^6 What hap...” more a^3 a^3 = a^6 What happens when you have a^3 + a^3?
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lk 10 years ago Posted 10 years ago. Direct link to lk's post “There are two ways you co...” more There are two ways you could approach this problem if you can't figure it out right away. First, you could think of the distributive property when you add (a^3) to (a^3). Second, you could substitute Y for (a^3) and solve.
(1) Use the distributive property.
a^3 + a^3 is equal to (1) x (a^3) + (1) x (a^3) since any number multiplied by positive 1 is itself. Therefore, the expression a^3 + a^3 can be written as (1) x (a^3) + (1) x (a^3).
Factor out (a^3) to get (a^3) x ( 1 + 1 ) = (a^3) x (2), or (2) x (a^3).
It's important to focus on the arithmetic side rather than on the exponential side of the problem. Arithmetic rules don't change when we add, multiply or factor out terms with exponents.
(2) Use Y instead.
Let Y be a^3. Then, the initial expression a^3 + a^3 is equal to Y + Y.
Y + Y = 2Y
Substitute in a^3 for Y to get (2) x (Y) = 2 x (a^3).
Hope this is useful.
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Aamyrah 2 years ago Posted 2 years ago. Direct link to Aamyrah's post “Why does every use "^"? e...” more Why does every use "^"? ex. (a^b)
What does it stand for?
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Mason Surfer 2 years ago Posted 2 years ago. Direct link to Mason Surfer's post “The "^" is a caret symbol...” more The "^" is a caret symbol on the keyboard which is a way to denote exponents while typing, which you can't do as correctly and that's why the "^" acts a substitute for the exponent on the higher end. For example: 5^3 = 5x5x5 = 125. So the last number after the "^" is the exponent. Hopefully that's helpful.
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shaban ali 2 years ago Posted 2 years ago. Direct link to shaban ali's post “can you use exponents to ...” more can you use exponents to show exponents? For example 3 to the 6th power and the 6th power is shown as 6 to the 1st power?
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Lucky 2 years ago Posted 2 years ago. Direct link to Lucky's post “yes, you can, 3^6^1 = 3^6...” more yes, you can, 3^6^1 = 3^6 = 729, for a better example, 2^2^3 = 2^8 = 256, but (2^2)^3 = 4^3 = 64
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Aamyrah 2 years ago Posted 2 years ago. Direct link to Aamyrah's post “This video is labelled "A...” more This video is labelled "Algebra". Im doing the 7th Grade math (FL B.E.S.T.) course so I do have 8th grade math and such. Could someone explain Algebra in some simple words for me to understand?
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Sanjanaa Vinod 2 years ago Posted 2 years ago. Direct link to Sanjanaa Vinod's post “Think of Algebra as a typ...” more Think of Algebra as a type of math. It's a category, like how you might have "Vocabulary" in Language Arts. It doesn't really have a definition of its own, other than being the category where equations and expressions come into play.
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𝓜𝓪𝓱𝓮𝓼𝓱 𝓜𝓪𝓱𝓮𝓷𝓭𝓻𝓪𝓴𝓪𝓻 6 years ago Posted 6 years ago. Direct link to 𝓜𝓪𝓱𝓮𝓼𝓱 𝓜𝓪𝓱𝓮𝓷𝓭𝓻𝓪𝓴𝓪𝓻's post “Will the property still h...” more Will the property still hold when ((a^x)^y)^z is the same as multiplying a^xyz ?
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gurushishya 2 years ago Posted 2 years ago. Direct link to gurushishya's post “Yes. Doing the parenthese...” more Yes. Doing the parentheses the way you show them is critical though. It would not be true without the nested parentheses.
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whitesugarcookie19 a year ago Posted a year ago. Direct link to whitesugarcookie19's post “What does "a to the b" m...” more What does "a to the b" mean?
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “When Sal says "a to the b...” more When Sal says "a to the b", he means a raised to the power of b, or a^b.
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Ritwik Ramsundar 4 months ago Posted 4 months ago. Direct link to Ritwik Ramsundar's post “Can an exponent be negati...” more Can an exponent be negative?
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JaniceHolz 4 months ago Posted 4 months ago. Direct link to JaniceHolz's post “Yes. That will be covere...” more Yes. That will be covered in the next lesson (Lesson 7).
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sb322363 2 years ago Posted 2 years ago. Direct link to sb322363's post “what would you get if you...” more what would you get if you had -(abcdefgh)^jklmnop
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “Your outer exponent would...” more Your outer exponent would be applied to all the variables inside the parentheses.
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arkrish.manigandan 2 years ago Posted 2 years ago. Direct link to arkrish.manigandan's post “how are you able to multi...” more how are you able to multiply powers ?
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questionExplainer a month ago Posted a month ago. Direct link to questionExplainer's post “If you're talking about s...” more If you're talking about something like a^3^2, I can walk you through.
a ^ 3 = aaa, and that to the power of 2 is (aaa)(aaa), which is the same as a^6, which is the product of 3 and 2. Hope this helps, if you have any questions, let me know
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AavyanS 2 years ago Posted 2 years ago. Direct link to AavyanS's post “Does (ab)^4 also equal ((...” more Does (ab)^4 also equal ((a^2)b)^4 (a(b^2))^4
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “I'm not sure where you go...” more I'm not sure where you got that from, since ((a^2)b)^4 (a(b^2))^4 evaluates to a^12b^12. The correct expansion to (ab)^4 is a^4b^4.
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Video transcript
And now I want to go over some of the other core exponent properties. But they really just fall out of what we already know about exponents. Let's say I have two numbers, a and b. And I'm going to raise it to-- I could do it in the abstract. I could raise it to the c power. But I'll do it a little bit more concrete. Let's raise it to the fourth power. What is that going to be equal to? Well that's going to be equal to-- I could write it like this. Copy and paste this, copy and paste. That's going to be equal to ab times ab times ab times ab times ab. But what is that equal to? Well when you just multiply a bunch of numbers like this it doesn't matter what order you're going to multiply it in. This right over here is going to be equivalent to a times a times a times a times-- We have four b's as well that we're multiplying together. Times b times b times b times b. And what is that equal to? Well this right over here is a to the fourth power. And this right over here is b to the fourth power. And so you see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent. And then taking their product. And here I just used the example with 4, but you could do this really with any arbitrary-- actually any exponent. This property holds. And you could satisfy yourself by trying different values, and using the same logic right over here. But this is a general property. That-- let me write it this way-- that if I have a to the b, to the c power, that this is going to be equal to a to the c times b to the c power. And we'll use this to throughout actually mathematics, when we try to simplify things or rewrite an expression in a different way. Now let me introduce you to another core idea here. And this is the idea of raising something to some power. And I'll just use example of 3. And then raising that to some power. What could this be simplified as? Well let's think about it. This is the same thing as a to the third-- let me copy and paste that-- as a to the third times a to the third. And what is a to the third times-- So this is equal to a to the third times a to the third. And that's going to be equal to a to the 3 plus 3 power. We have the same base, so we would add and they're being multiplied. They're being raised to these two exponents. So it's going to be the sum of the exponents, which of course is going to be equal to a-- that's a different color a-- it's going to be a to the sixth power. So what just happened over here? Well I took two a to the thirds. And I multiplied them together. So I took these two 3s and added them together. So this essentially right over here, you could view this as 2 times 3. That's how we got the 6. When I raise something to one exponent, and then raised it to another, that's the equivalent to raising the base to the product of those two exponents. I just did it with this example right over here. But I encourage you try other numbers to see how this works. And I could to do this in general. I could say a to the b power. And then-- let me copy and paste that-- and then I'm going to raise that to the c power. Well what is that going to give me? Well I'm essentially going to have to take c of these, so one, two, three. I don't know how large of a number c is, so I'll just do the dot, dot dot. So dot, dot, dot. I have c of these, right over here. So what is that going to be equal to? Well that is going to be equal to a to the-- well for each of these c, I'm going to have a b that I'm going to add together. So let me write this. So I'm going to have a b plus b plus b plus dot, dot, dot plus b. And now I have c of these b's, so I have c b's right over here. Or you could view this as a, this is equal to a to the c times b power. c or a, you could do a to the cb power. So very useful. So if someone were to say what is 35 to the third power, and then that raised to the seventh power? Well this is going to obviously be a huge number. But we could at least simplify the expression. This is going to be equal to 35 to the product of these two exponents. It's going to be 35 to the 3 times 7, or 35 to the 21, or to the 21st power.
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9138 | https://dummit.cos.northeastern.edu/docs/ringthy_3_homomorphisms_ideals_quotients.pdf | Ring Theory (part 3): Homomorphisms, Ideals, and Quotients (by Evan Dummit, 2018, v. 1.01) Contents 3 Homomorphisms, Ideals, and Quotients 1 3.1 Ring Isomorphisms and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 3.1.1 Ring Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 3.1.2 Ring Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 3.2 Ideals and Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 3.2.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 3.2.2 Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 3.2.3 Homomorphisms and Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 3.3 Properties of Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.3.1 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 3.3.2 Generation of Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 3.3.3 Maximal and Prime Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 3.3.4 The Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.4 Rings of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 3 Homomorphisms, Ideals, and Quotients In this chapter, we will examine some more intricate properties of general rings. We begin with a discussion of isomorphisms, which provide a way of identifying two rings whose structures are identical, and then examine the broader class of ring homomorphisms, which are the structure-preserving functions from one ring to another.
Next, we study ideals and quotient rings, which provide the most general version of modular arithmetic in a ring, and which are fundamentally connected with ring homomorphisms. We close with a detailed study of the structure of ideals and quotients in commutative rings with 1.
3.1 Ring Isomorphisms and Homomorphisms • We begin our study with a discussion of structure-preserving maps between rings.
3.1.1 Ring Isomorphisms • We have encountered several examples of rings with very similar structures.
• For example, consider the two rings R = Z/6Z and S = (Z/2Z) × (Z/3Z).
◦Here are the addition and multiplication tables in R: + 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 · 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 0 2 4 3 0 3 0 3 0 3 4 0 4 2 0 4 2 5 0 5 4 3 2 1 1 ◦Now compare those tables to the tables in S: + (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (1, 1) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (0, 2) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (1, 1) (1, 0) (1, 0) (0, 1) (1, 2) (0, 0) (1, 1) (0, 2) (0, 1) (0, 1) (1, 2) (0, 0) (1, 1) (0, 2) (1, 0) (1, 2) (1, 2) (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) · (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 1) (0, 0) (1, 1) (0, 2) (1, 0) (0, 1) (1, 2) (0, 2) (0, 0) (0, 2) (0, 1) (0, 0) (0, 2) (0, 1) (1, 0) (0, 0) (1, 0) (0, 0) (1, 0) (0, 0) (1, 0) (0, 1) (0, 0) (0, 1) (0, 2) (0, 0) (0, 1) (0, 2) (1, 2) (0, 0) (1, 2) (0, 1) (1, 0) (0, 2) (1, 1) ◦Notice that these tables look quite similar (especially given the artful reordering of the labels of the elements in S).
◦Indeed, if we relabel each entry n in the rst set of tables with the ordered pair corresponding to its reduction modulo 2 and 3 (so that 1 becomes (1, 1), 2 becomes (0, 2), and so forth) we will obtain the second set of tables!
• For another example, consider the rings R = (Z/2Z) × (Z/2Z) and S = F2[x]/(x2 + x).
◦Here are the addition and multiplication tables in R: + (0, 0) (1, 1) (1, 0) (0, 1) (0, 0) (0, 0) (1, 1) (1, 0) (0, 1) (1, 1) (1, 1) (0, 0) (0, 1) (1, 0) (1, 0) (1, 0) (0, 1) (0, 0) (1, 1) (0, 1) (0, 1) (1, 0) (1, 1) (0, 0) · (0, 0) (1, 1) (1, 0) (0, 1) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 1) (0, 0) (1, 1) (1, 0) (0, 1) (1, 0) (0, 0) (1, 0) (1, 0) (0, 0) (0, 1) (0, 0) (0, 1) (0, 0) (0, 1) ◦Now compare those tables to the tables in S: + 0 1 x x + 1 0 0 1 x x + 1 1 1 0 x + 1 x x x x + 1 0 1 x + 1 x + 1 x 1 0 · 0 1 x x + 1 0 0 0 0 0 1 0 1 x x + 1 x 0 x x 0 x + 1 0 x + 1 0 x + 1 ◦Here, if we relabel (0, 0) as 0, (1, 1) as 1, (1, 0) as x, and (0, 1) as x + 1, the rst pair of tables becomes the second set of tables.
• As a third example, consider the rings R = C = {a+bi : a, b ∈R} and S = a b −b a ∈M2×2(R) : a, b ∈R .
◦Notice that S is a subring of M2×2(R): we have a b −b a − c d −d c = a −c b −d −(b −d) a −c and a b −b a · c d −d c = ac −bd ad + bc −(ad + bc) ac −bd .
◦Compare these to the addition and multiplication operations in C: (a + bi) −(c + di) = (a −c) + (b −d)i and (a + bi) · (c + di) = (ac −bd) + (ad + bc)i.
◦Upon identifying the complex number a + bi with the matrix a b −b a , we see that the ring structure of S is precisely the same as the ring structure of C.
• Let us formalize the central idea in the examples above: in each case, we see that there is a way to relabel the elements of R using the elements of S in a way that preserves the ring structure.
◦The desired relabeling is a function ϕ : R →S with the property that ϕ is a bijection (so that each element of R is labeled with a unique element of S) and that ϕ respects the ring operations.
◦Explicitly, we require ϕ(r1 + r2) = ϕ(r1) + ϕ(r2) and ϕ(r1 · r2) = ϕ(r1) · ϕ(r2) for all elements r1 and r2 in R.
• De nition: Let R and S be rings. A ring isomorphism ϕ from R to S is a bijective1 function ϕ : R →S such that ϕ(r1 + r2) = ϕ(r1) + ϕ(r2) and ϕ(r1 · r2) = ϕ(r1) · ϕ(r2) for all elements r1 and r2 in R.
1Recall that a function ϕ : R →S is injective (one-to-one) if ϕ(x) = ϕ(y) implies x = y, and ϕ is surjective (onto) if for every s ∈S there exists an r ∈R with ϕ(r) = s. A bijective function is one that is both injective and surjective. Equivalently, ϕ is a bijection if it possesses a two-sided inverse function ϕ−1 : S →R with ϕ(ϕ−1(s)) = s and ϕ−1(ϕ(r)) = r for every r ∈R and s ∈S.
2 ◦We remark here that in both of the conditions ϕ(r1 + r2) = ϕ(r1) + ϕ(r2) and ϕ(r1 · r2) = ϕ(r1) · ϕ(r2), the operations on the left are performed in R while the operations on the right are performed in S.
◦Note: Isomorphisms arise in a variety of contexts (e.g., isomorphisms of vector spaces, isomorphisms of groups, etc.), and in some cases the rings we are considering may carry additional structure. We will simply say isomorphism rather than explicitly specifying ring isomorphism each time, unless there is a particular reason to do otherwise.
• Example: For R = Z/6Z and S = (Z/2Z)×(Z/3Z), the map ϕ : R →S de ned via ϕ(n) = (n mod 2, n mod 3) is an isomorphism.
◦Note that reducing a residue class in Z/6Z modulo 2 or modulo 3 is well-de ned, since 2 and 3 both divide 6, so ϕ is well-de ned.
◦It is then easy to see that ϕ is a bijection, and that ϕ(r1+r2) = ϕ(r1)+ϕ(r2) and ϕ(r1·r2) = ϕ(r1)·ϕ(r2) for any residue classes r1, r2 ∈Z/6Z. (It is also possible to compare the addition and multiplication tables as we did above.) ◦We therefore conclude that ϕ is an isomorphism.
• Example: For S = a b −b a ∈M2×2(R) : a, b ∈R , show that the map ϕ : C →S de ned via ϕ(a+bi) = a b −b a is an isomorphism.
◦First, we see that ϕ is a bijection since it has a two-sided inverse; namely, the map ϕ−1 : S →C de ned by ϕ−1 a b −b a = a + bi.
◦Furthermore, if z = a + bi and w = c + di, then ϕ(z + w) = ϕ((a + c) + (b + d)i) = a + c b + d −(b + d) a + c = a b −b a + c d −d c = ϕ(z) + ϕ(w) and also ϕ(zw) = ϕ((ac −bd) + (ad + bc)i) = ac −bd ad + bc −(ad + bc) ac −bd = a b −b a · c d −d c = ϕ(z) · ϕ(w).
◦Thus, ϕ satis es all the requirements, so it is an isomorphism.
• De nition: If there is an isomorphism ϕ : R →S, we say R and S are isomorphic, and write R ∼ = S.
◦Intuitively, isomorphic rings share the same structure, except that the elements and operations may be labeled dierently.
• Proposition (Properties of Isomorphisms): If R, S, T are any rings, the following hold: 1. The identity map I : R →R de ned by I(r) = r for all r ∈R is an isomorphism from R to R.
◦Proof: I is clearly a bijection and respects the ring operations.
2. If ϕ : R →S is an isomorphism, then the inverse map ϕ−1 : S →R is also an isomorphism.
◦Proof: Essentially by de nition, ϕ−1 is also a bijection.
◦Now suppose ϕ−1(s1) = r1 and ϕ−1(s2) = r2, so that ϕ(r1) = s1 and ϕ(r2) = s2.
◦Then ϕ(r1 +r2) = ϕ(r1)+ϕ(r2) = s1 +s2, meaning that ϕ−1(s1 +s2) = r1 +r2 = ϕ−1(s1)+ϕ−1(s2), and likewise for multiplication. Thus, ϕ−1 is also an isomorphism.
3. If ϕ : R →S and ψ : S →T are isomorphisms, then the composition ψϕ : R →T is also an isomorphism.
◦Proof: It is straightforward to see that the composition of two bijections is a bijection.
◦Furthermore, we have (ψϕ)(r1 + r2) = ψ(ϕ(r1 + r2)) = ψ(ϕ(r1) + ϕ(r2)) = ψϕ(r1) + ψϕ(r2), and likewise for multiplication. Thus ψϕ is an isomorphism.
3 4. If ϕ : R →S is an isomorphism, then ϕ(0R) = 0S, and if R has a 1, then so does S, and ϕ(1R) = 1S.
◦Proof: For any r ∈R, we have ϕ(r) = ϕ(r + 0R) = ϕ(r) + ϕ(0R): thus, by additive cancellation in S we see ϕ(0R) = 0S.
◦Likewise, if R has a 1, then let s ∈S be arbitrary and r = ϕ−1(s). Then s · ϕ(1R) = ϕ(r)ϕ(1R) = ϕ(r · 1R) = ϕ(r) = s, and likewise ϕ(1R) · s = s, so ϕ(1R) is a multiplicative identity in S.
• From the proposition, we immediately see that being isomorphic is an equivalence relation on any collection of rings. In general, it is not easy to determine whether two given rings are isomorphic, and even if two rings are isomorphic, there is no general method for constructing an isomorphism between them.
◦Two isomorphic rings have the same additive and multiplicative structures. Thus, any statement that only depends on the ring operations must be identical in two isomorphic rings.
◦Thus, for example, if ϕ : R →S is an isomorphism, then R is commutative if and only if S is commutative, and R has a 1 if and only if S has a 1. Likewise, R has zero divisors if and only if S has zero divisors, and the cardinalities of any two isomorphic rings (along with their sets of units) must be equal.
◦So, for example, we see that M2×2(R) is not isomorphic to the ring of real quaternions H, since the former has zero divisors and the latter does not.
◦Likewise, we see that none of the rings Z/mZ for m > 1 are isomorphic to one another, since they all have dierent cardinalities.
◦In a similar way, the ring R is not isomorphic to C since the polynomial equation x2 + 1 = 0 has no solutions in R, but does have solutions in C.
◦As a nal example, the rings Z/4Z and (Z/2Z) × (Z/2Z) are not isomorphic: there are two solutions to x2 = 0 in the rst ring (namely, 0 and 2) while there is only one solution to x2 = 0 in the second ring (namely, (0, 0)). Alternatively, the rst ring has 2 units, while the second ring has only 1.
• We also remark that there can exist nontrivial isomorphisms of a ring with itself. Such maps are known as automorphisms.
◦Remark (for those who like group theory): The set of automorphisms of a ring forms a group under function composition.
• Example: Show that the complex conjugation map ϕ(a + bi) = a −bi is an isomorphism from C to C.
◦It is easy to see that ϕ is a bijection, since it is its own inverse function.
◦Furthermore, it is a straightforward calculation that ϕ(z + w) = ϕ(z) + ϕ(w) and ϕ(zw) = ϕ(z)ϕ(w) for any complex numbers z and w, so ϕ is an isomorphism.
3.1.2 Ring Homomorphisms • We now study maps that respect the structure of ring operations without the requirement that they be bijections.
• De nition: A function ϕ : R →S is a ring homomorphism if ϕ(r1 + r2) = ϕ(r1) + ϕ(r2) and ϕ(r1 · r2) = ϕ(r1) · ϕ(r2) for all elements r1 and r2 in R.
◦Note of course that any isomorphism is a homomorphism, but the reverse is not typically true.
• Example: If m > 1, show that the map ϕ : Z →Z/mZ de ned by ϕ(a) = a, so that ϕ maps the integer a to its associated residue class a modulo m, is a ring homomorphism.
◦From our results on residue classes, we see ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b), and likewise ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b). Thus, ϕ is a homomorphism.
◦Notice that this map is surjective but not injective (since for example ϕ(0) = ϕ(m)), so it is not an isomorphism.
• In essentially the same way, we see that the reduction modulo p map inside F[x] is also a homomorphism: 4 • Example: Let F be a eld with R = F[x] and let p(x) ∈R be nonzero. Then the map ϕ : R →R/pR given by ϕ(a) = a, mapping the polynomial a to its associated residue class a modulo p, is a ring homomorphism.
◦From our results on residue classes, we see ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b), and likewise ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b). Thus, ϕ is a homomorphism.
◦In the next section, we will generalize the ideas in these two examples and describe a general procedure for constructing a quotient ring.
• Example: Let R be a commutative ring and a ∈R. Show that the evaluation at a map ϕa : R[x] →R de ned by ϕa(p) = p(a) is a ring homomorphism.
◦We have ϕa(p + q) = (p + q)(a) = p(a) + q(a) = ϕa(p) + ϕa(q) by the de nition of polynomial addition.
◦Likewise, we have ϕa(rbxb · rcxc) = rbrcab+c = (rbab)(rcac) = ϕa(rbxb)ϕa(rcxc) because R is commuta-tive.
◦Then for any polynomials p and q we see ϕa(pq) = ϕa(p)ϕa(q) by applying distributivity and the fact that ϕa respects multiplication of individual terms and addition.
• Example: Let R and S be any rings. The zero map Z : R →S given by Z(r) = 0S for every r ∈R is a ring homomorphism.
• Example: If S is a subring of R, the map ι : S →R given by ι(s) = s is a ring homomorphism. This map is called the inclusion map (since it simply re ects the set inclusion of S inside R).
• There exist numerous examples of maps that satisfy only one of the two requirements for being a homomor-phism.
◦Non-Example: The function f : Z →Z given by f(n) = 2n is not a homomorphism. Explicitly, although it satis es f(m + n) = 2(m + n) = f(m) + f(n), it is not multiplicative since f(1 · 1) = 2 while f(1) · f(1) = 4.
◦Non-Example: The function f : R →R given by f(x) = x2 is not a homomorphism. Explicitly, although it satis es f(xy) = (xy)2 = f(x)f(y), it is not additive since f(1 + 1) = 4 while f(1) + f(1) = 2.
• Here are a few more examples (and non-examples) of homomorphisms: • Example: Determine whether the map ϕ : M2×2(R) →R given by ϕ a b c d = b is a ring homomorphism.
◦We see that ϕ a1 b1 c1 d1 + a2 b2 c2 d2 = b1 + b2 = ϕ a1 b1 c1 d1 + ϕ a2 b2 c2 d2 .
◦However, ϕ a1 b1 c1 d1 · a2 b2 c2 d2 = a1b2 + b1d2 while ϕ a1 b1 c1 d1 · ϕ a2 b2 c2 d2 = b1b2, and these expressions are not equal in general. Thus, ϕ is is not a homomorphism .
• Example: Determine whether the map ϕ : (Z/15Z) →(Z/15Z) given by ϕ(a) = 10a is a ring homomorphism.
◦We have ϕ(a + b) = 10(a + b) = 10a + 10b = ϕ(a) + ϕ(b).
◦Likewise, ϕ(ab) = 10ab = 100ab = (10a)(10b) = ϕ(a)ϕ(b), since 10 ≡100 (mod 15).
◦Therefore, ϕ is a homomorphism .
• Example: Let R be the ring of in nitely dierentiable real-valued functions on R. Determine whether the derivative map D : R →R given by D(f) = f ′ is a ring homomorphism.
◦We have D(f + g) = (f + g)′ = f ′ + g′ = D(f) + D(g), so D is additive.
◦However, D does not respect ring multiplication, since for example D(x · x2) = 3x2 while D(x) · D(x2) = 2x. Therefore, ϕ is not a homomorphism .
5 • Example: Let R be any ring. Determine whether the map ϕ : R →R × R given by ϕ(r) = (r, r) is a ring homomorphism.
◦We have ϕ(r + s) = (r + s, r + s) = (r, r) + (s, s) = ϕ(r) + ϕ(s).
◦Likewise, ϕ(rs) = (rs, rs) = (r, r)(s, s) = ϕ(r)ϕ(s), so ϕ is a homomorphism .
• Like with isomorphisms, homomorphisms have a number of basic properties.
• Proposition (Properties of Homomorphisms): If R, S, T are any rings, the following hold: 1. If ϕ : R →S and ψ : S →T are homomorphisms, then the composition ψϕ : R →T is also a homomorphism.
◦Proof: Follows from the analogous calculation for isomorphisms.
2. If ϕ : R →S is a homomorphism, then ϕ(0R) = 0S, ϕ(−r) = −ϕ(r) for every r ∈R, and ϕ(r1 −r2) = ϕ(r1) −ϕ(r2) for every r1, r2 ∈R.
◦Proof: For any r ∈R, we have ϕ(r) = ϕ(r + 0R) = ϕ(r) + ϕ(0R): thus, by additive cancellation in S we see ϕ(0R) = 0S.
◦Then 0S = ϕ(0R) = ϕ(r + (−r)) = ϕ(r) + ϕ(−r) so by the uniqueness of additive inverses in S we conclude ϕ(−r) = −ϕ(r).
◦Finally, ϕ(r1 −r2) = ϕ(r1) + ϕ(−r2) = ϕ(r1) −ϕ(r2) by the above calculation.
3. If ϕ : R →S is a surjective homomorphism and R has a 1, then S also has a 1 and ϕ(1R) = 1S.
Furthermore, for any unit u ∈R, the value ϕ(u) is a unit in S whose inverse is ϕ(u−1).
◦Proof: Let s ∈S: then since ϕ is surjective there exists some r ∈R with ϕ(r) = s.
Then sϕ(1R) = ϕ(r)ϕ(1R) = ϕ(r1R) = ϕ(r) = s, and likewise ϕ(1R)s = s, so ϕ(1R) is a multiplicative identity in S.
◦For the other part, if u is a unit in R then 1S = ϕ(1R) = ϕ(u · u−1) = ϕ(u)ϕ(u−1), so ϕ(u) is a unit in S with inverse ϕ(u−1).
• Associated to a homomorphism are two fundamental objects: the kernel and image.
• De nition: If ϕ : R →S is a ring homomorphism, the kernel of ϕ, denoted ker ϕ, is the set of elements in R mapped to 0S by ϕ. In other words, ker ϕ = {r ∈R : ϕ(r) = 0}.
◦Intuitively, the kernel measures how close ϕ is to being the zero map: if the kernel is large, then ϕ sends many elements to zero, while if the kernel is small, ϕ sends fewer elements to zero.
◦Example: The kernel of the reduction homomorphism ϕ : Z →Z/mZ with ϕ(a) = a is the subring mZ.
◦Example: The kernel of the evaluation map ϕa : F[x] →F given by ϕa(p) = p(a) is the set of polynomials in F[x] with p(a) = 0, which is (equivalently) the set of polynomials divisible by x −a.
• De nition: If ϕ : R →S is a ring homomorphism, the image of ϕ, denoted im ϕ, is the set of elements in S of the form ϕ(r) for some r ∈R.
◦In the context of general functions, the image is often called the range of ϕ.
◦Intuitively, the image measures how close ϕ is to being surjective: indeed (by de nition) ϕ is surjective if and only if im ϕ = S.
• The kernel and image of a homomorphism are subrings of R and S respectively: • Proposition (Kernel and Image): Let ϕ : R →S be a ring homomorphism. Then 1. The image im ϕ is a subring of S.
◦Proof: Since ϕ(0R) = 0S, the image contains 0.
Furthermore, if s1 and s2 are in im ϕ so that ϕ(r1) = s1 and ϕ(r2) = s2 for some r1, r2 ∈R, then s1 −s2 = ϕ(r1 −r2) and s1s2 = ϕ(r1r2) are also in im ϕ.
◦Thus, im ϕ contains 0 and is closed under subtraction and multiplication, so it is a subring.
6 2. The kernel ker ϕ is a subring of R. In fact, if x ∈ker ϕ, then rx and xr are in ker ϕ for any r ∈R: in other words, ker ϕ is closed under multiplication by arbitrary elements of R.
◦Proof: Since ϕ(0R) = 0S, the kernel contains 0.
Furthermore, if r1 and r2 are in ker ϕ then ϕ(r1 −r2) = ϕ(r1) −ϕ(r2) = 0 and ϕ(r1r2) = ϕ(r1)ϕ(r2) = 0 · 0 = 0 ◦Thus, kerϕ contains 0 and is closed under subtraction and multiplication, so it is a subring.
◦Moreover, if x ∈ker ϕ then ϕ(rx) = ϕ(r)ϕ(x) = ϕ(r)0 = 0 and likewise ϕ(xr) = ϕ(x)ϕ(r) = 0ϕ(r) = 0.
3. The kernel is zero (i.e., ker ϕ = {0}) if and only if ϕ is injective. In particular, ϕ is an isomorphism if and only if ker ϕ = {0} and im ϕ = S.
◦Proof: If ϕ(a) = ϕ(b), then ϕ(a −b) = ϕ(a) −ϕ(b) = 0, so a −b ∈ker ϕ. Thus, if the only element in ker ϕ is 0, then we must have a −b = 0 so that a = b.
◦Conversely, if x ∈ker ϕ and ϕ is injective, then ϕ(x) = 0 = ϕ(0) implies x = 0.
◦The second statement follows from the facts that ker ϕ = {0} is equivalent to ϕ being injective and im ϕ = S is equivalent to ϕ being surjective.
3.2 Ideals and Quotient Rings • Our next task is to generalize the idea of modular arithmetic into general rings.
◦To motivate our discussion, recall the ideas behind the construction of Z/mZ and R/pR where R = F[x]: we rst de ned modular modular congruences and studied their properties, and then we constructed residue classes and showed that the collection of all residue classes had a ring structure.
• In both Z and F[x], we de ned modular congruences using divisibility, but let us take a broader approach: if I is a subset of R (whose properties we intend to characterize in a moment) let us say that two elements a, b ∈R are congruent modulo I if a −b ∈I.
◦This is a generalization of both types of congruence we have described thus far: for Z/mZ, the set I consists of the multiples of m, while for R/pR, the set I consists of the multiples of p.
◦We would like congruence modulo I to be an equivalence relation: this requires a ≡a (mod I), a ≡b (mod I) implies b ≡a (mod I), and a ≡b (mod I) and b ≡c (mod I) implies a ≡c (mod I).
◦It is easy to see that these three conditions require 0 ∈I, that I be closed under additive inverses, and that I be closed under addition. (Thus, I is in fact closed under subtraction.) ◦Furthermore, we would like the congruences to respect addition and multiplication: if a ≡b (mod I) and c ≡d (mod I), then we want a + c ≡b + d (mod I) and ac ≡bd (mod I).
◦In terms of ring elements, this is equivalent to the following: if b = a + r and d = c + s for some r, s ∈I, then we want (b+d)−(a+c) = r+s to be in I, and we also want bd−ac = (a+r)(c+s)−ac = as+rc+rs to be in I.
◦The rst condition clearly follows from the requirement that I is closed under addition. It is a bit less obvious how to handle the second condition, but one immediate implication follows by setting a = c = 0: namely, that rs ∈I.
◦Thus, I must be closed under multiplication, so it is in fact a subring of R.
◦But the well-de nedness of multiplication actually requires more: since 0 ∈I, we can set r = 0 to see that as ∈I, and we can also set s = 0 to see that rc ∈I.
◦So in fact, I must be closed under (left and right) multiplication by arbitrary elements of R, in addition to being a subring. It is then easy to see that this condition is also su cient to ensure that a ≡b (mod I) and c ≡d (mod I) imply a + c ≡b + d (mod I) and ac ≡bd (mod I).
◦Our last task is to de ne residue classes and then the ring operations: we de ne the residue class a (modulo I) to be the set of ring elements b congruent to a modulo I, which is to say, a = {a+r : r ∈I}.
◦Then we take the operations on residue classes to be a + b = a + b and a · b = a · b: then from our properties of congruences, we can verify that these operations are well-de ned and that the collection of residue classes forms a ring.
7 3.2.1 Ideals • Now that we have established the basic properties of the classes of the sets I we can use to construct congru-ences, we can run through the discussion more formally.
• De nition: A subring I of a ring R is called a left ideal of R if it is closed under arbitrary left multiplication by elements of R, and it is called a right ideal if it is closed under arbitrary right multiplication by elements of R.
◦Explicitly, I is a left ideal if I contains 0 and for any x, y ∈I and any r ∈R, the elements x −y and rx are in I, while I is a right ideal if I contains 0 and for any x, y ∈I and any r ∈R, the elements x −y and xr are in I.
• De nition: A subset I of a ring R that is both a left and a right ideal is called an ideal of R (or, for emphasis, a two-sided ideal).
◦Explicitly, I is an ideal if I contains 0 and for any x, y ∈I and any r ∈R, the elements x −y, rx, and xr are all in I.
◦If R is commutative, then left ideals, right ideals, and two-sided ideals are the same. (As we will mention below, when R is not commutative, there may exist left ideals that are not right ideals and vice versa.) • Here are a few basic examples of ideals: ◦Example: The subrings nZ are ideals of Z, since they are clearly closed under arbitrary multiplication by elements of Z.
◦Example: If R = F[x] and p is any polynomial, the subring pR of multiples of p is an ideal of F[x], since it is closed under arbitrary multiplication by polynomials in F[x].
◦Non-example: The subring Z of Q is not an ideal of Q, since it is not closed under arbitrary multiplication by elements of Q, since for example if we take r = 1 3 ∈Q and x = 4 ∈Z, the element rx = 4 3 is not in Z.
◦Example: For any ring R, the subrings {0} and R are ideals of R. We refer to {0} as the trivial ideal (or the zero ideal) and refer to any ideal I ̸= R as a proper ideal (since it is a proper subset of R).
• Here are a few more examples (and non-examples) of ideals.
• Example: In the polynomial ring Z[x], determine whether the set S of polynomials with even constant term (i.e., the polynomials of the form 2a0 + a1x + a2x2 + · · · + anxn for integers ai) forms an ideal.
◦It is easy to see that 0 ∈S and that S is closed under subtraction.
◦Furthermore, if q(x) is any other polynomial, and p(x) ∈S, then p(x)q(x) also has even constant term, so it is also in S.
◦Thus, S is closed under multiplication by elements of Z[x], so it is an ideal .
• Example: Determine whether the set S of upper-triangular 2 × 2 matrices is a left ideal or a right ideal of M2×2(R).
◦The upper-triangular matrices form a subring, so we need only determine whether they are closed under multiplication by arbitrary 2 × 2 matrices on the left and the right.
◦We can see that if r = 0 0 1 0 and x = 1 1 0 1 then x is upper-triangular but rx = 0 0 1 1 . Thus, S is not a left ideal .
◦Indeed, with the same choices, we have xr = 1 0 1 0 , so S also is not a right ideal .
• Example: Determine whether the set S = {0, 2, 4, 6} of even residue classes is an ideal of Z/8Z.
◦We have 0 ∈S, and it is a straightforward calculation to see that S is closed under subtraction, since the sum of two even residue classes modulo 8 will still be even.
8 ◦Furthermore, the product of any residue class with an even residue class will again be an even residue class (since 8 is even), so S is closed under multiplication by arbitrary elements of R. Thus, S is an ideal .
• Example: Determine whether the set S = {(2a, 3a) : a ∈Z} is an ideal of Z × Z.
◦We have 0 ∈S, and (2a, 3a) −(2b, 3b) = (2(a −b), 3(a −b)) so S is closed under subtraction.
◦But, for example, we can see that (1, 2) · (2, 3) = (2, 6) is not in S, even though (2, 3) is, so S is not closed under arbitrary multiplication by elements of Z × Z. Thus, S is not an ideal .
• Example: Determine whether the set S of matrices of the form a 0 b 0 for a, b ∈R is a left ideal or a right ideal of M2×2(R).
◦Clearly 0 ∈S and S is closed under subtraction. Furthermore, since a 0 b 0 · c 0 d 0 = ac 0 bc 0 , S is also closed under multiplication, so it is a subring.
◦Also, for r = e f g h and x = a 0 b 0 we have rx = ea + fb 0 ga + hb 0 and xr = ae af be bf .
◦Since rx ∈S for every r ∈R and x ∈S, but that xr is not always in S, we see that S is a left ideal but is not a right ideal (and hence not a two-sided ideal either).
◦Remark: By taking transposes, we can also see that the set of matrices of the form a b 0 0 is a right ideal of M2×2 that is not a left ideal.
• Several of the examples above are particular instances of a general class of ideals: • Proposition (Principal Ideals): If R is a commutative ring with 1, the set (a) = {ra : r ∈R} of all R-multiples of a forms a (two-sided) ideal of R, known as the principal ideal generated by a.
◦Proof: Since 0a = 0 we see 0 ∈(a). Furthermore, since ra −sa = (r −s)a we see that (a) is closed under subtraction.
◦Furthermore, if t ∈R then we have t(ra) = (tr)a, so since R is commutative, (a) is closed under multiplication by arbitrary elements of R. Thus, (a) is an ideal.
3.2.2 Quotient Rings • Now that we have discussed ideals, we can use them to study residue classes, and thereby discuss construct quotient rings.
• De nition: If I is an ideal of the ring R, then we say a is congruent to b modulo I, written a ≡b (mod I), if a −b ∈I.
◦As in Z and F[x], congruence modulo I is an equivalence relation that respects addition and multi-plication. The proofs are the same as in Z and F[x], once we make the appropriate translations from divisibility to containment in I.
• Proposition (Ideal Congruences): Let I be an ideal of R and a, b, c, d ∈R. Then the following are true: 1. a ≡a (mod I).
◦Proof: Since a −a = 0 ∈I, the statement is immediate.
2. a ≡b (mod I) if and only if b ≡a (mod I).
◦Proof: If a −b ∈I then −(a −b) = b −a ∈I since I is closed under additive inverses, and conversely if b −a ∈I then so is −(b −a) = a −b.
3. If a ≡b (mod I) and b ≡c (mod I), then a ≡c (mod I).
9 ◦Proof: We are given a−b ∈I and b−c ∈I, so since I is closed under addition, we see (a−b)+(b−c) = a −c ∈I.
4. If a ≡b (mod I) and c ≡d (mod I), then a + c ≡b + d (mod I).
◦Proof: We are given a−b ∈I and c−d ∈I, so since I is closed under addition, we see (a−b)+(c−d) = (a + c) −(b + d) ∈I.
5. If a ≡b (mod I) and c ≡d (mod I), then ac ≡bd (mod I).
◦Proof: We are given a −b ∈I and c −d ∈I. Then since I is closed under arbitrary left and right multiplication, we see that (a −b)c and b(c −d) are also in I. Hence ac −bd = (a −b)c + b(c −d) is also in I since I is closed under addition.
• Now we can de ne residue classes: • De nition: If I is an ideal of the ring R, then for any a ∈R we de ne the residue class of a modulo I to be the set a = a + I = {a + x : x ∈I}. This set is also called the coset of I represented by a.
◦We will use the notation a and a+I interchangeably. (The latter is intended to evoke the idea of adding a to the set I.) ◦We observe, as with our previous examples of residue classes, that any two residue classes are either disjoint or identical and that they partition R: speci cally, a = b if and only if a ≡b (mod I) if and only if a −b ∈I.
• All that remains is to verify that the residue classes form a ring, in the same way as in Z and F[x]: • Theorem (Quotient Rings): Let I be an ideal of the ring R. Then the collection of residue classes modulo I forms a ring, denoted R/I (read as R mod I), under the operations a + b = a + b and a · b = ab. (This ring is called the quotient ring of R by I.) If R is commutative then so is R/I, and likewise if R has a 1 then so does R/I.
◦Remark: The notation R/I is intended to emphasize the idea that I represents a single element (namely, 0) in the quotient ring R/I, and the other elements in R/I are translates of I. In this way, R/I is the ring obtained from R by collapsing or dividing out by I, whence the name quotient ring.
◦The proof of this fact is exactly the same as in the cases of Z and F[x], and only requires showing that the operations are well-de ned.
◦Proof: First we must show that the addition and multiplication operations are well-de ned: that is, if we choose dierent elements a′ ∈¯ a and b′ ∈¯ b, the residue class of a′ + b′ is the same as that of a + b, and similarly for the product.
◦To see this, if a′ ∈¯ a then a′ ≡a (mod I), and similarly if b′ ∈b then b′ ≡b (mod I).
◦Then a′ + b′ ≡a + b (mod I), so a′ + b′ = a + b. Likewise, a′b′ ≡ab (mod I), so a′b′ = ab.
◦Thus, the operations are well-de ned.
◦For the ring axioms [R1]-[R6], we observe that associativity, commutativity, and the distributive laws follow immediately from the corresponding properties in R: the additive identity in R/I is ¯ 0 and the additive inverse of a is −a.
◦Finally, if R is commutative then so will be the multiplication of the residue classes, and if R has a 1 then the residue class 1 is easily seen to be a multiplicative identity in R/I.
• This general description of quotient rings generalizes the two examples we have previously discussed: Z/mZ and R/pR where R = F[x].
◦To be explicit, Z/mZ is the quotient of Z by the ideal mZ, while F[x]/p is the quotient of the polynomial ring F[x] by the principal ideal (p) consisting of all multiples of p.
◦It is not hard to see that the integer congruence a ≡b (mod m), which we originally de ned as being equivalent to the statement m|(b −a), is the same as the congruence a ≡b (mod I) where I is the ideal mZ, since b −a ∈mZ precisely when b −a is a multiple of m.
10 • Here are some additional examples of quotient rings: • Example: If R is any ring, the quotient ring of R by the zero ideal, namely R/0, is (isomorphic to) R itself, while the quotient ring of R by itself, namely R/R, is (isomorphic to) the trivial ring {0}.
• Example: In R = Z[x], with I consisting of all multiples of x2 + 1, describe the structure of the quotient ring R/I.
◦It is easy to see that I is an ideal of R, since it is a subring that is closed under arbitrary multiplication by elements of R.
◦From our discussion of polynomial rings, we know that the residue classes in R/I are represented uniquely by residue classes of the form a + bx where a, b ∈Z. Note that in this quotient ring, we have x2 + 1 = 0, which is to say, x2 = −1.
◦The addition in this quotient ring is given by a + bx+c + dx = (a + c) + (b + d)x while the multiplication is given by a + bx · c + dx = (ac −bd) + (ad + bc)x, which follows from the distributive law and the fact that x2 = −1.
◦In this case, the quotient ring is isomorphic to the ring of Gaussian integers Z[i], with the isomorphism ϕ : R/I →Z[i] given by ϕ(a + bx) = a + bi.
• Example: In R = Z/8Z, with I = {0, 4}, describe the structure of the quotient ring R/I.
◦It is easy to see that I is an ideal of R, since it is a subring that is closed under arbitrary multiplication by elements of R. (Indeed, it is the principal ideal generated by 4.) ◦Since each residue class contains 2 elements, and R has 8 elements in total, there are four residue classes.
With this observation in hand, it is not hard to give a list: 0 = I = {0, 4}, 1 = 1 + I = {1, 5}, 2 = 2 + I = {2, 6}, and 3 = 3 + I = {3, 7}.
◦Notice, for example, that in the quotient ring R/I, we have 1 + 3 = 0, 2 · 2 = 0, and 2 · 3 = 2: indeed, we can see that the structure of R/I is exactly the same as Z/4Z (the labelings of the elements are even the same).
• Example: In the polynomial ring R = Z[x], with I consisting of the polynomials with even constant term (i.e., the polynomials of the form 2a0 + a1x + a2x2 + · · · + anxn for integers ai), describe the structure of the quotient ring R/I.
◦We observe that there are only two residue classes, namely 0 and 1: to see this observe that p(x) ∈0 when the constant term of p is even, and p(x) ∈1 when the constant term of p is odd.
◦Then it is fairly easy to see that the structure of this quotient ring is the same as Z/2Z (or more formally, it is isomorphic to Z/2Z), since 1 + 1 = 0.
3.2.3 Homomorphisms and Quotient Rings • Although homomorphisms and quotient rings may not immediately appear to be connected, in fact they are quite deeply related.
◦To begin, observe that if ϕ : R →S is a ring homomorphism, then the kernel of ϕ is an ideal of R.
◦In fact, we proved this fact earlier when we introduced the kernel, but let us remark again: if x ∈ker ϕ and r ∈R, then ϕ(rx) = ϕ(r)ϕ(x) = ϕ(r)0 = 0 and likewise ϕ(xr) = ϕ(x)ϕ(r) = 0ϕ(r) = 0.
◦Thus, we can use homomorphisms to construct new ideals.
◦Equally importantly, we can also do the reverse: we can use ideals to construct homomorphisms.
◦The key observation in this direction is that the map ϕ : R →R/I associating a ring element to its residue class (i.e., with ϕ(a) = a) is a ring homomorphism.
◦Indeed, the two parts of the de nition of homomorphism were precisely the properties we arranged for the residue classes modulo I to possess: ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b) and ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b).
11 ◦Furthermore, the kernel of this map ϕ is, by de nition, the set of elements in R with ϕ(r) = 0, which is to say, the set of elements r ∈I.
◦Thus, we see that kernels of homomorphisms and ideals are precisely the same things!
• Let us summarize these observations: • Proposition (Projection Homomorphisms): If I is an ideal of R, then the map ϕ : R →R/I de ned by ϕ(a) = a = a + I is a surjective ring homomorphism called the projection homomorphism from R to R/I.
◦Proof: We have ϕ(a + b) = a + b = a + b = ϕ(a) + ϕ(b) and ϕ(a · b) = a · b = a · b = ϕ(a) · ϕ(b), so ϕ is a homomorphism.
◦Furthermore, ϕ is surjective, essentially by de nition: any residue class in R/I is of the form a for some a ∈R, and then ϕ(a) = a.
• The next natural question to ask is: if ϕ : R →S is a homomorphism with kernel I, what can we say about the structure of R/I?
◦For example, if R = Q[x] and ϕ : R →R is de ned by ϕ(p) = p(0), then it is easy to see that ϕ is a homomorphism.
◦Furthermore, the kernel of ϕ is the ideal I of Q[x] consisting of the polynomials divisible by x, while the image of ϕ is the set of rational numbers.
◦Then it is easy to see (from our description of the kernel) that R/I is precisely the same as R/xR, and from the division algorithm for polynomials we know that the residue classes are represented by the polynomials of degree 0 in Q[x]; namely, the constant polynomials c for c ∈Q.
◦But now notice that the structure of R/I (namely, of Q) is exactly the same as the structure as the image of ϕ. More formally, these two rings are isomorphic, with an isomorphism given by identifying a residue class c with the rational number c.
◦This relabeling can, equivalently, be thought of as being done via the homomorphism ϕ: we associate the residue class c in R/I with the rational number ϕ(c) = c.
◦In other words: ϕ gives an isomorphism between R/ ker ϕ and the image im ϕ.
• Theorem (First Isomorphism Theorem): If ϕ : R →S is a homomorphism of rings, then R/ ker ϕ is isomorphic to im ϕ.
◦Intuitively, ϕ is a surjective homomorphism ϕ : R →imϕ. To turn it into an isomorphism, we must collapse its kernel to a single element: this is precisely what the quotient ring R/ ker ϕ represents.
◦Proof: Let I = ker ϕ. We use ϕ to construct a map ψ : R/I →im ϕ, and then show that it is injective and surjective.
◦The map is de ned as follows: for any residue class r ∈R/I, we de ne ψ(r) = ϕ(r).
◦We must verify that this map ψ is well-de ned, so suppose that r′ is some other representative of the residue class r: then r′ −r ∈I, so ϕ(r′ −r) = 0 and thus ϕ(r′) = ϕ(r).
◦Thus, ψ(r′) = ϕ(r′) = ϕ(r) = ψ(r), so the map ψ is well-de ned.
◦It is then easy to see ψ is a homomorphism, since ψ(r + s) = ϕ(r + s) = ϕ(r) + ϕ(s) = ψ(r) + ψ(s) and likewise ψ(r · s) = ϕ(r · s) = ϕ(r) · ϕ(s) = ψ(r) · ψ(s).
◦Next, we see that ψ(r) = 0 precisely when ϕ(r) = 0, which is to say r ∈ker(ϕ), so that r = 0. Thus, the only element in ker ψ is 0, so ψ is injective.
◦Finally, if s is any element of im ϕ, then by de nition there is some r ∈R with ϕ(r) = s: then ψ(r) = s, meaning that ψ is surjective.
◦Since ψ is a homomorphism that is both injective and surjective, it is an isomorphism.
• By using the rst isomorphism theorem, we can construct isomorphisms of rings.
◦In order to show that R/I is isomorphic to a ring S, we search for a surjective homomorphism ϕ : R →S whose kernel is I.
12 • Example: If R is any commutative ring, show that R[x]/(x) is isomorphic to R.
◦Let ϕ : R[x] →R be the evaluation at 0 homomorphism ϕ(p) = p(0). This map is clearly surjective since for any r ∈R we have ϕ(r) = r.
◦Furthermore, the kernel of this homomorphism is precisely the collection of polynomials p(x) = a0 + a1x + · · · + anxn with p(0) = 0, which is easily seen to be the ideal I = (x) consisting of polynomials divisible by x.
◦Thus, by the rst isomorphism theorem, for I = (x) we have R[x]/I ∼ = R.
• Example: Show that Z/12Z is isomorphic to (Z/3Z) × (Z/4Z).
◦We seek a surjective homomorphism ϕ : Z →(Z/3Z) × (Z/4Z) whose kernel is 12Z.
◦Once this idea is suggested, it is not hard to come up with a candidate, namely, ϕ(a) = (a mod 3, a mod 4).
◦It is easy to verify that map is a homomorphism (since the individual maps of reduction mod 3 and reduction mod 4 are homomorphisms) and it is likewise fairly easy to see that the map is surjective by checking that the images of 0, 1, ... , 11 represent all of the elements in (Z/3Z) × (Z/4Z).
◦Finally, the kernel of the map consists of all integers a with ϕ(a) = (0, 0), which is equivalent to saying a ≡0 (mod 3) and a ≡0 (mod 4), so that 3|a and 4|a: thus, the kernel is precisely 12Z.
◦Therefore, by the rst isomorphism theorem applied to this map ϕ, we conclude that Z/12Z is isomorphic to (Z/3Z) × (Z/4Z).
◦Remark: In fact, we could have avoided checking surjectivity explicitly by instead observing that the rst isomorphism theorem yields an injective homomorphism ψ : Z/12Z →(Z/3Z)×(Z/4Z), which must therefore also be surjective since there are 12 elements in both sets.
3.3 Properties of Ideals • Now that we have established basic properties of ideals, homomorphisms, and quotient rings, we embark on a deeper study of these topics.
3.3.1 The Isomorphism Theorems • We begin by discussing several fundamental theorems about rings, subrings, and ideals that are collectively known as the isomorphism theorems. We have already proven the rst one: • Theorem (First Isomorphism Theorem): If ϕ : R →S is a homomorphism of rings, then R/ ker ϕ is isomorphic to im ϕ.
• Theorem (Second Isomorphism Theorem): If A is a subring of R and B is an ideal of R, then A+B = {a+b : a ∈A, b ∈B} is a subring of A, A ∩B is an ideal of A, and (A + B)/B is isomorphic to A/(A ∩B).
◦Proof: Clearly A + B contains 0 and (a + b) −(a′ + b′) = (a −a′) + (b −b′) so it is also closed under subtraction. For multiplication, we observe (a + b)(a′ + b′) = aa′ + ba′ + ab′ + bb′: the rst term is in A since A is a subring, while the other three terms are in B (hence so is their sum) since B is an ideal.
◦For the last statement, consider the map ϕ : A →(A + B)/B de ned by ϕ(a) = a + B. This map is well-de ned and a homomorphism by the basic properties of quotient rings, and it is surjective since for any class r + B in (A + B)/B for some r = a + b ∈A + B, we have ϕ(a) = a + B = r + B.
◦The kernel of the map ϕ consists of all a ∈A with a + B = 0 + B, which is (by de nition) equivalent to saying a ∈B: thus, ker ϕ = A∩B. In particular, A∩B is an ideal since it is a kernel of a homomorphism.
◦Thus, by applying the rst isomorphism theorem to ϕ, we see that the rings A/(A ∩B) and (A + B)/B are isomorphic, as claimed.
• Theorem (Third Isomorphism Theorem): If I and J are ideals of R with I ⊆J, then J/I is an ideal of R/I and (R/I)/(J/I) is isomorphic to R/J.
13 ◦Proof: De ne the map ϕ : R/I →R/J given by setting ϕ(r + I) = r + J. This map is well-de ned because if r′ + I = r + I, then since J contains I, we also have r′ + J = r + J, and it is also surjective since for any class r + J in R/J, we clearly have ϕ(r + I) = r + J.
◦Furthermore, ϕ is a homomorphism by the basic properties of quotient rings, since for example ϕ((r1 + r2) + I) = (r1 + r2) + J = (r1 + J) + (r2 + J) = ϕ(r1 + I) + ϕ(r2 + I), which shows that ϕ is additive because (r1 + I) + (r2 + I) = (r1 + r2) + I.
◦Likewise, since (r1 + I)(r2 + I) = r1r2 + I, we see that ϕ(r1r2 + I) = r1r2 + J = (r1 + J)(r2 + J) = ϕ(r1 + I)ϕ(r2 + I) and so ϕ is multiplicative.
◦The kernel of the map ϕ consists of all r + I in R/I with the property that r + J = 0 + J, which is equivalent to saying r ∈J: thus, ker ϕ consists of the classes of the form r + I for r ∈J; this is simply another way of saying that ker ϕ = J/I.
◦Finally, by applying the rst isomorphism theorem to ϕ, we see that the rings (R/I)/(J/I) and R/J are isomorphic, as claimed.
• Example: Inside R = Z[x], let I be the ideal of all polynomials with zero constant term and J be the ideal of all polynomials with even constant term.
◦As we have already mentioned, both I and J are ideals of R, and clearly I ⊆J.
◦Furthermore, R/I is isomorphic to Z (per the division algorithm), and J/I is isomorphic to 2Z (the residue classes are represented by the even integers). Also, R/J is isomorphic to Z/2Z (since the residue classes are 0 and 1).
◦Then indeed (R/I)/(J/I) ∼ = Z/2Z ∼ = R/J, as claimed.
• Theorem (Fourth/Lattice Isomorphism Theorem): If I is an ideal of R, then there is an inclusion-preserving bijection between subrings A of R containing I and the subrings A = A/I of R/I. Furthermore, a subring A of R containing I is an ideal of R if and only if A/I is an ideal of R/I.
◦Proof: We showed during the proof of the second isomorphism theorem that if A contains I then I is an ideal of A, so the association of A with A = A/I is well-de ned. Conversely, if S is a subring of R/I, then the set A = {r ∈R : r + I ∈S} is the unique subring of R containing I with the property that A/I = S.
◦Furthermore, if B is a subring containing A, then B = B + I contains A = A + I, so the association preserves containment.
◦For the statements about ideals, we showed during the proof of the third isomorphism theorem that if J is an ideal containing I then J/I is an ideal of R/I. Conversely, if J/I is an ideal of R/I, then for any r ∈R and x ∈J we have r(x + I) ∈J/I, and this is equivalent to saying that rx ∈J: thus, J is an ideal of R (since it is already a subring, per the above).
• Example: For R = Z and I = 10Z, identify the ideals of R containing I and verify that they all yield ideals of R/I.
◦The ideals of R containing I are Z, 2Z, 5Z, and 10Z.
◦The corresponding ideals of R/I = Z/10Z are Z/10Z, 2Z/10Z = {0, 2, 4, 6, 8}, 5Z/10Z = {0, 5}, and 10Z/10Z = {0}.
◦As claimed, each of these is indeed an ideal of Z/10Z.
3.3.2 Generation of Ideals • In order to study the structure of ideals, we would like a simpler way to describe them. A convenient way is to describe ideals as being generated by subsets of a ring: ◦If R is a ring with 1 and A is a subset of R, we would like to de ne the ideal generated by A to be the smallest ideal containing A.
14 ◦A priori, it is not obvious that there is such a smallest ideal. However, since the intersection of any nonempty collection of ideals is also an ideal, and since A is contained in at least one ideal (namely the whole ring R), we can equivalently de ne (A) to be the intersection of all ideals containing A.
◦In a similar way, we could de ne the left ideal generated by A to be intersection of all left ideals containing A, and we could de ne the right ideal generated by A to be the intersection of all right ideals containing A.
◦However, although the above analysis clearly indicates that these de nitions are well-posed, we have not actually described what these ideals are.
◦If I is the left ideal generated by A, then if a1, a2, . . . , an are any elements of A, we see that I must contain the elements r1a1, r2a2, ... , rnan for any ri ∈R and hence also contain their sum.
◦On the other hand, if we let S be the set of elements of the form r1a1 + r2a2 + · · · + rnan for any ai ∈A and ri ∈R (and some n ≥0), then it is easy to see that S is a subring that is closed under left multiplication by elements of R, so S is a left ideal.
◦Furthermore, since R contains 1, S contains A, and so by the discussion above, we see that S is a left ideal containing A, hence must actually be the left ideal generated by A.
◦In a similar way, the right ideal generated by A consists of the elements of the form a1r1+a2r2+· · ·+anrn.
◦The two-sided ideal generated by A must contain all elements of both forms, but this is not su cient: indeed, the two-sided ideal must contain elements of the form ras for r, s ∈R, so the correct de nition in this case is the set of elements of the form r1a1s1 + r2a2s2 + · · · + rnansn for any ai ∈A and ri, si ∈R (and some n ≥0).
• Proposition (Generation of Ideals): Let R be a ring with 1 and A be a subset of R. Then the set RA = {r1a1 + · · · + rnan : ri ∈R and ai ∈A} is the smallest left ideal containing A, the set AR = {a1r1 + a2r2 + · · · + anrn : ri ∈R and ai ∈A} is the smallest right ideal containing A, and the set (A) = RAR = {r1a1s1 + r2a2s2 + · · · + rnansn : ri, si ∈R and ai ∈A} is the smallest ideal containing A.
◦We will also refer to RA, AR, and (A) as the left ideal generated by A, the right ideal generated by A, and the (two-sided) ideal generated by A, respectively.
◦Note of course that if R is commutative, then (A) = AR = RA = {r1a1+· · ·+rnan : ri ∈R and ai ∈A}.
◦Proof: As noted above, any left ideal containing A must contain RA, any right ideal containing A must contain AR, and any two-sided ideal containing A must contain (A).
◦Furthermore, since 1 ∈R, each of AR, RA, and (A) contains A.
◦Also, RA, AR, and (A) all contain 0 and are closed under subtraction and multiplication, so they are each subrings.
◦Furthermore, RA is closed under left multiplication, AR is closed under right multiplication, and (A) is closed under both, so they are a left ideal, a right ideal, and a two-sided ideal respectively.
◦Then since each of these is the appropriate type of ideal, by the rst observation, we conclude that RA is the smallest left ideal containing A, that AR is the smallest right ideal containing A, and that (A) is the smallest two-sided ideal containing A.
• The simplest class of ideals are those generated by a nite set, and (in particular) those generated by a single element: • De nition: If R is a ring with 1, we say an ideal I is nitely generated if I is generated by a nite set, and we say I is principal if I is generated by a single element. Thus, a nitely generated ideal has the form I = (a1, a2, . . . , an), while a principal ideal has the form I = (a).
◦Note that the de nition for principal ideal extends the one we gave before for commutative rings, since if R is commutative then (a) = Ra = {ra : r ∈R}.
◦If R is not commutative, however, then (a) is the set of elements of the form r1as1 + r2as2 + · · · + rnasn for ri, si ∈R. (Note in particular that (a) is not just the elements of the form ras for r, s ∈R, since the sum of two such elements need not also be of that form.) • Example: If R is any ring with 1, then R = (1) is principal. Likewise, the zero ideal 0 = (0) is also principal.
15 • Example: In Z, for any integer n we have (n) = nZ. Since every ideal of Z is of the form nZ, we see that every ideal of Z is principal.
◦Also, we remark that the notation nZ we have already used is consistent with the de nition above. (The same is true for the notation pR for R = F[x].) ◦We also remark that if a and b are integers with greatest common divisor d, then (a, b) = (d): this follows from the pair of observations that a and b are both contained in (d) so that (a, b) ⊆(d), and that d = xa + yb for some integers x and y by the Euclidean algorithm, so that d is contained in (a, b).
◦Indeed, as a re ection of this fact, many authors write (a, b) to denote the greatest common divisor of a and b.
• Since principal ideals are the easiest to describe, it is often useful to try to determine whether a particular ideal is principal (though this task is not always so easy!): • Example: Show that the ideal I = (2, x) in Z[x] is not principal.
◦Note that I = {2p(x) + xq(x) : p, q ∈Z[x]} is the collection of polynomials in Z[x] with even constant term.
◦If I were principal and generated by some polynomial r(x), then every polynomial in I would be divisible by r(x). Hence, in particular, r(x) would divide 2, so since 2 is a constant polynomial and a prime number, r(x) would have to be one of {±1, ±2}.
◦However, since r(x) must also divide x, the only possibility is that r(x) would be either 1 or −1. But it is easy to see that the ideal generated by 1 (or −1) is all of Z[x], so r(x) cannot be 1 or −1, since I ̸= Z[x].
◦Thus, there is no possible choice for r, so I is not principal . (Of course, it is still nitely generated!) • Example: Determine whether or not the ideal I = (2, 1 + √−5) in Z[√−5] is principal.
◦Suppose this ideal were principal with generator r = a + b√−5 in Z[√−5].
◦Then r would necessarily divide 2, meaning that 2 = rs for some s ∈Z[√−5]. By taking norms, we see that 4 = N(2) = N(r)N(s).
◦Likewise, since r divides 1+√−5, we would have 1+√−5 = rt for some t ∈Z[√−5], so by taking norms we would have 6 = N(1 + √−5) = N(r)N(t).
◦Since N(r) = a2 + 5b2 is a nonnegative integer, we see that N(r) must divide both 4 and 6, hence is either 1 or 2. However, it is easy to see that there are no integer solutions to a2 + 5b2 = 2, and the only elements of norm 1 are 1 and −1.
◦As in the examples above, the ideal generated by 1 (or −1) is all of Z[√−5], but (2, 1 + √−5) ̸= Z[√−5] since every element a + b√−5 in the ideal has a + b even.
◦Thus, I is not principal .
• Example: Determine whether the ideal I = (x3, x + 3) in Q[x] is principal.
◦In the same way as in the example above, if I were principal and generated by a polynomial r(x), then every polynomial in I would be divisible by r.
◦Here, since x3 and x + 3 are relatively prime, we can see that any generator would necessarily divide their gcd, which is 1.
◦In fact, 1 is a generator of I: via the Euclidean algorithm, we can see that 1 = −1 27x3 + (1 3 −1 9x + 1 27x2)(x + 3), and so since both x3 and x + 3 are in I, we see that 1 is also in I.
◦Then since 1 is in I, so is p(x) · 1 = p(x) for any p(x) ∈Q[x], meaning that in fact I = Q[x] and I indeed is principal (and generated by 1).
• We can in fact generalize the argument from the last example above: 16 • Proposition (Ideals of F[x]): If F is a eld, then every ideal in F[x] is principal.
◦Proof: Let I be an ideal of F[x]. If I is the zero ideal we are done, so assume I contains a nonzero element.
◦We claim that I = (d), where d is the monic greatest common divisor of all the elements in I. (Equiva-lently, d is the monic polynomial of largest degree dividing all the elements of I: such a polynomial must exist by the well-ordering axiom.) ◦If d divides every polynomial in I, then clearly I ⊆(d).
◦Conversely, since d is the gcd, by the Euclidean algorithm and the well-ordering axiom we can write d = x1p1 + x2p2 + · · · + xnpn for some polynomials xi ∈F[x] and pi ∈I: then we see that d ∈I, and hence (d) ⊆I. Thus, I = (d) is principal as claimed.
• As we also saw above, when R is a ring with 1, then 1 is a generator of R. We can likewise generalize this statement: • Proposition (Ideals and Units): If I is an ideal of the ring R with 1, then I = R if and only if I contains a unit.
◦Proof: If I = R then certainly I contains a unit (namely, 1).
◦Conversely, if u ∈I is a unit with ur = 1, then since I is an ideal we have 1 = ur ∈I, and then for any s ∈R, the element s = 1s is also in I, and so I = R.
• Since every nonzero element in a eld is a unit, we immediately see that the only nonzero ideal of a eld is the full ring. The converse is also true: • Corollary (Ideals of Fields): A commutative ring R with 1 is a eld if and only if the only ideals of R are 0 and R.
◦Proof: If F is a eld and I is any nonzero ideal, then I contains some nonzero element r. Since F is a eld, r is a unit, and so by the proposition above, I = R.
◦Conversely, if the only ideals of R are 0 and R, let r ∈R be any nonzero element. Then (r) contains r ̸= 0 so it cannot be the zero ideal, so we must have (r) = R.
◦By the previous proposition, this means (r) contains 1: then rs = 1 for some s ∈R, so r is a unit. Hence every nonzero element of R is a unit, so R is a eld as claimed.
◦Remark: In fact, the proof above shows that the only ideals of a division ring R are 0 and R. However, the converse direction does not hold: there exist noncommutative rings R with zero divisors whose only ideals are 0 and R. (One such ring is M2×2(R), although this is not completely trivial to prove.) 3.3.3 Maximal and Prime Ideals • An important class of ideals are those that are maximal under inclusion (i.e., which are not contained in any other ideal except the full ring): • De nition: If R is a ring, a maximal ideal of R is an ideal M ̸= R with the property that the only ideals of R containing M are M and R.
◦Example: If F is a eld, then since the only ideals of F are 0 and F, the zero ideal is a maximal ideal of F.
◦Example: In Z, the ideal mZ is contained in nZ precisely when n divides m. Accordingly, the maximal ideals of Z are precisely the ideals of the form pZ, where p is a prime.
◦Non-example: The ideal (x) is not a maximal ideal of Z[x] because it is contained in the proper ideal (2, x).
• A general ring need not possess any maximal ideals.
◦A trivial example is the zero ring, since its only ideal is itself.
17 ◦For a less trivial example, let R be the ring of rational numbers with trivial multiplication (i.e., so that ab = 0 for any a and b). Since multiplication is trivial, the ideals of R are precisely the sets containing 0 that are closed under subtraction.
◦Now suppose that I is any proper ideal of R, and let S = {n ∈N : 1 n ̸∈I}. If it were true that 1 n ∈I for every positive integer n, then since I is closed under addition and additive inverses, we would necessarily have I = R. Thus, S is a nonempty set of positive integers, so it contains some minimal element d.
◦Then de ne J = I +(1 d): it is not hard to verify that J is an ideal properly containing I. If 1 d2 were in J, then we could write 1 d2 = x + a d for some a ∈Z and x ∈I: multiplying through by d yields 1 d = dx + a, but since dx and a are both in I, this would imply 1/d is in I, which is impossible. Thus, J is a proper ideal, and so I is not maximal.
• However, it is true that a ring with 1 must have maximal ideals: • Theorem (Existence of Maximal Ideals): If R is a ring with 1, then any proper ideal of R is contained in a maximal ideal.
◦Like a number of other general existence theorems (e.g., the proof that every vector space has a basis), this proof requires the (in)famous axiom of choice from set theory. The version of the axiom of choice typically used in algebra is known as Zorn's lemma: if S is a nonempty partially ordered set with the property that every chain in S has an upper bound, then S contains a maximal element2.
◦Proof: Suppose R is a ring with 1 and I is a proper ideal of R.
◦Let S be the set of all proper ideals of R containing I, partially ordered under inclusion. Since I ∈S, S is nonempty.
◦If C is any nonempty chain in S, let J be the union of all ideals in C: then 0 ∈J since 0 is contained in any ideal in C.
◦Furthermore, if x, y ∈J and r ∈R, then by de nition x ∈Ii and y ∈Ij for some Ii and Ij in C. Since Ii ⊆Ij or Ij ⊆Ii since C is a chain, it follows that x −y, rx, and xr are all in one of Ii or Ij, hence in J. Thus, J is an ideal.
◦Also, if it were true that J = R, then the element 1 would be in J. But this is impossible, since by de nition J is the union of a collection of proper ideals of R, none of which therefore contains 1.
◦Therefore, J is an upper bound for S. Hence, by Zorn's lemma, J contains a maximal element, which is therefore a maximal ideal of R that contains I.
• It might initially appear to be di cult to detect whether a particular ideal is maximal. However, by using the isomorphism theorems, it is actually quite easy to detect maximal ideals in commutative rings: • Proposition (Maximal Ideals and Quotients): If R is a commutative ring with 1, then the ideal M is maximal if and only if R/M is a eld.
◦We will remark that this result is not true if we drop either of the assumptions on R (i.e., that it is commutative and has a 1).
◦Proof: By the lattice isomorphism theorem, the ideals of R/M are in bijection with the ideals of R containing M: therefore, M is maximal precisely when the only ideals of R/M are 0 and R/M.
◦Furthermore, if R is commutative with 1, then R/M is also a commutative ring with 1, so R/M is a eld if and only if the only ideals of R/M are 0 and R/M. Putting these two statements together yields the proposition.
• Corollary: If F is a eld, the maximal ideals of F[x] are precisely the principal ideals (p) where p is irreducible.
2A partial ordering on a set S a relation ≤such that for any x, y, z ∈S, (i) x ≤x (ii) x ≤y and y ≤x implies x = y, and (iii) x ≤y and y ≤z implies x ≤z. If S is a partially-ordered set, a subset C is a chain if for any x, y ∈C, either x ≤y or y ≤x, an upper bound for a subset B is an element w ∈B such that b ≤w for all b ∈B, and a maximal element of a subset B is an element m ∈B such that if x ∈B has m ≤x then m = x.
18 ◦Proof: Every ideal of F[x] is principal, and the quotient ring F[x]/(p) is a eld if and only if p is irreducible.
• Example: Determine whether the ideal I = (2, x) is a maximal ideal of R = Z[x].
◦As we have already shown, the quotient ring R/(2, x) is isomorphic to Z/2Z, which is a eld. Thus, I is a maximal ideal of R.
• Example: Determine whether the ideal I = (2) is maximal in R = Z[ √ 2].
◦In the quotient ring R/I, the residue class √ 2 + I is nonzero, but has the property that ( √ 2 + I)2 = 2 + I = 0 + I is equal to zero.
◦Thus, the quotient ring R/I has zero divisors hence is not a eld, meaning that I is not a maximal of R.
• In addition to maximal ideals, we have another important class of ideals in commutative rings: • De nition: If R is a commutative ring, a prime ideal of R is an ideal P ̸= R with the property that for any a, b ∈R with ab in P, at least one of a and b is in P.
◦Remark: There is also a de nition of prime ideal in a noncommutative ring, but it is more complicated (ultimately because the de nition above involves products of elements).
◦As naturally suggested by the name, prime ideals are a generalization of the idea of a prime number in Z: for n > 1, the ideal nZ is a prime ideal of Z precisely when ab ∈nZ implies a ∈nZ or b ∈nZ.
Equivalently (in the language of divisibility) this means n|ab implies n|a or n|b, and this is precisely the condition that n is either a prime number (or zero).
◦Example: The prime ideals of Z are (0) and the ideals pZ where p is a prime number.
◦A similar statement holds in R = F[x]: the ideal (p) is prime precisely when p is not a unit and p|ab implies p|a or p|b, and the latter condition is equivalent to saying that p is either irreducible or zero.
◦Example: The prime ideals of F[x] are (0) and the ideals (p) where p is an irreducible polynomial of positive degree.
• Like with maximal ideals, there is an easy way to test whether an ideal is prime using quotient rings: • Proposition (Prime Ideals and Quotients): If R is a commutative ring with 1, then the ideal P is prime if and only if R/P is an integral domain.
◦This proof is essentially just a restatement of the de nition of a prime ideal using residue classes in the quotient ring using the observation that r ∈P if and only if r = 0 in R/P.
◦Proof: If R is commutative with 1 and P ̸= R, then R/P is also commutative with 1, so we need only test for zero divisors.
◦If P is a prime ideal, then ab ∈P implies a ∈P or b ∈P. In the quotient ring, this says that ab = 0 implies a = 0 or b = 0, which is precisely the statement that R/P has no zero divisors.
◦Conversely, if R/P has no zero divisors, then ab = 0 implies a = 0 or b = 0, which is to say, ab ∈P implies a ∈P or b ∈P. Furthermore, since R/P is not the zero ring (since this possibility is excluded by the de nition of integral domain), we see P ̸= R, and therefore P is a prime ideal of R.
• Corollary: A commutative ring with 1 is an integral domain if and only if 0 is a prime ideal.
◦Proof: 0 is prime if and only if the quotient R/0 ∼ = R is an integral domain.
• Corollary: In a commutative ring with 1, every maximal ideal is prime.
◦Proof: If M is a maximal ideal, then R/M is a eld. Every eld is an integral domain, so M is a prime ideal.
• Example: Determine whether the ideals (x) and (x2) in Z[x] are prime ideals.
19 ◦Note that (x) is the kernel of the evaluation homomorphism ϕ : Z[x] →Z given by ϕ(p) = p(0), and this homomorphism is surjective.
◦Thus, by the rst isomorphism theorem, we see that Z[x]/(x) is isomorphic to Z. Since Z is an integral domain, we conclude that (x) is a prime ideal. (Note that it is not maximal, however, since Z is not a eld.) ◦On the other hand, by the division algorithm, we see that the residue classes in Z[x]/(x2) are of the form a + bx where a, b ∈Z. Since x · x = 0 but x ̸= 0, we see that Z[x]/(x2) has zero divisors, and so (x2) is not a prime ideal.
3.3.4 The Chinese Remainder Theorem • We now state an important theorem regarding quotient rings by products of ideals. We rst require a few preliminary de nitions: • De nition: If R is commutative with 1 and I and J are ideals of R, then the sum I +J = {a+b : a ∈I, b ∈J} is de ned to be the set of all sums of elements of I and J, and the product IJ = {a1b1 + · · · + anbn, : ai ∈ I, bi ∈J} is the set of nite sums of products of an element of I with an element of J.
◦It is not di cult to verify that I + J and IJ are both ideals of R, and that IJ contains the intersection I ∩J.
◦We can also speak of the product I1I2 · · · In of more than two ideals, de ned as the set of nite sums of products of an element from each of I1, I2, . . . , In.
• De nition: If R is commutative with 1, the ideals I and J are comaximal if I + J = R.
◦Note that aZ + bZ = Z precisely when a and b are relatively prime. (The appropriate notion in general rings is not primality but maximality, so we use the term comaximal rather than coprime.) • We can now state the theorem: • Theorem (Chinese Remainder Theorem): Let R be commutative with 1 and I1, I2, . . . , In be ideals of R. Then the map ϕ : R →(R/I1) × (R/I2) × · · · × (R/In) de ned by ϕ(r) = (r + I1, r + I2, . . . , r + In) is a ring homomorphism with kernel I1 ∩I2 ∩· · · ∩In. If all of the ideals I1, I2, . . . , In are pairwise comaximal, then ϕ is surjective and I1 ∩I2 ∩· · · ∩In = I1I2 · · · In, and thus R/(I1I2 · · · In) ∼ = (R/I1) × (R/I2) × · · · × (R/In).
◦Proof: First, ϕ is a homomorphism since ϕ(a + b) = (a + b + I1, . . . , a + b + In) = (a + I1, . . . , a + In) + (b + I1, . . . , b + In) = ϕ(a) + ϕ(b) and similarly ϕ(ab) = (ab + I1, . . . , ab + In) = (a + I1, . . . , a + In) · (b + I1, . . . , b + In) = ϕ(a)ϕ(b).
◦The kernel of ϕ is the set of elements r ∈R such that ϕ(r) = (0 + I1, . . . , 0 + In), which is equivalent to requiring r ∈I1, r ∈I2, ... , and r ∈In: thus, ker ϕ = I1 ∩I2 ∩· · · ∩In.
◦For the second statement, we will prove the results for two ideals and then deduce the general statement via induction.
◦So suppose I and J are ideals of R and ϕ : R →(R/I) × (R/J) has ϕ(r) = (r + I, r + J). We must show that if I + J = R, then I ∩J = IJ and ϕ is surjective.
◦If I + J = R then by de nition there exist elements x ∈I and y ∈J with x + y = 1.
◦Then for any r ∈I ∩J, we can write r = r(x + y) = rx + yr, and both rx and yr are in IJ: hence I ∩J ⊆IJ, and since IJ ⊆I ∩J we conclude IJ = I ∩J.
◦Furthermore, for any a, b ∈R we can write ay + bx = a(1 −x) + bx = a + (b −a)x so ay + bx ∈a + I, and likewise ay + bx = ay + b(1 −y) = b + (a −b)y ∈b + J.
◦Then ϕ(ay + bx) = (ay + bx + I, ay + bx + J) = (a + I, b + J), and therefore ϕ is surjective as claimed.
◦Finally, the statement that R/IJ ∼ = (R/I) × (R/J) then follows immediately by the rst isomorphism theorem. This establishes all of the results for two ideals.
◦For the general statement, we use induction on n: the base case n = 2 was done above, and for the inductive step, it is enough to show that the ideals I1 and I2 · · · In are comaximal, since then we may write R/(I1I2 · · · In) ∼ = (R/I1) × (R/I2 · · · In) and apply the induction hypothesis to R/I2 · · · In.
20 ◦If I1 and Ii are comaximal for 2 ≤i ≤n, then there exist elements xi ∈I1 and yi ∈Ii such that xi + yi = 1. Then 1 = (x2 + y2)(x3 + y3) · · · (xn + yn) ≡y2y3 · · · yn modulo I1. But since y2y3 · · · yn is in I2I3 · · · In, this means that I1 + I2I3 · · · In contains 1 and is therefore all of R, as required.
• The name of this theorem comes from its application inside Z to solving simultaneous modular congruences.
◦Explicitly, if m1, m2, . . . mn are relatively prime positive integers, then ϕ : Z →(Z/m1Z) × (Z/m2Z) × · · · × (Z/mnZ) given by ϕ(a) = (a mod m1, a mod m2, . . . , a mod mn) is a surjective homomorphism with kernel m1m2 · · · mnZ.
◦The fact that this map is surjective says that the system of simultaneous congruences x ≡a1 mod m1, x ≡a2 mod m2, ... , x ≡an mod mn always has a solution in Z. Furthermore, the characterization of the kernel says that the solution is unique modulo m1m2 · · · mn.
◦Systems of congruences of this form were studied by the ancient Chinese, whence the theorem's name.
• A useful application of the Chinese remainder theorem is to decompose Z/mZ as the direct product of other rings when m is composite (some examples of which we have already seen): • Corollary: If m is a positive integer with prime factorization m = pa1 1 pa2 2 · · · pan n , then Z/mZ ∼ = (Z/pa1 1 Z) × · · · × (Z/pan n Z). In particular, the number of units in Z/mZ is m(1 −1/p1)(1 −1/p2) · · · (1 −1/pn).
◦Proof: The rst statement follows from the Chinese remainder theorem along with the observation that if p and q are distinct primes, then the ideals paZ and qbZ are comaximal in Z.
◦The second statement follows from the observation that the number of units in Z/mZ is the same as the number of units in (Z/pa1 1 Z) × · · · × (Z/pan n Z).
◦Finally, an element of a direct product is a unit if and only if each of its components is a unit, and for any prime p, the units in Z/paZ are precisely the pa −pa−1 = pa(1 −1/p) residue classes that are not divisible by p.
◦Remark: The function ϕ(m) = m(1 −1/p1)(1 −1/p2) · · · (1 −1/pn) giving the the number of units in Z/mZ is called the Euler ϕ-function.
3.4 Rings of Fractions • Let R be a commutative ring. Our goal in this section is to discuss a construction for creating rings of fractions, which (as both a motivating example and a special case) includes the construction of the rational numbers Q from the integers Z.
◦A natural rst attempt is simply to construct symbols of the form a b where a, b ∈R, and then de ne addition and multiplication operations on these symbols.
◦However, even in the case of constructing Q from Z, complications already arise since it is not possible to divide by zero, and all rational numbers can be written in multiple forms (e.g., 1/2 = 3/6).
◦Indeed, we say that a/b and c/d are equal (as rational numbers) precisely when ad = bc.
◦To make this more precise, we can think of fractions as ordered pairs (a, b) of integers, with the rational numbers then being equivalence classes of these ordered pairs under the relation (a, b) ∼(c, d) when ad = bc, and then de ne (a, b)+(c, d) = (ad+bc, bd) and (a, b)·(c, d) = (ac, bd), per the usual arithmetic rules a b + c d = ad + bc bd and a b · c d = ac bd.
◦This approach also explains one reason why we should not allow 0 in denominators: if we did, then we would have (0, 0) ∼(a, b) for any integers a and b, and then ∼would not be an equivalence relation.
◦More generally, if we want ∼to be an equivalence relation, then we would need (a, b) ∼(c, d) and (c, d) ∼(e, f) to imply (a, b) ∼(e, f): thus we want ad −bc = 0 and cf −de = 0 to imply af −be = 0.
◦Since f(ad −bc) + b(cf −de) = (ad f −bcf) + (bcf −bde) = d(af −be), if d ̸= 0 we would be able to conclude that af −be = 0.
21 ◦This calculation suggests that, if we want to extend the construction of fractions to general rings, we should avoid having zero divisors (and zero) in our denominators. Indeed, if bd = 0, then we could write (b, 1) ∼(bd, d) ∼(0, d) ∼(0, 1), and so the associated fraction b/1 would be equal to 0/1 = 0.
◦Furthermore, the arithmetic rules clearly require the collection of denominators to be closed under multiplication, since we must be able to add and multiply fractions.
◦It turns out that these two restrictions are su cient to allow us to construct a ring of fractions with a given set of denominators.
• Theorem (Rings of Fractions): Let R be a commutative ring and D be a nonempty subset of R that is closed under multiplication and does not contain 0 or any zero divisors. Then there is a commutative ring D−1R with 1 consisting of elements of the form r·d−1 for r ∈R and d ∈D. This ring D−1R contains (an isomorphic copy of) R as a subring, and every element of D is a unit in D−1R.
◦Proof: Let S = {(r, d) : r ∈R and d ∈D} and de ne a relation on S via (r, d) ∼(s, e) precisely when re = sd.
◦First we observe that ∼is an equivalence relation on S: clearly (r, d) ∼(r, d) since rd −rd = 0, and (r, d) ∼(s, e) implies (s, e) ∼(r, d) since re −sd = 0 implies sd −re = 0.
◦Furthermore, if (r, d) ∼(s, e) and (s, e) ∼(t, f) then re −sd = sf −et = 0, and therefore we see that e(fr −dt) = f(re −sd) + d(sf −et) = 0. But since e ∈D, e is not a zero divisor, and therefore e(fr −dt) = 0 implies fr = dt, so (r, d) ∼(t, f).
◦Now let D−1R be the set of equivalence classes of S under the relation ∼, and write r d to represent the equivalence class of (r, d).
We de ne the addition and multiplication operations in D−1R to be a b + c d = ad + bc bd and a b · c d = ac bd.
◦In order to show that D−1R is a commutative ring with 1, we must verify that these operations are well-de ned and that they satisfy the ring axioms [R1]-[R8].
◦To see + is well-de ned: if a b = a′ b′ and c d = c′ d′ , then a′ b′ + c′ d′ = a′d′ + b′c′ b′d′ and we must show that this equals ad + bc bd . But (a′d′ + b′c′)(bd) −(ad + bc)(b′d′) = (a′b −ab′)dd′ −(c′d −cd′)bb′ = 0 since a′b −ab′ = c′d −cd′ = 0.
◦To see · is well-de ned: if a b = a′ b′ and c d = c′ d′ , then a′ b′ · c′ d′ = a′c′ b′d′ and we must show that this equals ac bd. But (a′c′)(bd) −(ac)(b′d′) = c′d(a′b −ab′) + ab′(c′d −cd′) = 0 since a′b −ab′ = c′d −cd′ = 0.
◦The ring axioms [R1]-[R8] are straightforward calculations (not even requiring the equivalence relation): the additive identity is 0 d for any d ∈D, the additive inverse of a b is −a b , and the multiplicative identity is d d for any d ∈D.
◦Furthermore, we can embed R in D−1R via the map ι : R →D−1R with ι(r) = dr d for any xed d ∈D (note that this embedding does not actually depend on d, since for any other d′ ∈D we have dr d = d′r d′ ): we have ι(a + b) = d(a + b) d = d2(a + b) d2 = da d + db d = ι(a) + ι(b) and likewise ι(ab) = d(ab) d = d2(ab) d2 = da d · db d = ι(a)ι(b), so ι is a homomorphism.
◦Furthermore, if ι(a) = 0 d then this means ad d = 0 d whence ad2 = 0 so that a = 0 (since d2 is not a zero divisor). Hence ι is injective, and so ι(R) is isomorphic to R.
◦Finally, for any e ∈D, we have de d · d de = d2e d2e = d d, so every element of D inside D−1R is a unit, and then any element r d ∈D−1R can be written as r · d−1 for r ∈R and d ∈D.
22 • Using this result, we can show that every integral domain can be viewed naturally as a subset of its eld of fractions: • Corollary (Fields of Fractions): If R is an integral domain, then R is a subring of its eld of fractions D−1R, where D = R{0}.
◦Proof: If R is an integral domain, then D = R{0} is a multiplicatively closed subset not containing zero or any zero divisors.
◦Then D−1R is a commutative ring with 1 in which every element of D is a unit, which is to say, in which every nonzero element of R is a unit.
◦But since the elements of D−1R are all of the form r/s for r, s ∈R and s ̸= 0, this means that every nonzero element of D−1R is a unit in D−1R, so it is a eld.
• Here are a few examples of rings and elds of fractions: ◦Example: The eld of fractions of Z is Q.
◦Example: The eld of fractions of Z[ √ D], or more generally the quadratic integer ring OQ( √ D) is Q( √ D).
◦Example: The eld of fractions of 2Z is the set of rational numbers of the form 2m 2n , hence is also Q. Notice in particular that although 2Z does not have a multiplicative identity, the fraction 2 2 is a multiplicative identity in its eld of fractions.
◦Example: If F is any eld, the eld of fractions of F is simply F itself.
◦Example: If F is any eld, the eld of fractions of F[x] consists of elements p(x) q(x) with q(x) ̸= 0: this is simply the eld of rational functions with coe cients in F.
◦Example: If R = Z and D = {1, p, p2, p3, . . . } where p is a prime number, the ring of fractions D−1R consists of the rational numbers whose denominator is a power of p. This ring is often denoted Z[1/p], since it is obtained by adjoining the number 1/p to Z (and indeed, it is not hard to see that it is the smallest subring of Q that contains 1/p).
• Inside D−1R, every element of D is a unit. In fact, D−1R is (in a fairly strong sense) the smallest ring in which this property holds: • Proposition (Minimality of D−1R): Let D be a multiplicatively closed subset of the ring R not containing 0 or any zero divisors. Suppose S is also a commutative ring with 1 and there is an injection ϕ : R →S with the property that ϕ(d) is a unit for all d ∈D. Then there is an injection Φ : D−1R →S such that Φ(r) = ϕ(r) for all r ∈R.
◦Equivalently, this result says that any ring that contains (an isomorphic copy of) R in which every element of D is a unit must actually contain (an isomorphic copy of) all of D−1R.
◦In the speci c case for elds of fractions, we obtain the following statement: if F is any eld that contains (an isomorphic copy of) an integral domain R and every element of R is a unit in F, then F contains (an isomorphic copy of) the eld of fractions of R.
◦Proof: Suppose ϕ : R →S is an injective ring homomorphism such that ϕ(d) is a unit for all d ∈D.
◦We then de ne Φ : D−1R →S by setting Φ(r/d) = ϕ(r)ϕ(d)−1.
◦To see that Φ is well-de ned, suppose that r′/d′ = r/d, so that r′d = rd′. Then ϕ(r′)ϕ(d) = ϕ(r′d) = ϕ(rd′) = ϕ(r)ϕ(d′), so multiplying by ϕ(d)−1ϕ(d′)−1 yields ϕ(r′)ϕ(d′)−1 = ϕ(r)ϕ(d)−1. Finally, we see Φ(r′/d′) = ϕ(r′)ϕ(d′)−1 = ϕ(r)ϕ(d)−1 = Φ(r/d).
◦Finally, Φ is a ring homomorphism, since Φ(r/d + s/e) = ϕ(re + ds)ϕ(de)−1 = [ϕ(r)ϕ(e) + ϕ(d)ϕ(s)] · ϕ(d)−1ϕ(e)−1 = ϕ(r)ϕ(d)−1 + ϕ(s)ϕ(e)−1 = Φ(r/d) + Φ(s/e) and Φ(r/d · s/e) = ϕ(rs)ϕ(de)−1 = ϕ(r)ϕ(s)ϕ(d)−1ϕ(e)−1 = [ϕ(r)ϕ(d)−1]·[ϕ(s)ϕ(e)−1] = Φ(r/d)·Φ(s/e), and Φ is injective since Φ(r/d) = 0 implies ϕ(r)ϕ(d)−1 = 0 so ϕ(r) = 0, and thus r = 0 since ϕ is injective.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2018. You may not reproduce or distribute this material without my express permission.
23 |
9139 | https://pmc.ncbi.nlm.nih.gov/articles/PMC136157/ | Ebola Virus VP40 Drives the Formation of Virus-Like Filamentous Particles Along with GP - PMC
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J Virol
. 2002 May;76(10):4855–4865. doi: 10.1128/JVI.76.10.4855-4865.2002
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Ebola Virus VP40 Drives the Formation of Virus-Like Filamentous Particles Along with GP
Takeshi Noda
Takeshi Noda
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
Find articles by Takeshi Noda
1,2, Hiroshi Sagara
Hiroshi Sagara
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
Find articles by Hiroshi Sagara
3, Emiko Suzuki
Emiko Suzuki
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
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3, Ayato Takada
Ayato Takada
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
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2, Hiroshi Kida
Hiroshi Kida
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
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1, Yoshihiro Kawaoka
Yoshihiro Kawaoka
Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
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2,4,
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Laboratory of Microbiology, Department of Disease Control, Graduate School of Veterinary Medicine, Hokkaido University, Sapporo 060-0818,1 Division of Virology, Department of Microbiology and Immunology,2 Fine Morphology Laboratory, Department of Basic Medical Science, Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan,3 Department of Pathobiological Science, School of Veterinary Medicine, University of Wisconsin-Madison, Madison, Wisconsin 53706 4
Corresponding author. Mailing address: Institute of Medical Science, University of Tokyo, Shirokanedai, Minato-ku, Tokyo 108-8639, Japan. Phone: 81-3-5449-5310. Fax: 81-3-5449-5408. E-mail: kawaoka@ims.u-tokyo.ac.jp
Received 2001 Nov 13; Accepted 2002 Feb 12.
Copyright © 2002, American Society for Microbiology
PMC Copyright notice
PMCID: PMC136157 PMID: 11967302
Abstract
Using biochemical assays, it has been demonstrated that expression of Ebola virus VP40 alone in mammalian cells induced production of particles with a density similar to that of virions. To determine the morphological properties of these particles, cells expressing VP40 and the particles released from the cells were examined by electron microscopy. VP40 induced budding from the plasma membrane of filamentous particles, which differed in length but had uniform diameters of approximately 65 nm. When the Ebola virus glycoprotein (GP) responsible for receptor binding and membrane fusion was expressed in cells, we found pleomorphic particles budding from the plasma membrane. By contrast, when GP was coexpressed with VP40, GP was found on the filamentous particles induced by VP40. These results demonstrated the central role of VP40 in formation of the filamentous structure of Ebola virions and may suggest an interaction between VP40 and GP in morphogenesis.
Ebola virus, a member of the family Filoviridae in the order Mononegavirales, causes severe hemorrhagic fever in humans and nonhuman primates, resulting in high mortality rates (27). This enveloped, nonsegmented, negative-strand RNA virus (27) has a filamentous appearance, but its shape may be branched, circular, U- or 6-shaped, or long and straight (4). Virions have a uniform diameter of approximately 80 nm but vary greatly in length.
Ebola virus particles consist of seven structural proteins. The glycoprotein (GP) of Ebola virus forms spikes of approximately 7 nm, which are spaced at 5- to 10-nm intervals on the virion surface (4, 27). GP is the only transmembrane protein of Ebola virus and is responsible for receptor binding and membrane fusion (31). Cells infected with recombinant vaccinia virus expressing the GP produced virosomes that varied in shape and diameter but uniformly possessed spike structures on their surface (35), although the effects of more than 80 vaccinia virus proteins (23) on the formation of particles are unknown. Similar virosomes are also released from Ebola virus-infected cells (35). These findings suggest that the GP contributes not only to an early stage of the viral infection cycle but also to viral budding.
The Ebola virus VP40 protein, equivalent to the matrix protein of other negative-strand RNA viruses, is the most abundant protein in virions and is located beneath the viral membrane, where it presumably maintains the structural integrity of the particle (4). In enveloped viruses, matrix proteins play important roles in virus assembly and budding. In vesicular stomatitis virus (VSV), for example, the matrix protein drives the formation of vesicles (16, 19). The matrix protein of human parainfluenza virus type 1, expressed in mammalian cells, assembles into virus-like particles that are released into culture medium (2). The retroviral Gag protein assembles into particles that bud from the cell surface (3, 7, 10). These observations indicate that the matrix proteins of these enveloped viruses have an intrinsic ability to form virus-like particles.
The matrix proteins of many enveloped viruses are associated with the plasma membrane and are thought to interact with the cytoplasmic tails of viral glycoproteins. Such interaction is believed to be important for virus assembly. In influenza viruses, the removal of the cytoplasmic tail of the hemagglutinin or neuraminidase glycoprotein alters virion morphology (14, 22). Although not essential for normal particle formation in rabies virus and VSV, glycoproteins enhance the efficiency of particle formation (20, 21, 30). However, little is known about the VP40-GP interaction in Ebola virion formation.
Using biochemical assays, Jasenosky et al. (12) and others (11, 33) recently showed that expression of VP40 in mammalian cells leads to the production of particles with a density corresponding to that of virions in the culture medium. To determine the morphological properties of these particles and to understand the VP40-GP interaction during virion morphogenesis, we used electron microscopy to examine cells expressing VP40 or GP alone or those expressing both proteins.
MATERIALS AND METHODS
Cells.
293T human embryonic kidney cells were maintained in Dulbecco's modified Eagle medium supplemented with 10% fetal calf serum, l-glutamine and penicillin-streptomycin-gentamicin solution (24). The cells were grown in an incubator at 37°C under 5% CO 2.
Plasmids.
Full-length cDNAs encoding the Ebola virus (species Zaire) VP40 or GP were cloned separately into a mammalian expression vector, pCAGGS/MCS (17, 25), which contains the chicken β-actin promoter. The resulting constructs were designated pCEboZVP40 and pCEboZGP, respectively.
Cell transfection for expression of VP40 and GP.
293T cells (10 6) were transfected with plasmids using the Trans IT LT-1 reagent (Panvera, Madison, Wis.) according to the manufacturer's instructions. Briefly, 1 μg of DNA in 0.1 ml of Opti-MEM (Gibco-BRL) and 3 μl of the transfection reagent were mixed, incubated for 10 min at room temperature, and added to the cells. Transfected cells were incubated at 37°C for 24 or 48 h.
Electron microscopy.
Ultrathin-section electron microscopy was performed as follows. Twenty-four hours posttransfection of 293T cells with plasmids, the cells were washed with phosphate-buffered saline (PBS) and fixed for 20 min with 2.5% glutaraldehyde (GLA) in 0.1 M cacodylate buffer (pH 7.4). They were scraped off the dish, pelleted by low-speed centrifugation, and then fixed for 30 min with the same fixative. Small pieces of fixed pellet were washed with the same buffer, postfixed with 2% osmium tetroxide in the same buffer for 1 h at 4°C, dehydrated with a series of ethanol gradients followed by propylene oxide, embedded in Epon 812 Resin mixture (TAAB), and polymerized at 70°C for 2 days. For immunoelectron microscopy, cells were fixed with 4% paraformaldehyde and 0.1% GLA, dehydrated, and embedded in LR White Resin (London Resin Company Ltd.). Thin sections were stained with uranyl acetate and lead citrate and examined with a JEM-1200EX electron microscope at 80 kV. For negative staining, culture media of 293T cells were collected at 24 h posttransfection onto a Formvar-coated copper grid, stained with 2% phosphotungstic acid solution (PTA), and examined with a JEM-1200 electron microscope at 80 kV. For immunoelectron microscopy, the samples were adsorbed to Formvar-coated nickel grids and washed with PBS containing 0.5% bovine serum albumin (PBS-BSA). The grids were then treated with mouse anti-GP monoclonal antibody (a mixture of ZGP12, ZGP42, and ZGP133 ; 1:150 in PBS-BSA) or rabbit anti-VP40 polyclonal antibody (1:300 in PBS-BSA) and rinsed six times with PBS, followed by incubation with a goat antimouse immunoglobulin conjugated to 15-nm gold particles (1:50 dilution; BB International) or a goat antirabbit immunoglobulin conjugated to 5-nm gold particles (1:100 dilution; BB International). After washing, the samples were fixed for 10 min in 2% glutaraldehyde and negatively stained with 2% PTA.
RESULTS
Pleomorphic particle formation by GP.
To determine the morphology of vesicles induced by Ebola virus GP expression, we analyzed GP-expressing cells and their supernatants by electron microscopy. The ultrathin sections of these cells showed particle-like structures with surface spikes budding from the plasma membrane (Fig. 1A); no such structures were observed using cells transfected with the expression vector alone (Fig. 1B). The spikes, which extended approximately 10 nm from the lipid membrane of the particle-like structure, did not seem to have a particularly ordered arrangement (Fig. 1A, inset). As previously observed with the recombinant vaccinia virus system (35), pleomorphic structures similar to virosomes with a range of diameters were apparent in the supernatants of GP-expressing cells (Fig. 2A and B). The spikes on the surface of the vesicles reacted with a mixture of anti-GP monoclonal antibodies (Fig. 2C and D), confirming the GP derivation of the structures.
FIG. 1.
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Budding of GP-associated particles from the plasma membrane. (A) 293T cells at 24 h posttransfection with a GP-expressing plasmid. (B) 293T cells transfected with an empty expression vector lack such particle formation. Bars, 100 nm.
FIG. 2.
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Pleomorphic particles resulting from GP expression. The supernatants of cells expressing GP were centrifuged through 20% sucrose, and the pelleted material was then negatively stained with 2% PTA. (A and B) Pleomorphic particles with surface spikes were observed. (C and D) Pelleted material was immunolabeled with a mixture of anti-GP monoclonal antibodies conjugated to 15-nm gold particles. Bars, 100 nm.
VP40 induces filamentous particle formation.
To determine how VP40 protein expressed in 293T cells is released into culture medium (11, 12, 33), we analyzed the VP40-expressing cells by transmission electron microscopy. The ultrathin sections of the cells expressing VP40 showed budding of filamentous structures (approximately 65 nm in diameter) on the cell surface (Fig. 3A). In some cells, the plasma membranes appeared ruffled and to consist of two bilayers (Fig. 3C). Aggregated ribosomes (Fig. 3E, arrows) were occasionally found in the cytoplasm of cells expressing VP40, as were electron-dense filamentous structures (approximately 45 nm in diameter; Fig. 3F, arrowheads), which were never seen in cells transfected with the expression vector alone. The budding particles and membrane ruffles reacted with rabbit anti-VP40 polyclonal antibody (Fig. 3B and D), confirming that VP40 had contributed to the generation of these structures. In studies to further determine the size and morphology of the VP40 particles released from cells, the supernatants of cells expressing this protein were centrifuged through 20% sucrose and the pelleted material was negatively stained with 2% PTA and analyzed by electron microscopy. Filamentous particles, which had uniform diameters of approximately 65 nm but varied lengths, were observed (Fig. 4A through C). These results indicate that VP40 alone can induce the formation of filamentous particles which bud from the cell surface.
FIG. 3.
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Morphological changes in 293T cells expressing VP40. At 24 h posttransfection of 293T cells with a VP40-expressing plasmid, filamentous particles budding from the plasma membrane (A), membrane ruffles and the adhering site of two bilayers (C, arrows), as well as aggregated ribosomes (E, arrows) were apparent. Intracellular electron-dense filamentous structures (F, arrowheads) were also observed. (B and D) The filamentous particles and membrane ruffles were immunolabeled with an anti-VP40 antibody conjugated with 5-nm gold particles. M, mitochondrion; mt, microtubule. Bars, 100 nm (panels A, B, C, D, and F) or 200 nm (panel E).
FIG. 4.
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To determine how GP expression affects VP40-driven particle formation, we transfected 293T cells with both VP40- and GP-expressing plasmids. In ultrathin sections of the tranfected cells, we observed filamentous particle-like structures of 80-nm external diameter that were budding from the plasma membrane (Fig. 5).
VP40-GP interaction in particle morphogenesis.
To determine how GP expression affects VP40-driven particle formation, we transfected 293T cells with both VP40- and GP-expressing plasmids. In ultrathin sections of the tranfected cells, we observed filamentous particle-like structures of 80-nm external diameter that were budding from the plasma membrane (Fig. 5).
FIG. 5.
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Filamentous, spiked particles budding from the plasma membrane at 24 h posttransfection of 293T cells with plasmids coexpressing VP40 and GP. Bars, 100 nm.
The structures possessed spikes of approximately 10 nm on their surface, in contrast to the structures observed in cells expressing VP40 alone (Fig. 3A). Also, in contrast to the arrangement of the spikes on the pleomorphic structures induced by GP expression, the spikes on the filamentous particles seemed to have an ordered arrangement (Fig. 5A, inset). Also, unlike the findings for expression of GP alone, few pleomorphic particles were observed. The particle structures were studied in more detail after negative staining of the particles in culture supernatants of cells expressing both VP40 and GP. Filamentous Ebola virus-like particles with surface spikes of approximately 85-nm in external diameter and lengths that ranged to 10 μm were observed (Fig. 6A through C). The spikes projected from the particle surface at 5- to 10-nm intervals and were morphologically indistinguishable from those on the Ebola virion surface (4, 27). Labeling the spikes with a mixture of anti-GP monoclonal antibodies conjugated with gold particles confirmed their identity as GP (Fig. 6D). Furthermore, when treated with 0.03% Triton X-100 and with both the anti-VP40 antibody conjugated to 5-nm gold particles and a mixture of anti-GP monoclonal antibodies conjugated to 15-nm gold particles, the filamentous particles became labeled with both antibodies, demonstrating that the Ebola virus-like particles contained GP as well as VP40 proteins (Fig. 6E). These results demonstrate GP incorporation into VP40-generated filamentous structures without affecting filamentous particle formation.
FIG. 6.
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Ebola virus-like particles produced by coexpression of VP40 and GP. The supernatants of cells coexpressing these two proteins were centrifuged through 20% sucrose, and the pelleted material was then negatively stained with 2% PTA. (A through C) Filamentous particles with surface spikes and varied lengths were observed. Pelleted material was immunolabeled with a mixture of anti-GP monoclonal antibodies conjugated to 15-nm gold particles (D, arrowheads), treated with 0.03% Triton X-100 at room temperature for 15 min, and then immunolabeled with a mixture of anti-GP antibodies conjugated to 15-nm gold particles (E, arrowheads) and an anti-VP40 antibody conjugated to 5-nm gold particles (E, arrows). Bars, 1 μm (panel A) or 100 nm (panels B through E).
DISCUSSION
A hallmark of Ebola viruses is their filamentous virions, as suggested by the family name Filoviridae. The shapes of enveloped viruses are determined by viral proteins of retroviruses (1, 5, 15) or by both viral RNA length and proteins of VSV (26). Because specific interactions among viral components are required for the formation of defined virion shapes, an understanding of such interactions can lead to the identification of targets for the development of antiviral compounds.
Here we showed by electron microscopy that the expression of VP40 in the absence of any other Ebola virus proteins leads to the formation of filamentous particles which resemble spikeless virions released into the supernatant of cultured Ebola virus-infected cells (6). Thus, our results suggest that the Ebola virus VP40 possesses structural information necessary and sufficient to induce the formation of filamentous particles, which then bud from the plasma membrane. Interestingly, some filamentous structures were observed in the cytoplasm of cells expressing VP40 as have been found in the cytoplasm of the cells infected with Ebola virus. Similar structures have also been observed in cells expressing the M1 protein of influenza virus or the Gag protein of retroviruses (3, 7, 9). However, the tubular structures observed upon expression of influenza virus M1 alone were not seen during normal viral infection or when M1 was coexpressed with other influenza virus proteins. Thus, VP40 may form intracellular filamentous structures by self-aggregation.
Membrane ruffles containing VP40 protein were observed in some VP40-expressing cells (Fig. 3C and D). The M protein of VSV induces similar double-layered membranes at the cell surface when expressed from recombinant Sendai virus (29). IpaC protein secreted by Shigella flexneri has also been linked to large-scale membrane extension in macrophages, including lamellipodia and membrane ruffles (18, 34), while Salmonella enterica serovar Typhimurium triggers the formation of host cell membrane ruffles in nonphagocytic cells (8, 36). These membrane ruffles are thought to result from interactions between the bacterial proteins, including IpaC, and the actin cytoskeletons of host cells (34, 36). In Ebola virus-infected cells, host cell plasma membranes proliferate extensively at the peak stage of viral budding (6), as observed in cells expressing VP40 alone. Thus, VP40 may interact with actin filaments during the assembly or budding of Ebola virus at the cell surface.
The impact of glycoprotein interaction with the matrix protein on virion morphology differs among viruses. For example, deletion of the cytoplasmic tails of the influenza virus hemagglutinin and neuraminidase alters virus morphology (14, 22), while the characteristic morphologies of rabies virus and VSV do not depend on glycoprotein-matrix protein interaction (20, 21, 28, 30). The Ebola virus GP, like that of VSV G, was incorporated into filamentous particles without affecting the morphology of the particles. However, such interaction may contribute to the efficiency of budding, as demonstrated by research with VSV (13, 21).
In conclusion, we demonstrated that VP40 induces VP40 containing-filamentous particle formation and that GP spikes are incorporated into VP40 induced-filamentous particles upon coexpression of GP and VP40, resulting in Ebola virus-like particles. This virus-like particle formation system will be useful to further elucidate the mechanism of Ebola virus particle formation, including the functional link between Ebola virus and cellular components.
Acknowledgments
We thank Krisna Wells and Martha McGregor for excellent technical assistance and John Gilbert for editing the manuscript.
This work was supported by Grants-in-Aid from the Ministry of Education, Culture, Sports, Science and Technology and the Ministry of Health, Labor and Welfare, Japan, and by a grant from the National Institute of Allergy and Infectious Diseases (NIH).
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9140 | https://www.ck12.org/flexi/cbse-math/perfect-cube-roots/how-do-i-calculate-the-cube-root-using-a-calculator/ | Flexi answers - How do I calculate the cube root using a calculator? | CK-12 Foundation
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Perfect Cube Roots
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How do I calculate the cube root using a calculator?
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To calculate the cube root of a number using a calculator, follow these steps: 1. Enter the number you want to find the cube root of. 2. Press the "root" or "∛" button. If your calculator doesn't have a cube root button, you can use the exponentiation button (usually marked as "^") and enter the fraction 1/3. 3. Press the "=" button to get the result. For example, to find the cube root of 8, you would enter "8", press the "∛" button (or "^" and then "1/3"), and then press "=". The result should be 2.
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9141 | https://www.math-aids.com/Ratios/ | Ratio Worksheets
Dynamically Created Ratio Worksheets
Here is a graphic preview for all of the Ratio Worksheets. You can select different variables to customize these Ratio Worksheets for your needs. The Ratio Worksheets are randomly created and will never repeat so you have an endless supply of quality Ratio Worksheets to use in the classroom or at home. Our Ratio Worksheets are free to download, easy to use, and very flexible.
These Ratio Worksheets are a great resource for children in Kindergarten, 1st Grade, 2nd Grade, 3rd Grade, 4th Grade, 5th Grade, 6th Grade, 7th Grade, and 8th Grade.
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Simple Ratio Worksheets These Ratio Worksheets will produce groups of symbols for the students to determine the ratio of the different symbols. These Ratio Worksheets are appropriate for 3rd Grade, 4th Grade, 5th Grade, 6th Grade, and 7th Grade.
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Ratios and Rates Word Problems Worksheets These Ratio Worksheets will produce eight ratio and rates word problems for the students to solve. These Ratio Worksheets are appropriate for 3rd Grade, 4th Grade, 5th Grade, 6th Grade, and 7th Grade. |
9142 | https://www.theopeneducator.com/doe/Regression/outlier-leverage-influential-points | Unusual Observations
Outlier, Leverage, and Influential Points
An observation could be unusual with respect to its y-value or x-value. However, rather than calling them x- or y-unusual observations, they are categorized as outlier, leverage, and influential points according to their impact on the regression model.
Outlier – an outlier is defined by an unusual observation with respect to either x-value or y-value. An x-outlier will make the scope of the regression too broad, which is usually considered less accurate. An x-outlier is uncommon, it may seriously affect the regression outcomes though. However, in an unplanned study, often the data is collected before putting much thought into it. In those situations, there could be a possibility of having x-outliers. The y-outliers are very common, and it is usually not as severe as the x-outlier. Nevertheless, the effects of the y-outliers must be investigated further to check whether it is just a simple data entry error, or some severe issue in the process, or just a random phenomenon. Figure 7 shows both x-outlier (left) and y-outlier (right). Both plots show that a better linear relationship will be possible without these outliers. In this situation, the x-outlier is rotating the line clockwise to change both the slope and the intercept of the relationship, while the y-outlier is moving the predicted line upward. The solid line shows the predicted relationship without the outliers.
Figure 7. Outlier with Respect to x-Value (Left) and y-Value (Right)
Leverage – a data point whose x-value (independent) is unusual, y-value follows the predicted regression line though (Figure 8). A leverage point may look okay as it sits on the predicted regression line. However, a leverage point will inflate the strength of the regression relationship by both the statistical significance (reducing the p-value to increase the chance of a significant relationship) and the practical significance (increasing r-square). Unfortunately, leverage points have no impact on the coefficients because the point follows the predicted regression line.
Practical significance of leverage point – think about a relationship between the muscle mass and the power. In the study, if most individuals weigh around 200 pounds and only one person weighs about 400 lbs. This 400-pound and his extreme y-value (power) will dictate the relationship more than all other individuals weighing near 200 pounds. Therefore, the conclusions for the study could be misleading. The Leverage points usually make the functional regression relationship too broad. Generally, a wider (too broad) model is conserved less accurate as compared to a shorter one. To improve the regression model accuracy, shorter models are recommended.
Figure 8. Leverage Point (Right) in a Regression Analysis
Influential – a data point that unduly influences the regression analyses outputs (Figure 9). A point is considered influential if its exclusion causes major changes in the fitted regression function. Depending on the location of the point, it may affect all statistics, including the p-value, r-square, coefficients, and intercept. Figure 9 shows the impact of an influential point on the regression statistics, including the r-square, slope, and the intercept.
Figure 9. Influential Point in a Regression Analysis
Statistical Diagnostic Tests for Unusual Observations
Any statistical software, including MS Excel will produce the diagnostic statistics results. Video 2 provides the diagnostic analysis using Minitab software. It also provides an explanation of the analysis results.
Video 2. How to Explain and Interpret the Linear Regression Diagnostics Analysis Explained Example in Minitab
Statistical Test
for
y-Outlier Point
Diagnostic analysis for each data point is provided in Table 2. An observation is generally considered an outlier if the absolute value of the residual (RESI) is higher. For example, the data point # 6 has a very high residual compared to any other data points of the data set. The absolute values for the other diagnostic statistics such as scaled or adjusted residuals, standardized residuals (SRES) and deleted residuals (TRES) are also observed to be higher for point # 6. Generally, higher absolute value for any of these diagnostic statistics for a point is considered an outlier.
Table 2
Regression Diagnostic Analysis: Detection of Outliers
Statistical Test
for
x-Outlier Point
An x-outlier is determined from the diagonal element of the hat matrix, HI. The diagonal elements of the hat matrix HI has some interesting properties, including
HI measures the weighted distance from the x-mean (mean of the independent variables).
The sum of all diagonal elements of the hat matrix, HI is equal to the sum of the total number of parameters and the intercept, p. In this example, there is one parameter and one intercept, which is equal to 2 = p. Therefore, the sum of HI column in Table 3 is equal to 2.
Therefore, any large value for the HI is considered an outlier with respect to the x-values.
Generally, any value exceeds twice the mean value of the HI (=2(p/n)) is considered an x-outlier.
Point #11 produces a value of 0.73 for the diagonal element of the hat matrix, HI, which is larger than the 2p/n (= 0.036). Therefore, this point #11 is considered an outlier with respect the x-value.
Table 3
Regression Diagnostic Analysis: Detection of x-Outlier and Leverage Points
Statistical Test
for
Leverage Point
A leverage point is determined by a point whose x-value is an outlier, while the y-value is on the predicted line (y-value is not an outlier). Therefore, this point is undetected by the y-outlier detection statistics, including the RESI, SRES, and TRES. For example, the RESI, SRES, and TRES values for the point # 11 are NOT considered large at all, rather they are very consistent with other points. Therefore, the point #11 is not considered an outlier with respect to the y-value. However, the value for the diagonal element of the hat matrix, HI is very large. Any point whose diagonal element of the hat matrix value exceeds 2p/n (22/11=0.36 for this example) is considered a leverage point. Therefore, the point # 11 is considered an x-outlier and it has high leverage on the regression analysis.
Statistical Test
for
Influential Point
DFIT and COOK distance is used to statistically determine the influential point. If the absolute value of DFIT exceeds 1 for small to medium data sets and for large data set, the point is considered as influential to the fitted regression. In this small data set example in Table 4, the absolute value of DFIT for point # 11 is observed to be 3.63 which exceeds 1 (one), and therefore, the point #11 is considered an influential point. While the DFIT measures the influence of the ith case on the fitted value for this case, Cook’s distance, COOK measures the influence of the ith case on all n fitted values. A very large Cook’s distance for a point indicates a potential influence on the fitted regression line. However, to statistically determine the influential point, the probability is calculated using the Cook’s distance as the value for the F-distribution. If the probability value for the Cook’s distance is 50% or more, the point has a major significant influence on the fitted regression line. Probability value between 10-20% indicates a very small influence, while 20-50% indicates moderate to high influence on the fitted regression. The probability for Cook’s distance is calculated using an F-distribution of p and n-p degrees freedom for the numerator and the denominator, respectively. For this example in Table 4, type /write/input = 1-FDIST(1.637,2,9) in MS Excel to calculate the p-value for the point # 11. The probability value calculated for point #11 is 75.2% indicating a major influence on the regression.
Table 4
Detection of Influential Point
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Tools & Reference>Anatomy
Testes and Epididymis Anatomy
Updated: Mar 11, 2025
Author: Todd M Hoagland, PhD; Chief Editor: Vinay K Kapoor, MBBS, MS, FRCSEd, FICS, FAMS more...;)
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Testes and Epididymis Anatomy
Sections Testes and Epididymis Anatomy
Overview
Development
Gross Anatomy
Microscopic Anatomy
Natural Variants
Pathophysiological Variants
Other Considerations
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Overview
Overview
The epididymis and the testes are integral components of the male reproductive system, performing both reproductive (spermatogenesis) and endocrine functions (hormone production or steroidogenesis). The testes are responsible for producing spermatozoa through spermatogenesis in the seminiferous tubules and synthesizing testosterone (by the interstitial endocrine cells [Leydig cells]), the primary male sex hormone. The testes are encapsulated by a dense fibrous layer called the tunica albuginea, which extends inward to form fibrous septa that divide the testis into lobules.
The epididymis is a highly coiled tubular structure located posterior and slightly lateral to the testis. It serves as a site for sperm maturation, storage, and transport to the vas deferens. Spermatozoa entering the epididymis from the seminiferous tubules are immature and nonmotile. During their transit through the epididymal duct, they acquire motility and fertilization capacity due to biochemical changes induced by epididymal secretions.
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Development
Initially, there is an undifferentiated gonad in the retroperitoneal area. Transcription of the SRY gene (sex-determining region Y gene), which is the testis-determining factor region on the Y chromosome, ultimately leads to sex differentiation. Without the SRY gene, the gonad would develop into an ovary. As the fetus develops, the functioning testis produces the male hormone testosterone to allow development of the male genitalia. Over the last 3 months' gestation, the testis must course its way down from its original retroperitoneal position to its final destination in the scrotum. During its journey, it must pass through the peritoneum, abdominal wall via the inguinal canal, and into the scrotal pouch.
An image depicting the testes and epididymis can be seen below.
Male reproductive organs, sagittal section.
View Media Gallery)
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Gross Anatomy
The testis is a paired, ovoid male reproductive organ that sits in the scrotum, separated from its mate by a scrotal septum. Described by some as being shaped and sized like a large olive or a small plum, the average volume of the adult testis is approximately 15-25 mL. Typically, it measures 3.5-5 cm in length, 2.5-3 cm in width and 3-4 cm in depth (anteroposterior diameter). [5, 2]
Smooth to palpation, the testis sits obliquely with its long axis mostly vertical with a slight anterior and lateral slant to the superior pole. Superiorly, it is suspended by the spermatic cord, with the left testis often sitting lower than the right testis. Inferiorly, the testis is anchored to the scrotum by the scrotal ligament, a remnant of the gubernaculum.
The tunica vaginalis testis (a remnant of the processus vaginalis) envelopes the testis in a double layer, except at the superior and posterior borders, where the spermatic cord and epididymis adhere to the testes.
The visceral layer of the tunica vaginalis testis is closely applied to the testis, epididymis, and ductus deferens. On the posterolateral surface of the testis, this layer invests a slit-like recess between the body of the epididymis and the testis called the sinus of epididymis.
The parietal layer of tunica vaginalis is adjacent to the internal spermatic fascia, is more extensive, and extends superiorly into the distal part of the spermatic cord.
Deep to the tunica vaginalis, the tunica albuginea is a tough, fibrous outer covering of the testis. On the posterior surface, it is reflected inwardly to form an incomplete vertical septum called the mediastinum testis.
The mediastinum testis extends from the superior to near the inferior portion of the gland. It narrows in width as it travels inferiorly. Anteriorly and laterally, numerous imperfect septa are given off, which radiate to the glands surface and are attached to the tunica albuginea. These divide the interior of the testis into numerous, cone-shaped spaces that have a wide base at the gland's surface and narrow as they converge to the mediastinum. In these spaces, the numerous lobules of glandular structures (the minute but long and highly coiled seminiferous tubules) are housed. The mediastinum supports the ducts and vessels as they pass to and from the glandular substance.
Internally, the tunica albuginea gives rise to septa that divide the testis into approximately 250 lobules. Each lobule contains 1-4 highly coiled seminiferous tubules, where spermatogenesis occurs. These tubules are lined by Sertoli cells and surrounded by interstitial tissue containing Leydig cells responsible for testosterone production.
The seminiferous tubules are lined with germ cells that produce sperm and nutrient fluid. The seminiferous tubules converge into straight tubules that lead to the rete testis, a network of interconnected channels located within the mediastinum testis. From here, sperm are transported via efferent ductules to the epididymis.
The epididymis is a comma-shaped, elongated structure composed of a single, fine tubular structure estimated up to 6 m (approximately 20 ft) in length. This tube is highly convoluted and tightly compressed (average size is ~5 cm) to the point of appearing solid. A deep groove, the sinus of the epididymis, is present laterally and it marks the boundary between the testis and epididymis. Located on the posterior border of the testis, it is composed of three parts, including:
The head (caput): About 8-12 efferent ductules from the superior pole of the testis drain into and form the head of the epididymis.
The body (corpora): Extends along the posterolateral aspect of the testis; this region is where the sperm undergo further maturation.
The tail (cauda): Located at the inferior pole, it serves as a storage site for mature sperm and transitions into the ductus deferens. The tail of the epididymis progressively tapers and becomes continuous with the convoluted portion of the ductus deferens.
Due to its length, the epididymal duct allows space for storage and maturation of sperm. (vas deferens; see the image below).
The epididymal duct is lined by pseudostratified columnar epithelium with stereocilia that facilitate fluid absorption and secretion of factors essential for sperm maturation. The tail contains smooth muscle layers that contract rhythmically during ejaculation to propel sperm into the vas deferens.
Male reproductive organs, sagittal section.
View Media Gallery)
The arterial supply to both testes is primarily from the testicular arteries, which arise from the anterolateral aspect of the abdominal aorta just inferior to the renal arteries. They travel retroperitoneally, cross over the ureters and the inferior parts of the external iliac arteries to pass through the deep inguinal ring to enter the inguinal canal and become one of the components of the spermatic cord. The testicular artery enters the testis through the posterior midportion. The testicular artery or one of its branches anastomoses with the artery of the ductus deferens.
Venous drainage from the testis and epididymis forms a network of 8-12 veins, called the pampiniform venous plexus, lying anterior to the ductus deferens and surrounding the testicular artery in the spermatic cord. This plexus plays a crucial role in thermoregulation via counter current heat exchange. The counterflowing arteries and veins are separated only by the thickness of their vascular walls. This permits the exchange of heat and small molecules and facilitates the maintenance of lower testicular temperatures. The veins converge superiorly, forming a testicular vein, after passing through the deep inguinal ring. The right testicular vein enters the inferior vena cava, and the left testicular vein drains into the left renal vein. This asymmetry in venous drainage has clinical significance, particularly in conditions such as varicocele, which is more common on the left side due to increased venous pressure.
Lymphatic drainage of the testis follows the testicular vessels (in the spermatic cord) to the right and left lumbar (caval/aortic) and preaortic lymph nodes at the second lumbar level. This pathway is distinct from that of scrotal lymphatics, which drains into superficial inguinal lymph nodes.
Autonomic innervations of the testis arise as the testicular plexus of nerves on the testicular artery, which contains vagal parasympathetic and visceral afferent fibers and sympathetic fibers from the T7 segment of the spinal cord.
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Microscopic Anatomy
The testis is composed of lobules of glandular tubules. These tubules are highly convoluted and held together by loose connective tissue with interspersed groups of "interstitial cells," which contain Leydig cells. The individual tubule consists of a basement membrane formed by laminated connective tissue with numerous elastic fibers with flattened cells between the layers and covered by an external layer of flattened epithelioid cells. Within the basement membrane are epithelial cells arranged in several irregular layers but may be separated into germ cells at the periphery and varying cells of spermatogenesis up to mature sperm cells as they advance toward the lumen. Also, interspersed in the layer are Sertoli cells, which project inward from the basement to the lumen and provide support to the developing sperm cells.
Testicular histology magnified 500 times. Leydig cells reside in the interstitium. Spermatogonia and Sertoli cells lie on the basement membrane of the seminiferous tubules. Germ cells interdigitate with the Sertoli cells and undergo ordered maturation, migrating toward the lumen as they mature.
View Media Gallery)
In the apices of the lobules, the tubules are less convoluted and converge into 20-30 larger, straight ducts (tubuli recti). These ducts merge into anastomosing tubes in the fibrous stroma, lined with flattened epithelium (rete testis). The tubes terminate into approximately 15 ducts that are initially straight in their course.
After piercing the tunica albuginea at the superior mediastinum, they enlarge and become increasingly convoluted. These convolutions are held together by fine areolar tissue and bands of fibrous tissue. They form a series of conical masses, the conic vasculosi. This series forms the head of the epididymis. These ducts are thicker and lined by ciliated columnar epithelium. Below this epithelium is muscular tissue arranged in a mostly circular fashion. As the tail of the epididymis merges with the ductus deferens, the microscopic anatomy demonstrates a thickened duct with increased muscular material, increased in diameter, and still lined with ciliated columnar epithelium.
The epithelial lining of the epididymal tubule contains principal, basal, apical, and clear cells. Principal cells are tall columnar cells with apical stereocilia that absorb excess fluid from testicular secretions and secrete glycoproteins essential for sperm maturation. Basal cells are small round cells located at the base of the epithelium that act as stem cells for epithelial renewal. Apical and clear cells are far less common than principal and basal cells. Apical cells have abundant mitochondria and are mainly found in the head of the epididymis. Clear cells are columnar and have few microvilli but numerous endocytic vesicles and lipid droplets . They are most abundant in the tail of the epididymis and facilitate the acidification of the luminal fluid. [2, 12, 4]
Beneath the epithelium lies a thin lamina propria followed by smooth muscle layers. In the head and body regions, smooth muscle is arranged circularly to facilitate gentle peristalsis. In the tail region, where sperm storage occurs, the muscular layer thickens to assist in powerful contractions during ejaculation. [2, 13]
Connecting the rete testis to the head of the epididymis are 10-15 efferent ductules. These ducts are lined with alternating patches of ciliated columnar cells (to propel sperm) and nonciliated cuboidal cells (for fluid absorption).
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Natural Variants
Two vestigial embryonic structures of no known physiological function are often found on the testis and epididymis as follows:
The appendix testis is a pear-shaped structure that is the vesicular remnant of the cranial end of the paramesonephric (Mullerian) duct. This is the embryonic genital duct that forms half the uterus in females. It is found in approximately 92% of all testes. Its typical location is at the superior testicular pole in the groove between the testis and the head of the epididymis.
The appendices of the epididymis are remnants of the cranial end of the mesonephric (Wolffian) duct, the embryonic duct that forms part of the ductus deferens in males. This is located in approximately 23% of testes. Its location may vary, but it usually projects from the head of the epididymis. Testicular and epididymal appendices are collectively referred to as hydatids. These structures may be sessile or pedunculated. Pedunculated appendices are more prone to torsion, which can mimic testicular torsion clinically.
In 6-7% of males, the epididymis is found on the anterior surface of the testis.
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Pathophysiological Variants
The testis may be arrested or delayed anywhere along its course of descent into the scrotum. When it is retained in the inguinal canal, it is often complicated by a congenital hernia. [7, 8, 15]
In premature infants, the testis may not be fully descended. This process is usually complete by 9 months' gestation. Special attention should be paid to an undescended testis, and this should be closely followed up. The testis may exit the superficial inguinal ring but slip down between the scrotum and thigh and come to rest in the perineum. This is known as perineal ectopia testis.
Androgen insensitivity may lead to the testes being found in the labia majora of a chromosomal male 46 XY but phenotypical female.
Another rare anomaly is polyorchidism, characterized by more than two testes, with triorchidism being the most common form. Management depends on factors such as drainage anatomy and associated complications such as torsion or malignancy. Preservation of functional supernumerary testes is often recommended when they share a common drainage system with normal testes and exhibit no malignancy on biopsy.
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Other Considerations
The testis begins as a retroperitoneal organ in the lumbar region. Its descent outside the abdominal cavity is imperative for normal testicular function and reproductive viability. By resting in the scrotum, the testis is kept at a temperature 2-3º C lower than the core body temperature. The countercurrent of blood flow with the venous plexus surrounding the testicular artery allows heat transfer to cool the blood flow. Further, the cremasteric reflex allows the testis to be elevated toward the body core to control testicular temperature.
Because of its intraabdominal origin, the testis have lymph drainage to the lumbar lymph nodes. Thus, infection of the epididymis or testicular carcinoma does not typically cause enlarged inguinal lymph nodes.
Cryptorchidism is important to recognize, not just because of sterility, but due to an increased incidence of testicular cancer in these males. Unless it is able to be surgically brought into the scrotum, where testicular surveillance may be performed, the undescended testis should be excised. Also, understanding the increased risk is extended to the contralateral testis as well is important. So, the individual should be instructed in careful, regular self-examinations.
Serous fluid may collect between the layers of the tunica vaginalis. This is termed hydrocele. This may be due to trauma, inflammation, or congenital due to persistent communication with the general peritoneal cavity.
Epididymitis is inflammation of the epididymis. It may be due to infectious process, commonly in men ages 19-35. In this age group, the treatment should cover gonorrhea and chlamydia. In young children and older men, E coli is the most common pathogens. Still, other causes include pathogens such as ureaplasma, Mycobacterium tuberculosis, and the drug amiodarone, which is commonly used for cardiac rate control.
Testicular torsion may occur if the testis twists on the suspending spermatic cord. This is a surgical emergency because the blood supply needs to be restored within 6 hours of initiation of symptoms. After 12 hours, the testis may be so badly damaged that it cannot be salvaged. The testis is fixed in the scrotum with suture to prevent recurrence, and the contralateral testis should be sutured in place as well. This is due to the increased incidence of testicular torsion recurrence.
Most acute presentations of scrotal pain and swelling can be attributed to epididymitis, testicular torsion, or torsion of a testicular appendage. In many cases, torsion of a testicular appendage, although a benign condition, may present identically to testicular torsion, a true urologic emergency. Ultrasound may be used for defining the problem, but clinical examination of a normal, nontender testis with the presence of a paratesticular nodule at the superior pole may point more to appendical torsion. Classically, a blue-dot appearance (blue dot sign) may be seen in the area of the injury, but this is only present in approximately 20% of cases.
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References
Gray H. XI. Splanchnology. 3c. The Male Genital. Anatomy of the Human Body. 20th ed. Philadelphia, New York: Lea & Febiger, 1918; Bartleby 2000; 2000. [Full Text].
Standring S. Gray's Anatomy: The Anatomical Basis of Clinical Practice. 42nd ed. 2021.
Mega Obukohwo O, Eze Kingsley N, Arientare Rume R, Victor E. The Concept of Male Reproductive Anatomy. IntechOpen; 2022. [Full Text].
James ER, Carrell DT, Aston KI, Jenkins TG, Yeste M, Salas-Huetos A. The Role of the Epididymis and the Contribution of Epididymosomes to Mammalian Reproduction. Int J Mol Sci. 2020 Jul 29. 21 (15):5377. [QxMD MEDLINE Link]. [Full Text].
Snell RS. Structures of the Anterior Abdominal Wall: Scrotum, Testis, and Epididymides. Clinical Anatomy for Medical Students. 6th ed. Philadelphia: Lippincott Williams & Wilkins; 2000. 153-157.
Swartz, MH. Male Genitalia and Hernias. Textbook of Physical Diagnosis: History and Examination. 5th ed. Philadelphia: Saunders Elsevier; 2006. 520-526.
Tintinalli JE, Kelen GD, Stapczynski JS. Testes. Emergency Medicine: A Comprehensive Study Guide. 6th ed. McGraw-Hill Professional: 2004.
Moore K, Daley A II. Inguinal Region. Clinically Oriented Anatomy. 5th ed. Philadelphia: Lippincott Williams & Wilkins; 2006. 214-230.
Pakurar AS, Bigbee JW. Digital Histology: An Interactive CD Atlas with Review Text. 2011. [Full Text].
De Grava Kempinas W, Klinefelter GR. Interpreting histopathology in the epididymis. Spermatogenesis. 2014 May-Aug. 4 (2):e979114. [QxMD MEDLINE Link]. [Full Text].
Kumar N, Swamy R, Patil J, Guru A, Aithal A, Shetty P. Presence of Arteriovenous Communication between Left Testicular Vessels and Its Clinical Significance. Case Rep Vasc Med. 2014. 2014:160824. [QxMD MEDLINE Link]. [Full Text].
Cornwall GA. New insights into epididymal biology and function. Hum Reprod Update. 2009 Mar-Apr. 15 (2):213-27. [QxMD MEDLINE Link]. [Full Text].
Gocht A, Merseburger AS, Ergün S, Roesch MC. The ductal network in the human testis and epididymis: What belongs to which?. Clin Anat. 2024 Nov 30. [QxMD MEDLINE Link].
Barbosa R, Favorito L, Sampaio F. Morphometric study applied to testicular and epididymis hydatids torsion. Sci Rep. 2024. 14:[Full Text].
Schneck FX, Bellinger MF. Abnormalities of the testes and scrotum and their surgical management. Wein AJ. Campbell-Walsh Urology. 9th ed. Philadelphia: Saunders Elsevier; 2007. chap 127.
Piro E, Abati L, Zocca V, Brugnoni M, D'Alessio A. [Triorchidism: which therapy?]. Pediatr Med Chir. 2017 Jun 23. 39 (2):141. [QxMD MEDLINE Link]. [Full Text].
Media Gallery
Male reproductive organs, sagittal section.
Testicular histology magnified 500 times. Leydig cells reside in the interstitium. Spermatogonia and Sertoli cells lie on the basement membrane of the seminiferous tubules. Germ cells interdigitate with the Sertoli cells and undergo ordered maturation, migrating toward the lumen as they mature.
Testicular and epididymal anatomy. Reprinted with permission, Cleveland Clinic Center for Medical Art & Photography © 2008-2016. All Rights Reserved.
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Author
Todd M Hoagland, PhD Professor Department of Medical Education Feinberg School of Medicine
Todd M Hoagland, PhD is a member of the following medical societies: Alpha Omega Alpha, American Physiological Society, American Association of Clinical Anatomists, American Association of Anatomists
Disclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: Elsevier Publishing, Netter's Atlas of Human Anatomy 6th-8th editions.
Specialty Editor Board
Disclosure: Nothing to disclose.
Chief Editor
Vinay K Kapoor, MBBS, MS, FRCSEd, FICS, FAMS Professor of Surgical Gastroenterology, Mahatma Gandhi Medical College and Hospital (MGMCH), Jaipur, India
Vinay K Kapoor, MBBS, MS, FRCSEd, FICS, FAMS is a member of the following medical societies: Association of Surgeons of India, Indian Association of Surgical Gastroenterology, Indian Society of Gastroenterology, Medical Council of India, National Academy of Medical Sciences (India), Royal College of Surgeons of Edinburgh
Disclosure: Nothing to disclose.
Additional Contributors
Edmund S Sabanegh, Jr, MD Chairman, Department of Urology, Glickman Urological and Kidney Institute, Cleveland Clinic Foundation
Edmund S Sabanegh, Jr, MD is a member of the following medical societies: American Medical Association, American Society of Andrology, Society of Reproductive Surgeons, Society for the Study of Male Reproduction, American Society for Reproductive Medicine, American Urological Association, SWOG
Disclosure: Nothing to disclose.
Martha K Terris, MD, FACS Professor and Chief of Urology, Witherington Distinguished Chair, Department of Surgery, Section of Urology, Director, Urology Residency Training Program, Medical College of Georgia at Augusta University; Professor, Department of Physician Assistants, Medical College of Georgia School of Allied Health; Chief, Section of Urology, Augusta Veterans Affairs Medical Center
Martha K Terris, MD, FACS is a member of the following medical societies: American Cancer Society, American College of Surgeons, American Institute of Ultrasound in Medicine, American Society of Clinical Oncology, American Urological Association, Association of Women Surgeons, New York Academy of Sciences, Society of Government Service Urologists, Society of University Urologists, Society of Urology Chairpersons and Program Directors, Society of Women in Urology
Disclosure: Nothing to disclose.
Thomas R Gest, PhD Professor of Anatomy, University of Houston College of Medicine
Thomas R Gest, PhD is a member of the following medical societies: American Association of Clinical Anatomists
Disclosure: Nothing to disclose.
Zachary W A Klaassen, MD Resident Physician, Department of Urology, Medical College of Georgia at Augusta University; Visiting Professor of Anatomy, St George’s University School of Medicine, Grenada
Disclosure: Nothing to disclose.
James H Wallace, PA-C Medical College of Georgia
James H Wallace, PA-C is a member of the following medical societies: American College of Physicians, Christian Medical and Dental Associations
Disclosure: Nothing to disclose.
Daniel J Greene, MD Resident Physician, Department of Urology, Cleveland Clinic Foundation
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Hepatocellular Adenoma (Hepatic Adenoma) Differential Diagnoses
Updated: Mar 06, 2025
Author: Michael H Piper, MD; Chief Editor: BS Anand, MDmore...;)
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Note that differentiation of a hepatocellular adenoma (HCA) (hepatic adenoma) from a high-grade hepatocellular carcinoma (HCC) can be difficult, [61, 62] if not impossible, so this should always be considered.
Also consider the following conditions in the evaluation of patients with suspected HCA:
Echinococcal cyst
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Media Gallery
Ultrasound in a patient with von Gierke disease (glycogen storage disease type 1) and several hepatic adenomas. This image shows a mass in the right lobe of the liver that is predominantly isoechoic relative to the liver parenchyma and contains a small, round, hypoechoic component. The 2 masses appear slightly hypoechoic. (Same patient as in the following image.)
T2-weighted fat-saturated fast spin-echo axial magnetic resonance image in a patient with von Gierke disease. This image shows 2 heterogeneous, hyperintense masses in the right lobe of the liver. (Same patient as in the following image.)
T1-weighted in-phase magnetic resonance image in a patient with von Gierke disease (same patient as in the following image). This image demonstrates normal hepatic signal intensity that is hyperintense relative to the spleen. Two heterogeneous masses that represent hepatic adenomas are seen in the right lobe and are slightly hypointense relative to the liver parenchyma.
T1-weighted out-of-phase magnetic resonance image in a patient with von Gierke disease (same patient as in the following image). This image shows abnormal low signal intensity of the liver, hypointense relative to the spleen, representing fatty infiltration of the liver. The hepatic adenomas are heterogeneous and slightly hyperintense relative to the fatty liver.
Single-shot fast spin-echo T2-weighted coronal magnetic resonance image in a patient with von Gierke disease (same patient as in the following image). This image shows a hyperintense mass in the right lobe of the liver and an additional hyperintense mass in the inferior tip of the liver, representing a third hepatic adenoma.
Fat-saturated 3-dimensional T1-weighted gradient-echo magnetic resonance image in a patient with von Gierke disease (same patient as in the following image). This image shows 2 heterogeneous, slightly hyperintense masses in the right lobe of the liver.
Gadolinium-enhanced fat-saturated 3-dimensional T1-weighted gradient-echo magnetic resonance image in a patient with von Gierke disease (same patient as in the following image). This image shows intense enhancement of the hepatic adenomas.
Gadolinium-enhanced fat-saturated 3-dimensional T1-weighted gradient-echo magnetic resonance image in a patient with von Gierke disease. This image shows that the hepatic adenoma remains hyperintense relative to the liver parenchyma. (Same patient as in the following image.)
Gadolinium-enhanced fat-saturated 3-dimensional T1-weighted gradient-echo magnetic resonance image in a patient with von Gierke disease (same patient in the previous image). This image shows that the hepatic adenoma remains hyperintense relative to the liver parenchyma.
Noncontrast computed tomography scan in a 41-year-old woman with a history of oral contraceptive use. This image demonstrates a heterogeneous, low-attenuation mass in the right lobe of the liver, a hepatic adenoma. (Same patient as in the following image.)
Contrast-enhanced computed tomography scan in the portal venous phase in a 41-year-old woman with a history of oral contraceptive use. This image demonstrates a heterogeneous, enhancing mass, a hepatic adenoma, predominantly isoattenuating relative to the liver with areas of low attenuation. (Same patient as in the following image.)
Technetium-99m (99mTc)–labeled red blood cell single-photon emission computed tomography scintigraphy in a 41-year-old woman with a history of oral contraceptive use. This image shows no demonstrable activity in the hepatic mass, indicating that it does not represent a hemangioma. (Same patient as in the previous image.)
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Contributor Information and Disclosures
Author
Michael H Piper, MD Clinical Assistant Professor, Department of Internal Medicine, Division of Gastroenterology, Wayne State University School of Medicine; Consulting Staff, Digestive Health Associates, PLC
Michael H Piper, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Gastroenterology, American College of Physicians, Michigan State Medical Society
Disclosure: Nothing to disclose.
Coauthor(s)
Janice M Fields, MD, FACG, FACP Assistant Professor of Internal Medicine, Oakland University William Beaumont School of Medicine; Consulting Staff, Department of Internal Medicine, Section of Gastroenterology, Providence Hospital, St John Macomb-Oakland Hosptial
Janice M Fields, MD, FACG, FACP is a member of the following medical societies: American College of Gastroenterology, American College of Physicians-American Society of Internal Medicine, American Gastroenterological Association, American Medical Association, American Society for Gastrointestinal Endoscopy, National Medical Association
Disclosure: Nothing to disclose.
Ryan R Kahl, DO Chief Fellow, Department of Gastroenterology, Ascension Providence-Providence Park Hospital, Michigan State University College of Human Medicine
Ryan R Kahl, DO is a member of the following medical societies: American College of Gastroenterology, American College of Physicians, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Specialty Editor Board
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Received salary from Medscape for employment.
Chief Editor
BS Anand, MD Professor, Department of Internal Medicine, Division of Gastroenterology, Baylor College of Medicine
BS Anand, MD is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, American Gastroenterological Association, American Society for Gastrointestinal Endoscopy
Disclosure: Nothing to disclose.
Additional Contributors
Karen Kodsi Garfield, MD Attending Physician in Body Imaging, Department of Radiology, St Luke's Hospital
Karen Kodsi Garfield, MD is a member of the following medical societies: American College of Radiology, American Medical Association, Radiological Society of North America
Disclosure: Nothing to disclose.
Tushar Patel, MB, ChB Professor of Medicine, Ohio State University Medical Center
Tushar Patel, MB, ChB is a member of the following medical societies: American Association for the Study of Liver Diseases, American Gastroenterological Association
Disclosure: Nothing to disclose.
Bradford A Whitmer, DO Fellow, Department of Gastroenterology, Providence Hospital
Bradford A Whitmer, DO is a member of the following medical societies: American College of Gastroenterology, American College of Physicians, American Osteopathic Association
Disclosure: Nothing to disclose.
Acknowledgements
Brian S Berk, MD Assistant Professor, Department of Medicine, Dartmouth Medical School; Director of End Stage Liver Disease, Section of Gastroenterology, Dartmouth Hitchcock Medical Center
Brian S Berk, MD is a member of the following medical societies: American Association for the Study of Liver Diseases, American College of Gastroenterology, and American Gastroenterological Association
Disclosure: Nothing to disclose. Kenneth Ingram, PAC Assistant Professor, Department of Medicine, Division of Gastroenterology and Hepatology, Oregon Health and Science University School of Medicine
Disclosure: Nothing to disclose.
Sandeep Mukherjee, MB, BCh, MPH, FRCPC Associate Professor, Department of Internal Medicine, Section of Gastroenterology and Hepatology, University of Nebraska Medical Center; Consulting Staff, Section of Gastroenterology and Hepatology, Veteran Affairs Medical Center
Disclosure: Merck Honoraria Speaking and teaching; Ikaria Pharmaceuticals Honoraria Board membership
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9145 | https://www.grc.nasa.gov/www/k-12/airplane/doppler.html | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- ---
| | | | --- | | + Text Only Site + Non-Flash Version + Contact Glenn | | | | | | | | | | --- --- --- | | | | | | | | | | | | | As any object moves through the air, the air near the object is disturbed. The disturbances are transmitted through the air at a distinct speed called the speed of sound. Sound is a sensation created in the human brain in response to small pressure fluctuations in the air. Sound moves through the air as a series of waves. When the waves pass our ears, a sound is detected. The distance between any two waves is called the wavelength and the time interval between waves passing is called the frequency . The wavelength and the frequency are related by the speed of sound; high frequency implies short wavelength and low frequency implies a long wavelength. The brain associates a certain musical pitch with each frequency; the higher the frequency, the higher the pitch. Similarly, shorter wavelengths produce higher pitches. The speed of transmission of the sound remains a constant regardless of the frequency or the wavelength. The speed of sound only depends on the state of the air (or gas) not on the characteristics of the generating source. Because the speed of sound depends only on the state of the gas, some interesting physical phenomena occur when a sound source moves through a uniform gas. You can study some of these phenomena by using the interactive sound wave simulator. As the source moves it continues to generate sound waves which move at the speed of sound. Since the source is moving slower than the speed of sound, the waves move out away from the source. Upstream (in the direction of the motion), the waves bunch up and the wavelength decreases. Downstream, the waves spread out and the wavelength increases. The sound that our ear detects will change in pitch as the object passes. This change in pitch is called a doppler effect. There are equations that describe the doppler effect. As the moving source approaches our ear, the wavelength is shorter, the frequency is higher and we hear a higher pitch. If we call the approaching frequency fa, the speed of sound a, the velocity of the approaching souce u, and the frequency of the sound at the source f, then fa = [f a] / [a - u] As the moving source leaves us, the wavelength is longer, the frequency is lower and the pitch is lower. Again. if the leaving frequency is called fl, then fl = [f a] / [a + u] Activities: Guided Tours Sound Waves: Sound Wave Simulator: Navigation .. Beginner's Guide Home Page | | | | | | | | --- --- | | | + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification | | Editor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn | | | |
9146 | https://metanumbers.com/1200 | 1200 (Number)
MetaNumbers
+-
?
| Max | 9223372036854775807 |
| 25835 | Positive integer |
| 0b10001100 | Binary number |
| 01771452 | Octal number |
| 0x129abf | Hex number |
| 5.012 10^7 | Scientific notation |
| 2^4 5 | Factorized form |
| | Random number |
More examples
Search
1200 (number)
1200 is an even four-digits composite number following 1199 and preceding 1201. In scientific notation, it is written as 1.2 × 10 3. The sum of its digits is 3. It has a total of 7 prime factors and 30 positive divisors. There are 320 positive integers (up to 1200) that are relatively prime to 1200.
Properties
Name
Notation
Prime Factorization
Divisors
Other Arithmetic Functions
Divisibility test
Base converter
Classification
Basic calculations
Geometrical shape
Hash Functions
Basic properties
Is Prime?no
Number parity even
Number length 4
Sum of Digits 3
Digital Root 3
Name
| Name | one thousand two hundred |
Notation
| Scientific notation | 1.2 × 10 3 |
| Engineering notation | 1.2 × 10 3 |
Prime Factorization of 1200
Prime Factorization2 4 × 3 × 5 2
Composite number
| ω | Distinct Factors | 3 | Total number of distinct prime factors |
| Ω | Total Factors | 7 | Total number of prime factors |
| rad | Radical | 30 | Product of the distinct prime numbers |
| λ | Liouville Lambda | -1 | Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) |
| μ | Mobius Mu | 0 | Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor |
| Λ | Mangoldt function | 0 | Returns log(p) if n is a power p k of any prime p (for any k>= 1), else returns 0 |
The prime factorization of 1200 is 2 4 × 3 × 5 2. Since it has a total of 7 prime factors, 1200 is a composite number.
Divisors of 1200
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 40, 48, 50, 60, 75, 80, 100, 120, 150, 200, 240, 300, 400, 600, 1200
30 divisors
| Even divisors | 24 |
| Odd divisors | 6 |
| 4k+1 divisors | 3 |
| 4k+3 divisors | 3 |
| τ | Total Divisors | 30 | Total number of the positive divisors of n |
| σ | Sum of Divisors | 3844 | Sum of all the positive divisors of n |
| s | Aliquot Sum | 2644 | Sum of the proper positive divisors of n |
| A | Arithmetic Mean | 128.133 | Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) |
| G | Geometric Mean | 34.641016151378 | Returns the nth root of the product of n divisors |
| H | Harmonic Mean | 9.3652445369651 | Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors |
The number 1200 can be divided by 30 positive divisors (out of which 24 are even, and 6 are odd). The sum of these divisors (counting 1200) is 3844, the average is 128.133.
Other Arithmetic Functions (n = 1200)
1 φ(n)n
| φ | Euler Totient | 320 | Total number of positive integers not greater than n that are coprime to n |
| λ | Carmichael Lambda | 40 | Smallest positive number such that a λ(n) ≡ 1 (mod n) for all a coprime to n |
| π | Prime Pi | ≈ 199 | Total number of primes less than or equal to n |
| r 2 | Sum of 2 squares | 0 | The number of ways n can be represented as the sum of 2 squares |
There are 320 positive integers (less than 1200) that are coprime with 1200. And there are approximately 199 prime numbers less than or equal to 1200.
Divisibility of 1200
| m | n mod m |
--- |
| 2 | 0 |
| 3 | 0 |
| 4 | 0 |
| 5 | 0 |
| 6 | 0 |
| 7 | 3 |
| 8 | 0 |
| 9 | 3 |
The number 1200 is divisible by 2, 3, 4, 5, 6 and 8.
Classification of 1200
By Arithmetic functions
Refactorable
Abundant
Expressible via specific sums
Polite
Practical
Other numbers
Frugal
Regular
Base conversion 1200
| Base | System | Value |
---
| 2 | Binary | 10010110000 |
| 3 | Ternary | 1122110 |
| 4 | Quaternary | 102300 |
| 5 | Quinary | 14300 |
| 6 | Senary | 5320 |
| 8 | Octal | 2260 |
| 10 | Decimal | 1200 |
| 12 | Duodecimal | 840 |
| 16 | Hexadecimal | 4b0 |
| 20 | Vigesimal | 300 |
| 36 | Base36 | xc |
Basic calculations (n = 1200)
Multiplication
n×y
| n×2 | 2400 |
| n×3 | 3600 |
| n×4 | 4800 |
| n×5 | 6000 |
Division
n÷y
| n÷2 | 600.000 |
| n÷3 | 400.000 |
| n÷4 | 300.000 |
| n÷5 | 240.000 |
Exponentiation
n y
| n 2 | 1440000 |
| n 3 | 1728000000 |
| n 4 | 2073600000000 |
| n 5 | 2488320000000000 |
Nth Root
y√n
| 2√n | 34.641016151378 |
| 3√n | 10.626585691826 |
| 4√n | 5.8856619127654 |
| 5√n | 4.1289179173334 |
1200 as geometric shapes
Circle
Radius = n
| Diameter | 2400 |
| Circumference | 7539.8223686155 |
| Area | 4523893.4211693 |
Sphere
Radius = n
| Volume | 7238229473.8709 |
| Surface area | 18095573.684677 |
| Circumference | 7539.8223686155 |
Square
Length = n
| Perimeter | 4800 |
| Area | 1440000 |
| Diagonal | 1697.0562748477 |
Cube
Length = n
| Surface area | 8640000 |
| Volume | 1728000000 |
| Space diagonal | 2078.4609690827 |
Equilateral Triangle
Length = n
| Perimeter | 3600 |
| Area | 623538.2907248 |
| Altitude | 1039.2304845413 |
Triangular Pyramid
Length = n
| Surface area | 2494153.1628992 |
| Volume | 203646752.98173 |
| Height | 979.79589711327 |
Cryptographic Hash Functions
| md5 | fe2d010308a6b3799a3d9c728ee74244 |
| sha1 | 73ee4958bdb5a056029ebd39b8abbaa3dbc0f333 |
| sha256 | 15197cf7214b58e67cae565e573ffd9aa44bcb81f8ee5d7185dfe8da0a16ef43 |
| sha512 | 3f981dcddcf1dcedd641a76c1ca0a4e996e6803a0775725efa23c6d95e959c7db52eb4a11a856cef4dc671e63c47005ee36d4f960632b16931789840832bbb51 |
| ripemd-160 | 14fdef6a95b7e4d85ffcdce5a3433ac0f170de6a |
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9147 | https://www.hmhco.com/blog/teaching-ratios-and-unit-rates-in-math?srsltid=AfmBOoqfi4e_KUYl63fjnYEeZdXt1DaO-ISH39giADjhdkmrl0cO4Wvl | From setup to support, HMH will ensure your back-to-school season runs smoothly. Let’s go.
Math
Teaching Ratios and Unit Rates in Math
The heart of middle school mathematics, and a key part of algebra readiness, is understanding ratios and rates. The overview and lessons below are tools to prepare students, usually in Grades 6 and up, who are ready to learn about these concepts. The lessons below will typically cover two days of instruction.
Ratios and Rates
A ratio is a comparison of two numbers or measurements. The numbers or measurements being compared are sometimes called the terms of the ratio. For example, if a store sells 6 red shirts and 8 green shirts, the ratio of red to green shirts is 6 to 8. You can write this ratio as 6 red/8 green, 6 red:8 green—or when writing fast or trying to make a point—simply 6/8 or 6:8. Both expressions mean that there are 6 red shirts “for every” 8 green shirts. Notice how you can rewrite 6/8 as 3/4, no different from any other time a math concept can appear as a fraction.
A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. This is not a ratio of two like units, such as shirts. This is a ratio of two unlike units: cents and ounces. The first term of the ratio (69¢) is measured in cents, and the second term (12) in ounces. You can write this rate as 69¢/12 ounces or 69¢:12 ounces. Both expressions mean that you pay 69¢ “for every” 12 ounces of corn, and similar to the shirt ratio, can enter calculations as the fraction 69/12. But notice that this time, a new unit is created: cents per ounce.
Rates are used by people every day, such as when they work 40 hours per week or earn interest every year at a bank. When rates are expressed as a quantity of 1, such as 2 feet per second (that is, per 1 second) or 5 miles per hour (that is, per 1 hour), they can be defined as unit rates. You can write any rate as a unit rate by reducing the fraction so it has a 1 as the denominator or second term. As a unit rate example, you can show that the unit rate of 120 students for every 3 buses is 40 students per bus.
120/3 = 40/1
You could also find the unit rate by dividing the first term of the ratio by the second term.
120 ÷ 3 = 40
When a price is expressed as a quantity of 1, such as $25 per ticket or $0.89 per can, it is called a unit price. If you have a non-unit price, such as $5.50 for 5 pounds of potatoes, and want to find the unit price, divide the terms of the ratio.
$5.50 ÷ 5 pounds = $1.10 per pound
The unit price of potatoes that cost $5.50 for 5 pounds is $1.10 per pound.
Rates in the Real World
Rate and unit rate are used to solve many real-world problems. Look at the following problem. “Tonya works 60 hours every 3 weeks. At that rate, how many hours will she work in 12 weeks?” The problem tells you that Tonya works at the rate of 60 hours every 3 weeks. To find the number of hours she will work in 12 weeks, write a ratio equal to 60/3 that has a second term of 12.
60/3 = ?/1260/3 = 240/12
Removing the units makes the calculation easier to see. However, it is important to remember the units when interpreting the new ratio.
Tonya will work 240 hours in 12 weeks.
You could also solve this problem by first finding the unit rate and multiplying it by 12.
60/3 = 20/1
20 × 12 = 240
When you find equal ratios, it is important to remember that if you multiply or divide one term of a ratio by a number, then you need to multiply or divide the other term by that same number.
Let's take a look at a problem that involves unit price. “A sign in a store says 3 Pens for $2.70. How much would 10 pens cost?” To solve the problem, find the unit price of the pens, then multiply by 10.
$2.70 ÷ 3 pens = $0.90 per pen
$0.90 × 10 pens = $9.00
Finding the cost of one unit enables you to find the cost of any number of units.
What Is a Unit Rate in Math?
Your students have no doubt encountered rates and ratios before (have they seen a speed limit sign?), but it may help them to review these concepts before solving problems that use them.
Standard: Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0. (6.RP.A.2)
Prerequisite Skills and Concepts: Students should have a basic understanding of ratios, how to write them, and an ability to simplify a ratio. Students should also have an ability to work with fractions and find equivalent fractions.
Developing the Concept: Rates
Now that students know how to find a unit rate, they will learn how to find an equivalent ratio using unit rates. Finding equivalent ratios uses the same thought process as finding equivalent fractions.
Standard: Use ratio and rate reasoning to find equivalent ratios and solve real-world problems (6.RP.A.3)
| | |
--- |
| Laps | Minutes |
| 1.5 | 1 |
| 3 | 2 |
| 6 | 4 |
| 12 | 8 |
| 18 | 12 |
| 24 | 16 |
| 30 | 20 |
Looking for more free math lessons and activities for elementary school students? Be sure to explore our Free Teaching Resources hub.
If you're in the market for a math curriculum that will unlock learning for students who struggle with mathematics, check out Math 180, our math intervention solution for Grades 5–12.
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9148 | https://www.wyzant.com/resources/answers/933519/related-rates-inverted-cone-problem | Related Rates Inverted Cone problem | Wyzant Ask An Expert
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Calculus
Felix W.
asked • 09/29/23
Related Rates Inverted Cone problem
Water is leaking out of an inverted conical tank at a rate of 7200 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 13 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 25 centimeters per minute when the height of the water is 1.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
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In order to find the rate at which water is being pumped into the tank, we need to find the net rate that the volume of water in the tank is increasing based the given information that the water level is rising at a rate of 25 cm/min when the height of the water is 1.0 m (100 cm).
We need to derive an equation that relates
the rate that the water level is rising, dh/dt
the rate that the volume of water is increasing, dV/dt
Step 1: express the volume of water (V) as a function of water level (h)
V = (1/3)πr 2 h
Since the tank dimensions are h = 13 cm and r = d/2 = 6.5/2 = 3.25 cm this means r = h/4
So V = (1/3)πr 2 h = (1/3)π(h/4)2 h = (1/48)πh 3
Step 2: Use the expression from step 1 to find an expression that relates dV/dt to dh/dt
dV/dt = (1/48)π(3h 2)(dh/dt) = (1/16)πh 2(dh/dt)
When h = 1m = 100 cm, dh/dt = 25 cm/min
This means dV/dt = (1/16)π(100)2(25) = 15,625π ≈ 49,087 cm 3/min
dV/dt = (dV/dt)in – (dV/dt)out where
(dV/dt)in is the rate that water is being pumped into the tank, which we are trying to find
(dV/dt)out is the rate that water is leaking out of the tank, which is given to be 7200 cm 3/min
dV/dt is the net rate that the volume of water in the tank is increasing
So the rate that water is being pumped into the tank is given by
(dV/dt)in = dV/dt + (dV/dt)out = 49,087 + 7200 = 56,287 cm 3/min
Answer (rounded to 2 significant figures)
The rate at which water is being pumped into the tank is 56,000 cm3/min
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9150 | https://www.econlib.org/library/Enc/IndustrialConcentration.html | Home /
Encyclopedia
/ Industrial Concentration
By William F. Shughart II,
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Industrial concentration” refers to a structural characteristic of the business sector. It is the degree to which production in an industry—or in the economy as a whole—is dominated by a few large firms. Once assumed to be a symptom of “market failure,” concentration is, for the most part, seen nowadays as an indicator of superior economic performance. In the early 1970s, Yale Brozen, a key contributor to the new thinking, called the profession’s about-face on this issue “a revolution in economics.” Industrial concentration remains a matter of public policy concern even so.
The Measurement of Industrial Concentration
Industrial concentration was traditionally summarized by the concentration ratio, which simply adds the market shares of an industry’s four, eight, twenty, or fifty largest companies. In 1982, when new federal merger guidelines were issued, the Herfindahl-Hirschman Index (HHI) became the standard measure of industrial concentration. Suppose that an industry contains ten firms that individually account for 25, 15, 12, 10, 10, 8, 7, 5, 5, and 3 percent of total sales. The four-firm concentration ratio for this industry—the most widely used number—is 25 + 15 + 12 + 10 = 62, meaning that the top four firms account for 62 percent of the industry’s sales. The HHI, by contrast, is calculated by summing the squared market shares of all of the firms in the industry: 252 + 152 + 122 + 102 + 102 + 82 + 72 + 52 + 52 + 32 = 1,366. The HHI has two distinct advantages over the concentration ratio. It uses information about the relative sizes of all of an industry’s members, not just some arbitrary subset of the leading companies, and it weights the market shares of the largest enterprises more heavily.
In general, the fewer the firms and the more unequal the distribution of market shares among them, the larger the HHI. Two four-firm industries, one containing equalsized firms each accounting for 25 percent of total sales, the other with market shares of 97, 1, 1, and 1, have the same four-firm concentration ratio (100) but very different HHIs (2,500 versus 9,412). An industry controlled by a single firm has an HHI of 1002 = 10,000, while the HHI for an industry populated by a very large number of very small firms would approach the index’s theoretical minimum value of zero.
Concentration in the U.S. Economy
According to the U.S. Department of Justice’s merger guidelines, an industry is considered “concentrated” if the HHI exceeds 1,800; it is “unconcentrated” if the HHI is below 1,000. Since 1982, HHIs based on the value of shipments of the fifty largest companies have been calculated and reported in the manufacturing series of the Economic Census.1 Concentration levels exceeding 1,800 are rare. The exceptions include glass containers (HHI = 2,959.9 in 1997), motor vehicles (2,505.8), and breakfast cereals (2,445.9). Cigarette manufacturing also is highly concentrated, but its HHI is not reported owing to the small number of firms in that industry, the largest four of which accounted for 89 percent of shipments in 1997. At the other extreme, the HHI for machine shops was 1.9 the same year.
Whether an industry is concentrated hinges on how narrowly or broadly it is defined, both in terms of the product it produces and the extent of the geographic area it serves. The U.S. footwear manufacturing industry as a whole is very unconcentrated (HHI = 317 in 1997); the level of concentration among house slipper manufacturers is considerably higher, though (HHI = 2,053.4). Similarly, although the national ready-mix concrete industry is unconcentrated (HHI = 29.4), concentration in that industry undoubtedly is much higher in specific cities and towns that typically are served by only a handful of such firms.
These examples suggest that concentration varies substantially across U.S. industries. Trends in concentration vary from industry to industry, but most changes in concentration proceed at a glacial pace. So, too, does aggregate concentration: the fifty largest U.S. companies accounted for 24 percent of manufacturing value added (revenue minus the costs of fuel, power, and raw materials) in 1997, the same percentage as in 1992 (and as in 1954, for that matter). On some measures—the percentages of total employment and total assets controlled by the nation’s 50, 100, or 200 largest firms—industrial concentration in the United States actually has declined since World War II.
Concentration indexes calculated for a particular year conceal the identities of the industry’s members. In reality, turnover among the nation’s leading firms is fairly regular over long time horizons, averaging between 2 and 5 percent annually. Success at one point in time does not guarantee survival: only three of the ten largest U.S. companies in 1909 made the top one hundred list in 1987. Available concentration indexes, which are based solely on domestic manufacturing data, also ignore the global dimensions of industrial production.
The Causes and Consequences of Industrial Concentration
Some industries are more concentrated than others because of technical properties of their production technologies or unique characteristics of the markets they serve. Economies of scale, which allow firms to reduce their average costs as they increase their rates of output, favor large-scale production over small-scale production. Thus, industries for which scale economies are important (e.g., auto manufacturing and petroleum refining) are expected to be more concentrated than others in which costs do not fall as rapidly as output expands (e.g., cut-and-sew apparel manufacturing). Similarly, concentration tends to be higher in industries, such as aircraft and semiconductor manufacturing, where learning curves generate substantial production-cost savings as additional units of the original model or design are made.
Owing to so-called network effects, some goods increase in value as more people use them. Computer operating systems, word-processing software, and video recorder-players are examples of such goods, as are literal networks such as railroads, commercial air transportation, and wire line telephony. Because standard technologies and protocols that provide compatible interconnections are critical to the realization of network effects— allowing faxes to be sent and received or computer users easily to exchange files—consumers rationally favor large networks over small ones. The necessity of building networks that accommodate critical masses of users means that only a few providers will achieve dominant positions, and therefore the industry will tend to be highly concentrated. Such domination is likely to be temporary, however, since consumers will switch networks when benefits outweigh costs, as illustrated by the replacement of Betaformatted video tapes by VHS formatted ones, which in turn are being replaced by DVDs.
Industrial concentration also is promoted by barriers to entry, which make it difficult for new firms to displace established firms. Barriers to entry are erected by government-conferred privileges such as patents, copyrights and trademarks, exclusive franchises, and licensing requirements. Existing firms may possess other advantages over newcomers, including lower costs and brand loyalty, which make entry more difficult.
The fundamental public policy question posed by industrial concentration is this: Are concentrated industries somehow less competitive than unconcentrated ones? Concentration would have adverse effects if it bred market power—the ability to charge prices in excess of costs—thereby increasing industry profits at consumers’ expense. In theory, industrial concentration can facilitate the exercise of market power if the members of the industry agree to cooperate rather than compete, or if the industry’s dominant firm takes the lead in setting prices that rivals follow. And, indeed, the evidence generated by hundreds of econometric studies suggests that concentrated industries are more profitable than unconcentrated ones. But that evidence begs the question. It does not tell us whether profits are higher in concentrated industries because of market power effects or because the firms in those industries use resources more efficiently (i.e., have lower costs).
Some economists have found that concentration leads to higher prices, but the link observed typically is both small (prices elevated by 1–5 percent) and statistically weak. A detailed econometric study by Sam Peltzman (1977) reaches the opposite conclusion. He reports that profits are higher in concentrated industries not because prices are higher, but because they do not decline as much as costs do as efficient firms expand their scales of operation. Analyses by Yale Brozen (1982), Harold Demsetz (1974), and others have found that the positive relation between industrial concentration and profits disappears altogether when firm size is taken into account. These results are consistent with the hypothesis that some industries are more concentrated than others because large firms have significant cost advantages over small firms. There is, in short, little unequivocal evidence that industrial concentration per se is worrisome. Just the reverse seems to be true.
Public Policies Toward Industrial Concentration
Consolidating production in the hands of fewer firms through mergers and acquisitions obviously is the most direct route to industrial concentration. Preventing transactions that, by eliminating one or more competitors, would lead to undue increases in concentration and the possible exercise of market power by the remaining firms is the mandate of the two federal antitrust agencies—the U.S. Department of Justice and the Federal Trade Commission—under section 7 of the Clayton Act (1914). That mandate was strengthened considerably by the Hart-Scott-Rodino Act (1978), which requires firms to notify the antitrust authorities of their intention to merge and then to hold the transaction in abeyance until it has been reviewed. Most transactions with summed firm values of fifteen million dollars or more had to file premerger notifications initially; in February 2001 that threshold was raised to fifty million dollars and indexed for inflation.
Two important factors that antitrust authorities consider in deciding whether to allow a proposed merger to proceed are the level of market concentration if the merger is consummated and the change in market concentration from its premerger level. (Note that the “market” considered relevant for merger analysis hardly ever corresponds to the “industry” defined by the Economic Census; antitrust markets may be defined more broadly or more narrowly; in practice, the definition of the relevant market usually is the key to whether a merger is lawful or not.) Concentration thresholds are laid out in the Justice Department’s merger guidelines, first promulgated in 1968, revised substantially in 1982, and amended several times since.
The guidelines state that proposed mergers are unlikely to be challenged if the postmerger market is unconcentrated (HHI remains below 1,000). However, mergers generally will not be approved if, following consummation, market concentration falls within the 1,000–1,800 range, and the HHI increases by more than 100 points or, if the postmerger HHI is 1,800 or more, concentration increases by more than 50 points.2 Exceptions are provided when the merging firms can demonstrate significant cost savings, when barriers to entry are low, or when one of the merger’s partners would fail otherwise. (In the European Union, by contrast, competition policy, including merger law enforcement, is shaped principally by fears of possible “abuses of dominant market positions” by large firms.)
Studies examining the enforcement of section 7 under the merger guidelines have found that they are not always followed closely. Mergers are, indeed, more likely to be challenged the greater the level of market concentration and the higher the barriers to entry are thought to be. But law enforcement also is found to be influenced significantly by political pressures on the antitrust authorities from groups that stand to lose if a merger is approved, including rivals worried that the transaction will create a more effective competitor. In fact, studies of stock-market reactions to news that a merger is likely to be challenged typically find competitors to be the main beneficiaries of such decisions.
About the Author
William F. Shughart II is F. A. P. Barnard Distinguished Professor of Economics at the University of Mississippi. He was special assistant to the director of the Federal Trade Commission’s Bureau of Economics during the Reagan administration and currently is editor in chief of Public Choice and associate editor of the Southern Economic Journal.
Further Reading
Introductory
Adams, Walter, and James Brock. The Structure of American Industry. 11th ed. Upper Saddle River, N.J.: Pearson/Prentice Hall, 2005.
Cabral, Luís M. B. Introduction to Industrial Organization. Cambridge: MIT Press, 2000.
Kwoka, John E. Jr., and Lawrence J. White. The Antitrust Revolution: Economics, Competition, and Policy. 4th ed. New York: Oxford University Press, 2004.
Pautler, Paul A. “Evidence on Mergers and Acquisitions.” Antitrust Bulletin 48 (Spring 2003): 119–221.
Shughart, William F. II. Antitrust Policy and Interest-Group Politics. New York: Quorum Books, 1990.
Shughart, William F. II. “Regulation and Antitrust.” In Charles K. Rowley and Friedrich Schneider, eds., The Encyclopedia of Public Choice. Vol. 1. Boston: Kluwer, 2004. Pp. 263–283.
Advanced
Brozen, Yale. Concentration, Mergers, and Public Policy. New York: Macmillan, 1982.
Carlton, Dennis W., and Jeffrey M. Perloff. Modern Industrial Organization. 3d ed. Reading, Mass.: Addison-Wesley, 2000.
Coate, Malcolm B., Richard S. Higgins, and Fred S. Mc-Chesney. “Bureaucracy and Politics in FTC Merger Challenges.” Journal of Law and Economics 33 (October 1990): 463–482.
Demsetz, Harold. “Two Systems of Belief About Monopoly.” In Harvey J. Goldschmid, H. Michael Mann, and J. Fred Weston, eds., Industrial Concentration: The New Learning. Boston: Little, Brown, 1974.
Goldschmid, Harvey J., H. Michael Mann, and J. Fred Weston, eds. Industrial Concentration: The New Learning. Boston: Little, Brown, 1974.
McChesney, Fred S., and William F. Shughart II, eds. The Causes and Consequences of Antitrust: The Public-Choice Perspective. Chicago: University of Chicago Press, 1995.
Peltzman, Sam. “The Gains and Losses from Industrial Concentration.” Journal of Law and Economics 20 (April 1977): 229–263.
Shy, Oz. The Economics of Network Industries. Cambridge: Cambridge University Press, 2001.
Stiglitz, Joseph E., and G. Frank Mathewson, eds. New Developments in the Analysis of Market Structure. Cambridge: MIT Press, 1986.
Footnotes
1.
The Economic Census has been conducted every five years since 1967, and before that for 1954, 1958, and 1963. Prior to 1997, it was known as the Census of Manufactures. That same year, industries began being categorized according to the North American Industry Classification System (NAICS), which replaced the Standard Industrial Classification (SIC) codes used until 1992. Industrial concentration also is reported by the Economic Census on the basis of value added. Industry concentration ratios and HHIs for the 1992 and 1997 economic censuses can be accessed online at: Information on industrial concentration is not readily available for sectors of the economy other than manufacturing.
2.
When firms with market shares of s1 and s2 merge, the HHI increases by (s1 + s2)2 − s12 − s22 = 2s1s2. So, for example, if a merger is proposed between the two largest firms in the hypothetical ten-firm industry described earlier, the HHI would increase by 2 × 25 × 15 = 750 points (from 1,366 to 2,116). According to the guidelines, that merger would in all likelihood be challenged.
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Crystal structure of Aldose Reductase with citrates bound in the active site
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9152 | https://engineering.stackexchange.com/questions/14122/what-area-should-i-use-when-calculating-induced-voltage | generator - What area should I use when calculating induced voltage? - Engineering Stack Exchange
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What area should I use when calculating induced voltage?
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\begingroup
I am trying to build a low power output generator with 12 pole pairs and 9 stator coils. The design will be similar to the one described in this link Basic Principles Of The Homebrew Axial Flux Alternator , but at a different scale/size. I am trying to find some way of estimating the number of turns I will need in the coils to produce an output voltage of 6.5V. The generator should produce this voltage at 100rpm. I am getting very confused about how to use Faraday's law to do my calculations.
\varepsilon = N\frac{d(BA)}{dt}
Faraday's Law as explained in this example where the area used is that of the magnet, while Faraday Law of Electromagnetic Induction states at the bottom that A is the area of the coil. Does this difference have to do with when the magnetic field is stationary while the coil is moving versus when the magnetic field is moving relative to a stationary coil? Also, should I be using \varepsilon = N\frac{d(BAcos(\theta))}{dt} instead? Thanks very much in advance, and apologies if I am missing something very obvious here.
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edited Mar 13, 2017 at 21:54
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asked Mar 8, 2017 at 0:59
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\begingroup
No expert, but the problem is interesting.
Assumptions:
Generate 6.5V_{RMS} at 100 rpm with 9 stator coils.
V_{MAX} = 9.2V
100rpm = 1.67rps
if radius to center of magnets/coils = 0.1m
C = 2 \pi r = 2 \pi \times 0.1m = 0.628m
at 1.67 rps, v = 1.05m/s
N48 Neodymium Bar Magnets 1 in x 1/2 in x 1/4 in
L = 0.0254m, W = 0.0127m, D = 0.00635m.
Separation z = 0.00635m.
Biggest problem is separation between magnets and windings. Closer they are the better.
From: How do you calculate the magnetic flux density?
From: Magnetic Properties of Sintered NdFeB Magnets
N48 has a Remanence field of B_r = 1.48T, which gives a flux density B = 0.164T, when you process the above equation.
V_{Ind} = N B l v
N = \frac {V_{Ind}} {B l v} = \frac {9.2V} {0.164T \times 0.0254m \times 1.05m/s} = 2108 turns
So that assumes you have one coil. Divide that by 3 (in series) gives 703 turns/coil (which is a lot). This coil has to be within length of magnet of 0.0254m. Whole area of coil has to fit within area of magnet for greatest effect.
The math is a bit distorted, but it should give you a ballpark to do your initial calculations. I'd guess, by the second or third prototype, you will have what you want. Not all three coils in series will generate maximum voltage at the same time, so some experimentation will have to come into play (i.e. more turns).
Realize that the coil will not experience a constant magnetic field. The further away windings are from the magnet the less the voltage induced. As already stated, biggest problem is air gap between windings and magnets. Air is a poor conductor of flux. I'd use a steel plate to attach your magnets.
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answered Mar 14, 2017 at 2:37
StainlessSteelRatStainlessSteelRat
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\begingroup Hi thanks for the reply! I thought about taking this approach, except I would assume the B field is constant, and then calculate dA/dt as you have. My point of contention though would be that vl gives you the area swept per second by the magnet. But is it not the rate of change of flux through the COIL's area? Currently I am using the length of the rectangular stator coil as the area, multiplied by the tangential velocity. Does this seem reasonable? I have a feeling this may depend on whether the magnets or coils are longer in length, but I'm not too sure.\endgroup masiewpao –masiewpao 2017-03-14 14:41:22 +00:00 Commented Mar 14, 2017 at 14:41
1 \begingroup As long as magnetic field changes, voltage will be induced. This gives you maximum voltage induced, which defines the sine wave. Ideally, full coil length fits within length of magnet because that is what produces the voltage.\endgroup StainlessSteelRat –StainlessSteelRat 2017-03-14 15:38:33 +00:00 Commented Mar 14, 2017 at 15:38
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9153 | https://thelawdictionary.org/prosecution/ | Prosecution
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In criminal law. A criminal action; a proceeding instituted and carried on by due course of law, before a competent tribunal, for the purpose of determining the guilt or innocence of a person charged with crime. See U. S. v. Reis-Inger, 12S U. S. 398, 9 Sup. Ct. 99, 32 L. Ed. 480; Tennessee v. Davis. 100 U. S. 257, 25 L. Ed. 648; Schulte v. Keokuk County, 74 Iowa, 292, 37 N. W. 376; Sigsbee v. State, 43 Fla. 524, 30 South. 816. By an easy extension of its meaning “prosecution” is sometimes used to designate the state as the party proceeding in a criminal action, or the prosecutor, or counsel; as when we speak of “the evidence adduced by the prosecution.”
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9154 | https://www.nature.com/articles/s41420-024-01979-4 | Skip to main content
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The emerging role of regulated cell death in ischemia and reperfusion-induced acute kidney injury: current evidence and future perspectives
Chenning Li1,2na1,
Ying Yu1,2na1,
Shuainan Zhu1,2,
Yan Hu1,2,
Xiaomin Ling1,2,
Liying Xu1,2,
Hao Zhang1,2 &
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Kefang Guo
ORCID: orcid.org/0000-0003-1146-513X1,2
Cell Death Discovery
volume 10, Article number: 216 (2024)
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Abstract
Renal ischemia‒reperfusion injury (IRI) is one of the main causes of acute kidney injury (AKI), which is a potentially life-threatening condition with a high mortality rate. IRI is a complex process involving multiple underlying mechanisms and pathways of cell injury and dysfunction. Additionally, various types of cell death have been linked to IRI, including necroptosis, apoptosis, pyroptosis, and ferroptosis. These processes operate differently and to varying degrees in different patients, but each plays a role in the various pathological conditions of AKI. Advances in understanding the underlying pathophysiology will lead to the development of new therapeutic approaches that hold promise for improving outcomes for patients with AKI. This review provides an overview of the recent research on the molecular mechanisms and pathways underlying IRI-AKI, with a focus on regulated cell death (RCD) forms such as necroptosis, pyroptosis, and ferroptosis. Overall, targeting RCD shows promise as a potential approach to treating IRI-AKI.
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Facts
Ischemia/reperfusion-induced damage in proximal tubular epithelial cells (TECs) is considered the main cause of AKI.
Various types of RCD have been linked to IRI, including necroptosis, apoptosis, pyroptosis, and ferroptosis.
RCD-targeted therapies have shown effectiveness in curing IRI-AKI.
Open questions
What kind of RCD is most closely associated with IRI-AKI?
Are there any other mechanisms through which RCD affects IRI-AKI? Do different types of RCD vary in their impact on IRI-AKI?
When it comes to translating RCD-targeted therapies from experimental mouse models to clinical trials, what are the key considerations and challenges that need to be addressed? How can the findings and insights gained from mouse studies be effectively applied in the context of human patients undergoing clinical trials for IRI-AKI?
Introduction
Acute kidney injury (AKI) is a complex syndrome that encompasses a range of conditions. It is characterized by a sudden decrease in kidney function that occurs within a period of 7 days, including an increase in serum creatinine levels and a reduction in urinary output . AKI is a significant complication in approximately 15% of hospitalized patients . AKI is even more common among patients in the intensive care unit (ICU) , with a prevalence of over 50%. Risk factors for AKI typically center around sepsis, major surgeries (including cardiac surgery) and drug toxicity . During hospitalizations, certain factors are often present that can lead to generalized or localized ischemia. Ischemia/reperfusion is a pathological condition in which an organ experiences a temporary restriction of blood supply, followed by the restoration of perfusion and reoxygenation. However, this restoration often exacerbates tissue injury and triggers a profound inflammatory response . Ischemia/reperfusion injury-related acute kidney injury (IRI-AKI) occurs when there is a mismatch between local tissue oxygen supply and demand. This results in sustained inadequate oxygen delivery, the accumulation of metabolites and ATP depletion, which can lead to ischemic acute tubular necrosis (ATN) and epithelial and endothelial injury if adequate renal perfusion is not restored. Ultimately, this causes activation of cell death programs and inflammatory processes, leading to renal dysfunction [6, 7]. Previous studies have suggested that tubular cell death by necrosis and apoptosis is the major mechanism associated with renal IRI . However, recent research has challenged traditional views of cell death by identifying new pathways in which cells die in a regulated manner, including pyroptosis , necroptosis, and ferroptosis . In this review, we summarize the current status of research on the molecular mechanisms and pathways of AKI induced by IRI. Advances in understanding the underlying pathophysiology will lead to the development of new therapeutic approaches that hold promise for improving outcomes for patients with AKI.
Cell injury and dysfunction in IRI-AKI
Epithelial cell injury
When blood flow to the kidneys decreases, the ATP levels in epithelial cells can drop, which may cause cellular injury or death. Although any part of the nephron can be affected, proximal tubular epithelial cells (TECs) are particularly vulnerable because of their high metabolic rate and limited ability to produce energy without oxygen. Ischemia-induced damage in proximal TECs is considered the main cause of AKI. Following injury, blood flow decreases, and congestion occurs in the outer stripe of the S3 segment, which can persist and contribute to ongoing ischemia. Ischemic cell injury causes the loss of the apical brush border of proximal TECs, exposing areas of the denuded tubular basement membrane, which leads to proximal tubular dilatation and the formation of distal tubule casts. Additionally, changes in the actin cytoskeleton and its dysfunction during ischemia alter cell polarity and function. Rapid depletion of ATP disrupts apical F-actin and tight junctions, resulting in increased permeability and back-leakage of the glomerular filtrate. Epithelial cells are also separated from the extracellular matrix due to integrin disruption, causing β-integrins to be relocated and viable cells to detach from the tubular basement membrane. The exfoliated cells then form cellular casts within the tubular lumen [6, 8]. Experiments have shown that mice with IRI-AKI exhibit distension, loss of cristae, and fragmentation of the mitochondria in transmission electron microscopy images. Additionally, there is higher expression of a marker of apoptosis (cleaved caspase 3) [11t
.")]. The mitochondrial dysfunction of TECs might be related to excessive formation of ROS and kidney damage in mouse models of IRI-AKI.
TECs play both victim and perpetrator roles in IRI-AKI. A study found that TECs experience complex molecular and cellular events that are particularly relevant to IRI, including oxidative damage and activation of the innate immune system through Toll-like receptors (TLRs), sphingosine-1-phosphate (S1P) receptors, and hypoxia-inducible factors (HIFs). This complexity provides various physiological advantages, such as molecular messaging, cellular specificity, and response amplification. Additionally, proinflammatory cascades triggered by TECs promote the recruitment of leukocytes and dilation of the vasculature, affecting both the injured TECs and their surrounding microenvironment (Fig. 1).
Endothelial dysfunction
Endothelial cells play a role in regulating vascular tone, blood flow, coagulation, inflammation, and permeability. Ischemia can affect the renal endothelium, resulting in microvascular dysregulation and further injury. This injury can manifest as vascular congestion, edema, diminished blood flow, and inflammation. Interpreting total kidney blood flow after injury is challenging due to the complexity of the kidney’s vascular beds . In postischemic kidney arterioles, vasoconstriction is amplified due to reduced production of vasodilatory substances by damaged endothelial cells and is augmented by vasoactive cytokines, causing them to constrict more than normal kidney vessels. Endothelial cells also contribute to IRI-AKI pathology by enhancing endothelium-leukocyte interactions, activating leukocytes, obstructing capillaries and postcapillary venules, and increasing microvascular permeability. In cases of renal hypoperfusion, epithelial cells may be damaged or killed via necrosis or apoptosis. In response to hypoxia, the functioning of adjacent injured or deceased TECs can be affected, leading to alterations in prostaglandin synthesis, the generation of reactive oxygen species, or the activation of inflammatory pathways. As a result, this can initiate inflammatory processes and injure the endothelium [7, 13]. Additionally, the number of microvessels in the inner stripe of the outer medulla declines after IRI, potentially leading to chronic hypoxia, increased tubular injury, and fibrosis .
Recent research shows that small membrane particles called extracellular vesicles (EVs) derived from endothelial colony-forming cells (ECFCs) may have the potential to treat AKI. EV treatment has advantages over stem or progenitor cells because it circumvents cell rejection, undesirable cells, and neoplasm formation . Furthermore, molecular manipulation of EVs with exogenous factors and specific surface molecules can be used for renal endothelial cell therapy, which can improve the prognosis of IRI-AKI .
Programmed cell death in IRI-AKI
Cell death, specifically the death of TECs, plays a significant role in the pathophysiology of IRI-AKI. Historically, cell death classifications were based on the morphological and structural details of individual tissues and cells. These physical characteristics have been used to classify cell death into three main types: apoptosis, autophagy, and necrosis . The Nomenclature Committee on Cell Death (NCCD) has updated the classification system for cell death based on functional aspects: accidental cell death (ACD) is an uncontrolled biological process, whereas regulated cell death (RCD) involves tightly structured signaling cascades and molecularly defined effector mechanisms . In addition to apoptosis, newly discovered forms of cell death, such as necroptosis, pyroptosis, and ferroptosis, are all examples of RCD. Recent studies suggest that different pathways for programmed cell death interact with each other (Fig. 2). The activation of these lytic cell death pathways leads to the release of inflammatory signals into the surrounding environment, which may prompt neighboring cells to die through different cell death modalities . Further studies are needed to determine which cells in the kidney are most susceptible to injury and how the various modalities of cell death interact to either cause or prevent IRI-AKI. Novel therapeutic approaches can improve outcomes in patients with AKI by targeting specific cell death pathways.
Apoptosis is the most well-established form of cell death that occurs in IRI-AKI. The caspase family of proteases is responsible for initiating and executing apoptosis. Both intrinsic and extrinsic apoptotic pathways are activated in AKI. Intrinsic apoptosis is marked by mitochondrial outer membrane permeabilization (MOMP), while extrinsic apoptosis is initiated by perturbations in the extracellular microenvironment detected by plasma membrane receptors and propagated by caspase-8. Both pathways activate executioner caspases, mainly caspase-3 . Effector caspases, such as caspase-3, -6, and -7, cleave numerous cellular proteins, which results in the classic apoptotic phenotype. Studies have shown that inhibiting caspase activity can be protective against injury . The balance between cell survival and intrinsic apoptotic cell death is determined by the relative concentrations of proapoptotic proteins (Bax, Bad, and Bid) and antiapoptotic proteins (Bcl2 and Bclxl) of the Bcl-2 protein family. Overexpression of proapoptotic proteins or underexpression of antiapoptotic proteins can create pores in the outer mitochondrial membrane and possibly other intracellular membranes . Other proteins involved in the apoptotic pathway include proximal-acting nuclear factors κB and p53. Kinases interact with signals from growth factors to mediate cellular responses involved in apoptosis, survival, and repair . Preventing apoptosis of endothelial cells and TECs may be a potential therapeutic strategy for preventing AKI to CKD progression [19, 20].
Previous studies on kidney and cell death have focused on apoptosis as a cause of AKI. Experimental evidence supports that proximal tubule epithelial cells are highly susceptible to apoptosis and that apoptosis plays a pathogenic role in AKI . However, early studies may not have clearly differentiated between apoptosis and regulated necrosis. These two processes share some of the same molecules and pathways. As a result, treatments targeting only apoptosis have been shown to be only partially effective .
Pyroptosis in IRI-AKI
Pyroptosis is an inflammatory form of regulated cell death characterized by the activation of proinflammatory caspases and the formation of plasma membrane pores, which eventually leads to cell lysis [16, 22]. Previous studies have shown that pyroptosis mediates a variety of downstream effects by activating caspase-1 through the canonical pathway or caspase-4/5/11 through the noncanonical pathway . In both pathways, gasdermin D (GSDMD) is the key effector of pyroptosis initiation. There is a growing body of evidence suggesting that pyroptosis may play a role in IRI. However, the role of pyroptosis and gasdermins in AKI has not been well studied and is not completely understood . In 2014, in one of the earliest studies, Yang et al. discovered a significantly increased presence of pyroptosis-related proteins, such as caspase-1, caspase-11, and IL-1, following renal IRI. This study also revealed that CHOP/caspase-11, triggered by overactivated endoplasmic reticulum (ER) stress, may be an essential pathway involved in pyroptosis of renal TECs .
In the canonical pathway, activated inflammasome sensors (NLRP1, NLRP3, NLRC4, and AIM2) form the inflammasome complex with the adapter apoptosis-associated speck-like protein containing a CARD (ASC) for caspase 1 activation. Caspase 1 cleaves pro-IL-1β, pro-IL-18, and GSDMD, releasing the amino-terminal domain of GSDMD (GSDMD-NT) that forms pores in the plasma membrane. Pore formation by GSDMD mediates pyroptosis, causing the release of cellular contents, mature IL-1β and IL-18, and lytic cell death .
In contrast, the noncanonical pathway directly activates caspase-4/5/11 through oligomerization by binding with intracellular bacterial lipopolysaccharide (LPS), which cleaves GSDMD. A study showed that disulfiram improved renal impairment after IR by blocking the upregulation of noncanonical pyroptosis pathway proteins, suggesting that it might reduce pyroptosis by antagonizing TLR4 and inhibiting the caspase-11/GSDMD pathway .
Although all of the studies mentioned above indicate that GSDMD is the key contributor to renal IRI, a recent study provided the first evidence that GSDMD-positive cells may function as a suppressor of AKI rather than being primarily involved in renal TEC death. Specifically, these cells constrain necroptotic cell death in the tubules through a previously unknown noncell autonomous crosstalk with the necroptosis machinery . Overall, most experimental studies investigating pyroptosis in AKI have been conducted using mouse models, and the results of these studies are quite inconsistent. Therefore, further research is needed to determine the role of this form of cell death in human AKI and its association with IRI.
Necroptosis in IRI-AKI
Necroptosis is a type of regulated cell death that leads to cell lysis and inflammation in nearby tissues, which is typically characterized by the morphological features of necrosis. This pathway is often associated with kidney disease and is defined at the molecular level by the involvement of three proteins: receptor-interacting serine/threonine protein kinase 3 (RIPK3), RIPK1, and the executioner pseudokinase mixed-lineage kinase domain-like protein (MLKL) . The necroptosis pathway induced by TNF-α has been extensively researched . Although necroptosis is commonly perceived as a highly proinflammatory mode of cell death due to its rapid lysis of cells and liberation of DAMPs, some researchers suggest that it might have anti-inflammatory effects in certain contexts. Specifically, it may reduce excessive TNF- or TLR-induced inflammatory cytokine production . IRI occurs when the blood and oxygen supply to the kidney is temporarily restricted, leading to depleted ATP and accumulated metabolites. Upon restoration of blood flow, necroptosis occurs, causing acute tubular necrosis and endothelial damage, which ultimately leads to inflammation .
There is a body of evidence indicating that a lack of RIPK3 and MLKL expression can reduce kidney damage due to IRI, and the necroinflammation driven by RIPK3/MLKL-dependent necroptosis may lead to the progression of IRI to CKD . Additionally, the accumulation of pMLKL was found in the necrotic tubules of human patients with AKI . Furthermore, Feng et al. discovered that in response to renal IRI, RIPK3 translocates into mitochondria and interacts with an inner mitochondrial protein (mitofilin). This interaction promotes the release of mitochondrial DNA (mtDNA) into the cytosol, which activates the cGAS/STING pathway, leading to increased nuclear transcription of proinflammatory markers, further exacerbating renal IRI . However, the interplay between RIPK3 and MLKL remains unclear and requires further study. The extent to which the effects of RIPK3 deficiency, MLKL deficiency and the combined deficiency on IRI differ is unclear [32, 33]. RIPK3 may worsen kidney injury even without the involvement of MLKL and necroptosis .
In addition, either apoptosis or necroptosis can be initiated by receptors such as TNFR and Fas, depending on the intracellular state of the target cell. Based on the significant amount of necrotic tubular cells, Linkermann et al. concluded that renal IRI primarily occurs through necroptosis, rather than apoptosis, and can be attenuated by inhibiting RIP1-mediated necroptosis by necrostatin-1 (nec-1) [34, 35]. However, subsequent studies have indicated that nec-1 could potentially inhibit ferroptosis and that ferroptosis might be more significant than necroptosis in renal IRI [18, 36]. In 2014, Linkermann et al. suggested that renal tubules are not sensitized to necroptosis by loss of FADD or caspase-8, and the RIPK1 inhibitor nec-1 does not protect freshly isolated tubules from hypoxic injury, while ferroptosis directly causes synchronized necrosis of renal tubules in models of IRI-AKI . Another point of view is that necroptosis and ferroptosis function independently in many pathologies that involve cell death, but that they are alternative pathways in murine renal IRI. This interconnection reflects that sensitization to one pathway can result in resistance to the other pathway, while compensation from one pathway can occur when the other is compromised . In conclusion, although most experimental studies investigating necroptosis in IRI-AKI have been conducted using mouse models, our understanding of the dynamics of necroptosis in human IRI-AKI remains inadequate. Additionally, further studies are needed to explore the interplay between necroptosis and other types of cell death.
Ferroptosis in IRI-AKI
Ferroptosis was first proposed by Dixon et al. in . It is an iron-dependent form of regulated necrotic cell death characterized by the accumulation of lipid ROS, which differs from other forms of cell death in terms of its morphology, biochemistry, and genetics . The most important feature underlying the distinction is lipid peroxidation in the intracellular microenvironment, which relies on the generation of ROS and the availability of iron . This process leads to membrane destruction but can be controlled by glutathione peroxidase 4 (GPX4) . The regulatory pathways of ferroptosis can be divided into three main groups: the GSH/GPX4 pathway, iron metabolism-related pathways, and lipid metabolism-related pathways (Fig. 3).
As a crucial regulator of ferroptosis, many studies have investigated the role of GPX4 in ferroptosis associated with IRI-AKI. If GPX4 activity is inhibited, lipid peroxides can accumulate. This buildup is a crucial factor in the occurrence of ferroptosis, and GPX4 is the primary regulator of this process . In 2014, genetic evidence showed that the knockout of GPX4 leads to ferroptosis, and by using inducible GPX4−/− mice, it was found that the glutathione/GPX4 axis plays an essential role in preventing lipid oxidation-induced AKI . In 2019, Huang et al. first found that I/R-induced kidney ferroptosis was mediated by augmenter of liver regeneration (ALR) via the GSH/GPX4 axis . A recent study suggested that GPX4 is a substrate of tripartite motif containing 21 (TRIM21) and can be degraded by TRIM21-mediated ubiquitination, suggesting that inhibiting TRIM21 attenuates ferroptosis. Loss of TRIM21 negatively regulates ferroptosis by upregulating GPX4, alleviates IRI-AKI, and improves renal function . Another member of the TRIM family, TRIM27, is involved in ubiquitin-specific peptidase 7 (USP7)-related ferroptosis. Inhibition of USP7 attenuated IRI-AKI by inhibiting ferroptosis through decreasing the ubiquitination of TRIM27-mediated TANK-binding kinase 1 (TBK1) and promoting DNA methyltransferase 1 (DNMT1)-mediated methylation of FMRP translational regulator 1 (FMR1) .
Another essential component for executing ferroptosis is acyl-CoA synthetase long-chain family member 4 (ACSL4), as it influences the necessary lipid composition in ferroptosis . In 2020, Zhao et al. first discovered ferroptosis in TECs of patients with clinical kidney disease, confirming the correlation between ACSL4 and kidney function decline. They also identified a potential ferroptosis inhibitor, XJB-5-131, which showed promising results by alleviating kidney injury, promoting TEC proliferation and repair, decreasing kidney inflammation, and inhibiting lipid peroxidation accumulation in the I/R model . In 2023, the same team showed that cytosolic high-mobility group box 1 (HMGB1) can cause ferroptosis by binding with ACSL4 after IRI. This study showed that inhibiting HMGB1 nucleocytoplasmic translocation pharmacologically and deleting HMGB1 in TECs in mice were effective in preventing IRI-AKI, tubular ferroptosis, and inflammation compared to control groups . Previous studies have demonstrated the significance of ferroptosis in AKI. Targeting specific genes or proteins involved in ferroptosis-related pathways could offer potential therapeutic targets for treating IRI-AKI.
New therapeutic targets: pyroptosis, necroptosis and ferroptosis
Renal IRI is a major contributor to AKI, for which an effective treatment has yet to be identified. We have compiled a list of clinically viable targeted drugs (Table 1) for RCD. Gasdermins and pyroptosis significantly impact tissue homeostasis, inflammation, and disease development. Approaches to attenuate renal IRI include inhibiting NLRP3 inflammasome activation and activating the Nrf2/NLRP3 signaling pathway. Other pathways, such as cholecalciferol pretreatment and Tisp40 overexpression, affect pyroptosis in IRI-AKI. Several novel inhibitors of necroptosis, such as Nec-1f, have shown promise in treating AKI in experimental models of renal IRI. The E-prostanoid 3 receptor (EP3) is another potential target for disrupting necroinflammation and improving ischemic AKI. Therapeutic strategies targeting Ferroptosis involve controlling lipid peroxidation, manipulating thiol metabolism, and regulating specific miRNAs and heme oxygenase-1. It is evident from the tables that current studies on drug therapy are being conducted in animal models. However, certain medications and small molecule inhibitors targeting regulated cell death pathways have shown promise in experimental models of renal IRI.
Therapeutic significance of pyroptosis in IRI-AKI
Although gasdermins and pyroptosis are not yet fully understood, they are thought to have a significant impact on tissue homeostasis, inflammation, and the development of various diseases. One approach is to prevent the assembly of caspase-1-activating inflammasomes. In response to various PAMPs and DAMPs, NLRP3 oligomerizes with the adapter protein ASC to form a high molecular weight complex that recruits and cleaves pro-caspase-1, activating it and driving the maturation of IL-1β and IL-18, GSDMD cleavage, and pyroptosis. Studies have found that there are several ways to ultimately attenuate renal IRI by inhibiting the activation of the NLRP3 inflammasome, such as using hydrogen sulfide (H2S) , β-hydroxybutyrate (β-OHB) treatment , inhibiting mTORC1 with activated protein C (aPC) , and promoting zinc ion influx in renal TECs with transient receptor potential-6 (TRPC6) . The cleavage of GSDMD by another caspase, caspase-11, promotes pyroptosis and IL-18 release in PTCs, playing a significant role in AKI. Therapeutically targeting both steps of Nlrp3 inflammasome with aPC and parmodulin-2 or small compound Nlrp3 inhibitors appears promising . The cleavage of GSDME by Caspase-3 is implicated in the formation of membrane pores, cell lysis, and the exacerbation of renal tubular injury and inflammation. Treatment with Z-VAD-FMK, a broad-spectrum caspase inhibitor, led to a notable decrease in the levels of GSDME-N. These observations suggest that partial inhibition of pro-pyroptotic factors could potentially ameliorate pyroptosis and associated tissue damage .
Moreover, nuclear factor erythroid 2 related factor 2 (Nrf2) is a significant regulator of redox balance that controls both proinflammatory and anti-inflammatory effects, but its detailed mechanism remains unknown. Nrf2 can inhibit NLRP3 inflammatory vesicle activation and prevent burn-induced inflammation by decreasing intracellular ROS. For example, salvianolic acid B (SalB) can inhibit caspase-1/GSDMD-mediated pyroptosis by activating the Nrf2/NLRP3 signaling pathway, resulting in alleviation of IRI in mice . In addition, Nrf2 is known to have an inhibitory effect on inflammation by inducing HO-1 expression. Inhibiting PRMT5 or administration of naringenin can activate the Nrf2/HO-1 pathway, which may alleviate pyroptosis and ameliorate renal IRI in mouse models. Further studies are necessary to clarify the regulation of inflammatory vesicles and burn injury by Nrf2.
Other pathways are also involved in pyroptosis in IRI-AKI. It appears that cholecalciferol pretreatment may reduce GSDMD-mediated pyroptosis by decreasing ROS production and NF-κB activation . Conversely, overexpression of Tisp40 increases TEC GSDMD-mediated pyroptosis and related protein expression via NF-κB signaling .
The study of gasdermins and pyroptosis in the kidney is a relatively new area of research. Therefore, it is premature to draw any firm conclusions about their roles in renal pathogenesis based on the literature. Currently there are no commercially available drugs that block the pyroptosis pathway to prevent AKI, but rather inhibitors of specific molecules. Further research is needed before gasdermins or pyroptosis can be considered potential treatment targets for human kidney disease. It is possible to start with existing drugs to discover their relationship with the pyroptosis in IRI-AKI. Viable treatment options have been reported in other causes of AKI, which might provide new directions for the research of IRI-AKI. Fibroblastic reticular cells (FRCs) derived exosomes (FRC-Exos) can improve kidney function in sepsis-induced acute kidney injury (S-AKI). FRC-Exos promote PINK1-dependent mitophagy and inhibit NLRP3 inflammasome activation in lipopolysaccharide (LPS)-stimulated primary kidney tubular cells (PKTCs) . It is important to validate any observations made from animal models in humans before clinical trials can be conducted to test treatments targeting gasdermins, pyroptosis or their upstream regulators in patients with kidney disease.
Therapeutic significance of necroptosis in IRI-AKI
Several novel inhibitors of necroptosis are effective in experimental models of renal IRI, making them promising targets for the treatment of AKI. Most of them inhibit necroptosis and ameliorate AKI by targeting RIPK1 and/or RIPK3 [61,62,63]. Tonnus et al. developed a combined small molecule inhibitor called Nec-1f, which can potently inhibit RIPK1 and mildly inhibit ferroptosis and has been shown to improve survival in models of IRI-AKI . Some inhibitors of AKI act through an MLKL-dependent mechanism. For example, repulsive guidance molecule (RGM)-b has been shown to inhibit MLKL membrane association and necroptosis in proximal TECs in mouse models . Another promising target is the E-prostanoid 3 receptor (EP3). Deficiency of this receptor on myeloid cells can ameliorate ischemic AKI by disrupting the autoamplification loop of necroinflammation .
The anticonvulsant Phenytoin, classified as a hydantoin, was identified as an RIPK1 inhibitor that shows promise in the inhibition of necroptosis, albeit with lower potency compared to other inhibitors . A novel small molecule Hsp90 inhibitor, C-316-1, attenuates acute kidney injury by suppressing RIPK1-mediated inflammation and necroptosis. The compound disrupts the Hsp90–Cdc37 protein–protein interactions, leading to a decrease in RIPK1 and reduced necroptosis. Despite the lack of approved Hsp90 inhibitors on the market due to toxicity concerns, C-316-1 shows promise due to its lower toxicity and better water solubility . Augmenter of liver regeneration (ALR), a multifunctional factor known for promoting hepatocyte proliferation, has been found to have antiapoptotic and anti-oxidative stress effects and shows promise as a protective agent against AKI. Exogenous injection of ALR has a protective effect against AKI, and overexpression of the ALR gene decreases mitochondrial damage .
While these medicines and small molecule inhibitors of necroptosis have demonstrated effectiveness in experimental models of renal IRI, our understanding of the dynamics of necroptosis in human IRI is limited and likely to hinder the clinical use of necroptosis inhibitors. Existing drugs are not yet ready for use, either because they are too toxic or because they are ineffective. Further exploration is warranted to fully understand its potential as an affordable and readily accessible treatment option, including the determination of in vivo concentrations in specific target organs. The potential correlation between the long-term side effects, such as gingival hyperplasia of phenytoin, should also be investigated. So, it is possible to start with what is already available and improve them.
Therapeutic significance of ferroptosis in IRI-AKI
Ferroptosis is implicated in the pathogenesis of kidney disease and could be targeted for therapy. This process occurs in cells with a redox imbalance and leads to cell death if other protective mechanisms fail. The excessive accumulation of certain products of phospholipid peroxidation triggers this process. Since iron and thiol imbalance can induce phospholipid oxidation through multiple pathways, there is potential for developing various regulators .
Among these therapeutic strategies, the control of lipid peroxidation might represent the most attractive strategy. ASCL4 plays a crucial role in synthesizing phospholipids that contain arachidonic acid, which are particularly vulnerable to oxidation during ferroptosis. Therefore, targeting ASCL4 could be a promising strategy for developing anti-ferroptosis therapies.
Dexmedetomidine, an α2-adrenergic receptor (α2-AR) agonist, has been shown to have renal protective effects against ferroptosis-mediated renal IRI and the inflammatory response through the suppression of ACSL4 . lncRNA TUG1 carried by human urine-derived stem cell (USC)-derived exosomes (USC-Exo) regulates ASCL4-mediated ferroptosis by interacting with SRSF1 and then protects against IRI-induced AKI . In addition, Shi et al. indicated that miR-20a-5p negatively regulated ACSL4 by targeting the 3’ UTR of ACSL4 mRNA, thereby inhibiting ACSL4-dependent ferroptosis and alleviating kidney IRI .
Another approach to targeting lipid peroxidation is through the use of radical-trapping antioxidants such as ferrostatin-1 or liproxstatin-1 . As previously mentioned, necrostatin-1, an inhibitor of RIPK1 and a crucial player in necroptosis, also hinders ferroptosis .
Targeting thiol metabolism can be an effective strategy. Small molecule compounds that boost GSH levels and GPX4 activity have been shown to inhibit ferroptosis, while depriving cells of cysteine and inhibiting GPX4 can trigger ferroptosis. However, directly administering GSH is not currently a practical option owing to its limited bioavailability and inability to pass through the plasma membrane [69, 75]. Some studies have found that other miRNAs can play a regulatory role in ferroptosis in IRI-AKI. According to Ding et al., there was an upregulation of miR-182-5p and miR-378a-3p following IRI, resulting in the activation of ferroptosis in renal injury through the downregulation of GPX4 and SLC7A11 . Moreover, heme oxygenase-1 (HMOX1) is significantly upregulated during the early stages of IRI in the kidneys. Inhibiting miR-3587 promotes upregulation of the HO-1 protein, which is encoded by HMOX1, and thus protects renal tissues from IR-induced ferroptosis .
Cyanidin-3-glucoside (C3G), an anthocyanin, can attenuate acute organ injury and modulate oxidation, which markedly ameliorated Era-induced ferroptosis, resulting in nearly equivalent effects as selective ferroptosis inhibitor Lip-1 . In a similar fashion, Entacapone, a medication commonly prescribed for Parkinson’s disease, has exhibited notable efficacy in preventing renal IRI by inhibiting ferroptosis. It achieves this by boosting antioxidant defenses through modulation of the p62-KEAP1-NRF2 pathway and the upregulation of SLC7A11 expression. Consequently, this intervention effectively suppresses oxidative stress and ferroptosis . Additionally, Legumain, a cysteine protease known for its role in promoting ferroptosis by interacting with the ferroptosis inhibitor GPX4, is being investigated as a potential therapeutic target and an early diagnostic marker for IRI-AKI .
The exact mechanism of ferroptosis in IRI-AKI remains to be determined experimentally. While ferroptosis-specific inhibitors, such as ferrostatin-1 (Fer-1) and liproxstatin-1 (Lip-1), have shown protective effects in scientific research, they are not yet ready for clinical use. Drug discovery is at a basic stage, and it is still very far from clinical application. Therefore, re-evaluation of marketed drugs is an important way to achieve breakthroughs rapidly. Further research should explore ways to regulate ferroptosis to improve AKI. These findings will provide evidence for clinical translation.
Conclusion and perspectives
The process of IRI-AKI involves multiple mechanisms of cell injury, cell dysfunction and various types of cell death. This paper outlines the preliminary understanding of the mechanisms underlying its incidence, characteristics, and targeted signaling pathways. Although most studies targeting IRI-AKI-related signaling pathways have been conducted in mouse models or in vitro experiments, exploring the role played by IRI in AKI and using potential targets to regulate AKI could provide new perspectives and therapeutic strategies to improve outcomes for AKI patients.
While current experimental techniques have been unable to determine the most significant form of cell death in IRI-AKI or whether they interact with each other, studies using animal models have provided increasing amounts of evidence of a correlation between RCD and IRI-AKI. The exact mechanism of RCD in the development and progression of IRI-AKI has yet to be determined experimentally. To do so, more functional studies that involve specific molecular markers and targeted inhibitors are necessary. In recent years, a significant amount of work has been done to identify biomarkers of cell death in vivo.
A novel form of cell death called PANoptosis has been discovered, involving multiple programmed cell death pathways such as pyroptosis, apoptosis, and necroptosis. Crosstalk among these pathways reveals new insights into the roles of inflammasome sensors and cell death complexes in disease. Caspase-8, traditionally linked to apoptosis, also plays a role in regulating the NLRP3 inflammasome and activating pyroptotic molecules. Caspase-3, a key player in apoptosis, can trigger pyroptosis as well, leading to both lytic and non-lytic cell death . This unique process, known as PANoptosis, involves the activation of various cell death mechanisms simultaneously, leading to rapid and irreversible loss of cell viability. Research is ongoing to investigate the underlying mechanisms and implications of PANoptosis, with a focus on the role of NLRP12 in initiating this form of cell death in conditions such as AKI and hemolytic disease . Enhancing the understanding of PANoptosis in the context of IRI-AKI has the potential to offer invaluable insights into the intricate interplay of cell death pathways in this condition. With the complexities surrounding IRI-AKI pathogenesis, delving into how PANoptosis interacts with conventional cell death mechanisms like apoptosis and necrosis could unveil innovative therapeutic targets for alleviating kidney damage and enhancing patient outcomes. Through unraveling the role of PANoptosis in IRI-AKI, researchers may discover new avenues for developing targeted interventions that target the specific mechanisms underlying renal injury and promote tissue regeneration.
When translating RCD-targeted therapies from experimental mouse models to clinical trials for IRI-AKI, there are several key considerations and challenges that need to be addressed: relevance of mouse models, dosing and pharmacokinetics, safety and toxicity, biomarkers and endpoint selection, patient heterogeneity, and regulatory approval. To effectively apply the findings and insights gained from mouse studies in the context of human patients undergoing clinical trials for IRI-AKI, collaboration between basic scientists, clinicians, and regulatory authorities is key. Robust preclinical data, including mechanistic insights and efficacy in relevant animal models, should guide the design of clinical trials. Additionally, close monitoring of patient responses, careful interpretation of clinical data, and adaptive trial designs are important strategies to optimize the translation of RCD-targeted therapies for IRI-AKI from bench to bedside.
Accurately identifying the form of RCD is of utmost importance, as it can serve as a solid diagnostic foundation for IRI-AKI and other related illnesses. It is imperative to conduct further research on how to effectively regulate RCD to enhance treatment strategies for IRI-AKI. These discoveries will ultimately offer supporting evidence for clinical translation.
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Funding
This work was supported by the Natural Science Foundation of Shanghai (22ZR1410900); Shanghai Municipal Health Commission Clinical Research Program (202140270) and Zhongshan Hospital Clinical Special Research Fund (2020ZSLC13).
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These authors contributed equally: Chenning Li, Ying Yu.
Authors and Affiliations
Department of Anesthesiology, Zhongshan Hospital, Shanghai, China
Chenning Li, Ying Yu, Shuainan Zhu, Yan Hu, Xiaomin Ling, Liying Xu, Hao Zhang & Kefang Guo
2. Shanghai Key Laboratory of Perioperative Stress and Protection, Shanghai, China
Chenning Li, Ying Yu, Shuainan Zhu, Yan Hu, Xiaomin Ling, Liying Xu, Hao Zhang & Kefang Guo
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Chenning Li
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HZ, and KG performed study concept and design; CL, YY and SZ performed literature investigation and wrote the manuscript. YH, XL, and LX edited and revised the paper. All authors have reviewed the paper and all approved of the final version.
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Li, C., Yu, Y., Zhu, S. et al. The emerging role of regulated cell death in ischemia and reperfusion-induced acute kidney injury: current evidence and future perspectives.
Cell Death Discov. 10, 216 (2024).
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9155 | https://www.chla.org/blog/experts/research-and-breakthroughs/improving-surgical-outcomes-children-rickets | Improving Surgical Outcomes for Children With Rickets | Children's Hospital Los Angeles
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Improving Surgical Outcomes for Children With Rickets
Research and Breakthroughs
Improving Surgical Outcomes for Children With Rickets
February 6, 2023
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Katie Sweeney
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A new study finds that metabolic control after surgery is associated with successful correction of leg deformities in children with hypophosphatemic rickets.
Orthopedic surgery plays a critical role in the treatment of angular deformities in children with severe forms of hypophosphatemic rickets, a rare disorder that leads to soft, painful bones and poor growth. But while some children respond well to surgery, others have leg deformities that continue to worsen.
Ian Marpuri, MD
Recently, a study led by Children’s Hospital Los Angeles revealed the importance of metabolic control of rickets on these varying outcomes. Early data turned up a surprising finding: Only metabolic control after surgery—not before—was associated with success.
Ian Marpuri, MD, a third-year pediatric endocrinology fellow at Children’s Hospital Los Angeles, presented early findings at the 2022 Pediatric Endocrine Society meeting, where the study was selected for a Presidential Abstract Award.
“Our original hypothesis was that children who went into surgery with well-controlled disease would have the best outcomes,” says Dr. Marpuri, who will join the CHLA faculty in the fall. “But it turns out that it’s the postoperative period that appears to have the biggest influence on success.”
Study findings
A disorder of poor bone mineralization, hypophosphatemic rickets is typically caused by genetic defects that lead to low levels of phosphate—one of the key building blocks of bone.
Children with severe forms of the disease can develop painful leg deformities, including bowed legs and knock knees. To correct these problems, many of these patients undergo hemiepiphysiodesis surgery. Also called guided growth surgery, the procedure uses strategically placed metal plates to direct bones to grow into a straighter position.
“Surgery does not treat the rickets itself, but it can be very helpful in decreasing pain and improving a child’s mobility and deformity,” says Anna Ryabets-Lienhard, DO, a senior investigator on the study and Medical Director of the Pediatric Bone Disorders Program at Children’s Hospital Los Angeles, one of the largest programs of its kind in the country.
Anna Ryabets-Lienhard, DO
“Unfortunately, some patients do not end up with a successful correction, or they develop additional deformities as their rickets continues,” adds Oussama Abousamra, MD—Director of the Limb Lengthening and Reconstruction Program in the Jackie and Gene Autry Orthopedic Center at CHLA. “Our goal was to better understand which patients are more likely to benefit from surgery and why.”
To investigate how metabolic control impacts surgical success, Dr. Marpuri, Dr. Ryabets-Lienhard and Dr. Abousamra collaborated with metabolic bone experts at Stollery Children’s Hospital in Alberta, Canada. Their retrospective study included 24 children with genetic forms of hypophosphatemic rickets who underwent guided growth surgery. The team reviewed results for each patient at one and two years after the procedure.
They then compared outcomes with patients’ pre- and postoperative metabolic control, which was measured by levels of alkaline phosphatase.
The researchers found that, two years after surgery:
Nearly 25% of affected limbs did not correct to normal alignment.
Postoperative metabolic control was significantly associated with deformities correcting back to normal alignment, signifying the importance of medical treatment of rickets.
However, pre-operative metabolic control was not linked to whether or not the leg abnormalities corrected.
“It didn’t matter what patients’ metabolic control was before surgery,” Dr. Marpuri says. “But if their rickets wasn’t well-controlled after surgery, then their bones wouldn’t grow in the way that the surgery intended.”
Oussama Abousamra, MD
A multicenter collaboration
The findings highlight the importance of optimizing metabolic control in these children. Traditional therapy has involved phosphate supplements and activated forms of vitamin D, but today, many patients are treated with an injectable monoclonal antibody called burosumab, which was approved by the Food and Drug Administration in 2018.
The team is continuing to analyze data for the study, which is part of a larger collaboration with Chelsey Grimbly, MD, at Stollery Children’s Hospital, and Leanne Ward, MD, at Children’s Hospital of Eastern Ontario. The multidisciplinary effort also includes investigations into other abnormalities associated with rickets—such as non-ossifying fibromas and craniosynostosis—and how those influence and predict outcomes for children.
“Hypophosphatemic rickets is a rare condition, and it’s important that we work together to advance knowledge and treatment,” Dr. Ryabets-Lienhard says. “The more we can heal the bone, the better we can improve bone health and growth and hopefully prevent deformities and pain. That is the ultimate goal.”
Learn more about the Pediatric Bone Disorders Program.
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9156 | https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_3?srsltid=AfmBOorLttxDxFP7iU5jRL7qJZXVAvCddDb3wYy22rqSFMaWQ_wmrwd6 | Art of Problem Solving
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2018 AIME II Problems/Problem 3
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 See Also
Problem
Find the sum of all positive integers such that the base- integer is a perfect square and the base- integer is a perfect cube.
Solution 1
The first step is to convert and into base-10 numbers. Then, we can write and . It should also be noted that .
Because there are less perfect cubes than perfect squares for the restriction we are given on , it is best to list out all the perfect cubes. Since the maximum can be is 1000 and • , we can list all the perfect cubes less than 2007.
Now, must be one of . However, will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to .
Because is a perfect square and is clearly divisible by 3, it must be divisible by 9, so is divisible by 3. Thus the cube, which is , must also be divisible by 3. Therefore, the only cubes that could potentially be now are and .
We need to test both of these cubes to make sure is a perfect square.
If we set so . If we plug this value of b into , the expression equals , which is indeed a perfect square.
If we set so . If we plug this value of b into , the expression equals , which is .
We have proven that both and are the only solutions, so
Solution 2
The conditions are: We can see is multiple is 3, so let , then . Substitute into second condition and we get . Now we know is both a multiple of 3 and odd. Also, must be smaller than 13 for to be smaller than 1000. So the only two possible values for are 3 and 9. Test and they both work. The final answer is . -Mathdummy
Solution 3
As shown above, let such that
Subtracting the equations we have
We know that and both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution must divide by the Rational Root Theorem.
We can instantly know since those will have negative solutions.
When we have , so then
When we have , so then
Therefore, the sum of all possible values of is
See Also
2018 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2Followed by
Problem 4
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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9157 | https://theengineeringmindset.com/thermodynamic-properties-refrigerant-r-134a/ | Thermodynamic properties of refrigerant R-134a
Thermodynamic properties of refrigerant R-134a.
Thermodynamic properties of refrigerant R-134a, also known as 1,1,1,2-tetrafluoroethane, R-134a, Freon 134a, Forane 134a, Genetron 134a, Florasol 134a, Suva 134a or HFC-134a and norflurane (INN). This is a haloalkane refrigerant with thermodynamic properties similar to R-12 (dichlorodifluoromethane) but with insignificant ozone depletion potential and a somewhat lower global warming potential (1,430, compared to R-12’s GWP of 10,900). It has the formula CH2FCF3 and a boiling point of −26.3 °C (−15.34 °F) at atmospheric pressure. This is a non-flammable gas used primarily as a “high-temperature” refrigerant for domestic refrigeration and automobile air conditioners.
Thermodynamic Properties of Saturated Refrigerant R134a
Pressure
Temperature
Specific volume (m^3/kg)
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
kPa
°C
Sat Liqvf
Sat Vapvg
Sat Liqhf
Sat Vaphg
Sat Liqsf
Sat Vapsg
60
-36.9
0.0007098
0.3112
3.9
227.8
0.0164
0.9645
80
-31.1
0.0007185
0.2376
11.3
231.5
0.0472
0.9572
100
-26.4
0.0007259
0.1926
17.3
234.5
0.0720
0.9519
120
-22.3
0.0007324
0.1621
22.5
237.0
0.0928
0.9478
140
-18.8
0.0007383
0.1402
27.1
239.2
0.1110
0.9446
160
-15.6
0.0007437
0.1235
31.2
241.1
0.1270
0.9420
180
-12.7
0.0007487
0.1104
35.0
242.9
0.1415
0.9397
200
-10.1
0.0007534
0.0999
38.5
244.5
0.1547
0.9378
220
-7.6
0.0007578
0.0912
41.7
245.9
0.1668
0.9361
240
-5.4
0.0007620
0.0839
44.7
247.3
0.1780
0.9347
260
-3.2
0.0007661
0.0777
47.5
248.6
0.1885
0.9333
280
-1.2
0.0007699
0.0724
50.2
249.7
0.1984
0.9322
300
0.7
0.0007737
0.0677
52.8
250.9
0.2077
0.9311
320
2.5
0.0007773
0.0636
55.2
251.9
0.2165
0.9301
340
4.2
0.0007808
0.0600
57.5
252.9
0.2248
0.9293
360
5.8
0.0007842
0.0567
59.8
253.8
0.2328
0.9284
400
8.9
0.0007907
0.0512
64.0
255.6
0.2477
0.9270
500
15.7
0.0008060
0.0411
73.4
259.3
0.2803
0.9241
600
21.6
0.0008200
0.0343
81.5
262.4
0.3081
0.9219
700
26.7
0.0008332
0.0294
88.8
265.1
0.3324
0.9200
800
31.3
0.0008459
0.0256
95.5
267.3
0.3541
0.9184
900
35.5
0.0008581
0.0227
101.6
269.3
0.3739
0.9170
1000
39.4
0.0008701
0.0203
107.4
271.0
0.3920
0.9157
1200
46.3
0.0008935
0.0167
117.8
273.9
0.4245
0.9131
1400
52.4
0.0009167
0.0141
127.3
276.2
0.4533
0.9106
1600
57.9
0.0009401
0.0121
136.0
277.9
0.4792
0.9080
1800
62.9
0.0009640
0.0106
144.1
279.2
0.5031
0.9051
2000
67.5
0.0009888
0.0093
151.8
280.1
0.5252
0.9020
2500
77.6
0.0010569
0.0069
169.7
280.9
0.5755
0.8925
3000
86.2
0.0011413
0.0053
186.6
279.2
0.6215
0.8792
Thermodynamic Properties of Superheated Refrigerant R134a
P=60 kPa (Tsat -36.9°C)
P=100 kPa (Tsat -26.4°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.3112
227.8
0.964
Sat.
0.1926
234.5
0.952
-20
0.3361
240.8
1.018
-20
0.1984
239.5
0.972
-10
0.3505
248.6
1.048
-10
0.2074
247.5
1.003
0
0.3648
256.5
1.077
0
0.2163
255.6
1.033
10
0.3789
264.7
1.107
10
0.2251
263.8
1.063
20
0.3930
272.9
1.135
20
0.2337
272.2
1.092
30
0.4071
281.4
1.164
30
0.2423
280.7
1.120
40
0.4210
290.0
1.192
40
0.2509
289.3
1.149
50
0.4350
298.7
1.219
50
0.2594
298.2
1.176
60
0.4488
307.7
1.246
60
0.2678
307.1
1.204
70
0.4627
316.8
1.273
70
0.2763
316.3
1.231
80
0.4765
326.0
1.300
80
0.2847
325.6
1.257
90
0.4903
335.4
1.326
90
0.2930
335.0
1.284
100
0.5041
345.0
1.352
100
0.3014
344.6
1.310
P=140 kPa (Tsat -18.8°C)
P=180 kPa (Tsat -12.7°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.1402
239.18
0.94462
Sat.
0.1104
242.9
0.940
-10
0.1461
246.4
0.972
-10
0.1119
245.2
0.948
0
0.1526
254.6
1.003
0
0.1172
253.6
0.980
10
0.1591
262.9
1.033
10
0.1224
262.0
1.010
20
0.1654
271.4
1.062
20
0.1275
270.6
1.040
30
0.1717
280.0
1.091
30
0.1325
279.3
1.069
40
0.1780
288.7
1.120
40
0.1374
288.1
1.098
50
0.1841
297.6
1.147
50
0.1423
297.0
1.126
60
0.1903
306.6
1.175
60
0.1472
306.1
1.153
70
0.1964
315.8
1.202
70
0.1520
315.3
1.181
80
0.2024
325.1
1.229
80
0.1567
324.6
1.207
90
0.2085
334.6
1.255
90
0.1615
334.1
1.234
100
0.2145
344.2
1.282
100
0.1662
343.8
1.260
P=200 kPa (Tsat -10.1°C)
P=240 kPa (Tsat -5.4°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0999
244.5
0.938
Sat.
0.0839
247.3
0.935
0
0.1048
253.1
0.970
0
0.0862
252.0
0.952
10
0.1096
261.6
1.001
10
0.0903
260.7
0.983
20
0.1142
270.2
1.030
20
0.0942
269.4
1.013
30
0.1187
278.9
1.060
30
0.0981
278.2
1.043
40
0.1232
287.7
1.088
40
0.1019
287.1
1.072
50
0.1277
296.7
1.116
50
0.1057
296.1
1.100
60
0.1321
305.8
1.144
60
0.1094
305.2
1.128
70
0.1364
315.0
1.171
70
0.1131
314.5
1.156
80
0.1407
324.4
1.198
80
0.1168
323.9
1.183
90
0.1451
333.9
1.225
90
0.1204
333.5
1.209
100
0.1493
343.6
1.251
100
0.1240
343.2
1.236
P=280 kPa (Tsat -1.2°C)
P=320 kPa (Tsat 2.5°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0724
249.7
0.932
Sat.
0.0636
251.9
0.930
10
0.0765
259.7
0.968
10
0.0661
258.7
0.954
20
0.0800
268.5
0.999
20
0.0693
267.7
0.986
30
0.0834
277.4
1.029
30
0.0723
276.7
1.016
40
0.0867
286.4
1.058
40
0.0753
285.7
1.045
50
0.0900
295.5
1.086
50
0.0782
294.9
1.074
60
0.0932
304.7
1.114
60
0.0811
304.1
1.102
70
0.0964
314.0
1.142
70
0.0839
313.5
1.130
80
0.0996
323.5
1.169
80
0.0868
323.0
1.157
90
0.1028
333.1
1.196
90
0.0895
332.6
1.184
110
0.1090
352.7
1.248
110
0.0950
352.3
1.237
120
0.1121
362.7
1.274
120
0.0978
362.4
1.263
P=400 kPa (Tsat 8.9°C)
P=500 kPa (Tsat 15.7°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0512
255.6
0.927
Sat.
0.0411
259.3
0.924
10
0.0515
256.6
0.931
20
0.0421
263.5
0.938
20
0.0542
265.9
0.963
30
0.0443
273.0
0.970
30
0.0568
275.1
0.994
40
0.0465
282.5
1.001
40
0.0593
284.3
1.024
50
0.0485
292.0
1.031
50
0.0617
293.6
1.053
60
0.0505
301.5
1.060
60
0.0641
301.0
1.081
70
0.0524
311.1
1.088
70
0.0664
312.4
1.109
80
0.0543
320.8
1.116
80
0.0687
322.0
1.137
90
0.0562
330.6
1.144
90
0.0710
331.7
1.164
100
0.0583
340.5
1.171
100
0.0735
341.6
1.191
110
0.0600
350.6
1.197
110
0.0755
351.5
1.217
120
0.0617
360.7
1.223
120
0.0777
361.6
1.243
130
0.0635
371.0
1.249
140
0.0653
381.5
1.275
P=600 kPa (Tsat 21.6°C)
P=700 kPa (Tsat 26.7°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0343
262.43
0.922
Sat.
0.0294
265.05
0.920
30
0.0360
270.8
0.950
30
0.0300
268.45
0.931
40
0.0379
280.6
0.982
40
0.0317
278.58
0.964
50
0.0397
290.3
1.012
50
0.0333
288.53
0.995
60
0.0414
300.0
1.042
60
0.0349
298.43
1.026
70
0.0431
309.7
1.071
70
0.0364
308.33
1.055
80
0.0447
319.6
1.099
80
0.0379
318.28
1.084
90
0.0463
329.5
1.126
90
0.0393
328.3
1.111
100
0.0479
339.5
1.154
100
0.0406
338.4
1.139
110
0.0495
349.6
1.180
110
0.0420
348.6
1.166
120
0.0510
359.8
1.207
120
0.0434
358.91
1.192
130
0.0525
370.2
1.233
130
0.0447
369.32
1.219
140
0.0540
380.7
1.258
140
0.0460
379.86
1.244
P=800 kPa (Tsat 31.3°C)
P=900 kPa (Tsat 35.5°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0256
267.3
0.918
Sat.
0.0227
269.3
0.917
40
0.0270
276.5
0.948
40
0.0234
274.2
0.933
50
0.0285
286.7
0.980
50
0.0248
284.8
0.966
60
0.0300
296.8
1.011
60
0.0261
295.1
0.998
70
0.0313
306.9
1.041
70
0.0274
305.4
1.028
80
0.0327
317.0
1.070
80
0.0286
<315.6/td>
1.057
90
0.0339
327.1
1.098
90
0.0298
325.9
1.086
100
0.0352
337.3
1.126
100
0.0310
336.2
1.114
110
0.0364
347.6
1.153
110
0.0321
346.6
1.141
120
0.0376
358.0
1.180
120
0.0332
357.0
1.168
130
0.0388
368.5
1.206
130
0.0342
367.6
1.195
140
0.0400
379.1
1.232
140
0.0353
378.2
1.221
150
0.0411
389.8
1.258
150
0.0363
389.0
1.247
160
0.0423
400.6
1.283
160
0.0374
399.9
1.272
P=1,000 kPa (Tsat 39.4°C)
P=1,200 kPa (Tsat 46.3°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0203
271.0
0.916
Sat.
0.0167
273.9
0.913
40
0.0204
271.7
0.918
50
0.0172
278.3
0.927
50
0.0218
282.7
0.953
60
0.0184
289.6
0.961
60
0.0231
293.4
0.985
70
0.0195
300.6
0.994
70
0.0243
303.9
1.016
80
0.0205
311.4
1.025
80
0.0254
314.3
1.046
90
0.0215
322.1
1.056
90
0.0265
324.7
1.075
100
0.0224
332.7
1.084
100
0.0276
335.1
1.103
110
0.0233
343.4
1.112
110
0.0286
345.5
1.131
120
0.0242
354.1
1.139
120
0.0296
356.1
1.158
130
0.0251
364.9
1.166
130
0.0306
366.7
1.185
140
0.0259
375.7
1.193
150
0.0325
388.2
1.237
160
0.0276
397.7
1.245
160
0.0335
399.2
1.262
170
0.0284
408.8
1.270
180
0.0292
420.1
1.295
P=1400 kPa (Tsat 52.4°C)
P=1600 kPa (Tsat 57.9°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0141
276.2
0.911
Sat.
0.0121
277.9
0.908
60
0.0150
285.5
0.939
60
0.0124
280.7
0.916
70
0.0161
297.1
0.973
70
0.0134
293.3
0.954
80
0.0170
308.3
1.006
80
0.0144
305.1
0.987
90
0.0179
319.4
1.036
90
0.0152
316.5
1.019
100
0.0188
330.3
1.066
100
0.0160
327.8
1.050
110
0.0196
341.2
1.095
110
0.0168
338.9
1.080
120
0.0204
352.1
1.123
120
0.0175
350.0
1.108
130
0.0212
363.0
1.150
130
0.0182
361.1
1.136
140
0.0219
374.0
1.177
140
0.0189
372.3
1.163
150
0.0226
385.1
1.204
150
0.0195
383.5
1.190
160
0.0234
396.2
1.230
160
0.0202
394.7
1.216
170
0.0241
407.4
1.255
170
0.0208
406.0
1.242
180
0.0247
418.8
1.281
180
0.0215
417.4
1.268
P=1800 kPa (Tsat 62.9°C)
P=2000 kPa (Tsat 67.5°C)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
Temperature
Volume
Enthalpy (kJ/kg)
Entropy (kJ/kg.K)
°C
m^3/kg
kJ/kg
kJ/kg.K
°C
m^3/kg
kJ/kg
kJ/kg.K
Sat.
0.0106
279.2
0.905
Sat.
0.0093
280.1
0.902
70
0.0113
288.9
0.934
70
0.0096
283.9
0.913
80
0.0123
301.5
0.970
80
0.0105
297.6
0.952
90
0.0131
313.5
1.003
90
0.0114
310.2
0.988
100
0.0139
325.1
1.035
100
0.0121
322.3
1.020
110
0.0146
336.5
1.065
110
0.0128
334.1
1.052
120
0.0152
347.9
1.094
120
0.0134
345.7
1.081
130
0.0159
359.2
1.123
130
0.0141
357.2
1.110
140
0.0165
370.5
1.150
140
0.0146
368.6
1.138
150
0.0171
381.6
1.177
150
0.0152
380.1
1.166
160
0.0177
393.2
1.204
160
0.0158
391.6
1.193
170
0.0183
404.6
1.230
170
0.0163
403.1
1.219
180
0.0189
416.1
1.256
180
0.0168
414.8
1.245
Thermodynamic Properties of Saturated Refrigerant R134a
Pressure | Temperature | Specific volume (m^3/kg) | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
kPa | °C | Sat Liqvf | Sat Vapvg | Sat Liqhf | Sat Vaphg | Sat Liqsf | Sat Vapsg
60 | -36.9 | 0.0007098 | 0.3112 | 3.9 | 227.8 | 0.0164 | 0.9645
80 | -31.1 | 0.0007185 | 0.2376 | 11.3 | 231.5 | 0.0472 | 0.9572
100 | -26.4 | 0.0007259 | 0.1926 | 17.3 | 234.5 | 0.0720 | 0.9519
120 | -22.3 | 0.0007324 | 0.1621 | 22.5 | 237.0 | 0.0928 | 0.9478
140 | -18.8 | 0.0007383 | 0.1402 | 27.1 | 239.2 | 0.1110 | 0.9446
160 | -15.6 | 0.0007437 | 0.1235 | 31.2 | 241.1 | 0.1270 | 0.9420
180 | -12.7 | 0.0007487 | 0.1104 | 35.0 | 242.9 | 0.1415 | 0.9397
200 | -10.1 | 0.0007534 | 0.0999 | 38.5 | 244.5 | 0.1547 | 0.9378
220 | -7.6 | 0.0007578 | 0.0912 | 41.7 | 245.9 | 0.1668 | 0.9361
240 | -5.4 | 0.0007620 | 0.0839 | 44.7 | 247.3 | 0.1780 | 0.9347
260 | -3.2 | 0.0007661 | 0.0777 | 47.5 | 248.6 | 0.1885 | 0.9333
280 | -1.2 | 0.0007699 | 0.0724 | 50.2 | 249.7 | 0.1984 | 0.9322
300 | 0.7 | 0.0007737 | 0.0677 | 52.8 | 250.9 | 0.2077 | 0.9311
320 | 2.5 | 0.0007773 | 0.0636 | 55.2 | 251.9 | 0.2165 | 0.9301
340 | 4.2 | 0.0007808 | 0.0600 | 57.5 | 252.9 | 0.2248 | 0.9293
360 | 5.8 | 0.0007842 | 0.0567 | 59.8 | 253.8 | 0.2328 | 0.9284
400 | 8.9 | 0.0007907 | 0.0512 | 64.0 | 255.6 | 0.2477 | 0.9270
500 | 15.7 | 0.0008060 | 0.0411 | 73.4 | 259.3 | 0.2803 | 0.9241
600 | 21.6 | 0.0008200 | 0.0343 | 81.5 | 262.4 | 0.3081 | 0.9219
700 | 26.7 | 0.0008332 | 0.0294 | 88.8 | 265.1 | 0.3324 | 0.9200
800 | 31.3 | 0.0008459 | 0.0256 | 95.5 | 267.3 | 0.3541 | 0.9184
900 | 35.5 | 0.0008581 | 0.0227 | 101.6 | 269.3 | 0.3739 | 0.9170
1000 | 39.4 | 0.0008701 | 0.0203 | 107.4 | 271.0 | 0.3920 | 0.9157
1200 | 46.3 | 0.0008935 | 0.0167 | 117.8 | 273.9 | 0.4245 | 0.9131
1400 | 52.4 | 0.0009167 | 0.0141 | 127.3 | 276.2 | 0.4533 | 0.9106
1600 | 57.9 | 0.0009401 | 0.0121 | 136.0 | 277.9 | 0.4792 | 0.9080
1800 | 62.9 | 0.0009640 | 0.0106 | 144.1 | 279.2 | 0.5031 | 0.9051
2000 | 67.5 | 0.0009888 | 0.0093 | 151.8 | 280.1 | 0.5252 | 0.9020
2500 | 77.6 | 0.0010569 | 0.0069 | 169.7 | 280.9 | 0.5755 | 0.8925
3000 | 86.2 | 0.0011413 | 0.0053 | 186.6 | 279.2 | 0.6215 | 0.8792
Thermodynamic Properties of Superheated Refrigerant R134a
P=60 kPa (Tsat -36.9°C) | | P=100 kPa (Tsat -26.4°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.3112 | 227.8 | 0.964 | | Sat. | 0.1926 | 234.5 | 0.952
-20 | 0.3361 | 240.8 | 1.018 | | -20 | 0.1984 | 239.5 | 0.972
-10 | 0.3505 | 248.6 | 1.048 | | -10 | 0.2074 | 247.5 | 1.003
0 | 0.3648 | 256.5 | 1.077 | | 0 | 0.2163 | 255.6 | 1.033
10 | 0.3789 | 264.7 | 1.107 | | 10 | 0.2251 | 263.8 | 1.063
20 | 0.3930 | 272.9 | 1.135 | | 20 | 0.2337 | 272.2 | 1.092
30 | 0.4071 | 281.4 | 1.164 | | 30 | 0.2423 | 280.7 | 1.120
40 | 0.4210 | 290.0 | 1.192 | | 40 | 0.2509 | 289.3 | 1.149
50 | 0.4350 | 298.7 | 1.219 | | 50 | 0.2594 | 298.2 | 1.176
60 | 0.4488 | 307.7 | 1.246 | | 60 | 0.2678 | 307.1 | 1.204
70 | 0.4627 | 316.8 | 1.273 | | 70 | 0.2763 | 316.3 | 1.231
80 | 0.4765 | 326.0 | 1.300 | | 80 | 0.2847 | 325.6 | 1.257
90 | 0.4903 | 335.4 | 1.326 | | 90 | 0.2930 | 335.0 | 1.284
100 | 0.5041 | 345.0 | 1.352 | | 100 | 0.3014 | 344.6 | 1.310
P=140 kPa (Tsat -18.8°C) | | P=180 kPa (Tsat -12.7°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.1402 | 239.18 | 0.94462 | | Sat. | 0.1104 | 242.9 | 0.940
-10 | 0.1461 | 246.4 | 0.972 | | -10 | 0.1119 | 245.2 | 0.948
0 | 0.1526 | 254.6 | 1.003 | | 0 | 0.1172 | 253.6 | 0.980
10 | 0.1591 | 262.9 | 1.033 | | 10 | 0.1224 | 262.0 | 1.010
20 | 0.1654 | 271.4 | 1.062 | | 20 | 0.1275 | 270.6 | 1.040
30 | 0.1717 | 280.0 | 1.091 | | 30 | 0.1325 | 279.3 | 1.069
40 | 0.1780 | 288.7 | 1.120 | | 40 | 0.1374 | 288.1 | 1.098
50 | 0.1841 | 297.6 | 1.147 | | 50 | 0.1423 | 297.0 | 1.126
60 | 0.1903 | 306.6 | 1.175 | | 60 | 0.1472 | 306.1 | 1.153
70 | 0.1964 | 315.8 | 1.202 | | 70 | 0.1520 | 315.3 | 1.181
80 | 0.2024 | 325.1 | 1.229 | | 80 | 0.1567 | 324.6 | 1.207
90 | 0.2085 | 334.6 | 1.255 | | 90 | 0.1615 | 334.1 | 1.234
100 | 0.2145 | 344.2 | 1.282 | | 100 | 0.1662 | 343.8 | 1.260
P=200 kPa (Tsat -10.1°C) | | P=240 kPa (Tsat -5.4°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0999 | 244.5 | 0.938 | | Sat. | 0.0839 | 247.3 | 0.935
0 | 0.1048 | 253.1 | 0.970 | | 0 | 0.0862 | 252.0 | 0.952
10 | 0.1096 | 261.6 | 1.001 | | 10 | 0.0903 | 260.7 | 0.983
20 | 0.1142 | 270.2 | 1.030 | | 20 | 0.0942 | 269.4 | 1.013
30 | 0.1187 | 278.9 | 1.060 | | 30 | 0.0981 | 278.2 | 1.043
40 | 0.1232 | 287.7 | 1.088 | | 40 | 0.1019 | 287.1 | 1.072
50 | 0.1277 | 296.7 | 1.116 | | 50 | 0.1057 | 296.1 | 1.100
60 | 0.1321 | 305.8 | 1.144 | | 60 | 0.1094 | 305.2 | 1.128
70 | 0.1364 | 315.0 | 1.171 | | 70 | 0.1131 | 314.5 | 1.156
80 | 0.1407 | 324.4 | 1.198 | | 80 | 0.1168 | 323.9 | 1.183
90 | 0.1451 | 333.9 | 1.225 | | 90 | 0.1204 | 333.5 | 1.209
100 | 0.1493 | 343.6 | 1.251 | | 100 | 0.1240 | 343.2 | 1.236
P=280 kPa (Tsat -1.2°C) | | P=320 kPa (Tsat 2.5°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0724 | 249.7 | 0.932 | | Sat. | 0.0636 | 251.9 | 0.930
10 | 0.0765 | 259.7 | 0.968 | | 10 | 0.0661 | 258.7 | 0.954
20 | 0.0800 | 268.5 | 0.999 | | 20 | 0.0693 | 267.7 | 0.986
30 | 0.0834 | 277.4 | 1.029 | | 30 | 0.0723 | 276.7 | 1.016
40 | 0.0867 | 286.4 | 1.058 | | 40 | 0.0753 | 285.7 | 1.045
50 | 0.0900 | 295.5 | 1.086 | | 50 | 0.0782 | 294.9 | 1.074
60 | 0.0932 | 304.7 | 1.114 | | 60 | 0.0811 | 304.1 | 1.102
70 | 0.0964 | 314.0 | 1.142 | | 70 | 0.0839 | 313.5 | 1.130
80 | 0.0996 | 323.5 | 1.169 | | 80 | 0.0868 | 323.0 | 1.157
90 | 0.1028 | 333.1 | 1.196 | | 90 | 0.0895 | 332.6 | 1.184
110 | 0.1090 | 352.7 | 1.248 | | 110 | 0.0950 | 352.3 | 1.237
120 | 0.1121 | 362.7 | 1.274 | | 120 | 0.0978 | 362.4 | 1.263
P=400 kPa (Tsat 8.9°C) | | P=500 kPa (Tsat 15.7°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0512 | 255.6 | 0.927 | | Sat. | 0.0411 | 259.3 | 0.924
10 | 0.0515 | 256.6 | 0.931 | | 20 | 0.0421 | 263.5 | 0.938
20 | 0.0542 | 265.9 | 0.963 | | 30 | 0.0443 | 273.0 | 0.970
30 | 0.0568 | 275.1 | 0.994 | | 40 | 0.0465 | 282.5 | 1.001
40 | 0.0593 | 284.3 | 1.024 | | 50 | 0.0485 | 292.0 | 1.031
50 | 0.0617 | 293.6 | 1.053 | | 60 | 0.0505 | 301.5 | 1.060
60 | 0.0641 | 301.0 | 1.081 | | 70 | 0.0524 | 311.1 | 1.088
70 | 0.0664 | 312.4 | 1.109 | | 80 | 0.0543 | 320.8 | 1.116
80 | 0.0687 | 322.0 | 1.137 | | 90 | 0.0562 | 330.6 | 1.144
90 | 0.0710 | 331.7 | 1.164 | | 100 | 0.0583 | 340.5 | 1.171
100 | 0.0735 | 341.6 | 1.191 | | 110 | 0.0600 | 350.6 | 1.197
110 | 0.0755 | 351.5 | 1.217 | | 120 | 0.0617 | 360.7 | 1.223
120 | 0.0777 | 361.6 | 1.243 | | 130 | 0.0635 | 371.0 | 1.249
| | | | | 140 | 0.0653 | 381.5 | 1.275
P=600 kPa (Tsat 21.6°C) | | P=700 kPa (Tsat 26.7°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0343 | 262.43 | 0.922 | | Sat. | 0.0294 | 265.05 | 0.920
30 | 0.0360 | 270.8 | 0.950 | | 30 | 0.0300 | 268.45 | 0.931
40 | 0.0379 | 280.6 | 0.982 | | 40 | 0.0317 | 278.58 | 0.964
50 | 0.0397 | 290.3 | 1.012 | | 50 | 0.0333 | 288.53 | 0.995
60 | 0.0414 | 300.0 | 1.042 | | 60 | 0.0349 | 298.43 | 1.026
70 | 0.0431 | 309.7 | 1.071 | | 70 | 0.0364 | 308.33 | 1.055
80 | 0.0447 | 319.6 | 1.099 | | 80 | 0.0379 | 318.28 | 1.084
90 | 0.0463 | 329.5 | 1.126 | | 90 | 0.0393 | 328.3 | 1.111
100 | 0.0479 | 339.5 | 1.154 | | 100 | 0.0406 | 338.4 | 1.139
110 | 0.0495 | 349.6 | 1.180 | | 110 | 0.0420 | 348.6 | 1.166
120 | 0.0510 | 359.8 | 1.207 | | 120 | 0.0434 | 358.91 | 1.192
130 | 0.0525 | 370.2 | 1.233 | | 130 | 0.0447 | 369.32 | 1.219
140 | 0.0540 | 380.7 | 1.258 | | 140 | 0.0460 | 379.86 | 1.244
P=800 kPa (Tsat 31.3°C) | | P=900 kPa (Tsat 35.5°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0256 | 267.3 | 0.918 | | Sat. | 0.0227 | 269.3 | 0.917
40 | 0.0270 | 276.5 | 0.948 | | 40 | 0.0234 | 274.2 | 0.933
50 | 0.0285 | 286.7 | 0.980 | | 50 | 0.0248 | 284.8 | 0.966
60 | 0.0300 | 296.8 | 1.011 | | 60 | 0.0261 | 295.1 | 0.998
70 | 0.0313 | 306.9 | 1.041 | | 70 | 0.0274 | 305.4 | 1.028
80 | 0.0327 | 317.0 | 1.070 | | 80 | 0.0286 | <315.6/td>1.057 | 1.057
90 | 0.0339 | 327.1 | 1.098 | | 90 | 0.0298 | 325.9 | 1.086
100 | 0.0352 | 337.3 | 1.126 | | 100 | 0.0310 | 336.2 | 1.114
110 | 0.0364 | 347.6 | 1.153 | | 110 | 0.0321 | 346.6 | 1.141
120 | 0.0376 | 358.0 | 1.180 | | 120 | 0.0332 | 357.0 | 1.168
130 | 0.0388 | 368.5 | 1.206 | | 130 | 0.0342 | 367.6 | 1.195
140 | 0.0400 | 379.1 | 1.232 | | 140 | 0.0353 | 378.2 | 1.221
150 | 0.0411 | 389.8 | 1.258 | | 150 | 0.0363 | 389.0 | 1.247
160 | 0.0423 | 400.6 | 1.283 | | 160 | 0.0374 | 399.9 | 1.272
P=1,000 kPa (Tsat 39.4°C) | | P=1,200 kPa (Tsat 46.3°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0203 | 271.0 | 0.916 | | Sat. | 0.0167 | 273.9 | 0.913
40 | 0.0204 | 271.7 | 0.918 | | 50 | 0.0172 | 278.3 | 0.927
50 | 0.0218 | 282.7 | 0.953 | | 60 | 0.0184 | 289.6 | 0.961
60 | 0.0231 | 293.4 | 0.985 | | 70 | 0.0195 | 300.6 | 0.994
70 | 0.0243 | 303.9 | 1.016 | | 80 | 0.0205 | 311.4 | 1.025
80 | 0.0254 | 314.3 | 1.046 | | 90 | 0.0215 | 322.1 | 1.056
90 | 0.0265 | 324.7 | 1.075 | | 100 | 0.0224 | 332.7 | 1.084
100 | 0.0276 | 335.1 | 1.103 | | 110 | 0.0233 | 343.4 | 1.112
110 | 0.0286 | 345.5 | 1.131 | | 120 | 0.0242 | 354.1 | 1.139
120 | 0.0296 | 356.1 | 1.158 | | 130 | 0.0251 | 364.9 | 1.166
130 | 0.0306 | 366.7 | 1.185 | | 140 | 0.0259 | 375.7 | 1.193
150 | 0.0325 | 388.2 | 1.237 | | 160 | 0.0276 | 397.7 | 1.245
160 | 0.0335 | 399.2 | 1.262 | | 170 | 0.0284 | 408.8 | 1.270
| | | | | 180 | 0.0292 | 420.1 | 1.295
P=1400 kPa (Tsat 52.4°C) | | P=1600 kPa (Tsat 57.9°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0141 | 276.2 | 0.911 | | Sat. | 0.0121 | 277.9 | 0.908
60 | 0.0150 | 285.5 | 0.939 | | 60 | 0.0124 | 280.7 | 0.916
70 | 0.0161 | 297.1 | 0.973 | | 70 | 0.0134 | 293.3 | 0.954
80 | 0.0170 | 308.3 | 1.006 | | 80 | 0.0144 | 305.1 | 0.987
90 | 0.0179 | 319.4 | 1.036 | | 90 | 0.0152 | 316.5 | 1.019
100 | 0.0188 | 330.3 | 1.066 | | 100 | 0.0160 | 327.8 | 1.050
110 | 0.0196 | 341.2 | 1.095 | | 110 | 0.0168 | 338.9 | 1.080
120 | 0.0204 | 352.1 | 1.123 | | 120 | 0.0175 | 350.0 | 1.108
130 | 0.0212 | 363.0 | 1.150 | | 130 | 0.0182 | 361.1 | 1.136
140 | 0.0219 | 374.0 | 1.177 | | 140 | 0.0189 | 372.3 | 1.163
150 | 0.0226 | 385.1 | 1.204 | | 150 | 0.0195 | 383.5 | 1.190
160 | 0.0234 | 396.2 | 1.230 | | 160 | 0.0202 | 394.7 | 1.216
170 | 0.0241 | 407.4 | 1.255 | | 170 | 0.0208 | 406.0 | 1.242
180 | 0.0247 | 418.8 | 1.281 | | 180 | 0.0215 | 417.4 | 1.268
P=1800 kPa (Tsat 62.9°C) | | P=2000 kPa (Tsat 67.5°C)
Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K) | | Temperature | Volume | Enthalpy (kJ/kg) | Entropy (kJ/kg.K)
°C | m^3/kg | kJ/kg | kJ/kg.K | | °C | m^3/kg | kJ/kg | kJ/kg.K
Sat. | 0.0106 | 279.2 | 0.905 | | Sat. | 0.0093 | 280.1 | 0.902
70 | 0.0113 | 288.9 | 0.934 | | 70 | 0.0096 | 283.9 | 0.913
80 | 0.0123 | 301.5 | 0.970 | | 80 | 0.0105 | 297.6 | 0.952
90 | 0.0131 | 313.5 | 1.003 | | 90 | 0.0114 | 310.2 | 0.988
100 | 0.0139 | 325.1 | 1.035 | | 100 | 0.0121 | 322.3 | 1.020
110 | 0.0146 | 336.5 | 1.065 | | 110 | 0.0128 | 334.1 | 1.052
120 | 0.0152 | 347.9 | 1.094 | | 120 | 0.0134 | 345.7 | 1.081
130 | 0.0159 | 359.2 | 1.123 | | 130 | 0.0141 | 357.2 | 1.110
140 | 0.0165 | 370.5 | 1.150 | | 140 | 0.0146 | 368.6 | 1.138
150 | 0.0171 | 381.6 | 1.177 | | 150 | 0.0152 | 380.1 | 1.166
160 | 0.0177 | 393.2 | 1.204 | | 160 | 0.0158 | 391.6 | 1.193
170 | 0.0183 | 404.6 | 1.230 | | 170 | 0.0163 | 403.1 | 1.219
180 | 0.0189 | 416.1 | 1.256 | | 180 | 0.0168 | 414.8 | 1.245
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3 COMMENTS
Hello Paul
I have enjoyed your youtube channel very much, thanks. I wander if you had any plans to publish the R410A thermodynamic property tables? I have found the R134A calculations very interesting and I wanted to do the same with R410A since I work with this with VRV and Split systems. I was looking to run through the same calculations and compare the theoretical COP and cooling loads etc with the real experimental data.
Thanks
Will
Thank you so much and I have one doubt that what is enthalphy where it was used
Hello paul,
Thank you because of your complete description For full description
I have a qaustion ,why do you start with 320 kpa Pressure in step one of r134a refrigeration ?
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9158 | https://mechanicsmap.psu.edu/websites/13_newtons_rigid_body/13-4_multi_body_motion/multi_body_motion.html | Multi-Body Kinetics
In cases where multiple connected rigid bodies are undergoing some sort of motion, we can extend our analysis of general planar motion to this multi-body situation. In these cases, which we will call multi-body kinetics problems, we will analyze each body independently as we did for general planar motion, but we will also need to pay attention to the Newton's Third Law pairs. Each body will have forces and/or moments exerted on it by the surrounding bodies, and it will exert equal and opposite forces/moments back through those same connections.
To analyze a multi-body system, we will start by drawing a free body diagram of each body in motion. Be sure to identify the center of mass, as well as identifying all known and unknown forces, and known and unknown moments acting on the body. When drawing forces and moments at connection points, be sure to make the forces and moments equal and opposite on the connected body to satisfy Newton's Third Law. It is also sometimes helpful to label any key dimensions as well as using dashed lines to identify any known accelerations or angular accelerations. Often, you will need to solve a kinematics problem using absolute motion analysis or relative motion analysis in order to determine the accelerations of the centers of mass and the angular accelerations for each body. Make sure all these accelerations are with respect to ground.
Next we move onto identifying the equations of motion for each body in the system. In two dimensions, we will use the same three equations we used for general planar motion. Be sure to find the the accelerations of all the centers of masses, find all moments about the center of mass, and take the mass moments of inertia about the center of mass of each body.
| |
| [\sum F_{x}=ma_{x}] |
| [\sum F_{y}=ma_{y}] |
| [\sum M_{G}=I_{G}\alpha] |
or (taking moments about the ICZV) ||| |
| [\sum \vec{M}_{IC}=I_{IC}\vec{\alpha}] |
or (taking moments about any point, P, on the body) ||| |
| [ \sum \vec{M}_{P}=I_{G}\vec{\alpha}+ \vec{r}_{G/P} \times m \vec{a}_G ] |
or ||| |
| [ \sum \vec{M}_{P}=I_{P}\vec{\alpha}+ \vec{r}_{G/P} \times m \vec{a}_P ] |
Plugging the known forces, moments, and accelerations into the above equations we can solve for up to three unknowns per body. If more than three unknowns exist in any one set of equations, you will need to start with an adjacent body. Once unknown forces and moments are determined on one body, they can become knowns on the connected body, reducing the number of unknowns to solve for.
Video Lecture
Worked Problems:
Question 1:
A robotic arm has two sections (OA and AB) with section OA having a mass of 10 kg and section AB having a mass of 7 kg. Treat each section as a slender rod. If we wish to accelerate member AB from a standstill at a rate of 3 rad/s2 and keep the right section stationary, what moments must we exert at joints O and A?
Solution:
Question 2:
An engineering student is testing a component of her vehicle for a design competition. The 5kg rectangular plate is pinned to a carriage at P. If the track is given an acceleration of 2m/s^2, determine the reaction forces at P and the angular acceleration of the plate. The height of the plate is h=2y and G is located a vertical distance y=0.8m from P. The plate has a length l=2m and point P is a horizontal distance x=0.6m from the edge.
Solution: |
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In a convex hexagon two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon is
A
5/12
B
7/12
C
2/5
D
none of these
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Text Solution
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The correct Answer is:A
To find the probability that two randomly drawn diagonals of a convex hexagon intersect at an interior point, we can follow these steps: 1. Identify the number of vertices in the hexagon: A convex hexagon has 6 vertices. 2. Calculate the number of diagonals in the hexagon: The formula for the number of diagonals in an n-sided polygon is given by: Number of diagonals=(n2)−n For a hexagon (n = 6): Number of diagonals=(62)−6 Calculating (62): (62)=6!2!(6−2)!=6×52×1=15 Therefore, the number of diagonals is: 15−6=9 3. Determine the total ways to choose 2 diagonals: The total ways to choose 2 diagonals from the 9 diagonals is given by: (92) Calculating (92): (92)=9!2!(9−2)!=9×82×1=36 4. Find the number of ways to choose 4 vertices: For two diagonals to intersect at an interior point, we need to choose 4 vertices from the 6 vertices of the hexagon. The number of ways to choose 4 vertices is: (64) Since (64)=(62), we already calculated: (64)=15 5. Calculate the probability: The probability that two diagonals intersect at an interior point is given by the ratio of the number of favorable outcomes to the total outcomes: P(intersect)=Number of ways to choose 4 verticesTotal ways to choose 2 diagonals=1536 Simplifying 1536: P(intersect)=512 Final Answer:The probability that the diagonals intersect at an interior point of the hexagon is 512. ---
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9160 | https://www.youtube.com/watch?v=jVQ2NS_7sYA | Geometry problems #7 (Radius of circumcircle)
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A problem on using the Law of Cosines to determine the radius of the circumcircle, given the side lengths of the triangle.
Transcript:
here's a geometry problem that might also give you some practice with algebra pause the video and give it a try okay here's one way to solve the problem we're trying to find the radius of this circle and the circle has a triangle inscribed in it now when you have angles inscribed in circles a useful theorem is the inscribed angle theorem for example if we pick this angle alpha then we know that the arc opposing it is 2 alpha in degrees which is the same thing as saying that this angle here is twice the angle alpha and that's useful because this triangle here is an isosceles triangle and both sides of it have length r since it's a circle of radius r so we have two triangles now that share the side length 15 and we know the angles being opposed alpha and 2 alpha so what we're going to want to use is the law of cosines so we're going to use it twice both for 15 here so this side length 15 can be made into a triangle using this big one or using this smaller one and we're going to do them both so first the big one the big one the law of cosines says the opposing side squared is equal to the sum of the squares of the two sides and if it were a right triangle that would be it but since it's not a right triangle we need to subtract off two times the side lengths times cosine of alpha that's for the bigger triangle now for the smaller one which has two alpha involved we again have 15 squared but now the opposing sides are these ones just the radius of the circle each one and so it's just r squared plus r squared minus 2 r squared cosine and now 2 alpha and so we have two equations for two unknowns alpha and r and it looks to me like the easiest way to get r which is what we're asked for is going to be first to solve the top one to get alpha so from this one we have 15 squared minus 10 squared minus 12 squared divided by minus 2 times 10 times 12 is equal to cosine alpha so alpha is just the arc cosine of this stuff in here and if you work it out and plug it into your calculator it comes out to about 1.5 radians so that's alpha and then from the bottom equation now that we know alpha we can just solve it for r so from this equation i see that the left hand side becomes 2 r squared times 1 minus cosine 2 alpha but alpha we know is 1.5 and that's all equal to 15 squared so finally we can solve for r squared is 15 squared over 2 times 1 minus cosine of approximately 3 radians or r is 15 over the square root of 2 1 minus cosine 3. and that's 3 radians and when you plug that into your calculator you'll find it's approximately 7 point |
9161 | https://middleschoolinthemitten.com/?p=1139 | A few years ago, I realized that my 7 th and 8 th grade students
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had to the work we put in to make referencing the page a routine in
our classroom. I hope this resource helps your students as much as it’s
helped mine! A few years ago, I realized that my 7 th and 8 th grade students
were lacking automaticity with integer rules, which was making pre -
algebra topics difficult and frustrating for them and for me! When my
students saw a negative sign in a problem, they either ignored it,
assumed their answer was negative, or wouldn’t even attempt the
problem. Maybe you are seeing similar behaviors in your classroom, too.
My coteacher and I quickly realized that we needed something
to help, so I created this simple reference page! I copied it on colorful
cardstock and added a multiplication chart on the back, since many of
my students didn’t have their math facts memorized either! (Have the
same issue? Here’s a link to the multiplication chart freebie that I use !)
In our intervention class, we passed out the note pages and filled in the
blanks together. We added examples in the extra space around the
rules. Each day, we reminded students to get that page out to help him,
no matter what topic we were working on. We modeled checking the
rules for integer problems and asked students to get out the pages
before we helped them during independent work time.
Over the next few weeks, students were using the pages all
the time! Even better, some of them were starting to remember the
rules and applying them independently. Students loved having a
reference page to look at and started gaining confidence in integer
operations. At the end of the year, many students asked if they could
keep the note page to help them in high school. Of course, the answer
was YES!
Now, my coteacher and I start the school year with these
reference pages in all of our classes. While I love using these as full
page reference charts for students, I’ve also enlarged them as anchor
chart posters or printed them several to a page for a pint sized
version! No matter the size you use, I attribute the success we have
had to the work we put in to make referencing the page a routine in
our classroom. I hope this resource helps your students as much as it’s
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School in the Mitten was to share engaging and
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Tag me on Instagram! – I love seeing my
products in classrooms around the world! After
using these pages in your classroom, snap a
photo and tag @middleschoolinthemitten! I can’t
wait to see!
Lastly, check out my website for more teacher
time savers, tips, and resources! While I focus on math and math intervention, I am still a special education
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By May 26, 2024, STUDENT will add, subtract, multiply, and divide rational
numbers with 75% accuracy as measured by samples of class work or
informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide integers
with 80% accuracy when given a multiplication chart as measured by
samples of classwork or informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide
fractions with 75% accuracy as measured by samples of classwork or
informal assessments.
Sample IEP Goal #2
By May 26, 2024, STUDENT will use inverse operations to solve two step
equations with 75% accuracy as measured by samples of class work or
informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide integers
with 80% accuracy when given a multiplication or reference chart as
measured by samples of classwork or informal assessments.
By May 26, 2024, STUDENT will use inverse operations to solve one step
equations with integers with 75% accuracy as measured by samples of
classwork or informal assessments.
Helpful Accommodations
Notes Provided: Simply print out the completed note page for students.
Use of Reference Page/Notes on Assignments/Assessments – Provides
students with a multiplication chart or integer rule page to help ensure
accuracy while solving problems on assessments or class work.
Extended time – provides students with extra time to use their integer
rule page as an accommodation. While I focus on math and math intervention, I am still a special education
teacher! Below are some sample IEP goals and IEP goal objectives that I
support with this resource, as well as some accommodations that will
benefit students with learning difficulties.
Sample IEP Goal #1
By May 26, 2024, STUDENT will add, subtract, multiply, and divide rational
numbers with 75% accuracy as measured by samples of class work or
informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide integers
with 80% accuracy when given a multiplication chart as measured by
samples of classwork or informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide
fractions with 75% accuracy as measured by samples of classwork or
informal assessments.
Sample IEP Goal #2
By May 26, 2024, STUDENT will use inverse operations to solve two step
equations with 75% accuracy as measured by samples of class work or
informal assessments.
By May 26, 2024, STUDENT will add, subtract, multiply, and divide integers
with 80% accuracy when given a multiplication or reference chart as
measured by samples of classwork or informal assessments.
By May 26, 2024, STUDENT will use inverse operations to solve one step
equations with integers with 75% accuracy as measured by samples of
classwork or informal assessments.
Helpful Accommodations
Notes Provided: Simply print out the completed note page for students.
Use of Reference Page/Notes on Assignments/Assessments – Provides
students with a multiplication chart or integer rule page to help ensure
accuracy while solving problems on assessments or class work.
Extended time – provides students with extra time to use their integer
rule page as an accommodation. |
9162 | https://brainly.com/question/29612049 | [FREE] Write a balanced half-reaction for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic - brainly.com
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Write a balanced half-reaction for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
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To determine the number of atoms of each element we need to multiply stoichiometry that is written in front of the molecule to the number that is written on the foot of the element. Hence, the balanced chemical equation for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic aqueous solution is NO₂ (g) + H₂O (l) → NO³⁻ (aq) + 2H⁺ + e⁻.
Balanced chemical equation
Balanced equation is defined as the reaction where the number of atoms of each species is same on reactant and product side. In balanced equation mass can neither be destroyed nor be created.
The skeletal equation can be written as:
NO₂ (g) + H₂O (l) → NO³⁻ (aq) + 2H⁺ + e⁻.
The number atoms of N on reactant and product side is 1 so N is balanced.
The number of atoms of H is 2 on reactant and on product side it is 2. So, H is balanced.
The number of atoms of O is 3 on reactant and on product side it is 3. So, O is balanced.
Hence, the balanced chemical equation for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic aqueous solution is:
NO₂ (g) + H₂O (l) → NO³⁻ (aq) + 2H⁺ + e⁻
Learn more about balanced chemical reaction from the link given below.
brainly.com/question/15133141
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The balanced half-reaction for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic solution is NO 3−(a q)+2 H+(a q)+2 e−→NO 2(g)+H 2O(l). The nitrogen atoms and charges are balanced by adding the appropriate number of water and hydrogen ions. This reaction illustrates the process of reduction in an acidic environment.
Explanation
To write a balanced half-reaction for the reduction of the nitrate ion (\text{NO}_3^-) to gaseous nitrogen dioxide (\text{NO}_2) in an acidic aqueous solution, we follow these steps:
Identify the species involved:
We have nitrate ion (\text{NO}_3^-) being reduced to nitrogen dioxide (\text{NO}_2).
Write the unbalanced half-reaction:
NO 3−→NO 2
Balance the nitrogen atoms:
There is one nitrogen atom on each side, so nitrogen is balanced.
Balance the oxygen atoms:
The reactant has three oxygen atoms, while the product has two oxygen atoms, so we need to add one water molecule to the product side:
NO 3−→NO 2+H 2O
Balance the hydrogen atoms:
We added two hydrogen atoms with water, so we will add two hydrogen ions (\text{H}^+) to the reactant side:
NO 3−+2 H+→NO 2+H 2O
Balance the charges:
The left side has a total charge of +1 (from 2H^+ and -1 from \text{NO}_3^-), and the right side has a charge of 0. To balance, we add two electrons to the left side:
NO 3−+2 H++2 e−→NO 2+H 2O
State the physical states:
\text{NO}_3^- is in aqueous solution (aq),
\text{NO}_2 is a gas (g),
\text{H}^+ is in aqueous solution (aq),
and \text{H}_2\text{O} is liquid (l).
Thus, the final balanced half-reaction is:
NO 3−(a q)+2 H+(a q)+2 e−→NO 2(g)+H 2O(l)
This reaction occurs in an acidic medium, making it essential for adequately balancing both mass and charge.
Examples & Evidence
For instance, when nitrate ions in acid react, they undergo reduction, showcasing how charge and mass balance is achieved through electrons and aqueous ions. An application of this reaction can be seen in processes like denitrification in the environment, where nitrates are converted back to nitrogen gas, influencing the nitrogen cycle.
The steps outlined in the balanced half-reaction follow fundamental principles of redox chemistry, which are standard and widely accepted in chemical education, particularly in studying reaction mechanisms and balancing chemical equations.
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To balance the reduction half-reaction for nitrate ion to nitrogen dioxide in acidic solution, balance each element and then charge, resulting in N O 3−(a q)+2 H+(a q)+e−→N O 2(g)+H 2O(l).
To write a balanced half-reaction for the reduction of nitrate ion to gaseous nitrogen dioxide in acidic solution, you should follow these steps:
Write the skeletal equation for the reduction process: N O 3−(a q)→N O 2(g).
Now, balance the oxygen atoms through addition of water molecules on the appropriate side of the derived equation: N O 3−(a q)→N O 2(g)+H 2O(l).
Then balance the (H) hydrogen atoms through addition of hydrogen ions (H+) on the appropriate side: N O 3−(a q)+H+→N O 2(g)+H 2O(l).
Finally, balance the charge by adding electrons (e−) to the side which is more positive: N O 3−(a q)+2 H++e−→N O 2(g)+H 2O(l).
The balanced half-reaction for the reduction of nitrate ion to nitrogen dioxide in acidic solution is:
N O 3−(a q)+2 H+(a q)+e−→N O 2(g)+H 2O(l)
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9163 | https://stats.stackexchange.com/questions/193990/approximate-e-using-monte-carlo-simulation | Approximate $e$ using Monte Carlo Simulation - Cross Validated
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Approximate e e using Monte Carlo Simulation
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I've been looking at Monte Carlo simulation recently, and have been using it to approximate constants such as π π (circle inside a rectangle, proportionate area).
However, I'm unable to think of a corresponding method of approximating the value of e e [Euler's number] using Monte Carlo integration.
Do you have any pointers on how this can be done?
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9 There are many, many, many ways to do this. That this is so might become evident by contemplating what the R command 2 + mean(exp(-lgamma(ceiling(1/runif(1e5))-1))) does. (If using the log Gamma function bothers you, replace it by 2 + mean(1/factorial(ceiling(1/runif(1e5))-2)), which uses only addition, multiplication, division, and truncation, and ignore the overflow warnings.) What might be of greater interest would be _efficient_ simulations: can you minimize the number of computational steps needed to estimate e e to any given accuracy? –whuber♦ Commented Feb 4, 2016 at 14:34
4 What a delightful question! I look forward to reading others' answers. One way that you could really draw attention to this question -- perhaps another half-dozen answers -- would be to revise the question and ask for efficient answers, as whuber suggests. That's like catnip for CV users. –Sycorax♦ Commented Feb 5, 2016 at 3:17
1 @EngrStudent I'm not sure the geometric analog exists for e e. It's simply not a naturally (pun intended) geometrical quantity like π π. –Aksakal Commented Feb 5, 2016 at 13:28
8 @Aksakal e e is an _exceptionally_ geometrical quantity. At the most elementary level it appears naturally in expressions for areas related to hyperbolas. At a slightly more advanced level it is intimately connected with all periodic functions, including trigonometric functions, whose geometric content is obvious. The real challenge here is that it just so _easy_ to simulate values related to e e! –whuber♦ Commented Feb 5, 2016 at 13:42
3 @StatsStudent: e e by itself is not interesting. However, if this leads to unbiased estimators of quantities like exp{∫x 0 f(y)d G(y)}exp{∫0 x f(y)d G(y)} this may prove most useful for MCMC algorithms. –Xi'an Commented Feb 12, 2016 at 18:58
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The simple and elegant way to estimate e e by Monte Carlo is described in this paper. The paper is actually about teaching e e. Hence, the approach seems perfectly fitting for your goal. The idea's based on an exercise from a popular Ukrainian textbook on probability theory by Gnedenko. See ex.22 on p.183
It happens so that E[ξ]=e E[ξ]=e, where ξ ξ is a random variable that is defined as follows. It's the minimum number of n n such that ∑n i=1 r i>1∑i=1 n r i>1 and r i r i are random numbers from uniform distribution on [0,1][0,1]. Beautiful, isn't it?!
Since it's an exercise, I'm not sure if it's cool for me to post the solution (proof) here :) If you'd like to prove it yourself, here's a tip: the chapter is called "Moments", which should point you in right direction.
If you want to implement it yourself, then don't read further!
This is a simple algorithm for Monte Carlo simulation. Draw a uniform random, then another one and so on until the sum exceeds 1. The number of randoms drawn is your first trial. Let's say you got:
r
0.0180
0.4596
0.7920
Then your first trial rendered 3. Keep doing these trials, and you'll notice that in average you get e e.
MATLAB code, simulation result and the histogram follow.
matlab
N = 10000000;
n = N;
s = 0;
i = 0;
maxl = 0;
f = 0;
while n > 0
s = s + rand;
i = i + 1;
if s > 1
if i > maxl
f(i) = 1;
maxl = i;
else
f(i) = f(i) + 1;
end
i = 0;
s = 0;
n = n - 1;
end
end
disp ((1:maxl)\f'/sum(f))
bar(f/sum(f))
grid on
f/sum(f)
The result and the histogram:
matlab
2.7183
ans =
Columns 1 through 8
0 0.5000 0.3332 0.1250 0.0334 0.0070 0.0012 0.0002
Columns 9 through 11
0.0000 0.0000 0.0000
UPDATE: I updated my code to get rid of the array of trial results so that it doesn't take RAM. I also printed the PMF estimation.
Update 2: Here's my Excel solution. Put a button in Excel and link it to the following VBA macro:
java
Private Sub CommandButton1\_Click()
n = Cells(1, 4).Value
Range("A:B").Value = ""
n = n
s = 0
i = 0
maxl = 0
Cells(1, 2).Value = "Frequency"
Cells(1, 1).Value = "n"
Cells(1, 3).Value = "# of trials"
Cells(2, 3).Value = "simulated e"
While n > 0
s = s + Rnd()
i = i + 1
If s > 1 Then
If i > maxl Then
Cells(i, 1).Value = i
Cells(i, 2).Value = 1
maxl = i
Else
Cells(i, 1).Value = i
Cells(i, 2).Value = Cells(i, 2).Value + 1
End If
i = 0
s = 0
n = n - 1
End If
Wend
s = 0
For i = 2 To maxl
s = s + Cells(i, 1) \ Cells(i, 2)
Next
Cells(2, 4).Value = s / Cells(1, 4).Value
Rem bar (f / Sum(f))
Rem grid on
Rem f/sum(f)
End Sub
Enter the number of trials, such as 1000, in the cell D1, and click the button. Here how the screen should look like after the first run:
UPDATE 3: Silverfish inspired me to another way, not as elegant as the first one but still cool. It calculated the volumes of n-simplexes using Sobol sequences.
matlab
s = 2;
for i=2:10
p=sobolset(i);
N = 10000;
X=net(p,N)';
s = s + (sum(sum(X)<1)/N);
end
disp(s)
2.712800000000001
Coincidentally he wrote the first book on Monte Carlo method I read back in high school. It's the best introduction to the method in my opinion.
UPDATE 4:
Silverfish in comments suggested a simple Excel formula implementation. This is the kind of result you get with his approach after about total 1 million random numbers and 185K trials:
Obviously, this is much slower than Excel VBA implementation. Especially, if you modify my VBA code to not update the cell values inside the loop, and only do it once all stats are collected.
UPDATE 5
Xi'an's solution #3 is closely related (or even the same in some sense as per jwg's comment in the thread). It's hard to say who came up with the idea first Forsythe or Gnedenko. Gnedenko's original 1950 edition in Russian doesn't have Problems sections in Chapters. So, I couldn't find this problem at a first glance where it is in later editions. Maybe it was added later or buried in the text.
As I commented in Xi'an's answer, Forsythe's approach is linked to another interesting area: the distribution of distances between peaks (extrema) in random (IID) sequences. The mean distance happens to be 3. The down sequence in Forsythe's approach ends with a bottom, so if you continue sampling you'll get another bottom at some point, then another etc. You could track the distance between them and build the distribution.
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edited Feb 23, 2023 at 23:45
community wiki
15 revs, 3 users 95%Aksakal
16
1 Wow, that's cool! Could you add a paragraph or two explaining why this works? –Sycorax♦ Commented Feb 5, 2016 at 4:10
12 (+1) Brilliant! The answer deserves the highest mark as it only relies on uniform simulations. And does not use any approximation but the one due to Monte Carlo. That it connects back to Gnedenko is a further perk. –Xi'an Commented Feb 5, 2016 at 15:10
3 Cool! Here is _Mathematica_ code for same, as a one-liner: …… Mean[Table[ Length[NestWhileList[(Random[]+#) &, Random[], #<1&]], {10^6}]] –wolfies Commented Feb 5, 2016 at 15:42
6 @wolfies The following direct translation of the R solution I posted in Xi'an's answer is twenty times faster: n=10^6; 1. / Mean[UnitStep[Differences[Sort[RandomReal[{0, n}, n + 1]]] - 1]] –whuber♦ Commented Feb 5, 2016 at 16:47
2 I have posted the "why is the mean e e?" question as a question in its own right; I suspect my sketch solution (which is what immediately came to mind as the "obvious" visualisation of the problem) isn't necessarily the way that Russian students were intended to do it! So alternative solutions would be very welcome. –Silverfish Commented Feb 6, 2016 at 18:04
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I suggest upvoting Aksakal's answer. It is unbiased and relies only on a method of generating unit uniform deviates.
My answer can be made arbitrarily precise, but still is biased away from the true value of e e.
Xi'an's answer is correct, but I think its dependence on either the log log function or a way of generating Poisson random deviates is a bit circular when the purpose is to approximate e e.
Estimating e e by Bootstrapping
Instead, consider the bootstrapping procedure. One has a large number of objects n n which are drawn with replacement to a sample size of n n. At each draw, the probability of _not_ drawing a particular object i i is 1−n−1 1−n−1, and there are n n such draws. The probability that a particular object is omitted from all draws is p=(1−1 n)n.p=(1−1 n)n.
Because I'm assuming we know that
exp(−1)=lim n→∞(1−1 n)n exp(−1)=lim n→∞(1−1 n)n
so we also can write
exp(−1)≈p^=∑i=1 m I i∈B j m exp(−1)≈p^=∑i=1 m I i∈B j m
That is, our estimate of p p is found by estimating the probability that a specific observation is omitted from m m bootstrap replicates B j B j across many such replicates -- i.e. the fraction of occurrences of object i i in the bootstraps.
There are two sources of error in this approximation. Finite n n will always mean that the results are approximate, i.e. the estimate is biased. Additionally, p^p^ will fluctuate around the true value because this is a simulation.
I find this approach somewhat charming because an undergraduate or another person with sufficiently little to do could approximate e e using a deck of cards, a pile of small stones, or any other items at hand, in the same vein as a person could estimate π π using a compass, a straight-edge and some grains of sand. I think it's neat when mathematics can be divorced from modern conveniences like computers.
Results
I conducted several simulations for various number of bootstrap replications. Standard errors are estimated using normal intervals.
Note that the choice of n n the number of objects being bootstrapped sets an absolute upper limit on the accuracy of the results because the Monte Carlo procedure is estimating p p and p p depends only on n n. Setting n n to be unnecessarily large will just encumber your computer, either because you only need a "rough" approximation to e e or because the bias will be swamped by variance due to the Monte Carlo. These results are for n=10 3 n=10 3 and p−1≈e p−1≈e is accurate to the third decimal.
This plot shows that the choice of m m has direct and profound consequences for the stability in p^p^. The blue dashed line shows p p and the red line shows e e. As expected, increasing the sample size produces ever-more accurate estimates p^p^.
I wrote an embarrassingly long R script for this. Suggestions for improvement can be submitted on the back of a $20 bill.
```
library(boot)
library(plotrix)
n <- 1e3
if p_hat is estimated with 0 variance (in the limit of infinite bootstraps), then the best estimate we can come up with is biased by exactly this much:
approx <- 1/((1-1/n)^n)
dat <- c("A", rep("B", n-1))
indicator <- function(x, ndx) xor("A"%in%x[ndx], TRUE) ## Because we want to count when "A" is not in the bootstrap sample
p_hat <- function(dat, m=1e3){
foo <- boot(data=dat, statistic=indicator, R=m)
1/mean(foo$t)
}
reps <- replicate(100, p_hat(dat))
boxplot(reps)
abline(h=exp(1),col="red")
p_mean <- NULL
p_var <- NULL
for(i in 1:10){
reps <- replicate(2^i, p_hat(dat))
p_mean[i] <- mean(reps)
p_var[i] <- sd(reps)
}
plotCI(2^(1:10), p_mean, uiw=qnorm(0.975)p_var/sqrt(2^(1:10)),xlab="m", log="x", ylab=expression(hat(p)), main=expression(paste("Monte Carlo Estimates of ", tilde(e))))
abline(h=approx, col='red')
```
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edited Feb 5, 2016 at 15:22
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Sycorax
13
4 +1 It makes a lot of sense. Any chance you can share your code if you wrote it? –Antoni Parellada Commented Feb 4, 2016 at 21:49
2 Even though this can be arbitrarily accurate, ultimately it is unsatisfactory because it only simulates an _approximation_ to e e rather than e e itself. –whuber♦ Commented Feb 5, 2016 at 13:44
1 Sure. You would just end with one replicate call inside of another, which is essentially the same as we have now. –Sycorax♦ Commented Feb 5, 2016 at 19:16
1 @whuber I don't really see the distinction between an arbitrarily accurate approximation to an arbitrarily accurate approximation to e e, and an arbitrarily accurate approximation to e e itself. –jwg Commented Feb 12, 2016 at 15:32
1 @jwg In addition to being conceptually important, it's also practically important because implementing an approximation to an approximation requires keeping track of how accurate each of the two approximations is. But I would have to agree that when both approximations are acceptably good, then the overall approach indeed is fine. –whuber♦ Commented Feb 12, 2016 at 16:47
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Solution 1:
For a Poisson P(λ)P(λ) distribution,
P(X=k)=λ k k!e−λ P(X=k)=λ k k!e−λ
Therefore, if X∼P(1)X∼P(1),
P(X=0)=P(X=1)=e−1 P(X=0)=P(X=1)=e−1
which means you can estimate e−1 e−1 by a Poisson simulation. And Poisson simulations can be derived from an exponential distribution generator (if not in the most efficient manner).
Remark 1: As discussed in the comments, this is a rather convoluted argument since simulating from a Poisson distribution or equivalently an Exponential distribution may be hard to imagine without involving a _log_ or an _exp_ function... But then W. Huber came to the rescue of this answer with a most elegant solution based on ordered uniforms. Which is an _approximation_ however, since the distribution of a uniform spacing U(i:n)−U(i−1:n)U(i:n)−U(i−1:n) is a Beta B(1,n)B(1,n), implying that
P(n{U(i:n)−U(i−1:n)}≥1)=(1−1 n)n P(n{U(i:n)−U(i−1:n)}≥1)=(1−1 n)n
which converges to e−1 e−1as n ngrows to infinity. As an other aside that answers the comments, von Neumann's 1951 exponential generator _only_uses uniform generations.
Solution 2:
Another way to achieve a representation of the constant e e as an integral is to recall that, when
X 1,X 2∼iid N(0,1)X 1,X 2∼iid N(0,1)
then
(X 2 1+X 2 2)∼χ 2 1(X 1 2+X 2 2)∼χ 1 2
which is also an E(1/2)E(1/2) distribution. Therefore,
P(X 2 1+X 2 2≥2)=1−{1−exp(−2/2)}=e−1 P(X 1 2+X 2 2≥2)=1−{1−exp(−2/2)}=e−1
A second approach to approximating e e by Monte Carlo is thus to simulate normal pairs (X 1,X 2)(X 1,X 2) and monitor the frequency of times X 2 1+X 2 2≥2 X 1 2+X 2 2≥2. In a sense it is the opposite of the Monte Carlo approximation of π π related to the frequency of times X 2 1+X 2 2<1 X 1 2+X 2 2<1...
Solution 3:
My Warwick University colleague M. Pollock pointed out another Monte Carlo approximation called Forsythe's method: the idea is to run a sequence of uniform generations u 1,u 2,...u 1,u 2,... until u n+1>u n u n+1>u n. The expectation of the corresponding stopping rule, N N, which is the number of time the uniform sequence went down is then e e while the probability that N N is odd is e−1 e−1! (Forsythe's method actually aims at simulating from any density of the form exp G(x)expG(x), hence is more general than approximating e e and e−1 e−1.)
This is quite parallel to Gnedenko's approach used in Aksakal's answer, so I wonder if one can be derived from the other. At the very least, both have the same distribution with probability mass 1/n!1/n! for value n n.
A quick R implementation of Forsythe's method is to forgo following precisely the sequence of uniforms in favour of larger blocks, which allows for parallel processing:
use=runif(n)
band=max(diff((1:(n-1))[diff(use)>0]))+1
bends=apply(apply((apply(matrix(use[1:((n%/%band)\band)],nrow=band),
2,diff)<0),2,cumprod),2,sum)
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edited Jan 24, 2018 at 14:36
answered Feb 4, 2016 at 12:36
Xi'anXi'an
109k 13 13 gold badges 195 195 silver badges 688 688 bronze badges
12
15 As long as one knows how to do Poisson simulation without knowing e e. –Glen_b Commented Feb 4, 2016 at 12:46
5 If I call R rpoiss() generator, I can pretend I do not know e e. More seriously, you can generate exponential variates E(1)E(1) [using a log log function rather than e e] until the sum exceeds 1 1 and the resulting number minus one is a Poisson P(1)P(1). –Xi'an Commented Feb 4, 2016 at 12:55
6 Computing log log is tantamount to computing exp exp, since they are inverses. You can avoid computing any such function in various ways. Here is one solution based on your first answer: n <- 1e5; 1/mean(n\diff(sort(runif(n+1))) > 1) It uses only elementary arithmetic. –whuber♦ Commented Feb 4, 2016 at 22:13
3 I believe that Forsythe's method is the same as Gnedenko's. Choosing a uniform x n x n such that ∑n x i∑n x i is less than 1 is the same as choosing x n x n smaller than 1−∑n−1 x i 1−∑n−1 x i, and if we are successful, 1−∑n x i 1−∑n x i is conditionally uniformly distributed between 1−∑n−1 x i 1−∑n−1 x i and 0. –jwg Commented Feb 12, 2016 at 8:31
3 I wasn't aware of Forsythe's approach. However, it's linked to something else very interesting. If instead of stopping at n+1 n+1 you keep sampling, then the expectation of the distance from n n to the next bottom is exactly 3. –Aksakal Commented Feb 17, 2016 at 14:40
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Not a solution ... just a quick comment that is too long for the comment box.
Aksakal
Aksakal posted a solution where we calculate the expected number of standard Uniform drawings that must be taken, such that their sum will exceed 1. In _Mathematica_, my first formulation was:
mrM := NestWhileList[(Random[] + #) &, Random[], #<1 &]
Mean[Table[Length[mrM], {10^6}]]
EDIT: Just had a quick play with this, and the following code (same method - also in Mma - just different code) is about 10 times faster:
Mean[Table[Module[{u=Random[], t=1}, While[u<1, u=Random[]+u; t++]; t] , {10^6}]]
Xian / Whuber
Whuber has suggested fast cool code to simulate Xian's solution 1:
R version: n <- 1e5; 1/mean(n\diff(sort(runif(n+1))) > 1)
Mma version: n=10^6; 1. / Mean[UnitStep[Differences[Sort[RandomReal[{0, n}, n + 1]]] - 1]]
which he notes is 20 times faster the first code (or about twice as fast as the new code above).
Just for fun, I thought it would be interesting to see if both approaches are as efficient (in a statistical sense). To do so, I generated 2000 estimates of e using:
Aksakal's method: dataA
Xian's method 1 using whuber code: dataB
... both in _Mathematica_. The following diagram contrasts a nonparametric kernel density estimate of the resulting dataA and dataB data sets.
So, while whuber's code (red curve) is about twice as fast, the method does not appear to be as 'reliable'.
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edited Oct 30, 2017 at 11:32
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4 revswolfies
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2 A vertical line at the location of the true value would vastly improve this image. –Sycorax♦ Commented Feb 5, 2016 at 19:17
1 It's a very interesting observation, thank you. Since the half-width will scale quadratically with the size of the simulation and the half-width of Xi'an's method is about twice that of Aksakal's, then running four times as many iterations will make them equally accurate. The question of how much effort is needed in each iteration remains: if one iteration of Xi'an's method takes less than one-quarter the effort, then that method would still be more efficient. –whuber♦ Commented Feb 5, 2016 at 19:29
2 I believe the situation becomes clear when you compare the numbers of realizations of random variables required in both methods rather than the nominal value of n n. –whuber♦ Commented Feb 5, 2016 at 20:01
1 @whuber wrote: running four times as many iterations will make them equally accurate ///// ..... Just had a quick play with this: increasing the number of sample points used in Xian's Method 1 from 10 6 10 6 to 6 x 10 6 10 6 (i.e. 6 times the number of points) produces a similar curve to Aksaksal. –wolfies Commented Feb 7, 2016 at 14:45
1 Well done with the code--it will be difficult to improve much on that. –whuber♦ Commented Feb 7, 2016 at 14:47
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Here is another way it can be done, though it is quite slow. I make no claim to efficiency, but offer this alternative in the spirit of completeness.
Contra Xi'an's answer, I will assume for the purposes of this question that you are able to generate and use a sequence of n n uniform pseudo-random variables U 1,⋯,U n∼IID U(0,1)U 1,⋯,U n∼IID U(0,1) and you then need to estimate e e by some method using basic arithmetic operations (i.e., you cannot use logarithmic or exponential functions or any distributions that use these functions).†† The present method is motivated by a simple result involving uniform random variables:
E(I(U i⩾1/e)U i)=∫1/e 1 d u u=1.E(I(U i⩾1/e)U i)=∫1/e 1 d u u=1.
Estimating e e using this result: We first order the sample values into descending order to obtain the order statistics u(1)⩾⋯⩾u(n)u(1)⩾⋯⩾u(n) and then we define the partial sums:
S n(k)≡1 n∑i=1 k 1 u(i)for all k=1,..,n.S n(k)≡1 n∑i=1 k 1 u(i)for all k=1,..,n.
Now, let m≡min{k|S(k)⩾1}m≡min{k|S(k)⩾1} and then estimate 1/e 1/e by interpolation of the ordered uniform variables. This gives an estimator for e e given by:
e^≡2 u(m)+u(m+1).e^≡2 u(m)+u(m+1).
This method has some slight bias (owing to the linear interpolation of the cut-off point for 1/e 1/e) but it is a consistent estimator for e e. The method can be implemented fairly easily but it requires sorting of values, which is more computationally intensive than deterministic calculation of e e. This method is slow, since it involves sorting of values.
Implementation in R: The method can be implemented in R using runif to generate uniform values. The code is as follows:
r
EST\_EULER <- function(n) { U <- sort(runif(n), decreasing = TRUE);
S <- cumsum(1/U)/n;
m <- min(which(S >= 1));
2/(U[m-1]+U[m]); }
Implementing this code gives convergence to the true value of e e, but it is very slow compared to deterministic methods.
r
set.seed(1234);
EST\_EULER(10^3);
2.715426
EST\_EULER(10^4);
2.678373
EST\_EULER(10^5);
2.722868
EST\_EULER(10^6);
2.722207
EST\_EULER(10^7);
2.718775
EST\_EULER(10^8);
2.718434
exp(1)
2.718282
†† I take the view that we want to avoid any method that makes use of any transformation that involves an exponential or logarithm. If we can use densities that use exponentials in their definition then it is possible to derive e e from these algebraically using a density call.
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edited Feb 23, 2023 at 23:47
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Method requiring an ungodly amount of samples
First you need to be able to sample from a normal distribution. Assuming you are going to exclude the use of the function f(x)=e x f(x)=e x, or look up tables derived from that function, you can produce approximate samples from the normal distribution via the CLT. For example, if you can sample from a uniform(0,1) distribution, then x¯12√n√∼˙N(0,1)x¯12 n∼˙N(0,1). As pointed out by whuber, to have the final estimate approach e e as the sample size approaches ∞∞, it would be required that the number of uniform samples used approaches ∞∞ as the sample size approaches infinity.
Now, if you can sample from a normal distribution, with large enough samples, you can get consistent estimates of the density of N(0,1)N(0,1). This can be done with histograms or kernel smoothers (but be careful not to use a Gaussian kernel to follow your no e x e x rule!). To get your density estimates to be consistent, you will need to let your df (number of bins in histogram, inverse of window for smoother) approach infinity, but slower than the sample size.
So now, with lots of computational power, you can approximate the density of a N(0,1)N(0,1), i.e. ϕ^(x)ϕ^(x). Since ϕ((√2))=(2 π)−1/2 e−1 ϕ((2))=(2 π)−1/2 e−1, your estimate for e=ϕ^(2–√)2 π−−√e=ϕ^(2)2 π.
If you want to go totally nuts, you can even estimate 2–√2 and 2 π−−√2 π using the methods you discussed earlier.
Method requiring very few samples, but causing an ungodly amount of numerical error
A completely silly, but very efficient, answer based on a comment I made:
Let X∼uniform(−1,1)X∼uniform(−1,1). Define Y n=|(x¯)n|Y n=|(x¯)n|. Define e^=(1−Y n)−1/Y n e^=(1−Y n)−1/Y n.
This will converge _very_ fast, but also run into extreme numerical error.
whuber pointed out that this uses the power function, which typically calls the exp function. This could be sidestepped by discretizing Y n Y n, such that 1/Y n 1/Y n is an integer and the power could be replaced with repeated multiplication. It would be required that as n→∞n→∞, the discretizing of Y n Y n would get finer and finer,and the discretization would have to exclude Y n=0 Y n=0. With this, the estimator _theoretically_ (i.e. the world in which numeric error does not exist) would converge to e e, and quite fast!
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edited Feb 5, 2016 at 18:04
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Cliff AB
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2 The CLT approach is less than satisfactory because ultimately you know these values are _not_ normally distributed. But there are plenty of ways to generate Normal variates without needing e e or logarithms: the Box-Muller method is one. That one, though, requires trig functions and (at a fundamental level) those are the same as exponentials. –whuber♦ Commented Feb 5, 2016 at 13:47
1 @whuber: I did not use the Box-Muller due to the required log transform too directly to exponential in my book. I would have reflexively allowed cos and sin, but that was only because I had forgotten about complex analysis for a moment, so good point. –Cliff AB Commented Feb 5, 2016 at 15:11
1 However, I would take argument with the idea that the generated normal approximation is the weak point of this idea; the density estimation is even weaker! You can think of this idea of having two parameters: n 1 n 1, the number uniforms used in your "approximated normal" and n 2 n 2 the number of approximated normals used estimate the density at ϕ(2–√)ϕ(2). As both n 1 n 1 and n 2 n 2 approach ∞∞, the estimator will approach e e. In fact, I'm very confident the convergence rate would be much more limited by n 2 n 2 than n 1 n 1; non-parametric density has a slow convergence rate! –Cliff AB Commented Feb 5, 2016 at 15:15
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If you do not have a calculator (ie you can not compute the exponential 'e' indirectly by using some related functions like computing a sample from a normal distribution or exponential distribution) and you have only coin flips or dice rolls available to you, then you could use the following puzzle to estimate the number e e:
The number e e appears in the expression for the expectation value for the frog problem with negative steps. We have E[J 1]=2 e−2 E[J 1]=2 e−2. So we could approximate e e with an approximate for E[J 1]=μ J 1 E[J 1]=μ J 1 using e^=0.5 μ^J 1+1 e^=0.5 μ^J 1+1
and dice rolls could be constructed from coin flips if you want to be more restrictive
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answered Dec 25, 2021 at 7:49
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Sextus Empiricus
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∫2 1 1 x d x=ln 2∫1 2 1 x d x=ln2
So if you draw uniformly from [1,2]2[1,2]2, the fraction of points whose product is less than 1 1 would converge to ln 2 ln2 by the LLN.
You can then get to e e by
2 1 ln 2=e 2 1 ln2=e
One might raise the issue that exponentiation might require knowledge of e e itself. I don’t have a good answer for that.
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answered Oct 11, 2022 at 14:48
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Calculon
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The Python version of this is the following if anyone is curious:
python
import random
print("Number of iterations: ", end="")
n = int(input())
sum\_total = 0
for \_ in range(n):
temp = 0
counter = 0
while temp < 1:
temp += random.random()
counter += 1
sum\_total += counter
print(sum\_total/n)
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edited Feb 23, 2023 at 23:46
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2 revs, 2 users 98%Andrew
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9164 | https://math.mit.edu/~sheffield/2018600/Lecture22.pdf | 18.600: Lecture 22 Sums of independent random variables Scott Sheffield MIT Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
▶Now let’s try to find FX+Y (a) = P{X + Y ≤a}.
Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
▶Now let’s try to find FX+Y (a) = P{X + Y ≤a}.
▶This is the integral over {(x, y) : x + y ≤a} of f (x, y) = fX(x)fY (y). Thus, Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
▶Now let’s try to find FX+Y (a) = P{X + Y ≤a}.
▶This is the integral over {(x, y) : x + y ≤a} of f (x, y) = fX(x)fY (y). Thus, ▶ P{X + Y ≤a} = Z ∞ −∞ Z a−y −∞ fX(x)fY (y)dxdy = Z ∞ −∞ FX(a −y)fY (y)dy.
Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
▶Now let’s try to find FX+Y (a) = P{X + Y ≤a}.
▶This is the integral over {(x, y) : x + y ≤a} of f (x, y) = fX(x)fY (y). Thus, ▶ P{X + Y ≤a} = Z ∞ −∞ Z a−y −∞ fX(x)fY (y)dxdy = Z ∞ −∞ FX(a −y)fY (y)dy.
▶Differentiating both sides gives fX+Y (a) = d da R ∞ −∞FX(a−y)fY (y)dy = R ∞ −∞fX(a−y)fY (y)dy.
Summing two random variables ▶Say we have independent random variables X and Y and we know their density functions fX and fY .
▶Now let’s try to find FX+Y (a) = P{X + Y ≤a}.
▶This is the integral over {(x, y) : x + y ≤a} of f (x, y) = fX(x)fY (y). Thus, ▶ P{X + Y ≤a} = Z ∞ −∞ Z a−y −∞ fX(x)fY (y)dxdy = Z ∞ −∞ FX(a −y)fY (y)dy.
▶Differentiating both sides gives fX+Y (a) = d da R ∞ −∞FX(a−y)fY (y)dy = R ∞ −∞fX(a−y)fY (y)dy.
▶Latter formula makes some intuitive sense. We’re integrating over the set of x, y pairs that add up to a.
Independent identically distributed (i.i.d.) ▶The abbreviation i.i.d. means independent identically distributed.
Independent identically distributed (i.i.d.) ▶The abbreviation i.i.d. means independent identically distributed.
▶It is actually one of the most important abbreviations in probability theory.
Independent identically distributed (i.i.d.) ▶The abbreviation i.i.d. means independent identically distributed.
▶It is actually one of the most important abbreviations in probability theory.
▶Worth memorizing.
Summing i.i.d. uniform random variables ▶Suppose that X and Y are i.i.d. and uniform on [0, 1]. So fX = fY = 1 on [0, 1].
Summing i.i.d. uniform random variables ▶Suppose that X and Y are i.i.d. and uniform on [0, 1]. So fX = fY = 1 on [0, 1].
▶What is the probability density function of X + Y ?
Summing i.i.d. uniform random variables ▶Suppose that X and Y are i.i.d. and uniform on [0, 1]. So fX = fY = 1 on [0, 1].
▶What is the probability density function of X + Y ?
▶fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy = R 1 0 fX(a −y) which is the length of [0, 1] ∩[a −1, a].
Summing i.i.d. uniform random variables ▶Suppose that X and Y are i.i.d. and uniform on [0, 1]. So fX = fY = 1 on [0, 1].
▶What is the probability density function of X + Y ?
▶fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy = R 1 0 fX(a −y) which is the length of [0, 1] ∩[a −1, a].
▶That’s a when a ∈[0, 1] and 2 −a when a ∈[1, 2] and 0 otherwise.
Review: summing i.i.d. geometric random variables ▶A geometric random variable X with parameter p has P{X = k} = (1 −p)k−1p for k ≥1.
Review: summing i.i.d. geometric random variables ▶A geometric random variable X with parameter p has P{X = k} = (1 −p)k−1p for k ≥1.
▶Sum Z of n independent copies of X?
Review: summing i.i.d. geometric random variables ▶A geometric random variable X with parameter p has P{X = k} = (1 −p)k−1p for k ≥1.
▶Sum Z of n independent copies of X?
▶We can interpret Z as time slot where nth head occurs in i.i.d. sequence of p-coin tosses.
Review: summing i.i.d. geometric random variables ▶A geometric random variable X with parameter p has P{X = k} = (1 −p)k−1p for k ≥1.
▶Sum Z of n independent copies of X?
▶We can interpret Z as time slot where nth head occurs in i.i.d. sequence of p-coin tosses.
▶So Z is negative binomial (n, p). So P{Z = k} = k−1 n−1 pn−1(1 −p)k−np.
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
▶What is the law of Z = Pn i=1 Xi?
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
▶What is the law of Z = Pn i=1 Xi?
▶We claimed in an earlier lecture that this was a gamma distribution with parameters (λ, n).
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
▶What is the law of Z = Pn i=1 Xi?
▶We claimed in an earlier lecture that this was a gamma distribution with parameters (λ, n).
▶So fZ(y) = λe−λy(λy)n−1 Γ(n) .
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
▶What is the law of Z = Pn i=1 Xi?
▶We claimed in an earlier lecture that this was a gamma distribution with parameters (λ, n).
▶So fZ(y) = λe−λy(λy)n−1 Γ(n) .
▶We argued this point by taking limits of negative binomial distributions. Can we check it directly?
Summing i.i.d. exponential random variables ▶Suppose X1, . . . Xn are i.i.d. exponential random variables with parameter λ. So fXi(x) = λe−λx on [0, ∞) for all 1 ≤i ≤n.
▶What is the law of Z = Pn i=1 Xi?
▶We claimed in an earlier lecture that this was a gamma distribution with parameters (λ, n).
▶So fZ(y) = λe−λy(λy)n−1 Γ(n) .
▶We argued this point by taking limits of negative binomial distributions. Can we check it directly?
▶By induction, would suffice to show that a gamma (λ, 1) plus an independent gamma (λ, n) is a gamma (λ, n + 1).
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
▶So fX(x) = λe−λx(λx)s−1 Γ(s) and fY (y) = λe−λy(λy)t−1 Γ(t) .
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
▶So fX(x) = λe−λx(λx)s−1 Γ(s) and fY (y) = λe−λy(λy)t−1 Γ(t) .
▶Now fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
▶So fX(x) = λe−λx(λx)s−1 Γ(s) and fY (y) = λe−λy(λy)t−1 Γ(t) .
▶Now fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶Up to an a-independent multiplicative constant, this is Z a 0 e−λ(a−y)(a−y)s−1e−λyyt−1dy = e−λa Z a 0 (a−y)s−1yt−1dy.
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
▶So fX(x) = λe−λx(λx)s−1 Γ(s) and fY (y) = λe−λy(λy)t−1 Γ(t) .
▶Now fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶Up to an a-independent multiplicative constant, this is Z a 0 e−λ(a−y)(a−y)s−1e−λyyt−1dy = e−λa Z a 0 (a−y)s−1yt−1dy.
▶Letting x = y/a, this becomes e−λaas+t−1 R 1 0 (1 −x)s−1xt−1dx.
Summing independent gamma random variables ▶Say X is gamma (λ, s), Y is gamma (λ, t), and X and Y are independent.
▶Intuitively, X is amount of time till we see s events, and Y is amount of subsequent time till we see t more events.
▶So fX(x) = λe−λx(λx)s−1 Γ(s) and fY (y) = λe−λy(λy)t−1 Γ(t) .
▶Now fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶Up to an a-independent multiplicative constant, this is Z a 0 e−λ(a−y)(a−y)s−1e−λyyt−1dy = e−λa Z a 0 (a−y)s−1yt−1dy.
▶Letting x = y/a, this becomes e−λaas+t−1 R 1 0 (1 −x)s−1xt−1dx.
▶This is (up to multiplicative constant) e−λaas+t−1. Constant must be such that integral from −∞to ∞is 1. Conclude that X + Y is gamma (λ, s + t).
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
▶We just need to compute fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
▶We just need to compute fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶We could compute this directly.
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
▶We just need to compute fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶We could compute this directly.
▶Or we could argue with a multi-dimensional bell curve picture that if X and Y have variance 1 then fσ1X+σ2Y is the density of a normal random variable (and note that variances and expectations are additive).
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
▶We just need to compute fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶We could compute this directly.
▶Or we could argue with a multi-dimensional bell curve picture that if X and Y have variance 1 then fσ1X+σ2Y is the density of a normal random variable (and note that variances and expectations are additive).
▶Or use fact that if Ai ∈{−1, 1} are i.i.d. coin tosses then 1 √ N Pσ2N i=1 Ai is approximately normal with variance σ2 when N is large.
Summing two normal variables ▶X is normal with mean zero, variance σ2 1, Y is normal with mean zero, variance σ2 2.
▶fX(x) = 1 √ 2πσ1 e −x2 2σ2 1 and fY (y) = 1 √ 2πσ2 e −y2 2σ2 2 .
▶We just need to compute fX+Y (a) = R ∞ −∞fX(a −y)fY (y)dy.
▶We could compute this directly.
▶Or we could argue with a multi-dimensional bell curve picture that if X and Y have variance 1 then fσ1X+σ2Y is the density of a normal random variable (and note that variances and expectations are additive).
▶Or use fact that if Ai ∈{−1, 1} are i.i.d. coin tosses then 1 √ N Pσ2N i=1 Ai is approximately normal with variance σ2 when N is large.
▶Generally: if independent random variables Xj are normal (µj, σ2 j ) then Pn j=1 Xj is normal (Pn j=1 µj, Pn j=1 σ2 j ).
Other sums ▶Sum of an independent binomial (m, p) and binomial (n, p)?
Other sums ▶Sum of an independent binomial (m, p) and binomial (n, p)?
▶Yes, binomial (m + n, p). Can be seen from coin toss interpretation.
Other sums ▶Sum of an independent binomial (m, p) and binomial (n, p)?
▶Yes, binomial (m + n, p). Can be seen from coin toss interpretation.
▶Sum of independent Poisson λ1 and Poisson λ2?
Other sums ▶Sum of an independent binomial (m, p) and binomial (n, p)?
▶Yes, binomial (m + n, p). Can be seen from coin toss interpretation.
▶Sum of independent Poisson λ1 and Poisson λ2?
▶Yes, Poisson λ1 + λ2. Can be seen from Poisson point process interpretation. |
9165 | https://www.investopedia.com/terms/n/netloss.asp | Skip to content
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Table of Contents
Table of Contents
What Is Net Loss?
Understanding Net Loss
Factors Contributing to a Net Loss
Examples
FAQs
Net Loss: Definition, Formula, and Examples
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Updated June 23, 2025
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Definition
A net loss is sometimes referred to as a net operating loss (NOL) because its cost of staying in business is greater than its earnings.
What Is Net Loss?
A net loss results when profits fall below the level of expenses and cost of goods sold (COGS) in a designated period. The most common factor that contributes to a net loss is a low revenue stream. A net loss may be contrasted with a net profit and is also known as after-tax income or net income.
Key Takeaways
A net loss occurs when the sum total of expenses exceeds the total income or revenue generated by a business, project, transaction, or investment.
Businesses would report a net loss on the income statement, effectively as a negative net profit.
Many factors can contribute to a net loss including low revenues, strong competition, unsuccessful marketing campaigns, and increased cost of goods sold (COGS).
Understanding Net Loss
For a business, net loss is sometimes referred to as a net operating loss (NOL). For tax purposes, net losses may be carried forward into future tax years to offset gains or profits in those years. A net loss appears on the company's bottom line or income statement. Net loss or net profit is calculated using the following formula:
Net Loss (or Net Profit) = Revenues - Expenses
Because revenues and expenses are matched during a set time, a net loss is an example of the matching principle, which is an integral part of the accrual accounting method. Expenses related to income earned during a set time are included in (or "matched to") that period regardless of when the expenses are paid.
Factors Contributing to a Net Loss
Strong competition, unsuccessful marketing programs, weak pricing strategies, not keeping up with market demands, and inefficient marketing staff contribute to decreasing revenues. Decreased revenues result in decreased profits. When profits fall below the level of expenses and cost of goods sold (COGS) in a given time, a net loss results.
COGS also affects net losses. Substantial production or purchase costs of products being sold are subtracted from revenue. The remaining money is used for covering expenses and creating profit. When COGS exceeds funding for expenses, a net loss occurs.
Expenses contribute to net losses as well. Even when targeted revenue is earned, and COGS remains within limits, unexpected expenses and overspending in budgeted areas may exceed gross profits.
Excessive carrying costs are a type of expense that can contribute to net losses. These are the costs a company pays for holding inventory in stock before it is sold to customers.
Fast Fact
Businesses that have a net loss do not necessarily go bankrupt immediately because they may opt to use their retained earnings or loans to stay afloat. This strategy, however, is only short-term, as a company without profits will not survive in the long term.
Net Loss Examples
Say that substantial refunds were expected as companies took advantage of outstanding tax credits previously issued as a way of retaining jobs in the state during the recession. As a result, the state treasurer anticipates a decrease of $99 million in revenue from the state’s principal business taxes. This prompts state officials to cut the current and upcoming fiscal year revenue projections by a significant amount and, unless they can cut expenditures as well, they will be operating at a net loss.
Another example would be if Company A has $200,000 in sales, $140,000 in COGS, and $80,000 in expenses. Subtracting $140,000 COGS from $200,000 in sales results in $60,000 in gross profit. However, because expenses exceed gross profit, a $20,000 net loss results.
Yet another example would be a company that sells frozen foods and needs to pay for refrigerated storage facilities, utility costs, taxes, employee expenses, and insurance. If sales are slow, the company will need to hold onto its inventory for a longer time, incurring additional carrying costs which could contribute to a net loss.
Can a Company With Positive Revenues Still Have a Net Loss?
Yes, even if a company has a large volume of sales, it can still end up losing money if the cost of goods or other expenses related to those sales (e.g., marketing) are too high. Other factors like taxes, interest expenses, depreciation and amortization, and one-time charges like a lawsuit can also take a company from a profit to a net loss.
What Is a Net Loss Carryforward?
The IRS allows certain net losses experienced in one tax period to be deducted from net profits earned in subsequent periods. The 2018 Tax Cuts and Jobs Act (TCJA) changed how businesses must account for net operating loss carryforwards. Check with your accountant for all tax matters
Is a Net Loss the Same As a Negative Profit?
A negative profit technically does not exist, since a profit, by definition, implies a gain in value. However, the term negative profit is used colloquially to describe a net loss.
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9166 | https://en.wikipedia.org/wiki/Interval_exchange_transformation | Jump to content
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1 Formal definition
2 Properties
3 Odometers
4 Higher dimensions
5 See also
6 Notes
7 References
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From Wikipedia, the free encyclopedia
In mathematics, an interval exchange transformation is a kind of dynamical system that generalises circle rotation. The phase space consists of the unit interval, and the transformation acts by cutting the interval into several subintervals, and then permuting these subintervals. They arise naturally in the study of polygonal billiards and in area-preserving flows.
Formal definition
[edit]
Let and let be a permutation on . Consider a vector of positive real numbers (the widths of the subintervals), satisfying
Define a map called the interval exchange transformation associated with the pair as follows. For let
Then for , define
if lies in the subinterval . Thus acts on each subinterval of the form by a translation, and it rearranges these subintervals so that the subinterval at position is moved to position .
Properties
[edit]
Any interval exchange transformation is a bijection of to itself that preserves the Lebesgue measure. It is continuous except at a finite number of points.
The inverse of the interval exchange transformation is again an interval exchange transformation. In fact, it is the transformation where for all .
If and (in cycle notation), and if we join up the ends of the interval to make a circle, then is just a circle rotation. The Weyl equidistribution theorem then asserts that if the length is irrational, then is uniquely ergodic. Roughly speaking, this means that the orbits of points of are uniformly evenly distributed. On the other hand, if is rational then each point of the interval is periodic, and the period is the denominator of (written in lowest terms).
If , and provided satisfies certain non-degeneracy conditions (namely there is no integer such that ), a deep theorem which was a conjecture of M.Keane and due independently to William A. Veech and to Howard Masur asserts that for almost all choices of in the unit simplex the interval exchange transformation is again uniquely ergodic. However, for there also exist choices of so that is ergodic but not uniquely ergodic. Even in these cases, the number of ergodic invariant measures of is finite, and is at most .
Interval maps have a topological entropy of zero.
Odometers
[edit]
The dyadic odometer can be understood as an interval exchange transformation of a countable number of intervals. The dyadic odometer is most easily written as the transformation
defined on the Cantor space The standard mapping from Cantor space into the unit interval is given by
This mapping is a measure-preserving homomorphism from the Cantor set to the unit interval, in that it maps the standard Bernoulli measure on the Cantor set to the Lebesgue measure on the unit interval. A visualization of the odometer and its first three iterates appear on the right.
Higher dimensions
[edit]
Two and higher-dimensional generalizations include polygon exchanges, polyhedral exchanges and piecewise isometries.
See also
[edit]
Odometer
Notes
[edit]
^ Keane, Michael (1975), "Interval exchange transformations", Mathematische Zeitschrift, 141: 25–31, doi:10.1007/BF01236981, MR 0357739.
^ Veech, William A. (1982), "Gauss measures for transformations on the space of interval exchange maps", Annals of Mathematics, Second Series, 115 (1): 201–242, doi:10.2307/1971391, MR 0644019.
^ Masur, Howard (1982), "Interval exchange transformations and measured foliations", Annals of Mathematics, Second Series, 115 (1): 169–200, doi:10.2307/1971341, MR 0644018.
^ Matthew Nicol and Karl Petersen, (2009) "Ergodic Theory: Basic Examples and Constructions", Encyclopedia of Complexity and Systems Science, Springer
^ Piecewise isometries – an emerging area of dynamical systems, Arek Goetz
References
[edit]
Artur Avila and Giovanni Forni, Weak mixing for interval exchange transformations and translation flows, arXiv:math/0406326v1,
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9167 | https://openoregon.pressbooks.pub/techmath/chapter/module-21-converting-units-of-area/ | Skip to content
Module 21: Converting Units of Area
You may use a calculator throughout this module.
Converting between units of area requires us to be careful because square units behave differently than linear units.
U.S. System: Converting Measurements of Area
Consider a square yard; the area of a square with sides yard long.
yard = feet, so we can divide the square into three sections vertically and three sections horizontally to convert both dimensions of the square from yards to feet. This forms a by grid, which shows us visually that square yard equals square feet, not square feet! The linear conversion ratio of to means that that the conversion ratio for the areas is to , or to .
Here’s another way to think about it without a diagram: , so . To remove the parentheses, we must square the number and square the units: .
More generally, we need to square the linear conversion factors when converting units of area. If the linear units have a ratio of to , the square units will have a ratio of to .
Exercises
1. An acre is defined as the area of a foot by foot rectangle. (That’s a furlong by a chain, if you were curious.) How many square feet are in acre?
2. How many square yards are in acre?
3. How many square inches equals square foot?
It should be no surprise that this module will be full of conversion ratios. As always, if you discover other conversion ratios that aren’t provided here, it would be a good idea to write them down so you can use them as needed.
An acre is defined as a unit of area; it would be wrong to say “acres squared” or put an exponent of on the units.
Exercises
4. A hallway is yards long and yards wide. How many square feet of linoleum are needed to cover the hallway?
5. A proposed site for an elementary school is feet by feet. Find its area, in acres.
Metric System: Converting Measurements of Area
A hectare is defined as a square with sides meters long. Dividing a square kilometer into ten rows and ten columns will make a by grid of hectares. As with acres, it would be wrong to say “hectares squared” or put an exponent of on the units.
Exercises
6. A hallway is meters long and meters wide. How many square centimeters of linoleum are needed to cover the hallway?
7. A proposed site for an elementary school is meters by meters. Find its area, in hectares.
Both Systems: Converting Measurements of Area
Converting between the U.S. and metric systems will involve messy decimal values. For example, because , we can square both numbers and find that . The conversions are rounded to three or four significant digits in the table below.
Exercises
8. The area of Portland is . Convert this area to square kilometers.
9. How many hectares is a acre ranch?
10. A sheet of paper measures inches by inches. What is the area in square centimeters?
11. A soccer field is meters long and meters wide. What is its area in square feet?
Areas of Similar Figures
Earlier in this module, it was stated that if the linear units have a ratio of to , the square units will have a ratio of to . This applies to similar figures as well.
If the linear dimensions of two similar figures have a ratio of to , then the areas will have a ratio of to .
This is true for circles, similar triangles, similar rectangles, similar hexagons, you name it. We’ll verify this in the following exercises.
Exercises
A personal pizza has a -inch diameter. A medium pizza has a diameter twice that of a personal pizza.
12. Determine the area of the medium pizza.
13. Determine the area of the personal pizza.
14. What is the ratio of the areas of the two pizzas?
Right triangle has legs and long. Right triangle has legs triple the length of ’s.
15. Determine the area of the larger triangle, .
16. Determine the area of the smaller triangle, .
17. What is the ratio of the areas of the two triangles?
License
Technical Mathematics Copyright © 2020 by Morgan Chase is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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9168 | https://math.stackexchange.com/questions/4426069/estimating-reaching-probabilities-of-a-simple-random-walk | Stack Exchange Network
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Estimating reaching probabilities of a simple random walk
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$\begingroup$
I'm currently reading the paper "Simple Random Walk" from Sven Erick Alm.
Nearly everything is clear, except for the one inequality.
A simple random walk started at $0$, with probability $p$ of going up ($+1$) and $q=1-p$ of going down ($-1$). He defined $P_k(n)$ as the probability to reach $x=k$ in the first $n$ steps and claims that conditioning on the first step gives $$ P_1(n) = p + qP_2(n-1) \le p+qP_1^2(n-1). $$
I am having problem with the inequality on the RHS. Is there an easy explanation for it?
I tried to prove it rigorously using induction, but so far I didn't get satisfying results.
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edited Apr 12, 2022 at 13:32
gt6989b
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asked Apr 12, 2022 at 13:25
user1047209user1047209
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$\endgroup$
8
$\begingroup$ Welcome to Math.SE! Seems like the only thing you need to prove is that $P_2(n-1) \le P_1^2(n-1)$, does that make sense? $\endgroup$
gt6989b
– gt6989b
2022-04-12 13:33:58 +00:00
Commented Apr 12, 2022 at 13:33
$\begingroup$ Also confused somewhat by that definition of $P_k(n)$. If the position of random walk is $S_n$, does that mean $S_n = k$ or that the first hitting time of $k$ is at most $n$? $\endgroup$
gt6989b
– gt6989b
2022-04-12 13:35:27 +00:00
Commented Apr 12, 2022 at 13:35
$\begingroup$ The definition is the probability, that the first hitting time of k is at most n. $\endgroup$
user1047209
– user1047209
2022-04-12 13:38:12 +00:00
Commented Apr 12, 2022 at 13:38
$\begingroup$ The first mentioned inequality isn't obvious for me, since the RHS is the probability of hitting 1 in less than n steps and from 1 hitting 2 in less than n steps, which is, for me like only one of the ways of getting from 0 to 2. $\endgroup$
user1047209
– user1047209
2022-04-12 13:41:50 +00:00
Commented Apr 12, 2022 at 13:41
$\begingroup$ I thought about that interpretation, that would imply that $P_2(n) \le P_1^2(n/2)$ though or something like that... more precisely, $$ P_2(n) = \sum_{k=1}^{n-1} P_1(k) P_1(n-k) $$ $\endgroup$
gt6989b
– gt6989b
2022-04-12 13:43:36 +00:00
Commented Apr 12, 2022 at 13:43
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$\begingroup$
We want to show $P_2(n-1) \le P_1(n-1)^2$, or in other words $P(\text{random walk hits $2$ in $n-1$ steps}) \le P(\text{random walk hits $1$ in $n-1$ steps})^2$.
Let $\tau$ be the first time the random walk $X$ hits $1$, and define $\hat X$ to be a random walk started at $1$ at time $\tau$ (basically thinking of restarting our random walk when it hits $1$ the first time). The strong Markov property implies $X$ is independent of $\hat X$. Then \begin{align}P(\text{$X$ hits $2$ in $n-1$ steps}) &= P(\text{$X$ hits $1$ in $\tau$ steps and $\hat X$ hits $1$ in $n-1-\tau$ steps}) \ &= P(\text{$X$ hits $1$ in $\tau$ steps}) P(\text{$\hat X$ hits $1$ in $n-1-\tau$ steps}) \ &\le P(\text{$X$ hits $1$ in $n-1$ steps}) P(\text{$\hat X$ hits $1$ in $n-1$ steps}) \ &= P(\text{$X$ hits $1$ in $n-1$ steps})^2 \ &= P_1(n-1)^2. \end{align}
That's not a rigorous proof or anything, but that's the basic idea: to hit $2$ in $n-1$ steps, you have to independently hit $1$ twice in less than $n-1$ steps. You can make this rigorous by being a little more careful about how you define $\hat X$.
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answered Apr 12, 2022 at 18:27
user6247850user6247850
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$\endgroup$
6
$\begingroup$ The definition of X^ won't be that hard to do formally, I think. Can I assume, that tau is w. l. o. g. <= n-1, since if it would be bigger, the LHS would be equal to 0? $\endgroup$
user1047209
– user1047209
2022-04-12 20:06:22 +00:00
Commented Apr 12, 2022 at 20:06
$\begingroup$ @user1047209 I think that handling the case where $\tau > n-1$ is where most of the difficulty in making this rigorous comes from. $\endgroup$
user6247850
– user6247850
2022-04-12 20:26:17 +00:00
Commented Apr 12, 2022 at 20:26
$\begingroup$ But if $\tau = n-1$ then we already have, that the LHS equals 0, since one cannot get from 1 to 2 in 0 steps; if its value is even bigger this makes even less sense, to get from 1 to 2 in $<0$ steps. $\endgroup$
user1047209
– user1047209
2022-04-13 05:17:34 +00:00
Commented Apr 13, 2022 at 5:17
$\begingroup$ @user1047209 The LHS is not random, so it can't equal $0$ depending on the value of the random variable $\tau$. It's correct that $P(\text{$X$ hits 2 in $n-1$ steps}, \tau \ge n-1) = 0$, but I think the justification for appealing to the strong markov property might get murky once you add in the extra event ${\tau < n-1}$ (maybe not, I might be misremembering here). Again, this all makes sense intuitively, and I think provides an explanation for the inequality, but rigorously justifying everything might be tricky. $\endgroup$
user6247850
– user6247850
2022-04-13 13:44:24 +00:00
Commented Apr 13, 2022 at 13:44
1
$\begingroup$ @user1047209 Yes, I think that works. $\endgroup$
user6247850
– user6247850
2022-04-22 13:28:41 +00:00
Commented Apr 22, 2022 at 13:28
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Possible clique numbers of a regular graph
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I have been stuck on a question
Let G G be a regular graph on n n vertices. Show that the possible clique numbers (the clique number being the maximal order of a complete subgraph in G G) are 1,2,...,⌊n 2⌋,n 1,2,...,⌊n 2⌋,n.
Note that this question was asked on comp sci stack exchange, however, the one answer only shows that one can always construct a regular graph with a clique number ⌊n 2⌋,n⌊n 2⌋,n and nothing intermediate. It does not show that one can definitely find a regular graph with a maximum complete graph of size 3, 4 ... say (with 1 and 2 it is trivial. For 1, you need the empty graph, and for 2, have a complete bipartite graph say if n n is even, or a cycle for any n n).
I'm not sure if this is completely obvious. I thought it might be by construction, for instance if I was to make two ⌊n−2 2⌋⌊n−2 2⌋ complete graphs and the remaining vertices I would give them the same degrees. How do I know that when I ad these remaining vertices and their associated edges, I don't end up necessarily making a larger clique, so that there are 'gaps' in the possible clique numbers?
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edited Sep 9, 2019 at 3:50
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To construct an n n-vertex regular graph with a clique of order k k (and no more), the easiest approach is to take a circulant graph in which we number the vertices 0,1,2,…,n−1 0,1,2,…,n−1 and make vertices i,j i,j adjacent if i−j mod n i−j mod n is one of {−k+1,−k+2,…,−1,1,2,…,k−1}{−k+1,−k+2,…,−1,1,2,…,k−1}. Here is an example with n=12 n=12 and k=4 k=4, and a clique of order 4 4 highlighted:
We can check that, provided k≤n 2 k≤n 2, no cliques of order more than k k are created. Take any clique, and without loss of generality, suppose that it contains vertex 0 0. Then at most the 2(k−1)2(k−1) other vertices {−k+1,−k+2,…,−1,1,2,…,k−1}{−k+1,−k+2,…,−1,1,2,…,k−1} can be in the clique. Moreover, these come in k−1 k−1 pairs {−k+1,1},{−k+2,2},…,{−1,k−1}{−k+1,1},{−k+2,2},…,{−1,k−1}, and at most one vertex from each pair can be in the clique (since the two vertices in a pair are not adjacent). This means there can be at most k−1 k−1 other vertices in the clique, so it has order at most k k.
As soon as k≥n+1 2 k≥n+1 2, this argument stops working, because then pairs such as {−1,k−1}{−1,k−1} are adjacent: although they are k k steps apart one way around the circle, they are k−1 k−1 steps apart or fewer the other way around. But as soon as k≥n+1 2 k≥n+1 2 but k<n k<n, there is no regular n n-vertex graph with clique number k k, anyway.
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answered Sep 7, 2019 at 22:53
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9170 | https://www.51jiaoxi.com/doc-17119209.html | 高中生物人教版 (新课标)选修1《生物技术实践》土壤中分解尿素的细菌的分离与计数教学设计-教案下载-教习网
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首页/高中生物/人教版 (新课标)/选修1《生物技术实践》/专题2 微生物的培养与应用/课题2 土壤中分解尿素的细菌的分离与计数
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》教案
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高中生物人教版 (新课标)选修1《生物技术实践》土壤中分解尿素的细菌的分离与计数教学设计
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这是一份高中生物人教版 (新课标)选修1《生物技术实践》土壤中分解尿素的细菌的分离与计数教学设计，共7页。教案主要包含了教材分析，学情分析，课题目标，课题重点与难点，教学策略，教学过程，板书设计，教学反思等内容，欢迎下载使用。
一、教材分析
“土壤中分解尿素的细菌的分离与计数”是选修1专题2《微生物的培养与应用》中的第2个课题，主要介绍了微生物培养过程中如何筛选菌株、统计菌落数目、设置对照、培养与观察并初步鉴定，既是课题1中所学基本技术在实践应用中的深化，同时也为课题3的学习打下基础。
学生通过学习，应当掌握微生物培养的一般方法，生产生活中关于微生物的分离、计数、培养等实际问题。在教材中占有重要的地位。
二、学情分析
在知识方面，学生已经掌握微生物实验室培养时所用到的技术——培养基的制备、接种技术和无菌技术，但对于微生物培养的生产生活方面的应用方面，尚缺乏一定的理论知识和操作技能。
在能力方面，高二学生思维敏捷，求知欲望强，思考问题有一定的深度，具备一定的自主学习及合作探究能力，但发现问题和创新意识还有待提高。
三、课题目标
知识与能力目标
通过实施“土壤中分解尿素的细菌的分离与计数”教学，让学生学会微生物的选择培养、分离、计数等技术的原理和方法，能够简述选择性培养基和鉴别性培养基的作用。
学科素养
（1）基础知识（选择培养基对微生物的选择作用；微生物数量的测定）；
（2）基本技能（通过分析研究思路的形成过程，提高实验设计的能力。通过对实验结果的分析，培养推理和计算能力）；
（3）基本思想（对生产生活中的尿素的介绍，认同“科学、技术、社会”息息相关）；
（4）基本活动经验（培养学生养他们保护、利用土壤微生物资源的意识）。
四、课题重点与难点
1.课题重点：对土样的选取和选择培养基的配制。
2.课题难点：对分解尿素的细菌的计数。
五、教学策略
采用课前和课中相结合的案例式教学策略。
课前，社团成员在老师的指导下，完成实验案例；课堂上，社团成员进行案例展示；老师引领学生以问题探讨的方式，拆分知识结构，细化知识内容，分析社团活动各个环节中存在的问题，以实践带动理论，深化理解，使得学生能够建立完整知识体系。
这种教学策略可以有效落实教学目标，破解理论脱离实践的教学困难，解决教师表述难与学生理解难之间教学矛盾。
六、教学过程
七、板书设计
课题2 土壤中分解尿素的细菌的分离与计数
（一）研究思路
1.筛选菌株 2.统计菌落数目 3.设置对照 4.培养与观察 5.初步鉴定
（二）实验流程
①准备工作 ②取样 ③样品稀释 ④涂布平板 ⑤培养 ⑥观察与记录
⑦数据处理 ⑧初步鉴定
八、教学反思
本课题的教学目标是研究培养基对微生物的选择作用并进行微生物数量的测定。教学内容为引导学生理解并掌握如何筛选菌株、统计菌落数目、设置对照、培养与观察并进行初步鉴定。
这些内容是微生物培养过程要解决的实验操作问题。知识难度不大，但理论知识抽象，学生不易理解和掌握。
为此，我注重课程资源的选择、整合和优化，采用课前和课中相结合的 “案例式”教学策略。
课前，社团成员在老师的指导下，完成实验案例；课堂上，社团成员进行案例展示；老师引领学生以问题探讨的方式，拆分知识结构，细化知识内容，分析社团活动各个环节中存在的问题，以实践带动理论，深化理解，使得学生能够建立完整知识体系。这种教学策略可以有效落实教学目标，破解理论脱离实践的教学困难，解决教师表述难与学生理解难之间教学矛盾。
本实验课题创新之处主要表现在三个方面：以学生社团活动成果展示的形式，将教学中要解决的问题与相应的实验成果联系，直观有效突破教学重难点。
实验过程中，将涂布器的灼烧灭菌、培养皿等玻璃器皿的干热灭菌，改为效果更为彻底的高压蒸汽灭菌；微量移液器、具塞试管的使用，使实验更加精准和便利。
本实验还存在一些不足，一方面，需要继续渗透“翻转课堂”模式在教学中的应用，将学生能自学的一定要放手，重点讲解学生不会的，或容易产生疑问的地方，老师角色从面面俱到的讲解到重点点播；另一方面，需要丰富实验教学手段，训练学生的思维能力。能力的形成能够避免学生死记硬背，更好的完成知识的内化。教学内容
教学活动
学生活动
1.情境激趣
展示图片1：农业工作者在劳作的场景，他们将大量的农业氮肥—尿素，洒向田间。
展示图片2：中国农大成果转化项目产品—微生物菌剂。
引导学生思考：本节课题的研究意义。如果我们能够将这类细菌分离出来，用相关技术，制成混合菌剂。施肥时，添加，既能提高尿素的利用率，降低农业生产成本，又能减少因过量使用尿素而造成的环境污染。
观察图片思考并讨论如何实现科学技术向科学成果的转化。
温故知新
温故知新1—培养基的制备
①技术展示：结合实验流程图片，学生回顾技术原理和相应操作。
②技术点评：老师对学生展示进行总结和评价。
温故知新2—接种技术（平板划线法）
①技术展示：结合自制实验微课，学生展示技术操作，讲解相应技术原理。
②技术点评：老师对学生展示进行总结和评价。
温故知新3—接种技术（稀释涂布平板法）
①技术展示：结合自制实验微课，学生展示技术操作，讲解相应技术原理。
②技术点评：老师对学生展示进行总结和评价。
温故知新4—无菌技术
①技术展示：学生回顾技术原理和相应方法。
②技术点评：老师对学生展示进行总结和评价。
①回顾、讨论、交流相关知识点；
②听取老师对知识点的讲解分析；
3、理论指导
结合自制微课，社团成员进行《校园土壤中真菌的分离与计数》研究展示。
理论指导1—筛选菌株
探究1：结合社团活动中马丁氏培养基的选择作用，探究设计选择性培养基的一般方法。
小结1：方法a加入某种化学物质；方法b改变培养基中营养成分；方法c改变微生物的培养条件。
探究2：结合实例PCR技术中用到的耐高温Taq酶，探究寻找实验样品方法。
小结2：在寻找目的菌株时，要根据它对生存环境的要求，到相应环境中去寻找。
归纳原理：人为提供有利于目的菌株生长的条件（包括营养、温度、pH等），同时抑制或阻止其他微生物生长。
结论：本课题中所用到的培养基应该以尿素作为唯一氮源，土壤样品应该到富含尿素的环境中去找采集（农田、厕所、化肥厂等）。
①探究选择性培养基的设计方法；
②探究寻找实验样品方法；
③讨论、交流、总结、归纳筛选菌株的方法；
3.理论指导
理论指导2—统计菌落数目
展示社团活动中取得的成果—每克校园土壤中真菌的数量为2.2×106株。
探究1：接种方法及原理
小结1：采用的接种技术是稀释涂布平板法，当样品的稀释度足够高时，在培养基表面形成单个菌落，一个菌落来源于样品稀释液中的一个活菌。
探究2：稀释度的确定
小结2：不同微生物采用不同稀释度，通常第一次选择范围宽泛，保证从中选择出菌落数目在30~300之间的平板。
探究3：数据的处理（结合教材资料）
小结3：每克样品中的菌落数=（C÷V）×M
其中，C代表某一稀释度下平板生长的平均菌落数，V代表涂布平板时所用的稀释液的体积（ml），M代表稀释倍数。
结论：本课题种稀释度应小于筛选细菌的稀释度，数据处理可采用公式（C÷V）×M。
①探究如何统计菌落数目；
②讨论并交流接种方及原理；
③资料分析，总结并归纳数据处理的方法；
3、理论指导
理论指导3—设置对照
展示社团活动中取得的成果—每克校园土壤中真菌的数量为2.2×106株。
探究：数据的可信度？是否被污染？如何证明马丁氏培养基具有选择作用？
小结1：设置空白对照。将未接种的平板和接种后的平板，放在相同且适宜的环境中培养，如果对照组有菌落生长，则说明被污染，反之则说明未被污染。
小结2：去除马丁氏培养基中的孟加拉红和链霉素。因失去对细菌和霉菌没有抑制作用，预期生长出来的菌落数目比马丁氏培养基上的数目多，从而说明马丁氏培养基的选择作用。
结论：本课题研究中应该设置空白对照和对比对照。
①探究如何设置对照；
②分析、归纳、总结设置对照的方法和原理。
3、理论指导
理论指导4—培养与观察。
展示社团活动中取得的成果。
探究1：参阅资料，接种后培养多久能形成菌落？
小结1：恒温培养箱中28℃培养3天。
探究2：结合社团活动成果（土壤真菌和教室空气中细菌），分析如何区分不同菌落？
小结2：除颜色、大小外，还可以从边缘、突起、形态等方面区分。
结论：本课题中细菌的培养温度为30—37℃，恒温培养1—2天。
①探究微生物的培养与观察；
②分析资料，讨论、交流并归纳出区分不同菌落的方法。
3、理论指导
理论指导5—初步鉴定 展示教师预实验成果：从厕所地面上，分离得到了3株分解尿素的细菌，接种到添加酚红指示剂的培养基中，结果两株红色，一株黄色。
探究：结合酚红指示剂的功能及教材中课题背景，分析实验结果。
小结：尿素被分解为氨，与水结合显示碱性，故为红色；氨被硝化细菌利用，转化为亚硝酸或硝酸，显示酸性，故为黄色。
结论：本课题可采用同样的方法，将分离得到的菌种做初步鉴定。
①案例分析，讨论并交流；
②归纳鉴定原理。
4.实验流程
在理清研究思路的基础上，探究本课题的研究流程应该是怎样的？
结论：①准备工作②取样③样品稀释④涂布平板⑤培养⑥观察与记录⑦数据处理⑧初步鉴定
讨论、交流、展示
实验流程
5.走近科学
倡议：科学理论服务于科学技术，技术服务于社会。各研究小组，秉承“追寻科学家足迹、铸就新科技梦想”理念，展开课题研究，期待大家在《成果展示课》上的出彩表现。
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及时进行课堂巩固
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课题2 土壤中分解尿素的细菌的分离与计数
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[精] 人教版新课标高中生物选修一 同步课件第二章 第二节 土壤中分解尿素的细菌的分离与计数 课件
[精] 高中生物 2.2土壤中分解尿素的细菌的分离与计数课件 新人教版选修1
高效课堂同步课件：2-2土壤中分解尿素的细菌的分离与计数（选修1）
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人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》复习课件
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》课件
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》课件
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》（第1课时）课件
高中生物 2.2土壤中分解尿素的细菌的分离与计数课件 新人教版选修1
高中生物 2.2土壤中分解尿素的细菌的分离与计数教案 新人教版选修1
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》（第1课时）教案
人教版 (新课标)高中生物选修1 2-2《土壤中分解尿素的细菌的分离与计数》教案
人教版高中生物选修一《生物技术实践》2.2 土壤中分解尿素的细菌的分离与计数导学案
生物：2.2《土壤中分解尿素的细菌的分离与计数》学案（3）（新人教版选修1）
生物：2.2《土壤中分解尿素的细菌的分离与计数》学案（1）（新人教版选修1）
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9171 | https://onesearch.neu.edu/discovery/fulldisplay?docid=alma9952563321801401&context=L&vid=01NEU_INST:NU&lang=en&adaptor=Local%20Search%20Engine&tab=Everything&query=sub%2Ccontains%2C%20%20Pharmacology%20%2CAND&mode=advanced&offset=0 | Full display result
E-book
Katzung's Basic & Clinical Pharmacology, 16e
; Vanderah, Todd W., editor.;
New York, N.Y. : McGraw Hill LLC 2023
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Title
Katzung's Basic & Clinical Pharmacology, 16e
Katzung's Basic & Clinical Pharmacology, 16e
Katzung's Basic & Clinical Pharmacology, 16e
Author
Vanderah, Todd W., editor.
Vanderah, Todd W., editor.
Vanderah, Todd W., editor.
Edition
Sixteenth Edition.
Sixteenth Edition.
Sixteenth Edition.
Rights
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
Licensed for use by Northeastern University faculty, staff, and currently enrolled students.
Subjects
Clinical pharmacology
Clinical pharmacology
Clinical pharmacology
Pharmacology
Pharmacology
Pharmacology
Pharmacology, Clinical
Pharmacology, Clinical
Pharmacology, Clinical
Description
The book is designed to provide a comprehensive, authoritative, and readable pharmacology textbook for students in the health sciences.
The book is designed to provide a comprehensive, authoritative, and readable pharmacology textbook for students in the health sciences.
The book is designed to provide a comprehensive, authoritative, and readable pharmacology textbook for students in the health sciences.
Publisher
New York, N.Y. : McGraw Hill LLC
New York, N.Y. : McGraw Hill LLC
New York, N.Y. : McGraw Hill LLC
Creation Date
2023
2023
2023
Bibliographical Notes
Includes bibliographical references and index.
Includes bibliographical references and index.
Includes bibliographical references and index.
Contents
SECTION I: BASIC PRINCIPLES -- Chapter 1: Introduction: The Nature of Drugs & Drug Development & Regulation -- Chapter 2: Drug Receptors & Pharmacodynamics -- Chapter 3: Pharmacokinetics & Pharmacodynamics: Rational Dosing & the Time Course of Drug Action -- Chapter 4: Drug Biotransformation -- Chapter 5: Pharmacogenomics -- SECTION II: AUTONOMIC DRUGS -- Chapter 6: Introduction to Autonomic Pharmacology -- Chapter 7: Cholinoceptor-Activating & Cholinesterase-Inhibiting Drugs -- Chapter 8: Cholinoceptor-Blocking Drugs -- Chapter 9: Adrenoceptor Agonists & Sympathomimetic Drugs -- Chapter 10: Adrenoceptor Antagonist Drugs -- SECTION III: CARDIOVASCULAR-RENAL DRUGS -- Chapter 11: Antihypertensive Agents -- Chapter 12: Vasodilators & the Treatment of Angina Pectoris & Coronary Syndromes -- Chapter 13: Drugs Used in Heart Failure -- Chapter 14: Agents Used in Cardiac Arrhythmias -- Chapter 15: Diuretic Agents -- SECTION IV: DRUGS WITH IMPORTANT ACTIONS ON SMOOTH MUSCLE -- Chapter 16: Histamine, Serotonin, Anti-Obesity Drugs, & the Ergot Alkaloids -- Chapter 17: Vasoactive Peptides -- Chapter 18: The Eicosanoids: Prostaglandins, Thromboxanes, Leukotrienes, & Related Compounds -- Chapter 19: Nitric Oxide -- Chapter 20: Drugs Used in Asthma & Chronic Obstructive Pulmonary Disease -- SECTION V: DRUGS THAT ACT IN THE CENTRAL NERVOUS SYSTEM -- Chapter 21: Introduction to the Pharmacology of CNS Drugs -- Chapter 22: Sedative-Hypnotic Drugs -- Chapter 23: The Alcohols -- Chapter 24: Antiseizure Medications -- Chapter 25: General Anesthetics -- Chapter 26: Local Anesthetics -- Chapter 27: Skeletal Muscle Relaxants -- Chapter 28: Pharmacologic Management of Parkinsonism & Other Movement Disorders -- Chapter 29: Antipsychotic Agents & Lithium -- Chapter 30: Antidepressant Agents -- Chapter 31: Opioid Agonists & Antagonists -- Chapter 32: Drugs of Abuse -- SECTION VI: DRUGS USED TO TREAT DISEASES OF THE BLOOD, INFLAMMATION, & GOUT -- Chapter 33: Agents Used in Cytopenias; Hematopoietic Growth Factors -- Chapter 34: Drugs Used in Disorders of Coagulation -- Chapter 35: Agents Used in Dyslipidemia -- Chapter 36: Nonsteroidal Anti-Inflammatory Drugs, Disease-Modifying Antirheumatic Drugs, Nonopioid Analgesics, & Drugs Used in Gout -- SECTION VII: ENDOCRINE DRUGS -- Chapter 37: Hypothalamic & Pituitary Hormones -- Chapter 38: Thyroid & Antithyroid Drugs -- Chapter 39: Adrenocorticosteroids & Adrenocortical Antagonists -- Chapter 40: The Gonadal Hormones & Inhibitors -- Chapter 41: Pancreatic Hormones & Glucose-Lowering Drugs -- Chapter 42: Agents That Affect Bone Mineral Homeostasis -- SECTION VIII: CHEMOTHERAPEUTIC DRUGS -- Section VIII: Chemotherapeutic Drugs -- Chapter 43: Beta-Lactam & Other Cell Wall- & Membrane-Active Antibiotics -- Chapter 44: Tetracyclines, Macrolides, Clindamycin, Chloramphenicol, Streptogramins, Oxazolidinones, & Pleuromutilins -- Chapter 45: Aminoglycosides & Spectinomycin -- Chapter 46: Sulfonamides, Trimethoprim, & Quinolones -- Chapter 47: Antimycobacterial Drugs -- Chapter 48: Antifungal Agents -- Chapter 49: Antiviral Agents -- Chapter 50: Miscellaneous Antimicrobial Agents; Disinfectants, Antiseptics, & Sterilants -- Chapter 51: Clinical Use of Antimicrobial Agents -- Chapter 52: Antiprotozoal Drugs -- Chapter 53: Pharmacology of the Antihelminthic Drugs -- Chapter 54: Cancer Chemotherapy -- Chapter 55: Immunopharmacology -- SECTION IX: TOXICOLOGY -- Chapter 56: Introduction to Toxicology: Occupational & Environmental -- Chapter 57: Heavy Metal Intoxication & Chelators -- Chapter 58: Management of the Poisoned Patient -- SECTION X: SPECIAL TOPICS -- Chapter 59: Special Aspects of Perinatal & Pediatric Pharmacology -- Chapter 60: Special Aspects of Geriatric Pharmacology -- Chapter 61: Dermatologic Pharmacology -- Chapter 62: Drugs Used in the Treatment of Gastrointestinal Diseases -- Chapter 63: Cannabinoid Drugs -- Chapter 64: Therapeutic & Toxic Potential of Over-the-Counter Agents -- Chapter 65: Dietary Supplements & Herbal Medications -- Chapter 66: Rational Prescribing & Prescription Writing -- Chapter 67: Important Drug Interactions & Their Mechanisms -- Appendix 1: Vaccines, Immune Globulins, & Other Complex Biologic Products -- Appendix 2: Additional Pharmacology Competencies According to Competency-Based -- Appendix 3: Pharmacology Competencies According to CBME Guidelines.
SECTION I: BASIC PRINCIPLES -- Chapter 1: Introduction: The Nature of Drugs & Drug Development & Regulation -- Chapter 2: Drug Receptors & Pharmacodynamics -- Chapter 3: Pharmacokinetics & Pharmacodynamics: Rational Dosing & the Time Course of Drug Action -- Chapter 4: Drug Biotransformation -- Chapter 5: Pharmacogenomics -- SECTION II: AUTONOMIC DRUGS -- Chapter 6: Introduction to Autonomic Pharmacology -- Chapter 7: Cholinoceptor-Activating & Cholinesterase-Inhibiting Drugs -- Chapter 8: Cholinoceptor-Blocking Drugs -- Chapter 9: Adrenoceptor Agonists & Sympathomimetic Drugs -- Chapter 10: Adrenoceptor Antagonist Drugs -- SECTION III: CARDIOVASCULAR-RENAL DRUGS -- Chapter 11: Antihypertensive Agents -- Chapter 12: Vasodilators & the Treatment of Angina Pectoris & Coronary Syndromes -- Chapter 13: Drugs Used in Heart Failure -- Chapter 14: Agents Used in Cardiac Arrhythmias -- Chapter 15: Diuretic Agents -- SECTION IV: DRUGS WITH IMPORTANT ACTIONS ON SMOOTH MUSCLE -- Chapter 16: Histamine, Serotonin, Anti-Obesity Drugs, & the Ergot Alkaloids -- Chapter 17: Vasoactive Peptides -- Chapter 18: The Eicosanoids: Prostaglandins, Thromboxanes, Leukotrienes, & Related Compounds -- Chapter 19: Nitric Oxide -- Chapter 20: Drugs Used in Asthma & Chronic Obstructive Pulmonary Disease -- SECTION V: DRUGS THAT ACT IN THE CENTRAL NERVOUS SYSTEM -- Chapter 21: Introduction to the Pharmacology of CNS Drugs -- Chapter 22: Sedative-Hypnotic Drugs -- Chapter 23: The Alcohols -- Chapter 24: Antiseizure Medications -- Chapter 25: General Anesthetics -- Chapter 26: Local Anesthetics -- Chapter 27: Skeletal Muscle Relaxants -- Chapter 28: Pharmacologic Management of Parkinsonism & Other Movement Disorders -- Chapter 29: Antipsychotic Agents & Lithium -- Chapter 30: Antidepressant Agents -- Chapter 31: Opioid Agonists & Antagonists -- Chapter 32: Drugs of Abuse -- SECTION VI: DRUGS USED TO TREAT DISEASES OF THE BLOOD, INFLAMMATION, & GOUT -- Chapter 33: Agents Used in Cytopenias; Hematopoietic Growth Factors -- Chapter 34: Drugs Used in Disorders of Coagulation -- Chapter 35: Agents Used in Dyslipidemia -- Chapter 36: Nonsteroidal Anti-Inflammatory Drugs, Disease-Modifying Antirheumatic Drugs, Nonopioid Analgesics, & Drugs Used in Gout -- SECTION VII: ENDOCRINE DRUGS -- Chapter 37: Hypothalamic & Pituitary Hormones -- Chapter 38: Thyroid & Antithyroid Drugs -- Chapter 39: Adrenocorticosteroids & Adrenocortical Antagonists -- Chapter 40: The Gonadal Hormones & Inhibitors -- Chapter 41: Pancreatic Hormones & Glucose-Lowering Drugs -- Chapter 42: Agents That Affect Bone Mineral Homeostasis -- SECTION VIII: CHEMOTHERAPEUTIC DRUGS -- Section VIII: Chemotherapeutic Drugs -- Chapter 43: Beta-Lactam & Other Cell Wall- & Membrane-Active Antibiotics -- Chapter 44: Tetracyclines, Macrolides, Clindamycin, Chloramphenicol, Streptogramins, Oxazolidinones, & Pleuromutilins -- Chapter 45: Aminoglycosides & Spectinomycin -- Chapter 46: Sulfonamides, Trimethoprim, & Quinolones -- Chapter 47: Antimycobacterial Drugs -- Chapter 48: Antifungal Agents -- Chapter 49: Antiviral Agents -- Chapter 50: Miscellaneous Antimicrobial Agents; Disinfectants, Antiseptics, & Sterilants -- Chapter 51: Clinical Use of Antimicrobial Agents -- Chapter 52: Antiprotozoal Drugs -- Chapter 53: Pharmacology of the Antihelminthic Drugs -- Chapter 54: Cancer Chemotherapy -- Chapter 55: Immunopharmacology -- SECTION IX: TOXICOLOGY -- Chapter 56: Introduction to Toxicology: Occupational & Environmental -- Chapter 57: Heavy Metal Intoxication & Chelators -- Chapter 58: Management of the Poisoned Patient -- SECTION X: SPECIAL TOPICS -- Chapter 59: Special Aspects of Perinatal & Pediatric Pharmacology -- Chapter 60: Special Aspects of Geriatric Pharmacology -- Chapter 61: Dermatologic Pharmacology -- Chapter 62: Drugs Used in the Treatment of Gastrointestinal Diseases -- Chapter 63: Cannabinoid Drugs -- Chapter 64: Therapeutic & Toxic Potential of Over-the-Counter Agents -- Chapter 65: Dietary Supplements & Herbal Medications -- Chapter 66: Rational Prescribing & Prescription Writing -- Chapter 67: Important Drug Interactions & Their Mechanisms -- Appendix 1: Vaccines, Immune Globulins, & Other Complex Biologic Products -- Appendix 2: Additional Pharmacology Competencies According to Competency-Based -- Appendix 3: Pharmacology Competencies According to CBME Guidelines.
SECTION I: BASIC PRINCIPLES -- Chapter 1: Introduction: The Nature of Drugs & Drug Development & Regulation -- Chapter 2: Drug Receptors & Pharmacodynamics -- Chapter 3: Pharmacokinetics & Pharmacodynamics: Rational Dosing & the Time Course of Drug Action -- Chapter 4: Drug Biotransformation -- Chapter 5: Pharmacogenomics -- SECTION II: AUTONOMIC DRUGS -- Chapter 6: Introduction to Autonomic Pharmacology -- Chapter 7: Cholinoceptor-Activating & Cholinesterase-Inhibiting Drugs -- Chapter 8: Cholinoceptor-Blocking Drugs -- Chapter 9: Adrenoceptor Agonists & Sympathomimetic Drugs -- Chapter 10: Adrenoceptor Antagonist Drugs -- SECTION III: CARDIOVASCULAR-RENAL DRUGS -- Chapter 11: Antihypertensive Agents -- Chapter 12: Vasodilators & the Treatment of Angina Pectoris & Coronary Syndromes -- Chapter 13: Drugs Used in Heart Failure -- Chapter 14: Agents Used in Cardiac Arrhythmias -- Chapter 15: Diuretic Agents -- SECTION IV: DRUGS WITH IMPORTANT ACTIONS ON SMOOTH MUSCLE -- Chapter 16: Histamine, Serotonin, Anti-Obesity Drugs, & the Ergot Alkaloids -- Chapter 17: Vasoactive Peptides -- Chapter 18: The Eicosanoids: Prostaglandins, Thromboxanes, Leukotrienes, & Related Compounds -- Chapter 19: Nitric Oxide -- Chapter 20: Drugs Used in Asthma & Chronic Obstructive Pulmonary Disease -- SECTION V: DRUGS THAT ACT IN THE CENTRAL NERVOUS SYSTEM -- Chapter 21: Introduction to the Pharmacology of CNS Drugs -- Chapter 22: Sedative-Hypnotic Drugs -- Chapter 23: The Alcohols -- Chapter 24: Antiseizure Medications -- Chapter 25: General Anesthetics -- Chapter 26: Local Anesthetics -- Chapter 27: Skeletal Muscle Relaxants -- Chapter 28: Pharmacologic Management of Parkinsonism & Other Movement Disorders -- Chapter 29: Antipsychotic Agents & Lithium -- Chapter 30: Antidepressant Agents -- Chapter 31: Opioid Agonists & Antagonists -- Chapter 32: Drugs of Abuse -- SECTION VI: DRUGS USED TO TREAT DISEASES OF THE BLOOD, INFLAMMATION, & GOUT -- Chapter 33: Agents Used in Cytopenias; Hematopoietic Growth Factors -- Chapter 34: Drugs Used in Disorders of Coagulation -- Chapter 35: Agents Used in Dyslipidemia -- Chapter 36: Nonsteroidal Anti-Inflammatory Drugs, Disease-Modifying Antirheumatic Drugs, Nonopioid Analgesics, & Drugs Used in Gout -- SECTION VII: ENDOCRINE DRUGS -- Chapter 37: Hypothalamic & Pituitary Hormones -- Chapter 38: Thyroid & Antithyroid Drugs -- Chapter 39: Adrenocorticosteroids & Adrenocortical Antagonists -- Chapter 40: The Gonadal Hormones & Inhibitors -- Chapter 41: Pancreatic Hormones & Glucose-Lowering Drugs -- Chapter 42: Agents That Affect Bone Mineral Homeostasis -- SECTION VIII: CHEMOTHERAPEUTIC DRUGS -- Section VIII: Chemotherapeutic Drugs -- Chapter 43: Beta-Lactam & Other Cell Wall- & Membrane-Active Antibiotics -- Chapter 44: Tetracyclines, Macrolides, Clindamycin, Chloramphenicol, Streptogramins, Oxazolidinones, & Pleuromutilins -- Chapter 45: Aminoglycosides & Spectinomycin -- Chapter 46: Sulfonamides, Trimethoprim, & Quinolones -- Chapter 47: Antimycobacterial Drugs -- Chapter 48: Antifungal Agents -- Chapter 49: Antiviral Agents -- Chapter 50: Miscellaneous Antimicrobial Agents; Disinfectants, Antiseptics, & Sterilants -- Chapter 51: Clinical Use of Antimicrobial Agents -- Chapter 52: Antiprotozoal Drugs -- Chapter 53: Pharmacology of the Antihelminthic Drugs -- Chapter 54: Cancer Chemotherapy -- Chapter 55: Immunopharmacology -- SECTION IX: TOXICOLOGY -- Chapter 56: Introduction to Toxicology: Occupational & Environmental -- Chapter 57: Heavy Metal Intoxication & Chelators -- Chapter 58: Management of the Poisoned Patient -- SECTION X: SPECIAL TOPICS -- Chapter 59: Special Aspects of Perinatal & Pediatric Pharmacology -- Chapter 60: Special Aspects of Geriatric Pharmacology -- Chapter 61: Dermatologic Pharmacology -- Chapter 62: Drugs Used in the Treatment of Gastrointestinal Diseases -- Chapter 63: Cannabinoid Drugs -- Chapter 64: Therapeutic & Toxic Potential of Over-the-Counter Agents -- Chapter 65: Dietary Supplements & Herbal Medications -- Chapter 66: Rational Prescribing & Prescription Writing -- Chapter 67: Important Drug Interactions & Their Mechanisms -- Appendix 1: Vaccines, Immune Globulins, & Other Complex Biologic Products -- Appendix 2: Additional Pharmacology Competencies According to Competency-Based -- Appendix 3: Pharmacology Competencies According to CBME Guidelines.
Related Titles
Available in other form: Online version : Katzung's Basic & Clinical Pharmacology, 16e. Sixteenth edition. New York, N.Y. : McGraw-Hill Education LLC., 2023
Available in other form: Online version : Katzung's Basic & Clinical Pharmacology, 16e. Sixteenth edition. New York, N.Y. : McGraw-Hill Education LLC., 2023
Available in other form: Online version : Katzung's Basic & Clinical Pharmacology, 16e. Sixteenth edition. New York, N.Y. : McGraw-Hill Education LLC., 2023
Series
McGraw-Hill's AccessScience.
McGraw-Hill's AccessScience.
McGraw-Hill's AccessScience.
Format
1 online resource (1347 pages) : illustrations
1 online resource (1347 pages) : illustrations
1 online resource (1347 pages) : illustrations
Identifier
ISBN : 9781260463316 (e-ISBN)
ISBN : 9781260463316 (e-ISBN)
ISBN : 9781260463316 (e-ISBN)
ISBN : 1260463311 (e-ISBN)
ISBN : 1260463311 (e-ISBN)
ISBN : 1260463311 (e-ISBN)
ISBN : 9781260463309
ISBN : 9781260463309
ISBN : 9781260463309
English
English
Library Catalog
Library Catalog
Electronic reproduction. New York, N.Y. : McGraw Hill, 2023. Mode of access: World Wide Web. System requirements: Web browser. Access may be restricted to users at subscribing institutions.
Electronic reproduction. New York, N.Y. : McGraw Hill, 2023. Mode of access: World Wide Web. System requirements: Web browser. Access may be restricted to users at subscribing institutions.
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9172 | https://math.stackexchange.com/questions/1279144/how-to-know-if-equation-can-be-solved-by-factorising-before-trying | Stack Exchange Network
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How to know if equation can be solved by factorising before trying?
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Asked
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$\begingroup$
So, I have core 1 test tomorrow and there is a lot of solving of quadratic equations without calculator and my weakest point is the time I waste in trying to factorise and equation but then it ends up it doesn't factorise or the opposite way in which I use equation while I could've done factorising. Is there any way you can know if equation factorise or not?
factoring
quadratics
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asked May 12, 2015 at 18:19
HannahHannah
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2
$\begingroup$ Check $b^2-4ac$ : If this is negative, there are no real roots; if it is not a square, you will not find two integer roots, (assuming the quadratic is of the form $ax^2+bx+c=0$). And if you need to use the quadratic formula, you'll need that number anyway. $\endgroup$
Joffan
– Joffan
2015-05-12 18:24:43 +00:00
Commented May 12, 2015 at 18:24
$\begingroup$ You can look at discriminant $D=b^2-4ac$, if it is a perfect square you can factorise the trinomial $ax^2+bx+c$ as a product of two factors with rational coefficients. If $D=b^2-4ac$ isn't a perfect square, then $ax^2+bx+c$ still can be factorised, but it includes factors with surds. $\endgroup$
Ángel Mario Gallegos
– Ángel Mario Gallegos
2015-05-12 18:24:53 +00:00
Commented May 12, 2015 at 18:24
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6 Answers 6
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$\begingroup$
Check the discriminant. If it is a perfect square, the radical expression in the quadratic formula becomes an integer (I'm assuming you've cleared fractions before beginning the problem, and all coefficients are integers after that).
Recall that the discriminant for $ax^2 + bx + c = 0$ is $\Delta = b^2-4ac$, and that the solution(s) are $$x = \frac{-b\pm\sqrt{\Delta}}{2a}$$
I'm also assuming you're only interested in factoring if you can do it without using radicals. For example, you could factor $x^2-2$ as $(x+\sqrt{2})(x-\sqrt{2})$, but I'm guessing that doesn't interest you.
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answered May 12, 2015 at 18:24
MPWMPW
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For any quadratic equation $ax^2+bx+c=0$, $a,b,c\in \mathbb{R}$, there exist solution(s) (i.e., you can factorise) if and only if the discriminant $$D=b^2-4ac$$ is higher than or equal to zero.
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answered May 12, 2015 at 18:24
Alberto DebernardiAlberto Debernardi
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Two ways. The first is particularly useful if the leading coefficient $a$ (as in $ax^2$) is $1$: If two factors $(x-m)(x-n)$ exist with $m$ and $n$ integers (or even rational) then $m$ and $n$ wil always be factors the constanc coefficient $c$. This means you need to try very few combinations, since depending on the sign of $c$ of the factors will will always be dictated by $b$.
For example, $$ x^2 +6x +10$$ could only have $m,n$ as $(1,10)$ or $(2,5)$; since neither of those add up to $6$ you can't factor this with easy factors.
The other way is to find $b^2-4ac$. If that is a perfect square, then the equation can be factored nicely. If not, then at least you are halfway toward finding the roots using the quadratic formula.
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answered May 12, 2015 at 18:27
Mark FischlerMark Fischler
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You can only factorise easily (without involving surds) if the discriminant is a perfect square. Mind you ,if you've already worked the discriminant out, you may as well go on and solve using the formula.
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answered May 12, 2015 at 18:26
David QuinnDavid Quinn
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You can check if their are easy factors by applying the rational root theorem, which in the special case of $$x^2+bx+c$$ says that the only rational roots can be factors of $c$, and in the more general case of $$ax^2+bx+c$$ says the only rational roots have to be of a form $\dfrac{\text{factor of }c}{\text{factor of }a}$.
Still, it seems that applying the quadratic formula isn't so hard to be fast at in simple cases.
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answered May 12, 2015 at 18:29
Peter WoolfittPeter Woolfitt
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For example, x2+6x+10 x2+6x+10 could only have m,nm,n as (1,10)(1,10) or (2,5)(2,5); since neither of those add up to 66 you can't factor this with easy factors.
The other way is to find b24acb24ac. If that is a perfect square, then the equation can be factored nicely. If not, then at least you are halfway toward finding the roots using the quadratic formula.
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answered Jan 25, 2017 at 1:45
Christopher BlantonChristopher Blanton
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25: The Binomial Theorem
Precalculus (Tradler and Carley)
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25.2: Binomial Expansion
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25.2: Binomial Expansion
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25.1: The Binomial Theorem
25.3: Exercises
Page ID 54485
Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Example 25.2.1
2. Observation: k th term of expansion
3. Example 25.2.2
4. Example 25.2.3
Using the binomial theorem, we can also expand more general powers of sums or differences.
Example 25.2.1
Expand the expression.
(x 2+2y 3)5
(2xy 2−4 y 2)3
(2+1)6
(i−3)4
Solution
We use the binomial theorem with a=x 2 and b=2y 3:
(x 2+2y 3)5=(x 2)5+(5 1)(x 2)4(2y 3)+(5 2)(x 2)3(2y 3)2+(5 3)(x 2)2(2y 3)3+(5 4)(x 2)(2y 3)4+(2y 3)5=x 10+5x 8⋅2y 3+10x 6⋅4y 6+10x 4⋅2 3y 9+5x 2⋅2 4y 12+2 5y 15=x 10+10x 8y 3+40x 6y 6+80x 4y 9+80x 2y 12+32y 15
For part (b), it is a=2xy 2 and b=−4 y 2.
(2xy 2−4 y 2)3=(2xy 2)3+(3 1)(2xy 2)2(−4 y 2)+(3 2)(2xy 2)(−4 y 2)2+(−4 y 2)3=2 3x 3y 6+3⋅2 2x 2y 4(−4 y 2)+3(2xy 2)(−1)24 2 y 4+(−1)34 3 y 6=8x 3y 6−48x 2y 2+96x⋅1 y 2−64⋅1 y 6
Similarly, for part (c), we now have a=2 and b=1:
(2+1)6=(2)6+(6 1)(2)5⋅1+(6 2)(2)4⋅1 2+(6 3)(2)3⋅1 3+(6 4)(2)2⋅1 4+(6 5)(2)⋅1 5+1 6=64+6⋅32+15⋅16+20⋅8+15⋅4+6⋅2+1=8+6⋅16⋅2+15⋅4+20⋅4⋅2+15⋅2+6⋅2+1=8+242+60+402+30+62+1=99+702
Note that the last expression cannot be simplified any further (due to the order of operations).
Finally, we have a=i and b=−3, and we use the fact that i 2=−1, and therefore, i 3=−i and i 4=+1:
(i−3)4=i 4+(4 1)⋅i 3⋅(−3)+(4 2)⋅i 2⋅(−3)2+(4 3)⋅i⋅(−3)3+(−3)4=1+4⋅(−i)⋅(−3)+6⋅(−1)⋅9+4⋅i⋅(−27)+81=1+12i−54−108i+81=28−96i
In some instances it is not necessary to write the full binomial expansion, but it is enough to find a particular term, say the k th term of the expansion.
Observation: k th term of expansion
Recall, for example, the binomial expansion of (a+b)6:
(6 0)a 6b 0+(6 1)a 5b 1+(6 2)a 4b 2+(6 3)a 3b 3+(6 4)a 2b 4+(6 5)a 1b 5+(6 6)a 0b 6 first second third fourth fifth sixth seventh term term term term term term term
Note that the exponents of the a’s and b’s for each term always add up to 6, and that the exponents of a decrease from 6 to 0, and the exponents of b increase from 0 to 6. Furthermore observe that in the above expansion the fifth term is (6 4)a 2b 4.
In general, we define the k th term by the following formula:
(25.2.1)The k th term in the expansion of(a+b)n is:(n k−1)a n−k+1b k−1
Note in particular, that the k th term has a power of b given by b k−1 (and not b k), it has a binomial coefficient (n k−1), and the exponents of a and b add up to n.
Example 25.2.2
Determine:
the 4 th term in the binomial expansion of (p+3q)5
the 8 th term in the binomial expansion of (x 3y−2x 2)10
the 12 th term in the binomial expansion of (−5a b 7−b)15
Solution
We have a=p and b=3q, and n=5 and k=4. Thus, the binomial coefficient of the 4 th term is (5 3), the b-term is (3q)3, and the a-term is p 2. The 4 th term is therefore given by
(5 3)⋅p 2⋅(3q)3=10⋅p 2⋅3 3q 3=270p 2q 3
In this case, a=x 3y and b=−2x 2, and furthermore, n=10 and k=8. The binomial coefficient of the 8 th term is (10 7), the b-term is (−2x 2)7, and the a-term is (x 3y)3. Therefore, the 8 th term is
(10 7)⋅(x 3y)3⋅(−2x 2)7=120⋅x 9y 3⋅(−128)x 14=−15360⋅x 23y 3
Similarly, we obtain the 12 th term of (−5a b 7−b)15 as
(15 11)⋅(−5a b 7)4⋅(−b)11=1365⋅5 4a 4 b 28⋅(−b 11)=1365⋅625⋅a 4⋅(−b 11)b 28=−853125⋅a 4 b 17
Here is a variation of the above problem.
Example 25.2.3
Determine:
the x 4y 12-term in the binomial expansion of (5x 2+2y 3)6
the x 15-term in the binomial expansion of (x 3−x)7
the real part of the complex number (3+2i)4
Solution
In this case we have a=5x 2 and b=2y 3. The term x 4y 12 can be rewritten as x 4y 12=(x 2)2⋅(y 3)4, so that the full term (n k−1)a n−k+1b k−1 (including the coefficients) is
(6 4)⋅(5x 2)2⋅(2y 3)4=15⋅25x 4⋅16y 12=6000⋅x 4y 12
The various powers of x in (x 3−x)7 (in the order in which they appear in the binomial expansion) are:
(x 3)7=x 21,(x 3)6⋅x 1=x 19,(x 3)5⋅x 2=x 17,(x 3)4⋅x 3=x 15,…
The last term is precisely the x 15-term, that is we take the fourth term, k=4. We obtain a total term (including the coefficients) of
(7 3)⋅(x 3)4⋅(−x)3=35⋅x 12⋅(−x)3=−35⋅x 15
Recall that i n is real when n is even, and imaginary when n is odd:
i 1=i i 2=−1 i 3=−i i 4=1 i 5=i i 6=−1⋮
The real part of (3+2i)4 is therefore given by the first, third, and fifth term of the binomial expansion:
real part=(4 0)⋅3 4⋅(2i)0+(4 2)⋅3 2⋅(2i)2+(4 4)⋅3 0⋅(2i)4=1⋅81⋅1+6⋅9⋅4i 2+1⋅1⋅16i 4=81+216⋅(−1)+16⋅1=81−216+16=−119
The real part of (3+2i)4 is −119.
This page titled 25.2: Binomial Expansion is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.
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9175 | https://math.stackexchange.com/a/3840904 | combinatorics - Proof of Generalized Pigeonhole Principle used in MIT OCW course 6.042 in Lecture 16. - Mathematics Stack Exchange
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Proof of Generalized Pigeonhole Principle used in MIT OCW course 6.042 in Lecture 16. [closed]
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The Generalized Pigeonhole principle states that :
If |X| >= K|Y|, then for all f:X -> Y, there exists K+1 different elements of X that are mapped to the same element in Y.
Prove this.
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asked Sep 26, 2020 at 6:57
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What have you tried ?user577215664 –user577215664 2020-09-26 06:58:04 +00:00 Commented Sep 26, 2020 at 6:58
@Aryadeva I was having trouble with this statement itself. I mean, there need not be necessarily only k+1 diff. elements, there can be more too. But this was the exact statement that was given.wormhole –wormhole 2020-09-26 07:01:48 +00:00 Commented Sep 26, 2020 at 7:01
@SoumyanilDas "there exists K+1 K+1 elements" is equivalent to "at least K+1 K+1 elements", it does not mean "exactly K+1 K+1 elements"angryavian –angryavian 2020-09-26 07:04:07 +00:00 Commented Sep 26, 2020 at 7:04
@angryavian I see. I did not know this. Thanks a lot. Now no worries.wormhole –wormhole 2020-09-26 07:06:14 +00:00 Commented Sep 26, 2020 at 7:06
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The proposition is false. For example,
let Y be a singleton and |X| = k.
If |X| > k|Y| and f:X -> Y, then
exists y with k > |f−1−1(x)|.
Otherwise, for all y, |f−1−1(y)| <= k;
X = ∪∪{ f−1−1(y) : y in Y };
|X| <= k|Y|, a contradiction.
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answered Sep 26, 2020 at 9:03
William ElliotWilliam Elliot
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The proposition is true for finite non-empty Y Y if you replace |X|≥K|Y|X|≥K|Y| with |X|>K|Y||X|>K|Y|.DanielWainfleet –DanielWainfleet 2020-09-26 10:52:06 +00:00 Commented Sep 26, 2020 at 10:52
Wake up @DanielWainfleet, that's exactly what I proved.William Elliot –William Elliot 2020-09-26 20:32:47 +00:00 Commented Sep 26, 2020 at 20:32
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9176 | https://edis.ifas.ufl.edu/publication/PI193 | Water pH and the Effectiveness of Pesticides
Frederick M. Fishel and J. A. Ferrell
Water pH and the Effectiveness of Pesticides
Introduction
Some pesticides lose their effectiveness when mixed with alkaline (high-pH) water. This document discusses the effects of alkalinity and presents some methods for preventing this reaction in pesticide mixes.
Water pH and Pesticides
What is pH?
The term pH, potential of hydrogen, refers to a measure of the concentration of hydrogen ion (H+) and hydroxide ion (OH-) in a solution. If hydrogen predominates, the solution is acidic; if hydroxide predominates, the solution is basic, or alkaline. A logarithmic scale of 0 to 14 is used to measure pH. A pH value of 7 indicates neutrality. Values below 7 indicate acidic conditions; pH values above 7 indicate alkaline conditions. Because a logarithmic scale is used in measuring pH, a pH of 6 is 10 times more acidic than a pH of 7, and a pH of 5 is 100 times more acidic than a pH of 7.
How does pH affect pesticides?
Some pesticides, particularly carbamate and organophosphate insecticides, undergo a chemical reaction in the presence of alkaline water (water that has a pH value greater than 7). The reaction is known as alkaline hydrolysis, and it reduces the effectiveness of the pesticide's active ingredient. The speed with which the breakdown occurs depends on the specific chemical properties of the pesticide, the pH of the mix water, and the length of time the pesticide is in contact with the water. Spray-mix water with a pH value between 8 and 9 can cause rapid hydrolysis to the point that the degree of pest control is greatly diminished or lost.
Chemical breakdown of a pesticide is commonly referred to in terms of its half-life. A half-life is the period of time it takes for one-half (50% hydrolysis) of the amount of pesticide in the water to degrade. The half-lives of some commonly used insecticides/miticides are presented in Table 1.
There are water sources in Florida that have pH values between 7 and 9 (from neutral to alkaline). This is the case wherever water comes from a limestone aquifer, such as the Floridan (majority of groundwater withdrawal in Florida) or Biscayne (south Florida), or from lakes or canals that cut into limestone. There is some variability in these values even if they are within the same hydrologic region of the state. Both surface and ground water pH values fluctuate over time and even seasonally. One factor that influences the pH of open water is the amount of resident plant life. In these systems, there are high concentrations of carbonate in the water. The pH of the water may rise in poorly buffered systems because carbonate leads to increases of pH. Therefore, in some Florida water bodies with high levels of healthy aquatic plants, it is possible for pH to reach a measurement of 9 or 10.
How can I determine the pH of the water I use to mix with pesticides?
A water test is the surest means of determining whether a pH problem exists. The UF/IFAS Extension Soil Testing Laboratory in Gainesville offers a water test to the public for $10.00 per sample. The form and instructions are available through all county UF/IFAS Extension offices or can be printed directly from the ESTL website ( A faster and cheaper way to determine the pH level of water is to use test paper. However, paper test strips can vary by as much as 2 pH points. A pH meter will also provide fast results with more reliable and consistent readings. Meters that measure pH within 0.2 points of accuracy are available commercially for as little as $50. More expensive models have greater precision and may have the ability to conduct additional measurements such as electrical conductivity.
When should a pesticide solution be acidified in the spray tank?
If you know that your mix water has a pH of 7.5 or greater, consider lowering the pH, especially if you are applying a pesticide that is sensitive to high pH. A pH of 4 to 7 is recommended for mixing most pesticides; a value of 5.5 to 6.5 is ideal. If your spray rig will be left to stand for several hours before the contents are applied, consider adding an acidifying agent to prevent alkaline hydrolysis.
Some product labels will direct you to avoid mixing the pesticide with alkaline water or other specific alkaline materials such as lime, lime sulfur, or Bordeaux mixtures (Figure 1). You may also see statements that the activity of the pesticide will be reduced under alkaline conditions. The directions will state that a buffering or acidifying agent should be added to the spray tank. There are a few pesticidal materials that should not be acidified under any circumstances: sprays containing fixed copper fungicides (Bordeaux mixture, copper oxide, basic copper sulfate, copper hydroxide, etc.) and lime and lime sulfur. Their labels will contain specific statements.
Acidifiers and Buffering Agents
Acidifiers and buffering agents are a type of pesticide spray mix adjuvant. All adjuvant materials are added to the chemical formulation or spray tank to make the application more effective, safer, or easier for the applicator. Various commercially available acidifying agents will lower the pH of spray solution. Buffers are capable of changing the pH of a water solution to a level which will be maintained even if the pH of the solution changes. Like pesticides, their labels should be read and followed closely. The amount to add will depend upon the initial pH, the volume of water, and the desired final results.
Flumioxazin—An Herbicide Case Study in Florida
Flumioxazin is an herbicide registered for use in a wide range of agricultural commodities and aquatic sites and sold under many trade names. Flumioxazin has been shown to provide consistent control of several broadleaf weed species, but this molecule is susceptible to alkaline hydrolysis. At pH 5, flumioxazin is very stable and will persist in water for several days. However, as pH increases to 7 the half-life decreases to approximately 24 hours, and at pH 9 the half-life is a mere 15 minutes. Hence, mixing flumioxazin with high pH water can cause this herbicide to degrade and completely lose its herbicidal activity before it can be applied. When using flumioxazin, it is important to know whether the available water should be acidified to enhance herbicidal activity. Otherwise, it is possible to lose herbicidal efficacy simply because high pH water was used to fill the sprayer.
Summary
Determining the pH of the spray mix water and adding an acidifier, if necessary, is inexpensive compared to the cost of losing a pesticide's effectiveness. There are water sources in Florida that are alkaline by nature, and the addition of an acidifying agent to the spray mix is an easy and economical way to guarantee maximum results from your pesticide applications.
Additional Information
Purdum, E. D. 2002. Florida Waters: A Water Resources Manual from Florida's Water Management Districts. 120 pages.
Half-lives of commonly used insecticides/miticides.
| Active ingredient | pH 6 | pH 7 | pH 8 | pH 9 |
--- ---
| Azinphos-methyl | | 10 days | | 12 hours |
| Captan | | 8 hours | 10 minutes | 2 minutes |
| Carbaryl | 100–150 days | 24–30 days | 2–3 days | 1–3 days |
| Carbofuran | 200 | 40 days | 5 days | 3 days |
| Chlorpyrifos | | 35 days | 22 days | |
| Diazinon | | 70 days | | 29 days |
| Dimethoate | 12 hours | | | 1 hour |
| Disulfoton | 32 hours | | | 7 hours |
| Malathion | 8 days | 3 days | 19 hours | |
| Methomyl | 54 weeks | 38 weeks | 20 weeks | |
| Phosmet | | 1 day | 4 hours (pH 8.3) | 1 minute (pH 10) |
| Propargite | 331 days | | | 1 day |
| Trichlorfon | 4 days | 6 hours | 1 hour | |
Active ingredient
pH 6
pH 7
pH 8
pH 9
Azinphos-methyl
10 days
12 hours
Captan
8 hours
10 minutes
2 minutes
Carbaryl
100–150 days
24–30 days
2–3 days
1–3 days
Carbofuran
200
40 days
5 days
3 days
Chlorpyrifos
35 days
22 days
Diazinon
70 days
29 days
Dimethoate
12 hours
1 hour
Disulfoton
32 hours
7 hours
Malathion
8 days
3 days
19 hours
Methomyl
54 weeks
38 weeks
20 weeks
Phosmet
1 day
4 hours (pH 8.3)
1 minute (pH 10)
Propargite
331 days
1 day
Trichlorfon
4 days
6 hours
1 hour
Related Pages
Contact IFAS Communications
P.O. Box 110810
Gainesville, FL 32611
edishelp@ifas.ufl.edu
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(352) 392-1761
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Joint And Combined Variation Word Problems
Related Pages:
Proportions
Proportion Word Problems
Direct Variation
Inverse Variation
More Algebra Lessons
In these lessons, we will learn to solve word problems where a quantity varies in relation to two or more other quantities.
Share this page to Google Classroom
In mathematics, variation describes how one quantity changes in relation to another. There are several types of variation, each with its own unique relationship between variables: Direct Variation, Inverse Variation,
Joint Variation, and Combined Variation.
The following diagrams give the different types of variation: Direct Variation, Inverse Variation,
Joint Variation, and Combined Variation. Scroll down the page for examples and solutions.
Ratio & Proportion Worksheets
Practice your skills with the following worksheets:
Printable & Online Proportion Worksheets
In this lesson, we will look at examples for Joint Variation and Combined Variation. Check out the following links for examples other types of variations.
Direct Variation Problems
Inverse Variation Problems
What Is Joint Variation And Combined Variation?
Joint variation is a variation where a quantity
varies directly as the product of two or more other quantities.
For example, the area of a rectangle varies whenever its length or its width varies. We say that
A ∝ lw, where A is the area, l is the length and w is the width.
Combined variation is a variation where a quantity depends
on two (or more) other quantities, and varies directly with some of them
and varies inversely with others.
Example 1: A quantity varies inversely as two or more other quantities.
The figure below shows a rectangular solid with a fixed volume. Express its width, w, as a
joint variation in terms of its length, l, and height, h.
Solution:
w∝1lh
In other words, the longer the length l or the height h, the narrower is the width w.
Example 2: A quantity varies directly as one quantity and inversely as another.
The speed, s, of a moving object varies directly as the distance traveled, d, and varies
inversely as the time taken, t. Express s as a joint variation in terms of d and t.
Solution:
s ∝ dt
In other words, the longer the distance or the shorter the time, the faster is the speed.
How To Solve Joint Variation Problems?
Example:
Suppose y varies jointly as x and z. What is y when x = 2 and z = 3, if y = 20 when x = 4 and z = 3?
Show Video Lesson
Joint Variation Problems
Example:
z varies jointly with x and y. when x = 3, y = 8, z = 6. Find z, when x = 6 and y = 4.
Show Video Lesson
How to solve Joint Variation Word Problems and Applications?
Example:
The energy that an item possesses due to its motion is called kinetic energy.
The kinetic energy of an object (which is measured in joules) varies jointly with the mass of the
object and the square of its velocity.
If the kinetic energy of a 3 kg ball traveling 12 m/s is 216 Joules, how is the mass of a ball that
generates 250 Joules of energy when traveling at 10 m/s?
Show Video Lesson
Direct, Inverse and Joint Variation
Example:
Determine whether the data in the table is an example of direct, inverse or joint variation. Then,
identify the equation that represents the relationship.
Show Video Lesson
What Is Combined Variation?
In Algebra, sometimes we have functions that vary in more than one element. When this happens,
we say that the functions have joint variation or combined variation. Joint variation is
direct variation to more than one variable (for example, d = (r)(t)). With combined variation,
we have both direct variation and indirect variation.
How to set up and solve combined variation problems?
Example:
Suppose y varies jointly with x and z.
When y = 20, x = 6 and z = 10. Find y when x = 8 and z =15.
Show Video Lesson
Lesson on combining direct and inverse or joint and inverse variation
Example:
y varies directly as x and inversely
as the square of z, and when x = 32, y = 6 and z = 4. Find x when y = 10 and z = 3.
Show Video Lesson
How to solve problems involving joint and combined variation?
Examples:
If t varies jointly with u and the square of v, and t is 1152 when u is 8
and v is 4, find t when v is 5 and u is 5.
The amount of oil used by a ship traveling at a uniform speed varies jointly with
the distance and the square of the speed. If the ship uses 200 barrels of oil in
traveling 200 miles at 36 miles per hour, determine how many barrels of oil are used
when the ship travels 360 miles at 18 miles per hour.
Designer Dolls found that its number of Dress-Up Dolls sold, N, varies directly
with their advertising budget, A, and inversely proportional with the price of each doll,
P. When $54,00 was spent on advertising and the price of the doll is $90, then 9,600
units are sold. Determine the number of dolls sold if the amount of advertising budget
is increased to $144,000.
Show Video Lesson
Combined Variation
Example:
y varies jointly as x and z and inversely as w, and y = 3/2, when x = 2, z =3 and w = 4. Find the equation of variation.
Show Video Lesson
Check out many other Algebra Word Problems
Age Word Problems, Average Word Problems, Coin Word Problems, Consecutive Integer Word Problems, Digit Word Problems, Distance Word Problems, Fraction Word Problems, Geometry Word Problems, Integer Word Problems, Interest Word Problems, Lever Word Problems, Mixture Word Problems, Money Word Problems, Motion & Distance Word Problems, Number Sequence Word Problems, Proportion Word Problems, Quadratic Equation Word Problems, Ratio Word Problems, Symbol Word Problems, Variation Word Problems, Work Word Problems.
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9178 | https://www.espressoenglish.net/stative-verbs-action-verbs-and-verbs-that-are-both/ | Skip to content
Stative Verbs, Action Verbs, and Verbs that are Both
Did you know that we have two types of verbs in English? They are action verbs (or dynamic verbs) and state verbs (also called “stative verbs” or “state of being verbs”).
There are even some verbs that can be BOTH action verbs and stative verbs, depending on the use and context! We’ve also got some special grammar rules about state verbs, and today you’ll learn all about them.
I’ve made the lesson guide available for you to download for free – just click below and enter your e-mail address to receive that free PDF download.
Download free lesson PDF
Let’s start with action verbs:
What are Action Verbs?
Action verbs (or dynamic verbs)are verbs that describe actions. We can use them in the simple or continuous verb tenses.
Action verbs: Examples
walk
Every day I walkhome from class.
I‘m walkingto the store right now.
read
I readmostly historical fiction.
I‘ve been readinga novel that takes place during colonial times.
help
My sister helpsme with my homework.
My father is helpingme learn how to drive.
watch
Bob watchesfour hours of TV every night.
Last night, he got angry at me because I changed the channel while he was watchinghis favorite show.
What are Stative Verbs?
Stative verbs(or state verbs) describe a status or qualityof something… NOT an action. Verbs of perception, opinion, the senses, emotion, possession, and state of being are often stative verbs.
Here are some examples:
Stative verbs of opinion / perception:
know, believe, understand, recognize, prefer, agree/disagree, approve/disapprove, suppose, suspect
I’ve knownmy best friend since childhood.
I‘ve been knowing my best friend since childhood.
We agreewith you.
We‘re agreeingwith you.
He doesn’t understandthe article.
He‘s not understandingthe article.
Stative verbs of possession:
have, own, belong, possess, include, owe
I havea bicycle.
I‘m havinga bicycle.
This book belongsto the teacher.
This book is belongingto the teacher.
Our tour includeda visit to the Modern Art Museum.
Our tour was includinga visit to the Modern Art Museum.
Stative verbs of the senses:
hear, smell, see, feel, appear, seem, resemble
I hearsome music playing.
I‘m hearingsome music playing.
This perfume smellslike roses.
This perfume is smellinglike roses.
He seemedupset last night.
He was seemingupset last night.
Stative verbs for emotional states:
love, hate, like, want, need, desire, wish
I loveice cream.
I‘m lovingice cream.
She has always hatedjazz.
She has always been hatingjazz.
They needsome help.
They‘re needingsome help.
Stative verbs of states/qualities:
weigh, contain, consist, measure, cost, exist, depend, deserve, involve, matter
This piece of meat weighstwo pounds.
This piece of meat is weighingtwo pounds.
The box containeda pair of earrings.
The box was containinga pair of earrings.
Success dependson how much effort you make.
Success is dependingon how much effort you make.
This class will involvelots of research.
This class will be involving lots of research.
Note: We do not use state verbs in continuous tenses when they’re the main verb… but they CAN be continuous when used as gerunds, in prepositional phrases, or after certain verbs that require -ING:
Main verb:
I have a bicycle.
Gerund:
Having a bicycle saves me a lot of money.
Main verb:
The box contained a pair of earrings.
Prepositional phrase:
The box containing a pair of earrings was stolen.
Main verb:
He felt unprepared for the presentation.
After verb requiring -ING:
I enjoy feeling well-prepared for every presentation.
Download free lesson PDF
Verbs that can be both action verbs and stative verbs
Some verbs can function as BOTH action verbs and stative verbs!
Here are some examples:
be
Stative verb:He isimmature. (he is always immature)
Action verb:
He is beingimmature. (he is temporarily acting immature)
have
Stative verb: possession
I havea car. He hasa dog.
Action verb: expressions with “have”
I’m havingbreakfast (eating breakfast).
He’s havingfun (experiencing fun).
see
Stative verb: perception with your eyes; understanding
I seesome birds.
I seewhat you mean.
Action verb: meet; have a relationship with
I’ll be seeingthe doctor tomorrow.
They’ve been seeingeach other for a month.
look
Stative verb: appearance
That cake looksdelicious!
Action verb: directing your eyes to something; phrasal verbs
He’s lookingat the computer screen.
She’s looking for(= seeking) a job.
They’re looking after(= taking care of) my dog for the weekend.
smell / taste
Stative verb: the quality of smell or taste possessed by something
The bar smells of smoke.
This meat tasteslike chicken.
Action verb: when a person uses their nose or mouth to test something
He’s smellingthe cookies.
She’s tastingthe soup to see if it needs more salt.
think / feel
Stative verb: when talking about your opinion
I thinkthat’s a great idea!
I feelthat this is not the best use of our time.
Action verb: when using your mind, or experiencing emotions or health issues
We’re thinkingabout moving to another city.
I’ve been feelingunusually tired lately.
weigh / measure
Stative verb: when talking about the quality possessed by something
The suitcase weighs20 pounds.
The surfboard measures2 meters by 55 centimeters.
Action verb: when a person performs the action of weighing/measuring something
The butcher is weighingthe meat on the scale.
The architects were measuringthe distance between the pillars.
Hope that helps you understand the difference between action verbs and stative verbs, as well as how we use them!
If you want to understand and really master English grammar so that you can use it successfully without thinking too much about all these rules, come join my Advanced English grammar course.
What’s great about this course is that not only will you learn grammar, but I’ll also help you put it into practice and use it yourself. It will help take your English language skills from “good” to GREAT!
Learn more: Linking verbs
Learn more about this course
You might also like...
Common Errors in English: See, Look, Watch
Intensifying Adverbs: Common Collocations
Difference between Present Perfect and Past Perfect in English |
9179 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5047698/ | COPI is essential for Golgi cisternal maturation and dynamics - PMC
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J Cell Sci
. 2016 Sep 1;129(17):3251–3261. doi: 10.1242/jcs.193367
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COPI is essential for Golgi cisternal maturation and dynamics
Midori Ishii
Midori Ishii
1 Live Cell Super-Resolution Imaging Research Team, RIKEN Center for Advanced Photonics, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
2 Department of Biological Sciences, Graduate School of Science, The University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan
Find articles by Midori Ishii
1,2, Yasuyuki Suda
Yasuyuki Suda
1 Live Cell Super-Resolution Imaging Research Team, RIKEN Center for Advanced Photonics, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
3 Laboratory of Molecular Cell Biology, Faculty of Medicine, University of Tsukuba, Tsukuba, Ibaraki 305-8575, Japan
Find articles by Yasuyuki Suda
1,3, Kazuo Kurokawa
Kazuo Kurokawa
1 Live Cell Super-Resolution Imaging Research Team, RIKEN Center for Advanced Photonics, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
Find articles by Kazuo Kurokawa
1,, Akihiko Nakano
Akihiko Nakano
1 Live Cell Super-Resolution Imaging Research Team, RIKEN Center for Advanced Photonics, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
2 Department of Biological Sciences, Graduate School of Science, The University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan
Find articles by Akihiko Nakano
1,2,
Author information
Article notes
Copyright and License information
1 Live Cell Super-Resolution Imaging Research Team, RIKEN Center for Advanced Photonics, 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
2 Department of Biological Sciences, Graduate School of Science, The University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan
3 Laboratory of Molecular Cell Biology, Faculty of Medicine, University of Tsukuba, Tsukuba, Ibaraki 305-8575, Japan
Authors for correspondence (kkurokawa@riken.jp; nakano@bs.s.u-tokyo.ac.jp)
Received 2016 Jun 2; Accepted 2016 Jul 15.
© 2016. Published by The Company of Biologists Ltd
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution and reproduction in any medium provided that the original work is properly attributed.
PMC Copyright notice
PMCID: PMC5047698 PMID: 27445311
ABSTRACT
Proteins synthesized in the endoplasmic reticulum (ER) are transported to the Golgi and then sorted to their destinations. For their passage through the Golgi, one widely accepted mechanism is cisternal maturation. Cisternal maturation is fulfilled by the retrograde transport of Golgi-resident proteins from later to earlier cisternae, and candidate carriers for this retrograde transport are coat protein complex I (COPI)-coated vesicles. We examined the COPI function in cisternal maturation directly by 4D observation of the transmembrane Golgi-resident proteins in living yeast cells. COPI temperature-sensitive mutants and induced degradation of COPI proteins were used to knockdown COPI function. For both methods, inactivation of COPI subunits Ret1 and Sec21 markedly impaired the transition from cis to medial and to trans cisternae. Furthermore, the movement of cisternae within the cytoplasm was severely restricted when COPI subunits were depleted. Our results demonstrate the essential roles of COPI proteins in retrograde trafficking of the Golgi-resident proteins and dynamics of the Golgi cisternae.
KEY WORDS: Golgi, COPI, Cisternal maturation
Highlighted Article: Knockdown of COPI function restricts retrograde recycling of Golgi-resident proteins and markedly impairs the transition from cis to medial and to trans cisternae, as demonstrated in living yeast cells.
INTRODUCTION
The Golgi is a central station of the membrane trafficking system in eukaryotic cells. It consists of flattened membrane-enclosed compartments, called cisternae, which are orderly differentiated in their functions and structures from cis to trans cisternae. The Golgi receives newly synthesized proteins from the endoplasmic reticulum (ER) at cis cisternae, and processes and glycosylates them as they proceed through the medial regions, and finally sorts them from trans cisternae and the trans-Golgi network (TGN) to their final destinations. These functions of the Golgi are evolutionarily conserved. However, the organization of the Golgi varies across cell types and species; higher animal and plant cells have stacked Golgi cisternae, whereas the budding yeast Saccharomyces cerevisiae has unstacked cisternae dispersed in the cytoplasm (Mowbrey and Dacks, 2009).
The mechanism of protein trafficking within the Golgi is a fundamental and intriguing question of cell biology. To explain anterograde transport of secretory cargo, the ‘cisternal maturation’ model is now widely accepted to explain the core mechanism for Golgi traffic (Glick and Luini, 2011; Glick and Nakano, 2009; Nakano and Luini, 2010). This model is based on the concept that Golgi cisternae progressively change their nature and work as anterograde carriers for secretory protein transport. The support for this model in mammalian cells was given by the observation that large aggregates of procollagen progressively moved through the Golgi stacks without leaving the lumen (Bonfanti et al., 1998). Maturation of the Golgi cisternae has also been directly observed in S. cerevisiae (Losev et al., 2006; Matsuura-Tokita et al., 2006). Individual early and late Golgi cisternae in S. cerevisiae were labeled with different fluorescent-protein-tagged Golgi-resident proteins and the colors of cisternae showed a unidirectional change from early to late under a confocal fluorescence microscope. In this view, Golgi-resident proteins should be transported from late to early cisternae by retrograde transport machineries. Furthermore, Rizzo et al. (2013) reported that artificial polymerization of Golgi-resident proteins to prevent recycling led to their progression through the Golgi stack, which also supports cisternal maturation (Rizzo et al., 2013). However, many questions remain as to its molecular mechanisms.
Coat protein complex I (COPI)-coated vesicles are the candidate retrograde transport carriers of Golgi-resident proteins. COPI coats are composed of seven subunits, known as α, β, β′, γ, δ, ε and ζ COP, which are classified into two groups: α, β′ and ε subunits forming the B trimeric adaptor complex, and β, γ, δ and ζ subunits forming the F tetrameric outer coat complex (Gabriely et al., 2007; Lee and Goldberg, 2010). A function of COPI-coated vesicles in anterograde cargo transport is still a matter of debate, however COPI has pivotal roles in Golgi–ER retrograde transport and probably also in the retrograde traffic between Golgi cisternae (Cosson et al., 2002; Emr et al., 2009; Martínez-Menárguez et al., 2001; Orci et al., 2000; Rabouille and Klumperman, 2005; Sato et al., 2001). Recruitment of COPI coat proteins to the Golgi membrane requires Arf GTPase (Serafini et al., 1991). A recent report has shown that disruption of Arf1 causes early Golgi cisternae to mature more slowly and less frequently, but does not alter the maturation of late Golgi cisternae in S. cerevisiae (Bhave et al., 2014).
We reported before that the α-COP temperature-sensitive mutant ret1-1, which is defective in COPI vesicle formation at the restrictive temperature, exhibited retarded but not completely blocked maturation from early to late cisternae at the restrictive temperature by 2D time-lapse observations (Matsuura-Tokita et al., 2006). From this result, we suggested that COPI is important for Golgi cisternal maturation, but other mechanisms might also operate. For technical reasons 3D observation was not performed, and we could not rule out the ambiguity of analysis from 2D data. To further clarify the role of COPI proteins in the cisternal maturation, we decided to revisit these results by using transmembrane Golgi-resident proteins as cis-, medial- and trans-cisternal markers and by a more-elaborate 3D observation by high-speed and high-resolution confocal microscopy method that we developed (super-resolution confocal live imaging microscopy, SCLIM) (Kurokawa et al., 2013). To inactivate COPI functions, not only temperature-sensitive defects of COPI proteins, but also a new method to induce their degradation by an auxin degron system (Nishimura et al., 2009) was employed. We found that disruption of COPI functions inhibited cisternal maturation and dynamic movement of cisternae in the cytoplasm. These results indicate that the retrograde transport of the Golgi-resident proteins is mediated through a COPI-dependent mechanism that plays a pivotal role in the cisternal maturation, and that COPI proteins also play some role in the Golgi dynamics.
RESULTS
Spatial distribution of transmembrane Golgi-resident proteins
A variety of fluorescent-tagged proteins have been used for live imaging of yeast Golgi cisternae (Matsuura-Tokita et al., 2006). Here, we employ as new markers, Golgi-resident proteins harboring transmembrane domains, which are Mnn9, a subunit of Golgi mamnosyltransferase complex, Gnt1, an N-acetylglucosaminyltransferase, and Sys1, a receptor for Arl3, for cis, medial, and trans cisternae, respectively (Behnia et al., 2004; Jungmann and Munro, 1998; Yoko-o et al., 2003). Mnn9 and Gnt1 are type II transmembrane glycosyltransferases and have a Vps74 recognition motif at their N-termini. Vps74 packs them into COPI-coated vesicles for recycling to early cisternae and controls their steady-state distributions (Tu et al., 2008). Sys1 has been shown to act as a transmembrane receptor for Arl3, but its retention is independent of Vps74 (Schmitz et al., 2008). These resident proteins and commonly used Golgi cisternae markers, Sed5, a t-soluble N-ethylmaleimide-sensitive factor attachment protein receptor (SNARE) molecule mostly reside in the cis cisternae (Hardwick and Pelham, 1992), and Sec7, the guanine-nucleotide exchange factor for Arf GTPase, which is peripherally associated with trans cisternae and TGN (Achstetter et al., 1988), were tagged with green or red fluorescent proteins. The degrees of colocalization of these Golgi marker proteins were examined by dual-color 3D observation (SCLIM). As shown in Fig.1A, the two cis markers GFP–Sed5 and Mnn9–mCherry showed a very high probability of colocalization. The percentage colocalization with Mnn9–mCherry for Gnt1–GFP, Sys1–GFP and Sec7–GFP was ∼77%, ∼26% and ∼5%, respectively, consistent with their medial- and trans-cisternal localization (Fig.1A,B). These results reflect the preferential distribution of Golgi-resident proteins to particular Golgi cisternae. Note that segregation of two markers is observed within a colocalizing cisterna by SCLIM (Fig.1C).
Fig. 1.
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Localization of Golgi-resident proteins. (A) Localization of Golgi marker proteins with the cis-Golgi-resident marker Mnn9–mCherry. Wild-type cells expressing GFP-tagged Sed5 (cis), Gnt1 (medial), Sys1 (trans) and Sec7 (trans) with Mnn9–mCherry were grown to a mid-logarithmic phase in synthetic medium at 30°C and observed by 3D confocal fluorescence microscopy. Dashed lines indicate the edge of the cells. (B) Bar graph showing the percentage of Mnn9–mCherry-positive cisternae containing GFP-tagged Golgi-resident proteins. Error bars indicate s.d. from 10 independent cells. (C) Images of indicated areas in A. Scale bars: 1 µm.
Visualization of cisternal maturation using transmembrane Golgi-resident proteins
We next examined cisternal maturation of the Golgi in detail by using these Golgi-resident markers. We conducted simultaneous dual-color 4D observations at time resolution of 8 frames/s by SCLIM (Kurokawa et al., 2013; Matsuura-Tokita et al., 2006). Similar to the case in the cell expressing mRFP–Sed5 and Sec7–GFP (Fig.2C and Movie 3), early cisternae labeled with Mnn9–mCherry changed their color to those of medial cisternae labeled with Gnt1–GFP or trans cisternae and TGN labeled with Sys1–GFP, giving evidence for cisternal maturation through recycling of Golgi-resident transmembrane proteins (Fig.2A,B; Movies 1 and 2). We determined the transition period between the intensity peaks for each pair of fluorescent markers on a single cisterna (peak-to-peak time) (Daboussi et al., 2012). The average peak-to-peak time from Mnn9–mCherry (cis) to Gnt1–GFP (medial) (38.6±11.6 s) was shorter than that from Mnn9–mCherry (cis) to Sys1–GFP (trans) (45.6±19.6 s) (mean±s.d., n>11; Fig.2A,B). The transition period from mRFP–Sed5 (cis) and Sec7–GFP (trans) (87.8±34.0 s) was longer than that from Mnn9–mCherry (cis) to Sys1–GFP (trans) (Fig.2C). Even though experimental variability was high, these results were consistent with the spatially ordered distribution of Golgi-resident proteins within the Golgi. High-resolution images during the transition of two fluorescent signals revealed that later markers of the Golgi cisternae, Gnt1–GFP, Sys1–GFP and Sec7–GFP, began to accumulate as small punctate structures on the early cisternae and increased their volume to cover entire regions on the membrane. By contrast, the earlier markers, Mnn9–mCherry and mRFP–Sed5 signals showed a gradual reduction (Fig.2A–C). In addition, we noticed that earlier and later Golgi markers did not completely overlap but rather showed segregation within the maturing cisternae, corroborating that functional domains exist within a Golgi cisterna (see also Fig.1C). We could not observe transition from Mnn9–mCherry-labeled cis cisternae to Sec7–GFP-labeled trans cisternae, indicating that Sec7–GFP begins to accumulate later than the disappearance of Mnn9–mCherry from cisternae. This is consistent with the above finding that Mnn9 rarely colocalized with Sec7 (Fig.1A,B; Fig.S1). The two trans markers Sys1 and Sec7 probably accumulate on cisternae at different maturation phases; Sec7 comes later than Sys1.
Fig. 2.
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4D observation of cisternal maturation. Wild-type cells expressing (A) Mnn9–mCherry (cis, magenta) and Gnt1–GFP (medial, green), (B) Mnn9–mCherry (cis, magenta) and Sys1–GFP (trans, green) and (C) mRFP–Sed5 (cis, magenta) and Sec7–GFP (trans, green) were grown to a mid-logarithmic phase in synthetic medium at 30°C and observed by SCLIM. Upper left panels show representative 3D images. Dashed lines indicate the edge of the cells. Upper right montages show 3D time-lapse images (4D) of the indicated squares. Lower left panels show the relative fluorescence intensities of green and red channels in the indicated cisterna. Lower right panels show magnified images of cells at selected time points. Scale bars: 1 µm. Representative images from at least 26 independent cisternae are shown.
4D observation of cisternal maturation in temperature-sensitive COPI mutant cells
In our previous report, we showed that cisternal maturation was slowed down but was still observed in α-COP temperature-sensitive mutant ret1-1 cells at the restrictive temperature (Matsuura-Tokita et al., 2006). Because this previous observation was made only in 2D time lapse, we decided to reinvestigate this issue using our much improved 4D imaging system, SCLIM, which has high sensitivity and high resolution.
Both in wild-type and ret1-1 cells at 25°C, transition of the fluorescent signals from mRFP–Sed5 to Sec7–GFP was observed within 120 s, indicating that efficient cisternal maturation occurred in the cells (Fig.3A,B). By contrast, in ret1-1 cells at the restrictive temperature, cis cisternae labeled with mRFP–Sed5 did not lose red fluorescence but kept it for more than 400 s, and never acquired Sec7–GFP signals (Fig.3B; Movie 4). The rate of successful Golgi maturation was greatly decreased (11%, 5 out of 55 cisternae) in ret1-1 cells at the restrictive temperature compared to the permissive temperature (62%, 13 out of 21 cisternae). No significant difference was observed for the wild-type upon temperature shift (Fig.3C). These observations indicate that COPI function is indispensable for cisternal maturation.
Fig. 3.
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4D observation reveals the defect of cisternal maturation in the α-COP mutant ret1-1 at the restrictive temperature. Wild-type (WT) and ret1-1 cells expressing mRFP–Sed5 (cis, magenta) and Sec7–GFP (trans, green) were grown to a mid-logarithmic phase in synthetic medium at 25°C. Wild-type cells cultured at 25°C and 38°C for 10 min (A) and ret1-1 cells cultured at 25°C (permissive) and 38°C (restrictive) temperature for 10 min (B) were observed by SCLIM. Representative 3D images are shown. Dashed lines indicate the edge of the cells. Right montages show 3D time-lapse (4D) images of the indicated squares. Scale bars: 1 µm. (C) The numbers of cisternae that had matured are shown. At least independent 17 cisternae were counted.
4D observation of cisternal maturation in cells depleted for COPI proteins
To further elucidate the role of COPI proteins in cisternal maturation, we developed cells in which COPI proteins were degraded through an auxin-inducible degron (AID) system. The AID system allows specific protein degradation in the presence of plant hormone auxin in non-plant cells (Nishimura et al., 2009). Arabidopsis thaliana E3 ubiquitin ligase TIR1 was constitutively expressed in yeast cells and the A. thaliana IAA17 sequence was added as a tag at the C-terminus of Ret1 (α-COP) or Sec21 (γ-COP). These cells are designated Ret1-aid and Sec21-aid cells, respectively, or COPI-aid cells, collectively. Expression of IAA17-tagged Ret1 and Sec21 could rescue the growth defects of corresponding temperature-sensitive mutants at the restrictive temperature (Fig.4A). COPI-aid cells exhibited strong growth defects in the presence of 0.5 mM 1-naphthaleneacetic acid (NAA, a synthetic auxin), whereas they grew normally as wild-type cells in the absence of NAA (Fig.4B). More than 80% of IAA17-tagged COPI proteins were degraded upon incubation with 1 mM NAA for 2 h (Fig.4D). Auxin-dependent degradation of COPI proteins was reversible, because the growth rates of COPI-aid cells after NAA washout were comparable to that of wild-type cells (Fig.4C). ER and Golgi precursor forms of a vacuolar protein, carboxypeptidase Y (CPY), accumulated in COPI-aid cells after NAA treatment for 2 h (Fig.4D).
Fig. 4.
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Degradation of the COPI proteins Ret1 and Sec21 by the AID system. (A) Expression of IAA17-tagged Ret1 and Sec21 from low-copy plasmids rescued the temperature-sensitive phenotypes of ret1-1 and sec21-1 or sec21-2 cells, respectively. (B) Growth of wild-type (WT), AtTIR1, Ret1-IAA17, Ret1-aid, Sec21-IAA17 and Sec21-aid cells on a YPD plate with or without 0.5 mM NAA at 30°C. (C) Growth of NAA-treated cells after NAA washout. Wild-type, Ret1-aid and Sec21-aid cells were cultured in YPD medium with or without 1 mM NAA for 2 h at 30°C. After washing out medium with water three times, cells were grown on a YPD plate at 30°C. (D) Ret1-aid and Sec21-aid cells were cultured in YPD medium with or without 1 mM NAA at 30°C. Cells were collected at 0, 1 and 2 h after NAA addition. Ret1-IAA17, Sec21-IAA17 and CPY proteins were detected by immunoblotting. The positions of the ER form (p1), Golgi form (p2) and mature (m) CPY are indicated on the right. The PGK lane shows the loading control. All experiments were performed three times and a representative example shown.
Using COPI-aid cells expressing Mnn9–mCherry and Gnt1–GFP, we next examined the effect of COPI protein removal on cisternal maturation. In COPI-aid cells treated with 1 mM NAA for 2 h, cisternae with the signal of Mnn9-mCherry were sustained and never acquired the Gnt1-GFP signal (Fig.5A–C; Movies 5 and 6). The rates of cisternal maturation in COPI-aid cells were significantly reduced by NAA treatment (Fig.5F). These observations again indicate that COPI function is required for the retrograde flux of Golgi-resident proteins and is essential for cisternal maturation of the Golgi. Intriguingly, functional domains labeled with Mnn9–mCherry and Gnt1–GFP were kept segregated within the cisterna in NAA-treated COPI-aid cells (Fig.5D,E), suggesting that COPI function is also necessary for domain dynamics. We next examined the effect of COPI depletion on the behaviors of the trans-Golgi and TGN marker Sys1–GFP. Transition from Mnn9–mCherry to Sys1–GFP was also disturbed in the presence of NAA in COPI-aid cells (Fig.6). Cis cisternae labeled with Mnn9–mCherry never lost their color (Fig.6A,B, boxed area 2; Movies 7 and 8). Cisternae harboring both Mnn9–mCherry and Sys1–GFP sustained both fluorescent signals for a prolonged time (Fig.6A,B, boxed area 1; Movies 7 and 8). These results indicate that the function of COPI is required over the whole maturation process of cis-to-trans cisternae within the Golgi.
Fig. 5.
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Depletion of COPI protein inhibits cis to medial cisternal maturation. (A) Wild-type (WT), (B) Ret1 - aid and (C) Sec21-aid cells expressing Mnn9–mCherry (cis, magenta) and Gnt1–GFP (medial, green) were grown to a mid-logarithmic phase in synthetic medium with or without 1 mM NAA at 30°C and observed by SCLIM. Representative 3D images of cells with (lower panels) or without (upper) NAA are shown. Dashed lines indicate the edge of cells. Right montages show 3D time-lapse images of the indicated areas. (D) Wild-type, Ret1 - aid and Sec21 - aid cells with NAA expressing Mnn9–mCherry (cis, magenta) and Gnt1–GFP (medial, green) were observed. Left panels show representative 3D images of cells. Dashed lines indicate the edge of cells. Right montages show 3D time-lapse images of the indicated areas. (E) Magnified images of selected time points in D are shown. Scale bars: 1 µm. (F) Bar graph shows the rate of maturation of cisternae from cis to medial. Error bars represent s.d. from at least five independent cells.
Fig. 6.
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Depletion of COPI protein inhibits cis to trans cisternal maturation. Ret1 - aid (A) and Sec21-aid (B) cells expressing Mnn9–mCherry (cis, magenta) and Gnt1–GFP (medial, green) were grown to a mid-logarithmic phase in synthetic medium with or without 1 mM NAA at 30°C and observed by SCLIM. Representative 3D images of cells with (lower panels) or without (upper) NAA are shown. Dashed lines indicate the edge of cells. Right panels show 3D time-lapse images of the indicated areas. Scale bars: 1 µm. Representative images from at least five independent cells are shown.
Reduced movements of the Golgi in COPI inactivated cells
In the course of our observation of cisternal maturation in ret1-1 cells and COPI-aid cells, we realized that the dynamics of Golgi cisternae was affected upon COPI disruption (Fig.7). S. cerevisiae has single layers of Golgi cisternae scattering and moving around in the cytoplasm. We have recently discovered that the cis cisternae of Golgi show repeated approach toward ER exit sites (ERES) (hug-and-kiss action) (Kurokawa et al., 2014) to capture cargo from the ER. In ret1-1 cells at restrictive temperature, the movement of Sed5-labeled cis cisternae was drastically restricted, whereas that of Sec7-labeled trans cisternae was less affected (Fig.7A; Movie 4). In NAA-treated COPI-aid cells, 3D tracking of cis and medial cisternae demonstrated remarkable inhibition of the cisternal movement (Fig.7B; Movies 5 and 6). The structures of the ER did not appear to be affected under these conditions (Fig.S2). Disruption of actin or microtubule cytoskeletons sensitive to latrunculin A and nocodazole, respectively, did not affect either cisternal maturation or movement (Fig.S3). These results suggest that COPI contributes to the dynamic behavior of Golgi cisternae in S. cerevisiae.
Fig. 7.
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Inhibition of COPI functions reduced cisternal movement. The z and time projections of the center of mass of each cisterna are shown. The colors of points indicate the time point shown at the bottom left. (A) Images of ret1-1 cells expressing mRFP–Sed5 (cis) and Sec7–GFP (trans) at 25°C and 38°C were processed. One z-stack was collected every 5 s. (B) Images of wild-type (WT), Ret1-aid and Sec21-aid cells expressing Mnn9–mCherry (cis) and Gnt1–GFP (medial) with or without 1 mM NAA for 2 h were processed. z-stack images were collected for every 5.7 s. Representative images from at least five independent cells are shown.
DISCUSSION
COPI-dependent recycling of Golgi-resident proteins drives cisternal maturation from cis to trans
Cisternal maturation explains a core mechanism for intra-Golgi traffic (Losev et al., 2006; Luini, 2011; Matsuura-Tokita et al., 2006). In a typical model, Golgi cisternae serve as anterograde carriers for secretory cargo, and the retrograde transport of Golgi-resident proteins executes the change in properties of cisternae called maturation. Direct evidence for cisternal maturation was provided by live imaging of yeast S. cerevisiae using fluorescent-protein-tagged Golgi-resident proteins (Losev et al., 2006; Matsuura-Tokita et al., 2006). We were aware, however, of a caveat in selection of fluorescent marker proteins. For example, a frequently used trans-Golgi marker, Sec7, is a peripheral membrane protein and its recycling between the cytosol and the membrane cannot be ruled out. SNARE proteins are also convenient markers but as they constitute important transport machinery their behaviors may not represent ‘resident’ Golgi proteins. In the present study, we have tried to exploit as new markers fluorescence-tagged Golgi-resident transmembrane proteins, such as enzymes, and confirmed that they are also good tools to analyze cisternal maturation. By using SCLIM imaging technology that we developed, we now provide 4D observation of these fluorescence Golgi cisternal markers at high spatiotemporal resolutions (Kurokawa et al., 2013). We addressed the roles of COPI in intra-Golgi transport processes, and the results indicate that the COPI functions are essential for fulfilling cisternal maturation. Both a temperature-sensitive defect (ret1-1 mutation) and auxin-induced degradation of Ret1 or Sec21 led to severe blockade of cisternal maturation in cis-to-medial and medial-to-trans steps.
Whether COPI executes its functions in the form of COPI vesicles is still an open question. COPI is also involved in tubule formation in mammalian cells (Yang et al., 2011). We sometimes see tubular connections between yeast Golgi cisternae (Matsuura-Tokita et al., 2006; Nakano and Luini, 2010). Our observation in the present study that small membrane puncta from later cisternae emerge on a maturing cisterna (discussed below) strongly suggests the presence of vesicles there, but at the moment we cannot conclude whether they are single vesicles or clusters of vesicles. Roles for COPI-coated tubules are also possible. We are now in the process of developing a second-generation SCLIM system, which has even better spatiotemporal resolutions, and we hope to clarify these questions in the near future.
Another important question is how COPI controls selective recycling of distinct Golgi-resident proteins to their own cisternae. One of the candidates for regulating the selectivity to specific cisternae would be golgin family proteins, because they are thought to specifically capture vesicles from different organelles, endosomes, the ER, and the Golgi. For example, Golgin-84 (also known as GOLGA5), GMAP210 (also known as TRIP11) and TMF1 have been shown to capture the Golgi-resident proteins in mammalian cells (Wong and Munro, 2014). In yeast, Rud3 and Sgm1, orthologs of GMAP210 and TMF1, localize to cis and trans cisternae, respectively (Munro, 2011). Whether selective degradation of these golgin family proteins by the AID system affects intra-Golgi trafficking of resident proteins would be a next question to be examined by our high-resolution 4D observation.
Domains are formed and maintained on Golgi cisternae during cisternal maturation
During cisternal maturation, segregation of early and late Golgi-resident proteins was observed in individual cisternae in our previous report (Matsuura-Tokita et al., 2006). Here, our 4D imaging with higher spatiotemporal resolutions reveals details of the compositional change in maturing individual cisternae (see Fig.2). At the beginning of the transition of cis cisternae, marked by Mnn9–mCherry, Gnt1–GFP coming from medial cisternae began to localize to small regions on the cis cisternae. These small puncta gradually expanded their area without extensive mixing with the cis component and eventually covered the whole cisternae. Similar events were also observed for the cases of Sys1–GFP and Sec7–GFP coming into the cis-cisternae. Furthermore, two cis-Golgi marker proteins, Sed5 and Mnn9, showed segregated distribution on a single cisterna (Fig.1C). Such segregation of membrane proteins might be an intrinsic property of the Golgi membrane and would be a key to understand their sorting and dynamic equilibrium. In COPI-depleted cells, domain segregation was still observed but in a fixed fashion (see Figs 5 and 6), suggesting the involvement of COPI in dynamic mixing of domains. The Rab family of GTPases regulate a variety of steps of membrane trafficking. They show phase-specific localization on the Golgi membrane, which is coordinately controlled by Rab GEF and Rab GAP cascades (Rivera-Molina and Novick, 2009; Suda et al., 2013). We previously showed that the cells disrupted for the Rab GAP cascade failed in proper Rab GTPase transition and thus in proper membrane segregation but somehow continued cisternal maturation (Suda et al., 2013). The significance of membrane domain segregation will need further investigation.
COPI also functions in dynamics of Golgi cisternae
Recently, Glick and colleagues showed that inactivation of COPI function by anchoring COPI subunits on mitochondria led to the formation of hybrid Golgi structures, in which both early (Vrg4–GFP) and late (Sec7–DsRed or Kex2–mCherry) Golgi-resident proteins were present (Papanikou et al., 2015). These structures fluctuated rapidly and moved around in the cytoplasm. They maintained the early Golgi marker, but showed repeated loss and gain of late Golgi and TGN makers. Anchored COPI proteins have potential to still interact to the Golgi. In our present study by contrast, the AID system directly removes COPI proteins from cells and COPI depletion brought about marked inhibition of maturation of cis-, medial- and trans-Golgi and TGN markers (Figs 5 and 6) as well as remarkable repression of the dynamic movement of Golgi cisternae (Fig.7). Cisternae labeled with both cis-Golgi and medial-Golgi or trans-Golgi and TGN markers might be deemed similar to the hybrid Golgi structures of Papanikou et al. (2015). However, they sustained these fluorescent signals and did not show repeated loss and gain behavior in our SCLIM observation. Colocalization of both early and late Golgi-resident proteins in a single cisterna was observed as only transient events during cisternal maturation in wild-type cells. These results suggest that in COPI-depleted cells the amount of COPI proteins is insufficient to recycle Golgi-resident proteins, resulting in intermediate cisternae becoming fixed in the maturation process.
Why the dynamics of cisternae is amazingly disturbed by COPI depletion remains a big question to be addressed. Perhaps dynamic release of COPI vesicles or projection of COPI tubules can affect the balance of individual cisternae. To test such a possibility mathematical modeling might be helpful. Recently, we have shown that cis-Golgi cisternae show repeated approach toward the COPII-coated area localized at ERES and capture cargo proteins there (hug-and-kiss action; Kurokawa et al., 2014). The dynamic behavior of Golgi cisternae must be a pivotal feature in S. cerevisiae that plays many roles in controlling anterograde and retrograde transport in secretory pathway. Analysis of COPI functions would be again a key to tackle the questions of Golgi dynamics.
MATERIALS AND METHODS
Yeast strains and plasmids
Yeast strains, plasmids and primers used in this study are listed in Tables S1, S2 and S3, respectively. ADE2+ cells were made by integration with pRS402 (Brachmann et al., 1998) digested by Stu I into the ade2 site. mRFP–Sed5 and Sec7–GFP were co-expressed under the control of the TDH3 promoter for mRFP–Sed5 and the ADH1 promoter for Sec7–GFP on the low-copy plasmid pRS316 (Matsuura-Tokita et al., 2006). Gnt1–GFP and Sys1–GFP were expressed under the control of ADH1 promoter and CMK1 terminator on the low-copy plasmids pRS316 or pRS314 (Sikorski and Hieter, 1989). These plasmids were constructed in several steps. First, the DNA fragment containing the ADH1 promoter or CMK1 terminator was obtained from yeast genomic DNA by PCR with an appropriate sets of primers, SacI-ADH1p-F and ADH1p-NotI-R, or XhoI-CMK1t-F and CMK1t-SacI-KpnI-R, was subcloned into the Sac I-Not I or Xho I-Kpn I sites of pRS316, respectively. Next, the GFP DNA fragment, obtained from pSKY5 (Sato et al., 2001) by PCR with primers SalI-GFP-F and GFP-XhoI-R, was subcloned into the Sal I-Xho I sites. Then, the GNT1 or SYS1 DNA fragments obtained from yeast genomic DNA by PCR with primers NotI-GNT1-F and GNT1-SalI-R or XbaI-SYS1-F and SYS1-HindIII-R were subcloned into the Not I-Sal I sites or Xba I-Hin dIII sites of pRS316 harboring the ADH1 promoter, GFP and the CMK1 terminator to develop pRS316-ADH1p-GNT1-GFP-CMK1t or pRS316-ADH1p-SYS1-GFP-CMK1t, respectively. pRS316-ADH1p-GNT1-GFP-CMK1t was digested by Sac I and subcloned into the Sac I site of pRS314 to produce pRS314-ADH1p-GNT1-GFP-CMK1t. pRS316-ADH1p-SYS1-GFP-CMK1t was digested by Pvu II and subcloned between the Pvu II sites of pRS314 to produce pRS314-ADH1p-SYS1-GFP-CMK1t. pRS304-SEC71TMD-GFP was constructed by subcloning the DNA fragment between Pvu II sites of pRS306-SEC71TMD-GFP (Sato et al., 2003) into pRS304 (Sikorski and Hieter, 1989). Strains expressing Sec71TMD-GFP were constructed by the integration with DNA fragment of pRS304-SEC71TMD-GFP digested by Bsu 36I into the trp1 site of yeast genome. Strains expressing fluorescent-protein-tagged Mnn9 were constructed by a PCR-based method described in a previous report (Kurokawa et al., 2014).
Strains expressing IAA17-tagged Ret1 and Sec21 at their C-termini were constructed by a PCR-based method using pMK43 plasmid as a template (Nishimura et al., 2009) and the primers listed in Table S3 (RET1-S2 and RET1-S3 for aid-tagged RET1 and SEC21-S2 and SEC21-S3 for aid-tagged SEC21). Arabidopsis thaliana E3 ubiquitin ligase TIR1 was expressed from pMK76 integrated into the ura3 site of each yeast genome by Stu I digestion (Nishimura et al., 2009).
Microscopy
Fluorescence microscopy was performed by super-resolution confocal microscopy (SCLIM). The system setup had an Olympus model IX-71 inverted fluorescence microscope with a UPlanSApo 100×NA 1.4 oil objective lens (Olympus, Japan), a high-speed spinning-disk confocal scanner (Yokogawa Electric, Japan), a custom-made spectroscopic unit, image intensifiers (Hamamatsu Photonics, Japan) with a custom-made cooling system, and two EM-CCD cameras (Hamamatsu Photonics, Japan) for green and red channels (Kurokawa et al., 2013). For microscopic observation, all strains were grown in selective medium (0.67% yeast nitrogen base without amino acids and 2% glucose) with appropriate supplements. ret1-1 mutant cells were cultured at the permissive temperature (25°C) and then incubated at the restrictive temperature (38°C) for 10 min on a thermo-controlled stage (Tokai Hit, Japan) whose temperature was kept at 38°C. Ret1-aid cells and Sec21-aid cells were grown with 1 mM NAA for 2 h. For 4D live imaging of wild-type cells, Ret1-aid cells and Sec21-aid cells, 32 optical slices spaced 0.1 µm apart were sequentially collected at 8 frames per second (fps), which takes 5.7 s in total including the time for data transfer. For 4D observation of ret1-1 mutant cells, 32 optical slices 0.1 µm apart were collected at 15 fps every 5 s. z-stack images were converted to 3D voxel data and processed by iterative deconvolution with Volocity (Perkin Elmer) using a theoretical point-spread function for spinning-disc confocal microscopy. Maximum intensity projection images from these deconvolved 3D images were used for the calculation of the relative fluorescent values of the green and red signals of the Golgi areas with Fiji software (Schindelin et al., 2012). The center of mass of each cisterna was calculated in a mask of the cisterna from raw 4D images by Fiji software. The projection of the center of mass was conducted by Temporal-Color Code of Fiji plugin with a thermal lookup table.
Immunoblotting
Cells expressing AID-tagged proteins were cultured in YPD (1% yeast extract, 2% peptone, and 2% glucose) medium at 30°C for overnight. Yeast culture was inoculated in fresh medium and grown until the optical density at 600 nm (OD 600) was 0.5. 0.5 OD 600 unit cells were collected for the loading sample. Cultures were incubated at 30°C for 1 or 2 h with or without 1 mM NAA, and were collected for loading samples at 1 or 2 h with or without NAA. 0.5 OD 600 unit cells were suspended in 100 µl Laemmli's sample buffer and then disrupted by vortexing with glass beads. The cell suspensions were boiled at 100°C for 5 min and cleared by centrifugation at 20,000 g for 5 min to prepare total cell lysates. Proteins corresponding to 5 µl of total cell lysates were analyzed by SDS/PAGE, followed by western blotting with anti-AID antibody (1:2000, BioROIS), anti-CPY antibody (1:3000, rabbit, polyclonal) (Yahara et al., 2001) and anti-PGK antibody (1:10,000, Invitrogen, 22C5). Bands were visualized by horseradish-peroxidase-conjugated sheep anti-mouse IgG and donkey anti-rabbit IgG antibodies (GE Healthcare).
Acknowledgements
We thank all members of the A.N. laboratory for helpful comments. We thank Dr Ryogo Hirata of Chemical Genetic Laboratory of RIKEN for valuable discussions and help.
Footnotes
Competing interests
The authors declare no competing or financial interests.
Author contributions
Conceptualization: Y.S., K.K., A.N.; formal analysis and investigation: M.I.; writing – original draft preparation: M.I., Y.S., K.K.; writing – review and editing: Y.S., K.K., A.N.; project administration: K.K.; supervision: K.K., A.N.
Funding
This work was supported by the Grants-in-Aid for Scientific Research from the Ministry of Education, Culture, Sports, Science, and Technology of Japan [grant numbers 25221103 to A.N., 25221103 to Y.S., 25221103 to K.K.]; the 4D measurements for Multilayered Cellular Dynamics Projects of RIKEN to A.N.; and also by the Takeda Science Foundation to Y.S. M.I. is a Research Fellow of Japan Society for the Promotion of Science (JSPS) and is supported by a Grant-in-Aid for JSPS fellows [grant number 25221103]. Deposited in PMC for immediate release.
Supplementary information
Supplementary information available online at
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How to Calculate Marginal Cost: A Step-by-Step Guide
7m Read
The concept of marginal cost can be difficult for business owners to understand. However, understanding how to calculate marginal cost is essential to good forecasting and business management. With that in mind, we’ve created a step-by-step guide detailing everything from the importance of marginal costs and formula examples.
Below, we’ll examine critical concepts involving the use of marginal cost. Highlights include the relation between variable and fixed costs, the related concept of marginal revenue, and how these are used to determine the optimal point at which a business can make the most profit. In addition, we’ll show you a formula that demonstrates how to find the marginal cost of goods.
The Importance of Marginal Cost
Knowing how to calculate marginal cost is important because every business should strive to expand to a point where marginal cost is equal to marginal value. So, let’s start by exploring what marginal cost is and how to find the marginal costs associated with your business.
Marginal cost is the added cost to produce an additional good. For example, say that to make 100 car tires, it costs $100. To make one more tire would cost $80. This is then the marginal cost: how much it costs to create one additional unit of a good or service.
The costs of production determine the marginal cost. These include fixed and variable costs. Fixed costs are things like monthly rent and utilities. Variable costs are things that can change over time, such as costs for labor and raw materials.
A good example is if demand for running shoes for a footwear company increases more machinery may be needed to expand production and is a one-off expense. However, it does need to be accounted for at the point the purchase takes place. Usually, marginal costs include all costs that vary with increases in production.
Short Run Cost Curves
A U-shaped short-run Average Cost (AC) curve. AVC is the Average Variable Cost, AFC the Average Fixed Cost, and MC the marginal cost curve crossing the minimum of both the Average Variable Cost curve and the Average Cost curve.
The Marginal Cost Curve
The marginal cost curve is the relation of the change between the marginal cost of producing a run of a product, and the amount of the product produced. In classical economics, the marginal cost of production is expected to increase until there is a point where producing more units would increase the per-unit production cost. Calculating marginal cost and understanding its curve is essential to determine if a business activity is profitable.
The Marginal Cost Formula
How do you find the marginal cost? There’s a mathematical formula that expresses the change in the total cost of a good or product that comes from one additional unit of that product. Knowing this formula is essential in learning how to calculate marginal cost. It is called the marginal cost equation or marginal cost formula.
Marginal Cost = (Changes in Costs)/(Changes in Quantity)
This is an important formula for cost projections and determining whether or not a business activity is profitable.
Change in Total Cost is the usual net fixed and variable costs that go into the production of goods. Variable costs include labor, raw materials, and so on.
Change in Quantity refers to an obvious increase in the number of goods produced.
A Marginal Cost Formula Example
Here is an example of how to calculate marginal cost:
Big Dynamo is a toy company that produces robot toys. Every month, they produce 2,000 robot toys for a total cost of $200,000. They expect to produce 4,000 robot toys next month for $250,000.
Marginal Cost
Current production amount 2000
Current production cost$200,000
Future production amount 4000
Future cost of production$250,000
Marginal Cost Formula$25
Since we know that Marginal Cost = (Change in Total Cost)/ (Change in Quantity), we have:
Change in total cost= (250,000-$200,000) = $50,000
Change in total units= (4000-2000) = 2000
So, the marginal cost equals $50,000/2000 = $25
Note that the marginal cost represents the change in the cost of a good, not the total cost of the good itself. This $25 represents the margin change.
Marginal Cost Pricing
Marginal cost pricing is a strategy that some businesses use to either regain market share or to increase cash flow. This where a company will lower the price on a product so that what is earned from a sale is what a product costs to produce, with no profit. Companies can do this for an enormous number of reasons, some of them actually rather shrewd, such as:
Generating cash to pay off an upcoming debt payment.
Lowering prices on a popular product so as to increase market share and win a larger portion of a market, with plans to slowly increase prices back to prior levels, or even increase them as the company approaches monopoly status.
Get rid of old stock and clear their distribution chain for new products. This can help a company by reducing transportation and inventory storage expenses.
Increase demand for a poorly performing product.
When used in conjunction with skilled planning and marketing, margin cost pricing can be an excellent tool to use in sales, increasing liquidity, and so on.
In addition to marginal cost pricing, it’s vital you create a competitivecash flow analysis. Doing so will allow you to forecast, and prepare for, a variety of financial scenarios for your business.
Synario has Your Solutions
Finance teams can run into trouble when forecasting marginal cost into the future. As your organization changes, your marginal cost formula may have to change with it. Updating that formula over time based on the completion or implementation of capital projects and initiatives can be a daunting task in a spreadsheet-based financial model.
Fortunately, Synario solves this challenging problem for CFOs and their finance teams. Custom formulas and ratios (like marginal cost) can be updated based on different factors or changed across different scenarios.
Synario’s proven suite offinancial modeling toolscan help you make intelligent business planning decisions. Contact us to discuss how our tools can help you more clearly understand the factors which comprise your business.
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9181 | https://www.chegg.com/homework-help/questions-and-answers/enthalpy-change-heating-nitrogen-600-k-1000-k-kj-kg-using-h-reference-given-nitrogen-table-q66443855 | Solved The enthalpy change during the heating of nitrogen | Chegg.com
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The enthalpy change during the heating of nitrogen from 600 K to 1000 K in kJ / kg, a) Using h reference given for nitrogen in Table 1, and find the error rate. b) Using the equation for temperature in Table 4, and find the error rate. c) Calculation using (Table 3), which gives specific heat value at average temperature, and find the error rate. d)
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Question: The enthalpy change during the heating of nitrogen from 600 K to 1000 K in kJ / kg, a) Using h reference given for nitrogen in Table 1, and find the error rate. b) Using the equation for temperature in Table 4, and find the error rate. c) Calculation using (Table 3), which gives specific heat value at average temperature, and find the error rate. d)
The enthalpy change during the heating of nitrogen from 600 K to 1000 K in kJ / kg,
a) Using h reference given for nitrogen in Table 1, and find the error rate.
b) Using the equation for temperature in Table 4, and find the error rate.
c) Calculation using (Table 3), which gives specific heat value at average temperature, and find the error rate.
d) Calculate by using the room Cp value given in Table 2, and find the error rate.
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To calculate the enthalpy change using the reference from Table 1 for nitrogen, integrate the specific heat capacity polynomial given for the temperature range from T 1=600 K to T 2=1,000 K. View the full answer Previous questionNext question
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Table 1 $ T к h kJ/kmol U kJ/kmol u kJ/kmol 5° kJ/(kmol.K) 0 220 230 240 250 260 270 280 290 298 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 0 6,391 6,683 6,975 7.266 7,558 7,849 8,141 8,432 8,669 8,723 9,014 9,306 9,597 9,888 10,180 10,471 10,763 11,055 11,347 11,640 11,932 12,225 12,518 12,811 13,105 13,399 13,693 13,988 14,285 14,581 14,876 15,172 15,469 15,766 16,064 16,363 16,662 16,962 17.262 h kJ/kmol 30,129 30,784 31,442 32,101 0 4,562 4.770 4,979 5,188 5,396 5,604 5,813 6,021 6,190 6,229 6,437 6,645 6,853 7,061 7,270 7,478 7,687 7,895 8,104 8,314 8,523 8,733 8,943 9,153 9,363 9,574 9,786 9,997 10,210 10,423 10,635 10.848 11,062 11,277 11,492 11,707 11,923 12,139 12.356 kJ/(kmol.K) 0 182.639 183.938 185.180 186.370 187.514 188.614 189.673 190.695 191.502 191.682 192.638 193.562 194.459 195.328 196.173 196.995 197.794 198.572 199.331 200.071 200.794 201.499 202.189 202.863 203.523 204.170 204.803 205.424 206.033 206.630 207.216 207.792 208.358 208.914 209.461 209.999 210.528 211.049 211.562 5 kJ/(kmol.K) 228.057 228.706 229.344 229.973 230.591 231.199 231.799 232.391 232.973 233.549 234.115 234.673 235.223 235.766 236.302 236.831 237.353 237.867 238.376 238.878 239.375 239.865 240.350 240.827 241.301 241.768 242.228 242.685 243.137 243.585 244.028 244.464 244.896 245.324 245.747 246.166 T K 600 610 620 630 640 650 660 670 680 690 700 710 720 730 740 750 760 770 780 790 800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990 h kJ/kmol 17,563 17,864 18,166 18,468 18,772 19,075 19,380 19,685 19,991 20,297 20,604 20,912 21,220 21,529 21,839 22,149 22,460 22,772 23,085 23,398 23,714 24,027 24,342 24,658 24,974 25,292 25,610 25,928 26,248 26,568 26,890 27,210 27,532 27,854 28,178 28,501 28,826 29,151 29,476 29.803 h kJ/kmol 56,227 56,938 57,651 58,363 59,075 59,790 60,504 61,220 61.936 62,654 63,381 64,090 64,810 66,612 68,417 70,226 72,040 73,856 75,676 77,496 79,320 81,149 82,981 84,814 86,650 88,488 90,328 92,171 94,014 95,859 97,705 99,556 101,407 103,260 105,115 106,972 12,574 12,792 13,011 13,230 13,450 13,671 13,892 14,114 14,337 14,560 14,784 15,008 15,234 15,460 15,686 15,913 16,141 16,370 16,599 16,830 17,061 17,292 17,524 17,757 17,990 18,224 18,459 18,695 18,931 19,168 19,407 19,644 19,883 20,122 20,362 20,603 20,844 21,086 21,328 21.571 212.066 212.564 213.055 213.541 214.018 214.489 214.954 215.413 215.866 216.314 216.756 217.192 217.624 218.059 218.472 218.889 219.301 219.709 220.113 220.512 220.907 221.298 221.684 222.067 222.447 222.822 223.194 223.562 223.927 224.288 224.647 225.002 225.353 225.701 226.047 226.389 226.728 227.064 227.398 227.728 S° kJ/(kmol.K) 247.396 247.798 248.195 248.589 248.979 249.365 249.748 250.128 250.502 250,874 251.242 251.607 251.969 252.858 253.726 254.578 255.412 256.227 257.027 257.810 T к U kJ/kmol T K U kJ/kmol 32,762 1000 1020 1040 1060 1080 1100 1120 1140 1160 1180 1200 1220 1240 1260 1280 1300 1320 1340 1360 1380 1400 1420 1440 1460 1480 33,426 34,092 34,760 35,430 36,104 36,777 37,452 38,129 38,807 39,488 40,170 40,853 41,539 42,227 42.915 43,605 44,295 44,988 45,682 46,377 47,073 47,771 48,470 49,168 49,869 50,571 51,275 51,980 52,686 53,393 54,099 21,815 22,304 22,795 23,288 23,782 24,280 24,780 25,282 25,786 26,291 26,799 27,308 27,819 28,331 28,845 29,361 29,378 30,398 30,919 31,441 31,964 32,489 33,014 33,543 34,071 34,601 35,133 35,665 36,197 36,732 37,268 37,806 38,344 38,884 39,424 39,965 1760 1780 1800 1820 1840 1860 1880 1900 1920 1940 1960 1980 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700 2750 2800 2850 41,594 42,139 42,685 43,231 43,777 44,324 44,873 45,423 45,973 46,524 47,075 47,627 48,181 49,567 50,957 52,351 53,749 55,149 56,553 57,958 59,366 60,779 62,195 63,613 65,033 66,455 67,880 69,306 70,734 72,163 73,593 75,028 76,464 77,902 79,341 80,782 258.580 259.332 260.073 260.799 261.512 1500 1520 262.213 262.902 263.577 264.241 264.895 1540 1560 1580 1600 1620 1640 1660 1680 1700 2900 2950 3000 3050 3100 3150 265.538 266.170 266.793 267.404 268.007 268.601 Table 2 CP Су Gas constant, R kJ/kg K kJ/kg K kJ/kg K k CO2 Perfect-gas specific temperatures of some known gases (a) At a temperature of 300 K Chemical Gas formula Air Argon Ar Butane C 4 H 10 Carbon dioxide Carbon monoxide CO Ethane C2H6 Ethylene C2H4 Helium He Hydrogen H2 Methane CH4 Neon What Nitrogen N2 Octane C 6 H 18 Oxygen 02 Propane C3HS H2O 0.2870 0.2081 0.1433 0.1889 0.2968 0.2765 0.2964 2.0769 4.1240 0.5182 0.4119 0.2968 0.0729 0.2598 0.1885 0.4615 1.005 0.5203 1.7164 0.846 1.040 1.7662 1.5482 5.1926 14.307 2.2537 1.0299 1.039 1.7113 0.918 1.6794 1.8723 0.718 0.3122 1.5734 0.657 0.744 1.4897 1.2518 3.1156 10.183 1.7354 0.6179 0.743 1.6385 0.658 1.4909 1.4108 1.400 1.667 1.091 1.289 1.400 1.186 1.237 1.667 1.405 1.299 1.667 1.400 1.044 1.395 1.126 1.327 Water vapor Note: unit of kJ/kg . K is equivalent to unit of kJ/kg-. Source: Chemical and Process Thermodynamics 3 / E by Kyle, BG, 2000. "Adapted with permission from Pearson Education, Inc., Upper Saddle River, NJ. Table 3 Perfect-gas specific temperatures of some known gases (Continued) (6) At different temperatures С у Су CP CP kJ/kg K kJ/kg K Ср kJ/kg K k kJ/kg K k kJ/kg K kJ/kg Kk Temperature, K Air Carbon dioxide, CO 2 Carbon monoxide, CO 250 300 350 400 450 500 550 600 650 700 750 800 900 one thousand 1.003 1.005 1.008 1.013 1.020 1.029 1.040 1.051 1.063 1.075 1.087 1.099 1.121 1.142 0.716 0.718 0.721 0.726 0.733 0.742 0.753 0.764 0.776 0.788 0.800 0.812 0.834 0.855 1.401 1.400 1.398 1.395 1.391 1.387 1.381 1376 1370 1.364 1.359 1.354 1.344 1.336 0.791 0.846 0.895 0.939 0.978 1.014 1.046 1.075 1.102 1.126 1.148 1.169 1.204 1.234 0.602 0.657 0.706 0.750 0.790 0.825 0.857 0.886 0.913 0.937 0.959 0.980 1.015 1.045 1.314 1.288 1.268 1.252 1.239 1.229 1.220 1.213 1.207 1.202 1.197 1.193 1.186 1.181 1.039 1.040 1.043 1.047 1.054 1.063 1.075 1.087 1.100 1.113 1.126 1.139 1.163 1.185 0.743 0.744 0.746 0.751 0.757 0.767 0.778 0.790 0.803 0.816 0.829 0.842 0.866 0.888 1.400 1.399 1.398 1.395 1.392 1.387 1.382 1.376 1.370 1.364 1.358 1.353 1.343 1.335 Hydrogen, H2 Nitrogen, N2 Oxygen, O2 250 300 350 400 450 500 550 600 650 700 750 800 900 one thousand 14.051 14.307 14.427 14.476 14.501 14.513 14.530 14.546 14.571 14.604 14.645 14.695 14.822 14.983 9.927 10.183 10.302 10.352 10.377 10.389 10.405 10.422 10.447 10.480 10.521 10.570 10.698 10.859 1.416 1.405 1.400 1.398 1.398 1.397 1.396 1.396 1.395 1.394 1.392 1.390 1.385 1.380 1.039 1.039 1.041 1.044 1.049 1.056 1.065 1.075 1.086 1.098 1.110 1.121 1.145 1.167 0.742 0.743 0.744 0.747 0.752 0.759 0.768 0.778 0.789 0.801 0.813 0.825 0.849 0.870 1.400 1.400 1.399 1.397 1.395 1.391 1.387 1.382 1.376 1.371 1.365 1.360 1.349 1.341 0.913 0.918 0.928 0.941 0.956 0.972 0.988 1.003 1.017 1.031 1.043 1.054 1.074 1.090 0.653 0.658 0.668 0.681 0.696 0.712 0.728 0.743 0.758 0.771 0.783 0.794 0.814 0.830 1.398 1.395 1.389 1.382 1.373 1.365 1.358 1.350 1.343 1.337 1.332 1.327 1.319 1.313 Table 4 Perfect-gas specific temperatures of some known gases (Continued) () As a function of temperature dT ср а от CT 2 (In TK and Cp kJ/kmol K units) Chemical % error Temperature range, K Matter formula a b с D Max. Cover. N2 0.59 0.34 28.90 25.48 02 1.19 0.28 0.1571 102 1.520 102 0.1967 102 0.1916 10 2 0.8081 10 0.7155 10 0.4802 103 0.4003 10 2.873 10 1.312 10 1.966 10 0.8704 104 273--1800 273--1800 273--1800 273--1800 28.11 0.72 0.33 0.26 H2 29.11 1.01 СО 28.16 0.1675 10 2 0.5372 10 2.222 10 273--1800 0.89 0.37 CO2 0.67 0.53 H2O 22.26 32.24 29.34 24.11 5,981 102 0.1923 10 2 0.09395 10 2 5.8632 102 3.501 10 1.055 105 0.9747 3.562 103 7.469 10 3.595 109 4.187 109 10.58 10 273--1800 273--1800 273--1500 273--1500 0.22 0.24 0.36 0.26 10 0.97 NO N20 0.59 NO2 22.9 0.18 27.568 5.715 102 2.5630 102 2.218 10 2 NH S2 3.52 10 0.99072 10 1.628 10 7.87 10 6.6909 109 3.986 109 0.46 0.91 273--1500 273--1500 273--1800 0.36 0.38 27.21 0.99 SO2 25.78 5.795 102 3.812 10 8,612 10 273--1800 0.45 0.24 Nitrogen Oxygen Air Hydrogen Carbon monoxide Carbon dioxide Water vapor Nitrous oxide Nitrous oxide Nitrogen dioxide Ammonia Sulfur Sulfur dioxide Sulfur trioxide Acetylene Benzene Methanol Ethanol Hydrogen chloride Methane Ethane Propane n-Butane 1-Butane n-Pentane n-Hexane Ethylene Propylene SO C2H2 C.HG CHO CHO 16.40 21.8 36.22 19.0 19.9 14.58 102 9.2143 102 48.475 10 2 9.152 102 20.96 102 11.20 10 6.527 10 31.57 10 1.22 10 10.38 103 32.42 10 18.21 10 77.62 109 8.039 10 20.05 10 273--1300 273--1500 273--1500 273--1000 273--1500 0.29 1.46 0.34 0.13 0.59 0.20 0.08 0.22 0.18 0.40 HCI 30.33 0.22 102 1.33 0.83 0.40 0.08 0.57 0.28 0.12 0.24 CH4 CH CH C4H10 CH10 CH 2 C.H 14 C2H4 CH 19.89 6,900 4.04 3.96 7.913 6.774 6.938 3.95 3.15 0.7620 10 2 5.024 17.27 102 30.48 102 37.15 10 2 41.60 10 2 45.43 102 55.22 102 15.64 102 23.83 102 1.327 103 1.269 10 6.406 10 15.72 103 18.34 10 23.01 103 22.46 10 28.65 10 8.344 10 4.338 109 11.01 10 9 7.285 10 31.74 10 35.00 109 49.91 109 42.29 10 57.69 109 17.67 10 24.62 109 273--1500 273--1500 273--1500 273--1500 273--1500 273--1500 273--1500 273--1500 273--1500 0.13 0.21 0.54 0.25 0.56 0.72 0.54 0.73 0.20 0.13 12.18 10 273--1500 0.17
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9182 | https://www.youtube.com/watch?v=AXLtE8zTRvE | Maximum And minimum value of aCos x + bSin x | trigonometry expression class 11
RKD CLASSES (Rajeeba Dash)
93800 subscribers
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Posted: 23 Dec 2020
Maximum And minimum value of
aCos x + bSin x | trigonometry expression class 11
In this video I explained about how to find the max. And Min. Value of trigonometric expression like
a Cos x +b Sin x.
12 comments
Transcript:
हेलो हेलो फ्रेंड्स वेलकम टू माय क्लास क्लास का संकल्प लें कि सीक्वेंस को सहलाना और ग्लैमर डुबो समर्थन मूल्य श्रीमंत इसने पिछला सॉन्ग सुनाओ डिफरेंशियल इक्वेशंस कर्नल अपने इंटरप्टेड मैक्सिमम एंड मिनिमम वेरी डिफरेंट फ्रॉम व्हाट इज द व्हाट इज द मीनिंग ऑफ दिस इज द फर्स्ट आपको कुछ पॉइंट्स को सबस्क्राइब करना न भूलें नुस्खे यदि मलिक रेस फर्स्ट है प्रयोग कौन फास्ट डेली पॉइंट मलिक प्ले साइन एक्स तो साइंस अकैडमी The Amazing नंबर सेट रहस्य को प्ले - 121 फॉलो बैक बैक इन थे वैल्यू आफ साइंस इन बिटवीन - 121 फर्स्ट नंबर टू प्ले को यह 90 यह ट्वीट 8 - इनिशिएटिव प्लस इनफील्ड अ कि तुम दिलीप राय नहीं प्ले ऑनलाइन या फिर यूं ही सम्राट तुम ही वाटरिंग भी मिली अपने हलवा जो प्रोग्राम्स आफ वरशिप लॉर्ड विघ्नेश्वरा तो देखी नहीं है नेक्स्ट 9 न्यूज रूम से - 121 हेलो हेलो एक एल्बम में ही पड़े तो गिरधर निकल डूम और फैक्ट ई लव ऊ मैं तेरा मूड चोट एक गुप्त कैमरा वर्क हिस्से को कॉकपिट गैलरी खोल दो कि कौशिक एक्सक्लूजन इक्वेडोर - उन्मुक्त स कि मुझे कुछ में मिले तो यह कह अलार्म करो चैप्टर थोड़ा कॉन्टिनेंट फंक्शन रोवर रेंजर प्ले फॉर स्कॉर्पियो बेस्ट कैमरा प्ले लिस्ट को सबस्क्राइब सब्सक्राइब करें कि सपोच कमला बिल्कुल एक्सप्रेस में कौन सा अम्ल एवं अरुचि टेक को सबस्क्राइब तब पर क्लिक करें तो चैनल को subscribe कि मुझे कुछ इन्हें दिखाओ यह चैप्टर कौन-कौन से क्वेश्चन को मैक्सिमम वॉल्यूम को सब्सक्राइब करें में बीएफ फुल स्किन एयरप्लेन आर सो स्वीट आ आ 210 प्ले आर को साफ आप आपको स्वीट आयुष को नंबर और 19 पोलर फॉर्म लक्ष्मी एवं अभिलेख मेडिकल बीइंग अब्जॉर्ब्ड इन बटन पर क्लिक करें प्ले लिस्ट प्ले लिस्ट प्ले करो आज तो दिन में कम खुली हुई एस प्लस बी स्क्वायर फीट स्पोर्ट्स प्लेयर्स पेंस पर अल्फा प्लस सब्सक्राइब टो कि चिकचिक उल्लू को लिखे अपने आर्ट टो हैव ए ग्रेट टाइम कुरान फालक हेल्पर है में आ झुलुआ निकले आखिरी कुल्लू 6S प्लस 20 मिनट पिंपले पिंपले आर इक्वल नंबर आफ इलेक्शन - वोट फॉर एडिफिकेशन रिमाइंडर प्ले सूची सब्सक्राइब टो है तो बिल्कुल आप ले टांग साइकिल शेंडी है है तो विल फाइनली Bigg Boss जो व्यक्ति नंबर चार कमरे में ताला बंद कर दिया अल्फ़ेज़ उत्तर प्रदेश में मिले तो यह लार अभिनव यूज करें अल्लू 300 यह क्वेश्चन लुट लुट लो में होने लग उसी लुट चेयरपर्सन आफ बललुट प्लेयर कौशल 5 मिनट पर अफवाहें चैनल पॉइंट क्वेश्चन रिलैक्स आर साइन फॉर कांटेक्ट प्लस वीरवार को सफेद 0r कौशिल्या सेंड ए के मुकाबले तक हसनपुर जैगुआर को ऑन कर लें तो 8 एंड वर्क फॉर फाइनल फॉर क्वालिफिकेशन फॉर साइंटिफिक तो विलियर्स मिले श्राइन आफ अंस प्लज लाइक थिस कि कल अप्लाइड साइंस और इंटर को मिला साइंटिफिक टेल मी द अमेज़िंग मैक्सिमम - 121 ए फुबल ए साइन आफ अल्फा प्लस एक्स इन बिटवीन - 121 डांस क्लास 9 मठ - 1898 प्लस माइनस माइनस सब्सक्राइब टो ए प्लेस इन कोल्ड यू आर साइनअप अल्फा प्लस एक्स विद्यूज लेफ्ट कि मैं यह फंक्शन है लगने से हुई थीं लेकिन एक वर्ल्ड जो आर्म होल लेग स्पिनर प्लस 2019 सब्सक्राइब लू लू लू लू कि मुझमें झड़प एस प्लस बी स्क्वायर को मिनिमम बेललू हो है - लौंग और ऑफ ए स्क्वायर प्लस बी एस वेल मैं उदयपुर से न भूलें तो यह लार मठरी लाभ ले पर SSC मुहल्ला अब देखो अमित जो क्वेश्चन अधेड़ उम्र की 2% तो यह YouTube चैनल को सबस्क्राइब करना न भूलें सब्सक्राइब डिफरेंस है के पास मित्र मंडल कला को शुक्रिया दिखाओ फिर क्वेश्चन फर्स्ट SIM में पिछ हुआ है हुआ है की कमी नहीं पड़ता घोटाला और कुछ प्ले लुटने ले पर क्वेश्चन है उसी फाइनल मैक्सिमम लक्षण खिलाफ एकदम मैक्सिमम एंड मिनिमम वैल्यू आफ एक्सप्रेशन मेज़ कॉन्टैक्ट उस ए प्लेस 406 हेलो कॉल बैक प्लान फॉर उस एंड टेक केयर लव यू ऑल सबस्क्राइब और शेयर और सब्सक्राइब जरूर सब्सक्राइब सब्सक्राइब कर दो है कि है 9th फंक्शन 9th फंक्शन अफेक्ट्स तरीकों से लें एक वैक्स प्लस बी टाउन अधिक कौन सी सब्सक्राइब टो ए गुय एंड को मैक्सिमम है टर्न ऑफ द फंक्शन मैं ऑफिस कौन ब्लुटूथ लैब्स है तो इस तरफ सब्सक्राइब कर लें और प्लस 6 एक क्वेश्चन एग्जाम बिल्कुल कम कर दो ना मिले तो लुट मिली है को मिनिमम बेंगलुरु ऑफ द फंक्शन ऑफ कनॉट प्लेस को सबस्क्राइब करना न भूलें और कि मैं एक ऑप्शन है गौरतलब है आलरेडी सिर्फ हमारा मतलब ले एक व्हाट्सएप प्लस 20 मिनट समर्थक तो आराम संस्पर्श लूप तक हुं हुं हुं हुं हुं क्वेश्चन नंबर टू फाइंड द वैल्यू ऑफ द फ्री फॉरेक्स तो प्लीज प्ले हाय टाइम टर्न ऑफ एक्स माइनस स्पाइडर छह अक्षर एक्स एक्स माइनस सब्सक्राइब माय चैनल को सबस्क्राइब करना न भूलें तो सिर्फ दो एक सलूशन है थे फंक्शन आफ वैक्सीन कोल्ड ऊ कैन गिव उस ए ट्रिक ओं सेट्स ए प्लस फाइव साइन आफ एक्स माइनस 5653 को सबस्क्राइब नाउ टू 9 9 थे कॉस्ट आफ बेसिक ए माइनस कॉस एक्स ए टाइनी प्ले पानी पिछले 6 उत्तर प्रदेश मणिपुर में बहुत रिक्वेस्ट तक एक ट्रैवलर प्लस अपने कंधों फेज़ सकूं सफेद छह पिसिस 3382 कि टीचर दाखिले के लिए कौन सी नेशनल कॉफी यह क्रॉस साइन अप कॉल अपने प्राइवेट 60 56001 बैक टू द बेल आइकॉन लक है - हां बेटू इंटर कॉलेज लैब्स MP3 - ही बैक टू राहुल को subscribe to subscribe to subscribe 21 अ कि अब हम लेंगे एक कांटेक्ट प्लस 20 सैनिक हमला स्लाइड तो यह है मैं एक उल्टा ऊंची हाफ एंड बीइंग क्लोज ब्लैक ₹5 बैक टू का रस का गाना प्ले मैक्सिमम करें मीडियम कर या कुछ ना पशुओं के तो मैक्सिमम मैक्सिमम मैक्सिमम सब्सक्राइब तो प्लीज सब्सक्राइब जरूर सब्सक्राइब करें और सब्सक्राइब सब्सक्राइब टो कि मुझे प्रणव प्ले यह लिखो नंबर ऑफ मैक्सिमम विल फाइंड ए प्लेस टो इंग्लिश मीनिंग - डफ सब्सक्राइब - 26 - 6a है तो यह सलाह और हाउ टू फाइंड मैक्सिमम एंड मिनिमम वैल्यू आफ फंक्शनिंग आफ इंपोर्टेंट रहना सब्सक्राइब करें तो पेज सब्सक्राइब करें अ क्वेश्चन विच विल हैव डाइड क्वेश्चन उत्तर पूर्व विद्यालय आईडिया हो रेड्युस का रूप चीज लाइनेक्स ए प्लस कौस एक्ट्स एस ए सिंगल कि रेसिंग गेम्स 10th 2.1 पॉसिबल है रकम गला टाइम ओनली थे 2.2 मुश्किल है कौन सा इन ओनली ओं ए रिप्लाई कैंसिल करवा द क्वेश्चन एकदम गुर्जर भला कौन दीजिए क्वेश्चन है क्वेश्चन राधे-राधे से आगे से रंग गुलाल उड़े एक संदेश पिछले सेम कमर टेक्स्ट संदेश से व्हाट्सएप खोल आप कमेंट करें प्रयोग reduce करना पड़ेगा इस एक्सप्रेस 126 और सिक्योरिटी मेडिकल कॉलेज को सिंगल पर्सन व्हो सैक्रिफिस विल नॉट प्ले 2.1 मल्टिप्लाई मल्टिप्लाई एंड डिवाइडिंग थे ए रुद्दर ऑफ स्पिनर प्लस बी स्क्वायर है कि मल्टीप्लाइंग हिंदी मेडिमिक्स कौन प्ले स्टोर टावर आफ ए स्क्वेयर इन थिस डिफिकल्ट कुछ सलूशन है कि अष्टम गोडसे को खोलकर सिद्ध प्रदर्शन करना अच्छा प्लैन क्वेश्चन कौन सी गिव मे न चला दो कि गिवर सिग्नल f5 फैक्ट इक्वेडोर रूट प्लीज साइन एक्स ए प्लस कॉस एक्स अक्षर एयरप्लेन मोड ऑन हाउ टू ड्रॉ ए स्पेशल ओं 5th कि एक लुट एंड बीइंग क्लोज ब्लैडर का मन है और सुनाओ थे रूट बर्फ एस प्लस बी स्क्वायर इक्वल टू अवर ऑफ रुष वृद्धि प्लस वन लॉर्ड फॉर तमिल टू टू टू टू टू टू टू सोनू निगम के लिए मुट्ठी टू मच नकुल वेट उंगली डिमांड कर ली थी तो लूट 382 साइन एक्स प्लस वन बाय टू को सेट में एक लव लव लव यू टू द ले रूप बैक टू डू सब्सक्राइब माय चैनल को सबस्क्राइब करना न भूलें और सब्सक्राइब करें इस चैनल सब्सक्राइब 61 कि ऐसी कौन लट्टू तो कार्यालय साइन ए प्लस बी प्लस को सेंड ए साइन आफ ए प्लस बी प्लस 6 में बैठो ब्लैक टॉप टेन फंक्शनल हद कन्वर्टेड ब्लैक ओनली टाइम फंक्शन बिजली सप्लाई ओं थे लास्ट डे दिखाओ ऑफिस नंबर टू बैक पंखुड़ी लाइक एंड कमेंट इट बैक टू टू टू टू ए प्लस वन बाय टू को सेट कि अब तक उन्हें इस फिल्म को को साइन रखा है से क्वेश्चन पेपर कौन है डीपी लगाओ 100 ग्राम साइन कर दो MP3 मिट्टू यूज मिले साइन अध्ययन कि यह कैमरा साइन ए प्लस न्यूज़ ब्लैक स्पॉट्स और 583 लेट्यूस प्ले कांटेक्ट से मेघा फ्रॉम लाल किला कौन से कॉस्बी प्लस साइनस एंड पुल हिम बैक हॉटस्पॉट ऑफ एनएफसी ऑफ कि मी टू इनटू पॉवरफुल स्प्लैशेद यू कैन एक्सप्रेस - 6a है तो को मिला प्ले दिस कौन एक्सप्रेस हलवा जो एक्स्ट्रा मतलब इलाकों में सोमवार को घने फर्स्ट टाइम फंक्शन आपको मिल जाए तो यह 125 तो यह अ कि अ 138 वाला लुटा थोड़ा समय दलों को संदेश हमको पे कॉन्टैक्ट - फॉर साइंस सीक्वेंस - 40 कि यह लव यू फंक्शन एंड समथिंग कन्वर्टेड इनटू साइन फंक्शन ओनली एंड क्वेश्चन फंक्शनालिटी लेकिन अधिक तो आदमी को मतलब लाल मिर्च हाउ टू फाइंड मैक्सिमम एंड मिनिमम घरलू ऑफिस तुम सपनों को डीप समिति बहुत अमीर विवाह निषेध हाउ टो कन्वर्ट इनटू ओनली और को फंक्शन प्ले लिस्ट में थैंक यू |
9183 | https://bxscience.edu/ourpages/auto/2017/9/5/54355963/CW009%20Sol%20-%20Intermediate%20Value%20Theorem.pdf | The Bronx High School of Science Mathematics Department Jean M. Donahue, Principal Vikram Arora, A.P. of Mathematics Ms. Perez (perez4@bxscience.edu) AP Calculus AB Intermediate Value Theorem Intermediate Value Theorem (IVT): If f is continuous on a closed interval [a,b] and u is any number between f(a) and f(b), inclusive, then there is at least one number c in the interval [a,b] such that f(c)=u. 1. Let f(x) be a continuous function on the closed interval [-‐3,6]. If f(-‐3)=-‐1 and f(6)=3, then which of the following must be true? a. f(0)=0 Not necessarily. There must be a y = 0, but it does not have to be when x= 0 b. −1≤f (x) ≤3 for al x between -‐3 and 6 Not necessarily. Function can be above and below c. f(c)=1 for at least one c between -‐3 and 6 YES d. f(c)=0 for at least one c between -‐1 and 3 No, only if the -‐3 is changed for 6 2. Show that p(x) = 2x3 −5x2 −10x + 5 has a root somewhere in the interval [-‐1,2]. p(−1) = 2(−1)3 −5(−1)2 −10(−1)+ 5 = 9 p(2) = 2⋅23 −5⋅22 −10⋅2 + 5 = −19 Since p is a polynomial, it is continuous. In particular, it is continuous on the closed interval −1,2 [ ]. Since p(−1) = 9 > 0 > −19 = p(2), by the IVT there exists a number c in −1,2 [ ] such that p(c) = 0 , so c is a root. 3. Sketch the graph of a function f that satisfies the stated conditions: a) f is continuous everywhere except at x=3, at which point it is continuous from the right. b) f has a two-‐sided limit at x=3, but it is not continuous at x=3. 4. A student parking lot at a university charges $2.00 for the first half hour (or any part) and $1.00 for each subsequent half hour (or any part) up to a daily maximum of $10.00. a. Sketch a graph of cost as a function of the time parked. b. Discuss the significance of the discontinuities in the graph to a student who parks there. 5. Show that there is a square with a diagonal length between r and 2r and an area that is half the area of a circle of radius r. 2x2 = d 2 2x = d x = 2d 2 r < d < 2r d = r →A(r) = r2 2 d = 2r →A(2r) = 2r2 Ao = πr2 r2 2 < πr2 2 < 2r2 By the IVT, such a square exists. 6. (Challenge) Prove that if a and b are positive, then the equation a x −1 + b x −3 = 0 has at least one solution in the interval (1,3). It is continuous on (1,3). lim x→1+ a x −1 + b x −3 = ∞,DNE lim x→3− a x −1 + b x −3 = −∞,DNE There must exist numbers c and d with 1< c < d < 3 such that f (c) < 0 and f (d) > 0 . Also, f is continuous on [c,d]. By the IVT, there is a root in [c,d]. The points of discontinuity are the times at which there is a price jump of $1, so every half an hour. These are the points at which the student is maximizing the value, because right after, the student would have to pay an extra $1. |
9184 | https://www.ijsrst.com/IJSRST2074161 | Water Molecule : Chemical and Physical Properties
Authors
Dr. Kumar Rajeev Kishore M.Sc., Ph.D. (Chemistry), B.R.A. Bihar University, Muzaffarpur (Bihar)
Keywords:
H2O, Chemical Compound, Polar Molecule, Extra-Terrestrial Life
Abstract
Water is a chemical compound and polar molecule, which is liquid at standard temperature and pressure. It has the chemical formula H2O, meaning that one molecule of water is composed of two hydrogen atoms and one oxygen atom. Water is found almost everywhere on earth and is required by all known life. About 70% of the Earth's surface is covered by water. Water is known to exist, in ice form, on several other bodies in the solar system and beyond, and proof that it exists (or did exist) in liquid form anywhere besides Earth would be strong evidence of extraterrestrial life.
References
"Bacterial cell structure." Wikipedia. June 24, 2015. Accessed July 5, 2015.
"Bent molecular geometry." UC Davis ChemWiki. Accessed July 5, 2015.
"Hydrogen bond." Wikipedia. October 2, 2015. Accessed October 4, 2015.
"Properties of water." Wikipedia. May 27, 2016. Accessed May 28, 2016.
Raven, P. H., G.B. Johnson, K. A. Mason, J. B. Losos, and S. R. Singer. "The nature of molecules and properties of water." In Biology, 17-30. 10th ed. AP ed. New York, NY: McGraw-Hill, 2014.
"Bacterial cell structure." Wikipedia. June 24, 2015. Accessed July 5, 2015.
"Bent molecular geometry." UC Davis ChemWiki. Accessed July 5, 2015.
"Hydrogen bond." Wikipedia. October 2, 2015. Accessed October 4, 2015.
"Properties of water." Wikipedia. May 27, 2016. Accessed May 28, 2016.
Raven, P. H., G.B. Johnson, K. A. Mason, J. B. Losos, and S. R. Singer. "The nature of molecules and properties of water." In Biology, 17-30. 10th ed. AP ed. New York, NY: McGraw-Hill, 2014.
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How to Cite
Dr. Kumar Rajeev Kishore "Water Molecule : Chemical and Physical Properties" International Journal of Scientific Research in Science and Technology(IJSRST), Online ISSN : 2395-602X, Print ISSN : 2395-6011,Volume 4, Issue 11, pp.468-473, November-December-2018.
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Online ISSN : 2395-602X | Print ISSN : 2395-6011 | UGC Approved Journal Number : 64011IJSRST @ Copyright 2025. All Rights Reserved. | Platform and Workflow by OJS/PKP |
9185 | https://artofproblemsolving.com/wiki/index.php/Factoring_Quadratics?srsltid=AfmBOoq97A9lr_VKV1VvblB_pvFAhaF3cOwUsyBhLU6qfVJm-jKS3ECr | Art of Problem Solving
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Factoring Quadratics
The purpose of factoring a quadratic is to turn the quadratic into a product of 2 binomials.
Contents
1 Method 1
1.1 Example
1.2 Limitations
2 Method 2
2.1 Example
2.2 Limitations
3 Also See
Method 1
Method 1 starts with factoring the product of the roots. Let the quadratic we are factoring be . When factored, it will be in the form of where and are the roots of the quadratic, and where and .
Example
Since the coefficient on the term is , we know are quadratic factors in the form of . We know that the factor pairs of 12 are and We can find that only and satisfy our equations and , so the factored form of is .
Limitations
This method cannot be used to factor quadratics with complex or irrational roots.
Method 2
Method 2 starts by using the sum. Let the quadratic we are factoring be . When factored, it will be in the form of where and are the roots of the quadratic, and where and .
Example
We know that , so we can set and . Then, we get that , giving us that , or . Because we have both and as our roots, it doesn't matter which one is plugged in, giving us that the factored form of is .
Limitations
None currently known.
Also See
Quadratic equation
Factoring
Vieta's Formulas
Retrieved from "
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Algebra
Quadratic equations
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9186 | https://blogs.sas.com/content/iml/?p=43059 | Blogs
Blogs
The variance of the sums of variables
Undergraduate textbooks on probability and statistics typically prove theorems that
show how the variance of a sum of random variables is related to the variance of the original variables and the covariance between them.
For example, the Wikipedia article on Variance
contains an equation for the sum of two random variables, X and Y:
(
\operatorname {Var} (X+Y)=\operatorname {Var} (X)+\operatorname {Var} (Y)+2\,\operatorname {Cov} (X,Y)
)
A SAS programmer wondered whether equations like this are also true for vectors of data.
In other words, if X and Y are columns in a data set, do the sample variance and covariance statistics
satisfy the same equation?
How can you use SAS to verify (or disprove) the equation for a sample of data?
This article shows how to use SAS to verify that the equation (and another similar equation)
is valid for vectors of data.
It is possible to verify the equation by using PROC CORR and the DATA step, but it is much simpler to use
SAS IML software because the IML language enables you to directly access cells of a variance-covariance matrix.
Create the sum of columns
Let's start by choosing some data to analyze.
The following DATA step renames some variables in the Sashelp.Iris data set to X1, X2, and X3.
The program also creates the variables
Y12 = X1 + X2,
Y13 = X1 + X3, and
Y23 = X2 + X3.
The program then reads the data into SAS IML matrices and computes the variance-covariance matrix
for the original variables (X) and their pairwise sums (Y):
| |
| / verify variance of a sum (or linear combination) of variables / data Want; set sashelp.iris(where=(Species="Versicolor")); X1 = PetalLength; X2 = PetalWidth; X3 = SepalLength; Y12 = X1 + X2; Y13 = X1 + X3; Y23 = X2 + X3; keep X1-X3 Y12 Y13 Y23; run; proc iml; use Want; read all var {'X1' 'X2' 'X3'} into X[c=XNames]; read all var {'Y12' 'Y13' 'Y23'} into Y; close; YNames = {'X1+X2' 'X1+X3' 'X2+X3'}; SX = cov(X); SY = cov(Y); print SX[c=XNames r=XNames F=best5. L='Cov(X)'], SY[c=YNames r=YNames F=best5. L='Cov(Sums)']; |
The variance of a sum
Let's use this information to verify the identity
(
\operatorname {Var} (X1+X2)=\operatorname {Var} (X1)+\operatorname {Var} (X2)+2\,\operatorname {Cov} (X1,X2)
)
From the displayed (rounded) values of the covariance matrices, you can mentally calculate that the equation could be true.
The variances of the original variables are along the diagonal of the first matrix, and the covariances are the off-diagonal elements.
The variance of the sum is the [1,1] cell of the second matrix.
A little mental arithmetic indicates that 40.6 ≈ 22 + 4 + 27.
The following SAS IML statements extract the relevant values from cells in the variance-covariance matrices.
The program then subtracts the right side of the equation from the left side. If the difference is 0,
then the equation is verified:
| |
| / Confirm that the sample variance of (x1+x2) satisfies Var(x1 +x2) = Var(x1) + Var(x2) + 2Cov(x1, x2) See / VarY12= SY[1,1]; VarX1 = SX[1,1]; VarX2 = SX[2,2]; Cov12 = SX[1,2]; Eqn1 = VarY12 - (VarX1 + VarX2 + 2Cov12); print Eqn1; |
Success! The left and right sides of the equation are equal to numerical precision.
This validates the equation for the data we are using.
Notice that this same technique could be used to analyze the variance of a general linear combination of other variables.
The covariance of sums
Let's verify one more equation.
The Wikipedia article about Covariance
states the following formula for four random variables, X, Y, Z, and W:
(
\operatorname {Cov} (X+Y,Z+W)=\operatorname {Cov} (X,Z)+\operatorname {Cov} (X,W)+\operatorname {Cov} (Y,Z)+\operatorname {Cov} (Y,W)
)
Since our example data only has three variables, let's simplify the equation by setting W=X. Then (changing variable names) the equation we want to verify is
(
\operatorname {Cov} (X1+X2,X1+X3)=\operatorname {Cov} (X1,X3)+\operatorname {Cov} (X1,X1)+\operatorname {Cov} (X2,X3)+\operatorname {Cov} (X1,X2)
)
The following SAS IML statements extract the relevant cells from the covariance matrices. The program subtracts the right side of the equation from the left side and prints the difference.
| |
| / Confirm that the sample covariance of (x1+x2) and (x1 + x3) satisfies Cov(x1+x2, x1+x3) = Cov(x1, x1) + Cov(x1, x2) + Cov(x1, x3) + Cov(x2, x3) See / CovY12_Y13 = SY[1,2]; Cov11 = SX[1,1]; / = Var(X1) / Cov12 = SX[1,2]; Cov13 = SX[1,3]; Cov23 = SX[2,3]; Eqn2 = CovY12_Y13 - (Cov11 + Cov12 + Cov13 + Cov23); print Eqn2; |
Once again, the difference is essentially zero, which validates the equation.
Summary
A SAS programmer asked whether she could use SAS to verify certain equations.
She wanted to verify that certain formulas for the variance and covariance of random variables are also true
for sample statistics for empirical data.
This article shows how to use SAS IML software to test certain equations that relate the variance and covariance of sums to the variances and covariances of a set of data variables.
About Author
Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of SAS/IML software. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
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9187 | https://www.youtube.com/watch?v=_XAzzuDqB7A | Evaluating sin(pi/2 - x)
Cowan Academy
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Description
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Posted: 22 Apr 2023
This video demonstrates how to express sin(pi/2 - x) in its simplest form.
By applying the trig identity sin(A+B), we can simplify sin(pi/2 - x) to cos(x).
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Transcript: |
9188 | https://bestpractice.bmj.com/topics/en-gb/815?locale=ko | Esophageal varices - Symptoms, diagnosis and treatment | BMJ Best Practice US
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Esophageal varices
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Last reviewed: 28 Aug 2025
Last updated: 26 Mar 2025
Summary
Esophageal varices are a direct consequence of portal hypertension as a progressive complication of cirrhosis.
The development of bleeding carries significant morbidity and mortality.
Nonselective beta-blockers and/or endoscopic ligation can prevent the development of variceal bleeding.
Acute hemorrhage can be managed with resuscitation, a vasoactive therapy, and endoscopic ligation. Additional management may include transjugular intrahepatic shunt therapy and prophylactic antibiotics.
Diagnosis and surveillance are important aspects of management.
Definition
Esophageal varices are dilated collateral blood vessels that develop as a complication of portal hypertension, usually in the setting of cirrhosis. They can be seen on endoscopy. In the US and Europe, the major cause of cirrhosis is alcoholic liver disease. Worldwide, hepatitis B virus infection and hepatitis C virus infection are the major causes of cirrhosis.Alter MJ. Epidemiology of hepatitis B in Europe and worldwide. J Hepatol. 2003;39 Suppl 1:S64-9. Kim WR, Brown RS Jr, Terrault NA, et al. Burden of liver disease in the United States: summary of a workshop. Hepatology. 2002 Jul;36(1):227-42. Once cirrhosis has developed, increasing hepatic vein pressure gradient and deteriorating liver function may result in the formation of esophageal varices.
Rupture of esophageal varices can cause life-threatening bleeding. The most important predictor of variceal hemorrhage is the size of varices, with the highest risk of first hemorrhage occurring in patients with large varices (15% per year).North Italian Endoscopic Club for the Study and Treatment of Esophageal Varices. Prediction of the first variceal hemorrhage in patients with cirrhosis of the liver and esophageal varices. A prospective multicenter study. N Engl J Med. 1988 Oct 13;319(15):983-9. Garcia-Tsao G, Bosch J. Management of varices and variceal hemorrhage in cirrhosis. N Engl J Med. 2010 Mar 4;362(9):823-32. Other important predictors of hemorrhage are decompensated cirrhosis (Child-Pugh B/C) and the endoscopic finding of red wale marks.North Italian Endoscopic Club for the Study and Treatment of Esophageal Varices. Prediction of the first variceal hemorrhage in patients with cirrhosis of the liver and esophageal varices. A prospective multicenter study. N Engl J Med. 1988 Oct 13;319(15):983-9. Kaplan DE, Ripoll C, Thiele M, et al. AASLD practice guidance on risk stratification and management of portal hypertension and varices in cirrhosis. Hepatology. 2024 May 1;79(5):1180-211.
History and exam
Key diagnostic factors
cirrhosis
severe liver disease
alcohol misuse
hepatitis B or C infection
ascites
spider angioma
caput medusa
jaundice
encephalopathy
hematemesis
melena
hematochezia
HIV coinfection
Full details
Other diagnostic factors
splenomegaly
Full details
Risk factors
portal hypertension
size of varices
red wale marks
decompensated cirrhosis
ascites
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Diagnostic tests
1st tests to order
hepatic venous pressure gradient (HPVG)
complete blood count
coagulation profile (INR/prothrombin time)
serum LFTs
BUN and creatinine
blood typing/cross-matching
hepatitis B surface antigen (HBsAg)
anti-hepatitis C virus IgG (anti-HCV IgG)
esophago-gastro-duodenoscopy (EGD)
liver stiffness measurement (LSM)
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Emerging tests
capsule endoscopy
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Treatment algorithm
ACUTE
decompensated cirrhosis with acute variceal hemorrhage (hepatic venous pressure gradient ≥10 mmHg)
decompensated cirrhosis with acute variceal hemorrhage and failed endoscopic/pharmacologic therapy
ONGOING
compensated cirrhosis with mild portal hypertension (hepatic venous pressure gradient >5 and <10 mmHg)
compensated cirrhosis with clinically significant portal hypertension (hepatic venous pressure gradient ≥10 mmHg): without gastroesophageal varices
compensated cirrhosis with clinically significant portal hypertension (hepatic venous pressure gradient ≥10 mmHg): with gastroesophageal varices (no bleeding)
previous variceal bleed
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Contributors
Expert advisers
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Savio John, MD, AGAF, FACG
Chief (Division of Gastroenterology), Director (Hepatology), Associate Professor
Department of Medicine
State University of New York Upstate Medical University
Syracuse
NY
Disclosures
SJ serves as editor of the hepatology section for Stat Pearls.
Kelita Singh, MD
Associate Professor of Medicine
Department of Gastroenterology
State University of New York Upstate Medical University
Syracuse
NY
Disclosures
KS declares that she has no competing interests.
Acknowledgements
Dr Savio John and Dr Keilita Singh would like to gratefully acknowledge Dr Grace E. Dolman, Dr Gennaro D'Amico, Dr Giuseppe Malizia, Dr Vikram Boolchand, and Dr Thomas Boyer, previous contributors to this topic.
Disclosures
GED, GDA, GM, VB, and TB declare that they have no competing interests.
Peer reviewers
VIEW ALL Peer reviewers
Shreyas Saligram, MD, MRCP, FACG, FASGE
Assistant Professor
Department of Gastroenterology
University of California
San Francisco
CA
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SS declares that he has no competing interests.
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BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic.
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Peer reviewer affiliations and disclosures pertain to the time of the review.
References
Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible.
Key articles
Kaplan DE, Ripoll C, Thiele M, et al. AASLD practice guidance on risk stratification and management of portal hypertension and varices in cirrhosis. Hepatology. 2024 May 1;79(5):1180-211.Full textAbstract
de Franchis R, Bosch J, Garcia-Tsao G, et al. Baveno VII - renewing consensus in portal hypertension. J Hepatol. 2022 Apr;76(4):959-74.Full textAbstract
Reference articles
A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice.
Differentials
Hiatal hernia
Gastric varices
Mallory-Weiss tear
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Guidelines
AASLD practice guidance on risk stratification and management of portal hypertension and varices in cirrhosis
American College of Gastroenterology–Canadian Association of Gastroenterology clinical practice guideline: management of anticoagulants and antiplatelets during acute gastrointestinal bleeding and the periendoscopic period
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9189 | https://math.stackexchange.com/questions/1969634/if-r-is-symmetric-must-s-x-y-in-pa-times-pa-forall-x-in-x-exists-y | elementary set theory - If R is symmetric, must $S={ (X,Y)\in P(A)\times P(A) |\forall x\in X \exists y\in Y (xRy)} $ be symmetric? - Mathematics Stack Exchange
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If R is symmetric, must S={(X,Y)∈P(A)×P(A)|∀x∈X∃y∈Y(x R y)}S={(X,Y)∈P(A)×P(A)|∀x∈X∃y∈Y(x R y)} be symmetric?
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Suppose R is a relation on A, and define a relation S on P(A) as follows:
S={(X,Y)∈P(A)×P(A)|∀x∈X∃y∈Y(x R y)}S={(X,Y)∈P(A)×P(A)|∀x∈X∃y∈Y(x R y)}
If R is symmetric, must S be symmetric? Prove or provide a counterexample.
A counterexample provided is as follows:
A={0,1,2}A={0,1,2}
R={(0,1),(1,0)}R={(0,1),(1,0)}
P(A)={{0},{1},{1,2},...}P(A)={{0},{1},{1,2},...}
X={0}X={0}
Y={1,2}Y={1,2}
({0},{1,2}∈S({0},{1,2}∈S because (0,1)∈R(0,1)∈R
({1,2},{0})∉S({1,2},{0})∉S because (2,0)∉R(2,0)∉R
I am confused about two thing:
What exactly should the conclusion we are supposed to show be? Is it ∀y∈Y∃x∈X(y R x)∀y∈Y∃x∈X(y R x)? Or ∀x∈X∃y∈Y(y R x)∀x∈X∃y∈Y(y R x)? And why? If it's the former, then indeed it seems that the proof does not work; but if it's the latter, I have actually worked out a proof, but what's wrong with it?
My proof is as follows.
Suppose R is symmetric, we need to assume ∀x∈X∃y∈Y(x R y)∀x∈X∃y∈Y(x R y) and prove ∀x∈X∃y∈Y(y R x)∀x∈X∃y∈Y(y R x). From the assumption we get x R z x R z. But since R is symmetric, we can use it to derive z R x z R x from x R z x R z. Then we can existential generalise z R x z R x to get ∃y∈Y(y R x)∃y∈Y(y R x). Since x x is arbitrary, this completes the universal proof.
What does the definition of S, ∀x∈X∃y∈Y(x R y)∀x∈X∃y∈Y(x R y) mean? Specifically, I am not sure what the ∃∃ means - I am having trouble understanding the last statement of the counterexample. (In fact I think my confusion is in part to blame for my 1st question for not knowing what the conclusion should be.)
Could anyone please help? (I understand that someone has always asked questions on exactly the same proof, but his concern seems to be different from mine)
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edited Jun 12, 2020 at 10:38
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asked Oct 15, 2016 at 15:13
Constantly confusedConstantly confused
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1 An arguably simpler counterexample: Let R R be the identity relation (which is clearly symmetric); then S S is the subset relation on P(A)P(A), which we know well is not symmetric.hmakholm left over Monica –hmakholm left over Monica 2016-10-15 15:20:41 +00:00 Commented Oct 15, 2016 at 15:20
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What's wrong with your proof is that you have unfolded the definition of S S wrong at a crucial step along the way. You want to prove
For all X X and Y Y, if X S Y X S Y then Y S X Y S X.
which unfolds to
For all X X and Y Y, if ∀x∈X∃y∈Y(x R y)∀x∈X∃y∈Y(x R y), then ∀x∈Y∃y∈X(x R y)∀x∈Y∃y∈X(x R y).
but what you're actually proving is
For all X X and Y Y, if ∀x∈X∃y∈Y(x R y)∀x∈X∃y∈Y(x R y), then ∀x∈X∃y∈Y(y R x)∀x∈X∃y∈Y(y R x).
The unfolding of Y S X Y S X is wrong here. When you unfold Y S X Y S X the universally quantified variable must be picked from Y Y, because that is what is to the left of the S S.
(3) is actually true and your proof of it seems to be correct, if a bit terse. But it's not what you need to prove here.
To see that Y S X Y S X means ∀y∈Y∃x∈X(y R x)∀y∈Y∃x∈X(y R x), it may be easier to do the renamings in two steps:
X S Y X S Y means ∀x∈X∃y∈Y(x R y)∀x∈X∃y∈Y(x R y).
Since the X X and Y Y in this definition are different from the X X and Y Y in Y S X Y S X, let's first temporarily rename the ones in the definition to P P and Q Q -- X X becomes P P and Y Y becomes Q Q:
P S Q P S Q means ∀x∈P∃y∈Q(x R y)∀x∈P∃y∈Q(x R y).
Now we can plug in Y Y for P P and X X for Q Q to get the definition as it applies to Y S X Y S X:
Y S X Y S X means ∀x∈Y∃y∈X(x R y)∀x∈Y∃y∈X(x R y).
Finally, just for tidiness we can switch the names of the dummy variables x x and y y (this changes nothing) to produce
Y S X Y S X means ∀y∈Y∃x∈X(y R x)∀y∈Y∃x∈X(y R x).
All of this has nothing in particular to do with the meaning of ∃∃ -- it's just a matter of substituting variable names in the expression.
For the last statement in the counterexample: In the penultimate line we have seen that {0}S{1,2}{0}S{1,2}. If S S were symmetric, we must therefore also have {1,2}S{0}{1,2}S{0}, and what this means is
∀x∈{1,2}∃y∈{0},(x R y)∀x∈{1,2}∃y∈{0},(x R y)
The example then implicitly argues: If this is true, then it must be true both for x=1 x=1 and x=2 x=2. But it is not true for x=2 x=2, because
∃y∈{0}(2 R y)∃y∈{0}(2 R y)
is false. The only value we can choose for y y is 0 0 and that does not make (2 R y)(2 R y) true, because (2,0)∉R(2,0)∉R.
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answered Oct 15, 2016 at 16:15
hmakholm left over Monicahmakholm left over Monica
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Thank you so much! Your explanation is incredibly clear!Constantly confused –Constantly confused 2016-10-17 14:01:45 +00:00 Commented Oct 17, 2016 at 14:01
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6S={(X,Y)∈P(A)×P(A)∣∀x∈X∃y∈Y(x R y)}.S={(X,Y)∈P(A)×P(A)∣∀x∈X∃y∈Y(x R y)}. If R is symmetric, must S be symmetric?
5How to prove relation is asymmetric if it is both anti-symmetric and irreflexive
2Prove that if R R is a symmetric, transitive relation on A A and the domain of R R is A A, then R R is reflexive on A A.
0Proving transitivity of a relation
0Let R R be a relation, R:A→A R:A→A where R R is both symmetric and transitive. Prove the following:
7Why is a symmetric relation defined: ∀x∀y(x R y⟹y R x)∀x∀y(x R y⟹y R x) and not ∀x∀y(x R y⟺y R x)∀x∀y(x R y⟺y R x)?
1Is the relation reflexive, symmetric and antisymmetric?
2Find a relation which is reflexive and symmetric but not transitive on integers
0If a relation is euclidean, is it necessarily asymmetric?
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9190 | https://books.google.com/books?id=ZtFwJCAiF3wC&printsec=frontcover | Phillips' Science of Dental Materials - eBook: Phillips' Science of Dental ... - Kenneth J. Anusavice - Google Books
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9191 | https://bitesizebio.com/163/10-ways-to-work-rnase-free/ | 10 Ways to Work RNase Free
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10 Ways to Work RNase Free
Written by: Dr Nick Oswald
last updated: April 10, 2025
Working with RNA? What fun! Those little, nearly indestructible RNases are everywhere – on your skin and mucous membranes, in the water and (some of the) enzymes you use, on lab surfaces, even in airborne microbes! Here are 10 ways to keep the RNases at bay, and keep your precious samples safe:
Clean everything; bench surfaces, pipettes, electrophoresis equipment and anything else you can think of with an RNase cleaning product, such as RNaseZap from Ambion (or 0.5% SDS followed by 3%H2O2). Establish a regular cleaning routine; a quick daily clean and a deeper weekly or monthly clean… and stick to it.
Treat your solutions. Good old DEPC is a fine way to keep your solutions RNase free. Use 0.5 mL DEPC/L, incubate for 2 hr, autoclave for 45 minutes minimum. DMPC can also be used and may be be safer than DEPC, which is a known carcinogen. Alternatively, many vendors offer certified nuclease-free water, which may be worth the investment. Note that ultrafiltered water is already RNase free so does not need DEPC treatment. Also, don’t use DEPC/DMPC on tris-based solutions.
Designate a workspace,and a set of pipettes, if possible, that are dedicated to RNase-free work.
Use barrier tips. Barrier tips stop cross-contamination of your reagents and samples by preventing aerosols reaching the barrel of your pipette. They are a must-have for RNA work.
Wear gloves and a lab coat. The obvious ones are the best. Gloves and a lab coat will stop you from contaminating your samples with your own RNases. Change both frequently (maybe once per week for lab coats). Also, when you have your gloves on don’t touch anything that is not decontaminated – door handles, taps, yourself… or other people (!).
Bake your glasswear. No enzyme can withstand baking for 300°C for 2 hours, but your glasswear can.
Isolate RNA using a method that eliminates endogenous RNAses, such as AquaRNA from Multitarget Pharmaceuticals, which is both clean and convenient.
Use RNase-free enzymes. Enzymes isolated from bacteria (e.g. DNase) can be full of RNase. Make sure you use certified RNase-free enzymes on your RNA samples where possible.
Use an RNase inhibitor when it’s not possible to keep things completely RNase-free. Roche’s Protector is a good example. Avoid high temperatures (above 60°C) or denaturing conditions that could deactivate the inhibitor!
Store RNA in ethanol at -80°C. Make aliquots if the sample is to be used a number of times to avoid freeze/thaw cycles. Before use, centrifuge to pellet the RNA, air dry then resuspend in an RNase-free buffer.
Be completely paranoid, work as far away from your colleagues as possible, and shower in RNaseZAP five times per day.Just kidding.
Dr Nick Oswald
Nick has a PhD from the University Dundee and is the Founder and Director of Bitesize Bio, Science Squared Ltd and The Life Science Marketing Society.
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CRC Handbook of Combinatorial Designs ===================================== Charles J. Colbourn CRC Press, Dec 12, 2010 - Mathematics - 784 pages From experimental design to cryptography, this comprehensive, easy-to-access reference contains literally all the facts you need on combinatorial designs. It includes constructions of designs, existence results, and properties of designs. Organized into six main parts, the CRC Handbook of Combinatorial Designs covers: More » Preview this book »
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VI 3
VII 41
VIII 47
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Charles J. Colbourn,Jeffrey Dinitz
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Projective Geometries Over Finite Fields
James William Peter Hirschfeld
No preview available - 1998
Principles and Practice of Constraint Programming - CP 2002: 8th ...
Pascal Van Hentenryck
No preview available - 2002
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Title CRC Handbook of Combinatorial Designs
Discrete Mathematics and Its Applications
EditorCharles J. Colbourn
Edition illustrated
Publisher CRC Press, 2010
ISBN 142004995X, 9781420049954
Length 784 pages
SubjectsComputers
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Operating Systems
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Computers / Operating Systems / General
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Mathematics / Combinatorics
Mathematics / Discrete Mathematics
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9193 | https://www.collinsdictionary.com/us/dictionary/english/cupidity | CUPIDITY definition in American English | Collins English Dictionary
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Definition of 'cupidity'
COBUILD frequency band
cupidity
(kyup ɪ dɪti)
uncountable noun
Cupidity is a greedy desire for money and possessions.
[formal]
avarice
His eyes gave him away, shining with cupidity.
Synonyms:avarice, greed, acquisitiveness, rapacityMore Synonyms of cupidity
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cupidity in American English
(kjuˈpɪdəti)
noun Origin: ME & Anglo-Fr cupidite< L cupiditas<cupidus: see Cupid
excessive desire for something, esp. for wealth; avarice; greed
Webster’s New World College Dictionary, 5th Digital Edition. Copyright © 2025 HarperCollins Publishers.
COBUILD frequency band
cupidity in American English
(kjuːˈpɪdɪti)
noun
eager or excessive desire, esp. to possess something; greed; avarice
SYNONYMS covetousness, avidity, hunger, acquisitiveness.
Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd
Derived forms
cupidinous (kjuːˈpɪdnəs)
adjective
Word origin
[1400–50; late ME cupidite (‹ MF) ‹ L cupiditās, equiv. to cupid(us) eager, desirous (cup(ere) to desire + -idus-id 4) + -itās-ity]
COBUILD frequency band
cupidity in British English
(kjuːˈpɪdɪtɪ)
noun
strong desire, esp for possessions or money; greed
Collins English Dictionary. Copyright © HarperCollins Publishers
Word origin
C15: from Latin cupiditās, from cupidus eagerly desiring, from cupere to long for
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In addition, its creator can instruct it to pile up treasures and watch over them and to simulate a form of cupidity.
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Incitement has a particularlybroad "actus reus"; it has been interpreted to include a suggestion, proposal, request, exhortation, gesture, argument, persuasion, inducement, goading or the arousal of cupidity.
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9194 | https://it.gauthmath.com/solution/1835993419441154/1-Indica-se-le-seguenti-parole-sono-maschili-M-femminili-F-o-invariabili-I-poi-f | Risolto:Indica se le seguenti parole sono maschili (M), femminili (F), o invariabili (I), poi for
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Indica se le seguenti parole sono maschili (M), femminili (F), o invariabili (I), poi forma il plurale di ognuna. 1. coche M_ 7. belga 2 atleta 8. problema _ 3. leche 9. modelo 10.viernes 4. director 5. pez _11. rey _ _12. actriz 6. televisión _
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coche (M) - coches; atleta (M) - atletas; leche (F) - leches; director (M) - directores; pez (M) - peces; televisión (F) - televisiones; belga (M/F) - belgas; problema (M) - problemas; modelo (M/F) - modelos; viernes (M) - viernes; rey (M) - reyes; actriz (F) - actrices.
Spiegazione
Identificaremos el género gramatical de cada palabra y luego formaremos su plural.
Aplicaremos las reglas de formación del plural en español. Para sustantivos masculinos que terminan en consonante, se añade "-es". Para sustantivos masculinos que terminan en vocal, se añade "-s". Para sustantivos femeninos que terminan en consonante, se añade "-es". Para sustantivos femeninos que terminan en vocal, se añade "-s". Hay excepciones, como "pez" que forma su plural en "peces".
Analizaremos cada palabra individualmente.
coche (M) - coches
atleta (M) - atletas
leche (F) - leches
director (M) - directores
pez (M) - peces
televisión (F) - televisiones
belga (M/F) - belgas (invariable en género, plural en -s)
problema (M) - problemas
modelo (M/F) - modelos (invariable en género, plural en -s)
viernes (M) - viernes (invariable en número)
rey (M) - reyes
actriz (F) - actrices
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9195 | https://zhidao.baidu.com/question/1755430522766389068.html | 在等差数列{an}中，已知S15=90,求a8.数学！谢谢！要过程_百度知道
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9196 | https://www-users.cse.umn.edu/~olver/ln_/odq.pdf | Nonlinear Ordinary Differential Equations by Peter J. Olver University of Minnesota 1.
Introduction.
These notes are concerned with initial value problems for systems of ordinary dif-ferential equations. Here our emphasis will be on nonlinear phenomena and properties, particularly those with physical relevance. Finding a solution to a differential equation may not be so important if that solution never appears in the physical model represented by the system, or is only realized in exceptional circumstances. Thus, equilibrium solu-tions, which correspond to configurations in which the physical system does not move, only occur in everyday situations if they are stable. An unstable equilibrium will not ap-pear in practice, since slight perturbations in the system or its physical surroundings will immediately dislodge the system far away from equilibrium.
Of course, very few nonlinear systems can be solved explicitly, and so one must typ-ically rely on a numerical scheme to accurately approximate the solution. Basic methods for initial value problems, beginning with the simple Euler scheme, and working up to the extremely popular Runge–Kutta fourth order method, will be the subject of the final section of the chapter. However, numerical schemes do not always give accurate results, and we briefly discuss the class of stiffdifferential equations, which present a more serious challenge to numerical analysts.
Without some basic theoretical understanding of the nature of solutions, equilibrium points, and stability properties, one would not be able to understand when numerical so-lutions (even those provided by standard well-used packages) are to be trusted. Moreover, when testing a numerical scheme, it helps to have already assembled a repertoire of nonlin-ear problems in which one already knows one or more explicit analytic solutions. Further tests and theoretical results can be based on first integrals (also known as conservation laws) or, more generally, Lyapunov functions. Although we have only space to touch on these topics briefly, but, we hope, this will whet the reader’s appetite for delving into this subject in more depth. The references [2, 9, 13, 15, 17] can be profitably consulted.
2.
First Order Systems of Ordinary Differential Equations.
Let us begin by introducing the basic object of study in discrete dynamics: the initial value problem for a first order system of ordinary differential equations. Many physical applications lead to higher order systems of ordinary differential equations, but there is a simple reformulation that will convert them into equivalent first order systems. Thus, we do not lose any generality by restricting our attention to the first order case throughout.
Moreover, numerical solution schemes for higher order initial value problems are entirely based on their reformulation as first order systems.
1/7/22 1 c ⃝2022 Peter J. Olver Scalar Ordinary Differential Equations As always, when confronted with a new problem, it is essential to fully understand the simplest case first. Thus, we begin with a single scalar, first order ordinary differential equation du dt = F(t, u).
(2.1) In many applications, the independent variable t represents time, and the unknown func-tion u(t) is some dynamical physical quantity. Throughout this chapter, all quantities are assumed to be real. (Results on complex ordinary differential equations can be found in .) Under appropriate conditions on the right hand side (to be formalized in the following section), the solution u(t) is uniquely specified by its value at a single time, u(t0) = u0.
(2.2) The combination (2.1–2) is referred to as an initial value problem, and our goal is to devise both analytical and numerical solution strategies.
A differential equation is called autonomous if the right hand side does not explicitly depend upon the time variable: du dt = F(u).
(2.3) All autonomous scalar equations can be solved by direct integration. We divide both sides by F(u), whereby 1 F(u) du dt = 1, and then integrate with respect to t; the result is Z 1 F(u) du dt dt = Z dt = t + k, where k is the constant of integration. The left hand integral can be evaluated by the change of variables that replaces t by u, whereby du = (du/dt) dt, and so Z 1 F(u) du dt dt = Z du F(u) = G(u), where G(u) indicates a convenient anti-derivative† of the function 1/F(u).
Thus, the solution can be written in implicit form G(u) = t + k.
(2.4) If we are able to solve the implicit equation (2.4), we may thereby obtain the explicit solution u(t) = H(t + k) (2.5) † Technically, a second constant of integration should appear here, but this can be absorbed into the previous constant k, and so proves to be unnecessary.
1/7/22 2 c ⃝2022 Peter J. Olver 0.5 1 1.5 2 -1 -0.5 0.5 1 1.5 2 Figure 1.
Solutions to u = u2.
in terms of the inverse function H = G−1. Finally, to satisfy the initial condition (2.2), we set t = t0 in the implicit solution formula (2.4), whereby G(u0) = t0 + k. Therefore, the solution to our initial value problem is G(u) −G(u0) = t −t0, or, explicitly, u(t) = H t −t0 + G(u0) .
(2.6) Remark: A more direct version of this solution technique is to rewrite the differential equation (2.3) in the “separated form” du F(u) = dt, in which all terms involving u, including its differential du, are collected on the left hand side of the equation, while all terms involving t and its differential are placed on the right, and then formally integrate both sides, leading to the same implicit solution formula: G(u) = Z du F(u) = Z dt = t + k.
(2.7) Before completing our analysis of this solution method, let us run through a couple of elementary examples.
Example 2.1. Consider the autonomous initial value problem du dt = u2, u(t0) = u0.
(2.8) To solve the differential equation, we rewrite it in the separated form du u2 = dt, and then integrate both sides: −1 u = Z du u2 = t + k.
1/7/22 3 c ⃝2022 Peter J. Olver Solving the resulting algebraic equation for u, we deduce the solution formula u = − 1 t + k .
(2.9) To specify the integration constant k, we evaluate u at the initial time t0; this implies u0 = − 1 t0 + k , so that k = −1 u0 −t0.
Therefore, the solution to the initial value problem is u = u0 1 −u0(t −t0) .
(2.10) Figure 1 shows the graphs of some typical solutions.
As t approaches the critical value t⋆= t0 + 1/u0 from below, the solution “blows up”, meaning u(t) →∞as t →t⋆. The blow-up time t⋆depends upon the initial data — the larger u0 > 0 is, the sooner the solution goes offto infinity. If the initial data is negative, u0 < 0, the solution is well-defined for all t > t0, but has a singularity in the past, at t⋆= t0 + 1/u0 < t0. The only solution that exists for all positive and negative time is the constant solution u(t) ≡0, corresponding to the initial condition u0 = 0.
In general, the constant equilibrium solutions to an autonomous ordinary differential equation, also known as its fixed points, play a distinguished role. If u(t) ≡u⋆is a constant solution, then du/dt ≡0, and hence the differential equation (2.3) implies that F(u⋆) = 0.
Therefore, the equilibrium solutions coincide with the roots of the function F(u). In point of fact, since we divided by F(u), the derivation of our formula for the solution (2.7) assumed that we were not at an equilibrium point. In the preceding example, our final solution formula (2.10) happens to include the equilibrium solution u(t) ≡0, corresponding to u0 = 0, but this is a lucky accident. Indeed, the equilibrium solution does not appear in the “general” solution formula (2.9). One must typically take extra care that equilibrium solutions do not elude us when utilizing this basic integration method.
Example 2.2.
Although a population of people, animals, or bacteria consists of individuals, the aggregate behavior can often be effectively modeled by a dynamical system that involves continuously varying variables. As first proposed by the English economist Thomas Malthus in 1798, the population of a species grows, roughly, in proportion to its size. Thus, the number of individuals N(t) at time t satisfies a first order differential equation of the form dN dt = ρN, (2.11) where the proportionality factor ρ = β −δ measures the rate of growth, namely the difference between the birth rate β ≥0 and the death rate δ ≥0. Thus, if births exceed deaths, ρ > 0, and the population increases, whereas if ρ < 0, more individuals are dying and the population shrinks.
In the very simplest model, the growth rate ρ is assumed to be independent of the population size, and (2.11) reduces to a simple linear ordinary differential equation whose 1/7/22 4 c ⃝2022 Peter J. Olver solutions satisfy the Malthusian exponential growth law N(t) = N0 eρt, where N0 = N(0) is the initial population size. Thus, if ρ > 0, the population grows without limit, while if ρ < 0, the population dies out, so N(t) →0 as t →∞, at an exponentially fast rate. The Malthusian population model provides a reasonably accurate description of the behavior of an isolated population in an environment with unlimited resources.
In a more realistic scenario, the growth rate will depend upon the size of the population as well as external environmental factors. For example, in the presence of limited resources, relatively small populations will increase, whereas an excessively large population will have insufficient resources to survive, and so its growth rate will be negative. In other words, the growth rate ρ(N) > 0 when N < N ⋆, while ρ(N) < 0 when N > N ⋆, where the carrying capacity N ⋆> 0 depends upon the resource availability. The simplest class of functions that satifies these two inequalities are of the form ρ(N) = µ(N ⋆−N), where µ > 0 is a positive constant. This leads us to the nonlinear population model dN dt = µN (N ⋆−N).
(2.12) In deriving this model, we assumed that the environment is not changing over time; a dynamical environment would require a more complicated non-autonomous differential equation.
Before analyzing the solutions to the nonlinear population model, let us make a pre-liminary change of variables, and set u(t) = N(t)/N ⋆, so that u represents the size of the population in proportion to the carrying capacity N ⋆. A straightforward computation shows that u(t) satisfies the so-called logistic differential equation du dt = λu(1 −u), u(0) = u0, (2.13) where λ = N ⋆µ, and, for simplicity, we assign the initial time to be t0 = 0. The logistic differential equation can be viewed as the continuous counterpart of the logistic map studied in my Notes on Nonlinear Systems. However, unlike its discrete namesake, the logistic differential equation is quite sedate, and its solutions easily understood.
First, there are two equilibrium solutions: u(t) ≡0 and u(t) ≡1, obtained by setting the right hand side of the equation equal to zero.
The first represents a nonexistent population with no individuals and hence no reproduction. The second equilibrium solution corresponds to a static population N(t) ≡N ⋆that is at the ideal size for the environment, so deaths exactly balance births. In all other situations, the population size will vary over time.
To integrate the logistic differential equation, we proceed as above, first writing it in the separated form du u(1 −u) = λ dt.
Integrating both sides, and using partial fractions, λt + k = Z du u(1 −u) = Z 1 u + 1 1 −u du = log u 1 −u , 1/7/22 5 c ⃝2022 Peter J. Olver 2 4 6 8 10 -1 -0.5 0.5 1 1.5 2 Figure 2.
Solutions to u′ = u(1 −u).
where k is a constant of integration. Therefore u 1 −u = ceλt, where c = ±ek.
Solving for u, we deduce the solution u(t) = ceλt 1 + ceλt .
(2.14) The constant of integration is fixed by the initial condition. Solving the algebraic equation u0 = u(0) = c 1 + c yields c = u0 1 −u0 .
Substituting the result back into the solution formula (2.14) and simplifying, we find u(t) = u0 eλt 1 −u0 + u0 eλt .
(2.15) The resulting solutions are illustrated in Figure 2. Interestingly, while the equilibrium solutions are not covered by the integration method, they reappear in the final solution formula, corresponding to initial data u0 = 0 and u0 = 1 respectively. However, this is a lucky accident, and cannot be anticipated in more complicated situations.
When using the logistic equation to model population dynamics, the initial data is assumed to be positive, u0 > 0. As time t →∞, the solution (2.15) tends to the equilibrium value u(t) →1 — which corresponds to N(t) →N ⋆approaching the carrying capacity in the original population model. For small initial values u0 ≪1 the solution initially grows at an exponential rate λ, corresponding to a population with unlimited resources.
However, as the population increases, the gradual lack of resources tends to slow down 1/7/22 6 c ⃝2022 Peter J. Olver the growth rate, and eventually the population saturates at the equilibrium value. On the other hand, if u0 > 1, the population is too large to be sustained by the available resources, and so dies offuntil it reaches the same saturation value. If u0 = 0, then the solution remains at equilibrium u(t) ≡0. Finally, when u0 < 0, the solution only exists for a finite amount of time, with u(t) − →−∞ as t − →t⋆= 1 λ log 1 −1 u0 .
Of course, this final case does appear in the physical world, since we cannot have a negative population!
The separation of variables method used to solve autonomous equations can be stra-ightforwardly extended to a special class of non-autonomous equations. A separable ordi-nary differential equation has the form du dt = a(t) F(u), (2.16) in which the right hand side is the product of a function of t and a function of u. To solve the equation, we rewrite it in the separated form du F(u) = a(t) dt.
Integrating both sides leads to the solution in implicit form G(u) = Z du F(u) = Z a(t) dt = A(t) + k.
(2.17) The integration constant k is then fixed by the initial condition. And, as before, one must properly account for any equilibrium solutions, when F(u) = 0.
Example 2.3. Let us solve the particular initial value problem du dt = (1 −2t) u, u(0) = 1.
(2.18) We begin by writing the differential equation in separated form du u = (1 −2t) dt.
Integrating both sides leads to log u = Z du u = Z (1 −2t) dt = t −t2 + k, where k is the constant of integration. We can readily solve for u(t) = cet−t2, where c = ±ek. The latter formula constitutes the general solution to the differential equation, and happens to include the equilibrium solution u(t) ≡0 when c = 0. The given initial condition requires that c = 1, and hence u(t) = et−t2 is the unique solution to the initial value problem. The solution is graphed in Figure 3.
1/7/22 7 c ⃝2022 Peter J. Olver 0.5 1 1.5 2 2.5 3 0.25 0.5 0.75 1 1.25 1.5 Figure 3.
Solution to the Initial Value Problem u = (1 −2t) u, u(0) = 1.
First Order Systems A first order system of ordinary differential equations has the general form du1 dt = F1(t, u1, . . . , un), · · · dun dt = Fn(t, u1, . . ., un).
(2.19) The unknowns u1(t), . . ., un(t) are scalar functions of the real variable t, which usually represents time. We shall write the system more compactly in vector form du dt = F(t, u), (2.20) where u(t) = ( u1(t), . . ., un(t) )T , and F(t, u) = ( F1(t, u1, . . ., un), . . ., Fn(t, u1, . . . , un) )T is a vector-valued function of n + 1 variables. By a solution to the differential equation, we mean a vector-valued function u(t) that is defined and continuously differentiable on an interval a < t < b, and, moreover, satisfies the differential equation on its interval of definition. Each solution u(t) serves to parametrize a curve C ⊂Rn, also known as a trajectory or orbit of the system.
In this chapter, we shall concentrate on initial value problems for such first order systems. The general initial conditions are u1(t0) = a1, u2(t0) = a2, · · · un(t0) = an, (2.21) or, in vectorial form, u(t0) = a (2.22) Here t0 is a prescribed initial time, while the vector a = ( a1, a2, . . ., an )T fixes the initial position of the desired solution. In favorable situations, as described below, the initial conditions serve to uniquely specify a solution to the differential equations — at least for nearby times. The general issues of existence and uniqueness of solutions will be addressed in the following section.
1/7/22 8 c ⃝2022 Peter J. Olver A system of differential equations is called autonomous if the right hand side does not explicitly depend upon the time t, and so takes the form du dt = F(u).
(2.23) One important class of autonomous first order systems are the steady state fluid flows.
Here F(u) = v represents the fluid velocity vector field at the position u. The solution u(t) to the initial value problem (2.23, 22) describes the motion of a fluid particle that starts at position a at time t0. The differential equation tells us that the fluid velocity at each point on the particle’s trajectory matches the prescribed vector field.
An equilibrium solution is constant: u(t) ≡u⋆for all t. Thus, its derivative must vanish, du/dt ≡0, and hence, every equilibrium solution arises as a solution to the system of algebraic equations F(u⋆) = 0 (2.24) prescribed by the vanishing of the right hand side of the system (2.23).
Example 2.4. A predator-prey system is a simplified ecological model of two species: the predators which feed on the prey. For example, the predators might be lions roaming the Serengeti and the prey zebra. We let u(t) represent the number of prey, and v(t) the number of predators at time t. Both species obey a population growth model of the form (2.11), and so the dynamical equations can be written as du dt = ρu, dv dt = σ v, (2.25) where the growth rates ρ, σ may depend upon the other species. The more prey, i.e., the larger u is, the faster the predators reproduce, while a lack of prey will cause them to die off. On the other hand, the more predators, the faster the prey are consumed and the slower their net rate of growth.
If we assume that the environment has unlimited resources for the prey, which, bar-ring drought, is probably valid in the case of the zebras, then the simplest model that incorporates these assumptions is the Lotka–Volterra system du dt = αu −δuv, dv dt = −β v + γ uv, (2.26) corresponding to growth rates ρ = α −δv, σ = −β + γ u. The parameters α, β, γ, δ > 0 are all positive, and their precise values will depend upon the species involved and how they interact, as indicated by field data, combined with, perhaps, educated guesses.
In particular, α represents the unrestrained growth rate of the prey in the absence of predators, while −β represents the rate that the predators die offin the absence of their prey. The nonlinear terms model the interaction of the two species: the rate of increase in the predators is proportional to the number of available prey, while the rate of decrease in the prey is proportional to the number of predators. The initial conditions u(t0) = u0, v(t0) = v0 represent the initial populations of the two species.
1/7/22 9 c ⃝2022 Peter J. Olver We will discuss the integration of the Lotka–Volterra system (2.26) in Section 4. Here, let us content ourselves with determining the possible equilibria. Setting the right hand sides of the system to zero leads to the nonlinear algebraic system 0 = αu −δuv = u(α −δv), 0 = −β v + γ uv = v(−β + γ u).
Thus, there are two distinct equilibria, namely u⋆ 1 = v⋆ 1 = 0, u⋆ 2 = β/γ, v⋆ 2 = α/δ.
The first is the uninteresting (or, rather catastrophic) situation where there are no animals — no predators and no prey. The second is a nontrivial solution in which both populations maintain a steady value, for which the birth rate of the prey is precisely sufficient to continuously feed the predators. Is this a feasible solution? Or, to state the question more mathematically, is this a stable equilibrium? We shall develop the tools to answer this question below.
Higher Order Systems A wide variety of physical systems are modeled by nonlinear systems of differential equations depending upon second and, occasionally, even higher order derivatives of the unknowns. But there is an easy device that will reduce any higher order ordinary differ-ential equation or system to an equivalent first order system. “Equivalent” means that each solution to the first order system uniquely corresponds to a solution to the higher order equation and vice versa. The upshot is that, for all practical purposes, one only needs to analyze first order systems. Moreover, the vast majority of numerical solution algorithms are designed for first order systems, and so to numerically integrate a higher order equation, one must place it into an equivalent first order form.
We have already encountered the main idea in our discussion of the phase plane approach to second order scalar equations d2u dt2 = F t, u, du dt .
(2.27) We introduce a new dependent variable v = du dt . Since dv dt = d2u dt2 , the functions u, v satisfy the equivalent first order system du dt = v, dv dt = F(t, u, v).
(2.28) Conversely, it is easy to check that if u(t) = ( u(t), v(t) )T is any solution to the first order system, then its first component u(t) defines a solution to the scalar equation, which establishes their equivalence. The basic initial conditions u(t0) = u0, v(t0) = v0, for the first order system translate into a pair of initial conditions u(t0) = u0, u(t0) = v0, specifying the value of the solution and its first order derivative for the second order equation.
Similarly, given a third order equation d3u dt3 = F t, u, du dt , d2u dt2 , 1/7/22 10 c ⃝2022 Peter J. Olver we set v = du dt , w = dv dt = d2u dt2 .
The variables u, v, w satisfy the equivalent first order system du dt = v, dv dt = w, dw dt = F(t, u, v, w).
The general technique should now be clear.
Example 2.5. The forced van der Pol equation d2u dt2 + (u2 −1) du dt + u = f(t) (2.29) arises in the modeling of an electrical circuit with a triode whose resistance changes with the current.
It also arises in certain chemical reactions and wind-induced motions of structures. To convert the van der Pol equation into an equivalent first order system, we set v = du/dt, whence du dt = v, dv dt = f(t) −(u2 −1)v −u, (2.30) is the equivalent phase plane system.
Example 2.6. The Newtonian equations for a mass m moving in a potential force field are a second order system of the form m d2u dt2 = −∇F(u) in which u(t) = ( u(t), v(t), w(t) )T represents the position of the mass, while F(u) = F(u, v, w) is the potential function. In components, m d2u dt2 = −∂F ∂u , m d2v dt2 = −∂F ∂v , m d2w dt2 = −∂F ∂w .
(2.31) For example, a planet moving in the sun’s gravitational field satisfies the Newtonian system for the gravitational potential F(u) = − α ∥u ∥= − α √ u2 + v2 + w2 , (2.32) where α depends on the masses and the universal gravitational constant. (This simplified model ignores any additional interplanetary forces.) Thus, the mass’ motion in such a gravitational force field follows the solution to the second order Newtonian system m d2u dt2 = −∇F(u) = −αu ∥u ∥3 = α (u2 + v2 + w2)3/2 u v w .
1/7/22 11 c ⃝2022 Peter J. Olver The same system of ordinary differential equations describes the motion of a charged particle in a Coulomb electric force field, where the sign of α is positive for attracting opposite charges, and negative for repelling like charges.
To convert the second order Newton equations into a first order system, we set v = u to be the mass’ velocity vector, with components p = du dt , q = dv dt , r = dw dt , and so du dt = p, dv dt = q, dw dt = r, (2.33) dp dt = −1 m ∂F ∂u (u, v, w), dq dt = −1 m ∂F ∂v (u, v, w), dr dt = −1 m ∂F ∂w (u, v, w).
One of Newton’s greatest acheivements was to solve this system in the case of the central gravitational potential (2.32), and thereby confirm the validity of Kepler’s laws of planetary motion.
Finally, we note that there is a simple device that will convert any non-autonomous system into an equivalent autonomous system involving one additional variable. Namely, one introduces an extra coordinate u0 = t to represent the time, which satisfies the el-ementary differential equation du0/dt = 1 with initial condition u0(t0) = t0. Thus, the original system (2.19) can be written in the autonomous form du0 dt = 1, du1 dt = F1(u0, u1, . . ., un), · · · dun dt = Fn(u0, u1, . . ., un).
(2.34) For example, the autonomous form of the forced van der Pol system (2.30) is du0 dt = 1, du1 dt = u2, du2 dt = f(u0) −(u2 1 −1)u2 −u1, (2.35) in which u0 represents the time variable.
3.
Existence, Uniqueness, and Continuous Dependence.
It goes without saying that there is no general analytical method that will solve all differential equations. Indeed, even relatively simple first order, scalar, non-autonomous ordinary differential equations cannot be solved in closed form. For example, the solution to the particular Riccati equation du dt = u2 + t (3.1) cannot be written in terms of elementary functions, although it can be solved in terms of Airy functions, . The Abel equation du dt = u3 + t (3.2) 1/7/22 12 c ⃝2022 Peter J. Olver fares even worse, since its general solution cannot be written in terms of even standard special functions — although power series solutions can be tediously ground out term by term.
Understanding when a given differential equation can be solved in terms of elementary functions or known special functions is an active area of contemporary research, .
In this vein, we cannot resist mentioning that the most important class of exact solution techniques for differential equations are those based on symmetry. An introduction can be found in the author’s graduate level monograph ; see also [5, 16].
Existence Before worrying about how to solve a differential equation, either analytically, qual-itatively, or numerically, it behooves us to try to resolve the core mathematical issues of existence and uniqueness. First, does a solution exist? If, not, it makes no sense trying to find one. Second, is the solution uniquely determined? Otherwise, the differential equation probably has scant relevance for physical applications since we cannot use it as a predictive tool. Since differential equations inevitably have lots of solutions, the only way in which we can deduce uniqueness is by imposing suitable initial (or boundary) conditions.
Unlike partial differential equations, which must be treated on a case-by-case basis, there are complete general answers to both the existence and uniqueness questions for initial value problems for systems of ordinary differential equations.
(Boundary value problems are more subtle.) While obviously important, we will not take the time to present the proofs of these fundamental results, which can be found in most advanced textbooks on the subject, including [2, 13, 15, 17].
Let us begin by stating the Fundamental Existence Theorem for initial value problems associated with first order systems of ordinary differential equations.
Theorem 3.1. Let F(t, u) be a continuous function. Then the initial value problem† du dt = F(t, u), u(t0) = a, (3.3) admits a solution u = f(t) that is, at least, defined for nearby times, i.e., when | t −t0 | < δ for some δ > 0.
Theorem 3.1 guarantees that the solution to the initial value problem exists — at least for times sufficiently close to the initial instant t0. This may be the most that can be said, although in many cases the maximal interval α < t < β of existence of the solution might be much larger — possibly infinite, −∞< t < ∞, resulting in a global solution.
The interval of existence of a solution typically depends upon both the equation and the particular initial data. For instance, even though its right hand side is defined everywhere, the solutions to the scalar initial value problem (2.8) only exist up until time 1/u0, and so, the larger the initial data, the shorter the time of existence. In this example, the only global solution is the equilibrium solution u(t) ≡0. It is worth noting that this short-term † If F(t, u) is only defined on a subdomain Ω⊂Rn+1, then we must assume that the point (t0, a) ∈Ωspecifying the initial conditions belongs to its domain of definition.
1/7/22 13 c ⃝2022 Peter J. Olver existence phenomenon does not appear in the linear regime, where, barring singularities in the equation itself, solutions to a linear ordinary differential equation are guaranteed to exist for all time.
In practice, one always extends a solutions to its maximal interval of existence. The Existence Theorem 3.1 implies that there are only two possible ways in which a solution cannot be extended beyond a time t⋆: Either (i) the solution becomes unbounded: ∥u(t) ∥→∞as t →t⋆, or (ii) if the right hand side F(t, u) is only defined on a subset Ω⊂Rn+1, then the solution u(t) reaches the boundary ∂Ωas t →t⋆.
If neither occurs in finite time, then the solution is necessarily global. In other words, a solution to an ordinary differential equation cannot suddenly vanish into thin air.
Remark: The existence theorem can be readily adapted to any higher order system of ordinary differential equations through the method of converting it into an equivalent first order system by introducing additional variables. The appropriate initial conditions guaranteeing existence are induced from those of the corresponding first order system, as in the second order example (2.27) discussed above.
Uniqueness and Smoothness As important as existence is the question of uniqueness. Does the initial value problem have more than one solution? If so, then we cannot use the differential equation to predict the future behavior of the system from its current state. While continuity of the right hand side of the differential equation will guarantee that a solution exists, it is not quite sufficient to ensure uniqueness of the solution to the initial value problem. The difficulty can be appreciated by looking at an elementary example.
Example 3.2. Consider the nonlinear initial value problem du dt = 5 3 u2/5, u(0) = 0.
(3.4) Since the right hand side is a continuous function, Theorem 3.1 assures us of the existence of a solution — at least for t close to 0. This autonomous scalar equation can be easily solved by the usual method: Z 3 5 du u2/5 = u3/5 = t + c, and so u = (t + c)5/3.
Substituting into the initial condition implies that c = 0, and hence u(t) = t5/3 is a solution to the initial value problem.
On the other hand, since the right hand side of the differential equation vanishes at u = 0, the constant function u(t) ≡0 is an equilibrium solution to the differential equation.
(Here is an example where the integration method fails to recover the equilibrium solution.) Moreover, the equilibrium solution has the same initial value u(0) = 0. Therefore, we have constructed two different solutions to the initial value problem (3.4). Uniqueness is not 1/7/22 14 c ⃝2022 Peter J. Olver 0.5 1 1.5 2 0.5 1 1.5 2 Figure 4.
Solutions to the Differential Equation u = 5 3 u2/5.
valid! Worse yet, there are, in fact, an infinite number of solutions to the initial value problem. For any a > 0, the function u(t) = 0, 0 ≤t ≤a, (t −a)5/3, t ≥a, (3.5) is differentiable everywhere, even at t = a. (Why?) Moreover, it satisfies both the differ-ential equation and the initial condition, and hence defines a solution to the initial value problem. Several of these solutions are plotted in Figure 4.
Thus, to ensure uniqueness of solutions, we need to impose a more stringent condition, beyond mere continuity. The proof of the following basic uniqueness theorem can be found in the above references.
Theorem 3.3.
If F(t, u) ∈C1 is continuously differentiable, then there exists one and only one solution† to the initial value problem (3.3).
Thus, the difficulty with the differential equation (3.4) is that the function F(u) = 5 3 u2/5, although continuous everywhere, is not differentiable at u = 0, and hence the Uniqueness Theorem 3.3 does not apply. On the other hand, F(u) is continuously differ-entiable away from u = 0, and so any nonzero initial condition u(t0) = u0 ̸= 0 will produce a unique solution — for as long as it remains away from the problematic value u = 0.
Blanket Hypothesis: From now on, all differential equations must satisfy the unique-ness criterion that their right hand side is continuously differentiable.
While continuous differentiability is sufficient to guarantee uniqueness of solutions, the smoother the right hand side of the system, the smoother the solutions. Specifically: Theorem 3.4. If F ∈Cn for n ≥1, then any solution to the system u = F(t, u) is of class u ∈Cn+1. If F(t, u) is an analytic function, then all solutions u(t) are analytic.
† As noted earlier, we extend all solutions to their maximal interval of existence.
1/7/22 15 c ⃝2022 Peter J. Olver The basic outline of the proof of the first result is clear: Continuity of u(t) (which is a basic prerequisite of any solution) implies continuity of F(t, u(t)), which means u is continuous and hence u ∈C1. This in turn implies F(t, u(t)) = u is a continuously differentiable of t, and so u ∈C2. And so on, up to order n. The proof of analyticity follows from a detailed analysis of the power series solutions, . Indeed, the analytic result underlies the method of power series solutions of ordinary differential equations, [2, 13].
Uniqueness has a number of particularly important consequences for the solutions to autonomous systems, i.e., those whose right hand side does not explicitly depend upon t. Throughout the remainder of this section, we will deal with an autonomous system of ordinary differential equations du dt = F(u), where F ∈C1, (3.6) whose right hand side is defined and continuously differentiable for all u in a domain Ω⊂Rn. As a consequence, each solution u(t) is, on its interval of existence, uniquely determined by its initial data. Autonomy of the differential equation is an essential hy-pothesis for the validity of the following properties.
The first result tells us that the solution trajectories of an autonomous system do not vary over time.
Proposition 3.5. If u(t) is the solution to the autonomous system (3.6) with initial condition u(t0) = u0, then the solution to the initial value problem e u(t1) = u0 is e u(t) = u(t −t1 + t0).
Proof : Let e u(t) = u(t −t1 + t0), where u(t) is the original solution. In view of the chain rule and the fact that t1 and t0 are fixed, d dt e u(t) = du dt (t −t1 + t0) = F(u(t −t1 + t0)) = F(e u(t)), and hence e u(t) is also a solution to the system (3.6). Moreover, e u(t1) = u(t0) = u0 has the indicated initial conditions, and hence, by uniqueness, must be the one and only solution to the latter initial value problem.
Q.E.D.
Note that the two solutions u(t) and e u(t) parametrize the same curve in Rn, differing only by an overall “phase shift”, t1 −t0, in their parametrizations. Thus, all solutions passing through the point u0 follow the same trajectory, irrespective of the time they arrive there. Indeed, not only is the trajectory the same, but the solutions have identical speeds at each point along the trajectory curve. For instance, if the right hand side of (3.6) represents the velocity vector field of steady state fluid flow, Proposition 3.5 implies that the stream lines — the paths followed by the individual fluid particles — do not change in time, even though the fluid itself is in motion. This, indeed, is the meaning of the term “steady state” in fluid mechanics.
1/7/22 16 c ⃝2022 Peter J. Olver One particularly important consequence of uniqueness is that a solution u(t) to an autonomous system is either stuck at an equilibrium for all time, or is always in motion.
In other words, either u ≡0, in the case of equilibrium, or, otherwise, u ̸= 0 wherever defined.
Proposition 3.6.
Let u⋆be an equilibrium for the autonomous system (3.6), so F(u⋆) = 0. If u(t) is any solution such that u(t⋆) = u⋆at some time t⋆, then u(t) ≡u⋆is the equilibrium solution.
Proof : We regard u(t⋆) = u⋆as initial data for the given solution u(t) at the initial time t⋆. Since F(u⋆) = 0, the constant function u⋆(t) ≡u⋆is a solution of the differential equation that satisfies the same initial conditions. Therefore, by uniqueness, it coincides with the solution in question.
Q.E.D.
In other words, it is mathematically impossible for a solution to reach an equilibrium position in a finite amount of time — although it may well approach equilibrium in an asymptotic fashion as t →∞; see Proposition 3.9 below for details. Physically, this obser-vation has the interesting and physically counterintuitive consequence that a mathematical system never actually attains an equilibrium position! Even at very large times, there is always some very slight residual motion. In practice, though, once the solution gets suffi-ciently close to equilibrium, we are unable to detect the motion, and the physical system has, in all but name, reached its stationary equilibrium configuration. And, of course, the inherent motion of the atoms and molecules not included in such a simplified model would hide any infinitesimal residual effects of the mathematical solution. Without uniqueness, the result is false. For example, the function u(t) = (t −t⋆)5/3 is a solution to the scalar ordinary differential equation (3.4) that reaches the equilibrium point u⋆= 0 in a finite time t = t⋆.
Continuous Dependence In a real-world applications, initial conditions are almost never known exactly. Rather, experimental and physical errors will only allow us to say that their values are approxi-mately equal to those in our mathematical model. Thus, to retain physical relevance, we need to be sure that small errors in our initial measurements do not induce a large change in the solution. A similar argument can be made for any physical parameters, e.g., masses, charges, spring stiffnesses, frictional coefficients, etc., that appear in the differential equa-tion itself. A slight change in the parameters should not have a dramatic effect on the solution.
Mathematically, what we are after is a criterion of continuous dependence of solutions upon both initial data and parameters. Fortunately, the desired result holds without any additional assumptions, beyond requiring that the parameters appear continuously in the differential equation. We state both results in a single theorem.
Theorem 3.7. Consider an initial value problem problem du dt = F(t, u, µ), u(t0) = a(µ), (3.7) 1/7/22 17 c ⃝2022 Peter J. Olver in which the differential equation and/or the initial conditions depend continuously on one or more parameters µ = (µ1, . . . , µk).
Then the unique† solution u(t, µ) depends continuously upon the parameters.
Example 3.8. Let us look at a perturbed version du dt = α u2, u(0) = u0 + ε, of the initial value problem that we considered in Example 2.1. We regard ε as a small perturbation of our original initial data u0, and α as a variable parameter in the equation.
The solution is u(t, ε) = u0 + ε 1 −α(u0 + ε)t .
(3.8) Note that, where defined, this is a continuous function of both parameters α, ε. Thus, a small change in the initial data, or in the equation, produces a small change in the solution — at least for times near the initial time.
Continuous dependence does not preclude nearby solutions from eventually becoming far apart. Indeed, the blow-up time t⋆= 1/ α(u0 + ε) for the solution (3.8) depends upon both the initial data and the parameter in the equation. Thus, as we approach the singularity, solutions that started out very close to each other will get arbitrarily far apart; see Figure 1 for an illustration.
An even simpler example is the linear model of exponential growth u = αu when α > 0. A very tiny change in the initial conditions has a negligible short term effect upon the solution, but over longer time intervals, the differences between the two solutions will be dramatic. Thus, the “sensitive dependence” of solutions on initial conditions already appears in very simple linear equations. For similar reasons, continuous dependence does not prevent solutions from exhibiting chaotic behavior. Further development of these ideas can be found in [1, 8] and elsewhere.
As an application, let us show that if a solution to an autonomous system converges to a single limit point, then that point is necessarily an equilibrium solution. Keep in mind that, owing to uniqueness of solutions, the limiting equilibrium cannot be mathematically achieved in finite time, but only as a limit as time goes to infinity.
Proposition 3.9. Let u(t) be a solution to the autonomous system u = F(u), with F ∈C1, such that lim t →∞u(t) = u⋆. Then u⋆is an equilibrium solution, and so F(u⋆) = 0.
Proof : Let v(t, a) denote the solution to the initial value problem v = F(v), v(0) = a.
(We use a different letter to avoid confusion with the given solution u(t).) Theorem 3.7 implies that v(t, a) is a continuous function of the initial position a. and hence v(t, u(s)) is a continuous function of s ∈R. Since lim s →∞u(s) = u⋆, we have lim s →∞v(t, u(s)) = v(t, u⋆).
† We continue to impose our blanket uniqueness hypothesis.
1/7/22 18 c ⃝2022 Peter J. Olver On the other hand, since the system is autonomous, Proposition 3.5 implies that v(t, u(s)) = u(t + s), and hence lim s →∞v(t, u(s)) = lim s →∞u(t + s) = u⋆.
Equating the preceding two limit equations, we conclude that v(t, u⋆) = u⋆for all t, and hence the solution with initial value v(0) = u⋆is an equilibrium solution.
Q.E.D.
The same conclusion holds if we run time backwards: if lim t →−∞u(t) = u⋆, then u⋆is also an equilibrium point. When they exist, solutions that start and end at equilibrium points play a particularly role in the dynamics, and are known as heteroclinic, or, if the start and end equilibria are the same, homoclinic orbits. Of course, limiting equilibrium points are but one of the possible long term behaviors of solutions to nonlinear ordinary differential equations, which can also become unbounded in finite or infinite time, or approach periodic orbits, known as limit cycles, or become completely chaotic, depending upon the nature of the system and the initial conditions. Resolving the long term behavior os solutions is one of the many challenges awaiting the detailed analysis of any nonlinear ordinary differential equation.
4.
Stability.
Once a solution to a system of ordinary differential equations has settled down, its limiting value is an equilibrium solution; this is the content of Proposition 3.9. However, not all equilibria appear in this fashion. The only steady state solutions that one directly observes in a physical system are the stable equilibria. Unstable equilibria are hard to sustain, and will disappear when subjected to even the tiniest perturbation, e.g., a breath of air, or outside traffic jarring the experimental apparatus. Thus, finding the equilibrium solutions to a system of ordinary differential equations is only half the battle; one must then understand their stability properties in order to characterize those that can be realized in normal physical circumstances.
We will focus our attention on autonomous systems u = F(u) whose right hand sides are at least continuously differentiable, so as to ensure the unique-ness of solutions to the initial value problem. If every solution that starts out near a given equilibrium solution tends to it, the equilibrium is called asymptotically stable.
If the solutions that start out nearby stay nearby, then the equilibrium is stable. More formally: Definition 4.1. An equilibrium solution u⋆to an autonomous system of first order ordinary differential equations is called • stable if for every (small) ε > 0, there exists a δ > 0 such that every solution u(t) having initial conditions within distance δ > ∥u(t0) −u⋆∥of the equilibrium remains within distance ε > ∥u(t) −u⋆∥for all t ≥t0.
• asymptotically stable if it is stable and, in addition, there exists δ0 > 0 such that whenever δ0 > ∥u(t0) −u⋆∥, then u(t) →u⋆as t →∞.
1/7/22 19 c ⃝2022 Peter J. Olver δ ε u⋆ u(t0) Stability δ0 u⋆ u(t0) Asymptotic Stability Figure 5.
Stability of Equilibria.
Thus, although solutions nearby a stable equilibrium may drift slightly farther away, they must remain relatively close. In the case of asymptotic stability, they will eventually return to equilibrium. This is illustrated in Figure 5 Example 4.2. As we saw, the logistic differential equation du dt = λu(1 −u) has two equilibrium solutions, corresponding to the two roots of the quadratic equation λ u(1 −u) = 0. The solution graphs in Figure 1 illustrate the behavior of the solutions.
Observe that the first equilibrium solution u⋆ 1 = 0 is unstable, since all nearby solutions go away from it at an exponentially fast rate. On the other hand, the other equilibrium solution u⋆ 2 = 1 is asymptotically stable, since any solution with initial condition 0 < u0 tends to it, again at an exponentially fast rate.
Example 4.3. Consider an autonomous (meaning constant coefficient) homogeneous linear planar system du dt = au + bv, dv dt = cu + dv, with coefficient matrix A = a b c d . The origin u⋆= v⋆= 0 is an evident equilibrium, solution, and, moreover, is the only equilibrium provided A is nonsingular. According to the results in [27; Section 9.3], the stability of the origin depends upon the eigenvalues of A: It is (globally) asymptotically stable if and only if both eigenvalues are real and negative, and is stable, but not asymptotically stable if and only if both eigenvalues are purely imaginary, or if 0 is a double eigenvalue and so A = O. In all other cases, the origin is an unstable equilibrium. Later, we will see how this simple linear analysis has a direct bearing on the stability question for nonlinear planar systems.
1/7/22 20 c ⃝2022 Peter J. Olver Stability of Scalar Differential Equations Before looking at any further examples, we need to develop some basic mathematical tools for investigating the stability of equilibria. We begin at the beginning. The stability analysis for first order scalar ordinary differential equations du dt = F(u) (4.1) is particularly easy. The first observation is that all non-equilibrium solutions u(t) are strictly monotone functions, meaning they are either always increasing or always decreas-ing. Indeed, when F(u) > 0, then (4.1) implies that the derivative u > 0, and hence u(t) is increasing at such a point. Vice versa, solutions are decreasing at any point where F(u) < 0. Since F(u(t)) depends continuously on t, any non-monotone solution would have to pass through an equilibrium value where F(u⋆) = 0, in violation of Proposition 3.6. This proves the claim.
As a consequence of monotonicity, there are only three possible behaviors for a non-equilibrium solution: (a) it becomes unbounded at some finite time: | u(t) | →∞as t →t⋆; or (b) it exists for all t ≥t0, but becomes unbounded as t →∞; or (c) it exists for all t ≥t0 and has a limiting value, u(t) →u⋆as t →∞, which, by Proposition 3.9 must be an equilibrium point.
Let us look more carefully at the last eventuality.
Suppose u⋆is an equilibrium point, so F(u⋆) = 0. Suppose that F(u) > 0 for all u lying slightly below u⋆, i.e., on an interval of the form u⋆−δ < u < u⋆. Any solution u(t) that starts out on this interval, u⋆−δ < u(t0) < u⋆must be increasing. Moreover, u(t) < u⋆for all t since, according to Proposition 3.6, the solution cannot pass through the equilibrium point. Therefore, u(t) is a solution of type (c). It must have limiting value u⋆, since by assumption, this is the only equilibrium solution it can increase to. Therefore, in this situation, the equilibrium point u⋆is asymptotically stable from below: solutions that start out slightly below return to it in the limit. On the other hand, if F(u) < 0 for all u slightly below u⋆, then any solution that starts out in this regime will be monotonically decreasing, and so will move downwards, away from the equilibrium point, which is thus unstable from below.
By the same reasoning, if F(u) < 0 for u slightly above u⋆, then solutions starting out there will be monotonically decreasing, bounded from below by u⋆, and hence have no choice but to tend to u⋆in the limit.
Under this condition, the equilibrium point is asymptotically stable from above.
The reverse inequality, F(u) > 0, corresponds to solutions that increase away from u⋆, which is hence unstable from above.
Combining the two stable cases produces the basic asymptotic stability criterion for scalar ordinary differential equations.
Theorem 4.4. A equilibrium point u⋆of an autonomous scalar differential equation is asymptotically stable if and only if F(u) > 0 for u⋆−δ < u < u⋆and F(u) < 0 for u⋆< u < u⋆+ δ, for some δ > 0.
1/7/22 21 c ⃝2022 Peter J. Olver u F(u) u⋆ u F(u) u⋆ u t u⋆ Stable Equilibrium u t u⋆ Unstable Equilibrium Figure 6.
Equilibria of Scalar Ordinary Differential Equations.
In other words, if F(u) switches sign from positive to negative as u increases through the equilibrium point, then the equilibrium is asymptotically stable. If the inequalities are reversed, and F(u) goes from negative to positive, then the equilibrium point is unstable.
The two cases are illustrated in Figure 6. An equilibrium point where F(u) is of one sign on both sides, e.g., the point u⋆= 0 for F(u) = u2, is stable from one side, and unstable from the other.
Example 4.5. Consider the differential equation du dt = u −u3.
(4.2) Solving the algebraic equation F(u) = u −u3 = 0, we find that the equation has three equilibria: u⋆ 1 = −1, u⋆ 2 = 0, u⋆ 3 = +1, As u increases, the graph of the function F(u) = u −u3 switches from positive to negative at the first equilibrium point u⋆ 1 = −1, which proves its stability. Similarly, the graph goes back from negative to positive at u⋆ 2 = 0, 1/7/22 22 c ⃝2022 Peter J. Olver -1 -0.5 0.5 1 -0.75 -0.5 -0.25 0.25 0.5 0.75 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0.5 1 1.5 2 Figure 7.
Stability of u = u −u3.
establishing the instability of the second equilibrium. The final equilibrium u⋆ 3 = +1 is stable because F(u) again changes from positive to negative there.
With this information in hand, we are able to completely characterize the behavior of all solutions to the system. Any solution with negative initial condition, u0 < 0, will end up, asymptotically, at the first equilibrium, u(t) →−1 as t →∞. Indeed, if u0 < −1, then u(t) is monotonically increasing to −1, while if −1 < u0 < 0, the solution is decreasing towards −1. On the other hand, if u0 > 0, the corresponding solution ends up at the other stable equilibrium, u(t) →+1; those with 0 < u0 < 1 are monotonically increasing, while those with u0 > 1 are decreasing. The only solution that does not end up at either −1 or +1 as t →∞is the unstable equilibrium solution u(t) ≡0.
Any perturbation of it, no matter how tiny, will force the solutions to choose one of the stable equilibria.
Representative solutions are plotted in Figure 7. Note that all the curves, with the sole exception of the horizontal axis, converge to one of the stable solutions ±1, and diverge from the unstable solution 0 as t →∞.
Thus, the sign of the function F(u) nearby an equilibrium determines its stability.
In most instances, this can be checked by looking at the derivative of the function at the equilibrium. If F ′(u⋆) < 0, then we are in the stable situation, where F(u) goes from positive to negative with increasing u, whereas if F ′(u⋆) > 0, then the equilibrium u⋆ unstable on both sides.
Theorem 4.6. Let u⋆be a equilibrium point for a scalar ordinary differential equa-tion u = F(u). If F ′(u⋆) < 0, then u⋆is asymptotically stable. If F ′(u⋆) > 0, then u⋆is unstable.
For instance, in the preceding example, F ′(u) = 1 −3u2, and its value at the equilibria are F ′(−1) = −2 < 0, F ′(0) = 1 > 0, F ′(1) = −2 < 0.
1/7/22 23 c ⃝2022 Peter J. Olver The signs reconfirm our conclusion that ±1 are stable equilibria, while 0 is unstable.
In the borderline case when F ′(u⋆) = 0, the derivative test is inconclusive, and further analysis is needed to resolve the status of the equilibrium point. For example, the equations u = u3 and u = −u3 both satisfy F ′(0) = 0 at the equilibrium point u⋆= 0. But, according to the criterion of Theorem 4.4, the former has an unstable equilibrium, while the latter’s is stable. Thus, Theorem 4.6 is not as powerful as the direct algebraic test in Theorem 4.4.
But it does have the advantage of being a bit easier to use. More significantly, unlike the algebraic test, it can be directly generalized to systems of ordinary differential equations.
Linearization and Stability In higher dimensional situations, we can no longer rely on simple monotonicity prop-erties, and a more sophisticated approach to stability issues is required. The key idea is already contained in the second characterization of stable equilibria in Theorem 4.6. The derivative F ′(u⋆) determines the slope of the tangent line, which is a linear approximation to the function F(u) near the equilibrium point.
In a similar fashion, a vector-valued function F(u) is replaced by its linear approximation near an equilibrium point. The basic stability criteria for the resulting linearized differential equation were established in [27; Section 9.2]. and, in most situations, the linearized stability or instability carries over to the nonlinear regime.
Let us first revisit the scalar case du dt = F(u) (4.3) from this point of view. Linearization of a scalar function at a point means to replace it by its tangent line approximation F(u) ≈F(u⋆) + F ′(u⋆)(u −u⋆) (4.4) If u⋆is an equilibrium point, then F(u⋆) = 0, and so the first term disappears. Therefore, we anticipate that, near the equilibrium point, the solutions to the nonlinear ordinary differential equation (4.3) will be well approximated by its linearization du dt = F ′(u⋆)(u −u⋆).
Let us rewrite the linearized equation in terms of the deviation v(t) = u(t) −u⋆of the solution from equilibrium. Since u⋆is fixed, dv/dt = du/dt, and so the linearized equation takes the elementary form dv dt = a v, where a = F ′(u⋆) (4.5) is the value of the derivative at the equilibrium point. Note that the original equilibrium point u⋆corresponds to the zero equilibrium point v⋆= 0 of the linearized equation (4.5).
We already know that the linear differential equation (4.5) has an asymptotically stable equilibrium at v⋆= 0 if and only if a = F ′(u⋆) < 0, while for a = F ′(u⋆) > 0 the origin is unstable. In this manner, the linearized stability criterion reproduces that established in Theorem 4.6.
1/7/22 24 c ⃝2022 Peter J. Olver The same linearization technique can be applied to analyze the stability of an equi-librium solution u⋆to a first order autonomous system u = F(u).
(4.6) We approximate the function F(u) near an equilibrium point, where F(u⋆) = 0, by its first order Taylor polynomial: F(u) ≈F(u⋆) + F′(u⋆)(u −u⋆) = F′(u⋆)(u −u⋆).
(4.7) Here, F′(u⋆) denotes its n × n Jacobian matrix at the equilibrium point. Thus, for nearby solutions, we expect that the deviation from equilibrium, v(t) = u(t)−u⋆, will be governed by the linearized system dv dt = Av, where A = F′(u⋆).
(4.8) Now, we already know the complete stability criteria for linear systems, [27; Section 9.2]. The zero equilibrium solution to (4.8) is asymptotically stable if and only if all the eigenvalues of the coefficient matrix A = F′(u⋆) have negative real part. In contrast, if one or more of the eigenvalues has positive real part, then the zero solution is unstable.
Indeed, it can be proved, [13, 15], that these linearized stability criteria are also valid in the nonlinear case.
Theorem 4.7. Let u⋆be an equilibrium point for the first order ordinary differential equation u = F(u). If all of the eigenvalues of the Jacobian matrix F′(u⋆) have negative real part, Reλ < 0, then u⋆is asymptotically stable. If, on the other hand, F′(u⋆) has one or more eigenvalues with positive real part, Re λ > 0, then u⋆is an unstable equilibrium.
Intuitively, the additional nonlinear terms in the full system should only slightly per-turb the eigenvalues, and hence, at least for those with nonzero real part, not alter their effect on the stability of solutions. The borderline case occurs when one or more of the eigenvalues of F′(u⋆) is either 0 or purely imaginary, i.e., Reλ = 0, while all other eigenval-ues have negative real part. In such situations, the linearized stability test is inconclusive, and we need more detailed information (which may not be easy to come by) to resolve the status of the equilibrium.
Example 4.8. The second order ordinary differential equation m d2θ dt2 + µ dθ dt + κ sin θ = 0 (4.9) describes the damped oscillations of a rigid pendulum that rotates on a pivot subject to a uniform gravitational force in the vertical direction. The unknown function θ(t) measures the angle of the pendulum from the vertical, as illustrated in Figure 8. The constant m > 0 is the mass of the pendulum bob, µ > 0 is the coefficient of friction, assumed here to be strictly positive, and κ > 0 represents the gravitational force.
1/7/22 25 c ⃝2022 Peter J. Olver θ Figure 8.
The Pendulum.
In order to study the equilibrium solutions and their stability, we must first convert the equation into a first order system. Setting u(t) = θ(t), v(t) = dθ dt , we find du dt = v, dv dt = −α sin u −β v, where α = κ m , β = µ m , (4.10) are both positive constants. The equilibria occur where the right hand sides of the first order system (4.10) simultaneously vanish, that is, v = 0, −α sin u −β v = 0, and hence u = 0, ±π, ±2π, . . . .
Thus, the system has infinitely many equilibrium points: u⋆ k = (kπ, 0) where k = 0, ±1, ±2, . . .
is any integer.
(4.11) The equilibrium point u⋆ 0 = (0, 0) corresponds to u = θ = 0, v = θ = 0, which means that the pendulum is at rest at the bottom of its arc. Our physical intuition leads us to expect this to describe a stable configuration, as the frictional effects will eventually damp out small nearby motions. The next equilibrium u⋆ 1 = (π, 0) corresponds to u = θ = π, v = θ = 0, which means that the pendulum is sitting motionless at the top of its arc. This is a theoretically possible equilibrium configuration, but highly unlikely to be observed in practice, and is thus expected to be unstable. Now, since u = θ is an angular variable, equilibria whose u values differ by an integer multiple of 2π define the same physical configuration, and hence should have identical stability properties.
Therefore, all the remaining equilibria u⋆ k physically correspond to one or the other of these two possibilities: when k = 2j is even, the pendulum is at the bottom, while when k = 2j + 1 is odd, the pendulum is at the top.
Let us now confirm our intuition by applying the linearization stability criterion of Theorem 4.7. The right hand side of the system (4.10), namely F(u, v) = v −α sin u −β v , has Jacobian matrix F′(u, v) = 0 1 −α cos u −β .
1/7/22 26 c ⃝2022 Peter J. Olver Figure 9.
The Underdamped Pendulum.
At the bottom equilibrium u⋆ 0 = (0, 0), the Jacobian matrix F′(0, 0) = 0 1 −α −β has eigenvalues λ = −β ± p β2 −4 α 2 .
Under our assumption that α, β > 0, both eigenvalues have negative real part, and hence the origin is a stable equilibrium. If β2 < 4 α — the underdamped case — the eigenvalues are complex, and hence, in the terminology of [27; Section 9.3], the origin is a stable focus.
In the phase plane, the solutions spiral in to the focus, which corresponds to a pendulum with damped oscillations of decreasing magnitude. On the other hand, if β2 > 4 α, then the system is overdamped. Both eigenvalues are negative, and the origin is a stable node.
In this case, the solutions decay exponentially fast to 0. Physically, this would be like a pendulum moving in a vat of molasses. The exact same analysis applies at all even equilibria u⋆ 2j = (2j π, 0) — which really represent the same bottom equilibrium point.
On the other hand, at the top equilibrium u⋆ 1 = (π, 0), the Jacobian matrix F′(0, 0) = 0 1 α −β has eigenvalues λ = −β ± p β2 + 4 α 2 .
In this case, one of the eigenvalues is real and positive while the other is negative. The linearized system has an unstable saddle point, and hence the nonlinear system is also unstable at this equilibrium point. Any tiny perturbation of an upright pendulum will dislodge it, causing it to swing down, and eventually settle into a damped oscillatory motion converging on one of the stable bottom equilibria.
The complete phase portrait of an underdamped pendulum appears in Figure 9. Note that, as advertised, almost all solutions end up spiraling into the stable equilibria. Solutions with a large initial velocity will spin several times around the center, but eventually the cumulative effect of frictional forces wins out and the pendulum ends up in a damped oscillatory mode.
Each of the the unstable equilibria has the same saddle form as its linearizations, with two very special solutions, corresponding to the stable eigenline of the linearization, in which the pendulum spins around a few times, and, in the t →∞limit, ends up standing upright at the unstable equilibrium position. However, like unstable 1/7/22 27 c ⃝2022 Peter J. Olver Figure 10.
Phase Portrait of the van der Pol System.
equilibria, such solutions are practically impossible to achieve in a physical environment as any tiny perturbation will cause the pendulum to sightly deviate and then end up eventually decaying into the usual damped oscillatory motion at the bottom.
A deeper analysis demonstrates the local structural stability of any nonlinear equi-librium whose linearization is structurally stable, and hence has no eigenvalues on the imaginary axis: Re λ ̸= 0.
Structural stability means that, not only are the stability properties of the equilibrium dictated by the linearized approximation, but, nearby the equilibrium point, all solutions to the nonlinear system are slight perturbations of solu-tions to the corresponding linearized system, and so, close to the equilibrium point, the two phase portraits have the same qualitative features. Thus, stable foci of the linearization remain stable foci of the nonlinear system; unstable saddle points remain saddle points, although the eigenlines become slightly curved as they depart from the equilibrium. Thus, the structural stability of linear systems, as discussed at the end of [27; Section 9.3] also carries over to the nonlinear regime near an equilibrium. A more in depth discussion of these issues can be found, for instance, in [13, 15].
, Example 4.9. Consider the unforced van der Pol system du dt = v, dv dt = −(u2 −1)v −u.
(4.12) that we derived in Example 2.5. The only equilibrium point is at the origin u = v = 0.
Computing the Jacobian matrix of the right hand side, F′(u, v) = 0 1 2uv −1 1 , and hence F′(0, 0) = 0 1 −1 1 .
1/7/22 28 c ⃝2022 Peter J. Olver Figure 11.
Phase Portrait for u = u(v −1), v = 4 −u2 −v2..
The eigenvalues of F′(0, 0) are 1 2 ± i √ 3 2 , and correspond to an unstable focus of the lin-earized system near the equilibrium point. Therefore, the origin is an unstable equilibrium for nonlinear van der Pol system, and all non-equilibrium solutions starting out near 0 eventually spiral away.
On the other hand, it can be shown that solutions that are sufficiently far away from the origin spiral in towards the center. So what happens to the solutions? As illustrated in the phase plane portrait sketched in Figure 10, all non-equilibrium solutions spiral towards a stable periodic orbit, known as a limit cycle for the system. Any non-zero initial data will eventually end up closely following the limit cycle orbit as it periodically circles around the origin. A rigorous proof of the existence of a limit cycle relies on the more sophisticated Poincar´ e–Bendixson Theory for planar autonomous systems, discussed in detail in .
Example 4.10. The nonlinear system du dt = u(v −1), dv dt = 4 −u2 −v2, has four equilibria: (0, ±2) and (± √ 3 , 1). Its Jacobian matrix is F′(u, v) = v −1 u −2u −2v .
A table of the eigenvalues at the equilibrium points and their stability follows.
These results are reconfirmed by the phase portrait drawn in Figure 11.
1/7/22 29 c ⃝2022 Peter J. Olver Equilibrium Point Jacobian matrix Eigenvalues Stability (0, 2) 1 0 0 −4 1, −4 unstable saddle (0, −2) −3 0 0 6 −3, 6 unstable saddle ( √ 3 , 1) 0 − √ 3 2 √ 3 −2 −1 ± i √ 5 stable focus (− √ 3 , 1) 0 − √ 3 2 √ 3 −2 −1 ± i √ 5 stable focus Conservative Systems When modeling a physical system that includes some form of damping — due to fric-tion, viscosity, or dissipation — linearization will usually suffice to resolve the stability or instability of equilibria. However, when dealing with conservative systems, when damp-ing is absent and energy is preserved, the linearization test is often inconclusive, and one must rely on more sophisticated stability criteria. In such situations, one can often exploit conservation of energy, appealing to our general philosophy that minimizers of an energy function should be stable (but not necessarily asymptotically stable) equilibria.
By saying that energy is conserved, we mean that it remains constant as the solution evolves. Conserved quantities are also known as first integrals for the system of ordinary differential equations. Additional well-known examples include the laws of conservation of mass, and conservation of linear and angular momentum. Let us mathematically formulate the general definition.
Definition 4.11. A first integral of an autonomous system u = F(u) is a real-valued function I(u) which is constant on solutions.
In other words, for each solution u(t) to the differential equation, I(u(t)) = c for all t, (4.13) where c is a fixed constant, which will depend upon which solution is being monitored.
The value of c is fixed by the initial data since, in particular, c = I(u(t0)) = I(u0). Or, to rephrase this condition in another, equivalent manner, every solution to the dynamical system is constrained to move along a single level set {I(u) = c} of the first integral, namely the level set that contains the initial data u0.
Note first that any constant function, I(u) ≡c0, is trivially a first integral, but this provides no useful information whatsoever about the solutions, and so is uninteresting. We will call any autonomous system that possesses a nontrivial first integral I(u) a conservative system.
1/7/22 30 c ⃝2022 Peter J. Olver How do we find first integrals? In applications, one often appeals to the underlying physical principles such as conservation of energy, momentum, or mass. Mathematically, the most convenient way to check whether a function is constant is to verify that its derivative is identically zero. Thus, differentiating (4.13) with respect to t and invoking the chain rule leads to the basic condition 0 = d dt I(u(t)) = ∇I(u(t)) · du dt = ∇I(u(t)) · F(u(t)).
(4.14) The final expression can be identified as the directional derivative of I(u) with respect to the vector field v = F(u) that specifies the differential equation. Writing out (4.14) in detail, we find that a first integral I(u1, . . ., un) must satisfy a first order linear partial differential equation: F1(u1, . . ., un) ∂I ∂u1 + · · · + Fn(u1, . . ., un) ∂I ∂un = 0.
(4.15) As such, it looks harder to solve than the original ordinary differential equation! Often, one falls back on either physical intuition, intelligent guesswork, or, as a last resort, a lucky guess. A deeper fact, due to the pioneering twentieth century mathematician Emmy Noether, cf. [24, 26], is that first integrals and conservation laws are the result of underlying symmetry properties of the differential equation. Like many nonlinear methods, it remains the subject of contemporary research.
Let us specialize to planar autonomous systems du dt = F(u, v), dv dt = G(u, v).
(4.16) According to (4.15), any first integral I(u, v) must satisfy the linear partial differential equation F(u, v) ∂I ∂u + G(u, v) ∂I ∂v = 0.
(4.17) This nonlinear first order partial differential equation can be solved by the method of characteristics. Consider the auxiliary first order scalar ordinary differential equation† dv du = G(u, v) F(u, v) (4.18) for v = h(u). Note that (4.18) can be formally obtained by dividing the second equation in the original system (4.16) by the first, and then canceling the time differentials dt. Suppose we can write the general solution to the scalar equation (4.18) in the implicit form I(u, v) = c, (4.19) † We assume that F(u, v) ̸≡0. Otherwise, I(u) = u is itself a first integral, and the system reduces to a scalar equation for v.
1/7/22 31 c ⃝2022 Peter J. Olver where c is a constant of integration. We claim that the function I(u, v) is a first integral of the original system (4.16). Indeed, differentiating (4.19) with respect to u, and using the chain rule, we find 0 = d du I(u, v) = ∂I ∂u + dv du ∂I ∂v = ∂I ∂u + G(u, v) F(u, v) ∂I ∂v .
Clearing the denominator, we conclude that I(u, v) solves the partial differential equation (4.17), which justifies our claim.
Example 4.12. As an elementary example, consider the linear system du dt = −v, dv dt = u.
(4.20) To construct a first integral, we form the auxiliary equation (4.18), which is dv du = −u v .
This first order ordinary differential equation can be solved by separating variables: v dv = −u du, and hence 1 2 u2 + 1 2 v2 = c, where c is the constant of integration. Therefore, by the preceding result, I(u, v) = 1 2 u2 + 1 2 v2 is a first integral. The level sets of I(u, v) are the concentric circles centered at the origin, and we recover the fact that the solutions of (4.20) go around the circles. The origin is a stable equilibrium — a center.
This simple example hints at the importance of first integrals in stability theory. The following key result confirms our general philosophy that energy minimizers, or, more generally, minimizers of first integrals, are stable equilibria.
Theorem 4.13. Let I(u) be a first integral for the autonomous system u = F(u).
If u⋆is a strict local extremum — minimum or mximum — of I, then u⋆is a stable equilibrium point for the system.
Remark: At first sight, the fact that strict maxima are also stable equilibria appears to contradict our intuition. However, energy functions typically do not have local maxima.
Indeed, physical energy is the sum of kinetic and potential contributions. While potential energy can admit maxima, e.g., the pendulum at the top of its arc, these are only unstable saddle points for the full energy function, since the kinetic energy can always be increased by moving a bit faster.
Proof : We first prove that u⋆is an equilibrium. Indeed, the solution u(t) with initial condition u(t0) = u⋆must maintain the value of I(u(t)) = I(u⋆). But, by definition of a strict local minimum, I(u) > I(u⋆) for all u near u⋆, and hence, by continuity, the solution has no choice but to remain at the point u⋆.
1/7/22 32 c ⃝2022 Peter J. Olver To prove stability, set M(r) = max { I(u) | ∥u −u⋆∥≤r } , m(r) = min { I(u) | ∥u −u⋆∥= r } .
Thus, M(r) is the maximum value of the first integral over a ball† of radius r centered at the minimum, while m(r) is the minimum over its boundary sphere of radius r. Since I is continuous, so are m and M. Since u⋆is a strict local minimum, M(r) ≥m(r) > m(0) = M(0) = I(u⋆) for any 0 < r < ε sufficiently small.
For each ε > 0, we can choose a δ > 0 such that M(δ) < m(ε). Then, whenever ∥u(t0) −u⋆∥≤δ, then I(u(t)) = I(u(t0)) ≤M(δ) < m(ε). Since m(ε) is the minimum possible value for I(u) when ∥u(t) −u⋆∥= ε, the solution u(t) cannot cross the sphere of radius ε at any t, and so ∥u(t) −u⋆∥< ε for all t ≥t0. Hence, we have fulfilled the stability criteria of Definition 4.1.
Q.E.D.
Example 4.14. Consider the specific predator-prey system du dt = 2u −uv, dv dt = −9v + 3uv, (4.21) modeling populations of, say, lions and zebra, and a special case of (2.26). According to Example 2.4, there are two possible equilibria: u⋆ 1 = v⋆ 1 = 0, u⋆ 2 = 3, v⋆ 2 = 2.
Let us first try to determine their stability by linearization. The Jacobian matrix for the system is F′(u, v) = 2 −v −u 3v 3u −9 .
At the first, trivial equilibrium, F′(0, 0) = 2 0 0 −9 , with eigenvalues 2 and −9.
Since there is one positive and one negative eigenvalue, the origin is an unstable saddle point. On the other hand, at the nonzero equilibrium, the Jacobian matrix F′(3, 2) = 0 −3 6 0 , has purely imaginary eigenvalues ± 3 √ 2 i .
So the linearized system has a stable center. However, as purely imaginary eigenvalues is a borderline situation, Theorem 4.7 cannot be applied. Thus, the linearization stability test is inconclusive.
It turns out that the predator-prey model is a conservative system. To find a first integral, we need to solve the auxiliary equation (4.18), which is dv du = −9v + 3uv 2u −uv = −9/u + 3 2/v −1 .
† We write as if the norm is the Euclidean norm, but any other norm will work equally well for this proof.
1/7/22 33 c ⃝2022 Peter J. Olver 2 4 6 8 1 2 3 4 5 6 Figure 12.
Phase Portrait and Solution of the Predator-Prey System.
Fortunately, this is a separable first order ordinary differential equation. Integrating, 2 log v −v = Z 2 v −1 dv = Z −9 u + 3 du = −9 log u + 3u + c, where c is the constant of integration. Writing the solution in the form (4.19), we conclude that I(u, v) = 9 log u −3u + 2 log v −v = c, is a first integral of the system. The solutions to (4.21) must stay on the level sets of I(u, v). Note that ∇I(u, v) = 9/u −3 2/v −1 , and hence ∇I(3, 2) = 0, which shows that the second equilibrium is a critical point. (On the other hand, I(u, v) is not defined at the unstable zero equilibrium.) Moreover, the Hessian matrix at the critical point, ∇2I(3, 2) = −3 0 0 −1 , is negative definite, and hence u⋆ 2 = ( 3, 2 )T is a strict local maximum of the first integral I(u, v). Thus, Theorem 4.13 proves that the equilibrium point is a stable center.
The first integral serves to completely characterize the qualitative behavior of the sys-tem. In the physically relevant region, i.e., the upper right quadrant Q = {u > 0, v > 0} where both populations are positive, all of the level sets of the first integral are closed curves encircling the equilibrium point u⋆ 2 = ( 3, 2 )T . The solutions move in a counter-clockwise direction around the closed level curves, and hence all non-equilibrium solutions in the positive quadrant are periodic. The phase portrait is illustrated in Figure 12, along with a typical periodic solution. Thus, in such an idealized ecological model, for any initial conditions starting with some zebra and lions, i.e., where u(t0), v(t0) > 0, the populations will maintain a balance over the long term, each varying periodically between maximum and minimum values. Observe also that the maximum and minimum values of the two 1/7/22 34 c ⃝2022 Peter J. Olver populations are not achieved simultaneously. Starting with a small number of predators, the number of prey will initially increase. The predators then have more food available, and so also start to increase in numbers. At a certain critical point, the predators are sufficiently numerous as to kill prey faster than they can reproduce. At this point, the prey population has reached its maximum, and begins to decline. But it takes a while for the predator population to feel the effect, and so it continues to increase. Eventually the increasingly rapid decline in the number of prey begins to affect the predators. After the predators reach their maximum number, both populations are in decline. Eventually, enough predators have died offso as to relieve the pressure on the prey, whose population bottoms out, and then slowly begins to rebound. Later, the number of predators also reaches a minimum, at which point the entire growth and decay cycle starts over again.
In contrast to a linear system, the period of the population cycle is not fixed, but depends upon how far away from the stable equilibrium the solution orbit lies.
Near equilibrium, the solutions are close to those of the linearized system which, in view of its eigenvalues ±3 i √ 2, are periodic of frequency 3 √ 2 and period √ 2 π/3. However, solutions that are far away from equilibrium have much longer periods, and so greater imbalances between predator and prey populations leads to longer periods, and more radically varying numbers. Understanding the mechanisms behind these population cycles is of increasing important in the ecological management of natural resources.
Example 4.15.
In our next example, we look at the undamped oscillations of a pendulum. When we set the friction coefficient µ = 0, the nonlinear second order ordinary differential equation (4.9) reduces to m d2θ dt2 + κ sin θ = 0.
(4.22) As before, we convert this into a first order system du dt = v, dv dt = −α sin u, (4.23) where u(t) = θ(t), v(t) = dθ dt , α = κ m .
The equilibria, u⋆ k = (kπ, 0) for k = 0, ±1, ±2, . . . , are the same as in the damped case, i.e., the pendulum is either at the top (k even) or the bottom (k odd) of the circle.
Let us see what the linearization stability test tells us. In this case, the Jacobian matrix of (4.23) is F′(u, v) = 0 1 −α cos u 0 .
At the top equilibria F′(u⋆ 2j+1) = F′ (2j + 1)π, 0 = 0 1 α 0 has real eigenvalues ± √α , 1/7/22 35 c ⃝2022 Peter J. Olver Figure 13.
The Undamped Pendulum.
and hence these equilibria are unstable saddle points, just as in the damped version. On the other hand, at the bottom equilibria F′(u⋆ 2j) = F′(2j π, 0) = 0 1 −α 0 , has purely imaginary eigenvalues ± i √α .
Without the benefit of damping, the linearization test is inconclusive, and the stability of the bottom equilibria remains in doubt.
Since we are dealing with a conservative system, the total energy of the pendulum, namely E(u, v) = 1 2 mv2 + κ (1 −cos u) = m 2 dθ dt 2 + κ (1 −cos θ) (4.24) should provide us with a first integral. Note that E is a sum of two terms, which represent, respectively, the kinetic energy due to the pendulum’s motion, and the potential energy† due to the height of the pendulum bob. To verify that E(u, v) is indeed a first integral, we compute dE dt = ∂E ∂u du dt + ∂E ∂v dv dt = (κ sin u)v + (mv)(−α sin u) = 0, since α = κ m .
Therefore, E is indeed constant on solutions, reconfirming the physical basis of the model.
The phase plane solutions to the pendulum equation will move along the level sets of the energy function E(u, v), which are plotted in Figure 13. Its critical points are the equilibria, where ∇E(u) = κ sin u mv = 0, and hence u = u⋆ k = ( kπ, 0 ), k = 0, ±1, ±2, . . . .
To characterize the critical points, we appeal to the second derivative test, and so evaluate the Hessian ∇2E(u, v) = κ cos u 0 0 m .
† In a physical system, the potential energy is only defined up to an additive constant. Here we have fixed the zero energy level to be at the bottom of the pendulum’s arc.
1/7/22 36 c ⃝2022 Peter J. Olver At the bottom equilibria, ∇2E(u⋆ 2j) = ∇2E(2j π, 0) = κ 0 0 m is positive definite, since κ and m are positive constants. Therefore, the bottom equilibria are strict local minima of the energy, and so Theorem 4.13 guarantees their stability.
Each stable equilibrium is surrounded by a family of closed oval-shaped level curves, and hence forms a center. Each oval corresponds to a periodic solution‡ of the system, in which the pendulum oscillates back and forth symmetrically around the bottom of its arc.
Near the equilibrium, the period is close to that of the linearized system, namely 2π/√α as prescribed by the eigenvalues of the Jacobian matrix. This fact underlies the use of pendulum-based clocks for keeping time, first recognized by Galileo. A grandfather clock is accurate because the amplitude of its pendulum’s oscillations is kept relatively small.
However, moving further away from the equilibrium point in the phase plane, we find that the periodic solutions with very large amplitude oscillations, in which the pendulum becomes nearly vertical, have much longer periods, and so would lead to inaccurate time-keeping.
The large amplitude limit of the periodic solutions is of particular interest. The pair of trajectories connecting two distinct unstable equilibria are known as the homoclinic orbits. Physically, a homoclinic orbit corresponds to a pendulum that starts out just shy of vertical, goes through exactly one full rotation, and eventually (as t →∞) ends up vertical again. The homoclinic orbits play an essential role in the analysis of the chaotic behavior of a periodically forced pendulum, [1, 8, 23].
Finally, the level sets lying above and below the “cat’s-eyes” formed by the homoclinic and periodic orbits are known as the running orbits. Since u = θ is a 2π periodic angular variable, a running orbit solution ( u(t), v(t) )T = (θ(t), θ(t))T , in fact, also corresponds to a periodic physical motion, in which the pendulum spins around and around its pivot.
The larger the total energy E(u, v), the farther away from the u–axis the running orbit lies, and the faster the pendulum spins.
In summary, the qualitative behavior of a solution to the pendulum equation is almost entirely characterized by its energy: • E = 0, stable equilibria, • 0 < E < 2κ, periodic oscillating orbits, • E = 2κ, unstable equilibria and homoclinic orbits, • E > 2κ, running orbits.
Example 4.16. The system governing the dynamical rotations of a rigid solid body around its center of mass are known as the Euler equations of rigid body mechanics, in honor of the prolific eighteenth century Swiss mathematician Leonhard Euler, cf. .
‡ More precisely, a family of periodic solutions indexed by their initial condition on the oval, and differing only by a phase shift: u(t −δ).
1/7/22 37 c ⃝2022 Peter J. Olver Figure 14.
The Rigid Body Phase Portrait.
The eigenvectors of the positive definite inertia tensor of the body prescribe its three mutually orthogonal principal axes. The corresponding eigenvalues I1, I2, I3 > 0 are called the principal moments of inertia. Let u1(t), u2(t), u3(t) denote the angular momenta of the body around its three principal axes. In the absence of external forces, the dynamical system governing the body’s rotations around its center of mass takes the symmetric form du1 dt = I2 −I3 I2 I3 u2u3, du2 dt = I3 −I1 I1 I3 u1u3, du3 dt = I1 −I2 I1 I2 u1u2.
(4.25) Th Euler equations model, for example, the dynamics of a satellite spinning in its orbit above the earth. The solution will prescribe the angular motions of the satellite around its center of mass, but not the overall motion of the center of mass as the satellite orbits the earth.
Let us assume that the moments of inertia are all different, which we place in increasing order 0 < I1 < I2 < I3. The equilibria of the Euler system (4.25) are where the right hand sides simultaneously vanish, which requires that either u2 = u3 = 0, or u1 = u3 = 0, or u1 = u2 = 0. In other words, every point on the three coordinate axes is an equilibrium configuration. Since the variables represent angular momenta, these equilibria correspond to the body spinning around one of its principal axes at a fixed angular velocity.
Let us analyze the stability of these equilibrium configurations. The linearization test fails completely — as it must whenever dealing with a non-isolated equilibrium. But the Euler equations turn out to admit two independent first integrals: E(u) = 1 2 u2 1 I1 + u2 2 I2 + u2 3 I3 , A(u) = 1 2 u2 1 + u2 2 + u2 3 .
(4.26) 1/7/22 38 c ⃝2022 Peter J. Olver The first is the total kinetic energy, while the second is the total angular momentum. The proof that dE/dt = 0 = dA/dt for any solution u(t) to (4.25) is left as an exercise for the reader.
Since both E and A are constant, the solutions to the system are constrained to move along a common level set C = {E = e, A = a}. Thus, the solution trajectories are the curves obtained by intersecting the sphere Sa = {A(u) = a} of radius √ 2a with the ellipsoid Le = {E(u) = e}. In Figure 14, we have graphed the solution trajectories on a fixed sphere. (To see the figure, make sure the left hand periodic orbits are perceived on the back side of the sphere.) The six equilibria on the sphere are at its intersections with the coordinate axes. Those on the x and z axes are surrounded by closed periodic orbits, and hence are stable equilibria; indeed, they are, respectively, local minima and maxima of the energy when restricted to the sphere. On the other hand, the two equilibria on the y axis have the form of unstable saddle points, and are connected by four distinct homoclinic orbits. We conclude that a body that spins around either of its principal axes with the smallest or the largest moment of inertia is stable, whereas one that spins around the axis corresponding to the intermediate moment of inertia is unstable. This mathematical deduction can be demonstrated physically by flipping a solid rectangular object, e.g., this book, up into the air. It is easy to arrange it to spin around its long axis or its short axis in a stable manner, but it will balk at attempts to make it rotate around its middle axis!
Lyapunov’s Method Systems that incorporate damping, viscosity and/or frictional effects do not typically possess non-constant first integrals. From a physical standpoint, the damping will cause the total energy of the system to be a decreasing function of time. Asymptotically, the system returns to a (stable) equilibrium, and the extra energy has been dissipated away. However, this physical principle captures important mathematical implications for the behavior of solutions. It leads to a useful alternative method for establishing stability of equilibria, even in cases when the linearization stability test is inconclusive. The nineteenth century Russian mathematician Alexander Lyapunov was the first to pinpoint the importance of such functions in dynamics.
Definition 4.17. A Lyapunov function for the first order autonomous system u = F(u) is a continuous real-valued function L(u) that is non-increasing on all solutions u(t), meaning that L(u(t)) ≤L(u(t0)) for all t > t0.
(4.27) A strict Lyapunov function satisfies the strict inequality L(u(t)) < L(u(t0)) for all t > t0, (4.28) whenever u(t) is a non-equilibrium solution to the system. (Clearly, the Lyapunov function must be constant on an equilibrium solution.) We can characterize continuously differentiable Lyapunov functions by using the ele-mentary calculus results that a scalar function is non-increasing if and only if its derivative is non-negative, and is strictly decreasing if its derivative is strictly less than 0. We can 1/7/22 39 c ⃝2022 Peter J. Olver compute the derivative of L(u(t)) by applying the same chain rule computation used to establish (4.14). As a result, we establish the basic criteria for Lyapunov functions.
Proposition 4.18. A continuously differentiable function L(u) is a Lyapunov func-tion for the system u = F(u) if and only if it satisfies the Lyapunov inequality d dt L(u(t)) = ∇L(u) · F(u) ≤0 for all solutions u(t).
(4.29) Furthermore, L(u) is a strict Lyapunov function if and only if d dt L(u(t)) = ∇L(u) · F(u) < 0 whenever F(u) ̸= 0.
(4.30) The main result on stability and instability of equilibria of a system that possesses a Lyapunov function follows.
Theorem 4.19. Let L(u) be a Lyapunov function for the autonomous system u = F(u). If u⋆is a strict local minimum of L, then u⋆is a stable equilibrium point. If L(u) is a strict Lyapunov function, then u⋆is an asymptotically stable equilibrium. On the other hand, any critical point of a strict Lyapunov function L(u) which is not a local minimum is an unstable equilibrium point.
In particular, local maxima of strict Lyapunov functions are not stable equilibria.
In outline, the proof relies on the fact that the Lyapunov function is non-increasing on solutions, and hence any solution that starts out sufficiently near a minimum value can-not go far away, demonstrating stability. Figuratively, if you start near the bottom of a valley and never walk uphill you stay near the bottom. In the strict case, the Lyapunov function must be strictly decreasing on the non-equilibrium solutions, which thus must go to the minimum value of L as t →∞. Again, starting near the bottom of a valley and always walking downhill takes you eventually to the bottom. On the other hand, if a non-equilibrium solution starts near an equilibrium point that is not a local minimum, then the fact that the Lyapunov function is steadily decreasing implies that the solution must move further and further away from the equilibrium, which suffices to demonstrate instability.
Unlike first integrals, which can, at least in principle, be systematically constructed by solving a first order partial differential equation, finding Lyapunov functions is much more of an art form, usually requiring some combination of physical intuition and inspired guesswork.
Example 4.20. Return to the planar system du dt = v, dv dt = −α sin u −β v, describing the damped oscillations of a pendulum, as in (4.10). Physically, we expect that the damping will cause a continual decrease in the total energy in the system, which, by (4.24), is E(u, v) = 1 2 mv2 + κ(1 −cos u).
1/7/22 40 c ⃝2022 Peter J. Olver Let us prove that E is, indeed, a Lyapunov function. We compute its time derivative, when u(t), v(t) is a solution to the damped system. Recalling that α = κ/m, β = µ/m, we find dE dt = ∂E ∂u du dt + ∂E ∂v dv dt = (κ sin u)v + (mv) (−α sin u −β v) = −µ v2 ≤0, since we are assuming that the frictional coefficient µ > 0. Therefore, the energy satisfies the Lyapunov stability criterion (4.29). Consequently, Theorem 4.19 re-establishes the stability of the energy minima u⋆ 2k = 2kπ, v⋆ 2k = 0, where the damped pendulum is at the bottom of the arc. In fact, since dE/dt < 0 except when v = 0, with a little more work, the Lyapunov criterion can be used to establish their asymptotic stability.
Hamiltonian and Poisson Systems One particularly important class of conservative systems were first introduced in the work of William Rowan Hamilton and Sim´ eon–Dennis Poisson in the early nineteenth century. Their research was concerned with methods for solving the conservative systems arising in classical mechanics.
Remarkably, Hamiltonian systems turned out to be the mathematical vehicle that unlocked the modern world of subatomic quantum mechanics.
Definition 4.21.
A Poisson system is a first order system of ordinary differential equations of the form u = J ∇H(u), where JT = −J (4.31) is a constant† skew-symmetric matrix. The real-valued function H(u) is known as the Hamiltonian function for the system.
The Hamiltonian function is automatically a first integral for the Poisson system (4.31). Indeed, to verify (4.14), we compute dH dt = ∇H · du dt = ∇H · F = ∇HT F = ∇HT J ∇H = 0.
The final equality follows from the fact that J is skew-symmetric, and hence vT J v = 0 for any vector v. In applications, the Hamiltonian function often represents the total energy of the system, and so a Poisson system necessarily conserves energy. Thus, friction and damping are not typically included in the Poisson framework.
Conservation of the Hamiltonian function means that the solutions of the Poisson system (4.31) move along its level sets { H(u) = c }.
In particular, if a level set is a single point u0, then the corresponding solution cannot move and hence is an equilibrium solution: u(t) ≡u0. If u0 is an isolated maximum or minimum of H, then it is necessarily stable, since the nearby level sets remain close to u0 and hence nearby solutions can never † Nonconstant Poisson matrices have additional, more subtle requirements, [26; Chapter 6].
1/7/22 41 c ⃝2022 Peter J. Olver Figure 15.
A Stable Planar Hamiltonian System.
get far away from the equilibrium solution. In Figure 15 we plot the elliptical level sets of a typical stable quadratic Hamiltonian in the plane — the solutions move periodically around the ellipses, all with the same period.
The simplest example is when J = 0 −1 1 0 is a 2 × 2 matrix. In this case, we set u(t) = ( p(t), q(t) )T , and the Poisson system (4.31) corresponding to a Hamiltonian function H(p, q) takes the form dp dt = −∂H ∂q , ∂q ∂t = ∂H ∂p .
(4.32) For example, consider the undamped pendulum equation (4.23).
Let us introduce the position variable q = θ representing the angular coordinate of the pendulum. Let p = m θ be its angular momentum. The pendulum energy function (4.24) is rewritten in terms of the position and momentum variables: H(p, q) = p2 2 m + κ (1 −cos q).
The resulting Hamiltonian system is (4.32) dp dt = κ sin q, ∂q ∂t = p m.
(4.33) If we convert back to u = q and velocity v = θ = p/m we immediately recover the phase plane form (4.23) of the pendulum equation. Thus, we recover the constancy of the energy first integral directly from the general Hamiltonian framework.
A Poisson system is called Hamiltonian if J is nonsingular: det J ̸= 0. Since J is a skew-symmetric matrix, this can only happen if its size, which is also the dimension 1/7/22 42 c ⃝2022 Peter J. Olver of the underlying space is even: n = 2k.
The most important case, generalizing the two-dimensional version (4.32), is when J = O −I I O where O denotes the k × k zero matrix and I denotes the k × k identity matrix. In this case, the variables are split into two vector components: u = ( p, q )T = ( p1, . . ., pk, q1, . . ., qk )T .
We find ∇H(p, q) = ∂H ∂p1 , . . . ∂H ∂pk , ∂H ∂q1 , . . . ∂H ∂qk T and so (4.31) takes on the explicit form dp1 dt = −∂H ∂q1 , . . .
dpk dt = −∂H ∂qk , dq1 dt = ∂H ∂p1 , . . .
dqk dt = ∂H ∂pk .
(4.34) In physical applications, the p variables typically represent momenta, while the q variables represent positions of the objects in the system; this was already noted in the pendulum example.
Conservative mechanical systems can always be represented in Hamiltonian form, .
The prototypical example is to take the Hamiltonian function of the particular form H(p, q) = p2 1 2 m1 + · · · + p2 k 2 mk + V (q1, . . ., qk), (4.35) where each mi > 0 is a positive constant. The canonical system (4.34) has the form dp1 dt = −∂V ∂q1 , . . .
dpk dt = −∂V ∂qk , dq1 dt = p1 m1 , . . .
dqk dt = pk mk , which we can rewrite in vectorial form M dq dt = p, dp dt = −∇V (q), (4.36) where M = diag (m1, . . ., mn). Eliminating the p variables, we find that (4.36) reduces to a second order system of ordinary differential equations in the Newtonian form M d2q dt2 = −∇V (q).
The vector q represents the position vector for the mechanical system, while V (q) is the potential energy. The constants mi are masses, and each pi = mi qi represents the momentum variable associated with the position variable qi. The Hamiltonian function is the total energy, the first term being the kinetic energy since the term p2 i 2mi = mi 2 dqi dt 2 is precisely one half mass times velocity squared.
1/7/22 43 c ⃝2022 Peter J. Olver Example 4.22. Consider a planetary system consisting of n bodies in three-dimensional space. We represent the planets as point masses, where the ith planet is concentrated at position qi(t) = ( xi(t), yi(t), zi(t) )T . The planet’s linear momentum vector is pi(t) = mi qi(t) = xi(t), yi(t), zi(t) T , and its kinetic energy is ∥pi ∥2 2mi = mi 2 ∥ qi ∥2 = mi 2 x2 i + y2 i + z2 i .
The Newtonian gravitational potential between two point masses is proportional to the inverse square of their distance; more specifically V (qi, qj) = G mi mj ∥qi −qj ∥2 (4.37) represents the gravitational potential between planets i and j, where mi, mj are their masses, and G the universal gravitational constant. The Hamiltonian of the system is the total kinetic plus potential energy, which is obtained by summing the individual contribu-tions from (pairs of) planets: H(p, q) = n X i=1 ∥pi ∥2 2mi + X 1≤i<j≤n G mi mj ∥qi −qj ∥2 .
The resulting Hamiltonian system is known as the n–body problem, and it has been the subject of intense research since the time of Newton. For a two body system, e.g., a planet and a sun, the motion is well understood; the smaller body moves around either an ellipse with the sum at a focus, or a parabola, or a hyperbola — the latter two cases apply to comets. Even today, there are fundamental unanswered questions about the nature of solutions to this basic physical system for 3 or more bodies. Recent developments have been the discovery of systems of planets in which one or more goes offto infinity in finite time, , and systems of planets that move around a figure 8 and even more complicated curves, [6, 22].
Remark: For systems in Hamiltonian form, Theorem 4.13 tells us that the strict lo-cal minima (and maxima) of the Hamiltonian function are necessarily stable equilibrium points! This reconfirms our general observation. Saddle points are typically unstable.
The Hamiltonian form of a classical mechanical system plays a profound role in the construction of its quantum mechanics version. There is a canonical quantization procedure that replaces the nonlinear Hamiltonian system of ordinary differential equations with a linear, quantum mechanics partial differential equation known as the Schr¨ odinger equation.
For example, the quantum mechanical system describing an electron orbiting a proton in a hydrogen atom is the quantized version of the classical 2–body problem describing the motion of a single planet around the sun. For details, we refer the reader to any basic text in quantum mechanics, e.g., [10, 19, 21].
1/7/22 44 c ⃝2022 Peter J. Olver 5.
Numerical Methods.
Since we have no hope of solving the vast majority of differential equations in explicit, analytic form, the design of suitable numerical algorithms for accurately approximating solutions is essential. The ubiquity of differential equations throughout mathematics and its applications has driven the tremendous research effort devoted to numerical solution schemes, some dating back to the beginnings of the calculus.
Nowadays, one has the luxury of choosing from a wide range of excellent software packages that provide reliable and accurate results for a broad range of systems, at least for solutions over moderately long time periods. However, all of these packages, and the underlying methods, have their limitations, and it is essential that one be able to to recognize when the software is working as advertised, and when it produces spurious results! Here is where the theory, particularly the classification of equilibria and their stability properties, as well as first integrals and Lyapunov functions, can play an essential role. Explicit solutions, when known, can also be used as test cases for tracking the reliability and accuracy of a chosen numerical scheme.
In this section, we survey the most basic numerical methods for solving initial value problems. For brevity, we shall only consider so-called single step schemes, culminating in the very popular and versatile fourth order Runge–Kutta Method. This should only serve as a extremely basic introduction to the subject, and many other important and useful methods can be found in more specialized texts, [12, 18]. It goes without saying that some equations are more difficult to accurately approximate than others, and a variety of more specialized techniques are employed when confronted with a recalcitrant system. But all of the more advanced developments build on the basic schemes and ideas laid out in this section.
Euler’s Method The key issues confronting the numerical analyst of ordinary differential equations already appear in the simplest first order ordinary differential equation. Our goal is to calculate a decent approximation to the (unique) solution to the initial value problem du dt = F(t, u), u(t0) = u0.
(5.1) To keep matters simple, we will focus our attention on the scalar case; however, all formulas and results written in a manner that can be readily adapted to first order systems — just replace the scalar functions u(t) and F(t, u) by vector-valued functions u and F(t, u) throughout. (The time t, of course, remains a scalar.) Higher order ordinary differential equations are inevitably handled by first converting them into an equivalent first order system, as discussed in Section 2, and then applying the numerical scheme.
The very simplest numerical solution method is named after Leonhard Euler — al-though Newton and his contemporaries were well aware of such a simple technique. Euler’s Method is rarely used in practice because much more efficient and accurate techniques can be implemented with minimal additional work. Nevertheless, the method lies at the core of the entire subject, and must be thoroughly understood before progressing on to the more sophisticated algorithms that arise in real-world computations.
1/7/22 45 c ⃝2022 Peter J. Olver Starting at the initial time t0, we introduce successive mesh points (or sample times) t0 < t1 < t2 < t3 < · · · , continuing on until we reach a desired final time tn = t⋆. The mesh points should be fairly closely spaced. To keep the analysis as simple as possible, we will always use a uniform step size, and so h = tk+1 −tk > 0, (5.2) does not depend on k and is assumed to be relatively small. This assumption serves to simplify the analysis, and does not significantly affect the underlying ideas. For a uniform step size, the kth mesh point is at tk = t0 + k h. More sophisticated adaptive methods, in which the step size is adjusted in order to maintain accuracy of the numerical solution, can be found in more specialized texts, e.g., [12, 18]. Our numerical algorithm will recursively compute approximations uk ≈u(tk), for k = 0, 1, 2, 3, . . . , to the sampled values of the solution u(t) at the chosen mesh points. Our goal is to make the error Ek = uk −u(tk) in the approximation at each time tk as small as possible.
If required, the values of the solution u(t) between mesh points may be computed by a subsequent interpolation procedure, e.g., cubic splines.
As you learned in first year calculus, the simplest approximation to a (continuously differentiable) function u(t) is provided by its tangent line or first order Taylor polynomial.
Thus, near the mesh point tk u(t) ≈u(tk) + (t −tk) du dt (tk) = u(tk) + (t −tk) F(tk, u(tk)), in which we replace the derivative du/dt of the solution by the right hand side of the governing differential equation (5.1). In particular, the approximate value of the solution at the subsequent mesh point is u(tk+1) ≈u(tk) + (tk+1 −tk) F(tk, u(tk)).
(5.3) This simple idea forms the basis of Euler’s Method.
Since in practice we only know the approximation uk to the value of u(tk) at the current mesh point, we are forced to replace u(tk) by its approximation uk in the preceding formula. We thereby convert (5.3) into the iterative scheme uk+1 = uk + (tk+1 −tk) F(tk, uk).
(5.4) In particular, when based on a uniform step size (5.2), Euler’s Method takes the simple form uk+1 = uk + hF(tk, uk).
(5.5) As sketched in Figure 16, the method starts offapproximating the solution reasonably well, but gradually loses accuracy as the errors accumulate.
Euler’s Method is the simplest example of a one-step numerical scheme for integrating an ordinary differential equation. This refers to the fact that the succeeding approximation, uk+1 ≈u(tk+1), depends only upon the current value, uk ≈u(tk), which is one mesh point or “step” behind.
1/7/22 46 c ⃝2022 Peter J. Olver t0 t1 t2 t3 u0 u1 u2 u3 u(t) Figure 16.
Euler’s Method.
To begin to understand how Euler’s Method works in practice, let us test it on a problem we know how to solve, since this will allow us to precisely monitor the resulting errors in our numerical approximation to the solution.
Example 5.1. The simplest “nontrivial” initial value problem is du dt = u, u(0) = 1, whose solution is, of course, the exponential function u(t) = et. Since F(t, u) = u, Euler’s Method (5.5) with a fixed step size h > 0 takes the form uk+1 = uk + huk = (1 + h) uk.
This is a linear iterative equation, and hence easy to solve: uk = (1 + h)ku0 = (1 + h)k is our proposed approximation to the solution u(tk) = etk at the mesh point tk = kh.
Therefore, the Euler scheme to solve the differential equation, we are effectively approxi-mating the exponential by a power function: etk = ek h ≈(1 + h)k When we use simply t to indicate the mesh time tk = kh, we recover, in the limit, a well-known calculus formula: et = lim h →0 (1 + h)t/h = lim k →∞ 1 + t k k .
(5.6) A reader familiar with the computation of compound interest will recognize this particular approximation. As the time interval of compounding, h, gets smaller and smaller, the amount in the savings account approaches an exponential.
How good is the resulting approximation? The error E(tk) = Ek = uk −etk 1/7/22 47 c ⃝2022 Peter J. Olver measures the difference between the true solution and its numerical approximation at time t = tk = kh. Let us tabulate the error at the particular times t = 1, 2 and 3 for various values of the step size h. The actual solution values are e1 = e = 2.718281828 . . . , e2 = 7.389056096 . . . , e3 = 20.085536912 . . . .
In this case, the numerical approximation always underestimates the true solution.
h E(1) E(2) E(3) .1 −.125 −.662 −2.636 .01 −.0134 −.0730 −.297 .001 −.00135 −.00738 −.0301 .0001 −.000136 −.000739 −.00301 .00001 −.0000136 −.0000739 −.000301 Some key observations: • For a fixed step size h, the further we go from the initial point t0 = 0, the larger the magnitude of the error.
• On the other hand, the smaller the step size, the smaller the error at a fixed value of t.
The trade-offis that more steps, and hence more computational effort† is required to produce the numerical approximation. For instance, we need k = 10 steps of size h = .1, but k = 1000 steps of size h = .001 to compute an approximation to u(t) at time t = 1.
• The error is more or less in proportion to the step size. Decreasing the step size by a factor of 1 10 decreases the error by a similar amount, but simultaneously increases the amount of computation by a factor of 10.
The final observation indicates that the Euler Method is of first order, which means that the error depends linearly on the step size h. More specifically, at a fixed time t, the error is bounded by | E(t)| = | uk −u(t) | ≤C(t) h, when t = tk = k h, (5.7) for some positive C(t) > 0 that depends upon the time, and the initial condition, but not on the step size.
Example 5.2. The solution to the initial value problem du dt = 1 −4 3 t u, u(0) = 1, (5.8) † In this case, there happens to be an explicit formula for the numerical solution which can be used to bypass the iterations. However, in almost any other situation, one cannot compute the approximation uk without having first determined the intermediate values u0, . . . , uk−1.
1/7/22 48 c ⃝2022 Peter J. Olver 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 h = .1 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 h = .01 Figure 17.
Euler’s Method for u = 1 −4 3 t u.
was found in Example 2.3 by the method of separation of variables: u(t) = exp t −2 3 t2 .
(5.9) Euler’s Method leads to the iterative numerical scheme uk+1 = uk + h 1 −4 3 tk uk, u0 = 1.
In Figure 17 we compare the graphs of the actual and numerical solutions for step sizes h = .1 and .01. In the former plot, we explicitly show the mesh points, but not in the latter, since they are too dense; moreover, the graphs of the numerical and true solutions are almost indistinguishable at this resolution.
The following table lists the numerical errors E(tk) = uk−u(tk) between the computed and actual solution values u(1) = 1.395612425 . . . , u(2) = .513417119 . . . , u(3) = .049787068 . . . , for several different step sizes: h E(1) E(2) E(3) .1000 .07461761 .03357536 −.00845267 .0100 .00749258 .00324416 −.00075619 .0010 .00074947 .00032338 −.00007477 .0001 .00007495 .00003233 −.00000747 As in the previous example, each decrease in step size by a factor of 10 leads to one additional decimal digit of accuracy in the computed solution.
1/7/22 49 c ⃝2022 Peter J. Olver Taylor Methods In general, the order of a numerical solution method governs both the accuracy of its approximations and the speed at which they converge to the true solution as the step size is decreased. Although the Euler Method is simple and easy to implement, it is only a first order scheme, and therefore of limited utility in serious computations. So, the goal is to devise simple numerical methods that enjoy a much higher order of accuracy.
Our derivation of the Euler Method was based on a first order Taylor approximation to the solution. So, an evident way to design a higher order method is to employ a higher order Taylor approximation.
The Taylor series expansion for the solution u(t) at the succeeding mesh point tk+1 = tk + h has the form u(tk+1) = u(tk + h) = u(tk) + h du dt (tk) + h2 2 d2u dt2 (tk) + h3 6 d3u dt3 (tk) + · · · .
(5.10) As we just saw, we can evaluate the first derivative term through use of the underlying differential equation: du dt = F(t, u).
(5.11) The second derivative term can be found by differentiating the equation with respect to t.
Invoking the chain rule†, d2u dt2 = d dt du dt = d dt F(t, u(t)) = ∂F ∂t (t, u) + ∂F ∂u (t, u) du dt = ∂F ∂t (t, u) + ∂F ∂u (t, u) F(t, u) ≡F (2)(t, u).
(5.12) This operation is known as the total derivative, indicating that that we must treat the second variable u as a function of t when differentiating.
Substituting (5.11–12) into (5.10) and truncating at order h2 leads to the Second Order Taylor Method uk+1 = uk + h F(tk, uk) + h2 2 F (2)(tk, uk) = uk + h F(tk, uk) + h2 2 ∂F ∂t (tk, uk) + ∂F ∂u (tk, uk) F(tk, uk) , (5.13) in which, as before, we replace the solution value u(tk) by its computed approximation uk. The resulting method is of second order, meaning that the error function satisfies the quadratic error estimate | E(t) | = | uk −u(t) | ≤C(t) h2 when t = tk = k h.
(5.14) † We assume throughout that F has as many continuous derivatives as needed.
1/7/22 50 c ⃝2022 Peter J. Olver Example 5.3.
Let us explicitly formulate the second order Taylor Method for the initial value problem (5.8). Here du dt = F(t, u) = 1 −4 3 t u, d2u dt2 = d dt F(t, u) = −4 3 u + 1 −4 3 t du dt = −4 3 u + 1 −4 3 t 2 u, and so (5.13) becomes uk+1 = uk + h 1 −4 3 tk uk + 1 2 h2 −4 3 uk + 1 −4 3 tk 2 uk , u0 = 1.
The following table lists the errors between the values computed by the second order Taylor scheme and the actual solution values, as given in Example 5.2.
h E(1) E(2) E(3) .100 .00276995 −.00133328 .00027753 .010 .00002680 −.00001216 .00000252 .001 .00000027 −.00000012 .00000002 Observe that, in accordance with the quadratic error estimate (5.14), a decrease in the step size by a factor of 1 10 leads in an increase in accuracy of the solution by a factor 1 100, i.e., an increase in 2 significant decimal places in the numerical approximation of the solution.
Higher order Taylor methods are obtained by including further terms in the expansion (5.10). For example, to derive a third order Taylor method, we include the third order term (h3/6)d3u/dt3 in the Taylor expansion, where we evaluate the third derivative by differentiating (5.12), and so d3u dt3 = d dt d2u dt2 = d dt F (2)(t, u) = ∂F (2) ∂t + ∂F (2) ∂u du dt = ∂F (2) ∂t + F ∂F (2) ∂u = ∂2F ∂t2 + 2 F ∂2F ∂t∂u + F 2 ∂2F ∂u2 + ∂F ∂t ∂F ∂u + F ∂F ∂u 2 ≡F (3)(t, u), (5.15) where we continue to make use of the fact that du/dt = F(t, u) is provided by the right hand side of the differential equation. The resulting third order Taylor method is uk+1 = uk + h F(tk, uk) + h2 2 F (2)(tk, uk) + h3 6 F (3)(tk, uk), (5.16) where the last two summand are given by (5.12), (5.15), respectively. The higher order expressions are even worse, and a good symbolic manipulation system is almost essential for accurate computation.
Whereas higher order Taylor methods are easy to motivate, they are rarely used in practice. There are two principal difficulties: 1/7/22 51 c ⃝2022 Peter J. Olver • Owing to their dependence upon the partial derivatives of F(t, u), the right hand side of the differential equation needs to be rather smooth.
• Even worse, the explicit formulae become exceedingly complicated, even for relatively simple functions F(t, u). Efficient evaluation of the multiplicity of terms in the Taylor approximation and avoidance of round offerrors become significant concerns.
As a result, mathematicians soon abandoned the Taylor series approach, and began to look elsewhere for high order, efficient integration methods.
Error Analysis Before pressing on, we need to engage in a more serious discussion of the error in a numerical scheme. A general one-step numerical method can be written in the form uk+1 = G(h, tk, uk), (5.17) where G is a prescribed function of the current approximate solution value uk ≈u(tk), the time tk, and the step size h = tk+1 −tk, which, for illustrative purposes, we assume to be fixed. (We leave the discussion of multi-step methods, in which G could also depend upon the earlier values uk−1, uk−2, . . . , to more advanced texts, e.g., [12, 18].) In any numerical integration scheme there are, in general, three sources of error.
• The first is the local error committed in the current step of the algorithm. Even if we have managed to compute a completely accurate value of the solution uk = u(tk) at time tk, the numerical approximation scheme (5.17) is almost certainly not exact, and will therefore introduce an error into the next computed value uk+1 ≈u(tk+1).
Round-offerrors, resulting from the finite precision of the computer arithmetic, will also contribute to the local error.
• The second is due to the error that is already present in the current approximation uk ≈ u(tk). The local errors tend to accumulate as we continue to run the iteration, and the net result is the global error, which is what we actually observe when comparing the numerical approximation with the exact solution.
• Finally, if the initial condition u0 ≈u(t0) is not computed accurately, this initial error will also make a contribution. For example, if u(t0) = π, then we introduce some initial error by using a decimal approximation, say π ≈3.14159.
The third error source will, for simplicity, be ignored in our discussion, i.e., we will assume u0 = u(t0) is exact. Further, for simplicity we will assume that round-offerrors do not play any significant role — although one must always keep them in mind when analyzing the computation. Since the global error is entirely due to the accumulation of successive local errors, we must first understand the local error in detail.
To measure the local error in going from tk to tk+1, we compare the exact solution value u(tk+1) with its numerical approximation (5.17) under the assumption that the current computed value is correct: uk = u(tk). Of course, in practice this is never the case, and so the local error is an artificial quantity. Be that as it may, in most circumstances the local error is (a) easy to estimate, and, (b) provides a reliable guide to the global accuracy of the numerical scheme. To estimate the local error, we assume that the step size h is small 1/7/22 52 c ⃝2022 Peter J. Olver and approximate the solution u(t) by its Taylor expansion† u(tk+1) = u(tk) + h du dt (tk) + h2 2 d2u dt2 (tk) + · · · = uk + h F(tk, uk) + h2 2 F (2)(tk, uk) + h3 6 F (3)(tk, uk) + · · · .
(5.18) In the second expression, we have employed (5.12, 15) and their higher order analogs to evaluate the derivative terms, and then invoked our local accuracy assumption to replace u(tk) by uk. On the other hand, a direct Taylor expansion, in h, of the numerical scheme produces uk+1 = G(h, tk, uk) = G(0, tk, uk) + h ∂G ∂h (0, tk, uk) + h2 2 ∂2G ∂h2 (0, tk, uk) + h3 6 ∂3G ∂h3 (0, tk, uk) + · · · .
(5.19) The local error is obtained by comparing these two Taylor expansions.
Definition 5.4. A numerical integration method is of order n if the Taylor expan-sions (5.18, 19) of the exact and numerical solutions agree up to order hn.
For example, the Euler Method uk+1 = G(h, tk, uk) = uk + hF(tk, uk), is already in the form of a Taylor expansion — that has no terms involving h2, h3, . . . .
Comparing with the exact expansion (5.18), we see that the constant and order h terms are the same, but the order h2 terms differ (unless F (2) ≡0). Thus, according to the definition, the Euler Method is a first order method. Similarly, the Taylor Method (5.13) is a second order method, because it was explicitly designed to match the constant, h and h2 terms in the Taylor expansion of the solution. The general Taylor Method of order n sets G(h, tk, uk) to be exactly the order n Taylor polynomial, differing from the full Taylor expansion at order hn+1.
Under fairly general hypotheses, it can be proved that, if the numerical scheme has order n as measured by the local error, then the global error is bounded by a multiple of hn. In other words, assuming no round-offor initial error, the computed value uk and the solution at time tk can be bounded by | uk −u(tk) | ≤M hn, (5.20) where the constant M > 0 may depend on the time tk and the particular solution u(t). The error bound (5.20) serves to justify our numerical observations. For a method of order n, decreasing the step size by a factor of 1 10 will decrease the overall error by a factor of about 10−n, and so, roughly speaking, we anticipate gaining an additional n digits of accuracy — † In our analysis, we assume that the differential equation, and hence the solution, has sufficient smoothness to justify the relevant Taylor approximation.
1/7/22 53 c ⃝2022 Peter J. Olver at least up until the point that round-offerrors begin to play a role. Readers interested in a more complete error analysis of numerical integration schemes should consult a specialized text, e.g., [12, 18].
The bottom line is the higher its order, the more accurate the numerical scheme, and hence the larger the step size that can be used to produce the solution to a desired accuracy.
However, this must be balanced with the fact that higher order methods inevitably require more computational effort at each step.
If the total amount of computation has also decreased, then the high order method is to be preferred over a simpler, lower order method. Our goal now is to find another route to the design of higher order methods that avoids the complications inherent in a direct Taylor expansion.
An Equivalent Integral Equation The secret to the design of higher order numerical algorithms is to replace the dif-ferential equation by an equivalent integral equation. By way of motivation, recall that, in general, differentiation is a badly behaved process; a reasonable function can have an unreasonable derivative.
On the other hand, integration ameliorates; even quite nasty functions have relatively well-behaved integrals. For the same reason, accurate numerical integration is relatively painless, whereas numerical differentiation should be avoided un-less necessary. While we have not dealt directly with integral equations in this text, the subject has been extensively developed by mathematicians, , and has many important physical applications.
Conversion of an initial value problem (5.1) to an integral equation is straightforward.
We integrate both sides of the differential equation from the initial point t0 to a variable time t. The Fundamental Theorem of Calculus is used to explicitly evaluate the left hand integral: u(t) −u(t0) = Z t t0 u(s) ds = Z t t0 F(s, u(s))ds.
Rearranging terms, we arrive at the key result.
Lemma 5.5. The solution u(t) to the the integral equation u(t) = u(t0) + Z t t0 F(s, u(s)) ds (5.21) coincides with the solution to the initial value problem du dt = F(t, u), u(t0) = u0.
Proof : Our derivation already showed that the solution to the initial value problem satisfies the integral equation (5.21).
Conversely, suppose that u(t) solves the integral equation. Since u(t0) = u0 is constant, the Fundamental Theorem of Calculus tells us that the derivative of the right hand side of (5.21) is equal to the integrand, so du dt = F(t, u(t)).
Moreover, at t = t0, the upper and lower limits of the integral coincide, and so it vanishes, whence u(t) = u(t0) = u0 has the correct initial conditions.
Q.E.D.
1/7/22 54 c ⃝2022 Peter J. Olver Left Endpoint Rule Trapezoid Rule Midpoint Rule Figure 18.
Numerical Integration Methods.
Observe that, unlike the differential equation, the integral equation (5.21) requires no additional initial condition — it is automatically built into the equation. The proofs of the fundamental existence and uniqueness Theorems 3.1 and 3.3 for ordinary differential equations are, in fact, based on the integral equation reformulation of the initial value problem; see [13, 15] for details.
The integral equation reformulation is equally valid for systems of first order ordinary differential equations.
As noted above, u(t) and F(t, u(t)) become vector-valued func-tions. Integrating a vector-valued function is accomplished by integrating its individual components. Complete details are left to the exercises.
Implicit and Predictor–Corrector Methods From this point onwards, we shall abandon the original initial value problem, and turn our attention to numerically solving the equivalent integral equation (5.21). Let us rewrite the integral equation, starting at the mesh point tk instead of t0, and integrating until time t = tk+1. The result is the basic integral formula u(tk+1) = u(tk) + Z tk+1 tk F(s, u(s))ds (5.22) that (implicitly) computes the value of the solution at the subsequent mesh point. Com-paring this formula with the Euler Method uk+1 = uk + h F(tk, uk), where h = tk+1 −tk, and assuming for the moment that uk = u(tk) is exact, we discover that we are merely approximating the integral by Z tk+1 tk F(s, u(s))ds ≈h F(tk, u(tk)).
(5.23) This is the Left Endpoint Rule for numerical integration — that approximates the area under the curve g(t) = F(t, u(t)) between tk ≤t ≤tk+1 by the area of a rectangle whose height g(tk) = F(tk, u(tk)) ≈F(tk, uk) is prescribed by the left-hand endpoint of the graph. As indicated in Figure 18, this is a reasonable, but not especially accurate method of numerical integration.
1/7/22 55 c ⃝2022 Peter J. Olver In first year calculus, you no doubt encountered much better methods of approximating the integral of a function. One of these is the Trapezoid Rule, which approximates the integral of the function g(t) by the area of a trapezoid obtained by connecting the two points (tk, g(tk)) and (tk+1, g(tk+1)) on the graph of g by a straight line, as in the second Figure 18.
Let us therefore try replacing (5.23) by the more accurate trapezoidal approximation Z tk+1 tk F(s, u(s))ds ≈1 2 h F(tk, u(tk)) + F(tk+1, u(tk+1)) .
(5.24) Substituting this approximation into the integral formula (5.22), and replacing the solution values u(tk), u(tk+1) by their numerical approximations, leads to the (hopefully) more accurate numerical scheme uk+1 = uk + 1 2 h F(tk, uk) + F(tk+1, uk+1) , (5.25) known as the Trapezoid Method. It is an implicit scheme, since the updated value uk+1 appears on both sides of the equation, and hence is only defined implicitly.
Example 5.6. Consider the differential equation u = 1 −4 3 t u studied in Exam-ples 5.2 and 5.3. The Trapezoid Method with a fixed step size h takes the form uk+1 = uk + 1 2 h 1 −4 3 tk uk + 1 −4 3 tk+1 uk+1 .
In this case, we can explicit solve for the updated solution value, leading to the recursive formula uk+1 = 1 + 1 2 h 1 −4 3 tk 1 −1 2 h 1 −4 3 tk+1 uk = 1 + 1 2 h −2 3 htk 1 −1 2 h + 2 3 h (tk + h) uk.
(5.26) Implementing this scheme for three different step sizes gives the following errors between the computed and true solutions at times t = 1, 2, 3.
h E(1) E(2) E(3) .100 −.00133315 .00060372 −.00012486 .010 −.00001335 .00000602 −.00000124 .001 −.00000013 .00000006 −.00000001 The numerical data indicates that the Trapezoid Method is of second order. For each reduction in step size by 1 10 , the accuracy in the solution increases by, roughly, a factor of 1 100 = 1 102 ; that is, the numerical solution acquires two additional accurate decimal digits.
The main difficulty with the Trapezoid Method (and any other implicit scheme) is immediately apparent. The updated approximate value for the solution uk+1 appears on both sides of the equation (5.25). Only for very simple functions F(t, u) can one expect to solve (5.25) explicitly for uk+1 in terms of the known quantities tk, uk and tk+1 = tk + h.
The alternative is to employ a numerical equation solver, such as the bisection algorithm 1/7/22 56 c ⃝2022 Peter J. Olver or Newton’s Method, to compute uk+1. In the case of Newton’s Method, one would use the current approximation uk as a first guess for the new approximation uk+1. The resulting scheme requires some work to program, but can be effective in certain situations.
An alternative, less involved strategy is based on the following far-reaching idea. We already know a half-way decent approximation to the solution value uk+1 — namely that provided by the more primitive Euler scheme e uk+1 = uk + h F(tk, uk).
(5.27) Let’s use this estimated value in place of uk+1 on the right hand side of the implicit equation (5.25). The result uk+1 = uk + 1 2 h F(tk, uk) + F(tk + h, e uk+1) = uk + 1 2 h F(tk, uk) + F tk + h, uk + h F(tk, uk) .
(5.28) is known as the Improved Euler Method. It is a completely explicit scheme since there is no need to solve any equation to find the updated value uk+1.
Example 5.7.
For our favorite equation u = 1 −4 3 t u, the Improved Euler Method begins with the Euler approximation e uk+1 = uk + h 1 −4 3 tk uk, and then replaces it by the improved value uk+1 = uk + 1 2 h 1 −4 3 tk uk + 1 −4 3 tk+1 e uk+1 = uk + 1 2 h 1 −4 3 tk uk + 1 −4 3 (tk + h) uk + h 1 −4 3 tk uk = h 1 −2 3 h2 1 + h 1 −4 3 tk + 1 2 h2 1 −4 3 tk 2 i uk.
Implementing this scheme leads to the following errors in the numerical solution at the indicated times. The Improved Euler Method performs comparably to the fully implicit scheme (5.26), and significantly better than the original Euler Method.
h E(1) E(2) E(3) .100 −.00070230 .00097842 .00147748 .010 −.00000459 .00001068 .00001264 .001 −.00000004 .00000011 .00000012 The Improved Euler Method is the simplest of a large family of so-called predictor– corrector algorithms. In general, one begins a relatively crude method — in this case the Euler Method — to predict a first approximation e uk+1 to the desired solution value uk+1.
One then employs a more sophisticated, typically implicit, method to correct the original prediction, by replacing the required update uk+1 on the right hand side of the implicit 1/7/22 57 c ⃝2022 Peter J. Olver scheme by the less accurate prediction e uk+1. The resulting explicit, corrected value uk+1 will, provided the method has been designed with due care, be an improved approximation to the true solution.
The numerical data in Example 5.7 indicates that the Improved Euler Method is of second order since each reduction in step size by 1 10 improves the solution accuracy by, roughly, a factor of 1 100. To verify this prediction, we expand the right hand side of (5.28) in a Taylor series in h, and then compare, term by term, with the solution expansion (5.18).
First†, F tk + h, uk + h F(tk, uk) = F + h Ft + F Fu + 1 2 h2 Ftt + 2F Ftu + F 2 Fuu + · · · , where all the terms involving F and its partial derivatives on the right hand side are evaluated at tk, uk. Substituting into (5.28), we find uk+1 = uk + h F + 1 2 h2 Ft + F Fu + 1 4 h3 Ftt + 2F Ftu + F 2 Fuu + · · · .
(5.29) The two Taylor expansions (5.18) and (5.29) agree in their order 1, h and h2 terms, but differ at order h3. This confirms our experimental observation that the Improved Euler Method is of second order.
We can design a range of numerical solution schemes by implementing alternative numerical approximations to the basic integral equation (5.22). For example, the Midpoint Rule approximates the integral of the function g(t) by the area of the rectangle whose height is the value of the function at the midpoint: Z tk+1 tk g(s) ds ≈h g tk + 1 2 h , where h = tk+1 −tk.
(5.30) See Figure 18 for an illustration. The Midpoint Rule is known to have the same order of accuracy as the Trapezoid Rule, . Substituting into (5.22) leads to the approximation uk+1 = uk + Z tk+1 tk F(s, u(s))ds ≈uk + h F tk + 1 2 h, u tk + 1 2 h .
Of course, we don’t know the value of the solution u tk + 1 2 h at the midpoint, but can predict it through a straightforward adaptation of the basic Euler approximation: u tk + 1 2 h ≈uk + 1 2 h F(tk, uk).
The result is the Midpoint Method uk+1 = uk + h F tk + 1 2 h, uk + 1 2 h F(tk, uk) .
(5.31) A comparison of the terms in the Taylor expansions of (5.18) and (5.31) reveals that the Midpoint Method is also of second order.
† We use subscripts to indicate partial derivatives to save space.
1/7/22 58 c ⃝2022 Peter J. Olver Runge–Kutta Methods The Improved Euler and Midpoint Methods are the most elementary incarnations of a general class of numerical schemes for ordinary differential equations that were first sys-tematically studied by the German mathematicians Carle Runge and Martin Kutta in the late nineteenth century. Runge–Kutta Methods are by far the most popular and powerful general-purpose numerical methods for integrating ordinary differential equations. While not appropriate in all possible situations, Runge–Kutta schemes are surprisingly robust, performing efficiently and accurately in a wide variety of problems. Barring significant complications, they are the method of choice in most basic applications. They comprise the engine that powers most computer software for solving general initial value problems for systems of ordinary differential equations.
The most general Runge–Kutta Method takes the form uk+1 = uk + h m X i=1 ci F(ti,k, ui,k), (5.32) where m counts the number of terms in the method. Each ti,k denotes a point in the kth mesh interval, so tk ≤ti,k ≤tk+1. The second argument ui,k ≈u(ti,k) can be viewed as an approximation to the solution at the point ti,k, and so is computed by a similar, but simpler formula of the same type. There is a lot of flexibility in the design of the method, through choosing the coefficients ci, the times ti,k, as well as the scheme (and all parameters therein) used to compute each of the intermediate approximations ui,k. As always, the order of the method is fixed by the power of h to which the Taylor expansions of the numerical method (5.32) and the actual solution (5.18) agree. Clearly, the more terms we include in the Runge–Kutta formula (5.32), the more free parameters available to match terms in the solution’s Taylor series, and so the higher the potential order of the method. Thus, the goal is to arrange the parameters so that the method has a high order of accuracy, while, simultaneously, avoiding unduly complicated, and hence computationally costly, formulae.
Both the Improved Euler and Midpoint Methods are instances of a family of two term Runge–Kutta Methods uk+1 = uk + h aF(tk, uk) + bF tk,2, uk,2 = uk + h aF(tk, uk) + bF tk + λh, uk + λhF(tk, uk) , (5.33) based on the current mesh point, so tk,1 = tk, and one intermediate point tk,2 = tk + λh with 0 ≤λ ≤1. The basic Euler Method is used to approximate the solution value uk,2 = uk + λhF(tk, uk) at tk,2. The Improved Euler Method sets a = b = 1 2 and λ = 1, while the Midpoint Method corresponds to a = 0, b = 1, λ = 1 2 . The range of possible values for a, b and λ is found by matching the Taylor expansion uk+1 = uk + h aF(tk, uk) + bF tk + λh, uk + λhF(tk, uk) = uk + h(a + b)F(tk, uk) + h2 bλ ∂F ∂t (tk, uk) + F(tk, uk) ∂F ∂u (tk, uk) + · · · .
1/7/22 59 c ⃝2022 Peter J. Olver (in powers of h) of the right hand side of (5.33) with the Taylor expansion (5.18) of the solution, namely u(tk+1) = uk + hF(tk, uk) + h2 2 [ Ft(tk, uk) + F(tk, uk)Fu(tk, uk) ] + · · · , to as high an order as possible. First, the constant terms, uk, are the same. For the order h and order h2 terms to agree, we must have, respectively, a + b = 1, bλ = 1 2.
Therefore, setting a = 1 −µ, b = µ, and λ = 1 2µ , where µ is arbitrary†, leads to the following family of two term, second order Runge–Kutta Methods: uk+1 = uk + h (1 −µ)F(tk, uk) + µF tk + h 2µ, uk + h 2µ F(tk, uk) .
(5.34) The case µ = 1 2 corresponds to the Improved Euler Method (5.28), while µ = 1 yields the Midpoint Method (5.31). Unfortunately, none of these methods are able to match all of the third order terms in the Taylor expansion for the solution, and so we are left with a one-parameter family of two step Runge–Kutta Methods, all of second order, that include the Improved Euler and Midpoint schemes as particular instances. The methods with 1 2 ≤µ ≤1 all perform more or less comparably, and there is no special reason to prefer one over the other.
To construct a third order Runge–Kutta Method, we need to take at least m ≥3 terms in (5.32).
A rather intricate computation (best done with the aid of computer algebra) will produce a range of valid schemes; the results can be found in [12, 18]. The algebraic manipulations are rather tedious, and we leave a complete discussion of the available options to a more advanced treatment. In practical applications, a particularly simple fourth order, four term formula has become the most used. The method, often abbreviated as RK4, takes the form uk+1 = uk + h 6 F(tk, uk) + 2F(t2,k, u2,k) + 2F(t3,k, u3,k) + F(t4,k, u4,k) , (5.35) where the times and function values are successively computed according to the following procedure: t2,k = tk + 1 2 h, u2,k = uk + 1 2 hF(tk, uk), t3,k = tk + 1 2 h, u3,k = uk + 1 2 hF(t2,k, u2,k), t4,k = tk + h, u4,k = uk + hF(t3,k, u3,k).
(5.36) The four term RK4 scheme (5.35–36) is, in fact, a fourth order method. This is confirmed by demonstrating that the Taylor expansion of the right hand side of (5.35) in powers of † Although we should restrict µ ≥1 2 in order that 0 ≤λ ≤1.
1/7/22 60 c ⃝2022 Peter J. Olver h matches all of the terms in the Taylor series for the solution (5.18) up to and including those of order h4, and hence the local error is of order h5. This is not a computation for the faint-hearted — bring lots of paper and erasers, or, better yet, a good computer algebra package! The RK4 scheme is but one instance of a large family of fourth order, four term Runge–Kutta Methods, and by far the most popular owing to its relative simplicity.
Example 5.8. Application of the RK4 Method (5.35–36) to our favorite initial value problem (5.8) leads to the following errors at the indicated times: h E(1) E(2) E(3) .100 −1.944 × 10−7 1.086 × 10−6 4.592 × 10−6 .010 −1.508 × 10−11 1.093 × 10−10 3.851 × 10−10 .001 −1.332 × 10−15 −4.741 × 10−14 1.932 × 10−14 The accuracy is phenomenally good — much better than any of our earlier numerical schemes. Each decrease in the step size by a factor of 1 10 results in about 4 additional decimal digits of accuracy in the computed solution, in complete accordance with its status as a fourth order method.
Actually, it is not entirely fair to compare the accuracy of the methods using the same step size. Each iteration of the RK4 Method requires four evaluations of the func-tion F(t, u), and hence takes the same computational effort as four Euler iterations, or, equivalently, two Improved Euler iterations. Thus, the more revealing comparison would be between RK4 at step size h, Euler at step size 1 4 h, and Improved Euler at step size 1 2 h, as these involve roughly the same amount of computational effort. The resulting errors E(1) at time t = 1 are listed in the following table.
Thus, even taking computational effort into account, the Runge–Kutta Method con-tinues to outperform its rivals. At a step size of .1, it is almost as accurate as the Im-proved Euler Method with step size .0005, and hence 200 times as much computation, while the Euler Method would require a step size of approximately .24 × 10−6, and would be 4, 000, 000 times as slow as Runge–Kutta! With a step size of .001, RK4 computes a solution value that is near the limits imposed by machine accuracy (in single precision arithmetic). The superb performance level and accuracy of the RK4 Method immediately explains its popularity for a broad range of applications.
h Euler Improved Euler Runge–Kutta 4 .1 1.872 × 10−2 −1.424 × 10−4 −1.944 × 10−7 .01 1.874 × 10−3 −1.112 × 10−6 −1.508 × 10−11 .001 1.870 × 10−4 −1.080 × 10−8 −1.332 × 10−15 1/7/22 61 c ⃝2022 Peter J. Olver 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Euler Method, h = .01 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Euler Method, h = .001 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 Improved Euler Method, h = .01 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 RK4 Method, h = .01 Figure 19.
Numerical Solutions of Predator–Prey Model.
Example 5.9. As noted earlier, by writing the function values as vectors uk ≈u(tk), one can immediately use all of the preceding methods to integrate initial value problems for first order systems of ordinary differential equations u = F(u). Consider, by way of example, the Lotka–Volterra system du dt = 2u −uv, dv dt = −9v + 3uv, (5.37) analyzed in Example 4.14. To find a numerical solution, we write u = ( u, v )T for the solution vector, while F(u) = ( 2u −uv, −9v + 3uv )T is the right hand side of the system.
The Euler Method with step size h is given by u(k+1) = u(k) + hF(u(k)), or, explicitly, as a first order nonlinear iterative system u(k+1) = u(k) + h(2u(k) −u(k) v(k)), v(k+1) = v(k) + h(−9v(k) + 3u(k) v(k)).
1/7/22 62 c ⃝2022 Peter J. Olver 5 10 15 20 25 -0.8 -0.6 -0.4 -0.2 Euler Method h = .001 10 20 30 40 50 -0.004 -0.003 -0.002 -0.001 Improved Euler Method h = .01 10 20 30 40 50 -0.00001 -7.5·10-6 -5·10-6 -2.5·10-6 2.5·10-6 5·10-6 RK4 Method h = .01 Figure 20.
Numerical Evaluation of Lotka–Volterra First Integral.
The Improved Euler and Runge–Kutta schemes are implemented in a similar fashion.
Phase portraits of the three numerical algorithms starting with initial conditions u(0) = v(0) = 1.5, and up to time t = 25 in the case of the Euler Method, and t = 50 for the other two, appear in Figure 19. Recall that the solution is supposed to travel periodically around a closed curve, which is the level set I(u, v) = 9 log u −3u + 2 log v −v = I(1.5, 1.5) = −1.53988 of the first integral.
The Euler Method spirals away from the exact periodic solution, whereas the Improved Euler and RK4 Methods perform rather well. Since we do not have an analytic formula for the solution, we are not able to measure the error exactly. However, the known first integral is supposed to remain constant on the solution trajectories, and so one means of monitoring the accuracy of the solution is to track the variation in the numerical values of I(u(k), v(k)).
These are graphed in Figure 20; the Improved Euler keeps the value within .0005, while in the RK4 solution, the first integral only experiences change in its the fifth decimal place over the indicated time period. Of course, the longer one continues to integrate, the more error will gradually creep into the numerical solution.
Still, for most practical purposes, the RK4 solution is indistinguishable from the exact solution.
In practical implementations, it is important to monitor the accuracy of the numer-ical solution, so to gauge when to abandon an insufficiently precise computation. Since accuracy is dependent upon the step size h, one may try adjusting h so as stay within a preassigned error. Adaptive methods, allow one to change the step size during the course of the computation, in response to some estimation of the overall error. Insufficiently ac-curate numerical solutions would necessitate a suitable reduction in step size (or increase in the order of the scheme). On the other hand, if the solution is more accurate than the application requires, one could increase the step size so as to reduce the total amount of computational effort.
How might one decide when a method is giving inaccurate results, since one presum-ably does not know the true solution and so has nothing to directly test the numerical approximation against? One useful idea is to integrate the differential equation using two different numerical schemes, usually of different orders of accuracy, and then compare the results. If the two solution values are reasonably close, then one is usually safe in as-suming that the methods are both giving accurate results, while in the event that they 1/7/22 63 c ⃝2022 Peter J. Olver differ beyond some preassigned tolerance, then one needs to re-evaluate the step size. The required adjustment to the step size relies on a more detailed analysis of the error terms.
Several well-studied methods are employed in practical situations; the most popular is the Runge–Kutta–Fehlberg Method, which combines a fourth and a fifth order Runge–Kutta scheme for error control. Details can be found in more advanced treatments of the subject, e.g., [12, 18].
StiffDifferential Equations While the fourth order Runge–Kutta Method with a sufficiently small step size will successfully integrate a broad range of differential equations — at least over not unduly long time intervals — it does occasionally experience unexpected difficulties. While we have not developed sufficiently sophisticated analytical tools to conduct a thorough analysis, it will be instructive to look at why a breakdown might occur in a simpler context.
Example 5.10. The elementary linear initial value problem du dt = −250 u, u(0) = 1, (5.38) is an instructive and sobering example. The explicit solution is easily found; it is a very rapidly decreasing exponential: u(t) = e−250t.
u(t) = e−250t with u(1) ≈2.69 × 10−109.
The following table gives the result of approximating the solution value u(1) ≈2.69×10−109 at time t = 1 using three of our numerical integration schemes for various step sizes: h Euler Improved Euler RK4 .1 6.34 × 1013 3.99 × 1024 2.81 × 1041 .01 4.07 × 1017 1.22 × 1021 1.53 × 10−19 .001 1.15 × 10−125 6.17 × 10−108 2.69 × 10−109 The results are not misprints! When the step size is .1, the computed solution values are perplexingly large, and appear to represent an exponentially growing solution — the complete opposite of the rapidly decaying true solution. Reducing the step size beyond a critical threshold suddenly transforms the numerical solution to an exponentially decaying function. Only the fourth order RK4 Method with step size h = .001 — and hence a total of 1, 000 steps — does a reasonable job at approximating the correct value of the solution at t = 1.
You may well ask, what on earth is going on? The solution couldn’t be simpler — why is it so difficult to compute it? To understand the basic issue, let us analyze how the Euler Method handles such simple differential equations. Consider the initial value problem du dt = λ u, u(0) = 1, (5.39) 1/7/22 64 c ⃝2022 Peter J. Olver which has an exponential solution u(t) = eλt.
(5.40) As in Example 5.1, the Euler Method with step size h relies on the iterative scheme uk+1 = (1 + λh) uk, u0 = 1, with solution uk = (1 + λh)k.
(5.41) If λ > 0, the exact solution (5.40) is exponentially growing. Since 1+λ h > 1, the numerical iterates are also growing, albeit at a somewhat slower rate. In this case, there is no inherent surprise with the numerical approximation procedure — in the short run it gives fairly accurate results, but eventually trails behind the exponentially growing solution.
On the other hand, if λ < 0, then the exact solution is exponentially decaying and positive. But now, if λ h < −2, then 1 + λ h < −1, and the iterates (5.41) grow expo-nentially fast in magnitude, with alternating signs. In this case, the numerical solution is nowhere close to the true solution; this explains the previously observed pathological behavior. If −1 < 1 + λ h < 0, the numerical solutions decay in magnitude, but continue to alternate between positive and negative values. Thus, to correctly model the qualitative features of the solution and obtain a numerically respectable approximation, we need to choose the step size h so as to ensure that 0 < 1 + λh, and hence h < −1/λ when λ < 0.
For the value λ = −250 in the example, then, we must choose h < 1 250 = .004 in order that the Euler Method give a reasonable numerical answer. A similar, but more complicated analysis applies to any of the Runge–Kutta schemes.
Thus, the numerical methods for ordinary differential equations exhibit a form of conditional stability. Paradoxically, the larger negative λ is — and hence the faster the solution tends to a trivial zero equilibrium — the more difficult and expensive the numerical integration.
The system (5.38) is the simplest example of what is known as a stiffdifferential equation. In general, an equation or system is stiffif it has one or more very rapidly decaying solutions. In the case of autonomous (constant coefficient) linear systems u = Au, stiffness occurs whenever the coefficient matrix A has an eigenvalue with a large negative real part: Re λ ≪0, resulting in a very rapidly decaying eigensolution. It only takes one such eigensolution to render the equation stiff, and ruin the numerical computation of even the well behaved solutions! Curiously, the component of the actual solution corresponding to such large negative eigenvalues is almost irrelevant, as it becomes almost instantaneously tiny. However, the presence of such an eigenvalue continues to render the numerical solution to the system very difficult, even to the point of exhausting any available computing resources. Stiffequations require more sophisticated numerical procedures to integrate, and we refer the reader to [12, 18] for details.
Most of the other methods derived above also suffer from instability due to stiffness of the ordinary differential equation for sufficiently large negative λ. Interestingly, stability for solving the trivial test scalar ordinary differential equation (5.39) suffices to characterize acceptable step sizes h, depending on the size of λ, which, in the case of systems, becomes the eigenvalue.
The analysis is not so difficult, owing to the innate simplicity of the 1/7/22 65 c ⃝2022 Peter J. Olver test ordinary differential equation (5.39). A significant exception, which also illustrates the test for behavior under rapidly decaying solutions, is the Trapezoid Method (5.25).
Let us analyze the behavior of the resulting numerical solution to (5.39). Substituting f(t, u) = λ u into the Trapezoid iterative equation (5.25), we find uk+1 = uk + 1 2 h λ uk + λ uk+1 , which we solve for uk+1 = 1 + 1 2 hλ 1 −1 2 hλ uk ≡µ uk.
Thus, the behavior of the solution is entirely determined by the size of the coefficient µ.
If λ > 0, then µ > 1 and the numerical solution is exponentially growing, as long as the denominator is positive, which requires h < 2/λ to be sufficiently small. In other words, rapidly growing exponential solutions require reasonably small step sizes to accurately compute, which is not surprising. On the other hand, if λ < 0, then | µ | < 1, no matter how large negative λ gets! (But we should also restrict h < −2/λ to be sufficiently small, as otherwise µ < 0 and the numerical solution would have oscillating signs, even though it is decaying, and hence vanishing small. If this were part of a larger system, such minor oscillations would not worry us because they would be unnoticeable in the long run.) Thus, the Trapezoid Method is not affected by very large negative exponents, and hence not subject to the effects of stiffness.
The Trapezoid Method is the simplest example of an A stable method. More precisely, a numerical solution method is called A stable if the zero solution is asymptotically stable for the iterative equation resulting from the numerical solution to the ordinary differential equation u = λ u for all λ < 0. The big advantage of A stable methods is that they are not affected by stiffness. Unfortunately, A stable methods are few and far between. In fact, they are all implicit one-step methods! No explicit Runge–Kutta Method is A stable; see for a proof of this disappointing result. Moreover, multistep methods also suffer from the lack of A stability and so are all prone to the effects of stiffness. Still, when confronted with a seriously stiffequation, one should discard the sophisticated methods and revert to a low order, but A stable scheme like the Trapezoid Method.
1/7/22 66 c ⃝2022 Peter J. Olver References Alligood, K.T., Sauer, T.D., and Yorke, J.A., Chaos. An Introduction to Dynamical Systems, Springer-Verlag, New York, 1997.
Birkhoff, G., and Rota, G.–C., Ordinary Differential Equations, Blaisdell Publ. Co., Waltham, Mass., 1962.
Bronstein, M., Symbolic integration I : Transcendental Functions, Springer–Verlag, New York, 1997.
Burden, R.L., and Faires, J.D., Numerical Analysis, Seventh Edition, Brooks/Cole, Pacific Grove, CA, 2001.
Cantwell, B.J., Introduction to Symmetry Analysis, Cambridge University Press, Cambridge, 2003.
Chenciner, A., and Montgomery, R., A remarkable periodic solution of the three-body problem in the case of equal masses, Ann. Math. 152 (2000), 881–901.
Courant, R., and Hilbert, D., Methods of Mathematical Physics, vol. I, Interscience Publ., New York, 1953.
Devaney, R.L., An Introduction to Chaotic Dynamical Systems, Addison–Wesley, Redwood City, Calif., 1989.
Diacu, F., An Introduction to Differential Equations, W.H. Freeman and Co., New York, 2000.
Dirac, P.A.M., The Principles of Quantum Mechanics, 3rd ed., Clarendon Press, Oxford, 1947.
Goldstein, H., Classical Mechanics, Second Edition, Addison–Wesley, Reading, Mass., 1980.
Hairer, E., Nørsett, S.P., and Wanner, G., Solving Ordinary Differential Equations, 2nd ed., Springer–Verlag, New York, 1993–1996.
Hale, J.K., Ordinary Differential Equations, Second Edition, R.E. Krieger Pub. Co., Huntington, N.Y., 1980.
Hille, E., Ordinary Differential Equations in the Complex Domain, John Wiley & Sons, New York, 1976.
Hirsch, M.W., and Smale, S., Differential Equations, Dynamical Systems, and Linear Algebra, Academic Press, New York, 1974.
Hydon, P.E., Symmetry Methods for Differential Equations, Cambridge Texts in Appl. Math., Cambridge University Press, Cambridge, 2000.
Ince, E.L., Ordinary Differential Equations, Dover Publ., New York, 1956.
Iserles, A., A First Course in the Numerical Analysis of Differential Equations, Cambridge University Press, Cambridge, 1996.
Landau, L.D., and Lifshitz, E.M., Quantum Mechanics (Non-relativistic Theory), Course of Theoretical Physics, vol. 3, Pergamon Press, New York, 1977.
1/7/22 67 c ⃝2022 Peter J. Olver Mather, J.N., and McGehee, R., Solutions of the collinear four body problem which become unbounded in finite time, Dynamical Systems, Theory and Applications; J. Moser, ed., Springer, Berlin, 1975, pp. 573–597..
Messiah, A., Quantum Mechanics, John Wiley & Sons, New York, 1976.
Montaldi, J., and Steckles, K., Classification of symmetry groups for planar n-body choreographies, preprint, 2013, arXiv:1305.0470v2.
Moon, F.C., Chaotic Vibrations, John Wiley & Sons, New York, 1987.
Noether, E., Invariante Variationsprobleme, Nachr. K¨ onig. Gesell. Wissen.
G¨ ottingen, Math.–Phys. Kl. (1918), 235–257. (See Kosmann-Schwarzbach, Y., The Noether Theorems. Invariance and Conservation Laws in the Twentieth Century, Springer, New York, 2011, for an English translation.) Olver, F.W.J., Lozier, D.W., Boisvert, R.F., and Clark, C.W., eds., NIST Handbook of Mathematical Functions, Cambridge University Press, Cambridge, 2010.
Olver, P.J., Applications of Lie Groups to Differential Equations, 2nd ed., Graduate Texts in Mathematics, vol. 107, Springer–Verlag, New York, 1993.
Olver, P.J., and Shakiban, C., Applied Linear Algebra, Second Edition, Undergraduate Texts in Mathematics, Springer, New York, 2018.
Whittaker, E.T., A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, Cambridge University Press, Cambridge, 1937.
1/7/22 68 c ⃝2022 Peter J. Olver |
9197 | http://tasks.illustrativemathematics.org/content-standards/8/G/A/1/tasks/1673 | Engage your students with effective distance learning resources. ACCESS RESOURCES>>
Grade 8
Domain Geometry
Cluster Understand congruence and similarity using physical models, transparencies, or geometry software.
Standard Verify experimentally the properties of rotations, reflections, and translations:
Task Reflections, Rotations, and Translations
Reflections, Rotations, and Translations
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Alignments to Content Standards: 8.G.A.1
Student View
Task
In this task, using computer software, you will apply reflections, rotations, and translations to a triangle. You will then study what happens to the side lengths and angle measures of the triangle after these transformations have been applied. In each part of the question, a sample picture of the triangle is supplied along with a line of reflection, angle of rotation, and segment of translation: the attached GeoGebra software will allow you to experiment with changing the location of the line of reflection, changing the measure of the angle of rotation, and changing the location and length of the segment of translation.
Below is a triangle $ABC$ and a line $\overleftrightarrow{DE}$:
Use the supplied GeoGebra application to reflect $\triangle ABC$ over $\overleftrightarrow{DE}$. Label the reflected triangle $A'B'C'$. What are the side lengths and angle measures of triangle $A'B'C'$? What happens when you change the location of one of the vertices of $\triangle ABC$? What happens when you change the location of line $\overleftrightarrow{DE}$?
2. Below is a triangle $ABC$ and a point $E$. Draw the rotation of $\triangle ABC$ about $E$ through an angle of 85 degrees in the counterclockwise direction.
Label the image of $\triangle ABC$ as $\triangle A'B'C'$. What happens to the side lengths and angle measures of $\triangle A'B'C'$ when you change the measure of the angle of rotation? What happens when you move the center of rotation $E$?
3. Below is a triangle $ABC$ and a directed line segment $\overline{ED}$.
Draw the translation of $\triangle ABC$ by $\overline{ED}$ and label it $\triangle A'B'C'$. What happens to the side lengths and angle measures of triangle $A'B'C'$ when you change one of the vertices, $A$, $B$, or $C$? What if you change the position, length, or direction of the directed line segment $\overline{ED}$?
IM Commentary
The goal of this task is to use technology to visualize what happens to angles and side lengths of a polygon (a triangle in this case) after a reflection, rotation, or translation. GeoGebra files are attached below with triangles already constructed as shown in the images. Students familiar with this or other technology (such as Geometer's Sketchpad) can construct their own polygons as well as their own lines of reflection, angles and centers of rotation, and segments of translation. The supplied GeoGebra files keep track of side lengths and angle measures, allowing the students to gain visualization skills as they study how the transformed polygons vary.
The emphasis for this problem is on MP5, Use Appropriate Tools Strategically. Applying transformations accurately to geometric shapes is a challenging endeavor. Dynamic geometry software, however, does this part of the work, allowing students to focus on how the shapes and their images change when one or more pieces of information is altered. It is important for teachers to use the technology to good effect, having students change the line of reflection, center and angle of rotation, and directed segment of translation. In addition they can change the given triangle and observe how its reflected, rotated, or translated image varies.
Though this is not the focus of the task, the teacher may wish to highlight that in each part of the problem, $\triangle ABC$ is congruent to $\triangle A'B'C'$ in order to familiarize students with this important word and concept.
Attached Resources
Geogebra Applet
Solution
Each picture below shows the requested reflection, rotation, and translation applied to the given triangles.
No matter where you move the vertices of the polygon or the polygon itself, the reflected image will have the same corresponding side lengths and angle measures as the original polygon. Moving the line of reflection does not change this either.
No matter where you move the vertices of the polygon or the polygon itself, the rotated image will have the same corresponding side lengths and angle measures as the original polygon. Moving the center of rotation or changing the measure of the angle of rotation does not change this either.
No matter where you move the vertices of the polygon or the polygon itself, the translated image will have the same corresponding side lengths and angle measures as the original polygon. Moving the position, length, or direction of the directed line segment does not change this either.
Reflections, Rotations, and Translations
In this task, using computer software, you will apply reflections, rotations, and translations to a triangle. You will then study what happens to the side lengths and angle measures of the triangle after these transformations have been applied. In each part of the question, a sample picture of the triangle is supplied along with a line of reflection, angle of rotation, and segment of translation: the attached GeoGebra software will allow you to experiment with changing the location of the line of reflection, changing the measure of the angle of rotation, and changing the location and length of the segment of translation.
Below is a triangle $ABC$ and a line $\overleftrightarrow{DE}$:
Use the supplied GeoGebra application to reflect $\triangle ABC$ over $\overleftrightarrow{DE}$. Label the reflected triangle $A'B'C'$. What are the side lengths and angle measures of triangle $A'B'C'$? What happens when you change the location of one of the vertices of $\triangle ABC$? What happens when you change the location of line $\overleftrightarrow{DE}$?
2. Below is a triangle $ABC$ and a point $E$. Draw the rotation of $\triangle ABC$ about $E$ through an angle of 85 degrees in the counterclockwise direction.
Label the image of $\triangle ABC$ as $\triangle A'B'C'$. What happens to the side lengths and angle measures of $\triangle A'B'C'$ when you change the measure of the angle of rotation? What happens when you move the center of rotation $E$?
3. Below is a triangle $ABC$ and a directed line segment $\overline{ED}$.
Draw the translation of $\triangle ABC$ by $\overline{ED}$ and label it $\triangle A'B'C'$. What happens to the side lengths and angle measures of triangle $A'B'C'$ when you change one of the vertices, $A$, $B$, or $C$? What if you change the position, length, or direction of the directed line segment $\overline{ED}$?
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9198 | https://energywavetheory.com/equations/epc/ | E=pc
Description
When a particle is in motion, its momentum (p) needs to be considered in an energy equation. In 1928, Paul Dirac extended Einstein’s mass-energy equivalence equation (E=mc2) to consider motion. The complete form of the energy-momentum relation equation is E2 = (mc2)2 + (pc)2. When mass isn’t considered, the energy is simply momentum times the speed of light (E=pc). Both versions will be derived on this page.
In energy wave theory, particles are formed from waves. When a particle is in motion, its frequency/wavelength changes. Its wavelength will be shorter in the direction of travel on its leading edge, and longer on its trailing edge, relative to the particle when it is at rest. To an observer, the particle experiences the Doppler effect and thus Doppler equations are used to find the leading edge and trailing frequencies. The particle’s frequency while in motion is the geometric mean of the lead and lag frequencies, which explains the use of the Lorentz factor and relativity in the equation. At relativistic speeds the Lorentz factor needs to be considered.
Derivation – Classical Constants
There is no derivation available for the energy-momentum equation using classical constants. The wave constant derivation is preferred as it describes waves in which wavelength changes with motion.
Derivation – Wave Constants
The energy-momentum equation is simply a change in wave frequency due to motion and it can derived from the base wave energy equation. Due to particle motion (velocity – v), it requires the complete form of the Longitudinal In- and Out-Wave Energy.
.
Proof
When velocity is zero, the sum of the in-wave and out-waves is the Longitudinal Energy Equation that accurately calculates rest energy and rest mass of the electron (the first part of the complete energy-momentum equation). See E=mc2. The addition of velocity into the equation correctly derives the Lorentz factor.
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9199 | https://www.uspharmacist.com/article/a-review-of-treatment-options-for-graves-disease | Published June 17, 2010
ENDOCRINOLOGY/METABOLISM
A Review of Treatment Options for Graves' Disease
Katrin Holt, BA, PharmD Innovations Medication Management, Inc. Bartlett, Tennessee
US Pharm. 2010;35(6)(Generic Drug Review):30-35.
Graves’ disease is an autoimmune disorder affecting approximately 0.3% of people living in the United States and 0.5% of the world population.1 Graves’ disease may include thyroid enlargement, ophthalmopathy, dermopathy, and hyperthyroidism. The most common cause of hyperthyroidism, this disorder occurs five to 10 times more often in women than in men, usually developing between the ages of 20 and 50 years.1,2 While the etiology of Graves’ disease is unknown, hereditary genetics has been implicated as predisposing an individual to developing the disorder later in life through environmental factors, such as smoking and stressful life events, or hormonal triggers like childbirth.2,3 This article will concentrate on the clinical presentation of Graves’ disease and currently available treatment options.
Thyroid Function in Graves’ Disease
Normally, the thyroid gland synthesizes, stores, and releases two kinds of hormones: thyroxine (T4) and triiodothyronine (T3). Iodine obtained through dietary intake is absorbed via the gastrointestinal (GI) tract; it is used as iodide by the thyroid follicular cells and converted to either monoiodotyrosine (MIT) or diiodotyrosine (DIT). MIT and DIT undergo coupling to form T3; T4 is formed by coupling of two DITs. The hypothalamic-pituitary negative feedback system signals the thyroid to stop hormone release when serum T4 and T3 levels are too high.1,4 In Graves’ disease, however, the negative feedback system is superseded by thyroid-stimulating antibodies activating thyroid receptor cells in the same way that thyroid-stimulating hormone (TSH) stimulates the thyroid gland.1,2 Autoantibody stimulation leads to thyroid enlargement by way of thyroid-cell hyperplasia and hypertrophy, along with an increase in serum T3 and T4 and suppression of TSH, leading to a thyrotoxic state.2
## Clinical Presentation
Upon physical examination, most patients exhibit an enlarged thyroid gland that is typically two to three times bigger than normal (>40 g). Extrathyroidal complications of ophthalmopathy and dermopathy also may be seen, and deep tendon reflexes usually are hyperactive.1 In a rare subset of individuals, deep tendon reflexes may be diminished as a result of hypokalemic periodic paralysis. More commonly seen in Asian and Hispanic patients, this phenomenon is related to an intracellular shift of potassium, usually after a period of strenuous exercise. Total body potassium remains normal. Periodic paralysis usually affects the group of muscles exercised and ranges from mild weakness to complete paralysis.1
Ophthalmopathy is apparent in about 30% of diagnosed individuals, with detection in more than 80% of patients assessed by orbital imaging.2,5 Photophobia is common, along with lid lag, periorbital edema, and proptosis. Other signs and symptoms include cardiovascular, central nervous system, and general mood problems, among others. See TABLE 1.1,2,4,6
## Diagnosis
In addition to gross signs and symptoms, abnormal laboratory values are used to make a clinical diagnosis of Graves’ disease (TABLE 2).1,7 Total T3, free T4, and TSH values usually reveal increases in T3 and T4 levels, with T3 being disproportionately higher than T4. TSH levels are extremely low or nondetectable owing to negative feedback from the pituitary gland.1,4,5,7 In patients not displaying overt signs and symptoms and in those experiencing recent onset of symptoms, a 24-hour radioactive iodine (RAI) uptake may be done to determine whether the thyroid gland is producing thyroid hormone (T4, T3, or both) when it is already in a thyrotoxic state. This procedure is performed only in nonpregnant individuals. Normal RAI uptake ranges from 10% to 30%; values higher than 30% indicate Graves’ disease.1,5
## Treatment
Once a diagnosis has been made, the patient’s treatment plan needs to be determined. Treatment options focus on the individual, taking into consideration the patient’s age, comorbidities, preference, and convenience. Three treatment options currently exist: thyroidectomy, antithyroid medications, and RAI.1-3,8
Thyroidectomy: A nonpharmacologic approach rarely used in the U.S. for Graves’ disease is surgical removal of the thyroid gland, which produces a euthyroid or hypothyroid state. Surgery is considered in patients who have a large goiter, have severe ophthalmopathy, are young, are pregnant or lactating, or have relapsed after a trial of antithyroid medications.1,2,4,9 With an 80% to 90% cure rate, surgical removal is considered safe and effective when performed by an experienced surgeon.1 However, the surgery comes with its own risks, including vocal-cord injury, hypoparathyroidism (1%-2%), transient hypocalcemia, bleeding, and infection. Permanent hypothyroidism is a direct result for the majority of patients who undergo surgery, and lifelong thyroid-hormone supplementation and monitoring are required.1,3,4
Antithyroid Medications: In Graves’ disease, propylthiouracil (PTU) and methimazole (MMI) are given to inhibit thyroid production. The primary objective of using antithyroid medications is to induce remission.1 Maximum remission rates of 30% to 50% are observed after 12 to 24 months of therapy. Remission is defined as a normal thyrotropin level while no medication is being taken. Patients who exhibit a smaller goiter (<50 g), are older (>40 years), have a short disease duration (<6 months), have taken antithyroid medication for 1 to 2 years, or have no history of relapse are more likely to experience remission.1-3
Both PTU and MMI are absorbed well from the GI tract, with respective half-lives of 1 to 2.5 hours and 6 to 9 hours. Peak serum concentrations are observed within 1 hour after ingestion.1,10,11 Only PTU inhibits peripheral conversion of T4 to T3; however, the effect is dose-related.1,3 Thionamides have been associated with decreasing thyroid-stimulating immunoglobulins in Graves’ disease; on the other hand, normalization of thyroid-hormone levels has shown a similar effect.3 Generally, normalization of hormone levels and symptom control occur 3 to 6 weeks after therapy is initiated. Loading doses, which range from 300 mg to 600 mg for PTU and 10 mg to 40 mg for MMI, are given for 4 to 6 weeks, followed by a dose reduction when symptoms resolve. For both PTU and MMI, the loading dose is normally given in 3 divided doses. The total daily maintenance dose ranges from 50 mg to 300 mg for PTU and 5 mg to 10 mg for MMI.1,3,6 PTU comes in 50-mg tablets and MMI is available in 5-mg and 10-mg tablets. MMI is typically taken once daily, whereas PTU is usually dosed 3 times daily.10,11
The side-effect profiles of PTU and MMI are similar. Arthralgias, rash, mild leukopenia (WBC <4,000 mm³), and urticaria occur in approximately 5% to 25% of patients taking thionamides. Arthralgias normally occur after 6 months of therapy. These effects are considered minor.1,3,8
Hepatotoxicity is a rare but severe side effect seen with both PTU and MMI. In June 2009 the FDA warned health care professionals about reports of PTU-induced liver failure.12 Currently, PTU-induced liver failure has been reported in 33 adults and 14 children. As reported by the United Network for Organ Sharing, from 1990 to 2007, 16 adults and 7 children received a liver transplant, for an average of two per year in the U.S. alone. MMI, also known for its hepatotoxicity, was not known to occur in patients receiving liver transplants during the same time period. Hepatotoxicity, reported with an average dose of 300 mg, usually occurred during the first 6 months of treatment. As a result, the American Thyroid Association (ATA) and the American Association of Clinical Endocrinologists (AACE) are drafting new hyperthyroidism guidelines that are slated to be published sometime in 2010. In the meantime, the ATA and AACE have recommended PTU as a second-line medication after MMI, except during the first trimester of pregnancy or in patients who are allergic to MMI. PTU should not be used in pediatric patients owing to their increased risk for liver damage. Patients should be encouraged to immediately report to their physician any symptoms of liver failure, such as yellowing of the eyes or skin, fatigue, malaise, pharyngitis, nausea, anorexia, or easy bruising.12,13
Agranulocytosis is a serious side effect occurring in 0.1% to 0.6% of patients taking thionamides. It is recommended that patients discontinue taking the thionamide and contact their physician if they experiencing fever, sore throat, mouth ulcers, or malaise. Onset tends to be sudden. Routine monitoring is not recommended.2,5
RAI (I131): Patients whose Graves’ disease is more severe and those who have relapsed after antithyroid therapy are candidates for RAI treatment. RAI given as I¹³¹ is the preferred drug for treating Graves’ disease in the U.S., as it is rapidly absorbed into and concentrated in thyroid follicular cells. Beta particles emitted from the absorbed I¹³¹ cause follicular-cell necrosis and destruction within weeks to years of treatment.1,4
RAI is administered in a clear liquid or tablet form. One approach to dosage is to give the patient a single fixed dose between 5 and 15 mCi; another approach is to calculate a specific dose based on age, gender, gland size, and thyroid uptake.3,14,15 Patients taking 30 mCi or more usually require hospitalization; patients taking less than 30 mCi usually are sent home and encouraged to take certain safety precautions for several days to minimize exposure to family and friends, such as staying 3 to 6 feet away from people at all times and sleeping in a separate bedroom. Precautions vary according to the dose given.3,6,14-16
Initially, the patient may get worse after treatment, as preformed thyroid hormone is released. Patients with cardiovascular complications and those who are elderly are usually prescribed thionamides, beta-blockers, or corticosteroids to manage these effects. Thionamides are generally given prior to I¹³¹ treatment to stabilize hormone production. Thionamides are stopped 4 to 6 days before treatment and then restarted approximately 4 days after therapy. Corticosteroid treatment tends to reduce T3 and T4 concentrations following RAI treatment.1,5
Normal thyroid function can occur as early as 6 to 8 weeks after treatment in 50% to 75% of patients. However, hypothyroidism results in more than 90% of patients within the first year after treatment, with a 2% to 3% rate each year thereafter.1,3,9 Lifelong thyroid supplementation and monitoring are necessary after treatment. Initially, monitoring occurs monthly; once a euthyroid state is obtained, monitoring is reduced to every 6 to 12 months. Mild thyroid-gland tenderness and dysphagia may occur. Reports of carcinomas or congenital defects are mixed; some studies report an increased risk, while others suggest no association. One study, however, reported an increased risk in all-cause mortality, including mortality from cardiovascular and cerebrovascular disease.1,2
Adjunctive Medications: Beta-blockers, calcium channel blockers, and centrally acting sympatholytics are used to control symptoms of hyperthyroidism such as anxiety, palpitations, heat intolerance, sweating, and tremors.1-3 These medications do not decrease thyroid-stimulating antibodies (TSAb) or prevent thyroid storm; propranolol and nadolol have been shown to partially inhibit conversion of T4 to T3, but the therapeutic effect is small.1,3 First-line treatment usually begins with a beta-blocker, particularly propranolol. The initial starting dose ranges from 20 mg to 40 mg four times daily, but may be up to 240 mg to 480 mg/day. Higher doses typically are used in patients who are younger or who are in a severely toxic state, as they tend to clear the drug faster.1,4 As with all beta-blockers, these medications are contraindicated in patients with decompensated heart failure, sinus bradycardia, or spontaneous hypoglycemia and in those who are taking monoamine oxidase inhibitors or tricyclic antidepressants.
More common side effects include nausea, vomiting, bradycardia, insomnia, anxiety, masked signs of hypoglycemia, and impotence. Nonselective beta-blockers should be used with caution in patients suffering from asthma or bronchospastic disorders. Calcium channel blockers such as diltiazem and centrally acting sympatholytic agents such as clonidine are used if beta-blockers are contraindicated.1,17
Iodide was the first drug used in Graves’ disease that is able to block thyroid hormone release, decreasing thyroid size and vascularity as well as inhibiting hormone synthesis. Effects may be seen as early as 2 to 7 days, but concentrations of T4 and T3 are reduced for only a few weeks. Iodides are now used to prepare patients for surgery or after RAI treatment.1,4
Potassium iodide (SSKI) or Lugol’s solution is typically used. SSKI contains between 35 and 50 mg of iodide per drop, and Lugol’s solution contains approximately 6 mg per drop. SSKI is usually given as 3 to 10 drops daily in water or juice and is administered 7 to 14 days before surgery. However, if the patient is undergoing RAI, SSKI is administered 3 to 7 days after treatment in order not to interfere with RAI being taken up into the thyroid. The most common side effects are salivary-gland swelling, hypersensitivity reactions, metallic taste, sore teeth and gums, stomach upset, diarrhea, and gynecomastia.1,4
Lithium has been used in the past to inhibit the release of thyroid hormone. The exact mechanism of action is unknown, but lithium is believed to act similarly to iodide. With its transient effect and side-effect profile, lithium has limited usefulness in the treatment of Graves’ disease.4
## Treatment During Pregnancy
Hyperthyroidism is suspected during pregnancy when there is persistent tachycardia along with an inability to gain weight despite a good appetite. No increase in maternal morbidity and mortality is seen. However, around 20% of patients experience thyroid storm during the postpartum period if left untreated. Other side effects, such as low fetal birthweight, premature delivery, fetal loss, and pre-eclampsia, may occur if the disease is left untreated. Fetal and neonatal hyperthyroidism may occur owing to TSAb crossing the placenta.1,3
RAI is contraindicated during pregnancy, and surgery is considered a second-line alternative. The treatment of choice is antithyroid medication. In the past, MMI was not recommended during pregnancy owing to rare complications like fetal aplasia cutis and fetal GI problems. However, both PTU and MMI have been shown to cross the placenta and may interact with the developing fetus, especially at higher doses. Lower doses are recommended, with PTU still being the preferred drug during the first trimester.1,8 The typical PTU dose is 300 mg or less daily to prevent fetal goiter or hypothyroidism. Normally, PTU is tapered to 50 mg to 150 mg daily after the first 4 to 6 weeks of treatment. Free T4 is usually kept near the upper limit of normal during pregnancy. Some patients go into remission during the last trimester, as TSAb tend to fall. Postpartum hyperthyroidism occurs in approximately 10% of women, however.1,3
## Summary
Graves’ disease is an autoimmune disorder that affects millions of people worldwide. A patient who is experiencing weight loss, dermopathy, ophthalmopathy, and/or cardiovascular complications needs to be referred to his or her primary care physician for thyroid-function tests. Diagnosis lies in overt signs and symptoms in combination with abnormal laboratory tests. There are currently three main treatment strategies, with RAI preferred during relapse and more severe disease, and antithyroid medications preferred during pregnancy. It is important for patients to have lifelong thyroid-function monitoring, as levels vary with age, during pregnancy, and even with the different treatment options.
REFERENCES
Sherman SI, Talbert RL. Thyroid disorders. In: DiPiro JT, Talbert RL, Yee GC, et al, eds. Pharmacotherapy: A Pathophysiologic Approach. 7th ed. New York, NY: McGraw-Hill Medical; 2008:1243-1263. 2. Brent GA. Graves’ disease. N Engl J Med. 2008;358:2594-2605. 3. Hegedüs L. Treatment of Graves’ hyperthyroidism: evidence-based and emerging modalities. Endocrinol Metab Clin N Am. 2009;38:355-371. 4. Streetman DD, Khanderia U. Diagnosis and treatment of Graves disease. Ann Pharmacother. 5. Pearce EN. Diagnosis and management of thyrotoxicosis. BMJ. 2006;332:1369-1373. 6. Iagaru A, McDougall IR. Treatment of thyrotoxicosis. J Nucl Med. 2007;48:379-389. 7. MedlinePlus. TSH. www.nlm.nih.gov/medlineplus/ency/article/003684.htm. Accessed March 9, 2010. 8. Cooper DS. Antithyroid drugs. N Engl J Med. 2005;352:905-917. 9. Stålberg P, Svensson A, Hessman O, et al. Surgical treatment of Graves’ disease: evidence-based approach. World J Surg. 2008;32:1269-1277. 10. Tapazole (methimazole) package insert. Bristol, TN: King Pharmaceuticals Inc; February 2009. 11. Propylthiouracil package insert. Pearl River, NY: Lederle; 2000. 12. FDA. Drug Safety and Availability. Information for healthcare professionals—propylthiouracil-induced liver failure. www.fda.gov/Drugs/DrugSafety/PostmarketDrugSafetyInformationforPatientsandProviders/DrugSafetyInformationforHeathcareProfessionals/ucm162701.htm. Accessed March 9, 2010. 13. Cooper DS, Rivkees SA. Putting propylthiouracil in perspective. J Clin Endocrinol Metab. 14. Kowalsky RJ, Falen SW. Radiopharmaceuticals in Nuclear Pharmacy and Nuclear Medicine. 2nd ed. Washington, DC: American Pharmacists Association; 2004. 15. Larose JH. Radionuclide therapy. In: Early PJ, Sodee DB, eds. Principles and Practice of Nuclear Medicine. 2nd ed. St. Louis, MO: Mosby; 1995:752. 16. Culver CM, Dworkin HJ. Radiation safety considerations for post-iodine-131 hyperthyroid therapy. J Nucl Med. 1991;32:169-173. 17. Propranolol hydrochloride package insert. Corona, CA: Watson Laboratories; June 2008.
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