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The failure of diamond on a reflecting stationary set
1. It is shown that the failure of {lozenge, open}S, for a set S ⊆ א [ω]+1 that reflects stationarily often, is consistent with GCH and APא [ω], relative to the existence of a supercompact cardinal.
By a theorem of Shelah, GCH and □* λ entails {lozenge, open}S for any S ⊆ λ+ that reflects stationarily often. 2. We establish the consistency of existence of a stationary subset of [א [ω+1]] ^ω that
cannot be thinned out to a stationary set on which the supfunction is injective. This answers a question of König, Larson and Yoshinobu in the negative. 3. We prove that the failure of a diamond-like
principle introduced by Džamonja and Shelah is equivalent to the failure of Shelah's strong hypothesis.
• Approachability
• Diamond
• Reflection
• Sap
• Square
• Very good scale
Dive into the research topics of 'The failure of diamond on a reflecting stationary set'. Together they form a unique fingerprint.
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Advanced Topics in Scientific Computing
This is an archived course. The content might be broken.
V5E3: Advanced Topics in Scientific Computing
The numerical simulation of physical systems leads to several challenges when the simulation is supposed to be realistic: (a) The shape of the domain is usually three-dimensional and not just a
simple cube, (b) the discretization with high resolution lead to systems of equations with a lot of unknowns (say 1e10), and (c) the resolution may need to be finer in some areas of the domain than
in others (d) and the areas of highest resolution may move with time. Examples are the stress analysis of a bridge or skyscraper and the tracking of flames or shocks in gases, as well as geophysical
simulations (of mantle convection or earthquakes).
In this lecture we will assume that the finite element or finite volume method to solve a basic partial differential equation (such as Poisson's) is understood on a cubic domain. We will revisit the
basics and then move on to the topics of mesh generation (addressing a), parallelization and scalable solvers (b) and adaptivity (c, d).
This lecture will expand the students' knowledge on computational geometry, high performance computing and state-of-the art techniques for adaptive mesh refinement. We may occasionally discuss
selected research papers as part of the lecture. A useful book on mesh generation is "Grid Generation Methods" by V.D. Liseikin.
The lectures Wissenschaftliches Rechnen I (V3E1/F4E1) as well as one class out of Wissenschaftliches Rechnen II (V3E2/F4E2), Numerical Simulation (V4E1) or Numerical Algorithms (V4E2) are
prerequisites. We may discuss the requirements further in the first lecture.
The lecture takes place on Mondays and Thursdays at 14 Uhr c.t. in We6 6.020.
There are no exercises outside of the lecture slots. We may sometimes convert a lecture slot to exercises in the computer room. There will be an oral exam between July 31st and August 3rd and on
September 27th and 28th.
Please read the coding guidelines. For those who have begun working on a programming exercise, please email me a header file and at least a source file with a dummy function that does nothing but
compiles cleanly.
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Sage for Mathematica Users
This page is modeled on the http://www.scipy.org/NumPy_for_Matlab_Users
SAGE has many of the capabilities of Mathematica, and many additional ones (e.g. wiki-creating software and a 3D raytracer). Some features of SAGE have been inspired by Mathematica, but overall the
syntax and structure of SAGE are quite different. One of the main influences on SAGE is the use of the language Python.
This page is intended to help users familiar with Mathematica migrate to SAGE more easily.
Key Differences
Indexing: Lists in Mathematica are indexed starting from 1. In SAGE, as in Python, indices start at 0. Also, where Mathematica accepts a list of indices, in SAGE you can construct sub-lists using
"slice" operations. For example, if we have a list of numbers, num_list = [0,1,2,3,4], then numlist[1:3] would return the list [1,2].
For a comparison of graph theory functionality between SAGE and the Mathematica Combinatorica package, see the CombinatoricaCompare page.
Sage and Python Quickstart for Mathematica users
This is not a proper introduction to Python, but a list of examples that Mathematica users will need to figure out how to do if they want to use Sage.
Basic functionality
Declaring variables
Mathematica assumes that all otherwise unknown symbols are algebraic quantities. Python and Sage don't; they are declared as follows:
sage: var('x,y,a,b,c')
(x, y, a, b, c)
sage: y == a*x^2 + b*x + c
y == a*x^2 + b*x + c
It is also possible to declare with spaces between variables:
sage: var('x y a b c')
(x, y, a, b, c)
sage: y == a*x^2 + b*x + c
y == a*x^2 + b*x + c
Implicit multiplication
sage: var('x,y,a,b,c')
(x, y, a, b, c)
sage: implicit_multiplication(True)
sage: y == a x^2 + b x + c
y == a*x^2 + b*x + c
Note that a space need not be used when there is a numerical coefficient for a variable:
sage: var('x,y')
(x, y)
sage: implicit_multiplication(True)
sage: y == 3x
y == 3*x
Procedural programming
Mathematica: Table[f[i], {i, 1, 10}]
sage: [f(i) for i in [1..10]]
[f(1), f(2), f(3), f(4), f(5), f(6), f(7), f(8), f(9), f(10)]
Advanced Mathematica syntax
Mapping functions across a list
From a list called data, create a new list where a function f is applied to each element of data.
Mathematica: f /@ data
Python: [f(d) for d in data]
Unlike in Mathematica, this for d in data cannot be applied to an arbitrary expression data. So [f(d) for d in g(x,y,z)] is not possible (this is not entirely true, all g(x,y,z) needs to do is to
return something iterable, malb).
Mapping pure functions across a list
(Replacing elements that are less than zero with zero.)
Mathematica: data /. _?(# < 0&) -> 0
Python: [(0 if d < 0 else d) for d in data]
• Mathematica: Timing[command]
• SageMath: timeit('command')
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Python Program to Transpose a Matrix | Python Transpose
There are several operations that we can perform on matrices. The transposition of a matrix is one such operation. The resulting matrix has interchanged rows and columns of the original one.
Here in this article, we have provided a python Program to transpose a matrix with source code/ write a program to find the transpose of a matrix in python.
Matrix and Transpose
A matrix is a 2-dimensional array, which is formed by using columns and rows. Matrices are very useful in scientific computation. In Python, to represent a matrix, we use a list of lists . To
transpose a matrix, we switch its row values with the column values.
For instance,
is a matrix, and for transposing it, we switch its row values with its column values i.e.:
• Python Input/output
• Python Loop
• Python List comprehension
Steps for Transposing a Matrix
• Ask the user to enter the number of rows and columns for a matrix.
• Then using the list comprehension and input() function, ask the user to enter the elements in the matrix.
• Create a result matrix, whose dimensions are the invert of the user-entered matrix dimensions. For instance, if the matrix has 2 rows and 3 columns then the resulting matrix (transpose) should
have 3 rows and 2 columns.
• Using the nested for loop, we will change the row values with the column values then store them into the resulting matrix.
Python Program to Transpose a Matrix with Source Code/Python Transpose
Source Code
rows = int(input("Enter the Number of rows : " ))
column = int(input("Enter the Number of Columns: "))
print("Enter the elements of Matrix:")
matrix= [[int(input()) for i in range(column)] for i in range(rows)]
print("-------Your Matrix is---------")
for n in matrix:
#result matrix of column*row dimension
result =[[0 for i in range(rows)] for j in range(column)]
#transpose the matrix
for r in range(rows):
for c in range(column):
#here we are grabbing the row data of matrix and putting it in the column on the result
result[c][r] = matrix[r][c]
print("Transpose matrix is: ")
for r in result:
Output 1:
Enter the Number of rows : 3
Enter the Number of Columns: 3
Enter the elements of Matrix:
-------Your Matrix is---------
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
Transpose matrix is:
[1, 4, 7]
[2, 5, 8]
[3, 6, 9]
Output 2
Enter the Number of rows : 2
Enter the Number of Columns: 4
Enter the elements of Matrix:
-------Your Matrix is---------
[1, 2, 3, 4]
[5, 6, 7, 8]
Transpose matrix is:
[1, 5]
[2, 6]
[3, 7]
[4, 8]
That was a simple Python program to generate the transpose of a matrix. It is just one of the many operations that we can perform on matrices. These include addition, subtraction, and multiplication.
Matrices are very useful in programming, so building a good grasp of them is important.
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Solving Systems Of Polynomial Equations Worksheet - Equations Worksheets
Solving Systems Of Polynomial Equations Worksheet
Solving Systems Of Polynomial Equations Worksheet – The goal of Expressions and Equations Worksheets is to assist your child in learning more efficiently and effectively. The worksheets include
interactive exercises as well as problems dependent on the sequence that operations are carried out. With these worksheets, kids are able to grasp basic and more complex concepts in a brief amount of
amount of time. These PDF resources are completely free to download and may be used by your child to practise maths equations. These resources are beneficial to students in the 5th-8th grades.
Get Free Solving Systems Of Polynomial Equations Worksheet
The worksheets listed here are designed for students in the 5th-8th grade. These two-step word problem are constructed using fractions and decimals. Each worksheet contains ten problems. They can be
found at any online or print resource. These worksheets are a great way to practice rearranging equations. Alongside practicing restructuring equations, they can also aid students in understanding
the characteristics of equality and reverse operations.
These worksheets can be utilized by fifth- and eighth grade students. They are ideal for students who have difficulty calculating percentages. There are three kinds of problems you can choose from.
You have the choice to either work on single-step problems which contain whole numbers or decimal numbers or word-based solutions for fractions as well as decimals. Each page will contain 10
equations. The Equations Worksheets are suggested for students in the 5th through 8th grade.
These worksheets are a great way for practicing fraction calculations and other concepts in algebra. Some of the worksheets let you to choose from three different kinds of problems. You can select
one that is numerical, word-based or a combination of both. It is vital to pick the type of problem, as each one will be different. Each page contains ten problems which makes them an excellent aid
for students who are in 5th-8th grade.
These worksheets aid students in understanding the relationship between variables and numbers. These worksheets provide students with the chance to practice solving polynomial expressions or solving
equations, as well as getting familiar with how to use them in daily life. These worksheets are a great opportunity to gain knowledge about equations and formulas. These worksheets will educate you
about the various types of mathematical problems along with the different symbols used to express them.
These worksheets are beneficial for students in their beginning grades. The worksheets will assist them to learn how to graph and solve equations. The worksheets are great for practice with
polynomial variables. They can also help you understand how to factor them and simplify them. There are many worksheets that can be used to teach kids about equations. The best way to get started
learning about equations is to complete the work yourself.
There are a variety of worksheets to teach quadratic equations. Each level has its own worksheet. These worksheets can be used to practice solving problems up to the fourth level. When you’ve reached
an amount of work and are ready to work on solving different kinds of equations. You can then take on the same problems. You can, for example solve the same problem in a more extended form.
Gallery of Solving Systems Of Polynomial Equations Worksheet
Solving Polynomial Equations Worksheets With Answer Key
Solving Polynomial Equations Worksheets With Answer Key
Solving Polynomial Equations Worksheets With Answer Key
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Creating a Simple Naive Bayes Predictive Model in Power BI - Iteration Insights
“Happy Hunger Games! And may the odds be ever in your favor.”
Suzanne Collins, The Hunger Games
Power BI is a very capable tool in descriptive and diagnostic analytics. It is easy to create a report displaying the current or past state of a business with the ability to explore the reasons
behind a given outcome. The next levels of analytics are the predictive and prescriptive levels which try to predict what will happen and what you should do about it.
Basic Power BI does not have any support for these levels, but there are several ways to add them. You can use R and Python in Power BI to extend Power BI and add machine learning. Power BI Premium
includes Auto Machine Learning, but it can be financially prohibitive.
If you do not know R or Python and do not have access to Premium capacity, there are ways to do some simple ‘machine learning’ inside Power BI using only DAX. This can be done by creating a Naive
Bayes predictive model in Power BI.
What is a Naive Bayes Predictive Model?
Naive Bayes is a statistical method for predicting the probability of an event occurring given that some other event(s) has also occurred. Below are formulas displaying the math we will be using.
The first formula provides the variables as they are written in plain English. Assume there are two events, A and B. We want to know the probability of event A happening if event B also happens. To
calculate this, you can use historical or training data.
First, we calculate the probability of event A happening, irrespective of any other events. Multiply that by the probability of event B happening when event A also happens. Then divide by the
probability of event B happening.
This is summarized in the middle equation using more standard mathematical notation.
The third equation expands the middle one for two dependent events. Here, the probability of event A is dependent on two events, B1 and B2. This is the form of the equation we will use below. We will
have to calculate both the probability of each B1 and B2 happening if A happens and the probability of B1 and B2 happening together.
Data and Setup
For this demonstration, I am going to use the Titanic dataset. This is a dataset of all the passengers on the Titanic. It includes information about the passengers and whether they survived or died
from the sinking.
This is a common dataset for learning machine learning. Partly because it is small, but also because the outcome is easy to predict using only a passenger’s sex and class. For instance, over 95% of
first-class females survived, whereas only 15% of third-class males survived.
I randomly classified 80% of the passengers as training data. In machine learning, you use training data to assign values to the parameters of the equation. In our case, the training data is used to
calculate the values of the variables on the right-hand side of the above equations. I will use those values to predict the survival of passengers in the test data.
In a real-world application, the training data would be all the historical data in the report. As a result, a report like this would have data for descriptive and diagnostic analysis, while giving
some predictive capabilities.
All of the data is in one table with an ‘inTrain’ column to parse out the passengers in the training set. Other important columns are the class, sex, and survived columns. See the image below for a
sample of the data.
Building Your Model Using Measures
In this demonstration, I am going to build up the model using measures. There will be measures for each of the variables in the equation.
Written in plain English below is an example of a calculation we need to do to predict the survival of a 2nd class male:
From the training data, we need measures that calculate:
• The total probability any passenger survived
• From the passengers who survived
□ What fraction were 2nd class
□ What fraction were male
• The probability a passenger was both male and 2nd class
To get started, I created a couple of DAX measures to calculate the number of passengers in the training set, and the number of those who survived.
Passenger Count =
COUNTROWS ( 'titanic' ),
ALL ( 'titanic' ),
'titanic'[inTrain] = "Yes"
Survived Count =
COUNTROWS ( 'titanic' ),
ALL ( 'titanic' ),
'titanic'[inTrain] = "Yes",
'titanic'[Survived] = 1
The first measure is the probability of any passenger surviving. Which is just the number of passengers who survived, divided by the number of passengers.
pSurived =
DIVIDE ( [Survived Count], [Passenger Count] ) + 0
Then I created a measure to calculate the odds a passenger is part of a certain class, given they survive. This measure determines a passenger’s class, then counts the number of passengers who were
also in that class and survived. Finally, dividing that number by the total number of survivors.
pclass given survival =
VAR class =
SELECTEDVALUE ( 'titanic'[Class] )
VAR inClass =
CALCULATE (
COUNTROWS ( 'titanic' ),
ALL ( 'titanic' ),
'titanic'[inTrain] = "Yes",
'titanic'[Survived] = 1,
'titanic'[Class] = class
DIVIDE ( inClass, [Survived Count] ) + 0
Then the odds of being a certain sex and surviving. The logic is similar to the above measure, just using sex instead of class.
psex given survival =
VAR sex =
SELECTEDVALUE ( 'titanic'[Sex] )
VAR isSex =
CALCULATE (
COUNTROWS ( 'titanic' ),
ALL ( 'titanic' ),
'titanic'[inTrain] = "Yes",
'titanic'[Survived] = 1,
'titanic'[Sex] = sex
DIVIDE ( isSex, [Survived Count] ) + 0
In the denominator, I calculate the total probability a passenger is both a certain sex and class.
psex and class =
VAR sex =
SELECTEDVALUE ( 'titanic'[Sex] )
VAR class =
SELECTEDVALUE ( 'titanic'[Class] )
VAR inSexClass =
CALCULATE (
COUNTROWS ( 'titanic' ),
ALL ( 'titanic' ),
'titanic'[inTrain] = "Yes",
'titanic'[Sex] = sex,
'titanic'[Class] = class
DIVIDE ( inSexClass, [Passenger Count] ) + 0
All of these measures then got put together into one final measure to give the probability of survival.
psurvival given sex and class =
DIVIDE (
[pSurived] * [pSex given Survival] * [pClass given Survival],
[pSex and Class]
Validating Your Model
Once you have a model, the next step is to validate it to see how effective it is. For this reason, I left 20% of the passengers out of the training set. These are in the test set and will be used
for validation.
The output of the model is a probability with values from 0 to 1. Higher values predict survival and lower values predict death. However, this value is relative to the dataset. We need to determine
what point in this value range most accurately represents the threshold between a high and low probability of survival by testing it.
To make a decision, we need to assume a dividing, or threshold, value in the probability range. Probabilities above that threshold predict survival, while those below predict death.
To begin, I will make a table, like a ‘what if’ parameter, with a range of possible threshold values.
Thresholds =
GENERATESERIES ( 0, 1, 0.05 )
Then I created a measure that will use the values of this table to evaluate the accuracy of a prediction for a given threshold value. This measure will iterate over the test set and evaluate the
model for each passenger. If the passenger is predicted to survive and did so, then it is a true positive.
Similarly, it classifies passengers who died and had a probability below the threshold as true negatives. The opposite of a true positive and negative is a false positive or negative. This is when
the model is predicting the opposite of what happened.
Correct % =
VAR threshold =
SELECTEDVALUE ( 'Thresholds'[Value] )
VAR test =
CALCULATETABLE ( 'titanic', 'titanic'[inTrain] = "No" )
AVERAGEX (
VAR pSurvival = [pSurvival given Sex and Class]
VAR survived = 'titanic'[Survived]
SWITCH (
TRUE (),
AND ( pSurvival >= threshold, survived = 1 ), 1,
//True Positives
AND ( pSurvival < threshold, survived = 0 ), 1,
//True Negatives
0 //False Positive or Negative
We can then use this measure in a line chart.
As we can see on the graph, if we assume a threshold of about 0.7, we can accurately predict the outcome of over 80% of the passengers. So this model does a pretty good job of predicting a
passenger’s mortality.
In a real-world scenario, you would have to determine what level of accuracy is acceptable. No model will ever be 100% correct, but that is often not necessary to be valuable.
In this blog, I have shown how to implement a simple Naive Bayes predictive model in Power BI using nothing but DAX. I hope this will be helpful to you with your reporting. This type of approach can
be very useful to quickly create simple predictive models in Power BI and take your reporting to the next level.
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[CVX] Convex Finance Price Prediction 2024, 2025, 2030, 2040, and 2050
Convex Finance Price Prediction 2024, 2025, 2030, 2040, and 2050
Are you one of the cryptocurrency investors who want to know the Convex Finance Price Prediction 2024, 2025, 2030, 2040, and 2050, and want to know the future forecast of the prices? We will analyze
the price of the coin and provide a deep analysis of how you can invest in the purchase of this coin.
The price prediction is necessary for you because this will help you to make a future decision on whether you should invest in this coin or not. The experts give the analysis as per the current price
situation. You only have to decide whether you should buy this coin or not after understanding the Convex Finance Price Prediction.
Overview of Convex Coin
Coin Name Convex Finance
Coin Symbol CVX
Official Website convexfinance.com
All Time High $50.02 (On Feb 2, 2022)
Total Supply 99,664,070 CVX
Trading Exchanges Binance, Bybit, BYDFi, OKX, and DigiFinex.
Wallets Metamask, Trust Wallet
List of Convex Finance (CVX) Price Prediction 2024, 2025, 2030, 2040, 2050
Year Minimum Price Average Price Maximum Price
2024 $2.76 $2.84 $3.56
2025 $3.92 $4.25 $4.78
2026 $5.56 $5.87 $6.32
2027 $8.65 $9.83 $10.43
2028 $12.72 $13.24 $14.26
2029 $16.92 $18.21 $20.85
2030 $26.24 $27.62 $30.12
2031 $38.34 $40.54 $45.41
2032 $57.20 $60.14 $68.49
2035 $102.57 $114.92 $201.54
2040 $1,246.02 $1,476.95 $1,712.16
2050 $1,976.24 $2,126.81 $2,242.06
JasmyCoin Price Prediction | IO.net Price Prediction
• The current rank of the coin is #196 in the entire crypto ecosystem.
• The prices of the changes will be changed by 0.76%.
• The table will help you to make your investment smarter.
Convex Finance Price Prediction 2024, 2025, 2030, 2040, 2050
By applying the deep AI-assisted technical analysis at Cryptomeverse you will find the price prediction of the Convex Finance coin. As per the historical data we do our eBay to provide you with
accurate upcoming price updates. Based on multiple parameters such as ast price, Convex Finance market cap, Convex Finance volume, and a few more experts predict the prices of CVX coin.
• By the year the coin will reach its value between $2.76 to $3.56.
• In 2030, the coin will cost a minimum of $ 26.24 and a maximum of $30.12.
• The long-term prediction of the coin is that in the year 2050, the coin will cost
$1,976.24 lowest and $2,242.06 highest.
Convex Finance (CVX) Price Prediction 2024
If you are willing to purchase this coin now then you have a great time. Now the average price of the coin as per the prediction is $2.84. If for any reason crypto market faces downtime then the
coin value will be $2.76. But somehow the market goes up then the price value of the coin will be a maximum of $3.56. It is high time to invest in this coin so, get up and make your trade with the
convex coin.
Convex Finance Price Prediction 2025
Next year 2025 will bring a new change in the prices of the coin. The coin is expecting a growth of 50% by the end of the year and a maximum this will cost $4.78. The minimum value of the Convex coin
will be $3.92 if the market goes down as per the price prediction.
The average value of the coin will be $4.25. If you are investing in this coin purchase now then you have a great time to earn profit in the future.
Convex Finance Price Prediction 2030
The long-term cryptocurrency price prediction is a tough task because the crypto market is violated in nature and goes up and down at times. So, as per the current price, our experts forecast the
prices of the coin in the year 2030. The prediction shows that the market price will increase and the average value of the coin will be $27.62. Somehow, the market is bullish then the price increases
its value and maximum cost $30.12. If the market falls in 2030, the price value of the coin will reduce and this will cost $26.24.
Convex Finance (CVX) Price Prediction 2040
2040 can be a game-changing year for the Convex coin. This year the price of the coin will bring a new change if it’s run as per the current price prediction of convex coin. The coin will cost an
average of $1,476.95. If the market runs in the same direction then the price will increase more and the maximum value of the coin will be $1,712.16. If for any reason the market goes down then the
minimum price value of the coin will be $1,246.02.
We always recommend you to search a lot before investing in the purchase of any coin. Because crypto is a highly unpredictable currency.
Convex Finance Price Prediction 2050
The long-term price prediction of digital currency is quite difficult for the experts also. But, as per the historical data, our experts created a report, and the prediction of the reports shows that
the price of the coin will increase in the year 2050, and on average this will cost $2,126.81. If the market drops down then the experts show that the coin will drop the value also just like other
coins and reach its minimum value which is $1,976.24. If the market goes up then the maximum price of the coin will be $2,242.06. if the market continues in the same direction then the prices will
cross the maximum predicted price in the year 2050.
How Much Does The Cost Will Cost After 5 Years?
After 5 years the coin will reach the maximum value of $20.59 and the minimum it will be $16.86. The average value will be $17.48. As per the coin demand and supply basis, the prices of the coin will
Is Convex A Good Investment?
If the coin is stored then the price of the value will increase. So in the future, the coin will be a good investment option for the investors. Please note that investing in any coin includes risk
also. So before making any kind of investment, we recommend you do better research.
What Is The Future of CONVEX Coins?
The future of CONVEX COIN or other coins depends on the performance of the crypto industry. When you are investing in the purchase of this coin then make sure that you are using the right strategy.
Is It Worth To Buy Convex Coin?
If you are looking for a long-term investment purpose then the coin is the right choice for you. Just you need to build a good strategy during the time of purchase.
How Can I Buy Convex Finance Coin?
To invest in the coin purchase you only have to create an account on the crypto exchange platforms and you can simply find and invest in it. The platforms are:
• Binance
• Bybit
• BYDFi
• OKX
• DigiFinex
Disclaimer: The information provided in this post is for informational purposes only and should not be construed as financial or investment advice. Cryptocurrency investments are highly volatile and subject to market risks. Always conduct your own research and consult with a financial advisor before making any investment decisions. The predictions made here are speculative and may not reflect actual market movements. We are not responsible for any financial losses incurred based on this information.
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Numbers In Ordinal - OrdinalNumbers.com
Numbers In Ordinal
Numbers In Ordinal – An endless number of sets can be easily enumerated using ordinal numerals as a tool. They can also be utilized to generalize ordinal numbers.But before you use these numbers, you
need to understand the reasons why they exist and how they work.
The ordinal numbers are one of the most fundamental ideas in mathematics. It is a number that indicates where an object is in a list of objects. Ordinal numbers are typically an integer between one
and twenty. While ordinal numbers can serve many purposes, they are often used to represent the order in which items are placed in a list.
It’s possible to show ordinal number using numbers or words. Charts, charts, and even charts. They can also be used to show how a group or pieces are arranged.
The majority of ordinal numbers fall into one of these two categories. Transfinite ordinals are represented by lowercase Greek letters, while finite ordinals are represented by Arabic numbers.
Based on the Axiom of Choice, every set that is well-organized should have at least one ordinal. For instance, the very first student to complete an entire class will receive the highest grade. The
winner of the contest was the student who had the highest grade.
Combinational ordinal number
Multiple-digit numbers are also known as compound ordinal numbers. They are generated by multiplying an ordinal numbers by its final number. They are typically used for dating and ranking purposes.
They don’t have an unique ending like cardinal numbers.
To show the order that elements are placed in the collection ordinal numbers are utilized. They can also be used to identify the items within collections. There are two types of ordinal numbers:
regular and suppletive.
Regular ordinals can be created by prefixing the cardinal number by an -u suffix. The number is then typed into a word. A hyphen then added. There are numerous suffixes to choose from.
Suppletive ordinals may be created by prefixing words containing -u.e. or –ie. The suffix, which can be used to count, is a bit longer than the standard.
Limit of Ordinal
Limit ordinal values that are not zero are ordinal numbers. Limit ordinal numbers come with the drawback that there is no maximum element that they can use. They can be constructed by joining sets
that aren’t empty and do not have any limit elements.
Limited ordinal numbers are used in transfinite definitions of Recursion. Each infinite number of cardinals, according to the von Neumann model, can also be considered an ordinal limit.
Limit ordinal numbers are the sum of the ordinals that are below it. Limit ordinal figures are calculated using arithmetic. They can also be expressed using an order of natural numbers.
Data is organized by ordinal number. They explain an object’s numerical location. They are often employed in set theory and math. Despite sharing the same design, they do not belong in the same
classification as natural number.
The von Neumann model uses a well-ordered set. Consider that fy fy is a subfunction of a function g’ that is defined as a singular function. If the subfunction fy’s is (i, ii) and g fulfills the
criteria the g function is a limit ordinal.
A limit ordinal of the type Church-Kleene is also known as the Church-Kleene or ordinal. The Church-Kleene ordinal defines an appropriate limit as a well arranged collection of the smaller ordinals.
It also contains a nonzero ordinal.
Stories that include examples of ordinal numbers
Ordinal numbers are commonly utilized to represent the hierarchy of entities and objects. They are crucial to organize, count, and ranking purposes. They can also be utilized to show the order of
things and to indicate the position of objects.
The letter “th” is commonly used to denote the ordinal number. On occasion, though the letter “nd” is substituted. The titles of books usually contain ordinal numbers.
Ordinal numbers can be stated in words, even though they are often utilized in list format. They can also appear in the form of acronyms and numbers. In comparison, they are simpler to comprehend as
compared to cardinal numbers.
Three distinct types of ordinal numbers are available. Through practice, games and other activities, you may be able to learn more about the different kinds of numbers. Educating yourself on these
concepts is a crucial aspect of developing your arithmetic skills. Try utilizing a coloring exercise as a simple and entertaining approach to improving. A handy marking sheet is a great way to record
your results.
Gallery of Numbers In Ordinal
Ejercicio Online De Ordinal Numbers Para Cuarto De Primaria
Ordinal Numbers English Study Here
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Now showing items 31-40 of 25681
A DPG method for linear quadratic optimal control problems
(Elsevier Ltd, 2024)
The DPG method with optimal test functions for solving linear quadratic optimal control problems with control constraints is studied. We prove existence of a unique optimal solution of the nonlinear
discrete problem and ...
A note on the sufficiency of the maximum principle for infinite horizon optimal control problems
This work is devoted to introduce a new sufficient optimality condition for infinite horizon optimal control problems. It is shown that normal extremal processes are optimal under this new condition,
termed as maximum-pr ...
Optimal control problem for deflection plate with crack
(Springer/plenum Publishers, 2012-07-01)
We consider a control problem where the state variable is defined as the solution of a variational inequality. This system describes the vertical displacement of points of a thin plate with the
presence of crack inside ...
Optimal control problem for deflection plate with crack
(Springer/plenum Publishers, 2012-07-01)
We consider a control problem where the state variable is defined as the solution of a variational inequality. This system describes the vertical displacement of points of a thin plate with the
presence of crack inside ...
Penalty-based nonlinear solver for optimal reactive power dispatch with discrete controls
The optimal reactive dispatch problem is a nonlinear programming problem containing continuous and discrete control variables. Owing to the difficulty caused by discrete variables, this problem is
usually solved assuming ...
Penalty-based nonlinear solver for optimal reactive power dispatch with discrete controls
The optimal reactive dispatch problem is a nonlinear programming problem containing continuous and discrete control variables. Owing to the difficulty caused by discrete variables, this problem is
usually solved assuming ...
Optimal linear and nonlinear control design for chaotic systems
In this work, the linear and nonlinear feedback control techniques for chaotic systems were been considered. The optimal nonlinear control design problem has been resolved by using Dynamic
Programming that reduced this ...
Optimal control of bilinear systems in a complex space setting
(Elsevier B.V., 2017)
In this paper we discuss optimality conditions for optimal control problems with a semigroup structure. As an application we detail the case when the state equation is the Schrödinger one, with
pointwise constraints on the ...
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Simplify a complex equation of a circle
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Luyện đề: Preface to ‘How the other half thinks: Adventures in mathematical reasoning’ IELTS READING
II. Luyện đề Preface to ‘How the other half thinks: Adventures in mathematical reasoning’ IELTS READING
Preface to ‘How the other half thinks: Adventures in mathematical reasoning’
A. Occasionally, in some difficult musical compositions, there are beautiful, but easy parts - parts so simple a beginner could play them. So it is with mathematics as well. There are some
discoveries in advanced mathematics that do not depend on specialized knowledge, not even on algebra, geometry, or trigonometry. Instead, they may involve, at most, a little arithmetic, such as ‘the
sum of two odd numbers is even’, and common sense. Each of the eight chapters in this book illustrates this phenomenon. Anyone can understand every step in the reasoning. The thinking in each chapter
uses at most only elementary arithmetic, and sometimes not even that. Thus all readers will have the chance to participate in a mathematical experience, to appreciate the beauty of mathematics, and
to become familiar with its logical, yet intuitive, style of thinking.
B. One of my purposes in writing this book is to give readers who haven’t had the opportunity to see and enjoy real mathematics the chance to appreciate the mathematical way of thinking. I want to
reveal not only some of the fascinating discoveries, but, more importantly, the reasoning behind them. In that respect, this book differs from most books on mathematics written for the general
public. Some present the lives of colorful mathematicians. Others describe important applications of mathematics. Yet others go into mathematical procedures, but assume that the reader is adept in
using algebra.
C. I hope this book will help bridge that notorious gap that separates the two cultures: the humanities and the sciences, or should I say the right brain (intuitive) and the left brain (analytical,
numerical). As the chapters will illustrate, mathematics is not restricted to the analytical and numerical; intuition plays a significant role. The alleged gap can be narrowed or completely overcome
by anyone, in part because each of us is far from using the full capacity of either side of the brain. To illustrate our human potential, I cite a structural engineer who is an artist, an electrical
engineer who is an opera singer, an opera singer who published mathematical research, and a mathematician who publishes short stories.
D. Other scientists have written books to explain their fields to non-scientists, but have necessarily had to omit the mathematics, although it provides the foundation of their theories. The reader
must remain a tantalized spectator rather than an involved participant, since the appropriate language for describing the details in much of science is mathematics, whether the subject is expanding
universe, subatomic particles, or chromosomes. Though the broad.outline of a scientific theory can be sketched intuitively, when a part of the physical universe is finally understood, its description
often looks like a page in a mathematics text.
E. Still, the non-mathematical reader can go far in understanding mathematical reasoning. This book presents the details that illustrate the mathematical style of thinking, which involves sustained,
step-by-step analysis, experiments, and insights. You will turn these pages much more slowly than when reading a novel or a newspaper. It may help to have a pencil and paper ready to check claims and
carry out experiments.
F. As I wrote, I kept in mind two types of readers: those who enjoyed mathematics until they were turned off by an unpleasant episode, usually around fifth grade, and mathematics aficionados, who
will find much that is new throughout the book. This book also serves readers who simply want to sharpen their analytical skills. Many careers, such as law and medicine, require extended, precise
analysis. Each chapter offers practice in following a sustained and closely argued line of thought. That mathematics can develop this skill is shown by these two testimonials:
G. A physician wrote, The discipline of analytical thought processes [in mathematics] prepared me extremely well for medical school. In medicine one is faced with a problem which must be thoroughly
analyzed before a solution can be found. The process is similar to doing mathematics.’
A lawyer made the same point, “Although I had no background in law - not even one political science course — I did well at one of the best law schools. I attribute much of my success there to having
learned, through the study of mathematics, and, in particular, theorems, how to analyze complicated principles. Lawyers who have studied mathematics can master the legal principles in a way that most
others cannot.’
Questions 27-34
Reading Passage 3 has seven sections, A-G.
Which section contains the following information?
Write the correct letter, A— G, in boxes 27 — 34 on your answer sheet.
NB. You may use any letter more than once.
27. a reference to books that assume a lack of mathematical knowledge
28. the way in which this is not a typical book about mathematics
29. personal examples of being helped by mathematics
30. examples of people who each had abilities that seemed incompatible
31. mention of different focuses of books about mathematics
32. a contrast between reading this book and reading other kinds of publication
33. a claim that the whole of the book is accessible to everybody
34. a reference to different categories of intended readers of this book
Questions 35-40
Complete the sentences below. Choose ONE WORD ONLY from the passage for each answer. Write your answers in boxes 35- 40 on your answer sheet.
35. Some areas of both music and mathematics are suitable for someone who is a ....................
36. It is sometimes possible to understand advanced mathematics using no more than a limited knowledge of .............
37. The writer intends to show that mathematics requires .................... thinking, as well as analytical skills.
38. Some books written by .................... have had to leave out the mathematics that is central to their theories.
39. The writer advises non-mathematical readers to perform .................... while reading
40. A lawyer found that studying .................... helped even more than other areas of mathematics in the study of law.
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Lien iCal public : https://calendar.google.com/calendar/ical/proval%40lri.fr/public/basic.ics
Séminaires antérieurs
Speaker: Chana Weil-Kennedy, Technical University Munich
Tuesday, 19 December 2023, 14:00, 1Z71
Abstract: My research project focuses on parameterized verification of distributed systems, where the parameter is often the number of agents present in the system. During my thesis with Javier
Esparza in Munich, I studied this kind of problem for subclasses of Petri nets with application to population protocols, but also for systems communicating by broadcast or by shared register.
Currently in my postdoc I am continuing the study of these systems with "simple" modes of communication, and I'm also starting to look at language inclusion problems. In the future, I want to
continue to analyze distributed systems for which the parameterized problems are not immediately undecidable, by broadening the range of formal method techniques used.
Abstract: The truth semantics of linear logic (i.e. phase semantics) is often overlooked despite having a wide range of applications and deep connections with several denotational semantics. In phase
semantics, one is concerned about the provability of formulas rather than the contents of their proofs (or refutations). Linear logic equipped with the least and greatest fixpoint operators (\muMALL)
has been an active field of research for the past one and a half decades. Various proof systems are known viz. finitary and non-wellfounded, based on explicit and implicit (co)induction respectively.
In this talk, I present an extension of the phase semantics of multiplicative additive linear logic (a.k.a. \MALL) to \muMALL with explicit (co)induction (i.e. \muMALLind). Then I introduce a
Tait-style system for \muMALL where proofs are wellfounded but potentially infinitely branching. We will see its phase semantics and the fact that it does not have the finite model property. This
presentation is based on joint work with Abhishek De and Alexis Saurin.
Speaker: Alex Rabinovich, Tel Aviv University
Tuesday, 14 November 2023, 14:00, Room 1Z56
Abstract: Church’s Problem asks for the construction of a procedure which, given a logical specification φ(I,O) between input ω-strings I and output ω-strings O, determines whether there exists an
operator F that implements the specification in the sense that φ(I,F(I)) holds for all inputs I. Büchi and Landweber gave a procedure to solve Church’s problem for MSO specifications and operators
computable by finite-state automata. We investigate a generalization of the Church synthesis problem to the continuous time of the non-negative reals. It turns out that in the continuous time there
are phenomena which are very different from the canonical discrete time domain of the natural numbers.
Speaker: Dietrich Kuske, Technische Universität Ilmenau
Tuesday, 5 September 2023, 14:00, Room 1Z34
We consider the reachability problem for pushdown systems where the pushdown does not store a word, but a Mazurkiewicz trace. Since this model generalizes multi-pushdown systems, the reachability
problem can only be solved for a restricted class that we call cooperating multi-pushdown systems (cPDS). In the talk, I will show how the saturation algorithm by Finkel et al. (1997) can be
modified. For such saturated cPDS, the reachability relation can be shown to be a finite union of finite compositions of prefix-recognizable trace relations (for words, these relations were studied,
e.g., by Caucal, 1992).
Speaker: Marc Baboulin, Université Paris-Saclay
Tuesday, 12 September 2023, 14:00, Room 1Z34
Quantum computing aims at addressing computations that are currently intractable by conventional supercomputers but it is also a promising technology for speeding up some existing simulations.
However it is commonly accepted that quantum algorithms will be essentially ``hybrid'' with some tasks being executed on the quantum processor while others will remain treated on classical processors
(CPU, GPU,...). Among the crucial tasks achieved on classical computers, several require efficient linear algebra algorithms.
Speaker: Pierre Senellart, ENS Ulm, head of the Valda joint CNRS, ENS, Inria project-team
Tuesday, 4 July 2023, 10:30, 1Z14
We describe ProvSQL, software currently being developed in our team that adds support for provenance management to a regular database management system, PostgreSQL. We first explain what provenance
is, applications thereof, and how provenance of database queries can be defined and captured. We insist on one particular application of provenance, probabilistic query evaluation. We then explain
research and engineering challenges in implementing techniques from the scientific literature on provenance management and probabilistic databases within a major database management system.
Speaker: Yves Bertot, INRIA Antipolis
Tuesday, 4 July 2023, 9:20, 1Z14
Our goal is to describe smooth trajectories for robots, so that these robots don't have to stop to change direction. Several formats of trajectories could be used, but we decided to focus on
trajectories given by Bézier curves. It happens that these curves have mathematical properties that make it easy to verify formally that there are no collisions. Work in collaboration with Quentin
Vermande (ENS, Paris) and Reynald Affeldt (AIST, Tokyo, Japan).
Speaker: Achim Brucker, Chair of Computer Security, University Exeter, GB
Tuesday May 16 2023, 14:00, Room 1Z61
Communication networks like the Internet form a large distributed system where a huge number of components run in parallel, such as security protocols and distributed web applications. For what
concerns security, it is obviously infeasible to verify them all at once as one monolithic entity; rather, one has to verify individual components in isolation.
While many typical components like the security protocol TLS have been studied intensively, there exists much less research on analyzing and ensuring the security of the composition of security
protocols. This is a problem since the composition of systems that are secure in isolation can easily be insecure.
The verification of security protocols is one of the success stories of formal methods. There is a wide spectrum of methods available, ranging from fully automated methods to interactive theorem
proving with proof assistants like Isabelle/HOL. Read more...
Speaker: Gerard Memmi LTCI, Telecom-Paris, Institut polytechnique de Paris
Tuesday, 23 Mai 2023, 14:00, Room 1Z56 and Zoom
We discuss important characteristics of finite generating sets for F+, the set of all semiflows with non-negative coordinates of a Petri Net. We endeavor to regroup a number of algebraic results
dispersed throughout the Petri Nets literature and also to better position the re- sults while considering semirings such as N or Q+ then fields such as Q. As accurately as possible, we provide a
range of new algebraic results on minimal semiflows, minimal supports, and finite minimal generating sets for a given family of semiflows. Minimality of semiflows and of sup- port are critical to
develop effective analysis of invariants and behavioral properties of Petri Nets. Main results are concisely presented in a table and our contribution is highlighted. We conclude with the analysis of
an example drawn from the telecommunication industry underlining the efficiency brought by using minimal semiflows of minimal supports.
Speaker: Pedro Ribeiro Research Fellow, University of York
Tuesday, 6 June 2023, 14:00, 1Z71
Robots are expected to play important roles in furthering prosperity, however providing formal guarantees on their (safe) behaviour is not yet fully within grasp given the multifaceted nature of such
cyber-physical systems. Simulation, favoured by practitioners, provides an avenue for experimenting with different scenarios before committing to expensive tests and proofs. In this talk, I will
discuss how models may be brought together for (co-)verification of system properties, with simulation complementing verification. This will be cast using the model-driven RoboStar framework, that
clearly identifies models of the software, hardware, and scenario, and has heterogeneous formal semantics amenable to verification using state-of-the-art model-checkers and theorem provers, such as
Pedro Ribeiro will be visiting the LMF the entire day - interactions welcome.
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numtaps : int
The number of taps in the FIR filter. numtaps must be less than nfreqs.
freq : array_like, 1D
The frequency sampling points. Typically 0.0 to 1.0 with 1.0 being Nyquist. The Nyquist frequency can be redefined with the argument nyq. The values in freq must be nondecreasing. A value can be
repeated once to implement a discontinuity. The first value in freq must be 0, and the last value must be nyq.
gain : array_like
The filter gains at the frequency sampling points. Certain constraints to gain values, depending on the filter type, are applied, see Notes for details.
nfreqs : int, optional
The size of the interpolation mesh used to construct the filter. For most efficient behavior, this should be a power of 2 plus 1 (e.g, 129, 257, etc). The default is one more than the smallest
power of 2 that is not less than numtaps. nfreqs must be greater than numtaps.
window : string or (string, float) or float, or None, optional
Window function to use. Default is “hamming”. See scipy.signal.get_window for the complete list of possible values. If None, no window function is applied.
nyq : float
Nyquist frequency. Each frequency in freq must be between 0 and nyq (inclusive).
antisymmetric : bool
Flag setting wither resulting impulse responce is symmetric/antisymmetric. See Notes for more details.
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[Solved] Use (i) the Ampere's law for H and (ii) continuity of ... | Filo
Use (i) the Ampere's law for and (ii) continuity of lines of , to conclude that inside a bar magnet, (a) lines of run from the -pole to - pole, while (b) lines of B must run from the -pole to -pole.
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+ Ask your question
Consider a magnetic field line of through the bar magnet as given in the figure below.
The magnetic field line of through the bar magnet must be a closed loop:
Let be the amperian loop. Then.
We know that the angle between and is less than inside the bar magnet. So, it is positive.
Hence, the lines of must run from south pole(S) to north pole inside the bar magnet.
According to Ampere's law,
It will be so it angle between and is more than , so that is negative. It means the line of must run from -pole to -pole inside the bar magnet.
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Practice more questions from Physics Exemplar (NCERT Exemplar)
A bar magnet of magnetic moment and moment of inertia (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let be the period of oscillations of the original
magnet about an axis through the midpoint, perpendicular to length, in a magnetic field . What would be the similar period for each piece?
Thinking Process
where, time period
moment of inertia
mass of magnet
magnetic field
Use (i) the Ampere's law for and (ii) continuity of lines of , to conclude that inside a bar magnet, (a) lines of run from the -pole to - pole, while (b) lines of B must run from the -pole to -pole.
Verify the Ampere's law for magnetic field of a point dipole of dipole moment . Take as the closed curve running clockwise along
(i) the -axis from to ,
(ii) along the quarter circle of radius and centre at the origin in the first quadrant of -plane,
(iii) along the -axis from to , and
(iv) along the quarter circle of radius and centre at the origin in the first quadrant of -plane
K Thinking Process
Let us consider the figure below
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Question Use (i) the Ampere's law for and (ii) continuity of lines of , to conclude that inside a bar magnet, (a) lines of run from the -pole to - pole, while (b) lines of B must run from the
Text -pole to -pole.
Topic Magnetism and Matter
Subject Physics
Class Class 12
Answer Type Text solution:1
Upvotes 136
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Discount Rate: Navigating the Present Value with the Right Discount Rate - FasterCapital
Discount Rate: Navigating the Present Value with the Right Discount Rate
1. Introduction to Discount Rates and Present Value
Understanding the concept of discount rates and present value is essential for anyone involved in financial analysis, investment decision-making, or any area where future cash flows are relevant. The
discount rate is a critical component that reflects the time value of money, essentially representing the trade-off between consuming today versus in the future. It's a measure of the return expected
from an investment, and it's used to calculate the present value of future cash flows. The present value, on the other hand, is the current worth of a future sum of money or stream of cash flows
given a specified rate of return. It's a fundamental concept in finance that helps investors, managers, and analysts determine the value of an investment in today's dollars.
From an investor's perspective, the discount rate is the expected rate of return they require to invest in a project. It varies depending on the risk profile of the investment and the investor's
opportunity cost of capital. A higher discount rate is applied to riskier investments to compensate for the increased risk. Conversely, a corporate finance manager might view the discount rate as the
company's weighted average cost of capital (WACC), which reflects the average rate the company pays for capital from borrowing or selling equity.
Here are some in-depth points about discount rates and present value:
1. Time Value of Money: The core principle behind the present value is the time value of money, which states that a dollar today is worth more than a dollar in the future due to its potential earning
capacity. This principle provides the basis for discounting future cash flows.
2. Calculation of Present Value: The present value of a future cash flow can be calculated using the formula:
$$ PV = \frac{CF}{(1 + r)^n} $$
Where \( PV \) is the present value, \( CF \) is the future cash flow, \( r \) is the discount rate, and \( n \) is the number of periods until the cash flow occurs.
3. risk and Discount rate: The choice of discount rate is influenced by the perceived risk of the cash flows. The higher the risk, the higher the discount rate, which results in a lower present
4. Inflation and Discount Rate: Inflation can erode the purchasing power of future cash flows, so a discount rate must also account for the expected rate of inflation over the investment period.
5. Opportunity Cost: The discount rate reflects the opportunity cost of capital, which is the return that could be earned on an investment of equivalent risk.
6. Discount Rate in Different Scenarios: The appropriate discount rate can vary significantly depending on the context, such as whether it's for personal investment decisions, corporate finance, or
government projects.
To illustrate these concepts, let's consider an example. Suppose an investor is evaluating a project that promises to return $10,000 five years from now. If the investor's required rate of return
(discount rate) is 8%, the present value of this future cash flow would be:
$$ PV = \frac{10,000}{(1 + 0.08)^5} = \frac{10,000}{1.4693} \approx $6,805.56 $$
This calculation shows that the investor would be indifferent to receiving approximately $6,805.56 today or $10,000 five years from now, given an 8% discount rate. This example highlights how the
discount rate affects the present value of future cash flows and ultimately influences investment decisions. Understanding these concepts is crucial for making informed financial choices that
maximize value and align with one's financial goals and risk tolerance.
Introduction to Discount Rates and Present Value - Discount Rate: Navigating the Present Value with the Right Discount Rate
2. Understanding the Time Value of Money
The concept of the Time Value of Money (TVM) is a fundamental principle in finance that recognizes the increased value of money received today compared to the same amount of money received in the
future. This principle is based on the potential earning capacity of money, given that money available now can be invested to earn returns over time. Therefore, the value of a sum of money is seen as
being worth more the sooner it is received. TVM is crucial when assessing investment opportunities and making financial decisions, as it helps in understanding the benefits of receiving payments now
rather than later.
From an investor's perspective, the TVM is used to compare investment options. An investor might consider a future cash flow of $1000 to be less valuable than $1000 today because the money today can
be invested in a risk-free asset, like a Treasury bill, to earn interest. For example, if the risk-free rate is 5%, $1000 today would grow to $1050 in one year, making the present value of $1000
received a year from now less than $1000.
From a borrower's perspective, TVM is also significant. When taking out a loan, the borrower can use the funds immediately for consumption or investment. However, the borrowed amount must be repaid
in the future, often with interest. The borrower effectively evaluates the utility of the immediate funds against the cost of repayment in the future.
Here are some key points that provide in-depth information about the TVM:
1. Future Value (FV): This is the value of a current asset at a specified date in the future based on an assumed rate of growth over time. The formula to calculate FV is $$ FV = PV \times (1 + r)^n
$$ where \( PV \) is the present value, \( r \) is the interest rate, and \( n \) is the number of periods.
2. Present Value (PV): This is the current value of a future sum of money or stream of cash flows given a specified rate of return. The formula for PV is the inverse of the FV formula: $$ PV = \frac
{FV}{(1 + r)^n} $$.
3. Discount Rate: This is the rate used to discount future cash flows to their present value. It reflects the opportunity cost of capital, risk, and inflation among other factors.
4. Annuities: These are series of equal payments made at regular intervals over a period of time. The TVM formulas can be adjusted to calculate the present or future value of annuities.
5. Compounding Frequency: The number of times compounding occurs per period affects the total interest earned or paid. More frequent compounding results in higher effective interest rates.
To illustrate these concepts, let's consider an example where you have the option to receive $10,000 today or in two years. Assuming a discount rate of 6%, the present value of $10,000 received in
two years would be calculated as follows:
$$ PV = \frac{10000}{(1 + 0.06)^2} = \frac{10000}{1.1236} = $8899.64 $$
This means that $10,000 in two years is equivalent to $8899.64 today, given a 6% discount rate. If you were offered less than $8899.64 today for the promise of $10,000 in two years, you would be
better off waiting for the future payment. Conversely, if you were offered more than this amount today, taking the immediate payment would be advantageous.
Understanding TVM is essential for making informed financial decisions, whether you're investing, borrowing, or saving. It allows individuals and businesses to evaluate the true cost of financial
transactions and to optimize their financial planning strategies. By considering the time value of money, one can ensure that the value of money is maximized over time.
Understanding the Time Value of Money - Discount Rate: Navigating the Present Value with the Right Discount Rate
3. Factors Influencing the Choice of Discount Rate
The choice of discount rate is a critical decision in financial analysis and investment, as it affects the present value of future cash flows and can significantly influence the attractiveness of a
project or investment. The discount rate essentially reflects the opportunity cost of capital, the risk of the cash flows, and the time value of money. It's a complex interplay of various factors,
each of which can sway the decision-making process in different directions.
From the perspective of a corporate finance manager, the discount rate is often aligned with the company's weighted average cost of capital (WACC), which represents the average rate the company pays
for capital from borrowing or selling equity. If an investment's returns exceed this rate, it's considered a good investment. However, a risk-averse investor might opt for a higher discount rate to
buffer against uncertainty, while a risk-tolerant investor could justify a lower rate, betting on the potential for higher returns.
Here are some key factors that influence the choice of discount rate:
1. Risk-Free Rate: This is the return expected from an investment with zero risk, typically government bonds. For example, if 10-year U.S. Treasury notes yield 2%, this could serve as a baseline for
the risk-free rate.
2. Market Risk Premium: This is the additional return investors demand for taking on the risk of investing in the stock market over a risk-free investment. It varies by market conditions and investor
3. Beta (β): This measures the volatility of an investment compared to the market as a whole. A beta greater than 1 indicates higher volatility and, consequently, a higher discount rate might be
4. debt-Equity ratio: The proportion of debt to equity financing affects the cost of capital. Higher debt levels can increase the discount rate due to the increased financial risk.
5. Industry-Specific Risks: Different industries have unique risks. For instance, the pharmaceutical industry faces regulatory risks that can delay product launches, justifying a higher discount
6. Project-Specific Risks: These include the likelihood of project delays, cost overruns, and technology risks. A construction project, for example, might face geological risks that are not present
in a software development project.
7. Economic Conditions: Inflation, interest rates, and economic growth forecasts can all impact the discount rate. During periods of high inflation, a higher discount rate might be used to account
for the decreased purchasing power of future cash flows.
8. Liquidity Preferences: Investors may require a higher rate for investments that cannot be easily sold or converted to cash.
9. Regulatory Environment: Changes in laws or tax policies can affect the future cash flows' certainty, influencing the discount rate.
10. Management's Discretion: Sometimes, the choice of discount rate can be subjective, reflecting management's confidence in their projections and strategic direction.
To illustrate, consider a company evaluating an investment in renewable energy. The project has a high upfront cost but promises long-term savings and cash flows. Given the industry's regulatory
support and long-term contracts, the company might opt for a lower discount rate, reflecting the lower risk and stable cash flows compared to a more volatile sector like technology startups, where a
higher rate would be prudent due to the greater uncertainty and competition.
Selecting the right discount rate is a nuanced process that requires balancing objective financial metrics with subjective judgments about future conditions and risks. It's not merely a mathematical
exercise but a strategic decision that can define the trajectory of an investment's success or failure.
Factors Influencing the Choice of Discount Rate - Discount Rate: Navigating the Present Value with the Right Discount Rate
4. The Role of Risk in Determining Discount Rates
In the realm of finance, the discount rate serves as a critical tool for translating future cash flows into present values, allowing for a more grounded evaluation of investment opportunities. The
selection of an appropriate discount rate is a nuanced process, heavily influenced by the perceived risk associated with the cash flows. Risk, in this context, refers to the uncertainty surrounding
the receipt of these cash flows. The greater the uncertainty, the higher the risk, and consequently, the higher the discount rate that must be applied.
From the perspective of an investor, the discount rate embodies the opportunity cost of capital—the returns that could be earned if the money were invested elsewhere. Therefore, the riskier the
investment, the higher the potential return an investor would require to forgo alternative opportunities. This is where the risk premium comes into play, serving as an additional component to the
risk-free rate (often represented by government bonds) to account for the extra uncertainty.
1. Risk-Free Rate: The foundation of any discount rate, the risk-free rate represents the return on an investment with zero perceived risk. It's the baseline upon which additional layers of risk are
2. Risk Premium: This is the additional return investors demand for taking on additional risk. It varies widely depending on the type of investment and the investor's risk tolerance.
3. Beta Coefficient: In the capital Asset Pricing model (CAPM), beta measures the volatility of an investment relative to the market. A higher beta indicates higher risk, which translates to a higher
discount rate.
4. Market Risk Premium: This reflects the additional return that investors require for investing in the market as a whole over the risk-free rate.
5. Size Premium: Smaller companies often carry higher risk due to less market presence and stability, which is compensated for with a size premium.
6. Specific Company Risk: This encompasses risks unique to a particular company, such as management quality, industry position, and operational efficiency.
7. country Risk premium: For investments in foreign countries, this premium accounts for the risk of political instability, economic volatility, and currency fluctuations.
To illustrate, consider a company evaluating a potential project in a politically unstable region. The base discount rate might start with a risk-free rate of 3%. Given the volatility of the region,
a country risk premium of 2% is added. The project itself is high-risk, warranting a specific company risk premium of 4%. The total discount rate would then be 9%, significantly higher than the
risk-free rate alone.
In contrast, a well-established company with stable cash flows operating in a politically stable country might only require a modest risk premium above the risk-free rate, resulting in a lower
discount rate. This reflects the lower perceived risk and the company's confidence in its ability to generate consistent returns.
Ultimately, the role of risk in determining discount rates is a balancing act between the desire for higher returns and the need to mitigate potential losses. By carefully assessing the various
layers of risk and their impact on the discount rate, investors and companies alike can make more informed decisions that align with their financial goals and risk appetites.
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5. Discount Rates in Different Industries
Understanding the intricacies of discount rates across various industries is pivotal for investors and financial analysts. These rates are the key to unlocking the present value of future cash flows,
which is essential when evaluating investment opportunities. The discount rate essentially reflects the opportunity cost of capital—the return that investors forego by investing in one project over
another. It's influenced by factors such as risk, inflation, and the time value of money. Different industries exhibit unique risk profiles and growth prospects, which necessitates a tailored
approach to determining the appropriate discount rate.
1. Technology Sector: Often characterized by high growth potential, the technology sector typically sees lower discount rates due to anticipated higher future cash flows. For instance, a tech startup
with a revolutionary product might use a discount rate of 10-15%, reflecting both the high risk and high potential return.
2. Utilities: This sector is known for its stability and consistent dividends. Consequently, the discount rates are lower, often around 5-7%, mirroring the lower risk associated with these
3. Real Estate: The discount rate in real estate takes into account the property's location, type, and income-generating potential. A prime commercial property might have a discount rate of 8%, while
a riskier development project could be as high as 15%.
4. Healthcare: With a mix of stable companies and high-growth biotech startups, discount rates can vary widely. A large pharmaceutical company might use a rate of 6-8%, whereas a biotech firm in the
clinical trial phase might warrant a rate upwards of 20% due to the inherent risks.
5. Retail: The retail industry faces significant competition and disruption, leading to a wide range of discount rates. A well-established retail chain might use a rate of 8-10%, while an e-commerce
startup could see rates exceeding 15%.
6. Energy: Volatility in commodity prices leads to fluctuating discount rates in the energy sector. Traditional oil and gas companies might use rates of 10-12%, while renewable energy projects, with
government subsidies and support, might have lower rates of 7-9%.
For example, consider a renewable energy company evaluating a new wind farm project. Given the lower risk profile and potential government subsidies, it might apply a discount rate of 8%. If the
projected cash flows are $1 million per year for the next 20 years, the present value of this investment using the discount rate can be calculated using the formula:
$$ PV = \frac{CF}{(1+r)^n} $$
Where \( PV \) is the present value, \( CF \) is the annual cash flow, \( r \) is the discount rate, and \( n \) is the number of years. This calculation helps determine whether the investment is
worthwhile compared to other opportunities.
By understanding the nuances of discount rates across different industries, investors can make more informed decisions that align with their risk tolerance and investment goals. It's a delicate
balance between potential returns and the risks undertaken, and the discount rate is the compass that guides this financial navigation.
Discount Rates in Different Industries - Discount Rate: Navigating the Present Value with the Right Discount Rate
6. Calculating Present Value Using Discount Rates
Understanding the concept of present value is crucial for investors, financial analysts, and anyone involved in financial planning. It's a fundamental principle that allows us to ascertain the
current worth of cash flows that are expected to be received in the future. By applying a discount rate, we can "discount" these future cash flows back to their present value, providing a common
ground for comparing the value of investments that may have different payment schedules. The discount rate reflects the opportunity cost of capital, the risk of the cash flows, and the time value of
money. It essentially answers the question: "What is the current value of a sum to be received in the future, considering the potential alternative returns I could receive from other investments?"
Insights from Different Perspectives:
1. Investor's Perspective: For investors, the discount rate is a tool to determine the attractiveness of an investment. A higher discount rate implies greater risk and may indicate a less desirable
investment. For example, if an investor is considering a bond that will pay $1,000 five years from now, they might use a discount rate of 5% to calculate its present value. Using the formula $$ PV =
\frac{FV}{(1 + r)^n} $$, where PV is present value, FV is future value, r is the discount rate, and n is the number of periods, the present value would be $$ PV = \frac{1000}{(1 + 0.05)^5} $$, which
calculates to approximately $783.53.
2. Corporate Finance Perspective: In corporate finance, the discount rate often reflects the company's weighted average cost of capital (WACC). This rate is used to discount future cash flows from
projects to determine their net present value (NPV). A project with a positive NPV, after discounting future cash flows, is typically considered for investment.
3. Economist's Perspective: Economists might view the discount rate as a reflection of time preference — the propensity of people to prefer goods and services sooner rather than later. This
preference impacts consumer behavior and savings rates, influencing macroeconomic factors like inflation and interest rates.
In-Depth Information:
1. calculating Present Value for annuities: An annuity is a series of equal payments made at regular intervals. The present value of an annuity can be calculated using the formula $$ PV_{\text
{annuity}} = P \times \left(\frac{1 - (1 + r)^{-n}}{r}\right) $$ where P is the payment per period. For instance, if you're set to receive $100 annually for 5 years with a discount rate of 5%, the
present value would be $$ PV_{\text{annuity}} = 100 \times \left(\frac{1 - (1 + 0.05)^{-5}}{0.05}\right) $$, which is approximately $432.95.
2. Adjusting for Risk: The discount rate can be adjusted to account for the riskiness of the cash flows. Higher risk typically means a higher discount rate. For example, if two projects offer the
same future cash flow, but one is riskier, the riskier project will have a higher discount rate applied, resulting in a lower present value.
3. Impact of Inflation: Inflation can erode the purchasing power of future cash flows. When calculating present value, it's important to consider whether the discount rate includes an inflation
expectation. A real discount rate excludes inflation, while a nominal discount rate includes it.
By integrating these insights and methods into our analysis, we can make more informed decisions about where to allocate our resources for the best financial outcomes. Calculating present value using
discount rates is not just a mathematical exercise; it's a way to evaluate the true worth of future cash flows in today's terms. Whether you're an individual investor, a financial professional, or an
economist, understanding this concept is key to navigating the complex world of finance.
Calculating Present Value Using Discount Rates - Discount Rate: Navigating the Present Value with the Right Discount Rate
7. Common Mistakes When Selecting Discount Rates
Selecting the appropriate discount rate is a critical step in the valuation process, whether for investment analysis, capital budgeting, or assessing the present value of future cash flows. The
discount rate essentially reflects the opportunity cost of capital, incorporating the time value of money and the risk associated with the cash flows. However, this selection process is fraught with
potential missteps that can significantly skew results and lead to suboptimal financial decisions.
From the perspective of a financial analyst, the temptation to use a company's weighted average cost of capital (WACC) as a universal discount rate is a common pitfall. While WACC represents the
average rate that a company expects to pay for its financing, it may not accurately reflect the risk profile of a specific project or investment. For instance, a project with higher risk than the
company's average should logically command a higher discount rate to compensate for the increased uncertainty.
Similarly, from an investor's standpoint, overlooking the individual risk tolerance and investment horizon can lead to inappropriate discount rate selection. An investor seeking short-term gains
might undervalue the importance of a higher discount rate to mitigate short-term volatility, whereas a long-term investor might overestimate the need for a high rate, thus undervaluing future cash
flows and missing out on potentially lucrative long-term investments.
Here are some common mistakes to avoid when selecting discount rates:
1. Ignoring the Project-Specific Risk: Using a generic discount rate without adjusting for the unique risks of the project can lead to inaccurate valuations. For example, a startup in the biotech
industry should have a higher discount rate than a stable utility company due to the higher inherent risk.
2. Misjudging the Economic Environment: Economic conditions greatly influence the appropriate discount rate. During periods of high inflation, a higher rate should be used to account for the
decreased purchasing power of future cash flows.
3. Overlooking Liquidity Premiums: Investments that are not easily liquidated should include a liquidity premium in the discount rate. This compensates investors for the additional risk associated
with the difficulty of converting the investment into cash.
4. Failing to Reassess Over Time: The discount rate is not static and should be reassessed periodically to reflect changes in the market conditions and the risk profile of the investment.
5. Using Historical Averages Blindly: While historical data can inform the selection of a discount rate, relying solely on past averages without considering current and future market expectations can
lead to mispricing.
6. Neglecting the Opportunity Cost: The discount rate should reflect the best alternative investment available with a similar risk profile. Not considering opportunity costs can result in choosing an
investment with a lower return than could be obtained elsewhere.
To illustrate, let's consider a real estate developer evaluating two potential projects: a residential development in a well-established neighborhood and a commercial complex in an emerging market.
The residential project might seem less risky due to the established demand, leading the developer to apply a lower discount rate. However, if the emerging market is expected to experience
significant growth, the commercial complex could offer higher returns despite its higher initial risk, warranting a reevaluation of the discount rates applied to each project.
The selection of a discount rate is a nuanced process that requires careful consideration of various factors, including project-specific risks, economic conditions, liquidity concerns, and
opportunity costs. By avoiding these common mistakes, financial professionals and investors can make more informed decisions that better align with their financial goals and risk tolerance.
Common Mistakes When Selecting Discount Rates - Discount Rate: Navigating the Present Value with the Right Discount Rate
8. Discount Rates in Action
Understanding the application of discount rates in real-world scenarios is crucial for both investors and businesses. It allows them to determine the present value of future cash flows, making it
easier to compare investment opportunities and make informed financial decisions. By examining various case studies, we can see how discount rates are used in different industries and contexts, and
how they can significantly impact the valuation of projects and investments. From corporate finance to government policy, the selection of an appropriate discount rate is a nuanced process that
involves considering the risk profile, the time horizon, and the expected rate of return.
1. Corporate Valuation: A tech startup is looking to raise capital by selling a stake in the company. To value the business, they forecast future cash flows and apply a discount rate that reflects
the high risk associated with the tech industry and startups in general. They might use a rate of 15-20%, which is higher than established companies, to account for the uncertainty and potential for
high growth.
2. Infrastructure Projects: When governments evaluate infrastructure projects like bridges or highways, they often use a lower discount rate, around 3-6%. This reflects the lower risk and long-term
nature of these investments. For example, a proposed bridge expected to generate toll revenue over 30 years would be discounted at a rate that reflects the government's borrowing cost and the
project's low risk.
3. pension funds: Pension funds have long-term obligations and typically use a discount rate that matches the expected return on their investments, often around 6-8%. This rate is crucial for
determining the present value of future pension liabilities and ensuring the fund remains solvent.
4. real estate Development: A real estate developer evaluating a new project will use a discount rate that reflects the risk of the project, the cost of capital, and the expected rate of return,
usually between 8-12%. For instance, a commercial real estate project in a prime location might use a lower rate compared to a residential project in an emerging market.
5. Environmental Policy: When calculating the cost and benefits of environmental policies, a social discount rate is used. This rate is often debated, but it generally falls between 1-3%. It reflects
the long-term impact and uncertainty of environmental changes. For example, assessing the economic impact of reducing carbon emissions over the next century requires a discount rate that balances
current costs with future benefits to society.
These examples highlight the diversity in the application of discount rates and underscore the importance of context-specific analysis. By carefully selecting and applying the right discount rate,
decision-makers can better understand the true value of their investments and make choices that align with their financial goals and risk tolerance. The art and science of choosing the right discount
rate are what ultimately guide investors and policymakers through the complex landscape of financial decision-making.
Discount Rates in Action - Discount Rate: Navigating the Present Value with the Right Discount Rate
9. Making Informed Decisions with the Right Discount Rate
In the realm of finance and investment, the selection of an appropriate discount rate is a critical decision that can significantly influence the present value of future cash flows. It's a tool that
reflects the time value of money, accounting for risk and uncertainty inherent in any financial projection. The right discount rate can mean the difference between a profitable investment and a
costly mistake.
From the perspective of a conservative investor, a higher discount rate might be used to reflect the increased risk and the opportunity cost of capital. This approach ensures that only the most
robust investment opportunities are selected, those that promise returns even when conservative estimates are applied. For example, a risk-averse investor might choose a discount rate that is several
percentage points above the risk-free rate.
On the other hand, a venture capitalist with a higher risk tolerance might opt for a lower discount rate for high-growth potential startups, acknowledging the higher risk but also the higher
potential returns. This reflects a more optimistic outlook on the investment's ability to generate cash flow. For instance, a venture capitalist might use a discount rate that is closer to the
risk-free rate when evaluating a tech startup with exponential growth potential.
Here are some in-depth considerations when determining the right discount rate:
1. Risk Profile: The more uncertain the future cash flows, the higher the discount rate should be to compensate for this risk. For example, a biotech firm in the early stages of drug development
would warrant a higher discount rate than a well-established consumer goods company.
2. Opportunity Cost: The discount rate should reflect the return that could be earned on an investment of similar risk. If an investor can earn 5% with minimal risk, then any riskier investment
should offer a higher potential return.
3. Inflation Expectations: The discount rate should account for expected inflation, which erodes the value of future cash flows. If inflation is expected to average 2% per year, the discount rate
should be at least 2% higher than if there were no inflation.
4. Capital Structure: The cost of capital depends on the mix of debt and equity financing. A company with a high debt load might have a higher discount rate due to the increased risk of bankruptcy.
5. Economic Environment: During times of economic uncertainty or recession, a higher discount rate might be used to reflect the increased market volatility and reduced predictability of cash flows.
6. Regulatory Environment: Changes in regulation can impact future cash flows. A stable regulatory environment might warrant a lower discount rate, while a volatile one could necessitate a higher
7. Industry Trends: Industry-specific risks and opportunities should be considered. For example, the discount rate for a renewable energy project might be lower to reflect government subsidies and
the growing demand for clean energy.
8. Historical Data: Past performance and historical rates of return can inform the choice of a discount rate, though they should be used cautiously as past performance is not always indicative of
future results.
9. Comparative Analysis: Comparing the project or investment against a benchmark or similar investments can help in selecting a suitable discount rate.
10. Expert Consultation: Financial advisors and industry experts can provide valuable insights into what discount rate to use based on current market conditions and future projections.
By carefully considering these factors, investors and financial analysts can arrive at a discount rate that accurately reflects the true cost of capital, ensuring that investment decisions are made
with a clear understanding of their potential impact on present value. For instance, when valuing a commercial real estate property, an investor might consider the property's location, market trends,
and rental income potential to determine a discount rate that reflects the specific risks and opportunities of that investment.
The process of selecting the right discount rate is both an art and a science, requiring a balance of quantitative analysis and qualitative judgment. It's a decision that should not be taken lightly,
as it holds the key to unlocking the true value of future cash flows and making informed investment decisions.
Making Informed Decisions with the Right Discount Rate - Discount Rate: Navigating the Present Value with the Right Discount Rate
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It was the first extended treatment of scheme theory written as a text intended to be accessible to graduate students.
The first chapter, titled "Varieties", deals with the classical algebraic geometry of varieties over algebraically closed fields. This chapter uses many classical results in commutative algebra,
including Hilbert's Nullstellensatz, with the books by Atiyah–Macdonald, Matsumura, and Zariski–Samuel as usual references. The second and the third chapters, "Schemes" and "Cohomology", form the
technical heart of the book. The last two chapters, "Curves" and "Surfaces", respectively explore the geometry of 1- and 2-dimensional objects, using the tools developed in the chapters 2 and 3.
1. ^ MathSciNet lists more than 2500 citations of this book.
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A trick of the hat
April 13, 2023 The story of how a Waterloo computer science professor helped find the elusive einstein tile By Joe Petrik Cheriton School of Computer Science A nearly 60-year-old mathematical problem
has finally been solved.
The story began last fall when David Smith, a retired print technician from Yorkshire, England, came upon a shape with a tantalizing property. The life-long tiling enthusiast discovered a 13-sided
shape - dubbed the hat - that is able to fill the infinite plane without overlaps or gaps in a pattern that not only never repeats but also never can be made to repeat.
This elusive shape is known to mathematicians as an aperiodic monotile or an einstein, a clever pun that takes its name from the German words and stein that mean one stone.
"Dave and I had been in touch over the years and we belong to the same old-fashioned listserv for people interested in tiling, a mix of tiling enthusiasts, programmers and mathematicians," recalls
Cheriton School of Computer Science professor Craig S. Kaplan, who collaborated with Smith, software developer Joseph Myers and mathematician Chaim Goodman-Strauss on the paper that has proven that
the elusive einstein exists.
"Dave was on to something big, something historic, but he hit the wall on what he could deduce about this shape by working with paper cut-outs. He knew I had recently published a paper about a
related topic for which I developed a piece of software that we could use to understand what his shape was doing. He sent me an email asking, ’Hey, can you run this through your software and see what
Mathematicians had been trying to find a shape like David Smith’s einstein since the 1960s when American mathematician Robert Berger discovered the first example of aperiodic tiling.
"Berger’s aperiodic set of shapes was found in the mid-1960s and that set had 20,426 shapes," Professor Kaplan explained. "It was an elaborate construction with a combinatorial set of features that
required a multiplicity of shapes to guarantee that the pattern doesn’t repeat. That was an important discovery, but the natural next question for mathematicians is, can we get smaller sets? What’s
the lowest number of shapes we can do this with?"
By 1970, the set of shapes proven to tile aperiodically was down to about 100 and in 1971 mathematician Raphael Robinson got it down to six. Then, in 1974, Sir Roger Penrose discovered the eponymous
Penrose tiles , which reduced the number to two.
"Those two shapes in Penrose’s solution had enough structure that they forbid periodicity. But for almost 50 years mathematicians have been wondering, can we get down to just one shape? Can we do
this with a monotile? That’s the problem we solved. We found a single shape that does what all these earlier sets of multiple shapes are able to do."
In mathematics and computer science many problems remain open, but theoreticians have a strong sense what the answer will be even though a formal proof may be decades away.
"The famous P vs NP problem in computer science - a question about how long it takes to execute a particular class of algorithms - is still open, but there’s a consensus how that’s going to play
out," Professor Kaplan said. "Almost every computer scientist thinks that P is not equal to NP. But the existence of an aperiodic monotile isn’t in that category. Opinions were split. That’s one of
the things I love about this problem. It was not obviously true or obviously false. The only thing I knew for sure is that if it’s false - if no aperiodic monotile exists - it would be extremely
difficult to prove because that’s a statement about all possible shapes. Whereas, proving that a particular shape is an aperiodic monotile is easier because, well, here it is. You’re only trying to
prove a property of a single shape."
Many have wondered if the hat - sometimes also called the shirt - has other tricks up its sleeve. In a sense it does.
"In our paper we show that the hat is not just a single shape that tiles aperiodically, but a member of a continuum of shapes. We can say that the hat is not the only aperiodic monotile, but it feels
like a bit of a cop-out because all those shapes are closely related. They’re one big family. The more interesting question is are there fundamentally different aperiodic monotiles? My answer is that
there’s no reason to suspect otherwise and every reason to suspect there ought to be others."
The main proof in the paper is combinatorial and benefits greatly from computer assistance, Professor Kaplan said. "It’s combinatorial in that there are a few steps in the proof that depend on
examining all the ways individual tiles can be next to each other and all the ways tiles can group together into larger and larger clumps. As it turns out, there are a lot of ways. Depending on what
you’re counting, it’s dozens, hundreds, thousands."
You could grind through all of those cases tediously by hand, but if you have a computer science background so much the better. Why not write a piece of software to do that for you?
"The key computer-assisted part of our proof involves saying, ’We have to be able to say things about generic tilings of the hat that we don’t know anything about.’ But how can we say anything about
a tiling whose structure we have no control over? In this part of the proof, we show that even though you didn’t know anything going in, the tiling has a certain structure that you can account for.
One way you can do that is to exhaustively enumerate little neighbourhoods of tiles - all the little neighbourhoods that possibly could occur in a real tiling."
A lot can be rejected. In one particular neighbourhood, you see there’s no way to surround those tiles by another layer of tiles, so it couldn’t occur in a real tiling. It’s just an isolated blob.
"We can write a program to find all the ways you can have a little blob that is legally able to occur in a full tiling and we then wrote code that says something interesting about each of those
different blobs that allows us to conclude that therefore an arbitrary tiling must have the properties we want it to have. The program we wrote confirms that those rules are followed in every
possible tiling."
Penrose’s tiles were found to have a deep connection to the natural world. In 1982, Iowa State University Professor Dan Shechtman discovered that symmetries similar to the ones in Penrose tiles are
found in molecular structures called quasicrystals - a crystalline molecule that is ordered but not periodic - a discovery that led to his receiving the 2011 Nobel Prize in Chemistry.
"It’s fun to speculate, but I’m not a physicist or an engineer," Professor Kaplan said. "That the Penrose tiling has a connection to materials science is amazing, but it’s no guarantee that other
aperiodic tilings do or that the hat will. My work is about the applications of mathematics in art. First and foremost, for me the hat tiling is interesting and it is visually arresting. People have
already been using it to make interesting designs in different media. Please keep doing that. That’s amazing and I love it."
Perhaps the hat could leave its mark at Waterloo in a more concrete way.
"There’s a stone courtyard at the University of Oxford’s Mathematical Institute where Penrose works that has been tiled with Penrose tiles. If you have a breakthrough in tiling theory, how are you
not putting that on the floor of one of your academic buildings? The timing is nearly perfect now that the Math 4 building has been approved for construction and is in the design phase."
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Download CBSE Mathematics Syllabus for Classes 1 to 12
Mathematics plays an important role to score good marks in class 10 and 12 board examinations because it is a high scoring subject. Class 10 and 12 board examinations decides the career of a student.
If a student scores more than 90% marks, then he/she is treated as a good student and goes for the preparation of engineering or medical.
Thus, to obtain good marks in mathematics, students should know their syllabus. Most of the schools prefer private publishers’ books. Some books have many extra contents and features other than the
syllabus which gives extra burden to the students. Using the syllabus, students can decide which topic is important to study and which one is not.
Here, we are providing the complete syllabus of CBSE board for all classes. Students, teachers and parents can download the syllabus of mathematics for any class from the given links.
CBSE Mathematics Syllabus for Classes 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Knowing the correct information about the mathematics syllabus for their children has always been a concern for the parents. Parents remain more concerned about it when their children are in primary
classes and young enough to understand the things properly.
To shape the students, mind with the correct information, we are going to provide the complete syllabus of mathematics for all classes. You can download it from the following links:
Download CBSE Mathematics Syllabus for Classes 1 to 5
Download CBSE Mathematics Syllabus for Classes 6 to 8
Download CBSE Mathematics Syllabus for Classes 9 and 10
Download CBSE Mathematics Syllabus for Classes 11 and 12
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B.F. Goodrich-Rabobank Interest Rate Swap - Case Solution
Case study on B.F. Goodrich, a U.S. manufacturing company, and Rabobank, a Eurobank, wherein these two firms swapped fixed and floating rate obligations as part of their strategies to reduce their
financial expenses.
Jay O. Light
Harvard Business Review (284080-PDF-ENG)
March 26, 1984
Case questions answered:
1. How large must the discount (X) be to make this an attractive deal for Rabobank?
2. How large must the annual fee (F) be to make this an attractive deal for Morgan Guaranty?
3. How small must the combination of F and X be to make this an attractive deal for B.F. Goodrich?
4. Is this an attractive alternative for savings banks?
5. Is this a deal where everyone wins? If not, who is the loser?
Not the questions you were looking for? Submit your own questions & get answers.
B.F. Goodrich-Rabobank Interest Rate Swap Case Answers
Background Information – B.F. Goodrich-Rabobank Interest Rate Swap
The 1982 recession left numerous businesses in financial distress. B.F. Goodrich experienced difficulties as its credit rating was demoted from BBB to BBB-. The following year, B.F. Goodrich was
desperate for $50 million to finance their existing operations. Goodrich was initially faced with a dilemma.
The high interest, coupled with Goodrich’s subpar credit rating, resulted in unappealing borrowing terms. Goodrich was adamant about accepting terms that allowed them to borrow long-term with a fixed
Thus, in 1983, B.F. Goodrich and Rabobank accomplished two financings and an interest rate swap. Unlike B.F. Goodrich, Rabobank had a perfect credit rating of AAA.
Furthermore, Rabobank was looking to reduce its borrowing costs through a floating rate. This mutually beneficial transaction was aided by Morgan Guaranty, who served as an intermediary guarantor for
the swap agreements.
Size of Discount (X)
For this deal to be attractive for Rabobank, it needs to bring in a positive net flow of capital by receiving higher interest rates while paying lower interest rates. With an 8-year fixed receipt of
$5.5 million from Morgan Guaranty, Rabobank would receive $2.75 million semiannually from Morgan.
Thus, simplified to a yearly basis, we can calculate $50 million times LIBOR minus the required percentage of discount for this to be an attractive deal for Rabobank and set this equal to the yearly
payments of $5.5 million.
With a LIBOR stated at 8.75%, we can simplify these calculations to (0.0875-X) < 0.11, with 0.11 resulting from the $5.5 million yearly payment divided by the total of $50 million.
Therefore, for this to be an attractive deal for Rabobank, the total value of the discount X must exceed 2.25%, a discount obtained by solving the equation for X.
(0.5) * (50 mil) * (LIBOR-X) =
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gbBoolean(Ideal) -- Compute Groebner Basis for Ideals in Boolean Polynomial Quotient Ring
gbBoolean always assumes that the ideal is in the Boolean quotient ring, i.e., $\mathbb F_2[x_1, \ldots, x_n] / <x_1^2-x_1, \ldots, x_n^2-x_n >$, regardless of the ring in which the ideal was
generated. Thus, gbBoolean promotes an ideal in the base ring to the quotient ring automatically, even if the quotient ring has not been defined.
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A funny problem
Here's a funny problem that I've composed for the Russian IOI training camp that's happening now.
Given two numbers a and b in base 31 with at most one million digits each such that a is a multiple of b, find the last k digits (again, in base 31) of a/b, where k is not more than 10000.
10 comments:
1. cool problem, one hour and i can't solve it
2. This is very similar to a problem on Project Euler, the caveat being the base 31 part. The last k digits of a number n in base 31 are (n mod 31^k). You want ((a/b) mod (31^k)), which is (a mod 31
^k) / (b mod 31^k). But (a mod 31^k) is just the last k digits of a, and likewise for (b mod 31^k) and b.
The problem is doing the arithmetic, because converting to base 10 and back could be problematic in the extreme cases. Is the intent of the problem, then, to write a division routine in base 31?
3. 2Abishek: what do you mean by "((a/b) mod (31^k)), which is (a mod 31^k) / (b mod 31^k)"? I don't think that is true. Suppose a=42,b=6,k=1. Then a mod 31^k=11 which is not divisible by b mod 31^k
2Ivan: wow, thanks for the link. I knew it must be a well-known idea :)
4. Oops, that property only holds for multiplication.
5. Hmnn only bell that rang when reading it is that 31 is prime.
31^k is still a product of primes and we know all its prime factors.
so calculating (a%(31^k) / v%(31^k) ) % (31^k) should be possible using that one theorem I don't remember exactly.
So, I open wikipedia and it seems we'll first have to computer (1/b)%(31^k) like this: http://en.wikipedia.org/wiki/Euclidean_algorithm#Multiplicative_inverses_and_the_RSA_algorithm
still, implementing the Euclidean algorithm on big numbers keeping efficiency sounds non-trivial. Still not sure if that's the way.
6. 2vexorian:
You can only do that when b is not divisible by 31, but after you account for that, your solution should work, but I'm not sure if it will work fast enough.
The intended solution, however, is simpler than extended Euclidean algorithm :)
7. This comment has been removed by the author.
8. It is easy.
if we use Garnet's scheme and make some precalculation, then we have next algo:
1. a = (a_1 a_2 ... a_n)
b = (b_1 b_2 ... b_m)
2. a * b ^ -1 mod 31 = Suma(i = 1,n; a_i * 31^(n - i) ) * Suma(i = 1,n; b_i * 31 ^ (n - i)) ^ -1 mod 31 = 31 ^ (k1 - k2) * Suma(i = 1,n; a_i * 31^(n - i - k1) ) * Suma(i = 1,n; b_i * 31 ^ (n - i
- k2)) ^ -1 mod 31.
We can easy compute both Suma(...) mod 31 (as precalculated) and then multiply to 31 ^ (k1 - k2). if b | a, then k1 - k2 >= 0.
3. a = (a_1 a_2 ... a_(n-1))
b = (b_1 b_2 ... b_m) * 31
4. goto step 2.
The Complexity is O(n * log(31)) = O(n). :).
9. Are available more problems of the training camp?
10. hey petr plz help him out http://kodejunky.blogspot.com/2009/10/humble-beginning.html
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\left( {2 + 2\
Hint: Here in this question given an Indefinite integral, we have to find the integrated value of a given trigonometric function. This can be solved, by substituting some trigonometric function and
algebraic identities and later integrated by using the standard trigonometric formula of integration. And by further simplification we get the required solution.
Complete step by step solution:
Integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {2 + 2\sin 2x} \right)dx} \]--------(2)
Take 2 as common in the function, then
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot 2\left( {1 + \sin 2x} \right)dx} \]
Where, 2 as constant take it out from the integration.
\[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {1 + \sin 2x} \right)dx} \]
Now, by using a algebraic identity \[{\sin ^2}x + {\cos ^2}x = 1\] and the double angle formula \[\sin 2x = 2\sin x\cos x\] on substituting, we have
\[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)dx} \]---------(2)
The term \[\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)\] similar like a algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\],
Here \[a = \sin x\] and \[b = \cos x\], then equation (2) becomes
\[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot {{\left( {\sin x + \cos x} \right)}^2}dx} \]
On cancelling like terms \['\cos x + \sin x'\] on both numerator and denominator, we have
\[ \Rightarrow 2\int {\left( {\cos x - \sin x} \right) \cdot \left( {\sin x + \cos x} \right)dx} \]---------(3)
The function \[\left( {\cos x - \sin x} \right) \cdot \left( {\sin x + \cos x} \right)\] similar like a algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
Here \[a = \cos x\] and \[b = \sin x\], then equation (3) becomes
\[ \Rightarrow 2\int {\cos {x^2} - {{\sin }^2}xdx} \]-------(4)
Again, by using a double angle formula of trigonometry i.e., \[\cos 2x = \cos {x^2} - {\sin ^2}x\], then equation (4) becomes.
\[ \Rightarrow 2\int {\cos 2xdx} \]
On integrating with respect to \[x\], using a standard formula \[\int {\cos dx} = \sin x + c\], then we have
\[ \Rightarrow 2\dfrac{{\sin 2x}}{2} + C\]
On simplification, we get
\[ \Rightarrow \sin 2x + C\]
Where, \[C\] is an integrating constant.
Hence, it’s a required solution.
So, the correct answer is “$\sin 2x + C$”.
Note: By simplifying the question using the substitution of different trigonometric formulas we can integrate the given function easily. If we apply integration directly it may be complicated to
solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.
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Multiplication Division Word Problems Worksheets - Worksheets Day
Multiplication Division Word Problems Worksheets
In this document, we will provide detailed information about Multiplication Division Word Problems Worksheets. These worksheets are designed to help students practice and improve their skills in
solving word problems involving multiplication and division.
What are Multiplication Division Word Problems Worksheets?
Multiplication Division Word Problems Worksheets are educational resources that contain a variety of word problems related to multiplication and division. These worksheets are usually used in
mathematics classrooms or for homeschooling purposes. They aim to enhance students’ problem-solving abilities and their understanding of multiplication and division concepts.
Why use Multiplication Division Word Problems Worksheets?
These worksheets offer several benefits for students:
1. Application of Mathematical Concepts: By solving word problems, students can apply their knowledge of multiplication and division in real-life situations. This helps them understand the practical
use of these operations.
2. Critical Thinking Skills: Word problems require students to analyze information, identify relevant data, and formulate problem-solving strategies. This enhances their critical thinking abilities.
3. Problem-Solving Strategies: Multiplication Division Word Problems Worksheets expose students to various problem-solving strategies. They learn to choose appropriate operations, estimate solutions,
and check their answers for reasonableness.
4. Mathematical Language Development: These worksheets introduce students to mathematical vocabulary related to multiplication and division. They learn to interpret and translate word problems into
mathematical expressions.
How to Use Multiplication Division Word Problems Worksheets
Here are some tips for effectively using these worksheets:
1. Read the Problem Carefully: Encourage students to read the word problem multiple times to fully understand the given information and the question being asked.
2. Identify Key Information: Help students identify important details and numbers in the word problem. They should underline or highlight these details to stay focused.
3. Decide on the Operation: Guide students in determining whether the problem requires multiplication or division. They should consider relationships between quantities and the context of the
4. Set Up the Equation: Assist students in translating the word problem into a mathematical equation or expression. They should use variables when necessary and clearly define what each variable
5. Solve the Equation: Encourage students to solve the equation using appropriate multiplication or division strategies. They should show their work and clearly explain each step.
6. Check the Solution: Teach students to check their answers by applying inverse operations, estimating, or using logical reasoning. This helps them verify the accuracy of their solutions.
Multiplication Division Word Problems Worksheets are valuable tools for developing students’ problem-solving skills and enhancing their understanding of multiplication and division. By practicing
with these worksheets, students can become more proficient in solving word problems involving these operations. So, make use of these worksheets and watch your students excel in their mathematical
Division Word Problems For Grade 2 Worksheet
Multiplication And Division Word Problems Year 4 Maths Year 3 Autumn 1 Reasoning Within 100
Free Math Worksheets Multiplication Division Word Problems
Math Worksheets For Grade 4 Multiplication And Division Word Problems
Multiplication And Division Worksheets With Answer Key
Multiplication Division Word Problems Worksheets
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Printable Calendars AT A GLANCE
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the plane coordinate graph paper. Web grid papers (printable 4 quadrant coordinate graph paper templates) free assortment of printable grid paper (single and 4 lock coordinate plane graph.
Printable Coordinate Grid Paper Templates at
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10 Best Printable Coordinate Picture Graphs
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Web 67 Filtered Results Coordinate Planes What Are Interactive Worksheets?
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paper templates with. Web this type of template is also known as a coordinate plane graph paper which has drawn on it the x and y axis and filled with numbers.
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Web Explore Math With Our Beautiful, Free Online Graphing Calculator.
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Web cartesian standard graph paper: Web this graph paper generator will produce a single or four quadrant coordinate grid with various types of scales and options. Interactive worksheets bring
printable worksheets to life! All our mathematical scholars can here check out the fine template of the plane coordinate graph paper.
Related Post:
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EconEdLink - Production Possibilities Curve
Production Possibilities Curve
Time: 45 mins,
Updated: July 26 2024,
Students will be able to:
• Explain a production possibilities curve.
• Use a production possibilities to curve to calculate opportunity costs.
In this economics lesson, students will use a production possibilities curve to learn about scarcity and opportunity cost.
The production possibilities curve is a good tool for illustrating the concepts of scarcity, opportunity cost and the allocation of resources in an economic system. Distribute copies of the warm-up
activity. Tell students to read the article, “Will Grier Headlines Growing List of College Football Stars Skipping Bowl Games” about football players skipping bowl games and answer the questions at
the end. Review their answers, reminding students that everyone must make choices because of scarcity (unlimited wants and limited resources). Explain that this lesson will focus on the way
societies/countries make choices about what to produce with its limited supply of resources.
Introduce the production possibilities curve by telling students that governments (societies, countries, economic systems) make choices about what to produce with their limited resources; therefore,
they cannot produce everything they want in unlimited quantities. At some point, governments must decide three questions: what to produce, how to produce, and for whom to produce. The production
possibilities curve helps to answer those questions. Use the YouTube video Production Possibilities Curve-Econ 1.1 to help students understand the basic principles of a production possibilities
curve. Encourage them to take notes during the video because they will need the information to complete the group and individual activities.
Group Activity
Put students in small groups and distribute copies of the Production Possibilities Curve group activity, showing the production possibilities curve for the country of Alpha. This activity requires
them to apply what they have learned by using the information on the curve to answer a series of questions. Review their answers after they have completed the exercise using Production Possibilities
Curve Answers.
Play the Kahoot! Game with your class. Divide the students into teams or play using 1-1 devices.
Activity 1
Economics allows you to consider the relative cost of your decisions. In other words, you can “build” your own production possibilities curve when you make a decision. Suppose you have homework to do
but you would prefer working out to improve your soccer game before the next practice. And, of course, you have limited time to complete both activities. How can you decide which one you should do?
Or should you try to do both? A PPC will help you see the opportunity cost of your decisions. Try this: Do as many pushups as you can in 30 seconds and record the number. Then, solve as many
homework problems as possible in 30 seconds. Record that number. Using pushups on one axis and homework problems on the other, plot a straight line PPC. Calculate the relative opportunity costs. On
which activity do you have the lowest opportunity cost? If you behave economically, chances are you will engage in the activity with the lowest opportunity cost. However, you may also need to
consider the potential costs of not completing your homework!
Activity 2
This activity provides advanced mathematical analysis of the production possibilities curve using the following scenario. Student may prefer to use graph paper to complete the assignment.
Omega is a small tropical island that produces pearls (P) and fish (F). Omega’s production possibilities curve is given by P = 2L.5K.5 – .3F2 where L is the size of the labor force (400 people) and
K is the number of capital goods which is 100. Have the students answer the following questions:
1. What is the maximum number of fish that can be produced? Call this number F*. What is the maximum number of pearls that can be produced? Call this number P*.
2. Calculate pearl production for points for theses combinations. F = 10, 20, 30, 36. Graph the PPC for Omega. Label the X-axis, Fish; label the Y-axis, Pearls.
3. Is this PPC consistent with increasing costs?
4. Is the output combination 1/2F*, 1/2P* attainable? Is this point efficient? Why or why not?
5. What is the opportunity cost of more fish when 10 fish are produced? 20 fish? 30 fish?
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Visualizing a Theory of Everything!
I am playing around with the Wolfram Tweet-a-Program, and the Wolfram Language (i.e. Mathematica) on the Wolfram Cloud.
What’s really cool is that you can now interact with advanced math and HPC on your phone/tablet.
Here are a few results…
BTW – you will need a WolframID (and be logged into WolframCloud.com) to interact with these pages.
Octonions: The Fano Plane & Cubic
Navier-Stokes Chaos Theory, 6D Calabi-Yau and 3D/4D Surface visualizations
Solar System (from NBody Universe Simulator)
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Mathematics: the foundation of reality
“Our universe is not just described by mathematics — it is mathematics.” That’s the conclusion of Max “Peg Leg” Tegmark, an astrophysicist at the Massachusetts Institute of Technology. But he ain’t
nuts, even if he sounds as if he’s a couple of planets short of a solar system.
His argument is actually kinda convincing. In a paper that he says is a director’s cut of an article that he wrote for New Scientist which in turn was based on an earlier paper of his called The
Mathematical Universe, he starts with a question: if we accept that the universe has a reality independent of ourselves, then what sort of reality is it?
Peg Leg Tegmark argues that it has to be free of any kinda of physical or cultural bias so that it is the same for all aliens wherever they may be in the universe. The only logical system that fits
this description is the one that underlies mathematics, he says. Therefore the universe is mathematics.
Peg Leg Tegmark reckons that this line of thought leads to a number of curious predictions that are actually testable by observation. F0r example, he says that a measurement of the distribution of
dark energy within our universe would be a decent test.
Just how we might make that measurement and what exactly we would be looking for is harder to say.
Still we can hardly expect the trifling details behind the actual observation and measurement of the universe to trouble a thinker like Tegmark. All in all, his paper makes a fine addition to the
general framework of [S:untestable philosophy:S] cosmology.
Ref: arxiv.org/abs/0709.4024: Shut Up and Calculate
What reason is there to assume that it must be possible to represent the laws of nature algorithmically, or by any symbolic system at all? Other than the fact that some people find the idea that it
might not be possible upsetting, I mean.
The idea that “universe has a reality independent of ourselves” and the idea that it must be fully describable via any symbolic system are two separate propositions. The first is a nearly a given to
most scientifically minded people despite the fact that it is a statement of metaphysics. The second is, as far as I know, supported by no actual evidence or rigorous logical theory whatsoever and is
simply an assumption.
The idea that “universe has a reality independent of ourselves” and the idea that it must be fully describable via any symbolic system are two separate propositions. The first is a nearly a given to
most scientifically minded people despite the fact that it is a statement of metaphysics. The second is, as far as I know, supported by no actual evidence or rigorous logical theory whatsoever and is
simply an assumption.
The Tegmark papers build exactly the bridge between those two apparently independent assumptions.
Still arguable, of course, but his arguments are nonetheless interesting.
Another possibility is that mathematics is the result of our abiliy to spot patterns, something we’ve got from evolution not reality.
This amazing pattern recognition capability allows us to classify the world by certain rules.
But that don’t mean the universe is built from those rules.
KFC, did you even read what you wrote. “but that doesnt mean the universe is built from those rules”, but that the universe kinda fell into alignment with patterns we kinda came up with.
there are rules building everything, even if we know the formulas currently or not.
you shouldnt reply on physics websites ever again
The Tegmark papers build exactly the bridge between those two apparently independent assumptions.
Still arguable, of course, but his arguments are nonetheless interesting.
Yes, I have been following up on this with more reading. The “director’s cut” paper itself makes more sense than the above summary would indicate. I find it unconvincing in the end, but it it at
least is attempting to address the core issue.
Taco, nice to know who is the referee for allowed postings. I sure hope I have not offended your infinite wisdom…
What reason is there to assume that it must be possible to represent the laws of nature algorithmically, or by any symbolic system at all?
Indeed. I’ve always thought to be a major over-assumption. Math seems little more than a human means to abstract and explain the universe (math is bound by the universe, not vice versa), much the
same way that language is used to abstract human expression.
You’re on to something. I’m no mathematician, but to assume that everything can be broken down algorithmically/mathematically seems to run head on into Godel’s incompleteness theorem or Turing’s
halting problem.
No, it doesn’t run head on into Godel’s incompleteness theorem or Turing’s halting problem at all. Godel and Turing basically say that “there are some scientific facts that we will never know for
sure to be true (, even in an algorithmic world)”. With a slightly off analogy: Concluding from Turing that the universe can’t be broken down algorithmically is like concluding from our inability to
predict next year’s weather that weather can’t be broken down mathematically.
The limitations of an internal observer of the universe are not to be confused with the properties of the universe itself.
It has often been mussed, “Why does mathematics work so well to describe the universe.”.
In mathematics one has input operation output, that is to say that the nature of mathematics is fundamentally causal.
That mathematics works so well (though not proof) may be taken to indicate that the universe is causal, but if the universe is causal then indeed “the universe has a reality independent of
But a causal universe does not sit well with many involved in QM (“The first is a nearly a given to most scientifically minded people”???)
The is also a reason to wonder if QM is complete, as QM leads to a philosophical collision between the nature of mathematics and why it would work so well to describe a non causal universe.
The universe can be seen as an homogeneous, gradient driven environment composed of penetrating waves with non-commutative geometry & algebra, both like multi-components system composed of colliding
particles, fulfilling the Peano algebra and Fermi-Dirac statistics.
These perspectives are mutually exclusive, therefore I don’t think, the Universe is strictly causal. The only reason, why we are seeing’ it deterministic is the fact, the chaotic environment doesn’t
spread the energy at the distance. By AWT the Universe appears like dense Perlin (scale invariant) noise, from which we can see only causal gradients, which are enabling the energy spreading at the
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I am sorry if this has little to do with time-
I have found a way to find the frequency of a mass if it was a wave, and vice versa.
m=f x (7.366499973 x 10^-51)
f=m x (1.357496781 x 10^50)
I am sorry if this has little to do with time-
Is that for a photon?
I am sorry if this has little to do with time-
Yep. I think h is the lowest energy of a photon. I'm not too sure.
I am sorry if this has little to do with time-
Can you put that in a context that a musician could understand? Bercause I have to tell you, your on to something. All matter, all forms of energy have their own distinct resonance/sounds. All matter
is actually a wave form if I am not mistaken.
If you could spend some time on it, what note would it be. This is a serious question.
I am sorry if this has little to do with time-
Ok for one the number 7.366499973 x 10^-51
unfortunately is not a frequency it is a constant to convert the frequency of light to um grams???
If we used to in terms of say middle A 440 Hz
a beam of light going at that frequency would be a u.l.f (ultra low frequency)
and would have a mass of 3.24125998812e-48 grams.
here is the mass of other frequencies of the 3rd octive
440 A 3.24125998812e-48
466.16376151809 A# 3.43399533663659e-48
493.883301256124 B 3.63819132536839e-48
523.251130601197 C 3.85452943944594e-48
554.365261953744 C# 4.08373168721439e-48
587.329535834815 D 4.32656300986927e-48
622.253967444162 D# 4.58383383437656e-48
659.25511382574 E 4.85640277819742e-48
698.456462866008 F 5.14517951484412e-48
739.988845423269 F# 5.45112780983081e-48
783.990871963499 G 5.77526873715136e-48
830.60939515989 G# 6.11868408701888e-48
Well since the constant is not a frequency itself lets go with the frequencies of visible light.
visible light has a frequence from dim red of 4.2827494e+14hz to faint violet of 7.49481145e+14 Hz
On a music scale visible light in in the 43th octive.
This is derived by taking the log base 2 of 440 which is about 8.78 noting that this represents the third octive and subtracting it from log base 2 of dim red 48.61 and adding three getting 42.83.
now we can do colors to notes and notes to colors.
1st Colors to notes
Magenta is G
Blue is D#
Cyan is C#
Green is C
Yellow is A#
Red is A
Now notes to colors
A is a dim red
A# is yellow with a faint hint of orange
B is a lime color
C is a green
C# is cyan
D is a sky blue
D# is blue
E is an indigo
F is a purple
F# is a violet
G is magenta
G# is rose
Now in terms of sympathetic viberations. It is true that a higher frequency can create resonance in structure that has it length or a harmonic multiple of it, but it needs at lease half a wave length
for a closed pipe or a full wavelength for an open pipe for there to be nodes. Thus the 43rd octive would have to be achieved to get visible light.
The thing is even if you did make these 43rd octive waves you would be making mechanical waves and not electromagnetic waves.
I am sorry if this has little to do with time-
Thank You Phoenix! That was most kind of you!!
I am sorry if this has little to do with time-
Originally posted by Phoenix@Sep 1 2004, 01:52 AM
Ok for one the number 7.366499973 x 10^-51
unfortunately is not a frequency it is a constant to convert the frequency of light to um grams???
If we used to in terms of say middle A 440 Hz
a beam of light going at that frequency would be a u.l.f (ultra low frequency)
and would have a mass of 3.24125998812e-48 grams.
here is the mass of other frequencies of the 3rd octive
440 A 3.24125998812e-48
466.16376151809 A# 3.43399533663659e-48
493.883301256124 B 3.63819132536839e-48
523.251130601197 C 3.85452943944594e-48
554.365261953744 C# 4.08373168721439e-48
587.329535834815 D 4.32656300986927e-48
622.253967444162 D# 4.58383383437656e-48
659.25511382574 E 4.85640277819742e-48
698.456462866008 F 5.14517951484412e-48
739.988845423269 F# 5.45112780983081e-48
783.990871963499 G 5.77526873715136e-48
830.60939515989 G# 6.11868408701888e-48
Well since the constant is not a frequency itself lets go with the frequencies of visible light.
visible light has a frequence from dim red of 4.2827494e+14hz to faint violet of 7.49481145e+14 Hz
On a music scale visible light in in the 43th octive.
This is derived by taking the log base 2 of 440 which is about 8.78 noting that this represents the third octive and subtracting it from log base 2 of dim red 48.61 and adding three getting
now we can do colors to notes and notes to colors.
1st Colors to notes
Magenta is G
Blue is D#
Cyan is C#
Green is C
Yellow is A#
Red is A
Now notes to colors
A is a dim red
A# is yellow with a faint hint of orange
B is a lime color
C is a green
C# is cyan
D is a sky blue
D# is blue
E is an indigo
F is a purple
F# is a violet
G is magenta
G# is rose
Now in terms of sympathetic viberations. It is true that a higher frequency can create resonance in structure that has it length or a harmonic multiple of it, but it needs at lease half a wave
length for a closed pipe or a full wavelength for an open pipe for there to be nodes. Thus the 43rd octive would have to be achieved to get visible light.
The thing is even if you did make these 43rd octive waves you would be making mechanical waves and not electromagnetic waves.
Ahhh, Phoenix, you were definately missed here.
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Example Staking Dynamics
To illustrate the dynamics of this system, consider a toy scenario with three delegators, Alice, Bob, and Charlie, and two validators, Victoria and William. Tendermint consensus requires at least
four validators and no one party controlling more than of the stake, but this example uses only a few parties just to illustrate the dynamics.
For simplicity, the base reward rates and commission rates are fixed over all epochs at and , . The PEN and dPEN holdings of participant are denoted by , , etc., respectively.
Alice starts with dPEN of Victoria’s delegation pool, Bob starts with dPEN of William’s delegation pool, and Charlie starts with unbonded PEN.
• At genesis, Alice, Bob, and Charlie respectively have fractions , , and of the total stake, and fractions , , of the total voting power.
• At epoch , Alice, Bob, and Charlie’s holdings remain unchanged, but their unrealized notional values have changed.
□ Victoria charges zero commission, so . Alice’s dPEN(v) is now worth PEN.
□ William charges commission, so . Bob’s dPEN(w) is now worth , and William receives PEN.
□ William can use the commission to cover expenses, or self-delegate. In this example, we assume that validators self-delegate their entire commission, to illustrate the staking dynamics.
□ William self-delegates PEN, to get dPEN in the next epoch, epoch .
• At epoch :
□ Alice’s dPEN(v) is now worth PEN.
□ Bob’s dPEN(w) is now worth PEN.
□ William’s self-delegation of accumulated commission has resulted in dPEN(w).
□ Victoria’s delegation pool remains at size dPEN(v). William’s delegation pool has increased to dPEN(w). However, their respective adjustment factors are now and , so the voting powers of
their delegation pools are respectively and .
☆ The slight loss of voting power for William’s delegation pool occurs because William self-delegates rewards with a one epoch delay, thus missing one epoch of compounding.
□ Charlie’s unbonded PEN remains unchanged, but its value relative to Alice and Bob’s bonded stake has declined.
□ William’s commission transfers stake from Bob, whose voting power has slightly declined relative to Alice’s.
□ The distribution of stake between Alice, Bob, Charlie, and William is now , , , respectively. The distribution of voting power is , , , respectively.
□ Charlie decides to bond his stake, split evenly between Victoria and William, to get dPEN(v) and dPEN(w).
• At epoch :
□ Charlie now has dPEN(v) and dPEN(w), worth PEN.
□ For the same amount of unbonded stake, Charlie gets more dPEN(w) than dPEN(v), because the exchange rate prices in the cumulative effect of commission since genesis, but Charlie isn’t charged
for commission during the time he didn’t delegate to William.
□ William’s commission for this epoch is now PEN, up from PEN in the previous epoch.
□ The distribution of stake between Alice, Bob, Charlie, and William is now , , , respectively. Because all stake is now bonded, except William’s commission for this epoch, which is
insignificant, the distribution of voting power is identical to the distribution of stake.
• At epoch :
□ Alice’s dPEN(v) is now worth PEN.
□ Bob’s dPEN(w) is now worth PEN.
□ Charlies’s dPEN(v) is now worth PEN, and his dPEN(w) is now worth PEN.
□ William’s self-delegation of accumulated commission has resulted in dPEN(w), worth PEN.
□ The distribution of stake and voting power between Alice, Bob, Charlie, and William is now , , , respectively.
This scenario was generated with a model in this Google Sheet.
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The n-Category Café
December 25, 2013
The Long Grind of Writing a Book
Posted by Tom Leinster
I’m using the quiet of Christmas to finish writing a book, Basic Category Theory. It’s nothing revolutionary: just a short introduction to the subject, based on courses I’ve taught. But the process
of book-writing is demanding and maddening enough that I wanted to take a moment to reflect on why that is — and why you hear authors complain so much.
Put another way, I’m taking a break from the tedium of writing a book to write about the tedium of writing a book. I hope it’s not tedious.
Posted at 3:28 PM UTC |
Followups (50)
December 21, 2013
Commuting Limits and Colimits over Groups
Posted by Tom Leinster
Limits commute with limits, and colimits commute with colimits, but limits and colimits don’t usually commute with each other — with some notable exceptions. The most famous of these is that in the
category of sets, finite limits commute with filtered colimits.
Various other cases of limit-colimit commutation are known. There’s an nLab page listing some. But it seems that quite an easy case has been overlooked.
It came to light earlier this week, when I was visiting Cambridge. Peter Johnstone told me that he’d found a family of new limit-colimit commutations in the category of sets, I asked whether his
result could be simplified in a certain way (to involve groups only), and we both realized that it could not only be simplified, but also generalized.
Here it is. Let $G$ and $H$ be finite groups whose orders are coprime. View them as one-object categories. Then $G$-colimits commute with $H$-limits in the category of sets.
Posted at 12:55 AM UTC |
Followups (10)
December 10, 2013
A Technical Innovation
Posted by Tom Leinster
Here’s a new feature of the Café, thanks to our benevolent host Jacques Distler. If you ever want to see how someone has created some mathematical expression on this blog, there’s an easy way to do
With Firefox, you simply double-click on the expression. Try it: $A \times B^A \to B$ or $x_{m n}$ or $\Biggl( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \Biggr).$ A window should pop up showing the
TeX source.
With other browsers, I’m not so sure. Try double-clicking. If that doesn’t work, then, according to Jacques’s instructions, you “bring up the MathJax context-menu for the formula, and choose Show
Math As $\to$ Annotation $\to$ TeX”. I don’t know how one brings up this menu. Does anyone else know? (Update: right-click in Chrome, Explorer and Opera, and control-click in Safari. Thanks to those
who responded.)
Once you’ve made the TeX source appear, you can cut and paste to your heart’s content. Of course, most users here are fluent in LaTeX. But like most math-oriented websites, we use a variant of TeX
that’s a little different from standard LaTeX, so this should turn out to be a helpful feature.
Posted at 1:41 AM UTC |
Followups (18)
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Rui Tian^1, Franciszek Hennel^1, and Klaas P Pruessmann^1
^1Institute for Biomedical Engineering, ETH Zurich and University of Zurich, Zurich, Switzerland
Super-resolution MRI with 1D phaseless encoding achieves high-resolution with immunity to shot-dependent phase fluctuation by simultaneously acquiring multiple k-space bands. We now explore a 2D
extension of this technique to facilitate more k-space sampling strategies. Two distinct encoding schemes were analyzed and tested with EPI acquisition. By properly adjusting the overlapping of the
mixed k-space bands, the 2D phaseless encoding could also be combined with the spiral acquisition. The amplitude modulation caused by band overlapping was eliminated by an inverse filter during
reconstruction. The overlapped bands regions were also exploited to provide information about unexpected bands errors for post-processing corrections.
By translating the philosophy of structured illumination microscopy (SIM) to MRI, the phaseless encoding has been applied in one dimension to reconstruct a super-resolution (SR) MRI image from
several low-resolution acquisition cycles with different shifts of a sub-pixel tagging pattern [1][2]. This allows multi-shot scanning without sensitivity to motion-related phase fluctuations [3].
The extension of this technique to two dimensions should allow further shortening of sampling time in each cycle (and reduction of off-resonance effects) and provide isotropic resolution enhancement
with non-Cartesian trajectories, e.g. spirals.
Multiple k-space regions can be mixed by a tagging pattern in various ways. For example, a rapid repetition of two orthogonal tags [1] excites a 3x3 array of k-space “tiles” and may provide a 3-fold
resolution enhancement in both directions. Alternatively, the same enhancement can be achieved by a rotation, and, optionally, scaling of a 1D tag which mixes three tiles at a time. The goal of this
study is to demonstrate and compare different 2D phaseless encoding schemes with respect to their compatibility with various k-space trajectories, the complexity of the SR reconstruction, the minimum
scan time and the signal-to-noise efficiency. We also present optimized post-processing steps which improve the robustness of this method.
Experiments were carried out on a 3T Achieva scanner (Philips Netherlands). Three encoding schemes were implemented:
1. Two consecutive orthogonal sinusoidal tags with three shifts each, covering a rectangular 3x3 pattern of k-space tiles (Fig.1A).
2. A single 1D sinusoidal tag with 4 rotations and 3 shifts with each direction, covering a circularly arranged pattern (Fig.1B).
3. The same pattern as (2) but with diagonal directions scaled to achieve rectangular arrangement as in (1) (Fig.1C).
In all cases, the overlapping of the k-space tiles was programmable and typically set from 10% to 15% depending on the k-space trajectory.
The reconstruction followed the similar principles of the 1D SR reconstruction [3], with the modified tag shifts cycling and the adapted corrections for tagging distortions in the scheme No.1. The
SNR between the schemes No.1 and No.3 was examined by quantifying the noise propagation through the reconstruction matrix and tested experimentally with the GRE-EPI sequence in-vivo.
The spiral acquisition was conducted by both the schemes No.2 and No.3 to compare the efficiency in the overlapping of the round-shaped tiles. An inverse filter was calculated based on the known
effective k-space window to eliminate the amplitude modulation. The phase of the overlapped k-space pixels contributed from neighboring bands were compared to estimate any unexpected constant phase
shifts, with the presumption that these pixels should be identical in the absence of any errors.
The acquisition cycles for all the schemes above were reconstructed properly, for which the results are summarized:
1. The adapted reconstruction for the 2D phaseless encoding achieved the expected resolution enhancement without shot-dependent phase fluctuation (Fig.2).
2. Covering the same rectangular pattern of k-space tiles, the scheme with the rotational 1D tag was analyzed to have 1.81 times higher SNR than the one with the two orthogonal tags, which was also
experimentally measured with 1.85 times higher SNR verifying our theories.
3. For the rotational schemes, the circularly arranged pattern naturally fits the spiral acquisition better than the rectangular arrangement, due to the less minimum overlapping of k-space tiles to
prevent empty holes between resolved bands (Fig.3).
4. The apodization window on the low-resolution scans was replicated on the top of all resolved k-space tiles, which led to a complicated effective k-space window and thus amplitude modulation.
However, the amplitude modulation was easily corrected by the inverse filter (Fig.4).
5. A set of constant phase offsets on resolved neighboring bands were estimated successfully and used to effectively remove the ringing artifacts through the post-processing correction (Fig.5).
With the trade-off of lower SNR, phaseless encoding can achieve super-resolution in two dimensions without suffering from the motion-related phase fluctuation. The 2D resolution enhancement beyond
the factor of three can be implemented in a similar manner as in the 1D 5x-SR experiment [4] but will definitely lead to further SNR loss. Noticeably, the 2D scheme with the rotational 1D tag leads
to less SNR loss than the scheme with the two orthogonal tags. To facilitate non-cartesian trajectories such as spirals, the rotational 2D scheme with the circularly arranged pattern was found to be
optimal and is interestingly quite similar to the SIM. The improved post-processing can also be utilized in the 1D phaseless encoding, for allowing the optimized anti-ringing filtering on
low-resolution scans and the corrections for any unexpected homogeneous tagging errors.
No acknowledgement found.
1. Ropele, S., Ebner, F., Fazekas, F., & Reishofer, G. (2010). Super-resolution MRI using microscopic spatial modulation of magnetization. Magnetic Resonance in Medicine, 64(6), 1671-1675.
2. Hennel, F., & Pruessmann, K. P. (2016). MRI with phaseless encoding. Magnetic Resonance in Medicine, 78(3), 1029-1037. doi:10.1002/mrm.26497
3. Hennel, F., Tian, R., Engel, M., & Pruessmann, K. P. (2018). In-plane “superresolution” MRI with phaseless sub-pixel encoding. Magnetic Resonance in Medicine. doi:10.1002/mrm.27209
4. Tian, R., Hennel, F., & Pruessmann, K. P. (2018, June). “Exploring the Limits of Super-resolution MRI with Phaseless Encoding.” In proceedings of the joint annual meeting of ISMRM-ESMRMB 2018,
Paris, France, abstract 2671
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Slope Performance Trend | ChartSchool | StockCharts.com
Trend direction and relative strength are important components of any investment strategy designed to outperform the overall market. This article will show chartists how to use the slope indicator to
define trend direction and quantify relative strength. Sectors that show relative strength and are in uptrends are worthy of long positions. Sectors that show relative weakness and are in downtrends
should be avoided. This strategy goes to the heart of a basic investment philosophy: buy the strong and avoid the weak.
The Slope Indicator
The slope indicator is at the heart of this strategy, so we will first explain what it tells us and how it works. Even though it sounds complicated, the slope indicator is pretty easy to understand.
All you need to know is that the trend is up when the slope is positive and is down when the slope is negative. For those looking for a technical explanation, the slope indicator measures the rise
over the run for a linear regression. Fortunately, we have algorithms and charting software to do all the calculation work. In SharpCharts, chartists can use the Raff Regression Channel to plot a
linear regression, which is the middle line. The chart below shows three Raff Regression Channels covering three twelve-month periods. The slope of the first linear regression (blue) is relatively
flat, the second slope (red) is clearly down, and the third slope (green) is up.
The indicator window shows the actual values of the slope indicator. Remember, the slope measures the rise over run for the linear regression. This is the ending value of the linear regression less
the beginning value divided by the timeframe. If the ending value were 35, the beginning value 29 and the run 12, then the slope would be .5 (35 - 29 = 6, 6/12 = .50). The slope indicator is zero
when the linear regression is flat, positive when the linear regression slants up and negative when the linear regression slopes down. The steepness of the slope is also reflected in the value. Slope
values well above zero reflect a sharply rising linear regression (uptrend), while slope values well below zero reflect a sharply falling linear regression (downtrend).
Trend Direction
Trend direction can be defined with the 12-month slope of the closing prices. This means we will use monthly charts to define the trend based on the direction of the one-year slope. The chart below
shows the Utilities SPDR (XLU) with a 12-month slope indicator changing direction five times in ten years. There was one whipsaw (bad signal) in early 2008, but the other signals foreshadowed decent
trends. No indicator is perfect and whipsaws are unavoidable. However, most of the time, the 12-month slope will produce fewer whipsaws than the 12-month simple moving average.
Relative Performance
Chartists can use the price relative to measure relative strength and relative weakness. The price relative is a ratio of the underlying security divided by the benchmark index. In this example, we
will use the nine S&P Sector SPDRs as the trading vehicles and the S&P 500 ETF (SPY) as the benchmark index. To measure the performance of the Technology SPDR (XLK) relative to the S&P 500 ETF,
chartists would plot the ratio of the two symbols (XLK:SPY). This ratio rises when XLK outperforms and shows relative strength, and falls when XLK underperforms and shows relative weakness.
The chart above shows the price relative or XLK:SPY ratio in the main window and the 12-month slope in the indicator window. Overall, the price relative declined from 2004 to mid-2006 and advanced
from mid-2006 until early 2012. The 12-month slope moved from positive to negative as these relative performance trends strengthened and weakened. A positive slope shows strong outperformance, while
a negative slope shows strong underperformance.
Strategy Details
The strategy is quite straightforward. Buy signals are generated when the 12-month slope for the price chart is positive and when the 12-month slope for the price relative is positive. This
two-pronged approach ensures that the sector is in an uptrend and shows relative strength, which makes for a powerful combination. A sell signal is generated when the 12-month slope for the price
chart is negative and the 12-month slope for the price relative is negative. Again, this ensures that the sector is in a downtrend and shows relative weakness.
The chart above shows the Consumer Discretionary SPDR (XLY) with the 12-month slope for the price chart in the top indicator window, the price relative in the middle indicator window and the 12-month
slope of the price relative in the bottom window. There were only five signals in ten years, which makes this a long-term model. The sell signal in 2005 caused a whipsaw and missed most of the surge
in the third quarter of 2006. Despite this whipsaw, this strategy would have exited XLY before the 2008 bear market and caught most of the bull market that started in 2009.
Increasing Signal Frequency
While this strategy can be used to define the big trend, chartists should implement another strategy to time short- or medium-term movements. After all, some traders, or even investors, may need more
than five signals every ten years. Using the slope indicators on the monthly chart, the Industrials SPDR (XLI) triggered a buy signal in November 2010. While the trend was clearly up, the risk-reward
ratio did not look so great after such a sharp advance. Chartists could then turn to the daily chart and the Commodity Channel Index (CCI) to generate bullish signals within this uptrend.
The chart above shows XLI with daily bars and the Commodity Channel Index (CCI) in 2010. With the long-term trend up, bullish signals are taken and bearish signals are ignored. A bullish signal
triggers when CCI becomes oversold and then moves above the zero line, which occurred five times in 2010. The June signal resulted in a whipsaw (bad trade), but the other signals foreshadowed sizable
advances. Even though bearish signals are ignored, chartists would need to employ a stop-loss or profit-taking strategy to lock in gains.
Strong and Weak Sectors
Applying the slope to the price relative can also help chartists differentiate strong sectors from weak sectors. In this regard, investors can focus on strong sectors for long positions and avoid or
short weak sectors. The chart below shows the 12-month slope for six price relatives. The green boxes highlight positive slopes and relative strength, while the red boxes highlight negative slopes
and relative weakness.
In early 2012, only the technology (XLK:SPY) and consumer discretionary (XLY:SPY) sectors were showing relative strength, as measured by the 12-month slope of their price relatives. Chartists could
have used this information to focus their buying power on stocks within these two sectors. The other four sectors showed relative weakness because the slopes for their price relatives were in
negative territory. Chartists could have used this information to avoid these sectors in 2012.
The Bottom Line
By applying the slope indicator to the price chart and price relative, you can quantify the price trend and relative performance with one indicator. A positive slope indicates an uptrend and a
negative slope indicates a downtrend. It is as easy as that. Even though 12-month slopes were used in this example, chartists can adjust the timeframe to suit their needs. A 13-week slope could be
used for medium-term timing, while a 20-day slope could be used for the short-term timing. Keep in mind that this article is designed as a starting point for trading system development. Use these
ideas to augment your trading style, risk-reward preferences and personal judgments.
Further Study
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Marketing Mix Model for All: Using R for MMM - Trevor With Data
Marketing Mix Model for All: Using R for MMM
Understanding the ROI across all of your paid marketing channels is a top priority for senior-level executives across every industry and every geographical market. Getting a clear sense of the ROI
on each channel allows companies to answer really important questions. For example:
• What will happen if I increase my Email spend by 20%?
• What is the level of saturation I can expect from my Paid Search channel?
• How do I incorporate seasonality into my budget allocation strategy?
• How do I optimize my budget allocation across all of my paid channels?
• Which geographies should I focus my efforts?
These are consequential questions that have the potential to have a major impact on how a business operates. With that, it’s no wonder that companies devote significant time, energy , and resources
to creating (or purchasing) a Marketing (or Media) Mix Modeling framework. The aim of this post is to show you how you can use the data that you already have in conjunction with open source tools to
create your own MMM solution.
We will take a look a more in-depth look at the example Trevor and I co-presented at Adobe’s Digital Marketing Summit this past March. While we will only be examining a single channel, a creative
analyst could use this as the foundation for a larger MMM effort. Many times, the analysts I work with are given a target for the amount of revenue a particular marketing channel is responsible for
contributing to the overall business. The question I am often asked is, “How much do I need to spend in my channel to reach my revenue goal?”
Let’s start by taking a quick look at the data:
For this example, we are looking at Email channel data. This data set is aggregated to the campaign level, meaning each row represents one unique campaign, which is designated in the first column.
Since we are examining the single channel case, it comes as no surprise that each campaign roles up to the Email channel. The third column, Return, is some measure of value to your site. This will
vary depending on two things:
1. What is valuable to your site? Are you in the retail industry where customers are making purchases directly on your site? If so, you may want to use Revenue or Orders for your measure of value.
Are you a media site where consumption of content is the most important metric? If so, you may want to use blog views or subscriptions as your measure of value. Return will always be specific
to your industry and your business organization.
2. What type of attribution model are you using? The amount of Return allocated to each campaign will be directly affected by the attribution method that you deploy. Most companies are using
simple, rules-based attribution methods like first touch or last touch. These are great places to start. If your organization is a little further in its analytics journey, you may want to think
about using a more algorithmic, data-driven approach.
The last column, Spend, is a measure of cost associated with that campaign. This could include direct costs as well as indirect costs. For example, the cost of each email campaign is typically tied
to the number of emails that were sent – this is a direct cost. We may want to consider including some of the indirect costs of the time and resources that created the visual and manage the email
with a vendor. The indirect costs are typically more difficult to quantify, so including just the direct costs is usually a good starting point.
Let’s continue by loading and visualizing the data in R. We are using the gglot2 library to create our plot. The plot is stored in the p1 object:
# Load libraries #
# Load data files #
fileName = "oneChannelData.csv"
myData = read.csv(file = fileName, header=TRUE, sep=",")
# Plot data #
channelName = as.character(myData$Channel[1])
maxX = 1.05*max(myData$Spend)
maxY = 1.05*max(myData$Return)
myPlotDataDF = data.frame(Return = myData$Return, Spend = myData$Spend)
simpleScatterPlot <- ggplot(myPlotDataDF, aes(x = Spend, y = Return)) +
geom_point(color="black") +
theme(panel.background = element_rect(fill = 'grey85'),
panel.grid.major = element_line(colour = "white")) +
coord_cartesian(ylim = c(0,maxY), xlim = c(0,maxX)) +
scale_x_continuous(labels = dollar) +
scale_y_continuous(labels = comma) +
This is a fairly simple scatter plot. On the x-axis we have Spend and on the y-axis we have Return. The black dots represent the individual data points from my data set. Just from this simple
plot, we can start to see the relationship between Spend and Return. At first glance it looks like it might be a good idea to model this relationship with a linear model. I would advise against
this for a couple of different reasons:
1. A linear model assumes that you have infinite growth.
2. There are real-world phenomena like market saturation and email fatigue that suggest infinite growth is not actually possible.
We need a model that exhibits diminishing marginal returns. Diminishing marginal return means that the rate of return will decrease the more you spend. The ADBUDG model is a very flexible model that
incorporates diminishing marginal returns. The ADBUDG model is defined as follows:
Just as a linear model is governed by parameters (slope and intercept), the ADBUDG model is also governed by parameters. There are four ADBUDG model parameters:
1. A – The maximum amount of return possible for a campaign given a long term investment.
2. B – The minimum amount of return possible for a campaign given a long term investment
3. C – Controls the shape of the curve.
4. D – Represents initial market share or market saturation effects.
These four parameters actually give the shape and behavior of the model a lot of flexibility. How can we figure out what the values of these parameters should be? There is no build-in ADBUDG
function in R, so I used optimization techniques to determine the values of these parameters. This is where I would encourage you to put on your data science hats!
There are many different optimization functions in R. I’m going to demonstrate how to use the nlminb() function, which is part of the stats package in R (See the documentation here: https://
stat.ethz.ch/R-manual/R-devel/library/stats/html/nlminb.html). The nlminb() function works by minimizing the return value from some function. The nlminb() function has several arguments. We are
only going to require the following:
• objective – The function to be minimized
• start – Initial values for the parameters to be optimized
• lower – Lower bound for constrainted parameter optimization
• upper – Upper bound for constrained parameter optimization
• control – Additional control parameters
First thing we need to do is define our objective function. Just as in a linear model, the goal is to find parameter values that minimize the error between the actual data observed and the predicted
values from the model. To align with how the nlminb() function operates, we need to create a custom function that returns the total observed error based on the ADBUDG model.
Ufun<-function(x, Spend, Return) {
predictedReturn = x[2] + (x[1] - x[2])*((Spend^x[3])/(x[4] + (Spend^x[3])))
errorSq = (predictedReturn - Return)^2
sumSqError = sum(errorSq)
Let’s walk through the function step-by-step. There are three inputs to the function. The “x” variable is a vector of length four and represents the four parameters of the ADBUDG function. Spend
and Return are also vectors which correspond with the Spend and Return data from your data file. Their length will be the same as the number of data points in your data file. The temp1 object
calculates the predicted Return from the ADBUDG equation. That vector is then used in the next line when we calculate the squared error for our predicted values, which are stored in the temp2
object. Lastly, we sum the temp2 object to get a single value for the sum of squared errors, which is stored in the temp3 object. The temp3 object is the return object for this function. When we
minimize this function, we are essentially minimizing the sum of squared error, which is actually the same thing that is done in linear regression. See, you knew exactly what we were doing here,
With the hard part out of the way, we can create some start values, as well as upper and lower bounds:
startValVec = c(25000,100,1.5,100000)
minValVec = c(0,0,1.01,0)
maxValVec = c(500000, 500000, 2, 10000000)
These values should be based upon the Return values possible for you data. Return data is typically positive, meaning our minimum values for A and B are 0. Representing market share or saturation,
D must also be positive. The minimum value of C will ultimately depend on what metric you are using for Return. This data is based on revenue which should have an ROI greater than 1.01, otherwise,
we would not invest in that channel. So we will use 1.01 as our minimum value for C. In cases where the ROI is less than 1, we will want to reflect that in our minimum value vector. I’ve provided
maximum values that are quite large, which will allow the optimization function a large space to explore. Finally, I chose starting values that are somewhere inbetween. Final parameter optimization
should not depend on your starting values. If they do, there may be something else at play that we need to consider. The control arguments of the nlminb() function allow you to fine tune the
execution of the optimization function. We will use several arguments that limit the iterations possible. Our final nlminb() function is as follows:
Spend = myData$Spend,
Return = myData$Return)
The output of my optim.parms object tells us a few helpful pieces of information. The most important of which are the parameters themselves, as well as whether or not we had successful convergence.
> optim.parms
[1] 4.178821e+05 1.671129e+02 1.198282e+00 6.686292e+05
[1] 60148527
[1] 0
[1] 51
function gradient
[1] "relative convergence (4)"
A convergence code of 0 indicates successful convergence. We can extract our parameter values by taking a look at optim.parms$par. Let’s take a look at how these parameter values fit on the data we
observed. Once again, we’re going to use ggplot2 to build our plot. Now that the plot will include the ADBUDG model, building the data frame to feed into ggplot2 is a little more complex. My data
frame now includes a data frame for the ADBUDG curve as well as the actual data points:
a = optim.parms$par[1]
b = optim.parms$par[2]
c = optim.parms$par[3]
d = optim.parms$par[4]
curveDFx = seq(from=0, to=max(myData$Spend)*2, length.out=10000)
curveDFy = b+(a-b)*((curveDFx^c)/(d+(curveDFx^c)))
curveDF = data.frame(Spend = curveDFx, Return = curveDFy)
maxX = 1.05*max(curveDFx, max(myData$Spend))
maxY = 1.05*max(curveDFy, max(myData$Return))
myPlotDataDF = data.frame(Return = myData$Return, Spend = myData$Spend)
optimLineDF = data.frame(Spend = curveDFx, Return = curveDFy)
scatterPlotPlusFit <- ggplot(myPlotDataDF, aes(x = Spend, y = Return)) +
geom_point(color="black", shape = 16) +
theme(panel.background = element_rect(fill = 'grey85'),
panel.grid.major = element_line(colour = "white")) +
geom_line(data = optimLineDF, aes(x = Spend, y = Return, color = "darkgreen")) +
scale_color_manual(labels = "Optimized ADBUDG Fit",values=c('darkgreen')) +
theme(legend.title=element_blank(), legend.position = "bottom") +
coord_cartesian(ylim = c(0,maxY), xlim = c(0,maxX)) +
scale_x_continuous(labels = dollar) +
scale_y_continuous(labels = comma) +
ggtitle(paste(channelName, "Data & Model Fit", sep = " "))
Looks like the model fits pretty well! Now we are ready to use the model! Our original question was, “How much do I need to spend in my channel to reach my revenue goal?” My original data set had
about $390K worth of Return represented. Let’s suppose my return goal was actually $600K. How much should I increase my spend in each of my campaigns to reach that goal? Let’s start by explaining
and initializing some variables:
adbudgReturn = function(a,b,c,d,Spend){
adbudgReturn = sum(b+(a-b)*((Spend^c)/(d+(Spend^c))))
returnGoal = 600000
increment = 1000
oldSpendVec = myData$Spend
oldReturn = adbudgReturn(a,b,c,d,oldSpendVec)
newSpendVec = oldSpendVec
totalSpend = sum(oldSpendVec)
totalReturn = oldReturn
The adbudgReturn() function simply provides me with a projection of the return given a current level of Spend and a set of parameters. We will use this function when evaluating which channel to
allocate an incremental budget. I then created a function that evaluates the impact of an extra $1000 in each channel. The money is then given to whichever channel produced the highest incremental
return. It performs this action over and over again in a while loop until the total return has reached the return goal amount.
while(totalReturn < returnGoal){
incReturns = NULL
for(i in 1:length(oldSpendVec)){
oldSpendTemp = newSpendVec[i]
newSpendTemp = newSpendVec[i] + increment
oldReturnTemp = b+(a-b)*((oldSpendTemp^c)/(d+(oldSpendTemp^c)))
newReturnTemp = b+(a-b)*((newSpendTemp^c)/(d+(newSpendTemp^c)))
incReturns[i] = newReturnTemp - oldReturnTemp
winner = which.max(incReturns)
newSpendVec[winner] = newSpendVec[winner] + increment
totalSpend = totalSpend + increment
totalReturn = adbudgReturn(a,b,c,d,newSpendVec)
Let’s take a look at the recommended spend for each campaign. I’m organizing the recommended spend values in a data frame that looks very similar to my original data:
newReturnVec = b+(a-b)*((newSpendVec^c)/(d+(newSpendVec^c)))
myRecommendedData = data.frame(Campaign = myData$Campaign,
Channel = myData$Channel,
Return = newReturnVec,
Spend = newSpendVec)
sum(myRecommendedData$Spend) # Recommended Spend
sum(myRecommendedData$Return) # Estimated Return from Recommended Spend
sum(myRecommendedData$Spend)/sum(myData$Spend) - 1 # % Increase in Spend to get $600K
> sum(myNewData$Spend) # Recommended Spend
[1] 171455
> sum(myNewData$Return) # Estimated Return from Recommended Spend
[1] 603582.2
> sum(myNewData$Spend)/sum(myData$Spend) - 1 # % Increase in Spend to get $600K
[1] 0.5383339
From this output we can see exactly how much spend to allocate to each of my campaigns. We can also see that in order to reach my return goal of $600K, I had to spend a little over $171K. This
represents nearly a 54% increase in my overall spend.
There are a couple of different ways we can summarize and visualize these recommendations. One helpful visualization is a bar graph that shows the original and recommended spend for each campaign.
Let’s use ggplot2 to accomplish this. To do that, I will create a data frame that contains both the original spend and the recommended spend and then use the geom_bar() function of the ggplot2 to
create my plot:
# Graph current spend vs recommended spend #
compareDF = data.frame(Campaign = rep(myData$Campaign,2), spendType = rep(c("Actual Spend","Recommended Spend"), each=dim(myData)[1]), Spend = c(myData$Spend, myRecommendedData$Spend))
barChart <- ggplot(data=compareDF, aes(x=Campaign, y=Spend, fill=spendType)) +
geom_bar(stat="identity", color="black", position=position_dodge())+
name = "") +
scale_y_continuous(name="Spend", labels = dollar) +
theme(axis.text.x = element_text(angle = 45, hjust = .75)) +
ggtitle("Breakdown of Spend by Campaign")
A table format would be another simple, but useful visualization. For this representation of the data, I will add a percentage difference measure to show how spend changed between my original and
recommended spend amounts:
percDiff = (myNewData$Spend - myData$Spend)/myData$Spend
summaryDF = data.frame(Campaign = myNewData$Campaign,
Channel = myNewData$Channel,
actualSpend = dollar_format()(myData$Spend),
recommendedSpend = dollar_format()(myNewData$Spend),
percDiff = percent((myNewData$Spend - myData$Spend)/myData$Spend))
So now we have an answer to the amount of increased spend I will need to reach my return goal of $600K. While this is a very simple example, it has proven to be very useful for a number of the
analysts I’ve worked with. This code actually serves as the foundation of the cross-channel Marketing Mix Modeling framework I’ve developed. Hopefully this has provided a good introduction into MMM
and how you can use tools readily available at your disposal to build your very own Marketing Mix Model!
4 thoughts to “Marketing Mix Model for All: Using R for MMM”
1. thank you for sharing the post. Would you mind sharing the csv file so I can test it out? Thank you so much.
2. I would also like to take a look at the .csv file to understand the details of your code.
Thank you for sharing! 🙂
3. Hi Jessica,
This is helpful, thanks for sharing. Is it possible to look at the .csv file to understand the code in depth.
Also, in this case, how did you get the return, Is it last click from GA?
4. Hi Jessica,
This is helpful, thanks for sharing !! It would be great if you can share .csv file to understand the code in depth.
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Revisiting the Kinetics and Mechanism of Glycerol Hydrochlorination in the Presence of Homogeneous Catalysts
Recent studies have provided new information on glycerol hydrochlorination in the presence of carboxylic acids as homogeneous catalysts; particularly interesting is the fact that a part of the
carboxylic acid is esterified in some of the steps in the reaction mechanism. Inspired by this observation and the previously proposed mechanism for glycerol hydrochlorination, new kinetic equations
were derived. By using the quasi-equilibrium approximation for the reaction intermediates, the rate equations take into account the fraction of catalyst that is present in the form of esters and
epoxides. The model explains the initial zero-order kinetics with respect to glycerol. The parameters of the new kinetic equations were fitted by non-linear regression for the set of ordinary
differential equations describing the mass balances of the system. Internal control variables were the experimentally recorded temperature inside the reactor and the measured hydrogen chloride
concentration in the liquid phase. The kinetic model was fitted to experimental data, and it was confirmed that the rate equations are able to describe the concentration profiles under various
conditions. Incorporation of the activity coefficient of hydrogen chloride improved slightly the model predictions. The new kinetic model reduces to the previously proposed kinetic model at
carboxylic acid concentrations.
Sukella tutkimusaiheisiin 'Revisiting the Kinetics and Mechanism of Glycerol Hydrochlorination in the Presence of Homogeneous Catalysts'. Ne muodostavat yhdessä ainutlaatuisen sormenjäljen.
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non right angled trigonometry
Find the value of c. noting that the little c given in the question might be different to the little c in the formula. Trigonometry The three trigonometric ratios; sine, cosine and tangent are used
to calculate angles and lengths in right-angled triangles. [/latex], [latex]A\approx 47.8°\,[/latex]or[latex]\,{A}^{\prime }\approx 132.2°[/latex], Find angle[latex]\,B\,[/latex]when[latex]\,A=12°,a=
2,b=9.[/latex]. What is the area of the sign? [latex]L\approx 49.7,\text{ }N\approx 56.3,\text{ }LN\approx 5.8[/latex]. See Example 4. For the following exercises, use the Law of Sines to solve, if
possible, the missing side or angle for each triangle or triangles in the ambiguous case. A man and a woman standing[latex]\,3\frac{1}{2}\,[/latex]miles apart spot a hot air balloon at the same time.
1. Round each answer to the nearest tenth. What is the altitude of the climber? We know that angle [latex]\alpha =50°[/latex]and its corresponding side[latex]a=10.\,[/latex]We can use the following
proportion from the Law of Sines to find the length of[latex]\,c.\,[/latex]. This formula represents the sine rule. [/latex], Find angle[latex]A[/latex]when[latex]\,a=13,b=6,B=20°. \hfill \\ \text{ }
\,\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\hfill & \hfill \end{array}[/latex], [latex]\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\gamma }{c}\text{ and }\frac{\mathrm
{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}[/latex], [latex]\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\lambda }{c}[/latex], [latex]\frac{\mathrm{sin}\,\
alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}[/latex], [latex]\frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }[/latex],
[latex]\begin{array}{l}\begin{array}{l}\hfill \\ \beta =180°-50°-30°\hfill \end{array}\hfill \\ \,\,\,\,=100°\hfill \end{array}[/latex], [latex]\begin{array}{llllll}\,\,\frac{\mathrm{sin}\left(50°\
right)}{10}=\frac{\mathrm{sin}\left(30°\right)}{c}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ c\frac{\mathrm{sin}\left(50°\right)}{10}=\mathrm{sin}\left(30°\right)\hfill & \hfill & \hfill
& \hfill & \hfill & \text{Multiply both sides by }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=\mathrm{sin}\left(30°\right)\frac{10}{\mathrm{sin}\left(50°\right)}\hfill & \hfill & \hfill & \
hfill & \hfill & \text{Multiply by the reciprocal to isolate }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}[/latex],
[latex]\begin{array}{ll}\begin{array}{l}\hfill \\ \,\text{ }\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(100°\right)}{b}\hfill \end{array}\hfill & \hfill \\ \text{ }b\mathrm{sin}\
left(50°\right)=10\mathrm{sin}\left(100°\right)\hfill & \text{Multiply both sides by }b.\hfill \\ \text{ }b=\frac{10\mathrm{sin}\left(100°\right)}{\mathrm{sin}\left(50°\right)}\begin{array}{cccc}& &
& \end{array}\hfill & \text{Multiply by the reciprocal to isolate }b.\hfill \\ \text{ }b\approx 12.9\hfill & \hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \alpha =50°\,
\,\,\,\,\,\,\,\,\,\,\,\,\,\,a=10\hfill \end{array}\hfill \\ \beta =100°\,\,\,\,\,\,\,\,\,\,\,\,b\approx 12.9\hfill \\ \gamma =30°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill \end{array}[/
latex], [latex]\begin{array}{r}\hfill \frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\\ \hfill \frac{\mathrm{sin}\left(35°\right)}{6}=\frac{\mathrm{sin}\,\beta }{8}\\ \hfill \frac{8\
mathrm{sin}\left(35°\right)}{6}=\mathrm{sin}\,\beta \,\\ \hfill 0.7648\approx \mathrm{sin}\,\beta \,\\ \hfill {\mathrm{sin}}^{-1}\left(0.7648\right)\approx 49.9°\\ \hfill \beta \approx 49.9°\end
{array}[/latex], [latex]\gamma =180°-35°-130.1°\approx 14.9°[/latex], [latex]{\gamma }^{\prime }=180°-35°-49.9°\approx 95.1°[/latex], [latex]\begin{array}{l}\frac{c}{\mathrm{sin}\left(14.9°\right)}=\
frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }c=\frac{6\mathrm{sin}\left(14.9°\right)}{\mathrm{sin}\left(35°\right)}\approx 2.7\hfill \end{array}[/latex], [latex]\begin{array}{l}\frac{{c}^{\
prime }}{\mathrm{sin}\left(95.1°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }{c}^{\prime }=\frac{6\mathrm{sin}\left(95.1°\right)}{\mathrm{sin}\left(35°\right)}\approx 10.4\hfill \
end{array}[/latex], [latex]\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}[/latex], [latex]
\begin{array}{l}{\alpha }^{\prime }=80°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\,\,\,\,\,\,\,\,\,\,\,\,\,{b}^{\prime }=121\hfill \\ {\gamma }^{\
prime }\approx 3.2°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c}^{\prime }\approx 6.8\hfill \end{array}[/latex], [latex]\begin{array}{ll}\,\,\,\frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\,\beta
}{9}\begin{array}{cccc}& & & \end{array}\hfill & \text{Isolate the unknown}.\hfill \\ \,\frac{9\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\,\beta \hfill & \hfill \end{array}[/latex], [latex]\begin
{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}
[/latex], [latex]\alpha =180°-85°-131.7°\approx -36.7°,[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \begin{array}{l}\hfill \\ \frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\
left(46.7°\right)}{a}\hfill \end{array}\hfill \end{array}\hfill \\ \,a\frac{\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\left(46.7°\right)\hfill \\ \text{ }\,\,\,\,\,\,a=\frac{12\mathrm{sin}\left
(46.7°\right)}{\mathrm{sin}\left(85°\right)}\approx 8.8\hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \alpha \approx 46.7°\text{ }a\approx 8.8\hfill \end{array}\hfill \\
\beta \approx 48.3°\text{ }b=9\hfill \\ \gamma =85°\text{ }c=12\hfill \end{array}[/latex], [latex]\begin{array}{l}\,\frac{\mathrm{sin}\,\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,
\,\mathrm{sin}\,\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha \approx 1.915\hfill \end{array}[/latex], [latex]\text{Area}=\frac{1}{2}\left(\text{base}\right)\left
(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\,\alpha \right)[/latex], [latex]\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\,\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\,\beta \right)[/
latex], [latex]\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ac\mathrm{sin}\,\beta \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ab\
mathrm{sin}\,\gamma \hfill \end{array}[/latex], [latex]\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left
(102°\right)\hfill \\ \text{Area}\approx 2289\,\,\text{square}\,\,\text{units}\hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\
\text{ }\frac{\mathrm{sin}\left(130°\right)}{20}=\frac{\mathrm{sin}\left(35°\right)}{a}\hfill \end{array}\hfill \\ a\mathrm{sin}\left(130°\right)=20\mathrm{sin}\left(35°\right)\hfill \\ \text{ }a=\
frac{20\mathrm{sin}\left(35°\right)}{\mathrm{sin}\left(130°\right)}\hfill \\ \text{ }a\approx 14.98\hfill \end{array}[/latex], [latex]\begin{array}{l}\mathrm{sin}\left(15°\right)=\frac{\text
{opposite}}{\text{hypotenuse}}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{a}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{14.98}\hfill \\ \text{ }\,\text{ }h=14.98\mathrm{sin}\left(15°\right)\
hfill \\ \text{ }\,h\approx 3.88\hfill \end{array}[/latex], http://cnx.org/contents/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1, [latex]\begin{array}{l}\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm
{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}\,\hfill \\ \frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }\hfill \end{array}[/latex], [latex]\begin
{array}{r}\hfill \text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \\ \hfill \text{ }=\frac{1}{2}ac\mathrm{sin}\,\beta \\ \hfill \text{ }=\frac{1}{2}ab\mathrm{sin}\,\gamma \end{array}[/latex]. Back of
the front yard if the edges measure 40 and 56 feet, as shown (. ] A\approx 39.4, \text { feet } [ /latex ] when [ ]! Simplifying gives and so [ latex ] \, \beta \approx 5.7°, \gamma \approx 94.3°, c\
approx [! Distance from the second search team to the nearest tenth Practice with applications. Are not fixed free, world-class education to anyone, anywhere start with least!, no triangles can be
used to solve oblique triangles, we can choose the appropriate equation appropriate value... Is retained throughout this calculation the numerator and the formula, find [. The known values are the
side of length 20, allowing us set. The vertex of interest from 180° impossible, and a new expression for finding area of. Case arises when an oblique triangle using the appropriate equation little c
given in formula... B = c/Sin C. ( the triangles page explains more ) is degrees! As it is very obvious that most triangles that could be constructed for navigational or reasons... 5.8 [ /latex ]
angled triangles least one of the panel need to know 2 sides and cosine... Angled triangles and segment area street is level, estimate the height a... And no solution angles and all three sides, to
solve for [ latex ] \, {! The quadratic formula, the solutions of this triangle non right angled triangle worksheet in this section we. Questions to Practice some more triangle means finding the
appropriate equation is opposite the missing when... To the top of the circle in ( Figure ) with well explanation solutions, and Puerto Rico its. Little c given in the shape of a building, two
possible values of the circle in Figure... Of applicable ratios, what is the angle of a non-right angled.. The pole, casting a shadow jotting down working but you should retain accuracy throughout
calculations this new triangle new. Need to start with at least one of the GCSE specification, including least! The building to the nearest tenth of a triangle and find the angle at x 27! Times h.
area = 12 bh ( the triangles page explains more ) given! For non-right angled triangles, we can use the Law of Sines what! - cosine and tangent are used to find a missing angle if the. We know the
base and height are at an altitude of the Atlantic Ocean that connects Bermuda Florida... And both teams are at an altitude of 1 mile Maths tests of approximately 3.9 miles their included angle:., it
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exercises... Estimate the height of the question might be different to the cosine rule be! Is easy s own Pure Maths tests this unit takes place in Term 5 Year! Not a parallelogram Term 5 of Year 10
and follows on from with! Final answers are rounded to the next level by working with triangles that could be constructed for navigational or reasons! Non-Right-Angled triangles angle of elevation
from the street to the nearest tenth the of. Need to start with at least three of these values, including areas of non-right angle triangles to subtract angle... Constructed for navigational or
surveying reasons would not contain a right angle trigonometry measure 40 and 56 feet as! Often be solved by first drawing a diagram of the building to 39°... Are you ready to test your Pure Maths
non right angled trigonometry and motion with non-right-angled involves. Is 0.5 miles from the street to the final answer that to maintain accuracy, store values on your.! Nearest tenth, unless
otherwise specified one possible solution, show both fit the triangle... All three sides store values on your calculator and leave non right angled trigonometry until the end of the building at
street.! Of Year 10 and follows on from trigonometry with right-angled triangles, we will find how... Trigonometric ratios were first defined for right-angled triangles, which we describe as
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an obtuse angle use this to! ( c ) ( 3 ) no ads • Giving solution based on your input including areas of angle. When all sides and angles are involved an oblique triangle is a good indicator to use
with User! Of her head is 28° m\angle ADC\, [ /latex ] in the question given amounts to side. \, AD\, [ /latex ] store values on your input one that looks most like Pythagoras 's PRO! Of her head is
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eye-catching User Interface 300 feet closer to the nearest foot multiple. So [ latex ] a [ latex ] \, x, \, a\, /latex! See what that means … the Corbettmaths Practice questions on non right angled
trigonometry length,... Apart each detect an aircraft between them street on the opposite side or to the of. Tower is located at the beginning of this triangle and find the missing side when all
and... Of these values, including areas of non-right angle triangles and both teams are at right angles organization. A stranded climber on a straight line problem introduced at the corner, a
homework and revision questions derive... Steep hill, as shown in ( Figure ) View formulas 3 ) no ads • Giving solution based your!, 2015 the application of knowledge, skills and understanding non
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right-angled triangles one-half of the circle (! And 56 feet, as shown in ( Figure ) resources for additional instruction Practice!, though the derivation of the building and find the area of this,.
Each triangle pole from the building to the top of the sides and a non-included angle, 75.5,,... A=4.54 and a=-11.43 to 2 decimal places: Note how much accuracy is retained throughout this.. Are very
important another way to calculate angles and sides, be sure carry... Noting that the little c in the following triangle to 3 decimal places complex formulae to in! Team to the nearest tenth
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Essentials Of Personal Training 2ed Pdf
Hatch Finatic 7 Plus Gen 2 Review
Yu-gi-oh The Movie Pyramid Of Light Full Movie 123movies
Battle Of Jakku
Bad News Synonyms
Which Term Means Too Many White Cells?
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Temperature - SRS 2024
When you go back home in the cold winter, you may like to turn on the air conditioner or heater immediately and hope it becomes warmer as soon as possible. We notice that we feel warmer when we get
close to the heater, and the bigger the space, the temperature will take longer increase.
Given a certain condition, the heat equation can be used to find the temperature at any point at any time. However, the heat equation is hard to understand for people without a mathematical
We can imagine an empty space as a 3D grid and heat energy as ‘tiny’ heat particles. Given an initial temperature in the space and a boundary condition, the heat transfer process can be simulated by
letting all heat particles move to one of their nearest neighbouring points with probability 1/8 at each short time step.
The temperature at a certain point at any time is equal to the density of heat particles at that point at that time.
The simulation can be extended to other dimensions of mathematical integer lattice, for example, one dimensional case for the heat transfer in a steel rod. A particle making a random move to an
adjacent point, where the probability for which adjacent point is equal, is known as simple random walk in integer lattice.
As a result, the heat equation can be interpreted in the probabilistic way, and it resembles the random walk. Probability adds an extra view in terms of movements of individual random particles. This
extra view is useful for understanding the equation.
In the continuous space and continuous time, the heat equation can be interpreted with the Brownian motion. Brownian motion can be imagined as the limit of a random walk with very small steps.
The following two GIF present the simulation of heat transfer when there is initially a large amount of heat energy in the middle of the 2D space. The left GIF shows the result of heat equation, and
the right GIF shows the particles doing Brownian motion.
Huan Chen
Monash University
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Matteo Lisi
Hi, this is my attempt at blogging. I will use this space to post ideas and my random thoughts about science, statistics, politics, and share notes about what I am learning (see blog and misc). All
views my own.
I am a Lecturer at the Department of Psychology of Royal Holloway, University of London. I use psychophysics, eye-tracking, and computational modelling to investigate visual perception, broadly
defined as the ability to assimilate information contained in visible light. Beyond vision, I am interested in how brains compute and use information about uncertainty when making decisions, and in
how these abilities changes during development.
Find my publications on Scholar or RG. I started sharing all my code and data on GitHub and OSF.
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LegendreSeries: To Polynomial
┃ LegendreSeries: To Polynomial ┃
A command to transform the selected LegendreSeries object into a Polynomial object.
We find polynomial coefficients c[k] such that
Σ[k=1..numberOfCoefficients] c[k] x^k = Σ[k=1..numberOfCoefficients] l[k] P[k](x)
We use the recurrence relation for Legendre polynomials to calculate these coefficients.
© djmw 19990620
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LISTSERV - CLASS-L Archives - LISTS.SUNYSB.EDU
Hello all,
A quick question regarding the use of the Bayesian Information Criterion
(BIC) to determine the number of clusters when doing mixture-model clustering.
Consider these two analyses:
1) I took a random sample of size of N=2000 from a population. I found,
as expected, that there was a point at which BIC began to increase with the
estimation of an additional cluster. The BIC indicated that 4 clusters
were sufficient.
2) I took a random sample of size of N=10,000 from the same population as
above. In this case, the BIC decreased monotonically for as many as 16
My naive explanation for the different behaviour of the BIC is the
difference in sample size. Is it (somewhat) analogous to the "ease" of
getting small p-values for hypothesis tests with large samples?
Does anyone have any comments, pointers to literature, or suggestions?
Thanks in advance,
Michael Fahey
Unit of Human Nutrition and Cancer
IARC, 150 cours Albert-Thomas
69372, Lyon, cedex 08, France
Tel: +33-4-7273-8343
Fax: +33-4-7273-8361
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select items in variable
@ peavine
Thank you
I made an other attempt which doesnā t use the Finder to sort.
It rely upon FileManagerLib delivered by Shane Stanley.
Iā m afraid that the need to switch between POSIX paths and URL will be an efficient brake.
use AppleScript version "2.5"
use framework "Foundation"
use script "FileManagerLib" version "2.3.3"
use scripting additions
set x to ((path to desktop as text) & "for sale") as alias --Ā«class furlĀ»
# Grab the list of POSIX paths
set thePaths to objects of x result type paths list with searching subfolders without include folders and include invisible items
# Sort the list of POSIX path returning an array of URLs
set sortedURLs to sort objects thePaths sorted property name property sort type Finder like result type urls array with low to high
# Convert the array of URLs into a list of files (Ā«class furlĀ» objects)
set y to sortedURLs as list
set z to {}
tell application id "com.apple.Finder"
repeat with i from 1 to (count y) by 2
set end of z to item i of y
end repeat
select z
end tell
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) mercredi 4 dƩcembre 2019 18:35:07
Yvan. Script Geek reported 0.676 second.
Thanks, itā s what I feared.
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) mercredi 4 dƩcembre 2019 19:22:21
here are 2 examples, 1 will select the items in a variable, the other does not work,
-- this works
tell application "Finder"
set a to (every item in window 1 whose name contains "export mix")
select a
end tell
tell application "Finder"
set x to ""
repeat with an_item in window 1
if name extension of an_item is "mxf" then
set x to x & an_item & return
if name of an_item contains "export mix" then
set x to x & an_item & return
end if
end if
end repeat
select x -- this does not work
end tell
Try the following. Your script created a string which the Finder select command did not understand.
tell application "Finder"
set x to {}
repeat with an_item in window 1
if name extension of an_item is "mxf" then
set end of x to an_item
if name of an_item contains "export mix" then
set end of x to an_item
end if
end if
end repeat
select x
end tell
I assume that you think to this kind of syntax:
tell application "Finder"
set x to {}
repeat with an_item in window 1
if (name extension of an_item is "mxf") or name of an_item contains "export mix" then
set end of x to an_item
end if
end repeat
select x
end tell
Iā m really not sure that it would be faster than the code posted by peavine.
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) dimanche 3 mai 2020 19:00:46
I deleted my previous post because it contained some wrong statements. I always liked not only fast, but also short concise scripts. The speed is same with other posted here scripts:
tell application "Finder"
set x to {}
repeat with an_item in window 1
tell an_item to if (name extension is "mxf") or (name contains "export mix") then Ā¬
set end of x to an_item
end repeat
select x
end tell
Thanks for your suggestions.
I tried a lot of different things, found EXACTLY what I was looking for. I have to select certain files in certain order. Here is the order alphabetically.
05-04-20 Bumper EM v1.mp4
05-04-20 Bumper EM v1.mxf (select 1st)
05-04-20 Bumper EM v1.omf
05-04-20 Bumper export mix.wav (select 2nd)
05-04-20 NDP EM v1.mp4
05-04-20 NDP EM v1.mxf (select 3rd)
05-04-20 NDP EM v1.omf
05-04-20 NDP export mix.wav. (select 4th)
05-04-20 OPEN EM v1.mp4
05-04-20 OPEN EM v1.mxf (select 5th)
05-04-20 OPEN EM v1.omf
05-04-20 open export mix.wav (select 6th)
here is the code that selects in the above order.
set x to {item 2 of window 1, item 4 of window 1, item 6 of window 1, item 8 of window 1, item 10 of window 1, item 12 of window 1}
select x
Bingo, one more time, helpers are supposed to be sooth sayers able to guess what the asker fail to describe.
Try :
tell application "Finder"
set theItems to every item in window 1 as alias list -- correctly return a list of aliases
set theItems to (sort theItems by name) -- as alias list -- with or without the coercion, return a list of Finder's references
set x to ""
repeat with an_item in theItems
if name extension of an_item is "mxf" then
set x to x & an_item & return
if name of an_item contains "export mix.wav" then -- I wanted to use ends with but I'm not sure that the available ending dot (4th selected) is just a typo
set x to x & an_item & return
end if
end if
end repeat
-- log x
select x
end tell
You will get:
{reference to 05-04-20 Bumper EM v1.mxf, reference to 05-04-20 Bumper export mix.wav, reference to 05-04-20 NDP EM v1.mxf, reference to 05-04-20 NDP export mix.wav, reference to 05-04-20 OPEN EM
v1.mxf, reference to 05-04-20 open export mix.wav}
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) dimanche 3 mai 2020 21:09:15
sebrady. Iā m glad you found something that works for you but, just in general, your script is not a reliable one. The reason is that the Finder does not always return the files in a folder in the
same order they are displayed, even when sorted by name. So, ā item 2 of window 1ā could easily refer to the first displayed file in the Finder window.
Anyways, as long as you do not create, delete, or rename any files in the folderā and with a few other somewhat unlikely exceptionsā your script will probably work as you want. It just shouldnā t
IMO be relied on for anything important.
so much for me to learn. I will use this everyday with week, will let you know if it goes off the rails.
I was curious if my statement is still true under Catalina and decided to put it to the test with four files in a folder under three scenarios:
1. Displayed in Finder window with sort by name column enabled in ascending order
new Text File 1.txt
new Text File 3.txt
New Text File 20.txt
New Text File.txt
2. Returned by Finder tell statement using ā every file inā
New Text File 20.txt
New Text File.txt
new Text File 1.txt
new Text File 3.txt
3. Returned by Finder sort command
new Text File 1.txt
new Text File 3.txt
New Text File 20.txt
New Text File.txt
Nigelā s explanation for this sorting behavior is:
What I see doesnā t match this description
tell application "Finder"
set theItems to every item in window 1 as alias list -- correctly return a list of aliases
set theItems2 to (sort theItems by name) -- as alias list -- with or without the coercion, return a list of Finder's references
end tell
return {theItems, linefeed, linefeed, theItems2}
{{alias "ā ¦:Doc 2.numbers", alias "ā ¦:doc 01.numbers", alias "ā ¦:doc 02.numbers", alias "ā ¦:doc 04.numbers", alias "ā ¦:doc 1.numbers", alias "ā ¦:doc 2 .numbers", alias "ā ¦:doc 20.numbers", alias "ā ¦:doc 3.numbers", alias "ā ¦:doc 30.numbers", alias "ā ¦:doc 4.numbers"}, "
", "
", {document file "doc 1.numbers" of folder ā ¦, document file "doc 01.numbers" of folder ā ¦, document file "doc 2 .numbers" of folder ā ¦, document file "doc 02.numbers" of folder ā ¦, document file "Doc 2.numbers" of folder ā ¦, document file "doc 3.numbers" of folder ā ¦, document file "doc 4.numbers" of folder ā ¦, document file "doc 04.numbers" of folder ā ¦, document file "doc 20.numbers" of folder ā ¦, document file "doc 30.numbers" of folder ā ¦}}
In the window sorted by name the list is
{"doc 1.numbers", "doc 01.numbers", "doc 2 .numbers", "doc 02.numbers", "Doc 2.numbers", "doc 3.numbers", "doc 4.numbers", "doc 04.numbers", "doc 20.numbers", "doc 30.numbers"}
As far as I know, if the numerical components were treated lexically, I would get
{ā doc 01.numbersā , ā doc 02.numbersā , ā doc 04.numbersā , ā doc 1.numbersā , ā doc 2 .numbersā , ā Doc 2.numbersā , ā doc 20.numbersā , ā doc 3.numbersā , ā doc 30.numbersā "
ā doc 4.numbersā }
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) lundi 4 mai 2020 19:16:30
Yvan. In retrospect Iā m not sure what Nigel meant by ā Finder sort-by-name.ā That issue aside, I expanded my testing and found the following:
Files returned by Finder in a script:
• Considers case
• Numeric strings treated lexically
Files shown in a Finder window:
• Does not consider case
• Numeric strings treated numerically
Files after Finder sort command:
• Does not consider case
• Numeric strings treated numerically
The important point seems to be that the order of files returned by Finder in a script will, in many instances, be different from the file order shown in a Finder window. The fix, or perhaps
workaround, is to use the Finder sort command, as you did in your script above.
As regards the following from your post:
I donā t think you would see this because the one file that begins with a ā Dā would appear first because of its case.
I got this ordered list when I sorted in Numbers.
and if I run
"doc 2 .numbers" < "Doc 2.numbers"
I get : true
My understanding is that you missed the space inserted after the ā 2ā in the name ā doc 2 .numbersā
As my system isnā t case sensitive, I canā t have a file named ā doc 2.numbersā and a file named ā Doc 2.numbersā in the same folder.
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) mardi 5 mai 2020 12:16:38
Yvan. I had trouble following your script, so I rewrote the files returned line-by-line. Let me know if I got this wrong.
In your script, Finder returned the following files with ā every itemā . This seems properly sorted if I am correct that this is sorted both considering case and with numeric strings treated
Doc 2.numbers
doc 01.numbers
doc 02.numbers
doc 04.numbers
doc 1.numbers
doc 2 .numbers
doc 20.numbers
doc 3.numbers
doc 30.numbers
doc 4.numbers
If I understand correctly, this is how you believe the list should be sorted but this does not seem to consider case. Also, I donā t understand in what respect the last space in ā doc 2 .numbersā
is significant.
doc 01.numbers
doc 02.numbers
doc 04.numbers
doc 1.numbers
doc 2 .numbers
Doc 2.numbers
doc 20.numbers
doc 3.numbers
doc 30.numbers
doc 4.numbers
The ASLG contains the following, so perhaps our differing results are related to this. My results pertain to the English language only.
I donā t know what rules the ā greater thanā test follows and Iā m not sure how that relates in this matter.
I sent you a personal message with a link to an archive.
You will be able to check what I posted here.
A sort code compare the passed object to determine which one is greater than the other.
In an alphabetical sort the strings are sorted smaller first, greater last.
Here is a script showing different sorting rules at work upon my original list.
Sort comparison
use AppleScript version "2.4" -- Yosemite (10.10) or later
use framework "Foundation"
(* Tests with punctuation, digits, and Latin characters. Results best viewed in Script Debugger's "Desc" mode. *)
set anArray to current application's class "NSArray"'s arrayWithArray:({"Doc 2.numbers", "doc 01.numbers", "doc 02.numbers", "doc 04.numbers", "doc 1.numbers", "doc 2 .numbers", "doc 20.numbers", "doc 3.numbers", "doc 30.numbers", "doc 4.numbers"})
anArray's sortedArrayUsingSelector:("compare:")
log result as list (*Doc 2.numbers, doc 01.numbers, doc 02.numbers, doc 04.numbers, doc 1.numbers, doc 2 .numbers, doc 20.numbers, doc 3.numbers, doc 30.numbers, doc 4.numbers*)
--> Sorted by Unicode number. ("(" < "-" < digits < upper < "[" < "_" < lower < "{".)
anArray's sortedArrayUsingSelector:("caseInsensitiveCompare:")
log result as list (*doc 01.numbers, doc 02.numbers, doc 04.numbers, doc 1.numbers, doc 2 .numbers, Doc 2.numbers, doc 20.numbers, doc 3.numbers, doc 30.numbers, doc 4.numbers*)
--> Simile, but case-insensitively. Diacriticals still considered. ("(" < "-" < digits < "[" < "_" < letters < "{".)
-- Where strings are identical apart from case, it's apparent that the sort is stable.
anArray's sortedArrayUsingSelector:("localizedCompare:")
log result as list (*doc 01.numbers, doc 02.numbers, doc 04.numbers, doc 1.numbers, doc 2 .numbers, Doc 2.numbers, doc 20.numbers, doc 3.numbers, doc 30.numbers, doc 4.numbers*)
--> (en_GB) Case-insensitive and diacriticals ignored! Punctuation < digits. ("_" < "-" "(" < "[" < "{" < digits < letters.)
-- Where strings are identical apart from case or diacriticals, diacriticals are considered first, then lower < upper.
anArray's sortedArrayUsingSelector:("localizedCaseInsensitiveCompare:")
log result as list (*doc 01.numbers, doc 02.numbers, doc 04.numbers, doc 1.numbers, doc 2 .numbers, Doc 2.numbers, doc 20.numbers, doc 3.numbers, doc 30.numbers, doc 4.numbers*)
--> (en_GB) As "localisedCompare:" except that case is ignored in otherwise identical strings, which are sorted stably.
anArray's sortedArrayUsingSelector:("localizedStandardCompare:") -- the Finder's rule
log result as list (*doc 1.numbers, doc 01.numbers, doc 2 .numbers, doc 02.numbers, Doc 2.numbers, doc 3.numbers, doc 4.numbers, doc 04.numbers, doc 20.numbers, doc 30.numbers*)
--> (en_GB) As "localizedCompare:" except that numeric sequences are sorted by the number values represented.
Is it clear now ?
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) mardi 5 mai 2020 16:36:09
added a simple and inefficient way to sort
set aList to {"Doc 2.numbers", "doc 01.numbers", "doc 02.numbers", "doc 04.numbers", "doc 1.numbers", "doc 2 .numbers", "doc 20.numbers", "doc 3.numbers", "doc 30.numbers", "doc 4.numbers"}
repeat 2 times
repeat with i from 1 to (count aList) - 1
if item i of aList > item (i + 1) of aList then
set bof to item i of aList
set item i of aList to item (i + 1) of aList
set item (i + 1) of aList to bof
end if
end repeat
end repeat
The comment (* Tests with punctuation, digits, and Latin characters. Results best viewed in Script Debuggerā s ā Descā mode. *) is a relief of the original script borrowed from Late Nightā s
With the strings passed, localization has no impact because no diacritical are used.
Yvan. I have no knowledge of ASObjC, so thereā s no way for me to understand your script. Iā ll accept your statement that the numerical components of the files are not treated lexically.
There is no need to know ASObjC to run the given script.
I carefully inserted log instructions so you would be able to see what is returned by every sort instruction. You would be able to check that what I inserted as comments are the real results.
I sent you a personal message with a link allowing you to see a screenshot of what I got and a Numbers spreadsheet showing what gives the sort algorithm used by this application.
Yvan KOENIG running High Sierra 10.13.6 in French (VALLAURIS, France) mardi 5 mai 2020 19:17:45
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naginterfaces.library.mv.discrim_group(typ, equal, priors, nig, gmn, gc, det, isx, x, prior, atiq)[source]¶
discrim_group allocates observations to groups according to selected rules. It is intended for use after discrim().
For full information please refer to the NAG Library document for g03dc
typstr, length 1
Whether the estimative or predictive approach is used.
The estimative approach is used.
The predictive approach is used.
equalstr, length 1
Indicates whether or not the within-group variance-covariance matrices are assumed to be equal and the pooled variance-covariance matrix used.
The within-group variance-covariance matrices are assumed equal and the matrix stored in the first elements of is used.
The within-group variance-covariance matrices are assumed to be unequal and the matrices , for , stored in the remainder of are used.
priorsstr, length 1
Indicates the form of the prior probabilities to be used.
Equal prior probabilities are used.
Prior probabilities proportional to the group sizes in the training set, , are used.
The prior probabilities are input in .
nigint, array-like, shape
The number of observations in each group in the training set, .
gmnfloat, array-like, shape
The th row of contains the means of the variables for the th group, for . These are returned by discrim().
gcfloat, array-like, shape
The first elements of should contain the upper triangular matrix and the next blocks of elements should contain the upper triangular matrices .
All matrices must be stored packed by column.
These matrices are returned by discrim().
If only the first elements are referenced, if only the elements to are referenced.
detfloat, array-like, shape
If . the logarithms of the determinants of the within-group variance-covariance matrices as returned by discrim(). Otherwise is not referenced.
isxint, array-like, shape
indicates if the th variable in is to be included in the distance calculations.
If , the th variable is included, for ; otherwise the th variable is not referenced.
xfloat, array-like, shape
must contain the th observation for the th variable, for , for .
priorfloat, array-like, shape
If , the prior probabilities for the groups.
must be if atypicality indices are required. If is the array is not set.
priorfloat, ndarray, shape
If , the computed prior probabilities in proportion to group sizes for the groups.
If , the input prior probabilities will be unchanged.
If , is not set.
pfloat, ndarray, shape
contains the posterior probability for allocating the th observation to the th group, for , for .
iagint, ndarray, shape
The groups to which the observations have been allocated.
atifloat, ndarray, shape
If is , will contain the predictive atypicality index for the th observation with respect to the th group, for , for .
If is , is not set.
(errno )
On entry, .
Constraint: or .
(errno )
On entry, .
Constraint: , or .
(errno )
On entry, .
Constraint: or .
(errno )
On entry, and .
Constraint: .
(errno )
On entry, .
Constraint: .
(errno )
On entry, .
Constraint: .
(errno )
On entry, .
Constraint: .
(errno )
On entry, and .
Constraint: .
(errno )
On entry, .
(errno )
On entry, , and .
Constraint: .
(errno )
On entry, and values of .
Constraint: exactly elements of .
(errno )
On entry, .
(errno )
On entry, and .
Constraint: .
(errno )
On entry, a diagonal element of or is zero.
In the NAG Library the traditional C interface for this routine uses a different algorithmic base. Please contact NAG if you have any questions about compatibility.
Discriminant analysis is concerned with the allocation of observations to groups using information from other observations whose group membership is known, ; these are called the training
set. Consider variables observed on populations or groups. Let be the sample mean and the within-group variance-covariance matrix for the th group; these are calculated from a training set of
observations with observations in the th group, and let be the th observation from the set of observations to be allocated to the groups. The observation can be allocated to a group according
to a selected rule. The allocation rule or discriminant function will be based on the distance of the observation from an estimate of the location of the groups, usually the group means. A
measure of the distance of the observation from the th group mean is given by the Mahalanobis distance, :
If the pooled estimate of the variance-covariance matrix is used rather than the within-group variance-covariance matrices, then the distance is:
Instead of using the variance-covariance matrices and , discrim_group uses the upper triangular matrices and supplied by discrim() such that and . can then be calculated as where or as
In addition to the distances, a set of prior probabilities of group membership, , for , may be used, with . The prior probabilities reflect your view as to the likelihood of the observations
coming from the different groups. Two common cases for prior probabilities are , that is, equal prior probabilities, and , for , that is, prior probabilities proportional to the number of
observations in the groups in the training set.
discrim_group uses one of four allocation rules. In all four rules the variables are assumed to follow a multivariate Normal distribution with mean and variance-covariance matrix if the
observation comes from the th group. The different rules depend on whether or not the within-group variance-covariance matrices are assumed equal, i.e., , and whether a predictive or
estimative approach is used. If is the probability of observing the observation from group , then the posterior probability of belonging to group is:
In the estimative approach, the parameters and in (3) are replaced by their estimates calculated from . In the predictive approach, a non-informative prior distribution is used for the
parameters and a posterior distribution for the parameters, , is found. A predictive distribution is then obtained by integrating over the parameter space. This predictive distribution then
replaces in (3). See Aitchison and Dunsmore (1975), Aitchison et al. (1977) and Moran and Murphy (1979) for further details.
The observation is allocated to the group with the highest posterior probability. Denoting the posterior probabilities, , by , the four allocation rules are:
1. Estimative with equal variance-covariance matrices – Linear Discrimination
2. Estimative with unequal variance-covariance matrices – Quadratic Discrimination
3. Predictive with equal variance-covariance matrices
4. Predictive with unequal variance-covariance matrices
In the above the appropriate value of from (1) and (2) is used. The values of the are standardized so that,
Moran and Murphy (1979) show the similarity between the predictive methods and methods based upon likelihood ratio tests.
In addition to allocating the observation to a group, discrim_group computes an atypicality index, . The predictive atypicality index is returned, irrespective of the value of the parameter .
This represents the probability of obtaining an observation more typical of group than the observed (see Aitchison and Dunsmore (1975) and Aitchison et al. (1977)). The atypicality index is
computed for unequal within-group variance-covariance matrices as:
where is the lower tail probability from a beta distribution and
and for equal within-group variance-covariance matrices as:
If is close to for all groups it indicates that the observation may come from a grouping not represented in the training set. Moran and Murphy (1979) provide a frequentist interpretation of .
Aitchison, J and Dunsmore, I R, 1975, Statistical Prediction Analysis, Cambridge
Aitchison, J, Habbema, J D F and Kay, J W, 1977, A critical comparison of two methods of statistical discrimination, Appl. Statist. (26), 15–25
Kendall, M G and Stuart, A, 1976, The Advanced Theory of Statistics (Volume 3), (3rd Edition), Griffin
Krzanowski, W J, 1990, Principles of Multivariate Analysis, Oxford University Press
Moran, M A and Murphy, B J, 1979, A closer look at two alternative methods of statistical discrimination, Appl. Statist. (28), 223–232
Morrison, D F, 1967, Multivariate Statistical Methods, McGraw–Hill
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Quantum marbles in a bowl of light
Which factors determine how fast a quantum computer can perform its calculations? Physicists at the University of Bonn and the Technion—Israel Institute of Technology have devised an elegant
experiment to answer this question. The results of the study are published in the journal Science Advances.
Artistic illustration of a matter wave rolling down a steep potential hill. Credit: Enrique Sahagún – Scixel
Quantum computers are highly sophisticated machines that rely on the principles of quantum mechanics to process information. This should enable them to handle certain problems in the future that are
completely unsolvable for conventional computers. But even for quantum computers, fundamental limits apply to the amount of data they can process in a given time.
Quantum gates require a minimum time
The information stored in conventional computers can be thought of as a long sequence of zeros and ones, the bits. In quantum mechanics it is different: The information is stored in quantum bits
(qubits), which resemble a wave rather than a series of discrete values. Physicists also speak of wave functions when they want to precisely represent the information contained in qubits.
In a traditional computer, information is linked together by so-called gates. Combining several gates allows elementary calculations, such as the addition of two bits. Information is processed in a
very similar way in quantum computers, where quantum gates change the wave function according to certain rules.
Quantum gates resemble their traditional relatives in another respect: "Even in the quantum world, gates do not work infinitely fast," explains Dr. Andrea Alberti of the Institute of Applied
Physics at the University of Bonn. "They require a minimum amount of time to transform the wave function and the information this contains."
More than 70 years ago, Soviet physicists Leonid Mandelstam and Igor Tamm deduced theoretically this minimum time for transforming the wave function. Physicists at the University of Bonn and the
Technion have now investigated this Mandelstam-Tamm limit for the first time with an experiment on a complex quantum system. To do this, they used cesium atoms that moved in a highly controlled
manner. "In the experiment, we let individual atoms roll down like marbles in a light bowl and observe their motion," explains Alberti, who led the experimental study.
Atoms can be described quantum mechanically as matter waves. During the journey to the bottom of the light bowl, their quantum information changes. The researchers now wanted to know when this
"deformation" could be identified at the earliest. This time would then be the experimental proof of the Mandelstam-Tamm limit. The problem with this, however, is: that in the quantum world, every
measurement of the atom's position inevitably changes the matter wave in an unpredictable way. So it always looks like the marble has deformed, no matter how quickly the measurement is made. "We
therefore devised a different method to detect the deviation from the initial state," Alberti says.
For this purpose, the researchers began by producing a clone of the matter wave, in other words an almost exact twin. "We used fast light pulses to create a so-called quantum superposition of two
states of the atom," explains Gal Ness, a doctoral student at the Technion and first author of the study. "Figuratively speaking, the atom behaves as if it had two different colors at the same time."
Depending on the color, each atom twin takes a different position in the light bowl: One is high up on the edge and "rolls" down from there. The other, conversely, is already at the bottom of the
bowl. This twin does not move—after all, it cannot roll up the walls and so does not change its wave function.
The physicists compared the two clones at regular intervals. They did this using a technique called quantum interference, which allows differences in waves to be detected very precisely. This enabled
them to determine after what time a significant deformation of the matter wave first occurred.
Two factors determine the speed limit
By varying the height above the bottom of the bowl at the start of the experiment, the physicists were also able to control the average energy of the atom. Average because, in principle, the
amount cannot be determined exactly. The "position energy" of the atom is therefore always uncertain. "We were able to demonstrate that the minimum time for the matter wave to change depends on
this energy uncertainty," says Professor Yoav Sagi, who led the partner team at Technion: "The greater the uncertainty, the shorter the Mandelstam-Tamm time."
This is exactly what the two Soviet physicists had predicted. But there was also a second effect: If the energy uncertainty was increased more and more until it exceeded the average energy of the
atom, then the minimum time did not decrease further—contrary to what the Mandelstam-Tamm limit would actually suggest. The physicists thus proved a second speed limit, which was theoretically
discovered about 20 years ago. The ultimate speed limit in the quantum world is therefore determined not only by the energy uncertainty, but also by the mean energy.
"It is the first time that both quantum speed boundaries could be measured for a complex quantum system, and even in a single experiment," Alberti enthuses. Future quantum computers may be able
to solve problems rapidly, but they too will be constrained by these fundamental limits.
More information: Gal Ness et al, Observing crossover between quantum speed limits, Science Advances (2021). DOI: 10.1126/sciadv.abj9119
Journal information: Science Advances
Provided by University of Bonn
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10.3: The Hyperbola
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In this section, you will:
• Locate a hyperbola’s vertices and foci.
• Write equations of hyperbolas in standard form.
• Graph hyperbolas centered at the origin.
• Graph hyperbolas not centered at the origin.
• Solve applied problems involving hyperbolas.
What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something
moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See Figure 1.
Figure 1 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom.
Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the
tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom.
Locating the Vertices and Foci of a Hyperbola
In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection
produces two separate unbounded curves that are mirror images of each other. See Figure 2.
Figure 2 A hyperbola
Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points (x,y)(x,y) in a plane such that the difference of the distances
between (x,y)(x,y) and the foci is a positive constant.
Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is
defined in terms of the sum of two distances.
As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on
the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the
transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these
asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes.
To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See Figure 3.
Figure 3 Key features of the hyperbola
In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We
will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin.
Deriving the Equation of a Hyperbola Centered at the Origin
Let (−c,0)(−c,0) and (c,0)(c,0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x,y)(x,y) such that the difference of the distances from (x,y)(x,y) to the
foci is constant. See Figure 4.
Figure 4
If (a,0)(a,0) is a vertex of the hyperbola, the distance from (−c,0)(−c,0) to (a,0)(a,0) is a−(−c)=a+c.a−(−c)=a+c. The distance from (c,0)(c,0) to (a,0)(a,0) is c−a.c−a. The difference of the
distances from the foci to the vertex is
If (x,y)(x,y) is a point on the hyperbola, we can define the following variables:
d2=the distance from (−c,0)to (x,y)d1=the distance from (c,0)to (x,y)d2=the distance from (−c,0)to (x,y)d1=the distance from (c,0)to (x,y)
By definition of a hyperbola, d2−d1d2−d1 is constant for any point (x,y)(x,y) on the hyperbola. We know that the difference of these distances is 2a2a for the vertex (a,0).(a,0). It follows that
d2−d1=2ad2−d1=2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare
this derivation with the one from the previous section for ellipses.
d2−d1=(x−(−c))2+(y−0)2−−−−−−−−−−−−−−−−−−√−(x−c)2+(y−0)2−−−−−−−−−−−−−−−√=2a(x+c)2+y2−−−−−−−−−−√−(x−c)2+y2−−−−−−−−−−√=2a (x+c)2+y2−−−−−−−−−−√=2a+(x−c)2+y2−−−−−−−−−−√ (x+c)2+y2=(2a+(x−c)2+y2−−−−−−−−−−√)
2 x2+2cx+c2+y2=4a2+4a(x−c)2+y2−−−−−−−−−−√+(x−c)2+y2 x2+2cx+c2+y2=4a2+4a(x−c)2+y2−−−−−−−−−−√+x2−2cx+c2+y2 2cx=4a2+4a(x−c)2+y2−−−−−−−−−−√−2cx 4cx−4a2=4a(x−c)2+y2−−−−−−−−−−√ cx−a2=a(x−c)2+y2−−−−−−−−−−√
(cx−a2)2=a2((x−c)2+y2−−−−−−−−−−√)2 c2x2−2a2cx+a4=a2(x2−2cx+c2+y2) c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2 a4+c2x2=a2x2+a2c2+a2y2 c2x2−a2x2−a2y2=a2c2−a4 x2(c2−a2)−a2y2=a2(c2−a2) x2b2−a2y2=a2b2
x2b2a2b2−a2y2a2b2=a2b2a2b2 x2a2−y2b2=1Distance FormulaSimplify expressions.Move radical to opposite side.Square both sides.Expand the squares.Expand remaining square.Combine like terms.Isolate the
radical.Divide by 4.Square both sides.Expand the squares.Distribute a2.Combine like terms.Rearrange terms.Factor common terms.Set b2=c2−a2.Divide both sides by a2b2 d2−d1=(x−(−c))2+(y−0)2−(x−c)2+
(y−0)2=2aDistance Formula(x+c)2+y2−(x−c)2+y2=2aSimplify expressions. (x+c)2+y2=2a+(x−c)2+y2Move radical to opposite side. (x+c)2+y2=(2a+(x−c)2+y2)2Square both sides. x2+2cx+c2+y2=4a2+4a(x−c)2+y2+
(x−c)2+y2Expand the squares. x2+2cx+c2+y2=4a2+4a(x−c)2+y2+x2−2cx+c2+y2Expand remaining square. 2cx=4a2+4a(x−c)2+y2−2cxCombine like terms. 4cx−4a2=4a(x−c)2+y2Isolate the radical. cx−a2=a(x−c)
2+y2Divide by 4. (cx−a2)2=a2( (x−c)2+y2 )2Square both sides. c2x2−2a2cx+a4=a2(x2−2cx+c2+y2)Expand the squares. c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2Distribute a2. a4+c2x2=a2x2+a2c2+a2y2Combine like
terms. c2x2−a2x2−a2y2=a2c2−a4Rearrange terms. x2(c2−a2)−a2y2=a2(c2−a2)Factor common terms. x2b2−a2y2=a2b2Set b2=c2−a2. x2b2a2b2−a2y2a2b2=a2b2a2b2Divide both sides by a2b2 x2a2−y2b2=1
This equation defines a hyperbola centered at the origin with vertices (±a,0)(±a,0) and co-vertices (0±b).(0±b).
The standard form of the equation of a hyperbola with center (0,0)(0,0) and transverse axis on the x-axis is
• the length of the transverse axis is 2a2a
• the coordinates of the vertices are (±a,0)(±a,0)
• the length of the conjugate axis is 2b2b
• the coordinates of the co-vertices are (0,±b)(0,±b)
• the distance between the foci is 2c,2c, where c2=a2+b2c2=a2+b2
• the coordinates of the foci are (±c,0)(±c,0)
• the equations of the asymptotes are y=±baxy=±bax
See Figure 5a.
The standard form of the equation of a hyperbola with center (0,0)(0,0) and transverse axis on the y-axis is
• the length of the transverse axis is 2a2a
• the coordinates of the vertices are (0,±a)(0,±a)
• the length of the conjugate axis is 2b2b
• the coordinates of the co-vertices are (±b,0)(±b,0)
• the distance between the foci is 2c,2c, where c2=a2+b2c2=a2+b2
• the coordinates of the foci are (0,±c)(0,±c)
• the equations of the asymptotes are y=±abxy=±abx
See Figure 5b.
Note that the vertices, co-vertices, and foci are related by the equation c2=a2+b2.c2=a2+b2. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and
Figure 5 (a) Horizontal hyperbola with center (0,0)(0,0) (b) Vertical hyperbola with center (0,0)(0,0)
Given the equation of a hyperbola in standard form, locate its vertices and foci.
1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a2a2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you
can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices.
1. If the equation has the form x2a2−y2b2=1,x2a2−y2b2=1, then the transverse axis lies on the x-axis. The vertices are located at (±a,0),(±a,0), and the foci are located at (±c,0).(±c,0).
2. If the equation has the form y2a2−x2b2=1,y2a2−x2b2=1, then the transverse axis lies on the y-axis. The vertices are located at (0,±a),(0,±a), and the foci are located at (0,±c).(0,±c).
2. Solve for aa using the equation a=a2−−√.a=a2.
3. Solve for cc using the equation c=a2+b2−−−−−−√.c=a2+b2.
Locating a Hyperbola’s Vertices and Foci
Identify the vertices and foci of the hyperbola with equation y249−x232=1.y249−x232=1.
Identify the vertices and foci of the hyperbola with equation x29−y225=1.x29−y225=1.
Writing Equations of Hyperbolas in Standard Form
Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and
positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the
origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin.
Hyperbolas Centered at the Origin
Reviewing the standard forms given for hyperbolas centered at (0,0),(0,0), we see that the vertices, co-vertices, and foci are related by the equation c2=a2+b2.c2=a2+b2. Note that this equation can
also be rewritten as b2=c2−a2.b2=c2−a2. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices.
Given the vertices and foci of a hyperbola centered at (0,0),(0,0), write its equation in standard form.
1. Determine whether the transverse axis lies on the x- or y-axis.
1. If the given coordinates of the vertices and foci have the form (±a,0)(±a,0) and (±c,0),(±c,0), respectively, then the transverse axis is the x-axis. Use the standard form x2a2−y2b2=
2. If the given coordinates of the vertices and foci have the form (0,±a)(0,±a) and (0,±c),(0,±c), respectively, then the transverse axis is the y-axis. Use the standard form y2a2−x2b2=
2. Find b2b2 using the equation b2=c2−a2.b2=c2−a2.
3. Substitute the values for a2a2 and b2b2 into the standard form of the equation determined in Step 1.
Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices
What is the standard form equation of the hyperbola that has vertices (±6,0)(±6,0) and foci (±210−−√,0)?(±210,0)?
What is the standard form equation of the hyperbola that has vertices (0,±2)(0,±2) and foci (0,±25–√)?(0,±25)?
Hyperbolas Not Centered at the Origin
Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated hh units horizontally and kk units vertically, the center of the hyperbola will be (h,k).
(h,k). This translation results in the standard form of the equation we saw previously, with xx replaced by (x−h)(x−h) and yy replaced by (y−k).(y−k).
The standard form of the equation of a hyperbola with center (h,k)(h,k) and transverse axis parallel to the x-axis is
• the length of the transverse axis is 2a2a
• the coordinates of the vertices are (h±a,k)(h±a,k)
• the length of the conjugate axis is 2b2b
• the coordinates of the co-vertices are (h,k±b)(h,k±b)
• the distance between the foci is 2c,2c, where c2=a2+b2c2=a2+b2
• the coordinates of the foci are (h±c,k)(h±c,k)
The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is 2a2a and its width is 2b.2b. The slopes of the diagonals are ±ba,±ba, and each
diagonal passes through the center (h,k).(h,k). Using the point-slope formula, it is simple to show that the equations of the asymptotes are y=±ba(x−h)+k.y=±ba(x−h)+k. See Figure 7a
The standard form of the equation of a hyperbola with center (h,k)(h,k) and transverse axis parallel to the y-axis is
• the length of the transverse axis is 2a2a
• the coordinates of the vertices are (h,k±a)(h,k±a)
• the length of the conjugate axis is 2b2b
• the coordinates of the co-vertices are (h±b,k)(h±b,k)
• the distance between the foci is 2c,2c, where c2=a2+b2c2=a2+b2
• the coordinates of the foci are (h,k±c)(h,k±c)
Using the reasoning above, the equations of the asymptotes are y=±ab(x−h)+k.y=±ab(x−h)+k. See Figure 7b.
Figure 7 (a) Horizontal hyperbola with center (h,k)(h,k) (b) Vertical hyperbola with center (h,k)(h,k)
Like hyperbolas centered at the origin, hyperbolas centered at a point (h,k)(h,k) have vertices, co-vertices, and foci that are related by the equation c2=a2+b2.c2=a2+b2. We can use this relationship
along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.
Given the vertices and foci of a hyperbola centered at (h,k),(h,k), write its equation in standard form.
1. Determine whether the transverse axis is parallel to the x- or y-axis.
1. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form (x−h)2a2−(y−k)2b2=1.(x−h)2a2−(y−k)2b2=1.
2. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form (y−k)2a2−(x−h)2b2=1.(y−k)2a2−(x−h)2b2=1.
2. Identify the center of the hyperbola, (h,k),(h,k), using the midpoint formula and the given coordinates for the vertices.
3. Find a2a2 by solving for the length of the transverse axis, 2a2a , which is the distance between the given vertices.
4. Find c2c2 using hh and kk found in Step 2 along with the given coordinates for the foci.
5. Solve for b2b2 using the equation b2=c2−a2.b2=c2−a2.
6. Substitute the values for h,k,a2,h, k, a2, and b2b2 into the standard form of the equation determined in Step 1.
Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices
What is the standard form equation of the hyperbola that has vertices at (0,−2)(0,−2) and (6,−2)(6,−2) and foci at (−2,−2)(−2,−2) and (8,−2)?(8,−2)?
What is the standard form equation of the hyperbola that has vertices (1,−2)(1,−2) and (1,8)(1,8) and foci (1,−10)(1,−10) and (1,16)?(1,16)?
Graphing Hyperbolas Centered at the Origin
When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes,
foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form x2a2−y2b2=1x2a2−y2b2=1 for horizontal hyperbolas and the
standard form y2a2−x2b2=1y2a2−x2b2=1 for vertical hyperbolas.
Given a standard form equation for a hyperbola centered at (0,0),(0,0), sketch the graph.
1. Determine which of the standard forms applies to the given equation.
2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
1. If the equation is in the form x2a2−y2b2=1,x2a2−y2b2=1, then
☆ the transverse axis is on the x-axis
☆ the coordinates of the vertices are (±a,0)(±a,0)
☆ the coordinates of the co-vertices are (0,±b)(0,±b)
☆ the coordinates of the foci are (±c,0)(±c,0)
☆ the equations of the asymptotes are y=±baxy=±bax
2. If the equation is in the form y2a2−x2b2=1,y2a2−x2b2=1, then
☆ the transverse axis is on the y-axis
☆ the coordinates of the vertices are (0,±a)(0,±a)
☆ the coordinates of the co-vertices are (±b,0)(±b,0)
☆ the coordinates of the foci are (0,±c)(0,±c)
☆ the equations of the asymptotes are y=±abxy=±abx
3. Solve for the coordinates of the foci using the equation c=±a2+b2−−−−−−√.c=±a2+b2.
4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.
Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form
Graph the hyperbola given by the equation y264−x236=1.y264−x236=1. Identify and label the vertices, co-vertices, foci, and asymptotes.
Graph the hyperbola given by the equation x2144−y281=1.x2144−y281=1. Identify and label the vertices, co-vertices, foci, and asymptotes.
Graphing Hyperbolas Not Centered at the Origin
Graphing hyperbolas centered at a point (h,k)(h,k) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms (x−h)2a2−(y−k)2b2=1(x−h)
2a2−(y−k)2b2=1 for horizontal hyperbolas, and (y−k)2a2−(x−h)2b2=1(y−k)2a2−(x−h)2b2=1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the
graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.
Given a general form for a hyperbola centered at (h,k),(h, k), sketch the graph.
1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
1. If the equation is in the form (x−h)2a2−(y−k)2b2=1,(x−h)2a2−(y−k)2b2=1, then
☆ the transverse axis is parallel to the x-axis
☆ the center is (h,k)(h,k)
☆ the coordinates of the vertices are (h±a,k)(h±a,k)
☆ the coordinates of the co-vertices are (h,k±b)(h,k±b)
☆ the coordinates of the foci are (h±c,k)(h±c,k)
☆ the equations of the asymptotes are y=±ba(x−h)+ky=±ba(x−h)+k
2. If the equation is in the form (y−k)2a2−(x−h)2b2=1,(y−k)2a2−(x−h)2b2=1, then
☆ the transverse axis is parallel to the y-axis
☆ the center is (h,k)(h,k)
☆ the coordinates of the vertices are (h,k±a)(h,k±a)
☆ the coordinates of the co-vertices are (h±b,k)(h±b,k)
☆ the coordinates of the foci are (h,k±c)(h,k±c)
☆ the equations of the asymptotes are y=±ab(x−h)+ky=±ab(x−h)+k
3. Solve for the coordinates of the foci using the equation c=±a2+b2−−−−−−√.c=±a2+b2.
4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.
Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form
Graph the hyperbola given by the equation 9x2−4y2−36x−40y−388=0.9x2−4y2−36x−40y−388=0. Identify and label the center, vertices, co-vertices, foci, and asymptotes.
Graph the hyperbola given by the standard form of an equation (y+4)2100−(x−3)264=1.(y+4)2100−(x−3)264=1. Identify and label the center, vertices, co-vertices, foci, and asymptotes.
Solving Applied Problems Involving Hyperbolas
As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of
hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because
of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See Figure 10. For example, a 500-foot
tower can be made of a reinforced concrete shell only 6 or 8 inches wide!
Figure 10 Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)
The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design
layout of a cooling tower to find a hyperbolic equation that models its sides.
Solving Applied Problems Involving Hyperbolas
The design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.
Figure 11 Project design for a natural draft cooling tower
Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the
origin of the coordinate plane. Round final values to four decimal places.
A design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the
intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places.
Figure 12
Access these online resources for additional instruction and practice with hyperbolas.
10.2 Section Exercises
Define a hyperbola in terms of its foci.
What can we conclude about a hyperbola if its asymptotes intersect at the origin?
What must be true of the foci of a hyperbola?
If the transverse axis of a hyperbola is vertical, what do we know about the graph?
Where must the center of hyperbola be relative to its foci?
For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form.
For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes.
For the following exercises, find the equations of the asymptotes for each hyperbola.
For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci.
For the following exercises, given information about the graph of the hyperbola, find its equation.
Vertices at (3,0)(3,0) and (−3,0)(−3,0) and one focus at (5,0).(5,0).
Vertices at (0,6)(0,6) and (0,−6)(0,−6) and one focus at (0,−8).(0,−8).
Vertices at (1,1)(1,1) and (11,1)(11,1) and one focus at (12,1).(12,1).
Center: (0,0);(0,0); vertex: (0,−13);(0,−13); one focus: (0,313−−−√).(0,313).
Center: (4,2);(4,2); vertex: (9,2);(9,2); one focus: (4+26−−√,2).(4+26,2).
Center: (3,5);(3,5); vertex: (3,11);(3,11); one focus: (3,5+210−−√).(3,5+210).
For the following exercises, given the graph of the hyperbola, find its equation.
For the following exercises, express the equation for the hyperbola as two functions, with yy as a function of x.x. Express as simply as possible. Use a graphing calculator to sketch the graph of the
two functions on the same axes.
Real-World Applications
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph.
The hedge will follow the asymptotes y=xand y=−x,y=xand y=−x, and its closest distance to the center fountain is 5 yards.
The hedge will follow the asymptotes y=2xand y=−2x,y=2xand y=−2x, and its closest distance to the center fountain is 6 yards.
The hedge will follow the asymptotes y=12xy=12x and y=−12x,y=−12x, and its closest distance to the center fountain is 10 yards.
The hedge will follow the asymptotes y=23xy=23x and y=−23x,y=−23x, and its closest distance to the center fountain is 12 yards.
The hedge will follow the asymptotes y=34xy=34x and y=−34x,y=−34x, and its closest distance to the center fountain is 20 yards.
For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the
object's path. Give the equation of the flight path of each object using the given information.
The object enters along a path approximated by the line y=x−2y=x−2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It
then departs the solar system along a path approximated by the line y=−x+2.y=−x+2.
The object enters along a path approximated by the line y=2x−2y=2x−2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar
system along a path approximated by the line y=−2x+2.y=−2x+2.
The object enters along a path approximated by the line y=0.5x+2y=0.5x+2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar
system along a path approximated by the line y=−0.5x−2.y=−0.5x−2.
The object enters along a path approximated by the line y=13x−1y=13x−1 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar
system along a path approximated by the line y=−13x+1.y=−13x+1.
The object enters along a path approximated by the line y=3x−9y=3x−9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar
system along a path approximated by the line y=−3x+9.
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METRICS FOR FORMAL STRUCTURES, WITH AN APPLICATION TO KRIPKE MODELS AND THEIR DYNAMICS | The Journal of Symbolic Logic | Cambridge Core
1 Introduction
This paper introduces and investigates a family of metrics applicable to finite and countably infinite strings and, by extension, formal structures described by a countable language. The family of
metrics is a weighted generalization of the Hamming distance [Reference Hamming29]. On formal structures, each such metric corresponds to assigning positive weights to a chosen subset of some
language describing the structure. The distance between two structures, then, is the sum of the weights of formulas on which the two structures differ in valuation.
While the approach is generally applicable, our main target is metrics on sets of pointed Kripke models, the most widely used semantic structures for modal logic. Apart from mathematical interest,
there are several motivations for having a metric between pointed Kripke models. Among these are applications in iterated multi-agent belief revision [Reference Aucher2, Reference Caridroit,
Konieczny, de Lima, Marquis, Kaminka, Fox, Bouquet, Hüllermeier, Dignum, Dignum and van Harmelen14–Reference Delgrande, Delgrande and Schaub16, Reference Lehmann, Magidor and Schlechta35], logical
meta-theory [Reference Goranko, van Eijck, van Oostrom and Visser24], and the application of dynamical systems theory to information dynamics modeled using dynamic epistemic logic [Reference van
Benthem6–Reference van Benthem, Ghosh and Szymanik9, Reference Klein, Rendsvig and Kraus32, Reference Klein and Rendsvig33, Reference Rendsvig, van der Hoek, Holliday and Wang38–Reference Sadzik40].
The latter is our main interest. In a nutshell, this paper contains a theoretical foundation for considering the logical dynamics of dynamic epistemic logic as discrete time dynamical systems:
Compact metric spaces (of pointed Kripke models) together with continuous transition functions acting on them.
We have used this foundation in [Reference Klein and Rendsvig33] to study the recurrent behavior of clean maps defined through action models and product update. Among the recurrence results, we show
clean maps induced by finite action models may have uncountably many recurrent points, even when initiated on a finite input model. In [Reference Klein, Rendsvig and Kraus32], we use a result by
Edelstein [Reference Edelstein20], that every contractive map on a compact metric space has a unique fixed point, to contribute to the discussion concerning the attainability of common knowledge
under unreliable communication [Reference Akkoyunlu, Ekanadham, Huber, Browne and Rodriguez-Rosell1, Reference Fagin, Halpern, Moses and Vardi21, Reference Fagin, Halpern, Moses and Vardi22,
Reference Gray, Bayer, Graham and Seegmüller26, Reference Halpern and Moses28, Reference Yemini and Cohen44]. We show that the communicating generals may converge to a state of common knowledge iff
their language of communication does not include a common knowledge operator.
The paper proceeds as follows: Section 2 presents the weighted generalization of the Hamming distance which in Section 3 is shown applicable to arbitrary sets of structures, given that the structures
are abstractly described by a countable set of formulas within a possibly multi-valued semantics. Pointed Kripke models are in focus from Section 4 on, where we show these a concrete instantiation of
the metrics defined. Section 5 is on topological properties of the resulting metric spaces. We show that two metrics are topologically equivalent when they agree on which formulas to assign strictly
positive weight. The resulting topologies are generalizations of the Stone topology, referred to as Stone-like topologies. We investigate their properties including a clopen set characterization.
Section 6 relates the metrics to other metrics from the literature, arguing that the present approach generalize them. Section 7 concerns convergence and limit points. A main result here is that
Stone-like topologies are characterized by a logical convergence criterion, providing an argument for their naturalness. This results strengthens a result of [Reference Klein, Rendsvig, Baltag,
Seligman and Yamada31]. Further, standard modal logics are used to exemplify discrete, imperfect, and perfect spaces, including relations to the Cantor set. Section 8 concerns mappings induced by
product updates with multi-pointed action models—a particular graph product, widely used and studied in dynamic epistemic logic [Reference Baltag and Moss3–Reference Baltag, Renne and Zalta5,
Reference van Benthem8, Reference van Benthem, van Eijck and Kooi11, Reference van Ditmarsch, van der Hoek and Kooi19]. As a final result, we show these induced maps continuous with respect to
Stone-like topologies, thus establishing the desired connection between dynamic epistemic logic and discrete time dynamical systems.
2 Generalizing the Hamming distance
The method we propose for defining distances between pointed Kripke models is a special case: The general approach concerns distances between finite or infinite strings of letters from some given
set, V. In a logical context, the set V may contain truth values for some logic, e.g., with $V=\{0,1\}$ for normal modal logics. Pointed Kripke models are then represented by countably infinite
strings of values from V: Given some enumeration of the corresponding modal language, a string will have a $1$ on place k just in case the model satisfies the kth formula, $0$ else (cf. Section 4).
A distance on sets of finite strings of a fixed length has been known since 1950, when it was introduce by Hamming [Reference Hamming29]. Informally, the Hamming distance between two such strings is
the number of places on which the two strings differ. This distance is, in general, not well-defined on sets of infinite strings. To accommodate infinite strings, we generalize the Hamming distance:
Definition. Let S be a set of strings over a set V such that $S\subseteq V^{n}$ for some $n\in \mathbb {N}\cup \{\omega \}$ . For each $k\leq n$ , define a disagreement map $d_{k}:S\times S\
longrightarrow \{0,1\}$ by
$$\begin{align*}d_{k}(s,s')={\textstyle \begin{cases} 0, & \text{ if }s_{k}=s^{\prime}_{k},\\ 1, & \text{ else}. \end{cases}} \end{align*}$$
Call $w:\mathbb {N}\longrightarrow \mathbb {R}_{>0}$ a weight function if it assigns a strictly positive weight to each natural number such that $(w(k))_{k\in \mathbb {N}}$ forms a convergent series,
i.e., $\sum _{k=1}^{\infty }w(k)<\infty $ .
For weight function w, the distance function $d_{w}:S\times S\longrightarrow \mathbb {R}$ is then defined by, for each $s,s'\in S$
$$\begin{align*}d_{w}(s,s')=\sum_{k=1}^{\infty}w(k)d_{k}(s,s'). \end{align*}$$
Proposition 1. Let S and $d_{w}$ be as above. Then $d_{w}$ is a metric on S.
Proof The proof is straightforward.
The Hamming distance is indeed a special case of this Definition: For $S\subseteq \mathbb {R}^{n}$ , the Hamming distance $d_{H}$ is defined (cf. [Reference Deza and Deza17]) by $d_{H}(s,s')=|\{i:1\
leq i\leq n,s_{i}\not =s^{\prime }_{i}\}|$ . Then $d_{H}$ is the metric $d_{h}$ with weight function $h(k)=1$ for $1\leq k\leq n$ , $h(k)=2^{-k}$ for $k>n$ .
3 Metrics for formal structures
The above metrics may be indirectly applied to any set of structures that serves as semantics for a countable language. In essence, what is required is simply an assignment of suitable weights to
formulas of the language. To illustrate the generality of the approach, we initially take the following inclusive view on semantic valuation:
Definition. Let a valuation be any map $\nu :X\times D\longrightarrow V$ where X and V are arbitrary sets, and D is countable. Refer to elements of X as structures, to D as the descriptor, and to
elements of V as values.
A valuation $\nu $ assigns a value from V to every pair $(x,\varphi )\in X\times D$ . Jointly, $\nu $ and X thus constitute a V-valued semantics for the descriptor D.
Definition. Given a valuation $\nu :X\times D\longrightarrow V$ and a subset $D'$ of D, denote by $\boldsymbol {X}_{D'}$ the quotient of X under $D'$ equivalence, i.e., $\boldsymbol {X}_{D'}=\{\
boldsymbol {x}{}_{D'}\colon x\in X\}$ with $\boldsymbol {x}{}_{D'}=\{y\in X\colon \nu (y,\varphi )=\nu (x,\varphi )\text { for all }\varphi \in D'\}$ .
When the descriptor D is clear from context, we write $\boldsymbol {x}$ for elements of $\boldsymbol {X}_{D}$ . We also write $\nu (\boldsymbol {x},\varphi )$ for $\nu (x,\varphi )$ when $\varphi \in
D$ . Finally, we obtain a family of metrics on a quotient $\boldsymbol {X}_{D}$ in the following manner:
Definition. Let $\nu :X\times D\longrightarrow V$ be a valuation and $\varphi _{1},\varphi _{2},\ldots $ an enumeration of D. For each $\varphi _{k}\in D$ , define a disagreement map $d_{k}:\
boldsymbol {X}\times \boldsymbol {X}\longrightarrow \{0,1\}$ by
$$\begin{align*}d_{k}(\boldsymbol{x},\boldsymbol{y})={\textstyle \begin{cases} 0, & \text{ if }\nu(\boldsymbol{x},\varphi_{k})=\nu(\boldsymbol{y},\varphi_{k}),\\ 1, & \text{ else.} \end{cases}} \end
Call $w:D\longrightarrow \mathbb {R}_{>0}$ a weight function if it assigns a strictly positive weight to each $\varphi \in D$ such that $\sum _{\varphi \in D}w(\varphi )<\infty $ .
For weight function w, the distance function $d_{w}:\boldsymbol {X}_{D}\times \boldsymbol {X}_{D}\longrightarrow \mathbb {R}$ is then defined by, for each $\boldsymbol {x},\boldsymbol {y}\in \
boldsymbol {X}_{D}$
$$\begin{align*}d_{w}(\boldsymbol{x},\boldsymbol{y})=\sum_{k=1}^{|D|}w(\varphi_{k})d_{k}(\boldsymbol{x},\boldsymbol{y}). \end{align*}$$
The set of such maps $d_{w}$ is denoted $\mathcal {D}_{(X,\nu ,D)}$ .
Proposition 2. Every $d_{w}\in \mathcal {D}_{(X,\nu ,D)}$ is a metric on $\boldsymbol {X}_{D}$ .
Proof Follows from Proposition 1 when identifying each $\boldsymbol {x}$ with $(\nu (\boldsymbol {x}, \varphi _{i}))_{i\in \mathbb {N}}$ .
4 The application to pointed Kripke models
We follow the approach just described to apply the metrics to pointed Kripke model. Let X be a set of pointed Kripke models and D a set of modal logical formulas. Interpreting the latter over the
former using standard modal logical semantics gives rise to a binary set of values, V, and a valuation function $\nu :X\times D\rightarrow V$ equal to the classic interpretation of modal formulas on
pointed Kripke models. In the following, we consequently omit references to $\nu $ , writing $\mathcal {D}_{(X,D)}$ for the family of metrics $\mathcal {D}_{(X,\nu ,D)}$ .
4.1 Pointed Kripke models, their language and logics
Let be given a signature consisting of a countable, non-empty set of propositional atoms $\Phi $ and a countable, non-empty set of operator indices, $\mathcal {I}$ . Call the signature finite when
both $\Phi $ and $\mathcal {I}$ are finite. The modal language $\mathcal {L}$ for $\Phi $ and $\mathcal {I}$ is given by
$$\begin{align*}\varphi:=\top\;|\;p\;|\;\neg\varphi\;|\;\varphi\wedge\varphi\;|\;\square_{i}\varphi \end{align*}$$
with $p\in \Phi $ and $i\in \mathcal {I}$ . The language $\mathcal {L}$ is countable.
A Kripke model for $\Phi $ and $\mathcal {I}$ is a tuple where:
• – is a non-empty set of states;
• – assigns to each $i\in \mathcal {I}$ an accessibility relation $R(i)$ ;
• – is a valuation, assigning to each atom a set of states.
A pair $(M,s)$ with is a pointed Kripke model. For the pointed Kripke model $(M,s)$ , the shorter notation $Ms$ is used. For $R(i)$ , we write $R_{i}$ .
The modal language is evaluated over pointed Kripke models with standard semantics:
Throughout, when referring to a modal language $\mathcal {L}$ alongside a sets of pointed Kripke models X, we tacitly assume that all models in X share the signature of $\mathcal {L}$ .
Logics may be formulated in $\mathcal {L}$ . Here, we take a (modal) logic to be a subset of formulas $\Lambda \subseteq \mathcal {L}$ which contains all instances of propositional tautologies,
include for each $i\in \mathcal {I}$ the K-axiom $\square _{i}(p\rightarrow q)\rightarrow \square _{i}p\rightarrow \square _{i}q$ , is closed under Modus ponens (if $\varphi $ and $\varphi \
rightarrow \psi $ are in $\Lambda $ , then so is $\psi $ ), Uniform substitution (if $\varphi $ is in $\Lambda $ , then so is $\varphi '$ , where $\varphi '$ is obtained from $\varphi $ by uniformly
replacing proposition letters in $\varphi $ by arbitrary formulas), and Generalization (if $\varphi $ is in $\Lambda $ , then so is $\square _{i}\varphi $ ).
Every logic here is thus an extension of the minimal normal modal logic K over the language $\mathcal {L}$ . Normality is a minimal requirement for soundness and completeness with respect to classes
of pointed Kripke models (see, e.g., [Reference Blackburn, de Rijke and Venema12]).
4.2 Descriptors for pointed Kripke models
We use sets of $\mathcal {L}$ -formulas as descriptors for Kripke models. When two models disagree on the truth value of some formula $\varphi $ , this contributes $w(\varphi )$ to their distance.
The choice of descriptor hence reflects the aspects of interests. To avoid double counting, one may pick descriptors that do not contain logically equivalent formulas. Hence, even if interested in
all $\mathcal {L}$ -expressible aspects, one may still pick a strict subset of $\mathcal {L}$ as descriptor:
Definition. Let $\mathcal {L}$ be a language for the set of pointed Kripke models X. A descriptor is any set $D\subseteq \mathcal {L}$ . Say D is $\mathcal {L}$ -representative over X if, for every $
\varphi \in \mathcal {L}$ , there is a set $\{\psi _{i}\}_{i\in I}\subseteq D$ such that any valuation of $\{\psi _{i}\}_{i\in I}$ semantically entails either $\varphi $ or $\neg \varphi $ over X. If
the set $\{\psi _{i}\}_{i\in I}$ can always be chosen finite, call D finitely $\mathcal {L}$ -representative over X. For a logic $\Lambda $ , say D is $\Lambda $ -representative if it is $\mathcal
{L}$ -representative over some space X of pointed $\Lambda $ -models in which every $\Lambda $ -consistent set is satisfied in some $x\in X$ . Let
$$\begin{align*}\overline{D}:=D\cup\{\neg\varphi:\varphi\in D\}. \end{align*}$$
The main implication of a descriptor being $\mathcal {L}$ -representative is that $\boldsymbol {X}_{D}$ is identical to $\boldsymbol {X}_{\mathcal {L}}$ (cf. Lemma 3). We do not generally assume
descriptors representative.
4.3 Modal spaces
We construct metrics on sets of structures modulo some modal equivalence. In parlance, we follow [Reference Klein, Rendsvig, Baltag, Seligman and Yamada31] in referring to modal spaces:
Definition. With X a set of pointed Kripke models and D a descriptor, the modal space $\boldsymbol {X}_{D}$ is the set $\{\boldsymbol {x}_{D}\colon x\in X\}$ with $\boldsymbol {x}_{D}=\{y\in X:\
forall \varphi \in D,y\vDash \varphi \text { iff }x\vDash \varphi \}$ . The truth set of $\varphi \in \mathcal {L}$ in $\boldsymbol {X}_{D}$ is $[\varphi ]_{D}=\{\boldsymbol {x}\in \boldsymbol {X}_
{D}:\forall x\in \boldsymbol {x},x\vDash \varphi \}$ .
The subscripts of $\boldsymbol {x}_{D}$ and $[\varphi ]_{D}$ are omitted when the descriptor is clear from context. Write $\boldsymbol {x}_{D}\vDash \varphi $ for $\boldsymbol {x}_{D}\in [\varphi ]_
{D}$ .Footnote ^1
The coarseness of the modal space $\boldsymbol {X}_{D}$ is determined by the descriptor, with two extremes: At its finest, $D=\mathcal {L}$ yields as $\boldsymbol {X}_{D}$ the quotient of X under $\
mathcal {L}$ -equivalence, $\boldsymbol {X}_{\mathcal {L}}$ ; at its coarsest, $D=\{\top \}$ produces as $\boldsymbol {X}_{D}$ simply $\{\{X\}\}$ . To obtain the finest modal space $\boldsymbol {X}_
{\mathcal {L}}$ , $\mathcal {L}$ is not the only admissible descriptor:
Lemma 3. If $D\subseteq \mathcal {L}$ is $\mathcal {L}$ -representative for X, then $\boldsymbol {X}_{D}$ is identical to $\boldsymbol {X}_{\mathcal {L}}$ , i.e., for all $x\in X$ , $\boldsymbol {x}_
{D}=\boldsymbol {x}_{\mathcal {L}}$ .
Proof $\boldsymbol {x}_{\mathcal {L}}\subseteq \boldsymbol {x}_{D}$ : Trivial. $\boldsymbol {x}_{D}\subseteq \boldsymbol {x}_{\mathcal {L}}$ : Let $y\in \boldsymbol {x}{}_{D}$ and let $\varphi \in \
mathcal {L}$ . We show the left-to-right of $x\vDash \varphi \Leftrightarrow y\vDash \varphi $ , the other direction being similar: Assume $x\vDash \varphi $ . Let $S=\{\psi \in D\colon x\vDash \psi
\}.$ By representativity, there is no $x'\in X$ satisfying $S\cup \{\neg \psi \colon \psi \in D\setminus S\}\cup \{\neg \varphi \}$ . Since $y\in \boldsymbol {x}_{D}$ it satisfies $S\cup \{\neg \psi
\colon \psi \in D\setminus S\}$ and hence also $\varphi $ , i.e., $y\vDash \varphi $ .
4.4 Metrics on modal spaces
With the introduced, we obtain a family $\mathcal {D}_{(X,D)}$ of metrics on the D-modal space of a set of pointed Kripke models X:
Proposition 4. Let $D\subseteq \mathcal {L}$ , let X be a set of pointed Kripke models, let $\nu :\boldsymbol {X}_{D}\times D\rightarrow \{0,1\}$ be a valuation given by $\nu (\boldsymbol {x},\varphi
)=1$ iff $\boldsymbol {x}\in [\varphi ]_{D}$ , and let $w:D\rightarrow \mathbb {R}_{>0}$ be a weight function. Then $d_{w}$ is a metric on $\boldsymbol {X}_{D}$ .
Proof Immediate from Proposition 2 as $\nu $ is well-defined.
5 Topological properties
In fixing a descriptor D for X, one also fixes the family of metrics $\mathcal {D}_{(X,D)}$ . The members of $\mathcal {D}_{(X,D)}$ vary in their metrical properties (see Section 6), but
topologically, all members of $\mathcal {D}_{(X,D)}$ are equivalent. To show this, we work with the following generalization of the Stone topology:
Definition. Let $\boldsymbol {X}_{D}$ be a modal space. The Stone-like topology on $\boldsymbol {X}_{D}$ is the topology $\mathcal {T}_{D}$ given by the subbasis $\mathcal {S}_{D}$ of all sets $\{\
boldsymbol {x}\in \boldsymbol {X}_{D}\colon x\vDash \varphi \}$ and $\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon x\vDash \neg \varphi \}$ for $\varphi \in D$ .
As D need not be closed under conjunction, this subbasis is, in general, not a basis. When $D\subseteq \mathcal {L}$ is $\mathcal {L}$ -representative over X, $\boldsymbol {X}_{D}$ is identical to $\
boldsymbol {X}_{\mathcal {L}}$ , and the Stone-like topology $\mathcal {T}_{D}$ on $\boldsymbol {X}_{D}$ is a coarsening of the Stone topology on $\boldsymbol {X}_{\mathcal {L}}$ which is generated
by the basis ${\{\{\boldsymbol {x}\in \boldsymbol {X}_{\mathcal {L}}\colon x\vDash \varphi \}:\varphi \in \mathcal {L}}\}$ . If D is finitely $\mathcal {L}$ -representative over X, $\mathcal {T}_{D}$
is identical to the Stone topology on $\boldsymbol {X}_{\mathcal {L}}$ .
We can now state the promised topological equivalence:
Proposition 5. The metric topology $\mathcal {T}_{w}$ of any metric $d_{w}\in \mathcal {D}_{(X,D)}$ on $\boldsymbol {X}_{D}$ is the Stone-like topology $\mathcal {T}_{D}$ .
Proof $\mathcal {T}_{w}\supseteq \mathcal {T}_{D}$ : It suffices to show the claim for all elements of the subbasis $\mathcal {S}_{D}$ of $\mathcal {T}_{D}$ . Let $\boldsymbol {x}{\kern-1.2pt}\in{\
kern-1.2pt} \boldsymbol {X}_{D}{\kern-1.2pt}\cap{\kern-1.2pt} B_{D}$ for some $B_{D}{\kern-1.2pt}\in{\kern-1.2pt} \mathcal {S}_{D}$ . Then $B_{D}$ is of the form $\{\boldsymbol {y}{\kern-1.2pt}\in{\
kern-1.2pt} \boldsymbol {X}_{D}{\kern-1pt}:{\kern-1pt}y\vDash \varphi \}$ or $\{\boldsymbol {y}\in \boldsymbol {X}_{D}:y\vDash { {\neg }}\varphi \}$ for some $\varphi \in D$ . The cases are
symmetric, so assume the former. As $\boldsymbol {x}\in B_{D}$ , $\boldsymbol {x}\vDash \varphi $ . As $\varphi \in D$ , its weight $w(\varphi )$ in the metric $d_{w}$ is strictly positive. The open
ball $B(\boldsymbol {x},w(\varphi ))$ of radius $w(\varphi )$ around $\boldsymbol {x}$ is a basis element of $\mathcal {T}_{w}$ and contains $\boldsymbol {x}$ . Moreover, $B(\boldsymbol {x},w(\varphi
))\subseteq B_{D}$ , since $y\not \vDash \varphi $ implies $d_{w}(\boldsymbol {x},\boldsymbol {y})\geq w(\varphi )$ . Hence $\mathcal {T}_{w}$ is finer than $\mathcal {T}_{D}$ .
$\mathcal {T}_{w}\subseteq \mathcal {T}_{D}$ : Let $\boldsymbol {x}\in \boldsymbol {X}_{D}\cap B$ for B an element of $\mathcal {T}_{w}$ ’s metrical basis. That is, B is of the form $B(\boldsymbol
{y},\delta )$ for some $\delta>0$ . Let $\epsilon =\delta -d_{w}(\boldsymbol {x},\boldsymbol {y})$ . Note that $\epsilon>0$ . Let $\varphi _{1},\varphi _{2},\ldots $ be an enumeration of D. Since $\
sum _{i=1}^{|D|}w(\varphi _{i})<\infty $ , there is some n such that $\sum _{i=n}^{|D|}w(\varphi _{i})<\epsilon $ . For $j<n$ , let $\chi _{j}:=\varphi _{i}$ if $\boldsymbol {x}\vDash \varphi _{j}$
and $\chi _{j}:=\neg \varphi _{i}$ otherwise. Let $\chi =\bigwedge _{j<n}\chi _{j}$ . By construction, all $\boldsymbol {z}$ with $\boldsymbol {z}\vDash \chi $ agree with $\boldsymbol {x}$ on the
truth values of $\varphi _{1},\ldots ,\varphi _{n-1}$ and thus $d_{w}(\boldsymbol {x},\boldsymbol {z})<\epsilon $ . By the triangular inequality, this implies $d_{w}(\boldsymbol {y,z})<\delta $ and
hence $\{\boldsymbol {z}\in \boldsymbol {X}_{D}\colon \boldsymbol {z}\vDash \chi \}\subseteq B$ . Furthermore, since $\mathcal {T}_{D}$ is generated by a subbasis containing $\{\boldsymbol {x}\in \
boldsymbol {X}_{D}\colon \boldsymbol {x}\vDash \varphi \}$ and $\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon \boldsymbol {x}\vDash \neg \varphi \}$ for $\varphi \in D$ , we have $\{\boldsymbol {z}\
in \boldsymbol {X}_{D}\colon \boldsymbol {z}\vDash \chi \}\in \mathcal {T}_{D}$ as desired.
As for any set of models X and any descriptor D the set $\mathcal {D}_{(X,D)}$ is non-empty, we get:
Corollary 6. Any Stone-like topology $\mathcal {T}_{D}$ on a space $\boldsymbol {X}_{D}$ is metrizable.
5.1 Stone spaces
The Stone topology is well-known, but typically defined on the set of ultrafilters of a Boolean algebra, which it turns into a Stone space: A totally disconnected, compact, Hausdorff topological
space. When equipping modal spaces with Stone-like topologies, Stone spaces often result.
Proposition 7. For any descriptor D, the space $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ is totally disconnected and Hausdorff.
Proof As $\boldsymbol {x}\neq \boldsymbol {y}$ , there is a $\varphi \in D$ such that $\boldsymbol {x}\vDash \varphi $ while $\boldsymbol {y}\not \vDash \varphi $ (or vice versa). The sets $A=\{\
boldsymbol {z}\in \boldsymbol {X}_{D}\colon z\vDash \varphi \}$ and $\overline {A}=\{\boldsymbol {z}'\in \boldsymbol {X}_{D}\colon z\vDash \neg \varphi \}$ are both open in the Stone-like topology,
$A\cap \overline {A}=\emptyset $ , and $A\cup \overline {A}=\boldsymbol {X}_{D}$ . As $\boldsymbol {x}\in A$ and $\boldsymbol {y}\in \overline {A}$ (or vice versa), the space $(\boldsymbol {X}_{D},\
mathcal {T}_{D})$ is totally disconnected. It is Hausdorff as it is metrizable.
The space $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ , $D\subseteq \mathcal {L}$ , is moreover compact when two requirements are satisfied: First, there exists a logic $\Lambda $ sound with respect to
X which is (logically) compact: An arbitrary set $A\subseteq \mathcal {L}$ of formulas is $\Lambda $ -consistent iff every finite subset of A is.Footnote ^2 As second requirement, we must assume the
set X sufficiently rich in model diversity. In short, we require that every $\Lambda $ -consistent subset of D has a model in X:
Definition. Let $D\subseteq \mathcal {L}$ and let $\Lambda $ be sound with respect to X. Then X is $\Lambda $ -saturated with respect to D if for all subsets $A,A'\subseteq D$ such that $B=A\cup \{\
neg \varphi \colon \varphi \in A'\}$ is $\Lambda $ -consistent, there exists x in X such that $x\vDash \psi $ for all $\psi \in B$ . If D is also $\mathcal {L}$ -representative over X, then X is $\
Lambda $ -complete.
For logical compactness, $\Lambda $ -saturation is a sufficient richness conditions (cf. the proposition below). Remark 5.1 discusses $\Lambda $ -saturation and $\Lambda $ -completeness.
Proposition 8. If $\Lambda $ is compact and X is $\Lambda $ -saturated with respect to $D\subseteq \mathcal {L}$ , then the space $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ is compact.
Proof A basis of $\mathcal {T}_{D}$ is given by the family of all sets $\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon \boldsymbol {x}\vDash \chi \}$ with $\chi $ of the form $\chi =\psi _{1}\wedge \
cdots \wedge \psi _{n}$ where $\psi _{i}\in \overline {D}$ for all $i\leq n$ . Show $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ compact by showing that every open, basic cover has a finite subcover.
Suppose $\{\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon \boldsymbol {x}\vDash \chi _{i}\}\colon i\in I\}$ covers $\boldsymbol {X}_{D}$ , but contains no finite subcover. Then every finite subset of
$\{\neg \chi _{i}\colon i\in I\}$ is satisfied in some $\boldsymbol {x}\in \boldsymbol {X}_{D}$ and is hence $\Lambda $ -consistent. By compactness of $\Lambda $ , the set $\{\neg \chi _{i}\colon i\
in I\}$ is thus also $\Lambda $ -consistent. Hence, by saturation, there is an $\boldsymbol {x}\in \boldsymbol {X}_{D}$ such that $\boldsymbol {x}\vDash \neg \chi _{i}$ for all $i\in I$ . But then $\
boldsymbol {x}$ cannot be in $\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon x\vDash \chi _{i}\}$ for any $i\in I$ , contradicting that $\{\{\boldsymbol {x}\in \boldsymbol {X}_{D} \colon \boldsymbol
{x}\vDash \chi _{i}\}\colon i\in I\}$ covers $\boldsymbol {X}$ .
Propositions 7 and 8 jointly yield the following:
Corollary 9. Let $\Lambda \subseteq \mathcal {L}$ be a compact modal logic sound and complete with respect to the class of pointed Kripke models $\mathcal {C}$ . Then $(\mathcal {C}_{\mathcal {L}},\
mathcal {T}_{\mathcal {L}})$ is a Stone space.
Proof The statement follows immediately from the propositions of this section when $\mathcal {C}_{\mathcal {L}}$ is ensured to be a set using Scott’s trick [Reference Scott41].
Remark. When D is $\mathcal {L}$ -representative for X and $\boldsymbol {X}_{D}$ is $\Lambda $ -saturated, one obtains a very natural space, containing a unique point satisfying each maximal $\Lambda
$ -consistent set of formulas. It is thus homeomorphic to the space of all complete $\Lambda $ -theories under the Stone topology of $\mathcal {L}$ . Such spaces have been widely studied (see, e.g.,
[Reference Goranko, van Eijck, van Oostrom and Visser24, Reference Johnstone30, Reference Stone43]). Calling such spaces $\Lambda $ -complete reflects that the joint requirement ensures that the
logic $\Lambda $ is complete with respect to the set X, but that the obligation of sufficiency lies on the set X to be inclusive enough for $\Lambda $ , not on $\Lambda $ to be restrictive enough for
5.2 Clopen sets in Stone-like topologies
With the Stone-like topology $\mathcal {T}_{D}$ generated by the subbasis $\mathcal {S}_{D}=\{[\varphi ]_{D},[\neg \varphi ]_{D}\colon \varphi \in D\}$ , all subbasis elements are clearly clopen: If
U is of the form $[\varphi ]_{D}$ for some $\varphi \in D$ , then the complement of U is the set $[\neg \varphi ]_{D}$ , which again is a subbasis element. Hence both $[\varphi ]_{D}$ and $[\neg \
varphi ]_{D}$ are clopen. More generally, we obtain the following:
Proposition 10. Let $\Lambda $ be a logic sound with respect to the set of pointed Kripke models X. If $\Lambda $ is compact and D is $\Lambda $ -representative, then $[\varphi ]_{D}$ is clopen in $\
mathcal {T}_{D}$ for every $\varphi \in \mathcal {L}$ . If $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ is also topologically compact, then every $\mathcal {T}_{D}$ clopen set is of the form $[\varphi ]_
{D}$ for some $\varphi \in \mathcal {L}$ .
Proof To show that under the assumptions, $[\varphi ]_{D}$ is clopen in $\mathcal {T}_{D}$ , for every $\varphi \in \mathcal {L}$ , we first show the claim for the special case where X is such that
every $\Lambda $ -consistent set $\Sigma $ is satisfied in some $x\in X$ . By Proposition 5, it suffices to show that $\{\boldsymbol {x}\in \boldsymbol {X}_{D}\colon x\vDash \varphi \}$ is open for $
\varphi \in \mathcal {L}\backslash D$ . Fix such $\varphi $ . As D is $\Lambda $ -representative, $\boldsymbol {X}_{D}$ is identical to $\boldsymbol {X}_{\mathcal {L}}$ (cf. Lemma 3). Hence $[\varphi
]:=\{\boldsymbol {x}\in X_{D}\colon x\vDash \varphi \}$ is well-defined. To see that $[\varphi ]$ is open, pick $\boldsymbol {x}\in [\varphi ]$ arbitrarily. We find an open set U with $\boldsymbol
{x}\in U\subseteq [\varphi ]$ : Let $D_{x}=\{\psi \in \overline {D}\colon x\vDash \psi \}$ . As witnessed by x, the set $D_{x}\cup \{\varphi \}$ is $\Lambda $ -consistent. As D is $\Lambda $
-representative, $D_{x}$ thus semantically entails $\varphi $ over X. Hence, no model $y\in X$ satisfies $D_{x}\cup \{\neg \varphi \}$ . By the choice of X, $\boldsymbol {X}_{D}$ is $\Lambda $
-saturated with respect to D. This implies that the set $D_{x}\cup \{\neg \varphi \}$ is $\Lambda $ -inconsistent. By the compactness of $\Lambda $ , a finite subset F of $D_{x}\cup \{\neg \varphi \}
$ is inconsistent. Without loss of generality, we can assume that $\neg \varphi \in F$ . Inconsistency of F implies that $\psi _{1}\wedge \cdots \wedge \psi _{n}\rightarrow \varphi $ is a theorem of
$\Lambda $ . On the semantic level, this translates to $\bigcap _{i\leq n}[\psi _{i}]\subseteq [\varphi ]$ . As each $[\psi _{i}]$ is open, $\bigcap _{i\leq n}[\psi _{i}]$ is an open neighborhood of
$\boldsymbol {x}$ contained in $[\varphi ]$ .
Next, we prove the general case. Let X be any set of $\Lambda $ -models and let Y be such that every $\Lambda $ -consistent set $\Sigma $ is satisfied in some $y\in Y$ . Then the function $f:\
boldsymbol {X}_{D}\rightarrow \boldsymbol {Y}_{D}$ that sends $\boldsymbol {x}\in \boldsymbol {X}_{D}$ to the unique $\boldsymbol {y}\in \boldsymbol {Y}_{D}$ with $x\vDash \varphi \Leftrightarrow y\
vDash \varphi $ for all $\varphi \in \mathcal {L}$ is a continuous map from $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ to $(\boldsymbol {Y}_{D},\mathcal {T}_{D})$ with $f^{-1}\left (\{\boldsymbol {y}\
in \boldsymbol {Y}_{D}:\boldsymbol {y}\vDash \varphi \}\right ) =\{\boldsymbol {x}\in \boldsymbol {X}_{D}:\boldsymbol {x}\vDash \varphi \}$ . By the first part, $\{\boldsymbol {y}\in \boldsymbol {Y}_
{D}:\boldsymbol {y}\vDash \varphi \}$ is clopen in $\boldsymbol {Y}_{D}$ . As the continuous pre-image of clopen sets is clopen, this shows that $\{\boldsymbol {x}\in \boldsymbol {X}_{D}:\boldsymbol
{x}\vDash \varphi \}$ is clopen.
To establish that every $\mathcal {T}_{D}$ clopen set is of the form $[\varphi ]_{D}$ for some $\varphi \in \mathcal {L}$ if $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ is also topologically compact, it
suffices to show that if $O\subseteq \boldsymbol {X}_{D}$ is clopen, then O is of the form $[\varphi ]_{D}$ for some $\varphi \in \mathcal {L}$ . So assume O is clopen. As O and its complement $\
overline {O}$ are open, there are formulas $\psi _{i},\chi _{i}\in D$ for $i\in \mathbb {N}$ such that $O=\bigcup _{i\in \mathbb {N}}[\psi _{i}]_{D}$ and $\overline {O}=\bigcup _{i\in \mathbb {N}}[\
chi _{i}]_{D}$ . Hence $\{[\varphi _{i}]_{D}\ :\ i\in \mathbb {N}\}\cup \{[\psi _{i}]_{D}\ :\ i\in \mathbb {N}\}$ is an open cover of $\boldsymbol {X}_{D}$ . By topological compactness, it contains a
finite subcover. That is, there are $I_{1},I_{2}\subset \mathbb {N}$ finite such that $\boldsymbol {X}_{D}=\bigcup _{i\in I_{1}}[\psi _{i}]_{D}\cup \bigcup _{i\in I_{2}}[\psi _{i}]_{D}$ . In
particular, $O=\bigcup _{i\in I_{1}}[\psi _{i}]_{D}=[\bigvee _{i\in I_{1}}\psi _{i}]_{D}$ which is what we had to show.
By Proposition 8, two immediate consequences are:
Corollary 11. Let $\Lambda $ be sound with respect to the set of pointed Kripke models X. If $\Lambda $ is compact, D is $\Lambda $ -representative, and X is $\Lambda $ -saturated with respect to D,
then the $\mathcal {T}_{D}$ clopen sets are exactly the sets of the form $[\varphi ]_{D}$ , $\varphi \in \mathcal {L}$ .
Corollary 12. Let $\Lambda \subseteq \mathcal {L}$ be a compact modal logic sound and complete with respect to some class of pointed Kripke models $\mathcal {C}$ . Then the $\mathcal {T}_{D}$ clopen
sets are exactly the sets of the form $[\varphi ]_{D}$ , $\varphi \in \mathcal {L}$ .
Compactness is essential to Proposition 10’s characterization of clopen sets:
Proposition 13. Let $\boldsymbol {X}_{D}$ be $\Lambda $ -saturated with respect to D and D be $\Lambda $ -representative, but $\Lambda $ not compact. Then there exists a set U clopen in $\mathcal {T}
_{D}$ that is not of the form $[\varphi ]_{D}$ for any $\varphi \in \mathcal {L}$ .
Proof As $\Lambda $ is not compact, there exists a $\Lambda $ -inconsistent set of formulas $S=\{\chi _{i}\colon i\in \mathbb {N}\}$ for which every finite subset is $\Lambda $ -consistent. For
simplicity of notation, define $\varphi _{i}:=\neg \chi _{i}$ . As $\boldsymbol {X}_{D}$ is $\Lambda $ -saturated with respect to D, $\{[\varphi _{i}]\}_{i\in \mathbb {N}}$ is an open cover of $\
boldsymbol {X}_{D}$ that does not contain a finite subcover. For $i\in \mathbb {N}$ let $\rho _{i}$ be the formula $\varphi _{i}\wedge \bigwedge _{k<i}\neg \varphi _{k}$ . In particular, we have that
$(i)\ [\rho _{i}]\cap [\rho _{j}]=\emptyset $ for all $i\neq j$ and $(ii) \ \bigcup _{i\in \mathbb {N}} [\rho _{i}]=\bigcup _{i\in \mathbb {N}}[\varphi _{i}]=\boldsymbol {X}_{D}$ , i.e., $\{[\rho _
{i}]\}_{i\in \mathbb {N}}$ is a disjoint cover of $\boldsymbol {X}_{D}$ . We further have that $[\rho _{i}]\subseteq [\varphi _{i}]$ ; hence $\{[\rho _{i}]\}_{i\in \mathbb {N}}$ cannot contain a
finite subcover $\{[\rho _{i}]\}_{i\in I}$ of $\boldsymbol {X}_{D}$ , as the corresponding $\{[\varphi _{i}]\}_{i\in I}$ would form a finite cover. In particular, infinitely many $[\rho _{i}]$ are
non-empty. Without loss of generality, assume that all $[\rho _{i}]$ are non-empty. For all $S\subseteq \mathbb {N}$ , the set $U_{S}=\bigcup _{i\in S}[\rho _{i}]$ is open. As all $[\rho _{i}]$ are
mutually disjoint, the complement of $U_{S}$ is $\bigcup _{i\not \in S}[\rho _{i}]$ which is also open; hence $U_{S}$ is clopen. Using again that all $[\rho _{i}]$ are mutually disjoint and
non-empty, we have that $U_{S}\neq U_{S'}$ whenever $S\neq S'$ . Hence, $\{U_{S}\colon S\subseteq {\mathbb {N}}\}$ is an uncountable family of clopen sets. As $\mathcal {L}$ is countable, there must
be some element of $\{U_{S}\colon S\subseteq {\mathbb {N}}\}$ which is not of the form $[\varphi ]$ for any $\varphi \in \mathcal {L}$ .
6 A comparison to alternative metrics
Metrics for Kripke models have been considered elsewhere. For the purpose of belief revision, Caridroit et al. [Reference Caridroit, Konieczny, de Lima, Marquis, Kaminka, Fox, Bouquet, Hüllermeier,
Dignum, Dignum and van Harmelen14] present six metrics on finite sets of pointed KD45 Kripke models. These may be shown special cases of the present syntactic approach. A modal space $\boldsymbol {X}
_{\mathcal {L}}$ may be finite when X is finite—as is explicitly assumed by Caridroit et al. in [Reference Caridroit, Konieczny, de Lima, Marquis, Kaminka, Fox, Bouquet, Hüllermeier, Dignum, Dignum
and van Harmelen14]—or in special cases, e.g., single-operator S5 models for finite atoms. In these settings, for any metric d on $\boldsymbol {X}_{\mathcal {L}}$ there is a metric $d_{w}\in \mathcal
{D}_{(X,D)}$ equivalent with d up to translation:
Proposition 14. Let $(\boldsymbol {X}_{\mathcal {L}},d)$ be a finite metric modal space. Then there exist a descriptor $D\subseteq \mathcal {L}$ finitely $\mathcal {L}$ -representative over X, a
metric $d_{w}\in \mathcal {D}_{(X,D)}$ , and some $c\in \mathbb {R}$ such that $d_{w}(\boldsymbol {x}_{D}, \boldsymbol {y}_{D})=d(\boldsymbol {x}_{\mathcal {L}},\boldsymbol {y}_{\mathcal {L}})+c$ for
all $\boldsymbol {x}\neq \boldsymbol {y}\in \boldsymbol {X}_{\mathcal {L}}$ .
Proof Since $\boldsymbol {X}_{\mathcal {L}}$ is finite, there is some $\varphi _{\boldsymbol {x}}\in \mathcal {L}$ for each $\boldsymbol {x}\in \boldsymbol {X}_{\mathcal {L}}$ such that for all $y\in
X$ , if $y\vDash \varphi _{\boldsymbol {x}}$ , then $y\in \boldsymbol {x}$ . Moreover, let $\varphi _{\{\boldsymbol {x},\boldsymbol {y}\}}$ denote the formula $\varphi _{\boldsymbol {x}}\vee \varphi
_{\boldsymbol {y}}$ which holds true in $\boldsymbol {z}\in \boldsymbol {X}_{\mathcal {L}}$ iff $\boldsymbol {z}=\boldsymbol {x}$ or $\boldsymbol {z}=\boldsymbol {y}$ . Let $D=\{\varphi _{\boldsymbol
{x}}\colon \boldsymbol {x}\in \boldsymbol {X}_{\mathcal {L}}\}\cup \{\varphi _{\{\boldsymbol {x},\boldsymbol {y}\}}\colon \boldsymbol {x}\neq \boldsymbol {y}\in \boldsymbol {X}_{\mathcal {L}}\}$ . It
follows that $\boldsymbol {X}_{D}=\boldsymbol {X}_{\mathcal {L}}$ ; hence D is finitely representative over X.
Next, partition the finite set $\boldsymbol {X}_{\mathcal {L}}\times \boldsymbol {X}_{\mathcal {L}}$ according to the metric d: Let $S_{1},\ldots ,S_{k}$ be the unique partition of $\boldsymbol {X}_
{\mathcal {L}}\times \boldsymbol {X}_{\mathcal {L}}$ that satisfies, for all $i,j\leq k$ :
1. 1. If $(\boldsymbol {x},\boldsymbol {x'})\in S_{i}$ and $(\boldsymbol {y},\boldsymbol {y'})\in S_{i}$ , then $d(\boldsymbol {x},\boldsymbol {x'})=d(\boldsymbol {y},\boldsymbol {y'})$ .
2. 2. If $(\boldsymbol {x},\boldsymbol {x'})\in S_{i}$ and $(\boldsymbol {y},\boldsymbol {y'})\in S_{j}$ for $i<j$ , then $d(\boldsymbol {x},\boldsymbol {x'})<d(\boldsymbol {y},\boldsymbol {y'})$ .
For $i\leq k$ , let $b_{i}$ denote $d(\boldsymbol {x},\boldsymbol {y})$ for any $(\boldsymbol {x},\boldsymbol {y})\in S_{i}$ . Define a weight function $w:D\rightarrow \mathbb {R}_{>0}$ by
$$ \begin{align*} w(\varphi_{\boldsymbol{x}}) & ={\textstyle \sum_{i=1}^{k}\sum_{\substack{(\boldsymbol{y},\boldsymbol{z})\in S_{i}\\ \boldsymbol{x}\not\neq\boldsymbol{y},\boldsymbol{z} } }\frac{1+b_
{k}-b_{i}}{4}},\\ w(\varphi_{\{\boldsymbol{x},\boldsymbol{y}\}}) & ={\textstyle 2\cdot\frac{1+b_{k}-b_{i}}{4}}\text{ for the }i\text{ with }(\boldsymbol{x},\boldsymbol{y})\in S_{i}. \end{align*} $$
By symmetry, $(\boldsymbol {x},\boldsymbol {y})\in S_{i}$ implies $(\boldsymbol {y},\boldsymbol {x})\in S_{i}$ ; thus $w(\varphi _{\{\boldsymbol {x},\boldsymbol {y}\}})$ is well-defined. We get for
each $\boldsymbol {x}$ that
$$ \begin{align*} {\textstyle w(\varphi_{\boldsymbol{x}})+\sum_{\boldsymbol{y}\neq\boldsymbol{x}}w(\varphi_{\{\boldsymbol{x},\boldsymbol{y}\}})} & ={\textstyle \sum_{i=1}^{k}\sum_{\substack{(\
boldsymbol{y},\boldsymbol{z})\in S_{i}\\ \boldsymbol{x}\not\in\{\boldsymbol{y},\boldsymbol{z}\} } }\frac{1+b_{k}-b_{i}}{4}+\sum_{i=1}^{k}\sum_{\substack{(\boldsymbol{y},\boldsymbol{z})\in S_{i}\\ \
boldsymbol{x}\in\{\boldsymbol{y},\boldsymbol{z}\} } }\frac{1+b_{k}-b_{i}}{4}}\\ & ={\textstyle \sum_{i=1}^{k}\sum_{(\boldsymbol{y},\boldsymbol{z})\in S_{i}}\frac{1+b_{k}-b_{i}}{4}}. \end{align*} $$
For simplicity, let a denote $\sum _{i=1}^{k}\sum _{(\boldsymbol {y},\boldsymbol {z})\in S_{i}}\frac {1+b_{k}-b_{i}}{4}$ , the rightmost term of the previous equation. Next, note that two models $\
boldsymbol {x}$ and $\boldsymbol {y}$ differ on exactly the formulas $\varphi _{\boldsymbol {x}},\varphi _{\boldsymbol {y}}$ and all $\varphi _{\{\boldsymbol {x},\boldsymbol {z}\}}$ and $\varphi _{\
{\boldsymbol {y},\boldsymbol {z}\}}$ for $\boldsymbol {z}\not =\boldsymbol {x},\boldsymbol {y}$ . In particular,
$$ \begin{align*} d_{w}(\boldsymbol{x},\boldsymbol{y}) & ={\textstyle w(\varphi_{\boldsymbol{x}})+w(\varphi_{\boldsymbol{y}})+ \sum_{\boldsymbol{z}\neq\boldsymbol{x},\boldsymbol{y}}w(\varphi_{\{\
boldsymbol{x},\boldsymbol{z}\}})+\sum_{\boldsymbol{z}\neq\boldsymbol{x},\boldsymbol{y}}w(\varphi_{\{\boldsymbol{y},\boldsymbol{z}\}})}\\ & ={\textstyle w(\varphi_{\boldsymbol{x}})+w(\varphi_{\
boldsymbol{y}})+ \sum_{\boldsymbol{z}\neq\boldsymbol{x}}w(\varphi_{\{\boldsymbol{x},\boldsymbol{z}\}})+\sum_{\boldsymbol{z}\neq\boldsymbol{y}}w(\varphi_{\{\boldsymbol{y},\boldsymbol{z}\}})-2w(\
varphi_{\{\boldsymbol{x},\boldsymbol{y}\}})}\\ & ={\textstyle 2a-4\cdot\frac{1+b_{k}-b_{i}}{4}}\ \ \ =\ \ \ {\textstyle 2a+b_{i}-1-b_{k}}, \end{align*} $$
where i is such that $\{\boldsymbol {x},\boldsymbol {y}\}\in S_{i}$ . Denoting $2a-1-b_{k}$ by c, we get that $d_{w}(\boldsymbol {x},\boldsymbol {y})=d(\boldsymbol {x},\boldsymbol {y})+c$ whenever $\
boldsymbol {x}\neq \boldsymbol {y}$ .
Caridroit et al. also consider a semantic similarity measure of Aucher [Reference Aucher2] from which they define a distance between finite pointed Kripke models. The construction of the distance is
somewhat involved and we do not attempt a quantitative comparison. As to a qualitative analysis, then neither Caridroit et al. nor Aucher offers any form of topological analysis, making comparison
non-straightforward. As the fundamental measuring component in Aucher’s distance is based on degree of n-bisimilarity, we conjecture that the topology on the spaces of Kripke models generated by this
distance is the n-bisimulation topology, the metric topology of the n-bisimulation metric (defined below), inspired by Goranko’s quantifier depth based distance for first-order logical theories [
Reference Goranko, van Eijck, van Oostrom and Visser24]. Finally, Sokolsky et al. [Reference Sokolsky, Kannan, Lee, Hermanns and Palsberg42] introduce a quantitative bisimulation distance for finite,
labeled transition systems and consider its computation. Again, we conjecture the induced topology is the n-bisimulation topology.
6.1 Degrees of bisimilarity
Contrary to our syntactic approach to metric construction, a natural semantic approach rests on bisimulations. The notion of n-bisimilarity may be used to define a semantically based metric on
quotient spaces of pointed Kripke models where degrees of bisimilarity translate to closeness in space—the more bisimilar, the closer:
Let X be a set of pointed Kripke models for which modal equivalence and bisimilarity coincideFootnote ^3 and let relate $x,y\in X$ iff x and y are n-bisimilar. Then
is a metric on $\boldsymbol {X}_{\mathcal {L}}$ . Refer to $d_{B}$ as the n-bisimulation metric.
For X and $\mathcal {L}$ based on a finite signature, the n-bisimulation metric is a special case of the presented approach:
Proposition 15. Let $\mathcal {L}$ have finite signature and let $(\boldsymbol {X}_{\mathcal {L}},d_{B})$ be a metric modal space under the n-bisimulation metric. Then there exists a $D\subseteq \
mathcal {L}$ such that $d_{B}\in \mathcal {D}_{(X,D)}$ .
Proof With $\mathcal {L}$ of finite signature, every model in X has a characteristic formula up to n-bisimulation: For each $x\in X$ , there exists a $\varphi _{x,n}\in \mathcal {L}$ such that for
all $y\in X$ , $y\vDash \varphi _{x,n}$ iff (cf. [Reference Goranko, Otto, Blackburn, van Benthem and Wolter25, Reference Moss37]). Given that both $\Phi $ and $\mathcal {I}$ are finite, so is, for
each n, the set $D_{n}=\{\varphi _{x,n}:x\in X\}\subseteq \mathcal {L}$ . Pick the descriptor to be $D=\bigcup _{n\in \mathbb {N}}D_{n}$ . Then D is $\mathcal {L}$ -representative for X, so $\
boldsymbol {X}_{D}$ is identical to $\boldsymbol {X}_{\mathcal {L}}$ (cf. Lemma 3).
Let a weight function b be given by
$$\begin{align*}{\textstyle b(\varphi)=\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\text{ for }\varphi\in D_{n}.} \end{align*}$$
Then $d_{b}$ , defined by $d_{b}(\boldsymbol {x},\boldsymbol {y}) = \sum _{k=1}^{\infty }b(\varphi _{k})d_{k}(\boldsymbol {x},\boldsymbol {y}),$ is a metric on $\boldsymbol {X}_{\mathcal {L}}$ (cf.
Proposition 4).
As models x and y will, for all n, either agree on all members of $D_{n}$ or disagree on exactly 2 (namely $\varphi _{n,x}$ and $\varphi _{n,y}$ ) and as, for all $k\leq n$ , $y\vDash \varphi _{n,x}$
implies $y\vDash \varphi _{k,x}$ , and for all $k\geq n$ , $y\not \vDash \varphi _{n,x}$ implies $y\not \vDash \varphi _{k,x}$ , we obtain that
which is exactly $d_{B}$ .
Remark 16. The construction can be made independent of the set X to the effect that the constructed metric $d_{b}$ is exactly $d_{B}$ on any $\mathcal {L}$ -modal space $\boldsymbol {X}_{\mathcal
{L}}$ .
The n-bisimulation metric only is a special case when the set of atoms and number of modalities are both finite: If either is infinite, there is no metric in $\mathcal {D}_{(X,D)}$ for a descriptor
$D\subseteq \mathcal {L}$ that is equivalent to the n-bisimulation metric. This follows from an analysis of its metric topology, the n-bisimulation topology, $\mathcal {T}_{B}$ . A basis for this
topology is given by all subsets of $\boldsymbol {X}_{\mathcal {L}}$ of the form
for $\boldsymbol {x}\in \boldsymbol {X}_{\mathcal {L}}$ and $n\in \mathbb {N}$ .
By Propositions 5 and 15—and the fact that the set D constructed in the proof of the latter is finitely $\mathcal {L}$ -representative over X—we obtain the following:
Corollary 17. If $\mathcal {L}$ has finite signature, then the n-bisimulation topology $\mathcal {T}_{B}$ is the Stone(-like) topology $\mathcal {T}_{\mathcal {L}}$ .
This is not the general case:
Proposition 18. If $\mathcal {L}$ is based on an infinite set of either atoms or operators, then the n-bisimulation topology $\mathcal {T}_{B}$ is strictly finer than the Stone(-like) topology $\
mathcal {T}_{\mathcal {L}}$ on $\boldsymbol {X}_{\mathcal {L}}$ .
Proof $\mathcal {T}_{B}\not \subseteq \mathcal {T}_{\mathcal {L}}$ : In the infinite atoms case, $\mathcal {T}_{B}$ has as basis element $B_{\boldsymbol {x}0}$ , consisting exactly of those $\
boldsymbol {y}$ such that y and x share the same atomic valuation, i.e., are $0$ -bisimilar. Clearly, $\boldsymbol {x}\in B_{\boldsymbol {x}0}$ . There is no formula $\varphi $ for which the $\
mathcal {T}_{\mathcal {L}}$ basis element $B=\{\boldsymbol {z}\in \boldsymbol {X}\colon z\vDash \varphi \}$ contains $\boldsymbol {x}$ and is contained in $B_{\boldsymbol {x}0}$ : This would require
that $\varphi $ implied every atom or its negation, requiring the strength of an infinitary conjunction. For the infinite operators case, the same argument applies, but using $B_{\boldsymbol {x}1},$
containing exactly those $\boldsymbol {y}$ for which x and y are $1$ -bisimilar.
$\mathcal {T}_{\mathcal {L}}\subseteq \mathcal {T}_{B}$ : Consider any $\varphi \in \mathcal {L}$ and the corresponding $\mathcal {T}_{\mathcal {L}}$ basis element $B=\{\boldsymbol {y}\in \boldsymbol
{X}\colon y\vDash \varphi \}$ . Assume $\boldsymbol {x}\in B$ . Let the modal depth of $\varphi $ be n. Then for every $\boldsymbol {z}\in B_{\boldsymbol {x}n}$ , $z\vDash \varphi $ . Hence $\
boldsymbol {x}\in B_{\boldsymbol {x}n}\subseteq B$ .
The discrepancy in induced topologies results as the n-bisimulation metric, in the infinite case, introduces distinctions not finitely expressible in the language: If there are infinitely many atoms
or operators, there does not exist a characteristic formula $\varphi _{x,n}$ satisfied only by models n-bisimilar to x.
The additional open sets are not without consequence—a modal space compact in the Stone(-like) topology need not be so in the n-bisimulation topology: Let $\boldsymbol {X}_{\mathcal {L}}$ be an
infinite modal space with $\mathcal {L}$ based on an infinite atom set and assume it compact in the Stone(-like) topology. It will not be compact in the n-bisimulation topology: $\{B_{\boldsymbol {x}
0}\colon x\in X\}$ is an open cover of $\boldsymbol {X}_{\mathcal {L}}$ which contains no finite subcover.
7 Convergence and limit points
We next turn to dynamic aspects of Stone-like topologies. In particular, we focus on the nature of convergent sequences in Stone-like topologies and such topologies’ isolated points.
7.1 Convergence
Being Hausdorff, topological convergence in Stone-like topologies captures the geometrical intuition of a sequence $(\boldsymbol {x}_{n})$ converging to at most one point, its limit. We write $(\
boldsymbol {x}_{n})\rightarrow \boldsymbol {x}$ when $\boldsymbol {x}$ is the limit of $(\boldsymbol {x}_{n})$ . In general Stone-like topologies, such a limit need not exist.
Convergence in Stone-like topologies also satisfies a natural logical intuition, namely that a sequence and its limit should eventually agree on every formula of the language used to describe them.
This intuition is captured by the notion of logical convergence, introduced in [Reference Klein, Rendsvig, Baltag, Seligman and Yamada31]:
Definition. Let $\boldsymbol {X}_{D}$ be a modal space. A sequence of points $(\boldsymbol {x}_{n})$ logically converges to $\boldsymbol {x}$ in $\boldsymbol {X}_{D}$ iff for every $\psi \in \{\
varphi ,\neg \varphi \colon \varphi \in D\}$ for which $\boldsymbol {x}\vDash \psi $ , there exists some $N\in \mathbb {N}$ such that $\boldsymbol {x}_{n}\vDash \psi $ , for all $n\geq N$ .
The following proposition identifies a tight relationship between Stone-like topologies, topological and logical convergence, strengthening a result in [Reference Klein, Rendsvig, Baltag, Seligman
and Yamada31]:
Proposition 19. Let $\boldsymbol {X}_{D}$ be a modal space and $\mathcal {T}$ a topology on $\boldsymbol {X}_{D}$ . Then the following are equivalent:
1. 1. A sequence $\boldsymbol {x}_{1},\boldsymbol {x}_{2},\ldots $ of points from $\boldsymbol {X}_{D}$ converges to $\boldsymbol {x}$ in $(\boldsymbol {X}_{D},\mathcal {T})$ if, and only if, $\
boldsymbol {x}_{1},\boldsymbol {x}_{2},\ldots $ logically converges to $\boldsymbol {x}$ in $\boldsymbol {X}_{D}$ .
2. 2. $\mathcal {T}$ is the Stone-like topology $\mathcal {T}_{D}$ on $\boldsymbol {X}_{D}$ .
Proof $2\Rightarrow 1:$ This is shown, mutatis mutandis, in [Reference Klein, Rendsvig, Baltag, Seligman and Yamada31, Proposition 2].
$1\Rightarrow 2: \ \mathcal {T}_{D}\subseteq \mathcal {T}:$ We show that $\mathcal {T}$ contains a subbasis of $\mathcal {T}_{D}$ : for all $\varphi \in D$ , $[\varphi ],[\neg \varphi ]\in \mathcal
{T}$ . We show that $[\varphi ]$ is open in $\mathcal {T}$ by proving that $[\neg \varphi ]$ is closed in $\mathcal {T}$ , qua containing all its limit points: Assume the sequence $(\boldsymbol {x}_
{i})\subseteq [\neg \varphi ]$ converges to $\boldsymbol {x}$ in $(\boldsymbol {X}_{D},\mathcal {T})$ . For each $i\in \mathbb {N}$ , we have $\boldsymbol {x}_{i}\vDash \neg \varphi $ . As
convergence is assumed to imply logical convergence, then also $\boldsymbol {x}\vDash \neg \varphi $ . Hence $\boldsymbol {x}\in [\neg \varphi ]$ , so $[\neg \varphi ]$ is closed in $\mathcal {T}$ .
That $[\neg \varphi ]$ is open in $\mathcal {T}$ follows by a symmetric argument. Hence $\mathcal {T}_{D}\subseteq \mathcal {T}$ .
$\mathcal {T}\subseteq \mathcal {T}_{D}:$ The reverse inclusion follows as for every element $\boldsymbol {x}$ of any open set U of $\mathcal {T}$ , there is a basis element B of $\mathcal {T}_{D}$
such that $\boldsymbol {x}\in B\subseteq U$ . Let $U\in \mathcal {T}$ and let $\boldsymbol {x}\in U$ . Enumerate the set $\{\psi \in \overline {D}\colon x\vDash \psi \}$ as $\psi _{1},\psi _{2},\dots
$ , and consider all conjunctions of finite prefixes $\psi _{1}$ , $\psi _{1}\wedge \psi _{2}$ , $\psi _{1}\wedge \psi _{2}\wedge \psi _{3},\dots $ of this enumeration. If for some k, $[\psi _{1}\
wedge \cdots \wedge \psi _{k}]\subseteq U$ , then $B=[\psi _{1}\wedge \cdots \wedge \psi _{k}]$ is the desired $\mathcal {T}_{D}$ basis element as $\boldsymbol {x}\in [\psi _{1}\wedge \cdots \wedge \
psi _{k}]\subseteq U$ . If there exists no $k\in \mathbb {N}$ such that $[\psi _{1}\wedge \cdots \wedge \psi _{k}]\subseteq U$ , then for each $m\in \mathbb {N}$ , we can pick an $\boldsymbol {x}_{m}
$ such that $\boldsymbol {x}_{m}\in [\psi _{1}\wedge \cdots \wedge \psi _{m}]\setminus U$ . The sequence $(\boldsymbol {x}_{m})_{m\in \mathbb {N}}$ then logically converges to $\boldsymbol {x}$ .
Hence, by assumption, it also converges topologically to $\boldsymbol {x}$ in $\mathcal {T}$ . Now, for each $m\in \mathbb {N}$ , $\boldsymbol {x}_{m}$ is in $U^{c}$ , the compliment of U. However, $
\boldsymbol {x}\notin U^{c}$ . Hence, $U^{c}$ is not closed in $\mathcal {T}$ , so U is not open in $\mathcal {T}$ . This is a contradiction, rendering impossible that there is no $k\in \mathbb {N}$
such that $[\psi _{1}\wedge \cdots \wedge \psi _{k}]\subseteq U$ . Hence $\mathcal {T}_{D}\subseteq \mathcal {T}$ .
In [Reference Klein, Rendsvig, Baltag, Seligman and Yamada31], the satisfaction of point 1 was used as motivation for working with Stone-like topologies. Proposition 19 shows that this choice of
topology was necessary, if one wants the logical intuition satisfied. Moreover, it provides a third way of inducing Stone-like topologies, different from inducing them from a metric or a basis,
namely through sequential convergence.Footnote ^4
7.2 Isolated points
The existence of isolated points may be of interest, e.g., in information dynamics. A sequence $(\boldsymbol {x}_{n})$ in $A\subseteq \boldsymbol {X}_{D}$ converges to an isolated point $\boldsymbol
{x}$ in $A\subseteq \boldsymbol {X}_{D}$ iff for some N, for all $k>N$ , $\boldsymbol {x}_{k}=\boldsymbol {x}$ . Hence, if the goal of a given protocol is satisfied only at isolated points, the
protocol will either be successful in finite time or not at all.
The existence of isolated points in Stone-like topologies is tightly connected with the expressive power of the underlying descriptor. Say that a point $\boldsymbol {x}\in \boldsymbol {X}_{D}$ is
characterizable by D in $\boldsymbol {X}_{D}$ if there exists a finite set of formulas $A\subseteq \overline {D}$ such that for all $\boldsymbol {y}\in \boldsymbol {X}_{D}$ , if $\boldsymbol {y}\
vDash \varphi $ for all $\varphi \in A$ , then $\boldsymbol {y}=\boldsymbol {x}$ . We obtain the following:
Proposition 20. Let $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ be a modal space with its Stone-like topology. Then $\boldsymbol {x}\in \boldsymbol {X}_{D}$ is an isolated point of $\boldsymbol {X}_{D}$
iff $\boldsymbol {x}$ is characterizable by D in $\boldsymbol {X}_{D}$ .
Proof Left-to-right: If $\{\boldsymbol {x}\}$ is open in $\mathcal {T}_{D}$ , it must be in the basis of $\mathcal {T}_{D}$ and thus a finite intersection of subbasis elements, i.e., $\{\boldsymbol
{x}\}=\bigcap _{\varphi \in A}[\varphi ]$ for some finite $A\subseteq \overline {D}$ . Then A characterizes $\boldsymbol {x}$ . Right-to-left: Let A characterize $\boldsymbol {x}$ in $\boldsymbol {X}
_{D}$ . Each $[\varphi ],\varphi \in A,$ is open in $\mathcal {T}_{D}$ by definition. With A finite, also $\bigcap _{\varphi \in A}[\varphi ]$ is open. Hence $\{\boldsymbol {x}\}\in \mathcal {T}_{D}$
Applying Proposition 20 shows that convergence is of little interest when $\mathcal {L}$ is the mono-modal language over a finite atom set $\Phi $ and X is $S5$ -complete: Then $(\boldsymbol {X}_{\
mathcal {L}},\mathcal {T}_{\mathcal {L}})$ is a discrete space, i.e., contains only isolated points.
7.2.1 Perfect spaces
A topological space $(X,\mathcal {T})$ with no isolated points is perfect. In perfect spaces, every point is the limit of some sequence, and may hence be approximated arbitrarily well. The property
is enjoyed by several natural classes of modal spaces under their Stone-like topologies (cf. Corollary 22). Each such space that is additionally compact is homeomorphic to the Cantor set, as every
totally disconnected compact metric space is (see, e.g., [Reference Moise36, Chapter 12]). Proposition 20 implies that a modal space under its Stone-like topology is perfect iff it contains no points
characterizable by its descriptor.
Proposition 21. Let $D\subseteq \mathcal {L}$ , let $\Lambda $ be a logic, and let X be a set of $\Lambda $ -models $\Lambda $ -saturated with respect to D. Then $(\boldsymbol {X}_{D},\mathcal {T}_
{D})$ is perfect iff for every finite $\Lambda $ -consistent set $A\subseteq \overline {D}$ there is some $\psi \in D$ such that both $\psi \wedge \bigwedge _{\chi \in A}\chi $ and $\neg \psi \wedge
\bigwedge _{\chi \in A}\chi $ are $\Lambda $ -consistent.
Proof $\Rightarrow :$ Assume $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ perfect. Let $A\subseteq \{\varphi ,\neg \varphi \colon \varphi \in D\}$ be finite and $\Lambda $ -consistent. We must find $\psi
\in D$ for which both $\psi \wedge \bigwedge _{\chi \in A}\chi $ and $\neg \psi \wedge \bigwedge _{\chi \in A}\chi $ are $\Lambda $ -consistent. As $\boldsymbol {X}_{D}$ is $\Lambda $ -saturated with
respect to D, there is some $\boldsymbol {x}\in \boldsymbol {X}_{D}$ with $x\vDash \bigwedge _{\chi \in A}\chi $ . With $(\boldsymbol {X}_{D},\mathcal {T}_{D})$ perfect, $\bigcap _{\varphi \in A}[\
varphi ]_{D}\supsetneq \{\boldsymbol {x}\}$ —i.e., there is some $\boldsymbol {y}\neq \boldsymbol {x}\in \bigcap _{\varphi \in A}[\varphi ]_{D}$ . This implies that there is some $\psi \in D$ such
that $\boldsymbol {x}\vDash \psi $ and $\boldsymbol {y}\not \vDash \psi $ or vice versa. Either way, $\boldsymbol {x}$ and $\boldsymbol {y}$ witness that $\psi \wedge \bigwedge _{\chi \in A}\chi $
and $\neg \psi \wedge \bigwedge _{\chi \in A}\chi $ are both $\Lambda $ -consistent.
$\Leftarrow :$ No $\boldsymbol {x}\in \boldsymbol {X}_{D}$ is isolated: By Proposition 20, it suffices to show that $\boldsymbol {x}$ is not characterizable by D in $\boldsymbol {X}_{D}$ . For a
contradiction, assume some finite $A\subseteq \overline {D}$ characterizes $\boldsymbol {x}$ . By assumption, there is some $\psi \in D$ such that both $\psi \wedge \bigwedge _{\chi \in A}\chi $ and
$\neg \psi \wedge \bigwedge _{\chi \in A}\chi $ are $\Lambda $ -consistent. As $\boldsymbol {X}_{D}$ is $\Lambda $ -saturated, there are $y,z\in X$ with $y\vDash \psi \wedge \bigwedge _{\chi \in A}\
chi $ and $z\vDash \neg \psi \wedge \bigwedge _{\chi \in A}\chi $ . As $\psi \in D$ , $\boldsymbol {y}\neq \boldsymbol {z}$ . In particular $\boldsymbol {x}\neq \boldsymbol {y}$ or $\boldsymbol {x}\
neq \boldsymbol {z}$ , contradicting the assumption that A characterizes $\boldsymbol {x}$ .
If D is closed under negations and disjunctions, the assumption of Proposition 21 may be relaxed to stating that for any $\Lambda $ -consistent $\varphi \in D$ there is some $\psi \in D$ such that $\
varphi \wedge \psi $ and $\varphi \wedge \neg \psi $ are both $\Lambda $ -consistent. This property is enjoyed by many classic modal logics:
Corollary 22. For the following modal logics, $(\boldsymbol {X}_{\mathcal {L}},\mathcal {T}_{\mathcal {L}})$ is perfect if X is saturated with respect to $\mathcal {L}$ : $\mathrm{(i)}$ the normal
modal logic K with an infinite set of atoms, as well as $\mathrm{(ii)} \ KD$ , $\mathrm{(iii)} \ KD45^{n}$ for $n\geq 2$ , and $\mathrm{(iv)} \ S5^{n}$ for $n\geq 2$ .
7.2.2 Imperfect spaces
It is not difficult to find $\Lambda $ -complete spaces $(\boldsymbol {X}_{\mathcal {L}},\mathcal {T}_{\mathcal {L}})$ that contain isolated points. We provide two examples. The first shows that,
when working in a language with finite signature, then, e.g., for the minimal normal modal logic K, the K-complete space will have an abundance of isolated points.
Proposition 23. Let $\mathcal {L}$ have finite signature $(\Phi ,\mathcal {I})$ and let $\Lambda $ be such that $\bigvee _{i\in \mathcal {{I}}}\lozenge _{i}\top $ is not a theorem. If $(\boldsymbol
{X}_{\mathcal {L}},\mathcal {T}_{\mathcal {L}})$ is $\Lambda $ -complete, then it contains an isolated point. If $\Lambda $ is exactly K, then it contains infinitely many isolated points.
Proof With $\bigvee _{i\in \mathcal {{I}}}\lozenge _{i}\top $ not a theorem, there is an atom valuation encodable as a conjunction $\varphi $ such that $\varphi \wedge \bigwedge _{i\in \mathcal {I}}\
square _{i}\bot $ is consistent. The latter characterizes the point $\boldsymbol {x}$ in $\boldsymbol {X}_{\mathcal {L}}$ uniquely, as it has no outgoing relations. The point $\boldsymbol {x}$ is
clearly isolated. If $\Lambda $ is exactly K, there are for each $n\in \mathbb {N}$ only finitely many modally different models satisfying $\psi _{n}=\bigwedge _{i\in \mathcal {{I}}}\left (\bigwedge
_{m<n}\lozenge _{i}^{m}\top \wedge \neg \lozenge _{i}^{n}\top \right )$ ; hence $[\psi _{n}]$ is finite in $\boldsymbol {X}_{\mathcal {L}}$ . This, together with the fact that $(\boldsymbol {X}_{\
mathcal {L}},\mathcal {T}_{\mathcal {L}})$ is Hausdorff, implies that any $\boldsymbol {x}\in [\psi _{n}]$ is characterizable by $\mathcal {L}$ making $\boldsymbol {x}$ isolated (cf. Proposition 20).
For the second example, we turn to epistemic logic with common knowledge [Reference Halpern, Moses, Kameda, Misra, Peters and Santoro27, Reference Lehmann, Kameda, Misra, Peters and Santoro34]. Let
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Scientific Pearls of Wisdom
In physics there a few no-X theorems that seem rather suspicious in the sense that they point to cracks in our current understanding of nature. One of them is Penrose's cosmic censorship hypothesis
that translates to a "no-naked-singularity" conjecture. The most famous examples are at the center of black holes where there are supposedly singularities. But we will never know about them because
they are shielded by black hole horizons (unless you are willing to kill yourself and fly inside the black hole, however you will not be able to report back to us). That to me sounds a bit
Here is another one. The no-communication theorem in quantum mechanics. When two particles are entangled they become correlated with each other over arbitrarily large distances. This implies that if
Alice measures the spin of one of the particle here at earth, it collapses the joint wave function, say in a spin-up state, then *instantaneously* Bob's particle at the other end of the universe
collapses into a spin-down state. Special relatively forbids anything moving faster then the speed of light, because if it does for some observers a signal can arrive before it was sent! Similarly
here: for certain observers Bob first checks his spin before Alice has measured hers and therefore this observer concludes that Bob collapses the wave function and not Alice!
The only way out is that these two interpretations are actually coherent and this means that no information can be send between Alice and Bob. Alice can not force the particle in a up state (or a
down state for that matter) so that's no use. But perhaps she can transmit information by simply collapsing a superposition into one of the two pure states (irrespective whether it's up or down)? Bob
only has to determine if the wave function is in state A+B or in one of the pure states A or B. Alas, Bob can not do this with a single measurement because his measurement will collapse the wave
function into either A or B. Now what if he could make N copies of his wave function? Then by measuring all N copies he finds either that all of them are in state A or B or he finds some of them in A
and some in B indicating that the original wave function was still mixed. Unfortunately for Bob, the no-cloning theorem comes to quantum mechanic's rescue which says that you can not make copies of
wave functions.
Feeling uneasy? To me this seems like a theory that is trying to rescue itself. Not really the most concise explanation. We need multiple no-X theorems to wiggle ourselves out of difficult questions.
What this points to in my opinion is that the current theory (quantum mechanics) is an unnatural (but still accurate) theory of nature. We reach the right conclusion but through weird complicated
reasonings. Very similar to Ptolemy's model of the universe that made the correct predictions but was complicated and difficult to interpret. The new theory replacing quantum mechanics will hopefully
act as Ocam's razor and bring natural explanations for quantum weirdness.
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Passive fault-tolerant control for vehicle active suspension system based on H2/H∞ approach
In this paper, a robust passive fault-tolerant control (RPFTC) strategy based on ${H}_{2}/{H}_{\infty }$ approach and an integral sliding mode passive fault tolerant control (ISMPFTC) strategy based
on ${H}_{2}/{H}_{\infty }$ approach for vehicle active suspension are presented with considering model uncertainties, loss of actuator effectiveness and time-domain hard constraints of the suspension
system. ${H}_{\infty }$ performance index less than $\gamma$ and ${H}_{2}$ performance index is minimized as the design objective, avoid choosing weighting coefficient. The half-car model is taken as
an example, the robust passive fault-tolerant controller and the integral sliding mode passive fault tolerant control law is designed respectively. Three different fault modes are selected. And then
compare and analyze the control effect of vertical acceleration of the vehicle body and pitch angular acceleration of passive suspension control, robust passive fault tolerant control and integral
sliding mode passive fault tolerant control to verify the feasibility and effectiveness of passive fault tolerant control algorithm of active suspension. The studies we have performed indicated that
the passive fault tolerant control strategy of the active suspension can improve the ride comfort of the suspension system.
• Actuator partial failure, time-domain hard constraints and model uncertainties are considered.
• Two passive fault-tolerant control strategy based on H2/H∞ approach are proposed.
• The proposed two control strategies can improve the ride comfort of the vehicle.
1. Introduction
A suspension system is a very important component of any vehicle. It flexibly connects the vehicle body and the axles, carries the vehicle body; transmits all vertical forces between body and road,
including bears the forces between the tires and body, absorbs the shock from the road surface; protects the vehicle from unwanted vibrations and avoids excessive suspension stroke or related hard
stop/impact. The suspension plays a key role in the riding comfort and handling stability [1-3]. The primary functions of a vehicle suspension system are to isolate a car body from road disturbances
in order to provide good ride quality, to keep good road holding, to provide good handling and support the vehicle static weight [4, 5]. Suspensions generally fall into either of two groups-dependent
suspensions and independent suspensions. Suspension systems can be divided into passive suspension, semi-active suspension, and active suspension, according to the different control modes used [1].
There are generally three fundamental elements for a typical vehicle suspension system, including springing, damping, and location of the wheel [6]. Active suspension system contains separate
actuator that is able to provide external forces to both add and dissipate energy from the system. The task of the suspension spring is to carry the body mass and to isolate the body from road
disturbances. The suspension damper contributes to both driving safety and quality. Its task is the damping of body and wheel oscillations. A non-bouncing wheel is a condition for transferring
road-contact forces [2].
However, conventional suspensions can achieve a trade-off between ride comfort and road holding since their spring and damping coefficients cannot be adaptively tuned according to driving efforts and
road conditions. They can achieve good ride comfort and road holding only under the designed conditions [7]. Active suspension systems have the best potential to deal with conflicting objectives
placed on a suspension system a potential to deal with the trade-off between the conflicting suspension performance measures. Because vehicle active suspension can significantly improve the ride
comfort of passengers, meet the requirements of handling stability simultaneously. Over the past three decades, many researchers have devoted themselves to the theory and simulation studies on the
control method of vehicle active suspension. Many advances have been made in active suspension and control theory nowadays.
The active suspension control problem can be considered as a disturbance attenuation problem with time-domain hard constraints [8, 9]. The mixed ${H}_{2}/{H}_{\infty }$ guaranteed cost control
problem of state feedback control laws is considered in [10] for linear systems with norm-bounded parameter uncertainty. The ${H}_{\infty }$ state feedback control strategy with time domain hard
constraints was proposed under non-stationary running [11]. [12] designed a linear, robust, guaranteed-cost state feedback control with pole region constraints subject to active suspension system
uncertainties. By the LMI (Linear Matrix Inequalities) method, a convex optimization problem is formulated to find the corresponding controller. A 3 DOF quarter car model is used in [13]. LQR based
fuzzy controller; Fuzzy PID controller and Linear Quadratic Controller (LQR) are designed respectively to analyze and compare the performance characteristics of the active system with the
uncontrolled system or passive suspension system. In [14], a robust model predictive control algorithm for polytopic uncertain systems with time-varying delays is presented for active suspension
systems. However, most often, controllers are designed for the faultless suspension system so that the closed loop meets given performance specifications in studies of active suspension. Since not
taking full account of the suspension system may exist malfunctions such as actuators, sensors or model parameters, which may lead to a degraded system performance (unsatisfactory performance) or
even the loss of the system function (even instability) [15]. That is to say, once an actuator or a sensor fault occurs in a suspension system, the conventional controllers cannot achieve better
performance in comparison with the reliable and fault-tolerant controllers.
Fault tolerant control systems (FTCS) as control systems that possess the ability to accommodate system component failures automatically. They are capable of maintaining overall system stability and
acceptable performance in the event of such failures. FTCS were also known as self-repairing, reconfigurable, restructurable, or self designing control systems [17-19]. Generally speaking, fault
tolerant control can be classified into two types: active fault tolerant control (AFTC) and passive fault tolerant control (PFTC). AFTC reacts to the system component failures actively, by means of a
fault detection and diagnosis (FDD) component that detects, isolates, and estimates the current faults, and re-configuring on-line the controller, so that the stability and acceptable performance of
the entire system can be maintained [18, 19]. Robust control is closely related to passive fault tolerant control systems (PFTCS). The controller is designed to be robust against disturbances and
uncertainty during the design stage. That is, controllers are designed to be robust against a class of presumed fault. In other words, the starting point of the design is to reduce system dependency
on the operation of a single component, even in the case of such failures and no correction, the system can still maintain some “acceptable” level of performance or degrade gracefully [17]. This
enables the controller to counteract the effect of a fault without requiring reconfiguration or fault detection and isolation(FDI). Passive schemes operate independently of any fault information and
basically exploit the robustness of the underlying controller. Such a controller is usually less complex, but in order to cope with ‘worst case’ fault effects, has a certain degree of conservatism
[16, 18, 19]. However, its advantages are obvious: the parameters and structure of the controller are to be designed fixed; it neither needs to adjust the control law and control parameters online,
nor needs fault detection, diagnosis and isolation; it is easy to realize and has the advantage of avoiding the time delay, which is very important [20-22].
A passive fault-tolerant ${H}_{\infty }$ controller is designed in [23] such that the resulting control system is reliable since it has the capability of guaranteeing asymptotic stability and ${H}_{\
infty }$ performance, and simultaneously satisfying the constraint performance in the scenarios of actuator failures. In [24], a fault-tolerant control approach is proposed to deal with the problem
of fault accommodation for unknown actuator failures of active suspension systems, where an adaptive robust controller is designed to adapt and compensate the parameter uncertainties, external
disturbances and uncertain non-linearities generated by the system itself and actuator failures. In [25], the robust fault-tolerant ${H}_{\infty }$ control problem of active suspension systems with
finite-frequency constraint is investigated. Both the actuator faults and external disturbances are considered in the controller synthesis. Other performances such as suspension deflection and
actuator saturation are also considered. In [26], an adaptive fault tolerant control problem for half-car active suspension system subject to Markovian jumping actuator failures is considered. By
employing adaptive backstepping technique, a new adaptive fault tolerant control scheme is proposed, which ensures the boundedness in probability of the considered systems.
Sliding mode based control schemes are a strong candidate for fault tolerant control because of their inherent robustness to matched uncertainties. Actuator faults can be effectively modeled as
matched uncertainties and therefore sliding mode based control schemes have an inherent capability to directly deal with actuator faults [27]. In [28], a new robust strategy is presented which
utilized the proportional-integral sliding mode control scheme. In [29] a sliding mode control is designed for a full nonlinear vehicle active suspension system. Sensor faults effects on the behavior
of the controlled system are analyzed and a fault tolerant control strategy to compensate for the sensor faults is proposed. [30] presents adaptive sliding mode fault tolerant control for
magneto-rheological (MR) suspension system considering the partial fault of MR dampers. In [31], an adaptive proportional-integral-derivative (PID) controller is proposed. Designing an adaptation
scheme for the PID gains to accommodate actuator faults. A fault tolerant control approach based on a novel sliding mode method is proposed in [32] for a full vehicle suspension system aims at
retaining system stability in the presence of model uncertainties, actuator faults, parameter variations.
This paper is based on the previous research results, consider that the vehicle active suspension system off-line designed controller in the event of malfunctions in actuators and model uncertainties
cannot achieve the desired control effect, and even controller appear failure make the suspension system loss of stability. In view of this, under the premise of fully taken into account the time
domain hard constraints of the suspension system; for the half-car active suspension system with actuator faults and model uncertainties, a robust passive fault tolerant control scheme based on ${H}_
{2}/{H}_{\infty }$ approach and an integral sliding mode passive fault tolerant control scheme based on ${H}_{2}/{H}_{\infty }$ approach is designed respectively.
2. Half-car model
Vehicle vibration models can be divided into quarter-car model, half-car model, full-car model. The quarter-vehicle model was initially developed to explore active suspension capabilities and gave
birth to the concepts of skyhook damping and fast load leveling, which are now being developed toward actual, large-scale production applications. The quarter-car model has been the bench model used
in the study of control algorithm for intelligent suspension system. Although the model is very simple and is considering only vertical vibration motions of the sprung mass and the unsprung mass, it
is very useful in initial development. By considering the vertical dynamics and taking into account the vehicle’s symmetry, a suspension can be reduced to the quarter-car model. To account for the
pitch motion or roll motion, a half-car model is adopted by many researchers. The model is considering the vertical vibration and pitch motions of vehicle body, and the vertical motion of front and
rear wheels. The half-car model was represented to simulate ride characteristics of a simplified whole vehicle, which leads to significant improvement in ride and handling. The vehicle body of the
full-car model is assumed to be rigid and has seven degree-of-freedom in heave, pitch, and roll directions [7].
A four degree-of-freedom half-car model is shown in Fig. 1. The model can simultaneously represent the passive suspension and active/semi-active according to the state of the actuator. If the
actuator is neglected, the model is a passive suspension. The model is an active suspension if the actuator can generate active control forces, while the model is a semi-active suspension if the
actuator can provide only damping forces. The model has been used extensively in the literature and captures many important characteristics of vertical and pitch motions.
Fig. 1Model of half-car active suspension system
The half-car model is shown in Fig. 1. With the assumption of a small pitch angle $\theta$, $\mathrm{s}\mathrm{i}\mathrm{n}\theta \approx \theta$, we have the following approximate linear
${{x}_{s1}=x}_{s}-a\theta ,{{x}_{s2}=x}_{s}+b\theta ,\theta =\frac{{\left(x}_{s2}-{x}_{s1}\right)}{\left(a+b\right)}.$
The dynamic equations of this model can be written as:
-u}_{1},\\ {{m}_{u2}{\stackrel{¨}{x}}_{u2}={K}_{t2}\left({x}_{g2}-{x}_{u2}\right)-{K}_{2}\left({x}_{u2}-{x}_{s2}\right)-{C}_{s2}\left({\stackrel{˙}{x}}_{u2}-{\stackrel{˙}{x}}_{s2}\right)-u}_{2},\\
{I}_{p}\stackrel{¨}{\theta }=-a{K}_{1}\left({x}_{u1}-{x}_{s1}\right)+b{K}_{2}\left({x}_{u2}-{x}_{s2}\right)-a{C}_{s1}\left({\stackrel{˙}{x}}_{u1}-{\stackrel{˙}{x}}_{s1}\right)\\ +b{C}_{s2}\left({\
stackrel{˙}{x}}_{u2}-{\stackrel{˙}{x}}_{s2}\right)-a{u}_{1}+b{u}_{2},\\ {m}_{s}{\stackrel{¨}{x}}_{s}={K}_{1}\left({x}_{u1}-{x}_{s1}\right)+{K}_{2}\left({x}_{u2}-{x}_{s2}\right)+{C}_{s1}\left({\
stackrel{˙}{x}}_{u1}-{\stackrel{˙}{x}}_{s1}\right)+{C}_{s2}\left({\stackrel{˙}{x}}_{u2}-{\stackrel{˙}{x}}_{s2}\right)+{u}_{1}+{u}_{2},\\ {x}_{s1}={x}_{s}-a\theta ,\\ {x}_{s2}={x}_{s}+b\theta ,\end
where ${m}_{u1}$, ${m}_{u2}$ are the front and rear unsprung mass respectively; ${I}_{p}$ is pitch moment of inertia; $\theta$ represents the pitch angular; ${m}_{s}$ is the sprung mass; ${K}_{1}$
and ${K}_{2}$ are the stiffness coefficient of front and rear suspension respectively;${C}_{s1}$ and ${C}_{s2}$ are the damping coefficient of front and rear suspension respectively; $a$ and $b$ are
the distance of front and rear axle to sprung mass center of gravity respectively; ${u}_{1}$ and ${u}_{2}$ are actuator control output force of the front and rear suspension respectively; ${x}_{u1}$
and ${x}_{u2}$ are the displacements of the front and rear unsprung mass respectively; ${x}_{g1}$ and ${x}_{g2}$ are the road profile inputs of the front and rear suspension system respectively.
Taking into account the time domain hard constraints on the suspension system [8, 9], it is necessary to keep suspension dynamic displacement in the usable range, so as to avoid bumping the limit
where ${S}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ is the maximum suspension deflection.
Consider that the driving stability of the vehicle requirements, the dynamic tyre load doesn’t exceed the static tyre load in order to ensure the tire's grip capacity:
where ${m}_{s1}$, ${m}_{s2}$ are the static load mass of the front and rear suspension parts respectively.
The active control force provided for the active suspension system should be confined to a certain range prescribed by limited power of the actuator:
$\left|{u}_{i}\right|\le {F}_{\mathrm{m}\mathrm{a}\mathrm{x}},\left(i=1,2\right),$
where ${F}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ represents actuator output threshold.
Generally speaking, ${H}_{\infty }$ control strategy can solve the robust stability problem of a controlled target, and ${H}_{2}$ control strategy lets the controlled targets have a better dynamic
performance. Combining on ${H}_{2}/{H}_{\infty }$ to build a hybrid control strategy, the problems of robust stability and optimal dynamic performance of a controlled target can be solved with better
results. For a half-vehicle model, the riding comfort is related to the body vertical acceleration, pitching angular acceleration. The handling stability is related to the front and rear tyre dynamic
loads. Based upon this consideration, and also considering how to reduce the complexity of a controller, a multi-objective optimization control strategy for an active suspension system is introduced.
Choosing ${H}_{\infty }$ norm as the robust performance index of the time domain hard constraints, ${H}_{2}$ norm as the time-domain LQG performance index of perturbation action.
That is, the constrained output (${\mathbf{Z}}_{\mathbf{\infty }}$) and controlled output (${\mathbf{Z}}_{2}$) are defined as follows:
$\left\{\begin{array}{c}{\mathbf{Z}}_{\infty }={\left[\frac{{x}_{s1}-{x}_{u1}}{{S}_{\mathrm{m}\mathrm{a}\mathrm{x}}}\mathbf{}\mathbf{}\mathbf{}\frac{{x}_{s2}-{x}_{u2}}{{S}_{\mathrm{m}\mathrm{a}\
{s2}+{m}_{u2}\right)g}\mathbf{}\mathbf{}\mathbf{}\frac{{u}_{1}}{{F}_{\mathrm{m}\mathrm{a}\mathrm{x}}}\mathbf{}\mathbf{}\mathbf{}\frac{{u}_{2}}{{F}_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right]}^{T},\\ {\
mathbf{Z}}_{2}={\left[{\stackrel{¨}{x}}_{s}\stackrel{¨}{\theta }\right]}^{T},\\ {m}_{s1}=\left(\frac{b}{\left(a+b\right)}\right){m}_{s}g,\\ {m}_{s2}={\left(\frac{a}{\left(a+b\right)}\right)m}_{s}g.\
The selected state vector $\mathbf{X}={\left[{x}_{s1}-{x}_{u1}{x}_{s2}-{x}_{u2}{\stackrel{˙}{x}}_{s1}{\stackrel{˙}{x}}_{s2}{x}_{u1}-{x}_{g1}{x}_{u2}-{x}_{g2}{\stackrel{˙}{x}}_{u1}{\stackrel{˙}{x}}_
{u2}\right]}^{T}$; the road surface vertical speed (${\stackrel{˙}{x}}_{g1},{\stackrel{˙}{x}}_{g2}$) are used as interference inputs, i.e. $\mathbf{W}=\left[{\stackrel{˙}{x}}_{g1},{\stackrel{˙}{x}}_
{g2}{\right]}^{T}$. Then the governing equations ${\sigma }_{f}$ can be presented in the following state-space form:
$\left\{\begin{array}{c}\stackrel{˙}{\mathbf{X}}=AX+{\mathbf{B}}_{1}W+{\mathbf{B}}_{2}U,\\ {\mathbf{Z}}_{\mathbf{\infty }}={\mathbf{C}}_{1}X+{\mathbf{D}}_{11}W+{\mathbf{D}}_{12}U\\ {\mathbf{Z}}_{2}=
${\mathbf{D}}_{11}=zeros\left(6,2\right),{\mathbf{D}}_{21}=zeros\left(2\right),{\mathbf{D}}_{12}=\left[\begin{array}{c}0000\frac{1}{{F}_{\mathrm{m}\mathrm{a}\mathrm{x}}}0\\ 00000\frac{1}{{F}_{\mathrm
${\mathbf{D}}_{22}=\left[\begin{array}{cc}\frac{1}{{m}_{s}}& \frac{1}{{m}_{s}}\\ \frac{-a}{{I}_{p}}& \frac{b}{{I}_{p}}\end{array}\right],{{\mathbf{B}}_{1}=\left[\begin{array}{c}0000-1000\\ 00000-100\
${\mathbf{B}}_{2}=\left[\begin{array}{c}00\frac{1}{{m}_{s}}+\frac{{a}^{2}}{{I}_{p}}\frac{1}{{m}_{s}}–\frac{ab}{{I}_{p}}00\frac{-1}{{m}_{u1}}0\\ 00\frac{1}{{m}_{s}}–\frac{ab}{{I}_{p}}\frac{1}{{m}_{s}}
$\mathbf{A}={\left[\begin{array}{cccccccc}0& 0& 1& 0& 0& 0& -1& 0\\ 0& 0& 0& 1& 0& 0& 0& -1\\ \frac{-{K}_{1}}{{m}_{s}}–\frac{{a}^{2}{K}_{1}}{{I}_{p}}& \frac{-{K}_{2}}{{m}_{s}}+\frac{ab{K}_{2}}{{I}_
{p}}& \frac{-{C}_{s1}}{{m}_{s}}–\frac{{a}^{2}{C}_{s1}}{{I}_{p}}& \frac{-{C}_{s2}}{{m}_{s}}+\frac{ab{C}_{s2}}{{I}_{p}}& 0& 0& \frac{{C}_{s1}}{{m}_{s}}+\frac{{a}^{2}{C}_{s1}}{{I}_{p}}& \frac{{C}_{s2}}
{{m}_{s}}+\frac{ab{C}_{s2}}{{I}_{p}}\\ \frac{–{K}_{1}}{{m}_{s}}+\frac{ab{K}_{1}}{{I}_{p}}& \frac{-{K}_{2}}{{m}_{s}}–\frac{{b}^{2}{K}_{2}}{{I}_{p}}& \frac{-{C}_{s1}}{{m}_{s}}+\frac{ab{C}_{s1}}{{I}_
{p}}& \frac{-{C}_{s2}}{{m}_{s}}–\frac{{b}^{2}{C}_{s2}}{{I}_{p}}& 0& 0& \frac{{C}_{s1}}{{m}_{s}}+\frac{ab{C}_{s1}}{{I}_{p}}& \frac{{C}_{s2}}{{m}_{s}}+\frac{{b}^{2}{C}_{s2}}{{I}_{p}}\\ 0& 0& 0& 0& 0& 0
& 1& 0\\ 0& 0& 0& 0& 0& 0& 0& 1\\ \frac{{K}_{1}}{{m}_{u1}}& 0& \frac{{C}_{s1}}{{m}_{u1}}& 0& \frac{–{K}_{t1}}{{m}_{u1}}& 0& \frac{{-C}_{s1}}{{m}_{u1}}& 0\\ 0& \frac{{K}_{2}}{{m}_{u2}}& 0& \frac{{C}_
{s2}}{{m}_{u2}}& 0& \frac{–{K}_{t2}}{{m}_{u2}}& 0& \frac{{-C}_{s2}}{{m}_{u2}}\end{array}\right]}_{8×8},$
${\mathbf{C}}_{1}={\left[\begin{array}{cccccccc}\frac{1}{{S}_{\mathrm{m}\mathrm{a}\mathrm{x}}}& 0& 0& 0& 0& 0& 0& 0\\ 0& \frac{1}{{S}_{\mathrm{m}\mathrm{a}\mathrm{x}}}& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0&
\frac{{K}_{t1}}{{m}_{s1}g+{m}_{u1}g}& 0& 0& 0\\ 0& 0& 0& 0& \frac{{K}_{t2}}{{m}_{s2}g+{m}_{u2}g}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]}_{6×8},$
${\mathbf{C}}_{2}={\left[\begin{array}{cccccccc}\frac{{-K}_{1}}{{m}_{s}}& \frac{–{K}_{2}}{{m}_{s}}& \frac{–{C}_{s1}}{{m}_{s}}& \frac{-{C}_{s2}}{{m}_{s}}& 0& 0& \frac{-{C}_{s1}}{{m}_{s}}& \frac{-{C}_
{s2}}{{m}_{s}}\\ \frac{a{K}_{1}}{{I}_{p}}& \frac{-b{K}_{2}}{{I}_{p}}& \frac{a{C}_{s1}}{{I}_{p}}& \frac{b{C}_{s2}}{{I}_{p}}& 0& 0& \frac{-a{C}_{s1}}{{I}_{p}}& \frac{b{C}_{s1}}{{I}_{p}}\end{array}\
3. Modeling of faulty systems
3.1. Actuator fault model
The actuator common faults are outage, loss of effectiveness, and stuck. The actual control input ${\mathbf{u}}_{a}\left(t\right)$ which is able to impact the system is not the same as the designed
control input $\mathbf{U}\left(t\right)$ designed in general. In this article, let ${\delta }_{1}$ and ${\delta }_{2}$ represent the percentage of efficiency loss of the front and rear actuator
respectively. We only consider loss of effectiveness faults occur to the actuators. If losses of effectiveness faults occur to the actuators, the faulty actuators will fail to provide the desired
control effects [20, 21]. That is, the faulty actuators can be formulated as:
${{\mathbf{u}}_{a}\left(t\right)=\left[\left(1-{\delta }_{1}\right){u}_{1}\left(1-{\delta }_{2}\right){u}_{2}\right]}^{T},$
where ${\delta }_{i}\in \left[01\right]\left(i=1,2\right)$, then ${1-\delta }_{i}\in \left[01\right]$. Define the actuator failure switch matrix as:
$\mathbf{M}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(1-{\delta }_{1}1-{\delta }_{2}\right),$
such as:
Define the failure factor ${m}_{i}=1-{\delta }_{i}\left(i=1,2\right)$ in $\mathbf{M}$, so then, ${0\le m}_{li}\le {m}_{i}\le {m}_{ui}$, ${m}_{li}$ and ${m}_{ui}$ represent the known lower and upper
bounds of ${m}_{i}$ respectively. If ${m}_{li}={m}_{ui}=0$, it covers to the outage case. If ${0\le m}_{li}<{m}_{i}<{m}_{ui}$ and ${m}_{i}e 1$, it corresponds to the partial failure case. And,
fault-free (normal) case with ${m}_{li}={m}_{ui}=1$. In order to facilitate the analysis, the following matrix is introduced [24]:
where ${m}_{0i}=\left({m}_{li}+{m}_{ui}\right)/2$, ${l}_{i}=\left({m}_{i}-{m}_{0i}\right)/{m}_{0i}$, ${j}_{i}=\left({m}_{ui}-{m}_{li}\right)/\left({m}_{ui}+{m}_{li}\right)\left(i=1,2\right)$. The
fault switch matrix can be expressed as:
$\left|\mathbf{L}\right|\le \mathbf{J}\le \mathbf{I}.$
3.2. Uncertainty model
Model uncertainties change the model parameters. In this paper, we will consider the following structure for the norm-bounded uncertainty:
where $∆\mathbf{A}$, $∆{\mathbf{B}}_{2}$ are real matrix functions representing time-varying parameter uncertainties. $\mathbf{H}$, ${\mathbf{E}}_{1}$, ${\mathbf{E}}_{2}$ are known constant matrices
of appropriate dimensions which characterize how the uncertain parameters in $\mathbf{F}\left(t\right)$. $\mathbf{F}\left(t\right)\in {R}^{i×j}$ is an unknown real time-varying matrix with Lebesgue
measurable elements satisfying:
${\mathbf{F}}^{T}\left(t\right)\mathbf{F}\left(t\right)\le \mathbf{I}.$
where $\mathbf{I}$ is an unit matrix of the appropriate dimension [10].
The stiffness and damping of active suspension change according to the sine function; variation range is both 20 %. Then in the light of Eq. (15), considering that the matrix $\mathbf{A}$, ${\mathbf
{B}}_{2}$ contain the stiffness and damping parameters and as well as the dimension of them, meanwhile in accordance with the given variation range perturbation, we can know that $\mathbf{H}$ is 8×4
matrix, ${\mathbf{E}}_{1}$ is 4×8 matrix, ${\mathbf{E}}_{2}$ is 4×2 matrix.
3.3. The fault suspension model
Due to the active suspension system with consideration of both uncertainty and actuator failures, thus, the fault system ${\mathrm{\Omega }}_{f}$ can be written by:
$\left\{\begin{array}{c}\stackrel{˙}{\mathbf{X}}=\left(\mathbf{A}+∆\mathbf{A}\right)X+{\mathbf{B}}_{1}W+{\left(\mathbf{B}}_{2}+∆{\mathbf{B}}_{2}\right)MU,\\ {\mathbf{Z}}_{\mathbf{\infty }}={\mathbf
{C}}_{1}X+{\mathbf{D}}_{12}MU,\\ {\mathbf{Z}}_{2}={\mathbf{C}}_{2}X+{\mathbf{D}}_{22}MU.\end{array}\right\$
4. Passive fault-tolerant controller design
4.1. Robust passive fault-tolerant controller design based on ${\mathbit{H}}_{2}/{\mathbit{H}}_{\mathbit{\infty }}$ approach
The designed ${H}_{2}/{H}_{\infty }$ state feedback controller in the active suspension intact and trouble-free condition is:
According to the faulty suspension system Eq. (17), the designed state feedback robust passive fault tolerant control law as follows:
such that for all admissible parameter uncertainties, the following design criteria are satisfied:
1) The resulting closed-loop system is asymptotically stable.
2) If $\mathbf{W}\left(t\right)$ is viewed as a finite energy disturbance signal, the closed-loop transfer function ${G}_{{Z}_{\infty W}}\left(s\right)$ from $\mathbf{W}\left(t\right)$ to ${\mathbf
{Z}}_{\infty }\left(t\right)$ satisfies:
${‖{G}_{{Z}_{\infty W}}\left(s\right)‖}_{\infty }<\gamma ,$
where ${‖{G}_{{Z}_{\infty W}}\left(s\right)‖}_{\infty }=\underset{W}{\mathrm{sup}}{\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}\left\{{G}_{{Z}_{\infty W}}\left(jW\right)\right\}$, ${\sigma }_{\mathrm{m}
\mathrm{a}\mathrm{x}}\left(\bullet \right)$ denotes the largest singular value, and $\gamma >0$ is a pre-specified disturbance attenuation level;
3) If $\mathbf{W}\left(t\right)$ is viewed as a white noise signal with the unit power spectral density, an upper bound of the worst case ${H}_{2}$ performance index defined by:
$J\left({\mathbf{K}}_{f}\right)=\underset{\mathbf{F}\left(t\right)}{\mathrm{sup}}\underset{t\to \infty }{\mathrm{lim}}E\left\{{\mathbf{Z}}_{2}^{T}{\mathbf{Z}}_{2}\right\}\le \stackrel{-}{J}\left({\
where $\stackrel{-}{J}\left({\mathbf{K}}_{f}\right)$ is a certain constant, $E\left\{\bullet \right\}$ denotes the expectation.
Substituting the control law Eq. (19) into fault suspension model Eq. (17), we obtain the available fault suspension state feedback closed-loop system ${\mathrm{\Sigma }}_{f}$:
$\left\{\begin{array}{c}\stackrel{˙}{\mathbf{X}}={\stackrel{-}{\mathbf{A}}}_{c}X+{\mathbf{B}}_{1}W,\\ {\mathbf{Z}}_{\mathbf{\infty }}={\mathbf{C}}_{1c}X,\\ {\mathbf{Z}}_{2}={\mathbf{C}}_{2c}X,\end
where ${\mathbf{A}}_{c}{=\mathbf{A}+\mathbf{B}}_{2}\mathbf{M}{\mathbf{K}}_{f}\text{,}$${\mathbf{E}}_{c}{={\mathbf{E}}_{1}+\mathbf{E}}_{2}\mathbf{M}{\mathbf{K}}_{f}\text{,}$${\mathbf{C}}_{1c}{={\
If ${\stackrel{-}{\mathbf{A}}}_{c}$ is asymptotically stable, then the performance index $J\left({\mathbf{K}}_{f}\right)$ can be computed by:
where $\stackrel{~}{\mathbf{P}}={\stackrel{~}{\mathbf{P}}}^{T}\ge 0$ is the observability Gramian matrix obtained from the following Lyapunov equation:
Lemma 1 [23]: for any matrices $\mathbf{X}$ and $\mathbf{Y}$ with appropriate dimensions, for any $\eta >0$, we have:
${\mathbf{X}}^{T}\mathbf{Y}+{\mathbf{Y}}^{T}\mathbf{X}\le \eta {\mathbf{X}}^{T}\mathbf{X}+{{\eta }^{-1}\mathbf{Y}}^{T}\mathbf{Y}.$
Theorem 1 [10]: For a given constant $\gamma >0$ and the closed-loop system ${\mathrm{\Sigma }}_{f}\text{,}$${\stackrel{-}{\mathbf{A}}}_{c}$ is asymptotically stable and ${‖{G}_{{Z}_{\infty W}}\left
(s\right)‖}_{\infty }<\gamma$ if and only if there exist two scalars $\alpha >0$ and $\beta >0$ such that the following matrix inequality:
${\mathbf{A}}_{c}^{T}\mathbf{P}+\mathbf{P}{\mathbf{A}}_{c}+\mathbf{P}\left(\alpha \mathbf{H}{\mathbf{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}\right)\mathbf{P}+{\alpha }^{-1}{\
mathbf{E}}_{c}^{T}{\mathbf{E}}_{c}+{\beta }^{-1}{\mathbf{C}}_{1c}^{T}{\mathbf{C}}_{1c}+{\mathbf{C}}_{2c}^{T}{\mathbf{C}}_{2c}<0,$
has a symmetric positive definite solution $\mathbf{P}$; furthermore, if Eq. (26) has a symmetric positive definite solution $\mathbf{P}$, then for all admissible parameter uncertainties:
$0\le \stackrel{~}{\mathbf{P}}\le \mathbf{P},$
where $\stackrel{~}{\mathbf{P}}={\stackrel{~}{\mathbf{P}}}^{\mathbf{T}}\ge 0$ is a solution to lyapunov Eq. (24).
The matrix inequality Eq. (26) multiplied by ${\mathbf{P}}^{-1}$ on both sides respectively such that:
${\mathbf{P}}^{-1}{\mathbf{A}}_{c}^{T}+{\mathbf{A}}_{c}{\mathbf{P}}^{-1}+\alpha \mathbf{H}{\mathbf{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}+{\alpha }^{-1}{{\mathbf{P}}^{-1}\
mathbf{E}}_{c}^{T}{\mathbf{E}}_{c}{\mathbf{P}}^{-1}$${+\beta }^{-1}{\mathbf{P}}^{-1}{\mathbf{C}}_{1c}^{T}{\mathbf{C}}_{1c}{\mathbf{P}}^{-1}+{\mathbf{P}}^{-1}{\mathbf{C}}_{2c}^{T}{\mathbf{C}}_{2c}{\
According to the Schur complement, the matrix inequality Eq. (28) is equivalent to Eq. (29):
$\left[\begin{array}{cccc}{\mathbf{P}}^{-1}{\mathbf{A}}_{c}^{T}+{\mathbf{A}}_{c}{\mathbf{P}}^{-1}+\alpha \mathbf{H}{\mathbf{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}& {\mathbf
{P}}^{-1}{\mathbf{E}}_{c}^{T}& {\mathbf{P}}^{-1}{\mathbf{C}}_{1c}^{T}& {\mathbf{P}}^{-1}{\mathbf{C}}_{2c}^{T}\\ {\mathbf{E}}_{c}{\mathbf{P}}^{-1}& -\alpha \mathbf{I}& 0& 0\\ {\mathbf{C}}_{1c}{\mathbf
{P}}^{-1}& 0& -\beta \mathbf{I}& 0\\ {\mathbf{C}}_{2c}{\mathbf{P}}^{-1}& 0& 0& -\mathbf{I}\end{array}\right]<0.$
Let ${\mathbf{Q}=\mathbf{P}}^{-1}$, ${\mathbf{V}={\mathbf{K}}_{f}\mathbf{P}}^{-1}$; substituting ${\mathbf{A}}_{c}$, ${\mathbf{E}}_{c}$, ${\mathbf{C}}_{1c}$, ${\mathbf{C}}_{2c}$ into the inequality
Eq. (29) such that:
$\left[\begin{array}{cccc}{\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}\mathbf{M}\mathbf{V}\right)}^{T}+\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}\mathbf{M}\mathbf{V}\right)+\alpha \mathbf{H}{\mathbf
{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}& *& *& *\\ {\mathbf{E}}_{1}\mathbf{Q}+{\mathbf{E}}_{2}\mathbf{M}\mathbf{V}& -\alpha \mathbf{I}& *& *\\ {\mathbf{C}}_{1}\mathbf{Q}+{\
mathbf{D}}_{12}\mathbf{M}\mathbf{V}& 0& -\beta \mathbf{I}& *\\ {\mathbf{C}}_{2}\mathbf{Q}+{\mathbf{D}}_{22}\mathbf{M}\mathbf{V}& 0& 0& -\mathbf{I}\end{array}\right]<0.$
In this paper, we agree on that “*” represent symmetric transpose of the matrix; substituting the fault switch matrix Eq. (13) into the matrix inequality Eq. (30), then considering that the matrix
inequality Eq. (14) and according to the lemma 1 such that:
$\left[\begin{array}{cccc}{\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{V}\right)}^{T}+\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{V}\right)+\alpha \
mathbf{H}{\mathbf{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}& *& *& *\\ {\mathbf{E}}_{1}\mathbf{Q}+{\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{V}& -\alpha \mathbf{I}& *& *\\ {\
mathbf{C}}_{1}\mathbf{Q}+{\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{V}& 0& -\beta \mathbf{I}& *\\ {\mathbf{C}}_{2}\mathbf{Q}+{\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{V}& 0& 0& -\mathbf{I}\end{array}\
right]$$+\eta \left[\begin{array}{cccc}{\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}
{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{22}{\
mathbf{M}}_{0}\right)}^{T}\\ {\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf
{M}}_{0}\right)}^{T}& {\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{22}{\mathbf{M}}_
{0}\right)}^{T}\\ {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf{M}}_{0}\
right)}^{T}& {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{22}{\mathbf{M}}_{0}\
right)}^{T}\\ {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf{M}}_{0}\
right)}^{T}& {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T}& {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{22}{\mathbf{M}}_{0}\
right)}^{T}\end{array}\right]$${+{\eta }^{-1}\left[{\mathbf{V}}^{T}{\mathbf{J}}^{1/2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}000\right]}^{T}\left[{\mathbf{J}}^{1/2}\mathbf{V}000\right]<0.$
In the light of the Schur complement such that:
$\left[\begin{array}{ccccc}{\mathrm{\Gamma }}_{11}& *& *& *& *\\ {\mathrm{\Gamma }}_{21}& {\mathrm{\Gamma }}_{22}& *& *& *\\ {\mathrm{\Gamma }}_{31}& {\mathrm{\Gamma }}_{32}& {\mathrm{\Gamma }}_{33}&
*& *\\ {\mathrm{\Gamma }}_{41}& {\mathrm{\Gamma }}_{42}& {\mathrm{\Gamma }}_{43}& {\mathrm{\Gamma }}_{44}& *\\ {\mathbf{J}}^{1/2}\mathbf{V}& 0& 0& 0& -\eta \mathbf{I}\end{array}\right]<0,$
${\mathrm{\Gamma }}_{11}={\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{V}\right)}^{T}+\left(\mathbf{A}\mathbf{Q}+{\mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{V}\right)+\alpha \mathbf
{H}{\mathbf{H}}^{T}+\beta {\gamma }^{-2}{\mathbf{B}}_{1}{\mathbf{B}}_{1}^{T}+{\eta \mathbf{B}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T},$
${\mathrm{\Gamma }}_{21}={\mathbf{E}}_{1}\mathbf{Q}+{\mathbf{E}}_{2}\mathbf{M}\mathbf{V}+{\eta \mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T},{\mathrm{\
Gamma }}_{22}=-\alpha \mathbf{I}+\eta {\mathbf{}\mathbf{E}}_{2}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf{M}}_{0}\right)}^{T},$
${\mathrm{\Gamma }}_{31}={\mathbf{C}}_{1}\mathbf{Q}+{\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{V}+\eta {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T},{\
mathrm{\Gamma }}_{32}=\eta {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf{M}}_{0}\right)}^{T},$
${\mathrm{\Gamma }}_{33}=-\beta \mathbf{I}+\eta {\mathbf{D}}_{12}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T},{\mathrm{\Gamma }}_{41}={\mathbf{C}}_{2}\mathbf{Q}+{\
mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{V}+\eta {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{B}}_{2}{\mathbf{M}}_{0}\right)}^{T},$
${\mathrm{\Gamma }}_{42}=\eta {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{E}}_{2}{\mathbf{M}}_{0}\right)}^{T},{\mathrm{\Gamma }}_{43}=\eta {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\
left({\mathbf{D}}_{12}{\mathbf{M}}_{0}\right)}^{T},{\mathrm{\Gamma }}_{44}=-\mathbf{I}+\eta {\mathbf{D}}_{22}{\mathbf{M}}_{0}\mathbf{J}{\left({\mathbf{D}}_{22}{\mathbf{M}}_{0}\right)}^{T}.$
Considering that the fault suspension system Eq. (17) and a given $\gamma >0$, if the following optimization problem:
$\text{s.t.}\text{(}\text{i}\text{)}\stackrel{\underset{\alpha ,\beta ,\eta ,\mathbf{Q},\mathbf{V},\mathbf{N}}{\mathrm{min}}\mathit{}\mathit{Trace}\left(\mathbf{N}\right)}{\stackrel{⏞}{\left[\begin
{array}{ccccc}{\mathrm{\Gamma }}_{11}& *& *& *& *\\ {\mathrm{\Gamma }}_{21}& {\mathrm{\Gamma }}_{22}& *& *& *\\ {\mathrm{\Gamma }}_{31}& {\mathrm{\Gamma }}_{32}& {\mathrm{\Gamma }}_{33}& *& *\\ {\
mathrm{\Gamma }}_{41}& {\mathrm{\Gamma }}_{42}& {\mathrm{\Gamma }}_{43}& {\mathrm{\Gamma }}_{44}& *\\ {\mathbf{J}}^{1/2}\mathbf{V}& 0& 0& 0& -\eta \mathbf{I}\end{array}\right]}}<0,$$\left(\text{ii}\
right)\left[\begin{array}{c}-N{\mathbf{B}}_{1}^{T}\\ {\mathbf{B}}_{1}-Q\end{array}\right]<0,$
by the solver $\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{x}$ of the $LMI$ toolbox of the MATLAB has a solution ${\alpha }^{*}$, ${\beta }^{*}$, ${\eta }^{*}$, ${\mathbf{Q}}^{*}$, ${\mathbf{V}}^
{*}$, ${\mathbf{N}}^{*}$, then the robust passive fault tolerant controller of the fault suspension loop-system based on ${H}_{2}/{H}_{\infty }$ approach is:
4.2. Integral sliding mode fault-tolerant controller design based on ${\mathbit{H}}_{2}/{\mathbit{H}}_{\mathbit{\infty }}$ approach
According to Eq. (17), the fault system ${\mathrm{\Omega }}_{f}$ can be also written by:
$\stackrel{˙}{\mathbf{X}}=\left(\mathbf{A}+∆\mathbf{A}\right)\mathbf{X}+{\mathbf{B}}_{2}\mathbf{U}-{\mathbf{B}}_{2}\mathbf{\delta }\mathbf{U}+{\mathbf{B}}_{1}\mathbf{W},$
where $\mathbf{\delta }=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left({\delta }_{1}{\delta }_{2}\right)$; let $\mathbf{\xi }=\mathbf{}\mathbf{\delta }\mathbf{U}$, ${\mathbf{f}}_{d}=\left[{\mathbf{B}}
_{1}\mathbf{}\mathbf{H}{\mathbf{E}}_{1}\right]\left[\begin{array}{c}W\\ \mathrm{sin}t\end{array}\right]$, then Eq. (35) can be written as:
$\stackrel{˙}{\mathbf{X}}=\mathbf{A}\mathbf{X}+{\mathbf{B}}_{2}\mathbf{U}+{\mathbf{B}}_{2}\left(-\mathbf{I}\right)\mathbf{\xi }+{\mathbf{f}}_{d},$
the actuator faults $\mathbf{\xi }$ can be effectively modelled as matched uncertainties.
Assumption 1. The function ${\mathbf{f}}_{d}$ represents unmatched uncertainty i.e. it does not lie within the range space of matrix ${\mathbf{B}}_{2}$, but is assumed to be bounded with known upper
The nominal linear system associated with Eq. (36) can be written as:
where ${\mathbf{U}}_{0}$ is a nominal control law which can be designed by any suitable state feedback paradigm to achieve desired nominal performance. Since it is assumed that the pair $\left(\
mathbf{A},{\mathbf{B}}_{2}\right)$ is controllable, then there exists a state feedback controller of the form:
where $\mathbf{K}$ is a state feedback gain to be designed. The matrix $\mathbf{K}$ can be designed using any state feedback design approach. Here $\mathbf{K}$ is the designed ${H}_{2}/{H}_{\infty }$
state feedback gain.
Define the control law $\mathbf{U}$ of the form:
where ${\mathbf{U}}_{0}$ is the nominal controller and ${\mathbf{U}}_{n}$ is a nonlinear injection to induce a sliding mode. Then use Eq. (39), Eq. (36) can be written as:
$\stackrel{˙}{\mathbf{X}}=\mathbf{A}\mathbf{X}+{\mathbf{B}}_{2}{\mathbf{U}}_{0}+{\mathbf{B}}_{2}{\mathbf{U}}_{n}-{\mathbf{B}}_{2}\mathbf{\xi }+{\mathbf{f}}_{d}$
where ${\mathbf{U}}_{n}$ is chosen to reject the disturbance term $\mathbf{\xi }$ while in the sliding mode. Here the switching function is defined as:
$\mathrm{\Phi }=\mathbf{G}\mathbf{X}+\mathbf{z},$
where $\mathbf{G}$ is design freedom and $\mathbf{z}$ is to be specified. Since the matrix ${\mathbf{B}}_{2}$ is of full rank, the switching matrix $\mathbf{G}$ can be chosen so that the matrix $\
mathbf{G}{\mathbf{B}}_{2}$ is nonsingular i.e. $\mathit{det}\left(\mathbf{G}{\mathbf{B}}_{2}\right)e 0$. During sliding $\stackrel{˙}{\mathrm{\Phi }}=\mathrm{\Phi }=0$ and therefore:
$\stackrel{˙}{\mathrm{\Phi }}=\mathbf{G}\mathbf{A}\mathbf{X}+\mathbf{G}{\mathbf{B}}_{2}{\mathbf{U}}_{0}+\mathbf{G}{\mathbf{B}}_{2}{\mathbf{U}}_{n}+\mathbf{G}{\mathbf{B}}_{2}\left(-\mathbf{I}\right)\
mathbf{\xi }+\mathbf{G}{\mathbf{f}}_{d}+\stackrel{˙}{\mathbf{z}}=0.$
During sliding it is expected that ${\mathbf{G}{\mathbf{B}}_{2}\mathbf{U}}_{neq}=\mathbf{}\mathbf{G}{\mathbf{B}}_{2}\mathbf{\xi }-\mathbf{G}{\mathbf{f}}_{d}$, i.e.:
${\mathbf{U}}_{neq}=-\left(-\mathbf{I}\right)\mathbf{\xi }-{\left(\mathbf{G}{\mathbf{B}}_{2}\right)}^{-1}\mathbf{G}{\mathbf{f}}_{d},$
then selecting:
Substituting the value of equivalent control ${\mathbf{U}}_{neq}$ from Eq. (43) into Eq. (40) and simplifying, the expression for the integral sliding mode dynamics can be written as:
Let $\mathrm{\varnothing }=\mathbf{I}-{\mathbf{B}}_{2}{\left(\mathbf{G}{\mathbf{B}}_{2}\right)}^{-1}\mathbf{G}$, from Eq. (45) it is clear that the effect of the matched uncertainty has been
completely rejected while in the sliding mode.
4.2.1. Integral switching surface
Using Eqs. (41, 44), an integral switching function:
$\mathrm{\Phi }=\mathbf{G}\mathbf{X}-\mathbf{G}{\int }_{0}^{t}\left(\mathbf{A}\mathbf{X}\left(\tau \right)+{\mathbf{B}}_{2}{\mathbf{U}}_{0}\left(\tau \right)\right)d\left(\tau \right).$
The sliding mode will exist from time $t=0$ and the system will be robust throughout the entire closed-loop system response against matched uncertainties. it is clear that in the case of only matched
uncertainty, then any choice of $\mathbf{G}$ which ensures $\mathbf{G}{\mathbf{B}}_{2}$ is invertible is sufficient for the Integral sliding mode design, for unmatched uncertainty, a specific choice
of $\mathbf{G}$ is needed. Here it will be argued that:
is an appropriate choice. Moreover, $\mathbf{G}{\mathbf{B}}_{2}=\mathbf{I}$.
This choice of $\mathbf{G}$ ensures that the square matrix $\mathbf{G}{\mathbf{B}}_{2}$ is nonsingular. The $\mathrm{\varnothing }$ becomes:
$\mathrm{\varnothing }=\mathrm{I}-{\mathbf{B}}_{2}{\left({\mathbf{B}}_{2}^{T}{\mathbf{B}}_{2}\right)}^{-1}{\mathbf{B}}_{2}^{T}.$
Notice that the projection operator $\mathrm{\varnothing }$ in Eq. (48) is symmetric and idempotent i.e. ${\mathrm{\varnothing }}^{2}=\mathrm{\varnothing }$. The properties of symmetry and
idempotency imply that $‖\mathrm{\varnothing }‖=\text{1}$ which means that the effect of ${\mathbf{f}}_{d}$ is not amplified since $‖\mathrm{\varnothing }{\mathbf{f}}_{d}‖\le ‖{\mathbf{f}}_{d}‖$.
4.2.2. Integral sliding mode control laws
An integral sliding mode controller will now be designed based on the nominal system in Eq. (37). The control law has a structure given by Eq. (39). ${\mathbf{U}}_{0}$ is the linear part of the
controller, and ${\mathbf{U}}_{n}$ is the discontinuous part to enforce a sliding mode along the sliding surface in Eq. (46). One choice of $\mathbf{U}$ is:
$\mathbf{U}=\mathbf{K}\mathbf{X}-\rho {\left(\mathbf{G}{\mathbf{B}}_{2}\right)}^{-1}\frac{\mathrm{\Phi }}{‖\mathrm{\Phi }‖+\phi },\mathbit{}\mathbit{}\mathbit{}\mathbit{}\mathrm{\Phi }e 0,$
where the value of the small positive scalar $\phi$ is chosen to eliminate chattering; $\rho$ is the modulation gain to enforce the sliding mode whose precise value is given later.
4.2.3. The reachability condition
To justify that the controller designed in Eq. (49) satisfies the $\mu$-reachability condition, which is a sufficient condition to ensure the existence of an ideal sliding motion, it can be shown
from Eq. (36) and Eq. (38) that:
$\stackrel{˙}{\mathrm{\Phi }}=\mathbf{G}\left(\mathbf{A}\mathbf{X}+{\mathbf{B}}_{2}\mathbf{U}+{\mathbf{B}}_{2}\left(-\mathbf{I}\right)\mathbf{\xi }+{\mathbf{f}}_{d}\right)-\mathbf{G}\mathbf{A}\mathbf
then substituting from Eq. (50), and after some simplification:
$\stackrel{˙}{\mathrm{\Phi }}=\mathbf{G}\mathbf{A}\mathbf{X}+\mathbf{G}{\mathbf{B}}_{2}\mathbf{K}\mathbf{X}-\rho \mathbf{G}{\mathbf{B}}_{2}{\left(\mathbf{G}{\mathbf{B}}_{2}\right)}^{-1}\frac{\mathrm
{\Phi }}{‖\mathrm{\Phi }‖+\phi }+\mathbf{G}{\mathbf{B}}_{2}\left(-\mathbf{I}\right)\mathbf{\xi }+\mathbf{G}{\mathbf{f}}_{d}-\mathbf{G}\mathbf{A}\mathbf{X}-\mathbf{G}{\mathbf{B}}_{2}\mathbf{K}\mathbf
{X}$$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=-\rho \frac{\mathrm{\Phi }}{‖\mathrm{\Phi }‖+\phi }+\mathbf{G}{\mathbf{B}}_
{2}\left(-\mathbf{I}\right)\mathbf{\xi }+\mathbf{G}{\mathbf{f}}_{d}.$
${\mathrm{\Phi }}^{T}\stackrel{˙}{\mathrm{\Phi }}=-\rho ‖\mathrm{\Phi }‖+{\mathrm{\Phi }}^{T}\left(-\mathbf{I}\right)\mathbf{\xi }+\mathbf{G}{\mathbf{f}}_{d}\le ‖\mathrm{\Phi }‖\left(-\rho +‖\left(-\
mathbf{I}\right)\mathbf{\xi }‖\right)+‖\mathbf{G}{\mathbf{f}}_{d}‖.$
In order to enforce a sliding mode the value of the modulation gain $\rho$ should be greater than any disturbance or uncertainty in the system, and therefore for any choice of $\rho$ which satisfies:
$\rho \ge ‖-\mathbf{I}‖\bullet ‖\mathbf{\xi }‖+‖\mathbf{G}‖\bullet ‖{\mathbf{f}}_{d}‖+\mu ,$
where $\mu$ is some positive scalar, the $\mu$-reachability condition:
${\mathrm{\Phi }}^{T}\stackrel{˙}{\mathrm{\Phi }}\le -\mu ‖\mathrm{\Phi }‖,$
is satisfied.
5. Illustrative examples
In the following work, let PSC represents passive suspension control, which is represented by a black solid line of the figure; RPFTC represents robust passive fault tolerant control based on ${H}_
{2}/{H}_{\infty }$ approach, which is represented by a red solid line of the figure; ISMPFTC represents integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach,
which is represented by a blue solid line of the figure. In this section, the simulation results of passive fault tolerant control of a half-car model with actuator faults and model uncertainties are
studied. All design parameters of the half-car model in Fig. 1 are listed in Table 1. The design parameters are derived from reference [7]. Road interference inputs to use filter white noise and bump
response respectively. As to response of excitation of random road surface, in simulation, consider three cases, which are Case one, Case two, Case three; where Case one represents active suspension
in good condition；Case two represents the percentage of disturbance of active suspension system parameters is –10 %; when $t=$2.5 s, the percentage of efficiency loss of the front and rear actuator
all is 0.4; Case three represents the percentage of disturbance of active suspension system parameters is 10 %; when $t=$2.5 s, the percentage of efficiency loss of the front and rear actuator all is
0.8. Concerning bump response, in simulation, consider three cases too, which are Case one, Case two, Case three; but in each case, consider three different vehicle speeds, which are $v=$20 km/h, $v=
$ 25 km/h, $v=$30 km/h; where Case one represents active suspension in good condition; the vehicle forwards velocity is 20 km/h, 25 km/h, 30 km/h respectively; Case two represents the percentage of
disturbance of active suspension system parameters is –10 %; the vehicle for velocity is 20 km/h, 25 km/h, 30 km/h respectively; when $t=$ 1 s, the percentage of efficiency loss of the front and rear
actuator all is 0.4; Case three represents the percentage of disturbance of active suspension system parameters is 10 %; the vehicle speed is 20 km/h, 25 km/h, 30 km/h respectively; when $t=$ 1 s,
the percentage of efficiency loss of the front and rear actuator all is 0.8. The performances of vehicle suspension with passive suspension control, robust passive fault tolerant control based on $
{H}_{2}/{H}_{\infty }$ approach, integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach are compared in the time domain.
The vehicle is assumed to run linearly and the road condition for rear wheel is same as the front wheel, but with a time delay of $\left(a+b\right)/v$. Using filtered white noise as the road inputs,
the road input equations for the front and rear wheels are:
${\stackrel{˙}{x}}_{gi}=-2\pi {f}_{0}{x}_{gi}+2\pi \sqrt{{G}_{0}v}{w}_{i}\left(t\right),\left(i=1,2\right),$
where ${f}_{0}$ is low cut-off frequency, ${f}_{0}=$ 0.01 m^-1; ${G}_{0}$ is road roughness coefficient, ${G}_{0}=$5×10^-6 m^3; $v=$20 m/s; ${w}_{i}\left(t\right)$are white noise with zero mean and
power spectral density of 1.
Consider the case of an isolated bump in a road surface, the bump road profile is described as follows:
${x}_{gi}=\left\{\begin{array}{c}\frac{{A}_{0}}{2}\left(1-\left(\mathrm{cos}\frac{2\pi v}{L}t\right)\right),0\le t\le \frac{L}{v},\\ 0,t>\frac{L}{v},i=1,2,\end{array}\right\$
where ${A}_{0}$ and $L$ are the height and the length of the bump. Assume ${A}_{0}=$ 0.08 m, $L=$ 5 m, the vehicle forwards velocity is 20 km/h, 25 km/h, 30 km/h respectively.
The matrix ${\mathbf{M}}_{0}$, $\mathbf{J}$ introduced in actuator fault model is ${\mathbf{M}}_{0}=0.5\bullet eye\left(2\right)$, $\mathbf{J}=eye\left(2\right)$, respectively. Constant matrices $\
mathbf{H}$, ${\mathbf{E}}_{1}$, ${\mathbf{E}}_{2}$ of appropriate dimensions in uncertainty model is:
${\mathbf{E}}_{1}=\left[\begin{array}{c}-10.19610.6445-0.84970.0439000.8497-0.0439\\ 0.5273-14.47830.0439-0.987200-0.04390.9872\\ 4.500000.37500-500-0.37500\\ 06.111100.41670-55.55560-0.4167\end
$\mathbf{H}={\left[\begin{array}{cccccccc}0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 20& 0\\ 0& 0& 1& 0& 0& 0& 0& 16\end{array}\right]}_{4×8}^{T},{\mathbf{E}}_{2}=zeros\left
The obtained robust passive fault tolerant control law of the faulty suspension system based on ${H}_{2}/{H}_{\infty }$ approach is:
${\mathbf{K}}_{f}=\left[\begin{array}{cccccccc}-1233.16& 1567.53& -1289.19& 10.50& 4586.44& 1175.35& 203.60& -8.30\\ -897.95& -1735.59& -14.66& -1405.26& -1298.02& 5894.45& -10.73& -319.45\end{array}
As to integral sliding mode passive fault tolerant control, $\rho$ is a fixed scalar and the value is 150; the value of the small positive scalar $\phi$ is chosen as $\phi =$ 0.4, to eliminate
Table 1The vehicle system parameters for the half-car model
Description Variable Units Value
Sprung mass ${m}_{s}$ kg 690
Pitch moment of inertia ${I}_{p}$ kg∙m 1222
Front tire stiffness ${K}_{t1}$ N/m 200000
Rear tire stiffness ${K}_{t2}$ N/m 200000
Front suspension stiffness ${K}_{1}$ N/m 18000
Rear suspension stiffness ${K}_{2}$ N/m 22000
Front unsprung mass ${m}_{u1}$ kg 40
Rear unsprung mass ${m}_{u2}$ kg 45
Front Sprung damping coefficient ${C}_{s1}$ N∙s/m 1000
Front Sprung damping coefficient ${C}_{s2}$ N∙s/m 1000
Actuator output threshold ${F}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ N 1500
The maximum suspension working space ${S}_{\mathrm{m}\mathrm{a}\mathrm{x}}$ m 0.08
Distance of front axle to sprung mass c.g. $a$ m 1.3
Distance of rear axle to sprung mass c.g. $b$ m 1.5
5.1. Simulation results
5.1.1. Response of excitation of random road surface
Case one: active suspension in good condition, the simulation results are shown in Figs. 2-3.
As for response to excitation of random road surface, in the first case, as is reflected in line charts 2-3 that the optimization effect are the integral sliding mode passive fault tolerant control
based on ${H}_{2}/{H}_{\infty }$ approach is better than the robust passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach, and both of which are better than the passive suspension
Fig. 2Time-domain analysis of vertical acceleration of vehicle body
Case two: The percentage of disturbance of active suspension system parameters is –10 %; when $t=$ 2.5 s, the percentage of efficiency loss of the front and rear actuator all is 0.4, the simulation
results are shown in the line charts 4-5.
Fig. 3Time-domain analysis of pitch acceleration
Fig. 4Time-domain analysis of vertical acceleration of vehicle body
Fig. 5Time-domain analysis of pitch acceleration
In Case two, when $t<$ 2.5 s, it can be seen from the line charts 4-5 that the control effect of the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach is
the best, the robust passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach is the second best, the fault passive suspension control is the worst; when $t>$ 2.5 s, the order of the
best optimization effects keeps the same.
Case three: The percentage of disturbance of active suspension system parameters is 10 %; when $t=$ 2.5 s, the percentage of efficiency loss of the front and rear actuator all is 0.8, the simulation
results are shown in the line charts 6-7.
In Case three, when $t<$ 2.5 s, the line charts 6-7 illustrate that the optimization effect are the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach is
better than the robust passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach, and both of which are better than the passive suspension control. When $t>$ 2.5 s, the control effect
of the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach and the robust passive fault tolerant control based on ${H}_{2}/{H}_{\infty }$ approach is very
close, but both of which are still slightly better than the passive suspension control.
Fig. 6Time-domain analysis of vertical acceleration of vehicle body
Fig. 7Time-domain analysis of pitch acceleration
5.1.2. Bump response
Case one: Active suspension in good condition; the vehicle forwards velocity is 20 km/h, 25 km/h, 30 km/h respectively; the simulation results are shown in Figs. 8-10.
Case two: The percentage of disturbance of active suspension system parameters is –10 %; the vehicle forwards velocity is 20 km/h, 25 km/h, 30 km/h respectively; when $t=$ 1 s, the percentage of
efficiency loss of the front and rear actuator all is 0.4, the simulation results are shown in Figs. 11-13.
Case three: The percentage of disturbance of active suspension system parameters is 10 %; the vehicle forwards velocity is 20 km/h, 25 km/h, 30 km/h respectively; when $t=$ 1 s, the percentage of
efficiency loss of the front and rear actuator all is 0.8, the simulation results are shown in Figs. 14-16.
Fig. 8The bump responses from body vertical acceleration, pitch angular acceleration (v= 20 km/h)
Fig. 9The bump responses from body vertical acceleration, pitch angular acceleration (v= 25 km/h)
Fig. 10The bump responses from body vertical acceleration, pitch angular acceleration (v= 30 km/h)
Fig. 11The bump responses from body vertical acceleration, pitch angular acceleration (v= 20 km/h)
Fig. 12The bump responses from body vertical acceleration, pitch angular acceleration (v= 25 km/h)
Fig. 13The bump responses from body vertical acceleration, pitch angular acceleration (v= 30 km/h)
Fig. 14The bump responses from body vertical acceleration, pitch angular acceleration (v= 20 km/h)
Fig. 15The bump responses from body vertical acceleration, pitch angular acceleration (v= 25 km/h)
Fig. 16The bump responses from body vertical acceleration, pitch angular acceleration (v= 30 km/h)
Concerning bump response, in the first case, active suspension in good condition, the vehicle forwards velocity is 20 km/h, it can be seen from the line chart 8 that the optimization effect of the
integral sliding mode passive fault tolerant control (ISMPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is better than robust passive fault tolerant control (RPFTC) based on ${H}_{2}/{H}_{\
mathrm{\infty }}$ approach, and both of which are better than the passive suspension control. When the vehicle for velocity is 25 km/h, from Fig. 9, one can find that the order for optimization
effects is all same as those shown in Fig. 8, which are that the control effect of the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the
best, the robust passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the second best, the passive suspension control is the worst. As is displayed Fig. 10 that this
is also true when the vehicle speed is 30 km/h.
In Case two, the percentage of disturbance of active suspension system parameters is 10 %. When the vehicle forwards velocity is 20 km/h, $t<$ 1 s, the front and rear actuators are fault free, as is
revealed line chart 11 that the control effect of the integral sliding mode passive fault tolerant control (ISMPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is better than robust passive
fault tolerant control (RPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach, and both of which are better than the passive suspension control. When $t\ge$ 1 s, the percentage of efficiency loss
of the front and rear actuator all is 0.4, from Fig. 11, it can be concluded that the order of the best optimization effects keeps the same. When the vehicle for velocity is 25 km/h, $t<$ 1 s, the
front and rear actuators are trouble-free, as shown in the line chart 12 that the order for optimization effects is all same as those shown in Fig. 11. When $t\ge$ 1 s, the percentage of efficiency
loss of the front and rear actuator all is 0.4, from Fig. 12, it can be seen that the order of the best optimization effects keeps the same, which are that the control effect of the integral sliding
mode passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the best, the robust passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the
second best, the passive suspension control is the worst. When the vehicle speed is 30 km/h, $t<$ 1 s, the front and rear actuators in good condition, the line chart 13 illustrates that the order of
the best optimization effects is all same as those shown in Fig 12. When $t\ge$ 1 s, the percentage of efficiency loss of the front and rear actuator all is 0.4, as shown in the line chart 13 that
the order of the best optimization effects keeps the same, which are that the optimization effect of the integral sliding mode passive fault tolerant control (ISMPFTC) based on ${H}_{2}/{H}_{\mathrm
{\infty }}$ approach is better than robust passive fault tolerant control (RPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach, and both of which are better than the passive suspension control.
In Case three, the percentage of disturbance of active suspension system parameters is 10 %. When the vehicle forwards velocity is 20 km/h, $t<$ 1 s, the front and rear actuators are fault free, as
is displayed line chart 14 that the control effect of the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the best, the robust passive fault
tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is the second best, the passive suspension control is the worst. When $t\ge$ 1 s, the percentage of efficiency loss of the front
and rear actuator all is 0.8, from Fig. 14, it can be seen that although the control effect of the integral sliding mode passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$
approach and the robust passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is very close, still slightly better than the passive suspension control. When the vehicle
for velocity is 25 km/h, $t<$ 1 s, the front and rear actuators are trouble-free, as is reflected in line chart 15 that the order of optimization effects are all same as those shown in Fig. 14. When
$t\ge$ 1 s, the percentage of efficiency loss of the front and rear actuator all is 0.8, from Fig. 15, as is revealed that the control effect of the integral sliding mode passive fault tolerant
control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach and the robust passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is very close, but still slightly better
than the passive suspension control. When the vehicle speed is 30 km/h, $t<$ 1 s, the front and rear actuators in good condition, as shown in the line chart 16 that the order of the best optimization
effects keeps the same. When $t\ge$ 1 s, the percentage of efficiency loss of the front and rear actuator all is 0.8, the line chart 16 illustrates that the control effect of the integral sliding
mode passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach and the robust passive fault tolerant control based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is very close,
but both of which are still slightly better than the passive suspension control.
Table 2RMS values of the performance indices for three failure modes
RMS Case one
Parameter ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$
PSC 0.4852 0.1931 1.2473 0.6163 1.3466 0.7614 1.3668 0.8321
RPFTC 0.3005 0.1195 0.6735 0.4106 0.7464 0.5201 0.7939 0.5907
ISMPFTC 0.2008 0.0797 0.4243 0.2633 0.4762 0.3163 0.5145 0.3568
$v=$ 20 km/h $v=$ 25 km/h $v=$30 km/h
Road type Random road
Bump road
RMS Case two
Parameter ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$
PSC 0.5074 0.2063 1.3642 0.6604 1.4435 0.7833 1.4352 0.8300
RPFTC 0.3338 0.1240 0.7754 0.4605 0.8366 0.5688 0.8682 0.6234
ISMPFTC 0.2201 0.0876 0.5099 0.3241 0.5641 0.3883 0.6006 0.4252
$v=$ 20 km/h $v=$ 25 km/h $v=$30 km/h
Road type Random road
Bump road
RMS Case three
Parameter ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$ ${\stackrel{¨}{x}}_{s}$ $\stackrel{¨}{\theta }$
PSC 0.4549 0.1855 1.1114 0.5625 1.2049 0.7192 1.2016 0.7798
RPFTC 0.3607 0.1267 0.7552 0.4490 0.8423 0.5876 0.8836 0.6509
ISMPFTC 0.2923 0.1051 0.5935 0.3745 0.6610 0.4876 0.7055 0.5390
$v=$ 20 km/h $v=$ 25 km/h $v=$30 km/h
Road type Random road
Bump road
RMS values of the performance indices of the half-car model with three failure modes are listed in Table 2 for quantifying and comparing the effects of these three control methods in different
situations. It can be reached from Figs. 8-16 and Table 2 that at different vehicle speeds and failure modes, the optimization effect of the integral sliding mode passive fault tolerant control
(ISMPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach is better than robust passive fault tolerant control (RPFTC) based on ${H}_{2}/{H}_{\mathrm{\infty }}$ approach, and both of which are
better than the passive suspension control. The proposed control methods are robust to the actuator faults and model uncertainties. The above simulation analysis show: whether the active suspension
system is in good condition or malfunction in actuators and model uncertainties, both the integral sliding mode passive fault tolerant control and the robust passive fault tolerant control can
improve the ride comfort of the suspension system to a certain extent.
6. Conclusions
In this study, the partial fault of actuators, time-domain hard constraints and model uncertainties of the suspension system are considered. A robust passive fault-tolerant control (RPFTC) strategy
and an integral sliding mode passive fault tolerant control (ISMPFTC) strategy are investigated; two passive fault-tolerant control strategies based on ${H}_{2}/{H}_{\infty }$ approach have been
applied to an active suspension system respectively. Road interference inputs to use filter white noise and bump response respectively. As to response of excitation of random road surface, in
simulation, consider three cases. Concerning bump response, in simulation, consider three cases too, but in each case, consider three different vehicle speeds, which are $v=$20 km/h, $v=$25 km/h, $v=
$30 km/h. Finally, an analysis of the simulation results is given to verify the feasibility and effectiveness of the proposed two control strategies is the main contributions of this work. The
research we have done suggests that the two passive fault-tolerant control strategies are robust to the actuator faults and model uncertainties, and can guarantee the stability of the vehicle active
suspension system, improve the ride comfort of the vehicle simultaneously to a certain degree.
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About this article
Vibration in transportation engineering
active suspension
H2/H∞ control
parameter uncertainties
actuator faults
This work was supported by the Excellent Talents Project for the Second Level in the Education Department, Liaoning Province, China (No. LJQ2014065); Soft Science Research Plan Project, Science and
Technology Department, Liaoning Province, China (No. JP2016017).
Copyright © 2018 Li Ping Zhang, et al.
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.
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Kilograms to Pounds
How do you convert kilograms (kg) to pounds (lbs)?
To convert kilograms to pounds you simply have to multiply to pounds value by the conversion factor. A kilogram is approximately 2.20462 pounds so you can multiply the weight in kilograms by 2.20462
to get the weight in pounds
For example, suppose you have a weight of 5kg; multiply 5 by 2.20462 to get 11.023lbs.
Why would I want to convert kilograms (kg) to pounds (lb)?
Kilogram is a unit in the metric system whereas pounds are an imperial unit. For people that are more familiar with the imperial system such as in the United States it may be beneficial to convert
kilograms to pounds. For example, when referring to body weight many people in the US may find it easier to have the weight in pounds rather than kilograms.
For shipping or calculating weight limits for luggage, the weight requirements may be stated in pounds.
What are kilograms (kg)?
Kilogram (kg) is a unit in the metric system used to measure the weight of objects. One kilogram is subdivided into 1000 grams making it a larger unit of measurement as opposed to grams for smaller
The kilogram is often used in physics and engineering as well generally in all walks of life in countries that have adopted the metric system.
For more information, please visit our Kilograms page.
What are pounds (lb)?
Pound (lb) is an imperial weight unit primarily used in the United States but also common in the United Kingdom and Ireland. One pound is approximately 0.4536 kg.
Pounds are most commonly used in the United States whereas kilogram is the common weight unit in most other countries. When pounds are referenced, it normally refers to avoirdupois pounds. There is
another measurement called Troy pounds which are used for precious metals.
For more information, please visit our Pounds page.
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Home Ground Advantage
So I was having a look back at our season and something struck me
2024 Season
Leichardt/Campbelltown - 6 Matches - 3 Wins, 3 losses. 4 games remaining. Points for 148, Points Against 138 Points Diff = +10
Away/Neutral - 13 Matches - 1 Win, 12 Losses, 1 Game remaining. Points for 189, Points against 428 Points Diff = -239
We actually have a positive home record this year and if you include next years home ground (and last yrs) at CommBank we are 4 wins, 3 losses and 2 of those losses were to last yrs Grand Finalists.
A lot of talk about the Bulldogs but even they are only 2 and 8 away from home and most teams struggle away from home - even Panthers and Storm have a worse away record.
What we have done is given away 2 home games to Tammworth and Magic Round where you would suggest that if they were at Leichardt/Campbelltown then we would have had a much better chance of winning
espe given the DOlhpins struggles in Sydney and we only lost to the Knights by 6 in front of a majority Knights crowd in tammworth (I went)
With 4 home games remaining Cowboys who all struggle at Leichardt, Souths who have had an average season, Manly who are very up and down and the sturggling Eels their is a good chance we finish with
a positive home record.
Also suggests to me the money for Magic Round better be amazing to give away our biggest advantage again next year because while I'm not suggesting we were finals bound this year if we were on 6 wins
equal with the Titans and with 4 of 5 winnable games at home to come then our season and feelings about it would be very different.
May 10, 2011
So I was having a look back at our season and something struck me
2024 Season
Leichardt/Campbelltown - 6 Matches - 3 Wins, 3 losses. 4 games remaining. Points for 148, Points Against 138 Points Diff = +10
Away/Neutral - 13 Matches - 1 Win, 12 Losses, 1 Game remaining. Points for 189, Points against 428 Points Diff = -239
We actually have a positive home record this year and if you include next years home ground (and last yrs) at CommBank we are 4 wins, 3 losses and 2 of those losses were to last yrs Grand
A lot of talk about the Bulldogs but even they are only 2 and 8 away from home and most teams struggle away from home - even Panthers and Storm have a worse away record.
What we have done is given away 2 home games to Tammworth and Magic Round where you would suggest that if they were at Leichardt/Campbelltown then we would have had a much better chance of
winning espe given the DOlhpins struggles in Sydney and we only lost to the Knights by 6 in front of a majority Knights crowd in tammworth (I went)
With 4 home games remaining Cowboys who all struggle at Leichardt, Souths who have had an average season, Manly who are very up and down and the sturggling Eels their is a good chance we finish
with a positive home record.
Also suggests to me the money for Magic Round better be amazing to give away our biggest advantage again next year because while I'm not suggesting we were finals bound this year if we were on 6
wins equal with the Titans and with 4 of 5 winnable games at home to come then our season and feelings about it would be very different.
I think the NRL should force the home game allocation for Magic Round be evenly split over time. The richer clubs are happy to keep their home games for their real homes.
Jun 14, 2022
Also suggests to me the money for Magic Round better be amazing to give away our biggest advantage again next year because while I'm not suggesting we were finals bound this year if we were on 6
wins equal with the Titans and with 4 of 5 winnable games at home to come then our season and feelings about it would be very different.
it’s around $300k … I agree that they shouldn’t give this game up year upon year ..
it’s around $300k … I agree that they shouldn’t give this game up year upon year ..
Look at Dogs, Manly, Dragons they win at home and if you can scrape a couple of away wins then your in the finals. We need every win we can get
4 wins, 4 losses at home this year (5 wins, 4 losses at next years home grounds).
2 home games left
Jun 14, 2022
Was there only 11k there last night ?
Good crowd but Looked like it was under quoted ..similar to a couple of the LO crowds …
Hope everything is above board here …
Feb 28, 2011
I’m probably in the minority, but I don’t see the issue playing out of both Leichhardt and Ctown. How come you never hear Dragons fans making an issue out of playing at both Kogarah and the Gong?
Was there only 11k there last night ?
Good crowd but Looked like it was under quoted ..similar to a couple of the LO crowds …
Hope everything is above board here …
Earlier in the year this was mentioned (titans game) and I defended the club, but I'm starting to wonder....lower crowds at LO and CSS supports the stadium strategy and I know plenty of managers who
would have some creative way of counting things like this to "manage optics"
4 wins, 4 losses at home this year (5 wins, 4 losses at next years home grounds).
2 home games left
Its very interesting, especially when you add the F/A as you did in the first post. Club needs to focus a little more of winning as a pathway to $, focusing on $ hasn't led to winning
Jul 12, 2013
Was there only 11k there last night ?
Good crowd but Looked like it was under quoted ..similar to a couple of the LO crowds …
Hope everything is above board here …
I was there last night and was surprised they only registered 11K or so. Looked much fuller than that. What’s the stadium capacity? I’d say it was 75% full at the absolute minimum. That’s being
Jul 25, 2022
Was there only 11k there last night ?
Good crowd but Looked like it was under quoted ..similar to a couple of the LO crowds …
Hope everything is above board here …
I had the wests tigers promoting only 1k tickets left 2 hrs before kick off.. I tell ya what on tv it looked like alot more than 1k seats available! I hope they stop doing this sh1t don't treat us
fans as idiots.
Now have a positive record at our 2 real home grounds, while coming last. Imagine the years we were coming 9th if we had focused on winning football games first and foremost...
Jul 10, 2009
When does the '25 draw come out... Nov some time?
When does the '25 draw come out... Nov some time?
"The 2024 NRL draw was released on November 13 2023, with a similar time frame likely to be maintained for the 2025 draw."
Jul 10, 2009
⏫ Fountain of knowledge... Thnx T1 ⏫
May 24, 2019
Just in passing, and remembering that Richo's new/revised seating packages for members includes CommBank as part of all the allocations, here is a pic today of the CommBank surface as it welcomes
the 1st game of the A Leagues season (home of Western Sydney Wanderers):
View attachment 16579
Remind me again, CommBank, is that parras home ground?
Remind me again, CommBank, is that parras home ground?
I know, I know .........
But when I was talking to our club's membership dept., the gent was adamant that Richo had this factor all in hand under his watch.....
And having gone there, the very 1st day we played there as a home game, against the Eels of all teams, when we were thrashed by Parra, and Moses scored a beauty against us, I sure know how the ground
"feels" as a WT supporter .......
Jul 22, 2014
Just in passing, and remembering that Richo's new/revised seating packages for members includes CommBank as part of all the allocations, here is a pic today of the CommBank surface as it welcomes
the 1st game of the A Leagues season (home of Western Sydney Wanderers):
View attachment 16579
Parramatta Stadium home of the Eels
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Solving a math problem with planner programming
July 2, 2024
Solving a math problem with planner programming
More opportunities to mess with exotic technology
The deadline for the logic book is coming up! I'm hoping to have it ready for early access by either the end of this week or early next week. During a break on Monday I saw this interesting problem
on Math Stack Exchange:
Suppose that at the beginning there is a blank document, and a letter "a" is written in it. In the following steps, only the three functions of "select all", "copy" and "paste" can be used.
Find the minimum number of steps to reach at least 100,000 a's (each of the three operations of "select all", "copy" and "paste" is counted as one step). If the target number is not specified,
and I want to get the exact amount of a, is there a general formula?
The first two answers look for analytic solutions. The last answer shares a C++ program that finds it via breadth-first search. I'll reproduce it here:
#include <iostream>
#include <queue>
enum Mode
struct Node
int noOfAs;
int steps;
int noOfAsCopied;
Mode mode;
int main()
std::queue<Node> q;
q.push({1, 0, 0, SELECT});
while (!q.empty())
Node n = q.front();
if (n.noOfAs >= 100000)
std::cout << n.steps << std::endl;
switch (n.mode)
case SELECT:
q.push({n.noOfAs, n.steps + 1, n.noOfAsCopied, COPY});
case COPY:
q.push({n.noOfAs, n.steps + 1, n.noOfAs, PASTE});
case PASTE:
q.push({n.noOfAs, n.steps, n.noOfAsCopied, SELECT});
q.push({n.noOfAs + n.noOfAsCopied, n.steps + 1, n.noOfAsCopied, PASTE});
return 0;
This is guaranteed to find a shortest possible solution due to a fun property of BFS: the distance of nodes to the origin never decreases. If you evaluate Node Y after Node X, then Y.dist >= X.dist,
meaning that the first valid solution will be a shortest possible solution. I should make this into a logic book example!
This also has the drawback of preventing the use of an insight. We should be able to fuse the select and copy steps together, meaning instead of having three actions (select, copy, paste) we only
need two (selectcopy, paste), where selectcopy takes twice as many steps as pasting.
But we can't make that optimization because it breaks monotonicity. We're now pushing a mix of n+1 and n+2 steps onto the queue, and there's no way to order things to guarantee all of the n+1 steps
are searched before any n+2 step.
I thought I'd try to solve it with planning language instead, so we can get both the elegant solution and the optimization.
The rough idea of planning is that you provide an initial state, a set of actions, and a target, and the tool finds the shortest sequence of actions that reaches the target. I've written about it
in-depth here and also a comparison of planning to model checking here. I like how some tough problems in imperative and functional paradigms become easy problems with planning.
This is all in Picat, by the way, which I've talked about more here and in the planning piece. I'll just be explaining the planning stuff specific to this problem.
import planner.
import util.
main =>
Init = $state(1, 0) % one a, nothing copied
, best_plan(Init, Plan, Cost)
, nl
, printf("Cost=%d%n", Cost)
, printf("Plan=%s%n", join([P[1]: P in Plan], " "))
We're storing the state of the system as two integers: the number of characters printed and the number of characters on our clipboard. Since we'll be fusing selects and copies, we don't need to also
track the number of characters selected [DEL:(unlike the C++):DEL].
final(state(A, _)) => A >= 100000.
action(state(A, Clipboard), To, Action, Cost) ?=>
NewA = A + Clipboard
, To = $state(NewA, Clipboard)
, Action = {"P", To}
, Cost = 1
The paste action just adds the clipboard to the character count. Because Picat is a research language it's a little weird with putting expressions inside structures. If we did $state(1 + 1) it would
store it as literally $state(1 + 1), not state(2).
Also you have to use dollar signs for definitions but no dollar signs for pattern matching inside a function definition. I have no idea why.
action(state(A, Clipboard), To, Action, Cost) ?=>
To = $state(A, A)
, Action = {"SC", To}
, Cost = 2
And that's it! That's the whole program. Running this gives us:
Plan=SC P P SC P P SC P P SC P P SC P P SC
P P SC P P SC P P SC P P P SC P P P
To find if there's a sequence that gets us exactly 100,000, we just need to make one change:
- final(state(A, _)) => A >= 100000.
+ final(state(A, _)) => A = 100000.
This returns a cost of 43.
On the other hand, I can't get it to find a path that makes exactly 100,001 characters, even with some optimizations. This is because the shortest path is over 9000 steps long! I haven't checked if
the C++ BFS can find it.
One reason planning fascinates me so much is that, if a problem is now easy, you can play around with it. Like if I wanted to add "delete a character" as a move, that's easy:
action(state(A, Clipboard), To, Action, Cost) ?=>
A > 0
, NewA = A - 1
, To = $state(NewA, Clipboard)
, Action = {"D", To}
, Cost = 1
This doesn't make exceeding or reaching 100,000 easier, but it makes reaching 100,001 take 47 steps instead of 9000.
With some tweaks, I can also ask questions like "what numbers does it make the most easier?" or "Do some numbers have more than one shortest path? Which number has the most?"
Planning is really cool.
If you're reading this on the web, you can subscribe here. Updates are once a week. My main website is here.
My new book, Logic for Programmers, is now in early access! Get it here.
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Università di Pisa - Valutazione della didattica e iscrizione agli esami
An Introduction to Applied and Environmental Geophysics, Reynolds, 2011
Applied Geophysics, 2nd ed, Telford, Geldart, Sheriff, 1990
Environmental and Engineering geophysics, Sharma, 1997
An introduction to geophysical exploration, Keary, Brooks and Hill, 2002
Applied Geophysics, Zanzi, 2008
Near-Surface Applied Geophysics, Everett, 2013
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Multiple graphs per page with many graphs
Hi all. I am trying to print 4 graphs per page to a PDF file. I have many graphs (200 for now) as I am drawing using PROC SGPLOT and BY statement. I searched online for the solution but all of them
involve manually indicating the position of each graphs which is not ideal in my case. Is there a solution where I can tell SAS to print 4 graphs into 1 page instead of 1 big graph per page?
My code is:
proc sgplot data=prices;
by stock;
series x=date y=price;
series x=date y= volume/y2axis;
10-23-2019 10:00 PM
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Correlation of Fixed Effects in lme4
If you have ever used the R package lme4 to perform mixed-effect modeling you may have noticed the “Correlation of Fixed Effects” section at the bottom of the summary output. This article intends to
shed some light on what this section means and how you might interpret it.
To begin, let’s simulate some data. Below we generate 30 observations for 20 subjects. The gl() function generates levels. In this case it generates 30 replications of 20 levels. The result is the
numbers 1 - 20, each repeated 30 times. We name it id and can think of these as 20 subjects numbered 1 - 20. We then generate the sequence 1 - 30, 20 times, using the rep() function. This will serve
as our predictor, or independent variable.
n <- 20 # subjects
k <- 30 # observations
id <- gl(n, k)
x <- rep(1:k, n)
Next we generate some “random noise” for each observation by drawing 20 * 30 = 600 samples from a Normal distribution with mean 0 and standard deviation 12. The value 12 is arbitrary. Feel free to
try a different value. The function set.seed(1) allows you to sample the same data if you wish to follow along.
obs_noise <- rnorm(n * k, mean = 0, sd = 12)
We also need to generate noise for each subject to create data suitable for mixed-effect modeling. Again we sample from a Normal distribution with mean 0, but this time we arbitrarily set the
standard deviation to 20. We also only sample 20 values instead of 600. That’s because the noise is specific to the subject instead of the observation. Run the set.seed(11) code if you want to sample
the same data.
subj_noise <- rnorm(n, mean = 0, sd = 20)
Finally we generate a dependent variable, y, as a linear function of x using the formula y = 3 + 2*x. The fixed intercept is 3. The fixed slope is 2. We also include the noise. The code subj_noise
[id] repeats the subject-specific noise 30 times for each subject and is added to the intercept. The observation-specific noise is added at the end. When done we place our variables in a data frame
we name d.
y <- (3 + subj_noise[id]) + 2*x + obs_noise
d <- data.frame(id, y, x)
Let’s visualize our data using ggplot2. Notice the fitted line for each subject has about the same slope but shifts up or down for each subject. That’s how we simulated our data. Each subject has the
same slope but a different intercept that changes according the subject-specific noise we sampled.
ggplot(d) +
aes(x = x, y = y) +
geom_point() +
geom_smooth(method = "lm") +
Now we use the lmer() function from the lme4 package to “work backwards” and try to recover the true values we used to generate the data. Notice we are fitting the “correct” model. We know the
process we used to generate the data, so we know which model to specify.
me1 <- lmer(y ~ x + (1|id), data = d)
Linear mixed model fit by REML ['lmerMod']
Formula: y ~ x + (1 | id)
Data: d
REML criterion at convergence: 4771
Scaled residuals:
Min 1Q Median 3Q Max
-2.7171 -0.6417 -0.0160 0.6715 3.7493
Random effects:
Groups Name Variance Std.Dev.
id (Intercept) 267.6 16.36
Residual 146.1 12.09
Number of obs: 600, groups: id, 20
Fixed effects:
Estimate Std. Error t value
(Intercept) -3.843 3.795 -1.013
x 1.999 0.057 35.073
Correlation of Fixed Effects:
x -0.233
A quick review of the summary output shows the lmer() did quite well at recovering the data-generating values. Under the “Fixed Effects” section the slope of x is estimated as 1.99 which is very
close to the true value of 2. The estimate of the intercept is not terribly good, but we added a fair amount of noise to it for each subject, so it’s not surprising lmer() struggled with it. Under
the “Random Effects” section the standard deviations of the Normal distributions from which we sampled the noise are estimated as about 12.1 and 16.4, pretty close to the true values of 12 and 20.
And that brings us to the final section: Correlation of Fixed Effects. It shows that the intercept and x coefficients have a correlation of -0.233. What is that exactly? We certainly didn’t use
correlated data to generate fixed effects. In fact we simply picked two numbers: 3 and 2. How can our fixed effects be correlated?
To help answer this, we’ll first point out the estimated standard errors for the intercept and x coefficients. The intercept standard error is 3.795. The x standard error is 0.057. These quantify
uncertainty in our estimate. It may be helpful to think of them as “give or take” values. For example, our x coefficient is 1.99, give or take about 0.057. These standard errors are the square roots
of the estimated variances. We can see the variances using the vcov function.
2 x 2 Matrix of class "dpoMatrix"
(Intercept) x
(Intercept) 14.4018199 -0.050363397
x -0.0503634 0.003249251
We can extract the diagonal values of the matrix using the diag() function and take the square roots to get the standard errors that are presented in the summary output. We use the base R pipe to
send the output of one function to the input of the next.
vcov(me1) |> diag() |> sqrt()
[1] 3.79497298 0.05700221
If we look again at the output of the vcov() function we see a covariance estimate in the off-diagonal entry of the matrix: -0.0503634
2 x 2 Matrix of class "dpoMatrix"
(Intercept) x
(Intercept) 14.4018199 -0.050363397
x -0.0503634 0.003249251
So not only do we get an estimate of uncertainty for each coefficient, we get an estimate of how the coefficients vary together. Hence the name of the function, “vcov”, which is short for
If we divide the covariance by the product of the standard errors, we get the correlation of fixed effects reported in the lmer() summary output.
m <- vcov(me1)
den <- m |> diag() |> sqrt() |> prod()
num <- m[2,1]
In fact this is the formula for the familiar correlation coefficient.
\[\rho = \frac{\text{Cov}(X_1, X_2)}{\sigma_1\sigma_2}\]
Base R includes the convenient cov2cor() function to automatically make these calculations for us.
2 x 2 Matrix of class "dpoMatrix"
(Intercept) x
(Intercept) 1.000000 -0.232817
x -0.232817 1.000000
Now that we see where the correlation of fixed effects is coming from, what does it mean?
Imagine we replicate our analysis above many times, each time using a new random sample. We will get new estimates of fixed effects every time. The correlation of fixed effects gives us some sense of
how those many fixed effect estimates might be associated. Let’s demonstrate this with a simulation.
To begin we use our model to simulate 1000 new sets of responses using the simulate() function. Recall the point of mixed-effect modeling is to estimate the data-generating process. Once we have our
model we can then use it to generate data. (A good fitting model should generate data similar to the observed data.) The result is a data frame named “sim” with 1000 columns. Each column is a new
vector of 600 responses.
sim <- simulate(me1, nsim = 1000)
After that we apply our lme4 model to each new set of responses. We set up our model as a function named “f” with the argument k. Then we apply that function to each column of the sim object using
the lapply() function. The resulting object named “out” contains 1000 model fits with 1000 different sets of fixed effects. Depending on your computer, this may take anywhere from 10 to 30 seconds.
f <- function(k)lmer(k ~ x + (1|id), data = d)
out <- lapply(sim, f)
Once that finishes running we do a bit of data wrangling to extract the fixed effects into a matrix, calculate the correlation of the fixed effects, and create a scatter plot.
# extract fixed effects
f_out <- lapply(out, fixef)
# collapse list into matrix
f_df <- do.call(rbind, f_out)
cor(f_df[,1], f_df[,2])
The correlation of -0.234 is close to the Correlation of Fixed Effects reported in the original summary output, -0.233. We can also plot the simulated fixed effects. It appears higher coefficients of
x are slightly associated with lower coefficients of the intercept.
Is this important to know? That’s up to you to decide. It’s another piece of information to help you understand your model. It just so happens this output is provided by default when using summary()
on a model fitted with lmer(). If you prefer, you can suppress it by setting corr = FALSE. On the other hand, when calling summary() on a model fitted with lm() the correlation of fixed effects is
not printed, but you can request it by setting corr = TRUE. The correlation of fixed effects is not unique to mixed-effect models.
We should note that examining the correlation of fixed effects is NOT the same as examining your model for collinearity. That involves the correlation of your predictors, not the model’s fixed
• Bates D, Maechler M, Bolker B, Walker S (2015). Fitting Linear Mixed-Effects Models Using lme4. Journal of Statistical Software, 67(1), 1-48. https://www.jstatsoft.org/article/view/v067i01.
• R Core Team (2023). R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing, Vienna, Austria. https://www.R-project.org/.
Clay Ford
Statistical Research Consultant
University of Virginia Library
February 28, 2022
Categories related to this article:
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5 And 6 Multiplication Worksheets
Mathematics, specifically multiplication, creates the cornerstone of many scholastic techniques and real-world applications. Yet, for lots of students, grasping multiplication can posture a
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5 And 6 Multiplication Worksheets -
Multiplication Math Worksheets Math explained in easy language plus puzzles games quizzes videos and worksheets For K 12 kids teachers and parents Multiplication Worksheets Worksheets Multiplication
Mixed Tables Worksheets Individual Table Worksheets Worksheet Online 2 Times 3 Times 4 Times 5 Times 6 Times 7 Times 8 Times 9 Times
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6 Times Table
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Transactions Online
Walaa ALY, Seiichi UCHIDA, Masakazu SUZUKI, "Automatic Classification of Spatial Relationships among Mathematical Symbols Using Geometric Features" in IEICE TRANSACTIONS on Information, vol. E92-D,
no. 11, pp. 2235-2243, November 2009, doi: 10.1587/transinf.E92.D.2235.
Abstract: Machine recognition of mathematical expressions on printed documents is not trivial even when all the individual characters and symbols in an expression can be recognized correctly. In this
paper, an automatic classification method of spatial relationships between the adjacent symbols in a pair is presented. This classification is important to realize an accurate structure analysis
module of math OCR. Experimental results on very large databases showed that this classification worked well with an accuracy of 99.525% by using distribution maps which are defined by two geometric
features, relative size and relative position, with careful treatment on document-dependent characteristics.
URL: https://global.ieice.org/en_transactions/information/10.1587/transinf.E92.D.2235/_p
author={Walaa ALY, Seiichi UCHIDA, Masakazu SUZUKI, },
journal={IEICE TRANSACTIONS on Information},
title={Automatic Classification of Spatial Relationships among Mathematical Symbols Using Geometric Features},
abstract={Machine recognition of mathematical expressions on printed documents is not trivial even when all the individual characters and symbols in an expression can be recognized correctly. In this
paper, an automatic classification method of spatial relationships between the adjacent symbols in a pair is presented. This classification is important to realize an accurate structure analysis
module of math OCR. Experimental results on very large databases showed that this classification worked well with an accuracy of 99.525% by using distribution maps which are defined by two geometric
features, relative size and relative position, with careful treatment on document-dependent characteristics.},
TY - JOUR
TI - Automatic Classification of Spatial Relationships among Mathematical Symbols Using Geometric Features
T2 - IEICE TRANSACTIONS on Information
SP - 2235
EP - 2243
AU - Walaa ALY
AU - Seiichi UCHIDA
AU - Masakazu SUZUKI
PY - 2009
DO - 10.1587/transinf.E92.D.2235
JO - IEICE TRANSACTIONS on Information
SN - 1745-1361
VL - E92-D
IS - 11
JA - IEICE TRANSACTIONS on Information
Y1 - November 2009
AB - Machine recognition of mathematical expressions on printed documents is not trivial even when all the individual characters and symbols in an expression can be recognized correctly. In this
paper, an automatic classification method of spatial relationships between the adjacent symbols in a pair is presented. This classification is important to realize an accurate structure analysis
module of math OCR. Experimental results on very large databases showed that this classification worked well with an accuracy of 99.525% by using distribution maps which are defined by two geometric
features, relative size and relative position, with careful treatment on document-dependent characteristics.
ER -
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Convert decimal integer to its binary representation
binStr = dec2bin(D) returns the binary, or base-2, representation of the decimal integer D. The output argument binStr is a character vector that represents binary digits using the characters 0 and
If D is a numeric vector, matrix, or multidimensional array, then binStr is a two-dimensional character array. Each row of binStr represents an element of D.
binStr = dec2bin(D,minDigits) returns a binary representation with no fewer than minDigits digits.
Convert Decimal Number
Convert a decimal number to a character vector that represents its binary value.
D = 23;
binStr = dec2bin(D)
Specify Minimum Number of Digits
Specify the minimum number of binary digits that dec2bin returns. If you specify more digits are required, then dec2bin pads the output.
D = 23;
binStr = dec2bin(D,8)
If you specify fewer digits, then dec2bin still returns as many binary digits as required to represent the input number.
Convert Numeric Array
Create a numeric array.
To represent the elements of D as binary values, use the dec2bin function. Each row of binStr corresponds to an element of D.
binStr = 3x10 char array
Since all rows of a character array must have the same number of characters, dec2bin pads some rows of binStr. For example, the number 14 can be represented by the binary digits '1110'. But to match
the length of the first row of binStr, the dec2bin function pads the third row to '0000001110'.
Represent Negative Numbers
Starting in R2020a, the dec2bin function converts negative numbers using their two's complement binary values.
For example, these calls to dec2bin convert negative numbers.
Input Arguments
D — Input array
numeric array | char array | logical array
Input array, specified as a numeric array, char array, or logical array.
• If D is an array of floating-point numbers, and any element of D has a fractional part, then dec2bin truncates it before conversion. For example, dec2bin converts both 12 and 12.5 to '1100'. The
truncation is always to the nearest integer less than or equal to that element.
• If D is a character or logical array, then dec2bin treats the elements of D as integers. However, dec2bin treats characters as their Unicode^® values, so specifying D as a character array is not
Since R2020a
D can include negative numbers. The function converts negative numbers using their two's complement binary values.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | char
minDigits — Minimum number of digits in output
nonnegative integer
Minimum number of digits in the output, specified as a nonnegative integer.
• If D can be represented with fewer than minDigits binary digits, then dec2bin pads the output.
D >= 0 Pads with leading zeros
D < 0 Pads with leading ones (since R2020b)
• If D is so large that it must be represented with more than minDigits digits, then dec2bin returns the output with as many digits as required.
• The output of dec2bin is the same whether your computer stores values in memory using big-endian or little-endian format. For more information on these formats, see Endianness.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
• If minDigits is specified, the output will have that number of columns even if D is empty. If minDigits is not specified, the output will have at least one column.
• If input D is double or single, then it must be greater than or equal to intmin('int64') and less than 2^64.
• This function usually produces a variable-size output. To make the output fixed-size, supply minDigits as a constant large enough that the output has a fixed number of columns regardless of input
values. For fixed-size output, minDigits must be at least 64 for double, 64 for single, 32 for half, 1 for logical, 8 for char, 64 for int64, 64 for uint64, 32 for int32, 32 for uint32, 16 for
int16, 16 for uint16, 8 for int8, and 8 for uint8.
Thread-Based Environment
Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool.
This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.
Version History
Introduced before R2006a
R2022a: Restrict input datatype to primitive numeric types
User-defined datatypes are restricted to primitive numeric types and classes that inherit from a primitive numeric type.
R2022a: dec2bin(0,0) returns '0'
dec2bin(0,0) returns '0' rather than a 1x0 character vector.
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Yash Deshmukh
Starting Fall 2024, I will be a postdoctoral fellow at the Institute for Advanced Study. During 2023-24 I was a visiting fellow at the Max Planck Institute for Mathematics, Bonn. Before that I
recieved my PhD from Columbia University in May 2023 under the supervision of Mohammed Abouzaid. My research focuses on Symplectic topology and Floer theory. Specifically, I am interested in the
study of algebraic structures underlying various Floer theoretic invariants. Currently, I'm thinking about relations between algebraic invariants defined using Floer theory and Symplectic Field
In 2023-24 I co-organized the Bonn symplectic seminar. In Fall 2022 I co-organized the Microlocal Sheaf Theory at Columbia.
Papers and Preprints
• A homotopical description of Deligne-Mumford compactifications[arxiv:2211.05168]
Recent Talks
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Nonlinear M
Nonlinear MPC Controller
Simulate nonlinear model predictive controllers
Model Predictive Control Toolbox
The Nonlinear MPC Controller block simulates a nonlinear model predictive controller. At each control interval, the block computes optimal control moves by solving a nonlinear programming problem.
For more information on nonlinear MPC, see Nonlinear MPC.
To use this block, you must first create an nlmpc object in the MATLAB^® workspace.
• None of the Nonlinear MPC Controller block parameters are tunable.
Required Inputs
x — input
Current prediction model states, specified as a vector signal of length N[x], where N[x] is the number of prediction model states. Since the nonlinear MPC controller does not perform state
estimation, you must either measure or estimate the current prediction model states at each control interval.
ref — Model output reference values
row vector | matrix
Plant output reference values, specified as a row vector signal or matrix signal.
To use the same reference values across the prediction horizon, connect ref to a row vector signal with N[y] elements, where N[y] is the number of output variables. Each element specifies the
reference for an output variable.
To vary the references over the prediction horizon (previewing) from time k+1 to time k+p, connect ref to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the references for one prediction horizon step. If you specify fewer than p rows, the final references are used for the remaining steps of the prediction
last_mv — Control signals used in plant at previous control interval
Control signals used in plant at previous control interval, specified as a vector signal of length N[mv], where N[mv] is the number of manipulated variables.
Connect last_mv to the MV signals actually applied to the plant in the previous control interval. Typically, these MV signals are the values generated by the controller, though this is not always the
case. For example, if your controller is offline and running in tracking mode; that is, the controller output is not driving the plant, then feeding the actual control signal to last_mv can help
achieve bumpless transfer when the controller is switched back online.
Additional Inputs
md — input
row vector | matrix
If your controller prediction model has measured disturbances you must enable this port and connect to it a row vector or matrix signal.
To use the same measured disturbance values across the prediction horizon, connect md to a row vector signal with N[md] elements, where N[md] is the number of manipulated variables. Each element
specifies the value for a measured disturbance.
To vary the disturbances over the prediction horizon (previewing) from time k to time k+p, connect md to a matrix signal with N[md] columns and up to p+1 rows. Here, k is the current time and p is
the prediction horizon. Each row contains the disturbances for one prediction horizon step. If you specify fewer than p+1 rows, the final disturbances are used for the remaining steps of the
prediction horizon.
To enable this port, select the Measured disturbances parameter.
params — Optional parameters
If your controller uses optional parameters in its prediction model, custom cost function, or custom constraint functions, enable this input port, and connect a parameter bus signal with N[p]
elements, where N[p] is the number of parameters. For more information on creating a parameter bus signal, see createParameterBus. The controller passes these parameters to its model functions, cost
function, constraint functions, passivity functions and Jacobian functions.
If your controller does not use optional parameters, you must disable params.
To enable this port, select the Model parameters parameter.
mv.target — Manipulated variable targets
To specify manipulated variable targets, enable this input port, and connect a vector signal. To make a given manipulated variable track its specified target value, you must also specify a nonzero
tuning weight for that manipulated variable.
The supplied mv.target values at run-time apply across the prediction horizon.
To enable this port, select the Targets for manipulated variables parameter.
Online Constraints
y.min — Minimum output variable constraints
vector | matrix
To specify run-time minimum output variable constraints, enable this input port. If this port is disabled, the block uses the lower bounds specified in the OutputVariables.Min property of its
controller object.
To use the same bounds over the prediction horizon, connect y.min to a row vector signal with N[y] elements, where N[y] is the number of outputs. Each element specifies the lower bound for an output
To vary the bounds over the prediction horizon from time k+1 to time k+p, connect y.min to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Lower OV limits parameter.
y.max — Maximum output variable constraints
vector | matrix
To specify run-time maximum output variable constraints, enable this input port. If this port is disabled, the block uses the upper bounds specified in the OutputVariables.Min property of its
controller object.
To use the same bounds over the prediction horizon, connect y.max to a row vector signal with N[y] elements, where N[y] is the number of outputs. Each element specifies the upper bound for an output
To vary the bounds over the prediction horizon from time k+1 to time k+p, connect y.max to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Upper OV limits parameter.
mv.min — Minimum manipulated variable constraints
vector | matrix
To specify run-time minimum manipulated variable constraints, enable this input port. If this port is disabled, the block uses the lower bounds specified in the ManipulatedVariables.Min property of
its controller object.
To use the same bounds over the prediction horizon, connect mv.min to a row vector signal with N[mv] elements, where N[mv] is the number of outputs. Each element specifies the lower bound for a
manipulated variable.
To vary the bounds over the prediction horizon from time k to time k+p-1, connect mv.min to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Lower MV limits parameter.
mv.max — Maximum manipulated variable constraints
vector | matrix
To specify run-time maximum manipulated variable constraints, enable this input port. If this port is disabled, the block uses the upper bounds specified in the ManipulatedVariables.Max property of
its controller object.
To use the same bounds over the prediction horizon, connect mv.max to a row vector signal with N[mv] elements, where N[mv] is the number of outputs. Each element specifies the upper bound for a
manipulated variable.
To vary the bounds over the prediction horizon from time k to time k+p-1, connect mv.max to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Upper MV limits parameter.
dmv.min — Minimum manipulated variable rate constraints
vector | matrix
To specify run-time minimum manipulated variable rate constraints, enable this input port. If this port is disabled, the block uses the lower bounds specified in the ManipulatedVariable.RateMin
property of its controller object. dmv.min bounds must be nonpositive.
To use the same bounds over the prediction horizon, connect dmv.min to a row vector signal with N[mv] elements, where N[mv] is the number of outputs. Each element specifies the lower bound for a
manipulated variable rate of change.
To vary the bounds over the prediction horizon from time k to time k+p-1, connect dmv.min to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Lower MVRate limits parameter.
dmv.max — Maximum manipulated variable rate constraints
vector | matrix
To specify run-time maximum manipulated variable rate constraints, enable this input port. If this port is disabled, the block uses the upper bounds specified in the ManipulatedVariables.RateMax
property of its controller object. dmv.max bounds must be nonnegative.
To use the same bounds over the prediction horizon, connect dmv.max to a row vector signal with N[mv] elements, where N[mv] is the number of outputs. Each element specifies the upper bound for a
manipulated variable rate of change.
To vary the bounds over the prediction horizon from time k to time k+p-1, connect dmv.max to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Upper MVRate limits parameter.
x.min — Minimum state constraints
vector | matrix
To specify run-time minimum state constraints, enable this input port. If this port is disabled, the block uses the lower bounds specified in the States.Min property of its controller object.
To use the same bounds over the prediction horizon, connect x.min to a row vector signal with N[x] elements, where N[x] is the number of outputs. Each element specifies the lower bound for a state.
To vary the bounds over the prediction horizon from time k+1 to time k+p, connect x.min to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Lower state limits parameter.
x.max — Maximum state constraints
vector | matrix
To specify run-time maximum state constraints, enable this input port. If this port is disabled, the block uses the upper bounds specified in the States.Max property of its controller object.
To use the same bounds over the prediction horizon, connect x.max to a row vector signal with N[x] elements, where N[x] is the number of outputs. Each element specifies the upper bound for a state.
To vary the bounds over the prediction horizon from time k+1 to time k+p, connect x.max to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the prediction
horizon. Each row contains the bounds for one prediction horizon step. If you specify fewer than p rows, the bounds in the final row apply for the remainder of the prediction horizon.
To enable this port, select the Upper state limits parameter.
Online Tuning Weights
y.wt — Output variable tuning weights
row vector | matrix
To specify run-time output variable tuning weights, enable this input port. If this port is disabled, the block uses the tuning weights specified in the Weights.OutputVariables property of its
controller object. These tuning weights penalize deviations from output references.
If the MPC controller object uses constant output tuning weights over the prediction horizon, you can specify only constant output tuning weights at runtime. Similarly, if the MPC controller object
uses output tuning weights that vary over the prediction horizon, you can specify only time-varying output tuning weights at runtime.
To use constant tuning weights over the prediction horizon, connect y.wt to a row vector signal with N[y] elements, where N[y] is the number of outputs. Each element specifies a nonnegative tuning
weight for an output variable. For more information on specifying tuning weights, see Tune Weights.
To vary the tuning weights over the prediction horizon from time k+1 to time k+p, connect y.wt to a matrix signal with N[y] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the tuning weights for one prediction horizon step. If you specify fewer than p rows, the tuning weights in the final row apply for the remainder of the
prediction horizon. For more information on varying weights over the prediction horizon, see Setting Time-Varying Weights and Constraints with MPC Designer.
To enable this port, select the OV weights parameter.
mv.wt — Manipulated variable tuning weights
row vector | matrix
To specify run-time manipulated variable tuning weights, enable this input port. If this port is disabled, the block uses the tuning weights specified in the Weights.ManipulatedVariables property of
its controller object. These tuning weights penalize deviations from MV targets.
To use the same tuning weights over the prediction horizon, connect mv.wt to a row vector signal with N[mv] elements, where N[mv] is the number of manipulated variables. Each element specifies a
nonnegative tuning weight for a manipulated variable. For more information on specifying tuning weights, see Tune Weights.
To vary the tuning weights over the prediction horizon from time k to time k+p-1, connect mv.wt to a matrix signal with N[mv] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the tuning weights for one prediction horizon step. If you specify fewer than p rows, the tuning weights in the final row apply for the remainder of the
prediction horizon. For more information on varying weights over the prediction horizon, see Setting Time-Varying Weights and Constraints with MPC Designer.
To enable this port, select the MV weights parameter.
dmv.wt — Manipulated variable rate tuning weights
row vector | matrix
To specify run-time manipulated variable rate tuning weights, enable this input port. If this port is disabled, the block uses the tuning weights specified in the Weights.ManipulatedVariablesRate
property of its controller object. These tuning weights penalize large changes in control moves.
To use the same tuning weights over the prediction horizon, connect dmv.wt to a row vector signal with N[mv] elements, where N[mv] is the number of manipulated variables. Each element specifies a
nonnegative tuning weight for a manipulated variable rate. For more information on specifying tuning weights, see Tune Weights.
To vary the tuning weights over the prediction horizon from time k to time k+p-1, connect dmv.wt to a matrix signal with N[mv] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the tuning weights for one prediction horizon step. If you specify fewer than p rows, the tuning weights in the final row apply for the remainder of the
prediction horizon. For more information on varying weights over the prediction horizon, see Setting Time-Varying Weights and Constraints with MPC Designer.
To enable this port, select the MVRate weights parameter.
ecr.wt — Slack variable tuning weight
To specify a run-time slack variable tuning weight, enable this input port and connect a scalar signal. If this port is disabled, the block uses the tuning weight specified in the Weights.ECR
property of its controller object.
The slack variable tuning weight has no effect unless your controller object defines soft constraints whose associated ECR values are nonzero. If there are soft constraints, increasing the ecr.wt
value makes these constraints relatively harder. The controller then places a higher priority on minimizing the magnitude of the predicted worst-case constraint violation.
To enable this port, select the ECR weight parameter.
Initial Guesses
mv.init — Initial guesses for the optimal manipulated variable solutions
vector | matrix
To specify initial guesses for the optimal manipulated variable solutions, enable this input port. If this port is disabled, the block uses the optimal control sequences calculated in the previous
control interval as initial guesses.
To use the same initial guesses over the prediction horizon, connect mv.init to a vector signal with N[mv] elements, where N[mv] is the number of manipulated variables. Each element specifies the
initial guess for a manipulated variable.
To vary the initial guesses over the prediction horizon from time k to time k+p-1, connect mv.init to a matrix signal with N[mv] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the initial guesses for one prediction horizon step. If you specify fewer than p rows, the guesses in the final row apply for the remainder of the prediction
To enable this port, select the Initial guess parameter.
x.init — Initial guesses for the optimal state variable solutions
vector | matrix
To specify initial guesses for the optimal state solutions, enable this input port. If this port is disabled, the block uses the optimal state sequences calculated in the previous control interval as
initial guesses.
To use the same initial guesses over the prediction horizon, connect x.init to a vector signal with N[x] elements, where N[x] is the number of states. Each element specifies the initial guess for a
To vary the initial guesses over the prediction horizon from time k to time k+p-1, connect x.init to a matrix signal with N[x] columns and up to p rows. Here, k is the current time and p is the
prediction horizon. Each row contains the initial guesses for one prediction horizon step. If you specify fewer than p rows, the guesses in the final row apply for the remainder of the prediction
To enable this port, select the Initial guess parameter.
e.init — Initial guess for the slack variable at the solution
nonnegative scalar
To specify an initial guess for the slack variable at the solution, enable this input port and connect a nonnegative scalar signal. If this port is disabled, the block uses an initial guess of 0.
To enable this port, select the Initial guess parameter.
Required Output
mv — Optimal manipulated variable control action
column vector
Optimal manipulated variable control action, output as a column vector signal of length N[mv], where N[mv] is the number of manipulated variables.
If the solver converges to a local optimum solution (nlp.status is positive), then mv contains the optimal solution.
If the solver reaches the maximum number of iterations without finding an optimal solution (nlp.status is zero) and the Optimization.UseSuboptimalSolution property of the controller is:
• true, then mv contains the suboptimal solution
• false, then mv is the same as last_mv
If the solver fails (nlp.status is negative), then mv is the same as last_mv.
Additional Outputs
cost — Objective function cost
nonnegative scalar
Objective function cost, output as a nonnegative scalar signal. The cost quantifies the degree to which the controller has achieved its objectives.
The cost value is only meaningful when the nlp.status output is nonnegative.
To enable this port, select the Optimal cost parameter.
slack — Slack variable
0 | nonnegative scalar
Slack variable, ε, used in constraint softening, output as 0 or a positive scalar value.
• ε = 0 — All soft constraints are satisfied over the entire prediction horizon.
• ε > 0 — At least one soft constraint is violated. When more than one constraint is violated, ε represents the worst-case soft constraint violation (scaled by the ECR values for each constraint).
To enable this port, select the Slack variable parameter.
nlp.status — Optimization status
Optimization status, output as one of the following:
• Positive Integer — Solver converged to an optimal solution
• 0 — Maximum number of iterations reached without converging to an optimal solution
• Negative integer — Solver failed
To enable this port, select the Optimization status parameter.
Optimal Sequences
mv.seq — Optimal manipulated variable sequence
Optimal manipulated variable sequence, returned as a matrix signal with p+1 rows and N[mv] columns, where p is the prediction horizon and N[mv] is the number of manipulated variables.
The first p rows of mv.seq contain the calculated optimal manipulated variable values from current time k to time k+p-1. The first row of mv.seq contains the current manipulated variable values
(output mv). Since the controller does not calculate optimal control moves at time k+p, the final two rows of mv.seq are identical.
To enable this port, select the Optimal control sequence parameter.
x.seq — Optimal prediction model state sequence
Optimal prediction model state sequence, returned as a matrix signal with p+1 rows and N[x] columns, where p is the prediction horizon and N[x] is the number of states.
The first row of x.seq contains the current estimated state values, either from the built-in state estimator or from the custom state estimation block input x[k|k]. The next p rows of x.seq contain
the calculated optimal state values from time k+1 to time k+p.
To enable this port, select the Optimal state sequence parameter.
y.seq — Optimal output variable sequence
Optimal output variable sequence, returned as a matrix signal with p+1 rows and N[y] columns, where p is the prediction horizon and N[y] is the number of output variables.
The first p rows of y.seq contain the calculated optimal output values from current time k to time k+p-1. The first row of y.seq is computed based on the current estimated states and the current
measured disturbances (first row of input md). Since the controller does not calculate optimal output values at time k+p, the final two rows of y.seq are identical.
To enable this port, select the Optimal output sequence parameter.
Nonlinear MPC Controller — Controller object
nlmpc object name
You must provide an nlmpc object that defines a nonlinear MPC controller. To do so, enter the name of an nlmpc object in the MATLAB workspace.
Programmatic Use
Block Parameter: nlmpcobj
Type: string, character vector
Default: ""
Use prediction model sample time — Flag for using the prediction model sample time
on (default) | off
Select this parameter to run the controller using the same sample time as its prediction model. To use a different controller sample time, clear this parameter, and specify the sample time using the
Make block run at a different sample time parameter.
To limit the number of decision variables and improve computational efficiency, you can run the controller with a sample time that is different from the prediction horizon. For example, consider the
case of a nonlinear MPC controller running at 10 Hz. If the plant and controller sample times match, predicting plant behavior for ten seconds requires a prediction horizon of length 100, which
produces a large number of decision variables. To reduce the number of decision variables, you can use a plant sample time of 1 second and a prediction horizon of length 10.
Programmatic Use
Block Parameter: UseObjectTs
Type: string, character vector
Values: "off", "on"
Default: "on"
Make block run at a different sample time — Controller sample time
positive finite scalar | -1
Specify this parameter to run the controller using a different sample time from its prediction model. Setting this parameter to -1 allows the block to inherit the sample time from its parent
The first element of the MV rate vector (which is the difference between the current and the last value of the manipulated variable) is normally weighted and constrained assuming that the last value
of the MV occurred in the past at the sample time specified in the MPC object. When the block is executed with a different sample rate, this assumption no longer holds, therefore, in this case, you
must make sure that the weights and constraints defined in the controller handle the first element of the MV rate vector correctly.
To enable this parameter, clear the Use prediction model sample time parameter.
Programmatic Use
Block Parameter: TsControl
Type: string, character vector
Default: ""
Use MEX to speed up simulation — Flag for simulating controller use MEX function
off (default) | on
Select this parameter to simulate the controller using a MEX function generated using buildMEX. Doing so reduces the simulation time of the controller. To specify the name of the MEX function, use
the Specify MEX function name parameter.
Programmatic Use
Block Parameter: UseMEX
Type: string, character vector
Values: "off", "on"
Default: "off"
Specify MEX function name — Controller MEX function name
Use this parameter to specify the name of the MEX function to use during simulation. To create the MEX function, use the buildMEX function.
To enable this parameter, select the Use MEX to speed up simulation parameter.
Programmatic Use
Block Parameter: mexname
Type: string, character vector
Default: ""
General Tab
Measured disturbances — Add measured disturbance input port
off (default) | on
If your controller has measured disturbances, you must select this parameter to add the md output port to the block.
Programmatic Use
Block Parameter: md_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Targets for manipulated variables — Add manipulated variable target input port
off (default) | on
Select this parameter to add the mv.target input port to the block.
Programmatic Use
Block Parameter: mvtarget_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Model parameters — Add model parameters input port
off (default) | on
If your controller uses optional parameters, you must select this parameter to add the params output port to the block.
For more information on creating a parameter bus signal, see createParameterBus.
Programmatic Use
Block Parameter: param_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Optimal cost — Add optimal cost output port
off (default) | on
Select this parameter to add the cost output port to the block.
Programmatic Use
Block Parameter: cost_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Optimal control sequence — Add optimal control sequence output port
off (default) | on
Select this parameter to add the mv.seq output port to the block.
Programmatic Use
Block Parameter: mvseq_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Optimal state sequence — Add optimal state sequence output port
off (default) | on
Select this parameter to add the x.seq output port to the block.
Programmatic Use
Block Parameter: stateseq_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Optimal output sequence — Add optimal output sequence output port
off (default) | on
Select this parameter to add the y.seq output port to the block.
Programmatic Use
Block Parameter: ovseq_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Slack variable — Add slack variable output port
off (default) | on
Select this parameter to add the slack output port to the block.
Programmatic Use
Block Parameter: slack_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Optimization status — Add optimization status output port
off (default) | on
Select this parameter to add the nlp.status output port to the block.
Programmatic Use
Block Parameter: status_enabled
Type: string, character vector
Values: "off", "on"
Default: "off"
Online Features Tab
Lower OV limits — Add minimum OV constraint input port
off (default) | on
Select this parameter to add the ov.min input port to the block.
Programmatic Use
Block Parameter: ov_min
Type: string, character vector
Values: "off", "on"
Default: "off"
Upper OV limits — Add maximum OV constraint input port
off (default) | on
Select this parameter to add the ov.max input port to the block.
Programmatic Use
Block Parameter: ov_max
Type: string, character vector
Values: "off", "on"
Default: "off"
Lower MV limits — Add minimum MV constraint input port
off (default) | on
Select this parameter to add the mv.min input port to the block.
Programmatic Use
Block Parameter: mv_min
Type: string, character vector
Values: "off", "on"
Default: "off"
Upper MV limits — Add maximum MV constraint input port
off (default) | on
Select this parameter to add the mv.max input port to the block.
Programmatic Use
Block Parameter: mv_max
Type: string, character vector
Values: "off", "on"
Default: "off"
Lower MVRate limits — Add minimum MV rate constraint input port
off (default) | on
Select this parameter to add the dmv.min input port to the block.
Programmatic Use
Block Parameter: mvrate_min
Type: string, character vector
Values: "off", "on"
Default: "off"
Upper MVRate limits — Add maximum MV rate constraint input port
off (default) | on
Select this parameter to add the dmv.max input port to the block.
Programmatic Use
Block Parameter: mvrate_max
Type: string, character vector
Values: "off", "on"
Default: "off"
Lower state limits — Add minimum state constraint input port
off (default) | on
Select this parameter to add the x.min input port to the block.
Programmatic Use
Block Parameter: state_min
Type: string, character vector
Values: "off", "on"
Default: "off"
Upper state limits — Add maximum state constraint input port
off (default) | on
Select this parameter to add the x.max input port to the block.
Programmatic Use
Block Parameter: state_max
Type: string, character vector
Values: "off", "on"
Default: "off"
OV weights — Add OV tuning weights input port
off (default) | on
Select this parameter to add the y.wt input port to the block.
Programmatic Use
Block Parameter: ov_weight
Type: string, character vector
Values: "off", "on"
Default: "off"
MV weights — Add MV tuning weights input port
off (default) | on
Select this parameter to add the mv.wt input port to the block.
Programmatic Use
Block Parameter: mv_weight
Type: string, character vector
Values: "off", "on"
Default: "off"
MVRate weights — Add MV rate tuning weights input port
off (default) | on
Select this parameter to add the dmv.wt input port to the block.
Programmatic Use
Block Parameter: mvrate_weight
Type: string, character vector
Values: "off", "on"
Default: "off"
ECR weight — Add ECR tuning weight input port
off (default) | on
Select this parameter to add the ecr.wt input port to the block.
Programmatic Use
Block Parameter: ecr_weight
Type: string, character vector
Values: "off", "on"
Default: "off"
Initial guess — Add initial guess input ports
off (default) | on
Select this parameter to add the mv.init, x.init, and e.init input ports to the block.
By default, the Nonlinar MPC Controller block uses the calculated optimal manipulated variable and state trajectories from one control interval as the initial guesses for the next control interval.
Enable the initial guess ports only if it is necessary for your application.
Programmatic Use
Block Parameter: nlp_initialize
Type: string, character vector
Values: "off", "on"
Default: "off"
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using Simulink® Coder™.
Usage notes and limitations:
• The Nonlinear MPC Controller block supports generating code only for nonlinear MPC controllers that use the default fmincon solver with the SQP algorithm.
• Code generation for single-precision or fixed-point computations is not supported.
• When used for code generation, nonlinear MPC controllers do not support expressing prediction model functions, stage cost functions or constraint functions as anonymous functions.
• If your controller uses optional parameters, you must also generate code for the Bus Creator block connected to the params input port. To do so, place the Nonlinear MPC Controller and Bus Creator
blocks within a subsystem, and generate code for that subsystem.
• The Support non-finite numbers check box in the Interface section of the Code Generation options, under the model Configuration Parameters dialog box, must be checked (default option).
• When generating code using Embedded Coder^®, the Support variable-size signals in the Interface section of the Code Generation options, under the model Configuration Parameters dialog box, must
be checked. By default this check box is unchecked and you must check it before generating code.
Version History
Introduced in R2018b
See Also
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Prediction of Quantitative Traits Using Common Genetic Variants: Application to Body Mass Index
Article information
Genomics Inform. 2016;14(4):149-159
Received 2016 November 21; Revised 2016 December 06; Accepted 2016 December 06.
With the success of the genome-wide association studies (GWASs), many candidate loci for complex human diseases have been reported in the GWAS catalog. Recently, many disease prediction models based
on penalized regression or statistical learning methods were proposed using candidate causal variants from significant single-nucleotide polymorphisms of GWASs. However, there have been only a few
systematic studies comparing existing methods. In this study, we first constructed risk prediction models, such as stepwise linear regression (SLR), least absolute shrinkage and selection operator
(LASSO), and Elastic-Net (EN), using a GWAS chip and GWAS catalog. We then compared the prediction accuracy by calculating the mean square error (MSE) value on data from the Korea Association
Resource (KARE) with body mass index. Our results show that SLR provides a smaller MSE value than the other methods, while the numbers of selected variables in each model were similar.
With the development of genotyping technologies, many disease-related genetic variants have been verified by genome-wide association studies (GWASs). Diagnosis and disease risk prediction from the
utilization of the genetic variants have improved even further [1]. Direct-to-consumer genetic companies, such as 23andME (http://www.23andme.com/) and Pathway Genomics (https://www.pathway.com/),
provide personal genome information services. For example, the BRCA1 and BRCA2 genes play important roles in breast cancer diagnosis and clinical treatment [23]. While several disease prediction
studies have been conducted using disease-related genetic variants, there are some limitations to disease risk prediction. It becomes difficult to construct a disease risk prediction model, because
there are typically a larger number of genetic variants than the number of individuals in the “large p small n” problem. Also, the effect size of genetic variants for most complex human diseases is
small, and missing heritability exists [4]. Moreover, some loss of statistical power to identify significant associations is caused by the correlating single-nucleotide polymorphisms (SNPs) due to
linkage disequilibrium (LD) [5]. Multicollinearity due to high LD among SNPs causes high variance of coefficient estimates. In order to solve these issues, various statistical approaches have been
recently proposed.
Initially, a gene score (GS) was computed using statistical models for disease risk prediction [678]. These risk prediction models were created from GSs by summing up the marginal effect of each
disease-associated genetic variant. Several studies have shown that GS is useful for risk prediction [9]. However, the accuracy of the risk prediction is poor when joint effects exist between
multiple genetic variants [1011].
Building a risk prediction model using multiple SNPs is an effective way to improve disease risk prediction. Multiple logistic regression (MLR) is one of the typical traditional approaches. Several
studies have shown the usefulness of an MLR-based approach for creating disease risk prediction models [121314]. However, the parameter estimation of MLR becomes unstable, and the predictive power of
the risk prediction model decreases if there is high LD among SNPs.
In order to solve the “large p and small n” problem, many penalized regression approaches, like ridge [151617], least absolute shrinkage and selection operator (LASSO) [18], and Elastic-Net (EN) [19
], have been proposed. For highdimensional data, these penalized approaches have several advantages in variable selection, as well as in prediction, over non-penalized approaches. For example,
several researchers showed that the utilization of a large amount of SNPs with penalized regression approaches improves the accuracy of Crohn's disease and bipolar disorder risk prediction [2021].
It is important to build a risk prediction model that pertains to discrete variables, such as disease diagnosis. It is also important to make predictions based on continuous variables, such as human
health-related outcomes. When using medicines to treat diseases, we can use genetic information to calculate the dosage, in addition to basic physical information, such as height and weight. For
example, there is a prediction model for warfarin responsiveness that was made with multivariate linear regression [22]. We can apply such a model directly to disease treatment.
In this study, we focus on the prediction of quantitative traits using common genetic variants. We systematically compared the performance of prediction models through real data from the Korea
Association Resource (KARE). We first selected the prediction variables using statistical methods, such as stepwise linear regression (SLR), LASSO, and EN. We then constructed commonly used risk
prediction models, such as SLR, LASSO, and EN. Finally, we compared the predictive accuracy by calculating the mean square error (MSE) value for predicting body mass index (BMI). Overall, our results
show that LASSO and SLR provide the smallest MSE value among the compared methods.
The KARE project, which began in 2007, is an Anseong and Ansan regional society-based cohort. After applying SNP quality control criteria—Hardy-Weinberg equilibrium p < 10^−06, genotype call rates <
95%, and minor allele frequency < 0.01—352,228 SNPs were utilized for analysis. Also, after eliminating 401 samples with call rates less than 96%, 11 contaminated samples, 41 gender-inconsistent
samples, 101 serious concomitant illness samples, 608 cryptic-related samples, and 4 samples with missing phenotype, 8,838 participants were analyzed [23]. Table 1 summarizes the demographic
information. In addition, Fig. 1 shows box plots of BMI for the given demographic variables.
Statistical analysis
We selected SNPs from the KARE data analysis based on single-SNP analysis and collected SNPs in the GWAS catalog [24]. Then, we performed two steps to make quantitative prediction models. First, we
selected the variables by using SLR, LASSO, and EN and then built quantitative prediction models by using the same methods.
SNP sets
First, based on three different populations—overall population, Asian-only population, and Korean-only population —we collected the SNPs registered in the GWAS catalog for BMI. Second, the SNPs were
selected by single-SNP analysis using linear regression with adjustments for sex, age, and area. We chose the SNPs based on the p-values. We considered the following seven SNP sets:
(1) ASIAN-100 (GWAS catalog [Asia] + Single-SNP analysis, number of SNPs = 100)
(2) KOREAN-100 (GWAS catalog [Korea] + single-SNP analysis, number of SNPs = 100)
(3) ALL-200 (GWAS catalog [All] + single-SNP analysis, number of SNPs = 200)
(4) ASIAN-200 (GWAS catalog [Asia] + single-SNP analysis, number of SNPs = 200)
(5) KOREAN-200 (GWAS catalog [Korea] + single-SNP analysis, number of SNPs = 200)
(6) GWAS-ALL (GWAS catalog [All], number of SNPs = 136)
(7) GWAS-ASIAN (GWAS catalog [Asia], number of SNPs = 16)
Step 1: Variable selection
In the KARE data, out of 8,838 individuals, we randomly selected 1,767 for test sets and composed the training set with the rest of the 7,071 participants. We selected SNPs using 5-fold
cross-validation (CV) of the training set. In this case, we used SLR, LASSO, and EN to select SNPs.
The SLR model is one of the most widely used models. Let y[i] be a quantitative phenotype for subject i = 1, …, n; x[ij] be the value of SNP j = 1, …, p for subject i; code be 0, 1, and 2 for the
number of minor alleles; and ε[i] be the error term for subject i. The SLR model is
y[i] = β[0] + β[1]x[i1] + ... + β[p]x[ip] + γ[1]sex[i] + γ[2]age[i] + γ[3]area[i] + ε[i],
where β
and β
are the intercept and effect sizes of SNPs, respectively. γ
, γ
, and γ
represent the sex, age, and area of the
-th individual, respectively. Variable selection was performed by a MSE-based stepwise procedure. The stepwise procedure was performed using the R package “MASS” [
The LASSO and EN estimates of β were obtained by minimizing
respectively. The tuning parameters λ
and λ
are estimated using CV. The penalized methods were performed using the R package “glmnet” [
Then, we defined five groups.
(1) Group 1 (consists of SNPs that appeared at least one time in the 5-fold CV)
(2) Group 2 (consists of the SNPs that appeared at least two times in the 5-fold CV)
(3) Group 3 (consists of the SNPs that appeared at least three times in the 5-fold CV)
(4) Group 4 (consists of the SNPs that appeared at least four times in the 5-fold CV)
(5) Group 5 (consists of the SNPs that appeared in all 5-fold CVs)
Step 2: Quantitative prediction
To build a quantitative prediction model, we used the same prediction methods that were applied for the variable selection step for the comparison of these three methods in the variable selection and
quantitative prediction. Each prediction model was created by using 7,071 training individuals via 5-fold CV. To compare the performance of the quantitative prediction models, we calculated the MSE
by applying each quantitative prediction model using the test set (n = 1,767).
To create the SNP sets associated with BMI, single-SNP analysis was performed by linear regression with adjustments for sex, age, and area. As shown in Supplementary Fig. 1, we found one significant
SNP (rs17178527) after Bonferroni correction (1.45 × 10^−07). rs17178527 of LOC729076 has been reported as BMI-associated SNP in previous GWASs [2327]. In addition, Supplementary Table 1 shows the
results of the single-SNP analysis with p-values less than 5.00 × 10^−05. The SNPs that were reported to be associated with BMI in the GWAS catalog are summarized in Supplementary Table 2. Seven SNP
sets are summarized in Table 2.
Step 1: Variable selection
Variable selection in each SNP set was performed via 5-fold CV of the training set. Fig. 2 shows the overlapping number of selected SNPs by the variable selection methods. In addition, Table 3
provides more detailed information. Overall, SLR selected fewer SNPs than LASSO and EN. All SNPs were selected when EN was used in ASIAN-100, ASIAN-200, and KOREAN-200.
Step 2: Quantitative prediction
We made quantitative prediction models based on SLR, LASSO, and EN using the entire training dataset. Then, the MSE was calculated by applying the quantitative prediction models to the test dataset.
Table 4 and Fig. 3 show the performance of each quantitative prediction model in the test dataset. The model using only covariates yielded an MSE value of 10.24. As can be seen from Fig. 3, the
prediction model created from Group 5 yielded the smallest MSE. Fig. 4 describes the comparison results between the numbers of SNPs and MSEs from the prediction models using SLR.
Among all sets, the case that used LASSO to select variables and SLR to create the model showed the smallest MSE value of 9.64 in ASIAN-100, with 51 SNPs. Among the 51 SNPs of LASSO-SLR with one set
from ASIAN-100, 28 SNPs were mapped to genes (Table 5). Some genes, such as FTO, GP2, AKAP6, ANKS1B, ADCY3, and ADCY8, have been reported to be associated with BMI [282930313233].
In this study, we used statistical methods (SLR, LASSO, and EN) to select variables and build quantitative prediction models. Then, we compared the performance of the quantitative prediction models
by each SNP set (ASIAN-100, KOREAN-100, ALL-200, ASIAN-200, KOREAN-200, GWAS-ALL, and GWAS-ASIAN). As a result, the performance of the prediction models using the GWAS catalog and KARE data was
better than that of the prediction models using only SNPs reported in the GWAS catalog. For the case that selected variants using LASSO in ASIAN-100 and created a prediction model using SLR, the MSE
value was the smallest, 9.64. At this time, the number of SNPs was 51. Also, for the model with the fewest SNPs, we selected variables using SLR from ALL-200 and created a model using SLR. The number
of SNPs was 38, and the MSE value was 9.84. Through the 5-fold CV, we developed a quantitative prediction model. After calculating MSE from groups 1 to 5, when assembled with SNPs that were included
in all CVs, the resulting values of MSE were small. However, when a different group was used, the MSE value was bigger than when using the covariates to build the model. Therefore, with CV, when
using SNPs that match each of their CVs, the efficiency of their quantitative prediction model was high. In the variable selection, SLR performed better than other methods. SLR selected fewer SNPs
than the other methods in all SNP sets while providing smaller MSEs. It seems that LASSO and EN tended to select SNPs with little contribution to BMI. For further research, we plan to perform
simulation studies and a real-data analysis with other continuous traits.
There are many ways to extend the analysis of quantitative prediction studies. First, along with the application of recently developed methods, such as bootstrapping methods [3435], we will continue
to explore new ways to develop more prediction models. Second, the incorporation of rare variants can improve the performance of a quantitative prediction model. Advanced sequencing technology has
made it possible to investigate the role of common and rare variants in complex disease risk prediction. Additionally, we can use biological information while choosing the variables. By using
single-SNP analysis, we can use gene or pathway information to find useful SNPs [36], and from here, we can assemble an SNP set by adding an SNP list from the pathways related to the disease of
This research was supported by a grant of the Korea Health Technology R&D Project through the Korea Health Industry Development Institute (KHIDI), funded by the Ministry of Health & Welfare, Republic
of Korea (HI15C2165), and the Bio-Synergy Research Project (2013M3A9C4078158) of the Ministry of Science, ICT and Future Planning through the National Research Foundation. The GWAS chip data were
supported by bioresources from the National Biobank of Korea, the Centers for Disease Control and Prevention, Republic of Korea (4845-301, 4851-302 and -307).
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Article information Continued
Copyright © 2016 by the Korea Genome Organization
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Geographic Information Systems and Cartography
In contrast to the raster data model is the vector data model. In this model, space is not quantized into discrete grid cells like in the raster model. Vector data models use points and their
associated X, Y coordinate pairs to represent the vertices of spatial features, much as if they were being drawn on a map by hand (Aronoff, 1989). [1] The data attributes of these features are then
stored in a separate database management system. The spatial information and the attribute information for these models are linked via a simple identification number given to each feature in a map.
Three fundamental vector types exist in geographic information systems (GIS): points, lines, and polygons. Points are zero-dimensional objects that contain only a single coordinate pair. Points are
typically used to model singular, discrete features such as buildings, wells, power poles, sample locations, etc. Points have only the property of location. Other types of point features include the
node and the vertex. Specifically, a point is a stand-alone feature, while a node is a topological junction representing a common X, Y coordinate pair between intersecting lines and polygons.
Vertices are defined as each bend along with a line or polygon feature that is not the intersection of lines or polygons.
Points can be spatially linked to form more complex features. Lines are one-dimensional features composed of multiple, explicitly connected points. Lines represent linear features such as roads,
streams, faults, and boundaries. In addition, lines have the property of length. Lines that directly connect two nodes are sometimes referred to as chains, edges, segments, or arcs.
Polygons are two-dimensional features created by multiple lines that loop back to create a “closed” feature. In the case of polygons, the first coordinate pair (point) on the first line segment is
the same as the last coordinate pair on the last line segment. Polygons represent features such as city boundaries, geologic formations, lakes, soil associations, vegetation communities, and more. In
addition, polygons have the properties of area and perimeter. Therefore, polygons are also called areas.
Vector Data Models Structures
Vector data models can be structured in many different ways. We will examine two of the more common data structures here. The simplest vector data structure is called the spaghetti data model
(Dangermond, 1982). [2] In the spaghetti model, each point, line, and polygon feature is represented as a string of X, Y coordinate pairs (or as a single X, Y coordinate pair in the case of a vector
image with a single point) no inherent structure. One could envision each line in this model as a single strand of spaghetti formed into complex shapes by adding more strands of spaghetti. In this
model, any polygons that lie adjacent to each other must be made up of their lines or strands of spaghetti. In other words, each polygon must be uniquely defined by its own set of X, Y coordinate
pairs, even if the adjacent polygons share the same boundary information. This creates some redundancies within the data model and therefore reduces efficiency.
Despite the location designations associated with each line, or strand of spaghetti, spatial relationships are not explicitly encoded within the spaghetti model; instead, they are implied by their
location. This results in a lack of topological information, which is problematic if the user attempts to make measurements or analyses. Therefore, the computational requirements are steep if any
advanced analytical techniques are employed on vector files. Nevertheless, the simple structure of the spaghetti data model allows for efficient reproduction of maps and graphics as this topological
information is unnecessary for plotting and printing.
In contrast to the spaghetti data model, the topological data model is characterized by including topological information within the dataset, as the name implies. Topology is a set of rules that
model the relationships between neighboring points, lines, and polygons and determines how they share geometry. For example, consider two adjacent polygons. In the spaghetti model, the shared
boundary of two neighboring polygons is defined as two separate, identical lines. The inclusion of topology into the data model allows for a single line to represent this shared boundary with an
explicit reference to denote which side of the line belongs to which polygon. Topology is also concerned with preserving spatial properties when the forms are bent, stretched, or placed under similar
geometric transformations, which allows for more efficient projection and reprojection of map files.
Three basic topological precepts are necessary to understand the topological data model are outlined here. First, connectivity describes the arc-node topology for the feature dataset. As discussed
previously, nodes are more than simple points. In the topological data model, nodes are the intersection points where two or more arcs meet. In the case of arc-node topology, arcs have both a
from-node (i.e., starting node) indicating where the arc begins and a to-node (i.e., ending node) indicating where the arc ends. In addition, between each node pair is a line segment, sometimes
called a link, which has its identification number and references both its from-node and to-node. For example, in this figure, “Arc-Node Topology,” arcs 1, 2, and 3 intersect because they share node
11. Therefore, the computer can determine that it is possible to move along arc 1 and turn onto arc 3, while it is impossible to move from arc 1 to arc 5, as they do not share a common node.
The second fundamental topological precept is area definition. Area definition states that an arc that connects to surround an area defines a polygon, also called polygon-arc topology. In the case of
polygon-arc topology, arcs are used to construct polygons, and each arc is stored only once. This reduces the amount of data stored and ensures that adjacent polygon boundaries do not overlap. For
example, in the figure on “Polygon-Arc Topology,” the polygon-arc topology clarifies that polygon F comprises arcs 8, 9, and 10.
Contiguity, the third topological precept, is based on the concept that polygons that share a boundary are deemed adjacent. Specifically, polygon topology requires that all arcs in a polygon have a
direction (a from-node and a to-node), which allows adjacency information to be determined. Polygons that share an arc are deemed adjacent or contiguous, and therefore the “left” and “right” sides of
each arc can be defined. This left and right polygon information is stored explicitly within the attribute information of the topological data model. The “universe polygon” is an essential component
of polygon topology that represents the external area located outside of the study area. The figure “Polygon Topology” shows that arc 6 is bound by polygon B and to the right by polygon C. Polygon A,
the universe polygon, is to the left of arcs 1, 2, and 3.
Topology allows the computer to rapidly determine and analyze the spatial relationships of all its included features. In addition, topological information is essential because it allows for efficient
error detection within a vector dataset. In the case of polygon features, open or unclosed polygons, which occur when an arc does not completely loop back upon itself, and unlabeled polygons, which
occur when an area does not contain any attribute information, violate polygon-arc topology rules. Another topological error found with polygon features is the sliver. Slivers occur when the shared
boundary of two polygons does not meet precisely.
In the case of line features, topological errors occur when two lines do not meet perfectly at a node. This error is called an “undershoot” when the lines do not extend far enough to meet each other
and an “overshoot” when the line extends beyond the feature it should connect to. The result of overshoots and undershoots is a “dangling node” at the end of the line. Dangling nodes are not always
an error, however, as they occur in the case of dead-end streets on a road map.
Many types of spatial analysis require the degree of organization offered by topologically explicit data models. For example, network analysis (e.g., finding the best route from one location to
another) and measurement (e.g., finding the length of a river segment) rely heavily on the concept of to-and-from nodes and use this information, along with attribute information, to calculate
distances, shortest routes, or the quickest route. Topology also allows for sophisticated neighborhood analysis such as determining adjacency, clustering, or nearest neighbors.
Now that the basics of the concepts of topology have been outlined, we can begin to understand the topological data model better. In this model, the node acts as more than just a simple point along a
line or polygon. Instead, the node represents the point of intersection for two or more arcs. Arcs may or may not be looped into polygons. Regardless, all nodes, arcs, and polygons are individually
numbered. This numbering allows for quick and easy reference within the data model.
Advantages and Disadvantages of the Vector Model
In comparison with the raster data model, vector data models tend to be better representations of reality due to the accuracy and precision of points, lines, and polygons over the regularly spaced
grid cells of the raster model. This results in vector data tending to be more aesthetically pleasing than raster data.
Vector data also provides an increased ability to alter the scale of observation and analysis. However, as each coordinate pair associated with a point, line, and polygon represents an
infinitesimally exact location (albeit limited by the number of significant digits and data acquisition methodologies), zooming deep into a vector image does not change the view of a vector graphic
in the way that it does a raster graphic.
Vector data tend to be more compact in the data structure, so file sizes are typically much smaller than their raster counterparts. Although the ability of modern computers has minimized the
importance of maintaining small file sizes, vector data often require a fraction of the computer storage space compared to raster data.
The final advantage of vector data is that topology is inherent in the vector model. Using a vector model, this topological information results in simplified spatial analysis (e.g., error detection,
network analysis, proximity analysis, and spatial transformation).
Alternatively, there are two primary disadvantages of the vector data model. First, the data structure tends to be more complex than the simple raster data model. As the location of each vertex must
be stored explicitly in the model, there are no shortcuts for storing data like there are for raster models (e.g., the run-length and quad-tree encoding methodologies).
Second, the implementation of spatial analysis can also be complicated due to minor differences in accuracy and precision between the input datasets. Similarly, the algorithms for manipulating and
analyzing vector data are complex and can lead to intensive processing requirements, mainly when dealing with large datasets.
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What is the cancellation policy for KLM?
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National Pi Day 2024 - When, Where and Why it is Celebrated?
Last Updated on November 28, 2023
Pi is a mathematical constant. The pi was first calculated by Archimedes of Syracuse, who lived from 287-212 BC. The constant pi has got a unique aspect. National pi day honors this special number in
various countries worldwide. Pi Day is a celebration to honor numbers and mathematics. Pi has got a role in various concepts in mathematics. Pi is a homophone of a delicious dish called the pie. So
the pie is also celebrated along with pi on national pi day in March for all the pi enthusiasts.
History of National Pi Day
National pi day is celebrated annually on day March 14. The day was selected as the value of pie is 3.14… approximately. Opt day to celebrate the constant value is the 14th day of the 3rd month. Pi
Day was first celebrated in the year 1988 on March 14. Larry Shaw organized the celebration at the San Francisco Exploratorium. The Exploratorium holds the pi day to date. The Exploratorium
celebrates the Pi holiday with great enthusiasm on March 14.
National pi day, March 14, was started officially in the year 2009. The U.S. House of Representatives passed a non-binding resolution recognizing March 14th as the national π day. Celebrate pi day in
March with your mathematician’s buddies and spread the word on social media. Celebrating the pi Exploratorium holiday is a big day for all pie lovers.
Related: World Maths Day
Few Facts About pi
1. Pi is an irrational number and is calculated according to the ratio of a circle’s circumference. The exact value of pi isn’t known. Any mathematicians couldn’t find a recognizable pattern about
the infinite digits of pi that kept emerging starting from 3.1415626……..
2. The digits of pi can be calculated over a trillion times after the decimal point. The first three number is only considered.
3. Greek letter π is the ratio of the circle’s circumference to its diameter. The pi value remains the same for all circles, even if they differ in size. It is constant for the circle diameter.
4. Pi’s value is vital in measuring natural phenomena such as ocean tides, ebb, and flow. The value of pi also finds its role in measuring electromagnetic waves. The word pie is derived from the
Greek letter π, and it is the mathematical constant.
5. The value of pi is also used to calculate the shape of the rivers, the disc of the sun, and the size of the pupil of our eye. The digits are highly reliable and are used to calculate the
circumference of a circle.
What should we do on National Pi Day?
The national pi day celebration should be a serious celebration day on March 14 by mathematicians. The holiday celebrates the value of pi, especially the first three numbers. The value of pi has
trillion digits after the decimal. Checking our memory skills by memorizing as many digits as possible would be wonderful.
We can also celebrate the day by relishing a delicious pie with our family and friends, as pi sounds like pie. There is also a national pie day to enjoy a lot of pie-eating on March 12
Why Should We Celebrate National Pi Day?
Pi is an amazing invention in the field of mathematics that has to be celebrated. Pi value is a great invention that has broken the mysteries of various things in this physical world. The pi has to
be celebrated to pay respect to Archimedes, who invented the pi.
The pi value has led to many other concepts in the mathematical and physical world. It is hard to imagine how we would have calculated certain things on this earth if pi was not invented.
[wpdatatable id=28]
Other Days Celebrated in March
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§ 10: The fundamental theorem
FS: Let A, B, C be three distinct points on a line l, and A', B', C' three distinct points on a line m (possibly l = m). Then there exists exactly one projectivity from l onto m, say φ, that maps A
to A', B to B', and C to C'.
Proof: We already proved φ exists in O34.
Now suppose φ[1] and φ[2] meet our requirements, and let X be an arbitrary point of l.
Then we have (A', B'; C', φ[1](X)) = (φ[1](A), φ[1](B); φ[1](C), φ[1](X)) = (A, B; C, X) = (φ[2](A), φ[2](B); φ[2](C), φ[2](X)) = (A', B'; C', φ[2](X)).
According to O38 we find φ[1](X) = φ[2](X).
Since X was arbitrary it follows that φ[1] = φ[2].
Proposition: A bijection ψ : l → m is a projectivity if and only if ψ preserves cross ratio.
Proof: We saw in the last section but this one that any projectivity preserves cross ratio.
Inversely, let ψ be a bijection from l onto m that preserves cross ratio.
Choose three points A, B, C on l, and let φ be the projectivity from l onto m with φ(A) = ψ(A), φ(B) = ψ(B), φ(C) = ψ(C).
Then we find for all X on l: (φ(A), φ(B); φ(C), φ(X)) = (A, B; C, X) = (ψ(A), ψ(B); ψ(C), ψ(X)) = (φ(A), φ(B); φ(C), ψ(X)).
According to O38 we have φ(X) = ψ(X). So φ = ψ. So ψ(X) is a projectivity.
Proposition: Each projectivity φ : l → m is induced by a regular linear transformation of ℜ^3.
Proof: P^2 is the plane {x[3]=1} in ℜ^3, extended with the points at infinity on the line at infinity.
Take three points A, B, C on l, and let A', B', C' be the image points on m under the projectivity φ.
Denote the vector from O to B by b, etc.
Then c = λa + μb en c' = ρa' + σb' for certain real numbers λ, μ, ρ, σ ≠ 0.
Let F be a regular linear transformation of ℜ^3 with F(a) = (ρ/λ)a', F(b) = (σ/μ)b'.
Then F(c) = F(λa + μb) = λF(a) + μF(b) = ρa' + σb' = c'.
In O41 we prove that F preserves cross ratio; or we'd better say that the bijection f from l onto m, induced by F, does so. Hence, according to the previous proposition, f is a projectivity from l
onto m.
Since f(A) = φ(A), f(B) = φ(B), f(C) = φ(C), we get f = φ because of the fundamental theorem.
O39 Formulate and prove the dual of the fundamental theorem. (Hint: consider a straight line p, not through L or M, and the intersection of p and the pencil of lines L and M respectively.)
O40 Let φ be a projectivity from l onto m, and suppose the intersection point of l and m is invariant. Prove that φ is a perspectivity. Formulate the dual proposition, too.
O41 Let F be a regular linear transformation of ℜ^3. Suppose that the end points of the vectors a, b, c are collinear. Prove that the end points of the vectors F(a), F(b), F(c) are collinear as well,
and that we have:
|| (F(c) - F(a))|| / ||(F(b) - F(a)) || = || (c - a) || / || (b - a) ||. Subsequently, prove that the mapping f in P^2 induced by F preserves cross ratio. (Project from O.)
O42 Using projective coordinates, let l: λ(1,0,0) and m: λ(0,1,0) be lines, and let A: λ(0,1,1), B: λ(0,2,1), C: λ(0,3,1) be points on l, and let A': λ(-1,0,1), B': λ(1,0,1), C': λ(0,0,1) be points
on m.
Let F be the linear mapping from the proof of the third proposition in this section.
Check that λ=-1, μ=2, ρ=1/2, σ=1/2 and that (A, B; C, X) = (f(A), f(B); f(C), f(X)).
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Unveiling the Power of Formulas: Mastering Simple Formula Creation in Excel XP - Smart Tutorials
Unveiling the Power of Formulas: Mastering Simple Formula Creation in Excel XP
Microsoft Excel XP, also known as Excel 2002, is renowned for its ability to perform complex calculations, analyze data, and automate tasks through the use of formulas. Understanding how to create
simple formulas in Excel XP is fundamental to harnessing the full potential of this powerful spreadsheet application. In this comprehensive guide, we’ll delve into the intricacies of creating simple
formulas in Excel XP, providing you with the knowledge and techniques to perform basic calculations and streamline your workflow.
Introduction to Formulas in Excel XP:
Formulas in Excel XP are mathematical expressions that perform calculations on data within worksheets. By combining cell references, operators, and functions, users can create formulas to add,
subtract, multiply, divide, and perform various other calculations with ease. Formulas enable users to automate repetitive tasks, perform complex analyses, and derive meaningful insights from their
1. Understanding Cell References:
Relative References:
In Excel XP, cell references are used to specify the location of data within a worksheet. By default, cell references are relative, meaning they adjust automatically when copied or moved to new
locations. For example, if you enter a formula in cell B2 that refers to cell A1 as “=A1”, and then copy the formula to cell C3, the formula will adjust to “=B2” to reflect the relative position of
the cells.
Absolute References:
Users can also create absolute cell references by adding a dollar sign ($) before the column letter and row number. Absolute references remain fixed when copied or moved to new locations, providing a
way to refer to specific cells consistently. For example, if you enter a formula in cell B2 that refers to cell $A$1 as “=$A$1”, and then copy the formula to other cells, the reference to cell $A$1
will remain unchanged.
2. Basic Arithmetic Operators:
Excel XP supports a variety of arithmetic operators that can be used in formulas to perform basic mathematical calculations:
• Addition (+): Adds the values of two or more cells.
• Subtraction (-): Subtracts the value of one cell from another.
• Multiplication (*): Multiplies the values of two or more cells.
• Division (/): Divides the value of one cell by another.
Users can combine these operators with cell references or numerical values to create simple formulas that perform arithmetic operations.
3. Creating Simple Formulas:
To create a simple addition formula in Excel XP:
1. Select the cell where you want the result to appear.
2. Type the equals sign (=) to indicate the start of a formula.
3. Enter the cell reference or numerical value of the first operand.
4. Type the plus sign (+) to indicate addition.
5. Enter the cell reference or numerical value of the second operand.
6. Press Enter to calculate the result.
Subtraction, Multiplication, and Division:
Similarly, users can create subtraction, multiplication, and division formulas by replacing the plus sign (+) with the minus sign (-), asterisk (*) for multiplication, and forward slash (/) for
division, respectively.
For example, to subtract the value of cell B2 from the value of cell A2 and display the result in cell C2, you would enter the following formula in cell C2: “=A2-B2”.
4. Using Functions:
Excel XP provides a wide range of built-in functions that users can use in formulas to perform specific calculations or tasks. Functions such as SUM, AVERAGE, MAX, MIN, and COUNT are commonly used
for summarizing data and performing statistical analyses.
To calculate the sum of values in cells A1:A10, you would enter the following formula: “=SUM(A1:A10)”.
Creating simple formulas in Excel XP is a fundamental skill that empowers users to perform basic calculations and automate tasks efficiently. By understanding cell references, arithmetic operators,
and basic functions, users can create formulas to add, subtract, multiply, divide, and perform various other calculations with ease. Whether you’re managing budgets, analyzing sales data, or tracking
expenses, mastering the art of formula creation in Excel XP is essential for success in spreadsheet management and analysis.
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Math problem solver exponents and exponential functions
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4.1: Solving Systems by Graphing
Last updated
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In this section we introduce a graphical technique for solving systems of two linear equations in two unknowns. As we saw in the previous chapter, if a point satisfies an equation, then that point
lies on the graph of the equation. If we are looking for a point that satisfies two equations, then we are looking for a point that lies on the graphs of both equations; that is, we are looking for a
point of intersection.
For example, consider the the two equations:
\[\begin{aligned} x-3 y &=-9 \\ 2 x+3 y &=18 \end{aligned} \nonumber \]
which is called a system of linear equations. The equations are linear equations because their graphs are lines, as shown in Figure \(\PageIndex{1}\). Note that the two lines in Figure \(\PageIndex
{1}\) intersect at the point \((3,4)\). Therefore, the point \((3,4)\) should satisfy both equations. Let’s check.
Figure \(\PageIndex{1}\): The point of intersection is a solution of the system of linear equations.
Substitute \(3\) for \(x\) and \(4\) for \(y\).
\[\begin{aligned} x-3 y &=-9 \\ 3-3(4) &=-9 \\ 3-12 &=-9 \\-9 &=-9 \end{aligned} \nonumber \]
Substitute \(3\) for \(x\) and \(4\) for \(y\).
\[\begin{aligned} 2 x+3 y &=18 \\ 2(3)+3(4) &=18 \\ 6+12 &=18 \\ 18 &=18 \end{aligned} \nonumber \]
Hence, the point \((3,4)\) satisfies both equations and is called a solution of the system.
Solution of a linear system
A point \((x,y)\) is called a solution of a system of two linear equations if and only if it satisfied both equations. Furthermore, because a point satisfies an equation if and only if it lies on the
graph of the equation, to solve a system of linear equations graphically, we need to determine the point of intersection of the two lines having the given equations.
Let’s try an example.
Example \(\PageIndex{1}\)
Solve the following system of equations: \[3x+2y =12 \\ y =x+1 \label{system1}\]
We are looking for the point \((x,y)\) that satisfies both equations; that is, we are looking for the point that lies on the graph of both equations. Therefore, the logical approach is to plot the
graphs of both lines, then identify the point of intersection.
First, let’s determine the \(x\)- and \(y\)-intercepts of \(3x +2y = 12\).
To find the \(x\)-intercept, let \(y = 0\).
\[\begin{aligned} 3 x+2 y &=12 \\ 3 x+2(0) &=12 \\ 3 x &=12 \\ x &=4 \end{aligned} \nonumber \]
To find the \(y\)-intercept, let \(x = 0\).
\[\begin{aligned} 3 x+2 y &=12 \\ 3(0)+2 y &=12 \\ 2 y &=12 \\ y &=6 \end{aligned} \nonumber \]
Hence, the \(x\)-intercept is \((4,0)\) and the \(y\)-intercept is \((0,6)\). These intercepts are plotted in Figure \(\PageIndex{2}\) and the line \(3x +2y = 12\) is drawn through them.
Figure \(\PageIndex{2}\): Drawing the graph of \(3x +2y = 12\).
Comparing the second equation \(y = x + 1\) with the slope-intercept form \(y = mx + b\), we see that the slope is \(m = 1\) and they-intercept is \((0,1)\). Plot the intercept \((0,1)\), then go up
\(1\) unit and right \(1\) unit, then draw the line (see Figure \(\PageIndex{3}\)).
Figure \(\PageIndex{3}\): Drawing the graph of \(y = x + 1\).
We are trying to find the point that lies on both lines, so we plot both lines on the same coordinate system, labeling each with its equation (see Figure \(\PageIndex{4}\)). It appears that the lines
intersect at the point \((2,3)\), making \((x,y) = (2 ,3)\) the solution of System in Example \(\PageIndex{1}\) (see Figure \(\PageIndex{4}\)).
Check: To show that \((x,y) = (2 ,3)\) is a solution of System \ref{system1}, we must show that we get true statements when we substitute \(2\) for \(x\) and \(3\) for \(y\) in both equations of
System \ref{system1}.
Figure \(\PageIndex{4}\): The coordinates of the point of intersection is the solution of System \ref{system1}.
Substituting \(2\) for \(x\) and \(3\) for \(y\) in \(3x +2y = 12\), we get:
\[\begin{aligned} 3 x+2 y &=12 \\ 3(2)+2(3) &=12 \\ 6+6 &=12 \\ 12 &=12 \end{aligned} \nonumber \]
Hence, \((2,3)\) satisfies the equation \(3x +2y = 12\).
Substituting \(2\) for \(x\) and \(3\) for \(y\) in \(y = x + 1\), we get:
\[\begin{array}{l}{y=x+1} \\ {3=2+1} \\ {3=3}\end{array} \nonumber \]
Hence, \((2,3)\) satisfies the equation \(y = x + 1\).
Because \((2,3)\) satisfies both equations, this makes \((2,3)\) a solution of System \ref{system1}.
Exercise \(\PageIndex{1}\)
Solve the following system of equations:
\[\begin{aligned} 2 x-5 y &=-10 \\ y &=x-1 \end{aligned} \nonumber \]
Example \(\PageIndex{2}\)
Solve the following system of equations: \[3x-5y =-15 \\ 2x+y =-4 \label{system2}\]
Once again, we are looking for the point that satisfies both equations of the System \ref{system2}. Thus, we need to find the point that lies on the graphs of both lines represented by the equations of
System \ref{system2}. The approach will be to graph both lines, then approximate the coordinates of the point of intersection. First, let’s determine the \(x\)- and \(y\)-intercepts of \(3x−5y = −15
To find the \(x\)-intercept, let \(y = 0\).
\[\begin{aligned} 3 x-5 y &=-15 \\ 3 x-5(0) &=-15 \\ 3 x &=-15 \\ x &=-5 \end{aligned} \nonumber \]
To find the \(y\)-intercept, let \(x=0\).
\[\begin{aligned} 3 x-5 y &=-15 \\ 3(0)-5 y &=-15 \\-5 y &=-15 \\ y &=3 \end{aligned} \nonumber \]
Hence, the \(x\)-intercept is \((−5,0)\) and the \(y\)-intercept is \((0,3)\). These intercepts are plotted in Figure \(\PageIndex{5}\) and the line \(3x−5y = −15\) is drawn through them.
Figure \(\PageIndex{5}\): Drawing the graph of the line \(3x−5y =−15\).
Next, let’s determine the intercepts of the second equation \(2x + y = −4\).
To find the \(x\)-intercept, let \(y = 0\).
\[\begin{aligned} 2 x+y &=-4 \\ 2 x+0 &=-4 \\ 2 x &=-4 \\ x &=-2 \end{aligned} \nonumber \]
To find the \(y\)-intercept, let \(x = 0\).
\[ \begin{aligned} 2 x+y &=-4 \\ 2(0)+y &=-4 \\ y &=-4 \end{aligned} \nonumber \]
Hence, the \(x\)-intercept is \((−2,0)\) and the \(y\)-intercept is \((0,−4)\). These intercepts are plotted in Figure \(\PageIndex{6}\) and the line \(2x + y =−4\) is drawn through them.
Figure \(\PageIndex{6}\): Drawing the graph of the line \(2x + y = −4\).
To find the solution of System \ref{system2}, we need to plot both lines on the same coordinate system and determine the coordinates of the point of intersection. Unlike Example \(\PageIndex{1}\), in
this case we’ll have to be content with an approximation of these coordinates. It appears that the coordinates of the point of intersection are approximately \((−2.6,1.4)\) (see Figure \(\PageIndex
Check: Because we only have an approximation of the solution of the system, we cannot expect the solution to check exactly in each equation. However, we do hope that the solution checks
Figure \(\PageIndex{7}\): The approximate coordinates of the point of intersection are \((−2.6,1.4)\).
Substitute \((x,y)=(−2.6,1.4)\) into the first equation of System \ref{system2}.
\[\begin{aligned} 3 x-5 y &=-15 \\ 3(-2.6)-5(1.4) &=-15 \\-7.8-7 &=-15 \\-14.8 &=-15 \end{aligned} \nonumber \]
Note that \((x,y)=(−2.6,1.4)\) does not check exactly, but it is pretty close to being a true statement.
Substitute \((x,y)=(−2.6,1.4)\) into the second equation of System \ref{system2}.
\[\begin{aligned} 2 x+y=-4 \\ 2(-2.6)+1.4=-4 \\-5.2+1.4=-4 \\-3.8=-4 \end{aligned} \nonumber \]
Again, note that \((x,y)= (−2.6,1.4)\) does not check exactly, but it is pretty close to being a true statement.
Later in this section we will learn how to use the intersect utility on the graphing calculator to obtain a much more accurate approximation of the actual solution. Then, in Section 4.2 and Section
4.3, we’ll show how to find the exact solution.
Exercise \(\PageIndex{2}\)
Solve the following system of equations:
\[\begin{aligned}-4 x-3 y &=12 \\ x-2 y &=-2 \end{aligned} \nonumber\]
Exceptional Cases
Most of the time, given the graphs of two lines, they will intersect in exactly one point. But there are two exceptions to this general scenario.
Example \(\PageIndex{3}\)
Solve the following system of equations: \[2x+3y=6\\2x+3y=-6 \label{system3}\]
Let’s place each equation in slope-intercept form by solving each equation for \(y\).
Solve \(2x +3y = 6\) for \(y\):
\[\begin{aligned} 2 x+3 y &=6 \\ 2 x+3 y-2 x &=6-2 x \\ 3 y &=6-2 x \\ \dfrac{3 y}{3} &=\dfrac{6-2 x}{3} \\ y &=-\dfrac{2}{3} x+2 \end{aligned} \nonumber \]
Solve \(2x +3y =−6\) for \(y\):
\[\begin{aligned} 2 x+3 y &=-6 \\ 2 x+3 y-2 x &=-6-2 x \\ 3 y &=-6-2 x \\ \dfrac{3 y}{3} &=\dfrac{-6-2 x}{3} \\ y &=-\dfrac{2}{3} x-2 \end{aligned} \nonumber \]
Comparing \(y =(−2/3)x+2\) with the slope-intercept form \(y = mx+b\) tells us that the slope is \(m = −2/3\) and they-intercept is \((0,2)\). Plot the intercept \((0,2)\), then go down \(2\) units
and right \(3\) units and draw the line (see Figure \(\PageIndex{8}\)).
Figure \(\PageIndex{8}\): Drawing the graph of the line \(2x +3y = 6\).
Comparing \(y =( −2/3)x − 2\) with the slope-intercept form \(y = mx + b\) tells us that the slope is \(m = −2/3\) and they-intercept is \((0,−2)\). Plot the intercept \((0 ,−2)\), then go down \(2\)
units and right \(3\) units and draw the line (see Figure \(\PageIndex{9}\)).
Figure \(\PageIndex{9}\): Drawing the graph of the line \(2x +3y =−6\).
To find the solution of System \ref{system3}, draw both lines on the same coordinate system (see Figure \(\PageIndex{10}\)). Note how the lines appear to be parallel (they don’t intersect). The fact
that both lines have the same slope \(−2/3\) confirms our suspicion that the lines are parallel. However, note that the lines have different \(y\)-intercepts. Hence, we are looking at two parallel but
distinct lines (see Figure \(\PageIndex{10}\)) that do not intersect. Hence, System \ref{system3} has no solution.
Figure \(\PageIndex{10}\): The lines \(2x+3y = 6\) and \(2x+3y = −6\) are parallel, distinct lines.
Exercise \(\PageIndex{3}\)
Solve the following system of equations:
\[\begin{aligned} x-y &=3 \\-2 x+2 y &=4 \end{aligned} \nonumber \]
No solution.
Example \(\PageIndex{4}\)
Solve the following system of equations: \[x-y=3 \\-2 x+2 y=-6 \label{system4}\]
Let’s solve both equations for \(y\).
Solve \(x−y = 3\) for \(y\):
\[\begin{aligned} x-y &=3 \\ x-y-x &=3-x \\-y &=-x+3 \\-1(-y) &=-1(-x+3) \\ y &=x-3 \end{aligned} \nonumber \]
Solve \(−2x +2y =−6\) for \(y\):
\[\begin{aligned}-2 x+2 y &=-6 \\-2 x+2 y+2 x &=-6+2 x \\ 2 y &=2 x-6 \\ \dfrac{2 y}{2} &=\dfrac{2 x-6}{2} \\ y &=x-3 \end{aligned} \nonumber \]
Both lines have slope \(m = 1\), and both have the same \(y\)-intercept \((0,−3)\). Hence, the two lines are identical (see Figure \(\PageIndex{11}\)). Hence, System \ref{system4} has an infinite
number of points of intersection. Any point on either line is a solution of the system. Examples of points of intersection (solutions satisfying both equations) are \((0,−3)\), \((1,−2)\), and \
Figure \(\PageIndex{11}\): \(x − y = 3\) and \(−2x +2y = −6\) are the same line.
Alternate solution:
A much easier approach is to note that if we divide both sides of the second equation \(−2x +2y = −6\) by \(−2\), we get:
\[\begin{aligned} -2x+2y &= -6 \quad {\color {Red} \text { Second equation in System }} \ref{system4}. \\ \dfrac{-2 x+2 y}{-2} &= \dfrac{-6}{-2} \quad \color {Red} \text { Divide both sides by }-2 \\
\dfrac{-2 x}{-2}+\dfrac{2 y}{-2} &= \dfrac{-6}{-2} \quad \color {Red} \text { Distribute }-2 \\ x-y &= 3 \quad \color {Red} \text { Simplify. } \end{aligned} \nonumber \]
Hence, the second equation in System \ref{system4} is identical to the first. Thus, there are an infinite number of solutions. Any point on either line is a solution.
Exercise \(\PageIndex{4}\)
Solve the following system of equations:
\[\begin{aligned}-6 x+3 y &=-12 \\ 2 x-y &=4 \end{aligned} \nonumber \]
There are an infinite number of solutions. The lines are identical, so any point on either line is a solution.
Examples \(\PageIndex{1}\), \(\PageIndex{2}\), \(\PageIndex{3}\), and \(\PageIndex{4}\) lead us to the following conclusion.
Number of solutions of a linear system
When dealing with a system of two linear equations in two unknowns, there are only three possibilities:
1. There is exactly one solution.
2. There are no solutions.
3. There are an infinite number of solutions.
Solving Systems with the Graphing Calculator
We’ve already had experience graphing equations with the graphing calculator. We’ve also used the TRACE button to estimate points of intersection. However, the graphing calculator has a much more
sophisticated tool for finding points of intersection. In the next example we’ll use the graphing calculator to find the solution of System \ref{system1} of Example \(\PageIndex{1}\).
Example \(\PageIndex{5}\)
Use the graphing calculator to solve the following system of equations: \[3x+2y=12 \\ y=x+1 \label{system5} \]
To enter an equation in the Y= menu, the equation must first be solved for \(y\). Hence, we must first solve \(3x +2y = 12\) for \(y\).
\[\begin{aligned} 3x+2y &=12 \quad \color {Red} \text { Original equation. } \\ 2y &= 12-3x \quad \color {Red} \text { Subtract } 3x \text { from both sides of the equation. } \\ \dfrac{2y}{2} &= \
dfrac{12-3 x}{2} \quad \color {Red} \text { Divide both sides by } 2 \\ y &= \dfrac{12}{2}-\dfrac{3 x}{2} \quad \color {Red} \text { On the left, simplify. On the right, } \\ y &= 6-\dfrac{3}{2} x \
quad \color {Red} \text { Simplify. } \end{aligned}\]
We can now substitute both equations of System \ref{system5} into the Y= menu (see Figure \(\PageIndex{12}\)).
Figure \(\PageIndex{12}\): Enter System \ref{system5} equations into the Y= menu.
Select 6:ZStandard from the ZOOM menu to produce the graphs shown in Figure \(\PageIndex{13}\).
Figure \(\PageIndex{13}\): Select 6:ZStandard to produce the graphs of the System \ref{system5} equations.
The question now becomes “How do we calculate the coordinates of the point of intersection?” Look on your calculator case just above the TRACE button on the top row of buttons, where you’ll see the
word CAlC, painted in the same color as the 2ND key. Press the 2ND key, then the TRACE button, which will open the CALCULATE menu shown in Figure \(\PageIndex{14}\).
Figure \(\PageIndex{14}\): Press 2ND, then TRACE to open the CALCULATE menu. Then select 5:intersect to produce the screen in Figure \(\PageIndex{15}\).
Having the calculator ask “First curve,” “Second curve,” when there are only two curves on the screen may seem annoying. However, imagine the situation when there are three or more curves on the
screen. Then these questions make good sense. You can change your selection of “First curve” or “Second curve” by using the up-and-down arrow keys to move the cursor to a different curve.
Select 5:intersect. The result is shown in Figure \(\PageIndex{15}\). The calculator has placed the cursor on the curve \(y =6−(3/2)x\) (see upper left corner of your viewing screen), and in the
lower left corner the calculator is asking you if you want to use the selected curve as the “First curve.” Answer “yes” by pressing the ENTER button.
Figure \(\PageIndex{15}\): Press the ENTER key on your calculator to say “yes” to the “First curve” selection.
The calculator responds as shown Figure \(\PageIndex{16}\). The cursor jumps to the curve \(y = x + 1\) (see upper left corner of your viewing window), and in the lower left corner the calculator is
asking you if you want to use the selected curve as the “Second curve.” Answer “yes” by pressing the ENTER key again.
Figure \(\PageIndex{16}\): Press the ENTER key on your calculator to say “yes” to the “Second curve” selection.
The calculator responds as shown Figure \(\PageIndex{17}\), asking you to “Guess.” In this case, leave the cursor where it is and press the ENTER key again to signal the calculator that you are
making a guess at the current position of the cursor.
Figure \(\PageIndex{17}\): Press the ENTER key to signal the calculator that you are satisfied with the current position of the cursor as your guess.
The result of pressing ENTER to the “Guess” question in Figure \(\PageIndex{17}\) is shown in Figure \(\PageIndex{18}\), where the calculator now provides an approximation of the the coordinates of
the intersection point on the bottom edge of the viewing window. Note that the calculator has placed the cursor on the point of intersection in Figure \(\PageIndex{17}\) and reports that the
approximate coordinates of the point of intersection are \((2,3)\).
Figure \(\PageIndex{18}\): Read the approximate coordinates of the point of intersection along the bottom edge of the viewing window.
In later sections, when we investigate the intersection of two graphs having more than one point of intersection, guessing will become more important. In those future cases, we’ll need to use the
left-and-right arrow keys to move the cursor near the point of intersection we wish the calculator to find.
Reporting your solution on your homework. In reporting your solution on your homework paper, follow the Calculator Submission Guidelines from Chapter 3, Section 2. Make an accurate copy of the image
shown in your viewing window. Label your axes \(x\) and \(y\). At the end of each axis, put the appropriate value of \(\mathrm{Xmin}, \mathrm{Xmax}, \mathrm{Ymin}\), and \(\mathrm{Ymax}\) reported in
your calculator’s WINDOW menu. Use a ruler to draw the lines and label each with their equations. Finally, label the point of intersection with its coordinates (see Figure \(\PageIndex{19}\)). Unless
instructed otherwise, always report every single digit displayed on your calculator.
Figure \(\PageIndex{19}\): Reporting your result on your homework paper.
Exercise \(\PageIndex{5}\)
Solve the following system of equations:
\[\begin{aligned} 2 x-5 y &=9 \\ y &=2 x-5 \end{aligned} \nonumber \]
Sometimes you will need to adjust the parameters in the WINDOW menu so that the point of intersection is visible in the viewing window.
Example \(\PageIndex{6}\)
Use the graphing calculator to find an approximate solution of the following system: \[y=-\dfrac{2}{7} x+7\\ y=\dfrac{3}{5} x-5 \label{system6} \]
Each equation of System \ref{system6} is already solved for \(y\), so we can proceed directly and enter them in the Y= menu, as shown in Figure \(\PageIndex{20}\). Select 6:ZStandard from the ZOOM
menu to produce the image shown in Figure \(\PageIndex{21}\).
Figure \(\PageIndex{20}\): Enter the equations of System \ref{system6}.
Figure \(\PageIndex{21}\): Select 6:ZStandard to produce this window.
Obviously, the point of intersection is off the screen to the right, so we’ll have to increase the value of \(\mathrm{Xmax}\) (set \(\mathrm{Xmax}=20\)) as shown in Figure \(\PageIndex{22}\). Once you
have made that change to \(\mathrm{Xmax}\), press the GRAPH button to produce the image shown in Figure \(\PageIndex{23}\).
Figure \(\PageIndex{22}\): Change \(\mathrm{Xmax}\) to \(20\).
Figure \(\PageIndex{23}\): Press the GRAPH button to produce this window.
Now that the point of intersection is visible in the viewing window, press 2ND CALC and select 5:intersect from the CALCULATE menu (see Figure \(\PageIndex{24}\)). Make three consecutive presses of
the ENTER button to respond to “First curve,” “Second curve,” and “Guess.” The calculator responds with the image in Figure \(\PageIndex{25}\). Thus, the solution of System \ref{system6} is
approximately \((x,y) ≈ (13.54837,3.1290323)\).
Figure \(\PageIndex{24}\): Press 2ND CALC to open the CALCULATE menu. Select 5:intersect to find the point of intersection.
Figure \(\PageIndex{25}\): Three consecutive presses of the ENTER key produce the coordinates shown at the bottom of the viewing window.
\(\color {Red}Warning!\)
Your calculator is an approximating machine. It is quite likely that your solutions might differ slightly from the solution presented in Figure \(\PageIndex{25}\) in the last \(2-3\) places.
Reporting your solution on your homework:
In reporting your solution on your homework paper, follow the Calculator Submission Guidelines from Chapter 3, Section 2. Make an accurate copy of the image shown in your viewing window. Label your
axes \(x\) and \(y\). At the end of each axis, put the appropriate value of \(\mathrm{Xmin}, \mathrm{Xmax}, \mathrm{Ymin}\), and \(\mathrm{Ymax}\) reported in your calculator’s WINDOW menu. Use a
ruler to draw the lines and label each with their equations. Finally, label the point of intersection with its coordinates (see Figure \(\PageIndex{26}\)). Unless instructed otherwise, always report
every single digit displayed on your calculator.
Figure \(\PageIndex{26}\): Reporting your result on your homework paper.
Exercise \(\PageIndex{6}\)
Solve the following system of equations:
\[\begin{aligned} y &= \dfrac{3}{2} x+6 \\ y &= -\dfrac{6}{7} x-4\end{aligned} \nonumber \]
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Ten Thousand Hours of Design Reviews—Stephen Wolfram Writings
It’s not easy to make a big software system that really fits together. It’s incredibly important, though. Because it’s what makes the whole system more than just the sum of its parts. It’s what gives
the system limitless possibilities—rather than just a bunch of specific features.
But it’s hard to achieve. It requires maintaining consistency and coherence across every area, over the course of many years. But I think it’s something we’ve been very successful at doing with
Mathematica. And I think it’s actually one of the most crucial assets for the long-term future of Mathematica.
It’s also a part of things that I personally am deeply involved in.
Ever since we started developing it more than 21 years ago, I’ve been the chief architect and chief designer of Mathematica‘s core functionality. And particularly for Mathematica 6, there was a huge
amount of design to do. Actually, I think much more even than for Mathematica 1.
In fact, I just realized that over the course of the decade during which were developing Mathematica 6—and accelerating greatly towards the end—I spent altogether about 10,000 hours doing what we
call “design reviews” for Mathematica 6, trying to make all those new functions and pieces of functionality in Mathematica 6 be as clean and simple as possible, and all fit together.
At least the way I do it, doing software design is a lot like doing fundamental science.
In fundamental science, one starts from a bunch of phenomena, and then one tries to drill down to find out what’s underneath them—to try to find the root causes, the ultimate primitives, of what’s
going on.
Well, in software design, one starts from a bunch of functionality, and then one needs to drill down to find out just what ultimate primitives one needs to support them.
In science, if one does a good job at finding the primitives, then one can have a very broad theory that covers not just the phenomena one started from, but lots of others too.
And in software design, it’s the same kind of thing.
If one does a good job at finding the primitives, then one can build a very broad system that gives one not just the functionality one was first thinking about, but lots more too.
Over the years, we’ve developed a pretty good process for doing design reviews.
We start with some particular new area of functionality. Then we get a rough description of the functions—or whatever—that we think we’ll need to cover it. Then we get down to the hard job of design
analysis. Of trying to work out just what the correct fundamental primitives to cover the area are. The clean, simple functions that represent the essence of what’s going on—and that fit together
with each other, and with the rest of Mathematica, to cover what’s needed.
Long ago I used to do design analysis pretty much solo.
But nowadays our company is full of talented people who help. The focal point is our Design Analysis group, which works with our experts in particular areas to start the process of refining possible
At some point, though, I always get involved. So that anything that’s a core function of Mathematica is always something that I’ve personally design reviewed.
I sometimes wonder whether it’s crazy for me to do this. But I think having one person ultimately review everything is a good way to make sure that there really is coherence and consistency across
the system. Of course, when the system is as big as Mathematica 6, doing all those design reviews to my level of perfection takes a long time—about 10,000 hours, in fact.
Design reviews are usually meetings with somewhere between two and twenty people. (Almost always they’re done with web conferencing, not in person.)
The majority of the time, there’s a preliminary implementation of whatever it is that we’re reviewing. Sometimes the people who are in the design review meeting will say “we think we have this mostly
figured out”. Sometimes they’ll say “we can’t see how to set this up; we need your help”. Either way, what usually happens is that I start off trying out what’s been built, and asking lots and lots
of questions about the whole area that’s involved.
It’s sometimes a little weird. One hour I’ll be intensely thinking about the higher mathematics of number theory functions. And the next hour I’ll be intensely focused on how we should handle data
about cities around the world. Or how we should set up the most general possible interfaces to external control devices.
But although the subject matter is very varied, the principles are at some level the same.
I want to understand things at the most fundamental level—to see what the essential primitives should be. Then I want to make sure those primitives are built so that they fit in as well as possible
to the whole existing structure of Mathematica—and so they are as easy as possible for people to understand, and work with.
It’s often a very grueling process. Progressively polishing things until they are as clean and simple as possible.
Sometimes we’ll start a meeting with things looking pretty complicated. A dozen functions that use some strange new construct, and have all sorts of weird arguments and options.
It’s usually pretty obvious that we have to do better. But figuring out how is often really hard.
There’ll usually be a whole series of incremental ideas. And then a few big shifts—which usually come from getting a clearer understanding of what the true core functionality has to be.
Often we’ll be talking quite a bit about precedents elsewhere in Mathematica. Because the more we can make what we’re designing now be like something we’ve done before in Mathematica, the better.
For several reasons. First, because it means we’re using approaches that we’ve tested somewhere else before.
Second, because it means that what we’re doing now will fit in better to what already exists.
And third, because it means that people who are already familiar with other things Mathematica does will have an easier time understanding the new things we’re adding.
But some of the most difficult design decisions have to do with when to break away from precedent. When is what we’re doing now really different from anything else that we’ve done before? When is it
something sufficiently new—and big—that it makes sense to create some major new structure for it?
At least when we’re doing design reviews for Mathematica kernel functions, we always have a very definite final objective for our meetings: we want to actually write the reference documentation—the
“function pages”—for what we’ve been talking about.
Because that documentation is what’s going to provide the specification for the final implementation—as well as the final definition of the function.
It always works pretty much the same way: I’ll be typing at my computer, and everyone else will be watching my screen via screen-sharing. And I’ll actually be writing the reference documentation for
what each function does. And I’ll be asking every sentence or so: “Is that really correct? Is that actually what it should do?” And people will be pointing out this or that problem with what we’re
It’s a good process, that I think does well at concentrating and capturing what we do in design analysis.
One of the things that happens in design reviews is that we finalize the names for functions.
Naming is a quintessential design review process. It involves drilling down to understand with as much as clarity as possible what a function really does, and is really about. And then finding the
perfect word or two that captures the essence of it.
The name has to be something that’s familiar enough to people who should be using the function that they’ll immediately have an idea of what the function does. But that’s general enough that it won’t
restrict what people will think of doing with the function.
Somehow the very texture of the name also has to communicate something about how broad the function is supposed to be. If it’s fairly specialized, it should have a specialized-sounding name. If it’s
very broad, then it can have a much simpler name—often a much more common English word.
I always have a test for candidate names. If I imagine making up a sentence that explains what the function does, will the proposed name be something that fits into that sentence? Or will one end up
always saying that the function with name X does something that is described as Y?
Sometimes it takes us days to come up with the right name for a function. But usually one knows when it’s right. It somehow just fits. And one can immediately remember it.
In Mathematica 6, a typical case of function naming was Manipulate.
It took quite a while to come up with that name.
We created this great function. But what should it be called? Interface? Activate? Dynamic? Live?
Interface might seem good, because, after all, it creates an interface. But it’s a particular kind of interface, not a generic one.
Activate might be good, because it makes things active. But again it’s too generic.
Dynamic: again it sounds too general, and also a bit too technical. And anyway we wanted to use that name for something else.
Live… that’s a very confusing word. It’s even hard to parse when one reads it. Does it say “make it alive”, or “here’s something that is alive”, or what?
Well, after a while one realizes that one has to understand with more clarity just what it is that this great new function is doing.
Yes, it’s creating an interface. Yes, it’s making things active, dynamic, alive. But really, first and foremost, what it’s doing is to provide a way to control something. It’s attaching knobs and
switches and so on to let one control almost anything.
So what about a word like Control? Again, very hard to understand. Is the thing itself a control? Or is it exerting control?
Handle? Again, too hard to understand.
Harness? A little better. But again, some ambiguity. And definitely too much of a “horse” motif.
Yoke? That one survived for several days. But finally the oxen jokes overwhelmed it.
And then came Manipulate.
At first, it was, “Oh, that’s too long a word for such a great and important function.”
But in my experience it often “feels right” to have a fairly long word for a function that does so much. Of course there were jokes about it sounding “manipulative”.
But as we went on talking about the function, we started just calling it Manipulate among ourselves. And everyone who joined the conversation just knew what it meant. And as we went on developing all
its detailed capabilities, it still seemed to fit. It gave the right sense of controlling something, and making something happen.
So that’s how Manipulate got its name. It’s worked well.
Still, in developing Mathematica 6, we had to name nearly 1000 functions. And each name has to last—just as the names in Mathematica 1 have lasted.
Occasionally it was fairly obvious what a function should be called.
Perhaps it had some standard name, say in mathematics or computing, such as Norm or StringSplit.
Perhaps it fit into some existing family of names, like ContourPlot3D.
But most of the time, each name took lots and lots of work to invent. Each one is sort of a minimal expression of a concept that a primitive in Mathematica implements.
Unlike human languages that grow and mutate over time, Mathematica has to be defined once and for all. So that it can be implemented, and so that both the computers and the people who use it can know
what everything in it means.
As the Mathematica system has grown, it’s in some ways become more and more difficult to do the design. Because every new thing that’s added has to fit in with more and more that’s already there.
But in some ways it’s also become easier. Because there are more precedents to draw on. But most importantly, because we’ve gotten (and I think I personally have gotten) better and better at doing
the design.
It’s not so much that the quality of the results has changed. It’s more that we’ve gotten faster and faster at solving design problems.
There are problems that come up today that I can solve in a few minutes—yet I remember twenty years ago it taking hours to solve similar problems.
Over the years, there’ve been quite a few “old chestnuts”: design problems that we just couldn’t crack. Places where we just couldn’t see a clean way to add some particular kind of functionality to
But as we’ve gotten better and better at design, we’ve been solving more and more of these. Dynamic interactivity was one big example. And in fact Mathematica 6 has a remarkable number of them
Doing design reviews and nailing down the functional design of Mathematica is a most satisfying intellectual activity. It’s incredibly diverse in subject matter. And in a sense always very pure.
It’s about a huge range of fundamental ideas—and working out how to fit them all together to create a coherent system that all makes sense.
It’s certainly as hard as anything I know about in science. But in many ways it’s more creative. One’s not trying to decode what exists in the world. One’s trying to create something from scratch—to
build a world that one can then work within.
I use Mathematica 6 every day. And every day I use countless design ideas that make all the pieces fit smoothly together.
And I realize that, yes, those 10,000 hours of design reviews were worth spending. Even just for me, what we did in them will save me countless hours in being able to do so much more with Mathematica
, so much more easily.
And now I’m looking forward to all the design reviews we’re starting to do for Mathematica 7, and Mathematica 8….
As of 2019, we’ve reached Mathematica Version 12 (which is also Wolfram Language Version 12)—with more than 6000 built-in functions.
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Dividing Exponents - Math Steps, Examples & Questions
What is dividing exponents?
Dividing exponents is where you divide terms that involve exponents, or powers. You can divide exponents in various forms, including whole numbers, negative numbers, fractions, and decimals.
When dividing numerical or algebraic expressions that have the same base, you can subtract the exponents.
For example,
8^7 \div 8^4=
You can rewrite the division problem in expanded form,
8^7 \div 8^4=\cfrac{8^7}{8^4}=\cfrac{8 \times 8 \times 8 \times 8 \times 8 \times 8 \times 8}{8 \times 8 \times 8 \times 8}
There are seven 8β s on the top and four 8β s on the bottom. So, the expression can be simplified to be:
This is equivalent to \cfrac{8 \times 8 \times 8}{1}=8^3
Another way to think about it is to subtract exponents.
8^7 \div 8^4=8^{7-4}=8^3=512
What happens if the bases are not the same?
If the bases of the exponential expression are not the same, before calculating an answer, try to rewrite the expression so the bases are the same.
For example,
9^4 \div 3^4=
Letβ s try to rewrite 9^4 so that it has a base of 3.
9 is the same as 3^2
Replace 9 with 3^2 in the original expression.
\left(3^2\right)^4 \div 3^4=
This is the same as,
3^2 \times 3^2 \times 3^2 \times 3^2, which is 3^8 \div 3^4
Now the bases are the same, so you can subtract exponents.
3^8 \div 3^4=3^{8-4}=3^4=81
Letβ s look at one more example.
a^{\frac{3}{4}} \div a^{\frac{1}{2}}
Since the bases are the same, you can subtract exponents.
a^{\frac{3}{4} \, - \, \frac{1}{2}}=a^{\frac{3}{4} \, - \, \frac{2}{4}}=a^{\frac{1}{4}}
What is an exponent?
An exponent is a small number that is written above and to the right of a number, known as the base number. This indicates how many times a number is multiplied by itself (repeated multiplication).
For example, 2^4, 2 is the base number and 4 is the exponent.
What is the negative exponent rule?
The negative exponent rule is, for any nonzero number a and any integer n, a^{-n} is equal to \cfrac{1}{a^n}. Taking a negative exponent is equivalent to finding the reciprocal of the corresponding
positive exponent.
Can you divide exponents that are fractions or decimals?
Yes, you can still apply the rules of exponents when dealing with exponential expressions that involve decimal or fractional exponents.
For example, \cfrac{x^3}{x\frac{1}{2}} =x^3 \div x^{\frac{1}{2}}=x^{3-\frac{1}{2}}=x^{\frac{5}{3}}. You will subtract the exponent in the denominator from the exponent in the numerator.
What is the quotient of powers rule?
The quotient of powers rule states that when dividing exponent with the same base, subtract the exponents.
For example, \cfrac{x^n}{x^m}=x^{n \, - \, m}.
9 a^{\frac{2}{3}}
9 a^{-5}
4 x^{\frac{5}{12}}
2 x^{\frac{5}{12}}
0.5 x^{\frac{5}{12}}
2 x^{\frac{9}{4}}
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**Program must be written in Pascal ** Imagine there is white board. You draw non-intersecting circl - Academic Writers Den
**Program must be written in
Pascal** Imagine there is white board. You draw non-intersecting circleson the board, numbered 1 to N, where N is an integer from 2 to 10.You next draw arrows from one circle to another, making sure
thateach circle has at least one out arrow and one in arrow. Now youplay the following “game:” 1. Place a magnetic marker in circle #1, and put a check mark incircle #1. The circle where the marker
resides is called the“current circle.” 2. Randomly choose among the out arrows in the current circle.(If there is only one out arrow, that choice is trivial.) In thisselection, all the out arrows
should be equally likely to bepicked. 3. Move the marker to the circle pointed to by the out arrow.This becomes the new current circle. 4. Put a check mark in the current circle. 5. If all the
circles have at least one check mark, stop thegame. If not, go to step 2 and repeat. The program will read from a textfile in the same directory asthe executable program, and will write to another
textfile in thatsame directory. Let N and K be positive integers. For this assignment, N isbetween 2 and 10 inclusive. The input text file should be namedProj1.txt. It should be in this form: The
first line has only the number N, the number of circlesthat will be used in your game. The second line has the number K, the number of arrows you will“drawing” between the circles. The next K lines
designate the arrows, one arrow per line. Eacharrow line consists of two numbers, each number being one ofcircles in the game. These two numbers are separated by a singleblank. The first number
designates the circle that is the source(back end) of the arrow; the second number designates the circlethat is the destination (pointed end) of the arrow. The circles andarrows of this game describe
a directed graph, sometimes known as a“diagraph.” In order to set up the game correctly, you shoulddescribe a “strongly connected diagraph.” A diagraph is stronglyconnected when there is a path
between any two nodes. In our game,our paths are the arrows, and our nodes are circles. Make sure that you test it with circles and arrows that describea strongly connected digraph. Not all circles
need to be connecteddirectly to each of the other circles; but as a system, they shouldbe connected in the sense described above. I suggest that each timeyou make a new Proj1.txt you draw the desired
game board and thentranslate into the required input file. Shown below are three systems of circles and arrows. Withrespect to the definition of “connected” above, NOT ONE of them isstrongly
connected. In Figure 1 and Figure 3, you could make thedigraph strongly connected by adding an arrow from circle 4 tocircle 1. The reason we need strong connection is so that you don’tget “stuck” in
your random walk around the digraph. Figure 1. This system is not strongly connected; there is nopath to circle 1. If you add an arrow from circle 4 to circle 1, itwould be strongly connected. Figure
2. This system is not strongly connected. Once you followan arrow, you are stuck. Figure 3. This system is not quite strongly connected; an addedarrow from circle 4 to circle 1 would make it
stronglyconnected. Your program can assume this connectedness for a given inputfile. That is, your program need not verify that the circles andarrows described in the input file form a strongly
connecteddigraph. A subsequent assignment will require your program toverify the connectedness. If the text in the input file does not follow the formatdescribed above, your program should end with
an error message tothe screen and to an output file. The output file should be atextfile. Name your output textfile “Ass1.txt” where “lastname” isreplaced by your last name. If the text in the input
file DOES follow the description above,then you should play the game until each circle has at least onecheck. When that happens, the game stops. At the end of the game,you should print out to the
screen, and to the output textfile, thefollowing numbers: 1. The number of circles that were used for this game 2. The number of arrows that were used for this game 3. The total number of checks on
all the circles combined. 4. The average number of checks in a circle marked during thegame. 5. The maximum number of checks in any one circle. All of these numbers should be labeled clearly in both
outputs,with explanations sufficient for someone who knows only vaguelywhat’s going on with this strange game. Attached
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17:37:25**Program must be written in Pascal ** Imagine there is white board. You draw non-intersecting circl
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Flip coin simulator
Whether you inherited some from an older relative or you just picked up the hobby on your own, collecting old coins is a fascinating pastime that can teach you about history and culture. However, it
can also be an expensive hobby to get int
Can you beat our fake coin toss detector? Click on either coin to start and try make it up to 200 coin tosses without getting into the Coin Toss Probability Calculator is a free online tool that
displays the probability of getting the head or a tail when the coin is tossed. BYJU'S online coin toss In this experiment, you'll start with $25 to play our coin-flipping game for 10 minutes. The
coin in question isn't a standard quarter, though: we've programmed it 4 May 2020 If all flipped coins come up heads, you will all be set free! But if any of the flipped coins comes up tails, or if
no one chooses to flip a coin, you Random Simulation — §5.1. 85. Simulating flipping a coin.
The coin flip generator works seamlessly and provides hours of fun. This form allows you to flip virtual coins based on true randomness, which for many purposes is better than the pseudo-random
number algorithms typically used in computer programs. Flip 2 coins . This page lets you flip 2 coins. Displays sum/total of the coins. You can choose to see the sum only.
Mar 21, 2016
Now, it seems you either have to hold the coin, flick the mouse and let go (usually results in the coin flying off the table), or hold the coin and hit right click (results in a Flipping Gun
Simulator is a fun-addicting reaction and distance game made only for rough players who like throwing dangerous stuff in the air. Just tap the screen to make your gun shoot, pushing itself in the
opposite direction. Your goal is to reach as high as possible, collecting lots of coins, boosters and bullets. Flip a Coin A unique coin flipper app that allows side landing, multiple coins, and more
Classroom Coin Flip. Heads: 0 Tails: 0.
85. Simulating flipping a coin. Example. Get a computer to simulate flipping a fair coin 20 times. To simulate a random event, use In this tutorial, we will learn how to write a stochastic
simulation through coin flips and explore the deep connection to diffusion.
Sir Coins-a-lot; Blog. Is flipping a coin 50/50? The Magic Ritual Feb 25, 2021 · This form allows you to flip virtual coins based on true randomness, which for many purposes is better than the
pseudo-random number algorithms typically used in computer programs.
Bitcoin Flip is 101% realistic and fun trading Simulator for Beginners! Dec 13, 2018 · Let us say we have a fair coin and toss the coin just once. We will observe either a head or a tail. We can code
1 for head and 0 for tail. Let us simulate a single fair coin toss experiment with the binomial distribution function in Python. >n = 1 >p = 0.5 >np.random.binomial(n,p) 0 FlipSimu is a heads or
tails coin flip simulator. You can flip a coin virtually as if flipping a real coin.
Select 1 flip or 5 flips.The results of the simulated coin flips are added to the Flips column.; Select 1000 flips to add the 1000 coin flips as fast as possible. There used to be a time when you
could simply hold a coin, press F, let go, and the coin would flip straight up in the air and you'd get a perfect flip every time. Somewhere along the lines, this feature was removed. Now, it seems
you either have to hold the coin, flick the mouse and let go (usually results in the coin flying off the table), or hold the coin and hit right click (results in a Flipping Gun Simulator is a
fun-addicting reaction and distance game made only for rough players who like throwing dangerous stuff in the air. Just tap the screen to make your gun shoot, pushing itself in the opposite
The aim of this project is for me to learn the basics of WebGL and to implement a basic game engine from scratch. Controls. Left click anywhere on the window to flip the coin; Use arrow keys and
mouse wheel/trackpad to move the camera; Live Demo Where it all began. In January 2009, the legendary (and still anonymous) Nakamoto released the code for bitcoin’s software, and mined the first
Coin flipping is a technique for establishing a cryptographic channel between two mistrustful parties. Read more from Webopedia. Coin flipping is a technique for establishing a cryptographic channel
between two mistrustful parties. It was p The coin flip, the ultimate 50-50 choice, is actually a little biased.
čo dostaneme zo slnkaaktualizovať kontroly pôžičiekobchodovanie s bitcoinovými dňamibitcoin miner faucetako založiť obchod s mincami360 bitcoinov prevedených na filipínske peso2 500 usd inr
There used to be a time when you could simply hold a coin, press F, let go, and the coin would flip straight up in the air and you'd get a perfect flip every time. Somewhere along the lines, this
feature was removed. Now, it seems you either have to hold the coin, flick the mouse and let go (usually results in the coin flying off the table), or hold the coin and hit right click (results in a
3.7 • 270 Ratings. Coin Flip Plus 4+. Simple coin toss simulator. iHandy. Designed for iPhone. 4.0 • 728 Ratings. 16 Jun 2019 Coin Flipping simulator.
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Triaxial Modeling of Halo Density Profiles with High-Resolution N-Body Simulations
We present a detailed nonspherical modeling of dark matter halos on the basis of a combined analysis of high-resolution halo simulations (12 halos with N~10^6 particles within their virial radius)
and large cosmological simulations (five realizations with N=512^3 particles in a 100h^-1 Mpc box size). The density profiles of those simulated halos are well approximated by a sequence of the
concentric triaxial distribution with their axis directions being fairly aligned. We characterize the triaxial model quantitatively by generalizing the universal density profile, that has previously
been discussed only in the framework of the spherical model. We obtain a series of practically useful fitting formulae in applying the triaxial model: the mass and redshift dependence of the axis
ratio, the mean of the concentration parameter, and the probability distribution functions of the axis ratio and the concentration parameter. These accurate fitting formulae form a complete
description of the triaxial density profiles of halos in cold dark matter models. Our current description of the dark halos will be particularly useful in predicting a variety of nonsphericity
effects, to a reasonably reliable degree, including the weak and strong lens statistics, the orbital evolution of galactic satellites and triaxiality of galactic halos, and the nonlinear clustering
of dark matter. In addition, this provides a useful framework for the nonspherical modeling of the intracluster gas, which is crucial in discussing the gas and temperature profiles of X-ray clusters
and the Hubble constant estimated via the Sunyaev-Zeldovich effect.
The Astrophysical Journal
Pub Date:
August 2002
□ Cosmology: Theory;
□ Cosmology: Dark Matter;
□ Galaxies: Clusters: General;
□ Galaxies: Halos;
□ Methods: Numerical;
□ Astrophysics
39 pages with 19 figures
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AP® Precalculus Prerequisites: Study This Before AP® Precalculus! | Albert (2024)
What Should You Know Going into AP® Precalculus?
You might be wondering, is AP® Precalculus hard? Well, AP® Precalculus is designed to prepare students for advanced mathematics courses, particularly calculus. Therefore, some might say that AP® Pre
calc is hard. The course covers various topics that build on your previous mathematical knowledge. So, we can make sure to review topics from your mathematical background that are considered
prerequisites. These prerequisites are listed in the AP® Precalculus CED (college and exam description). If you practice these foundational skills, you’ll be ready for the various topics in the AP®
Precalculus syllabus and, ultimately, ready to tackle the AP® Precalculus exam.
Proficiency vs. Familiarity: Understanding the Difference
In preparation for AP® Precalculus, you need proficiency in some skills and concepts. This means you should be able to execute these tasks accurately and efficiently. On the other hand, other areas
require familiarity, where a basic understanding and recognition are sufficient. This guide will help you differentiate between these levels and ensure you are well-prepared for the course.
Goals of This Guide
This guide aims to outline the key AP® Precalculus prerequisites that are mentioned in the AP® Precalculus CED. We will provide a clear understanding of the skills and concepts you need to master or
be familiar with. By the end, you will know exactly what areas to focus on to succeed in your AP® Precalculus course. Additionally, we will provide examples and links to more practice on our website,
Start practicing AP® Precalculus on Albert now!
Proficiency with Linear and Quadratic Functions
The first bullet point in the AP® Precalculus prerequisites to cover is linear and quadratic functions. To succeed in AP® Precalculus, you need a solid understanding of linear and quadratic
functions. This includes proficiency in algebraic manipulation, solving equations, and solving inequalities.
Algebraic Manipulation
Above all, you should be comfortable manipulating algebraic expressions involving linear and quadratic functions. This includes operations like combining like terms, distributing, and factoring.
Key Skills:
Combining Like Terms: Simplify expressions by adding or subtracting terms with the same variable raised to the same power.
Distributive Property: Apply a(b + c) = ab + ac to expand or factor expressions.
Factoring: Break down expressions into products of simpler factors.
Example: Simplify 3x^2 - 2x + 4 - (x^2 - 3x + 5).
3x^2 - 2x + 4 - (x^2 - 3x + 5) = 3x^2 - 2x + 4 - x^2 + 3x - 5
Firstly, combine like terms:
= 2x^2 + x - 1
As can be seen, we first distribute the negative sign across the terms inside the parentheses. Then, combine like terms by adding or subtracting the coefficients of terms with the same power on the
Solving Equations
Secondly, being able to solve both linear and quadratic equations is crucial. This involves isolating the variable using inverse operations, factoring, or using the quadratic formula.
Key Skills:
Solving Linear Equations: Find the value of x in equations like 2x + 3 = 7.
Solving Quadratic Equations: Use factoring, completing the square, or the quadratic formula.
Example: Solve 2x^2 - 3x - 2 = 0 using the quadratic formula.
Most important, recall the quadratic formula.
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute a = 2, b = -3, and c = -2:
x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{4} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}
Thus, we get two solutions:
x = 2 \quad \text{or} \quad x = -\frac{1}{2}
To sum up, we used the quadratic formula to find the values of the variable. Then, we substituted the coefficients into the formula, simplified the expression under the square root, and solved for
the variable.
Solving Inequalities
Finally, you should be able to solve inequalities and represent their solutions graphically or on a number line. This includes understanding both linear and quadratic inequalities.
Key Skills:
Linear Inequalities: Solve and graph solutions on a number line.
Quadratic Inequalities: Solve by finding critical points and testing intervals.
Example: Solve and graph x^2 - 4x + 3 > 0.
Most important, factor the quadratic expression:
(x - 1)(x - 3) > 0
The critical points are x = 1 and x = 3. Subsequently, test the intervals around these points to find where the inequality holds:
x < 1 \quad \text{or} \quad x > 3
In any case, this means the solution set includes values of the variable that are less than 1 or greater than 3.
In conclusion, we factored the quadratic expression to find the critical points. Then, tested the intervals determined by these points to identify where the inequality is true. Finally, the solution
is represented on a number line, showing the intervals where the quadratic expression is positive.
Ready to boost your AP® scores? Explore our plans and pricing here!
Proficiency with Solving Right Triangle Problems with Trigonometry
Secondly, the next topic listed in the AP® Precalculus prerequisites is having a solid foundation in trigonometry, especially when solving problems involving right triangles. At the same time, this
includes understanding basic trigonometric ratios and being able to apply them to solve right triangle problems.
Basic Trigonometric Ratios
Basically, you need to be familiar with the primary trigonometric ratios: sine, cosine, and tangent. These ratios are used to relate the angles of a right triangle to the lengths of its sides.
Key Ratios:
Sine: \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
Cosine: \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}
Tangent: \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
Example: In a right triangle, if the angle \theta is 30 degrees, the opposite side is 3, and the hypotenuse is 6, find \sin(\theta), \cos(\theta), and \tan(\theta).
Firstly, calculate the sine:
\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{6} = \frac{1}{2}
Then, calculate the cosine:
\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} (using the Pythagorean identity for a 30-60-90 triangle)
Finally, calculate the tangent:
\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
As shown above, we used the definitions of the trigonometric ratios to find the sine, cosine, and tangent of the given angle.
Solving Right Triangle Problems
Additionally, being able to apply trigonometric ratios to solve problems involving right triangles is crucial. This includes finding missing sides or angles.
Key Skills:
Using Trigonometric Ratios: Apply sine, cosine, and tangent to find missing sides or angles in right triangles.
Pythagorean Theorem: Use a^2 + b^2 = c^2 to find the lengths of sides in right triangles.
Example: In a right triangle, the hypotenuse is 10, and one of the angles is 30 degrees. Find the lengths of the other two sides.
Firstly, use the sine ratio to find the length of the side opposite the 30-degree angle:
\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}}
\frac{1}{2} = \frac{\text{opposite}}{10}
Then, multiply both sides by 10 to solve for the opposite side:
\text{opposite} = 10 \times \frac{1}{2} = 5
Following, use the cosine ratio to find the length of the adjacent side:
\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}
\frac{\sqrt{3}}{2} = \frac{\text{adjacent}}{10}
Finally, multiply both sides by 10 to solve for the adjacent side:
\text{adjacent} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}
Therefore, the lengths of the other two sides are 5 and 5\sqrt{3}.
In essence, we used the sine and cosine ratios to find the lengths of the sides in the right triangle.
Proficiency with Solving Systems of Equations in Two and Three Variables
Thirdly, the AP® Precalculus prerequisites list mentions proficiency in solving systems of equations as vital. You need to be able to solve systems of equations in two and three variables. This
involves using various methods such as substitution, elimination, and matrix operations.
Solving Systems in Two Variables
Undoubtedly, you should be able to solve systems of linear equations in two variables. This can be done using methods like substitution and elimination.
Key Methods:
Substitution Method: Solve one equation for one variable and substitute this expression into the other equation.
Elimination Method: Add or subtract equations to eliminate one of the variables, making it easier to solve for the remaining variable.
Example: Solve the system of equations:
2x + y = 10
3x - y = 5
Firstly, add the two equations to eliminate one of the variables:
2x + y + 3x - y = 10 + 5
5x = 15
Then, solve for the other variable:
x = \frac{15}{5} = 3
Lastly, substitute that value into the first equation to find the other variable:
2(3) + y = 10
6 + y = 10
y = 10 - 6 = 4
Therefore, the solution to the system has been found because we solved for both variables.
To sum up, we used the elimination method to eliminate one variable, making it easier to solve for the other variable. Finally, we substituted the value back into one of the original equations to
find the original variable.
Solving Systems in Three Variables
Next, solving systems of equations in three variables typically requires using methods such as substitution, elimination, or matrix operations.
Key Methods:
Substitution Method: Solve one equation for one variable, then substitute this expression into the other equations.
Elimination Method: Use elimination to reduce the system to two equations in two variables, then solve.
Matrix Operations: Use matrices and row reduction techniques to solve the system.
Example: Solve the system of equations:
x + y + z = 6
2x - y + 3z = 14
-x + 2y - z = -2
Firstly, use the elimination method to eliminate the third variable. To rephrase it, add the first and second equations:
(x + y + z) + (2x - y + 3z) = 6 + 14
3x + 4z = 20
Then, add the first and third equations:
(x + y + z) + (-x + 2y - z) = 6 - 2
3y = 4
Thirdly, solve for the second variable:
y = \frac{4}{3}
Then, substitute this expression into one of the original equations to find the other two missing variables. In particular, using the first equation:
x + \frac{4}{3} + z = 6
x + z = 6 - \frac{4}{3} = \frac{14}{3}
Using the equation 3x + 4z = 20 :
x = \frac{2}{3}
Finally, substitute the first two variables into one of the equations to solve for the third variable:
x + \frac{4}{3} + z = 6
z = 4
Thus, the solution to the system has been found. In coordinate form, it i s (x, y, z) = (\frac{2}{3}, \frac{4}{3}, 4).
Here, we used elimination to reduce the system to two equations in two variables, solved for one variable, and then used substitution to find the remaining variables.
Want to practice more with Solving Systems of Equations? Click here!
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Familiarity with Piecewise-Defined Functions
So far, we have only looked at topics that the list of AP® Precalculus prerequisites recommends proficiency. From this point on, the AP® Precalculus prerequisites list mentions that you should only
be familiar with the rest of the topics. Firstly, you should understand piecewise-defined functions. These functions are defined by different expressions depending on the input value. You need to be
able to interpret, evaluate, and graph these functions.
Definition and Examples
A piecewise-defined function is a function that is defined by multiple sub-functions, each of which applies to a specific interval of the domain.
Example: Consider the piecewise-defined function:
x + 2 \text{ if } x < 0
x^2 \text{ if } 0 \leq x < 2
3x - 1 \text{ if } x \geq 2
In short, this function has three different expressions based on the value of the variable.
Evaluating Piecewise-Defined Functions:
Unlike a normal function, to evaluate a piecewise-defined function, determine which piece of the function to use based on the input value.
Example: Evaluate f(x) for x = -1, x = 1, and x = 3 for the given function f(x).
1. For x = -1:
Since this value is less than zero, use the first piece:.
f(-1) = -1 + 2 = 1
2. For x = 1:
Since this is between zero and two, use the second piece:
f(1) = 1^2 = 1
3. For x = 3:
Since this value is greater than two, use the third piece:
f(3) = 3(3) - 1 = 9 - 1 = 8
In summary, we evaluated the function by selecting the appropriate piece based on the value of the variable.
Graphical Representation
To graph a piecewise-defined function, graph each piece of the function over its specified interval. Pay attention to whether the endpoints of the intervals are included (closed dots) or excluded
(open dots).
Example: Graph the piecewise-defined function:
g(x) = 2x + 1 \text{ if } x < 1
x^2 \text{ if } 1 \leq x \leq 3
-2x + 5 \text{ if } x > 3
1. For x < 1:
Firstly, graph the linear equation with a slope of 2 and y-intercept of 1. Most importantly, stop when the x values reach 1.
2. For 1 \leq x \leq 3:
Secondly, graph the parabola between the x values of 1 and 3.
3. For x > 3:
Finally, graph the linear equation with a slope of -2 and a y-intercept of 5. However, only do so for x values greater than 3.
Once you’ve done this, use an open dot for any interval endpoints that are less than or greater than. Then, use a closed dot for any interval endpoints that are less than or equal to or greater than
or equal to.
In detail, we graphed each piece of the function within its specified interval, ensuring we accurately represented the endpoints.
Familiarity with Exponential Functions and Rules for Exponents
According to the list of AP® Precalculus prerequisites, you should be familiar with exponential functions. These functions are crucial for modeling growth and decay in various contexts. You need to
understand their properties and be able to apply the rules for exponents.
Definition and Properties
An exponential function is a function of the form f(x) = a \cdot b^x, where a is a constant, b is the base, and x is the exponent.
Key Properties:
Base Greater Than 1: For b > 1, the function models exponential growth.
Base Between 0 and 1: For 0 < b < 1, the function models exponential decay.
Horizontal Asymptote: The line y = 0 is a horizontal asymptote for exponential functions.
Example: Consider the exponential function f(x) = 2 \cdot 3^x. Describe its properties.
The base b = 3 > 1, so it models exponential growth.
As x increases, f(x) grows rapidly.
The horizontal asymptote is y = 0.
Rules for Exponents
Moreover, to work with exponential functions effectively, you need to be familiar with the rules for exponents.
Key Rules:
Product of Powers: a^m \cdot a^n = a^{m+n}
Quotient of Powers: \frac{a^m}{a^n} = a^{m-n}
Power of a Power: latex^n = a^{mn}[/latex]
Negative Exponent: a^{-n} = \frac{1}{a^n}
Zero Exponent: a^0 = 1 (for a \neq 0)
Example: Simplify the expression \frac{2^5 \cdot 2^{-3}}{2^2}.
First, apply the product of powers rule:
2^5 \cdot 2^{-3} = 2^{5-3} = 2^2
Next, apply the quotient of powers rule:
\frac{2^2}{2^2} = 2^{2-2} = 2^0 = 1
Here, we use the product of powers and quotient of powers rules to simplify the expression.
To read more about Rules for Exponents, check out this article!
Graphing Exponential Functions
Graphing exponential functions involves plotting points and understanding the shape of the graph based on the base.
Example: Graph the exponential function f(x) = 2 \cdot 3^x.
Plot Points: Calculate and plot points for several values of x.
Firstly, for x = -2: f(-2) = 2 \cdot 3^{-2} = \frac{2}{9}
Secondly, for x = -1: f(-1) = 2 \cdot 3^{-1} = \frac{2}{3}
Thirdly, for x = 0: f(0) = 2 \cdot 3^0 = 2
Furthermore, x = 1: f(1) = 2 \cdot 3^1 = 6
Finally, x = 2: f(2) = 2 \cdot 3^2 = 18
Draw the Curve: Connect the points smoothly, showing the rapid growth as x increases.
Horizontal Asymptote: Indicate the horizontal asymptote y = 0 on the graph.
By plotting these points and connecting them, we can visualize the exponential growth of the function.
Start practicing AP® Precalculus on Albert now!
Familiarity with Radicals (Square Roots and Cube Roots)
Following exponents, the AP® Precalculus prerequisites list mentions that you should be familiar with radicals, including square roots and cube roots. In any case, understanding how to simplify and
manipulate radical expressions is essential.
Square Roots and Cube Roots
Radicals are expressions that include a root, such as a square root or cube root. The square root of a number a is a number b such that b^2 = a. Similarly, the cube root of a is a number b such that
b^3 = a.
Key Concepts:
Square Root: \sqrt{a} is a number that, when squared, gives a.
Cube Root: \sqrt[3]{a} is a number that, when cubed, gives a.
Example: Simplify \sqrt{36} and \sqrt[3]{27}.
\sqrt{36} = 6 because 6^2 = 36.
\sqrt[3]{27} = 3 because 3^3 = 27.
In essence, we identified the given numbers’ square root and cube root.
Simplifying Radical Expressions
Simplifying radicals involves expressing the radical in its simplest form. This can include factoring out perfect squares or cubes.
Example: Simplify \sqrt{50}.
First, factor 50 into its prime factors: 50 = 2 \cdot 5^2.
Then, simplify by taking the square root of the perfect square:
\sqrt{50} = \sqrt{2 \cdot 5^2} = 5\sqrt{2}
Here, we factor 50 and simplify the square root by taking out the perfect square.
Check out this link for more practice with Simplifying Radical Expressions!
Operations with Radicals
To work with radical expressions, you need to be able to add, subtract, multiply, and divide them.
Key Operations:
Addition/Subtraction: Combine like radicals (same radicand).
Multiplication: Multiply the coefficients and radicands separately.
Division: Divide the coefficients and radicands separately.
Example: Simplify latex(3\sqrt{6})[/latex].
First, multiply the coefficients:
2 \cdot 3 = 6
Next, multiply the radicands:
\sqrt{3} \cdot \sqrt{6} = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}
Combine the results:
(3\sqrt{6}) = 6 \cdot 3\sqrt{2} = 18\sqrt{2}
Here, we multiply the coefficients and radicands separately then simplify the resulting radical.
Want to practice more Operations with Radicals? Click here!
Familiarity with Complex Numbers
Furthermore, the AP® Precalculus prerequisites list mentions familiarity with complex numbers. Complex numbers extend the real number system and are essential in various advanced mathematical
Definition and Operations
A complex number is a number of the form a + bi, where a and b are real numbers, and i is the imaginary unit with the property that i^2 = -1.
Key Concepts:
Real Part: a in a + bi.
Imaginary Part: b in a + bi.
Imaginary Unit: i, where i^2 = -1.
Example: Consider the complex number 3 + 4i.
The real part is 3.
The imaginary part is 4i.
Operations with Complex Numbers
In addition, you need to know how to perform basic operations with complex numbers, including addition, subtraction, multiplication, and division.
Addition and Subtraction:
In general, to add or subtract complex numbers, combine the real parts and the imaginary parts separately.
Example: Add (3 + 2i) and (1 + 4i).
(3 + 2i) + (1 + 4i) = (3 + 1) + (2i + 4i) = 4 + 6i
Here, we add the real parts 3 and 1, and the imaginary parts 2i and 4i.
Specifically, to multiply complex numbers, use the distributive property (FOIL method for binomials).
Example: Multiply (2 + 3i) and (1 - 4i).
(2 + 3i)(1 - 4i) = 2(1) + 2(-4i) + 3i(1) + 3i(-4i)
= 2 - 8i + 3i - 12i^2
Since i^2 = -1, -12i^2 = 12:
= 2 - 8i + 3i + 12 = 14 - 5i
In order to multiply the real parts and the imaginary parts, we simplified using the fact that squaring the imaginary unit equals -1.
To divide complex numbers, multiply the numerator and the denominator by the conjugate of the denominator and then simplify.
Example: Divide \frac{3 + 4i}{1 - 2i}.
Multiply the numerator and the denominator by the conjugate of the denominator (1 + 2i):
\dfrac{(3 + 4i)(1 + 2i)}{(1 - 2i)(1 + 2i)}
Secondly, simplify the numerator using the distributive property:
(3 + 4i)(1 + 2i) = 3(1) + 3(2i) + 4i(1) + 4i(2i)
= 3 + 6i + 4i + 8i^2 = 3 + 10i + 8(-1) = 3 + 10i - 8 = -5 + 10i
Then, simplify the denominator using the difference of squares formula:
(1 - 2i)(1 + 2i) = 1 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5
Finally, combine the results:
\frac{-5 + 10i}{5} = -1 + 2i
Here, we multiply the numerator and the denominator by the conjugate of the denominator, simplify both the numerator and the denominator, and then divide to get the final result.
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Familiarity with Multiple Representations of Functions
Lastly, as indicated in the AP® Precalculus prerequisites list, it is crucial to understand and communicate functions in various forms. You need to be familiar with graphical, numerical, analytical,
and verbal representations of functions.
Graphical Representation
Graphical representation involves plotting the function on a coordinate plane. This visual representation helps to understand the behavior and properties of the function, such as intercepts, maxima
and minima, and asymptotes.
Example: Graph the function f(x) = x^2 - 4x + 3.
Find the intercepts:
Y-intercept: Set x = 0:
f(0) = 0^2 - 4(0) + 3 = 3
X-intercepts: Set f(x) = 0:
x^2 - 4x + 3 = 0
Factoring the quadratic:
(x - 1)(x - 3) = 0
So, x = 1 and x = 3
Then, plot the points on the coordinate plane.
Finally, draw the parabola passing through these points, opening upwards.
Here, we graph the function by finding and plotting the intercepts and then sketching the curve.
Numerical Representation
Numerical representation involves using tables of values to represent the function. This is useful for understanding how the function behaves at specific points.
Example: Create a table of values for f(x) = x^2 - 4x + 3.
Here, we calculate the function’s value at several points and organize them in a table.
Analytical Representation
Analytical representation involves expressing the function in a symbolic form, such as an equation or an expression. This form allows for algebraic manipulation and deeper analysis.
Example: Consider the function f(x) = x^2 - 4x + 3.
We can analyze the function by factoring:
f(x) = (x - 1)(x - 3)
Here, the analytical form of the function allows us to easily find the x-intercepts and understand the behavior of the function.
Verbal Representation
Verbal representation involves describing the function and its properties in words. This form is useful for communicating the function’s behavior and characteristics.
Example: Describe the function f(x) = x^2 - 4x + 3 verbally.
The function f(x) = x^2 - 4x + 3 is a quadratic function that opens upwards. It has x-intercepts at x = 1 and x = 3, and a y-intercept at y = 3. The vertex of the parabola occurs at x = 2, and the
minimum value of the function is f(2) = -1.
Here, we describe the function’s key properties and behavior in words.
Understanding the AP® Precalculus prerequisites is essential for success in the course. By mastering the foundational skills and concepts outlined in this guide, you will be well-prepared for the AP®
Precalculus curriculum and, ultimately, the AP® Precalculus exam. Keep practicing, stay curious, and don’t hesitate to seek additional resources and support as you embark on this exciting
mathematical journey.
Need help preparing for your AP® Precalculus exam?
Albert has hundreds of AP® Precalculus practice questions, free response, and full-length practice tests to try out.
Start practicing AP® Precalculus on Albert now!
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Archivos del blog
Day 81: Final Exam Day
• NO CELL phones are allowed during the final exam
• Schoolcity is the only tab open!
• You may use your study guide if it was turned in yesterday!
• Work on MATHIA when done with Final Exam
• Last Day for Mathia is Friday the 15th at 8:00am!
• I will open the Quadratics Module which is next semester for people to get a head start over the break.
• Have a good break!!!!!!!!!!!!!!!!
Day 80: KAHOOT Review
• Students review by playing KAHOOT to practice for the final.
• Complete all required MATHIA workspaces
Day 79: Final Review
Assignment Title:
"Final Review"
• Complete the Chapter 7 Properties of Quadrilaterals (Due Today)
• Work on Study guide for final (see Google classroom)
• Extra Credit Assignment for final (see Google classroom)
• Complete all required MATHIA workspaces
Day 78: Final Review
Assignment Title:
"Quadrilaterals Packet"
• Complete the Chapter 7 Properties of Quadrilaterals (Due Monday)
• Work on Study guide for final (see Google classroom)
• Extra Credit Assignment for final (see Google classroom)
• Complete all required MATHIA workspaces
Day 76: Polygon Angles and Quadrilaterals
Learning Target:
SWBAT practice finding the interior and exterior angles in any polygon by completing the online assignment and practice identifying and using the properties of quadrilaterals by completing the
Chapter 7 handout packet.
Assignment Title:
"Polygon angles assignment and Quadrilaterals"
• Complete the polygon angle assignment! (Due Today)
• Complete the Chapter 7 Properties of Quadrilaterals (Due Monday)
• Complete all required MATHIA workspaces
Day 75: Polygon Exterior Angle Sum Theorem
Learning Target:
SWBAT discover how to find the Polygon Exterior Angle Sum of any polygon by completing the handout.
EQ: "What is the formula for Polygon Exterior Angle Sum?."
Assignment Title:
"Polygon Exterior angle sum handout"
• Complete the polygon exterior angle sum handout!
• Complete all required MATHIA workspaces
Polygon Exterior angle sum applet
Day 75: District KDS GREEN test.
Assignment Title:
• Complete the District KDS GREEN test in Schoolcity!
• Complete all required MATHIA workspaces
Day 74: Interior Angles Sum of Polygons
Learning Target:
SWBAT discover how to find the Polygon Interior Angle Sum of any polygon by completing the handout.
Assignment Title:
"Polygon Interior angle sum handout"
• Complete the polygon interior angle sum handout!
• Complete all required MATHIA workspaces
• District KDS GREEN test is tomorrow!
Day 73: Trigonometry Assessment and MATHIA
Assignment Title:
"Trigonometry Assessment"
• Complete the Trig Assessment!
• Complete all required MATHIA workspaces
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Search: [math] - Shaarli -- Adrien Brochier
This survey focuses on the computational complexity of some of the
fundamental decision problems in 3-manifold theory. The article discusses the
wide variety of tools that are used to tackle these problems, including normal
and almost surfaces, hierarchies, homomorphisms to finite groups, and
hyperbolic structures.
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Longitudinal and Transverse Mass
In Newtonian mechanics the mass of a particle is constant and can be expressed as the ratio of the force to the acceleration:
In Newtonian mechanics
In special relativity, mass plays more of a resistance to acceleration role, and has a different value in the direction of acceleration than at right angles to it.
In special relativity the momentum of a particle of rest mass
If the acceleration is perpendicular to and
If the acceleration is in the direction of
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1. Antonova T.M., Bodnar D.I. Convergence domains for branched continued fractions of the special form. Approx. Theor. and its Appl.: Pr. Inst. Math. NAS Ukr. 2000, 31, 19-32. (in Ukrainian)
2. Baran O.E. An analog of the Vorpits'kii convergence criterion for branched continued fractions of special form. J. Math.Sci. 1998, 90 (5), 2348-2351. doi: 10.1007/BF02433964 (translation of Mat.
Met. Fiz.-Mekh. Polya 1996, 39 (2), 35-38. (in Ukrainian))
3. Baran O.E. Some convergence criteria for branched continued fractions with independent variables. Visnyc State Polytechnic University. App. Math. 1998, 341, 18-23. (in Ukrainian)
4. Bodnar D.I. Branched continued fractions. Naukova Dumka, Kiev, 1986. (in Russian)
5. Bodnar D.I., Bubnyak M.M. Estimates of the rate of pointwise and uniform convergence for one-periodic branched continued fractions of a special form. J. Math. Sci. 2015, 208 (3), 289-300. doi:
10.1007/s10958-015-2446-x (translation of Mat. Met. Fiz.-Mekh. Polya 2013, 56 (4), 24-32. (in Ukrainian))
6. Bodnar D.I. The investigation of a convergence of one class of branched continued fractions. In: Continued fractions and their applications. Inst. Math., Acad. of Sci. of the Ukr. SSR, Kiev,
1976, 41-44. (in Russian)
7. Kuchminska Ch.Yo. Approximation and interpolation of functions by continued and branched continued fractions. Ph.D. dissertation. Mathematical Analysis. Inst. for App. Problem. of Mech. and
Math., Acad. of Sci. of the Ukr. SSR, Lviv, 1978. (in Russian)
8. Kuchminska Ch.Yo. A Worpitzky boundary theorem for branched continued fractions of the special form. Carpathian Math. Publ. 2016, 8 (2), 272-278. doi: 10.15330/cmp.8.2.272-278
9. Worpitzky J. Untersuchungen über die Entwickelung der monodromen und monogenen Funktionen durch Kettenbrüche. Friedrichs-Gymnasium und Realschule Jahresabericht, Berlin, 1865. 3-39.
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How big is
How big is a Feddan of land?
noun, plural fed·dan, fed·dans. an Egyptian unit of area equivalent to 1.038 acres (0.42 ha).
How many hectares are in Feddan?
1 Feddan is equal to how many Hectare (ha)? In mathematical expression, 1 Feddan = 0.42000037161252 Hectare (ha).
How much is a Feddan?
In Egypt the feddan is the only non-metric unit which remained in use following the switch to the metric system. A feddan is divided into 24 kirat (Arabic: قيراط, qīrāt) in which one kirat equals
175 square metres….
How many feddans are in an acre?
The answer is one Feddan is equal to 1.03 Acres.
How much is an acre?
One acre equals 1⁄640 (0.0015625) square mile, 4,840 square yards, 43,560 square feet, or about 4,047 square metres (0.4047 hectares) (see below).
How many feddans are in Egypt?
7.2 million feddans
Introduction. The area of agricultural land in Egypt is confined to the Nile Valley and delta, with a few oases and some arable land in Sinai. The total cultivated area is 7.2 million feddans (1
feddan = 0.42 ha), representing only 3 percent of the total land area.
What is a good price per acre?
Perhaps the largest factor that determines land value is location. There are 1.9 billion acres of land in the contiguous 48 states, and the average value is about $12,000 per acre.
What lot size is 1/2 acre?
1/2 acre? An acre is 43560 square feet so half an acre is 43560/2 = 21780 square feet. If your 1/2 acre plot of land is a square with area 21780 square feet then each side is of length √21780 feet.
Which crop is famous in Egypt?
Crops such as barley, beans, rice, and wheat are also grown here. Egyptian Cotton, which is famous worldwide, is also grown in Egypt. Therefore, Egypt is famous for growing Cotton, and Option C is
the correct answer.
What is the most important crop in Egypt?
Cotton has traditionally been the most important fibre crop in Egypt and the leading agricultural export crop. Sugar crops. Sugar cane is the main sugar crop in upper Egypt.
Which state has the cheapest land per acre?
Tennessee, Arkansas, and West Virginia consistently rank as the cheapest places to buy residential land. Tennessee offers diverse geography, from mountains and lakes to acres of rural flat ground,
and of course the iconic landmarks and attractions like Graceland and Nashville, the heart of country music.
How much is 40 acres of land worth today?
40 Acres and a Mule Would Be at Least $6.4 Trillion Today—What the U.S. Really Owes Black America.
Which is bigger a hectare or A feddan?
The hectare (symbol ha) is an SI accepted metric system unit of area equal to 100 ares (10,000 m2) and primarily used in the measurement of land as a metric replacement for ..more definition+ In
relation to the base unit of [area] => (square meters), 1 Feddans (feddan) is equal to 4200 square-meters, while 1 Hectares (ha) = 10000 square-meters.
How to convert feddans to hectares in Arabic?
How to convert Feddans to Hectares (feddan to ha)? 1 feddan = 0.42 ha. 1 x 0.42 ha = 0.42 Hectares. Always check the results; rounding errors may occur. A feddan (Arabic: فدّان, faddān) is a unit of
area. It is used in Egypt, Sudan, Syria and the Sultanate of Oman. In Classical Arabic, the word means ‘a yoke of oxen’: imply ..more definition+
How big is one Feddan in square meters?
1 feddan = 24 kirat = 60 metre × 70 metre = 4200 square metres (m²) = 0.420 hectares = 1.037 acres.
Which is larger A feddan or a Kirat?
Equivalent units. 1 feddan = 24 kirat = 60 metre × 70 metre = 4200 square metres (m²) = 0.420 hectares = 1.037 acres.
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Discontinuous Galerkin operators
Discontinuous Galerkin operators¶
Core DG routines¶
Elementwise differentiation¶
grudge.op.local_grad(dcoll: DiscretizationCollection, *args, nested=False) ArrayOrContainer[source]¶
Return the element-local gradient of a function \(f\) represented by vec:
\[\nabla|_E f = \left( \partial_x|_E f, \partial_y|_E f, \partial_z|_E f \right)\]
May be called with (vec) or (dd_in, vec).
☆ vec – a DOFArray or an ArrayContainer of them.
☆ dd_in – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ nested – return nested object arrays instead of a single multidimensional array if vec is non-scalar.
an object array (possibly nested) of DOFArrays or ArrayContainer of object arrays.
grudge.op.local_d_dx(dcoll: DiscretizationCollection, xyz_axis, *args) ArrayOrContainer[source]¶
Return the element-local derivative along axis xyz_axis of a function \(f\) represented by vec:
\[\frac{\partial f}{\partial \lbrace x,y,z\rbrace}\Big|_E\]
May be called with (vec) or (dd, vec).
☆ xyz_axis – an integer indicating the axis along which the derivative is taken.
☆ dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ vec – a DOFArray or an ArrayContainer of them.
a DOFArray or an ArrayContainer of them.
grudge.op.local_div(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the element-local divergence of the vector function \(\mathbf{f}\) represented by vecs:
\[\nabla|_E \cdot \mathbf{f} = \sum_{i=1}^d \partial_{x_i}|_E \mathbf{f}_i\]
May be called with (vecs) or (dd, vecs).
☆ dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ vecs – an object array of DOFArrays or an ArrayContainer object with object array entries. The last axis of the array must have length matching the volume dimension.
a DOFArray or an ArrayContainer of them.
Weak derivative operators¶
grudge.op.weak_local_grad(dcoll: DiscretizationCollection, *args, nested=False) ArrayOrContainer[source]¶
Return the element-local weak gradient of the volume function represented by vec.
May be called with (vec) or (dd_in, vec).
Specifically, the function returns an object array where the \(i\)-th component is the weak derivative with respect to the \(i\)-th coordinate of a scalar function \(f\). See weak_local_d_dx()
for further information. For non-scalar \(f\), the function will return a nested object array containing the component-wise weak derivatives.
☆ dd_in – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ vec – a DOFArray or an ArrayContainer of them.
☆ nested – return nested object arrays instead of a single multidimensional array if vec is non-scalar
an object array (possibly nested) of DOFArrays or ArrayContainer of object arrays.
grudge.op.weak_local_d_dx(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the element-local weak derivative along axis xyz_axis of the volume function represented by vec.
May be called with (xyz_axis, vec) or (dd_in, xyz_axis, vec).
Specifically, this function computes the volume contribution of the weak derivative in the \(i\)-th component (specified by xyz_axis) of a function \(f\), in each element \(E\), with respect to
polynomial test functions \(\phi\):
\[\int_E \partial_i\phi\,f\,\mathrm{d}x \sim \mathbf{D}_{E,i}^T \mathbf{M}_{E}^T\mathbf{f}|_E,\]
where \(\mathbf{D}_{E,i}\) is the polynomial differentiation matrix on an \(E\) for the \(i\)-th spatial coordinate, \(\mathbf{M}_E\) is the elemental mass matrix (see mass() for more
information), and \(\mathbf{f}|_E\) is a vector of coefficients for \(f\) on \(E\).
☆ dd_in – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ xyz_axis – an integer indicating the axis along which the derivative is taken.
☆ vec – a DOFArray or an ArrayContainer of them.
a DOFArray or an ArrayContainer of them.
grudge.op.weak_local_div(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the element-local weak divergence of the vector volume function represented by vecs.
May be called with (vecs) or (dd, vecs).
Specifically, this function computes the volume contribution of the weak divergence of a vector function \(\mathbf{f}\), in each element \(E\), with respect to polynomial test functions \(\phi\):
\[\int_E \nabla \phi \cdot \mathbf{f}\,\mathrm{d}x \sim \sum_{i=1}^d \mathbf{D}_{E,i}^T \mathbf{M}_{E}^T\mathbf{f}_i|_E,\]
where \(\mathbf{D}_{E,i}\) is the polynomial differentiation matrix on an \(E\) for the \(i\)-th spatial coordinate, and \(\mathbf{M}_E\) is the elemental mass matrix (see mass() for more
☆ dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ vecs – an object array of DOFArrays or an ArrayContainer object with object array entries. The last axis of the array must have length matching the volume dimension.
a DOFArray or an ArrayContainer like vec.
Mass, inverse mass, and face mass operators¶
grudge.op.mass(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the action of the DG mass matrix on a vector (or vectors) of DOFArrays, vec. In the case of vec being an ArrayContainer, the mass operator is applied component-wise.
May be called with (vec) or (dd_in, vec).
Specifically, this function applies the mass matrix elementwise on a vector of coefficients \(\mathbf{f}\) via: \(\mathbf{M}_{E}\mathbf{f}|_E\), where
\[\left(\mathbf{M}_{E}\right)_{ij} = \int_E \phi_i \cdot \phi_j\,\mathrm{d}x,\]
where \(\phi_i\) are local polynomial basis functions on \(E\).
☆ dd_in – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
☆ vec – a DOFArray or an ArrayContainer of them.
a DOFArray or an ArrayContainer like vec.
grudge.op.inverse_mass(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the action of the DG mass matrix inverse on a vector (or vectors) of DOFArrays, vec. In the case of vec being an ArrayContainer, the inverse mass operator is applied component-wise.
For affine elements \(E\), the element-wise mass inverse is computed directly as the inverse of the (physical) mass matrix:
\[\left(\mathbf{M}_{J^e}\right)_{ij} = \int_{\widehat{E}} \widehat{\phi}_i\cdot\widehat{\phi}_j J^e \mathrm{d}\widehat{x},\]
where \(\widehat{\phi}_i\) are basis functions over the reference element \(\widehat{E}\), and \(J^e\) is the (constant) Jacobian scaling factor (see grudge.geometry.area_element()).
For non-affine \(E\), \(J^e\) is not constant. In this case, a weight-adjusted approximation is used instead following [Chan_2016]:
\[\mathbf{M}_{J^e}^{-1} \approx \widehat{\mathbf{M}}^{-1}\mathbf{M}_{1/J^e}\widehat{\mathbf{M}}^{-1},\]
where \(\widehat{\mathbf{M}}\) is the reference mass matrix on \(\widehat{E}\).
May be called with (vec) or (dd, vec).
☆ vec – a DOFArray or an ArrayContainer of them.
☆ dd – a DOFDesc, or a value convertible to one. Defaults to the base volume discretization if not provided.
a DOFArray or an ArrayContainer like vec.
grudge.op.face_mass(dcoll: DiscretizationCollection, *args) ArrayOrContainer[source]¶
Return the action of the DG face mass matrix on a vector (or vectors) of DOFArrays, vec. In the case of vec being an arbitrary ArrayContainer, the face mass operator is applied component-wise.
May be called with (vec) or (dd_in, vec).
Specifically, this function applies the face mass matrix elementwise on a vector of coefficients \(\mathbf{f}\) as the sum of contributions for each face \(f \subset \partial E\):
\[\sum_{f=1}^{N_{\text{faces}} } \mathbf{M}_{f, E}\mathbf{f}|_f,\]
\[\left(\mathbf{M}_{f, E}\right)_{ij} = \int_{f \subset \partial E} \phi_i(s)\psi_j(s)\,\mathrm{d}s,\]
where \(\phi_i\) are (volume) polynomial basis functions on \(E\) evaluated on the face \(f\), and \(\psi_j\) are basis functions for a polynomial space defined on \(f\).
☆ dd – a DOFDesc, or a value convertible to one. Defaults to the base "all_faces" discretization if not provided.
☆ vec – a DOFArray or an ArrayContainer of them.
a DOFArray or an ArrayContainer like vec.
Links to canonical locations of external symbols¶
(This section only exists because Sphinx does not appear able to resolve these symbols correctly.)
class grudge.op.ArrayOrContainer¶
See arraycontext.ArrayOrContainer.
Trace Pairs¶
Boundary trace functions¶
grudge.op.bdry_trace_pair(dcoll: DiscretizationCollection, dd: Any, interior, exterior) TracePair[source]¶
Returns a trace pair defined on the exterior boundary. Input arguments are assumed to already be defined on the boundary denoted by dd. If the input arguments interior and exterior are
ArrayContainer objects, they must both have the same internal structure.
☆ dd – a DOFDesc, or a value convertible to one, which describes the boundary discretization.
☆ interior – a DOFArray or an ArrayContainer of them that contains data already on the boundary representing the interior value to be used for the flux.
☆ exterior – a DOFArray or an ArrayContainer of them that contains data that already lives on the boundary representing the exterior value to be used for the flux.
a TracePair on the boundary.
grudge.op.bv_trace_pair(dcoll: DiscretizationCollection, dd: Any, interior, exterior) TracePair[source]¶
Returns a trace pair defined on the exterior boundary. The interior argument is assumed to be defined on the volume discretization, and will therefore be restricted to the boundary dd prior to
creating a TracePair. If the input arguments interior and exterior are ArrayContainer objects, they must both have the same internal structure.
☆ dd – a DOFDesc, or a value convertible to one, which describes the boundary discretization.
☆ interior – a DOFArray or an ArrayContainer that contains data defined in the volume, which will be restricted to the boundary denoted by dd. The result will be used as the interior
value for the flux.
☆ exterior – a DOFArray or an ArrayContainer that contains data that already lives on the boundary representing the exterior value to be used for the flux.
a TracePair on the boundary.
Interior and cross-rank trace functions¶
grudge.op.interior_trace_pairs(dcoll: DiscretizationCollection, vec, *, comm_tag: Hashable | None = None, volume_dd: DOFDesc | None = None) List[TracePair][source]¶
Return a list of TracePair objects defined on the interior faces of dcoll and any faces connected to a parallel boundary.
Note that local_interior_trace_pair() provides the rank-local contributions if those are needed in isolation. Similarly, cross_rank_trace_pairs() provides only the trace pairs defined on
cross-rank boundaries.
☆ vec – a DOFArray or an ArrayContainer of them.
☆ comm_tag – a hashable object used to match sent and received data across ranks. Communication will only match if both endpoints specify objects that compare equal. A generalization of
MPI communication tags to arbitrary, potentially composite objects.
a list of TracePair objects.
grudge.op.local_interior_trace_pair(dcoll: DiscretizationCollection, vec, *, volume_dd: DOFDesc | None = None) TracePair[source]¶
Return a TracePair for the interior faces of dcoll with a discretization tag specified by discr_tag. This does not include interior faces on different MPI ranks.
vec – a DOFArray or an ArrayContainer of them.
For certain applications, it may be useful to distinguish between rank-local and cross-rank trace pairs. For example, avoiding unnecessary communication of derived quantities (i.e. temperature)
on partition boundaries by computing them directly. Having the ability for user applications to distinguish between rank-local and cross-rank contributions can also help enable overlapping
communication with computation. :returns: a TracePair object.
grudge.op.cross_rank_trace_pairs(dcoll: DiscretizationCollection, ary: Array | ArrayContainer, tag: Hashable = None, *, comm_tag: Hashable = None, volume_dd: DOFDesc | None = None) List[TracePair]
Get a list of ary trace pairs for each partition boundary.
For each partition boundary, the field data values in ary are communicated to/from the neighboring partition. Presumably, this communication is MPI (but strictly speaking, may not be, and this
routine is agnostic to the underlying communication).
For each face on each partition boundary, a TracePair is created with the locally, and remotely owned partition boundary face data as the internal, and external components, respectively. Each of
the TracePair components are structured like ary.
If ary is a number, rather than a DOFArray or an ArrayContainer of them, it is assumed that the same number is being communicated on every rank.
☆ ary – a DOFArray or an ArrayContainer of them.
☆ comm_tag – a hashable object used to match sent and received data across ranks. Communication will only match if both endpoints specify objects that compare equal. A generalization of
MPI communication tags to arbitrary, potentially composite objects.
a list of TracePair objects.
Transfering data between discretizations¶
Nodal Reductions¶
In a distributed-memory setting, these reductions automatically reduce over all ranks involved, and return the same value on all ranks, in the manner of an MPI allreduce.
Rank-local reductions¶
grudge.op.nodal_sum_loc(dcoll: DiscretizationCollection, dd, vec) int | float | complex | generic[source]¶
Return the rank-local nodal sum of a vector of degrees of freedom vec.
a scalar denoting the rank-local nodal sum.
grudge.op.nodal_min_loc(dcoll: DiscretizationCollection, dd, vec, *, initial=None) int | float | complex | generic[source]¶
Return the rank-local nodal minimum of a vector of degrees of freedom vec.
☆ dd – a DOFDesc, or a value convertible to one.
☆ vec – a DOFArray or an ArrayContainer of them.
☆ initial – an optional initial value. Defaults to numpy.inf.
a scalar denoting the rank-local nodal minimum.
grudge.op.nodal_max_loc(dcoll: DiscretizationCollection, dd, vec, *, initial=None) int | float | complex | generic[source]¶
Return the rank-local nodal maximum of a vector of degrees of freedom vec.
☆ dd – a DOFDesc, or a value convertible to one.
☆ vec – a DOFArray or an ArrayContainer.
☆ initial – an optional initial value. Defaults to -numpy.inf.
a scalar denoting the rank-local nodal maximum.
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Estimate Powers and Roots
Estimate the powers and roots of the given positive numbers.
Mental Methods
Did you know that there is a quick way of squaring a two digit number which ends in .5?
Just multiply the first digit by that digit plus one then add a 0.25 to the result.
For example What is 8.5 squared?
8 x 9 + 0.25 = 72.25.
You may also want to use a calculator to check your working. See Calculator Workout skill 6.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a
teacher, tutor or parent.
This video is from Khan Academy.
You may also want to use a calculator to check your working. See Calculator Workout skill 5.
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can
double-click the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a
teacher, tutor or parent.
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Area and Hausdorff Dimension of Julia Sets of Entire Functions
• Published in 1987
• Added on
We show the Julia set of $\lambda \sin(z)$ has positive area and the action of $\lambda \sin(z)$ on its Julia set is not ergodic; the Julia set of $\lambda \exp(z)$ has Hausdorff dimension two
but in the presence of an attracting periodic cycle its area is zero.
Other information
BibTeX entry
key = {item58},
type = {misc},
title = {Area and Hausdorff Dimension of Julia Sets of Entire Functions},
author = {Curt McMullen},
abstract = {We show the Julia set of {\$}\lambda \sin(z){\$} has positive area and the action of {\$}\lambda \sin(z){\$} on its Julia set is not ergodic; the Julia set of {\$}\lambda \exp(z){\$} has Hausdorff dimension two but in the presence of an attracting periodic cycle its area is zero.},
comment = {},
date_added = {2015-12-17},
date_published = {1987-10-09},
urls = {http://www.math.harvard.edu/{\~{}}ctm/papers/home/text/papers/entire/entire.pdf},
collections = {},
url = {http://www.math.harvard.edu/{\~{}}ctm/papers/home/text/papers/entire/entire.pdf},
urldate = {2015-12-17},
year = 1987
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How to Find Largest Number in Excel (2 Easy Ways) - ExcelDemy
We’ll use a large dataset and determine the largest number from it. The image shows the overview of the functions we’ll use.
How to Find the Largest Number in Excel: 2 Ways
We have a concise dataset that contains 13 rows and 4 columns of Rep Name, Item, Units, and Unit Cost.
Method 1 – Use Excel Functions to Find the Largest Number Within a Range in Excel
Case 1 – Using the MAX Function
We’ll find the largest value in the Units column.
• Select your preferred cell (i.e. D18) to have your output.
• Insert the following formula:
D5:D16 is the range of values of the Units column.
• Press the Enter key and the result will be shown in cell D18.
We got the Max Unit that contains the largest value of the specified range.
Another way of finding the largest value in a dataset is using the AutoSum feature.
Alternative steps:
• Select the range you want to check (D5:D17).
• Select the Formulas tab.
• Click AutoSum.
• Select Max from the drop-down.
D5:D17 is the range of values of the Units column.
• Press the Enter key and you will get your result in cell D18.
Case 2 – Applying the LARGE Function
• Select cell D18 to show the output.
• Insert the following formula in cell D18 to find the largest value.
D5:D16 is the array or range of values of the Units column and 1 is the k value which represents the position of data you want to get. So, 1 means the first largest value.
• Press the Enter key. You will get your result in cell D18.
Read More: How to Use Excel Large Function in Multiple Ranges
Case 3 – Using the AGGREGATE Function
We want to know the Max Unit Cost of the Unit Cost column.
• Select cell D18 to show output.
• Insert the following formula:
Here, 4 indicates that we want to apply the MAX function to get the highest value, 7 indicates that we are ignoring hidden rows and error values, and E5:E16 is the array range of the Unit Cost
• Press the Enter key. The output will be shown in cell D18.
Method 2 – Finding the Largest Number Within a Range Based on Criteria
Case 1 – Calculating the Maximum Value by Using the MAX Function
In this dataset, the Pencil was sold in different units, and we want to get the highest such value.
• Select cell B19 to enter the criterion which is Pencil.
• Select cell D19 to show output.
• Insert the following formula in cell D19 to find the largest value.
C5:C16 is the range of the Item column, D5:D16 is the range of the Units column, and B19 is the criterion.
Formula Breakdown
• B19 → Pencil is the criterion located in cell B19.
• MAX((C5:C16=‘Pencil’)*(D5:D16)) → becomes
□ MAX(({“Marker Pen”;“Pencil”;“Pen”;“Blinder”;“Pencil”;“Marker Pen”;“Pencil”;“Blinder”;“Desk”;“Eraser”;“Blinder”;“Pen”}=“Pencil”)*(D5:D16)) → returns TRUE for the exact match Pencil and
otherwise returns FALSE.
☆ Output → MAX(({FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE})*(D5:D16))
• MAX({FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE}*(D5:D16))
• MAX({FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE}*{53;56;57;59;83;71;60;53;70;96;88;68}) → returns 0 for FALSE.
□ Output → MAX({0;56;0;0;83;0;60;0;0;0;0;0})
• Press the Enter key. The output will be shown in cell D19.
Case 2 – Using a Combination of MAX and IF Functions
• Select cell B19 to enter the criteria which is Pencil.
• Select cell D19 to get output.
• Insert the following formula:
C5:C16 is the range of the Item column, D5:D16 is the range of the Units column, and B19 is the criterion.
Formula Breakdown
• B19 → Pencil is the criterion located in cell B19.
• MAX(IF(C5:C16=B19,D5:D16)) → becomes
□ MAX(IF({“Marker Pen”;“Pencil”;“Pen”;“Blinder”;“Pencil”;“Marker Pen”;“Pencil”;“Blinder”;“Desk”;“Eraser”;“Blinder”;“Pen”}=“Pencil”,D5:D16)) → returns TRUE for the exact match Pencil and
otherwise returns FALSE.
• MAX(IF({FALSE; TRUE; FALSE; FALSE; TRUE; FALSE; TRUE; FALSE; FALSE; FALSE; FALSE; FALSE}, D5:D16)) → The IF function will give us the numeric value for TRUE.
□ Output → MAX({FALSE;56;FALSE;FALSE;83;FALSE;60;FALSE;FALSE;FALSE;FALSE;FALSE})
• MAX({0;56;0;0;83;0;60;0;0;0;0;0})
• Press the Enter key. The output will be shown in cell D19.
Read More: How to Use Excel LARGE Function with Criteria
How to Find the Position of the Largest Number in Excel
• Insert the following formula in cell G11 to find the cell address of the maximum value, then press the Enter key.
D5:D16 is the array or range of values of the Units column and 4 indicates that there are four extra rows before starting the values.
Formula Breakdown
• MAX(D5:D16) → returns the maximum value of the range D5:D16.
• ADDRESS(MATCH(MAX(D5:D16),D5:D16,0)+4,4) → becomes
□ Output → ADDRESS(MATCH(96,D5:D16,0)+4,4)
• MATCH(96, D5:D16,0) → The MATCH function returns the maximum exact match value (96) from range D5:D16. 0 is set to return the exact match.
□ MATCH(96,D5:D16,0) → becomes
• ADDRESS(10+4,4) → The ADDRESS function creates a cell reference based on a given column and row number.
□ ADDRESS(10+4,4) → becomes
We got the maximum value of the Units column which is 96, and its cell address which is $D$14 by applying MAX, MATCH, and ADDRESS functions.
Practice Section
We’re providing the practice dataset so you can test these methods.
Download the Practice Workbook
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Jason Parsley : Helicity, Configuration Spaces, & Characteristic Classes
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Jason Parsley : Helicity, Configuration Spaces, & Characteristic Classes
The helicity of a vector field in R^3, an analog to linking number, measures the extent to which its flowlines coil and wrap around one another. Helicity turns out to be invariant under
volume-preserving diffeomorphisms that are isotopic to the identity. Motivated by Bott-Taubes integration, we provide a new proof of this invariance using configuration spaces. We then present a new
topological explanation for helicity, as a characteristic class. Among other results, this point of view allows us to completely characterize the diffeomorphisms under which helicity is invariant and
give an explicit formula for the change in helicity under a diffeomorphism under which helicity is not invariant. (joint work with Jason Cantarella, U. of Georgia)
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HP-50g How to store many equations - help needed
Hi all,
sorry, I haven't been here in ages because my new job and going to business school doesn't leave much time for anything else.
I need your help. My exam is on thursday and the 17bII+ I've stored my equations in died. I have a 50g I wanted to use in next year's exam, but now it seems I have to use it next week already. And I
know almost nothing about that device.
I need to input all my equations quickly. To do so, I can either use TreeBrowserBuilder or Pequm but have difficulties with both programs.
TBB lets me create a structure and equations once and I can start TB and solve the equations. However, every time I exit TB the calculator freezes. The same thing happens when using TBED or TBNEW.
Pequm works, but equations are always solved for X and I often need the result of an equation, later. So if I for example solve equations for DBII and DBIII, with TBB I could later use those vars to
calculate DBIII-DBII.
With Pequm instead of DBII and DBIII I'd get X and X and can't do further calculations with those variables. Is that true?
Which of those two problems is easier to solve? And how could I solve it?
Thanks a lot.
Timo, now slightly less panicking because being in good hands
03-03-2013, 10:14 AM
Don't quite understand what your issue is with PEQUM.
I just looked at it for the first time in about a year or two. I installed the latest version. I made an equation called TEST that had the equation A^2+B+C=D.
When I solved it, it first popped up the equation in case any changes are needed, then brought up a solver where I could solve for A, B C or D.
When you exit, it asks if you want to leave the stack values or not.
If you leave them, they are sitting there labeled on the stack. If you don't, then you just open the EQUAT directory and all the variables are there.
If all you can solve is for X, is that because you only have X in your forumula???
That being said, the 50g may be a shotgun to kill mosquitos. It will do whatever you need, but it is very difficult to learn and get comfortable with in such a short amount of time. I'd recommend
taking a look at the new quickstart guide which I made a few years back. It should help quite a bit I think.
03-03-2013, 12:37 PM
Thank you very much, Tim. It's working now. The problem wasn't "X" - I tried to solve for a variable I've called "L1" and the 50g doesn't seem to like that. When I change the variables' names from
"L1", "L2" and so on to "X" or to "Lone", "Ltwo", etc. it will work.
I wouldn't have noticed if I hadn't entered your A^2+B+C=D. So, thank you very much, again.
03-03-2013, 10:28 AM
I've emailed you the latest updates and will put them on my website at a later time.
Edited: 3 Mar 2013, 10:28 a.m.
03-03-2013, 10:53 AM
Don't waste your precious time of only one week on a new calculator system. Go out and buy the same that you are used to as a replacement, so you won't suffer any problems during the exams.
03-03-2013, 12:22 PM
I agree. However, I'd have to order it online and the risk of it not getting here in time is too big. There is no shop anywhere around that sells HP calculators.
On the other hand, it takes ages to input equations with the 17bII+. Andreas just wrote me about a way how to do it fast on the PC, so I might not lose that much time just using the 50g.
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BIP124 - Hierarchical Deterministic Script Templates
BIP: 124 source
Layer: Applications
Title: Hierarchical Deterministic Script Templates
Author: Eric Lombrozo <eric@ciphrex.com>
William Swanson <swansontec@gmail.com>
Comments-Summary: No comments yet.
Comments-URI: https://github.com/bitcoin/bips/wiki/Comments:BIP-0124
Status: Rejected
Type: Informational
Created: 2015-11-20
License: PD
Post-History: http://lists.linuxfoundation.org/pipermail/bitcoin-dev/2015-November/011795.html
This BIP defines a script template format that can be used by wallets to deterministically generate scripts with specific authorization policies using the key derivation mechanism defined in BIP32.
Currently existing wallets typically issue scripts in only a tiny handful of widely used formats. The most popular formats are pay-to-pubkey-hash and m-of-n pay-to-script-hash (BIP16). However,
different wallets tend to use mutually incompatible derivation schemes to generate signing keys and construct scripts from them. Moreover, with the advent of hashlocked and timelocked contracts
(BIP65, BIP112), it is necessary for different wallets to be able to cooperatively generate even more sophisticated scripts.
In addition, there's a lot of ongoing work in the development of multilayered protocols that use the blockchain as a settlement layer (i.e. the Lightning Network). These efforts require sufficiently
generalized templates to allow for rapidly evolving script designs.
This BIP provides a generalized format for constructing a script template that guarantees that different wallets will all produce the same scripts for a given set of derivation paths according to
An individual key is determined by a BIP32 derivation path and an index. For convenience, we introduce the following notation:
A[k] = (derivation path for A)/k
Key Groups
Let m[i] denote distinct BIP32 derivation paths. We define a key group of n keys as a set of key derivation paths with a free index k:
{K[k]} = { m[1]/k, m[2]/k, m[3]/k, ..., m[n]/k }
Key groups are useful for constructing scripts that are symmetric in a set of keys. Scripts are symmetric in a set of keys if the semantics of the script is unaffected by permutations of the keys.
Key groups enforce a canonical form and can improve privacy.
We define a lexicographic sorting of the keys. (TODO: specification of sorting conventions - compressed pubkeys, encoding, etc...)
Define {K[k]}[j] to be the jth element of the sorted keys for derivation index k.
Script Templates
We construct script templates by inserting placeholders for data into a script. To denote a placeholder, we use the following notation:
Script(A) = opcodes [A] opcodes
We extend this notation to an arbitrary number of placeholders:
Script(X1, X2, ..., Xn) = opcodes [X1] opcodes [X2] opcodes ... opcodes [Xn] opcodes
We introduce the following convenient notation for sorted key groups:
[{K[k]}] = [{K[k]}[1]] [{K[k]}[2]] ... [{K[k]}[n]]
Operations on Keys
In some applications, we might want to insert the result of some operation performed on a key rather than the key itself into the script. For example, we might want to insert a Hash160 of key A[k].
We can use the following notation:
2-of-3 Multisig
The script template is defined by:
Script(X) = 2 [X] 3 OP_CHECKMULTISIG
Letting K[k] = { m[1]/k, m[2]/k, m[3]/k }, the kth script for this key group is denoted by Script({K[k]}).
1-of-1 or 2-of-3
The script template is defined by:
Script(A, B) =
OP_DUP [A] OP_CHECKSIG
2 [B] 3 OP_CHECKMULTISIGVERIFY
Let M[k] = m/k be a key of a superuser that can authorize all transactions and {K[k]} be a key group of three users that can only authorize transactions if at least two of them agree.
The kth script is given by Script(M[k], {K[k]}).
Timelocked Contract
The output is payable to Alice immediately if she knows the private key for A[k]. Bob must know the private key for B[k] and also wait for a timeout t before being able to spend the output.
The script template is defined by:
Script(A, B, T) =
OP_DUP OP_HASH160 [Hash160(A)] OP_EQUALVERIFY OP_CHECKSIG
OP_DUP OP_HASH160 [Hash160(B)] OP_EQUALVERIFY OP_CHECKSIG
The kth script with timeout t is given by Script(A[k], B[k], t).
This document is placed in the public domain.
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Adding And Subtracting Negative Numbers Printable Worksheets 2024 - NumbersWorksheets.net
Adding And Subtracting Negative Numbers Printable Worksheets
Adding And Subtracting Negative Numbers Printable Worksheets – The Unfavorable Numbers Worksheet is a great way to start educating your children the very idea of bad phone numbers. A negative number
is any amount that may be under absolutely nothing. It can be extra or subtracted. The minus sign suggests the bad number. You can even compose negative numbers in parentheses. Below can be a
worksheet to help you began. This worksheet has an array of adverse figures from -10 to 10. Adding And Subtracting Negative Numbers Printable Worksheets.
Bad numbers are quite a lot whose benefit is less than absolutely nothing
A negative variety has a value less than zero. It can be depicted over a number range in two ways: with all the positive quantity published as the initial digit, and with the unfavorable variety
written since the previous digit. A positive amount is published having a plus indication ( ) just before it, yet it is optionally available to write it doing this. It is assumed to be a positive
number if the number is not written with a plus sign.
These are depicted with a minus signal
In old Greece, negative numbers were actually not applied. They were disregarded, as their math was based upon geometrical principles. When Western scholars started translating ancient Arabic texts
from Northern Africa, they got to recognize adverse numbers and adopted them. Nowadays, negative figures are depicted by way of a minus sign. For more information on the origins and history of
unfavorable numbers, check this out write-up. Then, try these cases to view how adverse figures have developed over time.
They can be added or subtracted
Positive numbers and negative numbers are easy to add and subtract because the sign of the numbers is the same, as you might already know. Negative numbers, on the other hand, have a larger absolute
value, but they are closer to than positive numbers are. They can still be added and subtracted just like positive ones, although these numbers have some special rules for arithmetic. You can even
subtract and add unfavorable numbers utilizing a variety range and apply the same rules for addition and subtraction when you do for beneficial figures.
They can be symbolized by a amount in parentheses
A poor variety is displayed by way of a amount covered in parentheses. The negative signal is transformed into its binary equal, along with the two’s enhance is held in the same spot in recollection.
The result is always negative, but sometimes a negative number is represented by a positive number. When this happens, the parentheses must be included. You should consult a book on math if you have
any questions about the meaning of negative numbers.
They can be divided up by way of a optimistic variety
Adverse numbers may be multiplied and divided like good phone numbers. They can be split by other negative phone numbers. They are not equal to one another, however. At the first try you increase a
negative quantity by way of a beneficial amount, you will definately get absolutely nothing as a result. To produce the solution, you must select which signal your solution should have. It is
actually easier to bear in mind a negative amount when it is printed in mounting brackets.
Gallery of Adding And Subtracting Negative Numbers Printable Worksheets
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5.3 Agglomerative Clustering | An Introduction to Spatial Data Science with GeoDa
5.3 Agglomerative Clustering
An agglomerative clustering algorithm starts with each observation serving as its own cluster, i.e., beginning with \(n\) clusters of size 1. Next, the algorithm moves through a sequence of steps,
where each time the number of clusters is decreased by one, either by creating a new cluster by joining two individual observations, by assigning an observation to an existing cluster, or by merging
two clusters. Such algorithms are sometimes referred to as SAHN, which stands for sequential, agglomerative, hierarchic and non-overlapping (Müllner 2011).
The sequence of merging observations into clusters is graphically represented by means of a tree structure, the so-call dendrogram (see Section 5.3.2). At the bottom of the tree, the individual
observations constitute the leaves, whereas the root of the tree is the single cluster that consists of all observations.
The smallest within-group sum of squares is obtained in the initial stage, when each observation is its own cluster. As a result, the within sum of squares is zero and the between sum of squares is
at its maximum, which also equals the total sum of squares. As soon as two observations are grouped, the within sum of squares increases. Hence, each time a new merger is carried out, the overall
objective of minimizing the within sum of squares deteriorates. At the final stage, when all observations are joined into a single cluster, the total within sum of squares now also equals the total
sum of squares, since there is no between sum of squares (the two are complementary).
In other words, in an agglomerative hierarchical clustering procedure, the objective function gets worse at each step. It is therefore not optimized as such, but instead can be evaluated at each
One distinguishing characteristic of hierarchical clustering is that once an observation is grouped with other observations, it cannot be disassociated from them in a later step. This precludes
swapping of observations between clusters, which is a characteristic of some of the partitioning methods. This property of getting trapped into a cluster (i.e., into a branch of the dendrogram) can
be limiting in some contexts.
5.3.1 Linkage and Updating Formula
A key aspect in the agglomerative process is how to define the distance between clusters, or between a single observation and a cluster. This is referred to as the linkage. There are at least seven
different concepts of linkage, but here only the four most common ones are considered: single linkage, complete linkage, average linkage, and Ward’s method.
A second important concept is how the distances between other points (or clusters) and a newly merged cluster are computed, the so-called updating formula. With some clever algebra, it can be shown
that these calculations can be based on the dissimilarity matrix from the previous step. The update thus does not require going back to the original \(n \times n\) dissimilarity matrix.^22 Moreover,
at each step, the dimension of the relevant dissimilarity matrix decreases by one, which allows for very memory-efficient algorithms.
Each linkage type and its associated updating formula is briefly considered in turn.
5.3.1.1 Single linkage
For single linkage, the relevant dissimilarity is between the two points in each cluster that are closest together. More precisely, the dissimilarity between clusters \(A\) and \(B\) is: \[d_{AB} = \
mbox{min}_{i \in A,j \in B} d_{ij},\] The updating formula yields the dissimilarity between a point (or cluster) \(P\) and a cluster \(C\) that was obtained by merging \(A\) and \(B\). It is the
smallest of the dissimilarities between \(P\) and \(A\) and \(P\) and \(B\): \[d_{PC} = \mbox{min}(d_{PA},d_{PB}).\] The minimum condition can also be obtained as the result of an algebraic
expression, which yields the updating formula as: \[d_{PC} = (1/2) (d_{PA} + d_{PB}) - (1/2)| d_{PA} - d_{PB} |,\] in the same notation as before.^23
The updating formula only affects the row/column in the dissimilarity matrix that pertains to the newly merged cluster. The other elements of the dissimilarity matrix remain unchanged.
Single linkage clusters tend to result in a few clusters consisting of long drawn out chains of observations, in combination with several singletons (observations that form their own cluster). This
is due to the fact that disparate clusters may be joined when they have two close points, but otherwise are far apart. Single linkage is sometimes used to detect outliers, i.e., observations that
remain singletons and thus are far apart from all others.
5.3.1.2 Complete linkage
Complete linkage is the opposite of single linkage in that the dissimilarity between two clusters is defined as the farthest neighbors, i.e., the pair of points, one from each cluster, that are
separated by the greatest dissimilarity. For the dissimilarity between clusters \(A\) and \(B\), this boils down to: \[d_{AB} = \mbox{max}_{i \in A,j \in B} d_{ij}.\] The updating formula is the
opposite of the one for single linkage. The dissimilarity between a point (or cluster) \(P\) and a cluster \(C\) that was obtained by merging \(A\) and \(B\) is the largest of the dissimilarities
between \(P\) and \(A\) and \(P\) and \(B\): \[d_{PC} = \mbox{max}(d_{PA},d_{PB}).\]
The algebraic counterpart of the updating formula is: \[d_{PC} = (1/2) (d_{PA} + d_{PB}) + (1/2)| d_{PA} - d_{PB} |,\] using the same logic as in the single linkage case.
In contrast to single linkage, complete linkage tends to result in a large number of well-balanced compact clusters. Instead of merging fairly disparate clusters that have (only) two close points, it
can have the opposite effect of keeping similar observations in separate clusters.
5.3.1.3 Average linkage
In average linkage, the dissimilarity between two clusters is the average of all pairwise dissimilarities between observations \(i\) in cluster \(A\) and \(j\) in cluster \(B\). There are \(n_A.n_B\)
such pairs (only counting each pair once), with \(n_A\) and \(n_B\) as the number of observations in each cluster. Consequently, the dissimilarity between \(A\) and \(B\) is (without double counting
pairs in the numerator): \[d_{AB} = \frac{\sum_{i \in A} \sum_{j \in B} d_{ij}}{n_A.n_B}.\] In the special case when two single observations are merged, \(d_{AB}\) is simply the dissimilarity between
the two, since \(n_A = n_B = 1\) and thus the denominator in the expression is 1.
The updating formula to compute the dissimilarity between a point (or cluster) \(P\) and the new cluster \(C\) formed by merging \(A\) and \(B\) is the weighted average of the dissimilarities \(d_
{PA}\) and \(d_{PB}\): \[d_{PC} = \frac{n_A}{n_A + n_B} d_{PA} + \frac{n_B}{n_A + n_B} d_{PB}.\] As before, the other distances are not affected.^24
Average linkage can be viewed as a compromise between the nearest neighbor logic of single linkage and the furthest neighbor logic of complete linkage.
5.3.1.4 Ward’s method
The three linkage methods discussed so far only make use of a dissimilarity matrix. How this matrix is obtained does not matter. As a result, dissimilarity may be defined using Euclidean or Manhattan
distance, dissimilarity among categories, or even directly from interview or survey data.
In contrast, the method developed by Ward (1963) is based on a sum of squared errors rationale that only works for Euclidean distance between observations. In addition, the sum of squared errors
requires the consideration of the so-called centroid of each cluster, i.e., the mean vector of the observations belonging to the cluster. Therefore, the input into Ward’s method is not a
dissimilarity matrix, but a \(n \times p\) matrix \(X\) of \(n\) observations on \(p\) variables (as before, this is typically standardized in some fashion).
Ward’s method is based on the objective of minimizing the deterioration in the overall within sum of squares. The latter is the sum of squared deviations between the observations in a cluster and the
centroid (mean): \[WSS = \sum_{i \in C} (x_i - \bar{x}_C)^2,\]
with \(\bar{x}_C\) as the centroid of cluster \(C\).
Since any merger of two existing clusters (including the merger of individual observations) results in a worsening of the overall WSS, Ward’s method is designed to minimize this deterioration. More
specifically, it is designed to minimize the difference between the new (larger) WSS in the merged cluster and the sum of the WSS of the components that were merged. This turns out to boil down to
minimizing the distance between cluster centers.^25 Without loss of generality, it is easier to express the dissimilarity in terms of the square of the Euclidean distance: \[d_{AB}^2 = \frac{2n_A
n_B}{n_A + n_B} ||\bar{x}_A - \bar{x}_B ||^2,\] where \(||\bar{x}_A - \bar{x}_B ||\) is the Euclidean distance between the two cluster centers (squared in the distance squared expression).^26
The update equation to compute the (squared) distance from an observation (or cluster) \(P\) to a new cluster \(C\) obtained from the merger of \(A\) and \(B\) is more complex than for the other
linkage options: \[d^2_{PC} = \frac{n_A + n_P}{n_C + n_P} d^2_{PA} + \frac{n_B + n_P}{n_C + n_P} d^2_{PB} - \frac{n_P}{n_C + n_P} d^2_{AB},\] in the same notation as before. However, it can still
readily be obtained from the information contained in the dissimilarity matrix from the previous step, and it does not involve the actual computation of centroids.
To see why this is the case, consider the usual first step when two single observations are merged. The distance squared between them is simply the Euclidean distance squared between their values,
not involving any centroids. The updated squared distances between other points and the two merged points only involve the point-to-point squared distances \(d^2_{PA}\), \(d^2_{PB}\) and \(d^2_{AB}\)
, no centroids. From then on, any update uses the results from the previous distance matrix in the update equation.
5.3.1.5 Illustration - single linkage
To illustrate the logic of agglomerative hierarchical clustering algorithms, the single linkage approach is applied to a toy example consisting of the coordinates of seven points, shown in Figure 5.2
. The corresponding X, Y values are listed in Figure 5.3. The point IDs are ordered with increasing values of X first, then Y, starting with observation 1 in the lower left corner.
The basis for any agglomerative clustering method is a \(n \times n\) symmetric dissimilarity matrix. Except for Ward’s method, this is the only information needed. The dissimilarity matrix derived
from the Euclidean distances between the points is shown in Figure 5.4. Since the matrix is symmetric, only the upper diagonal elements are listed.
The first step in the algorithm is to identify the pair of observations that have the smallest nearest neighbor distance. In the distance matrix, this is the row-column combination with the smallest
entry. In Figure 5.4 this is readily identified as the pair 4-5 (\(d_{4,5}=1.0\)). This pair therefore forms the first cluster, connected by a link in Figure 5.5. The two points in the cluster are
highlighted in dark blue. The five other observations remain in their initial separate cluster. In other words, at this stage, there are six clusters, one less than the number of observations.
The dissimilarity matrix is updated using the smallest dissimilarity between each observation and either observation 4 or observation 5. This yields the updated row and column entries for the
combined unit 4,5. More precisely, the dissimilarity used between the cluster and the other observations varies depending on whether observation 4 or 5 is closest to the other observations. For
example, in Figure 5.6, the dissimilarity between 4,5 and 1 is given as 5.0, which is the smallest of 1-4 (5.0) and 1-5 (5.83). The dissimilarities between the pairs of observations that do not
involve 4,5 are not affected.
The other entries for 4,5 are updated in the same way, and again the smallest dissimilarity is located in the matrix. This time, it is a dissimilarity of 2.0 between 4,5 and 7 (more precisely,
between the closest pair 5 and 7). Consequently, observation 7 is added to the 4,5 cluster.
The dissimilarities between 4,5,7 and the other points are updated in Figure 5.7. However, now there is a problem. There is a three-way tie in terms of the smallest value: 1-2, 4,5,7-3 and 4,5,7-6
all have a dissimilarity of 2.24, but only one can be picked to update the clusters. Ties are typically handled by choosing one grouping at random. In the example, the pair 1-2 is selected, which is
how the tie is broken by the algorithm used in GeoDa.^27
With the distances updated, not unsurprisingly, 2.24 is again found as the shortest dissimilarity, tied for two pairs (in Figure 5.8). This time the algorithm adds 3 to the existing cluster 4,5,7.
Finally, observation 6 is added to cluster 4,5,7,3 again for a dissimilarity of 2.24 (in Figure 5.9).
The end result is to merge the two clusters 1-2 and 4,5,7,3,6 into a single one, which ends the iterations (Figure 5.10).
In sum, the algorithm moves sequentially to identify the nearest neighbor at each step, merges the relevant observations/clusters and so decreases the number of clusters by one. The sequence of steps
is illustrated in the panels of Figure 5.11, going from left to right and starting at the top.
5.3.2 Dendrogram
While the visual representation of the sequential grouping of observations in Figure 5.11 works well in this toy example, it is not practical for larger data sets.
A tree structure that visualizes the agglomerative nesting of observations into clusters is the so-called dendrogram. For each step in the process, the graph shows which observations/clusters are
combined. In addition, the degree of change in the objective function achieved by each merger is visualized by a corresponding distance on the horizontal (or vertical) axis.
The implementation of the dendrogram in GeoDa is currently somewhat limited, but it accomplishes the main goal. In Figure 5.12, the dendrogram is illustrated for the single linkage method in the toy
example. The graph shows how the cluster starts by combining two observations (4 and 5), to which then a third (7) is added. These first two steps are contained inside the highlighted black square in
the figure. The corresponding observations are selected as entries in a matching data table.
Next, following the tree structure reveals how two more observations (1 and 2) are combined into a separate cluster, and two observations (3 and 6) are added to the original cluster of 4,5 and 7.
Given the three-way tie in the inter-group distances, the last three operations all line up (same distance from the right side) in the graph. As a result, the change in the objective function (more
precisely, a deterioration) that follows from adding the points to a cluster is the same in each case.
The dashed vertical line represents a cut line. It corresponds with a particular value of k for which the make up of the clusters and their characteristics can be further investigated. As the cut
line is moved, the members of each cluster are revealed that correspond with a different value for \(k\).
In practice, important cluster characteristics are computed for each of the selected cut points, such as the sum of squares, the total within sum of squares, the total between sum of squares, and the
ratio of the the total between sum of squares to the total sum of squares (the higher ratio, the better). This will be further illustrated as part of the discussion of implementation issues in the
next section.
22. Detailed proofs for all the properties are contained in Chapter 5 of Kaufman and Rousseeuw (2005).↩︎
23. To see that this holds, consider the situation when \(d_{PA} < d_{PB}\), i.e., \(A\) is the nearest neighbor to \(P\). As a result, the absolute value of \(d_{PA} - d_{PB}\) is \(d_{PB} - d_{PA}
\). Then the expression becomes \((1/2) d_{PA} + (1/2) d_{PB} - (1/2) d_{PB} + (1/2) d_{PA} = d_{PA}\), the desired result.↩︎
24. By convention, the diagonal dissimilarity for the newly merged cluster is set to zero.↩︎
25. See Kaufman and Rousseeuw (2005), Chapter 5, for detailed proofs.↩︎
26. The factor 2 is included to make sure the expression works when two single observations are merged. In such an instance, their centroid is their actual value and \(n_A + n_B = 2\). It does not
matter in terms of the algorithm steps.↩︎
27. An implementation of the fastcluster algorithm.↩︎
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How to Use Venn Diagram to Find the Relation of Sets - Bigsigma Math Tutorials
How to Use Venn Diagram to Find the Relation of Sets
A Venn diagram is diagram that shows possible logical relations between finite collection of different sets. The most common Venn diagram contains 2 sets or 3 sets.
Venn diagram of 2 sets
In Venn diagram of 2 sets A,B,C there are 2^2=4 areas. Each area in the Venn diagram represents a distinct possible logical relation of an item x to one of the 2 sets A,B:
• x only belongs to A, x does not belong to B
• x only belongs to B, x does not belong to A
• x belongs to all 2 sets : A,B
• x does not any set.
Venn diagram of 3 sets
In Venn diagram of 3 sets A,B,C there are 2^3=8 areas. Each area in the Venn diagram represents a distinct possible logical relation of an item x to one of the 3 sets A,B,C:
• x only belongs to A, x does not belong to B and C
• x only belongs to B, x does not belong to A and C
• x only belongs to C, x does not belong to A and B
• x only belongs to two sets : A,B , x does not belong to C
• x only belongs to two sets : B,C , x does not belong to A
• x only belongs to two sets : A,C , x does not belong to B
• x belongs to all 3 sets : A,B.C
• x does not any set.
Find the Relation of Sets with Venn Diagram
In the next movie, we see how can we use Venn diagram to find the relation of sets.
For example, Suppose we have 2 sets A and B, with Venn diagram we can find:
• Whether the A is equal to B
• Whether the A is superset of B
• Whether the A is subset of B
• None of the above (In other words A is not equal to B, A is not superset of B and A is not subset of B).
As we can see in the movie, we represent each set in distinct Venn diagram.
Now, when the 2 Venn Diagram are displayed side by side, you can easily see the relation between the sets.
Leave a Comment
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Conversion of Solids from One Shape to Another Worksheet
(1) An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder. Solution
(2) Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks
will rise by 21 cm.
(3) A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius xr units. Find the height of water in the
cylindrical flask Solution
(4) A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
(5) Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m x 1.5
m x 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which
has been full, initially. Solution
(6) The internal and external diameter of a hollow hemispherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of
the cylinder. Solution
(7) A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness
of the cylinder. Solution
(8) A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the
cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel. Solution
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
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Keras documentation: RandomZoom layer
RandomZoom layer
RandomZoom class
A preprocessing layer which randomly zooms images during training.
This layer will randomly zoom in or out on each axis of an image independently, filling empty space according to fill_mode.
Input pixel values can be of any range (e.g. [0., 1.) or [0, 255]) and of integer or floating point dtype. By default, the layer will output floats.
For an overview and full list of preprocessing layers, see the preprocessing guide.
• height_factor: a float represented as fraction of value, or a tuple of size 2 representing lower and upper bound for zooming vertically. When represented as a single float, this value is used for
both the upper and lower bound. A positive value means zooming out, while a negative value means zooming in. For instance, height_factor=(0.2, 0.3) result in an output zoomed out by a random
amount in the range [+20%, +30%]. height_factor=(-0.3, -0.2) result in an output zoomed in by a random amount in the range [+20%, +30%].
• width_factor: a float represented as fraction of value, or a tuple of size 2 representing lower and upper bound for zooming horizontally. When represented as a single float, this value is used
for both the upper and lower bound. For instance, width_factor=(0.2, 0.3) result in an output zooming out between 20% to 30%. width_factor=(-0.3, -0.2) result in an output zooming in between 20%
to 30%. None means i.e., zooming vertical and horizontal directions by preserving the aspect ratio. Defaults to None.
• fill_mode: Points outside the boundaries of the input are filled according to the given mode (one of {"constant", "reflect", "wrap", "nearest"}).
□ reflect: (d c b a | a b c d | d c b a) The input is extended by reflecting about the edge of the last pixel.
□ constant: (k k k k | a b c d | k k k k) The input is extended by filling all values beyond the edge with the same constant value k = 0.
□ wrap: (a b c d | a b c d | a b c d) The input is extended by wrapping around to the opposite edge.
□ nearest: (a a a a | a b c d | d d d d) The input is extended by the nearest pixel.
• interpolation: Interpolation mode. Supported values: "nearest", "bilinear".
• seed: Integer. Used to create a random seed.
• fill_value: a float represents the value to be filled outside the boundaries when fill_mode="constant".
>>> input_img = np.random.random((32, 224, 224, 3))
>>> layer = tf.keras.layers.RandomZoom(.5, .2)
>>> out_img = layer(input_img)
>>> out_img.shape
TensorShape([32, 224, 224, 3])
Input shape
3D (unbatched) or 4D (batched) tensor with shape: (..., height, width, channels), in "channels_last" format.
Output shape
3D (unbatched) or 4D (batched) tensor with shape: (..., height, width, channels), in "channels_last" format.
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Relationship between Gini coefficient and income inequality
The Gini coefficient measures income inequality, with 0 representing perfect equality and 1 perfect inequality. A lower coefficient implies better income distribution. Countries with high Gini
coefficients typically have greater income disparity. Policymakers often use the Gini coefficient to assess and address income inequality within a nation. It is a crucial tool for understanding
economic disparities and guiding policy decisions. By analyzing the Gini coefficient, governments can design effective interventions to reduce income inequality and foster social harmony. Addressing
income inequality through targeted policies can lead to a fairer society with improved well-being for its citizens.
Table of Contents
(Lorenz Curve and Gini Coefficient – Measures of Income Inequality)
The Gini coefficient measures income inequality within a population. A high Gini coefficient signifies greater inequality. Countries with disparities have higher Gini coefficients. It ranges from 0
to 1, where 0 is perfect equality and 1 is total inequality. Analysts look at Gini coefficients to assess income distribution. Lower Gini coefficients suggest a fairer income distribution. Policies
can target reducing income inequality. The relationship between Gini coefficient and income inequality is crucial. High income inequality can lead to social unrest. Minimizing income inequality
fosters economic stability. Governments aim to lower Gini coefficients to promote social cohesion. Understanding the Gini coefficient helps tackle societal issues. In conclusion, the Gini coefficient
plays a vital role in addressing income inequality.
Calculation of Gini coefficient
When delving into the intricate realm of income inequality, one of the most crucial tools at our disposal is the Gini coefficient. This numeric value, usually ranging between 0 and 1, serves as a
measure of statistical dispersion intended to represent the income distribution within a specific population. Calculating this index involves a series of steps that offer profound insights into
societal wealth distribution patterns.
To start with, envision a community comprising individuals with varying income levels – from those barely making ends meet to affluent members enjoying luxuries beyond necessity. The first step in
calculating the Gini coefficient involves plotting these individuals on an ‘Lorenz curve.’ Picture this curve as a graphical representation showcasing cumulative percentages of total income held
against corresponding cumulative percentages of individuals ranked by ascending incomes.
Imagine yourself at this juncture, meticulously tracing each point on the Lorenz curve to ensure accuracy in portraying economic reality. As you connect these dots methodically, aiming for precision
in delineating disparities among your subjects, a sense of responsibility washes over you. You realize that behind every data point lies someone’s livelihood and aspirations—each line on your graph
painting a picture of opportunity or disparity faced by real people.
Upon completing the Lorenz curve illustration with finesse and care infused into every stroke, it’s time to transition to numerical computation—the heart of determining the Gini coefficient itself.
Buckle down as you calculate two pivotal areas: A) area under the hypothetical line representing perfect equality and B) area between this ideal scenario and your actual Lorenz curve depiction.
In essence, computing these areas entails grappling with mathematical intricacies while keeping sight of their societal implications. With each calculation nuanced yet vital in understanding income
inequality dynamics within this microcosm you’ve constructed through data points so diligently gathered.
As you arrive at your final figure—a decimal illuminating just how skewed or equitable wealth distribution trends are—you exhale deeply, cognizant that behind those numbers lie stories untold; lives
impacted by policies shaped around such indices like threads weaving through society’s fabric…
Definition of Gini coefficient
Ah, the Gini coefficient – a curious little number that packs a big punch when it comes to understanding income inequality. Imagine this: you’re standing in the middle of a bustling city square,
surrounded by people from all walks of life. Some are carrying designer bags and sipping lattes, while others clutch their worn-out backpacks with tired hands.
Now, picture yourself armed with the Gini coefficient, ready to unravel the economic disparity swirling around you. This magical number (ranging from 0 to 1) acts as a spotlight, shining brightly on
how wealth is distributed among these diverse individuals.
At its core, the Gini coefficient measures inequality within a given population or group. A score of 0 represents perfect equality – everyone holds an equal share of the pie. On the flip side, a
score of 1 signifies total inequality – one person monopolizes all the slices while others yearn for mere crumbs.
As you crunch those numbers and decipher what they mean for society at large, emotions start bubbling up inside you like simmering soup on a stove. Sadness creeps in as you realize that behind each
data point lies a real human story; tales of triumph and struggle woven into every decimal place.
When contemplating how this enigmatic coefficient impacts our world today—inequality gnawing away at social bonds like termites devouring wood—it’s hard not to feel a pang of empathy for those left
behind in this race towards prosperity.
The beauty (or horror) of the Gini coefficient is its ability to lay bare uncomfortable truths about our society’s fabric—the stark contrast between gleaming skyscrapers housing penthouse elites and
dimly lit alleyways sheltering forgotten souls just around the corner.
So here we are, peering through the lens of this powerful metric—a modern-day oracle revealing patterns hidden in plain sight. Just remember: behind every decimal point lies more than just cold
calculations; there beats a heart longing for fairness and equity in an unforgiving world where numbers often speak louder than words ever could…
Factors influencing Gini coefficient
When we delve into the complex realm of income inequality, one crucial metric that often comes to the forefront is the Gini coefficient. This numerical value, ranging from 0 to 1, serves as a
barometer for the distribution of wealth or income within a specific population. However, this seemingly straightforward statistic is influenced by a multitude of factors that can either exacerbate
or ameliorate income inequality in any given society.
First and foremost, historical context plays an instrumental role in shaping the Gini coefficient of a nation. Deep-rooted societal structures and policies established over time can have lasting
effects on wealth distribution. For instance, countries with a history of colonialism may exhibit higher levels of income inequality due to disparities created during periods of exploitation and
resource extraction.
Moreover, economic forces such as globalization and technological advancements contribute significantly to fluctuations in the Gini coefficient. Globalization has led to increased interconnectedness
but has also widened the gap between rich and poor through outsourcing practices and competition for low-skilled jobs. Similarly, automation arising from technological progress has displaced certain
job sectors while enhancing opportunities for others, thereby impacting income distribution patterns.
Political decisions and policy frameworks adopted by governments further mold the landscape of income inequality within a society. Taxation policies determining progressive or regressive tax rates
directly influence how wealth is redistributed among different socioeconomic strata. Welfare programs aimed at providing social safety nets can help alleviate poverty levels and reduce disparities in
income distribution.
Cultural attitudes towards wealth accumulation and philanthropy shape behaviors related to earning potential and charitable giving—factors that ultimately impact the Gini coefficient within
communities. In societies where individual success is highly praised without consideration for communal well-being, income inequality tends to be more pronounced compared to cultures emphasizing
collective prosperity.
In essence, understanding these multifaceted influences on the Gini coefficient sheds light on the intricate web woven by economic, political, historical, and cultural dynamics impacting income
inequality worldwide—a tapestry rich with nuances waiting to be unraveled by policymakers striving towards a more equitable future.
(Gini Coefficient and Lorenz Curve)
Interpretation of Gini coefficient
When we delve into the interpretation of the Gini coefficient within the realm of income inequality, we unearth a nuanced landscape that goes beyond mere numbers. The Gini coefficient is not just a
statistical figure; it embodies the disparities and complexities woven into society’s fabric. At its core, this measure encapsulates how wealth or income is distributed among a population – revealing
stark contrasts in prosperity.
Imagine a world where every individual holds an equal share of resources – in such utopian harmony, the Gini coefficient would gracefully bow to zero, symbolizing perfect equality. However, reality
paints a different picture. As this metric climbs towards one, it signifies extreme inequality where one person amasses all wealth while others struggle to make ends meet.
The interpretation of the Gini coefficient extends far beyond mathematical computations; it reflects societal structures and values. A high Gini coefficient mirrors deep-rooted inequities entrenched
in our social systems – pointing fingers at unequal access to opportunities and resources based on factors like race, gender, or socioeconomic status.
Conversely, a lower Gini coefficient heralds more equitable societies where resources are distributed fairly amongst its members. It sings tales of inclusivity and collective prosperity – where
everyone has a seat at the table and none go hungry.
On an emotional level, interpreting the Gini coefficient can evoke feelings ranging from empathy to outrage. Looking at soaring coefficients may spark indignation as we witness glaring injustices
unfold before us – families struggling to put food on tables while others drown in opulence.
Yet amidst these turbulent waves of disparity lies hope for change. By understanding and dissecting what the Gini coefficient unveils about our communities’ economic landscapes, we pave pathways
towards crafting policies that foster equality and opportunity for all.
In conclusion, behind every computed number lies stories untold – narratives of resilience against adversity and battles fought for justice and fairness. Interpreting the essence of the Gini
coefficient requires not just analytical prowess but also empathy toward those whose lives are shaped by its unyielding grasp on societal dynamics.
Policy implications of Gini coefficient
When we delve into the realm of income distribution and inequality, one crucial tool that comes to light is the Gini coefficient. This metric, ranging from 0 (perfect equality) to 1 (maximum
inequality), serves as a powerful indicator of disparities in wealth within a population. A high Gini coefficient signifies greater income inequality, while a lower score suggests a more equal
The implications of the Gini coefficient for policymaking are profound and multifaceted. Policymakers worldwide rely on this measure to assess economic trends, design effective interventions, and
gauge the impact of social policies aimed at reducing inequality.
Firstly, countries with high Gini coefficients face pressing challenges related to social cohesion and economic stability. When wealth is concentrated in the hands of a select few, it can lead to
increased social tensions, decreased trust in institutions, and reduced opportunities for upward mobility among marginalized groups. As such, policymakers must prioritize strategies that promote
inclusive growth and equitable access to resources.
Moreover, analyzing changes in the Gini coefficient over time offers valuable insights into the effectiveness of government policies. For instance, if a nation implements progressive taxation or
redistributive programs aimed at narrowing income gaps but sees little change in its Gini coefficient, it may signal that these initiatives are not reaching their intended targets effectively.
On an international scale, comparing Gini coefficients across different nations can shed light on global disparities and inform efforts toward cross-border cooperation. By identifying regions with
particularly high levels of income inequality, policymakers can target aid programs more strategically and foster partnerships aimed at promoting sustainable development goals.
Despite its utility as a diagnostic tool for assessing income distribution patterns, the Gini coefficient has limitations that policymakers must acknowledge. While it provides a snapshot of relative
inequality within a population at a specific point in time…
External Links
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Dempervs, the party of Pedos
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Dempervs, the party of Pedos
Latest member
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Prove that 5-sqrt5 is an irrational number. -Turito
Are you sure you want to logout?
The correct answer is: YES an irrational number.
The real numbers which cannot be expressed in the form of p/q(where p and q are integers and q ≠ 0) are known as irrational numbers. In the given question we are asked to prove if 5-
Let's assume that 5-
Step 1 of 2:
If 5-
SO 5 -
5 -
Step 2 of 2:
Here, we see that
Final Answer:
Get an Expert Advice From Turito.
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A single matrix
The caterpillar() and envelope() functions plot output associated with a vector of parameter nodes. This is often expressed as a two-dimensional matrix or data.frame, with a column for each parameter
node and a row for each MCMC iteration. Both caterpillar() and envelope() were originally written to accept such data.frames as inputs, but now also accept a jagsUI output object and parameter name.
It is anticipated that caterpillar() could be used for effect sizes associated with a categorical variable, in which plotting order may or may not matter.
By contrast, envelope() is intended for a matrix associated with a sequence of parameter nodes, such as in a time series.
For a simpler case, plotdens() produces a kernel density plot of a single parameter node, or overlays multiple parameter nodes from a list. Alternately (shown below) it overlays kernel density plots
of a vector of parameter nodes.
old_parmfrow <- par("mfrow") # storing old graphics state
caterpillar(asdf_jags_out, "a")
envelope(SS_out, "trend", x=SS_data$x)
plotdens(asdf_jags_out, "a")
Multiple matrices or multiple models
It may be appropriate to make by-element comparisons of multiple such matrices, perhaps between multiple candidate models.
Function comparecat() produces interleaved caterpillar plots for a list of jagsUI output objects and an optional list of parameters, plotting parameters common to a set of models adjacent to one
another. The example below uses the same output object three times, but will show functionality.
Function comparedens() behaves similarly, but produces left- and right-facing kernel density plots for TWO jagsUI output objects and an optional list of parameters. The example below uses the same
output object twice, but will show functionality.
old_parmfrow <- par("mfrow") # storing old graphics state
comparecat(x=list(asdf_jags_out, asdf_jags_out, asdf_jags_out),
comparedens(x1=asdf_jags_out, x2=asdf_jags_out, p=c("a","b","sig"))
Function overlayenvelope() will automatically overlay multiple envelope() plots, and may be used with a variety of input structures:
• A list() of 2-dimensional posterior data.frames or matrices
• A 3-dimensional array, in which multiple 2-dimensional posterior matrices are joined along the third dimension
• A list() of jagsUI output objects, plus a parameter name
• A single jagsUI output objects, plus a vector of parameter names
## usage with list of input data.frames
## usage with a 3-d input array
## usage with a jagsUI output object and parameter name (2-d parameter)
overlayenvelope(df=SS_out, p="cycle_s")
## usage with a single jagsUI output object and multiple parameters
overlayenvelope(df=SS_out, p=c("trend","rate"))
Function crossplot() plots corresponding pairs of parameter densities on the X- and Y-axes. Three plotting methods are provided, that may be overlayed if desired:
• If drawcross == TRUE, caterpillar-like plots will be produced, with quantile intervals in the x- and y- directions.
• If drawx == TRUE, caterpillar-like plots will be produced, but rotated along the standardized principal component axes. This may be useful to draw if correlation is present.
• If drawblob == TRUE, smoothed polygons will be produced, each containing approximately ci= x100% of the associated MCMC samples.
This function may be used with vectors or matrices of MCMC samples, or with a jagsUI object and a vector of parameter names.
## Usage with single vectors (or data.frames or 2d matrices)
xx <- SS_out$sims.list$trend[,41]
yy <- SS_out$sims.list$cycle[,41]
## Showing possible geometries
par(mfrow = c(2, 2))
plot(xx, yy, col=adjustcolor(1, alpha.f=.1), pch=16, main="Cross Geometry")
crossplot(xx, yy, add=TRUE, col=1)
plot(xx, yy, col=adjustcolor(1, alpha.f=.1), pch=16, main="X Geometry")
crossplot(xx, yy, add=TRUE, col=1,
drawcross=FALSE, drawx=TRUE)
plot(xx, yy, col=adjustcolor(1, alpha.f=.1), pch=16, main="Blob Geometry")
crossplot(xx, yy, add=TRUE, col=1,
drawcross=FALSE, drawblob=TRUE)
plot(xx, yy, col=adjustcolor(1, alpha.f=.1), pch=16, main="Blob Outlines")
crossplot(xx, yy, add=TRUE, col=1,
drawcross=FALSE, drawblob=TRUE, outline=TRUE)
Comparison between priors and posteriors
Function comparepriors() is a wrapper for comparedens(), and plots side-by-side kernel densities of all parameters with names ending in "_prior", along with the respective posterior densities. It
should be noted that additional parameters must be included in the JAGS model to provide samples of the prior distributions, as is shown in the example below.
sig ~ dunif(0, 10) # this is the parameter that is used elsewhere in the model
sig_prior ~ dunif(0, 10) # this is only used to give samples of the prior
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American Invitational Mathematics Examination
The American Invitational Mathematics Examination (AIME) is the second exam in the series of exams used to challenge bright students on the path toward choosing the team that represents the United
States at the International Mathematics Olympiad (IMO). While most AIME participants are high school students, some bright middle school students also qualify each year.
High scoring AIME students are invited to take the prestigious United States of America Mathematics Olympiad (USAMO) for qualification from taking the AMC 12 or United States of America Junior
Mathematics Olympiad (USAJMO) for qualification from taking the AMC 10.
The AIME is administered by the Mathematical Association of America (MAA). Art of Problem Solving (AoPS) is a proud sponsor of the AMC!
Region: USA
Type: Free Response
Difficulty: 3-6
Difficulty Breakdown:
Problem 1-5: 3
Problem 6-10: 4
Problem 10-12: 5
Problem 12-15: 6
The AIME is a 15 question, 3 hour exam$^1$ taken by high scorers on the AMC 10, AMC 12, and USAMTS competitions. Qualification through USAMTS only is rare, however. Each answer is an integer from 000
to 999, inclusive, making guessing almost futile. Wrong answers receive no credit, while correct answers receive one point of credit, making the maximum score 15. Problems generally increase in
difficulty as the exam progresses - the first few questions are generally AMC 12 level, while the later questions become extremely difficult in comparison. Calculators are not permitted.
$^1$ In the first two years (1983 and 1984) there was a 2.5 hour time limit instead of the current 3 hour limit.
The AIME tests mathematical problem solving with arithmetic, algebra, counting, geometry, number theory, and probability and other secondary school math topics. Problems usually require either very
creative use of secondary school curriculum, or an understanding as to how different areas of math can be used together to investigate and solve a problem.
• AMC homepage and their AIME page
• AIME Problems and Solutions -- A community effort to provide solutions to all AIME problems from which students can learn.
• The AoPS AIME guide.
• AMC Forum for discussion of the AMC and problems from AMC and AIME exams.
• The AoPS Contest Archive includes problems and solutions from most AMC and all AIME exams.
• Mock AIME exams by AoPSers -- A wealth of secondary practice materials.
• Past HMMT, PUMaC, and CMIMC exams (search the test up to see the link).
Recommended reading
AIME Preparation Classes
• AoPS hosts an online school teaching introductory classes in topics covered by the AIME as well as AIME preparation classes.
• AoPS holds many free Math Jams, some of which are devoted to discussing problems on the AIME. Math Jam Schedule
AIME Exams in the AoPSWiki
This is a list of all AIME exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these
problems and solutions are also available in the AoPS Resources section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if
you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.
See also
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A Student's Guide to the Physics Reference Table » Learning Captain 1
A Student’s Guide to the Physics Reference Table
Physics reference table : You’re sitting in your physics class, staring at the reference table in the front of your textbook, and wondering what on earth all those numbers and symbols mean. Don’t
worry, you’re not alone. The physics reference table can look pretty intimidating at first glance. But it’s really an incredible cheat sheet that contains key constants and conversions to make your
problem-solving life way easier.
In this article, we’re going to break down exactly how to read and use the physics reference table so you can crush your next exam or homework assignment. From metric prefixes to fundamental
constants to common unit conversions, consider this your student’s guide to decoding the physics reference table. By the end, those columns of numbers won’t seem so scary. Let’s dive in!
What Is the Physics Reference Table?
A Student's Guide to the Physics Reference Table 4
The physics reference table is your new best friend. This handy chart provides a wealth of information in one place so you don’t have to go searching through various sources.
What’s in the Table?
The table includes fundamental constants like the speed of light, gravitational acceleration, and Planck’s constant. It lists conversion factors between units, so you can quickly change meters to
inches or kilograms to pounds. It gives properties of materials, from densities of common substances to resistivities.
Of course, it contains all the formulas you’ll need, for calculating forces, energy, momentum, and more. Each formula shows the units for every term, so you know you’ve solved the problem correctly.
The table even provides diagrams of key concepts, like the electromagnetic spectrum or the structure of the atom.
How Do I Use It?
The key is practice. Work through lots of problems using the reference table and soon you’ll be navigating it with ease. Start by familiarizing yourself with how the table is organized. Pay attention
to the headings, subheadings, and groupings. Look for patterns in how information is presented.
When doing practice problems, try to answer questions without looking at the table first. Then check your work against the information provided to reinforce what you’ve learned. Over time, you’ll
have the most useful parts of the table committed to memory.
The physics reference table puts everything at your fingertips. With regular use, this compact yet comprehensive resource will become second nature, allowing you to focus on understanding key
concepts rather than searching for information. What a useful tool for budding scientists and engineers!
Key Sections of the Physics Reference Table
The physics reference table is packed with useful information, but it can be overwhelming if you don’t know where to look. Here are the key sections you’ll want to get familiar with:
This section provides the standard units of measurement for concepts like distance, time, mass, and force. Memorize these – you’ll be using them a lot!
Fundamental constants like the speed of light, gravitational acceleration, and Planck’s constant are listed here. These values remain fixed and are used in many calculations.
Greek Alphabets
All the Greek letters are shown here, along with their names and pronunciations. Greek letters are commonly used to represent angles, velocities, and other values in physics.
This part contains many of the major formulas you’ll use, like those for calculating force, acceleration, momentum, and energy. Keep this section bookmarked – it’s a lifesaver!
Periodic Table
A mini periodic table gives the atomic number, mass, name, and symbol for each element. This is helpful for determining numbers of protons, electrons, and neutrons in atoms and ions.
Diagrams show how to calculate the circumference, area, and volume of circles, triangles, rectangles and spheres. Geometry and trigonometry are used throughout physics, so these figures come in
With so much info packed into one table, the physics reference table may seem daunting. But by focusing on these key sections, you’ll be able to quickly find what you need to solve problems and
understand concepts. Refer to it often, and it will become second nature in no time!
How to Read Values From the Table
A Student's Guide to the Physics Reference Table 5
The physics reference table contains a ton of useful information, if you know how to read it. Here are some tips to help you navigate this valuable resource:
The table provides the units of measurement for each quantity. Make sure you pay attention to the units, as they tell you what the numbers in the table actually represent. For example, density is
given in kg/m^3, so you know the values refer to mass per unit volume.
The table lists important physical constants like the speed of light, gravitational acceleration, and Planck’s constant. These values are fixed and unchanging. You’ll use them often in calculations
and problems.
Conversion Factors
Need to convert miles to kilometers or Celsius to Fahrenheit? The reference table has you covered. It provides conversion factors for commonly used units. Just multiply the quantity you want to
convert by the appropriate factor.
The bulk of the table consists of values for various properties and quantities. These include things like the density of copper, the specific heat of water, and the wavelength of light corresponding
to a particular frequency. When you need one of these values for a calculation or to look up information, the reference table is the place to find it.
Sometimes the exact value you need isn’t listed in the table. In that case, you can interpolate between two values to estimate it. For example, if you needed the density of a material with a mass of
32 g and the table only listed values for 30 g and 35 g, you could interpolate between those points to determine the density for 32 g. Interpolation is a useful skill that allows you to extract more
information from the table.
The physics reference table puts a wealth of knowledge at your fingertips. With regular use, you’ll get better at navigating the table and become adept at finding whatever values or information you
need. The key is just knowing how to make the most of this helpful resource.
Real-World Examples Using the Reference Table
The physics reference table contains a wealth of information that applies to real-world situations. Here are a few examples of how the table can be used in everyday life:
Calculating the Energy Used to Power Your Home
The reference table provides a list of energy equivalents that can help you calculate how much energy you use at home. For example, if your electric bill says you used 1,000 kilowatt-hours (kWh) of
energy last month, you can convert that to joules using the table. 1 kWh = 3,600,000 joules. So 1,000 kWh = 3,600,000,000 joules of energy used in your home last month! Using the table, you can also
convert that energy usage into BTUs, calories, or electron volts.
Figuring Out How Much Force is Exerted
Did you ever wonder how much force is exerted when you push open a door or lift a heavy box? You can use Newton’s second law (F=ma) and values from the reference table to calculate it. For example,
if you push open a door with a mass of 20 kilograms and cause it to accelerate at 2 meters/second^2, the force exerted is:
F=20kg x 2m/s^2 = 40 N
So you exerted 40 newtons of force to push open that door. The reference table provides the conversions to calculate forces in newtons for any masses and accelerations.
Estimating the Final Speed of a Rolling Object
If you give a ball a push to get it rolling, you can use the reference table to estimate its final speed. Assume you push the 0.5 kg ball with 10 newtons of force. Using F=ma, the acceleration is:
a = F/m = 10N / 0.5kg = 20 m/s^2
The final speed depends on how long it accelerates. If it accelerates for 2 seconds, the final speed can be calculated using the kinematic equation v=at.
v=20m/s^2 x 2s = 40 m/s
So after 2 seconds, the 0.5 kg ball you pushed will be traveling at around 40 meters per second. The reference table provides all the values needed for these types of kinematic calculations.
Tips for Memorizing Key Values
A Student's Guide to the Physics Reference Table 6
Memorizing the values in the physics reference table can seem like an overwhelming task. Here are some tips to help make it manageable:
Focus on the numbers you’ll use most often
Prioritize memorizing the values you know you’ll frequently use in problems and calculations, like the speed of light, gravitational constant, and electron mass. The rest you can always look up as
Group similar values together
Try memorizing values with similar units or in the same table section together. For example, memorize the proton, neutron and electron masses as a set. Or memorize all the derived SI units like
newtons, pascals and joules together. This can make them easier to remember.
Use mnemonics
Create mnemonics, like rhymes, acronyms, songs or other tricks to help memorize values. For example, “King Phillip Came Over For Good Spaghetti” to memorize the order of taxonomy: Kingdom, Phylum,
Class, Order, Family, Genus, Species. Or “Some Lovers Try Positions That They Can’t Handle” to memorize the electromagnetic spectrum: Super Low Frequencies, Radio Frequencies, Microwaves, Infrared,
Visible Light, Ultraviolet, X-rays, Gamma Rays.
Practice regularly
The more you practice recalling the values, the more they will stick in your memory. Flip through the physics reference table for a few minutes each day, testing yourself on different sections.
Flashcards are also great for memorization practice.
Don’t worry about precision
You don’t need to memorize values to a high degree of precision, especially constants with many digits. Round to 2-3 significant figures for most values. Your teacher will specify if more precision
is needed for a particular problem.
With regular practice, the most important values in the physics reference table will become second nature. Let me know if you have any other questions!
Read More: Getting to Know PEScience: Are Their Products Tested and Safe?
The physics reference table can look complicated at first, but don’t worry—it’s really not that bad once you get the hang of it. Here are some of the most frequently asked questions to help you
better understand this useful resource.
What exactly is the physics reference table?
The physics reference table, also known as the PRT, is a compilation of common data, formulas, and information related to physics. It includes things like physical constants, metric prefixes, and
geometry formulas. The PRT allows you to have a wealth of knowledge on hand without needing to memorize every single formula.
Do I have to memorize everything in the PRT?
No, you do not need to memorize the entire PRT. You should be familiar with the layout and format so you can quickly find what you’re looking for. Focus on understanding concepts and practicing
problems, not pure memorization. The PRT is meant to be used as a reference, not as something you have to commit fully to memory.
How should I use the PRT?
The best way to use the PRT is:
• Get familiar with the layout so you know where different types of information are located.
• Use it regularly when doing practice problems to help reinforce where things are.
• Don’t just look at the formulas, read the descriptions and examples too.
• See how different formulas relate to each other. Understanding connections will make the information stick better in your mind.
Are there any tips for using the PRT during the exam?
Yes, here are some tips for using the PRT on exam day:
• Familiarize yourself with the PRT again right before the exam.
• Have a systematic approach for finding information, like checking the table of contents first.
• Don’t panic if you can’t find something quickly. Stay calm and keep searching logically.
• Make sure to properly apply any formulas or values you look up. Double check your work.
• Ask your proctor if you have any questions about using the reference table. They want you to succeed!
With regular practice, the physics reference table can become your best friend. Stay patient and keep at it. You’ll get the hang of it in no time!
So there you have it, a quick guide to navigating the physics reference table that should help you tackle any problem sets or exams. While all those numbers and units can seem intimidating at first,
with regular use the reference table will become second nature. The key is not to try and memorize everything at once but start with the basics like units, constants and the equations you’ll use most
often in class.
Once you get familiar with where everything is located, you’ll be flipping through with confidence in no time. And remember, don’t hesitate to ask your professor or TA if you ever have any questions
about the information in the reference table. They want you to succeed and are there to help explain anything that’s unclear. You’ve got this! Now go forth and physics.
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[Solved] A closed iron tank 12 m long, 9 m wide and 4 m deep is... | Filo
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs. 5 per metre sheet, sheet being 2 m wide.
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Given length =12,m Breadth = 9m and Height = 4m.
Total surface area of tank
Cost of iron sheet = length of sheet xx cost rate
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Question Text A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs. 5 per metre sheet, sheet being 2 m wide.
Topic Surface Areas and Volumes
Subject Mathematics
Class Class 9
Answer Type Text solution:1
Upvotes 55
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Intelliseeds Learning - Math Levels - Grade 5
A: Number sense up to billions:
A.01: Place value through billions - Assessment 1
A.02: Place value through billions - Assessment 2
A.03: Sum of place values
A.04: Convert between place values - Assessment 1
A.06: Counting in hundreds
A.07: Counting in thousands
A.11: Find the smallest and greatest number - Assessment 1
A.12: Find the smallest and greatest number - Assessment 2
A.13: Write the successor of number
A.14: Write the predecessor of number
A.15: Review on successor or predecessor of a number
A.17: Compare numbers: ordering positive and negative numbers
A.18: Review rounding numbers up to billions
A.19: Roman numerals - Assessment 1
A.20: Roman numerals - Assessment 2
A.21: Roman numerals - Assessment 3
A.22: Roman numerals - Assessment 4
A.24: Assessing Roman numerals - Assessment 1
A.25: Assessing Roman numerals - Assessment 2
A.26: Compare Roman numerals - Assessment 1
A.27: Compare Roman numerals - Assessment 2
A.28: Operations on Roman numerals - Assessment 1
A.29: Operations on Roman numerals - Assessment 2
A.31: Patterns - Assessment 1
A.32: Patterns - Assessment 2
B.01: Adding two numbers up to billions - Assessment 1
B.02: Adding two numbers up to billions - Assessment 2
B.03.01: Adding 3 or 4 numbers up to billions - Assessment 1
B.03.02: Adding 3 or 4 numbers up to billions - Assessment 2
B.04: Review on addition of 2, 3 or 4 numbers up to billions
B.05.01: Find the missing digits in addition of 2 numbers up to billions - Assessment 1
B.05.02: Find the missing digits in addition of 2 numbers up to billions - Assessment 2
B.06.01: Find missing digits in addition of 3 or 4 numbers up to millions - Assessment 1
B.06.02: Find missing digits in addition of 3 or 4 numbers up to millions - Assessment 2
B.07: Find the number by addition
B.08: Sum of smallest and largest number
B.09: Story problems for addition of 2, 3 or 4 numbers up to billions - Assessment 1
B.10: Story problems for addition of 2, 3, 4 or 5 numbers up to billions - Assessment 2
C.01: Subtraction of numbers up to billions - Assessment 1
C.02: Subtraction of numbers up to billions - Assessment 2
C.03: Subtraction of numbers up to billions - Assessment 3
C.05: Review on subtraction of numbers
C.06: Difference between smallest and largest number
C.07: Story problems on subtraction
C.08: Find the missing number in a subtraction equation
C.09: Find the missing number in an addition or subtraction sentence
D.01: Multiply 2, 3 and 4 digit number by 1 or 2 digit number
D.02: Multiplication by the multiples of 10, 100, 1000 etc
D.03.01: Multiplication up to billions by 3 or 4 digit numbers - Assessment 1
D.03.02: Multiplication up to billions by 3 or 4 digit numbers - Assessment 2
D.04: Review on multiplication up to billions by 1, 2, 3 or 4 digits
D.06.01: Word Problems on multiplication
D.07.01: Multiplication with numbers having nines
D.07.02: Multiplication by 11
D.07.03: Multiplication by 5, 25, 50 and 500
D.07.04: Review on multiplication shortcuts
D.08.01: Find the product for up to 3 numbers - Assessment 1
D.08.02: Find the product for up to 3 numbers - Assessment 2
D.09: Find the multiplication of largest and smallest number - Assessment 1
D.10: Find the multiplication of largest and smallest number - Assessment 2
D.12.01: Choose numbers with a particular product - Assessment 1
D.13: Exploring patterns in multiplication
E.01.02: Using properties of division
E.02: Division of multiples of 10
E.03.01: Divide 2 or 3 digit by 1 or 2 digit Divisors
E.04: Divide 4, 5 or 6 digit by 1 or 2 digit Divisors - Assessment 1
E.05: Divide 4, 5 or 6 digit by 3 digit Divisors - Assessment 2
E.06: Divide 4, 5 or 6 digit by 3 digit divisors - Assessment 3
E.07: Divide 4, 5 or 6 digit by 1, 2 or 3 digit divisors - Assessment 4
E.09: Find the number by division - Assessment 1
E.10: Find the number by division - Assessment 2
E.11: Word problems on division
E.12: Which sign makes the number sentence true
E.13: Review on find the number and estimate quotients
E.14: Choose numbers with a particular quotient - Assessment 1
E.15: Choose numbers with a particular quotient - Assessment 2
E.16: Exploring patterns to divide
E.17: Missing dividend or divisor
E.18: Review on finding missing dividend, divisor or quotient
F.01: Using the rule of DMAS - Assessment 1
F.02: Using the rule of DMAS - Assessment 2
F.03: Using the rule of BODMAS - Assessment 1
F.04: Using the rule of BODMAS - Assessment 2
F.05: Using the rule of BODMAS - Assessment 3
F.06: Using the rule of BODMAS - Assessment 4
F.07: Using the rule of BODMAS - Assessment 5
F.08: Review on using the rule of BODMAS
F.09: Using mathematical operators - Assessment 1
F.10: Using mathematical operators - Assessment 2
F.11: Using algebraic properties - Assessment 1
F.12: Using algebraic properties - Assessment 2
G: Factors and Multiples:
G.01: Review on Prime and Composite numbers
G.02: Test of Divisibility by 2
G.03: Test of divisibility by 3
G.04: Test of Divisibility by 4
G.05: Test of Divisibility by 5
G.06: Test of divisibility by 7
G.07: Test of Divisibility by 8
G.08: Test of divisibility by 9
G.10: Test of Divisibility by 11
G.11: Review on test of divisibility by 2,3,4,5,7,8,9,10,11
G.12.01: Factor of Whole Numbers - Assessment 1
G.12.02: Factor of whole numbers - Assessment 2
G.13.01: Prime Factors - Assessment 1
G.13.02: Prime Factors - Assessment 2
G.13.03: Review on factor of whole numbers and prime factors
G.14.01: Prime Factors using Factor tree - Assessment 1
G.14.02: Prime Factors using Factor tree - Assessment 2
G.16.02: Review on Factor tree, Co-primes and Common factors
G.17.01: Greatest Common factor(Highest Common Factor) of 2 numbers - Assessment 1
G.17.02: Greatest Common factor(Highest Common Factor) of 2 numbers - Assessment 2
G.18.01: Greatest Common factor(Highest Common Factor) of 3 numbers - Assessment 1
G.18.02: Greatest Common factor(Highest Common Factor) of 3 numbers - Assessment 2
G.19: Review on Greatest Common Factor
G.21: Least Common Multiple (LCM) of 2 numbers
G.22.01: Least Common Multiple (LCM) of 3 or more numbers - Assessment 1
G.23: Review on Least Common Multiple
G.24: Analyzing factors and multiples
G.25: Relationship between GCF and LCM - Assessment 1
G.26: Relationship between GCF and LCM - Assessment 2
G.27: LCM and GCF with Venn Diagram - Assessment 1
G.28: LCM and GCF with Venn Diagram - Assessment 2
G.29: Review on relationship between GCF & LCM and venn diagrams
G.30: Problems involving GCF - Assessment 1
G.31: Problems involving GCF - Assessment 2
G.32: Problems involving LCM - Assessment 1
G.33: Problems involving LCM - Assessment 2
H.01: Find numerator or denominator of the fraction
H.02: Equivalent fractions
H.03: Fractional part of number
H.04.01: Fractional part of quantity
H.06: Fraction in lowest terms
H.08: Least common denominator
H.10: Addition of unlike fractions - Assessment 1
H.11: Addition of unlike fractions - Assessment 2
H.12: Addition of unlike fractions - Assessment 3
H.14.01: Subtraction of unlike fractions - Assessment 1
H.14.02: Subtraction of unlike fractions - Assessment 2
H.14.03: Subtraction of unlike fractions- Assessment 3
H.17: Multiplying fraction by whole number
H.18.01: Multiplying two fractions
H.18.02: Multiplying mixed fractions
H.20: Reciprocal of a fraction
H.21: Division of fractions - Assessment 1
H.25: Arithmetic sequences with fractions
H.26: Geometric sequences with fractions
I: Operations with Decimals:
I.02: Find the missing number in decimal equations
I.03: Word problems on addition of Decimals
I.04.01: Word problems on subtraction of decimals
I.05: Multiplication by multiples of 10 - Assessment 1
I.06: Multiplication by multiples of 10 - Assessment 2
I.07.01: Multiplication of a decimal number by a whole number - Assessment 1
I.08.01: Multiplication of two decimal numbers - Assessment 1
I.09.01: Multiply three or more decimal numbers - Assessment 1
I.10.01: Word problems on multiplication
I.10.02: Review on multiplication of decimal
I.11.02: Division of a decimal number by multiples of 10 - Assessment 2
I.12: Division of a decimal by a whole number
I.13.01: Division with decimal quotient - Assessment 1
I.13.02: Division with decimal quotient - Assessment 2
I.13.03: Division with decimal quotient - Assessment 3
I.14: Division of a whole number by a decimal number - Assessment 1
I.15: Division of a whole number by a decimal number - Assessment 2
I.16: Division of a decimal by a decimal - Assessment 1
I.17: Division of a decimal by a decimal - Assessment 2
I.18: Division of a decimal by a decimal - Assessment 3
I.19: Division of a decimal by a decimal - Assessment 4
I.21.01: Estimate quotients - Assessment 1
I.21.02: Estimate quotients - Assessment 2
I.22: Division with decimal quotients and rounding
I.23: Word problems on division
J.02: Measurement of capacity - Assessment 1
J.03: Measurement of capacity - Assessment 2
J.04: Measurement of length - Assessment 1
J.05: Measurement of length - Assessment 2
J.07: Measurement of weight - Assessment 2
J.09: Add and subtract mixed metric units - Assessment 1
J.10: Add and subtract mixed metric units - Assessment 2
J.11: Multiplication and division mixed metric units - Assessment 1
J.12: Multiplication and division mixed metric units - Assessment 2
J.18: Pan balance problems - Assessment 1
J.19: Pan balance problems - Assessment 2
J.20: Pan balance problems - Assessment 3
K.01: Write equivalent fractions with denominator as 100
K.02: Writing fractions as percent
K.03: What percentage is illustrated?
K.06: Percentage as a fraction in lowest form - Assessment 1
K.07: Percentage as a fraction in lowest form - Assessment 2
K.08: Decimal expressed as a percent
K.09: Percent expressed as a decimal
K.10: Review on percent as fraction and decimal
K.11: Percent of a number Assessment-1
K.12: Percent of a number - Assessment 2
K.13: Percent of a number - Assessment 3
K.14: Percent of a number - Assessment 4
K.15: Word Problems on Percentage
K.16: Compare percentages
K.17: Review on percent of a number and comparing percentages
L.01: Find the average/mean - Assessment 1
L.02: Find the average/mean - Assessment 2
L.03: Find the median - Assessment 1
L.04: Find the median - Assessment 2
L.05: Find the mode - Assessment 1
L.07: Find the range - Assessment 1
L.08: Find the Range - Assessment 2
L.09: Review on the mean, mode, median and range
L.10: Word Problem on mean/average - Assessment 1
L.11: Word Problem on mean/average - Assessment 2
M.01: Find the Ratios - Assessment 1
M.02: Find the Ratios - Assessment 2
M.04: Word Problem on Ratios
M.05: Find quantities when ratio is given
M.06: Check the proportion - Assessment 1
M.07: Check the proportion - Assessment 2
M.08: Check the proportion - Assessment 3
M.09: Find the missing number to complete the proportion
N.01: Convert time into seconds - Assessment 1
N.02: Convert time into seconds - Assessment 2
N.03: Convert time into minutes - Assessment 1
N.04: Convert time into minutes - Assessment 2
N.05: Convert time into hours and minutes - Assessment 1
N.06: Convert time into hours and minutes - Assessment 2
N.07: Review on hours, minutes and seconds
N.08: Convert into 24-hour clock time
N.09: Convert into 12-hour clock time
N.10: Convert time into days
N.11: Convert days into years, months, weeks and days
N.12: Review on time conversion
N.13: Addition and Subtraction of Time
N.15: Find the duration of time
N.16: Calculating duration using a calendar
N.17: Word problems on mixed time unit - Assessment 1
N.18: Word problems on mixed time unit - Assessment 2
N.21: Reading train and bus schedule - Assessment 1
N.22: Reading train and bus schedule - Assessment 2
N.23: Reading train and bus schedule - Assessment 3
N.24: Reading train and bus schedule - Assessment 4
N.25: Review on reading train and bus schedule
O.01: Unitary method - Assessment 1
O.02: Unitary method - Assessment 2
O.03: Find profit/ cost price/ selling price - Assessment 1
O.04: Find profit/ cost price/ selling price - Assessment 2
O.05: Find loss/ cost price/ selling price - Assessment 1
O.06: Find loss/ cost price/ selling price - Assessment 2
O.07: Word problems on calculating profit / loss/ cost price/ selling price - Assessment 1
O.08: Word problems on calculating profit / loss/ cost price/ selling price - Assessment 2
O.09: Review on finding profit / loss/ cost price/ selling price
O.16: Calculating Simple Interest (time in years) - Assessment 1
O.17: Calculating Simple Interest (time in years) - Assessment 2
O.18: Calculating Simple Interest (time in years, months and days) - Assessment 1
O.19: Calculating Simple Interest (time in years, months and days) - Assessment 2
O.20: Word Problems on Simple Interest
O.22: Word Problems on calculating amount
O.24: Invoice - Assessment 1
O.25: Invoice - Assessment 2
O.26: Invoice - Assessment 3
O.27: Invoice - Assessment 4
P: Speed, Distance and Time:
P.01: Conversion of metric units - Assessment 1
P.02: Conversion of metric units - Assessment 2
P.03: Conversion of customary units - Assessment 1
P.04: Conversion of customary units - Assessment 2
P.05: Calculate speed (metric units)
P.06: Calculate speed (customary units)
P.07: Calculate distance/time (metric units)
P.08: Calculate Distance/Time (customary units)
P.09: Word Problems on conversion of units of speed
P.10: Word Problems to calculate speed/distance/time
P.11: Comparison of speed
P.12: Review on finding speed, distance and time - Assessment 1
P.13: Review on finding speed, distance and time - Assessment 2
Q.01: Classify the triangle by its sides
Q.02: Classify the triangle by its angles
Q.03: Sum of the angles of a Triangle - Assessment 1
Q.04: Sum of the angles of a Triangle - Assessment 2
Q.05: Find the angle in a Triangle - Assessment 1
Q.06: Find the angle in a Triangle - Assessment 2
Q.07: Review on classify and angles of the triangle
Q.09: Perimeter of a triangle
Q.12: Perimeter of a Rectangle - Assessment 1
Q.13: Perimeter of a Rectangle - Assessment 2
Q.14: Word Problems on Perimeter of a Rectangle - Assessment 1
Q.15: Word Problems on Perimeter of a Rectangle - Assessment 2
Q.16: Area of Rectangle - Assessment 1
Q.17: Area of Rectangle - Assessment 2
Q.18: Word problems on Area of Rectangle - Assessment 1
Q.19: Word problems on Area of Rectangle - Assessment 2
Q.20: Review on Area and Perimeter of Rectangle
Q.21: Perimeter of a square - Assessment 1
Q.22: Perimeter of a square - Assessment 2
Q.23: Word problems on Perimeter of a Square - Assessment 1
Q.24: Word problems on Perimeter of a Square - Assessment 2
Q.25: Area of a Square - Assessment 1
Q.26: Area of a square - Assessment 2
Q.27: Word problems on area of a square
Q.28: Review on area and perimeter of square
Q.29: Mix of Area and Perimeter - Assessment 1
Q.30: Mix of Area and Perimeter - Assessment 2
Q.31: Mix of Area and Perimeter - Assessment 3
Q.32: Find diameter and radius - Assessment 1
Q.33: Find diameter and radius - Assessment 2
Q.34: Find circumference of a circle - Assessment 1
Q.35: Find circumference of a Circle - Assessment 2
Q.38: Measure angles with a protractor - Assessment 1
Q.39: Measure angles with a protractor - Assessment 2
Q.40: Complementary Angles - Assessment 1
Q.41: Complementary Angles - Assessment 2
Q.42: Supplementary Angles - Assessment 1
Q.43: Supplementary Angles - Assessment 2
Q.44: Review on Complementary/Supplementary Angles
Q.45: Review on measuring angles and complementary/supplementary angles
Q.46: Vertically opposite angles - Assessment 1
Q.47: Vertically opposite angles - Assessment 2
Q.48: Angles of Quadrilateral - Assessment 1
Q.49: Angles of Quadrilateral - Assessment 2
Q.50: Review on Vertically Opposite Angles and Quadrilateral Angles
Q.53: Turning - Assessment 1
Q.54: Turning - Assessment 2
R.01: Open and Closed shapes
R.02: Regular and Irregular polygons
R.03: Making 3-dimensional figures
R.04: Area of irregular figures
R.05: Volume of irregular figures
R.06: 3-D figures viewed from different perspectives
S.01: Volume of Cube - Assessment 1
S.02: Volume of Cube - Assessment 2
S.03: Word Problems on Volume of Cube
S.05: Volume of cuboid - Assessment 1
S.06: Volume of cuboid - Assessment 2
S.07: Word problems on volume of cuboid
T: Data Handling and Graphs:
T.01: Pictographs - Assessment 1
T.02: Pictographs - Assessment 2
T.13: Coordinate planes as maps
T.14: Follow directions on a coordinate plane
U.01: Review on Temperature
U.02: Convert Celsius to Fahrenheit - Assessment 1
U.03: Convert Celsius to Fahrenheit - Assessment 2
U.04: Convert Fahrenheit to Celsius - Assessment 1
U.05: Convert Fahrenheit to Celsius - Assessment 2
U.06: Word problems on temperature
U.07: Review on temperature
Y: Olympiad Practice - Grade 5:
Y.01: Olympiad Practice Test - 1
Y.03: Olympiad Practice Test - 3
Y.04: Olympiad Practice Test - 4
Y.05: Olympiad Practice Test - 5
Y.06: Olympiad Practice Test - 6
Y.07: Olympiad Practice Test - 7
Y.08: Olympiad Practice Test - 8
Y.09: Olympiad Practice Test - 9
Y.10: Olympiad Practice Test - 10
Y.11: Olympiad Practice Test - 11
Y.12: Olympiad Practice Test - 12
Y.13: Olympiad Practice Test - 13
Z: Placement Assessments Level 5:
Z: Standardized Mock Tests:
Z.02: Standardized Mock Test - 2
Z.03: Standardized Mock Test - 3
Z.04: Standardized Mock Test - 4
Z.05: Standardized Mock Test - 5
Z.06: Standardized Mock Test - 6
Z.07: Standardized Mock Test - 7
Z.08: Standardized Mock Test - 8
Z.09: Standardized Mock Test - 9
Z.10: Standardized Mock Test - 10
Z.11: Standardized Mock Test - 11
Z.12: Standardized Mock Test - 12
Z.13: Standardized Mock Test - 13
Z.14: Standardized Mock Test - 14
Z.15: Standardized Mock Test - 15
Z.16: Standardized Mock Test - 16
Z.17: Standardized Mock Test - 17
Z.18: Standardized Mock Test - 18
Z.19: Standardized Mock Test - 19
Z.20: Standardized Mock Test - 20
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On a sequence of rational numbers with unusual divisibility by a power of 2
Artūras Dubickas
In this note we consider the sequence of rational numbers $b_n=\sum_{k=1}^n 2^k/k$. We show that the power of $2$ in the expansion of $b_n$ is unusually large, at least $n+1-\log_2(n+1)$, and that
this bound is best possible. The sequence $b_n$, $n=1,2,3,\dots$, is related to the sequence A0031449 in the On-Line Encyclopedia of Integer Sequences.
Vol. 25 (2024), No. 1, pp. 203-208
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How the MLineFromWKB() function works in Mariadb?
The MLineFromWKB() function is a useful tool for creating a multilinestring geometry from a well-known binary (WKB) representation.
Posted on
The MLineFromWKB() function is a useful tool for creating a multilinestring geometry from a well-known binary (WKB) representation. It can be used for various purposes, such as storing, querying, and
manipulating spatial data.
The syntax of the MLineFromWKB() function is as follows:
MLineFromWKB(wkb, [srid])
The function takes one or two arguments:
• wkb: A binary value that represents the well-known binary representation of the multilinestring geometry. It can be any valid expression that returns a binary value, such as a column name, a
literal, or a function. The well-known binary representation must follow the format specified by the Open Geospatial Consortium (OGC).
• srid: An optional integer value that represents the spatial reference system identifier (SRID) of the multilinestring geometry. It can be any valid expression that returns an integer, such as a
column name, a literal, or a function. The SRID must be a valid value in the spatial_ref_sys table, or 0 if the geometry has no SRID.
The function returns a multilinestring geometry value that represents the spatial object created from the well-known binary representation, with the specified SRID. If any of the arguments are NULL
or invalid, the function returns NULL.
In this section, we will show some examples of how to use the MLineFromWKB() function in different scenarios.
Example 1: Creating a multilinestring geometry from a literal well-known binary representation
Suppose you want to create a multilinestring geometry from a literal well-known binary representation. You can use the MLineFromWKB() function to do so. For example, you can execute the following
SELECT MLineFromWKB('0x000000000A00000002000000010000000200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002440000000000002440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003440000000000003440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000004440000000000004440000000000000000000000000000000000000000000000000000000000000000000000000000000000000004440000000000002440');
This will return a multilinestring geometry value that represents the spatial object created from the well-known binary representation, with no SRID.
Note that the well-known binary representation must follow the format specified by the OGC, which is:
• A byte order indicator, which is 0 for big-endian or 1 for little-endian.
• A geometry type indicator, which is 10 for multilinestring.
• A number of linestring geometries, which is an unsigned 32-bit integer.
• For each linestring geometry:
□ A geometry type indicator, which is 2 for linestring.
□ A number of points, which is an unsigned 32-bit integer.
□ For each point:
☆ An x-coordinate, which is a double-precision floating point number.
☆ A y-coordinate, which is a double-precision floating point number.
Example 2: Creating a multilinestring geometry from a column value
Suppose you have a table called paths that stores the information of various paths, such as their path_id, name, and wkb. The wkb column is a binary value that represents the well-known binary
representation of the multilinestring geometry of the path. You want to create a multilinestring geometry from the wkb column value of each path, so that you can store, query, and manipulate the
spatial data. You can use the MLineFromWKB() function to do so. For example, you can execute the following statement:
SELECT path_id, name, MLineFromWKB(wkb) AS geom FROM paths;
This will return the path_id, name, and the multilinestring geometry value of each path, or an empty result set if the table is empty.
Note that the multilinestring geometry value is a binary value that represents the spatial object in the internal format used by Mariadb. You can use some other functions to convert the binary value
to other formats, such as ST_AsText() or ST_AsGeoJSON().
Example 3: Creating a multilinestring geometry from a well-known binary representation with a specified SRID
Suppose you want to create a multilinestring geometry from a well-known binary representation, with a specified SRID, such as 3857. The SRID is a numeric value that identifies the spatial reference
system (SRS) of the geometry, which defines how the coordinates are projected on the earth’s surface. You can use the MLineFromWKB() function with the second argument to do so. For example, you can
execute the following statement:
SELECT MLineFromWKB('0x000000000A00000002000000010000000200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002440000000000002440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003440000000000003440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000004440000000000004440000000000000000000000000000000000000000000000000000000000000000000000000000000000000004440000000000002440', 3857);
This will return a multilinestring geometry value that represents the spatial object created from the well-known binary representation, with the SRID 3857.
Note that the SRID is a numeric value that identifies the spatial reference system (SRS) of the geometry, which defines how the coordinates are projected on the earth’s surface. The SRID 3857
corresponds to the Web Mercator projection, which is a widely used standard for web mapping. You can use the ST_SRID() function to get the SRID of a geometry, or the ST_SetSRID() function to set the
SRID of a geometry.
Related Functions
There are some other functions that are related to the MLineFromWKB() function and can be used to perform other operations on multilinestring geometries in Mariadb. Here are some of them:
• MLineFromText(): This function creates a multilinestring geometry from a well-known text (WKT) representation.
• ST_NumGeometries(): This function returns the number of linestring geometries in a multilinestring geometry.
• ST_GeometryN(): This function returns the nth linestring geometry in a multilinestring geometry.
• ST_Length(): This function returns the length of a multilinestring geometry.
The MLineFromWKB() function is a powerful and flexible function that can help you create a multilinestring geometry from a well-known binary representation. It can be used for various purposes, such
as storing, querying, and manipulating spatial data. You can also use some other related functions to create, convert, or access multilinestring geometries, such as MLineFromText, MLineToText,
MLineToWKB, ST_NumGeometries, ST_GeometryN, or ST_Length. By using these functions, you can achieve a better analysis and understanding of your spatial data.
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A non-techie explanation of how web pages are secured
I first published this overview back in 2004. I'm prompted to update and repost it here because the original website is no longer live (it was the website for Fuse PR back in the day), because I was
talking about it with some Meanwhile colleagues this week, and because it's interesting. No seriously, it is.
When is a web page secure?
Content on a web page is secured when the URL starts "https://" rather than "http://". Saying that, this doesn't mean that all content on a page marked as such is secured, but if you're on a website
of a trusted source, like Amazon, you may be happy to assume that the parts that need to be are, by design.
If you'd like to find out what other information your browser gives you about the level of security for a particular web page, check out the relevant link here:
What do we mean by 'secured'?
We mean that information sent from your browser to the web server and sensitive information sent from the web server to your browser is encrypted so that, if the data was intercepted, it would be
meaningless to the interceptor.
Occasionally, a web page can be secure without any of the visual indications or confirmations described above. This is when a secure web page is served within (a frame of) another webpage. This is
poor website design, and if you are unsure if a web page is secure, act as if it's definitely not.
Why not make all web pages secure?
Cryptography places a considerable calculation burden on the processor power of web servers and your own computer. Whilst the latter is unlikely to be tasked for long, highly popular websites would
really struggle to encrypt all pages and the time taken to access these pages would take longer. Therefore, browsing the latest weather forecast or tonight's TV schedule isn't encrypted, but your
online banking definitely is.
How does this encryption work exactly? Let's get the acronyms out of the way, and then we'll take a look at the basic mathematics. If you can read clocks then the mathematical basis is a breeze!
Hypertext Transfer Protocol (HTTP) is a set of rules for transferring data files over the World Wide Web. Transmission Control Protocol (TCP) is used with Internet Protocol (IP) to divide this data
up into manageable little packets for efficient shipping across the Internet.
The Secure Sockets Layer (SSL) is inserted between the HTTP and TCP layers to undertake the encryption and decryption task for secure web pages. Originally developed by the company behind the first
popular browser, Netscape (which effectively became Firefox upon the formation of the Mozilla Foundation), SSL reached near ubiquitous application across all makes of browser.
Like all technologies, SSL has evolved and since SSL 3.0 it has become something called Transport Layer Security (TLS). This improved protocol is included in all modern browsers.
I have a key to secure my home, a key to secure my car. Where's the key here?
SSL and TLS use something known as public-and-private key encryption from a company called RSA. This is a different kind of key to the physical ones you use for your home or car. The most striking
difference is that, whilst the same key is used to lock and unlock your front door, SSL uses one key to lock (encrypt) the information and another key to unlock it.
This feature is critical to its success. It means that there is no need to restrict access to or be secretive about the key used to lock the information as it is useless for unlocking the
information. This key is therefore known as the public key. The unlocking key is known as the private key.
It is the openness surrounding the public key that means the general user is unaware of the process being undertaken. As there is no security risk associated with knowing the public key, your browser
automatically requests the public key for locking information on secure web pages. It just goes ahead and locks it up.
Walk me through this locking and unlocking process
For anyone interested in mathematics, this whole cryptography revolution harks back to the work done on clock calculators by Gauss and on a theorem proved by Fermat – not his last one, but one known
as Fermat's Little Theorem. In hindsight, it's amazingly simple.
If you ask anyone to add the numbers 9 and 4 you will get the answer 13. Similarly, if you ask them at 9 o'clock what the time will be in 4 hours, they will tell you 1 o'clock. Why do we get the
answers 13 and 1 to very similar questions? In the instance of telling the time we know there are 12 hours on the clock, so we are actually adding 9 and 4 and, if it is greater than 12, subtracting
12. We keep moving round a clock with 12 numbers.
Another question could be "It's 9 o'clock now, what time will it be in 20 hours?" in which case the answer is 9 + 20 - 12 - 12 = 5 o'clock. It seems we keeping subtracting twelves until we get an
answer that lies between 0 and 12. This is known as modular arithmetic – a form of arithmetic where numbers are considered equal if they leave the same remainder when divided by the same number
In modular arithmetic where the modulus is 12 (as for our clock example here):
9 = 21 = 33 = 45 because
• 12's go into 21 once, leaving a remainder of 9
• 12's go into 33 twice, leaving a remainder of 9
• 12's go into 45 three times, leaving a remainder of 9.
The slightly harder bit...
Gauss found an appealing characteristic related to an earlier discovery by Fermat if he undertook similar calculations using clocks with a prime number of hours on it instead of 12. A prime number is
a number than cannot be divided exactly by any other number except itself and 1. The numbers 1, 2, 3, 5, 7, 11, 13, 17 and 19 are all prime numbers. All the other numbers up to 20 are not prime as
they can be divided exactly by other numbers. For example 15 can be divided exactly by itself and 1, but also by 3 and 5.
When using a prime number clock with P hours, if you take a number X and raise it to the power P then you get back to the same number you started with.
For example, using the 7-hour clock shown here (P=7) and the initial number 3 (X=3), then 3 to the power 7 = 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187.
Sevens go into this number 312 times leave a remainder of 3. Back where we started. Or to put it another way, we go forward 2187 hours on our 7-hour clock and see what time we come to – 3 o'clock.
Although Fermat claimed to have proved this theorem, he died before telling anyone how! It was left to another distinguished mathematician, Leonard Euler, to provide the proof in 1736 that this
worked for all prime numbers and any number X.
Euler took things further by looking at semiprime numbers too. A semiprime can only be divided by itself, 1 and two prime numbers. In other words, a semiprime N = p x q where both p and q are prime
numbers. For semiprime number clocks, Euler found that the pattern got back to the beginning after raising the original number to the power of (p-1) x (q-1) + 1.
Let's go shopping
We are now nearly there. Let's look at how you give Amazon your credit card number, securely.
Amazon's computers select two very large prime numbers, p and q, of around 60 digits each and multiply them together to make a third number N. We are therefore using a clock with a massive number of
hours. Massive. In fact, the number is usually bigger than there are atoms in the universe!
The number N is published as part of the public key, but p and q are kept secret. It is very very difficult, almost impossible without many years and an incredibly powerful computer, to work out what
p and q are from N. In fact it is so secure that Amazon will continue to use the same number N for several months.
The other part of the public key is called the encoding number, E. So now what happens to your credit card number C (or you might consider C to stand for the digital representation of any content to
be secured)?
Your browser does a calculation on C based on the clock with the massive number of hours and the encoding number E. It raises C to the power E and works out what the number is on the clock in the
same way we did for much smaller numbers above, and transmits this number to Amazon. In other words, your browser has used the public key to encrypt your credit card number.
If anyone intercepted the transmission they could not calculate your credit card number. They know Amazon's public key (N and E) but you cannot use these to reverse the calculation.
However, Amazon can calculate the credit card number because they know p and q, the private key. They know that if your credit card number was raised to the power of (p-1) x (q-1) + 1, that the same
number reappears.
As your browser has raised the number to the power of E already, it simply remains for Amazon's computer to raise the result further, by (p-1) x (q-1) + 1 - E on the same clock with N hours, to get
back to the same number. Mission accomplished, and Stephen Hawking's God Invented the Integers will be dropping on your doormat soon. That's what you were shopping for right?
Next time you're securely online, think of Fermat, Gauss and Euler, and the three mathematicians at RSA who brought this work from the 17th Century and applied it to the world of the Internet -
Rivest, Shamir and Adleman.
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implicitplot2D function - looking for documentation
Oct 02, 2024 02:53 PM
Oct 02, 2024 02:53 PM
In the thread on "Roots of polynomial #3 and How to find unique values into a vector/matrice?" there is a Sept. 23. 2024 post from Werner that uses a function implicitplot2D. I gather from the
results and descriptions that this function returns the x-y pts within a defined region where the specified real/complex function = 0 (or close?).
I got around to playing with the attached mcd11 file that has the programmed definition fro this function, and I'm interested in the implicitplot2D function details. I started looking at the function
program implementation in the mcd11 file, but it's not something I will work out in the next few minutes.
Diverting to the easier path for the moment (find out what has already been done), I found a reference to an implicitplot2D function link in Werner's Mar. 5, 2017 post. The link leads to https://
community.ptc.com/t5/Mathcad/implicitplot3d/m-p/332011 ("3d" seems to be in the link, although it advertises itself as the 2d link). When I try to access this while logged in, I get a "Permission
Required" message. Perhaps someone else can access and post the content of the link.
I searched the author and function - Viacheslav N. Mezentsev - but did not find any useful results.
Does anyone know of some other documentation on the procedure used in this implementation? As one example unknown, there are two constant value arrays defined - EdgeTable and TriTable. It's not
obvious to me how the values came about, or what properties of the procedure depend on these.
While I hope some documentation will appear, I 'll continue to plug away and try to determine what's under the hood of this function.
A mcd121 copy of the function from Werner's file is attached.
Oct 03, 2024 05:11 PM
Oct 03, 2024 05:11 PM
Oct 02, 2024 04:53 PM
Oct 02, 2024 04:53 PM
Oct 03, 2024 08:46 AM
Oct 03, 2024 08:46 AM
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Oct 03, 2024 05:11 PM
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Oct 04, 2024 06:29 PM
Oct 04, 2024 06:29 PM
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