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Pearson Correlation Coefficient Calculator Online | How to Find Pearson Coefficient? - Statisticscalculators.net
Pearson Correlation Coefficient Calculator
Pearson Correlation Coefficient Calculator OR pearson calculator OR pearson's correlation coefficient calculator OR pearson correlation calculator helps you determine the pearson correlation
coefficient between two variables easily. All you need to do is enter the input values in the input sections allotted and tap on the calculate button to avail the pearson correlation coefficient in a
matter of seconds.
What is meant by Pearson Correlation Coefficient?
Pearson Correlation Coefficient is one of the types to determine the correlation coefficient. It is used to measure the strength and direction of linear relation between two random variables.
Pearson's Correlation Coefficient is denoted by the symbol r and it is often called as Pearson's r.
Pearson Correlation Coefficient Formula and Properties
Mathematical Formula of Pearson Correlation Coefficient r is nothing but the covariance of two variables divided by the product of their respective standard deviations.
• We can understand that based on Cauchy–Schwarz inequality absolute value of correlation coefficient never exceeds 1.
• Correlation is symmetric i.e. relation between X and Y is same as Y and X.
• In case of variables being independent correlation is 0 but the converse is not true.
Interpretation of Pearson Correlation Coefficient
Sign of Pearson Correlation Coefficient determines the relationship direction. Go through below modules and see various cases
• If r is positive one variable increases the other also increases.
• If r is negative one variables decreases as the other one increases.
• Pearsons Value ranges from -1 to +-1
• The close it is to +-1 the stronger is the relation between the variables.
• If at all r equals -1 or +1 then all the data points lie on one line.
• If r is equal to 0 there will be no relation between the data existing.
How to find Pearson Correlation Coefficient Manually?
Go through the simple process listed below for finding the Pearson Correlation Coefficient r. They are along the lines
• Firstly, figure out the two data sets.
• Later find the mean of both the data sets x and y i.e. x[mean], y[mean]
• Now, find out the sums of squares x, y and their dot products.
• Apply all the values found in the formula of pearson correlation coefficient.
• Perform basic math calculations and compute the value of r.
How to use the Pearson Correlation Coefficient Calculator?
Follow the simple and easy steps listed below regarding the usage of pearson coefficient calculator. They are along the lines
• First and foremost step is to provide the data values x, y separated by commas in the input fields.
• Later click on the calculate button to get the result pearson correlation coefficient.
• Finally, you will get the pearson correlation coefficient displayed in a new window showing the work.
FAQs on Pearson Correlation Coefficient Calculator
1. What does Pearson's Correlation Coefficient tell you?
Pearson's Correlation Coefficient tells us the strength and direction of linear relation between two random variables.
2. What does a Pearson Coefficient of 0 mean?
Pearson Coefficient 0 tells us that there is no linear relationship between the data set present.
3. What is the formula for Pearson Correlation Coefficient?
The Formula for Pearson Correlation Coefficient is r = [n(∑xy) - (∑x)(∑y)] / √(n∑x² - (∑x)²) (n∑y² - (∑y)²)
4. Is there any website that offers the Pearson Correlation Coefficient Calculator for free?
statisticscalculators.net is the best portal that offers the Pearson Correlation Coefficient Calculator for free to aid your calculations. | {"url":"https://statisticscalculators.net/statistics/pearson-correlation-coefficient-calculator/","timestamp":"2024-11-11T09:29:13Z","content_type":"text/html","content_length":"30697","record_id":"<urn:uuid:e070ba68-2639-4a27-ac74-647ebe32e5a0>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00627.warc.gz"} |
Generating Unique Random Numbers In JavaScript Using Sets
Generating Unique Random Numbers In JavaScript Using Sets
Amejimaobari Ollornwi
JavaScript comes with a lot of built-in functions that allow you to carry out so many different operations. One of these built-in functions is the Math.random() method, which generates a random
floating-point number that can then be manipulated into integers.
However, if you wish to generate a series of unique random numbers and create more random effects in your code, you will need to come up with a custom solution for yourself because the Math.random()
method on its own cannot do that for you.
In this article, we’re going to be learning how to circumvent this issue and generate a series of unique random numbers using the Set object in JavaScript, which we can then use to create more
randomized effects in our code.
Note: This article assumes that you know how to generate random numbers in JavaScript, as well as how to work with sets and arrays.
Generating a Unique Series of Random Numbers
One of the ways to generate a unique series of random numbers in JavaScript is by using Set objects. The reason why we’re making use of sets is because the elements of a set are unique. We can
iteratively generate and insert random integers into sets until we get the number of integers we want.
And since sets do not allow duplicate elements, they are going to serve as a filter to remove all of the duplicate numbers that are generated and inserted into them so that we get a set of unique
Here’s how we are going to approach the work:
1. Create a Set object.
2. Define how many random numbers to produce and what range of numbers to use.
3. Generate each random number and immediately insert the numbers into the Set until the Set is filled with a certain number of them.
The following is a quick example of how the code comes together:
function generateRandomNumbers(count, min, max) {
// 1: Create a `Set` object
let uniqueNumbers = new Set();
while (uniqueNumbers.size < count) {
// 2: Generate each random number
uniqueNumbers.add(Math.floor(Math.random() * (max - min + 1)) + min);
// 3: Immediately insert them numbers into the Set...
return Array.from(uniqueNumbers);
// ...set how many numbers to generate from a given range
console.log(generateRandomNumbers(5, 5, 10));
What the code does is create a new Set object and then generate and add the random numbers to the set until our desired number of integers has been included in the set. The reason why we’re returning
an array is because they are easier to work with.
One thing to note, however, is that the number of integers you want to generate (represented by count in the code) should be less than the upper limit of your range plus one (represented by max + 1
in the code). Otherwise, the code will run forever. You can add an if statement to the code to ensure that this is always the case:
function generateRandomNumbers(count, min, max) {
// if statement checks that `count` is less than `max + 1`
if (count > max + 1) {
return "count cannot be greater than the upper limit of range";
} else {
let uniqueNumbers = new Set();
while (uniqueNumbers.size < count) {
uniqueNumbers.add(Math.floor(Math.random() * (max - min + 1)) + min);
return Array.from(uniqueNumbers);
console.log(generateRandomNumbers(5, 5, 10));
Using the Series of Unique Random Numbers as Array Indexes
It is one thing to generate a series of random numbers. It’s another thing to use them.
Being able to use a series of random numbers with arrays unlocks so many possibilities: you can use them in shuffling playlists in a music app, randomly sampling data for analysis, or, as I did,
shuffling the tiles in a memory game.
Let’s take the code from the last example and work off of it to return random letters of the alphabet. First, we’ll construct an array of letters:
const englishAlphabets = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
// rest of code
Then we map the letters in the range of numbers:
const englishAlphabets = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
// generateRandomNumbers()
const randomAlphabets = randomIndexes.map((index) => englishAlphabets[index]);
In the original code, the generateRandomNumbers() function is logged to the console. This time, we’ll construct a new variable that calls the function so it can be consumed by randomAlphabets:
const englishAlphabets = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
// generateRandomNumbers()
const randomIndexes = generateRandomNumbers(5, 0, 25);
const randomAlphabets = randomIndexes.map((index) => englishAlphabets[index]);
Now we can log the output to the console like we did before to see the results:
const englishAlphabets = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
// generateRandomNumbers()
const randomIndexes = generateRandomNumbers(5, 0, 25);
const randomAlphabets = randomIndexes.map((index) => englishAlphabets[index]);
And, when we put the generateRandomNumbers() function definition back in, we get the final code:
const englishAlphabets = [
'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M',
'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'
function generateRandomNumbers(count, min, max) {
if (count > max + 1) {
return "count cannot be greater than the upper limit of range";
} else {
let uniqueNumbers = new Set();
while (uniqueNumbers.size < count) {
uniqueNumbers.add(Math.floor(Math.random() * (max - min + 1)) + min);
return Array.from(uniqueNumbers);
const randomIndexes = generateRandomNumbers(5, 0, 25);
const randomAlphabets = randomIndexes.map((index) => englishAlphabets[index]);
So, in this example, we created a new array of alphabets by randomly selecting some letters in our englishAlphabets array.
You can pass in a count argument of englishAlphabets.length to the generateRandomNumbers function if you desire to shuffle the elements in the englishAlphabets array instead. This is what I mean:
generateRandomNumbers(englishAlphabets.length, 0, 25);
Wrapping Up
In this article, we’ve discussed how to create randomization in JavaScript by covering how to generate a series of unique random numbers, how to use these random numbers as indexes for arrays, and
also some practical applications of randomization.
The best way to learn anything in software development is by consuming content and reinforcing whatever knowledge you’ve gotten from that content by practicing. So, don’t stop here. Run the examples
in this tutorial (if you haven’t done so), play around with them, come up with your own unique solutions, and also don’t forget to share your good work. Ciao! | {"url":"http://fdswebdesign.com/index.php/2024/08/26/generating-unique-random-numbers-in-javascript-using-sets/","timestamp":"2024-11-14T12:10:35Z","content_type":"text/html","content_length":"47793","record_id":"<urn:uuid:c8aa4acc-a773-4f3b-a080-b3a14d09fbd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00279.warc.gz"} |
How To Compare The Size Of An Atom
When comparing atoms to larger objects — with a large disparity in size — orders of magnitude show how to quantify the size differences. Orders of magnitude allow you to compare the approximate value
of an extremely small object, such as the mass or diameter of an atom, to a much larger object. You can determine the order of magnitude using scientific notation to express these measurements and
quantify the differences.
TL;DR (Too Long; Didn't Read)
To compare the size of a large atom to a much smaller atom, the orders of magnitude allow you to quantify the size differences. Scientific notations help you to express these measurements and assign
a value to the differences.
The Tiny Size of Atoms
The average diameter of an atom is 0.1 to 0.5 nanometers. One meter contains 1,000,000,000 nanometers. Smaller units, such as centimeters and millimeters, typically used to measure small objects that
can fit within your hand, are still much larger than a nanometer. To carry this further, there are 1,000,000 nanometers in a millimeter and 10,000,000 nanometers in a centimeter. Researchers
sometimes measure atoms in ansgtoms, a unit that equals 10 nanometers. The size range of atoms is 1 to 5 angstroms. One angstrom equals 1/10,000,000 or 0.0000000001 m.
Units and Scale
The metric system makes it easy to convert between units because it is based on powers of 10. Each power of 10 is equal to one order of magnitude. Some of the more common units for measuring length
or distance include:
• Kilometer = 1000 m = 103 m
• Meter = 1 m = 101 m
• Centimeter = 1/100 m = 0.01 m = 10-2 m
• Millimeter = 1/1000 m = 0.001 m = 10-3 m
• Micrometer = 1/1,000,000 m = 0.000001 m = 10-6 m
• Nanometer = 1/1,000,000,000 m = 0.000000001 m = 10-9 m
• Angstrom = 1/10,000,000,000 m = 0.00000000001 m = 10-10 m
Powers of 10 and Scientific Notation
Express powers of 10 using scientific notation, where a number, such as a, is multiplied by 10 raised by an exponent, n. Scientific notation uses the exponential powers of 10, where the exponent is
an integer that represents the number of zeros or decimal places in a value, such as: a x 10n
The exponent makes large numbers with a lengthy series of zeros or small numbers with many decimal places much more manageable. After measuring two objects of vastly different sizes with the same
unit, express the measurements in scientific notation to make it easier to compare them by determining the order of magnitude between the two numbers. Calculate the order of magnitude between two
values by subtracting the difference between its two exponents.
For example, the diameter of a grain of salt measures 1 mm and a baseball measures 10 cm. When converted to meters and expressed in scientific notation, you can easily compare the measurements. The
grain of salt measures 1 x 10^-3 m and the baseball measures 1 x 10^-1 m. Subtracting -1 from -3 results in an order of magnitude of -2. The grain of salt is two orders of magnitude smaller than the
Comparing Atoms with Larger Objects
Comparing the size of an atom to objects large enough to see without a microscope requires much greater orders of magnitude. Suppose you compare an atom that has a diameter of 0.1 nm with a size AAA
battery that has a diameter of 1 cm. Converting both units to meters and using scientific notation, express the measurements as 10^-10 m and 10^-1 m, respectively. To find the difference in the
orders of magnitude, subtract the exponent -10 from the exponent -1. The order of magnitude is -9, so the diameter of the atom is nine orders of magnitude smaller than the battery. In other words,
one billion atoms could line up across the diameter of the battery.
The thickness of a sheet of paper is about 100,000 nanometers or 105 nm. A sheet of paper is about six orders of magnitude thicker than an atom. In this example, a stack of 1,000,000 atoms would be
the same thickness as sheet of paper.
Using aluminum as a specific example, an aluminum atom has a diameter of about 0.18 nm compared with a dime that has a diameter of about 18 mm. The diameter of the dime is eight orders of magnitude
greater than the aluminum atom.
Blue Whales to Honeybees
For perspective, compare the masses of two objects that can be observed without a microscope and are also separated by several orders of magnitude, such as the mass of a blue whale and a honeybee. A
blue whale weighs about 100 metric tons, or 10^8 grams. A honeybee weighs about 100 mg, or 10^-1 g. The whale is nine orders of magnitude more massive than the honeybee. One billion honeybees have
about the same mass as one blue whale.
Cite This Article
Mentzer, A.P.. "How To Compare The Size Of An Atom" sciencing.com, https://www.sciencing.com/compare-size-atom-7378966/. 23 April 2018.
Mentzer, A.P.. (2018, April 23). How To Compare The Size Of An Atom. sciencing.com. Retrieved from https://www.sciencing.com/compare-size-atom-7378966/
Mentzer, A.P.. How To Compare The Size Of An Atom last modified March 24, 2022. https://www.sciencing.com/compare-size-atom-7378966/ | {"url":"https://www.sciencing.com:443/compare-size-atom-7378966/","timestamp":"2024-11-08T21:06:33Z","content_type":"application/xhtml+xml","content_length":"76055","record_id":"<urn:uuid:391d2d7b-577c-481f-b030-4f12c9570915>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00092.warc.gz"} |
Best Scientific Calculator 2022 [Buying Guide] - mytechreviewer.com
The Best Scientific Calculators (2022 Reviews)
By Justin Stuart
This author has been vetted and has the necessary know-how or education to be able to write about this topic. Learn more on our about page.
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Whether you’re heading to high school or have a job which requires you to do advanced mathematical calculations, a scientific calculator will make your life a lot easier. To help you out, we’ve done
all of the research and legwork for you.
In this article, we’ll be reviewing some of the best scientific calculators on the market so that you can find one which fits your needs and budget perfectly.
Top 6 Best Scientific Calculators For 2022
Take a look at the table below; in it, you’ll find the name of each calculator we’ll be examining, as well as a couple of important specifications. We have a lot more to say about each of them,
though, so don’t make a decision just yet.
Let’s talk a little bit about our evaluation process. We’ll be looking for any strengths and weaknesses that each product might have, and at the very end, we’ll decide which is the best scientific
calculator overall based on our findings.
That said, let’s get started with the Casio FX-260 SOLAR.
Casio fx-260 SOLAR – Best Budget Scientific Calculator
If you just started looking for a scientific calculator, there’s a good chance you were shocked by how expensive some of them can be. Luckily, there are high-quality calculators out there which won’t
break the bank, like the Casio FX-260 SOLAR.
It costs around 6 USD, so it’s definitely affordable.
You also don’t have to worry about spending money on additional batteries since it’s powered by solar energy (hence the name). However, with such an inexpensive product, there are some downsides.
For instance, its display only shows a single line. Longer calculations can be shown, but you’ll have to scroll down to see anything after the first ten digits.
One area that hasn’t been compromised on, though, is the functionality.
This calculator can handle any high school level maths problems. It can calculate both square and cube roots of numbers, perform trigonometric functions, simplify fractions, calculate logarithms, and
display numbers in scientific notation.
This is just a small taste of what this product can do. Casio claims that it has over 140 different mathematical functions, and the good news is that you can even take it into the exam room with you.
It’s approved for use in PSAT/NMSQT, SAT and ACT entrance exams, and also AP tests.
Casio claims that this product has a long lifespan, but unfortunately, there’s no warranty for this product. This, regrettably, is to be expected for something which costs so little but to prolong
its life it comes with a slide-on hard case.
You can also choose to get a black or pink version, so there’s a little room for personalization here too.
The buttons are made of hard plastic instead of rubber, which again, is mostly due to the product’s low cost. This shouldn’t have any real downsides, but it does make the calculator seem lower
quality than it actually is.
We do like how small it is, though.
It’s just 6” tall, 4” wide and an inch thick, and weighs just 4.8 ounces. This makes it the perfect size to fit into a pocket or bag, and ensures that no matter how many textbooks you have to bring
with you, you’ll always have room for it.
All things considered, this is probably one of the best calculators you can find for under 10 USD. It has a wide range of different functions that can take care of even the most difficult high school
problems and conversions, so if you buy it when you first enroll, you’ll be covered until you graduate at the very earliest.
What We Liked
• Extremely affordable
• Can handle all high school level problems
• Comes in two colours
What We Didn’t Like
• No warranty
• Single line display
Texas Instruments TI-30X IIS – Best All-round Budget Scientific Calculator
Some scientific calculators focus only on the most advanced equations and neglect to address the simpler ones. The Texas Instruments TI-30X IIS is not one of these: it comes with the capability to
solve just about anything and several tweaks which make it a joy to use.
Firstly, let’s address the elephant in the room: it’s probably going to be expensive, right? Wrong. It retails for a little over 10 USD and stands shoulder to shoulder with some of the more expensive
calculators on the market.
If the standard black color scheme doesn’t do it for you, there are further eight colors to choose from.
We know that a calculator isn’t exactly a fashionable item, but at the very least it’ll help you stand out.
This calculator has a two-line display with a maximum of ten characters per line. Larger problems can be navigated a line at a time using the arrow keys and the last entered calculations are saved
for easy access.
It also has cut and paste functionality. Once an equation has been solved, you can go back and edit it, factoring in other numbers to see how it changes the answer. Additionally, you can quickly
switch between scientific and engineering notation.
The TI-30X ISS is primarily solar-powered, but we know that sometimes there just isn’t enough light in the classroom.
For this very reason, Texas Instruments have included a battery as a backup – this means that there’s no way the calculator can die at a vital moment, for example in the middle of an exam.
Make no mistake, you’ll want to bring this with you.
Whether you’re trying to convert something, simplify a fraction, solve a trigonometry problem, or wrap your head around algebra, this calculator can help. It even supports one and two-variable
statistical problems, which are outside of the bounds of a high school math class.
So what’s the catch? Well, it might not be up to scratch if you’re studying a college-level course.
For example, it can’t create graphs or tables, but really, for under 20 USD, you’d be hard-pressed to find a calculator that can.
Overall, the TI-30X IIS from Texas Instruments is a well-balanced calculator that can take you all the way through school. Its equation editing ability makes it a lot easier to learn exactly how to
solve difficult problems, and the low price tag makes it a very attractive product indeed.
It even comes with a one-year limited warranty, so you know it’s built to last.
What We Liked
• Wide range of functions
• Allows you to quickly edit and rerun calculations
• Comes in nine colours
What We Didn’t Like
• Not suitable for high-level maths courses
Casio fx-115ES PLUS – Best Advanced Scientific Calculator
This is the second Casio calculator we’ve covered so far, but there’s a reason for it. Casio is known for their high-quality products, and FX-115ES PLUS just proves how much bang you get for your
buck with this company.
It comes packed with over 280 different mathematical functions.
Fractions? No problem. Statistics? Great. Even more complex problems like linear regression and polar-rectangular conversions are a cinch for this model.
What’s more, it allows you to step backward through calculations for a better view of how exactly a problem was solved. You also have the option to edit these, which allows you to substitute other
numbers in – great for if you messed up several steps ago.
So how much does a calculator like this cost? Surprisingly little: it has a recommended retail price of less than 15 USD so it’s affordable no matter how small your budget is.
Some products allow you to convert between decimals and fractions, but that’s not enough for Casio.
This product allows you to convert numbers between forty different measurements so you’ll never find yourself struggling to work out how many inches are in a mile.
Like the TI-30X IIS, this product uses a hybrid solar power and battery power method to function.
It’s unlikely that you’ll have to rely on the battery, but if it does kick in, you can breathe a sigh of relief knowing that a lesser model would have failed you.
So does this calculator have any flaws? Well, yes, but they aren’t deal-breakers.
For a start, it comes with a grey and purple colour scheme which, to be honest, looks a little dorky. Its huge range of functionality may be a double-edged sword too – with so many options to choose
from, it can be a little overwhelming for someone with fairly simple needs.
So what’s our overall opinion of this product? Basically, we think it’s great for anyone heading into an advanced class, or who plans to study areas like engineering or computer science. However, if
you just need it for a calculus class, it’s probably overkilling.
What We Liked
• Almost 300 different mathematical functions to choose from
• Huge range of conversion types
• Has a battery backup power source
What We Didn’t Like
• May be too advanced for beginners
Texas Instruments TI-36X Pro – Best Mid-Budget Scientific Calculator
Have you ever heard the old saying “you get what you pay for”? A lot of the time it doesn’t hold up, but when it comes to scientific calculators, there is actually a grain of truth to it. Case in
point: The Texas Instruments TI-36X Pro is a little more expensive, but a lot stronger than its competitors.
Get this: it retails for a little under 20 USD, but it can handle just about any high school or college level problem. We say just about any because it doesn’t have any graphing functionality, but
any standard numerical or algebraic equations are fair game for this model.
One of the great things about the TI-36X Pro is that it displays equations and formulae the same way they appear in the textbooks. Fractions are stacked instead of being shown horizontally, and often
this can be the difference between total understanding and confusion.
Its screen has enough room for four lines of text, and each line holds up to thirteen digits (including three after a decimal point), so even the longest equations can be viewed on one screen. If you
do happen to enter something too long for a single screen, you can move up or down one line at a time using the arrow keys.
There are several different conversion settings (more than three dozen, in fact) that allow you to view a number in radians, floating-point form, or even hexadecimal. It can calculate the base
logarithm of any number too, which is excellent for computer science students.
Additionally, it allows you to calculate the standard deviation using one or two variables, so it’s a fantastic choice for anyone looking to study statistics or advanced mathematics. It may be a
little too advanced for people with simpler needs, though.
Here’s the deal: this scientific calculator is capable of solving most high school problems and even a good portion of college-level ones.
If you’re just looking for a standard calculator to help with trigonometry, you’d probably be better suited to one of the less expensive products.
That said, if you plan to pursue a career in a scientific field like engineering, this calculator will give you the tools that you need in order to succeed.
There are just two things that it can’t do: give you the willpower to make it to class, and graphs.
Are you looking for a graphing calculator? Check our guide about the best graphing calculators, this will help you to choose what to buy. Read it right here.
What We Liked
• Perfect for college-level students
• Shows equations as they appear in the textbooks
• Over 36 different conversions
What We Didn’t Like
• More expensive than most of its competitors
Sharp EL-W516TBSL – Best Scientific Calculator for College
Alright, now we’re getting onto the heavy duty scientific calculators. The Sharp EL-WT516TBSL is a model that can make even the most difficult maths problems run away with their tail between their
It costs a little over 20 USD, but you really get a lot of value for money with this product: some of the others we’ve seen so far can handle almost 300 functions, but this one goes above and beyond
with 640 different functions on offer.
Everything from cubic polynomials to imaginary numbers is accounted for. It can easily deal with quadratic and cubic equations, and even three-variable linear equations don’t pose much of a threat.
There’s even a dedicated equation mode that allows you to insert variables into your problems.
But there’s more: the calculator’s display can show four lines of text, with a huge 16 digits per line. Equations are shown as they are in the teaching material, too, with fractions appearing
vertically instead of horizontally.
You can also store up to four different formulas using the F keys.
Despite this feature, it still doesn’t count as a programmable calculator, so you’ll be able to take it into an exam room, no problem. It’s nice and small too, at just 6.6” tall, 3.1” wide, and 0.6”
When you’re paying a little more for a product, you expect it to be practical whenever you need it, which is why this calculator has a dual power system.
It primarily uses solar energy but switches to the battery backup when there’s too little light.
Can we address the elephant in the room for a second? This calculator actually looks great. It has a sleek black and silver colour scheme and the button labels are an interesting mint and orange
It actually looks and feels like a high-end product, even though it costs less than 25 USD.
So, our final thoughts: this calculator is aimed squarely at people with advanced mathematical needs. It can’t handle graphing, though, which is unfortunate, because that would have made this the
ultimate scientific calculator, hands down.
However, despite this limitation, it holds up very well. Regardless of whether you’re going to high school or university, this product will serve you well for years to come, particularly if your
chosen area of study is heavily mathematical.
What We Liked
• Can handle just about any problem other than graphs
• Can display several lines of text on screen simultaneously
• Very attractive casing
What We Didn’t Like
HP 35S – Best Premium Scientific Calculator
If you have the budget for it, and the need for a scientific calculator far more powerful than any of the others we’ve seen, the HP 35S may be the one for you. It costs significantly more, though, at
around 55 USD – considering this is a ten-year-old model, that’s a lot.
So how does this product justify its high price tag? Well, firstly, it has a high-quality construction. Its front panel is glossy black plastic, and each button has a nice colour-coded explanation of
its supplementary functions. It even comes with a genuine vinyl case, which is a nice touch.
More importantly, though, it has a functionality that the other products can’t even dream of matching. Simply, there are so many different functions available that it’s not even worth listing them
Suffice it to say: if you can dream it, you can do it (except for graphs).
Better still, if there’s something you want to do that doesn’t come as standard, you can add it since this is a programmable calculator.
It offers an additional degree of sophistication by allowing you to loop through equations multiple times, perform conditional tests to determine the result, and even create subroutines which can be
as many as 20 levels deep.
Now, this isn’t a solar-powered calculator. It relies upon its two batteries, but there’s a twist. These are wired in parallel which means that you can replace one at a time without losing all of the
functions that you’ve saved.
One of the only disadvantages of this model is its small screen size.
It shows just two lines of text at any one time and whilst it displays 14 digits per line, it really would have been nice to have a slightly larger display.
We did like that you can store up to 26 intermediate values at any one time though.
For more complex calculations, you’ll likely need a few of these, but it’s always good to have more than you need.
Calculators can be very confusing, and while you’re unlikely to read the entire 200-page digital manual, we’re glad that this calculator came with such in-depth documentation.
One thing is clear: it’s extremely versatile, extremely powerful, and although it costs a bit more, it justifies this by being the last non-graphing calculator that you’ll ever need to buy.
What We Liked
• Fully customisable – allows you to add your own functions
• Has a backup battery which can be replaced without turning the calculator off
• In-depth algorithm customisation options
What We Didn’t Like
How To Pick The Right Scientific Calculator For Me?
Hey, we get it. Maths is hard, and since most calculators brag about the advanced functions they offer, it can be difficult to accurately compare two models. However, once you know what you’re
looking for, it’s actually fairly simple.
To help you out, we’ve created a list with the aim of helping you find a scientific calculator that fits your needs, and more importantly, your budget.
First and foremost, you’ll have to consider how many of the functions offered are actually useful. For example, if you’re buying this calculator for high school level work, you’ll likely need one
which can solve quadratic equations and basic algebra problems. For an example of some of the problems you’ll need to solve, you can check out this sample curriculum.
You should also be keeping an eye out for one with a decently-sized display. Not only does this allow more information to be on the screen at one time, but it also allows them to display equations
the same way that they’re shown in textbooks, which makes them easier for learners to understand.
If you have higher ambitions, though, you’ll need a calculator that can deal with more advanced topics. Statistical analysis is a plus here, as is the ability to program in your own functions. Simply
put, you need a calculator that will last for your entire college career, not just for freshman year.
Other Considerations
There are a number of other things you should take into account. Take the screen, for instance: does it come with a protector, or is it prone to scratches? How large is the product? Is it heavy? Does
it come in a choice of colors or is it grey only?
Possibly the most important thing to think about is whether or not you’ll be allowed to take it into an exam room. There are specific models of calculator which are approved for certain tests, and
they usually mention this on the packaging, however you can also find out by looking up the specific exam online.
How is the calculator powered? Models that use solar energy are good because they don’t require additional batteries, however in dimly lit places, they can die without warning. If you have saved
functions or equations, these will disappear when this happens. Conversely, a product that uses batteries has ever so slightly higher running costs but is less likely to run out of power when you
need it most.
Some models even use a combination of these two power sources which gives you the best of both worlds. This allows you to change the battery without losing any of your stored information too since
the calculator will remain on even when the battery is removed as long as it’s in direct sunlight.
Choosing just one of these scientific calculators as the best overall was particularly difficult given how wildly people’s needs can vary. However, after careful consideration of the facts, we’ve
settled on the Casio fx-115ES PLUS.
It’s a very affordable product that offers a good level of depth and complexity whilst also providing tools to help you understand how to solve some of the more difficult equations.
It may not be a good choice for someone with a rudimentary understanding of mathematics, but for more advanced users, it’s perfect.
Now that you’ve read about the best scientific calculators. You may want to read our guide about the best portable tv, read it here.
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The Best Pop Filters (2022 Reviews)
Justin Stuart
Justin is the head writer for My Tech Reviewer and he ensures that readers always get what they are looking for. He's a hard worker and spends more hours writing for My Tech Reviewer than he works at
his part-time electrician job.
Justin Stuart
Justin is the head writer for My Tech Reviewer and he ensures that readers always get what they are looking for. He's a hard worker and spends more hours writing for My Tech Reviewer than he works at
his part-time electrician job. | {"url":"https://www.mytechreviewer.com/best-scientific-calculator/","timestamp":"2024-11-05T21:29:33Z","content_type":"text/html","content_length":"119826","record_id":"<urn:uuid:08e0d438-1fc3-4e8d-b190-3b50f3b183f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00564.warc.gz"} |
grinding mills in power plant
Engine, Turbine, and Power Transmission Equipment Manufacturing. Fabric Mills. Fiber, Yarn, and Thread Mills. Footwear Manufacturing. Forging and Stamping. Foundries. Fruit and Vegetable
Preserving and Specialty Food Manufacturing. Glass and Glass Product Manufacturing. Grain and Oilseed Milling.
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This project studied the relationship between PSD and power plant efficiency, emissions, and mill power consumption for lowrank highvolatilecontent Alaskan coal. The emissions studied were CO, CO
{sub 2}, NO {sub x}, SO {sub 2}, and Hg (only two tests). The tested PSD range was 42 to 81 percent passing 76 microns.
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This paper describes the simulation of the grinding process in vertical roller mills. It is based on actual experimental data obtained on a production line at the plant and from lab experiments.
Sampling and experiments were also carried out in a power plant that has four ballmill circuits used for coal grinding so that different equipment ...
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The power the mill consumes is a function of the ore and/or ball charge geometry and the mill speed. The motor only supplies the amount of power (or torque really) that the mill demands, and we
add some extra power to the motor to deal with 'upset' conditions like a change in ore density or a surge in ore level inside the mill.
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Power consumption of the grinding mill. The power consumption of the grinding mill is a critical parameter in the economics of the chromite beneficiation process. The ball mill consumes about
2530% of the total energy in the beneficiation plant, and hence any improvement will improve the overall economics of the plant.
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In this chapter an introduction of widely applied energyefficient grinding technologies in cement grinding and description of the operating principles of the related equipments and comparisons
over each other in terms of grinding efficiency, specific energy consumption, production capacity and cement quality are given.
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mal power plants running on ball drum mill grinding systems. In [2], it is stated that this share of energy is classified as very significant compared to the modern ones.
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A pulverizer or grinder is a mechanical device for the grinding of many different types of materials. For example, a pulverizer mill is used to pulverize coal for combustion in the
steamgenerating furnaces of coal power plants . Types of coal pulverizers Coal pulverizers may be classified by speed, as follows: [1] Low Speed Medium Speed High Speed
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Energy consumption represents a significant operating expense in the mining and minerals industry. Grinding accounts for more than half of the mining sector's total energy usage, where the
semiautogenous grinding (SAG) circuits are one of the main components. The implementation of control and automation strategies that can achieve production objectives along with energy efficiency
is a ...
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The production of fine and ultrafine limestone particles in grinding mills has an important role for the development of future products. Limestone as grinding material is used in ... gas
desulphurization is one of the most important processes in power plant operation and is dependent on efficient limestone grinding. Especially in the cement ...
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specific energy (kWh/t), plant size (t/d) and total milling process power. The required process power is divided into circuits and numbers of mills within a circuit, followed by the selection of
the mill sizes to fulfill the requirements. The optimal drive type can only be selected after determining the mill size, the need for variable speed ...
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A total of 120 t of Cristalino ore was prepared and sent to CIMM, where it was crushed and screened prior to grinding tests. The processing equipment included a m (8′) diameter by m (2′) length
AG/SAG mill equipped with a 20 kW motor, a cone crusher, a m (3′) diameter by m (4′) length ball mill equipped with a 15 kW ...
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The cost of optimization is minimal since inspecting the mill and the resulting modifications such as regrading the grinding media or moving the diaphragm are labor elements that can be handled
by the plant's maintenance crew. Upgrading the classifier and baghouse involves capital expenditure with a high benefit to cost ratio.
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Coal Mills in thermal power plant Free download as Powerpoint Presentation (.ppt), PDF File (.pdf), Text File (.txt) or view presentation slides online. for thermal powerplant familiarization ...
PRINCIPLES OF GRINDING Raw Coal is fed into the Mill for Grinding. Grinding takes place by Impact and attrition. MILL DETAILS. Design coal capacity ...
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Lower operating cost. Vertimill® is an energy efficient grinding machine. They tend to grind more efficiently than, for example, ball mills with feeds as coarse as 6 mm to products finer than 20
microns. This provides up to a 40% higher energy efficiency . With the Vertimill® simple and robust design, limited liner replacement is required.
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Power Distribution in Cement Plant. Grinding systems in the cement industry play a crucial role in particle size distribution and particle shape. This affects the reaction of the clinker and the
temperature dependence of the dehydrated gypsum which is also ground together with the clinker. ... The capacities of grinding mills range from 300 − ...
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Crushers, grinding mills and pulverizers are types of grinding equipment used to transform or reduce a coarse material such as stone, coal, or slag into a smaller, finer material. Grinding
equipment can be classified into to two basic types, crushers and grinders. Industrial crushers are the first level of size reducer; further granularization ...
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Hot air through the mill besides removing coal moisture, picks up the lighter particles and takes them through the classifier and drop down the higher size particles for further grinding. Fine
coal air mixture leaves the mill and enters the fuel piping system.
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Quantifying and improving the power efficiency of SAG milling circuits. Comminution circuits based on SAG (semiautogenous grinding) mills are not necessarily power efficient. So called critical
sized particles can build up in the SAG mill charge, limiting mill throughput, and increasing unit power consumption.
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Grinding mills in a largescale coalfueled power plant are formed of four core parts: the coal dryer, the coal handling and transportation system, the coal sieve, and the grinding mill [56,57].
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Balls and rods as the traditional grinding media are extensively used in pilot and fullscale mineral processing plants. ... Various power models for grinding mills have been used to prepare
powders in metallurgy, mining, cement or other industries and are now proposed to study the fundamental effect of load behavior on mill power (van Nierop ...
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The efficiency and the capacity of the plant can be increased as reduced. quantity of steam is required for the same power generation if high. pressure steam is used. 2. The forced circulation of
water through boiler tubes provides freedom in. the arrangement of furnace and water walls, in addition to the reduction in.
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This work concentrates on the energy consumption and grinding energy efficiency of a laboratory vertical roller mill (VRM) under various operating parameters. For design of experiments (DOE), the
response surface method (RSM) was employed with the VRM experiments to systematically investigate the influence of operating parameters on the energy consumption and grinding energy efficiency.
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Vertical roller mills (VRM) have found applications mostly in cement grinding operations where they were used in raw meal and finish grinding stages and in power plants for coal grinding. The
mill combines crushing, grinding, classification and if necessary drying operations in one unit and enables to decrease number of equipment in grinding ...
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The actual SAG mill power draw is 230370 kW lower. Mill runs 1 rpm lesser in speed on the average. The recirculation to the cone crusher is reduced by 110%, which means more efficient grinding of
critical size material is taking place in the mill.
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Small Pilot Plant Grinding Mill Sale! Grinding Mill for Metallurgical Pilot Testing of 10 to 150 Kilo/Hr US 50,000 40,000 The 911MPEPPGR426 is a small 300 kilo to ton per 24 hour day capacity
grinding mill acting primarily as a rod mill but can effortlessly be converted to a ball mill.
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increase of the surface area of a solid manufacturing of a solid with a desired grain size pulping of resources Grinding laws In spite of a great number of studies in the field of fracture
schemes there is no formula known which connects the technical grinding work with grinding results.
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Grinding mill performance (throughput or grind) is maximized using grind curves. • Grind curves map the essential performance measures of a grinding mill. • The need for a plant operator to
manually select operating conditions is reduced. • The grinding mill can be optimized without the need for a detailed process model.
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Coal mill pulverizer in thermal power plants Download as a PDF or view online for free. Submit Search. Upload. Coal mill pulverizer in thermal power plants ... The rolls do not touch the grinding
ring even when the mill is empty. 16. of CE Bowl Mills 17. Mill Grinding rolls 18. for 500 MW 1. ...
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After the closure of the mills and the old power plants not so long ago, the water circulation in this complex ... in 1919, the water mill was equipped with three millstones and produced 100 kg
of flour per day. The mill was used for grinding wheat and corn until the 1970s. Then, it was abandoned. In 2002, the water mill was renovated and re ...
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Modeling of wet stirred media mill processes is challenging since it requires the simultaneous modeling of the complex multiphysics in the interactions between grinding media, the moving internal
agitator elements, and the grinding fluid. In the present study, a multiphysics model of an HIG5 pilot vertical stirred media mill with a nominal power of kW is developed. The model is based on a
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Grind curves give the steadystate values of the performance variables throughput, power draw, and grind in terms of the mill filling and critical mill speed. The grind curves indicate the
operable region of the grinding mill. An analysis and dynamic simulation of the model show that the model captures the main dynamics of the grinding mill.
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Mill Power Pvt Ltd We are leading manufacturer of Atta Chakki, Double Stage Pulverisers, Flour Mill, Pulverisers since 1983. CALL OR WHATS APP +91 7435 92 6060. ... We bought a spice grinding
plant from Mill Power. Hitesh and his team helped us with right machine selection, optimal setup and made recommendations beyond their responsibilities
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Linear Algebra
Linear Algebra#
This module contains specialised linear algebra tools that are not currently available in the python standard scientific libraries.
Kronecker tools#
A kronecker matrix is matrix that can be written as a kronecker matrix of the individual matrices i.e.
\[\begin{split}K = K_0 \\otimes K_1 \\otimes K_2 \\otimes \\cdots\end{split}\]
Matrices which exhibit this structure can exploit properties of the kronecker product to avoid explicitly expanding the matrix \(K\). This module implements some common linear algebra operations
which leverages this property for computational gains and a reduced memory footprint.
kron_matvec(A, b) Computes the matrix vector product of a kronecker matrix in linear time.
kron_cholesky(A) Computes the Cholesky decomposition of a kronecker matrix as a kronecker matrix of Cholesky factors.
africanus.linalg.kron_matvec(A, b)[source]#
Computes the matrix vector product of a kronecker matrix in linear time. Assumes A consists of kronecker product of square matrices.
An array of arrays holding matrices [K0, K1, …] where \(A = K_0 \otimes K_1 \otimes \cdots\)
The right hand side vector
The result of A.dot(b)
Computes the Cholesky decomposition of a kronecker matrix as a kronecker matrix of Cholesky factors.
An array of arrays holding matrices [K0, K1, …] where \(A = K_0 \otimes K_1 \otimes \cdots\)
An array of arrays holding matrices [L0, L1, …] where \(L = L_0 \otimes L_1 \otimes \cdots\) and each Li = cholesky(Ki) | {"url":"https://codex-africanus.readthedocs.io/en/latest/linalg-api.html","timestamp":"2024-11-05T13:32:22Z","content_type":"text/html","content_length":"24712","record_id":"<urn:uuid:b7e0ff06-2130-4c09-a65a-009baf4c96d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00728.warc.gz"} |
The Absolute Center of a Network
This paper presents a new algorithm for finding an absolute center (minimax criterion) of an undirected network with n nodes and m arcs based on the concept of minimum-diameter trees. Local centers
and their associated radii are identified by a monotonically increasing sequence of lower bounds on the radii. Computational efficiency is addressed in terms of worst-case complexity and practical
performance. The complexity of the algorithm is 0(n^2 ℓg n + mn). In practice, because of its very rapid convergence, the algorithm renders the problem amenable even to manual solution for quite
large networks, provided that the minimal-distance matrix is given. Otherwise, evaluation of this matrix is the effective computational bottleneck. An interesting feature of the algorithm and its
theoretical foundations is that it synthesizes and generalizes some well-known results in this area, particularly Halpern's lower bound on the local radius of a network and properties of centers of
tree networks.
ASJC Scopus subject areas
• Software
• Information Systems
• Hardware and Architecture
• Computer Networks and Communications
Dive into the research topics of 'The Absolute Center of a Network'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/the-absolute-center-of-a-network","timestamp":"2024-11-02T05:12:48Z","content_type":"text/html","content_length":"54715","record_id":"<urn:uuid:949a9ea5-065a-427a-abde-5e6a55f81471>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00631.warc.gz"} |
Derivation of the Pauli Equation for a Charged Spin-1/2 Particle in a Weak Gravitational Field Using a Non-Relativistic Approximation
Derivation of the Pauli Equation for a Charged Spin-1/2 Particle in a Weak Gravitational Field Using a Non-Relativistic Approximation
Core Concepts
This paper presents a detailed derivation of the Pauli equation for a charged spin-1/2 particle in a weak gravitational field using a direct non-relativistic approximation of the Dirac equation,
comparing the results with previous approaches relying on Foldy-Wouthuysen transformations and highlighting discrepancies in the Newtonian limit.
• Bibliographic Information: Oliveira, S. W. P., Oyadomari, G. Y., & Shapiro, I. L. (2024). Pauli equation and charged spin-1/2 particle in a weak gravitational field. arXiv preprint
• Research Objective: To derive the Pauli equation for a charged spin-1/2 particle in a weak gravitational field using a direct non-relativistic approximation of the Dirac equation and compare the
results with previous approaches.
• Methodology: The authors employ a direct non-relativistic approximation of the Dirac equation in curved spacetime, expanding the metric tensor around a flat background and performing a power
series expansion in the inverse mass of the particle. They then impose a synchronous gauge condition to simplify the resulting equations and derive the Pauli equation and the equations of motion
for the particle. The results are compared with those obtained in previous works using Foldy-Wouthuysen transformations.
• Key Findings:
□ The authors successfully derive the Pauli equation for a charged spin-1/2 particle in a weak gravitational field using a direct non-relativistic approximation.
□ The derived equation is consistent with previous results obtained using perturbative and exact Foldy-Wouthuysen transformations in the case of a plane gravitational wave background.
□ However, discrepancies are found in the Newtonian limit compared to previous works employing Foldy-Wouthuysen transformations, particularly in the numerical coefficients of certain terms.
□ The authors suggest that these discrepancies might arise from the potential energy being proportional to the mass of the test particle in the Newtonian limit, leading to ambiguities in the 1/
m expansion.
• Main Conclusions: The direct non-relativistic approximation provides a consistent alternative approach to deriving the Pauli equation in curved spacetime. However, the discrepancies observed in
the Newtonian limit highlight potential ambiguities and scheme-dependent results when dealing with gravitational fields where the potential energy is proportional to the particle's mass.
• Significance: This work contributes to the understanding of the non-relativistic limit of quantum mechanics in curved spacetime and provides a valuable comparison of different approaches to this
problem. The highlighted discrepancies in the Newtonian limit raise important questions about the validity and interpretation of different approximation schemes in this regime.
• Limitations and Future Research: The study is limited to weak gravitational fields and a specific gauge condition. Further research could explore the non-relativistic limit in more general
spacetimes and gauge choices. Additionally, investigating the observed discrepancies in the Newtonian limit and their potential implications for experimental tests of gravity would be of
significant interest.
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Pauli equation and charged spin-1/2 particle in a weak gravitational field
Deeper Inquiries
How does the choice of gauge condition affect the form of the Pauli equation and the equations of motion in curved spacetime?
The choice of gauge condition significantly affects the form of the Pauli equation and the equations of motion in curved spacetime. This is a manifestation of the fundamental principle of general
covariance in general relativity, which states that the laws of physics should be independent of the choice of coordinates. Here's a breakdown of how the gauge condition influences the results:
Simplification of Equations: Choosing a specific gauge can significantly simplify the mathematical expressions involved. In the provided context, the authors employ the synchronous gauge condition
(h_0k = 0), which eliminates several terms in the derived equations. This simplification makes the analysis more manageable. Physical Interpretation: Different gauge choices correspond to different
choices of reference frames. While the underlying physics remains the same, the interpretation of individual terms in the equations can change. For instance, certain terms might be attributed to
inertial forces in one gauge and to gravitational forces in another. Comparison with Other Results: When comparing results derived using different gauge conditions, it's crucial to ensure
consistency. Direct comparison is only meaningful if the results are transformed to a common gauge. Limitations: Choosing a specific gauge might obscure certain physical effects or introduce
coordinate singularities. It's essential to be aware of the limitations imposed by the chosen gauge and consider alternative gauges if necessary. In summary, the choice of gauge condition is not
merely a mathematical convenience but has profound implications for the interpretation and analysis of the Pauli equation and equations of motion in curved spacetime.
Could the discrepancies observed in the Newtonian limit be resolved by employing a different regularization scheme or a more sophisticated treatment of the 1/m expansion?
The discrepancies observed in the Newtonian limit, particularly the differing numerical coefficients in the Hamiltonian and equations of motion, might indeed be resolved or at least better understood
by employing a different regularization scheme or a more sophisticated treatment of the 1/m expansion. Here's why: Ambiguities in 1/m Expansion: As the authors point out, the Newtonian limit presents
a unique challenge because the gravitational potential couples to the mass of the test particle, leading to terms proportional to mΦ. Since the non-relativistic expansion is in inverse powers of m,
this situation can introduce ambiguities. Regularization Scheme Dependence: The choice of regularization scheme, which essentially dictates how one handles divergent or ill-defined expressions during
the derivation, can influence the final result. Different schemes might lead to different finite contributions, potentially explaining the observed discrepancies in the coefficients. Higher-Order
Terms: The discrepancies might arise from neglecting higher-order terms in the 1/m expansion. A more accurate treatment might involve retaining these terms or employing resummation techniques to
capture their cumulative effect. Alternative Expansion Parameters: Instead of solely relying on the 1/m expansion, exploring alternative expansion parameters, such as the ratio of the particle's
Compton wavelength to the characteristic length scale of the gravitational field, might provide a more accurate description in the Newtonian limit. In conclusion, while the discrepancies observed in
the Newtonian limit are not uncommon in such derivations, exploring alternative regularization schemes and refining the 1/m expansion could lead to a more consistent and accurate description of the
non-relativistic limit of the Dirac equation in a weak gravitational field.
What are the potential experimental implications of the spin-gravity coupling terms derived in this paper, and how could they be tested in future experiments?
The spin-gravity coupling terms derived in the paper, while small, have intriguing potential experimental implications. These terms suggest that the spin of a particle is not merely a passive
property but dynamically interacts with the gravitational field. Here are some potential experimental avenues to explore: Precision Measurements of Spin Precession: The spin-gravity coupling could
manifest as subtle shifts in the precession frequency of particles with spin in a gravitational field. Experiments using ultra-sensitive atom interferometers or neutron interferometers could
potentially detect these minute variations. Spin-Dependent Gravitational Force: The equations of motion suggest a possible spin-dependent gravitational force. This force, though weak, might be
detectable by precisely measuring the motion of polarized masses or by looking for spin-dependent deflections of particles in a gravitational field. Macroscopic Quantum Effects: For macroscopic
objects with significant net spin, the spin-gravity coupling might lead to observable quantum effects. Experiments involving levitated micro- or nano-scale objects with controlled spin could probe
this regime. Astrophysical Observations: While challenging, astrophysical observations of the motion of spinning objects in strong gravitational fields, such as pulsars or black hole binaries, could
provide indirect evidence of spin-gravity coupling. Modified Gravity Tests: The spin-gravity coupling terms could be a signature of physics beyond general relativity. Experiments designed to test
alternative theories of gravity, such as those searching for Lorentz violation or torsion, might be sensitive to these effects. Realizing these experiments would require pushing the boundaries of
current technology, demanding exceptional sensitivity and control over experimental parameters. However, the potential payoff in terms of understanding the interplay between spin and gravity, and
possibly uncovering new physics, makes pursuing these experimental avenues a worthwhile endeavor. | {"url":"https://linnk.ai/insight/scientific-computing/derivation-of-the-pauli-equation-for-a-charged-spin-1-2-particle-in-a-weak-gravitational-field-using-a-non-relativistic-approximation-TtbSj9Q_/","timestamp":"2024-11-03T12:47:56Z","content_type":"text/html","content_length":"294527","record_id":"<urn:uuid:31615205-16d6-4ed0-a16a-ce9ab85e4530>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00724.warc.gz"} |
Differential Geometry/PDE Seminar
The UW Differential Geometry / Partial Differential Equations (DG/PDE) Seminar is held in Padelford C-38 on Wednesdays 4:00 p.m. unless otherwise noted.
Taking the DG/PDE Seminar for Credit
The Differential Geometry/PDE Seminar can be taken for graduate credit as Math 550A. You may obtain one credit by attending and participating in all but two of the seminar meetings, or two credits
for giving a talk in the seminar. If you have any questions about these requirements, feel free to contact Yu YUAN (yuan@math.washington.edu) or Jonathan Zhu (jonozhu@uw.edu).
Seminar Mailing List
Announcements of upcoming DG/PDE seminars are sent by e-mail to all interested participants. To be added to (or removed from) the DG/PDE seminar mailing list, send an e-mail to
yuan@math.washington.edu or jonozhu@uw.edu. | {"url":"https://math.washington.edu/events/series/differential-geometrypde-seminar","timestamp":"2024-11-11T08:42:11Z","content_type":"text/html","content_length":"77676","record_id":"<urn:uuid:b34f671e-131e-469b-9047-f9e925a345eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00488.warc.gz"} |
The following is a copy of my presentation to the founding members of the Academy for the Advancement of Post-Materialist Science, August 26, 2017
MATHEMATICS, PHYSICS AND CONSCIOUSNESS
A Presentation by Edward R. Close, August 2017
First, I want to thank Dr. Gary Schwartz, Dr. Marjorie Woollacott, Dr. Charles Tart, and all who have worked so hard to make the Academy for the Advancement of Post Materialist Sciences and this
meeting possible, including our anonymous benefactor. This meeting is the beginning of something I have dreamed of for many years.
I am struck by the similarities among the intellectual and psychic experiences of those gathered here today, but this should not be a surprise! It is evidence for what Erwin Schrӧdinger declared in
his wonderful little book “What is Life?” published by Cambridge University Press in 1967, when he said: “There is no evidence that consciousness is plural.” Many of us know that all things are
connected at a fundamental level, and, my friends, it is time for the first real scientific paradigm shift since relativity and quantum physics!
I want to start by sharing an experience I wrote about in my first book, “The Book of Atma”, published in 1977. It reveals the motivation that has propelled me throughout my life:
It was the summer of 1951. I was fourteen. I found a little book on analytical geometry written in German among some old books. Reading it, I had the distinct awareness that I already knew this
mathematics. It was as if I were remembering, not learning. Also, I had just discovered the work of Albert Einstein, which had opened a whole new world for me.
One evening, in the twilight just after sunset, I walked out of the little house on my parent’s farm in the Southern Missouri Ozarks, past a line of catalpa trees, to the bank of a pond. I had been
thinking about the “electrodynamics of moving objects” as described in Einstein’s special theory of relativity, and I had reached a point beyond which I could not go. Frustrated, I looked up at the
sky and complained: “God, I want to know everything!”
What followed was totally unexpected, but so real that I knew it was completely natural. Suddenly, I could “hear” the silence around me. My surroundings took on a glow, as if everything were alive.
My conscious mind seemed to melt, and the distinctions between my physical body and the surrounding landscape seemed to fade. I was filled with an all-pervading feeling of well-being. I knew I had
received my answer! I would be a theoretical physicist!
I could spend my twenty minutes describing the series of psychic experiences and epiphanies that led Dr. Vernon Neppe and me to develop the Triadic Dimensional Distinction Vortical Paradigm (TDVP),
and list the paradoxes it has resolved and the phenomena it has explained that are not explained by the current materialistic paradigm, but that would only scratch the surface. Instead, I want to
address Dr. Gary Schwartz’s last item in his list of important questions: “Do we need an expanded mathematics, as Close and Neppe propose, to advance Post Materialist Sciences?”
Of course my answer is yes; but let me illustrate and emphasize this answer with a short history of the development of the new mathematics that unites number theory, geometry, relativity, quantum
physics, some aspects of string theory, and the consciousness of the observer.
A paranormal experience in 1957 resulted in my discovery of the work of Pierre de Fermat. My College roommate, now Dr. David Stewart, and I were carrying out experiments in which we obtained
verifiable information not available to us by normal sensory means. One of the most successful of these experiments was submitted to Dr. J.B. Rhine at Duke University. During one of our early
experiments it was revealed that I had access to memories of the life of Pierre de Fermat. We obtained mathematical representations of concepts that far exceeded my training at the time, but were
verified by my physics professor.
In 1637, Fermat wrote in the margin of his copy of a book on Diophantine equations, that he had found a “marvelous” proof that the equation x^n + y^n = z^n has no integer solutions for n >2. But his
proof was never found. After receiving my degree in mathematics and physics in 1962, while teaching mathematics, I spent considerable time trying to access Fermat’s marvelous proof. Sometime during
that period, I realized that Fermat’s Last Theorem, considered by most to be nothing more than a hypothesis in pure number theory, had important implications for quantum physics if x, y and z
represent the radii of elementary particles that combine to form what we experience as ordinary physical reality.
This led to the realization that a quantum mathematics was urgently needed for describing the quantized reality we live in. The differential and integral calculus of Newton and Leibniz are
inappropriate for describing quantum phenomena because they depend on a continuity of the variables of measurement that does not exist in a quantized world. I believe that the inappropriate
application of Newtonian calculus to quantum phenomena gives rise to much of the ‘weirdness’ of quantum physics that physicists like to talk about.
I found the basis for the needed quantum mathematics in G. Spencer Brown’s calculus of indications published in his 1969 book “Laws of Form.” And it was obvious to me from the results of the Aspect
Experiment resolving the Einstein/Bohr debate, that we have to have a mathematics that incorporates the consciousness of the observer. I published the basic concepts of an adaptation of Brown’s
Calculus which I called the Calculus of Distinctions in my book, “Infinite Continuity,” in 1990. The Calculus of Distinctions is different from Brown’s Calculus of Indications in several ways that I
do not have time to go into here. Unfortunately, that book is now long out of print, but the basic logic is published in an appendix to my 1996 book, “Transcendental Physics.”
In those references, I show that the drawing of a distinction is comprised of a triad:
1. the object of distinction
2. the features distinguishing the object from everything else, and
3. the consciousness of the observer.
Thus, a distinction is inherently triadic, and the consciousness of the observer is implicit in the logic of the CoD. Therefore, application of these basic concepts inherently includes the
consciousness of the observer in the equations of science. I later adapted the CoD to reflect the multi-dimensional geometry of finite distinctions and the differentiation of existing distinctions
from conceptual distinctions in the Calculus of Dimensional Distinctions (CoDD).
With the help of Russian-born mathematician Vladimir Brandin in 2003, and Dr. Vernon Neppe, from 2008 to the present, application of the CoDD has allowed me to develop the definition of a true
quantum equivalence unit that I call the Triadic Rotational Unit of Equivalence (TRUE), and the discovery of the third form of the substance of reality, necessary for the stability of atomic
structure. This third form cannot be measured as mass or energy, but is detectable in the total angular momentum of any rotating physical system. Dr. Neppe proposed the name gimmel for the third form
for a variety of interesting reasons.
We decided to call the new paradigm TDVP: Triadic because that was the nature of the underlying structure of mass, energy and consciousness. Dimensional, because to be consistent, the mathematics had
to incorporate extra dimensions beyond three of space and one of time. Vortical, because of the spinning nature of elementary particles, and Paradigm to emphasize that it is a shift from the current
materialistic metaphysics of modern science.
Physicists talk about a “theory of everything”. But you can’t have a theory of everything if everything is not included in it. I see the discovery of gimmel as the fulfillment of my efforts over the
past 30 plus years to put consciousness into the equations of science. Gimmel has all the earmarks of consciousness, or at least of an agent of consciousness, acting through what I call the
conveyance equations, to bring the logic of the multi-dimensional substrate of Primary Consciousness into the 3 Spatial dimensions, 1 Time dimension, and 1 dimension of Consciousness, i.e., the
domain of physical observation.
The discovery of gimmel eliminates materialism as a viable metaphysical basis for science. It eliminates materialism because gimmel is inherently non-material, and because I have proved that it is
necessary for the stability of quarks and subatomic structure. Without it there would be no physical universe. The discovery of gimmel answers Gottfried Leibniz’s unanswered first priority question:
“Why is there something rather than nothing?”
I believe that gimmel is the manifestation of consciousness in physical reality. This view is justified in part because the elements and compounds supporting organic life forms prove to have the
highest levels of gimmel. TRUE units and gimmel provide the necessary basis to analyze and quantify consciousness working within our physical/spiritual/conscious reality.
Through the use of TRUE unit analysis and LHC data, and applying the principles of relativity and quantum physics, several unexplained phenomena have been explained quite elegantly by TDVP. Because
TDVP includes consciousness in the equations of science, and therefore is more comprehensive than materialistic theories, it can provide the mathematical basis for investigating and describing psi
phenomena like those experienced by virtually everyone in this room.
My answer to Gary’s question about whether the Academy needs an expanded math is this: It is my personal belief, based on over 50 years of explorations of mathematics, physics and consciousness
expansion techniques, that mathematics is not merely a tool, mathematics reflects the actual structure of reality. And if you look at the history of science, every real scientific paradigm shift of
the past has been accompanied by new mathematics. The paradigm shift to the primacy of consciousness can be no exception. It is my opinion that, in this case, a new mathematics is even more crucial
than ever before because of the magnitude of this shift. Post-Materialism Science cries out for a new more comprehensive mathematical paradigm, and in my opinion, that new paradigm is TDVP, and the
new math is the Calculus of Dimensional Distinctions.
46 comments:
1. As stated numerous times before, Ed, and appreciated by you, the above is certainly well suited to the 'One' of my mystically-inspired calculus formula, Y (Universal Intelligence) = X Squared
(Basic Evolutionary Elements: 0,1,2,3,4) + One (The Ultimate Force) - I hopefully still look forward to your sometime, sooner rather than later, introducing this concept to your peers to further
serve the fundamental nature of your own TDVP treatise.
The following URL, once again, provides full details: http://www.vigiltrust.lk/ultimate-forces-fundamental-calculus-revised-according-brian/
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Professor McGuckian's answer to Two-Sample t-test w Equal Var Question #25
I need help with question 18 from section 9.4. Could you please show how you get the test stat of 11.44, I get something different.
Thank You.
See the professor's answer below.
Hi Peggy,
I can create a video for this later, but for now, let me just explain here:
Because we assume equal variances for this problem, we first need to calculate Sp²
Sp² = [(n1-1)*S1² + (n2-1)*S2²]/(n1 + n2 - 2)
(27*4.51² + 28*3.98²)/(28+29-2) = 18.04934364 = Sp²
Then our test stat formula for this problem would be:
t = (Xbar1 - Xbar2) / SqrRoot(Sp²/n1 + Sp²/n2) = (38.28 - 25.4) / SqrRoot(18.04934/28 + 18.04934/29) = 11.443
Note: the denominator part of the test stat (the part under the squareroot) should be 1.125615643. Use that to check your calculation.
Hope that helps,
Hi Peggy,
Here is a video explanation of how to work out the test statistic: | {"url":"https://www.statsprofessor.com/solution/25","timestamp":"2024-11-14T05:14:50Z","content_type":"text/html","content_length":"48374","record_id":"<urn:uuid:59fb2e2a-62d6-4ac2-9d26-b65c8739d980>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00356.warc.gz"} |
1 googol is the number 1 followed by 100 zeros
Have you ever wondered about the largest number in existence? Well, let us introduce you to a mind-boggling quantity called “1 googol.” Represented by a one followed by a staggering 100 zeros, this
number is beyond imagination. In this article, we will delve into the fascinating world of googol and explore its incredible magnitude.
A googol was first conceptualized by mathematician Edward Kasner in the early 1900s. Kasner is also credited with coining the term “googol.” Its purpose was to demonstrate an unimaginably large
number that is far beyond human comprehension. The idea behind introducing this colossal quantity was to emphasize how vast mathematical possibilities could be.
To gain a better perspective, let’s compare a googol to other numbers. For instance, a million, usually considered a large number, is a mere 1 followed by six zeros, whereas a billion is 1 followed
by nine zeros. Ascending further, a trillion is 1 followed by twelve zeros. As we continue our journey, we reach a googol, which consists of an awe-inspiring hundred zeros!
To grasp the immensity of a googol, let’s consider some mind-blowing comparisons. If we were to count from one to a googol at a steady pace of one number per second, it would take us approximately
31.7 billion years to complete the task. This is nearly ten times the age of the universe! Moreover, if each second were represented by a particle of sand, counting to a googol would require more
grains of sand than the total number of atoms in the observable universe.
The concept of a googol also found its way into popular culture. Notably, it sparked the creation of the multinational technology company, Google Inc. Founded by Larry Page and Sergey Brin in 1998,
the name “Google” was derived from the term “googol.” Page and Brin wanted to symbolize the vast amount of information available on the internet that their search engine would explore and organize.
In essence, the existence of a googol showcases the vastness of our mathematical world. It serves as a reminder of how mathematics constantly stretches the boundaries of human understanding. Although
this colossal number might be inconceivable to comprehend fully, it serves as a testament to the infinite possibilities that mathematics uncovers.
In conclusion, a googol is an unimaginably large number represented by 1 followed by 100 zeros. Its magnitude surpasses our comprehension, and even counting to a googol would take billions of years.
Through its creation, mathematician Edward Kasner aimed to highlight the awe-inspiring potential of mathematics. The concept of a googol has also left its mark on popular culture, as evidenced by the
birth of Google Inc. Let the idea of a googol inspire you to explore the boundless realm of mathematics and embrace the wondrous possibilities it unveils.
Source: Wikipedia - Googol | {"url":"https://thefactbase.com/1-googol-is-the-number-1-followed-by-100-zeros/","timestamp":"2024-11-12T05:28:06Z","content_type":"text/html","content_length":"110446","record_id":"<urn:uuid:e0b2149b-fe50-4a60-a835-42f420e06dba>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00888.warc.gz"} |
Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r >
2 Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r
> 2
CAT 2021
Explanatory Answer :
we have given ,Region R is the triangle in the first quadrant with vertices (0,0) and (4,0) and (0,3) . The region R is the sample space. The area of the triangle is considered as it is a continuous
distribution of points in the region .
Area of region R is 6 units.
Further the event is r>2 with r,s being a point in the region R. So our event is basically the triangle with vertices (2,0) and (4,0) and (2,1.5)
It's the common area bounded by r>2 and the region R. The area of the event triangle is 1.5 units.
Hence the probability is ratio of area of the event and area of the region R which is the sample space.
Hence probability is 1/4
so ,A is the correct option | {"url":"https://www.mbarendezvous.com/question-answer/quant/coordinate-geometry/region-r-is-defined-as-the-region-in-the-first-quadrant/","timestamp":"2024-11-07T15:37:08Z","content_type":"text/html","content_length":"293644","record_id":"<urn:uuid:32a70041-465d-45d7-a04d-4beb56f1bb80>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00003.warc.gz"} |
Men with Hats | R-bloggersMen with Hats
[This article was first published on
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
Suppose N people (and their hats) attend a party (in the 1950s). For fun, the guests mix their hats in a pile at the center of the room, and each person picks a hat uniformly at random. What is the
probability that nobody ends up with their own hat?
EstimateProbability <- function(n.people, n.simulations=10000) {
NobodyGotTheirHat <- function(n.people) {
people <- 1:n.people
hats <- sample(people, size=n.people, replace=FALSE)
return(all(people != hats))
mean(replicate(n.simulations, NobodyGotTheirHat(n.people)))
CalculateProbability <- function(n.people) {
InclusionExclusionTerm <- function(i) {
return(((-1) ^ (i + 1)) * choose(n.people, i) *
factorial(n.people - i) / factorial(n.people))
1 - sum(sapply(1:n.people, InclusionExclusionTerm))
x.max <- 40
xs <- 1:x.max
dev.new(height=6, width=10)
plot(xs, sapply(xs, EstimateProbability), pch=4, ylim=c(0, 0.60),
main="Men with Hats: N hats uniformly assigned to N people",
ylab="probability that nobody ends up with their own hat")
lines(xs, sapply(xs, CalculateProbability), col="firebrick", lwd=2)
mtext("What is the probability that nobody ends up with their own hat?")
legend("topleft", "True probability", bty="n", lwd=2, col="firebrick")
legend("topright", "Estimate from 10k simulations", bty="n", pch=4)
# The probability converges to e^-1 as N -> Inf
exp(1) ^ -1
As in my earlier
puzzle post
, the solution is an application of the
principle. What’s fascinating about this particular puzzle is that the probability settles down not at zero or one, but instead converges to e^-1 as the number of people (and hats) grows large. I
don’t know about you, but I never would have guessed. | {"url":"https://www.r-bloggers.com/2011/07/men-with-hats/","timestamp":"2024-11-05T23:29:46Z","content_type":"text/html","content_length":"87349","record_id":"<urn:uuid:1bc18620-06e0-4112-b675-6589385a311a>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00655.warc.gz"} |
if this post gets 243 upvotes i will post again with 3 times as many triangles - tchncs
• If I’m counting right it’s actually X3+2 right? You are adding the inner black triangle and a new overarching triangle.
□ juliebean@lemm.ee English
2 months ago
not quite. if you were counting the black triangles in the previous iteration, then this should be three times as many triangles, minus three. each of the previous iteration has those two big
black triangles in the corner, but when you assemble them together, they merge into the two big corner triangles and the central triangle. | {"url":"https://discuss.tchncs.de/post/22121344/13103179","timestamp":"2024-11-03T13:09:27Z","content_type":"text/html","content_length":"65282","record_id":"<urn:uuid:0b1eac9b-285d-4151-adb9-9a35f7834fe7>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00617.warc.gz"} |
Simons Collaboration on Special Holonomy in Geometry, Analysis, and Physics
Ruobing Zhang: Lectures
September 9, 2021
TITLE: Collapsing geometry of hyperkaehler manifolds in dimension four
ABSTRACT: We will overview some recent developments on the degeneration theory of hyperkaehler 4-manifolds. As preliminaries, we will present some tools and conceptual ideas in understanding metric
geometric aspect of degenerating hyperkaehler metrics. After explaining the background, this talk will focus on the following classification results in the volume collapsed setting.
First, we will classify the collapsed limits of the bounded-diameter hyperkaehler metrics on the K3 manifold. More generally, we will precisely characterize the limiting singularities of the
hyperkaehler metrics with bounded quadratic curvature integral. We will also exhibit the classification of the complete non-compact hyperkaehler 4-manifolds with quadratically integrable curvature
which are called gravitational instantons and appear as bubbling limits of degenerating hyperkaehler metrics.
The above ingredients constitute a relatively complete picture of the collapsing geometry of hyperkaehler metrics in dimension 4.
April 12, 2018
TITLE: Gravitational collapsing of K3 surfaces II
ABSTRACT: We will exhibit some new examples of collapsed hyperkähler metrics on a K3 surface. This is my recent joint work with Hans-Joachim Hein, Song Sun and Jeff Viaclovsky. We will construct a
family of hyperkähler metrics on a K3 surface which are collapsing to a closed interval. Geometrically, each regular fiber is a Heisenberg manifold and each singular fiber is a singular circle
fibration over a torus. In our example, each bubble limit is either the Taub-NUT space or a complete hyperkähler space constructed by Tian-Yau. The regularity estimates in this example in fact
confirms a general picture given by the
April 9, 2018
TITLE: Quantitative nilpotent structure and regularity theorems of collapsed Einstein manifolds
ABSTRACT: This talk is on the new developments of the structure theory for collapsed Einstein manifolds. We will start with some motivating examples of collapsed Ricci-flat manifolds. Our main focus
is the
April 9, 2018
TITLE: Introduction to Ricci curvature and the convergence theory
ABSTRACT: The first talk is an overview of the convergence and regularity theory of the manifolds with Ricci and sectional curvature bounds. Specifically, we will review some both classical and new
structure theory such as the | {"url":"https://sites.duke.edu/scshgap/ruobin-zhang-lectures/","timestamp":"2024-11-03T10:38:48Z","content_type":"text/html","content_length":"46090","record_id":"<urn:uuid:287c60a6-96e0-43ed-9f75-59d90ba3b1f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00423.warc.gz"} |
The data are from the first 8 rows of the pollutant water treatment example in the book by Box, Hunter and Hunter, 2nd edition, Chapter 5, Question 19.
The 3 factors (C, T, and S) are in coded units where: C: -1 is chemical brand A; +1 is chemical brand B T: -1 is 72F for treatment temperature; +1 is 100F for the temperature S: -1 is No stirring; +1
is with fast stirring
The outcome variable is: y: the pollutant amount in the discharge [lb/day].
The aim is to find treatment conditions that MINIMIZE the amount of pollutant discharged each day, where the limit is 10 lb/day. | {"url":"https://www.rdocumentation.org/packages/pid/versions/0.50/topics/pollutant","timestamp":"2024-11-15T01:19:21Z","content_type":"text/html","content_length":"53821","record_id":"<urn:uuid:6bc05d29-3d2f-4554-979d-b64d63308065>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00336.warc.gz"} |
BurkeRatio: Burke ratio of the return distribution in guillermozbta/portafolio-master: Econometric tools for performance and risk analysis.
To calculate Burke ratio we take the difference between the portfolio return and the risk free rate and we divide it by the square root of the sum of the square of the drawdowns. To calculate the
modified Burke ratio we just multiply the Burke ratio by the square root of the number of datas.
R an xts, vector, matrix, data frame, timeSeries or zoo object of asset returns
Rf the risk free rate
modified a boolean to decide which ratio to calculate between Burke ratio and modified Burke ratio.
... any other passthru parameters
an xts, vector, matrix, data frame, timeSeries or zoo object of asset returns
a boolean to decide which ratio to calculate between Burke ratio and modified Burke ratio.
Modified Burke Ratio = (Rp - Rf) / (sqrt(sum(t=1..n)(Dt^2 / n)))
where n is the number of observations of the entire series, d is number of drawdowns, r_P is the portfolio return, r_F is the risk free rate and D_t the t^{th} drawdown.
Carl Bacon, Practical portfolio performance measurement and attribution, second edition 2008 p.90-91
1 data(portfolio_bacon)
2 print(BurkeRatio(portfolio_bacon[,1])) #expected 0.74
3 print(BurkeRatio(portfolio_bacon[,1], modified = TRUE)) #expected 3.65
5 data(managers)
6 print(BurkeRatio(managers['1996']))
7 print(BurkeRatio(managers['1996',1]))
8 print(BurkeRatio(managers['1996'], modified = TRUE))
9 print(BurkeRatio(managers['1996',1], modified = TRUE))
data(portfolio_bacon) print(BurkeRatio(portfolio_bacon[,1])) #expected 0.74 print(BurkeRatio(portfolio_bacon[,1], modified = TRUE)) #expected 3.65 data(managers) print(BurkeRatio(managers['1996']))
print(BurkeRatio(managers['1996',1])) print(BurkeRatio(managers['1996'], modified = TRUE)) print(BurkeRatio(managers['1996',1], modified = TRUE))
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/github/guillermozbta/portafolio-master/man/BurkeRatio.html","timestamp":"2024-11-09T13:03:41Z","content_type":"text/html","content_length":"43611","record_id":"<urn:uuid:d01a5b38-eb46-49f4-ab6c-a74af003fe1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00078.warc.gz"} |
Paired T-Test
The paired sample t-test, sometimes called the dependent sample t-test, is a statistical procedure used to determine whether the mean difference between two sets of observations is zero. In a paired
sample t-test, each subject or entity is measured twice, resulting in pairs of observations. Common applications of the paired sample t-test include case-control studies or repeated-measures designs.
Suppose you are interested in evaluating the effectiveness of a company training program. One approach you might consider would be to measure the performance of a sample of employees before and after
completing the program, and analyze the differences using a paired sample t-test.
Like many statistical procedures, the paired sample t-test has two competing hypotheses, the null hypothesis and the alternative hypothesis. The null hypothesis assumes that the true mean difference
between the paired samples is zero. Under this model, all observable differences are explained by random variation. Conversely, the alternative hypothesis assumes that the true mean difference
between the paired samples is not equal to zero. The alternative hypothesis can take one of several forms depending on the expected outcome. If the direction of the difference does not matter, a
two-tailed hypothesis is used. Otherwise, an upper-tailed or lower-tailed hypothesis can be used to increase the power of the test. The null hypothesis remains the same for each type of alternative
hypothesis. The paired sample t-test hypotheses are formally defined below:
• The null hypothesis (\(H_0\)) assumes that the true mean difference (\(\mu_d\)) is equal to zero.
• The two-tailed alternative hypothesis (\(H_1\)) assumes that \(\mu_d\) is not equal to zero.
• The upper-tailed alternative hypothesis (\(H_1\)) assumes that \(\mu_d\) is greater than zero.
• The lower-tailed alternative hypothesis (\(H_1\)) assumes that \(\mu_d\) is less than zero.
The mathematical representations of the null and alternative hypotheses are defined below:\(H_0:\ \mu_d\ =\ 0\)
\(H_1:\ \mu_d\ \ne\ 0\) (two-tailed)
\(H_1:\ \mu_d\ >\ 0\) (upper-tailed)
\(H_1:\ \mu_d\ <\ 0\) (lower-tailed)
Note. It is important to remember that hypotheses are never about data, they are about the processes which produce the data. In the formulas above, the value of \(\mu_d\) is unknown. The goal of
hypothesis testing is to determine the hypothesis (null or alternative) with which the data are more consistent.
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As a parametric procedure (a procedure which estimates unknown parameters), the paired sample t-test makes several assumptions. Although t-tests are quite robust, it is good practice to evaluate the
degree of deviation from these assumptions in order to assess the quality of the results. In a paired sample t-test, the observations are defined as the differences between two sets of values, and
each assumption refers to these differences, not the original data values. The paired sample t-test has four main assumptions:
• • The dependent variable must be continuous (interval/ratio).
• • The observations are independent of one another.
• • The dependent variable should be approximately normally distributed.
• • The dependent variable should not contain any outliers.
Level of Measurement
The paired sample t-test requires the sample data to be numeric and continuous, as it is based on the normal distribution. Continuous data can take on any value within a range (income, height,
weight, etc.). The opposite of continuous data is discrete data, which can only take on a few values (Low, Medium, High, etc.). Occasionally, discrete data can be used to approximate a continuous
scale, such as with Likert-type scales.
Independence of observations is usually not testable, but can be reasonably assumed if the data collection process was random without replacement. In our example, it is reasonable to assume that the
participating employees are independent of one another.
To test the assumption of normality, a variety of methods are available, but the simplest is to inspect the data visually using a tool like a histogram (Figure 1). Real-world data are almost never
perfectly normal, so this assumption can be considered reasonably met if the shape looks approximately symmetric and bell-shaped. The data in the example figure below is approximately normally
Histogram of an approximately normally distributed variable.
Outliers are rare values that appear far away from the majority of the data. Outliers can bias the results and potentially lead to incorrect conclusions if not handled properly. One method for
dealing with outliers is to simply remove them. However, removing data points can introduce other types of bias into the results, and potentially result in losing critical information. If outliers
seem to have a lot of influence on the results, a nonparametric test such as the Wilcoxon Signed Rank Test may be appropriate to use instead. Outliers can be identified visually using a boxplot
(Figure 2).
Boxplots of a variable without outliers (left) and with an outlier (right).
The procedure for a paired sample t-test can be summed up in four steps. The symbols to be used are defined below:
• \(D\ =\ \)Differences between two paired samples
• \(d_i\ =\ \)The \(i^{th}\) observation in \(D\)
• \(n\ =\ \)The sample size
• \(\overline{d}\ =\ \)The sample mean of the differences
• \(\hat{\sigma}\ =\ \)The sample standard deviation of the differences
• \(T\ =\)The critical value of a t-distribution with (\(n\ -\ 1\)) degrees of freedom
• \(t\ =\ \)The t-statistic (t-test statistic) for a paired sample t-test
• \(p\ =\ \)The \(p\)-value (probability value) for the t-statistic.
The four steps are listed below:
• 1. Calculate the sample mean.
• \(\overline{d}\ =\ \cfrac{d_1\ +\ d_2\ +\ \cdots\ +\ d_n}{n}\)
• 2. Calculate the sample standard deviation.
• \(\hat{\sigma}\ =\ \sqrt{\cfrac{(d_1\ -\ \overline{d})^2\ +\ (d_2\ -\ \overline{d})^2\ +\ \cdots\ +\ (d_n\ -\ \overline{d})^2}{n\ -\ 1}}\)
• 3. Calculate the test statistic.
• \(t\ =\ \cfrac{\overline{d}\ -\ 0}{\hat{\sigma}/\sqrt{n}}\)
• 4. Calculate the probability of observing the test statistic under the null hypothesis. This value is obtained by comparing t to a t-distribution with (\(n\ -\ 1\)) degrees of freedom. This can
be done by looking up the value in a table, such as those found in many statistical textbooks, or with statistical software for more accurate results.
• \(p\ =\ 2\ \cdot\ Pr(T\ >\ |t|)\) (two-tailed)
• \(p\ =\ Pr(T\ >\ t)\) (upper-tailed)
• \(p\ =\ Pr(T\ <\ t)\) (lower-tailed)
Determine whether the results provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
There are two types of significance to consider when interpreting the results of a paired sample t-test, statistical significance and practical significance.
Statistical Significance
Statistical significance is determined by looking at the p-value. The p-value gives the probability of observing the test results under the null hypothesis. The lower the p-value, the lower the
probability of obtaining a result like the one that was observed if the null hypothesis was true. Thus, a low p-value indicates decreased support for the null hypothesis. However, the possibility
that the null hypothesis is true and that we simply obtained a very rare result can never be ruled out completely. The cutoff value for determining statistical significance is ultimately decided on
by the researcher, but usually a value of .05 or less is chosen. This corresponds to a 5% (or less) chance of obtaining a result like the one that was observed if the null hypothesis was true.
Practical Significance
Practical significance depends on the subject matter. It is not uncommon, especially with large sample sizes, to observe a result that is statistically significant but not practically significant. In
most cases, both types of significance are required in order to draw meaningful conclusions.
Statistics Solutions can assist with your quantitative analysis by assisting you to develop your methodology and results chapters. The services that we offer include:
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*Clean and code dataset, and create composite scores
*Conduct descriptive statistics (i.e., mean, standard deviation, frequency and percent, as appropriate)
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Cómo calcular el coeficiente de variación utilizando Excel
Step 1: Prepare your data
When it comes to working with data, the first step is to prepare it in a format that is suitable for analysis. This involves cleaning and structuring the data in a way that allows for efficient
processing and meaningful insights.
To start the data preparation process, it is important to carefully assess the quality and completeness of the data. This includes checking for missing values, duplicate entries, and any
inconsistencies or errors that may affect the analysis results.
Once the data quality has been assessed and addressed, the next step is to organize the data into a structured format. This can include creating tables or spreadsheets, defining variables and their
types, and ensuring that the data is properly labeled and formatted.
Another important aspect of data preparation is data cleaning. This involves removing any outliers or anomalies in the data, correcting errors, and handling missing values. This process ensures that
the data is reliable and accurate for analysis.
Additionally, it may be necessary to transform or manipulate the data to make it suitable for analysis. This could involve performing calculations, aggregating data, or creating new variables based
on existing ones. These transformations help to derive meaningful insights and patterns from the data.
Lastly, it is important to ensure that the prepared data is properly documented. This includes documenting any assumptions made, the steps taken during data preparation, and any changes made to the
original data. Proper documentation helps to ensure transparency and reproducibility in the analysis process.
In conclusion, preparing data is an essential step in any data analysis project. It involves carefully assessing the data quality, organizing and structuring the data, cleaning and transforming it,
and documenting the entire process. By investing time and effort into data preparation, analysts can ensure the reliability and accuracy of their analysis results.
Step 2: Calculate the mean
Once you have collected the data you need, the next step is to calculate the mean. The mean, also known as the average, is a measure of central tendency that represents the typical value in a set of
To calculate the mean, you need to add up all the numbers in your data set and then divide the sum by the total number of values. Here are the steps to follow:
1. Add up all the numbers: Start by adding up all the numbers in your data set. This will give you a total sum.
2. Count the number of values: Determine how many values are in your data set. This will be the total number of observations or data points.
3. Divide the sum by the count: Take the total sum and divide it by the count. The result is the mean.
For example, let’s say you have the following data set:
5, 8, 3, 9, 2
To calculate the mean, you would:
1. Add up the numbers: 5 + 8 + 3 + 9 + 2 = 27
2. Count the number of values: There are 5 values in this data set.
3. Divide the sum by the count: 27 ÷ 5 = 5.4
Therefore, the mean of this data set is 5.4.
Calculating the mean is a fundamental step in many statistical analyses. It provides a representative value that can help summarize the data and make comparisons between different sets of numbers.
Remember, though, that the mean can be sensitive to outliers, so it’s important to consider other measures of central tendency in conjunction with the mean.
Step 3: Calculate the standard deviation
In statistics, the standard deviation measures the amount of variation or dispersion in a set of numbers. It tells us how tightly or loosely the data points are clustered around the mean. Calculating
the standard deviation gives us a sense of how spread out the data is.
To calculate the standard deviation, we follow these steps:
1. Find the mean of the data set. This is done by adding up all the values and dividing by the number of data points.
2. Subtract the mean from each data point and square the result.
3. Calculate the sum of all squared differences.
4. Divide the sum by the number of data points minus one.
5. Take the square root of the result to find the standard deviation.
Let’s go through an example to better understand how to calculate the standard deviation.
Suppose we have the following data set:
Step 1: Find the mean
Mean = (10 + 12 + 14 + 16 + 18) / 5 = 70 / 5 = 14
Step 2: Subtract the mean and square the result
• (10 – 14)^2 = 16
• (12 – 14)^2 = 4
• (14 – 14)^2 = 0
• (16 – 14)^2 = 4
• (18 – 14)^2 = 16
Step 3: Calculate the sum of squared differences
Sum = 16 + 4 + 0 + 4 + 16 = 40
Step 4: Divide the sum by the number of data points minus one
Divide by 5 – 1 = 4
Result = 40 / 4 = 10
Step 5: Take the square root
Standard Deviation = √10 ≈ 3.16
The standard deviation of this data set is approximately 3.16.
In conclusion, the standard deviation provides a measure of the spread of data around the mean. It is useful in analyzing the variability and distribution of a data set.
Step 4: Calculate the coefficient of variation
To calculate the coefficient of variation, follow these steps:
1. Calculate the mean (average) of the data set.
2. Calculate the standard deviation of the data set.
3. Divide the standard deviation by the mean.
4. Calculate the coefficient of variation using the formula:
Coefficient of Variation = (Standard Deviation / Mean) * 100
The coefficient of variation is a measure of relative variability. It is often used to compare the variability of different data sets, especially if the means of the data sets are different.
By calculating the coefficient of variation, you can determine which data set has a higher or lower variability relative to its mean. A lower coefficient of variation indicates that the data set has
less variability, while a higher coefficient of variation indicates greater variability.
Calculating the coefficient of variation is useful in various fields, such as finance, statistics, and quality control. It provides insight into the dispersion of data and can aid in decision-making
Step 5: Interpret the coefficient of variation
After calculating the coefficient of variation, it is important to interpret the result. The coefficient of variation is a statistical measure that shows the relative variability within a dataset.
What does the coefficient of variation tell us?
The coefficient of variation provides insight into the consistency or inconsistency of the data values. It is often used to compare the variability between different datasets or different samples
from the same population.
The coefficient of variation formula:
CV = (Standard Deviation / Mean) × 100
Interpreting the coefficient of variation:
1. If the coefficient of variation is low (less than 15%), it indicates a low relative variability and high consistency among the data values. This suggests that the dataset is relatively stable and
2. If the coefficient of variation is moderate (15% to 30%), it implies a moderate relative variability and moderate consistency. The dataset shows some level of variation, but it is not considered
highly unpredictable.
3. If the coefficient of variation is high (greater than 30%), it indicates a high relative variability and low consistency. The dataset has a large amount of variation, suggesting that it may be
unreliable or highly unpredictable.
Limitations of the coefficient of variation:
While the coefficient of variation is a useful measure for evaluating variability, it has some limitations. It is sensitive to outliers, meaning that extreme values can significantly impact the
result. Additionally, the coefficient of variation alone cannot provide information about the underlying distribution of the data.
In conclusion, the coefficient of variation is a valuable tool for assessing the relative variability within a dataset. By interpreting its value, we can gain insights into the consistency and
predictability of the data. However, it is important to consider its limitations and use it in conjunction with other statistical measures for a comprehensive analysis. | {"url":"https://matematizame.com/como-calcular-el-coeficiente-de-variacion-utilizando-excel/","timestamp":"2024-11-02T14:33:44Z","content_type":"text/html","content_length":"96160","record_id":"<urn:uuid:dc5e21da-0f6a-4e31-827a-674bd5018f1f>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00800.warc.gz"} |
Representing Numbers in Different Ways
Subject: Math
Lesson Length: 25 - 30 mins
Topic: Representing Numbers in Different Ways
Grade Level: 3, 4, 5
Standards / Framework:
Brief Description: Students will learn to represent numbers in different ways by creating a number comic.
Know Before You Start: Students should have a basic understanding of numbers up to 1000, and expanded notation.
• Read and discuss the sample comic. Is the student correct?
• In pairs, have students find other ways to show 822. They may use numbers or pictures. Have them share their answers with the class.
• Show students how to write 822 in expanded notation.
• List several numbers on the board.
• Using the sample comic as a guide, have students create a comic depicting one of the numbers in a different way.
• Share and discuss student comics with the class.
□ Are there any other ways to depict the numbers shown in each comic?
• Allow students to use the speech-to-text feature as needed.
• Allow students to use the voiceover feature to read their comics aloud.
• Provide a visual resource for students to represent numbers in a new way.
• Comic to print or display: Comic.
Suggested Story Starters: | {"url":"https://ideas.pixton.com/representing-numbers-in-different-ways","timestamp":"2024-11-02T08:38:47Z","content_type":"text/html","content_length":"32846","record_id":"<urn:uuid:9b1e853f-5daa-4dd0-9148-a3eb7e8f618e>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00576.warc.gz"} |
Divide Array in Sets of K Consecutive Numbers | CodingDrills
Divide Array in Sets of K Consecutive Numbers
Given an array of integers nums and an integer k, return true if it is possible to divide the array into some sets of k consecutive numbers.
Ada AI
I want to discuss a solution
What's wrong with my code?
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javascript (node 13.12.0) | {"url":"https://www.codingdrills.com/practice/divide-array-in-sets-of-k-consecutive-numbers","timestamp":"2024-11-08T05:47:35Z","content_type":"text/html","content_length":"13088","record_id":"<urn:uuid:57e51aaf-b434-41c7-974e-50e5811ed431>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00138.warc.gz"} |
Jeffrey C. Witt
I’ve been taking a calculus course of late and have been looking for applications. Besides math, I love to bike, and since biking involves lots of changes, it seems like a good place to experiment
with applications of my current calculus knowledge.
In this experiment I wanted to use calculus to answer some questions that often arise during my ride.
I often ride familiar routes, and for those familiar routes I often have a target “total miles per hour average” that I would like to reach. At given point in my route, I will look down at my simple
computer and learn that, thus far, I have been averaging 15mph.
But if my goal for the total route is 17mph, I’m usually curious about what it will take during the remainder of the proposed route in order to meet my goal.
Unfortunately, predicting what I will need to do (or whether it is even remotely possible) is not very intuitive. The rate of change in my overall average is dependent on a lot more than just my
current average. How quickly I can improve my overall average is significantly affected by how long I have been riding, how close my current average is to my goal average, and how much distance (or
time) remains in the overall route. Moreover, the rate of change is constantly in flux as the underlying parameters (overall average, distance traveled, distance remaining) are changing as I continue
my ride.
What I would like to do is be able to create a “bike computer” interface that (using its knowledge of my current distance, time, and average, and the amount distance or time remaining in my route)
constantly reports and updates the new average I need to maintain in order to meet my goal (as well as the distance or time I will require to meet my goal at my current pace).
Calculus (and integration in particular) will be particularly useful tools for building such an interface.
Let’s start with the general formula that will be needed to in order for my new computer to constantly perform these calculations.
Our overall goal is to reach a target speed. So this is a good place to start.
If our goal is to finish with an average of 17mph, we’re going to need to end with a distance and time that can give us this result.
\[Speed = \frac{Distance}{Time}\]
we know that we will need something like this:
\[17 = \frac{Distance}{Time}\]
But our challenge is to pick any point somewhere in the middle of the route and, based on the distance covered at that point in time, to pick a new speed that leads us to our desired overall average.
In this case, we know the starting distance (which we will call \(s\)) of the overall distance (\(d\)) (which we also know) and we know the time traveled (\(b\)) at \(T_1\).
But we want to discover the second part (which we will call \(r\) for remaining distance) of \(d\) at \(T_2\) and the additional time required to reach \(T_2\) (which we will call \(x\)) based on the
already completed distance \(s\) and initial time \(b\) and, most importantly, the speed required to cover that remaining distance (\(r\)) within the allotted amount of remaining time (\(x\)).
Let’s start by calculating the starting distance (\(s\)) at time \(T_1\) or \(b\).
This is an integral function:
\[\mathrm{d}y = \int_0^b f(x) \mathrm{d}t\]
Again \(b\) is the time (in hours) and \(f(x)\) is the function that describes the change in distance during \(\mathrm{d}t\). In our example \(f(x)\) will be very simple (i.e. the derivative of a
linear function, like 15).
But being general here will allow our calculations to not just work with linear functions but later with more elaborate functions (e.g. functions that might describe the average speed at \(T_1\) of a
rocket) and will allow our calculations to be even more accurate.
So let’s imagine I’ve been biking for 1 hours (\(T_1=1\) and \(b = 1\)) and my average over that 1 hour has been 15mph (\(f(x) = 15\)). We can compute the distance covered at this point (\(\mathrm{d}
y\)) by computing the definite integral: \(\int_0^1 15 \mathrm{d}t\) which becomes \(15(1)\) or 15 miles.
If we want to get to the speed over that hour, we can just divide \(\mathrm{d}y\) by how long I’ve been traveling, \(\mathrm{d}t\); in other words \(\frac{\mathrm{d}y}{\mathrm{d}t}\), or \(\frac{s}
{b}\) which equals \(\frac{15}{1}\) or 15 miles per hour.
Ok, but our final goal is \(\frac{17}{1}\) or 17 miles per hour.
So to reach our goal, we’re going to need an overall distance \(d\) (or \(dy\)) that, when divided by the overall time \(b + x\) (where x is the additional time traveled), gives us 17.
But we already know part of the overall function that is going to lead us to \(\mathrm{d}y\). So our question is really what do we need to add to get to our desired result? Or more concretely, how
fast do we need to travel over the additional amount of time \(x\). Let’s call this new and unknown rate of change \(g(x)\).
So getting to a \(\mathrm{d}y\) (\(d\), the overall distance) where \(\frac{\mathrm{d}y}{b+x}=17\) is a matter of adding another integral to the value of the already known integral.
\[\mathrm{d}y = \int_0^b f(x) \mathrm{d}t + \int_0^x g(x) \mathrm{d}t\]
And since we know our target average speed (17) is just the overall distance divided by the overall time \(\frac{\mathrm{d}y}{b + x}\) we have the following equation:
\[17 = \frac{\int_0^b f(x) \mathrm{d}t + \int_0^x g(x) \mathrm{d}t}{b+x}\]
Now we’re getting close. We can see our goal \(g(x)\), but before we can solve for \(g(x)\), we first need to find \(x\) or the additional amount of time anticipated in the planned route.
But this a little tricky because I don’t know the time yet. The amount of time it will take will depend on how fast I go or \(g(x)\) which is exactly what I’m trying to find.
However, because I know the route of my overall ride and thus the distance of the overall route, we can describe \(x\) in terms of \(g(x)\) and remaining distance which we have called \(r\).
Again, recall that:
\[Speed = \frac{Distance}{Time}\]
\[g(x) = \frac{r}{x}\]
So we just need to find \(r\) or the remaining distance.
And remaining distance will be the total distance of the route \(d\) minus the starting (already travelled) distance \(s\) at \(T_1\).
The starting distance \(s\) can be computed from the speed and time at \(T_1\) which is \(s = f(x)b\)
Putting all this together, we have:
\[r = d - f(x)b\]
And with the help of r, we can now replace all \(x\)’s (the remaining time) with \(\frac{d-f(x)b}{g(x)}\), and then we can solve for \(g(x)\).
Back to our above equation, summing the two integrals:
\[y = \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\int_0^b f(x) \mathrm{d}t + \int_0^x g(x) \mathrm{d}t}{b+x}\]
This can be now modified to:
\[y = \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\int_0^b f(x) \mathrm{d}t + \int_0^\frac{d - f(x)b}{g(x)} g(x) \mathrm{d}t}{b+\frac{d - f(x)b}{g(x)}}\]
It’s worth stopping here for a second to recognize this above equation as the important bit. All we have left to do is plug in our known quantities and solve for g(x). But the abstract equation here
helps us see that this equation should work in any scenario, even if the parameters are very different or f(x) is a very complicated function or even if g(x) needed to be variable function and not
just a constant.
With that in mind, let’s solve our original practical problem with our known quantities.
If my goal remains 17mph and the overall route is 30 miles and \(T_1 = 1\) and my average at \(T_1 = f(x) = 15\), then
\[17 = \frac{\int_0^1 15 \mathrm{d}t + \int_0^\frac{30-15(1)}{g(x)} g(x) \mathrm{d}t}{1+\frac{30-15(1)}{g(x)}}\]
which reduces to:
\[17 = \frac{15(1) + \frac{g(x)(30-15(1))}{g(x)}}{1 + \frac{30-15}{g(x)}}\]
As the \(g(x)\) in the top fraction cancels out and a few more sums can be simplified, we can further reduce to:
\[17 = \frac{15 + (30-15)}{1 + (\frac{30-15}{g(x)})}\] \[17 = \frac{15 + 15}{1 + \frac{30-15}{g(x)}}\] \[17 = \frac{30}{1 + \frac{15}{g(x)}}\]
Cross multiply to get:
\[17(1 + \frac{15}{g(x)}) = 30\]
\[17(1) + 17(\frac{15}{g(x)}) = 30\]
And then solve:
\[17(\frac{15}{g(x)}) = 30-17\] \[17(\frac{15}{g(x)}) = 13\] \[\frac{15}{g(x)} = \frac{13}{17}\] \[15 = \frac{13}{17}(g(x))\] \[\frac{15}{\frac{13}{17}} = g(x)\] \[15(\frac{17}{13}) = g(x)\] \[19.615
= g(x)\]
Thus after traveling 1 hour at an average of 15 miles per hour, with only 15 miles left, I would need to average 19.615 mph over the next 15 miles to reach my goal of an overall average 17 mph for
the entire trip.
The entire above set of calculations can be automated based on various input parameters and from there we get the BikeComputer Application which can provide a range of outputs depending on inputs.
Here you can see that given my average speed of 15 mph over 1 hour with 15 miles remaining, the speed needed to reach goal is calculated to be exactly what we concluded above 19.615.
But the computer shows more. It shows that if I persists at my current pace of 19mph over the next 15 miles, I will need 19 miles (not 15) to reach my goal, and that at the end of the remaining 15
miles I will only have achieved an average of 16.764 mph falling about 0.24 mph short of my goal. | {"url":"https://jeffreycwitt.com/2021/06/15/biking-with-calculus/","timestamp":"2024-11-11T13:30:07Z","content_type":"text/html","content_length":"16356","record_id":"<urn:uuid:4d58940f-d333-4125-8821-dee9c6113f41>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00647.warc.gz"} |
Write these fractions appropriately as additions or subtractions
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Write these fractions appropriately as additions or subtractions :
We will use the concepts of Fractions to solve this.
(a) Each rectangle has a total of 5 sections.
(b) Each circle has a total of 5 sections.
(c) There are a total of 6 dots in each rectangle.
You can also use the Fraction Calculator to check your answer.
NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.5 Question 1
Write these fractions appropriately as additions or subtractions :
(a) The image represents the addition of fractions, and the sum is 3/5. (b) The image represents the subtraction of fractions, and the difference is 2/5. (c) The image represents the addition of
fractions, and the sum is 5/6.
☛ Related Questions:
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/ncert-solutions/write-these-fractions-appropriately-as-additions-or-subtractions/","timestamp":"2024-11-11T07:47:38Z","content_type":"text/html","content_length":"209870","record_id":"<urn:uuid:11112d0b-027b-484c-9be7-b15117e11074>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00160.warc.gz"} |
A Wrinkle in Target: Why You Should Not Apply Actuarial Trend to Your GLM Target Variable | Published in Variance
1. Introduction
Creating the target response variable for an insurance loss cost predictive model has traditionally involved two main actuarial treatments: development and trending. Since claims can take a long time
before they are settled, development allows the actuary to estimate their ultimate (final) settlement values as of the time of analysis. Trending, which is the main subject of this paper, allows the
actuary to adjust for the general changes in loss costs over time particularly due to changes in prices of insurance cost factors such as business mix, inflation, technology, and tort. The interested
reader should see Basic Ratemaking by Werner and Modlin for a more in-depth discussion of these two methods (105-121). However, I will show that actuarial trend rates are not suitable for trending
target loss variables of actuarial predictive models. This is because they contemplate other dynamics—business mix—which are already contemplated in predictive models, and therefore introduce
extraneous modeling complications which bias predictions.
Such bias can be enough to eat away all if not a significant proportion of the underwriting profit margin (given the narrow profit margins of insurance books of business). Or equally regressive, it
can cause the book of business to shrink unnecessarily. A trend undershoot of 1% can eat away as much as 10% of margins, with the actual profit corrosion depending on size of profit provision
(assumed in this case to be 10%); and the impact of a 1% trend overshoot on the growth of your business will depend on elasticity of demand. Also, the trend bias problem, as I will show, may be so
elusive that it may cause an insurance book of business to bleed to its death—a condition I have called Pricing Hemorrhage. This paper discusses the distortionary effects of actuarial trends on
predictive modeling target loss variables (loss costs, frequency, severity, loss ratios or any other), and demonstrate ways to avoid them. Therefore, the purpose of this paper is central to driving
strong and consistent underwriting results for an insurance book of business. Its – the paper’s—close relationship to pricing precision also renders it relevant to any actuarial department which
seeks to promote the standards of pricing excellence.
The paper is arranged as follows: Part 2 is a theoretical exposition of how an inappropriate trend such as the actuarial trend rate can distort the target loss variable in a predictive modeling
framework; Part 3 presents a solution to the distortion; Part 4 uses simulated data to demonstrate Parts 2 and 3; Part 5 concludes by restating the major themes of the paper and their ramification on
the profitability of an insurance book of business.
2. The distortion of target loss costs by actuarial trends
Assume the ultimate loss cost of the ith risk in an insurance book of business in year t (denote evolves as follows^[1]:
is the model parameter associated with the constant term;
is the model parameter associated with the kth model variable;
is the kth model variable (such as insurance score) of the ith risk in exposure period t; and
the general inflation parameter, is the rate of change of loss costs over time due to changes in general economic and social climate such as monetary inflation, technology, insurance laws, and other
related factors; it does NOT however include changes in business mix or distribution as that is rather accounted for by the risk attributes, In this paper, we call such general inflation parameter
the econometric trend rate. It should be noted that even though I focus on loss costs in this section, all insights and conclusions apply freely to other target loss variables such as frequency,
severity and loss ratios.
The above is a generalized linear model (GLM) specification. The interested reader can consult with McCullagh and Nelder 1989 for a thorough treatment of the topic, or Duncan, Feldblum, Modlin,
Shirmacher, and Neeza 2004 for a casual treatment. It must however be pointed out that, an understanding of GLMs is not required to follow the rest of the paper. It is typical in actuarial predictive
modeling for the target loss costs to be trended to the level of the current year or a specified future year in which the model will be implemented (called trend-to year in actuarial argot). It is
also typical for an actuarial trend rate to be used for this purpose. This section shows that this culture of trending target loss variables with an actuarial trend rate creates a subtle wrinkle in
the target response variable that biases loss cost predictions. To see this, notice that the actuarial trend, which we will denote r, is a sum of and hence contemplates two main dynamics:
1. The changes in aggregate loss costs over time driven by general economic and social factors which we have denoted by and called the econometric trend rate above; and
2. The changes in aggregate loss costs driven by changes in business mix or distribution such as policy limits, deductibles, tenure and other risk characteristics over time. We will denote such
The reader should note that, in an actuarial predictive model, changes in business mix are already accounted for by the risk variables —We specifically build predictive models so that rates can
automatically change with risk characteristics. Hence, while applying to aggregate loss costs is in order, applying it to the granular (risk-level) target loss cost in a predictive model is
unnecessary and even distortionary. To explicitly show the distortionary effect of an actuarial trend on predictive model predictions, let us introduce a new term, , to denote the level of the target
loss cost of ith risk at period t when trended to period T with the actuarial trend rate. Mathematically,
Three things are worth pointing out to the reader about (1) and (2). The reader should remember that (1) provides us with the true values of loss cost for an insurance risk in any period and is hence
the reality the pricing actuary seeks to estimate.
i. (1) and (2) differ by two main parameters: The constant term and trend parameters In other words, if we fit the raw (non-trended) and trended target loss costs on the risk variables and trend
term, all model factors will be equal except for the ones associated with the constant and trend terms.
ii. The presence of a residual trend term, in (2) even after trending all loss costs to the same exposure year is concerning and highlights the elusive distortion, call it wrinkle, that is introduced
by the actuarial trend factor. A target variable, correctly trended to the same level, should have no correlation with time. Additionally, this residual trend term is not only elusive but also a bane
to modeling: As we will show in Part 4, when target loss costs are trended with the actuarial rate, model predictions will still be biased whether a trend term is included in or excluded from the
model. The modeler, unaware of the inappropriateness of the actuarial trend rate, will not expect any such nuisance trend term, let alone be inspired to rectify its distortions.
iii. (1) and (2) differ in loss cost predictions for all but exposure year T (the trend-to year). This implies that the predictions from (2) are almost always biased. Particularly, the bias factor,
measured as the ratio of predicted loss cost under (2) to that under (1), can be calculated as:
The following implications about the bias factor should also be noted:
I. If the actuarial trend rate, r, is positive, then the predictions under (2) will be overbiased for all exposure periods prior to the current exposure year (T), and underbiased for all exposure
periods afterwards; whereas if r is negative, it will be underbiased for all exposure periods prior to the current exposure year (T), and overbiased for all periods after. The level of bias is
proportional to the actuarial trend rate and the distance between the exposure and current (trend-to) years, implying the bias would continue to worsen over time if unchecked, a phenomenon I call, a
Pricing Hemorrhage. The relationship between the bias and exposure year is shown by the bias curve in Exhibit 1.
II. To eradicate the bias caused by trending, the loss cost predictions attained under (2) must be trended back to their exposure levels with the actuarial trend rate (i.e., multiplied by )—A
treatment that is elusive and will not be obvious to the modeler. This is one of the punchlines that the article seeks to make: That trending target loss costs with an actuarial trend rate in
predictive modeling presents an extraneous pricing trap which the pricing team can do without. It will therefore be in the modeler’s interest to rather model non-trended than trended loss costs as we
will demonstrate below.
3. The econometric trend— An efficient way to adjust for trends in predictive model target loss variables
A solution to the trend bias trap presented in Part 2 is to model non-trended ultimate loss costs (or frequency or severity) on an econometric trend term (time t variable) and other relevant risk
covariates As can be observed in equation (1) and is also shown with the simulations in Part 4, this provides unbiased loss cost estimates for all risks in all time periods. Because the econometric
trend effect is estimated simultaneously with the other risk covariates, it correctly measures the magnitude of the general inflation rate. In addition, the modeling team is saved from the burden of
having to trend loss costs and any corresponding costs of falling into the pricing trap discussed above.
A befitting question is why has the suboptimal trending culture—modeling target loss costs trended with actuarial trend rate—been so prevalent in actuarial predictive modeling? The first reason is
what I call, The Occasional Trap of Intuition (TOTI): The concept of trending target loss costs (just as we do with aggregate loss costs) sounds so intuitive that it has not attracted a serious
consideration or study, and thus, has maintained its popularity despite its sub-optimality. The other is the expertise asymmetry between the pricing actuary and predictive modeler in each other’s
body of work: The superficial knowledge of the predictive modeler about actuarial subject matters such as actuarial trends makes it hard for him or her to analyze their effects on predictive model
outcomes. I know of many fellow predictive modelers who used actuarial trends liberally but did not know what they contemplated or how they differed from an econometric trend. Likewise, most
actuaries are not statisticians and may hence not be able to see the subtle effects of actuarial applications in a predictive model framework. It takes a professional who has pursued a deep training
in both fields to catch such subtleties and arrange a healthy marriage between the products of the two fields.
4. Showing the trend bias with simulations
Assume we simulate a book of insurance business which follows the dynamics below:
1. Age group (young vs old) is the only relevant risk variable; and that a young insured is 3 times as risky as an old insured.
2. The proportion of young insureds increases by 5% every year, from 20% of the book in year 1 to 80% in year 13.
3. Loss severity increases exponentially by 3% per annum due to general inflation.
4. Loss severity for a given risk i in exposure year t, follows an exponential distribution, and per 1, 2, and 3, is defined as:
5. 40,000 claims are filed every year.
6. Exposure years 1 through 8 are the historic years whose data will be used for actuarial and predictive modeling analyses; and exposure years 9 through 13 are the prospective years in which the
proposed model will be implemented.
The reader should note that even though simple, the above simulation captures the major dynamics of an insurance book of business.
4a. Deriving the actuarial trend rate
An actuary derives his (her) trend rates by examining severity of ultimate claims at periodic intervals such as monthly, quarterly, or yearly. He (She) would then assess the magnitude and direction
of the periodic changes in ultimate claim severities to estimate a suitable trend rate he (she) believes will apply to future years. Using the simulated data above, Exhibit 2 shows the average
severity estimates and their natural logs for the historic exposure years.^[2]
In a log-link model specification, it is easier to rather estimate trends in the logged- than the raw-variable space. Exhibit 3 fits a trend line to log of loss severity to estimate the trend rate.
Per the above trend line, the actuary will select 8.7%—the coefficient of the time variable—as the trend rate for severity. As mentioned in Part 2 and evidenced here, it is different from the general
inflation rate because it also contemplates the continuous distribution changes in the book—increasing proportion of young insureds in this example. It therefore follows that the trend due to changes
in mix only, is roughly 5.7% (8.7%-3.0%). We show how the trend distortion and bias in predictions occur using the case analyses below.
4b. Case analyses
In this section, we analyze four different ways of dealing with trends in actuarial predictive models. For each case, we will present the GLM parameter estimates, and compare the means of predicted
and actual outcomes for the historic (t=1-8) and prospective (t=9-13) exposure years. These should help the reader build a vivid picture of the wrinkles caused by trending target loss variables with
the actuarial trend rate and how they can be unraveled. The following definitions will be handy:
Mean Actual Severity is the average value of the actual loss severities of the insured risks for the exposure year of interest. The actual loss severity for any risk in any exposure year is given by
equation (5) above.
Mean Modeled Severity is the average value of the modeled loss severities of the insured risks for the exposure year of interest. This is the mean of the dependent variable that was modeled. The
reader should note that, when we trend the target loss severity, the mean actual and mean modeled severities diverge.
Mean Predicted Severity is the average value of the predicted loss severities of the insured risks for the exposure year of interest.
It is also worthy to point out that the effect of the trend-induced biases on pricing in each case will depend on how the actuarial team uses the GLM output for ratemaking. In this article, we will
focus on two major approaches: In the traditional approach, the pricing team fetches only the GLM coefficients of the risk variables as relativities and combine them with the actuarially derived base
rate to calculate the proposed premiums. In the evolutionary approach (argued as superior and recommended in the paper, Ratemaking Reformed: The Future of Actuarial Indications in the Wake of
Predictive Analytics), the pricing team uses the risk’s GLM prediction—as the best estimate of the risk’s pure premium— and inflate with fixed and variable expense provisions to calculate the risk’s
proposed premium. Also, while traditional ratemaking exercise is performed at least once every year for most insurance products, actuarial GLMs can remain in production without major changes for 3-5
years. Therefore, defects in GLMs tend to persist for a longer period.
Case 1 (most common scenario): Target loss severity is trended with actuarial trend rate and a trend term is added to the set of covariates.
When we model actuarially trended loss severity with a trend term and a dummy for age group, the model parameter estimates are shown in Exhibit 4.
The results are consistent with what was discussed in Part 3: The parameter for young is unbiased (1.0977=ln(3)), and the trend term parameter is the negative of the distribution (business mix) trend
rate Also, as discussed above and shown in Exhibit 5, predictions of severity for both retrospective years and prospective years are biased (except for the year to which the target loss severity was
trended, year 8 in this case), and the biases are directly proportional to the absolute difference between the exposure and trend-to years.
The bias is vivid in the above table: Loss severity is overstated for historic years and understated for future years. The reader should see that, since the GLM risk relativities are unbiased in this
case, the traditional pricing approach will not be affected. However, the evolutionary pricing approach which rather uses the GLM predictions to make premiums will suffer the biases created by this
trending culture.
The above trend bias is not a defect to be taken lightly by the actuary. It can stealthily rupture the veins of an insurance book of business and cause it to bleed increasingly to death: A malady I
have named as Pricing Hemorrhage! The problem is that the pricing team (unaware of this distortionary effect) will not know that the target (modeled) severity is a contorted reality, and that even
though their predictions match this target, they deviate from actual severity. Sadly, the deviations worsen as time passes. This defect can go undetected for years. Even when the imprecision is
detected, it is likely to be misdiagnosed as a base rate inadequacy and consequently be treated, albeit ineffectively, with a constant base rate offset. But as can be inferred by the trending values
of the bias, the effective remedy should have a trend component. In fact, as noted in Part 2, we get the correct mean severity value for each exposure year by multiplying the GLM predictions by which
equals in this case, with t denoting the exposure year of interest. The need to remedy may also not be obvious to the pricing team because the presence of a trend term in the model creates a false
notion that the GLM estimates are already trended to their right exposure year levels!
Case 2: Target loss severity is trended with actuarial trend rate and a trend term is NOT added to the set of covariates.
Exhibit 6 shows the results when we trend our target loss severity with the actuarial trend rate but do not include a trend term in our predictive model.
As you can see, the age parameter is biased (1.0332=ln(2.81) < ln(3)). The reason for the bias is due to trending the target variable with trend rate which does not equal the econometric trend rate.
As shown in Part 3, the inappropriateness of the actuarial trend rate inhibits it from ridding the model of its trend effects and rather creates a distortionary residual trend effect; hence, when the
modeler excludes a trend term in his model, the residual trend effect is picked up by the intercept term and the age variable through a phenomenon which statisticians call the Omitted Variable Bias.
The intuition for this is simple: With the old segment being disproportionately larger in the earlier years, the application of an overstated trend rate exaggerates their—the old segment’s—
current-level severities, thereby dampening the disparity in severity of claims between the two groups.
The predicted estimates are also biased against the actual and even the modeled severities, as shown in Exhibit 7.
Unlike in case 1, both the traditional and evolutionary pricing approaches will be adversely impacted by the bias in this case since both the relativities and the predictions are biased.
In case 3, we see that these biases evaporate when we appropriately trend the target severities with the econometric trend rate.
Case 3: Target loss severity is trended with econometric trend rate and a trend term is included in the model.
As expected and observed in Exhibit 8, when the target loss severity is trended with the appropriate factor, all trend effects are flattened; thus, adding a trend term will be found to be
statistically insignificant as evidenced in the model output. The predictive modeler ideally should remove the trend term from the model for two main reasons. The first is to save a statistical
degree of freedom; but even more importantly, to eliminate any trace of false impression, albeit the almost zero trend coefficient, that trending has been already contemplated in the predictions.
That way, he or she will remember accordingly that, predictions are as of the trend-to year (i.e., year 8 in our example) and thus have to be de-trended back to their respective exposure years with
the same econometric trend rate:
Detrended Predicted = exp(0.03*(year-8)) * predicted
With respect to impact on the pricing approaches, the traditional approach will not be adversely affected since the relativities are unbiased and the evolutionary approach will also not be biased if
the detrended predictions are used.
This method, however, has application challenges. Finding the correct econometric trend rate to trend the target loss variable is difficult in practice. And as pointed out in Part 2, any inaccuracy
in the economic trend rate, like the actuarial trend rate, will lead to distortionary effects. Also, even though the trap presented here is less convoluted than case 1 (especially when the modeler
accordingly excludes the trend term), the extra step of detrending GLM predictions back to their exposure years is still a bane. It is a requirement induced by unnecessarily trending the target loss
severity to the same level of the trend-to year before modeling. Case 4 shows how we can accurately predict actual severity without trending the target loss variable.
Case 4: Target loss severity is NOT trended, and a trend term is added to the set of covariates.
This is the most preferred way for dealing with trends in an actuarial predictive modeling framework. No time or resource is wasted on trending the target loss variable before modeling; the model
parameters as well as target loss variable are accurately estimated; and no time or resource is wasted on detrending the GLM predictions back to their respective exposure years. See Exhibits 10 and
Since the relativities and the predictions are unbiased in this case, none of the two pricing approaches will be adversely affected.
Part 5: Concluding remarks
The three nonnegotiable metrics of business success —Underwriting Profit, Renewal Retention and New Business Production—have a special bond with prices (premiums) in insurance. Insurance underwriting
profit margin is one of the thinnest among industries, with even 10% being on the gracious side. Slight errors in premiums therefore have pronounced impact on profits. For instance, underpricing
customers by 1% can eat away as much as 10% of your profits, and even higher percentages with lower underwriting profit provisions. Also, given the low concentration (market power) of most insurance
industries, customer retention and demand can be highly elastic especially among the youthful and low-income segments; and lastly, the high cost of acquiring new business, ranging from 10% to 30% of
premiums, makes the loss of a renewal or quoted business expensive, especially when due to an evitable pricing bias. Pricing adequacy and accuracy, the two major goals in insurance pricing, are thus
It is therefore the obligation of the pricing actuary to espouse practices which allow him or her to climb up the rungs of pricing excellence and eschew the ones which bring him or her down that
ladder. Trending the predictive model target loss variable is one such regressive culture. Apart from being an extraneous toll on resources and time, it introduces a stealthy pricing trap which
produces biased predictions, with the biases worsening by the year after model implementation—A condition I have called Pricing Hemorrhage! The good news is that Pricing Hemorrhage is preventable:
Simply model non-trended target loss variable as a function of a trend term and model covariates.
1. The use of a log-link and a GLM framework is only for simplicity’s sake, and without a loss of generality.
2. Quarterly periods are the most used as it allows for seasonality effects to be controlled for. However, in this example, we would use yearly periods to keep it simple and avoid general trend
complications not relevant to our topic. | {"url":"https://variancejournal.org/article/37253","timestamp":"2024-11-05T18:39:04Z","content_type":"text/html","content_length":"347893","record_id":"<urn:uuid:415079d7-c5e0-4622-9cdb-ea36c30178b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00224.warc.gz"} |
VecLib Libraries
VecLib DSP Libraries
VecLib is a large set of routines offering high-performance, optimised implementations of vector elementwise functions.
The library contains 4576 functions of 2, 3, or 4 operands, real or complex, scalar or vector.
The functions have the following forms:
r[i] = a[i] op1 b[i], i = 1, ..., N
r[i] = a[i] op1 (b[i] op2 c[i]), i = 1, ..., N
r[i] = (a[i] op1 b[i]) op2 c[i], i = 1, ..., N
r[i] = (a[i] op1 b[i]) op2 (c[i] op3 d[i]), i = 1, ..., N, where
• each of op1, op2, and op3 can be +, -, × or ÷
• each of a, b, c, and d can be real or complex, scalar or vector
• the vectors can be strided independently.
Library Contents
Functions of 2 operands:
• Contains all possible combinations of real, complex, scalar or vector arguments.
• A total of 110 independent functions, 128 functions in all.
Real functions of 3 operands:
• Contains all possible combinations of real scalar or real vector arguments.
• A total of 149 independent functions, 336 functions in all.
Complex functions of 3 operands:
• Contains all possible combinations of complex scalar and complex vector arguments.
• A total of 1263 independent functions, 3024 functions in all.
Real vector functions of 4 operands:
• Contains all possible combinations of real vector arguments.
• A total of 62 independent functions, 64 functions in all.
Complex vector functions of 4 operands:
• Contains all possible combinations of complex vector arguments.
• A total of 805 independent functions, 1024 functions in all.
Naming Convention
VecLib is a large library, but using it is simplified by the systematic naming convention. Function names are made up of the components describing the function:
• brackets ( and ) are denoted b
• operations +, -, × and ÷ are denoted p, m, t, and d, respectively
• real vector or matrix operands are denoted r
• complex vector or matrix operands are denoted c
• conjugated complex vector or matrix operands are denoted j
• real scalar operands are denoted R
• complex scalar operands are denoted C
• conjugated complex scalars are denoted J
• functions for vectors start v
• functions for matrices start m
• transposed matrix inputs are denoted T
• transposed and conjugated matrix inputs are denoted H
• operands marked for cache removal have l appended to them.
Here are a few examples of function names: r, a, b, c are vectors, ß (beta) is a scalar.
┃Function Name│ Action │ Comment ┃
┃vrpbrmrb │r[i] = a[i] * (b[i] - c[i]) │all parameters real ┃
┃vcpbjmjb │r[i] = a[i] + (b[i]' - c[i]')│all parameters complex ┃
┃vcpbrtJb │r[i] = a[i] + (b[i] * ß') │r, a and beta are complex, b is real ┃
• ARM A53, A57, and A72
• PPC T2080
• PPC T2081
• PPC T4240
• Intel AVX512
• Intel Phi
• Intel AVX2
• Intel AVX
• Intel SSE | {"url":"https://www.nasoftware.co.uk/index.php/library-products/veclib","timestamp":"2024-11-12T13:27:14Z","content_type":"text/html","content_length":"29913","record_id":"<urn:uuid:54dcf84a-6806-4822-910a-16e2bd4af176>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00515.warc.gz"} |
F1 and F2 Values Calculation Formula
A simple model independent approach uses a difference factor (f1) and a similarity factor (f2) to compare dissolution profiles. The difference factor (f1) calculates the percent (%) difference
between the two curves at each time point and is a measurement of the relative error between the two curves:
The similarity factor (f2) is a logarithmic reciprocal square root transformation of the sum of squared error and is a measurement of the similarity in the percent (%) dissolution between the two
1. Determine the dissolution profile of two products (12 units each) of the test (postchange) and reference (prechange) products.
2. Using the mean dissolution values from both curves at each time interval, calculate the difference factor (f1) and similarity factor (f2) using the above equations.
3. For curves to be considered similar, f values should be close to 0, and f 1 2 values should be close to 100. Generally, f values up to 15 (0-15) and f values 1 2 greater than 50 (50-100) ensure
sameness or equivalence of the two curves and, thus, of the performance of the test (postchange) and reference (prechange) products.
Dissolution f2 Calculation Excel Sheet
In recent years, FDA has placed more emphasis on a dissolution profile comparison in the area of post-approval changes and biowaivers. Under appropriate test conditions, a dissolution profile can
characterize the product more precisely than a single point dissolution test.
A dissolution profile comparison between pre-change and post-change products for SUPAC related changes, or with different strengths, helps assure similarity in product performance.
As per USP,
For products that dissolve very rapidly (≥85% dissolution in 15 min), a profile comparison is not necessary (thus, f2 calculation is not required).
Read also:
Post a Comment | {"url":"https://www.pharmacalculation.com/2022/12/f1-and-f2-values-calculation-formula.html","timestamp":"2024-11-09T14:01:09Z","content_type":"application/xhtml+xml","content_length":"215992","record_id":"<urn:uuid:68954134-b05f-4127-9d01-7019e531ba37>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00717.warc.gz"} |
Introduction to Properties of Identity, Inverses, and Zero
What you’ll learn to do: Understand and apply the properties of identity, inverses, and zero when simplifying expressions
Sam is having a barbecue with his family to celebrate the Fourth of July. He’s grilling hamburgers for the adults and hot dogs for the children. He expects 14 adults and 8 children. After estimating
that each child will eat one hot dog, how many hot dogs should Sam grill? To figure this out, Sam can use the identity property of multiplication. You’ll learn about this property and a few
othersĀ in this section.
Before you get started, take this readiness quiz.
readiness quiz
If you missed this problem, review the video below.
If you missed this problem, review the following video.
If you missed this problem, review the following video. | {"url":"https://courses.lumenlearning.com/wm-developmentalemporium/chapter/properties-of-identity-inverses-and-zero/","timestamp":"2024-11-09T12:43:32Z","content_type":"text/html","content_length":"51482","record_id":"<urn:uuid:5ff7dbb2-db2d-47be-9f6d-5f63bf8c1071>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00273.warc.gz"} |
December 2007 – Page 6 – USABloggen.se
In statistics, a moving average (rolling average or running average) is a calculation to analyze data points by creating a series of averages of different subsets of the full data set. It is also
called a moving mean ( MM ) [1] or rolling mean and is a type of finite impulse response filter. 2013-05-06 · I have to run some reports and visualizations based on data that resides in SQL Server
DB. Some of the metrics I am pulling are rolling 7, 10 , 30 day averages for certain parameters. So, these parameters need to be refreshed on a daily basis -- capturing the last 7,10, 30 day data for
its calculations for any given day. If you have a SQL database in source, like Oracle or SQLServer you can use a partition by windowing function. Else, if you can't, maybe you can create a second
transaction date in your table which is 7 days after the real one. Then, with a interval match joining the table with itself you will be able to group all the transactions.
Cumulative. Monthly rate (averaged over prior 6 months). C Create a list of all dates using PROC SQL and the INTNX f In statistics, a moving average is a calculation to analyze data points by
creating a series of 5 Other weightings; 6 Moving median; 7 Moving average regression model; 8 See also; 9 Notes and references; 10 External links In an n- 25 May 2020 This example describes how to
use the rolling computational functions: EXAMPLE - Day of Functions ROLLINGAVERAGE - computes a rolling average from a window of rows before and 11/13/16, 7, 610, 2800, 400. 5 Jul 2016 In this
article, we'll look at how to calculate a moving average in PostgreSQL.
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Det här är 7 nov. 2017 — SQL Server 2012 och senare Moving Average. kan släta ut det genom att planera ett 7-dagars medelvärde ovanpå den underliggande data.
In the code below, the moving average window size is 15 (7 rows preceding the current row, plus the current row, plus the 7 following rows). The 7 period rolling average would be plotted in the
mid-week slot, starting at the 4th slot of seven, not the eight. Thereafter all would be the same. If you then plotted a curve through the smoothed data, it would help to identify upward/downward
trends, especially if the trends were small relative to the daily fluctuations.
If we want to calculate the moving average of 15 days, such that it can include 7 previous and 7 next days, we can just rewrite the query as follows. 1.
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In the code below, the moving average window size is 15 (7 rows preceding the current row, plus the current row, plus the 7 following rows). The 7 period rolling average would be plotted in the
mid-week slot, starting at the 4th slot of seven, not the eight. Thereafter all would be the same. If you then plotted a curve through the smoothed data, it would help to identify upward/downward
trends, especially if the trends were small relative to the daily fluctuations. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new
features Press Copyright Contact us Creators The chart below shows the same data seen above, arrayed on a daily interval, in a rolling 7 Day trend format. With this 7 Day Period Rolling View you can
see more than just the overall trend seen in the daily total line chart. rolling 7 days = VAR _max=max ('Table' [date]) VAR _min=_max-6 return if (_min=_min&&'Table' [date]<=_max)),7)) Did I answer
your question?
The first argument can be week, month, year and more. In statistics, a moving average (rolling average or running average) is a calculation to analyze data points by creating a series of averages of
different subsets of the full data set. It is also called a moving mean (MM) or rolling mean and is a type of finite impulse response filter. Variations include: simple, and cumulative, or weighted
forms (described below). Hi All, Need Help.. I want to create a 7 day rolling average but I need to use Month year on X-axis. I am using: rangeavg(above(sum(sales),0,7)) X - 1425633 2018-04-04 We can
apply the Average function to easily calculate the moving average for a series of data at ease.
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Some states may be calculating the positivity percentage for each day, and then doing the rolling 7-day average. Using Tableau's Superstore as the database, the SQL query: To add in a moving average,
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Windward Core - Recensioner 2021 - Capterra Sverige
Se hela listan på sqlservercentral.com To add a calculated moving average to a series on the chart. Right-click on a field in the Values area and click Add Calculated Series. The Calculated Series
Properties dialog box opens. Select the Moving average option from the Formula drop-down list. Specify an integer value for the Period that represents the period of the moving average. Se hela listan
på red-gate.com Moving Average: To add in a moving average, say 30 day moving average, the Tableau Calculation can be changed to Moving Calculation, the Summarise Values Using changed to ‘Average’
and the previous changed to ’29’. This Table Calculation then makes: How to calculate 7-day rolling average of tweets by each user for every date.
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XML, SQL, HTTP/HTTPS, and EAI processes. | {"url":"https://investeringartbvfm.netlify.app/21990/50703.html","timestamp":"2024-11-04T08:24:34Z","content_type":"text/html","content_length":"19175","record_id":"<urn:uuid:8b7d305a-b851-44d5-91b8-671e0d2f8e4b>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00803.warc.gz"} |
Math Problem Statement
DFF_NHL_cheatsheet_2024-10-08 (2).csv
3.31 KB
Given the attached dataset, list the top 5 players most likely to exceed their ppg_projection. The list must include at least one player from CHI. Also make sure the sum of total player salaries are
equal to 55000. Go!
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Math Problem Analysis
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Salary Sum Constraint: Total Salary = 55000
PPG Exceedance = L5_fppg_avg > ppg_projection
Suitable Grade Level
Grades 11-12 (High School, advanced optimization problems) | {"url":"https://math.bot/q/top-5-players-exceed-ppg-projection-salary-55000-FMOFPocG","timestamp":"2024-11-05T22:15:40Z","content_type":"text/html","content_length":"87156","record_id":"<urn:uuid:e5451d01-bc10-46bb-b2b6-b4431b0a055f>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00371.warc.gz"} |
Spring 2015
Title: A discrete to continuous framework for projection theorems
Speaker: Daniel Glasscock
Abstract: Projection theorems for planar sets take the following form: the image of a “large” set A⊆ℝ2 under “most” orthogonal projections πθ (to the line through the origin corresponding to θ∈S1) is
“large”; the set of those directions for which this does not hold is “small.” The first such projection theorem was given by J. M. Marstrand in 1954: assuming the Hausdorff dimension of A is less
than 1, the Hausdorff dimension of πθA is equal to that of A for Lebesgue-almost every θ∈S1.
Recent progress has been made on some fundamental problems in geometric measure theory by discretizing and using tools from additive combinatorics. In 2003, for example, J. Bourgain building on work
of N. Katz and T. Tao, used this approach to prove that a Borel subring of ℝ cannot have Hausdorff dimension strictly between 0 and 1 (a result shown independently by G. A. Edgar and C. Miller),
answering a question of P. Erd\H{o}s and B. Volkmann. The goal of my talk is to explain a discrete approach to continuous projection theorems. I will show, for example, how Marstrand’s original
theorem and a recent result of Bourgain and D. Oberlin can be obtained combinatorially through their discrete analogues. This discrete to continuous framework connects finitary combinatorial
techniques to continuous ones and hints at further parallels between the two regimes.
Seminar 3.31.15 Sharp
Title: Noncommutative geometry and measures on dynamical systems
Speaker: Richard Sharp (University of Warwick)
Abstract: Noncommutative geometry aims to describe a wide range of mathematical objects in terms of C^*-algebras, in particular through the notion of a spectral triple. We will discuss how to recover
important classes of invariant measures for certain dynamical systems on Cantor sets.
Seminar Schedule Spring 2015
Here is an overview of our current seminar schedule for this semester.
Feb 5: Brandon Seward (Michigan)
Feb 19: Dan Thompson (Ohio State)
Feb 26: Bill Mance (University of North Texas)
Mar 12: Sergey Bezuglyi (University of Iowa & Institute for Low Temperature Physics, Ukraine)
Tuesday Mar 31: Richard Sharp (Warwick) [Note unusual day]
Thursday Apr 2: Departmental Colloquium: Richard Sharp (Warwick)
Apr 9: Daniel Glasscock (Ohio State)
Seminar 3.12.15 Bezuglyi
Speaker: Sergey Bezuglyi (University of Iowa & Institute for Low Temperature Physics, Ukraine)
Title: Homeomorphic measures on Cantor sets and dimension groups
Abstract: Two measures, m and m’ on a topological space X are called homeomorphic if there is a homeomorphism f of X such that m(f(A)) = m'(A) for any Borel set A. The question when two Borel
probability non-atomic measures are homeomorphic has a long history beginning with the work of Oxtoby and Ulam: they found a criterion when a probability Borel measure on the n-dimensional cube [0,
1]^n is homeomorphic to the Lebesgue measure. The situation is more interesting for measures on a Cantor set. There is no complete characterization of homeomorphic measures. But, for the class of the
so called good measures (introduced by E. Akin), the answer is simple: two good measures are homeomorphic if and only if the sets of their values on clopen sets are the same.
In my talk I will focus on the study of probability measures invariant with respect to a minimal (or aperiodic) homeomorphism. These measures are in one-to-one correspondence with traces on the
corresponding dimension group. The technique of dimension groups allows us to apply new methods for studying good traces. A good trace is characterized by its kernel having dense image in the
annihilating set of affine functions on the trace space. A number of examples with seemingly paradoxical properties is considered.
The talk will be based on joint papers with D. Handelman and with O. Karpel.
Seminar 2.19.15 Thompson
Speaker: Dan Thompson (Ohio State)
Title: Unique equilibrium states for the robustly transitive diffeomorphisms of Mañé and Bonatti-Viana
Abstract: We establish results on uniqueness of equilibrium states for the well-known Mañé and Bonatti-Viana examples of robustly transitive diffeomorphisms. This is an application of machinery
developed by Vaughn Climenhaga and myself, which applies when systems satisfy suitably weakened versions of expansivity and the specification property. The Mañé examples are partially hyperbolic,
whereas the Bonatti-Viana examples are not partially hyperbolic but admit a dominated splitting. I’ll explain why these maps satisfy our hypotheses. This is joint work with Vaughn Climenhaga
(Houston) and Todd Fisher (Brigham Young).
Seminar 2.5.15 Seward
Speaker: Brandon Seward
Title: Krieger’s finite generator theorem for ergodic actions of countable groups
Abstract: The classical Krieger finite generator theorem states that if a free ergodic probability-measure-preserving action of Z has entropy less than log(k), then the action admits a
generating partition consisting of k sets. This was extended to actions of amenable groups independently by Rosenthal and Danilenko–Park. We introduce the notion of Rokhlin entropy which is defined
for actions of general countable groups. Rokhlin entropy may be viewed as a natural extension of classical entropy, as when the acting group is amenable the two notions coincide. Using this notion of
entropy, we prove Krieger’s finite generator theorem for actions of general countable groups. | {"url":"https://u.osu.edu/ergodictheory/category/seminar-spring-2015/","timestamp":"2024-11-03T06:20:27Z","content_type":"text/html","content_length":"57651","record_id":"<urn:uuid:afca68c0-77b3-472f-a0ca-900371b6d65b>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00674.warc.gz"} |
Bo Wahlberg
Publications by Bo Wahlberg
Peer reviewed
B. Wahlberg, B. Ninness and P. Van den Hof, "Chapter 1 : Introduction," in Modelling and Identification with Rational Orthogonal Basis Functions, Heuberger, Peter S.C.; Hof, Paul M.J. van den;
Wahlberg, Bo Ed., 1st ed. London : Springer London, 2005, pp. 1-14.
B. Wahlberg and T. E Oliveira Silva, "Construction and analysis," in Modelling and Identification with Rational Orthogonal Basis Functions, Peter S.C. Heuberger, Paul M.J. Van den Hof, Bo Wahlberg
Ed., : Springer London, 2005, pp. 15-39.
B. Wahlberg, "Transformation Analysis," in Modelling and Identification with Rational Orthogonal Basis Functions, Heuberger, Peter S.C.; Hof, Paul M.J. van den; Wahlberg, Bo Ed., 1st ed. : Springer
London, 2005, pp. 41-60.
B. Wahlberg and P. Lindskog, "Approximate modelling by means of orthonormal functions," in Modeling, estimation and control of systems with uncertainty, G.B. Di Masi, A. Gombani and A.B. Kurzhansky
Ed., : Birkhäuser Verlag, 1991, pp. 449-467.
H. Hjalmarsson, L. Ljung and B. Wahlberg, "Assessing model quality from data," in Modeling, estimation and control of systems with uncertainty, G.B. Di Masi, A. Gombani and A.B. Kurzhansky Ed., :
Birkhäuser Verlag, 1991, pp. 167-187.
Non-peer reviewed
Latest sync with DiVA:
2024-11-03 01:09:45 | {"url":"https://www.kth.se/profile/bo/publications/","timestamp":"2024-11-07T16:50:47Z","content_type":"text/html","content_length":"940867","record_id":"<urn:uuid:82fdb3db-1f93-4de4-8481-5ffe7afe89a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00463.warc.gz"} |
Mike Vermeulen's web
Last month, I posted a blog posted on approaches to creating an automatic solver for Wordle.
After that posting, there was also a Five Thirty Eight riddler question about the same topic by Zach Wissner Gross. In Zach’s original posting he referenced code and code and examples done by Laurent
Lessard. In Lessard’s work, including this blog post Lessard ran experiments trying several different techniques for creating an optimal approach, e.g.
• Picking the guess that created the most choices (most buckets)
• Picking the guess that minimized the size of the largest bucket
• Picking a guess that maximized entropy
Experimental results suggested that the first of these was slightly better than the other two alternatives.
A week later, Zach Wissner-Gross posted the best solution received a solution credited to Jenny Mitchell. Either from Jenny or from Zach, this solution posted also turned the problem into a lookup
table with each element of the tree having a designated solution. The declared victor was essentially a variation of the “maximize splits” approach
I was playing a little further with these solutions and trying to understand the search space. As discussed in previous post, each word splits the candidate mystery set into up to 238 unique buckets
(3^6 minus five invalid choices) and in practice up to 150 of these buckets populated using the word TRACE. As Jenny Michell’s solution and Laurent Lessard’s experiments show, this is indeed a very
good solution approach. However, I am skeptical whether it is the absolute best approach because of two reasons:
1. In the context of creating a lookup-table, it is possible that different approaches could create just slightly more optimal approaches for parts of the search subtree.
2. The splits themselves are also influenced by the choices of the valid guess words. There are quite a few, but it is not fully populated.
Zach made one statement in arguing for a maximizing splits approach that I am not convinced is true:
It didn’t matter how many words were in each constellation, just the total number of constellations.
To think of why, consider the following thought exercise of two hypothetical ways in which a a group of 20 candidates could be split by a guess into 10 buckets.
1. One method would be word just happened to split this into 10 buckets of two elements each. In this case, the expected number of guesses would be 2.5 – one for the initial split into 10 buckets
and then for each of these buckets one could either guess correctly (1 more guess) or incorrectly (2 more guesses) thus- 1 + (50% * 1) + (50% * 2) = 2.5
2. A second method would be a word that creates 9 buckets of one element each, and a remaining bucket of 11 elements. In this case the expected number of guesses would be one for the initial split
into 10 buckets and then all the buckets with one element could be guessed in the next guess – and the remaining ones however long it takes to split out the bucket of 16 – thus – 1 + 45% * 1) +
(55% * x).
There is no inherent reasons what these two calculations have to be the same. Some notion of entropy can still play a role, even if maximizing entropy is not the best solution. Hence, I believe
Zach’s statement above is a good description of the approach, but not guaranteed to find the absolute best solution.
So I created a metric that was a little more entropy based instead of purely the number of buckets or size of the largest bucket. I considered this as the “weighted size” of the bucket. In the
example above, 55% of the mystery words are in the bucket with 11 elements and 45% of the mystery words are in a bucket with 1 element. So the weighted size is (55% * 11 + 45% * 1) = 6.05. This is in
comparison to the weighted size of (100% * 2) = 2.0. So there is more entropy in the buckets that are all split out, than one that is still more concentrated in a larger bucket.
Note: That in general, I would guess that a lower weighted size should help – however, interestingly enough a bucket of size 2 is likely among one of the least efficient choices – so higher entropy
might not always be the best solution in some of these small cases.
When I print a list of the top 50 choices that minimize this weighted size, I can also print their number of buckets as well as the largest bucket sizes in the table below:
num word bucket max ones wsize eguess
1 roate 126 195 23 60.42 3.31
2 raise 132 168 28 61.00 3.28
3 raile 128 173 22 61.33 3.28
4 soare 127 183 22 62.30 3.24
5 arise 123 168 26 63.73 3.26
6 irate 124 194 14 63.78 3.28
7 orate 127 195 28 63.89 3.30
8 ariel 125 173 21 65.29 3.33
9 arose 121 183 23 66.02 3.30
10 raine 129 195 27 67.06 3.24
11 artel 128 196 27 67.50 3.27
12 taler 134 196 24 67.74 3.29
13 ratel 134 196 25 69.84 3.29
14 aesir 116 168 32 69.88 3.33
15 arles 108 205 14 69.89 3.34
16 realo 112 176 20 69.95 3.36
17 alter 128 196 25 69.99 3.31
18 saner 132 219 33 70.13 3.26
19 later 134 196 34 70.22 3.33
20 snare 132 219 32 71.10 3.18
21 oater 128 195 38 71.25 3.38
22 salet 148 221 34 71.27 3.22
23 taser 134 227 31 71.28 3.26
24 stare 133 227 23 71.29 3.19
25 tares 128 227 26 71.54 3.24
26 slate 147 221 29 71.57 3.22
27 alert 131 196 25 71.60 3.24
28 reais 114 168 24 71.61 3.35
29 lares 118 205 22 71.74 3.33
30 reast 147 227 29 71.77 3.15
31 strae 125 227 16 71.85 3.19
32 laser 123 205 27 72.12 3.33
33 saine 136 207 25 72.59 3.25
34 rales 117 205 21 72.80 3.34
35 urate 122 202 25 72.83 3.31
36 crate 148 246 30 72.90 3.17
37 serai 110 168 20 72.92 3.30
38 toile 123 204 20 73.04 3.23
39 seral 128 205 24 73.08 3.17
40 rates 118 227 24 73.33 3.30
41 carte 146 246 30 73.52 3.21
42 antre 134 226 31 73.94 3.25
43 slane 133 225 25 73.99 3.19
44 trace 150 246 32 74.02 3.16
45 coate 123 192 22 74.51 3.22
46 carle 144 249 37 74.68 3.23
47 carse 139 237 26 74.83 3.20
48 stoae 110 177 18 74.90 3.26
49 reals 116 205 21 74.94 3.27
50 terai 113 194 17 75.14 3.27
The word “TRACE” is 44th on this list, even though it has the highest value for the number of of buckets in column #3. Similarly, if you instead want to pick the words with lowest maximum bucket
size, a word like RAISE is lowest in column #4 but numbers in this column are not always the lowest expected weighted bucket size.
I am not convinced that a increase the entropy situation uniformly is best choice, any more than I am convinced a maximize buckets or maximize spread is optimal for the entire search tree. Instead, I
think it can be slightly influenced by subtleties of particular trees as well as counts.
Looking at weighted numbers of guesses, recursively
To look at potential approaches and expectations, I decided to look at this inductively. In the end, we are creating a search tree, where the tree is composed of nodes ranging from one or more
elements. What might be the expected number of guesses for these nodes and how can the be accumulated recursively?
• One of the bottom building blocks are nodes with one element. These are also listed in column #5 of the table above under the heading “ones”. There are not very many of them near the root of the
tree, only 32 of them at a first level guess of TRACE or ~1% of the 2315 cases. If you are fortunate enough to reach one of them, it takes one more guess to confirm. However, as you go to lower
levels of the tree, I also expect the percentage of buckets with one element to increase.
• Another building block can be an node with two elements. A described above, one can flip a coin get the right one ~50% of the time, so expected guesses are 1.5.
• Next are nodes with three elements. An average expectation could be to guess this in two guesses – using one of two possible approaches. Either you try the choices in this bucket one at a time
and (1/3 * 1) + (1/3 * 2) + (1/3 * 3) and have expectation of 2 guesses to solve – or you find something that splits the three choices into two and pick in the next guess. If you are lucky one of
the two guesses might split the other two if incorrect.
• Nodes with four or five elements might be even more “efficient” than nodes with three – since there is a larger probability of having a word that can split these into exactly four or five
choices. At least this is what I’ve observed.
• As the number of elements within a bucket increases, one gets into more situations where the problem instead gets solved recursively. Find the expected guesses and then break the problem up
So I created an example that created a search tree in this more dynamic fashion. When the node has only two elements, don’t find a pick that evenly splits it into two buckets to maximize the split
(or to minimize the bucket size) – but instead try guessing one of the choices and if that wasn’t correct – pick the other. Hence, using a strategy with a 1.5 expected count rather than 2 expected
I did this recursively, and when a node had too many elements in it, then rather than an exhaustive search for the absolute best candidate – I used a rough heuristic (wsize in my case, but it could
have been number of buckets or size of largest bucket) – and tried the best candidate. I think my rough heuristic can do a reasonable job – but I don’t claim to be absolutely optimal because there is
always a chance that further down on the list might just be slightly better.
Looking at the table above, I notice that TRACE does pretty well within this overall list if used as a first choice, even if followed up later with a different heuristic. I am also struck that there
is a word REAST at number 30 on the list that might be even slightly better as this first choice, even though it is not the best choice in terms of other metrics like maximal number of splits,
smallest bucket size or weighted bucket size.
Instead, REAST looks like it might have almost as many buckets as TRACE (247 vs 250) but then also a slightly smaller maximum bucket size (227 vs 246).
I am not claiming that I have found the best solution for solving Wordle.
Instead, I am suggesting that other claims of ideal might not as strong. With a lookup table approach, I do think one could find a case that optimizes. However, rather than a single heuristic for
creating all of that table, it may be the case that parts of the search tree use slightly different heuristics.
In the table above, the last column eguess is an estimated guess derived recursively and weighted by the number of elements in each node along the way. There is possibility my calculations are
slightly off, but I’ll also observe numbers slightly lower than those presented in FiveThirtyEight.
I expect the largest form of discrepancy is for nodes with two elements. A naive approach of “maximize spread” solves this in two guesses, whereas an approach of try one and if it isn’t right try the
other solves it on average in one and a half.
The second potential discrepancy isn’t as strong but comes from examining expectations for REAST vs. TRACE. These might suggest that it is possible to pick particular choices optimized for different
parts of the tree (in this case the initial guess, but it could also be lower down). Undoubtedly maximizing spread and minimizing largest buckets are good things to do and lead to strategies that
perform very well. It is with that step after that of claiming “ideal” that I am questioning.
Two general notes made in the last blog post also apply. First, this experiment was done in the context of a search tree for exactly the 2315 possible words in Wordle. I expect the general techniques
to still apply but results to be different if picking a larger set such as ~40,000 words in Google corpus that are five letters long.
Second note is that the way systems solve Wordle are different from how people solve Wordle. My observation is people often try to find initial letters to work from and go from there. Whereas a
system it is a combinatoric elimination problem. This was reinforced by Wordle for 14 February where chosing REATH followed by DOILY narrowed the space to a single word in a way I’m sure I would
still have no clue…
Wordle solvers – updated — No Comments | {"url":"http://mvermeulen.org/blog/2022/02/13/wordle-solvers-updated/","timestamp":"2024-11-03T06:03:41Z","content_type":"text/html","content_length":"69484","record_id":"<urn:uuid:04a03275-a65a-4efa-84fb-737506aa1a33>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00373.warc.gz"} |
Moss, Laura Willis South Shore Community Academy
The students will derive a general formula for the probability of a
given event given each possible outcome is equally likely.
Apparatus Needed:
6 dice
30 colored balls (1 each of five different colors)
6 coins
6 paper plate holders
6 plastic cups
6 paper bags
19 ping pong balls (numbered from 0 to 18)
Recommended strategy:
The students were told they had an opportunity to win a prize in
today's class by playing the lottery. They were to pick a 3 digit
number, a 4 digit number (repeats are allowed) and a combination of 4
numbers from 1 to 18 (no number could be chosen twice).
The students were divided into groups of 3 or 4. Each group received
a paper bag containing 1 die, 5 colored balls (1 of each color-blue,
green, yellow, orange and pink), 1 coin, 1 paper plate holder and 1
plastic cup. Each group had to conduct three experiments.
1. Flip the coin 50 times into the paper plate holder, record
the results (heads or tails).
2. Pull a ball from the bag, record its color, replace the
ball and repeat this process a total of 50 times.
3. Use the cup to shake the die. Roll the die into the paper
plate holder. Record the result and repeat the process
60 times.
As a class the data from each experiment was collected. The students
were asked questions such as: Were the results of each experiment
what you would have expected? What would you expect if we were to
repeat the process 1000 times? 10,000 times? 10,000,000 times?
The class then generated the formula for the probability of an event P(E)
P(E) = number of favorable outcomes
number of possible outcomes
Then the class discussed whether or not the formula will tell us
exactly what will happen for a given event. More examples were
The "lottery" ended the lesson. Students with a knowledge of
permutations and combinations can calculate the probability of each
P(3 digit number) = --- = ----
10^3 1000
P(4 digit number) = --- = ------
10^4 10,000
P(4 number from 1 to 18) = 1 = 1
18! 3060
Return to Mathematics Index | {"url":"https://smileprogram.info/ma8713.html","timestamp":"2024-11-09T03:44:39Z","content_type":"text/html","content_length":"3052","record_id":"<urn:uuid:3cba29e8-98c9-4663-9750-614aa7ea962e>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00157.warc.gz"} |
the S
Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Short Answer
Expert verified
The product of the lengths of the segments of one secant is equal to the product of the lengths of the segments of another secant: \(PA \times PD = BP \times PC\).
Step by step solution
Draw Diagram
Draw a circle with two secant lines intersecting at a point outside the circle. Label the point of intersection as point P. Label the points where the secants intersect the circle on both sides of
point P as A and B for the first secant, and C and D for the second secant. Then add an auxiliary line through points A and C inside the circle. Label the intersection point of this line with the
circle as E.
Identify Similar Triangles
Two pairs of similar triangles need to be identified. The first pair is \(\triangle PAB\) and \(\triangle PED\), the second pair is \(\triangle PBA\) and \(\triangle PEC\). They are similar by the
Angle-Angle criterion because they share an angle at P and have another pair of corresponding angles are equal (\(\angle APB = \angle CPD\) and \(\angle BPA = \angle DPC\)).
Apply the properties of the Similar Triangles
Since we identified that the triangles are similar, the ratios of the corresponding side lengths of these triangles should be in proportion. You can write following two proportions: \(\frac{PA}{PE} =
\frac{PE}{PD}\) (from triangles \(\triangle PAB\) and \(\triangle PED\)) and \(\frac{BP}{PE} = \frac{PE}{PC}\) (from triangles \(\triangle BPA\) and \(\triangle ECP\)).
Multiplication of Proportion
Rewrite these proportions to isolate PE and then multiply both results together to eliminate the PE in the fractions. This yields \( PA \times PD = BP \times PC\). This represents The Segments of
Secants Theorem.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Similar Triangles
Similar triangles are a fundamental concept in geometry that involve two or more triangles having the same shape but not necessarily the same size. For two triangles to be similar, all corresponding
angles must be equal, and the sides must be in proportion. This means that if you take the lengths of the sides of one triangle and multiply them by the same factor, you'll get the lengths of the
sides of the similar triangle.
In the context of the Segments of Secants Theorem, the similar triangles identified are useful for establishing relationships between the segment lengths of the secants. Understanding the properties
of similar triangles is critical for students to make sense of why the sides of the triangles are proportional and how this relates to the theorem in question.
Angle-Angle Criterion
The Angle-Angle (AA) Criterion is a way of determining whether two triangles are similar. According to this criterion, if two angles of one triangle are congruent to two angles of another triangle,
the triangles are similar. It's crucial for students to grasp this concept because it simplifies the complexity of proving triangle similarity. Just by confirming that two angles are the same, we can
conclude that the triangles share the same shape and that their sides are proportional.
In the proof for the Segments of Secants Theorem, the AA criterion is what allows us to confidently assert that the identified pairs of triangles are similar. Spotting corresponding congruent angles
is a critical thinking step that leads directly to the application of the theorem.
Proportions in Geometry
Proportions play a significant role in geometry, especially when it comes to similar figures like triangles. A proportion is an expression that indicates that two ratios are equal. When we have
similar triangles, the lengths of their corresponding sides are in proportion. This means that the ratios of the side lengths of one triangle are equal to the corresponding ratios of side lengths of
the other triangle.
By understanding proportions in geometry, students can solve for unknown lengths and apply theorems like the Segments of Secants Theorem. In the steps for solving the theorem, using proportions lets
us express the relationship between segments of secants in terms of equal products, thus proving the theorem.
Geometric Theorems
Geometric theorems are statements or conclusions about shapes, angles, and distances that have been proven to be true through logical deduction. These theorems are the building blocks for solving
complex problems in geometry. A well-known geometric theorem is the Segments of Secants Theorem, which relates the product of the lengths of one secant segment and its external segment to the product
of the lengths of another secant segment and its external segment.
Geometric theorems often involve a combination of known properties, like the similarity of triangles and the concept of proportions, to reach a logical conclusion. In teaching and learning geometry,
students should be encouraged to view theorems not just as formulas to memorize, but as logical sequences that build upon basic principles and previously established knowledge. | {"url":"https://www.vaia.com/en-us/textbooks/math/geometry-a-common-core-curriculum-2015-edition/chapter-10/problem-20-prove-the-segments-of-secants-theorem-theorem-101/","timestamp":"2024-11-03T17:24:49Z","content_type":"text/html","content_length":"249687","record_id":"<urn:uuid:34f93a61-5046-4679-87b5-8c37e9ff44cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00125.warc.gz"} |
Incremental consistent updates
A consistent update installs a new packet-forwarding policy across the switches of a software-defined network in place of an old policy. While doing so, such an update guarantees that every packet
entering the network either obeys the old policy or the new one, but not some combination of the two. In this paper, we introduce new algorithms that trade the time required to perform a consistent
update against the rule-space overhead required to implement it. We break an update in to k rounds that each transfer part of the traffic to the new configuration. The more rounds used, the slower
the update, but the smaller the rule-space overhead. To ensure consistency, our algorithm analyzes the dependencies between rules in the old and new policies to determine which rules to add and
remove on each round. In addition, we show how to optimize rule space used by representing the minimization problem as a mixed integer linear program. Moreover, to ensure the largest flows are moved
first, while using rule space efficiently, we extend the mixed integer linear program with additional constraints. Our initial experiments show that a 6-round, optimized incremental update decreases
rule space overhead from 100% to less than 10%. Moreover, if we cap the maximum rule-space overhead at 5% and assume the traffic flow volume follows Zipf's law, we find that 80% of the traffic may be
transferred to the new policy in the first round and 99% in the first 3 rounds.
Publication series
Name HotSDN 2013 - Proceedings of the 2013 ACM SIGCOMM Workshop on Hot Topics in Software Defined Networking
Other 2013 2nd ACM SIGCOMM Workshop on Hot Topics in Software Defined Networking, HotSDN 2013
Country/Territory China
City Hong Kong
Period 8/16/13 → 8/16/13
All Science Journal Classification (ASJC) codes
• Computer Networks and Communications
• Software
• Consistent network updates
• Frenetic
• Network programming languages
• OpenFlow
• Software-defined networking
Dive into the research topics of 'Incremental consistent updates'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/incremental-consistent-updates","timestamp":"2024-11-07T07:56:58Z","content_type":"text/html","content_length":"55722","record_id":"<urn:uuid:46d64319-4dfc-4336-b32e-e1a4995f9c0d>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00593.warc.gz"} |
The Mathematics of Linear Relations between Feynman Integrals
For the precise prediction of observables at particle colliders such as the LHC, the computation of a large number of Feynman integrals is indispensable. A crucial step in these computations is the
reduction to a small set of master integrals by exploiting linear relations.
In the standard approach such relations are derived from the famous integration-by-parts identities in momentum space and combined in Laporta's algorithm. Such reductions can easily grow too large to
be executed with reasonable computer resources. Motivated by this problem, many new ideas on efficient reductions of Feynman integrals were presented in the recent literature. This workshop is
dedicated to the mathematical techniques behind these developments, including finite field techniques, Gröbner bases, Milnor numbers, unitarity cuts and syzygy computations. In particular, techniques
from algebraic geometry and the theory of D-modules have provided new insights. In an attempt to boost the exchange of ideas and the development of advanced methods, this workshop brings together
mathematicians with relevant expertise, physicists with experience in Feynman integral reductions and specialists on computer algebra tools. | {"url":"https://indico.mitp.uni-mainz.de/event/179/timetable/?print=1&view=standard_numbered","timestamp":"2024-11-02T08:17:01Z","content_type":"text/html","content_length":"135859","record_id":"<urn:uuid:19a631aa-68b8-46a4-b2e9-f370ff97cd7b>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00569.warc.gz"} |
The principal of a high school needs to plan seating for commencement. The following two-way table displays data for the student
All categories
The principal of a high school needs to plan seating for commencement. The following two-way table displays data for the student
s who are involved in this year's commencement ceremony.
Grade level In choir Not in choir TOTAL
Junior 15 3 18
Senior 23 436 459
Other 3 0 3
TOTAL 41 439 480
For these students, are the events "senior" and "not in choir" mutually exclusive?
Find the probability that a randomly selected student from this group is a senior OR is not in choir.
1 answer:
Answer: 77/80= 0.9625
Step-by-step explanation:
You might be interested in
Hello from MrBillDoesMath!
a(n) = -3n , n>= 1
a(1) = -3
a(2) = -6
a(3) = -9
a(4) = -12
The pattern looks like a(n) = -3n
Thank you,
Add answer | {"url":"https://answer.ya.guru/questions/496-the-principal-of-a-high-school-needs-to-plan-seating-for.html","timestamp":"2024-11-04T18:02:51Z","content_type":"text/html","content_length":"57169","record_id":"<urn:uuid:5b691478-45fd-4513-83ed-e7ee19f4503c>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00349.warc.gz"} |
Symbolic Description of the Boundary Curves of the 2D Projections of the Unit Ball B_n - RISC - Johannes Kepler University
Symbolic Description of the Boundary Curves of the 2D Projections of the Unit Ball B_n
Date: 18/06/2019
Time: 14:00 - 15:00
Location: SP II, Lecture room S2 054
Abstract. In classical approximation theory the unit ball of the real univariate polynomials, B_n, that is, the set of polynomials of degree at most n with supremum norm less or equal than one on the
interval [-1,1] is widely investigated. This set is complicated for larger n's. Chebyshev famous result on the nice sharp bounds of the coefficients of such polynomials can be considered as a result
on the one dimensional projections of B_n along the coordinate axis in the standard monomial basis. However, less is known already about the exact 2D projections of B_n. In this talk we report how we
explored the boudary curves of some 2D projections with the aid of symbolic and numeric computational tools for small n's. It turns out that the all the boundary points of these curves correspond to
unit normed Zolotarev polynomials. Since for n<7 the latter polynomial families admit a univarite parametrization (see e.g. [Grasegger-Vo 2017], [Rack-Vajda 2019], there exact and relatively simple
description is possible. Some open problems will be also presented. [Grasegger 2017] G. Grasegger, N. Vo, An Algebraic-Geometric Method for Computing Zolotarev Polynomials, ISSAC '17 Proceedings of
the 2017 ACM on International Symposium on Symbolic and Algebraic Computation, 173-180. [Rack-Vajda 2019] An Explicit Univariate and Radical Parametrization of the Sextic Proper Zolotarev Polynomials
in Power Form, Dolomites Res. Notes Approx. 12 (2019), 43–50. | {"url":"https://www1.risc.jku.at/tk/symbolic-description-of-the-boundary-curves-of-the-2d-projections-of-the-unit-ball-b_n/","timestamp":"2024-11-03T09:56:17Z","content_type":"text/html","content_length":"21937","record_id":"<urn:uuid:7ed97a18-0c21-4dd8-a0c5-d5bbd6ea3869>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00146.warc.gz"} |
For the white screen buried point, I do this (front-end monitoring) - Moment For Technology
“White screen”, I as a cut to cut the hand of the cut to the cub, listen to the operation side of the feedback is really a lot; Sometimes early in the morning Shouting: “That Cheto boy, our
products again white screen.” ; At this point, MY mind is “WDNMD”. In fact, this problem is still quite disgusting, if detected in the test environment is ok, can be fixed in time; But in a
production environment, that’s the f * * k; Because you don’t know when it happened, whether it happened yesterday, a week ago… This will cause unnecessary losses. In the production environment,
our lovely users will think the “white screen” is loading, wait, wait until desperate, and then complain to the operation: “Your product is so rubbish, it hasn’t been loaded all day”. Finally,
the operation of YY front-end. Laozi, fed up, Laozi must be in the “white screen” the first time, eliminate it; So how do you know it’s there? It’s time to “plant a mine.” So how do you bury it?
And when?
How is buried?
Here should press own company’s product or own project to decide. Here are my three “burying points” for the company’s products that you might consider, but may not be right for you.
Vertical buried point
Because of our product, the main page is centered layout, so this “buried point” can meet 80% of our product pages. Vertical burial points, that is, along the X and Y axes. Look at the picture
Suppose you want to bury 10 points on the X axis and 10 points on the Y axis, each at the same distance, then the coordinates of the points on the X axis are (I /10 X the width of the screen, 1/2
X the height of the screen, and I represents the number of points. So the Y-axis is 1/2 the width of the screen, I / 10 the height of the screen, same thing with IT.
The coordinate formula of X and Y axis is obtained as follows:
$X = (\frac{I}{10}* width of screen, \frac{1}{2}* height of screen)$
$Y = (\frac{1}{2}* width of screen, \frac{I}{10}* height of screen)$
Buried cross point
Because, the vertical buried point can solve the center layout of the page, but we have some news pages, sometimes only in the first quadrant of the content, then into the blind area of this buried
point, will be misjudged as “white screen”. Look at the picture
Therefore, it is necessary to consider the cross-burying point scheme. The cross-burying point mainly involves burying “mines” on two diagonal lines of the screen. Look at the picture
These two diagonals are K1 and K2(X and Y are just auxiliary lines for easy understanding). We’re going to bury 10 points on each of them, and each point is the same distance apart, so how do we
pick their coordinates? It’s really easy, we just take I /10 screen width, I /10 screen height. K1 = (I /10 screen width, I /10 screen height), K2 is also easy, just reverse the coordinates of
K1, K2 = ((10-i)/10 screen width, 10-i /10 screen g height)
The coordinate formula of K1 and K2 is obtained as follows:
$K1 = (\frac{I}{10}* width of screen, \frac{I}{10}* height of screen)$
$K2 = (\ frac {10 – I} {10} * width of the screen, \ frac {10 – I} {10} * the height of the screen)$
Vertical cross burial point
As mentioned above, most of our pages are centered layout, if the content is less, and will enter the blind area of the cross buried point, look at the picture
Therefore, based on the above situation, we finally adopted a buried point scheme combining the two. Look at the picture
The formula is a combination of the two
$X = (\frac{I}{10}* width of screen, \frac{1}{2}* height of screen)$
$Y = (\frac{1}{2}* width of screen, \frac{I}{10}* height of screen)$
$K2 = (\frac{I}{10}* width of screen, \frac{I}{10}* height of screen)$
$K1 = (\ frac {10 – I} {10} * width of the screen, \ frac {10 – I} {10} * the height of the screen)$
How to code it
The rules of the buried point, we have found out, so it is very simple to implement in code.
function blankWhite(){
let point = 10
let warpTag = ['HTML'.'BODY']
let empty = 0
function isWarp (ele){
empty+=1}}for (let i = 1; i < point; i++) {
let x = document.elementsFromPoint(i / point * window.innerWidth, window.innerHeight / 2)
let y = document.elementsFromPoint( window.innerWidth / 2, i / point * window.innerHeight)
let k2 = document.elementsFromPoint(i/point * window.innerWidth, i / point * window.innerHeight)
let k1 = document.elementsFromPoint( (point-i)/point * window.innerWidth, (point-i)/point*window.innerHeight)
if(empty === 36) {// doimg someThings ....
// If the screen is blank, you can do something you want to do on the screen. For example, we will upload the error to the management background monitoring module of our company, so that we can know the occurrence of the screen in time and fix it}}Copy the code
So why is it empty === 36 instead of empty === 40? If you are careful, you will find that there are only 36 points buried (I < point). Why are there only 36 points buried instead of 40? Document.
ElementsFromPoint will return an empty array.
Also have homecoming questions document. ElementsFromPoint is a what? Take a look at the MDN hash point here
So when do you bury it?
This is a key point that we’ll bury after AJAX or after DOM loads.
if(document.readyState === 'complete'){
Copy the code
The complete code
if(document.readyState === 'complete'){
function blankWhite(){
let point = 10
let warpTag = ['HTML'.'BODY']
let empty = 0
function isWarp (ele){
empty+=1}}for (let i = 1; i < point; i++) {
let x = document.elementsFromPoint(i / point * window.innerWidth, window.innerHeight / 2)
let y = document.elementsFromPoint( window.innerWidth / 2, i / point * window.innerHeight)
let k2 = document.elementsFromPoint(i/point * window.innerWidth, i / point * window.innerHeight)
let k1 = document.elementsFromPoint( (point-i)/point * window.innerWidth, (point-i)/point*window.innerHeight)
if(empty === 36) {// doimg someThings ....
// If the screen is blank, you can do something you want to do on the screen. For example, we will upload the error to the management background monitoring module of our company, so that we can know the occurrence of the screen in time and fix it}}Copy the code
The last
This is my solution and implementation for the needs of my own company. It may not be suitable for you. You may have a better implementation.
Maybe you like the above buried like this, ha ha ha (🐺 save your life). | {"url":"https://dev.mo4tech.com/for-the-white-screen-buried-point-i-do-this-front-end-monitoring.html","timestamp":"2024-11-03T17:10:56Z","content_type":"text/html","content_length":"115242","record_id":"<urn:uuid:45927aa0-13a8-4354-b791-88407f4f54d7>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00177.warc.gz"} |
factor tree of 13
Factor tree of 13|Prime factor tree
Factors of 13 - Find Prime Factorization/Factors of 13
Factors of 13: Prime Factorization, Methods, Tree, and Examples
Factor Tree Worksheets page
How to Find the Factors of a Number | Find the Factors
Factor Tree Worksheets page
Factor Trees - Math Steps, Examples & Questions
Factor tree of 13|Prime factor tree - YouTube
Prime factor trees - finding HCF + LCM (N)
Factor Trees - Math Steps, Examples & Questions
1092 | Find the Factors
Factor tree of 13|Prime factor tree - YouTube
Factors of 13: Negative Factors, Factor Pairs, Sum, Factor Tree ...
Write the missing numbers in the following factor tree :
Factors of 52 - Find Prime Factorization/Factors of 52
How to Find the Factors of a Number | Find the Factors
Factor Tree Worksheets - 15 Worksheets.com
Number.factor tree | PPT
Using factor tree write as product of prime factorization 234.
Page 13 - 6-Math-3 FACTORS AND MULTIPLES
Factors, Primes, and Prime Factorization
Prime Numbers: Factorization & Factor Tree - Curvebreakers
Math Story : Prime Factorization Using Factor Tree Method - Fun2Do ...
Solved Complete the given factor tree and then give the | Chegg.com
Prime Factors - How to Find Prime Factors with Examples
Prime Factorisation Using Factor Tree - YouTube
Fill in the missing numbers in the Factor Tree | {"url":"https://worksheets.clipart-library.com/factor-tree-of-13.html","timestamp":"2024-11-03T07:47:45Z","content_type":"text/html","content_length":"20837","record_id":"<urn:uuid:5590692d-29ac-45e2-8a73-77cbaf68a9ca>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00870.warc.gz"} |
Meijster distance
A General Algorithm for Computing Distance Transforms in Linear Time
The Meijster distance computes a distance field from a boolean image. A boolean image is made up of zeros and ones, where zero means background (empty space) and one foreground. The resulting
distance field is an image whose pixels store distance to foreground.
A distance field can be used for stroking shapes, computing Voronoi diagrams, pathfinding, or performing mathematical morphology operators. The distance field changes depending on the distance
function used. Each distance function uses a different shape to compute distance, hold down option ⌥ to reveal it.
Euclidean distance
This is the most commonly used distance function. Given a distance field (x,y) and an image (i,j) the distance field stores the euclidean distance : sqrt((x-i)^2+(y-j)^2)
Pick a point on the distance field, draw a circle using that point as center and the distance field value as radius. The circle will hit the closest foreground point.
Manhattan distance (Taxicab geometry)
The distance field stores the Manhattan distance : abs(x-i)+abs(y-j)
Pick a point on the distance field, draw a diamond (rhombus) using that point as center and the distance field value as radius. The diamond will hit the closest foreground point.
Chessboard distance (Chebyshev)
The distance field stores the chessboard distance : max(abs(x-i),abs(y-j))
Pick a point on the distance field, draw a square using that point as center and the double distance field value as edge length. The square will hit the closest foreground point.
Left mouse button to paint, shift+left mouse button to clear.
Use each brush with its matching transform to recreate the pictures above.
• Circle matches Euclidean
• Diamond matches Manhattan
• Rectangle matches Chessboard
Using the Euclidean distance, paint points with a small brush and up contrast to get Voronoi lookalikes.
Image credits
Source is on | {"url":"https://parmanoir.com:443/distance/","timestamp":"2024-11-09T15:58:37Z","content_type":"text/html","content_length":"26762","record_id":"<urn:uuid:3cefc611-4656-4c19-96ca-906a3d879125>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00878.warc.gz"} |
AI Risk & Opportunity: Strategic Analysis Via Probability Tree — LessWrong
Part of the series AI Risk and Opportunity: A Strategic Analysis.
(You can leave anonymous feedback on posts in this series here. I alone will read the comments, and may use them to improve past and forthcoming posts in this series.)
There are many approaches to strategic analysis (Bishop et al. 2007). Though a morphological analysis (Ritchey 2006) could model our situation in more detail, the present analysis uses a simple
probability tree (Harshbarger & Reynolds 2008, sec. 7.4) to model potential events and interventions.
A very simple tree
In our initial attempt, the first disjunction concerns which of several (mutually exclusive and exhaustive) transformative events comes first:
• "FAI" = Friendly AI.
• "uFAI" = UnFriendly AI, not including uFAI developed with insights from WBE.
• "WBE" = Whole brain emulation.
• "Doom" = Human extinction, including simulation shutdown and extinction due to uFAI striking us from beyond our solar system.
• "Other" = None of the above four events occur in our solar system, perhaps due to stable global totalitarianism or for unforeseen reasons.
Our probability tree begins simply:
Each circle is a chance node, which represents a random variable. The leftmost chance node above represents the variable of whether FAI, uFAI, WBE, Doom, or Other will come first. The rightmost
chance nodes are open to further disjunctions: the random variables they represent will be revealed as we continue to develop the probability tree.
Each left-facing triangle is a terminal node, which for us serves the same function as a utility node in a Bayesian decision network. The only utility node in the tree above assigns a utility of 0
(bad!) to the Doom outcome.
Each branch in the tree is assigned a probability. For the purposes of illustration, the above tree assigns .01 probability to FAI coming first, .52 probability to uFAI coming first, .07 probability
to WBE coming first, .35 to Doom coming first, and .05 to Other coming first.
How the tree could be expanded
The simple tree above could be expanded "downstream" by adding additional branches:
We could also make the probability tree more actionable by trying to estimate the probability of desirable and undesirable outcomes given certain that certain shorter-term goals are met. In the
example below, "private push" means that a non-state actor passionate about safety invests $30 billion or more into developing WBE technology within 30 years from today. Perhaps there's a small
chance this safety-conscious actor could get to WBE before state actors, upload FAI researchers, and have them figure out FAI before uFAI is created.
We could also expand the tree "upstream" by making the first disjunction be not concerned with our five options for what comes first but instead with a series of disjunctions that feed into which
option will come first.
We could add hundreds or thousands of nodes to our probability tree, and then use the software to test for how much the outcomes change when particular inputs are changed, and learn what things we
can do now to most increase our chances of a desirable outcome, given our current model.
We would also need to decide which "endgame scenarios" we want to include as possible terminals, and the utility of each. These choices may be complicated by our beliefs about multiverses and
However, decision trees become enormously large and complex very quickly as you add more variables. If we had the resources for a more complicated model, we'd probably want to use influence diagrams
instead (Howard & Matheson 2005), e.g. one built in Analytica, like the ICAM climate change model. Of course, one must always worry that one's model is internally consistent but disconnected from the
real world (Kay 2012).
Oops, fixed.
Moving to a graph makes elicitation of the parameters a lot more difficult (to the extent that you have to start specifying clique potentials instead of conditional probabilities).
I think you can still get away with using conditional probabilities if you make the graph directed and acyclical, as I should've specified (my bad). The graph is still more complex than the tree, as
you said, but if we're using software for the tree, we might as well use one for the graph...
I'm rather pessimistic about that. I think that basically the tree branches into a very huge number of possibilities; one puts the possibilities into categories, but has no way of finding total
probabilities within the categories.
Furthermore, the categories themselves do not correspond to technological effort; there is the FAIs that resulted from regular AI effort via some rather simple insight by the scientist that came up
with AI, the insights that may be only visible up close when one is going over an AI and figuring out why the previous version killed half of itself repeatedly instead of self improving, or other
cases the probability of which we can't guess at without knowing how the AI is being implemented and what sort of great filters does the AI have to pass before it fooms. And there are uFAIs that
result from the FAI effort; those are the uFAIs of entirely different kind, with their own entirely different probabilities that can't be guessed at.
The intuitions are often very wrong, for example the intuitions about what 'random' designs do; firstly, our designs are not random, and secondarily, random code predominantly crashes, and the non
crashing space is utterly dominated by one or two simplest noncrash behaviours, and same may be true of the goal systems which pass the filters of not crashing, and recursively self improving over
enormous range. The filters are specific to the AI architecture.
The existence of filters, in my opinion, entirely thwarts any generic intuitions and generic arguments. The unknown , highly complex filters are an enormous sea between the logic and the probability
estimates, in the land of inferences. The illogic, sadly, does not need to pass through the sea, and rapidly suggests the numbers that are not, in any way, linked to the relevant issue-set.
I think that you are you are on a solid research path here. I think you have reached the bounds of business oriented software and it's time to look into something like apache mahout or RDF. Decision
tree implementations are available all over, just find a data structure and share them and run inference engines like owlim or pellet and see what you can see.
RDF is a good interim solution because you can start encoding things as structured data. I have some JSON->RDF stuff for inference if you get to that point.
Here is one way to represent these graphs as RDF.
Each edge becomes an edge to a blank node, that blank node has the label, arrival probability and could link to evidence supporting. Representing weighted graphs in RDF is fairly well studied.
The question is, what is your net goal of this from a computational artifact point of view?
I like the numbers plugged into that tree. Only slightly optimistic. :)
Teaching tree thinking through touch.
These experiments were done with video game trees showing evolutionary divergence, and this method of teaching outperformed traditional paper exercises. Perhaps a simple computer program would make
teaching probability trees easier, or the principles behind the experiments could be applied in another way to teach how to use these trees.
New Comment
11 comments, sorted by Click to highlight new comments since:
I'm not well-versed in these probability trees but something weird seems to be happening in the third one. The conditional probabilities under "private push" and those under no "private push" should
each separately sum to 1. Changing them to joint probabilities would resolve the problem. Alternatively, you could put a probability on "private push" and then put FAI, uFAI, and Other on each
This is kind of a weird article... It explains how to use decision trees, but then it just stops, without telling me what to expect, why I should care, or, assuming I did care, how to assign
probabilities to the nodes. So, the only feeling I'm left with at the end is, "all righty then, time for tea".
In addition, instead of saying "X | private push" and "X | no private push", it might be clearer to add the nodes "private push" and "no private push" explicitly, and then connect them to "FAI",
"uFAI", etc. An even better transformation would be to convert the tree into a graph; this way, you won't need to duplicate the terminal nodes all the time.
Moving to a graph makes elicitation of the parameters a lot more difficult (to the extent that you have to start specifying clique potentials instead of conditional probabilities). Global tasks like
marginalization or conditioning also become a lot harder.
But this is much too large a project for me to undertake now.
Too bad. I was excited about this post and thought it was a good sign that you took that path and that it would be highly promising to pursue it further.
Another worry is that putting so many made-up probabilities into a probability tree like this is not actually that helpful. I'm not sure if that's true, but I'm worried about it. | {"url":"https://www.lesswrong.com/posts/rvN6LrbtQDR2ee382/ai-risk-and-opportunity-strategic-analysis-via-probability","timestamp":"2024-11-10T04:59:00Z","content_type":"text/html","content_length":"286716","record_id":"<urn:uuid:6f51ae67-4c7e-4a97-9192-5e6d13bc4359>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00527.warc.gz"} |
Transactions Online
Satoshi TAYU, Mineo KANEKO, "Characterization and Computation of Steiner Routing Based on Elmore's Delay Model" in IEICE TRANSACTIONS on Fundamentals, vol. E85-A, no. 12, pp. 2764-2774, December
2002, doi: .
Abstract: As a remarkable development of VLSI technology, a gate switching delay is reduced and a signal delay of a net comes to have a considerable effect on the clock period. Therefore, it is
required to minimize signal delays in digital VLSIs. There are a number of ways to evaluate a signal delay of a net, such as cost, radius, and Elmore's delay. Delays of those models can be computed
in linear time. Elmore's delay model takes both capacitance and resistance into account and it is often regarded as a reasonable model. So, it is important to investigate the properties of this
model. In this paper, we investigate the properties of the model and construct a heuristic algorithm based on these properties for computing a wiring of a net to minimize the interconnection delay.
We show the effectiveness of our proposed algorithm by comparing ERT algorithm which is proposed in [2] for minimizing the maximum Elmore's delay of a sink. Our proposed algorithm decreases the
average of the maximum Elmore's delay by 10-20% for ERT algorithm. We also compare our algorithm with an O(n^4) algorithm proposed in [15] and confirm the effectiveness of our algorithm though its
time complexity is O(n^3).
URL: https://global.ieice.org/en_transactions/fundamentals/10.1587/e85-a_12_2764/_p
author={Satoshi TAYU, Mineo KANEKO, },
journal={IEICE TRANSACTIONS on Fundamentals},
title={Characterization and Computation of Steiner Routing Based on Elmore's Delay Model},
abstract={As a remarkable development of VLSI technology, a gate switching delay is reduced and a signal delay of a net comes to have a considerable effect on the clock period. Therefore, it is
required to minimize signal delays in digital VLSIs. There are a number of ways to evaluate a signal delay of a net, such as cost, radius, and Elmore's delay. Delays of those models can be computed
in linear time. Elmore's delay model takes both capacitance and resistance into account and it is often regarded as a reasonable model. So, it is important to investigate the properties of this
model. In this paper, we investigate the properties of the model and construct a heuristic algorithm based on these properties for computing a wiring of a net to minimize the interconnection delay.
We show the effectiveness of our proposed algorithm by comparing ERT algorithm which is proposed in [2] for minimizing the maximum Elmore's delay of a sink. Our proposed algorithm decreases the
average of the maximum Elmore's delay by 10-20% for ERT algorithm. We also compare our algorithm with an O(n^4) algorithm proposed in [15] and confirm the effectiveness of our algorithm though its
time complexity is O(n^3).},
TY - JOUR
TI - Characterization and Computation of Steiner Routing Based on Elmore's Delay Model
T2 - IEICE TRANSACTIONS on Fundamentals
SP - 2764
EP - 2774
AU - Satoshi TAYU
AU - Mineo KANEKO
PY - 2002
DO -
JO - IEICE TRANSACTIONS on Fundamentals
SN -
VL - E85-A
IS - 12
JA - IEICE TRANSACTIONS on Fundamentals
Y1 - December 2002
AB - As a remarkable development of VLSI technology, a gate switching delay is reduced and a signal delay of a net comes to have a considerable effect on the clock period. Therefore, it is required
to minimize signal delays in digital VLSIs. There are a number of ways to evaluate a signal delay of a net, such as cost, radius, and Elmore's delay. Delays of those models can be computed in linear
time. Elmore's delay model takes both capacitance and resistance into account and it is often regarded as a reasonable model. So, it is important to investigate the properties of this model. In this
paper, we investigate the properties of the model and construct a heuristic algorithm based on these properties for computing a wiring of a net to minimize the interconnection delay. We show the
effectiveness of our proposed algorithm by comparing ERT algorithm which is proposed in [2] for minimizing the maximum Elmore's delay of a sink. Our proposed algorithm decreases the average of the
maximum Elmore's delay by 10-20% for ERT algorithm. We also compare our algorithm with an O(n^4) algorithm proposed in [15] and confirm the effectiveness of our algorithm though its time complexity
is O(n^3).
ER - | {"url":"https://global.ieice.org/en_transactions/fundamentals/10.1587/e85-a_12_2764/_p","timestamp":"2024-11-05T22:16:35Z","content_type":"text/html","content_length":"62734","record_id":"<urn:uuid:fb733d47-3d23-42eb-9f5a-5cd52c0b569d>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00145.warc.gz"} |
Short Term Search Results
In this video, we learn how to improve short-term memory. There are many brain exercises that can help, that will jump start your abilities. Remember to focus your attention, take mental snapshots,
and connect your snapshots with memory. This will help you not only remember different things, but it will also help you to connect pictures and different details along with it. Just small things
like this while you are younger can help improve your short-term memory while you are both younger and ... | {"url":"https://www.wonderhowto.com/search/short%20term/","timestamp":"2024-11-12T10:48:41Z","content_type":"text/html","content_length":"184354","record_id":"<urn:uuid:7d857270-4905-409b-b709-f9530d6d0230>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00120.warc.gz"} |
Under the Hood of AdaBoost | HackerNoon
A short introduction to the AdaBoost algorithm
In this post, we will cover a very brief introduction to boosting algorithms, as well as delve under the hood of a popular boosting algorithm, AdaBoost. The purpose of this post is to provide a
gentle introduction to some of the key concepts of boosting and AdaBoost. This isn’t a definitive pros and cons of AdaBoost vs Gradient Boosting etc, but more of a summary of the key theory points to
understand the algorithm.
Real World Applications for AdaBoost
AdaBoost can be used to solve a variety of real-world problems, such as predicting customer churn and classifying the types of topics customers are talking/calling about. The algorithm is heavily
utilised for solving classification problems, given its relative ease of implementation in languages such as R and Python.
What are Boosting Algorithms?
Boosting algorithms fall within the broader family of ensemble modelling. Broadly speaking, there are two key approaches to model building within data science; building a single model and building an
ensemble of models. Boosting falls within the latter approach and with reference to AdaBoost, the models are constructed as follows: For each iteration, a new weak learner is introduced sequentially
and aims to compensate the “shortcomings” of the prior models to create a strong classifier. The overall goal of this exercise is to consecutively fit new models to provide more accurate estimations
of our response variable.
Boosting works from the assumption that each weak hypothesis, or model, has a higher accuracy than randomly guessing: This assumption is known as the “weak learning condition”.
What is AdaBoost?
The AdaBoost algorithm was developed by Freund and Schapire in 1996 and is still heavily used in various industries. AdaBoost reaches its end goal of a classifier by sequentially introducing new
models to compensate the “shortcomings” of prior models. Scikit Learn summarises AdaBoost’s core principle as that it “fits a sequence of weak learners on repeatedly modified versions of the data.”
This definition will allow us to understand and expand upon AdaBoost’s processes.
Getting Started
To begin with, a weak classifier is trained, and all of the example data samples are given an equal weight. Once the initial classifier is trained, two things happen. A weight is calculated for the
classifier, with more accurate classifiers being given a higher weight, and less accurate a lower weight. The weight is calculated based on the classifier’s error rate, which is the number of
misclassifications in the training set, divided by total training set size. This output weight per model is known as the “alpha”.
Calculating the Alpha
Each classifier will have a weight calculated, which is based on the classifier’s error rate.
For each iteration, the alpha of the classifier is calculated, with the lower the error rate = the higher the alpha. This is visualised as follows:
Image from Chris McCormick’s excellent AdaBoost tutorial
Intuitively, there is also a relationship between the weight of the training example and the alpha. If we have a classifier with a high alpha that misclassifies a training example, this example will
be given more weight than a weaker classifier which also misclassifies a training example. This is referred to as “intuitive”, as we can consider a classifier with a higher alpha as being a more
reliable witness; when it misclassifies something, we want to investigate that further.
Understanding Weights for Training Samples:
Secondly, the AdaBoost algorithm directs its attention to misclassified data examples from our first weak classifier, by assigning weights to each data sample, the value of which is defined by
whether the classifier correctly or incorrectly classified the sample.
We can break down a visualisation of weights per example, below:
Step 1: Our first model where wi=1/N
In this instance, we can see that each training example has an equal weight, and that the model has correctly and incorrectly classified certain examples. After each iteration, the sample weights are
modified, and those with higher weights (examples that have been incorrectly classified) are more likely to be included within the training sets. When a sample is correctly classified, it is given
less weightage in the next step of model building.
An example of a later model, where the weights have been changed:
The formula for this weight update is shown below:
Building the Final Classifier
Once all of the iterations have been completed, all of the weak learners are combined with their weights to form a strong classifier, as expressed in the below equation:
The final classifier is therefore built up of “T” weak classifiers, ht(x) is the output of the weak classifier, with at the weight applied to the classifier. The final output is therefore a
combination of all of the classifiers.
This is a whistle-stop tour of the theory of AdaBoost, and should be seen as an introductory exploration of the boosting algorithm. For further reading, I recommend the following resources:
Recommended Further Reading
Natekin, A. Knoll, A. (2018). Gradient boosting machines, a tutorial. Frontiers in Neurobotics. [online] Available at: https://core.ac.uk/download/pdf/82873637.pdf
Li, C. (2018). A Gentle Introduction to Gradient Boosting. [online] Available at: http://www.ccs.neu.edu/home/vip/teach/MLcourse/4_boosting/slides/gradient_boosting.pdf
Schapire, R. (2018). Explaining AdaBoost. [online] Available at: https://math.arizona.edu/~hzhang/math574m/Read/explaining-adaboost.pdf
YouTube. (2018). Extending Machine Learning Algorithms—AdaBoost Classifier | packtpub.com. [online] Available at: https://www.youtube.com/watch?v=BoGNyWW9-mE
Scikit-learn.org. (2018). 1.11. Ensemble methods—scikit-learn 0.20.2 documentation. [online] Available at: https://scikit-learn.org/stable/modules/ensemble.html#AdaBoost
McCormick, C. (2013). AdaBoost Tutorial · Chris McCormick . [online] Available at: http://mccormickml.com/2013/12/13/adaboost-tutorial/
Schapire, R. (2018). Explaining AdaBoost. [online] Available at: https://math.arizona.edu/~hzhang/math574m/Read/explaining-adaboost.pdf
Mohri, M. (2018). Foundations of Machine Learning [online] Available at: https://cs.nyu.edu/~mohri/mls/ml_boosting.pdf | {"url":"https://hackernoon.com/under-the-hood-of-adaboost-8eb499d78eab","timestamp":"2024-11-09T09:57:26Z","content_type":"text/html","content_length":"237007","record_id":"<urn:uuid:94f02a7a-0199-4b46-838e-263e94609032>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00368.warc.gz"} |
Application of Rapidly Exploring Random Tree Star-Smart and G2 Quintic Pythagorean Hodograph Curves to the UAV Path Planning Problem
Commenced in January 2007
Application of Rapidly Exploring Random Tree Star-Smart and G2 Quintic Pythagorean Hodograph Curves to the UAV Path Planning Problem
Authors: Luiz G. Véras, Felipe L. Medeiros, Lamartine F. Guimarães
This work approaches the automatic planning of paths for Unmanned Aerial Vehicles (UAVs) through the application of the Rapidly Exploring Random Tree Star-Smart (RRT*-Smart) algorithm. RRT*-Smart is
a sampling process of positions of a navigation environment through a tree-type graph. The algorithm consists of randomly expanding a tree from an initial position (root node) until one of its
branches reaches the final position of the path to be planned. The algorithm ensures the planning of the shortest path, considering the number of iterations tending to infinity. When a new node is
inserted into the tree, each neighbor node of the new node is connected to it, if and only if the extension of the path between the root node and that neighbor node, with this new connection, is less
than the current extension of the path between those two nodes. RRT*-smart uses an intelligent sampling strategy to plan less extensive routes by spending a smaller number of iterations. This
strategy is based on the creation of samples/nodes near to the convex vertices of the navigation environment obstacles. The planned paths are smoothed through the application of the method called
quintic pythagorean hodograph curves. The smoothing process converts a route into a dynamically-viable one based on the kinematic constraints of the vehicle. This smoothing method models the
hodograph components of a curve with polynomials that obey the Pythagorean Theorem. Its advantage is that the obtained structure allows computation of the curve length in an exact way, without the
need for quadratural techniques for the resolution of integrals.
Keywords: Path planning, path smoothing, Pythagorean hodograph curve, RRT*-Smart.
Digital Object Identifier (DOI): doi.org/10.5281/zenodo.1316688
Procedia APA BibTeX Chicago EndNote Harvard JSON MLA RIS XML ISO 690 PDF Downloads 896
[1] Farouki, Rida T. Pythagorean—hodograph Curves. Springer Berlin Heidelberg, 2008.
[2] Farouki, Rida T. The conformal map z → z2 of the hodograph plane. Computer Aided Geometric Design, v. 11, n. 4, p. 363-390, 1994.
[3] Farouki, Rida T. et al. Path planning with Pythagorean-hodograph curves for unmanned or autonomous vehicles. Proceedings of the Institution of Mechanical Engineers, Part G: Journal of Aerospace
Engineering, 2017.
[4] Islam, Fahad et al. RRT*-Smart: Rapid convergence implementation of RRT* towards optimal solution. In: Mechatronics and Automation (ICMA), 2012 International Conference on. IEEE, 2012. p.
[5] Lavalle, Steven M. Rapidly-exploring random trees: A new tool for path planning. 1998.
[6] Karaman, Sertac; Frazzoli, Emilio. Sampling-based algorithms for optimal motion planning. The international journal of robotics research, v. 30, n. 7, p. 846-894, 2011.
[7] Fabri, Andreas; Pion, Sylvain. CGAL: The computational geometry algorithms library. In: Proceedings of the 17th ACM SIGSPATIAL international conference on advances in geographic information
systems. ACM, 2009. p. 538-539.
[8] Dong, Bohan; Farouki, Rida T. Algorithm 952: PHquintic: A library of basic functions for the construction and analysis of planar quintic Pythagorean-hodograph curves. ACM Transactions on
Mathematical Software (TOMS), v. 41, n. 4, p. 28, 2015.
[9] Farouki, Rida T.; Sakkalis, Takis. Pythagorean hodographs. IBM Journal of Research and Development, v. 34, n. 5, p. 736-752, 1990. | {"url":"https://publications.waset.org/10008996/application-of-rapidly-exploring-random-tree-star-smart-and-g2-quintic-pythagorean-hodograph-curves-to-the-uav-path-planning-problem","timestamp":"2024-11-08T02:58:38Z","content_type":"text/html","content_length":"19623","record_id":"<urn:uuid:304a9e5c-2214-4c62-86f0-76244ea412ae>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00011.warc.gz"} |
On the entropy of distance coding
Let's look at a little math problem that I think is both amusing and illuminating, and is related to why LZ77 distance coding is fundamentally inefficient for sending a sequence of repeated tokens.
The toy problem we will study is this :
Given an array of random bytes, you for some silly reason decide to compress them by sending the distance to the most recent preceding occurance of that byte.
So to send 'A' you count backward from your current position P ; if there's an A at P-1, you send a 0, if it's at P-2 send a 1, etc.
What is the entropy of this distance encoding?
This "compressor" is silly, it expands (it will send more than 8 bits per byte). How much does it expand?
First of all, why are we looking at this. It's a boiled down version of LZ77 on a specific type of data. LZ77 sends repeated strings by sending the distance to the previous occurance of that string.
Imagine your data consists of a finite set of tokens. Each token is several bytes and imagine the tokens are drawn from a random source, and there are 256 of them. If you had the indices of the
tokens, you would just have a random byte sequence. But LZ77 does not have that and cannot convert the stream back to the token indexes. Instead the best it can do is to parse on token boundaries,
and send the distance to the previous occurance of that token.
This does occur in the real world. For example consider something like a palettized image. The palette indices are bytes that refer to a dictionary of 256 colors. If you are given the image to
compress after de-palettizing, you would see something like 32-bit RGBA tokens that repeat. Once you have seen the 256 unique colors, you no longer need to send anything but distances to previous
occurances of a color. English text is also a bit like this, with the tokens equal to a word+punction string of variable length.
So back to our toy problem. To get the entropy of the distance coding, we need the probability of the distances.
To find that, I think about coding the distance with a sequence of binary coding events. Start at distance = 0. Either the current byte matches ours, or it doesn't. Chance of matching is (1/256) for
random bytes. The chance of no-match is (255/256). So we multiply our current probability by (1/256) and the probability of all future distances by (255/256), then move to the next distance. (perhaps
it's easier to imagine that the preceding bytes don't exist yet; rather you are drawing a random number as you step back in distance; any time you get your own value you stop)
This gives you the probability distribution :
P(0) = (1/256)
P(1) = (255/256) * (1/256)
P(2) = (255/256)^2 * (1/256)
P(n) = (255/256)^n * (1/256)
an alternative way to get the same distribution is to look at distance D. First multiply by the probability that it is not a lower distance (one minus the sum of all lower distance probabilities).
Then the probability that it is here is (1/256). That's :
P(0) = (1/256)
P(1) = (1 - P(0)) * (1/256)
P(2) = (1 - P(0) - P(1)) * (1/256)
P(n) = (1 - P(0) - P(1) .. - P(n-1)) * (1/256)
which is equal to the first way.
Given this distribution we can compute the entropy :
H = - Sum_n P(n) * log2( P(n) )
starting at n = 0
let x = (255/256)
P(n) = (1-x) * x^n
log2( P(n) ) = log2( 1-x ) + n * log2( x )
H = - Sum_n (1-x) * x^n * ( log2( 1-x ) + n * log2( x ) )
terms that don't depend on n pull out of the sum :
H = - (1-x) * log2( 1-x ) * Sum_n x^n
- (1-x) * log2( x ) * Sum_n n * x^n
we need two sums :
G = Sum_n x^n
S = Sum_n n * x^n
G is just the geometric series
G = 1 / (1 - x)
recall the trick to find G is to look at G - x*G
we can use the same trick to find S
S - x*S = G - 1
the other standard trick to find S is to take the d/dx of G
either way you get:
S = x / (1-x)^2
H = - (1-x) * log2( 1-x ) * G
- (1-x) * log2( x ) * S
H = - (1-x) * log2( 1-x ) * ( 1 / (1-x) )
- (1-x) * log2( x ) * ( x / (1-x)^2 )
H = - log2( 1-x ) - log2( x ) * ( x / (1-x) )
putting back in x = (255/256)
1-x = 1/256
the first term "- log2( 1-x )" is just 8 bits, send a byte
H = 8 + 255 * log2( 256 / 255 )
H = 8 + 1.43987 bits
about 9.44 bits
or if we look at general alphabet size N now instead of source bytes
H = log2(N) + (N-1) * log2( N / (N-1) )
recalling of course log2(N) is just the number of bits it would take to code a random symbol of N possible values
we'll look at the expansion due to distance coding, H - log2(N)
if we now take the limit of N large
H - log2(N) -> (N-1) * log2( N / (N-1) )
H - log2(N) -> (N-1) * log2( 1 + 1 / (N-1) )
H - log2(N) -> log2( ( 1 + 1 / (N-1) ) ^ (N-1) )
H - log2(N) -> log2( ( 1 + 1/N ) ^ N )
( 1 + 1/N ) ^ N -> e !!
H - log2(N) -> log2( e ) = 1.44269 bits
for large alphabet, the excess bits sent due to coding distances instead of indices is log2(e) !!
I thought it was pretty entertaining for Euler to show up in this problem.
Why is distance coding fundamentally inefficient (vs just coding the indices of these repeated tokens) ? It's due to repetitions of values.
If our preceding bytes were "ABCB" and we now wish to code an "A" , it's at distance = 3 because we had to count the two B's. Our average distance is getting bumped up because symbols may occur
multiple times before we find our match. If we did an LZ77 coder that made all substrings of the match length unique, we would not have this inefficiency. (for example LZFG which sends suffix trie
node indices rather than distances does this)
We can see where this inefficiency appears in the probabilities:
if you are drawing a random number from 256 at each step
keep stepping until you get a match
each time you have to draw from all 256 possibilities (repeats allowed)
P(0) = (1/256)
P(1) = (255/256) * (1/256)
P(2) = (255/256)^2 * (1/256)
P(n) = (255/256)^n * (1/256)
(as above)
instead imagine drawing balls from a hat
once you draw a ball it is discarded so it cannot repeat
stop when you match your byte
first draw has the same 1/256 chance of match :
P(0) = (1/256)
as before multiply by 1-P to get the probability of continuing
but now we only have 255 balls left, so the chance of a match is (1/255)
P(1) = (255/256) * (1/255) = (1/256)
current P was (1/255) so multiply the next by (254/255)
now we have 254 , so we match (1/254) of the time :
P(2) = (255/256) * (254/255) * (1/254) = (1/256)
P(n) = (1/256)
it's just a flat probability.
decrementing the alphabet size by one each time makes a big difference.
This theoretical coding loss is also observed exactly in the real world.
experiment :
make 100,000 random bytes
make a palette of 256 32-bit random dwords
use the 100,000 random bytes to index the palette to output 400,000 bytes
what can LZ77 on these 400,000 bytes?
Our theoretical analysis says the best possible is 9.44 bits per palette index
plus we have to send 1024 bytes of the palette data as well
best = 100,000 * 9.44/8 + 1024 = 119,024
real world :
Leviathan Optimal5
random_bytes_depal.bin : 400,000 -> 119,034 = 2.381 bpb = 3.360 to 1
it turns out this theoretical limit is almost exactly achieved by Oodle Leviathan, only 10 bytes over due to headers and so on.
Leviathan is able to hit this limit (unlike simpler LZ77's) because it will entropy code all the match signals to nearly zero bits (all the matches will be "match len 4" commands, so they will have a
probability near 1 and thus entropy code to zero bits). Also the offsets are in multiples of 4, but Leviathan will see the bottom bits are always zero and not send them. The result is that Leviathan
sends the matches in this kind of data just as if it was sending a distance in tokens (as opposed to bytes) back to find the repeated token, which is exactly what our toy problem looked at.
We cannot do better than this with LZ77-style distance coding of matches. If we want to beat this we must induce the dictionary and send id references to whole words. Leviathan and similar LZ's
cannot get as close to the optimum on text, because we don't handle tokens of varying length as well. In this kind of depalettized data, the match length is totally redundant and should be coded in
zero bits. With text there is content-length correlation which we don't model at all.
Also note that we assume random tokens here. The loss due to sending distances instead of token id's gets even worse if the tokens are not equiprobable, as the distances are a poor way to capture the
different token probabilities.
Summary :
The coding loss due to sending repeated tokens by distance rather than by index is at least log2(e) bits for large alphabet. This theoretical limit acts as a real world limit on the compression that
LZ77 class algorithms can achieve on depalettized data.
2 comments:
Charles said...
Nice article; relatively easy to follow :). Let me see how much I got right here.
I think your argument boils down too:
Algorithms which say things like 'repeat that sequence we saw 2 tokens back' are less efficient than 'repeat that sequence at token 5'.
And you present some evidence centered on a dataset with 256 tokens of 4 bytes each.
An obvious argument; if we increase the size of the file in an unbounded way (so, 256 tokens of 4 bytes each, but 50GB of data), the distance between the sequence will, on average, remain the
same, while the absolute index value most recently seen would increase an unbounded (but much slower) way.
Unless, we cheat and assume we can reference the first value seen (the index of which will probably be quite small) for all instances of a value. You do seem to be making that assumption here,
and I don't really see a strong reason why you shouldn't do that, especially with the additional constraint 'your data consists of a finite set of tokens'. You call it out a bit in the hat
As an interesting extension to this; your assumptions here would not hold up as well if data was not random and tended towards 'chunking'. For example, if we had 4 billion bytes of 'black', then
4 billion of 'white', sending the index of a white pixel would use 4 bytes or so, where as sending the 'last reference' would be (on average) 1 byte.
So if there is a time component to the data (i.e., it changes from predominately one color to another over time), sending distance might work better. I wonder if that is more common in practice?
How strong does this tendency have to be to make distance more attractive?
PS; the lack of an open-source decompression library for Oodle makes me sad. I really want to extract images from one of my favorite games, but they use Oodle and cause me much sadness.
cbloom said...
No, you're not sending the index as in a position. The alternative to distance coding is indexing by *content*.
Instead of coding "repeat the thing at 200 bytes back" you say "repeat the 10th previous thing with unique contents of length 4"
It's really just about removing repeats from the indexing.
Another way to see it is, say you want to send length 4 tokens. As your file goes to 50 GB, the number of distances you can send goes up to 50 GB, a large number. But if you only count the number
of unique 4-byte strings that occur, that number goes up much more slowly, because some repeat.
How much more slowly depends on the entropy. On high entropy (random bytes) they go up at the same rate, on maximal low entropy (all bytes zero) the number of unique strings doesn't go up at all. | {"url":"https://cbloomrants.blogspot.com/2019/05/on-entropy-of-distance-coding.html","timestamp":"2024-11-12T13:41:11Z","content_type":"application/xhtml+xml","content_length":"74976","record_id":"<urn:uuid:4535e98c-fde5-414a-ad3b-030da328dad2>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00356.warc.gz"} |
Mental Math Archives - Page 2 of 4 - EducationUnboxed.com - Free Help for Homeschool
Please watch my previous free math tutoring videos on Mental Math before watching this video. Understanding mental math within twenty is foundational to understanding all mental math.
Subtracting Within 100 – Part One – Free Math Tutoring Video
Please watch my previous free math tutoring videos on Mental Math before watching this video. Understanding mental math within twenty is foundational to understanding all mental math.
Adding Within 100 – Free Math Tutoring Video
Please watch my previous free math tutoring videos on Mental Math before watching this video. Understanding mental math within twenty is foundational to understanding all mental math. | {"url":"http://www.educationunboxed.com/category/mental-math/page/2/","timestamp":"2024-11-09T01:00:50Z","content_type":"text/html","content_length":"47554","record_id":"<urn:uuid:c7d3e5ce-e911-45b7-be11-beba2dd43324>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00252.warc.gz"} |
Circle A has a radius of 1 and a center at (3 ,3 ). Circle B has a radius of 3 and a center at (6 ,4 ). If circle B is translated by <-3 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles? | HIX Tutor
Circle A has a radius of #1 # and a center at #(3 ,3 )#. Circle B has a radius of #3 # and a center at #(6 ,4 )#. If circle B is translated by #<-3 ,4 >#, does it overlap circle A? If not, what
is the minimum distance between points on both circles?
Answer 1
Yes they do, because the distance between the two circle centres $\mathbb{C} ' = 3$ is lower than the sum of both radii $R + R ' = 4$
The circle A has equation #(x-3)^2+(y-3)^2=1# with centre #C(3;3)# and radius #R=1#, whereas the circle B has equation #(x-6)^2+(y-4)^2=9# with radius #R'=3#. If we translate the second circle by the
vector #(-3,4)# the new equation of B circle is #(x-3)^2+y^2=9# and centre #C'(3;0)# and #R'=3# The distance between the two centres is #CC'=root2((3-3)^2+(3-0)^2)=3#. As the #CC'# is less than the
sum of the two radius we can deduce that the two circles overlap
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Answer 2
The translated center of circle B after the translation is (3, 8). To find if circle B overlaps circle A, we need to calculate the distance between the centers of the two circles and compare it to
the sum of their radii.
The distance between the centers of circles A and B after the translation is: [ \sqrt{(3 - 3)^2 + (3 - 8)^2} = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5 ]
The sum of the radii of circles A and B is: [ 1 + 3 = 4 ]
Since the distance between the centers of the circles (5) is greater than the sum of their radii (4), the circles do not overlap.
To find the minimum distance between points on both circles, we subtract the radii of circle A and circle B from the distance between their centers: [ \text{Minimum distance} = 5 - (1 + 3) = 5 - 4 =
1 ]
So, the minimum distance between points on both circles is 1.
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Answer 3
Circle B, after being translated by <-3, 4>, will have a new center at (6 - 3, 4 + 4) = (3, 8). The distance between the centers of Circle A and the translated Circle B is given by the distance
formula: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2) Plugging in the coordinates of the centers, we get: Distance = sqrt((3 - 3)^2 + (3 - 8)^2) = sqrt(0 + 25) = sqrt(25) = 5 The minimum distance
between points on both circles is equal to the difference between the radius of Circle B (3) and the distance between their centers (5), which is 3 - 5 = -2. This means that the circles do not
overlap, and there is a minimum distance of 2 units between them.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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: curious behavior of compoundProblem.pl
I wrote a WW problem using compoundProblem.pl. I didn’t want to rewrite the text to the first part in subsequent parts, so I formatted the problem as follows:
Does the following Diophantine equation have integer solutions?
\[ $a x + $b y = c \]
\{ $has_solution->menu() \} # $has_solution is a PopUp
END_TEXT Context->normalStrings;
ANS( $has_solution->cmp() );
if ( $part >= 2 ) {
Good! What is one possible solution?
\(x\equiv\) \{ ans_rule(5) \}
# and so forth
The key here is that
• I have no guard to print the introduction and its answer field for the first part; i.e., no if ($part == 1), and
• I have if ($part >= 2) rather than the documentation’s if ($part == 2) because there’s a third part, too, and I wanted the text in part 2 to stick around there, as well.
As I say, there were three parts total. The upshot is that the answer field for part 1 is counted 3 times (once for part 1, once again for part 2, and a third time for part 3), while the answer field
for part 2 was counted twice (once for part 2, and once again for part 3).
So the correct value of totalAnswers when declaring this compoundProblem was 6, not 3 as I originally expected. I say this because when the student gets the answers right, setting totalAnswers to 3
leads to a score of 200% and a mysterious-looking X on the student’s progress page, as well as problems with the student score, whereas setting totalAnswers to 6 leads to a 100% and the expected 100
on the student’s progress page.
I want to make sure this is the desired behavior. It took me a couple of hours to figure out what was going on, so I’d be happy to add a note to the Wiki for the other rare birds like myself who try
to do something like this. If this is not the desired behavior, but rather a bug, then of course I’d much prefer to change the problems now so that I do not have to worry about surprises in the
Hi John,
This isn't directly a reply to your question, but you should be aware that
compoundProblem.pl is deprecated (it is being kept around for backwards compatibility) and it is better to use scaffold.pl
which is more flexible and probably easier to use.
(There is also http://webwork.maa.org/wiki/CompoundProblem5 but again this is an older implementation of the scaffolding model.)
Hope this helps.
You aren’t kidding. I just wrote three problems using scaffold.pl, and it worked great! The only downside is that I now have to go back and rewrite my old problems.
That said, the natural place to go looking for help on this is (as far as I know) the Problem Techniques Wiki, and the only techniques listed that obviously apply to multi-part problems are “Compound
(Multi-part, sequential) Problems” and “Multi-Part, Sequential Problems”. Both link to a page that has nothing about scaffold.pl. I can edit that page with a note that the techniques listed there are
considered obsolete, adding a link to the page on Scaffolding. Is that OK? I’m leery about rushing in; for all I know, there’s a reason it hasn't already been done.
It's ok to edit -- in fact we encourage it. Pages are versioned so you can always back out if necessary. It also doesn't hurt to ask before or during edits if you have questions. That may save some
One template that might help you with the redirection task is this MediaWiki construction -- (you can google it for more information) or https://www.mediawiki.org/wiki/Category:Alert_templates
{{Historical|Foo [[test]] foo.}}
Which produces a banner about this being a historical document. The second part of the entry gives a link to follow for a more up-to-date information and some explanation.
The only reason this hasn't been done is lack of person-hours. Your help would be very useful.
There are also compoundProblems2.pl, etc as well as sequentialProblems.pl. The development of the right mechanism took many iterations of development over about 10 years.
The current version is quite powerful and could use a full length blog post about how to use it creatively. (That could be placed on the wiki or sometimes it gets more publicity if it's contributed
to the blog aggregator reached by the "blogs" link in the left margin of the wiki. ) It's most lack of time that keeps documentation from being higher priority. :-)
Glad this is now working well for you and I look forward to the problems you produce while experimenting with it. I think scaffolded.pl problems have a lot of potential that is only beginning to be
investigated. Thanks for letting us know how things turned out.
I’ve now modified the relevant page. But:
The template you showed doesn't appear on the page you gave, and doesn't work the way you suggested. That MediaWiki page also provided an Obsolete template, so I tried that, but it also doesn’t work
according to their documentation.
So I added both those tags/templates, as well as a separate paragraph redirecting the user. Finally, on the Problem Techniques page I added a “see also”.
Let me know if that was OK, or if any of it should change.
Thanks John. I admit I don't understand the MediaWiki templates very well either and some of their documentation is out of date. (So I guess WeBWorK is in good company in that regard.)
After much futzing around I've created a new template:
which accepts an additional parameter, any phrase you want.
{{OutdatedRedirect | You can write multi-part problems using the [[Scaffold|scaffold.pl]] macro file }}
gives you
IMPORTANT: The content of this page is outdated. Do not rely on the information
contained in this page.
You can write multi-part problems using the scaffold.pl macro file.
The template is at Template:OutdatedRedirect in the Altert_templates category:
Someone with more knowledge can got ahead and improve this template (and others in that category). Then we can update all the wiki pages to use OutdatedRedirect to point to the current information.
Thanks for the help.
-- Mike
Thank you. I see that you also modified the page for compound problems, as well, so it looks as if this is completely resolved.
Mike is right, the compundProblem.pl macros are very old and no longer supported, and the scaffold.pl macros are much better and easier to use. In particular, you don't have to keep track to the
number of problems, as you did with compoundProblem.pl.
You don't give enough of your problem code to really check what you are doing, but compoundProblem isn't designed to do what you are trying to do (you can't have the same answer rules show up in more
than one part, as you appear to be doing, or they will count as answers in both parts, and I'm not sure that will be graded correctly even if you do.
Fortunately, the scaffold macros are set up to provide more flexibility, and you can have previous parts open when later parts are being displayed.
Here is an example of what I think you are after:
Context->strings->are(yes=>{}, no=>{});
Scaffold::Begin(is_open => 'correct_or_first_incorrect');
$a = random(2,5,1);
$b = random(3,7,1);
$x = random(1,5,1);
$y = random(2,6,1);
$c = $a * $x + $b * $y;
$p = PopUp(["?","yes","no"],"yes");
$ma = MultiAnswer($x,$y)->with(
checker => sub {
my ($C, $S) = @_;
my ($x, $y) = @$S;
return $a * $x + $b * $y == $c && $x == int($x) && $y == int($y);
Section::Begin("Part 1: The Equation");
Does the following Diophantine equation have integer solutions?
\[ $a x + $b y = $c \]
\{ $p->menu \}
Section::Begin("Part 2: The Solution");
Good! What is one possible solution?
\(x\equiv\) \{ $ma->ans_rule(5) \} and \(y\equiv\) \{ $ma->ans_rule(5) \}
You should be able to start with this and modify it to your needs. See the scaffold.pl documentation for details.
Thank you! I didn't notice your code at first, but I ended up doing more or less what you have written here. | {"url":"https://webwork.maa.org/moodle/mod/forum/discuss.php?d=4454","timestamp":"2024-11-14T17:51:56Z","content_type":"text/html","content_length":"134788","record_id":"<urn:uuid:2b7e096a-55fe-41db-a0d9-601715246d7a>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00690.warc.gz"} |
Mathematics and engineering
Structure and Interpretation of Computer Programs is full of interesting footnotes. Here’s a good one about the Fermat test for prime numbers:
Numbers that fool the Fermat test are called Carmichael numbers, and little is known about them other than that they are extremely rare. There 255 Carmichael below 100,000,000. The smallest few
are 561, 1105, 1729, 2465, 2821 and 6601. In testing primality of very large numbers chosen at random, the chance of stumbling upon a value that fools the Fermat test is less than the chance that
cosmic radiation will cause the computer to make an error in carrying out a correct algorithm. Considering an algorithm to be inadequate for the first reason but not for the second illustrates
the difference between mathematics and engineering.
Taken from page 53. | {"url":"https://codeinthehole.com/tidbits/mathematics-and-engineering/","timestamp":"2024-11-04T08:19:08Z","content_type":"text/html","content_length":"6882","record_id":"<urn:uuid:9fa30d71-75bc-48a7-aa89-6b475f742a51>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00164.warc.gz"} |
ME 274: Basic Mechanics II
Any questions??
Four-step plan:
Step 1 - FBD: Draw a free body diagram of the crate.
Step 2 - Kinetics: Write down the Newton/Euler equations for the crate. Be careful with your moment equation. Since the crate has no fixed points, you need to sum moments about the center of mass of
the crate.
Step 3 - Kinematics: Crate does not tip, so α = 0.
Step 4 - Solve
Homework H5.B - Fa24
Problem statement
Solution video
Any questions??
For Part (a), the coefficient of static friction between the sphere and the ramp is sufficiently large that the sphere does not slip as it rolls. Here f ≠ μ[k]N.
For Part (b), the coefficient of friction is reduced such that slipping does occur. Here f = μ[k]N.
Can you see the difference in rotational motion of the sphere between the no-slip and slipping cases?
Four-step plan:
Step 1 - FBD: Draw a free body diagram of the sphere. Take care in getting the direction of the friction force on the sphere correct.
Step 2 - Kinetics: Write down the Newton/Euler equations for the sphere. Be careful with your moment equation - it is recommended that you use a moment equation about the center of mass of the
Step 3 - Kinematics: For Part (a), C is a no-slip point. Relate the angular acceleration of the sphere to the acceleration of G. For part (b), C is NOT a no-slip point. You cannot relate the angular
acceleration back to the acceleration of G through kinematics. Instead you need to use: f = μ[k]N.
Step 4 - Solve
Homework H4.U - Fa24
Problem statement
Solution video
Discussion and hints:
Recall that for oblique impact problems, it is recommended that you use three FBDs: A alone, B alone and A+B together. A+B together allows us to make the impact force internal (and not appear in the
linear impulse-momentum equation). FBDs for A and B individually allows us to determine the t-direction components of velocity each for A and B.
Step 1: FBDs
Draw FBDs of A, B and A+B.
Step 2: Kinetics (linear impulse/momentum)
Consider using the linear impulse-momentum equation for the t-direction for A and B individually, and in the n-direction for A+B. You need four equations to determine the two components of velocity
for each of the two particles. Consider the coefficient of restitution as your fourth equation.
Step 3: Kinematics
Step 4: Solve
Solve for the two components of velocity for each of the two particles. From these, the post-impact direction angles of motion can be found.
Any questions?
Homework H4.V - Fa24
Problem statement
Solution video
Discussion and hints:
For this problem, make your system big. In your FBD include A, B, P and link AOB. With this choice, you can simplify your angular impulse-momentum analysis since you do not need to deal with the
impact force between P and A since it will be internal to this system.
Step 1: FBDs
Draw an FBD of A, B, P and AOB together.
Step 2: Kinetics (angular impulse/momentum)
Consider the external forces acting on your system in your FBD. Which, if any, forces cause a moment about the fixed point O? Write down the angular momentum for each particle individually and add
together to find the angular momentum for your system: H[O] = 2m r[A/O ]x v[A] + m r[B/O ]x v[B ]+ m r[C/O] x v[C]. Is this momentum conserved? Also, consider using the coefficient of restitution
Step 3: Kinematics
Use the rigid body velocity equation to relate the velocities of A and B.
Step 4: Solve
Use your results from Steps 3 and 4 to solve for the angular speed of AOB.
Any questions?
Homework H4.S - Fa24
Problem statement
Solution video
Any questions??
Homework H4.T - Fa24
Problem statement
Solution video
Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.
You are asked to investigate the dynamics of this system during the short time of impact of P with A.
• It is suggested that you consider a system made up of A+P+bar (make the system "big").
• Draw a free body diagram (FBD) of this system.
• For this system, linear momentum is NOT conserved since there are non-zero reaction forces at O.
• Furthermore, energy is NOT conserved since there is an impact of P with A during that time.
• From your FBD of the system, you see that the moment about the fixed point O is zero. What does this say about the angular momentum of the system about O during impact? (Answer: It is conserved!)
STEP 1 - FBD: Draw a SINGLE free body diagram (FBD) of the system of A+P+bar.
STEP 2 - Kinetics: Consider the discussion above in regard to conservation of angular momentum about point O. Recall how to calculate the angular momentum about a point for a particle.
STEP 3 - Kinematics: At Instant 2, the P sticks to A: v[P2] = v[A2].
STEP 4 - Solve.
Homework H4.Q - Fa24
Problem statement
Solution video
Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.
HINT: This problem is an exercise in becoming familiar with calculating angular momenta. Also, you will observe a lot of cancelation in the calculations with you sum together the contributions from
the seven particles. Are you able to understand why these cancellations occur?
Homework H4.R - Fa24
Problem statement
Solution video
Any questions??
Step 1: FBD - Draw a free body diagram of the particle P and the bar. Are there any forces creating a moment about point O?
Step 2: Angular momentum - If you did not observe any moments about the fixed point O, then angular momentum about point O is conserved. This will provide you with one equation to find the angular
rotation rate of the bar at the second position. How do you find the radial component of velocity, R_dot? [HINT: Energy is also conserved.] This, along with the angular rotation rate of the bar, will
give you R_dot.
Step 3: Kinematics - Your kinematics all through the solution will be in terms of the polar components of velocity. Note that the velocity is NOT simply R*omega for the motion of P. Be careful!
Step 4: Solve
Homework H4.O - Fa24
Problem statement
Solution video
Any questions??
Step 1: FBD - Draw individual free body diagrams (FBDs) of each particle, along with an FBD of all three particles.
Step 2: Linear impulse/momentum - Carefully consider each FBD above. In which directions (if any) is linear momentum conserved for each FBD.
Step 3: Kinematics - None needed here.
Step 4: Solve - Use the conservation of linear momentum equations from Step 2, along with the coefficient of restitution equation in the "n"-directions for the two impacts.
Homework H4.P - Fa24
Problem statement
Solution video
Any questions??
Step 1: FBD - Draw individual free body diagrams (FBDs) of each particle, along with an FBD of both particles together.
Step 2: Linear impulse/momentum - Carefully consider each FBD above. In which directions (if any) is linear momentum conserved for each FBD.
Step 3: Kinematics - None needed here.
Step 4: Solve - Use the conservation of linear momentum equations from Step 2, along with the coefficient of restitution equation in the "n"-directions for the impact to solve for the post-impact
NOTE: Since e = 0, the "n"-components of velocity of A and B after impact should be equal. Can you see this in the animation above of the impact simulation? | {"url":"https://www.purdue.edu/freeform/me274/blog/page/2/","timestamp":"2024-11-07T12:40:32Z","content_type":"text/html","content_length":"62583","record_id":"<urn:uuid:0b1ce274-5ffd-4bd9-8ed7-7d42373adf72>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00334.warc.gz"} |
AmericanScience: A Team BlogSilver Linings and the Statistical Playbook
We asked historian of science Christopher J. Phillips, an expert on quantification in American public life, to reflect on the role of statistics—and Nate Silver—in the coverage of the 2012 election.
He was kind enough to write us the following guest post; you can find out more about his work here.
The 2012 election was a "Moneyball Election" and Nate Silver its big winner. Or so proclaimed the New Yorker's Adam Gopnik. He was certainly not alone. Deadspin's David Roher lamented the "braying
idiots" detracting from Silver's well-deserved limelight; President Obama jokingly praised Silver for having "nailed" the prediction of this year's Thanksgiving Turkey; and Wired's Angela Watercutter
perhaps gave the ultimate compliment by calling Silver a "Nerdy Chuck Norris."
Silver, for anyone who has spent the last few years under a rock, is the creator of the (mostly) political blog FiveThirtyEight. Picked up by the "New York Times" just before the 2010 midterm
elections, FiveThirtyEight has become one of the go-to sites for political junkies.
Source: https://pbs.twimg.com/media/A7Cbu_ZCQAE0m6o.jpg:large
An unlikely fate, to be sure, for an unknown consultant at the accounting firm KPMG a decade earlier. Silver tells it as an ersatz rags to riches story, a bored employee and mediocre online poker
player who designed a model for evaluating baseball players and then took on the pundits—mainly because in both baseball and politics the majority of "experts" knew close to nothing. This account is
too modest on the one hand—his evaluative system for baseball, PECOTA, became one of the central predictive tools of the respected Baseball Prospectus operation. On the other, it overstates his
statistical creativity—his overwhelming contribution has been to introduce clearer measures of confidence to poll predictions. As he explained to new readers in 2010, his blog was "devoted
to...rational analysis" and "prioritize[s] objective information over subjective information." Hardly a revolutionary statistical approach.
Notwithstanding these pedestrian goals, his message has certainly struck a chord. His 2012 book, The Signal and the Noise: Why So Many Predictions Fail—But Some Don't, was perfectly timed to capture
the election-season hype and indeed is still on the "New York Times" bestseller list. Few books could possibly be blurbed by both Bill James (of "Moneyball" fame) and Peter Orszag (of perhaps lesser
O.M.B. fame). Silver has, in effect, become a hero to thinking folks everywhere (particularly those with only passing statistical knowledge), who happily point to his columns as if to say to the
unwashed masses, "I told you so."
But what does this ascendance of "rational" thinking represent? For one thing, it is not about the spread of statistical knowledge. Silver himself almost never fully explains the mathematics behind
the models he uses. While he is footnote-happy in his book, he doesn't include any notes or appendices which would begin to teach others how to use the basic statistical concepts he deploys like
regression to the mean, probability distributions, or the "nearest neighbor" algorithm.
That's hardly criticism—each equation in a book is rumored to reduce readership, although to my knowledge no one's ever done the formal regression analysis—but it is telling that Silver is certainly
not trying to get more people to understand the models themselves.
Rather, he seems to be saying, "Trust me."
Source: http://www.newyorker.com/online/blogs/books/nate-silver.jpg
Like most in the futurology business, however, he is also saying, "Distrust others." Indeed, one lesson of his book, despite Silver's avowed optimism, is that predictive models are often worthless
and almost always fail to live up to their promoters' claims. Even well-defined rule-bound games like chess and poker turn out to be hard to predict in practice, a fact Silver acknowledges by going
Zen: "The closest approximation to a solution is to achieve a state of equanimity with the noise and the signal, recognizing that both are an irreducible part of our universe, and devote ourselves to
appreciating each for what it is."
Such concessions are not uncommon. Like nearly all modelers, Silver readily admits that models require active tinkering. He happily combines quantitative and qualitative information in his political
predictions, using the Cook Political Report to arbitrarily assign a code of +1 or 0 within his models, for example. And he emphasizes that Bayes's Theorem and conditional probability generally (the
probability of certain data given a hypothesis is not usually the same as the probability of a hypothesis given the same data but the probabilities are mathematically related) suggest the importance
of context and assumptions, although he perhaps overstates the theorem's importance.
Nevertheless, Silver doesn't really analyze the feedback loops of modeling that often interest historians of science. Following the language of Donald MacKenzie, models can act as a camera, giving a
snapshot of a particular process, but can also act as an engine, driving changes in the very thing being modeled.
Silver admits such loops occur in finance and fashion, and even in disease reporting, but does not address the possibility that all non-trivial models might work like this: more models mean more
noise, more noise will require more adjustments. Models are, after all, irreducibly human creations. And even after years of computer models analyzing "Big Data," the list of triumphs is strikingly
Source: http://mitpress.mit.edu/covers/9780262633673.jpg
Silver's rapid rise may ultimately represent a fear of declining discourse more than a triumph of the nerds. After all, this is an era in which Rep. Paul Ryan is considered a budget finance guru for,
apparently, basic arithmetic calculations. Silver's reputation has grown in no small part because he makes predictions in areas—sports, poker, politics—in which the loudest, wildest, crassest
"expert" opinions normally take center stage.
Maybe members of the "reality-based community" embraced Silver because they're tired of being told, "that's not the way the world really works" in the twenty-first century.
On the other hand, Silver predicted a Patriots-Seahawks Super Bowl in 2013. Oh well.
2 comments
Thanks for this Christopher. I enjoyed it.
Am I right in detecting an underlying question/critique of what Ted Porter talks about as the problem of "technicality": that too much science ends being reserved for the consideration of a
technically trained few, and not enough exists in a place that's truly susceptible to wide-spread discussion and debate? So Silver becomes a champion of "reality," but it's a reality that few others
can understand or recreate...
I think you're right about the large-scale effects of model proliferation. I was, still, generally impressed with Silver's 538 posts precisely because he did entertain (and discuss and investigate)
the performative aspects of modeling and polling. He mentioned often that polls tend to cluster around consensus numbers (which seems to denote fiddling the numbers on some pollsters part so as not
to stand out). And he also noted that his predictions were shaping the Intrade betting lines (and other betting markets too) that he sometimes used as a point of comparison.
On the math/equations line of discussion, I also take your point and I think it speaks more to the "technicality" of the situation that the book and 538 had so few particulars about mathematical
mechanics. Still, I often appreciated 538 most b/c of the very deep in the weeds discussions there of criteria for evaluating and rating data from a variety of sources. This struck me as old school
"statistics"---the sort of stuff that a pre-WWII statistician spent his or her time doing most often. And I sometimes think we have grown too enamored with fancy mathematics, to the detriment of
thinking about the collection and evaluation of data in the first place. In that way (but only that way), I appreciate the hidden mathematics.
Thanks again for providing so much food for thought.
Christopher Phillips January 28, 2013 at 2:02PM
Excellent thoughts.
I also appreciated the way Silver acts as a guide through the weeds and couldn't agree more on the fact that fancy doesn't imply quality or substitute for careful thought. It is not too hard to come
up with examples of complicated models that failed even in their basic assumptions, let along predictions. I definitely wanted to think more carefully, though, about the claim that Silver is showing
us the way, acting as a prophet with access to somehow deeper truths. Or even the idea that we are slowly but surely converging on some ultimate truth through these models. I think that's the problem
with his idea of "convergence" or with the number fiddling--that somehow with enough of it we're going to get to the bottom of things. The claim that he makes in his book that the truth stays
constant as the noise increases--so we just need to get better at filtering--comes across as a rather naive epistemology. | {"url":"http://americanscience.blogspot.com/2013/01/silver-linings-and-statistical-playbook.html","timestamp":"2024-11-05T12:27:04Z","content_type":"text/html","content_length":"107658","record_id":"<urn:uuid:01a0d8ab-251e-431b-ac7f-ffc9adc4a781>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00617.warc.gz"} |
The tipping points and early warning indicators for Pine Island Glacier, West Antarctica
Articles | Volume 15, issue 3
© Author(s) 2021. This work is distributed under the Creative Commons Attribution 4.0 License.
The tipping points and early warning indicators for Pine Island Glacier, West Antarctica
Mass loss from the Antarctic Ice Sheet is the main source of uncertainty in projections of future sea-level rise, with important implications for coastal regions worldwide. Central to ongoing and
future changes is the marine ice sheet instability: once a critical threshold, or tipping point, is crossed, ice internal dynamics can drive a self-sustaining retreat committing a glacier to
irreversible, rapid and substantial ice loss. This process might have already been triggered in the Amundsen Sea region, where Pine Island and Thwaites glaciers dominate the current mass loss from
Antarctica, but modelling and observational techniques have not been able to establish this rigorously, leading to divergent views on the future mass loss of the West Antarctic Ice Sheet. Here, we
aim at closing this knowledge gap by conducting a systematic investigation of the stability regime of Pine Island Glacier. To this end we show that early warning indicators in model simulations
robustly detect the onset of the marine ice sheet instability. We are thereby able to identify three distinct tipping points in response to increases in ocean-induced melt. The third and final event,
triggered by an ocean warming of approximately 1.2^∘C from the steady-state model configuration, leads to a retreat of the entire glacier that could initiate a collapse of the West Antarctic Ice
Received: 30 Jun 2020 – Discussion started: 04 Aug 2020 – Revised: 21 Jan 2021 – Accepted: 31 Jan 2021 – Published: 25 Mar 2021
The West Antarctic Ice Sheet (WAIS) is regarded as a tipping element in the Earth's climate system, defined as a major component of the Earth system susceptible to tipping-point behaviour (Lenton et
al., 2008). Its collapse, potentially driven by the marine ice sheet instability (MISI; Feldmann and Levermann, 2015), would result in over 3m of sea-level rise (Fretwell et al., 2013). Key to MISI
are the conditions at the grounding line – the transition across which grounded ice begins to float on the ocean forming ice shelves. In a steady state, ice flux across the grounding line balances
the surface accumulation upstream. If grounding-line retreat causes grounding-line flux to increase and this is not balanced by a corresponding increase in accumulation, the net mass balance is
negative and retreat will continue (Weertman, 1974; Schoof, 2007). Conversely, grounding-line advance leading to an increase in accumulation greater than the change in flux will lead to a continued
advance. In this regime, a small perturbation in the system can result in the system crossing a tipping point, beyond which a positive feedback propels the system to a contrasting state (Fig. 1c). A
complex range of factors can either cause or suppress MISI (Haseloff and Sergienko, 2018; Pegler, 2018; O'Leary et al., 2013; Gomez et al., 2010; Robel et al., 2016), and the difficulties in
predicting this behaviour are a major source of uncertainty for future sea-level-rise projections (Church et al., 2013; Bamber et al., 2019; Oppenheimer et al., 2019; Robel et al., 2019).
One area of particular concern is the Amundsen Sea region. Pine Island (PIG) and Thwaites glaciers, the two largest glaciers in the area, are believed to be particularly vulnerable to MISI (Favier et
al., 2014; Rignot et al., 2014). Palaeo-records and observational records of PIG show a history of retreat, driven by both natural and anthropogenic variability in ocean forcing (Jenkins et al.,
2018; Holland et al., 2019). One possible MISI-driven retreat might have happened when PIG unpinned from a submarine ridge in the 1940s (Jenkins et al., 2010; Smith et al., 2016). Recent modelling
studies indicate that a larger-scale MISI event may now be underway for both Pine Island and Thwaites glaciers that would lead to substantial and sustained mass loss throughout the coming centuries
(Favier et al., 2014; Jenkins et al., 2016; Joughin et al., 2010). Being able to identify a MISI-driven retreat and differentiate this from a retreat where a tipping point has not been crossed is
vital information for projections of future sea-level rise. One of the major hurdles in determining whether a tipping point has been crossed is that currently this necessitates time-consuming
steady-state simulations to calculate the hysteresis behaviour of an identified period of retreat (e.g. Garbe et al., 2020). An alternative methodology that can be applied directly to transient
simulations as a post-processing step would therefore be useful to the ice sheet modelling community. Such tools based on early warning indicators are presented in this paper.
The tipping behaviour of MISI is an example of a saddle-node (or fold) bifurcation in which three equilibria exist: an upper and lower stable branch and a middle unstable branch (Fig. 1c; Schoof,
2012). Starting on the upper stable branch, perturbing the system beyond a tipping point (x[1] in Fig. 1c) will induce a qualitative shift to the lower and contrasting stable state. Importantly (and
in contrast to a system such as that shown in Fig. 1a and b), in order to restore conditions to the state prior to a collapse it is not sufficient to simply reverse the forcing to its previous value.
Instead, the forcing must be taken back further (to point x[2]), which in some cases may be far beyond the parameter range that triggered the initial collapse. This type of behaviour is known as
hysteresis. A large change in response to a small forcing is not necessarily indicative of hysteresis, as shown in Fig. 1b. Tipping points are crossed in both Fig. 1c and Fig. 1d, and both cases are
often referred to as irreversible, although the two are distinct in that only Fig. 1d is irreversible for any change in the tested range of the control parameter. Hereafter we will refer to the
former as irreversible, in line with previous studies, and the latter as permanently irreversible, to differentiate the two. Diagnosing whether a tipping point has been crossed without some prior
knowledge of the system is not generally possible without reversing the forcing to see if hysteresis has occurred. An alternative approach to identify tipping points is based on a process known as
critical slowing down, which is known to precede saddle-node bifurcations of this type (Wissel, 1984; van Nes and Scheffer, 2007; Dakos et al., 2008; Scheffer et al., 2009). Critical slowing down is
a general feature of non-linear systems and refers to an increase in the time a system takes to recover from perturbations as a tipping point is approached (Wissel, 1984). We will explore both
hysteresis and critical slowing down as indicators of tipping points in our model simulations.
In Sect. 2, we explain critical slowing down and early warning indicators in the context of MISI. We then map out the stability regime of PIG using numerical model simulations. We force the model
with a slowly increasing ocean melt rate and identify three periods of rapid retreat with the methodology explained in Sect. 3.1. Using statistical tools from dynamical systems theory we find
critical slowing down preceding each of these retreat events and go on to demonstrate that these are indeed tipping points in Sect. 4. This is confirmed by analysing the hysteresis behaviour of the
glacier, showing the existence of unstable grounding-line positions. To our knowledge, this is the first time that the stability regime of PIG has been investigated in this detail and the first time
that tipping-point indicators have been applied to ice sheet model simulations. Our results reveal the existence of multiple tipping points leading to the collapse of PIG, rather than one single
event, that when crossed could easily be misidentified as simply periods of rapid retreat, with the irreversible and the self-sustained aspect of the retreat being missed.
2Critical slowing down and early warning indicators
As certain classes of complex systems approach a tipping point, they show early warning signals (i.e. specific changes in system behaviour as detailed below) which can allow us to anticipate or even
predict the onset of a tipping event by means of statistical tools called early warning indicators (EWIs; Wissel, 1984). Early warning signals have been found to precede, for example, collapse of the
thermohaline circulation (Held and Kleinen, 2014; Lenton, 2011), onset of epileptic seizures (Litt et al., 2001; McSharry and Tarassenko, 2003), crashes in financial markets (May et al., 2008; Diks
et al., 2018), onset of glacial terminations (Lenton, 2011) and wildlife population collapses (Scheffer et al., 2001). Although most commonly used to detect the onset of saddle-node bifurcations, of
which MISI is an example, they are not strictly limited to bifurcations of this type and have, for example, also been successfully used to indicate the onset of Hopf bifurcations (Chisholm and
Filotas, 2009).
2.1Critical slowing down preceding the marine ice sheet instability
Critical slowing down is one example of an early warning signal that has been used in the past for both model output and observational records such as palaeoclimate data, with the aim of detecting an
approaching bifurcation (Held and Kleinen, 2004; Livina and Lenton, 2007; Dakos et al., 2008; Lenton et al., 2009, 2012b). Critical slowing down is so called because, as a non-linear system is
gradually forced towards a bifurcation, that system will become more “sluggish” in its response to perturbations (see middle panel of Fig. 2). This can be shown mathematically, because the dominant
eigenvalue of the system tends to zero as a bifurcation point is approached (Wissel, 1984) or, equivalently, the recovery time (i.e. the time it takes for a system to return to a steady state after
small perturbations) tends to infinity. The response time of a glacier to external forcing has also been shown analytically to increase as a MISI bifurcation is approached (Robel et al., 2018). While
critical slowing down is a general characteristic behaviour of the dynamics underlying MISI, the question remains whether it can be reliably detected in the context of a complex glacier where many
other processes are at play.
As a first step to addressing this question, we model MISI in an idealised flow line setup of a marine ice sheet. In this setup, we determine the change in recovery time before a tipping point
directly through multiple stepwise perturbations of the control parameter (Appendix A). Our setup closely resembles the MISMIP experiments (Pattyn et al., 2012), and indeed hints of critical slowing
down can be identified in that paper (Fig. 2 in Pattyn et al., 2012). The results in Appendix A show that critical slowing down is easily identified preceding both MISI-driven advance and retreat
bifurcations. This demonstrates that there is at least the potential that critical slowing down could be found in a less simplified modelling framework. This is not clear a priori, and, for example,
adding noise to the bed topography reduces the ability to identify early warning, as detailed in the Appendix. Identifying critical slowing down in this stepwise perturbation manner is appealing
because it directly extracts the change in response time that we are searching for; however it is not practical for a realistic model forcing which would not normally take the form of a step
function. A more general approach, which we adopt for our simulation of PIG, is to use EWIs to analyse the recovery time of the system as it is forced with natural variability.
2.2Early warning indicators
As the field of EWIs has expanded, more methods have been developed for extracting critical-slowing-down information from model results and observational records. These methods seek to approximate
the system recovery time from some measure of the system state. The challenge is that, for most real-world applications, natural forcing does not take the form of a step function and the system is
continuously perturbed and so cannot return to a true steady state. However, if the recovery time of a system is indeed increasing, the response to a continual stochastic forcing could be detected as
a tendency for each measurement of the system state to become increasingly similar to the previous measurement, sometimes referred to as an increase in “memory” of small perturbations. This is shown
conceptually and with examples extracted from our PIG model in Fig. 2. One common way to measure this effect is by sampling the data at discrete time intervals and calculating the lag-1
auto-correlation, i.e. the correlation between values that are one time interval apart (examples given in Fig. 2). This measure, which we refer to hereafter as the ACF indicator, should increase as a
tipping point is approached (Dakos et al., 2008; Ives, 1995). Since recovery time tends to infinity as the bifurcation is approached, successive system states should become more and more similar and
the ACF indicator should tend to 1. This threshold of an indicator, corresponding to the point when a tipping point would be crossed as predicted by theory, is referred to hereafter as the critical
value. An alternative measure that also seeks to identify changes in recovery time is to use the detrended fluctuation analysis algorithm (Livina and Lenton, 2007; Lenton et al., 2012a, b). This
first calculates the mean-centred cumulative sum of the time series, splits the result into epochs of length n which are detrended and then calculates the rms F(n) for each epoch. This is repeated
for epochs of different length, and finally an exponent α can be fitted in log–log space such that F(n)∝n^α. This exponent yields information on the self-correlation of the original time series,
whereby a value of 0.5 corresponds to uncorrelated white noise and greater values indicate increasing “memory” up to a maximum of 1.5. To aid comparison with the ACF indicator and following Livina
and Lenton (2007), we rescale the exponent so that it reaches a critical value of 1 and call this the DFA indicator. These indicators can be supported by analysing the variance of the system state.
Variance can be shown to increase as a tipping point is approached, since perturbations to the system decay more slowly and thus large shifts from the mean state will persist for longer (Scheffer et
al., 2009). No critical value exists in this case, but a persistent positive trend in variance serves as additional evidence that a tipping point is being approached.
We conduct a quasi-steady modelling experiment whereby we subject PIG to slowly increasing rates of basal melt beneath its adjacent ice shelf (Fig. 3). Conducting a transient simulation with an
evolving basal melt that exactly tracks the equilibrium curve (Fig. 1c) is not computationally feasible or necessary for our purposes. Thus, we adopt this quasi-steady modelling approach in which the
forcing increases slowly enough that it approximates the steady-state behaviour but faster than the long response timescales the glacier would require to be truly in equilibrium. Quasi-steady-state
experiments have previously been successfully applied to identify the tipping point of the Greenland Ice Sheet with respect to the melt–elevation feedback (Robinson et al., 2012) and to identify
hysteresis of the Antarctic Ice Sheet (Garbe et al., 2020). In Garbe et al. (2020) it was shown that such transient experiments enable identification of hysteresis behaviour, while the exact shape of
the curve must be mapped out with equilibrium simulations. We accompany the quasi-steady simulations with simulations that run to a steady state for constant values of the control parameter at
discrete values (these simulations continue until the change in ice volume is approximately equal to zero). We use basal melt rate as the control parameter, i.e. the parameter that we will change to
drive the system towards a tipping point. We make this choice since erosion of ice shelves by the intrusion of warm ocean currents is widely accepted as the mechanism responsible for the considerable
changes currently observed in this region (Shepherd et al., 2004; Rignot et al., 2014; Rignot, 1998; Joughin et al., 2010; Park et al., 2013; Gudmundsson et al., 2019). Sub-ice-shelf melt rates are
increased linearly (with additional variability as explained below) from a value that generates a steady state for the present-day glacier configuration. Based on the numerical experiments we then
evaluate EWIs to test for critical slowing down.
3.1Model description
All simulations use the community Úa ice-flow model (Gudmundsson et al., 2012; Gudmundsson, 2013, 2020), which solves the dynamical equations for ice flow in the shallow-ice-stream approximation
(SSTREAM or SSA; Hutter, 1983). Bedrock geometry for the PIG domain is a combination of the R topo2 dataset (Schaffer et al., 2016) and, where available, an updated bathymetry of the Amundsen Sea
embayment (Millan et al., 2017). Surface ice topography is from CryoSat-2 altimetry (Slater et al., 2018). Depth-averaged ice density is calculated using a meteoric ice density of 917kgm^−3
together with firn depths obtained from the RACMO2.1 firn densification model (Ligtenberg et al., 2011). Snow accumulation is a climatological record obtained from RACMO2.1 and constant in time
(Lenaerts et al., 2012).
Viscous ice deformation is described by the Glen–Steinemann flow law $\stackrel{\mathrm{˙}}{\mathit{\epsilon }}=A{\mathit{\tau }}_{E}^{n}$ with exponent n=3, and basal motion is modelled using a
Weertman sliding law ${u}_{\mathrm{b}}=C{\mathit{\tau }}_{\mathrm{b}}^{m}$ with exponent m=3. The constitutive law and the sliding law use spatially varying parameters for the ice rate factor (A) and
basal slipperiness (C), respectively, to initialise the model with present-day ice velocities. These are obtained via optimisation methods using satellite observations of surface ice velocity from
the Landsat 8 dataset (Scambos et al., 2016; Fahnestock et al., 2016). An optimal solution is obtained by minimising a cost function that includes both the misfit between observed and modelled
velocities and regularisation terms. An additional term in the cost function penalises initial rates of ice thickness change in order to ensure that these are close to zero at the start of
simulations. This approach helps to provide a steady-state configuration of PIG from which we can conduct our perturbation experiments.
The Úa model solves the system of equations with the finite-element method on an unstructured mesh, generated with MESH2D (Engwirda et al., 2014). The mesh remains fixed throughout the simulation to
avoid contaminating the time series with errors resulting from remapping fields onto a new mesh. The mesh is refined in regions of high strain rate gradients and fast ice flow as well as around the
grounding line. The region of grounding-line mesh refinement, in which the average element size is ∼750m, extends upstream sufficiently far so that the grounding line always remains within this
region until after the final MISI collapse.
Basal melt rates are calculated using a widely used, local quadratic dependency on thermal forcing:
$\begin{array}{}\text{(1)}& M=f{\mathit{\gamma }}_{T}{\left(\frac{{\mathit{\rho }}_{\mathrm{w}}{c}_{p}}{{\mathit{\rho }}_{\mathrm{i}}{L}_{\mathrm{i}}}\right)}^{\mathrm{2}}\left({T}_{\mathrm{0}}-{T}_
where γ[T] is the constant heat exchange velocity, ρ[w] is seawater density, c[p] is the specific heat capacity of water, ρ[i] is ice density, L[i] is the latent heat of fusion of ice, T[0] is the
thermal forcing and T[f] is the freezing temperature (Favier et al., 2019). Melt rates are only applied beneath fully floating elements to ensure that no melting can possibly occur upstream of the
grounding line (Seroussi and Morlinghem, 2018). The initial melt rate factor (f) is chosen such that the model finds a steady state with a grounding line approximately coincident with its position as
given in Bedmap2 (Fretwell et al., 2013). This melt rate factor is the aforementioned control parameter that drives changes in the model, some of which may be identifiable as tipping points.
To effectively extract information about the system's recovery time using the statistical methods outlined in Sect. 2, we need to perturb the model in a way that has some measurable impact on the
system state. A slow and monotonically increasing forcing would make our chosen approach impractical and is arguably as unrealistic as a stepwise perturbation. We therefore add natural variability to
the linearly increasing melt rate factor (f). There is strong evidence that the inferred and observed changes in PIG over the last century can be linked to changes in thermocline depth of the
Amundsen Sea shelf, which in turn is influenced by an atmospheric Rossby wave train originating in the Pacific Ocean (Jenkins et al., 2018). Following Jenkins et al. (2018), we use a ∼130-year time
series of central tropical Pacific sea surface temperature anomalies as a proxy for relevant variability in our melt rate forcing. We create an autoregressive (AR) model-based surrogate from this
time series using the Yule–Walker method to fit the AR model and minimum description length to determine the maximum order of the model. This new surrogate time series has the same decadal
variability that would be expected for the melting beneath PIG and can be extended to any length required. As shown in more detail below, by superimposing this signal onto the linearly increasing
melt rate factor we ensure that the system response contains sufficient variability to extract information about critical slowing down and thereby enable the calculation of EWIs. Furthermore, using
natural variability enables us to test the versatility of EWIs if they were to be applied directly to observations.
3.2Detecting critical slowing down
We have already established the control parameter for our model, but another important decision to make is what model output should be used as a measure of the system state. One choice could be
changes in ice volume, since they can be related to sea-level rise and ice sheet model simulations tend to focus on this result. However, ice volume varies very smoothly over time, making it
difficult to detect changes in the system recovery time. Instead, we use the integrated grounding-line flux, which shows greater temporal variability and whose change is directly related to the MISI
mechanism. As with other studies of this type, the model output is processed prior to the calculation of EWIs. This consists of aggregating the output (i.e. data binning) to remove variability with a
frequency higher than that directly relevant to the internal ice dynamics considered here and thus not related to the system recovery time, together with detrending to remove non-stationarities
(detrending is included in the DFA algorithm and therefore not required before calculation of the DFA indicator). Detrending was carried out using a Gaussian kernel smoothing function that has been
shown to perform better than linear detrending (Lenton et al., 2012a). A smoothing bandwidth was selected that removed long-term trends without overfitting the model time series. Indicators are
calculated over a moving window with a length of 300 years. The optimal window length is further discussed in Sect. 4.3.
From the processed time series, we calculate three different EWIs:
1. Critical slowing down is measurable as an increase in the state variable auto-correlation. We measure this here using the lag-1 auto-correlation function (Dakos et al., 2008; Scheffer et al.,
2009; Held and Kleinen, 2004) applied to the grounding-line flux over a 300-year moving window preceding each tipping point (ACF indicator).
2. Similarly, DFA (Peng et al., 1994) measures increasing auto-correlation in a time series and we apply this with the same moving-window approach.
3. An additional consequence of critical slowing down is that variance will increase as a tipping point is approached (Scheffer et al., 2009). We calculate variance of grounding-line flux for each
moving window, and this can be used in conjunction with other indicators to increase robustness.
As described in Sect. 2, recovery time should tend to infinity as a tipping point is approached. This corresponds to the ACF and scaled DFA indicators reaching a critical value of 1. In practice, for
a complex model there are a wide variety of reasons why a tipping point might be crossed before the EWI reaches a critical value. For example, this can be a result of variability in the control
variable pushing the system over a tipping point despite its long-term mean still being some distance from its critical value. For this reason, most studies adopt an alternative approach of looking
for a consistent increase in the EWIs in the run up to a tipping event. This is often measured by calculating the nonparametric Kendall's τ coefficient, a measure of the ranking or ordering of a
variable, which equals 1 if the indicator is monotonically increasing with time (Dakos et al., 2008; Kendall, 1948). This single value enables a simple interpretation of our results, since $\mathrm
{0}<\mathit{\tau }\le \mathrm{1}$ means the EWI is tending to increase with time, suggesting an imminent tipping point. The closer to 1 the calculated τ coefficient is, the greater the tendency for
an indicator to be increasing with time, and conversely a τ coefficient close to zero suggests no clear tendency for an indicator to be changing with time. We present our results in terms of both
aforementioned criteria: whether an EWI reaches a critical value preceding the tipping point and whether the EWI is consistently increasing for a period of time before the tipping point.
The quasi-equilibrium simulation shows three potential tipping points with respect to the applied melt (Fig. 4). Upon crossing each threshold, indicated by the numbered blue dots in Fig. 4, PIG
undergoes periods of not only rapid but also (as we show below) self-sustained and irreversible mass loss. At this stage, relying only on a record of changes in ice volume resulting from an
increasing forcing (solid black line in Fig. 4), one can only speculate that these are indeed tipping points and more analysis is necessary to confirm this hypothesis, and we will address this point
in Sect. 4.2. The last of the three events causes a permanently irreversible collapse within the entire model domain (Fig. 4a). We focus our results on these three major changes in the glacier
configuration and ignore any possible smaller tipping points that do not result in significant grounding-line retreat or changes in ice volume. We increase basal melt rates gradually and in a
quasi-steady-state manner to ensure that successive retreat events can be isolated and their effects do not overlap during the simulation. A more rapidly increasing forcing could lead to one tipping
point cascading into the next and result in three individual tipping points being misinterpreted as only one event.
Grounding-line positions before each of these retreat events and after the final collapse are shown in Fig. 3. Events 1 and 2 each contribute approximately 20mm of sea-level rise, while event 3,
which arises after slightly more than doubling current melt rates, contributes approximately 100mm. The actual sea-level rise that would result from this third and largest event is likely to be
larger since in our simulation the effects stop at the domain boundary and in reality neighbouring drainage basins would be affected.
4.1Early warning for the marine ice sheet instability
The three periods of MISI-driven retreat, each following the crossing of an associated tipping point, can be identified clearly using EWIs (Fig. 5). The ACF indicator increases and tends to 1 as the
tipping points are approached (Fig. 5a–c), indicating a tendency towards an infinitely long recovery time as predicted by theory. We calculate Kendall's τ coefficient to identify trends in the
indicator, with a value of 1 representing a monotonic increase in the indicator with time. The positive Kendall's τ coefficient shows that in all three cases, the lag-1 auto-correlation increases
before the onset of unstable retreat. Furthermore, the ACF indicator reaches a critical value of 1 relatively close in time to when the MISI event begins.
These findings are supported by the DFA indicator, described in Sect. 2. As with the ACF indicator, Kendall's τ coefficient is positive and the DFA indicator trends towards a critical value of 1 as
each tipping point is approached. We show the change in normalised variance calculated over each time window, and in all cases this increases ahead of the tipping points being crossed with a positive
Kendall's τ coefficient. The increase in variance gives greater confidence to the findings of the other two EWIs, although variance cannot be used directly to predict when that threshold will be
crossed since it does not approach a critical value before a tipping point is crossed.
4.2Hysteresis of Pine Island Glacier
In order to verify that we have correctly identified tipping points using the EWIs, we run the model to a steady state for a given melt rate to search for hysteresis loops that indicate the presence
of unstable grounding-line positions. These simulations start from either the initial model setup (advance steady state) or the configuration just prior to the final tipping point (retreat steady
state). Each of these simulations samples a discrete mean melt factor between these two states that is held constant (but with the addition of the same natural variability as in the forward
simulations) and is run forward in time until the modelled ice volume reaches a steady state. The first two tipping events show relatively small but clearly identifiable hysteresis loops (Fig. 4b),
for which recovery of the grounding-line position requires reversing the forcing beyond the point at which retreat was triggered (i.e. as shown in Fig. 1c). The third event marks the onset of an
almost complete collapse of PIG (Fig. 4a). Unlike the previous two, this collapse cannot be reversed to regrow the glacier for any value of the control parameter. This is an example of a permanently
irreversible tipping point, as shown in Fig. 1d. Note that this permanent irreversibility is only true for the glacier modelled in isolation and by expanding the domain it would presumably be
possible for other catchments that may not have collapsed to enable this glacier to regrow.
4.3Robustness of the indicators
We carry out several tests to assess the robustness of the EWIs and their sensitivity to the pre-processing carried out on the model output prior to calculating each indicator. Two parameters in this
processing step are the bin size into which data are aggregated and the bandwidth of the smoothing kernel that removes long-term trends in the time series. To check that the increasing trends in our
indicators are a robust feature of our results, regardless of these choices, we conducted a sensitivity analysis. The parameters were varied by ±50%, and the indicators were recalculated for each
resulting time series. As before, we assess the utility of an indicator by whether it shows an increasing trend before each tipping point, as measured by a positive Kendall's τ coefficient. The
results of this sensitivity analysis are presented for each MISI event in Fig. 6. Kendall's τ coefficient is positive for all tested combinations of parameters and all MISI events, although MISI
event 2 is particularly insensitive to these parameter choices, whereas the spread in Kendall's τ coefficient is greater for the other two events.
In general, critical slowing down will only occur close to a tipping point. Determining how close to a tipping point a system must be in order to anticipate the approaching critical transition, i.e.
the prediction radius, is an important question and also informs the selection of palaeo-records that could be used to detect an upcoming MISI event. The results presented above are for a window size
of 300 years (i.e. a record length of 600 years), which is the shortest window size for which the DFA indicator provides a clear prediction for all tipping events. We explored the prediction radius
of our model by calculating Kendall's τ for the ACF and DFA indicators and the variance for a range of window lengths; see Fig. 7. For the main tipping event, preceded by the longest stable period,
the indicators gradually lose their ability to anticipate a tipping event. This is shown by Kendall's τ values approaching zero or in some cases becoming negative, meaning that the EWIs are not
robustly increasing before each tipping point as a result of more data being included further from the bifurcation. The same is true for the two smaller tipping events, but the drop-off is quicker
such that the indicators break down for window lengths >500 years. These results suggest that the prediction radius is relatively small, and thus window sizes that are too large, which hence include
data far from a tipping point, become less useful for the application of EWIs.
In addition to a sensitivity analysis, it is important to check that trends in the calculated indicators are statistically significant and not the result of random fluctuations. We follow the method
originally proposed by Dakos et al. (2012) and produce surrogate datasets from the model time series that have many of the same properties but should not contain any critical-slowing-down trends. We
generate 1000 of these datasets using an autoregressive AR(1) process-based surrogate. For each of these datasets we calculate the ACF and DFA indicators and variance in the same way as with the
model time series and then estimate the trend with values of Kendall's τ coefficient. We calculate the probability of our results being a result of chance for each indicator and for all three
combined as the proportion of cases for which the surrogate dataset was found to have a higher correlation than the model time series. We find that P<0.1 in all but one instance for the ACF and DFA
indicators but variance trends were generally less significant (Table 1). However, the combined probability that all three indicators would be equally positive as a result of chance was less than
0.02 for the first MISI event and less than 0.005 for the second and third events.
The indicators we have tested provide early warning of tipping points as they are approached in our transient simulation with gradually increasing melt rates. Tipping points driven by MISI represent
potential “high-impact” shifts in the Earth climate system, since they may lead to considerable changes in the configuration of the Antarctic Ice Sheet that are effectively irreversible on human
timescales. Computational models are frequently used to forecast future changes in the Antarctic Ice Sheet in response to various greenhouse gas emission and warming scenarios. Predictive studies of
this kind sometimes label periods of rapid retreat as “unstable” without further analysis of the type performed here (e.g. Joughin et al., 2014; Ritz et al., 2015; Favier et al., 2014) or avoid
making this diagnosis altogether (DeConto and Pollard, 2016). Here, we have demonstrated that EWIs robustly approach critical thresholds preceding tipping points driven by MISI. Our results show that
EWIs can be used as a method to identify instabilities without the need for the aforementioned modelling approach based on computationally expensive equilibrium simulations.
It is important to clearly understand what critical threshold is identified by the EWIs. In Fig. 4 the simulated steady states show the crossing of the tipping point earlier than identified by the
indicators in the transient simulation. Since the timescales of ice flow are longer than the forcing timescale, the ice sheet system modelled here does not evolve along the steady-state branch (as
shown schematically in Fig. 1c). Relaxation to a steady state takes centuries to millennia in the simulations. This means that while technically the critical value of the control parameter (basal
melt rate) might have already been crossed, the glacier could return to its previous state in the transient simulation at that point if the basal melt rate was reduced below the critical threshold.
This is true until the system state variable crosses its critical value (point x[t] in Fig. 1c) – and this is the point identified by the EWIs. This complication in interpreting EWIs is inherent to
ice dynamics because of its long response timescales.
We find that both the ACF and the DFA indicators not only increase as a tipping point is approached, as shown by positive Kendall coefficients, but also generally approach the critical value of 1,
although with varying degrees of precision (Fig. 5). This enhances their predictive power, since by extending a positive trend line it is possible to approximate what value of the control parameter
will eventually cause a tipping point to be crossed. While our experiments in Appendix A showed that critical slowing down can accurately predict onset of tipping points in an idealised setup,
applying this method to a more complex case study may fail, and in this context our finding that these indicators largely retain their predictive power is very encouraging. One area of additional
complexity in our model of PIG compared to the setup in Appendix A is the bed geometry, which is obtained from observations and so is much less smooth than the synthetic retrograde bed used in the
MISMIP experiments. We explored how the addition of “bumpiness” to bed geometry affects the performance of EWIs and found that it reduces how clearly we can resolve the change in response time
(Appendix A). This effect may account for the fact that EWIs do not precisely reach a value of 1 at the bifurcation point, but confirming this would require further testing.
There are several important caveats to the use of EWIs as presented here. Firstly, and as explained above, the tipping point identified is that of the transient system not in a steady state. Although
the transient behaviour is arguably of greater societal relevance and an ice sheet is unlikely to ever truly be in a steady state, this is an important distinction to make. Secondly, the predictive
power of this method decreases as the distance to tipping increases and must eventually break down altogether. This effect can be clearly seen in Fig. 7 as Kendall's τ coefficient decreases with
increasing window length. Thirdly, there is a risk of so-called “false alarms” and “missed alarms” (Lenton, 2011). False alarms, whereby a positive trend in an indicator that is incorrectly
interpreted as a tipping point being imminent, can occur for a wide variety of reasons. First and foremost, interpreting EWIs requires robust statistical analysis and judicious data processing to
ensure that the response time being measured is that of the critical mode (Lenton, 2011). It is possible that rising auto-correlation is a result of other processes, and using more than one indicator
together with changes in variance can help mitigate this risk (Ditlevsen and Johnsen, 2010). It is also possible for a tipping point to be crossed with no apparent warning i.e. missed alarms. This
could happen if the internal variability in a system is high so that it changes state before a bifurcation point is reached or similarly if the forcing is too sudden. This last point is particularly
pertinent, since we intentionally perturb our model slowly and do not explore how a change in forcing rate affects the performance of our chosen EWIs. Increasing the forcing rate might present
further difficulties in identifying tipping points by leading to multiple tipping points being crossed coincidentally. Since in our methodology the control parameter is not held constant after a
tipping point is crossed, then if that parameter changes sufficiently during the time it takes for one tipping event to conclude, it might reach a second threshold while the first event is still
underway, disguising the fact that two distinct tipping points have been crossed. Changing the control parameter very slowly alleviates this issue, since it will only have altered slightly during the
time it takes for a tipping event to happen. This issue, along with the related issue of cascading tipping points, is one that we try to avoid in our experiments to simplify the analysis but is known
to influence EWI performance (Dakos et al., 2015; Brock and Carpenter, 2010). Despite our use of a very slow forcing rate, it is possible that more than three tipping points exist in our model
configuration. Finding all possible tipping points would necessitate infinitesimally small changes in the control parameter, in either the steady-state or the transient simulations, greatly
increasing computational cost but with little benefit in terms of detecting tipping events that constitute substantial mass loss.
In this paper we have presented an application of EWIs to model output to anticipate tipping points. This is a useful approach in and of itself, since it could be used in model studies to detect
bifurcations in the system with minimal computational expense or to check whether a model might be on a trajectory to cross a tipping point at some point in time beyond the simulation. Alternatively,
it may be possible to use this method on observational data, palaeo-records or some combination thereof. This raises the question of what data might qualify as useful for the application of EWIs,
which can be broken down further into (1) the type of data needed and (2) the length of record necessary. As mentioned previously, ice volume or related measures of an ice sheet's size do not show
sufficient variability for information on the recovery time to be extracted. Ice speed however can change significantly over very short timescales; for example many ice streams show large variability
over timescales as short as tidal periods (Anandakrishnan and Alley, 1997; Gudmundsson, 2006; Minchew et al., 2017). Ice flux was chosen in this study since it is closely related to the MISI
mechanism and because flux is proportional to velocity, but it is possible that other metrics related to ice velocity might also exhibit critical slowing down in a similar way. With regards to record
length, we find in this study that early warning of tipping points becomes less reliable (with a low or even negative Kendall's τ coefficient) for a moving-window size shorter than 200–300 years.
However, this does not mean that this represents the minimum window size in general and is likely sensitive to a number of the choices in our methodology. For example, this value is likely to be
sensitive to the rate of forcing applied to the system. In the limiting case of a forcing rate approaching zero, the necessary window length must increase since EWIs are only expected to work
relatively close to the tipping point. Both of these points require further study in order to establish suitable datasets for prediction of MISI onset.
Conducting quasi-steady numerical experiments, whereby the underside of the PIG ice shelf is forced with a slowly increasing ocean-induced melt, we have established the existence of at least three
distinct tipping points. Crossing each tipping point initiates periods of irreversible and self-sustained retreat of the grounding line (MISI) with significant contributions to global sea-level rise.
The tipping points are identified through critical slowing down, a general behavioural characteristic of non-linear systems as they approach a tipping point. EWIs have been successfully applied to
detect critical slowing down in other complex systems. We show here that they robustly detect the onset of the marine ice sheet instability in the simulations of the realistic PIG configuration which
is promising for application of early warning to further cryospheric systems and beyond. While the possibility of PIG undergoing unstable retreat has been raised and discussed previously, this is to
our knowledge the first time the stability regime of PIG has been mapped out in this fashion. The first and second tipping events are relatively small and could be missed without careful analysis of
model results but nevertheless are important in that they lead to considerable sea-level rise and would require a large reversal in ocean conditions for recovery. The third and final tipping point is
crossed with an increase in sub-shelf melt rates equivalent to a +1.2^∘C increase in ocean temperatures from initial conditions and leads to a complete collapse of PIG. Long-term warming and
shoaling trends in Circumpolar Deep Water (Holland et al., 2019), in combination with changing wind patterns in the Amundsen Sea (Turner et al., 2017), can expose the PIG ice shelf to warmer waters
for longer periods of time and make temperature changes of this magnitude increasingly likely.
Appendix A:Flow line experiments
MISI has been a major focus of modelling efforts within the glaciological community in recent years. In an effort to assess how ice-flow models capture this behaviour, a model inter-comparison
experiment was performed to calculate the hysteresis loop of advance and retreat of a marine ice sheet on a retrograde slope, known as MISMIP experiment 3 (referred to as EXP 3 hereafter; Pattyn et
al., 2012). As a first step to establishing whether critical slowing down can be observed prior to MISI, we undertook a slightly modified version of this experiment using the Úa ice-flow model
(Gudmundsson, 2012, 2013; see Methods). In our modified experiment, the marine ice sheet is forced towards tipping points through step perturbations in the control parameter as before but with
smaller steps and the additional constraint that the model must be in a steady state after each perturbation before moving on to the next. In this experiment the chosen control parameter is the ice
rate factor, a parameter linked to ice viscosity and temperature.
Following each perturbation in the ice rate factor, we analyse the e-folding relaxation time (T[R]) of the state variable (in this case, grounding-line position) to directly extract the recovery
time of the model as it approaches each tipping point (both advance and retreat). Theory predicts that T[R]→∞ close to a tipping point and that the point at which ${T}_{\mathrm{R}}^{-\mathrm{2}}$ (as
plotted versus the control parameter) reaches 0 thus identifies the critical value of the control parameter, beyond which a tipping point is crossed (Wissel, 1984). We show this plot for both the
advance and retreat scenarios of EXP 3 in Fig. A1. In both cases the relaxation time increases (${T}_{\mathrm{R}}^{-\mathrm{2}}$ decreases) as predicted by theory, even far from the tipping point. A
linear fit through the last six perturbations yields a good agreement with theory and accurately predicts the critical value of the control parameter when compared to the analytical solution (red
arrows in Fig. A1) given by Schoof (2007). Critical slowing down still occurs outside of this range (equivalent to a change in ice temperature of >5^∘C), but using these more distant points to
forecast the tipping point would yield a less accurate prediction. These results therefore provide some insight into how far from the tipping point we can expect the predicted linear response.
One major simplification in this idealised experiment is the bed geometry, which is synthetic and arguably unrealistically smooth. To test whether the addition of bumpiness to the bed affects how
accurately the critical value of the control parameter can be predicted, we conducted further experiments in which the bed was made successively less smooth. One simple but flexible way to generate
the desired roughness is to add Perlin noise to the bed. Perlin noise is a commonly used method in terrain generation that adds noise at a number of levels with successively smaller wavelengths and
amplitudes. The number of levels is denoted by the octave; the rate at which each octave changes frequency is the lacunarity, and the rate at which each octave changes amplitude is the persistence.
We made the common choice of a lacunarity greater than 1 and a persistence less than one, meaning that each octave adds noise of a higher frequency and lower amplitude. For a starting octave
amplitude of 25m the difference between the analytical solution and linear fit is less than 1%, but this grows to ∼5% with an amplitude of 50m (note the change in height from peak to trough in
the retrograde region of the smooth bed is ∼120m). This suggests that more realistic bed geometries with increased roughness might make the task of predicting tipping points more challenging than
it is in this simplified case.
The source code of the Úa ice-flow model is available from https://github.com/ghilmarg/UaSource (last access: 30 June 2020, https://doi.org/10.5281/zenodo.3706624, Gudmundsson, 2020), and raw model
output is available from the authors upon request.
SHRR and RR conceived the study, SHRR conducted the modelling experiments; JFD contributed to the statistical analysis and surrogate time series; JDR provided an initial model setup. SHRR and RR
wrote the manuscript with contributions from all authors.
The authors declare that they have no conflict of interest.
We are grateful to the anonymous reviewer, Alexander Robel and the editor Pippa Whitehouse for their comments, which have greatly improved our paper.
This research has been supported by the Natural Environment Research Council (grant nos. NE/L013770/1 and NE/S006745/1), the Deutsche Forschungsgemeinschaft (grant no. WI4556/3-1) and Horizon 2020
(TiPACCs (grant no. 820575)).
This paper was edited by Pippa Whitehouse and reviewed by Alexander Robel and one anonymous referee.
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doi.org/10.1073/pnas.1904822116, 2019.
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bathymetry, Earth Syst. Sci. Data, 8, 543–557, https://doi.org/10.5194/essd-8-543-2016, 2016.
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Cryosphere, 12, 1551–1562, https://doi.org/10.5194/tc-12-1551-2018, 2018.
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Wissel, C.: A universal law of The characteristic return time near thresholds, Oecologia, 65, 101–107, https://doi.org/10.1007/BF00384470, 1984. | {"url":"https://tc.copernicus.org/articles/15/1501/2021/","timestamp":"2024-11-08T20:27:00Z","content_type":"text/html","content_length":"282602","record_id":"<urn:uuid:15bde89c-ad9a-4031-bd12-1563dc37d0b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00398.warc.gz"} |
Counting Sort Implementation in Java - Daily Java Concept
Counting Sort Implementation in Java
Counting Sort is a non-comparison-based sorting algorithm that efficiently sorts integers within a specified range. Unlike traditional comparison-based sorting algorithms like Quicksort or Merge Sort
, Counting Sort Implementation in Java exploits the specific characteristics of the input elements to achieve linear time complexity.
Here’s a brief overview of how Counting Sort works:
1. Counting Occurrences: The algorithm begins by counting the number of occurrences of each unique element in the input array.
2. Creating a Counting Array: Based on the counts, it creates an auxiliary array, often called the “counting array,” to store the cumulative count of elements up to each unique value.
3. Cumulative Counting Array: The counting array is then modified to represent the cumulative count of elements. This step helps determine the correct position of each element in the sorted array.
4. Sorted Array Construction: Using the information from the counting array, a sorted array is constructed by placing elements in their correct positions based on their cumulative counts.
5. Copying Back to Original Array: Finally, the sorted array is copied back to the original array, completing the sorting process.
Counting Sort is particularly efficient when the range of possible values in the input array is small compared to the number of elements. Its linear time complexity, O(n + k), where n is the number
of elements and k is the range, makes it well-suited for certain scenarios, such as sorting integers within a limited range.
Java Implementation
Counting Sort operates by counting the number of occurrences of each element in the input array and using that information to construct a sorted array. It is especially efficient when dealing with a
relatively small range of integer values.
Java Program – Counting Sort
import java.util.Arrays;
public class CountingSort {
public static void main(String[] args) {
int[] array = {4, 2, 2, 8, 3, 3, 1};
System.out.println("Original Array: " + Arrays.toString(array));
// Call the sort method to perform Counting Sort
System.out.println("Sorted Array: " + Arrays.toString(array));
public static void sort(int[] array) {
int n = array.length;
// Find the maximum element to determine the counting array size
int max = Arrays.stream(array).max().orElse(0);
// Create a counting array to store the count of each element
int[] count = new int[max + 1];
// Populate the counting array with element counts
for (int i : array) {
// Update the counting array to represent the cumulative count
for (int i = 1; i <= max; i++) {
count[i] += count[i - 1];
// Create the sorted array based on the counting array information
int[] sortedArray = new int[n];
for (int i = n - 1; i >= 0; i--) {
sortedArray[count[array[i]] - 1] = array[i];
// Copy the sorted array back to the original array
System.arraycopy(sortedArray, 0, array, 0, n);
Understanding the Logic
Counting Array:
• Create a counting array to store the count of each element.
• Populate the counting array with the occurrences of each element in the input array.
Cumulative Count:
• Update the counting array to represent the cumulative count of elements.
• This step helps determine the correct positions of elements in the sorted array.
Sorted Array Construction:
• Create a sorted array based on the counting array information.
• Populate the sorted array by decrementing the count for each element.
Original Array Update:
• Copy the sorted array back to the original array, completing the sorting process.
Example Usage
Consider an example array: {4, 2, 2, 8, 3, 3, 1}.
Counting Array:
• The counting array would represent the occurrences of each element: {1, 2, 2, 0, 1, 0, 0, 1}.
Cumulative Count:
• After updating, the counting array becomes {1, 3, 5, 5, 6, 6, 6, 7}.
Sorted Array:
• The sorted array is constructed using the counting array information: {1, 2, 2, 3, 3, 4, 8}.
Output – Counting Sort
Original Array: [4, 2, 2, 8, 3, 3, 1]
Sorted Array: [1, 2, 2, 3, 3, 4, 8]
Complexity Best Case Average Case Worst Case
Time Complexity O(n + k) O(n + k) O(n + k)
Space Complexity O(k) O(k) O(k)
Counting Sort is a powerful algorithm for sorting integers within a specific range efficiently. Understanding its implementation in Java provides valuable insights into its application and
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Mathematics Methods Yr 11
Watch Activity Tutorial
If A = 1, B = 2 ... your name could be converted into numbers and described as a function, your Personal Polynomial. What does your polynomial look like? Students find their own personal polynomial
and then study its properties. They set up and use simultaneous equations to find their polynomial, the bisection method to locate x-axis intercepts and transformations to compare others. Palindromic
names create polynomials with an axis of symmetry. Is it possible for two names to generate the same polynomial, Alex(x) compared with Alexander(x)? A guided exploration task that will run over
several lessons. | {"url":"https://education.ti.com/en-au/seniornspiredcurriculum/aus-nz/qld-84/methods-yr-11","timestamp":"2024-11-11T11:27:59Z","content_type":"text/html","content_length":"55384","record_id":"<urn:uuid:e359c809-bb5c-45de-a782-9ff4ad9df226>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00546.warc.gz"} |
4.6 Representing Relationships Among Variables
Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
□ 4.6 Representing Relationships Among Variables
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Digging Deeper into Group Models
• segmentChapter 9 - Models with a Quantitative Explanatory Variable
• segmentPART III: EVALUATING MODELS
• segmentChapter 10 - The Logic of Inference
• segmentChapter 11 - Model Comparison with F
• segmentChapter 12 - Parameter Estimation and Confidence Intervals
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
list High School / Advanced Statistics and Data Science I (ABC)
4.6 Representing Relationships Among Variables
We’ve learned a lot of ways to visualize the relationship between an outcome variable and an explanatory variable, using color and fill, using gf_facet_grid(), using gf_point(), gf_jitter(), and
There are many ways to represent the relationship between the outcome and explanatory variables. We’ve learned how to make visualizations and write code that represents this relationship. We will
introduce even more ways to represent these relationships. In this section, we will show you two: path diagrams, and word equations.
Building off the example we are now very familiar with, here’s a way to represent the relationship between Sex and Thumb:
In path diagram convention, the explanatory variable “points to” the outcome variable. The direction of the arrow identifies Thumb as the outcome variable and Sex as the explanatory variable. So, we
can read this diagram like this: “The variation in Thumb length is explained, at least in part, by variation in Sex.” It’s good to practice making diagrams like this. The same relationship is
represented in the R code Thumb ~ Sex.
These path diagrams can be used to represent a relationship we have found in our data, or to one we hypothesize is true but that we have not yet investigated. Assuming we do have data to support it,
we still can read it in two senses: as a relationship that exists in our data, or as a relationship that we believe exists in the world beyond our data, in the DGP.
A second way of representing relationships is with word equations:
Thumb = Sex
We can read this equation in the same way as a path diagram: “Variation in Thumb length is explained by variation in Sex.” A word equation is another way of representing relationships, real or
hypothesized, in our data or in the DGP. By convention, the outcome variable, Thumb, is written on the left of the equal sign and the explanatory variable, Sex, is written to the right.
Note that this kind of equation is not the same as a mathematical equation. It doesn’t mean, for example, that thumb length and sex are the same thing, or that they represent the same quantity. It
also doesn’t mean that the variation in thumb length is equal to the variation in sex. It simply means that some of the variation in thumb length is explained by variation in sex.
You may also notice that the equation is, in a sense, not true. Variation in thumb length is not fully explained by variation in sex. In the analyses we did above, we saw that even though some of the
variation in thumb length appeared to be explained by sex, there is still plenty of overlap in the distributions of thumb length for males and females.
Another way of putting this is: just knowing the sex of a person will not enable you to perfectly predict their thumb length. There is a lot of variability within each group. So, obviously, sex is
not the only variable that could explain variation in thumb length.
To represent this idea more clearly—and it is an important idea, as you will see later—we can add something to our word equation:
Thumb = Sex + Other Stuff
We can read this as: “Variation in thumb length is explained by variation in sex plus variation in other stuff.” In the next section we will discuss what this “other stuff” might be. But for now,
it’s useful to think of the total variation in our outcome variable as the sum of variation due to an explanatory variable, plus variation due to other stuff.
We may also refer to these word equations as models. We will talk more about why we call them models later, but it’s helpful to start thinking of them as models. | {"url":"https://coursekata.org/preview/book/f84ca125-b1d7-4288-9263-7995615e6ead/lesson/6/5","timestamp":"2024-11-05T12:34:11Z","content_type":"text/html","content_length":"93268","record_id":"<urn:uuid:421988e4-196c-4087-9b54-c9adf1a9fd80>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00295.warc.gz"} |
Description 1 online resource (116 pages)
Contents Contents; Preface to the first edition; Preface to the second edition; 1 Strategies in problem solving; 2 Examples in number theory; 3 Examples in algebra and analysis; 4 Euclidean
geometry; 5 Analytic geometry; 6 Sundry examples; References; Index
Summary Authored by a leading name in mathematics, this engaging and clearly presented text leads the reader through the tactics involved in solving mathematical problems at the Mathematical
Olympiad level. With numerous exercises and assuming only basic mathematics, this text is ideal for students of 14 years and above in pure mathematics
Notes Description based upon print version of record
Print version record
Subject Mathematical analysis
Solving problems
Form Electronic book
ISBN 9780191568695 (electronic bk.)
0191568694 (electronic bk.) | {"url":"https://library.deakin.edu.au/search~S1?/dSolvents+--+Handbooks%2C+manuals%2C+etc./dsolvents+handbooks+manuals+etc/-3%2C1%2C1%2CB/frameset&FF=dsolving+problems&0%2C%2C0","timestamp":"2024-11-09T17:24:45Z","content_type":"text/html","content_length":"20678","record_id":"<urn:uuid:25aeb88b-0a70-4848-9253-a7e43d17d5a9>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00719.warc.gz"} |
Using numpy hstack() to horizontally stack arrays - Data Science Parichay
In this tutorial, we will look at how to use the numpy hstack method to horizontally stack (or concat) numpy arrays with the help of some examples.
How to concatenate numpy arrays horizontally?
You can use the numpy hstack() function to stack numpy arrays horizontally. It concatenates the arrays in sequence horizontally (column-wise). The following is the syntax.
import numpy as np
# tup is a tuple of arrays to be concatenated, e.g. (ar1, ar2, ..)
ar_h = np.hstack(tup)
It takes the sequence of arrays to be concatenated as a parameter and returns a numpy array resulting from stacking the given arrays. This function is similar to the numpy vstack() function which is
also used to concatenate arrays but it stacks them vertically.
Let’s look at some examples of how to use the numpy hstack() function.
1. Horizontally stack two 1D arrays
Let’s stack two one-dimensional arrays together horizontally.
import numpy as np
# create two 1d arrays
ar1 = np.array([1, 2, 3])
ar2 = np.array([4, 5, 6])
# hstack the arrays
ar_h = np.hstack((ar1, ar2))
# display the concatenated array
[1 2 3 4 5 6]
Here, we created two 1D arrays of length 3 and then horizontally stacked them with the hstack() function. The resulting array is also one-dimensional since we are concatenating them horizontally.
You can also stack more than two arrays at once with the numpy hstack() function. Just pass the arrays to be stacked as a tuple. For example, let’s stack three 1D arrays horizontally at once.
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# create two 1d arrays
ar1 = np.array([1, 2, 3])
ar2 = np.array([4, 5, 6])
ar3 = np.array([7, 8, 9])
# hstack the arrays
ar_h = np.hstack((ar1, ar2, ar3))
# display the concatenated array
[1 2 3 4 5 6 7 8 9]
Here we concatenated three arrays of length 3 horizontally. The resulting array is one-dimensional and has length 9. Similarly, you can stack multiple arrays, just pass them in the order you want as
a sequence.
2. Horizontally stack a 1D and a 2D array
Now let’s try to stack a 1D array with a 2D array horizontally.
# create a 1d array
ar1 = np.array([1, 2])
# create a 2d array
ar2 = np.array([[0, 0, 0],
[1, 1, 1]])
# hstack the arrays
ar_h = np.hstack((ar1, ar2))
# display the concatenated array
ValueError Traceback (most recent call last)
<ipython-input-6-5f5f4f1cdb7b> in <module>
7 # hstack the arrays
----> 8 ar_h = np.hstack((ar1, ar2))
9 # display the concatenated array
10 print(ar_h)
<__array_function__ internals> in hstack(*args, **kwargs)
~\anaconda3\envs\dsp\lib\site-packages\numpy\core\shape_base.py in hstack(tup)
341 # As a special case, dimension 0 of 1-dimensional arrays is "horizontal"
342 if arrs and arrs[0].ndim == 1:
--> 343 return _nx.concatenate(arrs, 0)
344 else:
345 return _nx.concatenate(arrs, 1)
<__array_function__ internals> in concatenate(*args, **kwargs)
ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 1 dimension(s) and the array at index 1 has 2 dimension(s)
We get an error. This is because the dimensions of the arrays are such that you cannot stack them together horizontally.
If you want to stack the two arrays horizontally, they need to have the same number of rows.
# create an array of shape (2, 1)
ar1 = np.array([[1], [2]])
# create a 2d array
ar2 = np.array([[0, 0, 0],
[1, 1, 1]])
# hstack the arrays
ar_h = np.hstack((ar1, ar2))
# display the concatenated array
[[1 0 0 0]
[2 1 1 1]]
3. Horizontally stack two 2D arrays
You can also horizontally stack two 2D arrays together in a similar way.
# create a 2d array
ar1 = np.array([[1, 2],
[3, 4],
[5, 6]])
# create a 2d array
ar2 = np.array([[0, 0],
[1, 1],
[2, 2]])
# hstack the arrays
ar_h = np.hstack((ar1, ar2))
# display the concatenated array
[[1 2 0 0]
[3 4 1 1]
[5 6 2 2]]
Here we stacked two 2d arrays of shape (3, 2) horizontally resulting in an array of shape (3, 4).
For more on the numpy hstack() function, refer to its documentation.
With this, we come to the end of this tutorial. The code examples and results presented in this tutorial have been implemented in a Jupyter Notebook with a python (version 3.8.3) kernel having numpy
version 1.18.5
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What is The Basic Measure of Electricity?
Everything worth measuring has a unit of measurement associated with it. In the US, we use inches and feet to measure how tall something is, pounds and ounces to measure how much something weighs and
Fahrenheit to measure the temperature of something. But what about electricity? What units of measurement or used to talk about electricity?
Before we talk about how to measure electricity, we first need to understand what it is. At a base level, electricity is the movement of electrons. Your computer, your lights, your television, your
fridge, etc all operate using the same basic power source - the movement of electrons.
Whenever we talk about the power of electricity, we’re actually talking about the charge created by moving electrons.
The basic units of measurement for electricity are current, voltage and resistance.
Current (I)
Current, measured in amps, is the rate at which charge is flowing - how fast the electrons are moving. Amps, or amperes, are the basic unit for measuring electricity and measure how many electrons
move past a point every second. One amp is equal to 6.25 x 1018 electrons per second.
Voltage (V)
Voltage, measured in volts, is the difference in charge between two points. Put simply, it is the difference in the concentration of electrons between two points.
Resistance (R)
Resistance is a material’s tendency to resist the flow of charge (current). It is measured in ohms.
The Water Pipe Analogy
Now, let’s put these ideas to work. The most common analogy used to understand these ideas is water in a pipe. When thinking about how fast water can move through a pipe, there are three main
components that need to be considered: water pressure, flow rate and pipe size. To combine these two ideas, voltage is equivalent to water pressure, current is the flow rate and resistance is the
pipe size.
So, when we talk about these values, we’re really describing the movement of charge, and thus, the behavior of electrons. A circuit is a closed loop that allows charge to move from one place to
another. Components in the circuit allow us to control this charge and use it to do work.
Ohm’s Law
Ohm’s Law is a basic, and highly important, equation that is used to determine how current, voltage and resistance interact. It says that the current is equal to the voltage divided by the resistance
or I = V/R. Ohm’s Law can be used to accurately describe the conductivity of most electrically conductive materials. As long as you know two of the values, it is possible to determine the third. The
three variations of this equation are: I = V/R, V=IR, R=V/I
There is another term you may have heard used in relation to electricity: Watts. Watts measure the rate at which energy is used or transferred and are not just used for electronics. A Watt is the
basic unit of electric, mechanical, or thermal power. One Watt is equal to one amp under the pressure of one volt. (Watt = Amps x Volts)
For a more in-depth look at voltage, current, resistance, & Ohm's Law, check out this post. | {"url":"https://blog.sparkfuneducation.com/what-is-the-basic-measure-of-electricity","timestamp":"2024-11-12T12:32:55Z","content_type":"text/html","content_length":"63077","record_id":"<urn:uuid:db3f4b47-571c-4c0b-89fe-beb38d4f904e>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00081.warc.gz"} |
ST952-15 An Introduction to Statistical Practice
Introductory description
This module runs in Term 1 and is core for students on an MSc in Statistics. It is not available for undergraduate students.
Module aims
Students on the Diploma and MSc often had diverse academic backgrounds. This course complements ST903 Statistical Methods in giving a common starting point to the programme, with an emphasis on
learning skills in practical statistics.
Outline syllabus
This is an indicative module outline only to give an indication of the sort of topics that may be covered. Actual sessions held may differ.
• Exploratory data analysis (numerical and graphical measures)
• A hands-on introduction to R, exercises to learn basics of R.
• Simpson's paradox, Regression to the mean, Correlation vs causation
• Simple linear regression; Correlation coefficient, SD line, Regression Line
• Multiple linear regression; Diagnostic plots, Hypothesis testing, ANOVA
• Structured Data (coming from simple experimental designs)
• Generalised Linear Models; Poisson and Binomial data
• Resampling methods such as the BootStrap
Learning outcomes
By the end of the module, students should be able to:
• Computational skills: Basic use of R, search for commands in help files and understand them, dealing with data (collecting, typing in, downloading, storing, sharing etc.)
• Descriptive statistics and Explorative Data Analysis (EDA): Data structures, appropriateness of data (relevance to the scientific question(S), completeness, quality etc.), representation of data
(choice of the form, optimal layout, misleading representation etc.), strategies to explore certain aspects of the data
• Modelling and analysis: choice of model, discussion of model assumptions, fitting models, validation and comparison of models, prediction, sensitivity analysis (in respect to assumptions and
sample data), simulation
• Context: translating scientific queries into statistical questions, classification of investigations, drawing scientific conclusions from statistical analysis
• Communication skills: listening, asking questions, explaining analysis, approach and delivering results to a non-statistician, writing a report
Indicative reading list
View reading list on Talis Aspire
Subject specific skills
-Data structures, appropriateness of data (relevance to the scientific question(s), completeness, quality etc.), representation of data (choice of the form, optimal layout, misleading representation
etc.), strategies to explore certain aspects of the data
-choice of model, discussion of model assumptions, fitting models, validation and comparison of models, prediction, sensitivity analysis (in respect to assumptions and sample data), simulation
-translating scientific queries into statistical questions, classification of investigations, drawing scientific conclusions from statistical analysis
Transferable skills
-Basic use of R, search for commands in help files and understand them, dealing with data (collecting, typing in, downloading, storing, sharing etc.)
-listening, asking questions, explaining analysis approach and delivering results to a non-statistician, writing a report.
Study time
Type Required
Lectures 20 sessions of 1 hour (13%)
Practical classes 10 sessions of 2 hours (13%)
Private study 36 hours (24%)
Assessment 74 hours (49%)
Total 150 hours
Private study description
Weekly revision of lecture notes and materials, wider reading, practice exercises, learning to code in R and preparing for examination.
No further costs have been identified for this module.
You must pass all assessment components to pass the module.
Students can register for this module without taking any assessment.
Assessment group C4
Weighting Study time Eligible for self-certification
Assignment 1 & 2 50% 72 hours No
You will work as part of a small group to carry out analysis of a dataset and provide a written report in response to a set of prompt questions.
500 words is equivalent to one page of text, diagrams, formula or equations.
You will work as part of a small group to carry out analysis of a dataset and provide a written report in response to a set of prompt questions.
500 words is equivalent to one page of text, diagrams, formula or equations.
In-person Examination 50% 2 hours No
The examination paper will contain four questions, of which the best marks of THREE questions will be used to calculate your grade.
• Answerbook Pink (12 page)
• Students may use a calculator
• Cambridge Statistical Tables (blue)
Feedback on assessment
Feedback for reports will be available within 20 working days.
Cohort level feedback and solutions will be provided for the examination.
Post-requisite modules
If you pass this module, you can take:
• ST409-15 Medical Statistics with Advanced Topics
• ST955-60 Dissertation
This module is Core for:
• Year 1 of TSTA-G4P1 Postgraduate Taught Statistics | {"url":"https://courses.warwick.ac.uk/modules/2024/ST952-15","timestamp":"2024-11-09T23:41:52Z","content_type":"text/html","content_length":"18167","record_id":"<urn:uuid:03044dda-2015-4814-8228-cbc3d5375b38>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00132.warc.gz"} |
The values to be attached to the probability tables, - Writer Scholar
The values to be attached to the probability tables,
This document will cover two aspects:
1. The values to be attached to the probability tables, and
2. The formulae to be used to compute the probabilities that are required.
Probability table specification In total there are 16 nodes for the Car diagnosis Bayesian network. This means that 16 probability tables are
required. These tables must be assigned values according to the specifications below.
1. Battery age table
ba _y 0.2 ba_n 0.8
2. Alternator broken table
ab_y 0.1 ab_n 0.9
3. Fanbelt broken table
fb_y 0.3 fb_n 0.7
4. Battery dead table
ba_y_bd_y 0.7 ba_y_bd_n 0.3 ba_n_bd_y 0.3 ba_n_bd_n 0.7
Note that this table represents conditional probabilities. Thus for example, the row ba_y_bd_y=0.2 should be
interpreted as the Pr(battery being dead|battery is aged) is 0.2. All other tables below which have more than two
rows also happen to represent conditional probabilities.
5. No charging table
ab_y_fb_y_nc_y 0.75
ab_y_fb_n_nc_y 0.4
ab_n_fb_y_nc_y 0.6 ab_n_fb_n_nc_y 0.1 ab_y_fb_y_nc=n 0.25 ab_y_fb_n_nc=n 0.6 ab_n_fb_y_nc=n 0.4 ab_n_fb_n_nc=n 0.9
6. Battery meter table
bd_y_bm_y 0.9 bd_y_bm_n 0.1 bd_n_bm_y 0.1 bd_n_bm_n 0.9
7. Battery flat table
bd_y_nc_y_bf_y 0.95
bd_y_nc_n_bf_y 0.85
bd_n_nc_y_bf_y 0.8 bd_n_nc_n_bf_y 0.1 bd_y_nc_y_bf_n 0.05 bd_y_nc_n_bf_n 0.15 bd_n_nc_y_bf_n 0.2 bd_n_nc_n_bf_n 0.9
8. No oil table
no _y 0.05 no_n 0.95
9. No gas table
ng _y 0.05 ng_n 0.95
10. Fuel line blocked table
fb _y 0.1 fb_n 0.9
11. Starter broken table
sb _y 0.1 sb_n 0.9
12. Lights table
l_y_bf_y 0.9 l_n_bf_y 0.3 l_y_bf_n 0.1 l_n_bf_n 0.7
13. Oil lights table
bf_y_no_y_ol_y 0.9
bf_y_no_n_ol_y 0.7
bf_n_no_y_ol_y 0.8 bf_n_no_n_ol_y 0.1 bf_y_no_y_ol_n 0.1 bf_y_no_n_ol_n 0.3 bf_n_no_y_ol_n 0.2 bf_n_no_n_ol_n 0.9
14. Gas gauge table
bf_y_ng_y_gg_y 0.95
bf_y_ng_n_gg_y 0.4
bf_n_ng_y_gg_y 0.7 bf_n_ng_n_gg_y 0.1 bf_y_ng_y_gg_n 0.05 bf_y_ng_n_gg_n 0.6 bf_n_ng_y_gg_n 0.3 bf_n_ng_n_gg_n 0.9
15. Car won’t start table
This table has 64 rows. There are 3 cases to consider:
1. For every combination of bf, no, ng, fb, sb with at least one of these variables taking the value y, the
probability is 0.9 for the cs_n outcome.
2. For the case when all 5 variables bf, no, ng, fb, sb take the value n with cs_n. In this case the probability is 0.1
3. The remaining case is when cs_N. The probabilities are now defined as the complement of the probabilities
of the first 32 rows. That is if the probability is p for the first row then it is (1-p) for the 33rd row, if it is q for
row 2 then it is (1-q) for row 34 and so on.
Note this table must be encoded in the graph with 64 rows and each row should have a probability as specified
16. Dipstick low table
no_y_dl_y 0.95 no_n_dl_y 0.3 no_y_dl_n 0.05 no_n_d_n 0.7
Formulae for computation of probabilities In addition to the discussion below you are strongly advised to refer to the class notes on Bayesian learning.
Let us illustrate the computation of the probabilities by taking R2 as an example.
For R2 you are asked to compute P(-cs, +ab, +fb) – this is the joint probability that the car does not start whenever
both the alternator and fan belt are not functioning at the same time.
To understand how this is done, let us first look at a simpler situation.
P(+c, +a, +b) = P(+c|+a, +b)*P(+a)*P(+b)
Using this as a guide we can now work out P(-cs, +ab, +fb)
You didn't find what you were looking for? Upload your specific requirements now and relax as your preferred tutor delivers a top quality customized paper
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A1 Quadratic functions and equations | Five Minute Math
top of page
Quadratic functions and equations. The student applies the mathematical process standards when using properties of quadratic functions to write and represent in multiple ways, with and without
technology, quadratic equations. The student is expected to:
(A) determine the domain and range of quadratic functions and represent the domain and range using inequalities;
(B) write equations of quadratic functions given the vertex and another point on the graph, write the equation in vertex form (f(x) = a(x - h)2+ k), and rewrite the equation from vertex form to
standard form (f(x) = ax2+ bx + c); and
(C) write quadratic functions when given real solutions and graphs of their related equations.
Quadratic functions and equations. The student applies the mathematical process standards when using graphs of quadratic functions and their related transformations to represent in multiple ways and
determine, with and without technology, the solutions to equations. The student is expected to:
(A) graph quadratic functions on the coordinate plane and use the graph to identify key attributes, if possible, including x-intercept, y-intercept, zeros, maximum value, minimum values, vertex, and
the equation of the axis of symmetry;
(B) describe the relationship between the linear factors of quadratic expressions and the zeros of their associated quadratic functions; and
(C) determine the effects on the graph of the parent function f(x) = x2 when f(x) is replaced by af(x), f(x) + d, f(x - c), f(bx) for specific values of a, b, c, and d.
Quadratic functions and equations. The student applies the mathematical process standards to solve, with and without technology, quadratic equations and evaluate the reasonableness of their
solutions. The student formulates statistical relationships and evaluates their reasonableness based on real-world data. The student is expected to:
(A) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula; and
(B) write, using technology, quadratic functions that provide a reasonable fit to data to estimate solutions and make predictions for real-world problems.
A.6 Quadratic functions and equations
A.7 Quadratic Functions and Equations
A.8 Quadratic Functions and Equations
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AUMC Managerial Finance the Stock of Matrix Computing & Cost of Capital Worksheet - Research Wire
Attached Files:
Question 1
Start with the partial model in the attached.
The stock of Matrix Computing sells for $65, and last year s dividend was $2.53. Security analysts are projecting that the common dividend will grow at a rate of 9% a year. A flotation cost of 12%
would be required to issue new common stock. Matrix s preferred stock sells for $42.00, pays a dividend of $3.32 per share, and new preferred stock could be sold with a flotation cost of 10%. The
firm has outstanding bonds with 25 years to maturity, a 15% annual coupon rate, semiannual payments, $1,000 par value. The bonds are trading at $1,271.59. The tax rate is 20%. The market risk premium
is 5.5%, the risk-free rate is 7.0%, and Matrix s beta is 1.2. In its cost-of-capital calculations, Matrix uses a target capital structure with 40% debt, 10% preferred stock, and 50% common equity.
a. Calculate the cost of each capital component in other words, the after-tax cost of debt, the cost of preferred stock (including flotation costs), and the cost of equity (ignoring flotation costs).
Use both the CAPM method and the dividend growth approach to find the cost of equity.
b. Calculate the cost of new stock using the dividend growth approach.
c. Assuming that Matrix will not issue new equity and will continue to use the same tar-get capital structure, what is the company s WACC
Question 2
Start with the partial model in the attached.
Pinto.com has developed a powerful new server that would be used for corporations Internet activities. It would cost $25 million at Year 0 to buy the equipment necessary to manufacture the server.
The project would require net working capital at the beginning of each year in an amount equal to 12% of the year s projected sales; for example, NWC[0] = 12%(Sales[1] ). The servers would sell for
$21,000 per unit, and Pinto believes that variable costs would amount to $15,000 per unit. After Year 1, the sales price and variable costs will increase at the inflation rate of 2.5%. The company s
nonvariable costs would be $1.5 million at Year 1 and would increase with inflation. The server project would have a life of 4 years. If the project is undertaken, it must be continued for the entire
4 years. Also, the project s returns are expected to be highly correlated with returns on the firm s other assets. The firm believes it could sell 2,000 units per year. The equipment would be
depreciated over a 5-year period, using MACRS rates. The estimated market value of the equipment at the end of the project s 4-year life is $1 million. Pinto.com s federal-plus-state tax rate is 20%.
Its cost of capital is 10% for average-risk projects, defined as projects with a coefficient of variation of NPV between 0.8 and 1.2. Low-risk projects are evaluated with an 8% project cost of
capital and high-risk projects at 13%.
1. Develop a spreadsheet model, and use it to find the project s NPV, IRR, and payback.
2. Now conduct a sensitivity analysis to determine the sensitivity of NPV to changes in the sales price, variable costs per unit, and number of units sold. Set these variables values at 10% and 20%
above and below their base-case values.
3. Now conduct a scenario analysis. Assume that there is a 25% probability that best-case conditions, with each of the variables discussed in Part b being 20% better than its base-case value, will
occur. There is a 25% probability of worst-case conditions, with the variables 20% worse than base, and a 50% probability of base-case conditions.
4. If the project appears to be more or less risky than an average project, find its risk-adjusted NPV, IRR, and payback.
5. On the basis of information in the problem, would you recommend the project should be accepted?
Question 3
You are a financial analyst for the Waffle Company. The director of capital budgeting has asked you to analyze two proposed capital investments, Projects A and B. Each project has a cost of $50,000,
and the cost of capital for each is 10%.
The projects expected net cash flows are as follows:
Expected Net Cash Flows
Year Project A Project B
0 ($50,000) ($50,000)
1 25,000 15,000
2 20,000 15,000
3 10,000 15,000
4 5,000 15,000
5 5,000 15,000
1. Calculate each project s payback period, net present value (NPV), internal rate of return (IRR), modified internal rate of return (MIRR), and profitability index (PI).
2. Which project will you select if your decision was based solely on the project s payback period?
3. Which project or projects should be accepted if they are independent?
4. Which project should be accepted if they are mutually exclusive?
5. d. How might a change in the cost of capital produce a conflict between the NPV and IRR rankings of these two projects? Would this conflict exist if r were 6%? (Hint: Plot the NPV profiles.)
Question 4
Marvin Industries must choose between an electric-powered and a coal-powered forklift machine for its factory. Because both machines perform the same function, the firm will choose only one. (They
are mutually exclusive investments.) The electric-powered machine will cost more, but it will be less expensive to operate; it will cost $102,000, whereas the coal-powered machine will cost $69,500.
The cost of capital that applies to both investments is 10%. The life for both types of machines is estimated to be 6 years, during which time the net cash flows for the electric-powered machine will
be $26,150 per year, and those for the coal-powered machine will be $20,000 per year. Annual net cash flows include depreciation expenses.
1. Calculate the NPV and IRR for each type of machine, and decide which to recommend
Submit your answers in a Word document. | {"url":"https://researchwire.blog/2024/03/aumc-managerial-finance-the-stock-of-matrix-computing-cost-of-capital-worksheet/","timestamp":"2024-11-08T08:55:51Z","content_type":"text/html","content_length":"163624","record_id":"<urn:uuid:121c5721-d1d3-4272-bac7-461495286086>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00224.warc.gz"} |
Research Guides: MAT392H5 Ideas of Mathematics (Winter 2022): Finding Resources
The Concise Oxford Dictionary of Mathematics by Christopher Clapham; James Nicholson
ISBN: 0199679592
Publication Date: 2014-07-22
Authoritative and reliable, this A-Z provides jargon-free definitions for even the most technical mathematical terms. With over 3,000 entries ranging from Achilles paradox to zero matrix, it covers
all commonly encountered terms and concepts from pure and applied mathematics and statistics,for example, linear algebra, optimisation, nonlinear equations, and differential equations. In addition,
there are entries on major mathematicians and on topics of more general interest, such as fractals, game theory, and chaos.Using graphs, diagrams, and charts to render definitions as comprehensible
as possible, entries are clear and accessible. Almost 200 new entries have been added to this edition, including terms such as arrow paradox, nested set, and symbolic logic. Useful appendices follow
the A-Z dictionary andinclude lists of Nobel Prize winners and Fields' medallists, Greek letters, formulae, and tables of inequalities, moments of inertia, Roman numerals, a geometry summary,
additional trigonometric values of special angles, and many more. This edition contains recommended web links, which areaccessible and kept up to date via the Dictionary of Mathematics companion
website.Fully revised and updated in line with curriculum and degree requirements, this dictionary is indispensable for students and teachers of mathematics, and for anyone encountering mathematics
in the workplace. | {"url":"https://guides.library.utoronto.ca/c.php?g=728685&p=5226946","timestamp":"2024-11-05T01:08:03Z","content_type":"text/html","content_length":"30626","record_id":"<urn:uuid:c19f3358-9d7b-4d88-bd2c-0662b346450d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00650.warc.gz"} |
32-bit Single-Precision Floating Point in Details - ByteScout (2024)
In modern days, programming languages tend to be as high-level as possible to make the programmer’s life a little bit easier. However, no matter how advanced programming language is, the code still
has to be converted down to the machine code, via compilation, interpretation, or even virtual machine such as JVM. Of course, at this stage, rules are different: CPU works with addresses and
registers without any classes, even «if» branches look like conditional jumps. One of the most important aspects of this execution is the arithmetic operation, and today we will be talking about one
of these «cornerstones»: floating-point numbers and how they may affect your code.
A brief introduction to the history
The need for processing large or small values was present since the very first days of computers: even first designs of Charles Babbage’s Analytical Engine sometimes included floating-point
arithmetic along with usual integer arithmetic. For a long time, the floating-point format was used primarily for scientific research, especially in physics, due to the large variety of data. It is
extremely convenient that distance between Earth and Sun can be expressed in the same amount of bits as the distance between hydrogen and oxygen atoms in water molecules with the same relative
precision and, even better, values of different magnitudes may be freely multiplied without large losses in precision.
Almost all the early implementations of floating-point numbers were software due to the complexity of the hardware implementations. Without this common standard, everybody had to come up with their
own formats: this is how Microsoft Binary Format and IBM Floating Point Architecture were born; the latter is still used in some fields such as weather forecasting, although it is extremely rare by
Intel 8087 coprocessor, announced in 1980, also used its own format called «x87». It was the first coprocessor specifically dedicated to floating-point arithmetic with aims to replace slow library
calls with the machine code. Then, based on x87 format, IEEE 754 was born as the first and successful attempt to create a universal standard for floating-point calculations. Soon, Intel started to
integrate IEEE 754 into their CPUs, and nowadays almost every system except some embedded ones supports the floating-point format.
Theory and experiments
In IEEE 754 single-precision binary floating-point format, 32 bits are split into 1-bit sign flag, 8-bit exponent flag, and 23-bit fraction part, in that order (bit sign is the leftmost bit). This
information should be enough for us to start some experiments! Let us see how number 1.0 looks like in this format using this simple C code:
union { float in; unsigned out;} converter; converter.in = float_number; unsigned bits = converter.out;
Of course, after getting the bits variable, we only need to print it. For instance, this way:
1.0 | 1 | S: 0 E: 01111111 M: 00000000000000000000000
Common sense tells that 1 can be expressed in binary fluting-point form as 1.0 * 2<sup>0</sup>, so exponent is 0 and significand is 1, while in IEEE 754 exponent is 1111111 (127 in decimal) and
significand is 0.
The mystery behind exponent is simple: the exponent is actually shifted. A zero exponent is represented as 127; exponent of 1 is represented as 128 and so on. Maximum value of exponent should be 255
– 127 = 128, and minimum value should be 0 – 127 = -127. However, values 255 and 0 are reserved, so the actual range is -126…127. We will talk about those reserved values later.
The significand is even simpler to explain. Binary significand has one unique property: every significand in normalized form, except for zero, starts with 1 (this is only true for binary numbers).
Next, if a number starts with zero, then it is not normalized. For instance, non-normalized 0.000101 * 10<sup>101</sup> is the same as normalized 1.01 * 10<sup>1</sup>. Due to that, there is no need
to write an initial 1 for normalized numbers: we can just keep it in mind, saving space for one more significant bit. In our case, the actual significand is 1 and 23 zeroes, but because 1 is skipped,
it is only 23 zeroes.
Let us try some different numbers in comparison with 1.
1.0 | 1 | S: 0 E: 01111111 M: 00000000000000000000000
-1.0 | -1 | S: 1 E: 01111111 M: 00000000000000000000000
2.0 | 2 | S: 0 E: 10000000 M: 00000000000000000000000
4.0 | 4 | S: 0 E: 10000001 M: 00000000000000000000000
1 / 8 | 0.125 | S: 0 E: 01111100 M: 00000000000000000000000
As we can see, a negative sign just inverts sign flag without touching the rest (this seems obvious, but it is not always the case in computer science: for integers, a negative sign is much more
complex than just flipping one bit!). Changing the exponent by trying different powers of two works as expected.
1.0 | 1 | S: 0 E: 01111111 M: 00000000000000000000000
3.0 | 3 | S: 0 E: 10000000 M: 10000000000000000000000
5.0 | 5 | S: 0 E: 10000001 M: 01000000000000000000000
0.2 | 0.2 | S: 0 E: 01111100 M: 10011001100110011001101
It is easy to see that numbers 3 and 5 are represented as 1.1 and 1.01 with aproper exponent. 0.2 should not differ much from them, but it is. What happened?
It is easier to explain on decimals. 0.2 is the same number as 1/5. At the same time, not each radical can be represented as a decimal floating-point number: for example, 2/3 is 0.666666… It happens
because 3 does not have any non-trivial common divisors with 10 (10 = 2 * 5, neither of them is 3). In the same time, 2/3 can be easily represented in base 12 as 0.8 (12 = 2 * 2 * 3). The same trick
is true for a binary system: 2 does not have any common divisors with 5, so 0.2 can only be represented as infinitely long 0.00110011001100… At the same time, we only have 23 significant bits! So, we
are inevitably losing precision.
Let us try with some multiplications.
1.0 | 1 | S: 0 E: 01111111 M: 00000000000000000000000
0.2^2*25 | 1 | S: 0 E: 01111111 M: 00000000000000000000001
25*0.2^2 | 1 | S: 0 E: 01111111 M: 00000000000000000000000
Both 1 and 0.2 * 0.2 * 25 are printed as 1, but they are actually different! Due to the precision loss, 0.2 * 0.2 * 25 is not the same as 1, and the expression (0.2f * 0.2f * 25.0f == 1.0f) is
actually false. At the same time, if we execute 25 * 0.2 first, then the result is actually correct. It means that the rule (a * b) * c = a * (b * c) is not always true for floating-point numbers!
Special numbers
Remember about the fact that zero can never be written in the normalized form because it does not contain any 1s in its binary representation? Zero is a special number.
0 | 0 | S: 0 E: 00000000 M: 00000000000000000000000
-0 | -0 | S: 1 E: 00000000 M: 00000000000000000000000
For zero, IEEE 754 uses an exponent value of 0 and a significand value of 0. In addition, as you can see, there are actually two zero values: +0 and -0. In terms of comparison, (0.0f == -0.0f), is
actually true, sign just does not count. +0 and -0 loosely correspond to the mathematical concept of the infinitesimal, positive and negative.
Are there any special numbers with an exponent value of 0? Yes. They are called «denormalized numbers». Those numbers can represent extremely small values, lesser than the minimum normalized number
(which should be a little larger than 1 * 2<sup>-127</sup>). Examples:
2^-126 | 1.17549e-38 | S: 0 E: 00000001 M: 00000000000000000000000
2^-127 | 5.87747e-39 | S: 0 E: 00000000 M: 10000000000000000000000
2^-128 | 2.93874e-39 | S: 0 E: 00000000 M: 01000000000000000000000
2^-149 | 1.4013e-45 | S: 0 E: 00000000 M: 00000000000000000000001
2^-150 | 0 | S: 0 E: 00000000 M: 00000000000000000000000
A denormalized number has the virtual exponent value of 1, but, at the same time, they do not have omitted 1 as their first omitted digit. The only consequence is that denormalized numbers quickly
lose precision: to store numbers between 2<sup>-128</sup> and 2<sup>-127</sup>, we are only using 21 digits of information instead of 23.
It is easy to see that zero is the special case of the denormalized number. Moreover, as we can see, the least possible single-precision floating-point number is actually 2<sup>-149</sup>, or
approximately 1.4013 * 10<sup>-45</sup>.
Numbers with the exponent of 11111111 are reserved for the «other end» of the number scale: Infinity and the special value called «Not a Number».
1 / 0 | inf | S: 0 E: 11111111 M: 00000000000000000000000
1 / -0 | -inf | S: 1 E: 11111111 M: 00000000000000000000000
2^128 | inf | S: 0 E: 11111111 M: 00000000000000000000000
As with zeroes, infinity can be either positive or negative. It can be achieved by dividing any non-zero number to zero or by getting any number larger than maximum allowed (which is a little less
than 2<sup>128</sup>). Infinity is processed as follows:
Infinity > Any number
Infinity = Infinity
Infinity > -Infinity
Any number / Infinity = 0 (sign is set properly)
Infinity * Infinity = Infinity (again, sign is set properly)
Infinity / Infinity = NaN
Infinity * 0 = NaN
Not a Number, or NaN, is, perhaps, the most interesting floating-point value. It can be obtained in multiple ways. First, it is the result of any indeterminate form:
Infinity * 0
0 / 0 or Infinity / Infinity
Infinity – Infinity or –Infinity + Infinity
Secondly, it can be the result of some non-trivial operations. Power function may return NaN in case of any of those indeterminate forms: 0<sup>0</sup>, 1<sup>Infinity</sup>, Infinity<sup>0</sup>.
Any operation which can result in complex number may return NaN in this case: log(-1.0f), sqrt(-1.0f), sin(2.0f) are the examples.
Lastly, any operation involving NaN as any of the operands always returns NaN. Because of that, NaN can sometimes quickly “spread” through data like a computer virus. The only exception is min or
max: those functions should return the non-NaN argument. NaN is never equal to any other number, even itself (it can be used to test numbers against NaN).
Actual contents of NaN are implementation-defined; IEEE 754 only requires that exponent should be 11111111, significand should be non-zero (zero is reserved for the infinity) and the sign does not
0/0 | -nan | S: 1 E: 11111111 M: 10000000000000000000000
IEEE 754 differentiates two types of NaN: quiet NaN and signaling NaN. Their only difference is that signaling NaN generates interruption while quiet NaN does not. Again, the application decides if
it generates quiet NaN or signaling NaN. For instance, the GCC C compiler always generates quiet NaN unless explicitly specified to behave the other way around.
What can we learn from all the facts and experiments above? In any language operating with the floating-point data type, beware of the following:
– You should almost never directly compare two floating-point numbers unless you know what you are doing! A better way to do it is to compare it with some precision.
if (a == b) – wrong!
if (fabsf(a – b) < epsilon) – correct!
– Floating-point numbers lose precision even when you are just working with such seemingly harmless numbers as 0.2 or 71.3. You should be extra careful when working with a large amount of
floating-point operations over the same data: errors may build up rather quickly. If you are getting unexpected results and you suspect rounding errors, try to use a different approach, and minimize
– In the world of floating-point arithmetic, multiplication is not associative: a * (b * c) is not always equal to (a * b) * c.
– Additional measures should be taken if you are working with either extremely large values, extremely small numbers, and/or numbers close to zero: in case of overflow or underflow those values will
be transformed into +Infinty, -Infinity or 0. Numeric limits for single-precision floating-point numbers are approximately 1.175494e-38 to 3.402823e+38 (1.4013e-45 to 3.402823e+38 if we also count
denormalized numbers)а.
– Beware if your system generates «quiet NaN». Sometimes, it may help you to not crash the application. Sometimes, it may spoil program execution beyond recognition.
Nowadays, floating-point numbers operations are extremely fast, with speed comparable to the usual integer arithmetic: a number of floating-point operations per second, or FLOPS, is perhaps the most
well-known measure of computer performance. The only downside is that the programmer should be aware of all the pitfalls regarding the precision and special floating-point values.
About the Author
ByteScout Team of WritersByteScout has a team of professional writers proficient in different technical topics. We select the best writers to cover interesting and trending topics for our readers. We
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How many combination of 3 numbers out of 5 numbers? - Answers
How many combinations are there in 39 numbers?
It depends on how many numbers are in each combination. If there are 5 numbers in each combination, then you have 575,757 possibilities. If there are 2 in each combination, there you have 741
possibilities. If there's only 1 number in each combination, then there are only 39 possibilities. | {"url":"https://math.answers.com/math-and-arithmetic/How_many_combination_of_3_numbers_out_of_5_numbers","timestamp":"2024-11-12T20:17:34Z","content_type":"text/html","content_length":"163112","record_id":"<urn:uuid:e34aaf10-f394-4fd3-9c60-f4689e4d636a>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00660.warc.gz"} |
Prosper IPMT mode to help you calculate notice part of a loan fee - RokaFlex | Policarbonat | Trape de fum
Prosper IPMT mode to help you calculate notice part of a loan fee
New course shows ways to use the IPMT setting when you look at the Do well to find the focus part of an intermittent commission to your a beneficial mortgage or financial.
When you remove financing, whether it is a mortgage, financial otherwise car finance, you pay right back the amount your in the first place borrowed and you may notice on top of it. Basically, desire
‘s the cost of using somebody’s (always an excellent bank’s) currency.
The interest part of a loan commission will be computed by hand from the multiplying the period’s interest of the kept equilibrium. However, Microsoft Do just fine provides another type of mode for
this – the newest IPMT setting. Inside example, we are going to come in-depth describing their syntax and you may bringing genuine-lifestyle formula advice.
Prosper IPMT function – sentence structure and you can basic spends
IPMT try Excel’s attract commission form. It productivity the attention level of that loan fee during the good given several months, and if the speed while the total amount of a payment try ongoing
in most periods.
For example, if you make annual payments with the that loan that have a yearly interest rate out of 6 per cent, play with six% otherwise 0.06 having price.
If you make each week, monthly, otherwise every quarter repayments, separate brand new yearly speed by the quantity of percentage symptoms for every single season, while the revealed contained in
this analogy. State, if one makes every quarter money towards a loan having a yearly rate of interest from six percent, fool around with six%/cuatro having rate.
• For each (required) – that point the place you need to estimate the eye. It should be an integer about may include step one to nper.
• Nper (required) – the entire quantity of money when you look at the longevity of the borrowed funds.
• Sun (required) – today’s property value the loan otherwise resource. Simply put, simple fact is that financing dominant, we.elizabeth. the amount you borrowed from.
• Fv (optional) – the future really worth, i.e. the necessary balance following past commission is done. When the excluded, it’s implied to be no (0).
• Sort of (optional) – specifies if payments try due:
• 0 or omitted – money were created after for every single period.
• step 1 – payments are produced at the beginning of each months.
Instance, for people who received financing regarding $20,one hundred thousand, you need to pay of in annual installments within the 2nd 36 months with an annual interest out of 6%, the attention
part of the step 1 st year percentage might be computed with this particular formula:
Instead of giving the amounts into an algorithm, you can type in her or him in a number of predetermined structure and refer to the individuals cells such as for example shown regarding the
screenshot lower than.
In accordance with the income indication summit, as a result, came back because the a bad matter since you shell out out that it money. Automagically, it’s showcased into the yellow and you can
closed in the parenthesis (Currency structure having bad quantity) once the shown about left the main screenshot below. Off to the right, you will find caused by a comparable algorithm in the
Standard style.
If you would as an alternative rating notice since a confident number, put a without signal prior to both the entire IPMT function or the fresh new photo voltaic dispute:
Types of having fun with IPMT algorithm in the Excel
Now you be aware of the basics, let us see how to make use of the IPMT setting to get the number of interest for various wavelengths out of fee, as well as how changing the loan criteria transform
the possibility attention.
Prior to we plunge inside the, it ought to be listed one to IPMT formulas should be getting put after the PMT function that exercise the total amount of an occasional percentage (notice + principal).
To discover the notice part of financing payment proper, it is best to http://www.loansavesolutions.com/installment-loans-mt/ convert the fresh new yearly rate of interest into the related period’s
rates and lifetime into final number away from fee attacks:
• To the price disagreement, separate the fresh new annual interest from the level of payments for each season, while aforementioned is equal to just how many compounding attacks per year.
• Into nper dispute, proliferate how many age because of the quantity of money each seasons.
Such as, let us find the quantity of attention you will need to shell out on the same mortgage however in some other fee frequencies:
The bill following the history commission is going to be $0 (the latest fv argument excluded), and the payments is due at the end of for each and every period (the sort argument omitted).
Looking at the screenshot less than, you could observe that the eye count reduces with each subsequent period. Simply because one percentage leads to reducing the financing prominent, hence reduces
the left harmony about what interest rates are calculated.
Plus, delight observe that the total amount of attention payable for the same financing differs getting yearly, semi-annual and every quarter installment payments:
Full-form of your own IPMT mode
Within analogy, we’re going to calculate notice for similar financing, an identical commission volume, however, various other annuity versions (regular and you can annuity-due). For it, we need to
use the full form of your IPMT means.
Note. If you plan to use the latest IPMT algorithm for more than one several months, please notice the fresh cellphone references. All sources with the type in cells should be pure (towards the
dollars signal) so they really was secured to people cells. This new per dispute must be a member of family telephone reference (without having any buck signal such as for example A9) because is to
changes in line with the cousin standing out of an excellent row that this new algorithm are copied.
Thus, we enter the significantly more than algorithm into the B9, drag it down towards the left attacks, and have the second results. If you examine brand new amounts regarding the Attention articles
(normal annuity to your left and annuity-due to the right), you will find that appeal is a little lower when you shell out at the beginning of months.
Do well IPMT mode not working
Which is the way you make use of the IPMT setting from inside the Excel. Having a close look from the algorithms talked about inside tutorial, you are invited to download our Do well IPMT form take
to workbook. We thank you for learning and you will hope to view you towards the all of our blog next week! | {"url":"http://rokaflex.ro/prosper-ipmt-mode-to-help-you-calculate-notice/","timestamp":"2024-11-06T05:19:20Z","content_type":"text/html","content_length":"31332","record_id":"<urn:uuid:907efb77-125d-4f5b-8a5f-237a49c6052b>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00470.warc.gz"} |
acronymforme wrote:
I bought a CE a few weeks ago and I am about ready to throw it into Tampa Bay. Now I carry both the CE and the Silver with me. Why look like a fool when I go into class and tests with 2 calculators?
It's because the CE removed the ability to answer questions automatically in fraction format. Here's an example. On the CE type in the fraction 2/3. It will convert to the decimal .6666666667.
Awesome. Now - try to convert that back into a fraction. You can't. You can't until the end of time. So, tell me, how awesome is that? To have all of your answers in decimal format where you might or
most likely will not be able to convert back into a logical and recognizable value? Calculating function graph intercepts and values in Calculus are also real fun when 2 decimal points are dropped
from the x values displayed as decimals that I will never be able to convert into a fraction on this device. This calculator should come with a crystal ball to help me guess. What makes it a real
party favor is the fact that Pearson Edu Math applications require all the answers on homework, quizzes, tests, midterms and finals to be entered as fractions! This thing is an overpriced piece of
useless junk. Sure it has pretty color and a lot of features - but the basics, the basic taken for granted things you depend on to work are not there. That makes the CE undependable. I chalk it up to
lazy programming which was most likely outsourced to save a few bucks.
If you press [math] and select the >Frac option, it will convert the answer back into fractional form.
acronymforme wrote:
I bought a CE a few weeks ago and I am about ready to throw it into Tampa Bay. Now I carry both the CE and the Silver with me. Why look like a fool when I go into class and tests with 2 calculators?
It's because the CE removed the ability to answer questions automatically in fraction format. Here's an example. On the CE type in the fraction 2/3. It will convert to the decimal .6666666667.
Awesome. Now - try to convert that back into a fraction. You can't. You can't until the end of time. So, tell me, how awesome is that? To have all of your answers in decimal format where you might or
most likely will not be able to convert back into a logical and recognizable value? Calculating function graph intercepts and values in Calculus are also real fun when 2 decimal points are dropped
from the x values displayed as decimals that I will never be able to convert into a fraction on this device. This calculator should come with a crystal ball to help me guess. What makes it a real
party favor is the fact that Pearson Edu Math applications require all the answers on homework, quizzes, tests, midterms and finals to be entered as fractions! This thing is an overpriced piece of
useless junk. Sure it has pretty color and a lot of features - but the basics, the basic taken for granted things you depend on to work are not there. That makes the CE undependable. I chalk it up to
lazy programming which was most likely outsourced to save a few bucks.
I don't believe that anything you've described has changed for the CE.
You can view answers as fractions, just as you've always been able to, by using the ►Frac conversion:
If you really need more than 8 digits of precision from coordinates on the graph screen, you can view the coordinate values on the home screen, just as you've always been able to:
Basically, I'm researching this for a high school math student. I'm deciding between the TI-84 Plus and the TI-84 Plus CE.
Can anyone comment on the protective cover that comes with the TI-84 Plus CE? I've read multiple reviews that state that it leaves a small gap at the top of the screen, which can allow a pencil or
something else to fall in and scratch the screen. If that's true, how do you overcome that?
I've also read multiple reviews of engineering students and math tutors saying that, if they key too quickly, the keystrokes are not always registered, causing them to have to check and recheck that
the buttons were all registered before relying on the result.
From what I understand, the basic black 84Plus has the advantage of being built like a tank and being less likely to get damaged than the CE. In terms of computer connectivity, I researched before
buying, but TI put out a new OS for the CE very recently (5.2). 5.1.5 was supposedly compatible with an older Mac running 10.6.8, but 5.2 is only compatible with Macs running 10.10 or 10.11.
According to TI's website, the TI-84 Plus is compatible with Mac 10.6.8, but is that actually true? (Mac discussion threads don't seem to support that idea.)
The one obvious advantage of the CE is that the pixels seem crisper and sharper and the graphing function appears inordinately easier to read than the basic 84Plus. The CE also appears to have a
rechargeable battery, though it seems like problems with the "deep sleep mode" have come up.
Any advice anyone can offer will be welcome ... and will be shared with my entire school community, all of which are using older computers at home than probably anyone on this board. Thank you!
About the issue with fractions on the TI-84 Plus CE ....
I thought that the CE eliminated the option to format answers as "FRAC-APPROX", which means you can't choose a mode that auto-converts all answers from decimals back to fractions with the CE, the way
you could with the TI-84 Plus C. (The C version allowed for AUTO, DEC or FRAC-APPROX.)
It seems like the CE allows you to press [math] and select >Frac, and convert it manually that way, but there isn't an auto-setting for that anymore. I'm not sure why not.
Has this annoyed anyone using this for school?
indywriter13 wrote:
Can anyone comment on the protective cover that comes with the TI-84 Plus CE? I've read multiple reviews that state that it leaves a small gap at the top of the screen, which can allow a pencil or
something else to fall in and scratch the screen. If that's true, how do you overcome that?
You'd have to have a pretty small pencil to get in that gap. It's super small, at least on my calculator.
I've also read multiple reviews of engineering students and math tutors saying that, if they key too quickly, the keystrokes are not always registered, causing them to have to check and recheck that
the buttons were all registered before relying on the result.
Its more of a problem in the programs than the actual calculator. However, yes sometimes the calculator misses some keypress', but that's usually because of the calculator still turning on and things
like that and when you're just typing too fast. Type slower. Big deal.
From what I understand, the basic black 84Plus has the advantage of being built like a tank and being less likely to get damaged than the CE. In terms of computer connectivity, I researched before
buying, but TI put out a new OS for the CE very recently (5.2). 5.1.5 was supposedly compatible with an older Mac running 10.6.8, but 5.2 is only compatible with Macs running 10.10 or 10.11.
According to TI's website, the TI-84 Plus is compatible with Mac 10.6.8, but is that actually true? (Mac discussion threads don't seem to support that idea.)
2 things.
1. Don't drop it. It's not super fragile, but you just shouldn't.
2. Compatibility is something that is always going down. Apple does that all the time themselves. Like iOS 10 isn't compatible with the Ipod 5 when it's the second newest Ipod.
The one obvious advantage of the CE is that the pixels seem crisper and sharper and the graphing function appears inordinately easier to read than the basic 84Plus. The CE also appears to have a
rechargeable battery, though it seems like problems with the "deep sleep mode" have come up.
That is a super big advantage. Another is that there is more RAM, and development for it is on the rise above all other calculators. TI and 3rd party developers are now focusing on the CE and
improving that.
I thought that the CE eliminated the option to format answers as "FRAC-APPROX", which means you can't choose a mode that auto-converts all answers from decimals back to fractions with the CE, the way
you could with the TI-84 Plus C. (The C version allowed for AUTO, DEC or FRAC-APPROX.)
It seems like the CE allows you to press [math] and select >Frac, and convert it manually that way, but there isn't an auto-setting for that anymore. I'm not sure why not.
Has this annoyed anyone using this for school?
Yes, however, we get over it. Maybe they'll release it in a newer update.
For the issue with fractions on the TI-84 Plus CE, you can always make an asm program that forces fractions.
Try this
And run it from the homescreen with the Asm() token
This will force fractional results when it can until you turn the calc off or press [2nd] and [mode]
Once a Zstart-like program is released for the CE, you wil be able to make it run this command on startup and always get fractional results when you turn it on. | {"url":"https://dev.cemetech.net/forum/viewtopic.php?t=11434&postdays=0&postorder=asc&start=80","timestamp":"2024-11-12T05:16:52Z","content_type":"text/html","content_length":"49868","record_id":"<urn:uuid:e40e6b5e-03ce-42f5-8485-8744829c0c82>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00293.warc.gz"} |
The Last Ride Together | Blablawriting.com
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Question 1:
Write a program to input a start limit S (S>0) and the last limit L (L>0). Print all the prime triplets between S and L (both inclusive), if S<=L otherwise your program should ask to re-enter the
values of S and L again with a suitable error message.
* Start
* To input the lower limit
* To input the upper limit
* To run the outer loop
* To run inner loop
* To calculate total number of prime numbers between lower and upper limits
* To declare array with it’s number of elements as ‘s’
* To run the outer loop
* To run the inner loop
* To store the prime numbers in array a[]
* To run a loop for every position of array a[]
* If condition matches for the number for prime triplets
* Continue till all Prime Triplets are printed
Question 2:
A unique digit integer is a positive integer (without leading zeros) with no duplicate digits. For example 7, 135, 214 are all unique digit integers whereas 33, 3121, 300 are not. Given two positive
integers m and n, where m<n, write a program to determine how many unique digit integers are there in the range between m and n (both inclusive) and output them.
* Start
* To input the starting limit
* To input the last limit
* To run the outer loop
* To store the value of ‘i’ as a string
* To run the inner loop
* To run a nested loop of the inner loop
* To check for repetition of any digit in the number
* To store all the unique digit integers in a string
* To store the frequency of unique digit integers
* To print the unique digit integers and their frequency
Question 3:
Write a program which inputs Natural numbers N and M followed by integer arrays A[ ] and B[ ], each consisting of N and M numbers of elements respectively. Sort the arrays A[ ] and B[ ] in Ascending
order of magnitude. Use the sorted arrays A[ ] and B[ ] to generate a merged array C[ ] such that the elements of A[ ] and B[ ] appears in C[ ] in Ascending order without any duplicate elements.
Sorting of array C[ ] is not allowed.
* Start
* To enter the limit of first array (<21)
* To enter the limit of the second array (<21)
* To run a loop
* To enter the two arrays
* To sort the two arrays using bubble sort technique
* To copy the two arrays into a single array
* To sort the merged array using bubble sort technique
* To run a loop
* To print the elements of the i-th position
* To go to the next index of the next number
Question 4:
Write a program to input and store n integers (n > 0) in a single subscripted variable and print each number with their frequencies of existence. The output should contain number, asterisk symbol and
its frequency and be displayed in separate lines.
* Start
* To enter capacity
* To enter the numbers in an array
* To run outer loop
* To run inner loop
* To sort the numbers in the array
* To run outer loop
* To run inner loop
* To transfer the values of the array in another array
* To check frequency and print
Question 5:
Write a program to input an arithmetic expression in String form which contains only one operator between two numeric operands. Print the output in the form of number. ( If more than one operators
are present, an error output message “ INVALID EXPRESSION” should appear).
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Raw College Application Material
I don’t want to go into the scientific community with a sense of competition, competing for spots at a college, but with a sense of ambition for discovery.
If not first, then at least independently. I always jump to ask questions, moreover I enjoy answering questions I don't think anyone has asked and more rarely nobody has been able to answer.
I don’t believe that I’m driven by an extrinsic desire to just be different, but rather I am intrinsically different. This is corroborated when I’m asked to solve a problem and I do so in an unconventional manner.
I’m very visual. I have an abstraction threshold that, when a concepts exceeds, I must understand it visually. Not any proof will work for me, I can’t just see and follow the math to accept the derived concept or rule; I have to understand things intuitively. It’s not that I was instructed to understand this way, (in fact, there was little mention of an ‘intuitive understanding’ in many of my classes) it’s that I feel uneasy employing an equation that I don’t understand intuitively. I’ve found that inferences come extremely easily with an intuitive understanding of something. Because of this need of an intuitive understanding, I’ve had difficulty with concepts like electromagnetism, some advanced integration techniques, centripetal motion, etc. This is likely because the proofs of these are often heavily mathematical. Also, of course, this makes the questions that lead to an intuitive understanding of things (Why? How?) nearly impossible to apply to laws in physics and mathematics: Why does every object attract every other object? Why is there a magnetic field created in the movement of charge? etc.
If I can't be the first to explore something, then I much prefer to figure it out myself than have it explained. It's more challenging that way and it's personal and hopefully original. If the exact same viewpoint on a subject persisted through time and the same questions persisted with them, it may take a new interpretation to figure it out. Perhaps it's better to be wrong and original than right and mainstream.
I try to apply this ideology to everything I do: this makes learning a difficult process but I think I cull an enormous amount more from it than if I were to accept or request a basic, common viewpoint. To come to a realization realization/derivation independently is an enormously pleasant experience. It's even more pleasant when you find that the methods of derivation are NOT mainstream. For this reason, I learn with my own methodology, but I am by no means resistant. Learning naturally leads to more learning; it's a recursive process. It’s quite common that I’ll encounter a concept that I’m hesitant to accept in Physics or Mathematics. When I do, I’ll spend some time ruminating on it and, if I can’t at that point see action I can take to understand it on my own, I’ll research the topic further. Finally, if I still can’t come to an understanding, I’ll look for instruction (often online) for an intuitive understanding, Khan academy, explanations for everyone, etc. I’ve found that simply going through this process, even if I’m not successful, can help me understand many things along the way. If I am successful, I try to prove the concept using thought experiments, programming, or a graphical representation. This allows me to expand on the topic further (as the process of proving often provokes more questions) and also to present the proof to others and my own personal reference.
Many times, my proofs are not efficient, but they are quite cognitively accessible.
There's a certain discomfort I feel when I know something but don't understand it. My entire life has been and I predict will be to satiate this feeling.
In my high school, I’ve found that much of the work suppresses this desire
When looking through a high school textbook in say, Physics, I’m perplexed by the brevity with which the authors explain each topic. I’m stressed to find instructors whose basis of instruction is intuition. I think this may be because understanding something intuitively takes a lot of time and effort, for instance, permutations. It’s difficult to imagine the Birthday paradox intuitively, and is much easier for a teacher to regurgitate an equation and show an example to demonstrate that it works. But, with some effort, one can easily see that permutations and combinations can be understood visually and intuitively. Insert permutations here
Accepting and forgetting is one of the most dreadful experiences in the school system and, tragically, it's encountered often. What’s even more tragic is that the results of this process (grades) are used in the college application process.
Seeing a proof or demonstration of a concept isn't enough for me, I have to feel it.
Underneath all learning is the building of intuition, so the best approach is an intuitive one.
I wrote a program to analyze Facebook. It took learning two new languages and 30 hours of work to complete, but doing so was the most satisfying work I’ve ever had.
I’ve found that the stuff that I really care about, the work I’m actually proud of completing, is the work I asked myself to do. The reason why I’m so proud is that I know this work is unique. It’s not the essay that the past 5 generations have written, it’s not the project everyone in my grade completed, it was thought up, executed, and completed by only me. Because I find this kind of work innately pleasureful, I’m disposed to putting priority on it. I’m always prudent to preserve this wonder and inquisitiveness that sometimes thwarts my progress in standard instruction (I go on a tangent I find interesting and build continually). But, I find that I gain far more from my personal endeavors.
Put the 400 hours of lab work on my college apps and get scholarships for having 400 hours of community service. | {"url":"https://joepucc.io/notes/raw-college-application-material.php","timestamp":"2024-11-14T05:17:32Z","content_type":"text/html","content_length":"8396","record_id":"<urn:uuid:45117844-9431-45b3-9007-9b23bad32bd3>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00203.warc.gz"} |
Math, Grade 6, Fractions and Decimals, Fractions and Division in Word Problems
Fractions and Division in Word Problems
Math Mission
Solve word problems involving fractions.
Dog Play Area
Work Time
Dog Play Area
Denzel wants to fence in a rectangular play area for his dog. The play area will extend the entire length of his backyard, which is $8\frac{1}{3}$ yards.
• If he wants his dog to have 50 square yards to play in, how wide does he need to make the play area?
Ask yourself:
• What is the formula for the area of a rectangle?
• What measurements does the problem provide?
• What is the unknown measurement?
Walking to School
Work Time
Walking to School
Martin walks $\frac{4}{5}$ of a mile to school each day. The distance Emma walks to school is $\frac{5}{6}$ of the distance Martin walks.
Ask yourself:
Can you use a number line to model the situation?
Work Time
The male hippopotamus at the zoo weighs 2 tons. He weighs $1\frac{1}{2}×f=2$ times as much as the female hippopotamus weighs.
• How much does the female hippopotamus weigh?
Does the female hippopotamus weigh more or less than the male?
Prepare a Presentation
Work Time
Prepare a Presentation
Choose one of the problems you solved. Prepare a solution that you can share with your classmates. Include a drawing and an explanation of what you did to solve the problem.
Challenge Problem
Write two word problems.
• Write one word problem that uses multiplication and the numbers $4\frac{2}{3}$ and $\frac{1}{3}$.
• Write a second word problem that uses division and the numbers $4\frac{2}{3}$ and $\frac{1}{3}$.
• Solve both problems.
Make Connections
Performance Task
Ways of Thinking: Make Connections
Take notes on your classmates’ approaches to solving and writing fraction word problems.
As your classmates present, ask questions such as:
• How are the quantities in the problem related?
• What is the unknown quantity in the problem?
• How does your equation represent the problem situation?
• Is your answer reasonable? How do you know?
• Where are the known and unknown quantities in the problem you wrote?
• How do you know that your problem is a multiplication or a division situation?
Reciprocals and Dividing by Fractions
Formative Assessment
Summary of the Math: Reciprocals and Dividing by Fractions
Read and Discuss
• The fractions $\frac{a}{b}$ and $\frac{b}{a}$, with numerators and denominators inverted, are called reciprocals of each other.
• The key property of reciprocals is that their product is always 1.
• Reciprocals are also called multiplicative inverses. This name refers to the fact that if you multiply a fraction by a number, and then multiply the result by the reciprocal of the fraction, the
result is “undone.” For example:
• You can use the properties of reciprocals to help you divide by fractions. The general rule is: To divide by a fraction, multiply by the reciprocal of the fraction. Algebraically, the rule looks
like this: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}×\frac{d}{c}=\frac{ad}{bc}$
Can you:
• Solve a problem that involves dividing a fraction by a fraction?
• Determine which operation is needed to solve a word problem?
• Explain what reciprocals are and how to use them to help you solve division problems involving fractions?
Reflect On Your Work
Work Time
Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.
An example of an everyday situation that involves fraction division is … | {"url":"https://openspace.infohio.org/courseware/lesson/2099/student/?section=9","timestamp":"2024-11-12T05:49:40Z","content_type":"text/html","content_length":"39729","record_id":"<urn:uuid:4d41e349-aaa2-46be-8759-d6d94e445b3d>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00124.warc.gz"} |
Higher Order Functions Exercises | R-bloggersHigher Order Functions Exercises
Higher Order Functions Exercises
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Higher order functions are functions that take other functions as arguments or return functions as their result. In this set of exercises we will focus on the former. R has a set of built-in higher
order functions: Map, Reduce, Filter, Find, Position, Negate. They enable us to complete complex operations by using simple single-purpose functions as their building blocks. In R this is especially
helpful in cases where we cannot depend on vectorization and have to utilize control statements like for loops. In such scenarios higher order functions help us by: a) simplifying and shortening the
syntax, b) getting rid of counter indices and c) getting rid of temporary storage values.
Exercises in this section will have to be solved by using one or more of the higher order functions mentioned above. It might be useful reading their help page before continuing.
Answers to the exercises are available here.
If you obtained a different (correct) answer than those listed on the solutions page, please feel free to post your answer as a comment on that page.
Exercise 1
You are working on 3 datasets all at once:
multidata <- list(mtcars, USArrests, rock)
summary(multidata[[1]]) will return the summary information for a single dataset.
Obtain summary information for every dataset in the list.
Exercise 2
cumsum(1:100) returns the cumulative sums of a vector of numbers from 1 to 100.
Do the same using sum and an appropriate higher order function.
Exercise 3
You have a vector of numbers from 1 to 10. You want to multiply all those numbers first by 2 and then by 4. Why the following line does not work and how to fix it?
Map(`*`, 1:10, c(2,4))
Exercise 4
Expression sample(LETTERS, 5, replace=TRUE) obtains 5 random letters.
Generate a list with 10 elements, where first element contains 1 random letter, second element 2 random letters and so on.
Note: use a fixed random seed: set.seed(14)
Exercise 5
Library spatstat has a function is.prime() that checks if a given number is a prime.
Find all prime numbers between 100 and 200.
Exercise 6
We have a vector containing all the words of the English language –
words <- scan("http://www-01.sil.org/linguistics/wordlists/english/wordlist/wordsEn.txt", what="character")
a. Using a function that checks if a given words contains any vowels:
containsVowel <- function(x) grepl("a|o|e|i|u", x)
find all words that do not contain any vowels.
b. Using a function is.colour() from the spatstat library find the index of the first word inside the words vector corresponding to a valid R color.
Exercise 7
a. Find the smallest number between 10000 and 20000 that is divisible by 1234.
b. Find the largest number between 10000 and 20000 that is divisible by 1234.
Exercise 8
Consider the babynames dataset from the babynames library.
Start with a list containing the used names for each year:
library(babynames); namesData <- split(babynames$name, babynames$year)
a. Obtain a set of names that were present in every year.
b. Obtain a set of names that are only present in year 2014
Exercise 9
Using the same babynames dataset and a function that checks if word has more than 3 letters: moreThan3 <- function(x) nchar(x) > 3
Inside each year list leave only the names that have 3 letters or less.
Exercise 10
Using the same babynames dataset:
a. Split each name to a list of letters.
b. Join each list of letters by inserting an underscore “_” after each letter.
Note: if you have a word x <- "exercise" you can split it with x2 <- strsplit(x, "")
and join using underscores with paste(x2[[1]], collapse="_") | {"url":"https://www.r-bloggers.com/2016/08/higher-order-functions-exercises/","timestamp":"2024-11-10T15:51:12Z","content_type":"text/html","content_length":"91742","record_id":"<urn:uuid:191275a4-c085-4fe0-9376-a7f5c3c7a317>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00719.warc.gz"} |
1049 - One is an Interesting Number
Numbers are interesting, but some are inherently more interesting than others, by various criteria. Given a collection of numbers, you are to find the most interesting ones.
A number X is more interesting than another number Y if it has more attributes than Y. For the purposes of this problem, the attributes that are interesting are:
Attribute Name Description Example Numbers
prime:The number is prime (not divisible by numbers other than itself and 1). eg:2, 113,
square: The number is the second power of an integer. eg:4, 225, 1089
cube: The number is the third power of an integer. eg:8, 3375, 35937
quad: The number is the fourth power of an integer. eg:16, 50625, 1185921
sum-multiple: The number is a multiple of the sum of its digits. eg:1, 24, 100
multiple-multiple:The number is a multiple of the number made when multiplying its digits together. eg:1, 24, 315
Note that 0 has no multiples other than itself, and 1 is not prime.
In addition to the above attributes, there are also those which depend on the other numbers in a given collection:
Attribute Name Description
factor: The number is a factor of another number in the collection.
multiple: The number is a multiple of another number in the collection.
other-square: The number is the second power of another number in the collection.
other-cube: The number is the third power of another number in the collection.
other-quad: The number is the fourth power of another number in the collection.
other-sum-multiple: The number is a multiple of the sum of digits of another number in the collection.
other-multiple-multiple: The number is a multiple of the number made when multiplying the digits of another number in the collection together.
This makes for a total of thirteen possible attributes. Note that meeting the criteria for a particular attribute in multiple ways (1 is the factor of all other numbers, for example) still only
counts as a single instance of an attribute.
Given a collection of numbers, you are to determine which numbers in that collection are most interesting.
Input to this problem will begin with a line containing a single integer N (1 ≤ N ≤ 100) indicating the number of data sets. Each data set consists of the following components:
* A line containing a single integer M (1 ≤ M ≤ 100) indicating how many numbers are in the collection;
* A series of M lines, each with a single integer X (1 ≤ X ≤ 1000000). There will be no duplicate integers X within the same data set.
For each data set in the input, output the heading "DATA SET #k" where k is 1 for the first data set, 2 for the second, and so on. For each data set, print the number or numbers that are most
interesting in the collection. If more than one number ties for "most interesting," print them in ascending order, one to a line.
sample input
sample output
DATA SET #1
DATA SET #2 | {"url":"http://hustoj.org/problem/1049","timestamp":"2024-11-13T15:24:26Z","content_type":"text/html","content_length":"10495","record_id":"<urn:uuid:fb3aaa99-3fbf-47fe-a3a2-24f5f81cd790>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00455.warc.gz"} |
Mushrooms 3932 - math word problem (3932)
Mushrooms 3932
Peter Milan and Valika collected a total of 88 mushrooms. Valika collected the most - up to 34 mushrooms. How many mushrooms did Peter and Milan collect together?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
You need to know the following knowledge to solve this word math problem:
Grade of the word problem:
Related math problems and questions: | {"url":"https://www.hackmath.net/en/math-problem/3932","timestamp":"2024-11-06T08:18:09Z","content_type":"text/html","content_length":"47057","record_id":"<urn:uuid:0169517e-8426-4fca-b9f4-73f4b1a01945>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00208.warc.gz"} |
Derivative-Calculator.org – Constant Multiple Rule for finding Derivatives
The Constant Multiple Rule, also known as the Constant Coefficient Rule, is a rule in calculus used for differentiating functions that are multiplied by a constant. This rule simplifies the process
of finding derivatives for functions involving constant multiples. The Constant Multiple Rule states that the derivative of a constant times a function is equal to the constant times the derivative
of the function.
Constant Multiple Rule Formula and Formal Definition
The Constant Multiple Rule formula is as follows:
where $c$ is a constant and $f\left(x\right)$ is a function of $x$.
For a function $g\left(x\right)=c·f\left(x\right)$, where $c$ is a constant and $f\left(x\right)$ is a function of $x$, the derivative of $g\left(x\right)$ with respect to $x$ is given by:
${g}^{\prime }\left(x\right)=\frac{\text{d}}{\text{d}x}\left(c·f\left(x\right)\right)=c·\frac{\text{d}}{\text{d}x}\left(f\left(x\right)\right)=c·{f}^{\prime }\left(x\right)$
Intuitive Understanding of the Constant Multiple Rule
To understand the Constant Multiple Rule intuitively, consider the following example. Let $f\left(x\right)={x}^{2}$ (plotted in orange below), and let’s multiply this function by a constant $c=3$ to
get $c·g\left(x\right)=3·\left({x}^{2}\right)=3{x}^{2}$ (in blue below).
If we think about the graph of $g\left(x\right)$, it will have the same shape as the graph of $f\left(x\right)$, but it will be vertically stretched by a factor of 3. This means that for any change
in $x$, the change in $g\left(x\right)$ will be 3 times the change in $f\left(x\right)$.
Now, recall that the derivative of a function at a point is the slope of the tangent line to the function’s graph at that point. Since the graph of $g\left(x\right)$ is stretched vertically by a
factor of 3, the slope of the tangent line to $g\left(x\right)$ at any point will be 3 times the slope of the tangent line to $f\left(x\right)$ at the corresponding point.
Therefore, the derivative of $g\left(x\right)$ will be 3 times the derivative of $f\left(x\right)$, which is exactly what the Constant Multiple Rule states.
Steps to Apply the Constant Multiple Rule
1. Identify the constant coefficient: Determine the constant $c$ that is multiplying the function $f\left(x\right)$.
2. Find the derivative of the inner function: Calculate $\frac{\text{d}}{\text{d}x}\left(f\left(x\right)\right)$, which is the derivative of the function being multiplied by the constant.
3. Multiply the constant and the derivative: Multiply the constant $c$ from step 1 and the derivative $\frac{\text{d}}{\text{d}x}\left(f\left(x\right)\right)$ from step 2 to obtain the final result.
Proof of the Constant Multiple Rule
To prove the Constant Multiple Rule, we can use the definition of the derivative:
${g}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{g\left(x+h\right)-g\left(x\right)}{h}$
Step 1: Substitute $g\left(x\right)=cf\left(x\right)$ into the definition of the derivative.
${g}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{c·f\left(x+h\right)-c·f\left(x\right)}{h}$
Step 2: Factor out the constant $c$.
${g}^{\prime }\left(x\right)=c·{lim}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
Step 3: Recognize that ${lim}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}={f}^{\prime }\left(x\right)$, which is the definition of the derivative of $f\left(x\right)$.
${g}^{\prime }\left(x\right)=c·{f}^{\prime }\left(x\right)$
Thus, we have proven that the derivative of a constant times a function is equal to the constant times the derivative of the function.
1. Find the derivative of $f\left(x\right)=3{x}^{2}+5x$.
Using the Constant Multiple Rule and the Power Rule, we get:
${f}^{\prime }\left(x\right)=3·\frac{\text{d}}{\text{d}x}\left({x}^{2}\right)+5·\frac{\text{d}}{\text{d}x}\left(x\right)=3·2x+5·1=6x+5$
2. Find the derivative of $g\left(x\right)=-2\mathrm{sin}\left(x\right)$.
Using the Constant Multiple Rule and the derivative of sine, we get:
${g}^{\prime }\left(x\right)=-2·\frac{\text{d}}{\text{d}x}\left(\mathrm{sin}\left(x\right)\right)=-2·\mathrm{cos}\left(x\right)$
3. A particle’s position is given by the function $s\left(t\right)=4{t}^{3}-2t$, where $s$ is measured in meters and $t$ is measured in seconds. Find the particle’s velocity and acceleration at time
To find the velocity, we take the derivative of the position function using the Constant Multiple Rule and the Power Rule:
$v\left(t\right)={s}^{\prime }\left(t\right)=4·\frac{\text{d}}{\text{d}t}\left({t}^{3}\right)-2·\frac{\text{d}}{\text{d}t}\left(t\right)=4·3{t}^{2}-2·1=12{t}^{2}-2$
To find the acceleration, we take the derivative of the velocity function:
$a\left(t\right)={v}^{\prime }\left(t\right)=12·\frac{\text{d}}{\text{d}t}\left({t}^{2}\right)=12·2t=24t$
So, the particle’s velocity is $v\left(t\right)=12{t}^{2}-2$ meters per second, and its acceleration is $a\left(t\right)=24t$ meters per second squared.
4. Find the marginal cost function if the total cost function is given by $C\left(x\right)=100x+500$, where $C\left(x\right)$ is the total cost in dollars and $x$ is the number of units produced.
The marginal cost function is the derivative of the total cost function. Using the Constant Multiple Rule, we get:
$MC\left(x\right)={C}^{\prime }\left(x\right)=100·\frac{\text{d}}{\text{d}x}\left(x\right)+500·\frac{\text{d}}{\text{d}x}\left(1\right)=100·1+500·0=100$
So, the marginal cost is constant at 100 dollars per unit.
5. Solve the differential equation ${y}^{″}-4{y}^{\prime }+4y=0$.
This is a second-order linear differential equation with constant coefficients. To solve it, we first find the characteristic equation by replacing ${y}^{″}$ with ${r}^{2}$, ${y}^{\prime }$ with
$r$, and $y$ with $1$:
Factoring this equation, we get:
So, the characteristic equation has a double root at $r=2$. This means that the general solution to the differential equation is:
where ${C}_{1}$ and ${C}_{2}$ are arbitrary constants determined by initial conditions.
Note that the Constant Multiple Rule was used implicitly when we replaced ${y}^{″}$ with ${r}^{2}$, ${y}^{\prime }$ with $r$, and $y$ with $1$ in the characteristic equation, as this is
equivalent to finding the derivatives of ${e}^{rx}$ and multiplying by the constant coefficients. | {"url":"https://derivative-calculator.org/rules/constant-multiple-rule/","timestamp":"2024-11-04T10:17:23Z","content_type":"text/html","content_length":"39165","record_id":"<urn:uuid:5e66f615-2a92-4ac4-aa6e-78557b71aec8>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00718.warc.gz"} |
MTH 3300 Homework 2 - Population, T
MTH 3300 Homework 2 - Population, Triangle and Binary
1) population.py
Suppose that you are a demographist, who wants to model the growth of the population of a nation over time. A simple model for this growth is the standard exponential model
P = P0*e^rt
where P0 is the initial population at time t = 0, r is the relative growth rate in percent per year (expressed as a decimal), t is the time elapsed in years, and P is the population at time t. e, of
course, is the base of the natural logarithm (e ≈ 2.718).
Write a program that prompts the user to enter a value for the initial population. The program should then do the following three times: it will ask for a number of years and a relative growth rate;
the program will then compute the new population after that many years have elapsed, using the population growth formula provided above. This should be done cumulatively – that is, the end population
for the first iteration should be used as the initial population for the second iteration, and the end population for the second iteration should be used as the initial population for the third
So, for example: suppose that the user enters an initial population of 300, a first time period of 4 years, and a first growth rate of 1.2%. Then the population at the end of the first time period
would be 314.751, since
300 · e(0.012)(4) = 314.751.
Suppose then that the second time period was 2.5 years, and the second growth rate is 5%; then the population at the end of the second time period will be
314.751 · e(0.05)(2.5) = 356.66.
Finally, suppose that the user enters 1 year for the last time period, and 2.1% for the last growth rate; then the population at the end of the last time period will be
356.66 · e(0.021)(1) = 364.229.
That last number, 364.229, is what the program should display as the final population. (Don’t worry about that fact that many of these numbers aren’t integers.)
So, when you run your program, it should look something like this:
Enter initial population: 300
Enter first time period (in years): 4 Enter first growth rate (in percent): 1.2 Enter second time period (in years): 2.5 Enter second growth rate (in percent): 5 Enter third time period
(in years): 1 Enter third growth rate (in percent): 2.1 The final population is 364.2288848868699
(The numbers after each :, like 300 and 4 and 1.2, are user entries; the rest should be produced by the program.)
Hints: make sure you try calculating with my sample values by hand before you start programming, to make sure you understand the task! Finally, in case you are tempted to figure out a way to get
Python to “repeat itself three times” – don’t do that, just write similar code three times over (we’ll learn about repetition soon enough).
Specifications: your program must
ˆ clearly ask the user to enter the initial population.
ˆ ask the user three times to enter a number of years, and a growth rate. The program should accept the growth rates input as percents (but input without the percent sign – so “3.5%” will be input to
the program as just 3.5).
ˆ compute the final population as described above, and clearly display it. You are not required to round this value to an integer.
Challenge: try to write this program using only three variables.
2) triangle.py
Write a program that will ask the user to input three positive numbers in descending order. The program will then – using a specific procedure! – compute the three angles of the triangle that has
those three numbers as sidelengths, in degrees.
You will do this by using the Law of Cosines to find the largest angle, then using the Law of Sines to find the middle angle, and then subtract to find the smallest angle. (There are other ways to
proceed – you might say they are better ways! However, I am insisting that you do it in this manner.)
I strongly suggest you look up the Law of Sines and the Law of Cosines in case you’ve forgotten them. Also, don’t forget about converting from radians to degrees. In addition, you should Google the
trigonometric and inverse trigonometric functions that you will need for your calculations.
When you run your program, a sample run might look like this (where the user inputs 5, 4 and 3):
Enter largest side length: 5 Enter middle side length: 4 Enter smallest side length: 3 The angles are:
or this (where the user inputs 4.1, 2.6 and 2.4):
Enter largest side length: 4.1 Enter middle side length: 2.6 Enter smallest side length: 2.4 The angles are:
Hints: the largest angle is opposite the largest side, the middle angle is opposite the middle side, and the smallest angle is opposite the smallest side. Also, be very careful about order of
Specifications: your program must
ˆ ask the user to enter three side lengths in descending order (two consecutive equal sides is valid also), each of which can be any positive number – even one with decimals. You may assume that the
user obeys – if the user enters sides out of order, or negative numbers, then your program does not need to work. You may also assume that the user enters numbers that are the sidelengths of a real
triangle! That is, you don’t have to worry about the user entering 100, 2 and 1, because no real triangle has those three sidelengths (remember the triangle inequality?).
ˆ correctly calculate the three angles in the triangle, in degrees, under the assumptions mentioned above.
ˆ use the strategy I outlined above: Law of Cosines to find the largest angle, Law of Sines to find the middle angle, and subtracting to find the smallest angle.
ˆ print out the three angles in degrees, each on a separate line. You don’t have to worry about how many decimal places they are output with.
Challenge: if the user is disobedient – that is, if they enter side lengths that are either not all positive, aren’t entered in descending order, or don’t satisfy the triangle inequality – make the
program print out a message saying so instead of trying to perform the computations (which could result in an error). You might need to read ahead to do this
3) binary.py
Recall that we briefly discussed the binary system (or base two) in class. A number is represented in binary by a sequence of 1’s and 0’s (also known as bits), which have place values given by powers
of 2 instead of powers of 10. In an eight-digit binary number, the left digit would be the 128’s digit, the next digit would be the 64’s digit, the third would be the 32’s digit, followed by the
16’s, 8’s, 4’s, 2’s, and 1’s digits.
For example, 01011101 in binary would represent the (base-10) number 93, because this number has, reading from the left, no 128, one 64, no 32, one 16, one 8, one 4, no 2, and one 1, and 64 + 16 + 8
+ 4 + 1 = 93. And 00000111 would represent
the (base-10) number 7, since this number has no 128, no 64, no 32, no 16, no 8, one 4, one 2, and one 1.
Write a program that takes no input from the user, but prints a random 8-digit binary number, and its base-10 equivalent. Specifically, your program should use the random module to pick 8 random
numbers that should each be either 0 or 1. The program should then display those 8 digits in a row, together with the equivalent decimal number.
When you run your program, a sample run might look like this:
Here’s a random example of binary!
The binary number 1 0 0 0 1 1 1 0 is the same as the decimal number 142.
or this:
Here’s a random example of binary!
The binary number 0 0 1 1 0 0 1 1 is the same as the decimal number 51.
Every time I run the program, I should get a different number (between 0 and 255).
You should be able to complete this program (and the challenge) only using things we have discussed already in the class: don’t use lists or tuples (these are sequences of variables enclosed in
either [] or ()), don’t use loops, and definitely don’t use any of Python’s functions for binary (like bin). If/else statements are acceptable, but unnecessary. Instead, use mathematics –
multiplication, exponentiation, and maybe // and % (the last two might helpful for the challenge).
Specifications: your program must
ˆ accept NO user input.
ˆ use random to generate eight different random numbers – each should be 0 or 1.
ˆ display the eight generated digits, as well as the equivalent decimal number, in the manner shown above. In particular, I want to see binary numbers listed like 1 0 1 0 1 1 0 0, not like (1,
0, 1, 0, 1, 1, 0, 0). (10101100, without the spaces in between, is acceptable.)
ˆ display different numbers each time the program is run.
ˆ only use techniques that we have discussed in class thus far – in particular, no use of the bin function, no use of loops, and no use of lists or tuples.
Challenge: after you answer this question, have the program continue by asking the user to type in a number between 0 and 255 in decimal, and then printing out the binary equivalent. For example, a
run might look like this (where the user enters the value 24):
Now, enter a decimal number between 0 and 255: 24
This number is equivalent to the binary number 0 0 0 1 1 0 0 0
Need a custom answer at your budget?
This assignment has been answered 2 times in private sessions. | {"url":"https://codifytutor.com/marketplace/mth-3300-homework-2-population-triangle-and-binary-1be369ab-5737-40a4-9574-203e4614312c","timestamp":"2024-11-02T12:22:55Z","content_type":"text/html","content_length":"29726","record_id":"<urn:uuid:40f1c092-c874-48ec-9722-82465371952f>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00171.warc.gz"} |
: list solutions to periodic equations
Having had a look at
for how to formulate question so that a student can give any answer to periodic equation. I was wondering if anyone has authored or has any insight into authoring trig questions where the answer(s)
would be a list of solutions for a particular interval of the function argument. For example
calculate all the solutions to cos(\theta) = 1/2 for -2\pi < \theta < 2\pi
Thanks in advance for any help
Hi Zak,
One example of what you're looking for is done using if-then-else code blocks:
I am sure there are other examples like this in Chapter 6 and Chapter 7 of Library/LoyolaChicago/Precalc.
Best regards,
Paul Pearson | {"url":"https://webwork.maa.org/moodle/mod/forum/discuss.php?d=3223","timestamp":"2024-11-14T17:50:04Z","content_type":"text/html","content_length":"74627","record_id":"<urn:uuid:f7edbdae-3cde-4798-bf29-22177a70bdbe>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00383.warc.gz"} |
Audiogon Discussion Forum
Bryan, it's not wrong, but they're assuming you're going into a phono preamp meant for MC cartridges. And that you might want to adjust the output impedance of the transformer to fit a lower MC phono
preamp input. This is not necessary when going into a 47K MM phonostage IMO.
I'm afraid there's a lot of confused information being posted here.
The resistor sockets on the secondary side of your BentAudio Mu's are not there simply to show your phono inputs a particular impedance. Nor is BentAudio assuming you're using an MC phono stage. If
you were, you probably wouldn't need a transformer in the first place. Bent is assuming you're using a MM input with 47K ohms input impedance. The page you linked is correct.
1) Resistance on the secondary side of a transformer IS reflected back to the primary side. Your transformer's primary is connected with the cartridge coils in a closed AC circuit. The resistance
reflected from the secondary is seen as impedance by the cartridge. Adjusting that is the purpose of the Mu's resistor sockets.
2) Impedance on any AC circuit does act as a frequency tone control, as Raul said, but in this application it also acts as impedance on the electro-mechanical motors in the cartridge. If circuit
impedance is too high the cantilever will be under-damped and excessive HFs will result. (You may also get higher output, which is probably what you were hearing before you inserted the resistors.)
If impedance is too low the cantilever will be over-damped. HF response and overall dynamics will suffer.
2) The formula "25 times coil resistance" is for MC's feeding a voltage gain stage. It has nothing to do with transformers. When playing through stepup trannies most MC's like to see a reflected
impedance somewhere between 1x and 5x the coil resistance. IME low output ZYX's usually prefer something between 1.5x and 2x their 4 ohm coil resistance, ie, 6-8 ohms. (You're actually in this range
now, stay with me.)
3) When using a tranny the impedance "seen" by the primary = the resistance on the secondary side divided by the SQUARE of the tranny's turns ratio. Your 26db trannies have a 1:20 turns ratio, so
those 3.3K resistors are "seen" by the cartridge as an impedance of 3300/20^2 = 8.25 ohms.
4) Any resistor you place in the Mu's socket is in parallel with the resistance of your phono inputs. With a 47K ohm MM phono stage, the actual resistance on your secondary side can be calculated
with the formula for resistors wired in parallel:
1/R = 1/ra + 1/rb...
R = total resistance on secondary side
ra = resistance value "a" (eg, phono inputs)
rb = resistance value "b" (eg, resistor in Mu socket)
For your present setup, the formula is:
1/R = 1/47K + 1/3.3K
R = 3083 ohms
Dividing this figure by 400 (the square of your turns ratio) gives us 7.71 ohms. That is the actual impedance your ZYX is presently seeing (not counting the trivial additional impedance of your phono
cable and the Mu's wiring).
5) IME, MC's playing through a stepup are EXTREMELY sensitive to VERY small changes in reflected impedance. A change of 0.1 ohms will be more audible than a change of 25 ohms when using an MC gain
stage. In one demo I took a ZYX UNIverse from dull to perfect to bright by going from 6.54 to 6.65 to 6.71 ohms of reflected impedance.
This sensitivity is both a blessing and a curse of course.
How do you make these tiny impedance adjustments? Simple. You put two resistors in each socket of the Mu. You can calculate the total impedance on the secondary side by extending the formula above
1/R = 1/ra + 1/rb + 1/rc
Divide R by 400 to get the impedance seen by the cartridge.
For my low output ZYX I'm current using a 680 ohm and a 27Kohm resistor in each socket of my (20db, 1:10) Mu's. That works out like this:
1/R = 1/47K + 1/680 + 1/27K
R = 654 ohms
654/10^2 = 6.54 ohms
In my system with this cartrige, 6.50 ohms is too low, 6.62 ohms is too high. That's how touchy this is. I'd suggest buying some cheap resistors and experimenting. IME the same cartridge will need
slightly different values in different systems.
Unfortunately, asking the manufacturer will get you nowhere. Few MC makers offer more than general impedance guidelines, and even fewer offer any guidelines at all when using a tranny. How could
they? The turns ratios of trannies vary all over the lot, and as you can see from the formula above that has a major impact on the impedance actually seen by the cartridge.
Hope this helps more than confuses,
I have my vinyl rig sounding better then ever.
It was always better then my digital rig, but now it is light years ahead.
The performer is now in my room.
Thanks again.
Bryan P | {"url":"https://ag-forum.herokuapp.com/discussions/loading-my-xyz-3-s/post?postid=311635#311635","timestamp":"2024-11-03T00:01:32Z","content_type":"text/html","content_length":"70787","record_id":"<urn:uuid:cf9f0bf8-7233-419a-b5e2-c36e8ed87167>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00173.warc.gz"} |
EXAMPLE 1 Use a formula High-speed Train The Acela train travels between Boston and Washington, a distance of 457 miles. The trip takes 6.5 hours. What. - ppt download
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A singularity in an unsteady free-convection boundary layer
We consider the evolution of the boundary layer formed by the sudden heating of a horizontal circular cylinder in air, assuming that the Grashof number is large. To begin with the tangential velocity
vanishes at the upper generator, from symmetry considerations, but this condition is violated after a finite time and a collision then occurs. The transition between the two types of boundary layers
is manifested by a singularity and we elucidate its structure by numerical and analytic studies. It is characterised by a central region in which the inertia forces dominate, the velocity gradient in
the tangential direction becomes large and the temperature of the air is almost constant. On either side of this central region the boundary layer is of a more conventional type and formally free of
singular behavior. It is claimed that the structure proposed is also relevant to two other unsteady boundary-layer problems, in which singularities dominated by inertia forces have been found to
Quarterly Journal of Mechanics and Applied Mathematics
Pub Date:
August 1982
□ Aerothermodynamics;
□ Boundary Layer Stability;
□ Free Convection;
□ Thermal Boundary Layer;
□ Unsteady Flow;
□ Circular Cylinders;
□ Heat Transfer Coefficients;
□ Skin Friction;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1982QJMAM..35..291S/abstract","timestamp":"2024-11-14T16:00:16Z","content_type":"text/html","content_length":"36230","record_id":"<urn:uuid:2b9a2f66-4d28-418e-b448-22ccb374111f>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00576.warc.gz"} |
Maths - Dependent Types
A dependent type is modeled mathematically as a fibre bundle. We discuss fibre bundles in topology context on the page here and in the category theory context on the page here.
The theory of fibres is conventionally explained in terms of sets but here we are working with types. There is a fundamental difference between a set and a type which is that the elements (terms) of
a type don't exist outside that type. This makes working with the concept of subtype more problematic.
One of the ways of describing dependent types (on the page here) is as a type indexed over a set. The theory of fibres gives us a way to generalise this set to a more general type.
So, as an example, a vector depends on its dimension:
In set theory each element of set A corresponds to a set but, on this page, we are concerned with types, so in this diagram B (Base) and E[n] are types.
Fibre Product
We can take the product of these fibres so we have two bases (indexes - in the diagram B and C) over the same space.
An example might be matrices as a vector of vectors.
This change of indexes is related to the concept of substitution.
More information about fibre product as pullback on this page.
A fibre is a generalisation of a fibre bundle where the elements of the base space don't have to be members of the same type.
An example is one type defined 'over' another type.
We can again use vectors as an example where we could have vectors over reals or vectors over complex numbers.
Geometric Example
Shown is a common example where a spiral projects onto a circle. To reverse the direction of this projection we have to create a set (type) of points.
In this example of a spiral, it can be projected onto a circle since any local section corresponds to a section of a circle. Locally a section of the spiral looks like a section of the circle,
globaly the circle and the spiral are different. | {"url":"https://euclideanspace.com/maths/discrete/types/dependent/fibre/index.htm","timestamp":"2024-11-10T12:43:25Z","content_type":"text/html","content_length":"17139","record_id":"<urn:uuid:7891490e-e3f6-47b5-9df0-a0b649ec9263>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00708.warc.gz"} |
Sunday, June 18, 2023 - 17:15
There has been a bit of progress since the last post.
The module videos don't really give a good sense of how the app works, so this time I've tried to give some idea of how things are bolted together.
The "Map Tools" section of the app deals with ECU firmware. The flow through the modules in this section is input -> information -> output.
Input options are: - Map Finder: new firmware from database - Open File: read firmware from file (Nanocom .map, 16kb bin, 256kb bin, 256kb Kess bin, MEMS3 Flasher Map and Firmware formats) - Read
from ECU
Output options are: - Write file: write to any of the above formats. - Write ECU: map programming.
This allows the app to work as a "map wizard", file format convertor and programmer.
The video shows the Map Finder's option to display maps that suit the connected ECU.
The programming module defaults to programming the tune only if the variant code already exists on the ECU. In these cases "Overwrite Variant" gives the option to program the variant
As a bonus there is a quick tour of the Injector programming module. There isn't much to see - it edits and programs codes.
The app is currently running on macOS and Windows 10+.
I'm having some issues with PyInstaller built .exe files being identified as malware. These are false positives and its a known issue which seems to require an expensive code-signing signature to | {"url":"https://discotd5.com/","timestamp":"2024-11-09T03:37:32Z","content_type":"text/html","content_length":"52694","record_id":"<urn:uuid:9667cac9-f370-4b9d-94ee-e616fdc965bf>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00885.warc.gz"} |
Encoding zooms (scaling) with a diagonal matrix
Encoding zooms (scaling) with a diagonal matrix#
If I want to express the fact that I am expanding or contracting a coordinate along the x axis, then I multiply the x coordinate by some scalar \(p\):
\[\begin{split} \begin{bmatrix} x'\\ y'\\ z'\\ \end{bmatrix} = \begin{bmatrix} p x\\ y\\ z\\ \end{bmatrix} \end{split}\]
In general if I want to scale by \(p\) in \(x\), \(q\) in \(y\) and \(r\) in \(z\), then I could multiply each coordinate by the respective scaling:
\[\begin{split} \begin{bmatrix} x'\\ y'\\ z'\\ \end{bmatrix} = \begin{bmatrix} p x\\ q y\\ r z\\ \end{bmatrix} \end{split}\]
We can do the same thing by multiplying the coordinate by a matrix with the scaling factors on the diagonal:
\[\begin{split} \begin{bmatrix} x'\\ y'\\ z'\\ \end{bmatrix} = \begin{bmatrix} p x\\ q y\\ r z\\ \end{bmatrix} = \begin{bmatrix} p & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & r \\ \end{bmatrix} \begin{bmatrix} x
\\ y\\ z\\ \end{bmatrix} \end{split}\]
You can make these zooming matrices with np.diag:
import numpy as np
zoom_mat = np.diag([3, 4, 5])
array([[3, 0, 0],
[0, 4, 0],
[0, 0, 5]]) | {"url":"https://textbook.nipraxis.org/diagonal_zooms.html","timestamp":"2024-11-06T08:09:17Z","content_type":"text/html","content_length":"29334","record_id":"<urn:uuid:f8e38694-8385-40fb-9381-6f98a0e7b9d9>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00532.warc.gz"} |
What is the state in the WRT TQFT associated to a handlebody?
1792 views
Let $Y^3$ be a handlebody with boundary $\Sigma$. By definition, there is some associated vector $v_{WRT}(Y^3)\in Z(\Sigma)$, the (finite dimensional) Hilbert space associated to $\Sigma$ by the
Witten-Reshetikhin-Turaev TQFT. I'd like to understand what this vector is.
In short, $Z(\Sigma)$ is a space of sections of a line bundle over the $SU(2)$ character variety of $\Sigma$. I am hoping that the section $v_{WRT}(Y^3)$ achieves its maximum value (with respect to
the canonical inner product on the line bundle) on the Lagrangian submanifold of the character variety consisting of those representations which extend to $Y^3$. [EDIT: there is a good reason to
believe this holds, since then high powers of the section will concentrate on this Lagrangian, giving Volume Conjecture-like convergence to the classical Lagrangian intersection theory as the level
of the TQFT goes to infinity]
In more detail, let's discuss an explicit description of $Z(\Sigma)$. There is a natural line bundle $\mathcal L$ over the character variety $X:=\operatorname{Hom}(\pi_1(\Sigma),SU(2))/\\!/SU(2)$.
There is a natural symplectic form on $X$, and choosing a complex structure on $\Sigma$ equips $X$ with a complex structure which together with the symplectic form makes $X$ a Kahler manifold. Then
$Z(\Sigma)$ is the Hilbert space of square integrable holomorphic sections of $\mathcal L$ ($\mathcal L$ carries a natural inner product, and the curvature form of the induced connection coincides
with the natural symplectic form on $X$).
My question is then: how can one describe $v(Y^3)\in Z(\Sigma)$? Does the corresponding section achieve its maximum value on the Lagrangian subvariety of $X$ comprised of those characters of $\pi_1(\
Sigma)$ extending to characters of $\pi_1(Y)$?
A comment: answering this question for an arbitrary $3$-manifold $Y^3$ seems unlikely to yield a clean answer, since it includes as a special case calculating the value of the WRT TQFT applied to $Y$
(and the description of this requires the introduction of a whole bunch of extra stuff, e.g. surgery diagrams for $Y^3$, etc.). This is why I am restricting to the case that $Y^3$ is a handlebody, in
hopes that in this special case, there is a clean answer to this question.
This post imported from StackExchange MathOverflow at 2014-09-04 08:37 (UCT), posted by SE-user John Pardon
You can define it up to phase. The idea Is to see your setting as a fiber Bundle over Teichmuller space. There Is a projectively flat connection That relates state spaces over different Points.
Complete with stable curves. Over A surface that has been pinched down to a collection of spheres with three singular points there is a canonical vector, drag it back.
This post imported from StackExchange MathOverflow at 2014-09-04 08:37 (UCT), posted by SE-user Charlie Frohman
It is even a little more complicated than Charlie says. Not only does defining the invariant of a 3-fold require extra info (framing), but vector space associated with $\Sigma$ requires extra info to
define (they are all isomorphic, but to find a natural basis in which to specify $Z(y^3)$ you will have to address this. I can speak about all of this precisely in surgery / 4 fold terms, but no idea
how to relate it to $SU(2)$ character varieties. If you want my spiel let me know.
This post imported from StackExchange MathOverflow at 2014-09-04 08:37 (UCT), posted by SE-user Steve Sawin
One way to figure this out is to perform the path integral
$\int DA e^{iS}$
over the handlebody with boundary conditions specified by fixing the (flat) connection along the boundary Riemann surface. This number is the value of the wavefunction you want on the boundary
configuration specified. In the abelian case this should actually only depend on $g$ numbers describing the holonomies of the boundary gauge field around the cycles which bound discs in your
handle-body. In general it is a section of the line bundle you mention on the character variety.
This path integral can be (presumably) performed using the state-sum representation of the RT invariant. | {"url":"https://www.physicsoverflow.org/23199/what-is-the-state-in-the-wrt-tqft-associated-to-a-handlebody","timestamp":"2024-11-09T22:50:29Z","content_type":"text/html","content_length":"125520","record_id":"<urn:uuid:bab8d06d-a978-460c-bfbb-c2692d37fac5>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00155.warc.gz"} |
Dummy Variables and Omitted Variable Bias
Dummy Variables and Omitted Variable Bias
Dummy Variables and Omitted Variable Bias
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
2. <strong>Omitted</strong> <strong>Variable</strong> <strong>Bias</strong>So far we have assumed that the linear regression model is the correct specification of therelationship between the
dependent <strong>and</strong> the explanatory variables. But suppose it is not.If it is not the correct specification, the model is then termed “mis-specified”. There arenumber of kinds of
mis-specification <strong>and</strong> each kind has different consequences forestimation <strong>and</strong> hypothesis testing. Here we will deal with one of the most commonkinds; the omission of
an explanatory variable.The linear regression model can, of course, contain a number of explanatory variables.The number is limited by the number of observations. In general, the number
ofobservations should be several times greater than the number of explanatory variables.Nevertheless, it is still possible for an explanatory variable to be omitted either becauseits influence on the
dependent variable is unknown or because it is difficult or impossibleto find data on such a variable. We are interested in the consequences for the leastsquares estimators of this omission.We will
take the simplest possible case. Suppose the true model is;y t =a 1 x t + 2 z ta + u tt=1,2,… T (2)where u t is an unobserved r<strong>and</strong>om variable, E( ut⏐x t , z t ) = 0.However for
whatever reason the second explanatory variable z t is omitted <strong>and</strong> theeconometrician assumes that the correct model is;y t =a 1 x t + u tt=1,2,… T (3)The least squares estimate of a
1 will beâ 1 =∑ x y∑ xt t2t(4)What are the properties of â 1?As before (see Notes 3), we take the expression for â 1 (4) <strong>and</strong> substitute for y t . On thisoccasion we substitute not
from the false model (3), but from the true model (2). Thusâ 1 =∑ x (a x + a z + ut1t∑ x2 t t2t)= a 1 +a ∑ x z2∑ xt t2t+∑ x u∑ xt t2t(5)We now wish to examine whether â 1 is biased or unbiased. To do
this we takeexpectations of (5).⎛E( â 1 ) = E( a 1 ) + E⎜a2∑x z⎝ ∑ xt t2t⎞ ⎛⎟ + E⎜∑x tu2⎠ ⎝ ∑ x tt⎞⎟⎠4
e positive for theoretical reasons as well), the sign of the omitted variable bias is, in thisexample, negative.The consequence of this bias is that the estimated coefficient of unemployment in (8)
willbe smaller that it should be. You will be able to confirm that the least squares estimate ofa 1 in (8) is smaller (a larger negative number) than −0.509. The comparatively largenegative effect of
unemployment on wages which was found by estimating (8) is nowconsiderably reduced.In addition the estimated coefficient of unemployment (unlike that of inflation) is notsignificant in (9). We can
tell this by calculating the absolute value of its t ratio;0.509/0.501 = 1.016This has a t distribution with 19 degrees of freedom. We cannot reject the null hypothesisthat the coefficient of
unemployment is zero (for the 95% critical value see above).Notice that the st<strong>and</strong>ard error here (0.501) is comparatively large. The 95% confidenceinterval goes from −1.56 to 0.54.
This includes some relatively large negative numbersas well as some smaller positive ones. It is still possible that there maybe a fairlysubstantial effect of unemployment on percentage wage changes.
It would be unwise toconclude from these estimates that there is no unemployment effect. (see Tests ofSignificance in Notes 4.)It is interesting to compare the results in (9) with those when the
dummy variable is2included as in (1). Equation (1) has the higher R <strong>and</strong> thus (1) is the better “fit” to thedata. The estimates of the unemployment coefficient differ by a fairly wide
margin(− 1.709 , − 0.509). Such a difference could have important policy implications.Which is the better equation will depend on a number of factors, such as the pattern of theresiduals, which we
have not discussed. However it would seem that there are goodtheoretical reasons for including retail price inflation in the regression. It appears tocapture whatever political effects occurred in
1975 <strong>and</strong> 1980 (the residuals in (9) for 1975<strong>and</strong> 1980 are not particularly large). Where an economic explanatory variable isavailable it is to be preferred to the “ad
hoc” dummy variable. <strong>Dummy</strong> variables shouldreally be reserved for events which are completely non-economic. Thus I would preferthe estimates in (9) to those in (1).One problem with
(9), concerns causation. Was price inflation driving up wages in thisperiod, or were wage changes driving up prices? Or were both variables influencing eachother? The answers to these <strong>and</
strong> other questions will be dealt with in subsequent units inEconometrics.David WinterApril 20007 | {"url":"https://www.yumpu.com/en/document/view/40796215/dummy-variables-and-omitted-variable-bias","timestamp":"2024-11-15T01:24:59Z","content_type":"text/html","content_length":"128551","record_id":"<urn:uuid:93aade73-686e-4dbe-abf7-80aa18cb0772>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00548.warc.gz"} |
Futurebasic/Language/Reference/atanh - Wikibooks, open books for an open world
✔ Appearance ✔ Standard ✔ Console
result# = atanh( expr )
Returns the inverse hyperbolic tangent of expr. This is the inverse of the tanh function, so that atanh(tanh(x)) equals x. atanh always returns a double-precision result.
No special notes. | {"url":"https://en.m.wikibooks.org/wiki/Futurebasic/Language/Reference/atanh","timestamp":"2024-11-09T12:35:58Z","content_type":"text/html","content_length":"25961","record_id":"<urn:uuid:ec202e18-294a-4959-b3ac-f6f4fc867f50>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00155.warc.gz"} |
Synovec Corp. is experiencing rapid growth. Dividends are
expected to grow at 24 percent per year...
Synovec Corp. is experiencing rapid growth. Dividends are expected to grow at 24 percent per year...
Synovec Corp. is experiencing rapid growth. Dividends are expected to grow at 24 percent per year during the next three years, 14 percent over the following year, and then 8 percent per year,
indefinitely. The required return on this stock is 10 percent and the stock currently sells for $86 per share. What is the projected dividend for the coming year? (Do not round intermediate
calculations and round your answer to 2 decimal places, e.g., 32.16.)
Let x be the dividend paid next year. So, dividend paid for year 2 and 3 will be 1.24x and 1.24^2*x. for Year 4 will be 1.24^2*1.14*x and 1.24^2*1.14*1.08 in Year 5.
Using Dividend Discount Model (DDM), Price of this stock can be calculated using the formula: D1/(1+r)+D2/(1+r)^2+D3/(1+r)^3+D4/(1+r)^4+(D5/(r-g))/(1+r)^4; where D1 to D5 are dividends paid from Year
1 to Year 5, r is required rate of return and g is constant growth rate of dividend.
So, Price of the stock= 86= x/1.1+(1.24x)/1.1^2+(1.24^2*x)/1.1^3+(1.24^2*1.14*x)/1.1^4+((1.24^2*1.14*1.08*x)/(10%-8%))/1.1^4
86= 68.9367*x
x= $1.25
So, Projected dividend for coming year is $1.25 | {"url":"https://justaaa.com/finance/25053-synovec-corp-is-experiencing-rapid-growth","timestamp":"2024-11-09T23:44:40Z","content_type":"text/html","content_length":"41353","record_id":"<urn:uuid:5ec64cb1-49a4-4a38-9b48-370a220aab32>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00612.warc.gz"} |
Electric Potential in a Uniform Electric Field
Learning Objectives
By the end of this section, you will be able to:
• Describe the relationship between voltage and electric field.
• Derive an expression for the electric potential and electric field.
• Calculate electric field strength given distance and voltage.
In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric
field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. (See Figure 1.)
Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field.
From a physicist’s point of view, either ΔV or E can be used to describe any charge distribution. ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar
quantity and has no direction, while E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by E
below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from point A to point B.
But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an
interesting special case.
The work done by the electric field in Figure 1 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is
W = −ΔPE = −qΔV.
The potential difference between points A and B is
−ΔV = −(V[B] − V[A]) = V[A] − V[B] = V[AB].
Entering this into the expression for work yields W = qV[AB].
Work is W = Fd cos θ; here cos θ = 1, since the path is parallel to the field, and so W = Fd. Since F = qE, we see that W = qEd. Substituting this expression for work into the previous equation
gives qEd = qV[AB].
The charge cancels, and so the voltage between points A and B is seen to be
[latex]\begin{cases}V_{\text{AB}}&=&Ed\\E&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (uniform E − field only)
where d is the distance from A to B, or the distance between the plates in Figure 1. Note that the above equation implies the units for electric field are volts per meter. We already know the units
for electric field are newtons per coulomb; thus the following relation among units is valid: 1 N/C = 1 V/m.
Voltage between Points A and B
[latex]\begin{cases}V_{\text{AB}}&=&Ed\\E&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (uniform E − field only)
where d is the distance from A to B, or the distance between the plates.
Example 1. What Is the Highest Voltage Possible between Two Plates?
Dry air will support a maximum electric field strength of about 3.0 × 10^6 V/m. Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge
or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?
We are given the maximum electric field E between the plates and the distance d between them. The equation V[AB] = Ed can thus be used to calculate the maximum voltage.
The potential difference or voltage between the plates is
V[AB] = Ed.
Entering the given values for E and d gives
V[AB] = (3.0 × 10^6 V/m)(0.025 m) 7.5 × 10^4 V or V[AB] = 75 kV.
(The answer is quoted to only two digits, since the maximum field strength is approximate.)
One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between
conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks
down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.
Example 2. Field and Force inside an Electron Gun
1. An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates?
2. What force would this field exert on a piece of plastic with a 0.500 μC charge that gets between the plates?
Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression [latex]E=\frac{V_{\text{AB}}}{d}\\[/latex]. Once the electric field
strength is known, the force on a charge is found using F = qE. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F = qE.
Solution for Part 1
The expression for the magnitude of the electric field between two uniform metal plates is
Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for V[AB] and the plate separation of 0.0400 m, we obtain
[latex]E=\frac{25.0\text{ kV}}{0.0400\text{ m}}=6.25\times10^5\text{ V/m}\\[/latex]
Solution for Part 2
The magnitude of the force on a charge in an electric field is obtained from the equation F = qE.
Substituting known values gives
F = (0.500 × 10^−6 C)(6.25 × 10^5 V/m) = 0.313 N.
Note that the units are newtons, since 1 V/m = 1 N/C. The force on the charge is the same no matter where the charge is located between the plates. This is because the electric field is uniform
between the plates.
In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of E
and also in the direction of lower potential V. Furthermore, the magnitude of E equals the rate of decrease of V with distance. The faster V decreases over distance, the greater the electric field.
In equation form, the general relationship between voltage and electric field is
where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the
gradient (as in grade or slope) of the electric potential.
Relationship between Voltage and Electric Field
In equation form, the general relationship between voltage and electric field is
where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the
gradient (as in grade or slope) of the electric potential.
For continually changing potentials, ΔV and Δs become infinitesimals and differential calculus must be employed to determine the electric field.
Section Summary
• The voltage between points A and B is
[latex]\begin{cases}V_{\text{AB}}&=&Ed\\E&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (uniform E − field only)
where d is the distance from A to B, or the distance between the plates.
• In equation form, the general relationship between voltage and electric field is
where Δs is the distance over which the change in potential, ΔV, takes place. The minus sign tells us that E points in the direction of decreasing potential.) The electric field is said to be the
gradient (as in grade or slope) of the electric potential.
Conceptual Questions
1. Discuss how potential difference and electric field strength are related. Give an example.
2. What is the strength of the electric field in a region where the electric potential is constant?
3. Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.
Problems & Exercises
1. Show that units of V/m and N/C for electric field strength are indeed equivalent.
2. What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50 × 10^4 V?
3. The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50 × 10^4 V/m. (a) What is the potential difference between the plates? (b) The plate with the lowest
potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?
4. How far apart are two conducting plates that have an electric field strength of 4.50× 10^3 V/m between them, if their potential difference is 15.0 kV?
5. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air (3.0 × 10^6 V/m) if the plates are separated by 2.00 mm and a potential
difference of 5.0 × 10^3 V is applied? (b) How close together can the plates be with this applied voltage?
6. The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are
discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms.) You may assume a uniform electric field.
7. Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What
is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.
8. Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero
volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?
9. Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be 3.0 × 10^6 V/m.
10. A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the
11. An electron is to be accelerated in a uniform electric field having a strength of 2.00 × 10^6 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over
what distance would it have to be accelerated to increase its energy by 50.0 GeV?
scalar: physical quantity with magnitude but no direction
vector: physical quantity with both magnitude and direction
Selected Solutions to Problems & Exercises
3. (a) 3.00 kV; (b) 750 V
5. (a) No. The electric field strength between the plates is 2.5 × 10^6 V/m, which is lower than the breakdown strength for air (3.0 × 10^6 V/m}); (b) 1.7 mm
7. 44.0 mV
9. 15 kV
11. (a) 800 KeV; (b) 25.0 km | {"url":"https://courses.lumenlearning.com/suny-physics/chapter/19-2-electric-potential-in-a-uniform-electric-field/","timestamp":"2024-11-11T07:04:19Z","content_type":"text/html","content_length":"61637","record_id":"<urn:uuid:595f6af6-cba7-46a4-90af-d78c0741419f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00538.warc.gz"} |
Interaction contrast after I got the estimation of parameter by PROC GLM
I intend to test a interaction between two variables, but not via multiplicative interaction variable (e.g., Y = beta0 + beta1*X1 + beta2*X2 + beta3*X1*X2).
My GLM model is Y = beta0 + beta1*X1 + beta2*X2. After the PROC GLM I got the estimation of beta1 and beta2 with their 95% confidence intervals. Both X1 and X2 are binary variables, means they either
can be 0 or 1, respectively.
Now I want to estimate the interaction contrast (additive interaction instead of multiplicative one), defined as,
Interaction contrast = Y11 - Y01 - Y10 + Y00, and to compare it with ZERO. It would be better to have its 95% confidence interval. However, I don't know how to. Thanks.
04-03-2021 06:28 AM | {"url":"https://communities.sas.com/t5/Statistical-Procedures/Interaction-contrast-after-I-got-the-estimation-of-parameter-by/m-p/731122","timestamp":"2024-11-10T21:19:18Z","content_type":"text/html","content_length":"251868","record_id":"<urn:uuid:2e442e7f-216e-4c16-8c3f-7dc26d7bf865>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00834.warc.gz"} |
Alessandro Mininno, PhD
Researcher in Theoretical High Energy Physics
I am a postDoc researcher, from October 2024 at the Department of Physics at the University of Wisconsin-Madison.
Here you can find information about my research interests, my studies and how to contact me.
My research focuses on String Theory, String Phenomenology, but also Superconformal Field Theories in various dimensions.
In the context of String Phenomenology, I am mainly interested in the so-called Swampland Program, aimed to find constraints that any Effective Field Theory coupled to gravity must satisfy in order
to be compatible with Quantum Gravity. Most of the time, such constraints are expressed in terms of conjecture, and I focus on various versions of Weak Gravity Conjecture that, together with my
collaborators, we tested in different set-ups, such as type-II / F-theory and M-theory compactifications.
I am also interested in formal aspects of String compactifications, such study of supersymmetric vacua with minimal supersymmetry.
My other research interest are Argyres-Douglas Theories, and their compactifications in 3d. In the past years, we studied generalized global symmetries of these theories, such as 1-form and 2-group
symmetries, as well as non-invertible symmetries.
One can find selected papers per year in my Publications section, as well as my complete list of publications.
2024 - ...
University of Wisconsin-Madison
Postdoctoral Researcher in Theoretical High Energy Physics
Group Leader: Prof. G. Shiu
2021 - 2024
II. Institut für Theoretische Physik, UHH
Postdoctoral Researcher in Theoretical High Energy Physics
Group Leader: Prof. T. Weigand
2023 - 2024
Introduction to Solid State Physics
Teaching assistant at University of Hamburg
Professors: Prof. Nils Huse and Prof. Christian Schroer
General Relativity
Teaching Assistant at University of Hamburg
Professor: Prof. Gudrid Moortgat-Pick
2022 - 2023
Quantum Field Theory 1
Teaching Assistant at University of Hamburg
Professor: Prof. Timo Weigand
Mathematical Foundations of Physics
Teaching Assistant at University of Hamburg
Professor: Prof. Gleb Arutyunov
2021 - 2022
Quantum Field Theory 1
Teaching assistant at University of Hamburg
Professor: Prof. Timo Weigand
Quantum Field Theory 2
Teaching assistant at University of Hamburg
Professor: Prof. Timo Weigand | {"url":"https://www.amininno.com/home","timestamp":"2024-11-11T01:58:46Z","content_type":"text/html","content_length":"113883","record_id":"<urn:uuid:71f4c5e4-642c-4a72-9e50-96c6b940a2c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00081.warc.gz"} |
Math in Motion | Texas Instruments
Students write their own programs in the Math in Motion Plus activities to control Rover, or students can explore math concepts using pre-made programs in the Math in Motion activities. Either way,
students will experience math in a whole new way!
Move the Cone
In this Math in Motion Plus activity, students will write their own program for the TI-Innovator™ Rover. They will use several commonly used Rover and basic calculator commands while exercising their
knowledge of polygons, the coordinate plane, and equations of lines to write code to have the Rover complete a series of challenges. Students will progress through a series of challenges to build
skills and then attempt a more complex final challenge. Student “Challenge” cards have been provided and can be printed/cut into individual task cards, and distributed to students (optional).
Material List:
• 3’x3’ Butcher paper
• 3 colors of sticky dots
• Miniature traffic cones (or some other object(s) to move)
Math Topic: Algebra 1 — the coordinate plane, collinear points, equations of lines
TI-84 Plus CE activity files TI-84 Plus CE Python activity files TI-Nspire™ activity files TI-Nspire™ CX II Python activity files
Navigate “Math-hattan” Challenge
In this Math in Motion Plus activity, students will write their own program for the TI-Innovator™ Rover. Students will use measurement of distance and angles to write code to explore a set of
preliminary challenges. Students will then apply their knowledge of the relationships among distance, rate and time to write equations to optimize routes through a “city” (Math-hattan) or maze with
obstacles. Student “Challenge” cards have been provided and can be printed/cut into individual task cards, and distributed to students (optional).
Material List:
• “City Scape”- can be constructed of various materials (see activity sheet for options)
• Masking tape
• Meter sticks or measuring tapes and protractors
• Miniature traffic cones to temporarily block paths, and challenge students as needed, once activity has begun
Math Topic: Algebra I — Linear Equations and Distance, Rate and Time
TI-84 Plus CE activity files TI-84 Plus CE Python activity files TI-Nspire™ activity files TI-Nspire™ CX II Python activity files
Drive the Line Challenge
In this Math in Motion Plus activity, students will write their own program for the TI-Innovator™ Rover. They will explore multiple representations of position, velocity and time to write code that
will navigate a set of challenges. Students will apply their knowledge of the relationships between position vs. time graphs and velocity vs. time graphs, and their verbal descriptions, to write
programs for Rover to carry out the motion described. The “Challenge” cards provided can be printed and/or cut into individual task cards and distributed to students (optional).
Material List:
• Number line for positioning Rover for each challenge (scale should use 10 cm units as shown in the activity sheet)
Math Topic: Algebra I — Linear Equations modeling position as a function of time
TI-84 Plus CE activity files TI-84 Plus CE Python activity files TI-Nspire™ activity files TI-Nspire™ CX II Python activity files
Driving Inequalities Challenge
In this Math in Motion Plus activity, students will write their own program for the TI-Innovator™ Rover. Students will explore inequalities and the number line while writing code to navigate a set of
challenges. They will apply their knowledge of compound inequalities to write programs for Rover to demonstrate the inequalities on the number line. This activity is appropriate for students familiar
with graphing inequalities on a number line diagram. Student “Challenge” cards have been provided and can be printed/cut into individual task cards, and distributed to students (optional).
Material List:
• Number line for position. Use a 2 meter course, from -10 units to +10 units. (1 Rover unit = 10 cm)
Math Topic: Middle Grades — inequalities, number lines
TI-84 Plus CE activity files TI-84 Plus CE Python activity files TI-Nspire™ activity files TI-Nspire™ CX II Python activity files
Students write their own programs in the Math in Motion Plus activities to control Rover, or students can explore math concepts using pre-made programs in the Math in Motion activities. Either way,
students will experience math in a whole new way!
Design a Path
Students will use the TI-Innovator™ Rover and the provided file to design and measure a segmented path. They will then verify their measurements and directions using Rover with the provided file.
Material List:
• Meter sticks
• Protractors
• Rulers
• Butcher paper
• Tape (painters tape is recommended)
Math Topic: Geometry — Angle Measurements
TI-84 Plus CE activity files TI-Nspire™ activity files
Rover Rate
Students will use the TI-Innovator™ Rover and the provided files to explore the relationship between distance, rate and time. The students will relate their findings to the actions of Rover and
understand they have the ability to apply this relationship to control how it drives a given path.
Material List:
• Meter stick
• Stopwatch (One per student group)
• 1.5 by 2.5 meters of floor or table space
• Tape (painters tape is recommended)
Math Topic: Middle Grades — Distance, Rate and Time
TI-84 Plus CE activity files TI-Nspire™ activity files
Two Rovers Leave the Station ...
Students will use the TI-Innovator™ Rover and the provided file to help illustrate an intersection point of two lines. Students will crash two Rovers into each other. The crash will use the
horizontal distance from the origin as the dependent variable and time as the independent variable. The velocity is the slope of the linear equation, and the starting distance from the origin is the
Material List:
• Meter stick
• Tape (painters tape is recommended)
Math Topic: Algebra I — Intersection of Lines, System of Equations
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Watts to Lux Calculator | Convert Light Power to Illuminance
Watts to Lux Calculator
What are Watts and Lux?
Watts (W) is a unit of power that measures the rate of energy transfer. In lighting, it represents the amount of electrical power consumed by a light source. Lux (lx) is the SI unit of illuminance,
which measures the amount of light that falls on a surface area.
The Watts to Lux Formula
The conversion from watts to lux involves two steps:
1. Convert watts to lumens: \[\text{Lumens} = \text{Watts} \times \text{Efficacy}\] Where efficacy is measured in lumens per watt (lm/W).
2. Convert lumens to lux: \[\text{Lux} = \frac{\text{Lumens}}{\text{Surface Area}}\] Where surface area is measured in square meters (m²).
Combining these steps, we get:
\[\text{Lux} = \frac{\text{Watts} \times \text{Efficacy}}{\text{Surface Area}}\]
• Lux (lx) = Illuminance
• Watts (W) = Power of the light source
• Efficacy (lm/W) = Luminous efficacy of the light source
• Surface Area (m²) = Area illuminated by the light source
Step-by-Step Watts to Lux Calculation
1. Identify the power of the light source in watts (W).
2. Determine the luminous efficacy of the light source in lumens per watt (lm/W).
3. Measure or calculate the surface area being illuminated in square meters (m²).
4. Multiply the watts by the efficacy to get lumens.
5. Divide the lumens by the surface area to get lux.
Example Calculation
Let's calculate the illuminance for a 100W LED light with an efficacy of 100 lm/W, illuminating an area of 10 m²:
1. Power = 100 W
2. Efficacy = 100 lm/W
3. Surface Area = 10 m²
4. Lumens = 100 W × 100 lm/W = 10,000 lm
5. Lux = 10,000 lm ÷ 10 m² = 1,000 lx
Visual Representation
This diagram illustrates how a 100W LED light source with an efficacy of 100 lm/W produces an illuminance of 1,000 lux over a surface area of 10 m². The red point represents the light source, and the
yellow circle shows the illuminated area. | {"url":"https://www.thinkcalculator.com/light/watt-to-lux-calculator.php","timestamp":"2024-11-08T21:44:42Z","content_type":"text/html","content_length":"36396","record_id":"<urn:uuid:e7a92a19-2e92-40ee-addd-dbe0df5a4df7>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00237.warc.gz"} |
How do I ensure that my isometric drawing assignment meets all requirements? | AutoCAD Assignment Help
How do I ensure that my isometric drawing assignment meets all requirements? My student was told by the teachers in my class that I should be able to do the graduation from the student’s exam and if
he can access my class web page, but since that was not the best way to do so, I was concerned. I found a way for him to see what the goals have been and found if they are similar. They didn’t use a
formula to meet the criteria and they did some basic maths & trigonometry see it here This is my first time teaching Math and trigonometry, so hopefully it will help. My school year: 2009. What is
your model school/work experiences? Is math curriculum your school/work topic? I would like to know more details about how you think your college/work experience will or should be shaped since it can
be a subject which you are most satisfied with? A: The school at your school has one of the widest set of subjects to consider and the only school that has the most out of every subject is the Maths.
The Math that you teach will include all basic math will be very much in the spirit of Math teacher guide. A good math teacher provides methods to your Math system which you can implement if given
the opportunity. Some basic Math is important to any student. Some general understanding of basic math may be important as a minor. Note that if it is you who are likely to be teaching, you should
really be targeting at the highest level possible, it is just as much for students as it is for teachers from the beginning. Regarding the other way to think about what you are doing, there is a lot
of research which isn’t available in one place. It is also a great chance for you to look into the subject of the teacher. If the Math that is taught is general, you may also focus on specific skills
or knowledge to do with skills that will be considered the same skill as the Math teacher. About go mathematics style In my experience, the Math that is taught has a very very sensitive discipline.
First you will want to know the context of which part of the world is going to be made up, and the history. Then you will want some sort of analogy and generalize it to other parts of the world. The
first thing to do is to ask yourself the same question for the start and end of the Math. On some research and reading that would help, I’ve found some relevant knowledge to do that and it is quite
enlightening. How do I ensure that my isometric drawing assignment meets all requirements? I have tried the following, but I dont have an idea of what I can do.
Is The Exam Of Nptel In Online?
I don’t need the “paper” being a point on the page, and I know the paper object really does not have “the page” element. I just want the paper an element to “stand”. What is the approach you
suggested? I’m checking whether the paper should “stand” for my isometric drawing test. I’m not sure who is making this decision (maybe I just tried too many papers beforehand, or was hoping for
something a little more simple?). I would like to know more about why that is, before I waste the precious while’s. Thanks in advance….. A: Create a new Sheet and store your isometric signature on
the page. Then you add something to create the paper being drawn. You can use paintComponent to draw the signature to the page, then repeat. I would expect it to be very easy, I have no experience
with paper writing, so you can refer here. How do I ensure that my isometric drawing assignment meets all requirements? As explained herein, it is important to note that any attempt to image an
isometric drawing is not just that, the drawing of the object is itself and its relationship must be the same throughout a series of drawing operations. How it is described in this article can also
vary and vary too on how the objects they are described are designed, the size and number of points that need to be captured being presented. I have not presented you with any specific instructions
and without further information, I will merely review other related material. As with any field of research with representation and theory, I will not go into detail until I have read the article
published here. Rounding out with the Nested Loop One of the most important developments in this field takes place in this article, section 7, which presents an architectural framework which allows
for the building of many objects using nested loops. I have chosen this structure because it is similar to one used by Lego so I will refer to it as such in this chapter for simplicity.
Best Online Class Help
You can see from the following picture that the nested loop allows for an unlimited set of instructions to repeatedly execute, at a set of steps. This can be achieved only just by using the
JavaScript to get the original bounding box of a given object and then performing the addition to that object. This method is called a conditional loop function squareRect(innerWidth, innerHeight) {
// If outerWidth > innerHeight > outerWidth set outerHeight to zero. If innerWidth < outerHeight set innerHeight to 2.0 seconds, at a time offset from outerHeight. if (innerWidth > outerWidth ) { //
If innerHeight > innerWidth set innerLength to 0 milliseconds. parent.innerVisible = false if (outerHeight > innerHeight ) { // If innerHeight – innerWidth < outerHeight - innerHeight, return an
infinite loop. return 1.0 - (innerHeight < outerHeight ? 1.0 - 2.0 : ((innerHeight > outerHeight)? 1.0 : 0) : 5.0) // – 45.6 seconds } } parent.innerVisible = true // Inside of child of current
block. Use this to get the global pointer to the window. go to website 0, 1); var window = $(‘#draw’).
Easy E2020 Courses
val(); // Get the global window object that the rest of the commands look for. // Now you can apply your window commands to your object. if (window && window.prototype.ins()!= window.prototype.dom) {
// Advance forward. var stack = arrayOf.slice(arrayWidth, arrayHeight) arrayWidth = stack.length arrayHeight = stack.length // Check if line-width-relative to the node box should give a different
line-width-relative value. var lineWidth = stack.sub(0).indexOf(10)!== -1 if ((–stack.length) > 1) { | {"url":"https://autocadhelp.net/how-do-i-ensure-that-my-isometric-drawing-assignment-meets-all-requirements","timestamp":"2024-11-10T20:37:52Z","content_type":"text/html","content_length":"87332","record_id":"<urn:uuid:33e2f7f5-f1e9-40b4-ad4a-55133e9d0192>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00333.warc.gz"} |
(3) Calculate the perimeter and area of the triangle in the pic... | Filo
Question asked by Filo student
(3) Calculate the perimeter and area of the triangle in the picture. is a natural number or a fraction. 1) (i) iiin (i)) x) Division the product can also be writt In general, for any natural number
or fraction , we as and the division as and
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(3) Calculate the perimeter and area of the triangle in the picture. is a natural number or a fraction. 1) (i) iiin (i)) x) Division the product can also be writt In general, for any
Question Text natural number or fraction , we as and the division as and
Updated On Dec 8, 2022
Topic All topics
Subject Mathematics
Class Class 9
Answer Type Video solution: 1
Upvotes 132
Avg. Video 4 min | {"url":"https://askfilo.com/user-question-answers-mathematics/3-calculate-the-perimeter-and-area-of-the-triangle-in-the-33323330363434","timestamp":"2024-11-12T12:10:08Z","content_type":"text/html","content_length":"316875","record_id":"<urn:uuid:efda0e8e-9e1e-40ed-95f0-4f6bc9d1111a>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00584.warc.gz"} |
Cioppa's Nearly Orthogonal Latin Hypercube Designs
nolhDesign {DiceDesign} R Documentation
Cioppa's Nearly Orthogonal Latin Hypercube Designs
This function generates a NOLH design of dimension 2 to 29 and normalizes it to the selected range. The design is extracted from Cioppa's NOLHdesigns list.
nolhDesign(dimension, range = c(0, 1))
dimension number of input variables
range the scale (min and max) of the inputs. Range (0, 0) and (1, 1) are special cases and call integer ranges (-m, m) and (0, 2m). See the examples
A list with components:
n the number of lines/experiments
dimension the number of columns/input variables
design the design of experiments
T.M. Cioppa for the designs. P. Kiener for the R code.
See Also
Cioppa's list NOLHdesigns. Other NOLH and OLH designs: nolhdrDesign, olhDesign.
## Classical normalizations
nolhDesign(8, range = c(1, 1))
nolhDesign(8, range = c(0, 0))
nolhDesign(8, range = c(0, 1))
nolhDesign(8, range = c(-1, 1))
## Change the dimnames, adjust to range (-10, 10) and round to 2 digits
xDRDN(nolhDesign(8), letter = "T", dgts = 2, range = c(-10, 10))
## A list of designs
lapply(5:9, function(n) nolhDesign(n, range = c(-1, 1))$design)
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Get Started with Problem-Based Optimization and Equations
Get started with problem-based setup
To solve a problem using the problem-based approach, perform these steps.
• Create an optimization problem using optimproblem or an equation-solving problem using eqnproblem.
• Create optimization variables using optimvar.
• Create expressions using the optimization variables representing the objective, constraints, or equations. Place the expressions into the problem using dot notation, such as
prob.Objective = expression1;
probl.Constraints.ineq = ineq1;
• For nonlinear problems, create an initial point x0 as a structure, with the names of the optimization variables as the fields.
• Solve the problem by calling solve.
To improve your setup, increase performance, or learn details about problem-based setup, see Improve Problem-Based Organization and Performance.
For parallel computing in Optimization Toolbox™, see the last section; for parallel computing in Global Optimization Toolbox, see How to Use Parallel Processing in Global Optimization Toolbox (Global
Optimization Toolbox).
Create Variables and Problem
eqnproblem Create equation problem
optimproblem Create optimization problem
optimvalues Create values for optimization problem (Since R2022a)
optimvar Create optimization variables
show Display information about optimization object
showbounds Display variable bounds
write Save optimization object description
writebounds Save description of variable bounds
Create Expressions, Constraints, and Equations
fcn2optimexpr Convert function to optimization expression
optimconstr Create empty optimization constraint array
optimeq Create empty optimization equality array
optimineq Create empty optimization inequality array
optimexpr Create empty optimization expression array
show Display information about optimization object
write Save optimization object description
Solve and Analyze
evaluate Evaluate optimization expression or objectives and constraints in problem
findindex Find numeric index equivalents of named index variables
infeasibility Constraint violation at a point
issatisfied Constraint satisfaction of an optimization problem at a set of points (Since R2024a)
paretoplot Pareto plot of multiobjective values (Since R2022a)
prob2struct Convert optimization problem or equation problem to solver form
show Display information about optimization object
solve Solve optimization problem or equation problem
solvers Determine default and valid solvers for optimization problem or equation problem (Since R2022b)
varindex Map problem variables to solver-based variable index
write Save optimization object description
EquationProblem System of nonlinear equations
OptimizationConstraint Optimization constraints
OptimizationEquality Equalities and equality constraints
OptimizationExpression Arithmetic or functional expression in terms of optimization variables
OptimizationInequality Inequality constraints
OptimizationProblem Optimization problem
OptimizationValues Values for optimization problems (Since R2022a)
OptimizationVariable Variable for optimization
Live Editor Tasks
Optimize Optimize or solve equations in the Live Editor (Since R2020b)
• Variables with Duplicate Names Disallowed
Learn how to solve a problem that has two optimization variables with the same name.
• Expression Contains Inf or NaN
Optimization expressions containing Inf or NaN cannot be displayed, and can cause unexpected results.
Tune and Monitor Solution Process
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Slow is Smooth. Smooth is Fast.
Ok, here’s the deal:
I feel like the way most of us currently assess and view mathematical fluency and ability (timed tests) is complete garbage.
Now that I have that out of the way, let me explain myself.
I took German in high school. I can’t speak German, but I can recognize words and phrases, and I can usually tell if someone is speaking German. I can spout off phrases and words that I’ve committed
to memory, but I wouldn’t consider myself even close to being fluent in German.
Math fluency is a lot like learning to speak a foreign language. Simply committing a series of facts or procedural steps to memory is no more a sign of math fluency (or numeracy) than my ability to
speak German.
In English, I can carry an intelligible (perhaps intelligent even) conversation and think of many ways to express myself so that I will be understood by others. I can also understand what is being
shared with me in the same language when it’s communicated in a variety of contexts, accents, dialects, and some colloquialisms or slang. If I’m communicating with someone who has a lesser command of
the English language, I employ more simplistic words and phrases. Similarly, I can dip into a deep well of rich vocabulary when conversing with someone who is well-versed in my native tongue. Quite
clearly, I am fluent in English.
Likewise, when individuals are fluent with math, they can speak about concepts in an intelligible manner and communicate their understanding in a variety of contexts for a wide range of audiences.
Those with an understanding of the properties of numbers and the operations performed on those numbers are able to choose from multiple pathways to solve problems. People who are mathematically
fluent are not limited to any one singular method or algorithm but are able to choose an efficient strategy based on the context of the problem.
Another important consideration is that speed is a byproduct of fluency. Too often, we focus on the speed with which an individual can solve a mathematical task or recall a fact from memory,
mistaking it as the telltale marker of true mathematical ability or fact fluency. Yes, a reasonable rate of speed is desirable, but oftentimes, the cart is put before the horse; fluency begets speed.
While I was in the Marine Corps, we had a phrase, “Slow is smooth. Smooth is fast.” Once you understand a concept and can carry it out accurately and proficiently, speed can be increased without
additional stress or missteps taking place. We ought to work on building speed from conceptual understanding and repeated practice over time.
Some might assume that if we can just get a student to memorize their facts or the steps to a procedure, increasing the speed with which they solve problems, they’ll make sense of it later. NCTM
(National Council of Teachers of Mathematics), however, would disagree.
Research suggests that once students have memorized and practiced procedures that they do not understand, they have less motivation to understand their meaning or the reasoning behind them.
NCTM Standards and Positions: Procedural Fluency in Mathematics
Just as wise individuals in meaningful conversations often pause to measure the weight of their words before letting them loose from their lips, or a skilled writer allows themselves to wait briefly
before putting pen to paper, being sure of the same, so also those who are fluent mathematically may take a brief moment before being sure they have calculated correctly.
Computational fluency refers to having efficient and accurate methods for computing. Students exhibit computational fluency when they demonstrate flexibility in the computational methods they
choose, understand and can explain these methods, and produce accurate answers efficiently. The computational methods that a student uses should be based on mathematical ideas that the student
understands well, including the structure of the base-ten number system, properties of multiplication and division, and number relationships.
NCTM. Principles and Standards for School Mathematics, p. 152
Clearly, there is much more to being fluent in mathematics than computing quickly. Simply committing a series of math facts or procedural steps to memory does not make one mathematically fluent.
Mathematically fluent individuals, like those fluent in a language, possess the ability to think flexibly and communicate their reasoning effectively, in a variety of contexts, for wide ranges of
I feel we need to assess students for true understanding and not just speed. Are they able to communicate their understanding in an intelligible manner? Can they use effective and efficient
strategies to arrive at correct solutions? Are they limited to using one method or strategy to solve a particular type of problem, or do they have options? Let’s not mistake memorization of facts and
procedures or speed as a true measure of mathematical ability of fluency.
What’s you take on math fluency? How are you assessing fluency in your classroom or with your student(s)?
2 Comments
1. Hi Shawn,
Nice article. I use Cathy Fosnot’s work, including her number strings almost daily. Her assessment app has something called “2 pen” assessments which are great assessments for fluency. Have you
used her work?
2. Thanks, Emily! I recently purchased Conferring with Young Mathematicians at Work, but I haven’t got around to reading it yet. Is this where students use two colors and switch colors after time
runs out? If so, I’ve done that in the past. | {"url":"https://shawnseeley.com/slow-is-smooth-smooth-is-fast/","timestamp":"2024-11-14T04:16:28Z","content_type":"text/html","content_length":"90261","record_id":"<urn:uuid:48eb225c-4108-4906-96c7-7606e81d0362>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00311.warc.gz"} |
Rydberg Atoms for a Quantum Computer
Rydberg Atoms for a Quantum Computer
Experimental investigations of atoms in highly excited states are performed at the Institute of Semiconductor Physics SB RAS. It is expected that these atoms can be used as fast logic gates to a
quantum computer
The origins of quantum information theory are associated with Russian mathematician Yury Manin and American physicist Richard Feynman, who, more than thirty years ago, suggested using quantum systems
(groups of microparticles with a discrete set of states) to calculate molecular dynamics and quantum effects.
As a basic example of a quantum computer, let us consider a system, which consists of a number of quantum objects. Each of them can be found only in two states, which are associated with the logical
values “true” and “false.” These quantum objects are termed quantum bits or qubits, and the whole quantum system is termed a quantum register. A classical bit has to be either in the “true” state or
in the “false” state at each particular moment in time, whereas a quantum bit can be in a superposition of both states at the same time. Hence, we can only speak about a probability to find a qubit
in each of the two states.
If qubits are interacting with each other, their quantum states are entangled; therefore, measuring the quantum state of one of the qubits will instantly change the quantum state of the other qubits.
The consequent features of quantum computing (for example, quantum parallelism) lead to great advantages of the quantum computer in solving certain mathematical problems, which are considered
incomputable on classical computers. This includes factorization of very large numbers, which is used in modern cryptography, and search in very large unsorted databases.
The simplest quantum computations were demonstrated on organic molecules using nuclear magnetic resonances in liquids. On the other hand, a solid-state quantum computer seems to be more natural and
easier for implementation. Today, the most advantageous results in implementing simple quantum algorithms are obtained with single ions in electrostatic ion traps. However, it is difficult to
construct a large quantum register using this method, since the charged ions always strongly interact with each other and with the neighborhood. That leads to strong decoherence, which means
destruction of the quantum state of the register.
Single neutral atoms can be more promising as qubits of a quantum computer. A pair of long-lived hyperfine sublevels of the ground state of an atom can be considered as the two states of a qubit,
which are appropriate for storing information for a long period. The quantum state of a qubit can be controlled by sequences of laser pulses; and the quantum register of arbitrary size can be
constructed from the neutal atoms, trapped in the optical lattice created by laser radiation. The greatest advantage of neutral atoms is the possibility to control their interaction via laser
excitation of highly-excited energy levels. Highly-excited, or Rydberg, atoms have numerous unique properties, including large lifetimes and the ability to interact at large distances. Their
interaction can be switched on and off by laser excitation and de-excitation of atoms.
Recently, success in implementating quantum logical operations with single atoms was reported by a group of researchers from Wisconsin-Madison University, USA. They have demonstrated controllable
inversion of the quantum state of an atom, which corresponds to the “Controlled logical NOT” operation, the most important operation for a universal quantum computer. The sequence of the laser pulses
required to perform this operation lasted 7 microseconds. However, the full experimental cycle, including preparation of the qubits and detection of the final quantum states, lasted about a second.
So, a classical computer remains faster yet.
In Russia, implementation of quantum computing based on neutral atoms is studied by researchers at the Institute of Semiconductor Physics SB RAS. They are performing experiments with ultracold
rubidium atoms, which are excited into Rydberg states by pulsed lasers with large repetition frequency. One of the most important achievements of the laboratory is observation of the
electrically-controlled resonant dipole interaction of cold Rydberg atoms. An original method of rapid determinating of the number of atoms excited by each laser pulse was also developed here. The
method will help to observe the effect of dipole blockade: in an ensemble of strongly interacting atoms only one of them can be excited into the Rydberg state by laser radiation. This effect can be
used for developing fast quantum logic gates; consequently, detailed investigation of dipole blockade is being planned.
The next step will be implementing the two-qubit quantum phase gate in spatially separated optical dipole traps. Inversion of the phase of the composite wavefunction is expected to occur, which is
required for a “Controlled logical Not” quantum gate.
Quantum computing is one of the most challenging tasks for scientists all over the world. Solving this problem will lead to significant progress not only in physics, mathematics, an information
technology, but in academic research overall.
Jaksh D., Cirac J. I., Zoller P., Rolston S. L. et al. Quantum Gates for Neutral Atoms // Phys. Rev. Lett. 2000. V. 85. P. 2208. Manin Yu. I. Computable and Incomputable (In Russian). M.: Soviet
Radio, USSR, 1980.
Ryabtsev I. I., Tretyakov D. B., Beterov I. I., Entin V. M. Observation of the Stark-tuned Forster resonance between two Rydberg atoms // Phys. Rev. Lett. 2010. V. 104. P. 073003. Valiev K. A.
Quantum computers and quantum computations // PHYS-USP, 2005. 48 (1). P. 1—36.
This work was supported by the Presidential grant MK-6386.2010.2,
RFBR grants 10-02-92624, 09-02-90427, 10-02-00133,
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