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Divisibility Worksheet Answer Key - Divisonworksheets.com
Divisibility Worksheet Answer Key
Divisibility Worksheet Answer Key – Let your child learn about division with division worksheets. There are numerous types of worksheets available and you are able to design your own. These
worksheets are amazing because they are available for download at no cost and modify the exact layout you desire them. These worksheets are ideal for first-graders and kindergarteners.
Two people can perform huge quantities of work
It is crucial for kids to work on division worksheets. Many worksheets only allow for two, three or even four different divisors. Your child won’t need worry about forgetting to divide the big number
or making mistakes in their tables of times because of this method. You can find worksheets on the internet, or download them onto your computer to help your youngster in developing the mathematical
skills required.
Multi-digit division worksheets are an excellent method for kids to practice and build their understanding. This ability is essential for maths that are complex and everyday calculations. These
worksheets include engaging questions and activities that strengthen the understanding.
It can be difficult to divide huge numbers for students. These worksheets generally employ the same algorithm and follow step-by–step directions. It is possible for students to not have the knowledge
required. Utilizing base ten blocks to illustrate the process is one technique to teach long division. The steps to learn should make long division easier for students.
The pupils can work on the division of large numbers by using a variety exercises and worksheets. In the worksheets, you will also find fractional results which are written in decimals. There are
worksheets that allow you to determine hundreds ofths. This is especially useful when you need to divide large sums of money.
Divide the numbers into smaller ones.
It isn’t easy to arrange a number into small groups. While it looks great on paper, many facilitators in small groups are averse to the process. It genuinely reflects how the human body develops and
it can aid in the Kingdom’s unending growth. Additionally, it inspires others to reach out to the undiscovered and new leadership to take the helm.
It is useful for brainstorming. It is possible to form groups of people who share the same traits and experience. You could think of creative solutions using this method. Reintroduce yourself to each
person once you’ve created your groups. It’s a great way to encourage innovation and fresh thinking.
The most fundamental operation in arithmetic is division is to split big amounts into smaller numbers. It’s a good option when you need equal items for multiple groups. It is possible to break down a
large class into groups of five students. This gives you the 30 pupils that were in the first group.
Keep in mind that when you divide numbers, there’s a divisor as well as an Quotient. Dividing one number by another produces “ten/five,” while divising two by two yields the similar result.
In large numbers, a power of 10 should not be used.
To help us compare huge numbers, we could divide them into powers of 10. Decimals are an essential part of shopping. You can find them on receipts, price tags and food labels. They are used by fuel
pumps to show the cost per gallon and the amount of fuel transported via a sprayer.
There are two methods to divide a large number into its powers of ten. One is by moving the decimal point to the left and using a multiplier of 10-1. The second option is to use the power of ten’s
association feature. Once you’ve learned how to utilize the power of ten’s associative feature you can split huge numbers into smaller powers.
The first method employs mental computation. Divide 2.5 by 10 to find patterns. The decimal point shifts to the left as the power of 10 rises. This principle is simple to grasp and is applicable to
every situation regardless of how complex.
Mentally breaking large numbers down into powers is the third method. It is easy to express massive numbers by using the scientific notation. When writing in scientific notation, huge numbers must be
expressed as positive exponents. For instance, by shifting the decimal point five spaces to the left, you could write 450,000 into 4.5. To split a large number into smaller powers 10, you could make
use of exponent 5, or divide it into smaller powers 10 until it’s 4.5.
Gallery of Divisibility Worksheet Answer Key
Divisibility Rules Worksheets Grade 5 27 Divisibility Rules Worksheet
Single Digit Division With No Remainders Worksheets For Grade 5 Answer
Divisibility Rules Worksheets With Answer Key Pdf Grade 5 Worksheetpedia
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What does Binary mean?
What does Binary mean?
Binary is a numeral system that uses only two symbols, 0 and 1, to represent numbers. It is the foundation of digital electronics and computers, as computers use binary to process and store
information. In binary, each digit represents a switch that can either be on or off, and these binary digits, or bits, are used to perform complex operations through simple binary calculations. For
example, the binary number 10011011 can represent the decimal number 155, allowing computers to perform mathematical operations with decimal values by converting them to binary. Binary is also used
in coding and data compression, where it is used to represent text, images, and other data in a compact and efficient manner. The use of binary allows for efficient processing and storage of data, as
well as easy transmission between computers and other digital devices. In short, binary is a crucial part of the digital world, providing a simple and efficient way to represent and manipulate data
in computers and other digital devices. | {"url":"https://www.ascii-code.com/glossary/binary","timestamp":"2024-11-13T19:23:53Z","content_type":"text/html","content_length":"20056","record_id":"<urn:uuid:c70b8de6-a44e-4019-ace9-7ac22ae72dd6>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00590.warc.gz"} |
Linear Regression
• Linear regression builds a model which establishes a relationship between features and targets.
• For simple linear regression, the model has two parameters w and b whose values are fit using training data. f(X) = wX + b
• Once a model’s parameters have been determined, the model can be used to make predictions on new data.
• Linear regression with one variable ==> univariate linear regression
• Linear function generates best fit line
To train the model:
You feed the training set (both the input features and the output targets) to your learning algorithm.
Then your supervised learning algorithm will produce some function (f). f is called the model.
Function takes a new input x and
Estimates or makes prediction (y hat) for y.
import numpy as np
import matplotlib.pyplot as plt
x_train = np.array([1.0, 2.0, 3.0,4.0, 5.0])
y_train = np.array([300.0, 500.0, 600.0, 700.0, 800.0])
m = x_train.shape[0]
# Try with different w and b values
def compute_model_output(x,w,b):
m = x.shape[0]
f_wb = np.zeros(m)
for i in range(m):
f_wb[i] = w*x[i] + b
return f_wb
prediction = compute_model_output(x_train,w,b)
plt.plot(x_train, prediction, c="b", label = "Our Prediction")
plt.scatter(x_train, y_train, c="r", marker="x", label = "Actual Values")
plt.title("House Pricing")
Linear Regression
Here, we randomly gave the values w and b that fits the linear line. Now that we have a model, we can use it to make our original prediction : f(x) = 160 * x + 100
x_new = 3.5
f_wb_new = w* x_new + b
print(f"${f_wb_new:0f} thousands dolar")
# prediction : $660.000000 thousands dolar
How to find w and b?
The more fit w and b are found, the closer the prediction for y^ is to the true target.
How to measure how well a line fits the training data?
TO DO THAT, construct a cost function! | {"url":"https://datasciencemynotes.com/linear-regression/","timestamp":"2024-11-04T21:30:38Z","content_type":"text/html","content_length":"54338","record_id":"<urn:uuid:5a1d9eea-6f81-476f-8613-136ed2881027>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00583.warc.gz"} |
[Solved] Statement I ∣f∣a+b∣=∣a−b∣, then a and b are perpendicu... | Filo
Statement I , then and are perpendicular to each other. Statement II If the diagonals of a parallelogram are equal in magnitude, then the parallelogram is a rectangle
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Sol. (a) are the diagonals of a parallelogram whose sides are and . Thus, diagonals of the parallelogram have the same length. So, the parallelogram is a rectangle, i.e. .
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Question Text Statement I , then and are perpendicular to each other. Statement II If the diagonals of a parallelogram are equal in magnitude, then the parallelogram is a rectangle
Topic Vector Algebra
Subject Mathematics
Class Class 12
Answer Type Text solution:1
Upvotes 111 | {"url":"https://askfilo.com/math-question-answers/statement-i-f-mathbfamathbfb-mathbfa-mathbfb-mid-then-mathbfa-and-mathbfb-are","timestamp":"2024-11-03T09:54:38Z","content_type":"text/html","content_length":"565958","record_id":"<urn:uuid:09659adf-b0fe-444f-979a-6bdfbb771260>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00229.warc.gz"} |
Recommended Age: 5+ | Mathematical Topic: Shapes & Geometry
What You Need
Straws cut to various lengths; playdough rolled into many small balls.
What To Do
Have your child build different shapes by using the playdough to attach the straws at the corners. For example: To make a square, use four straws for each side of the shape and 4 balls of playdough
to connect the straws at the corners. Use this activity as an opportunity to discuss basic shape features as you build together.
Moving On
When your child understands how to build the shapes well and understands the shapes they’re making, use the shapes they build to talk about more advanced shape features like right (90 degrees), acute
(less than 90 degrees), obtuse (greater than 90 degrees) angles, and other aspects of the shapes they construct. | {"url":"https://becomingamathfamily.uchicago.edu/activities/46","timestamp":"2024-11-06T21:02:56Z","content_type":"text/html","content_length":"19457","record_id":"<urn:uuid:92f68b7e-2918-4492-84e1-cd57b4f52a27>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00691.warc.gz"} |
What Is Leverage Ratio, And What Is Its Significance?
What Is Leverage Ratio?
A leverage ratio is a financial metric used to measure the level of debt a company uses to finance its operations. It is a ratio of the company’s total debt to its assets or equity. The leverage
ratio is an important indicator of a company’s financial health, and investors, analysts, and lenders use it to assess the risk associated with a company’s debt level.
Leverage Ratio Formula
Leverage ratio = Total Debt / Total Assets
Types Of Leverage Ratios
A company can use three types of leverage ratios to measure its debt usage, these are:
1. Debt-to-equity ratio
2. Debt-to-assets ratio
3. Interest coverage ratio
Let’s dive into each type in more detail, along with their formulas:
Debt-to-Equity Ratio
This ratio compares a company’s total debt to the equity shareholders have invested. It helps to determine the proportion of financing that is being provided by creditors versus shareholders. A high
debt-to-equity ratio indicates the company is highly leveraged, which may pose a greater risk to investors.
Formula: Debt-to-equity ratio = Total liabilities / Shareholders’ equity
Debt-to-Assets Ratio
With the help of this ratio, it measures the percentage of a company’s assets financed by debt. A high debt-to-assets ratio indicates that a significant portion of the company’s assets is financed by
debt, which may increase financial risk.
Formula: Debt-to-assets ratio = Total liabilities / Total assets
Interest Coverage Ratio
This ratio measures a company’s ability to pay interest on its outstanding debt. It helps to define whether the business generates enough earnings to meet its interest obligations. A high-interest
coverage ratio indicates that the company is generating enough earnings to cover its interest payments, which may signal a lower risk to investors.
Formula: Interest coverage ratio = Earnings before interest and taxes (EBIT) / Interest expense
How Risks of High Operating Leverage Different from High Financial Leverage?
High operating leverage occurs when a company has high fixed costs relative to its variable costs. This means that small changes in sales revenue can significantly impact on the company’s profits.
High operating leverage can be risky because if sales decline, the company’s profits can decline even more sharply, leading to potential losses. Similarly, if the company needs more revenue to cover
its fixed costs, it may be unable to remain in business.
High financial leverage occurs when a company relies heavily on debt financing to fund its operations. This means that the company has a high level of debt relative to its equity. High financial
leverage can be risky because if the company’s profits decline, it may not be able to meet its debt obligations. This can lead to defaults, bankruptcy, or a decrease in the company’s credit rating.
Additionally, if interest rates rise, the company’s interest expenses may increase, which could further strain its financial position.
How to Calculate the Leverage Ratio?
The leverage ratio is calculated by dividing a company’s total debt by its equity or total assets. Here’s a step-by-step guide on how to calculate the leverage ratio:
Step 1: Determine the Total Debt of the Company
The first step in calculating the leverage ratio is to determine the company’s total debt. This includes both short-term and long-term debt. You can find this information in the company’s balance
Step 2: Determine the Total Equity of the Company
The second step is to determine the total equity of the company. This includes the value of all the company’s assets minus its liabilities. You can find this information in the company’s balance
sheet as well.
Step 3: Calculate the Leverage Ratio
Once you have determined the company’s total debt and equity, you can calculate the leverage ratio by dividing the total debt by the total equity. The formula is as follows:
Leverage Ratio = Total Debt / Total Equity
Alternatively, you can calculate the leverage ratio by dividing the total debt by the total assets. The formula for this is as follows:
Leverage ratio = Total Debt / Total Assets
Step 4: Interpret the Results
The leverage ratio measures a company’s debt relative to its assets or equity. A higher leverage ratio indicates that the company has more debt than its assets or equity, which can cause concern for
investors and creditors. On the other hand, a lower leverage ratio indicates that the company has less debt relative to its assets or equity, which can be a positive sign.
Example: Let’s say a company has total debt of $500,000 and total equity of $1,000,000. To calculate the leverage ratio using the debt-to-equity formula, we divide the total debt by the total
Leverage Ratio = $500,000 / $1,000,000 = 0.5
So, the leverage ratio for this company is 0.5. The company has $0.50 of debt for every $1.00 of equity.
What is the Significance of the Leverage Ratio?
A leverage ratio is used by investors, creditors, and regulators to assess a company’s ability to meet its financial obligations and manage its financial risk.
A high leverage ratio indicates that a company relies heavily on debt to finance its operations, making it more vulnerable to economic downturns or changes in interest rates. On the other hand, a low
leverage ratio indicates that a company has a lower level of debt and may be better able to weather financial challenges.
From a regulatory perspective, the leverage ratio is often used to ensure that banks and other financial institutions have a sufficient cushion of capital to absorb losses in the event of financial
stress. A higher leverage ratio requirement for these institutions can help mitigate the risk of systemic financial instability.
In conclusion, the leverage ratio is a financial metric providing insight into a company’s debt levels and ability to meet its financial obligations. It measures the proportion of a company’s debt to
its equity, and a high ratio indicates that a company has a higher level of debt and is thus at a greater risk of defaulting on its debts. The significance of the leverage ratio lies in its ability
to provide investors, lenders, and other stakeholders with valuable information about a company’s financial health and risk profile. By understanding the leverage ratio and its implications,
investors can make more informed decisions about which companies to invest in.
Read Also: The 5 Most Important Profitability Ratios You Need for Your Small Business
Farwah Jafri is a financial management expert and Product Owner at Monily, where she leads financial services for small and medium businesses. With over a decade of experience, including a
directorial role at Arthur Lawrence UK Ltd., she specializes in bookkeeping, payroll, and financial analytics. Farwah holds an MBA from Alliance Manchester Business School and a BS in Computer
Software Engineering. Based in Houston, Texas, she is dedicated to helping businesses better their financial operations. | {"url":"https://monily.com/m/blog/what-is-leverage-ratio/","timestamp":"2024-11-09T13:07:52Z","content_type":"text/html","content_length":"255116","record_id":"<urn:uuid:60a1b6ad-2717-46b2-a8fb-4ddb126f8d47>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00789.warc.gz"} |
Product Rule: Definition, Examples
What is the Product Rule?
The product rule is used to differentiate many functions where one function is multiplied by another. The formal definition of the rule is:
(f * g)′ = f′ * g + f * g′.
While this looks tricky, you’re just multiplying the
of each function by the other function. Recognizing the functions that you can differentiate using the
product rule
in calculus can be tricky. Working through a few examples will help you recognize when to use the product rule and when to use other rules, like the
chain rule
Product Rule Example 1: y = x^3 ln x
The derivative of x^3 is 3x^2, but when x^3 is multiplied by another function—in this case a natural log (ln x), the process gets a little more complicated.
Step 1: Name the first function “f” and the second function “g.” Go in order (i.e. call the first function “f” and the second “g”).
Step 2: Rewrite the equation using the new function names f and g you started using in Step 1:
Multiply f by the derivative of g, then add the derivative of f multiplied by g. You don’t need to actually differentiate at this point: just rewrite the equation.
y’= x^3 D (ln x) + D (x^3) ln x
Step 3: Take the derivative of the two functions in the equation you wrote in Step 2. Leave the two other functions in the sequence alone.
y′= x^3 (1/x) + (3x^2 ln x).
Step 4: Use algebra to simplify the result. This step is optional, but it keeps things neat and tidy.
y′= x^2 + 3x^2 ln x.
That’s it! If you differentiate y=x^3 ln 3, the answer is y’= x^2 + 3x^2 ln x.
Graph of y=x^3 ln x (red line). When you differentiate y=xx^3 ln x you get x^2 + 3x^2 ln x (red line).
Product Rule Example 2: y = (x^3 + 7x – 7)(5x + 2)
Step 1: Label the first function “f” and the second function “g”.
• f = (x^3 + 7x – 7)
• g = (5x + 3)
Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The
tick marks mean “derivative” but we’ll use “D” instead.
y′ = (x^3 + 7x – 7) D(5x + 3) + D(x^3 + 7x – 7)(5x + 3)
Step 3: Take the derivative of the two functions identified in the equation you wrote in Step 2.
y′ = (x^3 + 7x – 7) (5) + (3x^2 + 7)(5x + 3)
Step 4: Use algebra to multiply out and neaten up your answer:
y′ = 5x^3 + 35x – 35 + 15x^3 + 9x^2 + 35x + 21 = 20x^3 + 9x^2 + 70x – 14
That’s it!
Product Rule Example 3: y = x^-3 (17 + 3x^-3)
Example problem: Differentiate y = x^-3(17 + 3x^-3) using the product rule.
Step 1: Name the functions so that the first function is “f” and the second function is “g.” In this example, we have:
• f = x^-3 and
• g = (17 + 3x^-3)
Step 2: Rewrite the equation: Multiply f by the derivative of g, added to the derivative of f multiplied by g.
f′ = x^-3 D (17 + 3x^-3) + D(x^-3) (17 + 3x^-3).
Step 3: Take the two derivatives of the equation from Step 2:
f′ = x^-3 (-9x^-4) + (-3x^-4) (17 + 3x^-3).
Step 4: Use algebra to expand and simplify the equation:
f′ = -9x^-7-51x^-4-9x^-7 = -18x^-7-51x^-4.
That’s it!
Product Rule Example 4: y = 6x^3/2 cot x.
Step 1: Label the first function “f” and the second function “g”.
f = 6x^3/2
g = cot x
Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The
tick(‘) in the formal definition means “derivative” but we’ll use “D” instead.
y′ = (6x^3/2)* D (cot x) + D(6x^3/2)* cot x
Step 3: Take the derivative of the two functions from Step 2.
y′ = (6x^3/2)* (– csc^2 x) + (6^(3/2)x^1/2)* cot x
Step 4: Use algebra to multiply out and neaten up your answer:
f′ = 6x^3/2 – csc^2 x + 9x^1/2cot x
= 3x^1/2(2x – csc^2 x + 3 cot x)
That’s it!
Tip: Don’t be tempted to skip steps, especially when multiplying out algebraically. Although you might think you’re in calculus (and therefore know it all when it comes to algebra!), common mistakes
usually happen in differentiation not by the actual differentiating process itself, but when you try and multiply out “in your head” instead of being careful to multiply out piece-wise.
Product Derivative Theorem
The product derivative theorem states that if two functions f and g are differentiable at some point x = a, then f * g is also differentiable at z. In other words, if two different functions have
derivatives at a point, then their product inherits the differentiable property (Swann & Johnson, 2014).
The theorem also tells us that, for some point a:
D(f * g) = f(a)Dg + g(a)Df
In words, that’s: the derivative of f * g at some point a is equal to:
• The function f’s value at a, multiplied by the derivative of g at that point, plus
• The function g’s value at a, multiplied by the derivative of f at that point.
Swann, H. & Johnson, J. (2014). Prof. E. McSquared’s Calculus Primer. Expanded Intergalactic Version! Dover Publications. | {"url":"https://www.statisticshowto.com/derivatives/product-rule/","timestamp":"2024-11-02T23:59:08Z","content_type":"text/html","content_length":"66823","record_id":"<urn:uuid:4c105ddb-e108-451c-ad62-e584c6feb6f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00659.warc.gz"} |
Pattern Charts
Years 2 - 6
This investigation has several levels of challenge. You don't have to do it all. Do as much as you can then stop. If you keep good journal notes you can come back any time and know what you were
up to.
□ Print this Professor Morris Poster. (Artist: Rob Mullarvey)
□ If you are a teacher using a Poster Problem Clinic you will also need this Professor Morris Slide.
Allow full screen when asked by the slide. Use Esc to return to menu view.
□ Write the title of this challenge and today's date on a fresh page in your maths journal.
Patterns In The Puzzle
□ Professor Morris has hidden some numbers. Write in the missing numbers.
□ If your numbers are correct there will be pattern.
□ Stick the puzzle sheet in your journal. You can colour it first if you want to.
□ There's lots of patterns in this puzzle.
- Show someone else the patterns you see.
- Can they see any others?
- Write and draw about the patterns you find.
Numbers In The Puzzle
□ Make a list of the numbers Professor Morris used. Just write each number once.
□ Make a list of the numbers you used. Just write each number once.
□ One of you used Odd numbers and one of you used Even numbers.
- Who used which numbers?
- How do you know?
□ Pretend Professor Morris has drawn the chart on the driveway with chalk.
Choose any odd number on the chart and pretend to stand on it.
- If you jump left or right one square, what sort of number do you land on?
- If you jump forward or backward one square, what sort of number do you land on?
- If you jump diagonally forward or backward one square, what sort of number do you land on?
□ It doesn't matter which odd number you stand on, the same thing always happens. Can you explain why?
□ What happens if you start on an even number?
Send A Text
Professor Morris loved the patterns in his chart and he wanted to share it with his friend Mini. He sent her this text.
To make my puzzle you draw 5 rows of 6 squares.
Start with number 1 in the bottom right corner.
To fill in the rest of the numbers you ...
Uh oh! The text went off the screen. What did Professor Morris write next? Write what you think in your journal.
Have fun exploring Pattern Charts.
Walking On The Puzzle
Suppose six (6) children were in the driveway with one person in each box of the bottom row.
They start with the number they are standing on and walk to the top adding each number they step on.
□ Calculate and record each total.
Hint: look for ways to group each person's numbers to help you add them quickly.
□ What is the pattern in the totals?
Try to explain the pattern?
□ Mini-Challenge: Find the total of the totals WITHOUT adding them all up
In your journal record how you did it.
Suppose five (5) children were in the driveway with one person in each box of the right hand column.
They start with the number they are standing on and walk to the left side adding each number they step on.
□ Calculate and record each total.
Hint: can you use knowledge you already have?
□ What is the pattern in the totals?
Try to explain the pattern?
□ Mini-Challenge: Find the total of the totals WITHOUT adding them all up
Professor Morris got a surprise when he first did this.
Why do think he was surprised?
In your journal explain why both 'totals of totals' get the same answer.
Use diagrams to explain if you want to.
Then Ali noticed something:
Look. If I stand on this 6 and I walk this way, then this way, I get to ten. That means its 6 + 1 + 3 = 10 and I did it in two moves.
When Ali says 'move' she means she walks in a straight line until she turns a corner or reaches her finish square.
□ Where did Ali stand and how did she walk?
□ Find another place Ali could start and do two (2) moves to get to 10. Write its equation.
□ Find five (5) more.
□ One of the kids said they did this three (3) move walk: 1 + 3 + 4 + 2 =10. Show how they walked.
□ Make up your own three move walk without looking at the chart. Does it work?
Then Ali said:
Okay, let's try getting to 1 in two moves. Suppose I start on 6 again. I can do this: 6 - 2 - 3 = 1
□ How did Ali walk?
□ Find all the ways Ali could walk to 1 from a 6 and record them.
□ One 6 is special. Why?
□ Explore more two move and three move walks to get to 1.
Another kid came up with a new game.
Hey guys. I know. I'll give you a total and you have to walk a straight line from edge to edge horizontally, vertically or diagonally to make that total. First to get it wins a point.
Try these questions and write the answers in your journal.
□ One of the totals was 20. How would you do that one?
□ Is there another way?
□ How many solutions are there?
□ How do you know when you have found them all?
A mathematician might ask these questions:
□ What is the lowest total I can walk in this game?
□ What is the highest total I can walk in this game?
□ Can I make all the totals between the highest and the lowest?
Frames On The Chart
The children found an old picture frame in the garage. □ Use the frame horizontally or vertically.
It exactly fitted around two (2) squares. □ Explore.
They put it around lots of pairs of numbers on the board. □ Report on what you discover with this frame.
1 row of 3 Extra Challenges
1 row of 4 1. Choose one of the other frames on the left.
See what you can find out about the numbers inside.
2 rows of 2
2. Print these two Extra Challenges.
Professor Morris has made his chart bigger and you have to fill in ALL of it.
Extra Challenge 1 (could be a bit easy) ... Extra Challenge 2 (could be a bit hard)
3. Print this Chart Paper and create your own Professor Morris puzzle.
□ You choose the size.
□ You choose the starting point.
□ You choose the starting number.
□ You choose the pattern to build the chart.
Investigate your own puzzle and report.
Just Before You Finish
For this part you need your maths journal and your Working Like A Mathematician page.
□ How did you work like a mathematician today? Record 2 ways.
□ What do you know now that you didn't know when you started Pattern Charts?
Send any comments or photos about this activity and we can start a gallery here.
Maths At Home is a division of Mathematics Centre | {"url":"http://www.mathematicscentre.com/mathsathome/challenges/patchart.htm","timestamp":"2024-11-08T23:58:13Z","content_type":"text/html","content_length":"10254","record_id":"<urn:uuid:1d056af4-2e55-4e3f-9683-f20b74cd6584>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00320.warc.gz"} |
How many mL are there in half a Litre?
How many mL are there in half a Litre?
There are 1000 milliliters (ml) in 1 Liter. So 250ml is equivalent to one quarter of a Liter. And 500ml is equivalent to half a Liter.
Is 500ml a half liter?
One Liter is more than 500 mL since one Liter is equal to 1000 mL.
How many Millilitres does it take to make 1 l?
1000 ml
To convert liters to milliliters, we multiply the given value by 1000 because 1 liter = 1000 ml.
What is half of 1 litre?
the half of 1 litre is 500 ml.
How many mL is two and a half?
mL and cups conversion chart
Milliliters Cups Cups (fraction approx)
500 mL 2.11 Cups 2 and 1/10 cups
550 mL 2.32 Cups 2 and 1/3 cups
600 mL 2.54 Cups 2 and 1/2 cups
650 mL 2.75 Cups 2 and 3/4 cups
How many tbsp are in a liter?
Liter to Tablespoon Conversion Table
Liters Tablespoons
1 l 67.63 tbsp
2 l 135.26 tbsp
3 l 202.88 tbsp
4 l 270.51 tbsp
What is bigger 1 mL or 1 L?
In the metric system, the prefix m stands for “milli”, which means “1/1,000 of”. So 1 ml (milliliter) is only 1/1,000 of 1 l (liter). Therefore, 1 ml is smaller than 1 l.
How many mg are in a Gramm?
Answer: It takes 1000 milligrams to make a gram.
What is a liter?
Litre (l), also spelled liter, unit of volume in the metric system, equal to one cubic decimetre (0.001 cubic metre). From 1901 to 1964 the litre was defined as the volume of one kilogram of pure
water at 4 °C (39.2 °F) and standard atmospheric pressure; in 1964 the original, present value was reinstated.
How many mL is 2 litres?
Liters to Milliliters table
Liters Milliliters
2 L 2000.00 mL
3 L 3000.00 mL
4 L 4000.00 mL
5 L 5000.00 mL
How do you convert from liters to milliliters?
Multiply the number of liters (L) by 1,000 to find the number of milliliters (mL). There are 1,000 times as many milliliters as there are liters. For instance, say you have 3 liters. Simply multiply
3 liters by 1,000 to get 3,000 milliliters.
How many ounces are in a liters?
33.814 fl oz
There are 33.814 fl oz in a liter. There are two different values for ounces in UK and US measurements: For US, 1 liter = 33.814 fluid ounces.
How many milliliters is a half liter of milk?
It would be to the power of ten for each capacity amount of one type of measurement. Half a liter would be 500 mm because 1 liter is equal to 1000 milliliter. Half of 1000 would be 500 milliliters.
How many milliliters are in 1.5 liters?
To convert liters to mL, multiply the liter value by 1000. For example, to find out how many milliliters in a liter and a half, multiply 1.5 by 1000, that makes 1500 mL in 1.5 liters. 1 Liter = 1000
How many mL in one leter?
1 Liter = 1000 Milliliters. In metric system, the prefix “milli” means “one thousandth of”, that makes easier to remember how to convert from liters to milliliters, because if you replace the “milli”
with “1000” in word “milliliter”, 1 milliliter means 1000 of a liter (or 1 L = 1000 mL).
How to find out how many liters are 500 ml?
For example, to find out how many liters is 500 mL, divide 500 by 1000, that makes 0.5 L is 500 mL. You may also use this volume units conversion calculator tool to convert between liters,
milliliters and all other volume units. What is Liter/Litre? Liter (litre) is a metric system volume unit. 1 L = 1000 mL. | {"url":"https://sage-answer.com/how-many-ml-are-there-in-half-a-litre/","timestamp":"2024-11-08T18:47:56Z","content_type":"text/html","content_length":"142064","record_id":"<urn:uuid:419a07d5-3934-47a1-aff8-4413dea12b06>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00659.warc.gz"} |
**vegan** FAQ ============= This document contains answers to some of the most frequently asked questions about R package **vegan**. > This work is licensed under the Creative Commons Attribution 3.0
> License. To view a copy of this license, visit > or send a letter to > Creative Commons, 543 Howard Street, 5th Floor, San Francisco, > California, 94105, USA. > > Copyright © 2008-2016 vegan
development team ------------------------------------------------------------------------ Introduction ------------ ------------------------------------------------------------------------ ### What
is **vegan**? **Vegan** is an R package for community ecologists. It contains the most popular methods of multivariate analysis needed in analysing ecological communities, and tools for diversity
analysis, and other potentially useful functions. **Vegan** is not self-contained but it must be run under R statistical environment, and it also depends on many other R packages. **Vegan** is [free
software](https://www.gnu.org/philosophy/free-sw.html) and distributed under [GPL2 license](https://www.gnu.org/licenses/gpl.html).
------------------------------------------------------------------------ ### What is R? R is a system for statistical computation and graphics. It consists of a language plus a run-time environment
with graphics, a debugger, access to certain system functions, and the ability to run programs stored in script files. R has a home page at . It is [free software](https://www.gnu.org/philosophy/
free-sw.html) distributed under a GNU-style [copyleft](https://www.gnu.org/copyleft/copyleft.html), and an official part of the [GNU](https://www.gnu.org/) project (“GNU S”).
------------------------------------------------------------------------ ### How to obtain **vegan** and R? Both R and latest release version of **vegan** can be obtained through [CRAN](https://
cran.r-project.org). Unstable development version of **vegan** can be obtained through [GitHub](https://github.com/vegandevs/vegan). The github page gives further instructions for obtaining and
installing development versions of **vegan**. ------------------------------------------------------------------------ ### What R packages **vegan** depends on? **Vegan** depends on the **permute**
package which will provide advanced and flexible permutation routines for **vegan**. The **permute** package is developed together with **vegan** in [GitHub](https://github.com/gavinsimpson/permute).
Some individual **vegan** functions depend on packages **MASS**, **mgcv**, **parallel**, **cluster** and **lattice**. **Vegan** dependence on **tcltk** is deprecated and will be removed in future
releases. These all are base or recommended R packages that should be available in every R installation. **Vegan** declares these as suggested or imported packages, and you can install **vegan** and
use most of its functions without these packages. **Vegan** is accompanied with a supporting package **vegan3d** for three-dimensional and dynamic plotting. The **vegan3d** package needs **tcltk**
and non-standard packages **rgl** and **scatterplot3d**. ------------------------------------------------------------------------ ### What other packages are available for ecologists? CRAN [Task
Views](https://cran.r-project.org/web/views/) include entries like `Environmetrics`, `Multivariate` and `Spatial` that describe several useful packages and functions. If you install R package
**ctv**, you can inspect Task Views from your R session, and automatically install sets of most important packages. ------------------------------------------------------------------------ ### What
other documentation is available for **vegan**? **Vegan** is a fully documented R package with standard help pages. These are the most authoritative sources of documentation (and as a last resource
you can use the force and the read the source, as **vegan** is open source). **Vegan** package ships with other documents which can be read with `browseVignettes("vegan")` command. The documents
included in the **vegan** package are - **Vegan** `NEWS` that can be accessed via `news()` command. - This document (`FAQ-vegan`). - Short introduction to basic ordination methods in **vegan**
(`intro-vegan`). - Introduction to diversity methods in **vegan** (`diversity-vegan`). - Discussion on design decisions in **vegan** (`decision-vegan`). - Description of variance partition procedures
in function `varpart` (`partitioning`). Web documents outside the package include: - : development page. - : **vegan** homepage.
------------------------------------------------------------------------ ### Is there a Graphical User Interface (GUI) for **vegan**? Roeland Kindt has made package **BiodiversityR** which provides a
GUI for **vegan**. The package is available at [CRAN](https://cran.r-project.org/package=BiodiversityR). It is not a mere GUI for **vegan**, but adds some new functions and complements **vegan**
functions in order to provide a workbench for biodiversity analysis. You can install **BiodiversityR** using `install.packages("BiodiversityR")` or graphical package management menu in R. The GUI
works on Windows, MacOS X and Linux. ------------------------------------------------------------------------ ### How to cite **vegan**? Use command `citation("vegan")` in R to see the recommended
citation to be used in publications. ------------------------------------------------------------------------ ### How to build **vegan** from sources? In general, you do not need to build **vegan**
from sources, but binary builds of release versions are available through [CRAN](https://cran.r-project.org/) for Windows and MacOS X. If you use some other operating systems, you may have to use
source packages. **Vegan** is a standard R package, and can be built like instructed in R documentation. **Vegan** contains source files in C and FORTRAN, and you need appropriate compilers (which
may need more work in Windows and MacOS X). ------------------------------------------------------------------------ ### Are there binaries for devel versions? Binaries can be available from R
Universe: see for instructions. ------------------------------------------------------------------------ ### How to report a bug in **vegan**? If you think you have found a bug in **vegan**, you
should report it to **vegan** maintainers or developers. The preferred forum to report bugs is [GitHub](https://github.com/vegandevs/vegan/issues). The bug report should be so detailed that the bug
can be replicated and corrected. Preferably, you should send an example that causes a bug. If it needs a data set that is not available in R, you should send a minimal data set as well. You also
should paste the output or error message in your message. You also should specify which version of **vegan** you used. Bug reports are welcome: they are the only way to make **vegan** non-buggy.
Please note that you shall not send bug reports to R mailing lists, since **vegan** is not a standard R package. ------------------------------------------------------------------------ ### Is it a
bug or a feature? It is not necessarily a bug if some function gives different results than you expect: That may be a deliberate design decision. It may be useful to check the documentation of the
function to see what was the intended behaviour. It may also happen that function has an argument to switch the behaviour to match your expectation. For instance, function `vegdist` always calculates
quantitative indices (when this is possible). If you expect it to calculate a binary index, you should use argument `binary = TRUE`.
------------------------------------------------------------------------ ### Can I contribute to **vegan**? **Vegan** is dependent on user contribution. All feedback is welcome. If you have problems
with **vegan**, it may be as simple as incomplete documentation, and we shall do our best to improve the documents. Feature requests also are welcome, but they are not necessarily fulfilled. A new
feature will be added if it is easy to do and it looks useful, or if you submit code. If you can write code yourself, the best forum to contribute to vegan is [GitHub](https://github.com/vegandevs/
vegan). ------------------------------------------------------------------------ Ordination ---------- ------------------------------------------------------------------------ ### I have only numeric
and positive data but **vegan** still complains You are wrong! Computers are painfully pedantic, and if they find non-numeric or negative data entries, you really have them. Check your data! Most
common reasons for non-numeric data are that row names were read as a non-numeric variable instead of being used as row names (check argument `row.names` in reading the data), or that the column
names were interpreted as data (check argument `header = TRUE` in reading the data). Another common reason is that you had empty cells in your input data, and these were interpreted as missing
values. ------------------------------------------------------------------------ ### Can I analyse binary or cover class data? Yes. Most **vegan** methods can handle binary data or cover abundance
data. Most statistical tests are based on permutation, and do not make distributional assumptions. There are some methods (mainly in diversity analysis) that need count data. These methods check that
input data are integers, but they may be fooled by cover class data. ------------------------------------------------------------------------ ### Why dissimilarities in **vegan** differ from other
sources? Most commonly the reason is that other software use presence–absence data whereas **vegan** used quantitative data. Usually **vegan** indices are quantitative, but you can use argument
`binary = TRUE` to make them presence–absence. However, the index name is the same in both cases, although different names usually occur in literature. For instance, Jaccard index actually refers to
the binary index, but **vegan** uses name `"jaccard"` for the quantitative index, too. Another reason may be that indices indeed are defined differently, because people use same names for different
indices. ------------------------------------------------------------------------ ### Why NMDS stress is sometimes 0.1 and sometimes 10? Stress is a proportional measure of badness of fit. The
proportions can be expressed either as parts of one or as percents. Function `isoMDS` (**MASS** package) uses percents, and function `monoMDS` (**vegan** package) uses proportions, and therefore the
same stress is 100 times higher in `isoMDS`. The results of `goodness` function also depend on the definition of stress, and the same `goodness` is 100 times higher in `isoMDS` than in `monoMDS`.
Both of these conventions are equally correct. ------------------------------------------------------------------------ ### I get zero stress but no repeated solutions in `metaMDS` The first (try 0)
run of `metaMDS` starts from the metric scaling solution and is usually good, and most sofware only return that solution. However, `metaMDS` tries to see if that standard solution can be repeated, or
improved and the improved solution still repeated. In all cases, it will return the best solution found, and there is no burning need to do anything if you get the message tha the solution could not
be repeated. If you are keen to know that the solution really is the global optimum, you may follow the instructions in the `metaMDS` help section "Results Could Not Be Repeated" and try more. Most
common reason is that you have too few observations for your NMDS. For `n` observations (points) and `k` dimensions you need to estimate `n*k` parameters (ordination scores) using `n*(n-1)/2`
dissimilarities. For `k` dimensions you must have `n > 2*k + 1`, or for two dimensions at least six points. In some degenerate situations you may need even a larger number of points. If you have a
lower number of points, you can find an undefined number of perfect (stress is zero) but different solutions. Conventional wisdom due to Kruskal is that you should have `n > 4*k + 1` points for `k`
dimensions. A typical symptom of insufficient data is that you have (nearly) zero stress but no two convergent solutions. In those cases you should reduce the number of dimensions (`k`) and with very
small data sets you should not use `NMDS`, but rely on metric methods. It seems that local and hybrid scaling with `monoMDS` have similar lower limits in practice (although theoretically they could
differ). However, higher number of dimensions can be used in metric scaling, both with `monoMDS` and in principal coordinates analysis (`cmdscale` in **stats**, `wcmdscale` in **vegan**).
------------------------------------------------------------------------ ### Zero dissimilarities in isoMDS Function `metaMDS` uses function `monoMDS` as its default method for NMDS, and this
function can handle zero dissimilarities. Alternative function `isoMDS` cannot handle zero dissimilarities. If you want to use `isoMDS`, you can use argument `zerodist = "add"` in `metaMDS` to handle
zero dissimilarities. With this argument, zero dissimilarities are replaced with a small positive value, and they can be handled in `isoMDS`. This is a kluge, and some people do not like this. A more
principal solution is to remove duplicate sites using R command `unique`. However, after some standardizations or with some dissimilarity indices, originally non-unique sites can have zero
dissimilarity, and you have to resort to the kluge (or work harder with your data). Usually it is better to use `monoMDS`. ------------------------------------------------------------------------ ###
I have heard that you cannot fit environmental vectors or surfaces to NMDS results which only have rank-order scores Claims like this have indeed been at large in the Internet, but they are based on
grave misunderstanding and are plainly wrong. NMDS ordination results are strictly metric, and in **vegan** `metaMDS` and `monoMDS` they are even strictly Euclidean. The method is called “non-metric”
because the Euclidean distances in ordination space have a non-metric rank-order relationship to community dissimilarities. You can inspect this non-linear step curve using function `stressplot` in
**vegan**. Because the ordination scores are strictly Euclidean, it is correct to use **vegan** functions `envfit` and `ordisurf` with NMDS results.
------------------------------------------------------------------------ ### Where can I find numerical scores of ordination axes? Normally you can use function `scores` to extract ordination scores
for any ordination method. The `scores` function can also find ordination scores for many non-**vegan** functions such as for `prcomp` and `princomp` and for some **ade4** functions. In some cases
the ordination result object stores raw scores, and the axes are also scaled appropriate when you access them with `scores`. For instance, in `cca` and `rda` the ordination object has only so-called
normalized scores, and they are scaled for ordination plots or for other use when they are accessed with `scores`. ------------------------------------------------------------------------ ### How the
RDA results are scaled? The scaling or RDA results indeed differ from most other software packages. The scaling of RDA is such a complicated issue that it cannot be explained in this FAQ, but it is
explained in a separate pdf document on “Design decision and implementation details in vegan” that you can read with command `browseVignettes("vegan")`.
------------------------------------------------------------------------ ### I cannot print and plot RDA results properly If the RDA ordination results have a weird format or you cannot plot them
properly, you probably have a name clash with **klaR** package which also has function `rda`, and the **klaR** `print`, `plot` or `predict` functions are used for **vegan** RDA results. You can
choose between `rda` functions using `vegan::rda()` or `klaR::rda()`: you will get obscure error messages if you use the wrong function. In general, **vegan** should be able to work normally if
**vegan** was loaded after **klaR**, but if **klaR** was loaded later, its functions will take precedence over **vegan**. Sometimes **vegan** namespace is loaded automatically when restoring a
previously stored workspace at the start-up, and then **klaR** methods will always take precedence over **vegan**. You should check your loaded packages. **klaR** may be also loaded indirectly via
other packages (in the reported cases it was most often loaded via **agricolae** package). **Vegan** and **klaR** both have the same function name (`rda`), and it may not be possible to use these
packages simultaneously, and the safest choice is to unload one of the packages if only possible. See discussion in [vegan github issues](https://github.com/vegandevs/vegan/issues/277).
------------------------------------------------------------------------ ### Ordination fails with “Error in La.svd” Constrained ordination (`cca`, `rda`, `dbrda`, `capscale`) will sometimes fail
with error message `Error in La.svd(x, nu, nv): error code 1 from Lapack routine 'dgesdd'.` It seems that the basic problem is in the `svd` function of `LAPACK` that is used for numerical analysis in
R. `LAPACK` is an external library that is beyond the control of package developers and R core team so that these problems may be unsolvable. Reducing the range of constraints (environmental
variables) helps sometimes. For instance, multiplying constraints by a constant \< 1. This rescaling does not influence the numerical results of constrained ordination, but it can complicate further
analyses when values of constraints are needed, because the same scaling must be applied there. The reports on the problems are getting rare and it may that this problem is fixed in R and `LAPACK`.
------------------------------------------------------------------------ ### Variance explained by ordination axes. In general, **vegan** does not directly give any statistics on the “variance
explained” by ordination axes or by the constrained axes. This is a design decision: I think this information is normally useless and often misleading. In community ordination, the goal typically is
not to explain the variance, but to find the “gradients” or main trends in the data. The “total variation” often is meaningless, and all proportions of meaningless values also are meaningless. Often
a better solution explains a smaller part of “total variation”. For instance, in unstandardized principal components analysis most of the variance is generated by a small number of most abundant
species, and they are easy to “explain” because data really are not very multivariate. If you standardize your data, all species are equally important. The first axes explains much less of the “total
variation”, but now they explain all species equally, and results typically are much more useful for the whole community. Correspondence analysis uses another measure of variation (which is not
variance), and again it typically explains a “smaller proportion” than principal components but with a better result. Detrended correspondence analysis and nonmetric multidimensional scaling even do
not try to “explain” the variation, but use other criteria. All methods are incommensurable, and it is impossible to compare methods using “explanation of variation”. If you still want to get
“explanation of variation” (or a deranged editor requests that from you), it is possible to get this information for some methods: - Eigenvector methods: Functions `rda`, `cca`, `dbrda` and `capscale
` give the variation of conditional (partialled), constrained (canonical) and residual components. Function `eigenvals` extracts the eigenvalues, and `summary(eigenvals(ord))` reports the proportions
explained in the result object `ord`, and also works with `decorana` and `wcmdscale`. Function `RsquareAdj` gives the R-squared and adjusted R-squared (if available) for constrained components.
Function `goodness` gives the same statistics for individual species or sites. In addition, there is a special function `varpart` for unbiased partitioning of variance between up to four separate
components in redundancy analysis. - Nonmetric multidimensional scaling. NMDS is a method for nonlinear mapping, and the concept of of variation explained does not make sense. However, 1 - stress\^2
transforms nonlinear stress into quantity analogous to squared correlation coefficient. Function `stressplot` displays the nonlinear fit and gives this statistic.
------------------------------------------------------------------------ ### Can I have random effects in constrained ordination or in `adonis`? No. Strictly speaking, this is impossible. However,
you can define models that respond to similar goals as random effects models, although they strictly speaking use only fixed effects. Constrained ordination functions `cca`, `rda` and `dbrda` can
have `Condition()` terms in their formula. The `Condition()` define partial terms that are fitted before other constraints and can be used to remove the effects of background variables, and their
contribution to decomposing inertia (variance) is reported separately. These partial terms are often regarded as similar to random effects, but they are still fitted in the same way as other terms
and strictly speaking they are fixed terms. Function `adonis2` can evaluate terms sequentially. In a model with right-hand-side `~ A + B` the effects of `A` are evaluated first, and the effects of `B
` after removing the effects of `A`. Sequential tests are also available in `anova` function for constrained ordination results by setting argument `by = "term"`. In this way, the first terms can
serve in a similar role as random effects, although they are fitted in the same way as all other terms, and strictly speaking they are fixed terms. All permutation tests in **vegan** are based on the
**permute** package that allows constructing various restricted permutation schemes. For instance, you can set levels of `plots` or `blocks` for a factor regarded as a random term. A major reason why
real random effects models are impossible in most **vegan** functions is that their tests are based on the permutation of the data. The data are given, that is fixed, and therefore permutation tests
are basically tests of fixed terms on fixed data. Random effect terms would require permutations of data with a random component instead of the given, fixed data, and such tests are not available in
**vegan**. ------------------------------------------------------------------------ ### Is it possible to have passive points in ordination? **Vegan** does not have a concept of passive points, or a
point that should only little influence the ordination results. However, you can add points to eigenvector methods using `predict` functions with `newdata`. You can first perform an ordination
without some species or sites, and then you can find scores for all points using your complete data as `newdata`. The `predict` functions are available for basic eigenvector methods in **vegan**
(`cca`, `rda`, `decorana`, for an up-to-date list, use command `methods("predict")`). ------------------------------------------------------------------------ ### Class variables and dummies You
should define a class variable as an R `factor`, and **vegan** will automatically handle them. R (and **vegan**) knows both unordered and ordered factors. Unordered factors are internally coded as
dummy variables, but one redundant level is removed or aliased. With default contrasts, the removed level is the first one. Ordered factors are expressed as polynomial contrasts. Both of these
contrasts explained in standard R documentation. ------------------------------------------------------------------------ ### How are environmental arrows scaled? The printed output of `envfit` gives
the direction cosines which are the coordinates of unit length arrows. For plotting, these are scaled by their correlation (square roots of column `r2`). You can see the scaled lengths of `envfit`
arrows using command `scores`. The scaled environmental vectors from `envfit` and the arrows for continuous environmental variables in constrained ordination (`cca`, `rda`, `dbrda`) are adjusted to
fill the current graph. The lengths of arrows do not have fixed meaning with respect to the points (species, sites), but they can only compared against each other, and therefore only their relative
lengths are important. If you want change the scaling of the arrows, you can use `text` (plotting arrows and text) or `points` (plotting only arrows) functions for constrained ordination. These
functions have argument `arrow.mul` which sets the multiplier. The `plot` function for `envfit` also has the `arrow.mul` argument to set the arrow multiplier. If you save the invisible result of the
constrained ordination `plot` command, you can see the value of the currently used `arrow.mul` which is saved as an attribute of `biplot` scores. Function `ordiArrowMul` is used to find the scaling
for the current plot. You can use this function to see how arrows would be scaled: ```{r eval=FALSE} sol <- cca(varespec) ef <- envfit(sol ~ ., varechem) plot(sol) ordiArrowMul(scores(ef, display=
"vectors")) ``` ------------------------------------------------------------------------ ### I want to use Helmert or sum contrasts `vegan` uses standard R utilities for defining contrasts. The
default in standard installations is to use treatment contrasts, but you can change the behaviour globally setting `options` or locally by using keyword `contrasts`. Please check the R help pages and
user manuals for details. ------------------------------------------------------------------------ ### What are aliased variables and how to see them? Aliased variable has no information because it
can be expressed with the help of other variables. Such variables are automatically removed in constrained ordination in **vegan**. The aliased variables can be redundant levels of factors or whole
variables. **Vegan** function `alias` gives the defining equations for aliased variables. If you only want to see the names of aliased variables or levels in solution `sol`, use `alias(sol,
names.only=TRUE)`. ------------------------------------------------------------------------ ### Plotting aliased variables You can fit vectors or class centroids for aliased variables using `envfit`
function. The `envfit` function uses weighted fitting, and the fitted vectors are identical to the vectors in correspondence analysis.
------------------------------------------------------------------------ ### Restricted permutations in **vegan** **Vegan** uses **permute** package in all its permutation tests. The **permute**
package will allow restricted permutation designs for time series, line transects, spatial grids and blocking factors. The construction of restricted permutation schemes is explained in the manual
page `permutations` in **vegan** and in the documentation of the **permute** package. ------------------------------------------------------------------------ ### How to use different plotting
symbols in ordination graphics? The default ordination `plot` function is intended for fast plotting and it is not very configurable. To use different plotting symbols, you should first create and
empty ordination plot with `plot(..., type="n")`, and then add `points` or `text` to the created empty frame (here `...` means other arguments you want to give to your `plot` command). The `points`
and `text` commands are fully configurable, and allow different plotting symbols and characters. ------------------------------------------------------------------------ ### How to avoid cluttered
ordination graphs? If there is a really high number of species or sites, the graphs often are congested and many labels are overwritten. It may be impossible to have complete readable graphics with
some data sets. Below we give a brief overview of tricks you can use. Gavin Simpson’s blog [From the bottom of the heap](https://fromthebottomoftheheap.net) has a series of articles on “decluttering
ordination plots” with more detailed discussion and examples. - Use only points, possibly with different types if you do not need to see the labels. You may need to first create an empty plot using
`plot(..., type="n")`, if you are not satisfied with the default graph. (Here and below `...` means other arguments you want to give to your `plot` command.) - Use points and add labels to desired
points using interactive `identify` command if you do not need to see all labels. - Add labels using function `ordilabel` which uses non-transparent background to the text. The labels still shadow
each other, but the uppermost labels are readable. Argument `priority` will help in displaying the most interesting labels (see [Decluttering blog, part 1](https://fromthebottomoftheheap.net/2013/01/
12/decluttering-ordination-plots-in-vegan-part-1-ordilabel/)). - Use `orditorp` function that uses labels only if these can be added to a graph without overwriting other labels, and points otherwise,
if you do not need to see all labels. You must first create an empty plot using `plot(..., type="n")`, and then add labels or points with `orditorp` (see [Decluttering blog](https://
fromthebottomoftheheap.net/2013/01/13/decluttering-ordination-plots-in-vegan-part-2-orditorp/)). - Use `ordipointlabel` which uses points and text labels to the points, and tries to optimize the
location of the text to minimize the overlap (see [Decluttering blog](https://fromthebottomoftheheap.net/2013/06/27/decluttering-ordination-plots-in-vegan-part-3-ordipointlabel/)). - Ordination `text
` and `points` functions have argument `select` that can be used for full control of selecting items plotted as text or points. - Use interactive `orditkplot` function (**vegan3d** package) that lets
you drag labels of points to better positions if you need to see all labels. Only one set of points can be used (see [Decluttering blog](https://fromthebottomoftheheap.net/2013/12/31/
decluttering-ordination-in-vegan-part-4-orditkplot/)). - Most `plot` functions allow you to zoom to a part of the graph using `xlim` and `ylim` arguments to reduce clutter in congested areas.
------------------------------------------------------------------------ ### Can I flip an axis in ordination diagram? Use `xlim` or `ylim` with flipped limits. If you have model `mod <- cca(dune)`
you can flip the first axis with `plot(mod, xlim = c(3, -2))`. ------------------------------------------------------------------------ ### Can I zoom into an ordination plot? You can use `xlim` and
`ylim` arguments in `plot` or `ordiplot` to zoom into ordination diagrams. Normally you must set both `xlim` and `ylim` because ordination plots will keep the equal aspect ratio of axes, and they
will fill the graph so that the longer axis will fit. Dynamic zooming can be done with function `orditkplot` in CRAN package **vegan3d**. You can directly save the edited `orditkplot` graph in
various graphic formats, or you can export the graph object back to R session and use `plot` to display the results. ------------------------------------------------------------------------ Other
analysis methods ---------------------- ------------------------------------------------------------------------ ### Is there TWINSPAN? TWINSPAN for R is available in [github](https://github.com/
jarioksa/twinspan). ------------------------------------------------------------------------ ### Why restricted permutation does not influence adonis results? The permutation scheme influences the
permutation distribution of the statistics and probably the significance levels, but does not influence the calculation of the statistics.
------------------------------------------------------------------------ ### How is deviance calculated? Some **vegan** functions, such as `radfit` use base R facility of `family` in maximum
likelihood estimation. This allows use of several alternative error distributions, among them `"poisson"` and `"gaussian"`. The R `family` also defines the deviance. You can see the equations for
deviance with commands like `poisson()$dev` or `gaussian()$dev`. In general, deviance is 2 times log.likelihood shifted so that models with exact fit have zero deviance. | {"url":"https://cran.fhcrc.org/web/packages/vegan/vignettes/FAQ-vegan.Rmd","timestamp":"2024-11-14T23:33:21Z","content_type":"text/plain","content_length":"33629","record_id":"<urn:uuid:fae34caa-f596-478a-8001-68d21bf1d77b>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00387.warc.gz"} |
50G: calling Stat49Pro
I would like to be able to embed a Stat49Pro command inside of an algebraic equation that will subsequently be used with the numerical solver. Here is a simplified version of what I want to do:
Result=ZALPHA(x/y), where ZALPHA is a command from the Stat49Pro library. I haven't a whole lot of experience using external library calls in my equations, so I don't really know how to proceed.
Any and all help would be much appreciated.
(real name: Tod)
09-17-2008, 01:53 AM
Just a side comment, and then I'll let someone else answer your direct question... ;-)
StatPro is one of the BEST programs for the 50g that I have ever come across. It truly brings the statistics capabilities of the 50g above and beyond those of all the TI-8x series. I wish that HP
would just buy the source code and slam it in the ROM somewhere -- it would probably make the 50g the best calculator out there for stats.
Just MHO.
09-18-2008, 06:54 AM
Excuse my igorance, but what is Stat49Pro and where do you find it?
Sorry I can't help you with original question.
Regards Stuart.
09-18-2008, 07:21 AM
Quote: Excuse my igorance, but what is Stat49Pro and where do you find it?
Sorry I can't help you with original question.
Regards Stuart.
09-18-2008, 10:14 AM
I've downloaded the lib to my 50g and installed it in port 2. How can I run the commands from the prompt instead of the STAT menu? Do I need to enter a specific directory?
09-18-2008, 10:28 AM
Quote: I've downloaded the lib to my 50g and installed it in port 2. How can I run the commands from the prompt instead of the STAT menu? Do I need to enter a specific directory?
[RightShift] [2] softkey [statp].
09-19-2008, 03:53 AM
I had to enter 1043 ATTACH once. Now it's there.
09-22-2008, 07:32 AM
Thanks Damir. I have located it and will try it out.
09-19-2008, 04:08 AM
I've played around a little and came to the following solution:
'ZALPHA(x)' does not work but you can embed the command in a program object:
<< -> x << x ZALPHA >> >> 'ZA' STO
Now, ZA is a function to be used in an algebraic expression.
09-21-2008, 01:49 PM
Thx for the reply, I'll try this today! | {"url":"https://archived.hpcalc.org/museumforum/thread-140862-post-141038.html#pid141038","timestamp":"2024-11-12T04:16:40Z","content_type":"application/xhtml+xml","content_length":"50346","record_id":"<urn:uuid:8d687582-6dd5-4b2b-8777-3016a7f007f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00203.warc.gz"} |
Matlab Cross Product | Learn How to Implement Cross Product in Matlab?
Updated March 27, 2023
Introduction to Matlab Cross Product
In this article, we will learn about Matlab Cross Product. Vectors are defined as the quantity that has both magnitude and direction. They are mainly used in the field of mathematics and physics to
demonstrate or define any mathematical quantity. There are various operations that can be performed using vectors, one such function is the multiplication of different vectors. There are two ways in
which we can multiply vectors with other vectors, which are known as Cross Product and Dot Product. While dealing with the dot product, the resultant is always scalar whereas in Cross Product the
resultant quantity is a vector.
How does Cross Product Work in Matlab?
The Following are the Working with Cross Product in Matlab with Syntax and Examples:
Cross product is defined as the quantity, where if we multiply both the vectors (x and y) the resultant is a vector(z) and it is perpendicular to both the vectors which are defined by any right-hand
rule method and the magnitude is defined as the parallelogram area and is given by in which respective vector spans. In Matlab, the cross product is defined by using the cross () function and serves
the same purpose as the normal cross product in mathematics.
Please find the below syntax which is used in Matlab to define the cross product:
• Z=cross (x, y): This returns the cross product of x and y which is Z, where x and y are vectors and they should have a length of three. If x and y are multidimensional arrays or matrices, then
they should be of the same size.
• Z=cross (x, y, dimension): This returns the cross product of x and y along the defined dimension which is given by “dimension” in the syntax. It should be noted that the size of x and y should be
the same, where size (x, dimension) and size (y, dimension) must have a length of three.
The input x and y can be input arrays which are generally numeric in nature. The data types that are supported are single and double. It also supports complex numbers. Another input i.e.dimension,
it’s valued should be a positive value. If there is no value given to it, then by default it considers the dimension of the first array. If we mention the function as a cross (x, y,1) then it gives
the cross product of x and y across the columns, where the inputs are treated as vectors. If we mention the function as a cross (x, y,2) then it gives the cross product of x and y across the rows,
where the inputs are treated as vectors.
Examples to Implement in Matlab Cross Product
Below are the examples to implement in Matlab Cross Product:
Example #1
a. To find the cross product of the two vectors and check whether the resultant is perpendicular to the inputs using the dot product:
x = [5 -2 2];
y = [2 -1 4];
Z = cross(x,y)
b. To check whether the resultant is perpendicular to the inputs i.e. x and y, we have used the dot product.
p=dot(Z,x)==0 & dot(Z,y)==0
In the above example, Z is the cross product of the two input arrays x and y. p=1, which means that the output is perpendicular to the inputs x and y. If it is not perpendicular, then p will be 0.
Example #2
To find the cross product of the two matrices where integers are arranged randomly:
x = randi(12,3,4)
y = randi(15,3,4)
Z= cross(x,y)
There are various properties associated with the cross product, some of them are described below:
• Cross product of two vectors length is defined as ||x*y||=|x||y|sin α, where α is the angle of the defined vectors. The cross product is given by the length of each vector x and y and the angle
between them. If the two vectors x and y are parallel to each other, then the result of the cross product is zero.
• It follows the property of anticommutativity, which means the result is the negative of the input. It is given by ||x*y||= -x*y.
• If we multiply any scalar quantity to the vectors, then it is in the below format: (cx)*y=c(x*y) =x*(cy), where c is the scalar quantity, x and y are the vectors.
• It follows the property of Distributivity, which is one of the most important properties in Mathematics. It follows the format as: x*(y+z) =x*y+x*z
• If x, y, and z are the vectors, then the scalar triple product of these vectors will be in the form of x+(y*z) =(x*y) +z
• If x, y, and z are the vectors, then the vector triple product of these vectors will be in the form of x*(y*z) =(x+z) y –(x+y) z
Example #3
a. To find the cross product of the multi-dimensional arrays which contain the random integers. Random Integers in Matlab can be defined by the “randi” function which generates the random numbers
within a given range.
x = randi(15,3,2,2);
y = randi(10,3,2,2);
Z= cross(x,y)
b. If we consider the cross product along the rows, then we have to specify “2”, which is meant for rows in the command. Here resultant will be an accumulation of all the row vectors.
x = randi(15,3,3,2);
y = randi(10,3,3,2);
Z= cross(x,y,2)>/code>
c. If we want to find the cross product along the third dimension of the matrix, then we can mention it in the command.
x = randi(15,3,3,3);
y = randi(10,3,3,3);
Z= cross(x,y,3)
Cross Product is an important aspect while dealing with vector quantities. They are mainly used in mathematical sectors to determine the cross product of the vectors. They are also used to find
whether the two vectors are orthogonal in nature. So, learning the use of cross products is important if we are dealing with any vector related work.
Recommended Articles
This is a guide to Matlab Cross Product. Here we discuss basic concept, how does it works and examples to implement in Matlab Cross Product. You can also go through our other related articles to
learn more – | {"url":"https://www.educba.com/matlab-cross-product/","timestamp":"2024-11-07T22:11:02Z","content_type":"text/html","content_length":"315913","record_id":"<urn:uuid:f268ce55-2f9d-4484-96a3-e3c445142c24>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00512.warc.gz"} |
Is the int8 type ciphertext realized by only a single LWE ciphertext?
I wonder that if the LWE ciphertext support the int8 type arithmetic? If so, how large is the ring dimension N used in blind rotation?
hello @GuLu_GuLu it can in shortint, but the parameters we have for it don’t allow to make much computations on it except bootstrapping IIRC because there is no margin for noise for leveled
pub const PARAM_MESSAGE_8_CARRY_0_KS_PBS_GAUSSIAN_2M64: ClassicPBSParameters =
ClassicPBSParameters {
lwe_dimension: LweDimension(1110),
glwe_dimension: GlweDimension(1),
polynomial_size: PolynomialSize(32768),
lwe_noise_distribution: DynamicDistribution::new_gaussian_from_std_dev(StandardDev(
glwe_noise_distribution: DynamicDistribution::new_gaussian_from_std_dev(StandardDev(
pbs_base_log: DecompositionBaseLog(15),
pbs_level: DecompositionLevelCount(2),
ks_base_log: DecompositionBaseLog(2),
ks_level: DecompositionLevelCount(11),
message_modulus: MessageModulus(256),
carry_modulus: CarryModulus(1),
max_noise_level: MaxNoiseLevel::new(1),
log2_p_fail: -64.011,
ciphertext_modulus: CiphertextModulus::new_native(),
encryption_key_choice: EncryptionKeyChoice::Big,
1 Like
Thanks for the reply. My understanding is: to support any binary arithmetic for int8 type, the ring dimension N should at least (8 + e) -bits, where the noise part e mainly comes from the modulus
switching, for example, e can be 9-bit if the LWE dimension is 1024. Is that right?
Hello @GuLu_GuLu
the ring dimension N should at least (8 + e) -bits, where the noise part e mainly comes from the modulus switching
you are right. In practice e is composed of the input noise and some additional noise due to the modulus switch. This additional noise is drawn from a gaussian distribution with a variance ~ n/24. So
for an LWE dimension n of 1024, the modulus switching part of the noise e will be written on 5-6 bits.
1 Like
Thanks, @Sam .
I got what you mean. I want to know more details about the implementation of the FHEint8 in TFHE.rs. As you described above, the modulus switching will incur about 5~6 bits of noise, as a result, to
enable the look-up table, the ring dimension N should be at least 2^(8 + 6) = 16384 ?
I have also have an another question, how does the multiplication performed about the FHEint8 type? Is it similar to the ciphertext multiplication in BFV? | {"url":"https://community.zama.ai/t/is-the-int8-type-ciphertext-realized-by-only-a-single-lwe-ciphertext/1357","timestamp":"2024-11-05T22:25:17Z","content_type":"text/html","content_length":"25515","record_id":"<urn:uuid:c04d7264-eade-4586-8b27-ecd9ed483875>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00242.warc.gz"} |
A quick picture of current implementation:
The LargeIntegers in Squeak are accelerated by using a multi-layer mechanism:
• The first layer is in VM Integer primitives (for example primitive 29 for LargePositiveInteger>>#*). If the Integer parameters fit on 64 bits and so does the result, then they are handled quite
fast at this level, especially in COG VM, the most common primitives being directly inlined into JIT.
• The second layer is using the LargeIntegersPlugin (named LargeIntegers) - see Integer>>#* and #digitMultiply:neg: for example which uses primDigitMultiplyNegative.
• The third layer when above two failed is to fall back to Smalltalk code iterating on 8 bits digits (found at end of Integer>>#digitMultiply:neg: for example)
Currently, the 3rd layer is not used, because the plugin is integrated internally in every distributed VM. But it suggests that the LargeIntegersPlugin is also performing operations on 8-bits digits
which is quite dated... Most arithmetic units handle at least 32 or 64 bits nowadays.
A new 32-bit implementation:
So what performance could we gain by using 32-bits digits? To do so, I thought of two options:
• add a new Integer variableWordSubclass: #LargePositiveInteger32, and its Negative32 counterpart, plus a dedicated LargeInteger32Plugin with new primitives
• change the LargeIntegersPlugin to operate on 32-bits digits
The first option is completed by adding byteAt: byteAt:put: and byteLength primitives to preserve a 8-bits digit based fallback code (this is required because fallback must operate on SmallInteger).
The second option would be possible, because the allocated size is always a multiple of 4 bytes, even for Byte Arrays. But it has two drawbacks:
• LargeInteger are necessary for a modern Squeak image to work, so each mistake would cost me a lot of debugging pain, and I would first have to learn how to use the VM simulator...
• It would not be a great option on big-endian machines due to necessary swapping of each word (unless of course we use native word ordering and cheat for digitAt: and digitAt:put: fallback code).
So I opted for a new LargeInteger32Plugin solution tested in a 32bits VM. All my operations will be defined using unsigned int, those requiring 64 bits (* + - quo:) will use unsigned long long. Note
that 64 bits operations are emulated by gcc which does not use native 64 bits capabilities of ALU in this configuration.
Micro-benchmark results:
So here are the first results on my old 2.26GHz Intel core 2 duo Mac-mini for a hacked COG VM derivated from version 2640. They are presented in term of number of operations executed per second
(8bits on left, 32bits on middle, percentage of improvement on right column) for various operands bit lengths.
Number of operations per second in 8-bit vs 32-bit Large Integer Plugin
The main improvement is for *, around 7x at kilobit is already something. Note the primitive still implements a naïve O(N²) product, we know from other experiments that a simple Karatsuba algorithm
programmed at image side already improves things a bit above kilobit length in a COG VM (see optimizing-product-of-prime-powers). Then come the score of rem: but it is not fair... I cheated to use
the same primitive as quo: which in fact computes both quotient and remainder. It is not used in 8-bit version because performing operations with Integer primitives in case of 64 bits or less is much
much faster than using the LargeIntegersPlugin. Otherwise, we would have something similar to \\ score, which is already interesting thanks to usage of *... The improvment of + and - are low, those
are only O(N), probably we mainly measure the overhead because internal loops are too simple...
Micro-benchmark source code:
Here are the gory details: #(100 1000 10000) collect: [:n |
p1 := 1<<n - 1.
p2 := p1 sqrtFloor*p1+1.
q1 := LargePositiveInteger32 digitLength: p1 digitLength.
1 to: p1 digitLength do: [:i | q1 digitAt: i put: (p1 digitAt: i)].
q2 := LargePositiveInteger32 digitLength: p2 digitLength.
1 to: p2 digitLength do: [:i | q2 digitAt: i put: (p2 digitAt: i)].
n printString , ' bits' -> {
'+' -> {
[p1+p2] bench.
[q1+q2] bench}.
'-' -> {
[p2-p1] bench.
[q2-q1] bench}.
'*' -> {
[p1*p2] bench.
[q1*q2] bench}.
'//' -> {
[p2//p1] bench.
[q2//q1] bench}.
'\\' -> {
[p2\\p1] bench.
[q2\\q1] bench}.
'quo:' -> {
[p2 quo: p1] bench.
[q2 quo: q1] bench}.
'rem:' -> {
[p2 rem: p1] bench.
[q2 rem: q1] bench}.
'quo: 10' -> {
[p1 quo: 10] bench.
[q1 quo: 10] bench}.
'rem: 10' -> {
[p1 rem: 10] bench.
[q1 rem: 10] bench}.
'<<53' -> {
[p1 << 53] bench.
[q1 << 53] bench}.
'>>35' -> {
[p2 >> 35] bench.
[q2 >> 35] bench}.
This benchmark is very naive, because distribution of 1-bit and 0-bit used above is not fair. Since the probability of having a null 8-bit digit is higher than that of having a null 32-bit digit, and
since primDigitMultiplyNegative eliminates N loops at each null digits, this is probably unfair for the 8-bit version, but for a first idea of performance, we don't care - only 1 byte out of 256 is
null without prior knowledge of distribution.
Basic unit tests:
If results are false, then the benchmark is worth nothing, so I performed some quick sanity tests with involved pairs of Integers. Since I can't mix the 8-bit and 32-bit Integers (no interest by
now), the tests are quite low level. I disabled logDebuggerStackToFile Preferences, I don't use traditional assert:, and only evaluate from a Workspace, because a Debugger trying to print the 32-bit
LargeIntegers may involve such mixed arithmetic and crash the image. IMO, the Squeak Debugger is much too clever and performs too many message sends on the debuggee...
assert :=
[p digitLength = q digitLength ifFalse: [self halt: 'different length'].
1 to: (p digitLength min: q digitLength) do: [:i | (p digitAt: i) = (q digitAt: i) ifFalse: [self halt: 'different digit at ' , i printString]]]. #(100 1000 10000) do: [:n |
p1 := 1<<n - 1.
p2 := p1 sqrtFloor*p1+1.
q1 := LargePositiveInteger32 digitLength: p1 digitLength.
1 to: p1 digitLength do: [:i | q1 digitAt: i put: (p1 digitAt: i)].
q2 := LargePositiveInteger32 digitLength: p2 digitLength.
1 to: p2 digitLength do: [:i | q2 digitAt: i put: (p2 digitAt: i)]. p := p1. q := q1. assert value. p := p2. q := q2. assert value. p := p1*3. q := q1*3. assert value. p := p1+5. q := q1+5. assert
value. p := p1-3. q := q1-3. assert value. p := p1+p2. q := q1+q2. assert value. p := p1-p2. q := q1-q2. assert value. p := p2-p1. q := q2-q1. assert value. p := p1*p2. q := q1*q2. assert value. p :=
p2<<53. q := q2<<53. assert value. p := p2>>35. q := q2>>35. assert value. p := p2 quo: p1. q := q2 quo: q1. assert value. p := p2 rem: p1. q := q2 rem: q1. assert value. p := p2//p1. q := q2//q1.
assert value. p := p2\\p1. q := q2\\q1. assert value].
Source code and other details:
I went thru complications for either adding my new LargePositive/NegativeInteger32 classes to Smalltalk specialObjectsArray or using addGCRoot: mechanism. I finally used specialObjectsArray for the
bench to be fair. Note that unlike 8-bit LargeInteger, my 32-bit LargeInteger classes are not compact, but I have no idea of the influence on above micro-benchmark. The LargePositiveInteger32 and
LargeInteger32Plugin code is unpublished yet. I have to clean it up and increase testing before publishing. And I don't really know yet where to publish this stuff. But it will of course be available
as MIT.
The volume is also too high for commenting it in this too long post, but this might feed a few more posts if there is any interest... | {"url":"https://smallissimo.blogspot.com/2012/","timestamp":"2024-11-06T18:21:27Z","content_type":"text/html","content_length":"223491","record_id":"<urn:uuid:657cc4fe-4ead-4e9b-82ba-487984d33dff>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00705.warc.gz"} |
The Geometry of Complex Numbers
Any nonzero complex number modulus and argument. If you plot z in the complex plane (where the x axis is the real part and the y axis is the imaginary part) at modulus of z is the distance, r, from
the origin to P. The argument of z is the angle,
Using basic trigonometry, we can write z as
This last expression is often abbreviated
You might notice that zero, which is
Now if you multiply
The Rule to Remember: When you multiply complex numbers, you multiply moduli and add arguments.
This is actually the way many mathematicians remember the trig identities. Once you know this general rule for multiplying complex numbers, you don't need to memorize the details of the two
identities we used
De Moivre's formula, named after Abraham de Moivre (1667 - 1754), follows from this more general rule. Given any nonzero complex number z and any integer n, the
Related Links:
Dave's Short Course on Complex Numbers provides a thorough and accessible introduction to complex numbers, their meaning, geometry, and operations.
Geometry and Complex Numbers offers a text with exercises on complex numbers and trigonometry.
Complex Applet provides an interactive tool for studying the geometry of complex numbers.
John and Betty's Journey Through Complex Numbers is a picture book introduction to complex numbers. | {"url":"https://www2.edc.org/makingmath/mathtools/complex/complex.asp","timestamp":"2024-11-01T19:03:53Z","content_type":"text/html","content_length":"11287","record_id":"<urn:uuid:887679c0-5b15-4cb5-8f65-4619cd031b7e>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00601.warc.gz"} |
Physics Project Class 12 ISC for Students 2024 | MCQTUBE
Physics Project Class 12 ISC
Physics Project Class 12 ISC. We covered all the Physics Project Class 12 ISC MCQs in this post for free so that you can practice well for the exam.
Install our MCQTUBE Android app from the Google Play Store and prepare for any competitive government exams for free.
These types of competitive MCQs appear in exams like SSC, Railway, Bank, Delhi Police, UPSSSC, UPSC (Pre), State PCS, CDS, NDA, Assistant Commandant, and other Competitive Examinations, etc.
We created all the competitive exam MCQs into several small posts on our website for your convenience.
You will get their respective links in the related posts section provided below.
Related Posts:
Physics Project Class 12 ISC Objective for Students
If the work done on the system or by the system is zero, which of the following statements for a system kept at a certain temperature is correct?
(a) Change in internal energy of the system is equal to the flow of heat in or out of the system.
(b) Change in internal energy of the system is less than heat transferred.
(c) Change in the internal energy of the system is more than the heat flow.
(d) Cannot be determined.
Option a – Change in internal energy of the system is equal to the flow of heat in or out of the system
Blowing air with an open pipe is an example of ( Physics Project Class 12 MCQs )
(a) isothermal process
(b) isochoric process
(c) isobaric process
(d) adiabatic process
Option c – isobaric process
A cycle tire bursts suddenly. This represents an
(a) isothermal process
(b) adiabatic process
(c) isochoric process
(d) isobaric process
Option b – adiabatic process
Who codified the first two laws of thermodynamics and deduced that the absolute zero of temperature is -273.15°C, leading to the naming of the Kelvin temperature scale?
(a) William Crookes
(b) William Thomson
(c) Luis Alvarez
(d) Robert Hooke
Option b – William Thomson
The statement that ‘heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature’ is known as ( Physics Project Class 12 MCQs )
(a) Zeroth Law of Thermodynamics.
(b) First Law of Thermodynamics.
(c) Second Law of Thermodynamics.
(d) Third Law of Thermodynamics.
The greenhouse effect is the heating up of the Earth’s atmosphere due to
(a) the ultraviolet rays
(b) gamma rays
(c) the infrared rays
(d) X-rays
Option c – the infrared rays
Zeroth Law of Thermodynamics leads to
(a) the concept of temperature.
(b) the concept of specific heat.
(c) the concept of internal energy.
(d) None of the above
Option a – the concept of temperature
The first law of thermodynamics is simply the case of ( Physics Project Class 12 MCQs )
(a) Charles’s Law
(b) Newton’s Law of Cooling
(c) The Law of Heat Exchange
(d) The Law of Conservation of Energy
Option d – The Law of Conservation of Energy
A weightless string can bear tension up to 3.7 kgwt. A stone of mass 0.5 kg is tied to it and rotated in a vertical circular path of radius 4 m then the maximum angular velocity of the stone :
(A) 2 rad/s
(B) 2.7 rad/s
(C) 3 rad/s
(D) 4 rad/s
A motorcycle is going on over the bridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the bridge, The normal force on it ( Physics Project Class 12 MCQs )
(A) Increases
(B) Remains constant
(C) Decrease
(D) fluctuates
A frictionless track ends in a circular loop of radius small body slides down the track from a 3 cm. A height h then the minimum value of ‘h’ point ‘p’ at for a body to complete a circular loop :
(A) 3 cm
(B) 7.5 cm
(C) 12.5 cm
(D) 14 cm
A vertical section of the flyover bridge is in the form of an arc of radius 19.5 m, the truck crosses the bridge without losing contact, and the center of gravity of the truck is 0.5 m above the
surface. The maximum speed upto which the truck can be driven is : ( Physics Project Class 12 MCQs )
(A) 7 m/s
(B) 14 m/s
(C) 21 m/s
(D) 28 m/s
A stone attached to a rope of length 1 = 80 cm is rotated at a speed of 240 rpm. At the moment when the velocity is directed upward the rope breaks, to what height does the stone rise further?
(A) 10.3 m
(B) 41.2 m
(C) 20.6 m
(D) 24.9 m
A 4 kg ball swings in a vertical circle at the end of the chord 1 m long. What maximum speed at which it can swing if the chord can sustain maximum tension of 163.6 N? ( Physics Project Class 12 MCQs
(A) 5.57 m/s
(B) 4 m/s
(C) 10.2 m/s
(D) 31.1 m/s
The string of a pendulum of mass m and length 1 is displaced through 90°. The minimum strength of the string to withstand the tension will be :
(A) mg
(B) 2 mg
(C) 3 mg
(D) 4 mg
A small body of mass 0.1 kg swings in a vertical circle, at the end of the chord of length 1 m. If the speed is 2 m/s when the chord makes an angle of 30° with vertical. Find the tension in the chord
: (g=9.8 m/s²) ( Physics Project Class 12 MCQs )
(A) 0.4 N
(B) 0.85 N
(C) 0.98 N
(D) 1.25 N
A particle of mass 1 kg is moving in the xy plane parallel to the y-axis, with a uniform speed of 5 m/s, 2 m away from the origin. Angular momentum of particle about origin :
(A) 10kg m² s
(B) 5kg m²/s
(C) 2.5kg m²/s
(D) 2kgm²/s
A bob of mass m attached to an inextensible string of length I is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed of rad/s about the vertical. About
the point of suspension : ( Physics Project Class 12 MCQs )
(A) angular momentum changes in direction but not in magnitude
(B) angular momentum changes both in direction and magnitude
(C) angular momentum is conserved
(D) angular momentum changes in magnitude but not in direction
Option a – angular momentum changes in direction but not in magnitude
If the radius of the earth is suddenly expanded by ‘n’ times its present value without change in its mass then the length of the day is :
(A) 24 hours
(B) 24/n² hours
(C) n² 24 hours
(D) n 24 hours
A wheel is rotating with an angular frequency of 500 revolutions per minute on a shaft. The second wheel whose mass and radius are half of the first is coupled with the first wheel then the angular
speed of rotation becomes : ( Physics Project Class 12 MCQs )
(A) 500 rpm
(B) 450 rpm
(C) 250 rpm
(D) 125 rpm
A particle of mass m is rotating in a plane in a circular path of radius Y’. Its angular momentum is L. The centripetal force acting on the particle is :
(A) L²/mr³
(B) L²m/r
(C) L²/mr²
(D) L²m/r²
If the ice of polar caps melts the duration of the day will : ( Physics Project Class 12 MCQs )
(A) Increase
(B) 20 gm cm²
(C) Remains the same
(D) cannot be predicted
A particle performs UCM with an angular momentum of L. If the frequency of the particle in motion is doubled and Its KE is halved, the angular momentum becomes :
(A) L/4
(B) L/2
(C) L
(D) 4L
A body starts from rest and completes 20 revolutions in 5 minutes then its number of revolutions in the next 5 minutes : ( Physics Project Class 12 MCQs )
(A) 20 revolutions
(B) 40 revolutions
(C) 60 revolutions
(D) 80 revolutions
Option c – 60 revolutions
A flywheel of MI 15 kgm² slows down from 90 rad/ s to 40 rad/s in 20 seconds by a constant torque than the number of revolutions during this time interval :
(A) 100 rev
(B) 207 rev
(C) 215 rev
(D) 1300 rev
A 0.5 kW motor acts for 10 seconds on an initially non-rotating wheel with a moment of inertia of 1 kg m². What is the angular velocity developed in the wheel neglecting friction? ( Physics Project
Class 12 MCQs )
(A) 50 rad/s
(B) 120 rad/s
(C) 70 rad/s
(D) 160 rad/s
A wheel of the moment of inertia 50 kg m² is rotating at a uniform angular velocity of 5 rad/ s. The torque required to stop it in 2 s has a magnitude :
(A) 125 Nm
(B) 250 Nm
(C) 500 Nm
(D) 1000 Nm
A motor running at a rate of 1200 rpm can supply a torque of 80 Nm. What power does it develop?
(A) 1.6 π kW
(B) 3.2 π kW
(C) 0.8 π kW
(D) 4.8 π kW
Starting from rest a fan takes 5 seconds to attain the maximum speed of 400 rpm. Assuming constant acceleration, the time taken by the fan in attaining half the maximum speed is :
(A) 20 s
(B) 10 s
(C) 2.5 s
(D) 2.0 s
A motor of an engine is rotating about its axis with an angular velocity of 100 rev/min. It comes to rest in 15 seconds after being switched off. A number of revolutions made by the motor before
coming to rest :
(A) 25 rev
(B) 25 π rev
(C) 12.5 rev
(D) 50 π rev
The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle ‘0’ without slipping and slipping down the incline without rolling is :
(A) 5 : 7
(B) 2 : 5
(C) 2 : 3
(D) 7 : 5
A body rolls down an inclined plane. If its kinetic energy of rotational motion is 40% of its kinetic energy of translational then the body is :
(A) a Disc
(B) a Hollow sphere
(C) a Ring
(D) a Solid sphere
Option d – a Solid sphere
The solid cylinder is raised to a certain height on the inclined plane and then rolls down with a velocity of 7 m/s. What is the height to which the cylinder is raised?
(A) 1.2 m
(B) 4.9 m
(C) 3 m
(D) 3.75 m
An inclined plane makes an angle of 30° with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has linear acceleration equal to
(A) 3g
(B) 5g
(C) 5g/14
(D) 5g/7
A solid sphere of mass 2 kg rolls up a 30° incline with an initial speed of 10 m/s. The maximum height reached by the sphere is : (g = 10 m/s²)
(A) 3.5 m
(B) 7 m
(C) 10.5 m
(D) 14 m
A solid sphere rolls down an inclined plane the percentage of total energy which is rotational K.E. is :
(A) 28%
(B) 72%
(C) 100%
(D) 75%
A solid sphere rolls down an inclined plane, and the percentage of total energy which is translational K.E. is :
(A) 28%
(B) 72%
(C) 100%
(D) 25%
The speed of a solid sphere after rolling from a height of 14 m horizontally is (g = 9.8 m/s²) :
(A) 14 m/s
(B) 7 m/s
(C) 9.8 m/s
(D) 10 m/s
A small spherical liquid drop is moving in a viscous medium. The viscous force does not depend on
(A) the nature of the medium
(B) the density of the medium
(C) the instantaneous speed of the spherical drop
(D) the radius of the spherical drop
Option b – the density of the medium
If the temperature rises, the coefficient of viscosity of a liquid :
(A) decreases
(B) increases
(C) remains unchanged
(D) increases for some liquids and decreases for others
A metal plate 48 cm² in area rests horizontally on a layer of oil 1 mm thick. A force of 0.25 N applied to the plate horizontally keeps it moving with a uniform speed of 2.5 cm/s. The coefficient of
viscosity of the oil is :
(A) 1.083 Ns/m²
(B) 3.083 Ns/m²
(C) 2.083 Ns/m²
(D) 4.083 Ns/m²
A plate of area 100 cm² is lying on the upper surface of a 3 mm thick oil film. If the coefficient of viscosity of the oil is 15.5 poise then the horizontal force required to move the plate with a
velocity of 3 cm/s will be
(A) 0.155 N
(B) 15.5 N
(C) 1.55 N
(D) 155 N
Stoke’s law is applicable only to :
(A) púre liquids
(B) solutions
(C) non-viscous liquids
(D) viscous liquids
Option d – viscous liquids
The SI unit of the coefficient of viscosity is :
(A) m/kg-s
(B) kg/m-s²
(C) m-s/kg²
(D) kg/m-s
The coefficient of viscosity of a liquid does not depend upon
(A) the density of the liquid
(B) the temperature of the liquid
(C) the pressure of the liquid
(D) the nature of liquid
Option a – the density of the liquid
1 centipoise is equal to :
(A) 0.001 kg/m-s
(B) 1 kg/m-s
(C) 0.1 kg/m-s
(D) 1000 kg/m-s
One poise is equivalent to :
(A) 0.1 Pa-s
(B) 0.001 Pa-s
(C) 0.01 Pa-s
(D) 0.0001 Pa-s
Two sound waves of wavelengths 0.87 m and 0.885 m produce 7 beats per second. The velocity of sound in the air will be near about :
(A) 359 m/s
(B) 340 m/s
(C) 280 m/s
(D) 320 m/s
Eleven tuning forks are arranged in ascending order of frequencies. Each fork produces 8 beats per second with the other. The last fork is an octave of the first. The frequency of the 10th fork is :
(A) 80 Hz
(B) 152 Hz
(C) 172 Hz
(D) 184 Hz
A tuning fork C sounded together with a tuning fork D of frequency 256 Hz emits 2 beats. On loading the tuning fork D the number of beats, heard is 1 per second. The frequency of tuning fork C is :
(A) 257 Hz
(B) 256 Hz
(C) 258 Hz
(D) 254 Hz
Two bodies have frequencies of 252 and 256 vibrations per second respectively. The beat frequency produced when they vibrate together is :
(A) 16
(B) 12
(C) 20
(D) 4
When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the
number of beats is two per second. The natural frequency of the second tuning fork is :
(A) 335 Hz
(B) 339 Hz
(C) 343 Hz
(D) 347 Hz
Two tuning forks A and B produce 8 beats per second when sounded together. When B is slightly loaded with wax the beats are reduced to 4 per second. If the frequency of A is 512 Hz, the frequency of
B is :
(A) 508 Hz
(B) 516 Hz
(C) 504 Hz
(D) 520 Hz
A source of the sound of frequency 450 cycle/sec is moving, towards a stationary observer with 34 m/s. If the speed of sound is 340 m/s, then the apparent frequency will be :
(A) 410 cycle/sec
(B) 500 cycle/sec
(C) 550 cycle/sec
(D) 450 cycle/sec
Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats per second. The highest frequency is twice the lowest. The possible
highest and lowest frequencies are :
(A) 80 Hz and 40 Hz
(B) 100 Hz and 50 Hz
(C) 44 Hz and 22 Hz
(D) 72 Hz and 36 Hz
Option d – 72 Hz and 36 Hz
Two tuning forks have frequencies of 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be :
(A) 1/4 second
(B) 1/2 second
(C) 1 second
(D) 2 second
A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of
the first tuning fork?
(A) 95 Hz
(B) 100 Hz
(C) 105 Hz
(D) 110 Hz
Two tuning forks of frequency 256 and 258 vibrations per second are sounded together, then the time interval between consecutive maxima heard by the observer is :
(A) 2 sec
(B) 0.5 sec
(C) 250 sec
(D) 252 sec
Two tuning forks A and B vibrating simultaneously produce 5 beats. The frequency of B is 512 Hz. It is seen that if one arm of A is filed, then the number of beats increases. The frequency of A will
be :
(A) 502 Hz
(B) 507 Hz
(C) 517 Hz
(D) 522 Hz
We covered all the physics project class 12 isc mcqs above in this post for free so that you can practice well for the exam.
Check out the latest MCQ content by visiting our mcqtube website homepage.
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Units to measure capacity, mass and length | Stage 3 Maths | HK Secondary S1-S3
How we measure things
When we measure things, the unit of measurement we use depends on what it is we are measuring. We are going to look at three different types of measurement here, as well as the words we use to
describe what it is we are measuring.
Capacity is a word to describe how much something holds. We often use this when we think about liquids, though you could use it to describe how many people a sports ground holds, perhaps. Today,
we'll be looking at liquids, and comparing some different containers, and how much they hold. The units of measurement we use for capacity are:
• millilitres (mL)
• litres (L)
• kilolitres (kL) and
• megalitres (ML)
If we measure length, we're thinking about how far away one end of something is from the other. We could also be talking about how tall someone (or something) is, or how wide something is. The units
of measurement we use for length are:
• millimetres (mm)
• centimetres (cm)
• metres (m) and
• kilometres (km)
When we want to know how heavy something is, we refer to its mass. Some objects may weigh a lot, even though they are very small. Other objects may be very big but weigh very little. Imagine a big
bag of feathers- it takes up quite a bit of space but it probably doesn't weight very much. On the other hand, an elephant is big and definitely weights a lot! Often, we call of an object's mass its
In this video, we'll look at all capacity, length and mass and see examples of the units of measurement for each.
Did you know?
'kilo' is used at the start of a word to suggest 'thousand' - a kilogram is $1000$1000grams
'milli' is used at the start of a word to suggest 'thousandth' - a millilitre is $0.001$0.001 of a litre
Worked examples
Question 1
Look at the picture of the scale to answer the following questions.
1. What do we measure with a scale like this one?
2. Which of these units can we measure with this scale?
Question 2
Which unit is best to measure how far a person has jumped in the long jump?
Question 3 | {"url":"https://mathspace.co/textbooks/syllabuses/Syllabus-98/topics/Topic-4531/subtopics/Subtopic-58730/","timestamp":"2024-11-11T21:26:11Z","content_type":"text/html","content_length":"420701","record_id":"<urn:uuid:2bc3130c-1500-483b-ab8f-405195da1c44>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00447.warc.gz"} |
agged bit patterns
Snippets tagged bit patterns
• A function that returns a sequence of subsets generated by the power set of a specified set. The function use bit patterns to generate sets. for example the power set generated by a set with 3
elements set [1;2;3;] has 2^3 sets. each set in the power set is represented by the set bits in each of the integer from 0 to (2^3) -1
8 people like this
Like the snippet!
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Ch. 1 Key Terms - College Algebra 2e | OpenStax
Key Terms
algebraic expression
constants and variables combined using addition, subtraction, multiplication, and division
associative property of addition
the sum of three numbers may be grouped differently without affecting the result; in symbols, $a+( b+c )=( a+b )+c a+( b+c )=( a+b )+c$
associative property of multiplication
the product of three numbers may be grouped differently without affecting the result; in symbols, $a⋅( b⋅c )=( a⋅b )⋅c a⋅( b⋅c )=( a⋅b )⋅c$
in exponential notation, the expression that is being multiplied
a polynomial containing two terms
any real number $a i a i$ in a polynomial in the form $a n x n +...+ a 2 x 2 + a 1 x+ a 0 a n x n +...+ a 2 x 2 + a 1 x+ a 0$
commutative property of addition
two numbers may be added in either order without affecting the result; in symbols, $a+b=b+a a+b=b+a$
commutative property of multiplication
two numbers may be multiplied in any order without affecting the result; in symbols, $a⋅b=b⋅a a⋅b=b⋅a$
a quantity that does not change value
the highest power of the variable that occurs in a polynomial
difference of squares
the binomial that results when a binomial is multiplied by a binomial with the same terms, but the opposite sign
distributive property
the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, $a⋅( b+c )=a⋅b+a⋅c a⋅( b+c )=a⋅b+a⋅c$
a mathematical statement indicating that two expressions are equal
in exponential notation, the raised number or variable that indicates how many times the base is being multiplied
exponential notation
a shorthand method of writing products of the same factor
factor by grouping
a method for factoring a trinomial in the form $a x 2 +bx+c a x 2 +bx+c$ by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring
out the GCF of the entire expression
an equation expressing a relationship between constant and variable quantities
greatest common factor
the largest polynomial that divides evenly into each polynomial
identity property of addition
there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols, $a+0=a a+0=a$
identity property of multiplication
there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols, $a⋅1=a a⋅1=a$
the number above the radical sign indicating the nth root
the set consisting of the natural numbers, their opposites, and 0: ${ …,−3,−2,−1,0,1,2,3,… } { …,−3,−2,−1,0,1,2,3,… }$
inverse property of addition
for every real number $a, a,$ there is a unique number, called the additive inverse (or opposite), denoted $−a, −a,$ which, when added to the original number, results in the additive identity, 0;
in symbols, $a+( −a )=0 a+( −a )=0$
inverse property of multiplication
for every non-zero real number $a, a,$ there is a unique number, called the multiplicative inverse (or reciprocal), denoted $1 a , 1 a ,$ which, when multiplied by the original number, results in
the multiplicative identity, 1; in symbols, $a⋅ 1 a =1 a⋅ 1 a =1$
irrational numbers
the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers
leading coefficient
the coefficient of the leading term
leading term
the term containing the highest degree
least common denominator
the smallest multiple that two denominators have in common
a polynomial containing one term
natural numbers
the set of counting numbers: ${ 1,2,3,… } { 1,2,3,… }$
order of operations
a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations
perfect square trinomial
the trinomial that results when a binomial is squared
a sum of terms each consisting of a variable raised to a nonnegative integer power
principal nth root
the number with the same sign as $a a$ that when raised to the nth power equals $a a$
principal square root
the nonnegative square root of a number $a a$ that, when multiplied by itself, equals $a a$
the symbol used to indicate a root
radical expression
an expression containing a radical symbol
the number under the radical symbol
rational expression
the quotient of two polynomial expressions
rational numbers
the set of all numbers of the form $m n , m n ,$ where $m m$ and $n n$ are integers and $n≠0. n≠0.$ Any rational number may be written as a fraction or a terminating or repeating decimal.
real number line
a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left.
real numbers
the sets of rational numbers and irrational numbers taken together
scientific notation
a shorthand notation for writing very large or very small numbers in the form $a× 10 n a× 10 n$ where $1≤| a |<10 1≤| a |<10$ and $n n$ is an integer
term of a polynomial
any $a i x i a i x i$ of a polynomial in the form $a n x n +...+ a 2 x 2 + a 1 x+ a 0 a n x n +...+ a 2 x 2 + a 1 x+ a 0$
a polynomial containing three terms
a quantity that may change value
whole numbers
the set consisting of 0 plus the natural numbers: ${ 0,1,2,3,… } { 0,1,2,3,… }$ | {"url":"https://openstax.org/books/college-algebra-2e/pages/1-key-terms","timestamp":"2024-11-13T08:28:24Z","content_type":"text/html","content_length":"353495","record_id":"<urn:uuid:57ba9405-c798-4253-b4df-9772e079f8be>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00553.warc.gz"} |
ball mill simulation software download
The ball mill grinding process is mainly composed of a ball mill, a hydrocyclone cluster, a sump, a slurry pump and an ore bin. The schematic diagram of this process is shown in Fig. 1. Fig. 1
Schematic diagram of ball mill grinding process. The variables of this process shown in Fig. 1 are explained as follows: W
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When the test ball mill is run in open circuit the design feed rate is tonnes/hour. Figure 4 compares the predicted size distributions from the open circuit ball mill with feed rates of, and
tonnes/hour respectively. As might be expected, the larger the feed rate the coarser the grinding mill discharge product.
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For instance, high energy ball milling is a topdown method using planetary ball mills. To obtain optimized milling parameters in a planetary ball mill, many trials are needed. ... (2017) Grinding
of classF fly ash using planetary ball mill: a simulation study to determine the breakage kinetics by directand backcalculation method. S Afr J ...
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A method for simulating the motion of balls in tumbling ball mill under wet condition is investigated. The simulation method is based on the threedimensional discrete element method (DEM) and
takes into account the effects of the presence of suspension,, drag force and buoyancy. The impact energy on balls' collision, which enables us to provide useful information for predicting the
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%free Downloads. 2151 "ball mill" 3D Models. Every Day new 3D Models from all over the World. Click to find the best Results for ball mill Models for your 3D Printer.
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Fig. 1 shows the spreadsheet which includes the graphical view of various flowsheets. For example to simulate a tumbling ball mill in an open circuit, the corresponding button, No. 1, must be
selected and pressed. Then, automatically another spreadsheet will be opened to enter simulation data ( Fig. 2, Fig. 3 ).
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Contribute to chengxinjia/sbm development by creating an account on GitHub.
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The submodels of comminution and classification for vertical spindle mill (VSM) presented in Part 1 of this paper have been integrated in the VSM simulation models for the Emill, MPS mill and CKP
mill. Plant survey data from an Emill (ballrace) and MPS mill (rollerrace), both including internal streams and external sampling, and the CKP ...
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1. Fill the container with small metal balls. Most people prefer to use steel balls, but lead balls and even marbles can be used for your grinding. Use balls with a diameter between ½" (13 mm)
and ¾" (19 mm) inside the mill. The number of balls is going to be dependent on the exact size of your drum.
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The simulation model for tumbling ball mills proposed by Austin, Klimpel and Luckie (AKL) was used to simulate wet grinding in ball mills, and it gave good agreement with experimental results
from ...
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Abstract. The multisegment ball mill model developed by Whiten and Kavetsky has been used together with an extensive range of data from operating mills to establish the parameters of a new ball
mill model suitable for simulation and design of coarse grinding ball mills ( mills containing some plus 2mm particles in the mill discharge).
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The simulation is applied over all the PID sets aiming to find the parameter region that provides the minimum integral of absolute error, which functions as a performance ... The ball cement mill
(CM) is fed with raw materials. The milled product is fed via a recycle elevator to a dynamic separator. The high fineness stream of the separator
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to the ineficiency of ball mills for fine grind applications. The difficulty encountered in fine grinding is the increased resistance to comminute small particles compared to coarse particles. As
a result, increased energy inputs are necessary to raise the number of collisional events in a mill ... Vertical Stirred Mill Simulation Using a ...
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Planetary ball mill is a powerful tool, which has been used for milling various materials for size reduction. The discrete element method (DEM) was used to simulate the dynamics of particle ...
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The simulation started with the small ball mill with partially filled monosized spherical particles (Fig. 1 (a)). Periodic boundary condition was applied along the axial direction to avoid the
wall effect. The mill size D is 280 mm, particle diameter d is mm, mill fill level fraction M* is, and critical speed fraction N * is at ...
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The planetary ball mill is promising in that it makes grinding to submicron sizes possible by imparting high energy to the ground powder. In this context, there is a need to understand the
dynamics of ultrafine grinding within the mill. ... Simulation of grinding in different devices was studied extensively in the literature [10,11,12,13 ...
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The objective of this work is to investigate the effect of milling parameters, including shape of powder particles, rotation speed, and balltopowder diameter (BPDR) on DEM simulation results in a
planetary ball mill and to develop a method to minimize the calculation cost during simulation. 2. DEM Model and Simulation Conditions.
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The ball mill is the important equipment in the mining industry, with the development of the macroscale ball mill, and is more difficult of liner's wearing,installation and taking apart. The
service life which raises a liner will prolong the production period that the ball mill, influencing an economic benefit of the concentrating mill.
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Abstract Talc powder samples were ground by three types of ball mill with different sample loadings, W, to investigate rate constants of the size reduction and structural change into the
amorphous state. Ball mill simulation based on the particle element method was performed to calculate the impact energy of the balls, E i, during grinding. Each rate constant correlated with the
specific ...
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Correspondingly, the ball mill has to replace the worn lifters. Therefore, determining the impact of mill speed and lifter on the load behavior of iron ore particles in a ball mill not only
potentially facilitates the improvement in highperformance liners but also optimizes the mill speed in the predesign stage.
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A ball mill, which is used to finely grind materials, causes high levels of vibration and sound during grinding operations. The vibration and sound of mills provide significant information about
the internal conditions and can be used to estimate the status of the ground material. We developed a simulation model for the vibration of a mill wall ...
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What is the Vector graphics and uses of Vector graphics
Vector graphics and it uses
Vector graphics is the creation of digital images through a sequence of commands or mathematical statements. It places lines and shapes in a given two-dimensional or three-dimensional space.
In physics, a vector is a representation of both a quantity and a direction at the same time. In vector graphics, the file that results from a graphic artist’s work is created and saved as a sequence
of vector statements. For example, instead of containing a bit in the file for each bit of a line drawing, a vector graphic file describes a series of points to be connected. One result is a much
smaller file. | {"url":"https://www.servercake.blog/vector-graphics/","timestamp":"2024-11-12T09:05:27Z","content_type":"text/html","content_length":"72480","record_id":"<urn:uuid:046c27b5-3513-40fd-9891-8fd1c0a56059>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00527.warc.gz"} |
Brownian motion Archives – H. Paul Keeler
This probability blog came up in my news feed:
It seems to focus on stochastic processes such as Brownian motion and friends.
Wiener or Brownian (motion) process
One of the most important stochastic processes is the Wiener process or Brownian (motion) process. In a previous post I gave the definition of a stochastic process (also called a random process) with
some examples of this important random object, including random walks. The Wiener process can be considered a continuous version of the simple random walk. This continuous-time stochastic process is
a highly studied and used object. It plays a key role different probability fields, particularly those focused on stochastic processes such as stochastic calculus and the theories of Markov
processes, martingales, Gaussian processes, and Levy processes.
The Wiener process is named after Norbert Wiener, but it is called the Brownian motion process or often just Brownian motion due to its historical connection as a model for Brownian movement in
liquids, a physical phenomenon observed by Robert Brown. But the physical process is not true a Wiener process, which can be treated as an idealized model. I will use the terms Wiener process or
Brownian (motion) process to differentiate the stochastic process from the physical phenomenon known as Brownian movement or Brownian process.
The Wiener process is arguably the most important stochastic process. The other important stochastic process is the Poisson (stochastic) process, which I cover in another post. I have written that
and the current post with the same structure and style, reflecting and emphasizing the similarities between these two fundamental stochastic process.
In this post I will give a definition of the standard Wiener process. I will also describe some of its key properties and importance. In future posts I will cover the history and generalizations of
this stochastic process.
In the stochastic processes literature there are different definitions of the Wiener process. These depend on the settings such as the level of mathematical rigour. I give a mathematical definition
which captures the main characteristics of this stochastic process.
Definition: Standard Wiener or Brownian (motion) process
A real-valued stochastic process \(\{W_t:t\geq 0 \}\) defined on a probability space \((\Omega,\mathcal{A},\mathbb{P})\) is a standard Wiener (or Brownian motion) process if it has the following
1. The initial value of the stochastic process \(\{W_t:t\geq 0 \}\) is zero with probability one, meaning \(P(W_0=0)=1\).
2. The increment \(W_t-W_s\) is independent of the past, that is, \(W_u\), where \(0\leq u\leq s\).
3. The increment \(W_t-W_s\) is a normal variable with mean \(o\) and variance \(t-s\).
In some literature, the initial value of the stochastic process may not be given. Alternatively, it is simply stated as \(W_0=0\) instead of the more precise (probabilistic) statement given above.
Also, some definitions of this stochastic process include an extra property or two. For example, from the above definition, we can infer that increments of the standard Wiener process are stationary
due to the properties of the normal distribution. But a definition may include something like the following property, which explicitly states that this stochastic process is stationary.
4. For \(0\leq u\leq s\), the increment \(W_t-W_s\) is equal in distribution to \(W_{t-s}\).
The definitions may also describe the continuity of the realizations of the stochastic process, known as sample paths, which we will cover in the next section.
It’s interesting to compare these defining properties with the corresponding ones of the homogeneous Poisson stochastic process. Both stochastic processes build upon divisible probability
distributions. Using this property, Lévy processes generalize these two stochastic processes.
The definition of the Wiener process means that it has stationary and independent increments. These are arguably the most important properties as they lead to the great tractability of this
stochastic process. The increments are normal random variables, implying they can have both positive and negative (real) values.
The Wiener process exhibits closure properties, meaning you apply certain operations, you get another Wiener process. For example, if \(W= \{W_t:t\geq 0 \}\) is a Wiener process, then for a scaling
constant \(c>0\), the resulting stochastic process \(\{W_{ct}/\sqrt{c}:t \geq 0 \}\)is also a Wiener process. Such properties are useful for proving mathematical results.
Two realizations of a Wiener (or Brownian motion) process. The sample paths are continuous (but non-differentiable) almost everywhere.
Properties such as independence and stationarity of the increments are so-called distributional properties. But the sample paths of this stochastic process are also interesting. A sample path of a
Wiener process is continuous almost everywhere. (The term almost everywhere comes from measure theory, but it simply means that the only region where the property does not hold is mathematically
negligible.) Despite the continuity of the sample paths, they are nowhere differentiable. (Historically, it was a challenge to find such a function, but a classic example is the Weierstrass function
The standard Wiener process has the Markov property, making it an example of a Markov process. The standard Wiener process \(W=\{ W_t\}_{t\geq 0}\) is a martingale. Interestingly, the stochastic
process \(W=\{ W_t^2-t\}_{t\geq 0}\) is also a martingale. The Wiener process is a fundamental object in martingale theory.
There are many other properties of the Brownian motion process; see the Further reading section for, well, further reading.
Playing a main role in the theory of probability, the Wiener process is considered the most important and studied stochastic process. It has connections to other stochastic processes and is central
in stochastic calculus and martingales. Its discovery led to the development to a family of Markov processes known as diffusion processes.
The Wiener process also arises as the mathematical limit of other stochastic processes such as random walks, which is the subject of Donsker’s theorem or invariance principle, also known as the
functional central limit theorem.
The Wiener process is a member of some important families of stochastic processes, including Markov processes, Lévy processes, and Gaussian processes. This stochastic process also has many
applications. For example, it plays a central role in quantitative finance. It is also used in the physical sciences as well as some branches of social sciences, as a mathematical model for various
random phenomena.
Generalizations and modifications
For the Brownian motion process, the index set and state space are respectively the non-negative numbers and real numbers, that is \(T=[0,\infty)\) and \(S=[0,\infty)\), so it has both continuous
index set and state space. Consequently, changing the state space, index set, or both offers an ways for generalizing or modifying the Wiener (stochastic) process.
A single realization of a two-dimensional Wiener (or Brownian motion) process. Each vector component is an independent standard Wiener process.
The defining properties of the Wiener process, namely independence and stationarity of increments, results in it being easy to simulate. The Wiener can be simulated provided random variables can be
simulated or sampled according to a normal distribution. The main challenge is that the Wiener process is a continuous-time stochastic process, but computer simulations run in a discrete universe.
I will leave the details of sampling this stochastic process for another post.
Further reading
A very quick history of Wiener process and the Poisson (point) process is covered in this talk by me.
There are books almost entirely dedicated to the subject of the Wiener or Brownian (motion) process, including:
Of course the stochastic process is also covered in any book on stochastic calculus:
More advanced readers can read about the Wiener process, its descrete-valued cousin, the Poisson (stochastic) process, as well as other Lévy processes:
• Kyprianou, Fluctuations of Lévy Processes with Applications;
• Bertoin, Lévy Processes;
• Applebaum, Lévy Processes and Stochastic Calculus.
On this topic, I recommend the introductory article:
• 2004, Applebaum, Lévy Processes – From Probability to Finance and Quantum Groups.
The Wiener process is of course also covered in general books on stochastic process such as: | {"url":"https://hpaulkeeler.com/tag/brownian-motion/","timestamp":"2024-11-02T02:58:07Z","content_type":"text/html","content_length":"70936","record_id":"<urn:uuid:a3a46611-3c29-4f4b-8c8a-179ca5b14fea>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00279.warc.gz"} |
Monash topology talk on sensitivity conjecture and Clifford algebras, July 2019, Daniel Mathews
On 31 July 2019 I gave at talk at Monash University in the topology seminar.
The sensitivity conjecture, induced subgraphs of cubes, and Clifford algebras
Recently, Hao Huang gave an ingenious short proof of a longstanding conjecture in computer science, the Sensitivity Conjecture. Huang proved this conjecture by establishing a result about the maximal
degree of induced subgraphs of cube graphs. In very recent work, we gave a new version of this result, and slightly generalise it, by connecting it to the theory of Clifford Algebras, algebraic
structures which arise naturally in geometry, topology and physics.
Monash topology talk on sensitivity conjecture and Clifford algebras, July 2019 | {"url":"https://www.danielmathews.info/2019/07/31/monash-topology-talk-on-sensitivity-conjecture-and-clifford-algebras-july-2019/","timestamp":"2024-11-07T19:41:40Z","content_type":"text/html","content_length":"38969","record_id":"<urn:uuid:a510829b-84fd-4e79-a54f-44cae5f4ae8a>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00021.warc.gz"} |
Fingeprinting, Polynomial Identities, Matchings and Isolation Lemma | Rafael Oliveira
Fingeprinting, Polynomial Identities, Matchings and Isolation Lemma
It is hard to overstate the importance of algebraic techniques in computer science. Algebraic techniques are used in many areas of computer science, including randomized algorithms (hashing, today’s
lecture), parallel algorithms (also this lecture), efficient proof/program verification (PCPs), coding theory, cryptography, and complexity theory.
We begin with a basic problem: suppose Alice and Bob each maintain the same large database of information (think of each as being a server from a comapany that deals with a lot of data). Alice and
Bob want to check if their databases are consistent. However, they do not want to reveal their entire database to each other (as that would be too expensive).
So, sending the entire database to each other is not an option. What can they do? Deterministic consistent checking requires sending the entire database to each other. However, if we use randomness
we can do much better, using a technique called fingerprinting.
The problem above can be more succinctly stated as follows: if Alice’s version of the database is given by string $a = (a_0, a_1, \ldots, a_{n-1})$ and Bob’s is given by $b = (b_0, b_1, \ldots, b_
{n-1})$, then given two strings $a, b \in \{0,1\}^n$, how can we check if they are equal?
Fingerprinting Mechanism
Let $\alpha := \sum_{i=0}^{n-1} a_i \cdot 2^i$ and $\beta := \sum_{i=0}^{n-1} b_i \cdot 2^i$. Let $p$ be a prime number and let $F_p(x) := x \bmod p$ be the function that maps $x$ to its remainder
modulo $p$. This function is called the fingerprinting function.
Now, we can describe the fingerprinting mechanism/protocol as follows:
1. Alice picks a random prime $p$ and sends $(p, F_p(\alpha))$ to Bob.
2. Bob checks if $F_p(\beta) \equiv F_p(\alpha) \bmod p$, and sends to Alice $$\begin{cases} 1 & \text{if } F_p(\beta) \equiv F_p(\alpha) \bmod p \\ 0 & \text{otherwise} \end{cases}$$
In the above algorithm, the total number of bits communicated is $O(\log p)$. And it is easy to see that if $a = b$, the protocol always outputs $1$. What happens when $a \neq b$?
Verifying String Inequality
If $a \neq b$, then $\alpha \neq \beta$. For how many primes $p$ is it true that $F_p(\alpha) \equiv F_p(\beta)$? (i.e., the protocol will fail) Note that $F_p(\alpha) \equiv F_p(\beta) \bmod p$ if
and only if $p \mid \alpha - \beta$. This leads us to the following claim:
Claim: If a number $M \in {-2^n, \ldots, 2^n}$, then the number of distinct primes $p$ such that $p \mid M$ is less than $n$.
Proof: each prime divisor of $M$ is $\geq 2$, so if $M$ has $k$ distinct prime divisors, then $|M| > 2^k$. Since $|M| \leq 2^n$, we have $k < n$.
By the above claim, the number of primes $p$ such that $p \mid \alpha - \beta$ is at most $n$. By the prime number theorem, we know that there are $m/\log m$ primes among the first $m$ positive
integers. Choosing our prime $p$ among the first $tn \log(tn)$ positive integers, we have that the probability that $p \mid \alpha - \beta$ is at most $\dfrac{n}{tn \log(tn) /\log(tn \cdot \log tn)}
= \tilde{O}(1/t)$.
Thus, the number of bits sent is $\tilde{O}(\log(tn))$. Choosing $t = n$ gives us a protocol which works with high probability.
Polynomial Identity Testing
The technique of fingerprinting can be used to solve a more general problem: given two polynomials $f(x), g(x) \in \mathbb{F}[x]$ (where $\mathbb{F}$ is a field), how can we check if $f(x) = g(x)$?
Two polynomials are equal if and only if their difference is the zero polynomial. Hence, the problem reduces to checking if a polynomial is the zero polynomial. Since a polynomial of degree $d$ is
uniquely determined by its values at $d+1$ points, we can check if a polynomial is the zero polynomial by checking if it is zero at $d+1$ points. If we want to turn this into a randomized algorithm,
we can simply sample one point uniformly at random from a set $S \subseteq \mathbb{F}$ with $|S| = 2(d+1)$ and check if the polynomial is zero at that point. By the above argument, the probability
that a nonzero polynomial evaluates to zero most $1/2$.
If we want to increase the success probability, there are two ways to do it: either we can increase the number of points we check, or we can repeat the above procedure multiple times.
The above problem as well as the approach can be generalized to polynomials in many variables.
The general problem is known as polynomial identity testing, which we now formally state:
Polynomial Identity Testing (PIT): Given a polynomial $f(x_1, \ldots, x_n) \in \mathbb{F}[x_1, \ldots, x_n]$, is $f(x_1, \ldots, x_n) \equiv 0$?
What do we mean by “given a polynomial”? This can come in many forms, but in this class we will only assume that we have access to an oracle that can evaluate the polynomial at any point in $\mathbb
Generalizing the above approach yields the following lemma, that can be used in a randomized algorithm for polynomial identity testing.
Lemma 1 (Ore-Schwartz-Zippel-de Millo-Lipton): Let $f(x_1, \ldots, x_n) \in \mathbb{F}[x_1, \ldots, x_n]$ be a nonzero polynomial of degree $d$. Then, for any set $S \subseteq \mathbb{F}$, we have $$
\Pr_{a_1, \ldots, a_n \in S}[f(a_1, \ldots, a_n) = 0] \leq \dfrac{d}{|S|}$$
Proof: We prove the lemma by induction on $n$. The base case $n = 1$ follows from the argument above.
For the inductive step, we assume that the lemma holds for $n-1$ variables. Let $f(x_1, \ldots, x_n) = \sum_{i=0}^d f_i(x_1, \ldots, x_{n-1}) x_n^i$. Since $f$ is non-zero, it must be the case that
$f_i$ is non-zero for some $i$. Let $k$ be the largest index such that $f_k$ is non-zero. Then, we have that $f_k(x_1, \ldots, x_{n-1})$ is a nonzero polynomial of degree $d-k$ in $n-1$ variables. By
the inductive hypothesis, we have that $$\Pr_{a_1, \ldots, a_{n-1} \in S}[f_k(a_1, \ldots, a_{n-1}) = 0] \leq \dfrac{d-k}{|S|}$$
Now, we have $$ \Pr_{a_1, \ldots, a_n \in S}[f(a_1, \ldots, a_n) = 0] = $$ $$\Pr_{a_1, \ldots, a_n \in S}[f(a_1, \ldots, a_{n-1}, a_n) = 0 \mid f_k(a_1, \ldots, a_{n-1}) \neq 0] \cdot \Pr_{a_1, \
ldots, a_{n-1} \in S}[f_k(a_1, \ldots, a_{n-1}) \neq 0] + $$ $$\Pr_{a_1, \ldots, a_n \in S}[f(a_1, \ldots, a_{n-1}, a_n) = 0 \mid f_k(a_1, \ldots, a_{n-1}) = 0] \cdot \Pr_{a_1, \ldots, a_{n-1} \in S}
[f_k(a_1, \ldots, a_{n-1}) = 0] \leq $$ $$\dfrac{k}{|S|} \cdot \Pr_{a_1, \ldots, a_{n-1} \in S}[f_k(a_1, \ldots, a_{n-1}) \neq 0] + $$ $$\Pr_{a_1, \ldots, a_n \in S}[f(a_1, \ldots, a_{n-1}, a_n) = 0
\mid f_k(a_1, \ldots, a_{n-1}) = 0] \cdot \dfrac{d-k}{|S|} \leq $$ $$\dfrac{k}{|S|} \cdot 1 + 1 \cdot \dfrac{d-k}{|S|} = \dfrac{d}{|S|}$$
where in the second to last inequality we simply applied the inductive hypothesis for the cases of 1 variable and $n-1$ variables. In the last inequality, we simply used the fact that any probability
is upper bounded by $1$.
Randomized Matching Algorithms
We now use the above lemma to give a randomized algorithm for the perfect matching problem. We begin with the problem of deciding whether a bipartite graph $G = (L \cup R, E)$ has a perfect matching.
Input: A bipartite graph $G = (L \cup R, E)$.
Output: YES if $G$ has a perfect matching, NO otherwise.
Let $n = |L| = |R|$ and let $X \in \mathbb{F}[x_{11}, x_{12}, \ldots, x_{nn}]^{n \times n}$ be the symbolic adjacency matrix of $G$. That is, $X_{ij} = x_{ij}$ if $(i,j) \in E$ and $X_{ij} = 0$
Since $$\det(X) = \sum_{\sigma \in S_n} (-1)^{\sigma} \prod_{i=1}^n X_{i, \sigma(i)}$$ and since each permutation corresponds to a perfect matching, we have that $\det(X) \not\equiv 0$ (as a
polynomial) if and only if $G$ has a perfect matching.
Thus, we can use Lemma 1 to give a randomized algorithm for the perfect matching problem! In other words, the perfect matching problem for bipartite graphs is a special case of the polynomial
identity testing problem.
Thus, our algorithm is simply to evaluate the polynomial $\det(X)$ at a random point in $\mathbb{F}^{n \times n}$. The analysis is the same as the one in the previous section.
Isolation Lemma
Often times in parallel algorithms, when solving a problem with many possible solutions, it is important to make sure that different processors are working towards the same solution.
For this, we need to single out (i.e. isolate) a specific solution without knowing any element of the solution space. How can we do this?
One way to do this is to implicitly define a random ordering on the solution space and then pick the first solution (i.e. lowest order solution) in this ordering. This approach also has applications
in distributed computing, where we want to pick a leader among a set of processors, or break deadlocks. We can also use this approach to compute a minimum weight perfect matching in a graph (see
references in slides).
We now state the isolation lemma:
Lemma 2 (Isolation Lemma): given a set system over $[n] := {1, 2, \dots, n}$, if we assign a random weight function $w: [n] \rightarrow [2n]$, then the probability that the minimum weight set is
unique is at least $1/2$.
Example: Suppose $n = 4$, and our set system is given by $S_1 = {1, 4}, S_2 = {2, 3}, S_3 = {1, 2, 3}$.
Then a random weight function $w: [4] \rightarrow [8]$ might be $w(1) = 3, w(2) = 5, w(3) = 8, w(4) = 4$. Then, the minimum weight set is $S_1$. However, if we had instead chosen $w(1) = 5, w(2) = 1,
w(3) = 7, w(4) = 3$, then we will have two sets with minimum weight, i.e., $S_1, S_2$.
Remark: The isolation lemma can be quite counter-intuitive. A set system can have $\Omega(2^n)$ sets, and on average, there are $\Omega(2^n/2n^2)$ sets of a given weight, as the max weight is $2n^2$.
The isolation lemma tells us that even though there are exponentially many sets, the probability that the minimum weight set is unique is still at least $1/2$.
Proof of Isolation Lemma: Let $\mathcal{S}$ be a set system over $[n]$, let $v \in [n]$, and for each $A \in \mathcal{S}$, let $w(A)$ be the weight of $A$. Also, let $\mathcal{S}_v \subset \mathcal
{S}$ be the family of sets in $\mathcal{S}$ that contain $v$, and similarly define $\mathcal{N}_v := \mathcal{S} \setminus \mathcal{S}_v$, that is, the family of sets in $\mathcal{S}$ that do not
contain $v$.
Let $$ \alpha_v := \min_{A \in \mathcal{N}_v} w(A) - \min_{B \in \mathcal{S}_v} w(B \setminus {v})$$
Note that
• $\alpha_v < w(v) \Rightarrow v$ does not belong to any minimum weight set.
• $\alpha_v > w(v) \Rightarrow v$ belongs to every unique minimum weight set.
• $\alpha_v = w(v) \Rightarrow v$ belongs to some but not all minimum weight sets (so this is an ambiguous case).
Since the weight function $w$ is chosen uniformly at random, we have that $\alpha_v$ is independent of $w(v)$. Hence, we have that $$\Pr[\alpha_v = w(v)] \leq 1/2n \Rightarrow \Pr[\text{there is an
ambiguous element}] \leq 1/2$$ where the last inequality follows from the union bound.
Note that if we have two sets $A, B$ of minimum weight, then any element $v \in A \Delta B$ is ambiguous. But as we saw above, the probability that there is an ambiguous element is at most $1/2$. | {"url":"https://cs.uwaterloo.ca/~r5olivei/courses/2024-spring-cs466/lecture-notes/lecture7/","timestamp":"2024-11-03T16:12:49Z","content_type":"text/html","content_length":"31257","record_id":"<urn:uuid:2e5e7f24-f8e5-452e-a711-ac1e764da92d>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00301.warc.gz"} |
An airway tree-shape model for geodesic airway branch labeling
We present a mathematical airway tree-shape framework
where airway trees are compared using geodesic distances. The framework
consists of a rigorously dened shape space for treelike shapes,
endowed with a metric such that the shape space is a geodesic metric
space. This means that the distance between two tree-shapes can be realized
as the length of the geodesic, or shortest deformation, connecting
the two shapes. By computing geodesics between airway trees, as well as
the corresponding airway deformation, we generate airway branch correspondences.
Correspondences between an unlabeled airway tree and a set
of labeled airway trees are combined with a voting scheme to perform
automatic branch labeling of segmented airways from the challenging
EXACT'09 test set. In spite of the varying quality of the data, we obtain
robust labeling results.
Originalsprog Engelsk
Titel MFCA 2011 : 3rd MICCAI Workshop on Mathematical Foundations of Computational Anatomy
Redaktører Xavier Pennec, Sarang Joshi, Mads Nielsen
Antal sider 12
Publikationsdato 2011
Sider 123-134
Status Udgivet - 2011
3rd MICCAI Workshop on Mathematical Foundations of Computational Anatomy - Westin Harbour Castle, Toronto, Canada
Begivenhed Varighed: 22 sep. 2011 → 22 sep. 2011
Konferencens nummer: 3
Konference 3rd MICCAI Workshop on Mathematical Foundations of Computational Anatomy
Nummer 3
Lokation Westin Harbour Castle
Land/Område Canada
By Toronto
Periode 22/09/2011 → 22/09/2011 | {"url":"https://researchprofiles.ku.dk/da/publications/an-airway-tree-shape-model-for-geodesic-airway-branch-labeling","timestamp":"2024-11-02T09:31:48Z","content_type":"text/html","content_length":"47345","record_id":"<urn:uuid:296affab-84dd-4ae6-9671-f467a8799163>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00063.warc.gz"} |
Using Python Splitting Colored Image in to Red, Blue & Green Component
Splitting Colored Image in to Red, Blue & Green Component¶
Re integrating the Red, Blue and Green Component into Original Image¶
Image is nothing but a numeric matrix stored in the memory. Each colored Image is a combination of three componenets i.e. Red, Green & Blue. Suppose we have an image of lets say 200 X 300 resolution.
The corresponding to that image three matrixes of size 200 X 300 will be stored in the memory each representing Red, Green & Blue component of the image. The values in the matrixes are in the range 0
to 255 wherein 0 represents least intensity of color and 255 represents maximum intensity of the color.
Below are the steps for writing the code to read the colored image and display it using matplotlib.
Step 1 : Import the required libraries. Here we will be using three python libraries cv2 (OpenCV), numpy and matplotlib
Step 2 : Read the image in to 3 dimesnional matrix using opencv imread function
Step 3 : Since, OpenCV takes BGR format and matplotlib takes RGB format. So, if we want to view the image read by OpenCV then we need to convert BGR format in to RGB Format
Step 4: View the RGB Format Image by using imshow() function of matplotlib
In [40]:
import cv2
import numpy as np
from matplotlib import pyplot as plt
img = cv2.imread('C:/My/Video/standard_test_images/standard_test_images/peppers_color.tif')
b,g,r = cv2.split(img) # get b,g,r
rgb_img = cv2.merge([r,g,b])
<matplotlib.image.AxesImage at 0x27844644160>
Now we will break down the three dimensional matrix rgb_img in to three different components Red, Green and Blue.Below are the steps to do the same:
Step 1 : Get the size of three dimsnional matrix rgb_img
Step 2 : Declare three new matrixes corresponding to each color with size as of original image or matrix
Step 3 : Iterate for ‘x’ and ‘y’ axis (2 Dimesnions) to populate values in respective matrixes
Step 4 : Populate first Value in Red
Step 5 : Populate second value in Green
Step 6 : Populae third value in Blue
Step 7 : Display Each Matrix or Image using matplotlib imshow() function
In [46]:
x,y,z = np.shape(img)
red = np.zeros((x,y,z),dtype=int)
green = np.zeros((x,y,z),dtype=int)
blue = np.zeros((x,y,z),dtype=int)
for i in range(0,x):
for j in range(0,y):
red[i][j][0] = rgb_img[i][j][0]
green[i][j][1]= rgb_img[i][j][1]
blue[i][j][2] = rgb_img[i][j][2]
<matplotlib.image.AxesImage at 0x2784469a780>
<matplotlib.image.AxesImage at 0x278446f6390>
<matplotlib.image.AxesImage at 0x27844751080>
Now we Will try to re create the orignal image from the Red, Green A=and Blue Component of the image.
Steps for the same are as follows:
Step 1 : Declare a three dimensional matrix of type integer of the size of original image
Step 2 : Again iterate through ‘x’and ‘y’ axis (2 Dimesnions)
Step 3 : In the first index populate the value in Red Matrix
Step 4 : In the second index populate the value in Green Matrix
Step 5 : In the third index populate the value in Blue Matrix
Step 6 : Use the imwrite() function of OpenCV to save the image back to disk
Step 7 : Use the imshow() funciton of matplotlib to view the re created image
In [47]:
#Now we will again create the original image from these Red, Blue and Green Images
retrack_original = np.zeros((x,y,z),dtype=int)
for i in range(0,x):
for j in range(0,y):
retrack_original[i][j][0] = red[i][j][0]
retrack_original[i][j][1] = green[i][j][1]
retrack_original[i][j][2] = blue[i][j][2]
<matplotlib.image.AxesImage at 0x27844804d68>
Leave a Comment | {"url":"https://www.instrovate.com/using-python-splitting-colored-image-in-to-red-blue-green-component/","timestamp":"2024-11-12T22:08:39Z","content_type":"text/html","content_length":"720483","record_id":"<urn:uuid:9a7b1d6a-1893-4d10-901c-cb554a707a3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00149.warc.gz"} |
NIPS 2017
Mon Dec 4th through Sat the 9th, 2017 at Long Beach Convention Center
Reviewer 1
Summary: This paper proposes a new optimization algorithm, Online Lazy Newton (OLN), based on Online Newton Step (ONS) algorithm. Unlike ONS which tries to utilize curvature information within convex
functions, OLN aims at optimizing general convex functions with no curvature. Additionally, by making use of low rank structure of the conditioning matrix, the authors showed that OLN yields better
regret bound under certain conditions. Overall, the problem is well-motivated and the paper is easy to follow. Major Comments: 1. The major difference between OLN and ONS is that ONS introduces a
lazy evaluation step, which accumulates the negative gradients at each round. The authors claimed in Lines 155-157 that this helps in decoupling between past and future conditioning and projections
and it is better in the case when transformation matrix is changing between rounds. It would be better to provide some explanations. 2. Lines 158-161, it is claimed that ONS is not invariant to
affine transformation due to its initialization. In my understanding, the regularization term is added partly because it allows for an invertible matrix and can be omitted if Moore-Penrose
pseudo-inverse is used as in the FTAL. 3. Line 80 and Line 180, it is claimed that O(\sqrt(r T logT)) is improved upon O(r \sqrt(T)). The statement will hold under the condition that r/logT = O(1),
is this always true? Minor: 1. Lines 77-79, actually both ONS and OLN utilizes first-order information to approximate second-order statistics. 2. Line 95, there is no definition of the derivative on
the right-hand-side of the equation prior to the equation. 3. Line 148-149, on the improved regret bound. Assuming that there is a low rank structure for ONS (similar to what is assumed in OLS),
would the regret bound for OLS still be better than ONS? 4. Line 166, 'In The...' -> 'In the...' 5. Line 181, better to add a reference for Hedge algorithm. 6. Line 199, what is 'h' in '... if h is
convex ...'? 7. Line 211-212, '...all eigenvalues are equal to 1, except for r of them...', why it is the case? For D = I + B defined in Line 209, the rank for B is at most r, then at least r of the
eigenvalues of D are equal to 1. The regret bound depends on the low rank structure of matrix A as in Theorem 3, another direction that would be interesting to explore is to consider the low rank
approximation to the matrix A and check if similar regret bound can be derived, or under which conditions similar regret bound can be derived. I believe the proposed methods will be applicable to
more general cases along this direction.
Reviewer 2
This paper considers online learning problem with linear and low-rank loss space, which is an interesting and relative new topic. The main contribution lies in proposing an online newton method with
a better regret than [Hazan et. al. 2016], namely O(\sqrt(rT logT) vs. O(r\sqrt(T)). And the analysis is simple and easy to follow. There are a few concerns listed below. 1. The better regret bound
is arguable. Assuming low rank, ‘r’ is typical small while ‘T’ could be very large in online setting. Thus, comparing with existing work, which establishes regrets of O(r\sqrt(T)) and O(\sqrt(rT)
+logNlogr)), the achieved result is not very exciting. 2. Algorithm 1 requires a matrix inverse and solving a non-linear programming per iteration, and a sub-routine appears inevitable for most
problems (e.g. the low rank expert example with domain `simplex’). Such complexity prevents the algorithm from real online applications and limits it in analysis. 3. The reviewer appreciates strong
theoretical work without experiments. However, the presented analysis of this paper is not convincing enough under NIPS criterion. A empirical comparison with AdaGrad and [Hazan et. al. 2016] would
be a nice plus. 5. `A_t^{-1}’ in Algorithm 1 is not invertible in general. Is it a Moore–Penrose pseudoinverse? And does a pseudoinverse lead to a failure in analysis? 6. Though the authors claim
Algorithm 1 is similar to ONS, I feel it is closer to `follow the approximate leader (ver2)’ in [Hazan et.al. 2006]. Further discussion is desirable. Overall, this work makes a theoretical step in
special online setting and may be interesting to audiences working in this narrow direction. But it is not very exciting to general optimization community and appears too expensive in practice.
************* I read the rebuttal and removed the comment on comparison with recent work on arxiv. Nevertheless, I'm still feel that O(\sqrt(rT logT) is not a strong improvement over O(r\sqrt(T)),
given that T is much larger than r. An ideal answer to the open problem should be O(\sqrt(rT) or a lower bound showing that O(\sqrt(rT logT) is inevitable.
Reviewer 3
The paper analyzes a particular variant of online Newton algorithm for online linear(!) optimization. Using second-order algorithm might sound non-sensical. However, the point of the paper is to
compete with the best preconditioning of the data. The main reason for doing this is to solve the low rank expert problem. (In the low-rank expert problem, the whole point is to find a basis of the
loss matrix.) The main result of the paper is an algorithm for low rank expert problem that has regret within sqrt(log T) of the lower bound for the low rank experts problem. In particular, it
improves sqrt{r} factor on the previous algorithm. Large parts of the analysis are the same as in Vovk-Azoury-Warmuth forecaster for online least squares (see e.g. the book Gabor Lugosi & Nicolo
Cesa-Bianchi), the second-order Perceptron ("A second-order perceptron " by Nicolo Cesa-Bianchi, Alex Conconi, and Claudio Gentile), or the analysis of Hazan & Kale for exp-concave functions. I
suggest that authors reference these papers. The second-order perceptron paper is particularly relevant since it's a classification problem, so there is no obvious second-order information to use,
same as in the present paper. Also, the paper "Efficient Second Order Online Learning by Sketching" by Haipeng Luo, Alekh Agarwal, Nicolo Cesa-Bianchi, John Langford is worth mentioning, since it
also analyzes online Newton's method that is affine invariant i.e. the estimate of the Hessian starts with zero (see https://arxiv.org/pdf/1602.02202.pdf Appendix D). The paper is nicely written.
I've spot-checked the proofs. They look correct. At some places (Page 7, proof of Theorem 1), inverse of non-invertible matrix A_t is used. It's not a big mistake, since a pseudo-inverse can be used
instead. However, this issue needs to be fixed before publication. | {"url":"https://papers.nips.cc/paper_files/paper/2017/file/347665597cbfaef834886adbb848011f-Reviews.html","timestamp":"2024-11-12T23:15:39Z","content_type":"text/html","content_length":"8190","record_id":"<urn:uuid:b2d49fb3-8118-4484-b8f6-9f0ca033359a>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00540.warc.gz"} |
[Solved] Since its inception in 1967, the Super Bo | SolutionInn
Since its inception in 1967, the Super Bowl is one of the most watched events on television
Since its inception in 1967, the Super Bowl is one of the most watched events on television in the United States every year. The number of U.S. households that tuned in for each Super Bowl, reported
by Nielson.com, is provided in the data set SuperBowlRatings.
a. Construct a time series plot for the data. What type of pattern exists in the data? Discuss some of the patterns that may have resulted in the pattern exhibited in the time series plot of the
b. Given the pattern of the time series plot developed in part (a), do you think the forecasting methods discussed in this chapter are appropriate to develop forecasts for this time series? Explain.
c. Use simple linear regression analysis to find the parameters for the line that minimizes MSE for this time series.
Fantastic news! We've Found the answer you've been seeking! | {"url":"https://www.solutioninn.com/since-its-inception-in-1967-the-super-bowl-is-one","timestamp":"2024-11-02T14:59:08Z","content_type":"text/html","content_length":"81199","record_id":"<urn:uuid:aa42b11a-0088-4b26-9b73-a04909d4cfc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00877.warc.gz"} |
Implementation of Sorted Arrays - Learn Coding Online - CodingPanel.comImplementation of Sorted Arrays - Learn Coding Online - CodingPanel.com
As the name suggests, a sorted array has all the elements sorted in a specific order, such as numerical or alphabetical order. One of the main advantages of the sorted array is that it makes the
search operation very efficient. We can perform the lookup in O(logn) time by applying the binary search algorithm. However, the time complexity of insertion and deletion is O(n) as we have to shift
How to Implement a Sorted Array?
def binarySearch(array, x, start, end):
if (end < start):
return -1 # x is not found
mid = (start+end)//2 #find the mid
if (x==array[mid]): #if x is equal to middle element
return mid
elif (x> array[mid]): # x lies in the upper half
return binarySearch(array, x, mid+1, end)
else: # x lies in the lower half
return binarySearch(array, x, start, mid-1)
def insert(array, x):
n = len(array)
if x>=array[n-1]: #add element at the end
return array
for i in range(0, n):
if x < array[i]:
sliced_array = array[i:] #copy the elements greater than x
array[i] = x #add x
array[i+1:]=sliced_array #put the remaining elements
return array
def delete(array, x):
n = len(array)
index = binarySearch(array, x, 0, n-1)
if index == -1: #element not found
print("Element not found")
return -1
if index == n-1: #deleting the last element
return array[0: n-1]
elif index ==0: #deleting the first element
return array[1:n]
left = array[0:index] #items before the index
right = array[index+1:] #items after the index
left.extend(right) #combining
return left
Operations on Sorted Arrays
As already mentioned, we perform the lookup operation using the binary search algorithm. The method checks if the given element is equal to the middle value, and in that case, its index gets
returned. If it is greater than the middle value, it means that the element can lie in the upper half, so only that part gets considered. If it is less than the middle value, it can exist in the
lower half.
scores = [50, 75.5, 85, 90, 91.5]
n = len(scores)
index = binarySearch(scores, 90, 0, n-1)
if index!=1:
print(f"Element found at index {index}")
Element found at index 3
The insert() method checks whether an element is greater than the last value or not. If it is, it will get appended to the list, and no shifting is required. Otherwise, it iterates through the array
until it finds the greater than x. The x will get inserted at that index, and the rest of the elements will get shifted. Therefore, we copy the rest of the items and break from the loop. Finally, we
add x at its position and put the remaining elements after that.
scores = [50, 75.5, 85, 90, 91.5]
n = len(scores)
scores = insert(scores, 100)
scores = insert(scores, 49.6)
scores = insert(scores, 70)
[49.6, 50, 70, 75.5, 85, 90, 91.5, 100]
The delete() method finds the index of the input element if it exists in the array, then removes it by slicing the array depending upon the item’s position.
scores = [50, 75.5, 85, 90, 91.5]
n = len(scores)
scores = delete(scores, 85)
[50, 75.5, 90, 91.5]
Concluding, sorted arrays are useful in those cases where the frequency of the lookup operation is more than insertion and deletion operations. | {"url":"https://www.codingpanel.com/lesson/implementation-of-sorted-arrays/","timestamp":"2024-11-11T10:18:44Z","content_type":"text/html","content_length":"67020","record_id":"<urn:uuid:0423f0ac-5804-48cd-aac4-a1d3b5425ca0>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00298.warc.gz"} |
Healing with Frequencies : explanations
Author's page :
http://altered-states.net/barry/newsletter420/Healing with Frequencies
To start, let us say, that everything is vibration. From the chair that you may be sitting in to the paper or the mouse you are holding, everything is in a state of vibration. This is not a new idea.
Your ancient mystics have known this for many a millennium, but now your scientists are beginning to understand this and agree. It is a wonderful start.
From the electrons spinning around the nucleus of an atom, to the planets spinning around suns in the galaxy, everything is in movement. Everything is in vibration a frequency.
"You are a digital, bioholographic, precipitation, crystallization, miraculous manifestation, of Divine frequency vibrations, forming harmonically in hydro-space."
Dr. Leonard Horowitz, author, investigator and speaker
What is illness? “
Emotional issues that are unresolved block the healing vibrations or cause the disease state to return
.” R Gordon
Every object has a natural vibratory rate. This is called resonance. One of the basis principles of using frequency as a transformative and healing modality is to understand the idea that part of the
body is in a state of vibration : every organ, every bone, every tissue, every system... all are in a state of vibration. Now, when we are in a state of health, the body puts out an overall harmonic
of health. However, when a frequency that is counter to our health sets itself up in some portion of the body, it creates a disharmony that we call
I learned to listen to my body with an inner concentration like meditation, to get guidance as to when to exercise and when to rest. I learned that healing and cure are active processes in which I
myself needed to participate.
- Rollo May
What is the mechanism for healing ? Resonance !
“When two systems are oscillating at different frequencies, there is an impelling force called resonance that causes the two to transfer energy from one to another. When two similarly tuned systems
vibrate at different frequencies, there is another aspect of this energy transfer called entrainment, which causes them to line up and to vibrate at the same frequency.” (Richard Gordon)
As Bob Beck has said, the only thing keeping the devices and procedures mentioned in these pages from working is if a person doesn't use them. Too many times I have listened to people complain
repeatedly about their health problems, declaring they are willing to do almost anything to get relief. Yet these same people somehow never quite manage to actually do anything.
In 1929
George Lakhovsky
, a Russian engineer, published a book called, 'The Secret Life' and
"waves that heal"
which gave birth to an innovative new concept in healing, Radiobiology. In another book titled, 'The Cancer Conspiracy' by Barry Lynes, reviewer Theresa Welsh of The Seeker Books website stated,
Lakhovsky maintained all living cells, from people to parasites, produce and radiate oscillations at high frequencies, and they respond to oscillations of different frequencies from outside sources.
The world today is bombarded with electro-magnetic impulses from cell phones to microwaves and researchers fear this may be the cause of increased cancer risks. But what happens when outside
oscillations concur with the frequency of internal cell oscillations? According to Lakhovsky, and even some modern scholars, the living being grows stronger.
The Law Of Vibration
Just as a pebble creates vibrations that appear as ripples, which travel outward in a body of water, your thoughts create vibrations that travel outward into the Universe, and attract similar
vibrations that manifest as circumstances in your life.
Consider oxygen. It is something that we use everyday, and each of us realize how crucial it is to our survival, yet we aren't able to experience it with touch, taste, smell, hearing or feelings. The
fact that we can't experience it with these senses certainly doesn't mean it doesn't exist. We know it does. The reason that we are unable to sense it with our physical sensory perception is because
it's rate of vibration is outside of our physical ability to do so.
It's interesting that the latest quantum physics theory, born only a decade or so ago, arrives at a similar conclusion. It is called String Theory and it basically suggests that the physical universe
is built out of sound vibrations, kind of like everything is the result of some huge cosmic guitar being played somewhere. It's a mind-blowing concept that is held by some of the sharpest minds in
the physics community, including Steven Hawking.Video not working ?
Watch it
>> HERE << Michio Kaku Explains String TheorySolfeggio Harmonics - 528 HZ - Miracle Frequency
The 528 hz frequency is known as, the "528 Miracle," because it has the remarkable capacity to heal and repair DNA within the body and is the exact frequency that has been used by genetic
"528 cycles per second is literally the core creative frequency of nature. It is love," proclaims renowned medical researcher Dr. Leonard G. Horowitz.
Video not working ?
Watch it
>> HERE << Sacred Harmony Resonator
Low frequencies, and frequencies that are out of balance, cause illness. By using frequency healing tools, you can help correct those imbalances, even before they create disease. Frequency healing
tools are complimentary to each other and modern medicine and have no negative side effects.
Watch Parasites Die from FrequenciesAny positive emotion causes a cell to vibrate at a higher frequency and negative vibrations cause the cells to vibrate at lower frequency. The negative emotion is
nothing but an incompletely experienced emotion. These emotions when stored in the cells of the body are the diseases
In essence, everything in the world is made up of energy. We are all constantly vibrating masses of microscopic particles that are always in motion. Every object, person and organ has a healthy
vibration rate called resonance. If that vibration is out of resonance, disease results. These imbalances can be treated with frquencies...
By: Barbara Hero(1996)
Personality C+ 264
Circulation,Sex C# 586
Adrenals,Thyroid & Parathyroid B 492.8 *
Kidneys Eb 319.88 *
Liver Eb 317.83 *
Bladder F# 352 *
Small Intestine C# 281.6 *
Lungs A 220 *
Colon F# 176 *
Gall Bladder E 164.3 *
Pancreas C# 117.3
Stomach A 110 *
Spleen B 492
Blood Eb 321.9
Fat Cells C# 295.8
Muscles E 324
Bone Ab 418.3
Dr. Rife made incredible progress in this field that has unfortunately not been picked up on and continued with by our modern medical society. His research eventually documented 52 specific
frequencies which could be used to treat many common health maladies, including tuberculosis and cancer. His laboratory work showed that he could safely destroy these bad cells and microbes by simply
increasing the intensity of the frequency until they disintegrated from the pressure. He documented successful results in both the laboratory environment and in humans. The human body's cell
structure and good bacteria were unaffected by these treatments. That's because those cells resonate at entirely different frequencies and are naturally insulated from potentially harmful radio
Q&A on our rife units
Some Example here :
(use Parasite general, roundworm, and ascaris if these don't work
long term) - 414, 464, 877, 866, 886, 254.2, 381, 661, 762, 742, 1151, 450
20, 35, 465, 6.8, 440, 484, 660, 727, 787, 800, 803, 880, 1850, 2008, 2127, 2000, 2003, 2013, 2050, 2080 for 3 min, 5000 for 15 min.
Fungus and mold, general
- 728, 880, 784, 464, 886, 866, 414, 254, 344,
2411, 321, 555, 942, 337, 766, 1823, 524, 374, 743, 132, 866
(aches and respiratory) - 440, 512, 683, 728, 784, 787, 800, 875,
880, 885, 2050, 2720, 5000 for 5 min, 7760, 7766 for 10 min, 304 for 3 min
and more frequencies
Rife Tools hereNew Way to Kill Viruses: Shake Them to Death
Quantum physics proved that all matter, both physical and chemical, is comprised of sub atomic particles with positive and negative electrical charge. Therefore, we are electrical beings and so is
our universe and everything in it.
Through this discovery, it was determined that every form of chemical or physical matter has a specific, measurable frequency. This includes everything that makes up who we are: organs, blood, the
neuropeptides and neurotransmitters that we experience as emotions or thoughts, amino acids that construct or DNA, hormones that control and regulate or bodies, minerals, vitamins, and fatty acids
that feed our metabolism, etc. Electrical energy is our life force.
There appears to be a correlation between a specific frequency and the atomic weight of the elements. For instance, if the note of "C" is low in a person's voice, chances are the element of the zinc
is also low in the body. The frequency of the note of "C" at the second octave is 65.40 cycles per second (hertz), and the atomic weight of the element of zinc is 65.37. So by listening to the
frequency of the zinc the cells of the body will receive the vibration; and when the person eats foods that contain zinc, the body will resonate with this vibration and absorb the zinc. Not only will
the body become more balanced, but the voice will improve; for it will produce all the notes in a more harmonious way.
" Measurement of standing waveforms from electrical storms confirmed what he had suspected, that the earth had a resonant frequency and could therefore be used as a wave carrier to transmit signals.
He established that lightning storms as they swooped down the Rockies and then rumbled across the plains into Kansas were resonating at a frequency of 7.68-7.82 cycles per second, or “Hertz” (Hz)
This natural phenomenon was rediscovered in the 1960s by researcher W.O. Schumann while working for the Navy on ways to broadcast nuclear war orders to submerged submarines "
360Hz = The Balance Frequency (add the numbers! What do you get?) is derived from the Golden Section and is a harmonic that naturally brings sensations of joy and healing. Vibrational Medicine
science assert that the Golden section tones as well as Fibonacci sequence music brings balance to health. Even more amazing, NASA astronauts have long proven that the earth creates a tone in space
of 360Hz !!
The ancient Chinese knowledge of 172 Hz as the fundament harmonic frequency of nature.
172.06 - Resonates with the Platonic year {about 26,000 years} (Note=F) The great tone of nature in China known as the Kung is the musical note F, while in Tibet, the notes A, F, and G are the sounds
of power. The Emporer of China kept the peace by travelling once a year with his entourage to each province to tune the notes of the scale. This procedure maintained peace for thousands of years.
(Color=purple-violet) (Effects=joyful, cheerful, spiritual effect) [PSI]; The Frequency Of The Platonic Year (Color=red-violet {purple}) (Tempo=80.6 BPM) (Chakra=Sahasrar/Crown chakra) (Effects=
cheerfulness, clarity of spirit, cosmic unity on highest levels) (Medicinal=antidepressive) (Other=F is considered the tone of the spirit, and had a lot of significance to the Chinese) * Other
sources [PM] disagree about the tone F being associated with the Crown chakra, which is how HC/Planetware connects this frequency to the crown chakra. [PM] considers the crown chakra to be associated
with the B note, and not F.
For a photograph of a 65 bell ensemble of ancient Chinese bells using a norm tone (2 millenia before such a concept was instituted in Europe) of 345 Hz (344 Hz would be a harmonic of 172 Hz).441Hz =
The King's Chamber Frequency. Like the Balance Frequency, the King's Chamber acts towards preservation and equilibrium. Play a 441Hz tone in a chaotic room and people will find themselves mellowing
down. Ed Skilling designed a unit to output 728 Hz which is the Rife frequency considered to be the most healing. This frequency is carried on a radio frequency wave to transport it to the body. This
works in the same way a radio transmitter carries the signal for a particular radio station so it can be received by a radio in any given area. As with Lakhovsky's work, the cells can then pick up
their resonant healthy frequency. The immune system can then gradually strengthen.Reported Diseases Affected by 727 -728 hzWhen an artist expresses visual information which is originated in a certain
dimension, whatever passes through to the viewer is not only a picture, but its energetic essence - energetic frequency as well.Electricity for health in the 21st century
The human body is a symphony of sounds. Every chakra, every organ, every bone, every tissue, every cell has its own resonant frequency, its own sound. Together, they create a unified or composite
frequency, with its own sound, like the instruments of an orchestra coming together. Ideally, the individual sounds and frequencies comprise a harmonious whole. That is when the body is functioning
as it should, in health. However, when an organ is out of time or out of tune with the rest, then the entire body is affected. This disharmony leads to states of disease and disintegration.
What secret is there in music which attracts all those who listen to it? It is the rhythm which is being created. It is the tone of that music which tunes a soul and raises it above depression and
despair of everyday life in this world. And if one knew what rhythm was needed for a particular individual in his trouble and despair, what tone was needed, and to what pitch that person's soul
should be raised, one would then be able to heal him with music. (healing music
University of California at Los Angeles nanotechnologist Jim Gimzewski is pioneering a new science he calls sonocytology, the study of cell sounds. His first experiments began with yeast cells, using
a nanotechnology tool called an atomic force microscope to detect sound-generating vibrations and then using a computer to enhance the volume. The yeast cells were heard to produce harmonics, around
1,000 cps. In musical terms, they were "singing" in the range of C-sharp to D above middle C. Killing the yeast cells with alcohol, the pitch rose dramatically as if the cells were screaming.
Cellular harmonics were also affected by temperature, speeding them up or slowing them down, genetic mutations were found to make a slightly different sound than normal cells. Dead cells emitted a
low rumbling like radio static. Distinguishing between the sound signatures of healthy and diseased cells may be a part of the medicine of the future.
Dissonance and Rhythm
Living in a city, unfortunately, means living with noise. The etymology of "noise" derives from the Latin "nausea." We are bombarded by these upsetting, stress-inducing sounds -- road traffic,
subways, airplanes, emergency vehicle sirens, garbage trucks, car alarms, construction equipment, cell phones, workplace machinery, lawn mowers, leaf blowers, hair dryers, boom boxes, the din of
chatter in crowded restaurants and coffee shops, and on and on. Noise pollution is among the most pervasive pollutants to which we are exposed.
Toxic noise is literally poisoning to our health and well-being. When hair cells in the ear, the sensory organs that allow us to hear, are injured by noise, they cannot be regenerated. The result is
hearing damage and, in some cases, permanent hearing loss. Noise-induced hearing loss can be caused by a one-time exposure to loud sound, such as an explosion, or by repeated exposure to sounds at
various loudness levels over an extended period of time. Problems related to noise include hearing loss, stress, high blood pressure, peptic ulcers, degradation of the immune system, sleep loss and
fatigue, distraction and poor work performance, impairment of learning, increased aggression, depression, withdrawal, and a general reduction in the quality of life and opportunities for tranquility.
Dissonant sounds create disharmony -- rifts between the individual and her environment, as well as within the body's own frequencies. If 10 tuning forks tuned to the same frequency are lined up
together and one is struck, they will all begin to reverberate together. This is resonance. However, if you strike a tuning fork of a different frequency and place it near the others, they will all
stop. This is dissonance. When you're feeling irritable or "not yourself" and you don't quite know why, pay attention to your environment. Quite often you'll find that nearby is some sound --
machinery, music, voices -- that is creating discord in your own frequency. If the offending sound is not something that can be eliminated, try to create a stronger vibration that has a positive
resonance. One on-the-fly solution is humming or the
Schumman Resonator
. It doesn't need to be loud, but just enough to feel its vibrations in your own body. You will find the resonant frequencies that will make you feel better, and the dissonant sound you can't escape
from will cease to bother you.
Source"Many years ago the Author was enthralled by the sight of a certain genus of flowering plant in a remote mountainous area, being pollinated by bees called by the plants by their emission of a
distinct humming sound. After recording the television documentary, I checked the frequency of the fundamental frequency generated by the plants and found it to be 432 Hertz, or cycles per second.
This prompted me to place small battery powered sound generators in the flower beds on my farm where I kept bee hives, and to discovering a whole new world of plant and bee intelligence."
www.hinduism.co.za/anahata.htmScientists Discover Healing Frequency In Animal SoundsWatch Parasites Die from Frequencies
Any number of things can easily jostle our Frequencies and cause these frequencies to become out of tune. Whether it be a traumatic experience, a drop in temperature, or even a stressful incident at
work. The balance of our system is so fragile. Just the same as how a guitar needs to be tuned from time to time our system is the same way.
Each body has it’s own unique frequency, when the interaction of these frequencies are balanced, you feel peaceful and at perfect harmony with yourself as well as towards other people. When they are
off balance, it can have significant negative effects. An unbalanced body, is unable to fulfill it’s energy contribution to the system. This can have negative psychological and physical consequences
on an individual. Anger, depression, constipation,lack of concentration, and sexual dysfunctions are just a few examples of symptoms due to unbalanced internal frequencies.
In an article in Radio News Magazine in February 1925, Lakhovsky wrote: "In conclusion I wish to call attention of the reader to the fact that I have obtained very conclusive results not only with a
wavelength of two meters, but with longer and shorter wavelengths. The main thing is to produce the greatest number of harmonics possible." [Also see Electricity for Health in the 21st Century.]
Healing Frequencies
When examining Healing Frequencies we can say that every biotic organism, or a biotic object, resonate in a particular frequency. The major aspect of of frequency is the pace of repetition. What
repeats itself is the physical resonance of an object. That resonance or vibration that can be measured on a molecular level. Molecules that constitute every physical object are constantly moving in
a certain frequency, with relatively permanent repetitive nature.
A great deal of the frequency measurement is being detected and measured on the electronic level, which is exponentially smaller than the molecular. It is much more accurate to measure frequency (as
part of the Healing Frequencies) on an electro magnetic and light spectrum level.
The electromagnetic resonance of any object fluctuate between high and low values at different frequency levels. This fluctuation nature can be absorbed and measured by certain devices and by the
senses as sound, light and vibratory response of the touch sense.
The Planet: The cure for cancer was covered up ?The newspaper article provided here was included in a newspaper called The Planet and published February 1986 in the Washington, D.C.. It was delivered
to every member of the U.S. House of Representatives and every member of the United States Senate. Not one representative, senator or staff assistant was motiviated sufficiently to investigate
further. The newspaper was also provided free to the George Washington University Medical School students and professors. Again, not one was motivated to investigate further.More .
The healing nature of some of the frequency spectrum lies in the ability of a living organism to absorb very precise and particular set or range frequencies that can physically create a healing
effect (and thus compose a set of frequencies) on certain organs and organic systems. In many cases the healing effect of frequencies is achieved by the ability of the frequency to create a very
accurate affect on specific bacteria & viruses by sending highly matched Healing Frequencies capable of neutralizing their chemical structure.
"Put a cat and a bunch of broken bones in the same room" some veterinary schools joke, "and the bones will heal." Only two years ago scientists discovered that vibrations between 20-140 Hz (at low
db) are anabolic for bone growth and will also help to heal fractures, mend torn muscles and ligaments, reduce swelling, and relieve pain. Fauna have found that a cat's purr not only matches this
vibration, but its dominant frequencies are 25 and 50 Hz - the optimum frequencies for bone growth and fracture healing. All cats, including larger ones such as pumas, ocelots and lions, have further
sets of strong harmonics at the exact hertz (number of cycles per second) that generate muscle strength, increase joint mobility and provide therapeutic pain relief.
Richard Gerber, M.D. states in his book "Vibrational Medicine" :
When viral and chemical environmental stressors are introduced into the human biological system, the place where they will cause the most damage will be partially determined by the weakest link in
the physiologic/subtle energy chain.
From an energetic standpoint, the human body, when weakened or shifted from equilibrium, oscillates at a different and less harmonious frequency than when healthy. This abnormal frequency reflects a
general state of cellular energetic imbalance within the physical body. When a weakened individual is unable to shift their energetic mode to the needed frequency a certain amount of subtle energetic
help may be needed. When supplied with a dose of the needed energetic frequency, it allows the cellular bioenergetic systems to resonate in the proper vibrational mode, thereby throwing off the
toxicities of the illness.
Create a Symphony
We are the instruments, We are the orchestra, We are the music.
Each cell takes part in the symphony of our body. Our role as a conductor is to orchestrate harmony. When a musician (organ or system), produces a sour note, we bring them back into harmony by
helping them to retune their instrument, or refocus their attention. We don't cover up their disharmony or remove them from the orchestra. Each musician (or part of the body), is important in its
Divine Expression for the creation of the symphony.
Different frequencies, tones, and sounds -- through drumming, chanting, toning, or the use of Rife or Frequencie tools like the MWO-- can induce different states to promote healing for the body,
mind, emotions, and spirit. On a molecular level, our bodies are systems of vibrating atomic particles. We are living receivers and transmitters of vibration. We can use frequencies to vibrate matter
and promote healing and regeneration of the different body systems. These frequencies also shift etheric patterning to heal the emotional and mental causes of disease.
Kondaa (Barry) Kapke, ACST,
"Royal Raymond Rife, a researcher in San Diego in the early part of the 20th century, sucessfully eliminated Dis-ease using an electronic device he invented that emitted specific frequencies..."
Rife Tools here
Dr. Hulda Clark, Ph.D., N.D., and author of The Cure For All Cancers, "studied the work of Rife and learned that every living creature has a vibration..."
See Dr. Hulda Clark Zappers hereRife machines and Multiwave oscillators are claimed to complement each other based on the principle that life forms absorb energy. A multiwave Oscillator uses this
principle to strengthen cells within the body to resist disease while a Rife machine uses this principle to destroy microorganisms with an overdose of frequency energy.
May/June 1997 Leading Edge Newspaper
" ...Every cell has its own frequency. When you offer, through technology, a harmonic opportunity to the cell it can choose that frequency and become established to its ideal resonance and become
recharged to its normal energy state...When we apply this technology we are affecting the intelligence of each individual cell. Every cell is a hologram for the entire body. This area of cellular
resonance is the fundamental aspect of vibrational technology and has been vastly overlooked."-By Keich Frick
See the MWOLIVE HAPPY ! | {"url":"https://www.neeeeext.com/index.php?PHPSESSID=b0a3effbd35e129920c6e51d2c1cd00f&topic=508.msg918","timestamp":"2024-11-13T21:07:04Z","content_type":"application/xhtml+xml","content_length":"73462","record_id":"<urn:uuid:c19e6136-acbf-4c16-a8d5-a9cf5eb25c97>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00581.warc.gz"} |
Serialize and Deserialize
Problem Statement:
Serialization is converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed
later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search
tree can be serialized to a string, and this string can be deserialized to the original tree structure.
The encoded string should be as compact as possible.
Example 1:
Input: root = [2,1,3]
Output: [2,1,3]
Example 2:
Input: root = []
Output: []
I would highly encourage you to pay special attention to the implementation of deserialization as here we have implemented a very simple yet powerful approach using the most fundamental
characteristic of a
Binary Search Tree
for every node N1 all the nodes in the left subtree have values less than or equal to the value of node N1, and all the nodes in the right subtree have values greater than the value of node N1. It is
for this characteristic that the INORDER Traversal of a Binary Search Tree happens to be always SORTED.
This implementations could be reused in solving various other
Binary Search Tree
problems of all difficulty levels.
We already talked about Inorder Traversal for
Binary Search Tree
above. Now let's concentrate on the other two tree traversals,Preorder Traversal and Postorder Traversal, for a minute. In Preorder Traversal we
1. visit a node.
2. then traverse left subtree,
3. at the end, we traverse right subtree.
/ \
n2 n3
For the above node the Preorder Traversal would give n1 -> n2 -> n3
If you are well versed in Preorder Traversal and recursion, just by giving a little thoughts you would realize that
if you are given the result of Preorder Traversal of a Binary Search Tree
you could reconstruct the tree in the below way by leverage the basic characteristic of a Binary Search Tree:
public TreeNode reconstructBstFomPreorder(ArrayDeque<Integer> preorder) {
if (preorder.isEmpty()) {
return null;
return reconstructBST(preorder, Integer.MIN_VALUE, Integer.MAX_VALUE);
public TreeNode reconstructBST(ArrayDeque<Integer> preorder, Integer lower, Integer upper) {
if (preorder.isEmpty()) {
return null;
int val = preorder.getFirst();
if (val < lower || val > upper) {
return null;
TreeNode root = new TreeNode(val);
root.left = reconstructBST(preorder, lower, val);
root.right = reconstructBST(preorder, val, upper);
return root;
We already know that any Binary Tree could be reconstructed if you are given the result of inorder traversal and either preorder or postorder.
Inorder traversal of BST is an array sorted in the ascending order: inorder = sorted(preorder) OR sorted(postorder).
This means that BST structure is already encoded in the preorder or postorder traversal and we do not need to be given the result of the inorder traversal separately.
Using the discussion above we could say that Binary Search Tree could be constructed from preorder or postorder traversal only, as we just saw in the implementation above.
If we use Postorder instead of Preorder, the above implementation would look like below:
public TreeNode reconstructBstFomPostorder(ArrayDeque<Integer> postorder) {
if (postorder.isEmpty()) {
return null;
return reconstructBST(postorder, Integer.MIN_VALUE, Integer.MAX_VALUE);
public TreeNode reconstructBST(ArrayDeque<Integer> postorder, Integer lower, Integer upper) {
if (postorder.isEmpty()) {
return null;
int val = postorder.getLast(); // we need to process from end for Postorder
if (val < lower || val > upper) {
return null;
postorder.removeLast(); // Remove the last element so that the second last element becomes available for processing
TreeNode root = new TreeNode(val);
root.right = reconstructBST(postorder, val, upper);
root.left = reconstructBST(postorder, lower, val);
return root;
From the above discussion the below two things are clear:
1. Preorder or Postorder traversal result of a Binary Search Tree uniquely encodes a BST. So we could use either of Preorder or Postorder for serialization. The below implementation uses Preorder
but you could use Postorder as well as shown above.
2. We decode the encoded BST (either Preorder or Postorder) to get deserialized BST as shown above. The below code implements the same.
Login to Access Content
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A common activity on a graph is visiting each vertex of it in a given order. We will start by introducing the breadth-first search, and then follow with depth-first search. Both of these techniques
form the archetype for many important graph algorithms, as we will see later with the cycle detection and Dijkstra's algorithm for single-source shortest paths.
Given a graphG = (V, E)and a source vertexs, breadth-first search explores the edges ofGsystematically to discover every vertex that is reachable froms. While doing so, it computes the smallest
number of edges fromsto each reachable vertex, making it suitable to solve the single-source shortest path problem on unweighted graphs, or graphs whose edges all have the same weight.
Breadth-First Search (BFS)is named so because it expands the frontier between discovered and undiscovered vertices uniformly across the breadth of the frontier. In that sense, the algorithm first
explores vertices at distancekfrom... | {"url":"https://subscription.packtpub.com/book/programming/9781789537178/6/ch06lvl1sec30/prime-numbers-in-algorithms","timestamp":"2024-11-08T07:38:49Z","content_type":"text/html","content_length":"168503","record_id":"<urn:uuid:6d917996-021b-44dc-b030-2c5b306be56b>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00715.warc.gz"} |
How do i find a function such as f(a)f(b)f(c) = f(sqrt(a^2+b^2+c^2))f^2(0) ? | Socratic
How do i find a function such as #f(a)f(b)f(c) = f(sqrt(a^2+b^2+c^2))f^2(0)# ?
1 Answer
$f \left(x\right) = p {k}^{{x}^{2}}$ for constants $p \in \mathbb{R}$, $k > 0$
I will assume that by ${f}^{2} \left(0\right)$ you mean ${\left(f \left(0\right)\right)}^{2}$ rather than $f \left(f \left(0\right)\right)$ or ${f}^{\left(2\right)} \left(0\right)$
$\textcolor{w h i t e}{}$
f(x) is an even function
First note that if $f \left(0\right) = 0$ then ${\left(f \left(a\right)\right)}^{3} = 0$ for all $a$, hence $f \left(a\right) = 0$ for all $a$. So one option for $f \left(x\right)$ is the constant
function $f \left(x\right) = 0$.
Otherwise, if we let $b = c = 0$ then we find:
$f \left(a\right) f \left(0\right) f \left(0\right) = f \left(\sqrt{{a}^{2} + {0}^{2} + {0}^{2}}\right) f \left(0\right) f \left(0\right)$
and hence:
$f \left(a\right) = f \left(\sqrt{{a}^{2}}\right) = f \left(\left\mid a \right\mid\right)$
So we can deduce that $f \left(x\right)$ is an even function.
$\textcolor{w h i t e}{}$
Any constant function is a solution
Suppose $f \left(x\right) = k$ for all $x \in \mathbb{R}$
$f \left(a\right) f \left(b\right) f \left(c\right) = {k}^{3} = f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right) f \left(0\right) f \left(0\right)$
$\textcolor{w h i t e}{}$
Are there any non-constant solutions?
Suppose $f \left(0\right) = 1$ and $f \left(1\right) = k$ for some constant $k > 0$
Notice that:
$f \left(\sqrt{2}\right) = f \left(\sqrt{{1}^{2} + {1}^{2} + {0}^{0}}\right) f \left(0\right) f \left(0\right) = f \left(1\right) f \left(1\right) f \left(0\right) = {k}^{2}$
$f \left(\sqrt{3}\right) = f \left(\sqrt{{1}^{2} + {1}^{2} + {1}^{2}}\right) f \left(0\right) f \left(0\right) = f \left(1\right) f \left(1\right) f \left(1\right) = {k}^{3}$
Observing this pattern, we can define
$f \left(x\right) = {k}^{{x}^{2}}$
To find:
$f \left(a\right) f \left(b\right) f \left(c\right) = {k}^{{a}^{2}} {k}^{{b}^{2}} {k}^{{c}^{2}}$
$= {k}^{{a}^{2} + {b}^{2} + {c}^{2}}$
$= {k}^{{\left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right)}^{2}}$
$= f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right)$
$= f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right) f \left(0\right) f \left(0\right)$
Note that if $p \in \mathbb{R}$ is any constant then:
$f \left(x\right) = p {k}^{{x}^{2}}$
will also be a solution.
The case $k = 1$ then covers the previously identified constant solution.
Impact of this question
2103 views around the world | {"url":"https://api-project-1022638073839.appspot.com/questions/how-do-i-find-a-function-such-as-f-a-f-b-f-c-f-sqrt-a-2-b-2-c-2-f-2-0","timestamp":"2024-11-11T03:11:17Z","content_type":"text/html","content_length":"37658","record_id":"<urn:uuid:4336554a-7a46-408f-9ab6-0029902d50ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00089.warc.gz"} |
The heat capacity at constant volume of a sample of a monatomic gas is 35J
Hint: Firstly, we will write the 1st law of thermodynamics $dQ = dU + dW$ . Then we will see that constant volume change in heat becomes equal to the change in internal energy when volume remains
constant. After that we will write the value of molar specific heat at constant volume and in constant pressure. Then we will find out the ratio between value of molar specific heat at constant
volume and in constant pressure. Then we will calculate the molar heat capacity at constant pressure for 1 mol of gas after putting all the values.
Complete step by step solution
We know that the first law of thermodynamics says that when system takes some amount of heat from its surroundings, it is used up in two ways: A part of it increases the internal energy of the system
and the remaining part is conserved into some external work done by the system that is heat absorbed= rise in internal energy external work done.
So from the first law of thermodynamics we can write if a small amount of heat dQ changes the internal energy of the system by dU and an external work dW is done then, $dQ = dU + dW$
At constant volume, $dV = 0$ so $dW = pdV = 0$ .
So, $dQ = dU$
So, we can write that the heat gained or lost at constant volume for a temperature change dT of 1 mol of a gas is $dQ = {C_v}dT$
Here, ${C_v}$ is the molar specific heat at constant volume
${C_v} = \dfrac{{dQ}}{{dT}} = \dfrac{{dU}}{{dT}}$
The total internal energy of 1 mol of an ideal gas is $U = \dfrac{3}{2}RT$
So, ${C_v} = \dfrac{{dU}}{{dT}} = \dfrac{3}{2}R....(i)$
The molar specific heat at constant pressure is ${C_p} = {C_v} + R = \dfrac{3}{2}R + R = \dfrac{5}{2}R.....(ii)$
The ratio between the two specific heats is called heat capacity ratio
$\gamma = \dfrac{{{C_p}}}{{{C_V}}} = \dfrac{{\dfrac{5}{2}R}}{{\dfrac{3}{2}R}} = \dfrac{5}{3}....(iii)$
\Rightarrow {C_P} = {C_V} \times \gamma \\
\Rightarrow {C_P} = \dfrac{5}{3}{C_V} = 58.33\dfrac{J}{{mol - K}} \\
Hence the required solution is 58.33J/mol-K
Additional Information: The average molecular kinetic energy of any substance is equally shared among the degrees of freedom; the average kinetic energy of a single molecule associated with each
degree of freedom is $\dfrac{1}{2}kT$ , here T=absolute temperature and k= Boltzmann constant. The number of degrees of freedom of an ideal gas molecule is 3. So, from the equipartition of energy we
can write the average kinetic energy of a molecule $ = 3 \times \dfrac{1}{2}kT = \dfrac{3}{2}kT$ . The molecule has no potential energy so average total energy $e = \dfrac{3}{2}kT$ and the total
internal energy of 1 mol of an ideal gas is $U = \dfrac{3}{2}RT$
Note: The molecules of monoatomic gas can move in any direction in space so it has three independent motions and hence 3 degrees of freedom. So one may put the wrong value of degrees of freedom for a
monatomic gas. Especially every 3 dimensional motion has 3 degrees of motion.
Alternate method:
The minimum number of independent coordinates necessary to specify the instantaneous position of a moving body is called the degree of freedom of the body.
We know that for monoatomic gas, degree of freedom is 3
From equation (i) we can write ${C_v} = \dfrac{f}{2}R[\because f = 3]$
And ${C_P} = \dfrac{f}{2}R + R = \dfrac{{f + 2}}{2}R$
So, from equation (iii) we can write $\gamma = \dfrac{{{C_p}}}{{{C_V}}} = \dfrac{{f + 2}}{f} = 1 + \dfrac{2}{f}$
Given that ${C_V} = 35\dfrac{J}{k}$
So, we can write, ${C_P} = \gamma {C_V} = \dfrac{{f + 2}}{f}{C_V} = \dfrac{5}{3}{C_V} = 58.33\dfrac{J}{{mol - K}}$ | {"url":"https://www.vedantu.com/question-answer/the-heat-capacity-at-constant-volume-of-a-sample-class-11-physics-cbse-5f8a54be5db3b03fe6de6be1","timestamp":"2024-11-05T20:19:06Z","content_type":"text/html","content_length":"170904","record_id":"<urn:uuid:dbab5408-fc71-4e2d-9430-78b1ede1aca8>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00194.warc.gz"} |
Tom Bertalan
I am a Research Software Engineer in the Chemical and Biomolecular Engineering department at The University of Massachusetts at Lowell. I received my PhD from Princeton University's department of
Chemical and Biological Engineering, with a Graduate Certificate in Computational and Information Science. Before working at UML, I was a postdoc at MIT and JHU.
My research interests are in data mining, dimensionality reduction, and system identification (using neural networks) for high-dimensional dynamical systems, with applications in robotic perception
and planning, and computational neuroscience.
Visual Scene Grammars
Foundation models can analyze and generate both images and (noisy) grammars. Let's use that to do some visual scene understanding.
Build an Ackermann robot with RGBD as its primary sense.
Robotics Simulators
Mostly about using Farm Simulator, GTA V, and other games for training both perception and control systems.
Scan for SyncThing Conflicts
Cross-platform Python app to quickly compare conflict files created by SyncThing
Equal Space
Nature Comm. 2022. Use nonlinear manifold learning to discover automatically both the true dimensionality and the underlying spatial coordinates that define a high-dimensional simulation trajectory.
Local Neural Text-to-Speech App
A one-pyfile PyTorch+tk GUI for local neural text-to-speech synthesis.
Faster RNN warmup via manifold learning
Use diffusion maps to skip the warmup phase of RNN inference; demonstrated witha chemical model system.
Learning ODEs from Patchy Observations
Extract Neural ODEs from data whose channels are observed at different times and frequencies.
Certified Invertibility in NNs via MILP
Explore excessive NN invariance in various contexts, with methods for certifying invertibility pointwise across input space.
ANOVA and PCE for Biological Neural Networks
Use ANOVA to perform integrals for polynomial chaos expansions.
Representation Learning
PNAS 2020. Unsupervised learning methods to transform data into a form that's somehow more useful.
Iterative ANNs
Neural networks build on various numerical iterative algorithms.
CHO Neural ODEs
Fit a neural ordinary differential equation to Chinese hamster ovary metabolism data, with a grey-box structure including internal constraints.
Learning stochastic DEs from data
Suggest alterative methods for learning stochastic differential equations from data as neural networks.
Hamiltonian Neural Networks
Learn dynamics with constrained quantities.
Build a differential-drive robot with LIDAR as its primary sense.
Meta-learning of ODE integrators
Rather than learning the RHS of an ODE, learn the parameters of the integrator itself.
Learning for Multiphase Flow
After some dimension reduction by PCA and autoencoder, learn an ODE for the slow dynamics of the Navier-Stokes equations in a multiphase flow setup.
Project Opener Menu
A little TK menu for quickly getting to my project directories.
GPT3 for Seminar Announcements
Use OpenAI's API to generate ics files from email text.
Next Task Decider
Process task list and decide what I should do next.
Cat Wrangler
A feline surveillance bot using the guts of an iRobot Braava.
Boston AV Group Robocar
Teach a one-week workshop to high school students on building and programming a small autonomous car.
Hierarchy Formation
Simulate the formation of dominance hierarchies through social combat.
Circadian Rhythms
Simulate circadian rhythms in the suprachiasmatic nucleus of the hypothalamus. | {"url":"https://www.tomsb.net","timestamp":"2024-11-02T02:29:57Z","content_type":"text/html","content_length":"20318","record_id":"<urn:uuid:84ba5358-eafe-44da-acef-6b7c6100a82a>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00369.warc.gz"} |
Command-line reference
Command-line reference¶
GROMACS includes many tools for preparing, running and analyzing molecular dynamics simulations. These are all structured as part of a single gmx wrapper binary, and invoked with commands like gmx
grompp. mdrun is the only other binary that can be built; in the normal build it can be run with gmx mdrun. Documentation for these can be found at the respective sections below, as well as on man
pages (e.g., gmx-grompp(1)) and with gmx help command or gmx command -h.
If you’ve installed an MPI version of GROMACS, by default the gmx binary is called gmx_mpi and you should adapt accordingly.
Command-line interface and conventions¶
All GROMACS commands require an option before any arguments (i.e., all command-line arguments need to be preceded by an argument starting with a dash, and values not starting with a dash are
arguments to the preceding option). Most options, except for boolean flags, expect an argument (or multiple in some cases) after the option name. The argument must be a separate command-line
argument, i.e., separated by space, as in -f traj.xtc. If more than one argument needs to be given to an option, they should be similarly separated from each other. Some options also have default
arguments, i.e., just specifying the option without any argument uses the default argument. If an option is not specified at all, a default value is used; in the case of optional files, the default
might be not to use that file (see below).
All GROMACS command options start with a single dash, whether they are single- or multiple-letter options. However, two dashes are also recognized (starting from 5.1).
In addition to command-specific options, some options are handled by the gmx wrapper, and can be specified for any command. See wrapper binary help for the list of such options. These options are
recognized both before the command name (e.g., gmx -quiet grompp) as well as after the command name (e.g., gmx grompp -quiet). There is also a -hidden option that can be specified in combination with
-h to show help for advanced/developer-targeted options.
Most analysis commands can process a trajectory with fewer atoms than the run input or structure file, but only if the trajectory consists of the first n atoms of the run input or structure file.
Handling specific types of command-line options¶
boolean options
Boolean flags can be specified like -pbc and negated like -nopbc. It is also possible to use an explicit value like -pbc no and -pbc yes.
file name options
Options that accept files names have features that support using default file names (where the default file name is specific to that option):
□ If a required option is not set, the default is used.
□ If an option is marked optional, the file is not used unless the option is set (or other conditions make the file required).
□ If an option is set, and no file name is provided, the default is used.
All such options will accept file names without a file extension. The extension is automatically appended in such a case. When multiple input formats are accepted, such as a generic structure
format, the directory will be searched for files of each type with the supplied or default name. When no file with a recognized extension is found, an error is given. For output files with
multiple formats, a default file type will be used.
Some file formats can also be read from compressed (.Z or .gz) formats.
enum options
Enumerated options (enum) should be used with one of the arguments listed in the option description. The argument may be abbreviated, and the first match to the shortest argument in the list will
be selected.
vector options
Some options accept a vector of values. Either 1 or 3 parameters can be supplied; when only one parameter is supplied the two other values are also set to this value.
selection options
Commands by topic¶
Trajectory analysis¶
Calculate angles
Calculate distances between pairs of positions
Calculate free volume
Calculate pairwise distances between groups of positions
Calculate radial distribution functions
Compute solvent accessible surface area
Print general information about selections
Generating topologies and coordinates¶
Edit the box and write subgroups
Generate a primitive topology from coordinates
Solvate a system
Insert molecules into existing vacancies
Multiply a conformation in ‘random’ orientations
Generate monoatomic ions on energetically favorable positions
Generate position restraints or distance restraints for index groups
Convert coordinate files to topology and FF-compliant coordinate files
Running a simulation¶
Make a run input file
Perform a simulation, do a normal mode analysis or an energy minimization
Make a modifed run-input file
Viewing trajectories¶
Generate a virtual oscillating trajectory from an eigenvector
View a trajectory on an X-Windows terminal
Processing energies¶
Extract an energy matrix from an energy file
Writes energies to xvg files and display averages
(Re)calculate energies for trajectory frames with -rerun
Converting files¶
Convert and manipulates structure files
Convert energy files
Convert c6/12 or c6/cn combinations to and from sigma/epsilon
Concatenate trajectory files
Convert and manipulates trajectory files
Convert XPM (XPixelMap) matrices to postscript or XPM
Analyze data sets
Interpolate and extrapolate structure rotations
Frequency filter trajectories, useful for making smooth movies
Estimate free energy from linear combinations
Interpolate linearly between conformations
Estimate the error of using PME with a given input file
Compute free energies or other histograms from histograms
Calculate the spatial distribution function
Plot x, v, f, box, temperature and rotational energy from trajectories
Time mdrun as a function of PME ranks to optimize settings
Perform weighted histogram analysis after umbrella sampling
Check and compare files
Make binary files human readable
Make index files
Generate index files for ‘gmx angle’
Order molecules according to their distance to a group
Convert XPM (XPixelMap) matrices to postscript or XPM
Distances between structures¶
Cluster structures
Fit two structures and calculates the RMSD
Calculate RMSDs with a reference structure and RMSD matrices
Calculate atomic fluctuations
Distances in structures over time¶
Calculate the minimum distance between two groups
Calculate residue contact maps
Calculate static properties of polymers
Calculate atom pair distances averaged with power -2, -3 or -6
Mass distribution properties over time¶
Calculate the radius of gyration
Calculates mean square displacements
Calculate static properties of polymers
Calculate radial distribution functions
Calculate the rotational correlation function for molecules
Plot the rotation matrix for fitting to a reference structure
Compute small angle neutron scattering spectra
Compute small angle X-ray scattering spectra
Plot x, v, f, box, temperature and rotational energy from trajectories
Compute Van Hove displacement and correlation functions
Analyzing bonded interactions¶
Calculate distributions and correlations for angles and dihedrals
Generate index files for ‘gmx angle’
Structural properties¶
Cluster structures from Autodock runs
Analyze bundles of axes, e.g., helices
Calculate size distributions of atomic clusters
Analyze distance restraints
Compute and analyze hydrogen bonds
Compute the order parameter per atom for carbon tails
Calculate principal axes of inertia for a group of atoms
Calculate radial distribution functions
Compute salt bridges
Analyze solvent orientation around solutes
Analyze solvent dipole orientation and polarization around solutes
Kinetic properties¶
Calculate free energy difference estimates through Bennett’s acceptance ratio
Calculate dielectric constants and current autocorrelation function
Analyze density of states and properties based on that
Extract dye dynamics from trajectories
Calculate principal axes of inertia for a group of atoms
Calculate viscosities of liquids
Plot x, v, f, box, temperature and rotational energy from trajectories
Compute Van Hove displacement and correlation functions
Calculate velocity autocorrelation functions
Electrostatic properties¶
Calculate dielectric constants and current autocorrelation function
Calculate frequency dependent dielectric constants
Compute the total dipole plus fluctuations
Calculate the electrostatic potential across the box
Analyze solvent dipole orientation and polarization around solutes
Generate monoatomic ions on energetically favorable positions
Protein-specific analysis¶
Assign secondary structure and calculate solvent accessible surface area
Calculate everything you want to know about chi and other dihedrals
Calculate basic properties of alpha helices
Calculate local pitch/bending/rotation/orientation inside helices
Compute Ramachandran plots
Plot helical wheels
Analyze bundles of axes, e.g., helices
Calculate the density of the system
Calculate 2D planar or axial-radial density maps
Calculate surface fluctuations
Compute the orientation of water molecules
Compute tetrahedrality parameters around a given atom
Compute the order parameter per atom for carbon tails
Calculate the electrostatic potential across the box
Covariance analysis¶
Analyze the eigenvectors
Calculate and diagonalize the covariance matrix
Generate input files for essential dynamics sampling
Normal modes¶
Analyze the normal modes
Diagonalize the Hessian for normal mode analysis
Generate a virtual oscillating trajectory from an eigenvector
Generate an ensemble of structures from the normal modes
Make a run input file
Find a potential energy minimum and calculate the Hessian
Special topics¶
The information in these topics is also accessible through gmx help topic on the command line.
Selection syntax and usage¶
Command changes between versions¶
Starting from GROMACS 5.0, some of the analysis commands (and a few other commands as well) have changed significantly.
One main driver for this has been that many new tools mentioned below now accept selections through one or more command-line options instead of prompting for a static index group. To take full
advantage of selections, the interface to the commands has changed somewhat, and some previous command-line options are no longer present as the same effect can be achieved with suitable selections.
Please see Selection syntax and usage additional information on how to use selections.
In the process, some old analysis commands have been removed in favor of more powerful functionality that is available through an alternative tool. For removed or replaced commands, this page
documents how to perform the same tasks with new tools. For new commands, a brief note on the available features is given. See the linked help for the new commands for a full description.
This section lists only major changes; minor changes like additional/removed options or bug fixes are not typically included.
Version 2016¶
Analysis on arbitrary subsets of atoms¶
Tools implemented in the new analysis framework can now operate upon trajectories that match only a subset of the atoms in the input structure file.
gmx insert-molecules¶
gmx insert-molecules has gained an option -replace that makes it possible to insert molecules into a solvated configuration, replacing any overlapping solvent atoms. In a fully solvated box, it is
also possible to insert into a certain region of the solvent only by selecting a subset of the solvent atoms (-replace takes a selection that can also contain expressions like not within 1 of ...).
gmx rdf¶
The normalization for the output RDF can now also be the radial number density.
gmx genconf¶
Removed -block, -sort and -shuffle.
Version 5.1¶
Symbolic links from 5.0 are no longer supported. The only way to invoke a command is through gmx <command>.
gmx pairdist¶
gmx pairdist has been introduced as a selection-enabled replacement for gmx mindist (gmx mindist still exists unchanged). It can calculate min/max pairwise distances between a pair of selections,
including, e.g., per-residue minimum distances or distances from a single point to a set of residue-centers-of-mass.
gmx rdf¶
gmx rdf has been rewritten for 5.1 to use selections for specifying the points from which the RDFs are calculated. The interface is mostly the same, except that there are new command-line options to
specify the selections. The following additional changes have been made:
• -com and -rdf options have been removed. Equivalent functionality is available through selections:
□ -com can be replaced with a com of <selection> as the reference selection.
□ -rdf can be replaced with a suitable set of selections (e.g., res_com of <selection>) and/or using -seltype.
• -rmax option is added to specify a cutoff for the RDFs. If set to a value that is significantly smaller than half the box size, it can speed up the calculation significantly if a grid-based
neighborhood search can be used.
• -hq and -fade options have been removed, as they are simply postprocessing steps on the raw numbers that can be easily done after the analysis.
Version 5.0¶
Version 5.0 introduced the gmx wrapper binary. For backwards compatibility, this version still creates symbolic links by default for old tools: e.g., g_order <options> is equivalent to gmx order
<options>, and g_order is simply a symbolic link on the file system.
This tool has been removed in 5.0. A replacement is gmx distance.
You can provide your existing index file to gmx distance, and it will calculate the same distances. The differences are:
• -blen and -tol options have different default values.
• You can control the output histogram with -binw.
• -aver and -averdist options are not present. Instead, you can choose between the different things to calculate using -oav (corresponds to -d with -averdist), -oall (corresponds to -d without
-averdist), -oh (corresponds to -o with -aver), and -oallstat (corresponds to -l without -aver).
You can produce any combination of output files. Compared to g_bond, gmx distance -oall is currently missing labels for the output columns.
This tool has been removed in 5.0. A replacement is gmx distance (for most options) or gmx select (for -dist or -lt).
If you had index groups A and B in index.ndx for g_dist, you can use the following command to compute the same distance with gmx distance:
gmx distance -n index.ndx -select 'com of group "A" plus com of group "B"' -oxyz -oall
The -intra switch is replaced with -nopbc.
If you used -dist D, you can do the same calculation with gmx select:
gmx select -n index.ndx -select 'group "B" and within D of com of group "A"' -on/-oi/-os/-olt
You can select the output option that best suits your post-processing needs (-olt is a replacement for g_dist -dist -lt)
gmx distance¶
gmx distance has been introduced as a selection-enabled replacement for various tools that computed distances between fixed pairs of atoms (or centers-of-mass of groups). It has a combination of the
features of g_bond and g_dist, allowing computation of one or multiple distances, either between atom-atom pairs or centers-of-mass of groups, and providing a combination of output options that were
available in one of the tools.
gmx gangle¶
gmx gangle has been introduced as a selection-enabled replacement for g_sgangle. In addition to supporting atom-atom vectors, centers-of-mass can be used as endpoints of the vectors, and there are a
few additional angle types that can be calculated. The command also has basic support for calculating normal angles between three atoms and/or centers-of-mass, making it a partial replacement for gmx
angle as well.
gmx protonate¶
This was a very old tool originally written for united atom force fields, where it was necessary to generate all hydrogens after running a trajectory in order to calculate e.g. distance restraint
violations. The functionality to simply protonate a structure is available in gmx pdb2gmx. If there is significant interest, we might reintroduce it after moving to new topology formats in the
gmx freevolume¶
This tool has been introduced in 5.0. It uses a Monte Carlo sampling method to calculate the fraction of free volume within the box (using a probe of a given size).
This tool has been rewritten in 5.0, and renamed to gmx sasa (the underlying surface area calculation algorithm is still the same).
The main difference in the new tool is support for selections. Instead of prompting for an index group, a (potentially dynamic) selection for the calculation can be given with -surface. Any number of
output groups can be given with -output, allowing multiple parts of the surface area to be computed in a single run. The total area of the -surface group is now always calculated.
The tool no longer automatically divides the surface into hydrophobic and hydrophilic areas, and there is no -f_index option. The same effects can be obtained by defining suitable selections for
-output. If you want output that contains the same numbers as with the old tool for a calculation group A and output group B, you can use
gmx sasa -surface 'group "A"' -output '"Hydrophobic" group "A" and charge {-0.2 to 0.2}; "Hydrophilic" group "B" and not charge {-0.2 to 0.2}; "Total" group "B"'
Solvation free energy estimates are now calculated only if separately requested with -odg, and are written into a separate file.
Output option -i for a position restraint file is not currently implemented in the new tool, but would not be very difficult to add if requested.
This tool has been removed in 5.0. A replacement is gmx gangle (for angle calculation) and gmx distance (for -od, -od1, -od2).
If you had index groups A and B in index.ndx for g_sgangle, you can use the following command to compute the same angle with gmx gangle:
gmx gangle -n index.ndx -g1 vector/plane -group1 'group "A"' -g2 vector/plane -group2 'group "B"' -oav
You need to select either vector or plane for the -g1 and -g2 options depending on which one your index groups specify.
If you only had a single index group A in index.ndx and you used g_sgangle -z or -one, you can use:
gmx gangle -n index.ndx -g1 vector/plane -group1 'group "A"' -g2 z/t0 -oav
For the distances, you can use gmx distance to compute one or more distances as you want. Both distances between centers of groups or individual atoms are supported using the new selection syntax. | {"url":"https://manual.gromacs.org/2016.6/user-guide/cmdline.html","timestamp":"2024-11-07T15:57:15Z","content_type":"application/xhtml+xml","content_length":"81579","record_id":"<urn:uuid:90c686d8-ebb5-48ae-963f-b88f16672d3c>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00761.warc.gz"} |
Random Number Generator
You may have wondered why predictable computers can generate randomness. In fact, most random numbers used in computer programs are pseudorandom, meaning that they are generated in a predictable way
using a mathematical formula. This is great for many purposes, but it can’t be considered random, as you’d expect if you’re used to dice rolls and lottery draws.
This version of the generator creates a random integer
Generating a random, non-repeating number
A random number generator is a device that can generate one or more random numbers from a specific area. Random number generators can be hardware or pseudo-random.
A pseudorandom number generator is an algorithm for generating a sequence of numbers whose properties approximate the properties of sequences of random numbers. Computer random number generators are
almost always pseudo-random number generators. However, the numbers generated by pseudo-random number generators are not really random. Similarly, our generators above are also pseudo-random number
generators. Generated random numbers are sufficient for most applications, but they should not be used for cryptographic purposes. True random numbers are based on physical phenomena such as
atmospheric phenomena, temperature, and other quantum phenomena. Methods that generate true random numbers also include compensation for potential distortions caused by the measurement process. | {"url":"https://calculators.vip/random-number-generator/","timestamp":"2024-11-06T23:53:47Z","content_type":"text/html","content_length":"43719","record_id":"<urn:uuid:124b3adb-0f0a-40ab-a839-468857a9820f>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00345.warc.gz"} |
Bayesian Stochastic Search Variable Selection
This example shows how to implement stochastic search variable selection (SSVS), a Bayesian variable selection technique for linear regression models.
Consider this Bayesian linear regression model.
${y}_{t}=\sum _{k}{\beta }_{k}{x}_{tk}+{\epsilon }_{t}.$
• The regression coefficients ${\beta }_{k}|{\sigma }^{2}\sim N\left({\mu }_{j},{\sigma }^{2}{V}_{k}\right)$.
• $k=0,...,p$.
• The disturbances ${\epsilon }_{t}\sim N\left(0,{\sigma }^{2}\right)$.
• The disturbance variance ${\sigma }^{2}\sim IG\left(A,B\right)$, where $IG\left(A,B\right)$ is the inverse gamma distribution with shape A and scale B.
The goal of variable selection is to include only those predictors supported by data in the final regression model. One way to do this is to analyze the ${2}^{p}$ permutations of models, called
regimes, where models differ by the coefficients that are included. If $p$ is small, then you can fit all permutations of models to the data, and then compare the models by using performance
measures, such as goodness-of-fit (for example, Akaike information criterion) or forecast mean squared error (MSE). However, for even moderate values of $p$, estimating all permutations of models is
A Bayesian view of variable selection is a coefficient, being excluded from a model, has a degenerate posterior distribution. That is, the excluded coefficient has a Dirac delta distribution, which
has its probability mass concentrated on zero. To circumvent the complexity induced by degenerate variates, the prior for a coefficient being excluded is a Gaussian distribution with a mean of 0 and
a small variance, for example $N\left(0,0.{1}^{2}\right)$. Because the prior is concentrated around zero, the posterior must also be concentrated around zero.
The prior for the coefficient being included can be $N\left(\mu ,V\right)$, where $V$ is sufficiently away from zero and $\mu$ is usually zero. This framework implies that the prior of each
coefficient is a Gaussian mixture model.
Consider the latent, binary random variables ${\gamma }_{k}$, $k=0,...,p$, such that:
• ${\gamma }_{k}=1$ indicates that ${\beta }_{k}\sim N\left(0,{\sigma }^{2}{V}_{1k}\right)$ and that ${\beta }_{k}$ is included in the model.
• ${\gamma }_{k}=0$ indicates that ${\beta }_{k}\sim N\left(0,{\sigma }^{2}{V}_{2k}\right)$ and that ${\beta }_{k}$ is excluded from the model.
• ${\gamma }_{k}\sim Bernoulli\left({g}_{k}\right)$.
• The sample space of ${\gamma }_{k}$ has a cardinality of ${2}^{\mathit{p}+1}$, and each element is a $p+1$-D vector of zeros or ones.
• ${V}_{2k}$ is a small, positive number and ${V}_{1k}>{V}_{2k}$.
• Coefficients ${\beta }_{j}$ and ${\beta }_{k}$, $je k$ are independent, a priori.
One goal of SSVS is to estimate the posterior regime probabilities ${g}_{k}$, the estimates that determine whether corresponding coefficients should be included in the model. Given ${\beta }_{k}$, $
{\gamma }_{k}$ is conditionally independent of the data. Therefore, for $k=0,...,p$, this equation represents the full conditional posterior distribution of the probability that variable k is
included in the model:
$P\left({\gamma }_{k}=1|\beta ,{\sigma }^{2},{\gamma }_{e k}\right)\propto {g}_{k}\varphi \left({\beta }_{k};0,{\sigma }^{2}{V}_{1k}\right),$
where $\varphi \left(\mu ,{\sigma }^{2}\right)$ is the pdf of the Gaussian distribution with scalar mean $\mu$ and variance ${\sigma }^{2}$.
Econometrics Toolbox™ has two Bayesian linear regression models that specify the prior distributions for SSVS: mixconjugateblm and mixsemiconjugateblm. The framework presented earlier describes the
priors of the mixconjugateblm model. The difference between the models is that $\beta$ and ${\sigma }^{2}$ are independent, a priori, for mixsemiconjugateblm models. Therefore, the prior variance of
${\beta }_{k}$ is ${V}_{k1}$ (${\gamma }_{\mathit{k}}=1$) or ${V}_{k2}$ (${\gamma }_{\mathit{k}}=0$).
After you decide which prior model to use, call bayeslm to create the model and specify hyperparameter values. Supported hyperparameters include:
• Intercept, a logical scalar specifying whether to include an intercept in the model.
• Mu, a (p + 1)-by-2 matrix specifying the prior Gaussian mixture means of $\beta$. The first column contains the means for the component corresponding to ${\gamma }_{k}=1$, and the second column
contains the means corresponding to ${\gamma }_{k}=0$. By default, all means are 0, which specifies implementing SSVS.
• V, a (p + 1)-by-2 matrix specifying the prior Gaussian mixture variance factors (or variances) of $\beta$. Columns correspond to the columns of Mu. By default, the variance of the first component
is 10 and the variance of the second component is 0.1.
• Correlation, a (p + 1)-by-(p + 1) positive definite matrix specifying the prior correlation matrix of $\beta$ for both components. The default is the identity matrix, which implies that the
regression coefficients are uncorrelated, a priori.
• Probability, a (p + 1)-D vector of prior probabilities of variable inclusion (${g}_{k}$, k = 0,...,_p_) or a function handle to a custom function. ${\gamma }_{j}$ and ${\gamma }_{k}$, $je k$, are
independent, a priori. However, using a function handle (@functionname), you can supply a custom prior distribution that specifies dependencies between ${\gamma }_{j}$ and ${\gamma }_{k}$. For
example, you can specify forcing ${x}_{2}$ out of the model if ${x}_{4}$ is included.
After you create a model, pass it and the data to estimate. The estimate function uses a Gibbs sampler to sample from the full conditionals, and estimate characteristics of the posterior
distributions of $\beta$ and ${\sigma }^{2}$. Also, estimate returns posterior estimates of ${g}_{k}$.
For this example, consider creating a predictive linear model for the US unemployment rate. You want a model that generalizes well. In other words, you want to minimize the model complexity by
removing all redundant predictors and all predictors that are uncorrelated with the unemployment rate.
Load and Preprocess Data
Load the US macroeconomic data set Data_USEconModel.mat.
The data set includes the MATLAB® timetable DataTimeTable, which contains 14 variables measured from Q1 1947 through Q1 2009; UNRATE is the US unemployment rate. For more details, enter Description
at the command line.
Plot all series in the same figure, but in separate subplots.
for j = 1:size(DataTimeTable,2)
All series except FEDFUNDS, GS10, TB3MS, and UNRATE appear to have an exponential trend.
Apply the log transform to those variables with an exponential trend.
hasexpotrend = ~ismember(DataTimeTable.Properties.VariableNames,...
["FEDFUNDS" "GD10" "TB3MS" "UNRATE"]);
DataTimeTableLog = varfun(@log,DataTimeTable,'InputVariables',...
DataTimeTableLog = [DataTimeTableLog ...
DataTimeTableLog is a timetable like DataTimeTable, but those variables with an exponential trend are on the log scale.
Coefficients that have relatively large magnitudes tend to dominate the penalty in the lasso regression objective function. Therefore, it is important that variables have a similar scale when you
implement lasso regression. Compare the scales of the variables in DataTimeTableLog by plotting their box plots on the same axis.
h = gcf;
h.Position(3) = h.Position(3)*2.5;
title('Variable Box Plots');
The variables have fairly similar scales.
To tune the prior Gaussian mixture variance factors, follow this procedure:
1. Partition the data into estimation and forecast samples.
2. Fit the models to the estimation sample and specify, for all $k$, ${V}_{1k}=\left\{10,\phantom{\rule{0.2777777777777778em}{0ex}}50,\phantom{\rule{0.2777777777777778em}{0ex}}100\right\}$ and ${V}_
3. Use the fitted models to forecast responses into the forecast horizon.
4. Estimate the forecast MSE for each model.
5. Choose the model with the lowest forecast MSE.
George and McCulloch suggest another way to tune the prior variances of $\beta$ in [1].
Create estimation and forecast sample variables for the response and predictor data. Specify a forecast horizon of 4 years (16 quarters).
fh = 16;
y = DataTimeTableLog.UNRATE(1:(end - fh));
yF = DataTimeTableLog.UNRATE((end - fh + 1):end);
isresponse = DataTimeTable.Properties.VariableNames == "UNRATE";
X = DataTimeTableLog{1:(end - fh),~isresponse};
XF = DataTimeTableLog{(end - fh + 1):end,~isresponse};
p = size(X,2); % Number of predictors
predictornames = DataTimeTableLog.Properties.VariableNames(~isresponse);
Create Prior Bayesian Linear Regression Models
Create prior Bayesian linear regression models for SSVS by calling bayeslm and specifying the number of predictors, model type, predictor names, and component variance factors. Assume that $\beta$
and ${\sigma }^{2}$ are dependent, a priori (mixconjugateblm model).
V1 = [10 50 100];
V2 = [0.05 0.1 0.5];
numv1 = numel(V1);
numv2 = numel(V2);
PriorMdl = cell(numv1,numv2); % Preallocate
for k = 1:numv2
for j = 1:numv1
V = [V1(j)*ones(p + 1,1) V2(k)*ones(p + 1,1)];
PriorMdl{j,k} = bayeslm(p,'ModelType','mixconjugateblm',...
PriorMdl is a 3-by-3 cell array, and each cell contains a mixconjugateblm model object.
Plot the prior distribution of log_GDP for the models in which V2 is 0.5.
for j = 1:numv1
[~,~,~,h] = plot(PriorMdl{j,3},'VarNames',"log_GDP");
title(sprintf("Log GDP, V1 = %g, V2 = %g",V1(j),V2(3)));
h.Tag = strcat("fig",num2str(V1(j)),num2str(V2(3)));
The prior distributions of $\beta$ have the spike-and-slab shape. When V1 is low, more of the distribution is concentrated around 0, which makes it more difficult for the algorithm to attribute a
high value for beta. However, variables the algorithm identifies as important are regularized, in that the algorithm does not attribute a high magnitude to the corresponding coefficients.
When V1 is high, more density occurs well away from zero, which makes it easier for the algorithm to attribute non-zero coefficients to important predictors. However, if V1 is too high, then
important predictors can have inflated coefficients.
Perform SSVS Variable Selection
To perform SSVS, estimate the posterior distributions by using estimate. Use the default options for the Gibbs sampler.
PosteriorMdl = cell(numv1,numv2);
PosteriorSummary = cell(numv1,numv2);
rng(1); % For reproducibility
for k = 1:numv2
for j = 1:numv1
[PosteriorMdl{j,k},PosteriorSummary{j,k}] = estimate(PriorMdl{j,k},X,y,...
Each cell in PosteriorMdl contains an empiricalblm model object storing the full conditional posterior draws from the Gibbs sampler. Each cell in PosteriorSummary contains a table of posterior
estimates. The Regime table variable represents the posterior probability of variable inclusion (${g}_{k}$).
Display a table of posterior estimates of ${g}_{k}$.
RegimeTbl = table(zeros(p + 2,1),'RowNames',PosteriorSummary{1}.Properties.RowNames);
for k = 1:numv2
for j = 1:numv1
vname = strcat("V1_",num2str(V1(j)),"__","V2_",num2str(V2(k)));
vname = replace(vname,".","p");
tmp = table(PosteriorSummary{j,k}.Regime,'VariableNames',vname);
RegimeTbl = [RegimeTbl tmp];
RegimeTbl.Var1 = [];
RegimeTbl=15×9 table
V1_10__V2_0p05 V1_50__V2_0p05 V1_100__V2_0p05 V1_10__V2_0p1 V1_50__V2_0p1 V1_100__V2_0p1 V1_10__V2_0p5 V1_50__V2_0p5 V1_100__V2_0p5
______________ ______________ _______________ _____________ _____________ ______________ _____________ _____________ ______________
Intercept 0.9692 1 1 0.9501 1 1 0.9487 0.9999 1
log_COE 0.4686 0.4586 0.5102 0.4487 0.3919 0.4785 0.4575 0.4147 0.4284
log_CPIAUCSL 0.9713 0.3713 0.4088 0.971 0.3698 0.3856 0.962 0.3714 0.3456
log_GCE 0.9999 1 1 0.9978 1 1 0.9959 1 1
log_GDP 0.7895 0.9921 0.9982 0.7859 0.9959 1 0.7908 0.9975 0.9999
log_GDPDEF 0.9977 1 1 1 1 1 0.9996 1 1
log_GPDI 1 1 1 1 1 1 1 1 1
log_GS10 1 1 0.9991 1 1 0.9992 1 0.9992 0.994
log_HOANBS 0.9996 1 1 0.9887 1 1 0.9763 1 1
log_M1SL 1 1 1 1 1 1 1 1 1
log_M2SL 0.9989 0.9993 0.9913 0.9996 0.9998 0.9754 0.9951 0.9983 0.9856
log_PCEC 0.4457 0.6366 0.8421 0.4435 0.6226 0.8342 0.4614 0.624 0.85
FEDFUNDS 0.0762 0.0386 0.0237 0.0951 0.0465 0.0343 0.1856 0.0953 0.068
TB3MS 0.2473 0.1788 0.1467 0.2014 0.1338 0.1095 0.2234 0.1185 0.0909
Sigma2 NaN NaN NaN NaN NaN NaN NaN NaN NaN
Using an arbitrary threshold of 0.10, all models agree that FEDFUNDS is an insignificant or redundant predictor. When V1 is high, TB3MS borders on being insignificant.
Forecast responses and compute forecast MSEs using the estimated models.
yhat = zeros(fh,numv1*numv2);
fmse = zeros(numv1,numv2);
for k = 1:numv2
for j = 1:numv1
idx = ((k - 1)*numv1 + j);
yhat(:,idx) = forecast(PosteriorMdl{j,k},XF);
fmse(j,k) = sqrt(mean((yF - yhat(:,idx)).^2));
Identify the variance factor settings that yield the minimum forecast MSE.
minfmse = min(fmse,[],'all');
[idxminr,idxminc] = find(abs(minfmse - fmse) < eps);
bestv1 = V1(idxminr)
Estimate an SSVS model using the entire data set and the variance factor settings that yield the minimum forecast MSE.
XFull = [X; XF];
yFull = [y; yF];
EstMdl = estimate(PriorMdl{idxminr,idxminc},XFull,yFull);
Method: MCMC sampling with 10000 draws
Number of observations: 201
Number of predictors: 14
| Mean Std CI95 Positive Distribution Regime
Intercept | 29.4598 4.2723 [21.105, 37.839] 1.000 Empirical 1
log_COE | 3.5380 3.0180 [-0.216, 9.426] 0.862 Empirical 0.7418
log_CPIAUCSL | -0.6333 1.7689 [-5.468, 2.144] 0.405 Empirical 0.3711
log_GCE | -9.3924 1.4699 [-12.191, -6.494] 0.000 Empirical 1
log_GDP | 16.5111 3.7131 [ 9.326, 23.707] 1.000 Empirical 1
log_GDPDEF | 13.0146 2.3992 [ 9.171, 19.131] 1.000 Empirical 1
log_GPDI | -5.9537 0.6083 [-7.140, -4.756] 0.000 Empirical 1
log_GS10 | 1.4485 0.3852 [ 0.680, 2.169] 0.999 Empirical 0.9868
log_HOANBS | -16.0240 1.5361 [-19.026, -13.048] 0.000 Empirical 1
log_M1SL | -4.6509 0.6815 [-5.996, -3.313] 0.000 Empirical 1
log_M2SL | 5.3320 1.3003 [ 2.738, 7.770] 0.999 Empirical 0.9971
log_PCEC | -9.9025 3.3904 [-16.315, -2.648] 0.006 Empirical 0.9858
FEDFUNDS | -0.0176 0.0567 [-0.125, 0.098] 0.378 Empirical 0.0269
TB3MS | -0.1436 0.0762 [-0.299, 0.002] 0.026 Empirical 0.0745
Sigma2 | 0.2891 0.0289 [ 0.238, 0.352] 1.000 Empirical NaN
EstMdl is an empiricalblm model representing the result of performing SSVS. You can use EstMdl to forecast the unemployment rate given future predictor data , for example.
[1] George, E. I., and R. E. McCulloch. "Variable Selection Via Gibbs Sampling." Journal of the American Statistical Association. Vol. 88, No. 423, 1993, pp. 881–889.
See Also
estimate | sampleroptions
Related Topics | {"url":"https://it.mathworks.com/help/econ/implement-bayesian-variable-selection.html","timestamp":"2024-11-04T21:19:59Z","content_type":"text/html","content_length":"116779","record_id":"<urn:uuid:279566f1-7cac-4da2-b6d8-1340071a7e2b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00322.warc.gz"} |
Estimating Jones polynomials is a complete problem for one clean qubit
Title Estimating Jones polynomials is a complete problem for one clean qubit
Publication Journal Article
Year of 2008
Authors Shor, PW, Jordan, SP
Journal Quantum Information & Computation
Volume 8
Issue 8
Pages 681-714
Date 2008/09/01
It is known that evaluating a certain approximation to the Jones polynomial for the plat closure of a braid is a BQP-complete problem. That is, this problem exactly captures the power of
the quantum circuit model. The one clean qubit model is a model of quantum computation in which all but one qubit starts in the maximally mixed state. One clean qubit computers are
believed to be strictly weaker than standard quantum computers, but still capable of solving some classically intractable problems. Here we show that evaluating a certain approximation to
Abstract the Jones polynomial at a fifth root of unity for the trace closure of a braid is a complete problem for the one clean qubit complexity class. That is, a one clean qubit computer can
approximate these Jones polynomials in time polynomial in both the number of strands and number of crossings, and the problem of simulating a one clean qubit computer is reducible to
approximating the Jones polynomial of the trace closure of a braid.
URL http://dl.acm.org/citation.cfm?id=2017011.2017012 | {"url":"https://www.quics.umd.edu/publications/estimating-jones-polynomials-complete-problem-one-clean-qubit","timestamp":"2024-11-06T07:40:06Z","content_type":"text/html","content_length":"21695","record_id":"<urn:uuid:b424c1e7-f16b-4a78-9d93-34c22696c80e>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00791.warc.gz"} |
Problem F: Failed in Linear Algebra
One day, Wavator is taking his Linear algebra course. He hates calculating the expression of matrix so he wants to develop a calculator to help him. But, he got 59 in last years’ DSAA course, so he
turns you for help.
n square matrices of size m are given, and we defines an operation like “(1+2)*1” which means the matrix 1 plus matrix 2 and then multiplies matrix 1.
Wavator only wants to calculate “+” and “-” and “*”, so he denotes that “+” means A + B = C where C[i][j] = A[i][j] + B[i][j] .The rule of “-” is similar with "+".
Notice that in matrix multiplication, a*b and b*a is not the same.
Since the number may be too large during the calculation process, in each step you should mod 1000000007. | {"url":"https://acm.sustech.edu.cn/onlinejudge/problem.php?cid=1033&pid=5","timestamp":"2024-11-04T23:44:50Z","content_type":"text/html","content_length":"10224","record_id":"<urn:uuid:46fb3fda-5f47-46a7-9888-773e222febc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00590.warc.gz"} |
Stability of a mixed type cubic and quartic functional equation in fuzzy Banach spaces
J. Math. Comput. Sci. 7 (2017), No. 5, 821-831 ISSN: 1927-5307
Department of Mathematics, Tianjin University of Technology, Tianjin 300384, P.R. China
Copyright c2017 Zhu Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unre-stricted use, distribution, and reproduction in any medium,
provided the original work is properly cited.
Abstract. In this paper, we generalized Ulam-Hyers stability of the mixed type cubic and quartic functional equation in fuzzy Banach space.
Keywords:fuzzy normed spaces; stability of quartic and cubic mapping; Banach space. 2010 AMS Subject Classification:Primary 46S40, secondary 39B52, 26E50.
1. Introduction
In 1940, Ulam [13] posed the first stability problem concerning group homomorphisms. In the next year, Hyers [7] gave an affirmative answer to the question of Ulam in Banach spaces. Aoki [14]
generalized Hyers result for additive mappings. For additive mapping involving dif-ferent powers of norms [18,20]. This stability is also investigates by Park [6]. In 1984, Katsaras [1] constructed a
fuzzy vector topological structure on the linear space. Later, some mathe-maticians considered some other type fuzzy norms and some properties of fuzzy normed linear spaces [5,15]. Recently, several
various fuzzy versions stability problem concerning quadratic,
E-mail address: [email protected] Received May 17, 2017
cubic and quartic function equation have been considered [2,3].
Hyers [8] was the first person to point out the direct method for studying the stability of func-tional equation. In 2003, Radu [19] proposed the fixed point alternative method to solve the Ulam
problem. Subsequently, Mihet [9] applied the fixed point alternative method to solve fuzzy stability of Jensen functional equation in fuzzy normed space.
Jun and Kim [12] introduced cubic function equation and they established the solution of Hyers-Ulam-Rassias stability for the functional equation (1.1), and this equation is called the cubic function
equation, if the cubic function f(x) =cx3satisfies (1.1). The quartic functional equa-tion was introduced by Rassias [19] in 2000
f(2x+y) +f(2x−y) =4f(x+y) +4f(x−y) +24f(x)−6f(y) (1.2) It is easy to show that the function f(x) =cx4 satisfies the functional equation (1.2). In this paper, we establish a fuzzy version stability
for following functional equation:
f(x+2y) +f(x−2y) =4(f(x+y) +4f(x−y))−24f(y)−6f(x) +3f(2y) (1.3) in fuzzy Banach space and the function f(x) =ax3+bx4is a solution of the functional equation (1.3). We using the fixed point
alternative method to establish fuzzy stability.
2. Preliminaries
We start our works with basic definition using in this paper.
Definition 2.1. [16]Let X be a real linear space. A fuzzy subset N of X×[R]is called afuzzy normon X if and only if
(N1)For all t ∈[R]with t≤0, N(x,t) =0;
(N2)For all t ∈[R]with t>0, N(x,t) =1if and only if x=0; (N3)For allλ ∈[R]withλ 6=0, N(λx,t) =N(x,t/|λ|);
(N4)For all s,t∈[R], N(x+y,s+t)≥min{(N(x,s),N(y,t)};
(N5)N(x,·)is a non-decreasing function on[R]andlimt→∞N(x,t) =1;
Example 2.2.[4]Let(X,k · k)be a normed space. For every x∈X , we define
N(x,t) =
t+kxk, when t>0,
0, when t≤0.
Then(X,N)is a fuzzy normed linear space.
A sequence{x[n]}inX is called Cauchy if for eachε >0 and eacht>0 there existsn0such
that for alln≥n0and all p>0, we haveN(xn+p−xn,t)>1−ε. If every Cauchy sequence is
convergent, then the fuzzy normed space space is called a fuzzy Banach space.
3.fuzzy stability of Cubic and Quartic Function Equation Using Direct
In this section, for given f :X→Y, we define operatorD f :X×X →Y by
D f(x,y) = f(x+2y) + f(x−2y)−4[f(x+y) + f(x−y)]−3f(2y) +24f(y) +6f(x) Theorem 3.1.(The fixed point alternative theorem,[17]) Let(Ω,d)be a complete generalized metric space and T :Ω→Ωbe a strictly
contractive mapping with Lipschitz constant L, that is
d(T x,Ty)≤Ld(x,y), ∀x,y∈Ω.
Then for each given x∈Ω,either
d(Tnx,Tn+1y) =∞, ∀n≥0,
or there exists a natural number n[0]such that
(1)d(Tnx,Tn+1y)<∞, ∀n≥0,
(2)The sequence Tnis convergent to a fixed point y∗of T .
(3)y∗is the unique fixed point of T in the set4={y∈Ω:d(Tn0[x][,][y)][<][∞][}][.]
(4)d(y,y∗)≤ 1
1−Ld(y.Ty)for all y∈ 4.
mapping from X×X →Z such that
(α ϕ(x,y),t)
for all x∈X,t>0, and
lim k→∞
N0(ϕ(2kx,2ky),16kt) =1
for all x,y∈X,t>0,k≥0. If f :X→Y be an even function and f(0) =0. In the sense that
N(D f(x,y),t)≥N0(ϕ(x,y),t) (3.1)
for all x,y∈X,t>0. Then there exists a unique quartic mapping C:X→Y such that
for all x∈X,t>0. Moreover,
C(x) = lim n→∞
16n for all x∈X .
Proof.We assume that 0<α <16. Let
Ω={g:g:X →Y,g(0) =0}
and introduce the generalized metricd onΩby
d(g,h) =in f{β ∈(0,∞):N(g(x)−h(x),βt)≥N
We know that(Ω,d)is complete generalized metric onΩ. We now defined a mappingT :Ω→ Ωby
T g(x) = 1 16g(2x)
We now proveT is a strictly contractive mapping with the Lipschitz constant α
16. Giveng,h∈Ω,
setε ∈(0,∞)be an arbitrary constant withd(g.h)<ε. Then
N(T g−T h,α εt
16 ) =N( 1
16g(2x)− 1
α εt
16 ) =N(g(2x)−h(2x),α εt)
≥N0(α ϕ(0,x),16αt)
Hence ,we can conclude that
d(T g,T h)≤ α ε
16 Hence
d(g,h)<ε⇒d(T g,T h)≤ α ε
16, g,h∈Ω. That is
d(T g,T h)≤ α
16d(g,h) Putx=0 in (3.1), then replace y by x, we obtain
16 −f(x),t)≥N
for allx∈X,t>0, it follows thatd(T f,f)≤1. From the fixed point alternative theorem, we can conclude that, there exists a fixed pointCofT inΩsuch that
C(2x) =16C(x),∀x∈X.
Moreover, we have limn→∞d(Tnf,C)→0, which implies
N(lim n→∞
16n −C(x),t) =0. By the fixed point alternative, we conclude that
d(f,C)≤ 1
1−Ld(T f,f) Then
d(f,C)≤ 16
This means that
for allx∈X,t>0. The uniqueness ofC follows from the fact thatCis the unique fixed point ofT with the property that there existsk∈(0,∞)such that
N(C(x)− f(x),kt)≥N0(ϕ(0,x),t), ∀x∈X,t>0.
This completes the proof of this theorem.
Corollary 3.3. Let(X,k · k)be a normed space,(Y,N)be a fuzzy Banach space and(Z,N0)be a fuzzy normed space, u,v,γ,s be non-negative real numbers satisfies u+v,γ,s<4.If f :X→Y
be a mapping such that for some u0∈Z
N(D f(x,y),t)≥N0((kxkukykv+kxkγ[+][k][y][k]s[)u]
for all x,y∈X,t>0. Then there exists a unique quartic mapping C:X→Y such that N(f(x)−C(x),t)≥N0(kxksu0,(16−α)t)
Proof.We defineϕ:X×X →Zby
ϕ(x,y) = (kxkukykv+kxkγ+kyks)u0.
for allx,y∈X. It follows the conditions of Theorem 3.2, then completes the proof.
Theorem 3.4. Let X be a linear space,(Y,N)and(Z,N0)be a fuzzy Banach space and a fuzzy normed linear space respectively. Suppose that α is a constant satisfies0<|α|<8, ϕ is a
mapping from X×X →Z such that
(α ϕ(x,y),t)
for all x∈X,t>0, and
lim k→∞
N0(ϕ(2kx,2ky),8kt) =1
for all x,y∈X,t>0,k≥0. If f :X→Y be an odd function and f(0) =0. In the sense that N(D f(x,y),t)≥N0(ϕ(x,y),t) (3.2)
for all x∈X,t>0. Moreover,
C(x) = lim n→∞
f(2nx) 8n for all x∈X .
Proof.Similar to the proof of Theorem 3.2. We can assume that 0<α<8. Let Ω={g:g:X →Y,g(0) =0}
and introduce the generalized metricd onΩby
d(g,h) =in f{β ∈(0,∞):N(g(x)−h(x),βt)≥N
We know that(Ω,d)is complete generalized metric onΩ. We now defined a mappingT :Ω→ Ωby
T g(x) =1 8g(2x)
We now proveT is a strictly contractive mapping with the Lipschitz constantα
8. Giveng,h∈Ω,
setε ∈(0,∞)be an arbitrary constant withd(g.h)<ε. Then
(ϕ(0,x),24t), ∀x∈X,t>0
N(T g−T h,α εt
8 ) =N( 1
8g(2x)− 1 8h(2x),
α εt
8 ) =N(g(2x)−h(2x),α εt)
≥N0(α ϕ(0,x),24αt)
Hence ,we can conclude that
d(T g,T h)≤ α ε
8 Hence
d(g,h)<ε⇒d(T g,T h)≤ α ε
8 , g,h∈Ω. That is
d(T g,T h)≤ α
Putx=0 in(3.2), then replace y by x, we obtain
8 −f(x),t)≥N
for allx∈X,t>0, it follows thatd(T f,f)≤1. From the fixed point alternative theorem, we can conclude that, there exists a fixed pointCofT inΩsuch that
C(2x) =8C(x),∀x∈X
Moreover, we have limn→∞d(Tnf,C)→0, which implies
N(lim n→∞
8n −C(x),t) =0. By the fixed point alternative, we can conclude that
d(f,C)≤ 1
1−Ld(T f,f) Then
d(f,C)≤ 8
This means that
for allx∈X,t>0. The uniqueness ofC follows from the fact thatCis the unique fixed point ofT. This completes the proof of this theorem.
Corollary 3.5. Let(X,k · k)be a normed space,(Y,N)be a fuzzy Banach space and(Z,N0)be a fuzzy normed space,u,v,γ,s be non-negative real numbers satisfies u+v,γ,s<3. If f :X→Y
be a mapping such that for some u0∈Z
N(D f(x,y),t)≥N0((kxkukykv+kxkγ[+][k][y][k]s[)u]
for all x,y∈X,t>0. Then there exists a unique cubic mapping C:X→Y such that N(f(x)−C(x),t)≥N0(kxks[u]
Theorem 3.6. Let X be a linear space,(Y,N)and(Z,N0)be a fuzzy Banach space and a fuzzy normed linear space respectively. Suppose that α is a constant satisfies0<|α|<8, ϕ is a
mapping from X×X →Z such that
(α ϕ(x,y),t)
for all x∈X,t>0, and
lim k→∞
N0(ϕ(2kx,2ky),8kt) =1
for all x,y∈X,t>0,k≥0. If f :X→Y be a function such that f(0) =0. In the sense that N(D f(x,y),t)≥N0(ϕ(x,y),t)
for all x,y∈X,t>0. Then there exists a unique cubic mapping C:X →Y and a unique quartic mapping Q:X →Y such that
2 t), 0<α ≤4, N0(ϕ(0,x),3(8−α)
2 t), 4<α <8. for all x∈X,t>0. Moreover,
C(x) = lim n→∞
8n ,Q(x) =[n→]lim[∞]
16n for all x∈X .
Proof. We assume that 0<α <8. Let f0(x) = 1[2](f(x)− f(−x))for all x∈X. Then f0(0) =
0,f0(−x) =−f0(x)and
Let f[1](x) = 1[2](f(x) + f(−x))for allx∈X. Then f[1](0) =0,f[1](−x) = f[1](x)and N(D(f[1](x,y),t)≥min{N0(ϕ(x,y),t),N
Using the proof Theorem 3.2 and 3.4, we get unique cubic mapping C and unique quartic mappingQsatisfying
N(f[0](x)−C(x))≥N0(ϕ(0,x),3(8−α)t),N(f1(x)−Q(x))≥N 0
2),N(f1(x)−Q(x), t 2)}
2 t),N
2 t)}. This means that
2 t), 0<α ≤4, N0(ϕ(0,x),3(8−α)
2 t), 4<α <8. This completes the proof of this theorem.
Conflict of Interests no conflict of interest.
[1] A. K. Katsaras, Fuzzy topological vector spacesΠ. Fuzzy Sets Syst. 12 (1984), 143-154.
[2] A. K. Mirmostafaee, M. S. Moslehian, Fuzzy almost quadratic functions. Result. Math., 52 (2008), 161-177. [3] A. K. Mirmostafaee, M. S. Moslehian, Fuzzy approximately cubic mappings.Inf. Sci. 178
(2008), 3791-3798. [4] A. K. Mirmostafaee, M. S. Moslehian,: Fuzzy approximately cubic mappings, Inf. Sci. 178 (2008),
[5] C. Felbin, Finite dimensional fuzzy normed linear space. Fuzzy Sets Syst. 48 (1992), 239-248.
[6] Ch. Park, Hyers-Ulam-Rassias stability of homomorphisms in quasi-Banach algebras. Bull. Sci .Math., 132 (2008), 87-96.
[7] D. H. Hyers, On the stability of the linear functional equation, Proc. Natl. Acad. Sci. USA 27 (1941), 222-224. [8] D. H. Hyers, On the stability of the linear functional equation. Proc. Nat.
Acad. Sci. U.S.A., 27 (1941),
[9] D. Mihet, The fixed point method for fuzzy stability of the Jensen functional equation. Fuzzy Sets Syst. 160 (2009), 1663-1667.
[10] D. Mihet, V. Radu, On the stability of the additive Cauchy functional equation in random normed spaces, J. Math. Anal. Appl. 343 (2008), 567-572.
[11] J. M. Rassias, Solution of the Ulam stability problem for quartic mappings, J. Ind. Math. Soc. 67 (2000), 169-178.
[12] K. W. Jun, H. M. Kim, The generalized Hyers-Ulam-Rassias stability of a cubic functional equation, J. Math. Anal. Appl. 274 (2002), 267-278.
[14] T. Aoki, On the stability of the linear transformation in Banach spaces. J. Math. Soc. Japan, 2 (1950), 64-66. [15] T. Bag, Samanta, S. K.: Fuzzy bounded linear operators. Fuzzy Sets Syst. 151
(2005), 513-547.
[16] T.Bag and S. K.Samanta,: Finite dimensional fuzzy normed linear spaces. J. Fuzzy Math., 11(3) (2003), 687-705.
[17] T.Z.Xu, On fuzzy approximately cubic type mapping in fuzzy Banach spaces,Inf. Sci. 278 (2014), 56-66. [18] T. Aoki, On the stability of the linear transformation in Banach spaces, J. Math. Soc.
Japan, 2 (1950), 64-66. [19] V. Radu,:The fixed point alternative and the stability of functional equations. Sem. Fixed Point Theory, 4
(2003), 91-96. | {"url":"https://1library.net/document/yn46vvpz-stability-mixed-cubic-quartic-functional-equation-banach-spaces.html","timestamp":"2024-11-14T05:45:49Z","content_type":"text/html","content_length":"155774","record_id":"<urn:uuid:c746581e-0a2c-42de-91e9-ede1ccf47e63>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00528.warc.gz"} |
On Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order Stochastic Differential Equations
Ganiyu A. A.;^1 , Kayode S. J.;^2 , Augustine A. C.;^2 & & Fakunle I.^3
^1Adeyemi Federal University of Education, Ondo State, Nigeria
*Corresponding Author Email: ganiyuaa@aceondo.edu.ng …
This paper examines the approximate solution of general first order stochastic differential equations (SDEs). Two different methods of solution were considered. They include Drift-implicit
Euler-Maruyama method (DIEMM) and full-implicit Euler-Maruyama method (FIEMM). The two methods were adapted from Explicit Euler-Maruyama method (EEMM). Two problems in the form of first order SDEs
considered are Black-Scholes option price model (BSOPM) with a drift and without a drift function. The absolute errors were calculated using the exact solution and numerical solution for stepsizes 2^
-4, 2^-5, 2^-6, 2^-7, 2^-8, 2^-9. Comparison in performance of the methods was achieved using mean absolute error criterion. The mean absolute error was then used to determine the order of
convergence for each method. The order of convergence obtained for DIEMM and FIEMM was compared with that obtained using EEMM. The results showed that the performance of EEMM was better than DIEMM
while the performance of FIEMM was better than that of EEMM and DIEMM for first problem. It was noted that the order of convergence of EEMM and DIEMM are approximately the same in the second problem.
This can be associated with the absence of Drift function in second problem. However, FIEMM outperformed EEMM and DIEMM. This is because its order of convergence was less than that of EEMM and DIEMM
for second problem. Also, the graphical solutions were constructed for each method can also be identified for stepsize 2^-4.
Keywords: Stochastic Differential Equations, Itô Lemma, Explicit Euler-Maruyama Method, Drift-Implicit Euler-Maruyama method, Full-Implicit Euler-Maruyama method, Wiener process, Black Scholes Option
Price Model, Mean Absolute Error, Order of Convergence.
1. Introduction
Modeling physical systems using Ordinary Differential Equations (ODEs) overlooks stochastic effects. Incorporating random elements into these equations results in Stochastic Differential Equations
(SDEs), with the term “stochastic” referring to noise (Rezaeyan and Farnoosh, 2010). A first-order Stochastic Differential Equation is an equation of the specified form.
where f:[0,T]×〖R^n⟶R〗^n is a drift function.
Equation (1.0) can be written as
Where g:[0,T]×〖R^n⟶R〗^(n×m)is the diffusion function. The noise in equation (1.1) is generally called Gaussian white noise. It is expressed as , where is the Wiener process. For the properties of
Wiener process see Higham (2001) and Williams (2006) in Ganiyu et al (2015).
Equation (1.1) can be written as
Integrating (1.2) from 0 to t we have
The first integral on the right-hand side of equation (1.3) is known as the Riemann integral, while the second is referred to as the Itô integral or stochastic integral. Numerous researchers have
conducted studies on Stochastic Differential Equations (SDEs) of the type described in form (1.2). Among them are Platen (1992), Oksendal (1998), Higham (2001), Burrage et al. (2000), Burrage (2004),
Richardson (2009), Anna (2010), Razaaeyan and Farnoosh (2010), Fadugba et al. (2013), Bokor (2003), Sauer (2013), Kayode and Ganiyu (2015), Kayode et al. (2016), Ganiyu et al. (2018), and Ganiyu et
al. (2021a).
The aim of this paper is to find the numerical solution of stochastic differential equations using two methods. The methods are drift-implicit Euler-Maruyama and full-implicit Euler-Maruyama methods.
The objectives are; to use each of the method aforementioned to determine the approximate solution of two stochastic differential equations used in option pricing, obtain the absolute error for each
of the method from the corresponding exact solution and numerical solution for stepsizes 2^-4, 2^-5, 2^-6, 2^-7, 2^-8, 2^-9. to compare the performance of the methods using mean absolute error
criterion and to determine the accuracy of each method using strong order of convergence property.
2. Research Methodology
Numerous methodologies exist for solving SDE (1.2), including the Euler-Maruyama method, Milstein method, explicit strong order one Runge-Kutta method, Heun method, and others. For solving SDE (1.1),
the Drift-implicit Euler-Maruyama method (DIEMM) and Full-implicit Euler-Maruyama method (FIEMM) were employed, as utilized by Wang and Liu (2009). Their outcomes were benchmarked against results
from the Explicit Euler-Maruyama method (EEMM), referenced in the studies by Kayode et al. (2016) and Ganiyu et al. (2018). Additionally, Higham (2001) applied the EEMM in the context of an
autonomous system of first-order stochastic differential equations.
The EEMM derived by Kayode et al (2016) is of the form
According to Wang and Liu (2009), the EEMM in equation (2.1) can made implicit by introducing implicitness in the term δtf (ꞇ[j], X[j]) giving rise to drift-implicit Euler-Maruyama method (DIEMM)
By introducing implicitness in the second function at the right hand side of equation (2.2), this gives full-implicit Euler-Maruyama method (FIEMM).
2.1 Implementation of the Methods
The method in equation (2.1) was considered by Higham (2001) for EEMM using backward difference. In this paper, we shall apply the two methods (2.2) and (2.3) to SDE (1.2) using discritised interval
[0,Ͳ] as 0< ꞇ[1]< ꞇ[2]<…<ꞇ[j+1]=Ͳ. Let δt=Ͳ/L be the stepsize defined as δt=ꞇ[j+1],- ꞇ[j], where N are some integer and ꞇ[j]=jδt. The δt-space path increment dW[j]=W[j]– W[j-1] will be approximated
by summing the underlying dt -space increments as established by Higham (2001) using
Wiener increment dW will be generated in MATLAB over the space intervals by using dW:=sqrt (dt)*rand (1,N). For computational purpose, we shall assume that R=1, Dt=R*dt and L=N/R The exact and
numerical solution will be obtained using MATLAB software program.
2.2 Mean Absolute Error Criterion
In assessing the accuracy of any numerical method, it is essential to consider the properties of the solution produced by such a method. A key property of SDEs is the convergence of the solution
method employed. The convergence issue of SDEs has been explored by numerous researchers, including Higham (2001), Burrage (2004), Beretta et al. (2000), Lactus (2008), Sauer (2013), Fadugba et al.
(2013), Kayode and Ganiyu (2015), Kayode et al. (2016), Ganiyu et al. (2018), Ganiyu et al. (2021a), and Ganiyu et al. (2021b), among others.
The convergence of a method of solution of SDEs depends on the magnitude of the mean or expected value of the absolute error being measured. This can be defined as follows. Suppose a stochastic
differential equation of the form (1.2) is given. Suppose further that X(t) and X^h(t) (where h=δt is the stepsize) represent the true solution and numerical approximation of the SDE respectively,
the absolute error ‘E’ is defined by
The absolute error in any experiment shall be investigated by chosen the stepsize δt=2^-4
The mean absolute error (MAE) E^h is defined by
where E represent the mean or the expected value.
The Strong order of convergence (SOC) of each method is determined by considering the mean absolute error for six stepsizes, 2^-4, 2^-5, 2^-6, 2^-7, 2^-8, 2^-9.
Remark 2.1
The accuracy of a solution method is gauged by its Mean Absolute Error (MAE). From a list comparing the MAE values of different methods, the one with the lowest MAE is considered to be the most
accurate. Additionally, the performance and accuracy of a solution method can be evaluated by identifying which method exhibits the lowest Strong Order of Convergence (SOC).
3. Solution of First Order Stochastic Differential Equations Using Drift-Implicit and Full Implicit Euler-Maruyama Methods
In this section, two problems in the form of first order stochastic differential equation (1.1) will be considered. The two methods (2.2) and (2.3) will be applied to find the approximate solution of
the SDEs. The targeted problems are stated below.
Problem 1
where σ=0.0002 and μ=0.0001 are arbitrary values.
The exact solution of the SDE (3.1) is
Problem 1 is the Black-Scholes option price model with a drift function μX(t) and diffusion function σX(t). The problem was also used by Higham (2001) and Sauer (2013).
Problem 2
where σ=0.0001 is an arbitrary value.
The exact solution of the SDE (3.3) is
Problem 2 is the Black-Scholes option price model without a drift function but with diffusion function σX(t). In carrying out our numerical experiment, the stepsizes considered are 2^-4, 2^-5, 2^-6,
2^-7, 2^-8, 2^-9. It is assumed that X[0]=1 for each of the problem.
The numerical solution of the given problems in (3.1) and (3.3) can then be determined using DIEMM and FIEMM. For the study of solution to first order Stochastic Differential Equations using Explicit
Euler-Maruyama Method (EEMM) for Problem 1 as well as simulation curve associated with the method see Ganiyu et al (2018).
3.1 Solution of First Order Stochastic Differential Equations using Drift Implicit Euler-Maruyama Method (DIEMM) For Problem 1
Applying DIEMM of equation (2.2) to problem1 gives
Table 1: Result of using DIEMM (3.5) for solution of Problem 1 with h= 2^-4
The mean absolute error (E^h) is 7.484290742709731e-010.
Table 1 shows the exact solution and numerical solution of problem 1 using DIEMM with stepsize 2^-4. The mean absolute error for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be similarly
Figure 1 shows the sample path of the exact solution and numerical solution of problem 1 using DIEMM. The graphical solution for other stepsizes, 2^-6, 2^-7, 2^-8, 2^-9 can be obtained in a similar
3.2 Solution of First Order Stochastic Differential Equations using Full-Implicit Euler-Maruyama Method (FIEMM) for Problem 1
Applying FIEMM of equation (2.3) to problem 1 gives
Table 2: Result of using FIEMM (3.6) for solution of Problem 1 with h= 2^-4
The mean absolute error (E^h) is 4.074086257244148e-009.
Table 2 shows the exact solution and numerical solution of problem 1 using FIEMM with stepsize 2^-4. The mean absolute error for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be similarly
Figure 2 shows the sample path of the exact solution and numerical solution of problem 1 using FIEMM. The graphical solution for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be obtained in a
similar manner.
3.3 Solution of First Order Stochastic Differential Equations using Drift Implicit Euler-Maruyama Method (DIEMM) For Problem 2
Applying DIEMM of equation (2.2) to problem 2 gives
Table 3: Result of using DIEMM (3.7) for solution of Problem 2 with h= 2^-4
The mean absolute error (E^h) is 1.162394613896112e-009.
Table 3 shows the exact solution and numerical solution of problem 2 using DIEMM with stepsize 2^-4. The mean absolute error for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be similarly
Figure 3 shows the sample path of the exact and numerical solution of problem 2 using DIEMM. The graphical solution for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be obtained in a similar
Remark 3.1
The result obtained using DIEMM and EEMM (See Ganiyu et al (2018) for problem 2 are the same, the reason for this is that there exist no drift function to be made implicit in the method.
3.4 Solution of First Order Stochastic Differential Equations using Full-Implicit Euler-Maruyama Method (FIEMM) For Problem 2
Applying FIEMM of equation (2.3) to problem 2 gives
Table 4: Result of using FIEMM (3.8) for solution of Problem 2 with h= 2^-4
The mean absolute error (E^h) is 4.588471202993105e-009.
Table 4 above shows the exact and numerical solution of problem 2 using FIEMM with stepsize 2^-4. The mean absolute error for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be similarly determined.
Figure 4 shows the sample path of the exact and numerical solution of problem 2 using FIEMM. The graphical solution for other stepsizes 2^-5, 2^-6, 2^-7, 2^-8, 2^-9 can be obtained in a similar
4. Comparison of Mean Absolute Error (MAE) of Explicit, Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order SDEs of Problem 1 (P1)
Stepsize Explicit EMM P1 Drift Implicit EMM P1 Full Implicit EMM P1
Kayode and Ganiyu (2015)
2^-4 6.36669439e-010 7.48429074e-010 4.07408626e-009
2^-5 1.04921388e-009 1.04294593e-009 4.48671372e-009
2^-6 9.69581149e-010 9.70649827e-010 4.40708724e-009
2^-7 6.30707042e-010 5.55700275e-010 4.06850471e-009
2^-8 2.52163668e-010 2.21823759e-010 3.67923086e-009
2^-9 1.14792709e-010 9.61350666e-011 3.54315376e-009
4.1 Comparison of Strong Order Convergence of Explicit, Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order SDEs of Problem 1.
Method Order of Convergence P1 Residual P1
Explicit EMM Kayode and Ganiyu (2015) 0.54710253 1.09749078
Drift Implicit EMM 0.63736669 1.03290638
Full Implicit EMM 0.05660868 0.13288578
4.2 Comparison of Mean Absolute Error (MAE) of Explicit, Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order SDEs of Problem 2 (P2).
Stepsize Explicit EMM P2 Ganiyu et al (2018) Drift Implicit EMM P2 (New Method1) Full Implicit EMM P2 (New Method2)
2^-4 1.16239446e-9 1.16239461e-009 4.58847120e-009
2^-5 1.25768166e-9 1.25768170e-009 4.69481000e-009
2^-6 1.07798138e-9 1.07798177e-009 4.51515269e-009
2^-7 6.09374184e-10 6.09373751e-010 4.04685462e-009
2^-8 2.23838359e-10 2.23838781e-010 3.62954590e-009
2^-9 1.05161935e-10 1.05163089e-010 3.52669949e-009
4.3 Comparison of Strong Order Convergence of Explicit, Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order SDEs of Problem 2
Method Order of Convergence P2 Residual P2
Explicit EEMM Ganiyu et al (2018) 0.73216346 0.85947327
Drift Implicit EMM 0.73216104 0.85946835
Full Implicit EMM 0.09057947 0.09205844
5. Discussion
In this paper, we have explored two methods for solving general first-order stochastic differential equations (SDEs): the two methods are Drift Implicit Euler Maruyama method (DIEMM) and full
implicit Euler Maruyama method (FIEMM). Each of the method was used to determine the numerical solution of two problems used in option pricing. The first problem is with a drift function and the
second is without a drift function. The exact solutions of the given stochastic differential equations were determined. This provided the opportunity to obtain the absolute error ‘E’ at time t=[0,Τ]
, where T=1. The mean absolute error E^h of each method was calculated to compare the performance of the methods. The mean absolute errors were used to determine the order of convergence of each
method. This had provided the opportunity to determine the accuracy of the methods. The results obtained were used to compare the results obtained using explicit Euler Maruyama method (EEMM) by
Kayode and Ganiyu (2015) and Ganiyu et al (2018). Our analysis revealed that the order of convergence of the EEMM is less than that of the DIEMM, while the FIEMM’s order of convergence is lower than
both the EEMM and DIEMM for the problem 1. However, for the problem 2, the order of convergence of the EEMM closely matches that of the DIEMM, with the FIEMM’s order of convergence still being the
lowest. Graphical representations of each method’s solutions were also generated for stepsize 2^-4.
We have analyzed two scenarios involving first-order stochastic differential equations (SDEs). The first scenario addresses the Black-Scholes option pricing model incorporating a drift function,
while the second scenario considers the same model but without a drift function. To solve these SDEs, we applied two distinct methodologies: the drift-implicit Euler-Maruyama method (DIEMM) and the
full-implicit Euler-Maruyama method (FIEMM). The effectiveness of these methods was evaluated by calculating the absolute errors between the exact solutions and the numerical solutions derived using
the aforementioned methods. To compare the performance of the methods, the mean absolute error for each method was obtained using stepsizes 2^-4, 2^-5, 2^-6, 2^-7, 2^-8, 2^-9. The results showed that
the accuracy of EEMM is better than that of DIEMM because the order of convergence of EEMM is less than that of DIEMM, while that of FIEMM is better than that of EEMM and DIEMM because its order of
convergence (OOC) is less than that of EEMM and DIEMM for problem 1. However, the order of convergence of EEMM is approximately the same as that of DIEMM, while that of FIEMM is more accurate than
EEMM and DIEMM because the OOC is less than that of EEMM and DIEMM for problem 2. It can be concluded that the performance of EEMM is better than that of DIEMM, while the performance of FIEMM is
better than that of EEMM and DIEMM for problem 1. The order of convergence of EEMM and DIEMM being approximately equal can be associated with the absence of Drift function in problem 2. However,
FIEMM outperformed EEMM and DIEMM since its order of convergence was less than that of EEMM and DIEMM for problem 2.
The graphical solutions of each method were constructed for stepsize 2^-4. It can be observed that the graphical solutions of DIEMM and FIEMM are almost the same for problem 1. Similarly, that of
DIEMM and FIEMM is almost the same too for problem 2. This confirms the statement credited to Wang and Liu (2009) that there is no simple stochastic counterpart of the Eular method; that is, the
method fails because, for example Ε|(1-μδt-σδW[j] )^-1|=+∞ for a linear SDE in equation (3.1), nevertheless, the treatment could be to look at a higher-order explicit Strong method like the Milstein
method and try to introduce implicitness there (this is for future research).
Conflict of Interest
The research was completed with no conflict of interest.
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About this Article
Cite this Article
Ganiyu A. A., Kayode S. J., Augustine A. C., & Fakunle I. (2024). On Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order Stochastic Differential Equations. In K. S.
Adegbie, A. A. Akinsemolu, & B. N. Akintewe (Eds.), Exploring STEM frontiers: A festschrift in honour of Dr. F. O. Balogun. SustainE.
Ganiyu A. A., Kayode S. J., Augustine A. C., & Fakunle I. 2024. “On Drift-Implicit and Full-Implicit Euler-Maruyama Methods for Solution of First Order Stochastic Differential Equations.” In
Exploring STEM Frontiers: A Festschrift in Honour of Dr. F.O. Balogun, edited by Adegbie K.S., Akinsemolu A.A., and Akintewe B.N., SustainE.
Corresponding Author Email: famuagunks@aceondo.edu.ng
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An analysis of evolutionary-based sampling methodologies
A common approach for solving simulation-driven engineering problems is by using metamodel-assisted optimization algorithms, namely, in which a metamodel approximates the computationally expensive
simulation and provides predicted values at a lower computational cost. Such algorithms typically generate an initial sample of solutions which are then used to train a preliminary metamodel and to
initiate optimization process. One approach for generating the initial sample is with the design of experiment methods which are statistically oriented, while the more recent search-driven sampling
approach invokes a computational intelligence optimizer such as an evolutionary algorithm, and then uses the vectors it generated as the initial sample. Since the initial sample can strongly impact
the effectiveness of the optimization process, this study presents an extensive comparison and analysis between the two approaches across a variety of settings. Results show that evolutionary-based
sampling performed well when the size of the initial sample was large as this enabled a more extended and consequently a more effective evolutionary search. When the initial sample was small the
design of experiments methods typically performed better since they distributed the vectors more effectively in the search space.
Original language English
Title of host publication New Developments in Evolutionary Computation Research
Publisher Nova Science Publishers, Inc.
Pages 183-213
Number of pages 31
ISBN (Electronic) 9781634635257
ISBN (Print) 9781634634939
State Published - 1 Jan 2015
• evolutionary algorithms
• expensive optimization problems
• metamodelling
• sampling methods
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A non-Euclidean characterization of convexity
A set X in a real vector space is convex if for every x, y in X the segment xy = {λx + (1−λ)y : 0 ≤ λ ≤ 1} is included in X. There are ways to extend the concept of convexity from vector spaces to
metric spaces based on the translation of the "segment" concept: for instance, a set X in some suitable metric spaces is geodesically convex if for any pair of points x, y in X the geodesic between x
and y is also in X. We explore an alternate extension of the concept of convexity that does not involve any kind of segment construct. The following considerations motivate the definition.
Consider some set X in R^n (say R^2 for simplicity of display) composed of several irregular components. Imagine now that this set X is made of some radioactive material so that it creates a
radiation wavefront propagating from X outwards:
If seen from a sufficiently long distance, the radiation wavefront is indistinguishable from that of a convex set Y with the same exterior shape as X:
So, the convex hull of X (or something very similar to it, as we will see later) can be reconstructed from its radiation wavefront at some distance r by retropropagating the wavefront towards X. Let
us see a graphical example where X consists of three points and the wavefront w is considered at a distance r larger than the diameter of X:
Retropropagating w is equivalent to considering the wavefront at distance r generated by the exterior of w:
The resulting shape includes X and approximates the convex hull of X (in this case, the triangle with vertices in X) as r grows. So, we have a characterization of convexity resorting only to our
wavefront construction, which can be formalized by means of metric space balls:
Definition. Let S be some metric space and X a set of points in S. The wavefront hull of X, H[wave](X), is defined as
H[wave](X) = U[r≥0]B[r](B[r](X) \ S) \ S,
) is defined as the union of all
closed balls
of radius
centered at points in
Theorem. For any X in R^n
H[wave](X) = X U int(H[convex](X)),
where int(A) denotes the interior set of A and H[convex](X) is the convex hull of X.
Corollary. cl(H[convex](X)) = ∩[r][>0]H[wave](B[r](X)), where cl(A) denotes the topological closure of A.
We will prove the theorem and corollary in a later entry.
1 comment :
1. Ya he visto tu blog, muy interesante.
Por cierto, el otro día estuve pensando en el teorema López-García de desgajamiento de yogures y se cumple también en las tres dimensiones(nº de cortes = nº total de elementos -1), ponte a ver si
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Probability Chart
PLEASE GO BACK AND USE THE BIG BLUE 'PRINT' BUTTON ON THE PAGE TO PRINT THE WORKSHEET CORRECTLY!
Sorry for the trouble! The browser won't print the embedded worksheet PDF directly using the normal 'Print' command in the file menu, so you need to click the big 'Print' button to send just the
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Probability Chart
Probability anchor chart for word problem reference! This illustrated chart describes scenarios with coins, dice and playing cards. It includes odds for most likely and least likely outcomes.
Probability Chart for Word Problems
Probability problems can be tricky for kids, and many of the devices we use to communicate probability may initially be unfamiliar to kids. We take for granted that kids understand the idea of
tossing a coin or picking a specific card from a deck of playing cards, however many kids will be unfamiliar with some of these common activities.
Probability for Coin Tosses
A common type of probability word problem involves calculating the odds of results from multiple coin tosses. The probability chart on this page breaks down how many possible outcomes there are from
a given number of coin tosses and gives the odds of a specific sequence of heads or tails outcomes occurring. It also discusses probabilities where a series of coin tosses might generate an outcome
regardless of the order of the results.
Probability for Rolling Dice
Many word problems involve calculating the probability of rolling a pair of six sided game dice. This probability chart gives the probability of all of the sums you can roll with a pair of dice. The
chart illustrates the dice as a single white die and a single red die to emphasize the way the probability is calculated. Some students may be confused by the idea that rolling a three and a four is
distinct from rolling a four and three (and therefore increases the odds of rolling a sum of seven.) To help explain this, the two colored die can help kids understand that order of the rolls and the
individual die play a role (pardon the pun) in how the probabilities stack up.
You can further elaborate on this by explaining how getting a result on one die, and then rolling the second die, creates a distinct set of outcomes.
Probability for Playing Cards
Playing cards are a common feature of many probability word problems, but you may be surprised at how many kids haven't been exposed to a traditional deck of 52 cards at home. This probability chart
breaks down the composition of a deck of cards can gives probability for individual cards, face cards, suits and more. | {"url":"https://dadsworksheets.com/printables/probability-chart.html","timestamp":"2024-11-10T21:29:16Z","content_type":"text/html","content_length":"98061","record_id":"<urn:uuid:cc1c1259-e052-45d5-8047-353c0001327a>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00367.warc.gz"} |
Partnership Business - ConceptEra
Q1. I and my friend Mala together have started a business with capitals of ₹ 15000 and ₹ 25000 respectively. If the make a profit of ₹ 16,800 in a year, let us see what the profit share shall we each
Q2. Priyam Supriya and Bulu have opened a small shop of grocery shop with capitals of ₹ 15000, ₹ 10000, and ₹ 25000 respectively. But after a year there was a loss of ₹ 3000. Let us write by
calculating what each must pay to make up the loss.
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to pay to make up that loss to undisturb their capitals. Let us write by calculating the amount they have to pay.
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of the profit equally among themselves and the remaining in the ratio of their capital, let us find the profit share of each.
Q6. Three friends have started a business by investing ₹ 8000, ₹ 10000 and ₹ 12000 respectively. They also took an amount as bank loan. At the end of year, they made a profit of ₹ 13400. After paying
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capitals and repaid back their individual loans amount to the bank. Let us write by calculating the amount of their individual share and the ratio of their shares.
Q8. Three friends invested ₹ 1200, ₹ 15000 and ₹ 11000 respectively to purchase a bus. The first person is a driver and the other two are conductors. They decided to divide 2/5^th of the profit among
themselves in the ratio of 3:2:2 according to their work and the remaining in the ratio of their capitals. If they earn ₹ 29260 in one month, let us find the share of each of them.
Q9. Pradipbabu and Aminabibi started a business by investing ₹ 24,000 and ₹ 30,000 respectively at the beginning of a year. After 5 months Pradipbabu invested the capital of ₹ 4000 more. If the
yearly profit was ₹ 27716, let us write by calculating the share of each of them.
Q10. Niyamat chacha and karabi didi have started a partnership business together by investing ₹ 3000 and ₹ 50000 respectively. After 6 months Niyamat chacha has invested of ₹ 4000 more but Karabi
didi ha a withdrawn ₹ 10000 for personal need. If the profit at the end of the year is ₹ 19000, Let u write by calculating the profit share of each of them.
Q11. Srikant and Saiffuddin invested ₹ 40,000 and ₹ 300.000 respectively at the beginning of the year to purchase a mini bus to run it on a route. After 4 months, their friend peter joined them with
a capital of ₹ 81000. Srikant and Soiffuddin have withdrawn that money in the ratio of their capital. Let uc write by calculating the share of each if they make a profit of ₹ 39150 at the end of the
Q12. Arun and Ajoy started a business jointly by investing ₹ 24000 and ₹ 30000 respectively at the beginning of the year. But after a few months Arun invested ₹ 12000 more. After a year, the profit
was ₹ 14030 and Arun received the profit share of ₹ 7130. Let us find after how months did does Arun invest money in that business.
Q13. Three clay modelers from Kumartuli collectively took a loan of ₹ 100000 from a co-operative bank to set up a modelling workshop. They made a construct that after paying back the annual back
instalment of ₹ 28100, they would divide half of the profit among themselves in terms of the number of working days and the other half will be equally divided among them. Last year they worked 300
days, 275 days and 350 days respectively and made a profit of ₹ 1391oo. Let us write by calculating the share of each in this profit.
Q14. Two friends invested ₹ 40000 and ₹ 50000 respectively to start a business. They made a contract that they would divide 50% of the profit equally among themselves and the remaining the profit in
the ratio of their capitals. Let us write the share of profit of the first friend if it is ₹ 800 less than that of the 2^nd friend.
Q15. Puja, Uttam and Meher started a partnership business with a capital of ₹ 5000, ₹ 7000 and ₹ 10000 respectively with the conditions that
(i)Monthly expense the running business is ₹ 125 (ii) Puja and Uttam each will get ₹ 200 the keeping the accounts. If the profit is ₹ 6960 at the end of the year, let us write by calculating the
profit share each would get. | {"url":"https://conceptera.in/partnership-business/","timestamp":"2024-11-06T20:58:29Z","content_type":"text/html","content_length":"202450","record_id":"<urn:uuid:53ed428a-23ec-42b6-a40f-fd54e0c3cd84>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00774.warc.gz"} |
Outline of a Shape
The algorithm for drawing the Outline of a Shape is as follows:
1) The Destination starts filled with whatever background we need to draw on top of. The distance buffer is assumed to be initially filled with 'infinite' (in practice, any sufficiently large value
will do).
2) The algorithm continues by following the Trajectory specified by the outline. This is done in discrete steps or jumps. At each step, the pen is moved to a new point in the trajectory, jumping by a
fraction of a pixel. A step of about 1/4 of a pixel is small enough. This is done using float (or fixed point) arithmetic.
3) For each step, the pen is at some (float) point. The possibly affected pixels are in the circle defined by the pen outer diameter (w+r). For each of these pixels, the Euclidean Distance from the
pixel center to the pen center is computed. This distance is stored in the distance buffer if it is less that the value already there.
4) When the pen has finished following the prescribed trajectory, the distance buffer contains, for each pixel, the minimum distance d to the trajectory. The opacity value is obtained by evaluating
the opacity function of the filtering pen at distance d. This opacity value is used to alpha-blend the shape color over the destination frame buffer. Afterwards, the distance buffer is filled with
infinite, to be ready for the next call.
[…] Figure 5. Each pixel is painted with the opacity obtained by taking its distance to the path, and using it as the argument to the prefiltering pen function. To illustrate, for some pixels, the
small black arrows show this distance to the path. As a further example, for the two arrows closest to the bottom of the figure, the distance is projected (dashed lines) over the filtering pen
function, to show how to obtain the opacity values used to paint the pixels. The same procedure is used for all pixels.
Here's a possible implementation of the algorithm in JavaScript:
function drawOutline(destination, trajectory, penOuterDiameter, filteringPen) { const width = destination.width; const height = destination.height; const distanceBuffer = []; for (let i = 0; i <
width * height; i++) { distanceBuffer[i] = Number.POSITIVE_INFINITY; } for (let i = 0; i < trajectory.length - 1; i++) { const start = trajectory[i]; const end = trajectory[i + 1]; const step = (end
- start) / 4; let current = start; while (current <= end) { const x = Math.round(current.x); const y = Math.round(current.y); const r = penOuterDiameter / 2; for (let i = Math.max(0, x - r); i <=
Math.min(width - 1, x + r); i++) { for (let j = Math.max(0, y - r); j <= Math.min(height - 1, y + r); j++) { const distance = Math.sqrt((i - x) ** 2 + (j - y) ** 2); if (distance < distanceBuffer[i +
j * width]) { distanceBuffer[i + j * width] = distance; } } } current = current.add(step); } for (let i = 0; i < width * height; i++) { const x = i % width; const y = Math.floor(i / width); const
opacity = filteringPen(distanceBuffer[i]); const index = (x + y * width) * 4; destination.data[index + 3] = opacity * 255; } }
The graphics engines commonly used to draw vector graphics apply the antialiasing technique known as Pixel Coverage.
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Normal Distribution: Meaning, Examples and Uses in Economics (2024)
Did you know that nearly 99.7% of data points in a normal distribution are within three standard deviations of the mean? This fundamental aspect of the normal distribution, or Gaussian distribution,
highlights its significance in statistical analytics and econometric models. Also known for its symmetric bell-shaped curve, this distribution is key for understanding natural occurrences.
The essence of the normal distribution lies in its mean and standard deviation. The mean showcases the data’s central point, whereas the standard deviation reflects how spread out the values are.
This setup allows it to closely fit diverse data types, serving as a bedrock for economists, econometricians and data scientists.
Therefore, a grasp of the normal distribution is indispensable for professionals dealing with analytical data or economic studies. It stands as a cornerstone in the realm of statistics and
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What is Normal Distribution?
A normal distribution, also known as a Gaussian distribution, encompasses a probability framework symmetric about the mean. This symmetry implies that data points clustering around the mean are more
common. Its significance in statistical analysis is underpinned by its frequent occurrence in nature. Moreover, its fundamental properties aid in simplifying the interpretation of complex data sets.
A normal distribution represents a continuous probability structure for real-valued random variables in the realm of probability theory and statistics. The standard variant, or standard normal
distribution, is characterized by a central mean of 0, a variance and a standard deviation of 1. Furthermore, a mere two parameters, namely the mean and standard deviation, succinctly capture the
essence of a normal distribution. Moreover, the central limit theorem underscores that the averages of independently and identically distributed variables tend to approximate a normal distribution.
The figure above shows the shape of a Standard Normal Distribution which has a mean 0 and standard deviation of 1.
The hallmark of a normal distribution is its symmetry, ensuring zero skewness. It has a kurtosis value of 3.0, depicting a mesokurtic form. Distributions deviating from this value exhibit varying
kurtosis characteristics, being either leptokurtic or platykurtic. Such symmetry allows the mean and standard deviation to precisely identify the data’s central tendencies and variabilities. Notably,
roughly 68.27% of data fall within one standard deviation, 95.45% within two, and 99.73% within three standard deviations from the mean in normal distribution.
Its application spans diverse arenas, notably in technical financial analysis, owing to its symmetrical and simplistic nature. It also finds wide use in Statistics and Econometric techniques.
Characteristics of Normal Distribution
The normal distribution is defined by its mean and standard deviation. They elucidate the probability distribution along with the center and spread of the data set.
The mean serves as the central spindle of the normal distribution, encapsulating where data concentrates. Therefore, it essentially defines the middle point. Adjusting the mean shifts the curve
without altering its symmetry as shown in the figure below. In the context of a standard normal distribution shown in maroon colour with a mean of zero, half of the data points fall below it. This
feature is pivotal for assessing central tendencies. The Normal Distributions in the figure have the same standard deviation of 1, but different means of -2, 0 and 2.
Standard Deviation
Standard deviation measures how spread out data points are from the mean, thus influencing the curve’s breadth. About 68.27% of the data falls within one standard deviation from the mean, with
increasing coverage at further deviations. In the figure below, we can observe 3 different normal distributions which have the same mean of 0. However, their standard deviations are different at 0.5,
1 and 2 which also changes their shape. The Standard Normal Distribution is shown in maroon with a mean of 0 and a standard deviation of 1.
Bell Curve
The bell curve, an iconic graphical depiction, serves to outline the normal distribution’s attributes. This symmetrical plot, with its mean-centered shape, is widely lauded for its capacity to
represent statistical data with clarity. Its utility extends across various fields, also aiding in the understanding of data distribution.
Visual Representation
The alignment of the curve’s peak with the mean, median, and mode allows for the division of data into mirror-image sections. By doing so, it simplifies the comprehension of both the spread and
central values of a dataset. In the context of the standard normal distribution, specific percentages of data concentrate within defined standard deviations from the mean. In the figures shown above,
each normal distribution curve is characterised by mean = median = mode.
Normal Distribution Formula: Probability Density Function
The normal distribution formula or normal probability density function lies at the heart of statistical analysis, serving as a key method for computing probabilities within economic distributions. It
is characterized by a probability function utilizing the mean (μ) and standard deviation (σ) to determine the likelihood f(x) of a variable x. Mathematically, this relationship is expressed as:
Furthermore, through z-score standardization, analysts simplify the use of normal distributions. This process mitigates the need for intricate cumulative probability computations, enhancing
The versatility of the normal distribution formula also extends to various transformation applications. For example, it facilitates the conversion of any normal distribution into the standard form
using Z = (X – μ) / σ. This standardization enables expedited cumulative probability determination for specific value intervals through reference to standardized tables.
Furthermore, the formula’s utility includes approximating the binomial distribution conditionally. Such approximations also simplify otherwise complex statistical analyses, rendering them more
accessible and relevant in economic contexts. Mastery of the normal distribution formula empowers economists and financial experts to derive significant insights from raw data. This significantly
boosts the accuracy and impact of their analytical and decision-making processes.
Normal Cumulative Distribution Function
The normal cumulative distribution function (Normal CDF) is fundamental in statistical analysis, portraying the likelihood of a normally distributed random variable falling within a specific
interval. It is a key element in a myriad of statistical contexts, facilitating extensive analysis.
The normal cumulative distribution function (CDF) emerges from the integral of the probability density function (PDF) of a normal distribution. Specifically, for the standard normal distribution,
Normal CDF or Φ(x) signifies the probability that the standard normal variable Z is less than or equal to a given value of x. This is expressed mathematically by the equation:
The first figure shows the Normal CDF of a Standard Normal Distribution which has a mean of 0 and standard deviation of 1. In the second figure, we have shown CDFs of Normal distributions with
different means and standard deviations. The one in orange is the CDF of Standard Normal Distribution. The Normal CDF in dark green has a mean of -1 and a standard deviation of 0.5. Finally, the plot
in dark purple has a mean of 0 same as the Standard Normal, but a higher standard deviation of 2.
This Normal CDF also has its applications in Statistics, Economics and Econometrics. For example, Normal CDF is used as a link function in the Probit Model with a qualitative dependent variable.
The Central Limit Theorem
The Central Limit Theorem (CLT), a cornerstone in probability theory, asserts that the sample means’ distribution transforms into a normal distribution with expanding sample sizes. This phenomenon of
convergence is agnostic to the original distribution of the population and is contingent upon independent sample selection and a finite population variance, crucial for diverse statistical
investigations in disciplines like economics and finance.
Essentially, the CLT posits that large, random samples from any population culminate in the sample means assuming a normal distribution, despite the variance in original population distributions. Its
historical lineage traces back to early 19th-century rudimentary forms, evolving into a definitive structure by 1920, further amalgamating classical and modern probability theory. A critical tenet of
the CLT elucidates how random fluctuations, surrounding a fixed parameter, trend towards a Gaussian distribution as the sample size increases.
The Empirical Rule in Normal Distribution
68-95-99.7 Rule
The empirical rule, known as the 68-95-99.7 rule, also aids in comprehending data spread in a normal distribution. It articulates that a significant portion of data falls within certain deviations
from the mean: approximately 68.27% within one standard deviation, 95.45% within two, and 99.73% within three standard deviations. Moreover, its application extends to predictive analytics and
thorough risk evaluation activities.
The 68-95-99.7 rule ensures, therefore, that a vast majority of the dataset is encapsulated within three standard deviations of the average. It significantly simplifies the determination of event
probabilities and their associated likelihoods, bolstering its role in analytical scenarios.
Extensively employed in sectors reliant on meticulous risk assessment and foresight analytics, the rule is indispensable. In the field of finance, it also facilitates the assessment of market
instability through the computation of standard deviations. For logistics, it is further instrumental in approximating delivery timelines. Its widespread application exemplifies its pivotal role in
performing efficient data analysis.
While financial market data might not adhere strictly to a normal distribution, the use of standard deviation remains paramount in the estimation of financial risks and market fluctuations. Moreover,
proficiency in applying these empirical principles equips professionals with the tools they need. This includes making data-informed decisions, maintaining quality surveillance, and forecasting
results based on probabilistic evaluations.
Skewness in Normal Distribution
Skewness in a normal distribution signifies the extent of its asymmetry from the ideal bell curve symmetry. A normal distribution serves as a paramount case of symmetry, therefore, holding a skewness
value of zero. This also implies equilibrium around its central mean.
Conversely, within distinct distributions, we also unearth disparate skewness characterizations. For example, the double exponential’s skewness registers around 0, stipulating symmetry whereas an
exponential distribution has a skewness of 2.
Skewness wields critical implications in data analysis, especially within the economic and financial realms. A symmetric, zero-skewness normal distribution stands as a foundational model for economic
prediction under complete symmetry. Real data items, however, typically manifest skewness, positive or negative, affecting a distribution’s tail lengths asymmetrically. This departure from ideal
symmetry impacts financial prognosis and peril evaluation models, hence, necessitating skewness consideration within data scrutiny.
Kurtosis in Normal Distribution
Kurtosis delves into the distribution’s tail characteristics, offering insight into the data’s extremities and outliers. These characteristics include mesokurtic, platykurtic, and leptokurtic, each
illuminating different data density profiles. Mesokurtic distributions, for example, mirror the normal curve with a kurtosis of approximately 3.
Leptokurtic distributions, with kurtosis above 3, signify an increased amount of outliers and thicker tails. Therefore, instances like the Student’s t-distribution highlight this phenomenon.
Conversely, the term platykurtic describes distributions with kurtosis below 3, implying slender tails with fewer extreme values.
For investors, understanding kurtosis is pivotal to accurately gauge the risk in their investments, with high kurtosis further signalling a greater risk of extreme price movements. In the financial
sector, professionals may rely on kurtosis for assessing the probability of these extreme events, also aiding in the formation of risk mitigation strategies. Analyzing a distribution’s kurtosis
enables these experts to foresee the presence of fat tails accurately and make informed risk management decisions.
Examples of Normal Distribution
Grasping the manifestation of normal distribution in our day-to-day experiences also significantly enhances the linkage between theoretical and practical contexts. Hence, let us consider 2 examples
of a normal distribution:
Height and Weight
Classic examples of normal distribution include human height and weight variables. Suppose, the typical human height average approximates 175 cm. A normal distribution indicates that nearly 99.73% of
individuals fall within three standard deviations from this average. For illustration, in a hypothetical town with adult heights conforming to a normal distribution, where the mean is 175 cm and the
standard deviation is 10 cm, one would expect a minute fraction of the population, about 0.135%, to surpass 205 cm. If the town has a population of 330,000, this translates to roughly 446 individuals
above 205 cm. Such insights further underscore the robustness and suitability of normal distribution in capturing the variances in human physical traits.
IQ Scores
Intelligence Quotient (IQ) scores represent another paramount domain for normal distribution analysis. Suppose, the standard IQ mean stands at 100, encapsulating a majority of the global intelligence
metric. Approximately 68.27% of IQ scores encompass 1 standard deviation range from the mean, with 95.45% contained within a 2 standard deviations range, and nearly 99.73% lying within three
deviations. The culminated distribution depicts a recognizable bell curve, hence, embodying the application of normal distribution in the understanding of human cognitive prowess.
Role of Normal Distribution in Economics
Asset Prices
The normal distribution also plays a central role in economic analysis, specifically in market evaluation and asset pricing appraisal. Through its utilization, economists can apply the bell curve to
discern significant deviations in asset prices from the average. This method further unveils potential overvaluation or undervaluation scenarios, critical in investment decision-making.
Underpinning asset pricing mechanisms in financial markets is the supposition of returns conforming to a normal distribution. This paradigm simplifies return modelling and strategizing investments.
Within a normal distribution, the majority of readings reside close to the mean, outlining a standard by which to evaluate asset value.
Market Analysis
Crucial to market analysis is the normal distribution, applied to project future market shifts and trends. However, it is imperative to acknowledge the inadequacy of this assumption alone, as
real-world market returns often exhibit skewness and fat tails, signalling a departure from the norm. While foundational, reliance solely on the normal distribution proves insufficient. This further
necessitates a blend of approaches and models to account for these irregularities in market dynamics and asset valuations.
Normal Distribution in Econometrics
Econometrics is also deeply grounded in the Gaussian distribution, essential for evaluating economic relationships and hypothesis testing. It operates under the assumption that regression model
errors conform to a normal distribution, facilitating more precise economic parameter inferences and reliable future trend predictions. The Gaussian distribution’s symmetries and the empirical rule’s
application cement its utility in econometric analyses.
Hence, the normal distribution is vital in econometrics for handling random variables in economic models. This simplification comes from the central limit theorem, which states that the average of
large random samples with a finite mean and variance will tend to a normal distribution as the sample size increases. This theorem is foundational, also supporting a multitude of econometric methods
and analyses.
Econometric techniques often employ normal distribution tables for probability calculations and inferences. These tables match Z-values with associated probabilities, further easing complex
statistical work. The distribution’s unique features, like a symmetrical peak with the mean, median, and mode aligned, further enhance its utility and practicality in econometrics.
In assessing regression residuals, economists leverage the Gaussian distribution framework. If the residuals’ normality is confirmed, it verifies the model’s assumptions and the validity of the
economic relationships under scrutiny. Hence, this step is pivotal in ensuring trustworthy and meaningful economic forecasts and policy decisions.
Normal Distribution in Other Fields
Economics and Econometrics are not the only fields where normal distribution is extensively employed. Normal distribution also forms the basis for hypothesis testing in clinical trials, an essential
part of medicine and biostatistics. Moreover, in population studies, it is used to model attributes like height, weight, and blood pressure. This enables accurate forecasts of various demographic
Hence, the normal distribution is a fundamental aspect of statistical analysis and economic modelling. Its symmetrical, bell-shaped curve, characterized by mean and standard deviation, provides a
powerful means for understanding and predicting data trends. Economists also rely on this distribution to model various economic phenomena, given its nature where the majority of observations cluster
around the mean.
An essential feature of the normal distribution is its adherence to the 68-95-99.7 rule. This rule stipulates that a significant portion of data, around 68.27%, falls within one standard deviation of
the mean. Moreover, 95.45% lie within two standard deviations, and an overwhelming 99.73% within three standard deviations. Such predictability simplifies decision-making processes for economists and
data analysts in various fields.
Nonetheless, the utility of the normal distribution faces challenges, particularly in finance. Market prices’ fluctuation can lead to distributions with skewness and kurtosis beyond its capability.
In light of these inaccuracies, there’s a requirement for economists and financial analysts to explore alternative modelling frameworks, to ensure a better fit for their datasets.
In summary, it serves as a critical backbone for economics and econometric modelling. Its foundational nature simplifies the interpretation of complex data, enhancing the clarity of economic theories
and forecasts. Yet, due to its limitations, which are often glaring in the financial sector, a nuanced and cautious approach is crucial for its application in decision-making and analysis.
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Ecological Equivalence: A Realistic Assumption for Niche Theory as a Testable Alternative to Neutral Theory
Hubbell's 2001 neutral theory unifies biodiversity and biogeography by modelling steady-state distributions of species richness and abundances across spatio-temporal scales. Accurate predictions have
issued from its core premise that all species have identical vital rates. Yet no ecologist believes that species are identical in reality. Here I explain this paradox in terms of the ecological
equivalence that species must achieve at their coexistence equilibrium, defined by zero net fitness for all regardless of intrinsic differences between them. I show that the distinction of realised
from intrinsic vital rates is crucial to evaluating community resilience.
Principal Findings
An analysis of competitive interactions reveals how zero-sum patterns of abundance emerge for species with contrasting life-history traits as for identical species. I develop a stochastic model to
simulate community assembly from a random drift of invasions sustaining the dynamics of recruitment following deaths and extinctions. Species are allocated identical intrinsic vital rates for neutral
dynamics, or random intrinsic vital rates and competitive abilities for niche dynamics either on a continuous scale or between dominant-fugitive extremes. Resulting communities have steady-state
distributions of the same type for more or less extremely differentiated species as for identical species. All produce negatively skewed log-normal distributions of species abundance, zero-sum
relationships of total abundance to area, and Arrhenius relationships of species to area. Intrinsically identical species nevertheless support fewer total individuals, because their densities impact
as strongly on each other as on themselves. Truly neutral communities have measurably lower abundance/area and higher species/abundance ratios.
Citation: Doncaster CP (2009) Ecological Equivalence: A Realistic Assumption for Niche Theory as a Testable Alternative to Neutral Theory. PLoS ONE 4(10): e7460. https://doi.org/10.1371/
Editor: Stephen J. Cornell, University of Leeds, United Kingdom
Received: May 6, 2009; Accepted: September 25, 2009; Published: October 14, 2009
Copyright: © 2009 C. Patrick Doncaster. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original author and source are credited.
Funding: This work was supported by grant NE/C003705/1 from the UK Natural Environment Research Council. The funders had no role in study design, data collection and analysis, decision to publish, or
preparation of the manuscript.
Competing interests: The author has declared that no competing interests exist.
Hubbell's 2001 neutral theory (HNT) unifies the disciplines of biodiversity and biogeography by modelling steady-state distributions of species richness and relative species abundance across
spatio-temporal scales [1]. Surprisingly accurate predictions have issued from its core premise that all species are exactly identical in their vital rates. As a null hypothesis to explain what
should be observed if all species were perfectly equal with respect to all ecologically relevant properties, it has proved hard to refute [2]. Yet no ecologist, including Hubbell, believes that
species are equivalent in reality [3], [4]. The challenge presented by HNT is to justify invoking anything more complex than ecological drift to define community structure [5]. Its extravagant
simplicity has had an explosive impact on ecology (>1100 citations, rising exponentially), because it appears to discount 100 years of traditional conventions on niche differentiation. If
biodiversity encompasses the great richness of differently attributed species that constitutes the natural world, how can ecological equivalence yield such predictive power about the numbers of
species [6]? If HNT is based on a ludicrous assumption [7], then our conceptual understanding is thrown into disarray by its fit to empirical patterns [8]. Here I explain this paradox in terms of the
ecological equivalence realised by coexisting species at demographic equilibrium. Analyses and simulations of coexistence equilibria demonstrate the emergent property of ecological equivalence
amongst species with a rich diversity of attributes, leading to novel predictions for a quantifiable gradation in species-area relationships between neutral and niche models.
A neutral model of empirical relationships eliminates “the entire set of forces competing for a place in the explanation of the pattern” [9]. Accordingly, HNT assumes that all species behave
identically in a zero-sum game such that the total density of individuals in a trophically similar community remains constant regardless of species composition. The defining image of this ecological
equivalence is a tropical forest canopy, with remarkably constant total densities of trees regardless of large regional variations in constituent species [1]. Interpretations of zero-sum equivalence
routinely omit to distinguish between the equal vital rates realized at the system carrying capacity approximated in this image (and most datasets), and the intrinsic vital rates that define the
heritable character traits of each species. Models of HNT consistently prescribe identical intrinsic rates and niche dimensions. Hubbell [1] anticipated the disjuncture between realized and intrinsic
rates by comparing ecological equivalence to the fitness invariance achieved at carrying capacity, allowing for different trade-off combinations in life-history traits. The prevailing convention,
however, remains that ecological equivalence explicitly requires symmetric species with identical per capita vital rates, thereby promulgating the notion that HNT is built on an unrealistic
foundation [3].
Theoretical studies have sought various ways to reconcile neutral patterns with niche concepts. Intrinsically similar species can coexist under niche theory [7], and niches add stabilizing mechanisms
that are absent under the fitness equivalence of intrinsic neutrality [10]. Comparisons of niche to neutral simulations in a saturated system of fixed total abundance have shown that they can predict
similar species-abundance distributions and species-area relationships [11], demonstrating that neutral patterns need not imply neutral processes [12]. Even neutral processes of intraspecific
competition and dispersal limitation cannot be distinguished in principle for species-abundance predictions [13]–[16]. Here I use an analysis and simulation of Lotka-Volterra dynamics to model
zero-sum ecological drift as an emergent property of stochastic niche structures at dynamic equilibrium. I explain its appearance in the steady-state distributions even of extremely dissimilar
species in terms of the trivial expectation that species must achieve ecological equivalence at their coexistence equilibrium, which is defined by equal realised fitness for all. Although the
predictions are standards of Lotka-Volterra analysis for a homogeneous environment, they drive a simulation that for the first time spans across dispersal-limited neutral to stochastic niche
scenarios without fixing the total abundance of individuals.
The neutral simulation developed here is consistent with the models of Solé et al. [17] and Allouche & Kadmon [18] in having total species, S, abundance of individuals, N, and zero-sum dynamics as
emergent properties (in contrast to refs [1], [11], [12], [19]). The S species are identical in all respects including interspecific interactions equal to intraspecific (in contrast to refs [13],
[16]). Non-neutral simulations developed here extend the model of Chave et al. [11] by allowing competitive differences to vary stochastically on a continuous scale, as in Purves & Pacala [12]. They
extend both these models by allowing pre-emptive recruitment and emergent zero-sum dynamics, and the model of Calcagno et al. [20] by adding dispersal limitation. They are consistent with Tilman's
niche theory [21], [22] in their population abundances being a function of species-specific vital rates.
These simulations confirm the previously untested prediction [12] that colonization-competition trade-offs with stochastic colonization will exhibit zero-sum ecological drift and produce rank
abundance curves that resemble neutral drift. Truly neutral dynamics should nevertheless sustain a lower total density of individuals at density-dependent equilibrium. This is because intrinsically
identical species must interact as strongly between as within species. They therefore experience no competitive release in each others' presence, contrasting with the net release to larger
populations obtained by segregated niches. The simulations demonstrate this fundamental difference, and I discuss its use as a signal for dynamic processes when predicting species-area relationships.
Analysis of abundance patterns for two-niche communities
Species characterized by extremely different intrinsic attributes can achieve ecological equivalence in a zero-sum game played out at dynamic equilibrium. Take for example a two-species community
comprising a dominant competitor displacing the niche of a fugitive (e.g., [23]). The fugitive survives even under complete subordination, provided it trades competitive impact for faster growth
capacity [24]. Figure 1 illustrates the equal fitness, zero-sum outcome at density-dependent equilibrium under this most extremely asymmetric competition. The carrying capacity of each species is a
function of its intrinsic lifetime reproduction (detailed in Methods Equation 1), and equilibrium population sizes are therefore a function of the species-specific vital rates. Regardless of
variation in the ratio of dominant to fugitive carrying capacities, 0≤k[D] / k[F]≤1, the system density of individuals is attracted to the stable equilibrium at N=n[F] + n[D]=k[F]. Knocking out
the fugitive reduces N to the smaller k[D], but only until invasion by another fugitive. This may be expected to follow rapidly, given the fugitive characteristic of fast turnover. The steady-state
scenario is effectively neutral by virtue of the dominant and fugitive realising identical vital rates and constant total density at their coexistence equilibrium despite contrasting intrinsic
(heritable) rates. The reality that species differ in their life history traits therefore underpins the assumption of ecological equivalence, which then permits fitting of intrinsically neutral
models with vital rates set equal to the realised rates. In the next section, these predictions are extended to simulate the drift of species invasions that sustains the dynamics of recruitment
following deaths and extinctions amongst multiple species of dominants and fugitives.
With competition coefficients α[DF]=0, α[FD]=1, the fugitive persists provided it has the greater carrying capacity: k[F]/k[D]>1. (A) Lotka-Volterra phase plane with steady-state abundance at the
intersection of the isoclines for the fugitive (dashed line) and the dominant (solid line). (B) Equilibration of abundances over time given by Runge-Kutta solutions to Equation 1, with a 20% drop in
the dominant's intrinsic death rate, d[D], imposed at t=3 (equivalent to a rightward shift in its isocline) to illustrate the constancy of N=n[F]+n[D].
The same principle of trade-offs in character traits conversely allows a sexually reproducing species to withstand invasion by highly fecund asexual mutants [25], [26]. A two-fold advantage to the
mutant in growth capacity resulting from its production of female-only offspring is cancelled by even a small competitive edge for the parent species (Fig. 2). Sexual and asexual types coexist as
ecological equivalents to the extent that each invades the other's population to symmetric (zero) net growth for all. Although the dynamics are not zero-sum if the mutant has some competitive impact
on the parent species, they approach it the higher the impact of parent on mutant and the faster its growth capacity (albeit half the mutant's). Attributes such as these accommodate greater
similarity between the types in their carrying capacities and competitive abilities, which aligns the two isoclines. A consequently reduced stability of the coexistence equilibrium may result in the
sexual parent ousting the asexual mutant over time, for example if the latter accumulates deleterious mutations [26], [27].
With the mutant having identical vital rates except for twice the intrinsic propagation rate per capita: b[M]=2⋅b[P], the parent species persists if α[PM]<k[P]/k[M]. (A) Phase plane. (B)
Equilibration of abundances over time given by Equation 1, with a 50% drop in the parent's intrinsic death rate imposed at t=3 to illustrate approximate constancy of N=n[M]+n[P].
These local-scale dynamics apply equally at the regional scale of biogeography, reconfiguring individual death as local extinction, and birth as habitat colonization [24]. Equally for regional as for
local scales, rate equations take as many dimensions as species in the community, with their coupling together defining niche overlap [24], [28]. Coexistence of the species that make up a community
is facilitated by their different heritable traits, which is a fundamental premise of niche theory. Ecological equivalence, and hence modelling by neutral theory is nevertheless possible by virtue of
the coexistence equilibrium levelling the playing field to zero net growth for all.
The above examples of dominant versus fugitive and sexual versus asexual were illustrated with models that gave identical realised rates of both birth and death at coexistence equilibrium. Fitness
invariance and zero-sum dynamics, however, require only that species have identical net rates of realised birth minus death. The simulations in the next section show how neutral-like dynamics are
realised for communities of coexisting species with trade-offs in realized as well as intrinsic vital rates.
Comparison of simulated neutral and multi-niche communities with drift
Figure 3 illustrates the species-abundance distributions and species-area relationships of randomly assembled S-species systems under drift of limited immigration and new-species invasions (protocols
described in Simulation Methods). From top to bottom, its graphs show congruent patterns between an intrinsically neutral community with identical character traits for all species (equivalent to
identically superimposed isoclines in Figs-1 and -2 models), and communities that trade growth capacity against competitive dominance increasingly starkly. The non-neutral communities sustain more
total individuals and show greater spread in their responses, reflecting their variable life-history coefficients. Their communities nevertheless follow qualitatively the same patterns as those of
neutral communities. For intrinsically neutral and niche-based communities alike, Fig. 3 shows species-abundance distributions negatively skewed from log-normal (all P<0.05, every g[1]<0), and an
accelerating decline in rank abundances of rare species (cf. linear for Fisher log-series) that is significantly less precipitous than predicted by broken-stick models of randomly allocated
abundances amongst fixed S and N; Fig. 4 shows constant densities of total individuals regardless of area (unambiguously linear), and Arrhenius relationships of species richness to area
(unambiguously linear on log-log scales).
From top to bottom, graphs show average patterns for intrinsically neutral, Lotka-Volterra, and dominant-fugitive communities. SADs each show mean ± s.e. of six replicate communities with carrying
capacity K=1000 habitable patches. Frequencies are compared to log-normal (left-hand column) and MacArthur's broken-stick (right-hand column). See Methods for input parameter values and the process
of random species assembly.
SARs each show mean ± s.e. of three replicate communities. See Methods for input parameter values and the process of random species assembly.
The extended tail of rare species seen in the Fig.-3 species-abundance distributions is caused by single-individual invaders replacing random extinctions of n-individual species. Further trials
confirm that reduced dispersal limitation exacerbates the negative skew from the log-normal distribution, while sustaining a higher total density of individuals. The extinction-invasion imbalance
sets the equilibrium species richness, S, as a power function of total population density, N. This can be expressed as the Arrhenius relationship: S=cK^ z (Fig. 4 right-hand column) by virtue of
the zero-sum relation of N to K (Fig. 4 left-hand column). Supporting Text S1 provides a full analysis of the departure from MacArthur's broken-stick model, and the derivations of the Arrhenius c and
z. Further simulations show that reduced dispersal limitation raises c and reduces z, and a higher rate of new-species invasions raises c (though not z, in contrast to predictions from spatially
explicit neutral models [29]).
The closely aligned proportionality of total individuals to habitable area for all communities illustrates emergent zero-sum dynamics for neutral and non-neutral scenarios (Fig. 4 left-hand column).
Despite sharing this type of pattern, and rather similar densities of species (Fig. 4 right-hand column), the non-neutral communities sustain more than double the total individuals. This difference
is caused by a more than halving of their competition coefficients on average (all α[ij]=1 for neutral, mean α[ij] (i≠j)=0.45 for Lotka-Volterra, mean ratio of 0∶1 values=58∶42 for
dominant-fugitive). The zero-sum gradient of N against K is simply the equilibrium fraction of occupied habitat, which is 1–1/R for a closed neutral scenario, where R is per capita lifetime
reproduction before density regulation (b/d in Methods Equation 1 [23], [24]). The closed dominant-fugitive scenario modelled in Fig. 1 has a slope of k[F]/K=(1–1/R)/α, where R and α are system
averages. Further simulation trials show the slope increasing with immigration, for example by a factor of 1.9 between closed and fully open (dispersal unlimited) Lotka-Volterra communities.
Dispersal limitation therefore counterbalances effects of the net competitive release obtained in niche scenarios from α[ij]<1 (as also seen in models of heterogeneous environments [19]).
The less crowded neutral scenario sustains a somewhat higher density of species than non-neutral scenarios (comparing Fig. 4 z-values for right-hand graphs), and consequently it maximizes species
packing as expressed by the power function predicting S from N in Fig. 5. With no species intrinsically advantaged in the neutral scenario, its coefficient of power is higher than for pooled
non-neutral scenarios (0.594 and 0.384 respectively, log-log covariate contrasts: F[1,42]=122.72, P<0.001). The lower coefficients of Lotka-Volterra and dominant-fugitive scenarios are further
differentiated by competitive asymmetry (0.412 and 0.355 respectively, F[1,42]=7.24, P<0.01). In effect, the neutral scenario has the lowest average abundance of individuals per species, n, for a
community of size K with given average R, which is also reflected in the modal values in Fig. 3 histograms for K=1000 patches.
Each point shows the mean ± s.e. of the three replicate communities in Fig. 4, and regression lines on the means are the power functions for intrinsically neutral (top) Lotka-Volterra (middle) and
dominant-fugitive (lower) scenarios.
The lower N and n predicted for the intrinsically neutral scenario point to a detectable signal of steady-state intrinsically neutral dynamics: α=1 for all, because intrinsically identical species
cannot experience competitive release in each others' presence (cf. α[ij]<1 in niche models). These interactions may be measurable directly from field data as inter-specific impacts of equal
magnitude to intra-specific impacts; alternatively, Lotka-Volterra models of the sort described here can estimate average competition coefficients at an observed equilibrium N, given an average R (a
big proviso, as field data generally measure realised rather than intrinsic vital rates). This distinction of intrinsically neutral from non-neutral dynamics has been masked in previous theory by the
convention for neutral models either to fix N [1], [11], [12] or to set zero interspecific impacts [13,16). By definition, identical species cannot be invisible to each other unless they are
invisible to themselves, which would require density independent dynamics. Simulations of non-interacting species under density-dependent regulation therefore embody an extreme version of niche
theory whereby each species occupies a unique niche, somehow completely differentiated by resource preferences rather than partially by trade-offs in vital rates. These models fit well to species
abundance distributions in rainforests and coral reefs [13]–[16], though without providing any explanation for what attributes would allow each species to be invisible to all others (in contrast to
the trade-off models). Indeed the condition is unrealistic at least for mature trees that partition a homogeneous environment by each making their own canopy. This so-called neutral scenario ([13],
[16], more appositely a neutral-niche scenario) has no steady state outcomes in the analyses and simulations described here, because setting all α[ij]=0 (i≠j) allows indefinite expansion of S and
hence also of N. A slightly less extreme neutral-niche community is modelled by setting all interspecific impacts to a common low value. Simulations at α[ij]=0.1 for all i≠j give a zero-sum
relation N=4.026K, which has >4-fold steeper gradient than that for the Lotka-Volterra scenario (Fig. 4) reflecting its >4-fold reduction in α and consistent with its representation of a highly
niched scenario.
Although intrinsic identity is clearly not a necessary condition of ecological equivalence or of zero-sum abundances at dynamic equilibrium, only neutral models sustain these outcomes over all
frequencies. It is their good fit to steady-state patterns of diversity and abundance even for communities subject to species turnover in ecological drift that has argued powerfully for niche
differences having a limited role in community structure. The Fig.-3 simulations reveal these types of patterns to be equally well represented by niche models, however, despite constituent
individuals and species achieving fitness equivalence only at dynamic equilibrium. Non-neutral dynamics of a mature community express the community-wide average of fluctuations either side of
equilibrium. Outcomes regress to the equilibrium mean for a random assembly of species undergoing stochastic extinctions of rare members, regulated by spatially autocorrelated immigration, and
replacement by initially rare invaders. The predicted power of neutral theory can be taken as evidence for ecological equivalence at the coexistence equilibrium of species with more or less different
intrinsic attributes.
Modelling zero-sum ecological drift as an emergent property reveals a key distinguishing feature of truly neutral communities. Their intrinsically identical species self-regulate to a lower total
density as a result of inter-specific impacts equalling intra-specific impacts. Any empirical test for competitive release is therefore also a test for niche structure. For example, removing habitat
is predicted to give a relative or absolute advantage to species towards the fugitive end of a dominant-fugitive spectrum, which may be picked up in correlated life-history traits for winners or
losers under habitat loss or degradation [23], [24]. In contrast, neutral dynamics lead to sudden biodiversity collapse at a system-wide extinction threshold of habitat [17]. The extinction threshold
of habitat for a resource-limited metapopulation is set by the fraction 1/R [30], [31]. The value of R is thus an important yardstick of resilience in conservation planning. A neutral model fitted to
empirical zero-sum abundances will overestimate their community-wide R, and hence overestimate community resilience, if α[ij] are overvalued by setting all to unity. Likewise, a neutral model that
sets all α[ij]=0 (i≠j) will underestimate R, and hence resilience, if the α[ij] are undervalued by setting all to zero.
Ecological equivalence is a much more permissive requirement for neutrality than is currently acknowledged in theoretical developments on HNT. Coexistence equilibria largely achieve the
neutrality-defining mission, to eliminate all of the forces competing for a place in explanations of pattern. It remains an open question whether they do so best amongst species with most or least
competitive release in each others' presence (e.g., Fig. 1 versus Fig. 2 respectively, and Fig. 3 dominant-fugitive versus Lotka-Volterra respectively; [7], [10], [32]). Models need to incorporate
the ecologically realistic dynamics of interspecific interactions simulated here in order to explore the true nature of competitive release between extreme scenarios of niches that are all
intrinsically identical (HNT [1]) and intrinsically unique [13], [16]. Simulations of niches distributed along environmental gradients have found emerging groups of intrinsically similar species over
evolutionary timescales [33]. For the spatially homogeneous environments modelled here, competition-recruitment trade-offs will always sustain species differences. In their absence, however,
homogenous environments will tend to favour fast-recruiting competitive dominants. This species type may eventually prevail, with runaway selection checked by other forces such as predation, disease,
mutation accumulation and environmental variability. These systems would merit further study because many of their attributes could be those of intrinsically neutral dynamics.
The following protocols apply to simulations of single and multi-niche communities with density-dependent recruitment and density-independent loss of individuals. They produce the outcomes
illustrated in Figs 3– 5 from input parameters specified at the end of this Section. The general model has species-specific vital rates; the intrinsically neutral and dominant-fugitive scenarios are
special cases of this model, with constrained parameter values.
The community occupies a homogenous environment represented by a matrix of K equally accessible habitat patches within a wider meta-community of K[m] patches. The dynamics of individual births and
deaths are modelled at each time step by species-specific probability b of each resident, immigrant, and individual of new invading species producing a propagule, and species-specific probability d
of death for each patch resident. Recruitment to a patch is more or less suppressed from intrinsic rate b by the presence there of other species according to the value of α[ij], the impact of species
j on species i relative to i on itself, where the intraspecific impact α[ii]=1 always. A patch can be occupied by only one individual of a species, and by only one species unless all its resident α
[ij]<1. Conventional Lotka-Volterra competition is thus set in a metapopulation context by equating individual births and deaths to local colonisations and extinctions (following [24], consistent
with [20]). A large closed metapopulation comprising S species has rates of change for each species i in its abundance n[i] of individuals (or equally of occupied patches) over time t approximated
by:(1)This is the rate equation that also drives the dynamics of Figs 1 and 2, where k[i]=(1–d[i]/b[i])K. Coexistence of any two species to positive equilibrium n[1], n[2] requires them to have
intrinsic differences such that k[1]>α[12] k[2] and k[2]>α[21] k[1].
Each time-step in the simulation offers an opportunity for one individual of each of two new species to attempt invasion (regardless of the size of the meta-community). Each new species i has
randomly set competitive impacts with respect to each other resident species j, of α[ij] received and α[ji] imposed. It has randomly set b[i], and an intrinsic lifetime reproduction R[i]=b[i]/d[i]
that is stratified in direct proportion to its dominance rank amongst residents, obtained from its ranked mean α-received minus mean α-imposed. For example, an invader with higher dominance than all
of three resident species will have random R[i] stratified in the bottom quartile of set limits R[min] to R[max]. Communities are thereby structured on a stochastic life-history trade-off between
competitive dominance and population growth capacity. This competition-growth trade-off is a well-established feature of many real communities, which captures the fundamental life-history principle
of costly adaptations [11], [17], [21]. Its effect on the community is to prevent escalations of growth capacity or competitive dominance amongst the invading species. Neutral communities are a
special case, with identical values of b and R for all species and α=1 for all.
At each time step, new invaders and every resident each have species-specific probability b of producing a propagule. Each propagule has small probability ν of speciation (following [1]). The sample
community additionally receives immigrant propagules of its resident species that arrive from the wider meta-community in proportion to their expected numbers out there ([K[m]/K–1]n[i]), assuming the
same density n[i]/K of each species i as in the sample community, and in proportion to their probability (K/K[m]) of landing within the sample community, and modified by a dispersal limitation
parameter ω. In effect, for each resident species in the community, [(1–K/K[m]) n[i]]^1–ω external residents each produce an immigrating propagule with probability b[i]. Thus if K[m]≫K and ω=0, a
colonist is just as likely to be an immigrant from outside as produced from within the sample community (no dispersal limitation, following [1]). This likelihood reduces for ω>0, and also for smaller
K[m]. None of the propagules generated within the sample community emigrate out into the meta-community, making K a sink if smaller than K[m] (sensu [34]), or a closed community if equal to K[m]. The
simulation is thus conceptually equivalent to randomly assembled S-species systems previously studied (e.g., [35]), except that it additionally accommodates a random drift of invasions to sustain the
dynamics of recruitment following deaths and extinctions.
Each propagule lands on a random patch within the sample community and establishes there only if (a) its species is not already present, and (b) it beats each probability α[ij] of repulsion by each
other resident species j, and (c) it either beats the odds on repulsion by all other propagules simultaneously attempting to colonise the patch, or benefits from the random chance of being the first
arrival amongst them. Each pre-established resident risks death with species-specific probability d[i]=b[i]/R[i] at each time step. Each patch has probability X of a catastrophic hazard at each
time step that extirpates all its occupants. The model thus captures the principles of stochastic niche theory [21], [22] and pre-emptive advantage [20].
Each of the replicate communities contributing to distributions and relationships in Figs 3– 5 is represented by values averaged over time-steps 401–500, long after the asymptote of species richness.
For all graphs in Figs 3– 5, meta-community carrying capacity K[m]=10^6, dispersal limitation parameter ω=0.5, speciation probability per resident propagation event ν=10^−12, two invasion
attempts per time-step (setting Hubbell's [1] fundamental biodiversity number θ∼4 independently of K[m]), probability of catastrophe per patch X=0.01. For neutral communities, all species take
competition coefficients α=1, individual intrinsic propagation probability b=0.5, individual intrinsic lifetime reproduction R=1.5 (so lifespan R/b=3); for Lotka-Volterra communities, each
species i takes random 0≤α[ij]≤1, random 0≤b[i]≤1, R[i] between 1.2 and 1.8 and proportional to dominance rank; dominant-fugitive communities are as Lotka-Volterra except for random binary α[ij]=0
or 1. All scenarios are thereby sampled from a large meta-community with moderate dispersal limitation, low extrinsic mortality, and sufficient invasions to sustain a reasonably high asymptote of
species richness from the starting point of two species each occupying 5 patches. Skew in the lognormal distribution of species abundances (Fig. 3) was measured for each replicate in its
dimensionless third moment about the mean, g[1] [36], and confidence limits for the sample of six values were tested against H[0]: g[1]=0.
I thank Simon J. Cox for valuable discussions, and David Alonso and an anonymous referee for helping to sharpen my focus.
Author Contributions
Conceived and designed the experiments: CPD. Performed the experiments: CPD. Analyzed the data: CPD. Contributed reagents/materials/analysis tools: CPD. Wrote the paper: CPD. | {"url":"https://journals.plos.org:443/plosone/article?id=10.1371/journal.pone.0007460","timestamp":"2024-11-05T13:23:20Z","content_type":"text/html","content_length":"185630","record_id":"<urn:uuid:83f66c1d-13e5-4721-a815-6577ddfe7ded>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00704.warc.gz"} |
incomplete gamma function
The incomplete gamma function Γ(a,x) is a variation on the gamma function.
An alternative incomplete gamma function can be defined as the integral from 0 to x. It may be clear that this is the same function, apart from a constant.
The function can be normalized to the regularized incomplete gamma function Q(a,x) defined by: Γ(a,x) / Γ(a).
The incomplete gamma function and its inverse are used in statistics. | {"url":"https://www.2dcurves.com/gamma/gammagi.html","timestamp":"2024-11-13T19:34:56Z","content_type":"text/html","content_length":"2982","record_id":"<urn:uuid:9d715591-ea07-46ef-b944-88e62189eb19>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00351.warc.gz"} |
Introduction to Number Patterns - School Games
Introduction to Number Patterns
Fullscreen Mode
Number pattern activities.
A pattern of numbers is A series of numbers that follow A pattern. So let’s look at an example: 20 15 10 5 0. The numbers are reduced 5 Each this is the pattern.
Number patterns are sequences of numbers which follow a particular rule or pattern. They can be used to help students develop understanding of mathematical concepts including algebra and arithmetic.
In many different areas of mathematics Number patterns are found such as geometry trigonometry and calculus.
For identification and extension of number patterns it is important to understand the terms used to describe them. In a number pattern The first term is usually called an initial term and subsequent
terms are generated by applying a consistent rule or formula. The difference between consecutive terms is called The common difference in The pattern in some patterns The ratio of consecutive terms
is called The common difference.
Number pattern can be represented by different methods including tables graphs and algebraic expressions. Using The study of number patterns students can develop problem solving skills logical
reasoning and critical thinking skills. Students can apply this knowledge to solve complex mathematical problems in a variety of fields By understanding the underlying principles of number patterns. | {"url":"https://www.schoolgames.io/edu/introduction-to-number-patterns/3619400/","timestamp":"2024-11-03T10:25:38Z","content_type":"text/html","content_length":"157280","record_id":"<urn:uuid:e9b387d8-8f16-4a9f-b462-bc04340628a1>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00647.warc.gz"} |
Basic Geometry 1st Grade Math Practice Test | [Difficulty - Easy]
Basic Geometry Practice Test for 1st Grade – [Easy]
Updated on September 26, 2023
Welcome to Brighterly’s introduction to basic geometry for young mathematicians! Geometry isn’t just about complicated formulas and high-level calculations. For our 1st graders, it’s all about
recognizing and understanding the wonderful world of shapes that surrounds us every day.
Introduction to Shapes
The first thing you’ll notice when looking around you – from your favorite toys to the building blocks you play with – is that everything has a shape.
• Circles: Like the pizza you enjoy on weekends or the clock on the wall.
• Squares: Think of your beloved board games or certain puzzle pieces.
• Triangles: Seen in the pyramids in Egypt, if you’ve ever flipped through an math worksheets.
Recognizing these shapes and more helps us describe the world and even solve everyday problems!
Exploring Angles
Even in 1st grade, we can begin to understand the idea of angles. Think of an angle as a twist or a turn.
• When you open a book halfway, you form an angle.
• When you completely open a toy box’s lid, another angle is formed!
By understanding these tiny turns and twists, we can start building the foundation for more advanced geometry in the future. Wondering how? Check out Brighterly’s advanced geometry for older kids!
Why Geometry Matters
You might wonder, why should 1st graders bother with shapes and angles? Well, geometry is everywhere!
1. Building Skills: When you stack blocks or fit puzzle pieces together, you’re using geometric understanding.
2. Describing Things: It helps you describe things better. Instead of saying “that thing over there”, you can say “that round ball” or “that square box”.
3. Problem Solving: Believe it or not, geometry can help solve problems. For example, knowing how big a space is can help you figure out if your big teddy bear will fit in it!
Basic Geometry Practice Test
Get ready for math lessons with Brighterly! This easy-level assessment is specially designed for our budding mathematicians to confidently recognize and understand fundamental geometric concepts.
From circles to squares and angles in between, this test promises a blend of fun and learning.
1 / 20
Which shape looks like a stretched circle?
2 / 20
A flat surface of a 3D shape is called a:
3 / 20
Which of these shapes has no corners?
4 / 20
If a shape has 6 sides, it is called a:
5 / 20
A soccer ball is shaped like a:
6 / 20
Which shape has more sides: a triangle or a rectangle?
7 / 20
What shape is the base of a pyramid?
8 / 20
Which shape does NOT have straight sides?
9 / 20
Which shape has all equal sides and angles?
10 / 20
How many long sides does a rectangle have?
11 / 20
If you put two triangles together, you can make a:
12 / 20
Which shape rolls easily?
13 / 20
How many vertices does a rectangle have?
14 / 20
What is another name for a corner of a shape?
15 / 20
Which of these is round and not a polygon?
16 / 20
How many sides does a pentagon have?
17 / 20
What shape is like a flat ring?
18 / 20
How many corners does a square have?
20 / 20
Which shape looks like a box?
Poor Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Mediocre Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Needs Improvement
Start practicing math regularly to avoid your child`s math scores dropping to C or even D.
High Potential
It's important to continue building math proficiency to make sure your child outperforms peers at school. | {"url":"https://brighterly.com/tests/basic-geometry-practice-test-for-1st-grade-easy/","timestamp":"2024-11-06T09:26:45Z","content_type":"text/html","content_length":"233529","record_id":"<urn:uuid:f79c7e70-3349-4042-8a2c-f1e10229734e>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00045.warc.gz"} |
Free Space Path Loss
Click here to go to our decibel page
Click here to go to our main antenna page
New for May 2008! Let us know if this Rule of Thumb works for you, or if we screwed it up when we edited it.
Free space loss can be estimated in your head by:
22 dB for the first wavelength,
plus an additional 20xlog(number of wavelengths in the path)
According to Mike, another way of stating the rule is: 22 dB for the first wavelength plus 6 dB for each doubling of distance. Estimate how many doublings of the first wavelength are needed to arrive
at the required distance, then multiply this by 6 and add 22. Got that?
Frequency: 1 GHz
Distance: 1 km
Wavelength at 1 GHz is 30 cm. Take 22 dB for the first wavelength.
There's 100 cm in a meter and 1000 meters in a kilometer, so the distance is 100,000 cm
Number of wavelengths for a distance of 1 km:
100,000/30=3300 wavelengths
Now take "normal" dB of 3300
(actually, 10xlog of 1/3 is -4.77 dB but this is an approximation, right?)
Because of the 20 log.... the 35 dB becomes 70 dB.
Sum the two contributions to arrive at the estimated path loss at 1 GHz for 1km:
22 dB + 70 dB = 92 dB
With some decibel numbers in memory this estimate can be pretty easy.
Factor 2 = 3 dB
Factor 3 = 5 dB (4.77 dB)
Factor 10 = 10 dB
For a more accurate calculation of FSPL, go to our download center and loof for Tim's spreadsheet on this topic. | {"url":"https://www.microwaves101.com/encyclopedias/free-space-path-loss","timestamp":"2024-11-08T18:41:56Z","content_type":"application/xhtml+xml","content_length":"32094","record_id":"<urn:uuid:ecc18770-a7ae-459e-b83d-89f4b183b7e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00838.warc.gz"} |
Unscramble ABOITEAUX
How Many Words are in ABOITEAUX Unscramble?
By unscrambling letters aboiteaux, our Word Unscrambler aka Scrabble Word Finder easily found 98 playable words in virtually every word scramble game!
Letter / Tile Values for ABOITEAUX
Below are the values for each of the letters/tiles in Scrabble. The letters in aboiteaux combine for a total of 20 points (not including bonus squares)
• A [1]
• B [3]
• O [1]
• I [1]
• T [3]
• E [1]
• A [1]
• U [1]
• X [8]
What do the Letters aboiteaux Unscrambled Mean?
The unscrambled words with the most letters from ABOITEAUX word or letters are below along with the definitions.
• aboiteaux () - Sorry, we do not have a definition for this word | {"url":"https://www.scrabblewordfind.com/unscramble-aboiteaux","timestamp":"2024-11-06T23:17:33Z","content_type":"text/html","content_length":"57219","record_id":"<urn:uuid:624d6719-a859-4bd8-b8a4-ecdeb392a62c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00531.warc.gz"} |
CIE May 2020 9709 Mechanics Paper 41
CIE May 2020 9709 Mechanics Paper 41 (pdf)
1. Three coplanar forces of magnitudes 100 N, 50 N and 50 N act at a point A, as shown in the diagram. The value of cos α is 4/5.
Find the magnitude of the resultant of the three forces and state its direction
2. A car of mass 1800 kg is towing a trailer of mass 400 kg along a straight horizontal road. The car and trailer are connected by a light rigid tow-bar. The car is accelerating at 1.5 m s^−2. There
are constant resistance forces of 250 N on the car and 100 N on the trailer.
(a) Find the tension in the tow-bar
(b) Find the power of the engine of the car at the instant when the speed is 20 m s^−1
3. A particle P is projected vertically upwards with speed 5 m s^−1 from a point A which is 2.8 m above horizontal ground.
(a) Find the greatest height above the ground reached by P.
(b) Find the length of time for which P is at a height of more than 3.6 m above the ground
4. The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod. The rod is rough and the coefficient of friction between the ring and the rod is 0.8. A force of magnitude T N acts on
the ring in a direction at 30° to the rod, downwards in the vertical plane containing the rod. Initially the ring is at rest.
(a) Find the greatest value of T for which the ring remains at rest
(b) Find the acceleration of the ring when T = 3.
5. A child of mass 35 kg is swinging on a rope. The child is modelled as a particle P and the rope is modelled as a light inextensible string of length 4 m. Initially P is held at an angle of 45Å to
the vertical (see diagram).
(a) Given that there is no resistance force, find the speed of P when it has travelled half way along the circular arc from its initial position to its lowest point.
(b) It is given instead that there is a resistance force. The work done against the resistance force as P travels from its initial position to its lowest point is X J. The speed of P at its
lowest point is 4 m s^−1.
Find X.
6. A particle moves in a straight line AB. The velocity vm s^−1 of the particle ts after leaving A is given by v = k(t^2 − 10t + 21), where k is a constant. The displacement of the particle from A,
in the direction towards B, is 2.85 m when t = 3 and is 2.4 m when t = 6.
(a) Find the value of k. Hence find an expression, in terms of t, for the displacement of the particle from A
(b) Find the displacement of the particle from A when its velocity is a minimum
7. A particle P of mass 0.3 kg, lying on a smooth plane inclined at 30° to the horizontal, is released from rest. P slides down the plane for a distance of 2.5 m and then reaches a horizontal plane.
There is no change in speed when P reaches the horizontal plane. A particle Q of mass 0.2 kg lies at rest on the horizontal plane 1.5 m from the end of the inclined plane (see diagram). P
collides directly with Q.
(a) It is given that the horizontal plane is smooth and that, after the collision, P continues moving in the same direction, with speed 2 m s^−1.
Find the speed of Q after the collision
(b) It is given instead that the horizontal plane is rough and that when P and Q collide, they coalesce and move with speed 1.2 m s^−1.
Find the coefficient of friction between P and the horizontal plane
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"https://www.onlinemathlearning.com/may-2020-9709-41.html","timestamp":"2024-11-06T02:14:52Z","content_type":"text/html","content_length":"38143","record_id":"<urn:uuid:4088c1db-ea39-47e5-bc2a-47410ef22ac0>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00095.warc.gz"} |
Torque Calculator
Last updated:
Torque Calculator
This torque calculator helps you find the torque arising in a rotating object. What exactly is this torque? Imagine an object that can rotate around some point called the pivot point. If you exert a
force at some distance from the pivot point, then even though the force will act along a straight line, the object will begin to rotate. Continue reading if you want to learn how to calculate torque
and have the torque formula explained in detail.
If you're after torque in the context of the automotive industry, then the torque to hp calculator might be for you!
Torque equation
The torque (tendency of an object to rotate) depends on three different factors:
τ = r F sin(θ)
• r — Lever arm — the distance between the pivot point and the point of force application;
• F — Force acting on the object;
• θ — Angle between the force vector and lever arm. Typically, it is equal to 90°; and
• τ — Torque, whose units are newton-meters (symbol: N⋅m).
Imagine that you try to open a door. The pivot point is simply where the hinges are located. The closer you are to the hinges, the larger the force you must use. If you use the handle, though, the
lever arm will increase, and the door will open with less force exerted.
💡 Do not confuse this concept with the centrifugal force — the centrifugal force is an inertial force that pulls away from the pivot point, parallel to the lever arm. Such a force doesn't cause
torque (you can check it by substituting an angle of 0° into the torque formula).
How to calculate torque
1. Start with determining the force acting on the object. Let's assume that F = 120 N.
2. Decide on the lever arm length. In our example, r = 0.5 m.
3. Choose the angle between the force vector and lever arm. We assume θ = 90°, but if it is not equal to the default 90°, you can change its value.
4. Enter these values into our torque calculator. It uses the torque equation: τ = rFsin(θ) = 0.5 × 120 × sin(90°) = 60 N⋅m.
5. The torque calculator can also work in reverse, finding the force or lever arm if torque is given.
If you want to learn more about the concept of force and Newton's second law, try the acceleration calculator and the Newton's second law calculator.
How do I calculate torque?
To calculate torque, follow the given instructions:
1. Find out the magnitude of the applied force, F.
2. Measure the distance, r, between the pivot point and the point the force is applied.
3. Determine the angle θ between the direction of the applied force and the vector between the point the force is applied to the pivot point.
4. Multiply r by F and sin θ, and you will get the torque.
What is the SI unit of torque?
The SI unit of torque is the newton meter or N⋅m. To express torque in imperial units, we use pound-force foot or lbf⋅ft.
What is the dimensional formula for torque?
The magnitude of torque is equal to the product of the magnitude of the force and the lever arm. The dimensional formula for force is [M¹L¹T⁻²] while for a lever arm it is [L]. Hence, the dimensional
formula for torque is [M¹L²T⁻²].
How do I convert torque to lbf⋅ft from N⋅m?
As we know that, 1 pound-force (lbf) = 4.448 newton (N) and 1 foot (ft) = 0.3048 meter (m). Hence, To convert torque from N⋅m to lbf⋅ft divide by 1.355818 or multiply by 0.737562. | {"url":"https://www.omnicalculator.com/physics/torque","timestamp":"2024-11-03T22:30:23Z","content_type":"text/html","content_length":"426035","record_id":"<urn:uuid:3054061f-0c7f-4b6b-9c4d-702cff8b5852>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00212.warc.gz"} |
Introduction to Cryptography and Coding Theory
Practice Problems
This page generates arbitrarily many random practice problems for you!
Warning: These are the most basic computational problems. These are good for making sure you have your basic arithmetic and algorithms down. But they don’t (and can’t) test conceptual
understanding, and test problems may NOT resemble these.
Modular arithmetic.
This box will generate a random addition and multiplication problem.
This box will show the answers.
Complex Numbers
Miller-Rabin Primality Testing
These are not designed to be easy by hand, because the first successive squaring part might be annoying; you can use Sage to do the modular arithmetic for you.
RSA Factoring
This box produces RSA moduli that are good for testing your factoring methods on.
The following box will output the answer.
Elliptic curve addition.
This box will generate a random elliptic curve point addition problem.
This box will show the answer.
Elliptic Curve Diffie-Hellman Key Exchange
This box will provide you a key exchange question.
This box will show the answer.
Elliptic Curve Factoring
This box will generate an example where an elliptic curve computation will help you factor a number. It will typically ask you to multiply a point by 3 or 6. Think about how to do this efficiently!
(For 6, don’t just add the point to itself 5 times.)
The following box will output the answer.
Single Qubit States
This box will provide the answer. | {"url":"https://crypto.katestange.net/practice-problems/","timestamp":"2024-11-13T12:34:51Z","content_type":"text/html","content_length":"48955","record_id":"<urn:uuid:10bb304c-d15d-40f6-84de-620b60041ec1>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00529.warc.gz"} |
Calculate elastic constants
8.3.5. Calculate elastic constants
Elastic constants characterize the stiffness of a material. The formal definition is provided by the linear relation that holds between the stress and strain tensors in the limit of infinitesimal
deformation. In tensor notation, this is expressed as
\[s_{ij} = C_{ijkl} e_{kl}\]
where the repeated indices imply summation. \(s_{ij}\) are the elements of the symmetric stress tensor. \(e_{kl}\) are the elements of the symmetric strain tensor. \(C_{ijkl}\) are the elements of
the fourth rank tensor of elastic constants. In three dimensions, this tensor has \(3^4=81\) elements. Using Voigt notation, the tensor can be written as a 6x6 matrix, where \(C_{ij}\) is now the
derivative of \(s_i\) w.r.t. \(e_j\). Because \(s_i\) is itself a derivative w.r.t. \(e_i\), it follows that \(C_{ij}\) is also symmetric, with at most \(\frac{7 \times 6}{2}\) = 21 distinct
At zero temperature, it is easy to estimate these derivatives by deforming the simulation box in one of the six directions using the change_box command and measuring the change in the stress tensor.
A general-purpose script that does this is given in the examples/ELASTIC directory described on the Examples doc page.
Calculating elastic constants at finite temperature is more challenging, because it is necessary to run a simulation that performs time averages of differential properties. There are at least 3 ways
to do this in LAMMPS. The most reliable way to do this is by exploiting the relationship between elastic constants, stress fluctuations, and the Born matrix, the second derivatives of energy w.r.t.
strain (Ray). The Born matrix calculation has been enabled by the compute born/matrix command, which works for any bonded or non-bonded potential in LAMMPS. The most expensive part of the calculation
is the sampling of the stress fluctuations. Several examples of this method are provided in the examples/ELASTIC_T/BORN_MATRIX directory described on the Examples doc page.
A second way is to measure the change in average stress tensor in an NVT simulations when the cell volume undergoes a finite deformation. In order to balance the systematic and statistical errors in
this method, the magnitude of the deformation must be chosen judiciously, and care must be taken to fully equilibrate the deformed cell before sampling the stress tensor. An example of this method is
provided in the examples/ELASTIC_T/DEFORMATION directory described on the Examples doc page.
Another approach is to sample the triclinic cell fluctuations that occur in an NPT simulation. This method can also be slow to converge and requires careful post-processing (Shinoda). We do not
provide an example of this method.
A nice review of the advantages and disadvantages of all of these methods is provided in the paper by Clavier et al. (Clavier).
(Ray) J. R. Ray and A. Rahman, J Chem Phys, 80, 4423 (1984).
(Shinoda) Shinoda, Shiga, and Mikami, Phys Rev B, 69, 134103 (2004).
(Clavier) G. Clavier, N. Desbiens, E. Bourasseau, V. Lachet, N. Brusselle-Dupend and B. Rousseau, Mol Sim, 43, 1413 (2017). | {"url":"https://docs.lammps.org/Howto_elastic.html","timestamp":"2024-11-12T17:08:01Z","content_type":"text/html","content_length":"16711","record_id":"<urn:uuid:568e747d-97d9-4a9e-b6b2-c30b672d674b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00402.warc.gz"} |
Memory: How to Develop, Train and Use It
Chapter XIV
How to Remember Numbers
The faculty of Number–that is the faculty of knowing, recognizing and remembering figures in the abstract and in their relation to each other, differs very materially among different individuals. To
some, figures and numbers are apprehended and remembered with ease, while to others they possess no interest, attraction or affinity, and consequently are not apt to be remembered. It is generally
admitted by the best authorities that the memorizing of dates, figures, numbers, etc., is the most difficult of any of the phases of memory. But all agree that the faculty may be developed by
practice and interest. There have been instances of persons having this faculty of the mind developed to a degree almost incredible; and other instances of persons having started with an aversion to
figures and then developing an interest which resulted in their acquiring a remarkable degree of proficiency along these lines.
Many of the celebrated mathematicians and astronomers developed wonderful memories for figures. Herschel is said to have been able to remember all the details of intricate calculations in his
astronomical computations, even to the figures of the fractions. It is said that he was able to perform the most intricate calculations mentally, without the use of pen or pencil, and then dictated
to his assistant the entire details of the process, including the final results. Tycho Brahe, the astronomer, also possessed a similar memory. It is said that he rebelled at being compelled to refer
to the printed tables of square roots and cube roots, and set to work to memorize the entire set of tables, which almost incredible task he accomplished in a half day–this required the memorizing of
over 75,000 figures, and their relations to each other. Euler the mathematician became blind in his old age, and being unable to refer to his tables, memorized them. It is said that he was able to
repeat from recollection the first six powers of all the numbers from one to one hundred.
Wallis the mathematician was a prodigy in this respect. He is reported to have been able to mentally extract the square root of a number to forty decimal places, and on one occasion mentally
extracted the cube root of a number consisting of thirty figures. Dase is said to have mentally multiplied two numbers of one hundred figures each. A youth named Mangiamele was able to perform the
most remarkable feats in mental arithmetic. The reports show that upon a celebrated test before members of the French Academy of Sciences he was able to extract the cube root of 3,796,416 in thirty
seconds; and the tenth root of 282,475,289 in three minutes. He also immediately solved the following question put to him by Arago: “What number has the following proportion: That if five times the
number be subtracted from the cube plus five times the square of the number, and nine times the square of the number be subtracted from that result, the remainder will be 0?” The answer, “5” was
given immediately, without putting down a figure on paper or board. It is related that a cashier of a Chicago bank was able to mentally restore the accounts of the bank, which had been destroyed in
the great fire in that city, and his account which was accepted by the bank and the depositors, was found to agree perfectly with the other memoranda in the case, the work performed by him being
solely the work of his memory.
Bidder was able to tell instantly the number of farthings in the sum of £868, 42s, 121d. Buxton mentally calculated the number of cubical eighths of an inch there were in a quadrangular mass
23,145,789 yards long, 2,642,732 yards wide and 54,965 yards in thickness. He also figured out mentally, the dimensions of an irregular estate of about a thousand acres, giving the contents in acres
and perches, then reducing them to square inches, and then reducing them to square hair-breadths, estimating 2,304 to the square inch, 48 to each side. The mathematical prodigy, Zerah Colburn, was
perhaps the most remarkable of any of these remarkable people. When a mere child, he began to develop the most amazing qualities of mind regarding figures. He was able to instantly make the mental
calculation of the exact number of seconds or minutes there was in a given time. On one occasion he calculated the number of minutes and seconds contained in forty-eight years, the answer:
“25,228,800 minutes, and 1,513,728,000 seconds,” being given almost instantaneously. He could instantly multiply any number of one to three figures, by another number consisting of the same number of
figures; the factors of any number consisting of six or seven figures; the square, and cube roots, and the prime numbers of any numbers given him. He mentally raised the number 8, progressively, to
its sixteenth power, the result being 281,474,976,710,656; and gave the square root of 106,929, which was 5. He mentally extracted the cube root of 268,336,125; and the squares of 244,999,755 and
1,224,998,755. In five seconds he calculated the cube root of 413,993,348,677. He found the factors of 4,294,967,297, which had previously been considered to be a prime number. He mentally calculated
the square of 999,999, which is 999,998,000,001 and then multiplied that number by 49, and the product by the same number, and the whole by 25–the latter as extra measure.
The great difficulty in remembering numbers, to the majority of persons, is the fact that numbers “do not mean anything to them”–that is, that numbers are thought of only in their abstract phase and
nature, and are consequently far more difficult to remember than are impressions received from the senses of sight or sound. The remedy, however, becomes apparent when we recognize the source of the
difficulty. The remedy is: _Make the number the subject of sound and sight impressions._ Attach the abstract idea of the numbers to the sense of impressions of sight or sound, or both, according to
which are the best developed in your particular case. It may be difficult for you to remember “1848” as an abstract thing, but comparatively easy for you to remember the _sound_ of “eighteen
forty-eight,” or the _shape and appearance_ of “1848.” If you will repeat a number to yourself, so that you grasp the sound impression of it, or else visualize it so that you can remember having
_seen_ it–then you will be far more apt to remember it than if you merely think of it without reference to sound or form. You may forget that the number of a certain store or house is 3948, but you
may easily remember the sound of the spoken words “thirty-nine forty-eight,” or the form of “3948” as it appeared to your sight on the door of the place. In the latter case, you associate the number
with the door and when you visualize the door you visualize the number.
Pages: Page 1, Page 2, Page 3, Page 4, Page 5, Page 6, Page 7, Page 8, Page 9, Page 10, Page 11, Page 12, Page 13, Page 14, Page 15, Page 16, Page 17, Page 18, Page 19, Page 20, Page 21, Page 22,
Page 23, Page 24, Page 25, Page 26, Page 27, Page 28, Page 29, Page 30, Page 31, Page 32, Page 33, Page 34, Page 35, Page 36
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Precision in Construction: Tips for Using Concrete Calculators Effectively
In the dynamic world of construction, precision is paramount. When it comes to concrete, accurate calculations are crucial for minimizing waste, controlling costs, and ensuring project success. This
is where concrete calculators emerge as invaluable tools for contractors, architects, and engineers.
This comprehensive guide dives into the functionalities of concrete calculators, explores their benefits, and equips you with essential tips for utilizing them effectively in your construction
What is a Concrete Calculator?
A concrete calculator is an online tool designed to estimate the volume of concrete required for a specific construction element. It typically requires you to input the dimensions (length, width,
depth) of the concrete structure you're planning, such as a slab, foundation, wall, or column. Some calculators allow for additional factors like shape (rectangular, circular) and waste percentage.
Based on your input, the calculator provides an estimate of the total concrete volume needed in cubic yards or meters.
Benefits of Using Concrete Calculators
Integrating concrete calculators into your construction workflow offers a multitude of advantages:
• Improved Accuracy: Manual calculations are prone to errors. Concrete calculators minimize human error, leading to more precise volume estimates and reducing the risk of under-ordering or
over-ordering concrete.
• Efficient Material Ordering: Accurate volume estimates ensure you order the exact amount of concrete required for the project. This eliminates unnecessary material costs and prevents delays
caused by insufficient concrete on-site.
• Effective Cost Control: By minimizing concrete waste, you control material costs and optimize your budget. Concrete calculators empower you to anticipate potential costs based on volume
• Faster Project Planning: Quickly determine concrete requirements for different project phases, facilitating efficient planning and scheduling of concrete pours.
• Simplified Waste Reduction: Concrete waste can significantly impact your budget and environmental footprint. Calculators help you minimize waste by providing accurate volume estimates and
allowing you to factor in a waste percentage.
Types of Concrete Calculators
There are various concrete calculators available online, each catering to specific needs:
• Basic Concrete Calculators: These offer a simple interface for calculating the volume of basic shapes like rectangular slabs or square footings.
• Advanced Concrete Calculators: These provide functionalities for calculating complex shapes like circular columns, tapered walls, or sloped slabs.
• Custom Concrete Calculators: Some calculators allow users to define custom shapes with specific dimensions for highly unique project elements.
Choosing the Right Concrete Calculator
Selecting the most appropriate concrete calculator depends on the complexity of your project and your specific needs. Consider these factors:
• Project Complexity: For simple slabs or footings, a basic calculator suffices. For complex shapes, opt for an advanced option with relevant functionalities.
• Level of Detail: Some calculators provide basic volume estimates, while others offer detailed breakdowns of cubic yards per area or separate calculations for different project phases. Choose a
level of detail that aligns with your needs.
• Additional Features: Do you need functionalities like waste percentage calculation or integration with concrete mix design tools? Select a calculator that offers the features you require.
Using a Concrete Calculator Effectively
Here are some essential tips for maximizing the benefits of a concrete calculator:
• Accurate Measurements: Precise project measurements are crucial for accurate volume estimates. Double-check your length, width, and depth measurements before inputting them into the calculator.
• Understanding Waste Percentage: Factor in a realistic waste percentage to account for spillage, cutting errors, and leftover concrete. This helps ensure you order sufficient material.
• Consider Formwork Volume: While the calculator estimates concrete volume, remember to account for the additional volume occupied by formwork (molds) when calculating the total material needed.
• Consult Project Specifications: Refer to project specifications and structural drawings to ensure you're calculating the volume for the correct elements and factoring in any design complexities.
• Double-Checking Calculations: While calculators minimize errors, it's good practice to double-check complex calculations, especially for large projects. Consider performing manual calculations
alongside the calculator output for verification.
• Utilizing Advanced Features: Explore advanced features like waste percentage calculation, formwork volume estimation, or integration with concrete mix design tools offered by some calculators.
These features can significantly streamline your concrete planning and ordering process.
• Communicate Effectively: Once you have the concrete volume estimate, communicate it clearly to your concrete supplier, project team, and subcontractors involved in the pouring process. This
ensures everyone is on the same page regarding material requirements and avoids confusion during project execution.
Concrete calculators empower construction professionals to estimate concrete volume accurately, optimize material ordering, minimize waste, and ensure project success. By leveraging these tools
effectively alongside best practices and additional resources, you can build with confidence, maintain tight control over costs, and contribute to a more sustainable construction process.
Remember, concrete calculators are a valuable addition to your construction toolkit, but they should not replace sound judgment and a thorough understanding of project specifications. Embrace the
technology, utilize it strategically, and watch your concrete projects reach new heights of precision and efficiency.
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ax-addf - Intuitionistic Logic Explorer
Description: Addition is an operation on the complex numbers. This deprecated axiom is provided for historical compatibility but is not a bona fide axiom for complex numbers (independent of set
theory) since it cannot be interpreted as a first- or second-order statement (see https://us.metamath.org/downloads/schmidt-cnaxioms.pdf). It may be deleted in the future and should be avoided for
new theorems. Instead, the less specific addcl 7872 should be used. Note that uses of ax-addf 7869 can be eliminated by using the defined operation
This axiom is justified by Theorem axaddf 7803. (New usage is discouraged.) (Contributed by NM, 19-Oct-2004.) | {"url":"https://us.metamath.org/ilegif/ax-addf.html","timestamp":"2024-11-13T22:32:37Z","content_type":"text/html","content_length":"9962","record_id":"<urn:uuid:78adfcba-d089-46fc-8a4f-3e1d81e509a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00443.warc.gz"} |
CBSE Solutions For Class 10 Mathematics Chapter 9 Some Applications of Trigonometry - CBSE School Notes
CBSE Solutions For Class 10 Mathematics Chapter 9 Some Applications of Trigonometry
Class 10 Maths Some Applications of Trigonometry
1. The length of the Shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height. Show that the angle of elevation of the Sun is 60°.
Let PQ be a vertical pole whose height is h. Its Shadow is OQ whose height is \(\frac{h}{\sqrt{3}}\)
Let the angle of elevation of the Sun is ∠POQ = 0
In APOQ,
⇒ \(\tan \theta=\frac{P Q}{O Q}=\frac{h}{h / \sqrt{3}}=\sqrt{3}=\tan 60^{\circ}\)
θ = 60°
∴ The angle of elevation of Sun = 60°
Read and Learn More Class 10 Maths
2. If a tower 30m high, casts a shadow \(10 \sqrt{3} \mathrm{~m}\) long on the ground, then what is the angle of elevation of the Sun?
It is given that AB = 30m be the tower and BC = \(10 \sqrt{3} \mathrm{~m}\) m be its shadow on the ground.
Let θ be the angle of elevation.
In a right triangle,
tan θ = \(\frac{AB}{BC}\)
⇒ \(\frac{30}{10 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\)
= tan 60°
θ = 60°
∴ Hence, the angle of elevation θ = 60°
3. A ladder 15 meters long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
Let PR be a ladder of length 15m and QR, a wall of height h.
Given that ∠PRQ = 60°
In ΔPQR,
Cos 60° = \(\frac{h}{PR}\) = \(\frac{1}{2}\) = \(\frac{h}{15}\)
⇒ h = \(\frac{15}{2}\)m
∴ Height of the wall = \(\frac{15}{2} m\)
4. A Circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the Pole, if the angle made by the rope with the
ground level is 30°.
In ΔABC,
Sin 30°= \(\frac{A B}{A C}\)
⇒ \(\frac{1}{2}=\frac{A B}{20}\)
AB = 10
∴ Height of pole = 10m
5. A tree breaks due to stom and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top
touches the ground is 8m. Find the height of the tree.
Let the part CD of the tree BD broken in air and touches the ground at point A.
According to the problem,
AB = 8M
and ∠BAC = 30°
In ΔABC,
tan 30° = \(\frac{BC}{AB}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{B C}{8}\)
⇒ \(B=\frac{8}{\sqrt{3}} \mathrm{~m}\)
and Cos 30° = \(\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{A C}\)
AC = \(\frac{16}{\sqrt{3}} m\)
CD = \(\frac{16}{\sqrt{3}} m\) (∵ AC = CD)
Now, the height of tree = BC + CD
⇒ \(\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8 \sqrt{3} \mathrm{~m}\)
6. The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is 30°: Find the height of the tower.
Let AB be the tower.
The angle of elevation of the top of the tower from Point C, 30m away from A is 30°
∴ In ABAC,
tan 30°= \(\frac{A B}{A C}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{A C}\)
AB = \(\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m}\)
∴ Height of the tower = \(10 \sqrt{3} \mathrm{~m}\)
7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building and 60° respectively. Find the height of the tower.
Let, CD be the height of the transmission tower.
Here, the height of the building
BC = 20m
In ΔABC,
tan 45° = \(\frac{B C}{A B}\)
1 = \(\frac{20}{A B}\)
AB = 20m
In ΔABD,
tan 60° = \(\frac{B D}{A B} \Rightarrow \sqrt{3}=\frac{B D}{20}\)
⇒ \(B D=20 \sqrt{3} \mathrm{~m}\)
⇒ \(B C+C D=20 \sqrt{3}\)
⇒ \(20+C D=20 \sqrt{3}\)
⇒ \(C D=20(\sqrt{3}-1) m\)
∴ Height of transmission tower = \(20(\sqrt{3}-1) m\)
8. The Shadow of a tower Standing on a level plane is found to be much longer when the Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Let AB be a tower of height ‘h’ meters and BD and BC be its shadows when the angles of elevation of the sun are 30° and 60° respectively.
∴ ∠ADB =30°, ∠ACB = 60° and CD = 50m
Let BC = X meters.
In ΔABC
tan 60° = \(\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x}\)
⇒ \(x=\frac{h}{\sqrt{3}}\)
In ΔABD
tan 30° = \(\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50}\)
⇒ \(\sqrt{3} h=x+50 \Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50\)
2h= \(50 \sqrt{3}\)
h = \(25 \sqrt{3}\)
∴ Height of the tower = \(25 \sqrt{3}\)m
9. The angle of elevation of the top of a tower from a point on the ground is 30: After walking nom towards the tower, the angle of elevation becomes 60° Find the height of the tower.
Let AB be a tower of height ‘h’ meters. From points D and c on the ground, the angle of elevation of top A of the tower is 30° and 60° respectively.
Given that CD = 40m
let BC = x meters
In ΔABC
tan 60° = \(\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x}\)
⇒ \(x=\frac{h}{\sqrt{3}}\)
In ΔABD
tan 30° =\(\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40+x}\)
⇒ \(\sqrt{3} h=40+x\)
⇒ \(\sqrt{3} h=40+\frac{h}{\sqrt{3}}\) [from (1)]
⇒ \(3 h=40 \sqrt{3}+h\)
⇒ \(2 h=40 \sqrt{3}\)
⇒ \(h=20 \sqrt{3}\)
∴ Height of the tower = \(20 \sqrt{3} \mathrm{~m}\) m
Leave a Comment | {"url":"https://cbseschoolnotes.com/cbse-solutions-for-class-10-mathematics-chapter-9/","timestamp":"2024-11-06T12:02:42Z","content_type":"text/html","content_length":"152929","record_id":"<urn:uuid:6fb0e71d-7de9-4cc9-9830-a74e00e9a150>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00175.warc.gz"} |
Development of the Math Strands - Connected Mathematics Project
Development of the Math Strands
The “Connected” in Connected Mathematics has several meanings. First, there are contexts that connect to the world in which students live. Second, there are mathematical ideas that serve as unifying
themes to connect Units and strands together. Lastly, goals are developed in symbiotic tandem with each other, and over Units and grade levels. The result is a coherent whole.
Within each CMP Unit, the Problems are carefully sequenced to address important goals. This might imply that the goals are a discrete, linear sequence, but goals are often developed in parallel, as
well as in sequence.
While exploring relationships among variables in Variables and Patterns, students are simultaneously beginning to develop strategies for solving equations, two prominent goals for CMP’s Algebra and
Functions strand.
Likewise, organizing Units by mathematical strands does not imply that all the goals for each Unit are related to the same strand. A Unit might be listed in the one strand but also carry key
mathematical goals for another strand. For example, Looking For Pythagoras, while primarily about the Pythagorean Theorem, also carries forward the development of the Number and Operations strand by
introducing students to irrational numbers and the set of real numbers. The Pythagorean Theorem also leads naturally to the equation of a circle and other important ideas.
Not only does goal development transcend the boundaries of a strand, but some mathematical ideas are so powerful that they permeate several strands and serve as unifying themes. Two of these
overarching themes are proportional reasoning and mathematical modeling.
This section provides an overview to the development of the four mathematical strands, Number, Operations, Rates, and Ratio, Geometry and Measurement, Data and Probability, and Algebra and Functions
and two of the unifying themes.
As you study the goals, the development of the mathematics in each strand, and in the collection of Units that comprise each grade-level course, it will be helpful to ask:
• What are the big ideas of the strand and key objectives of each Unit in the strand?
• How are the key concepts of the strand developed in depth and complexity over time?
• What connections are made between the Units of this strand and those of other strands which are interpreted in the sequence of Units?
• How are the unifying themes reflected in the strands? Units? | {"url":"https://connectedmath.msu.edu/the-math/development-of-the-math-strands/index.aspx","timestamp":"2024-11-13T09:37:10Z","content_type":"text/html","content_length":"52557","record_id":"<urn:uuid:75da4cc9-55b8-4176-b6dd-052d52e5294e>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00463.warc.gz"} |
Math Colloquia - Homogeneous dynamics and its application to number theory
Homogeneous dynamics, the theory of flows on homogeneous spaces, has been proved useful for certain problems in Number theory.
In this talk, we will explain what kind of geometry and dynamics we need to solve certain number theoretic questions such as counting matrices of integer entries, or some problems in Diophantine
approximation. The appropriate manifold can often be seen as a space of lattices, and its asymptotic geometry is governed by the smallest length of a non-zero vector in a given lattice, which is also
the backbone of post-quantum cryptography.
We will then explain how (partial) solutions of Oppenheim conjecture and Littlewood conjecture were obtained using homogeneous dynamics. We will also survey some recent results and remaining | {"url":"http://my.math.snu.ac.kr/board/index.php?mid=colloquia&page=1&document_srl=1266013&l=en&sort_index=room&order_type=asc","timestamp":"2024-11-02T18:53:03Z","content_type":"text/html","content_length":"43461","record_id":"<urn:uuid:b862c53c-8c0c-47af-80a4-08d6f7c00e0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00443.warc.gz"} |
OpenStax College Physics for AP® Courses, Chapter 30, Problem 2 (Problems & Exercises)
(a) Calculate the mass of a proton using the charge-to- mass ratio given for it in this chapter and its known charge. (b) How does your result compare with the proton mass given in this chapter?
Question by
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CC BY 4.0
Final Answer
a. $1.67\times 10^{-27}\textrm{ kg}$
b. This is the same as the modern accepted value for the mass of a proton.
Solution video
OpenStax College Physics for AP® Courses, Chapter 30, Problem 2 (Problems & Exercises)
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Video Transcript
This is College Physics Answers with Shaun Dychko. Given the charge to mass ratio for a proton is 9.58 times 10 to the 7 coulombs per kilogram, we can figure out the mass of the proton knowing its
charge; it has an elementary charge of 1.602 times 10 to the minus 19 coulombs for every proton and we multiply that by the reciprocal of this ratio which is 1 kilogram for every 9.58 times 10 to the
7 coulombs and we see that the coulombs cancel leaving us with kilograms per proton and that is 1.67 times 10 to the minus 27 kilograms and this is the same as the modern accepted value for the mass
of a proton. | {"url":"https://collegephysicsanswers.com/openstax-solutions/calculate-mass-proton-using-charge-mass-ratio-given-it-chapter-and-its-known-0","timestamp":"2024-11-08T18:53:49Z","content_type":"text/html","content_length":"191319","record_id":"<urn:uuid:e3a47162-9371-4fdc-b0e6-b9d3c779ffeb>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00210.warc.gz"} |
Introducing the Newest Hitter xBABIP
It’s been a looooooooong journey toward understanding what underlying skills drive a hitter’s BABIP ability. No matter how much understanding we have gained over the years, it has been a struggle to
develop an equation that produced an R-squared much over 0.50. That’s not terrible, but when my hitter xHR/FB equation spits out an impressive 0.826 R-Squared, I continue to strive for better. I
shared my last hitter xBABIP equation almost exactly five years ago, and since, I have yet to see a better one.
After the 2021 season, I went back to the drawing board, as I usually do for my xMetrics, to figure out if there was something I was missing. Or perhaps there was some time-consuming data gathering
task I could do that would improve the equation, maybe one I had been too lazy to do in the past. Sure enough, I suddenly had several epiphanies and excitedly pulled the data I needed and ran a
regression. Then I stared at a lower R-squared than my 2017 equation and my excitement waned. Huh? I was about to give up and just stick with my latest equation until I remembered something.
Statcast already calculates a hitter’s expected batting average, or xBA! They do so by using the holy grail — individual batted ball data. Rather than use season totals and averages like all my
xMetrics, Statcast’s xBA calculates a hit probability for every single batted ball and then totals those probabilities up for an expected hit total, which is used to calculate xBA. That’s a far
superior method. So why reinvent the wheel if I could just drop the idea of creating a better xBABIP and simply go with what Statcast is selling?
Well, Statcast’s xBA metric only accounts for exit velocity, launch angle, and as its glossary states “on certain types of batted balls, Sprint Speed.” What’s missing here and makes an obvious
difference is horizontal direction, as in pull, center, or opposite. Statcast includes a filter for that in its search, but doesn’t incorporate that data into its calculation. The side effect of the
missing variables is that xBA also does not account for defensive shifts, which we know have had a dramatic effect on some hitters as the strategy has become more common over the years.
Knowing xBA’s shortcomings, I figured it might be prudent to attempt to improve upon the metric by incorporating what’s missing. First though, I needed to calculate an implied Statcast xBABIP,
because it doesn’t calculate it for us. How, you ask? Let’s go over the steps:
1. Go to this custom leaderboard for all the stats you need.
2. Let’s take Marcus Semien as an example: In Excel, or in your math ninja head, multiply Semien’s xBA of 0.245 by his AB of 652 to calculate his expected hits or xH. So, xH = Statcast xBA * AB. In
this case, xH = 159.74, versus 173 actual hits.
3. Then, Statcast xBABIP = (Statcast xH – HR)/(AB – HR – SO + SF). For Semien, Statcast xBABIP = (159.74 – 45) / (652 – 45 – 146 + 3) = 0.247. That compares to an actual BABIP of .276, if you were
Just like that, you have officially calculated an implied Statcast xBABIP!
Once I calculated Statcast’s xBABIP, I wanted to compare how well it correlated with actual BABIP during the entire Statcast era from 2015-2021. Shockingly, its R-squared was just 0.462. If you
recall from the intro, I mentioned developing an equation with an R-squared of just over 0.50, as my 2017 xBABIP sported an R-squared of 0.538. So Statcast’s xBABIP, which uses individual batted
ball data, explained actual BABIP worse than my equation that used season totals and averages. That was confirmation that Statcast’s xBABIP was ripe for improvement. The hope then became that I
could develop a version of xBABIP that included Statcast xBABIP as a variable and it would produce a better R-squared than 0.538, and hopefully much better.
I decided to do some horizontal direction research by using Statcast search to find out the league average BABIP on various combinations of batted ball variables. Below are the results:
Batted Ball Type BABIP
Batted Ball Type BABIP
Opposite Shift IF Alignment GB 0.527
Opposite GB 0.376
Opposite Standard IF Alignment GB 0.356
Opposite Strategic IF Alignment GB 0.353
Straightaway Strategic IF Alignment GB 0.264
Strategic IF Alignment GB 0.256
Standard IF Alignment GB 0.252
Ground Ball 0.247
Straightaway GB 0.241
Straightaway Standard IF Alignment GB 0.239
Straightaway Shift IF Alignment GB 0.236
Pull Standard IF Alignment GB 0.233
Pull Strategic IF Alignment GB 0.215
Pull GB 0.214
Shift IF Alignment GB 0.213
Pull Shift IF Alignment GB by R 0.166
Pull Shift IF Alignment GB 0.127
Pull Shift IF Alignment GB by L 0.109
The Ground Ball row highlighted in yellow in the middle is the control. The league average BABIP on all grounders has been .247. I then chose to highlight Opposite GB, or grounders hit to the
opposite field, and three buckets of pulled grounders into the shift that appear at the bottom. I included the BABIP on all pulled grounders while the infield alignment was shifted, as well as
that same BABIP broken out by batter handedness. You’ll notice that left-handed hitters have been hurt more than right-handed hitters when pulling grounders into the shift. This was an aha!
I knew immediately that Opposite GB% would become a variable and given the stark BABIP difference between right-handed and left-handed batters on pulled grounders into the shift, I would include
both Pull Shift IF Alignment GB As R% and Pull Shift IF Alignment GB As L%. But I wasn’t done quite yet.
Despite the Statcast glossary page quoted above that mentions accounting for Sprint Speed, I have continued to find that it’s not accounted for enough. When I have sorted hitters by BABIP –
xBABIP differential, the underperforming group had a significantly higher Sprint Speed than the overperforming group. That discovery confirmed that I still needed to add a speed variable. My
preference was to add HP to 1B, for obvious reasons. Unfortunately, it’s not available for every player, and I wasn’t going to use a different xBABIP equation for hitters with no HP to 1B data.
So I settled on using Sprint Speed, as I had no other choice. That became the fourth variable I would add to Statcast xBABIP for my new and (hopefully) improved Pod xBABIP.
After running my regression, I was thrilled — R-squared improved and even jumped over my 2017 equation. However, there was a problem when looking at individual seasons, as my 2015 to 2019 league
xBABIP marks came close to actual league BABIP marks, but my 2020 and 2021 marks did not. Perhaps you could figure out why from this table:
League BA vs Statcast xBA
BA Statcast xBA Diff
2015 0.259 0.244 0.015
2016 0.259 0.248 0.012
2017 0.259 0.251 0.009
2018 0.252 0.244 0.008
2019 0.256 0.248 0.009
2020 0.246 0.245 0.001
2021 0.248 0.246 0.002
From 2015 to 2019, Statcast’s xBA consistently sat well below actual BA. That’s pretty odd, as an xMetric should come pretty close to the actual metric for the entire league during a season. On
an individual player basis, it’s going to be all over the map, but it shouldn’t be for the league in aggregate. But then beginning in 2020 and continuing in 2021, Statcast’s xBA was suddenly very
close to actual BA.
So I reached out to Mike Petriello to find out if he had an explanation, and this was our short Twitter convo:
today of all days, Mike
I don't know the precise answer off-hand but my guess (emphasis on guess here) would be the switch to much better tracking hardware starting in 2020 limited the number of fill-in-the-gaps
that had to be done.
— Mike Petriello (@mike_petriello) December 2, 2021
So it would seem as if there was an actual change that led to improved Statcast xBA calculations beginning in 2020. No wonder my 2020 and 2021 calculations were off! Those seasons were using the
same equation as the 2015 to 2019 seasons when xBA was further away from actual BA and needed to be corrected. But the 2020 and 2021 xBA marks closely matched actual BA and did not need to be
corrected. So I decided to create two separate regression equations — one to use for 2015 to 2019 and another for 2020 to 2021 and all future seasons. That solved the issue and I was back in
It’s now time to reveal the equations:
Pod xBABIP 2015-2019 = -0.01876 + (Statcast xBABIP * 0.84139) + (Sprint Speed * 0.00276) + (Pull Shift IF Alignment GB As R% * -0.08450) + (Pull Shift IF Alignment GB As R% * -0.12089) +
(Opposite GB% * 0.14197)
Pod xBABIP 2020 & Beyond = -0.02373 + (Statcast xBABIP * 0.93377)+ (Sprint Speed * 0.00175) + (Pull Shift IF Alignment GB As R% * -0.11485) + (Pull Shift IF Alignment GB As R% * -0.11195) +
(Opposite GB% * 0.11621)
Here is a table of adjusted R-squared comparisons:
Comparison of Adjusted R-Squared With BABIP
Seasons Pod xBABIP Statcast xBABIP
2015-2019 0.538 0.459
2020-2021 0.593 0.542
Overall 0.551 0.462
Those are big improvements from the Pod xBABIP over Statcast. The 2020-2021 marks are higher because the sample size of player seasons was much smaller.
Note that while these marks aren’t significantly higher than my 2017 equation, it’s not a true apples-to-apples comparison. I used a minimum of 200 non-home run balls in play for my equations
this time, versus 400 at-bats back then. If you could believe it, the fewest at-bats a hitter recorded while still putting 200 non-homers in play was 224. Not only does using a balls in play
minimum versus an at-bat minimum make far more sense for an xBABIP equation, but the smaller sample size of balls in play I used for this new equation gives it a disadvantage versus the 2017. If
I used a 400 at-bat minimum for my 2015-2019 equation instead of the 200 balls in play minimum, my adjusted R-squared would rise to 0.564, from the 0.538 in the above table. So, it’s another
confirmation that this latest Pod xBABIP is superior to my 2017 version.
Now for a quick explanation on pulling the data for the four variables used with Statcast xBABIP in the equations:
Sprint Speed
Pull Shift IF Alignment GB As R% & Pull Shift IF Alignment GB As L%
1. Perform a Statcast search by filtering Batted Ball Direction = Pull, IF Alignment = Shift, Batted Ball Type = Ground Ball, and Batter Handedness = Right or Left
2. Click the disk button at the top right of the search results that when hovered over says “Download Results Comma Separated Values File”.
3. Column A, “pitches”, is your value. It’s the total number of pulled ground balls hit into a shifted infield.
4. Perform the same search, but switch the Batter Handedness filter to the other hand to ensure you download the data for each handedness.
Opposite GB%
1. Perform a Statcast search by filtering Batted Ball Direction = Opposite and Batted Ball Type = Ground Ball.
2. Click the disk button at the top right of the search results that when hovered over says “Download Results Comma Separated Values File”.
3. Column A, “pitches”, is your value. It’s the total number of opposite field ground balls hit.
Once you have these totals, you will need to calculate what percentage of balls in play (which excludes home runs) these batted ball buckets represent. Calculate balls in play (BIP) using the
stats you have already downloaded as:
BIP = AB – HR – SO + SF
Now simply divide each of the three batted ball bucket totals by BIP and you have your percentages to use in the Pod xBABIP equation.
That’s a wrap for today. We’ll dive into the fun part next, looking at hitters whose Pod xBABIP most differs from Statcast xBABIP, underperformers and overperformers, and perhaps some
leaderboards in each of the batted ball bucket rates.
Mike Podhorzer is the 2015 Fantasy Sports Writers Association Baseball Writer of the Year. He produces player projections using his own forecasting system and is the author of the eBook Projecting X
2.0: How to Forecast Baseball Player Performance, which teaches you how to project players yourself. His projections helped him win the inaugural 2013 Tout Wars mixed draft league. Follow Mike on
Twitter @MikePodhorzer and contact him via email.
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Differential inequalities for one component of solution vector for systems of linear functional differential equations
The method to compare only one component of the solution vector of linear functional differential systems, which does not require heavy sign restrictions on their coefficients, is proposed in this
paper. Necessary and sufficient conditions of the positivity of elements in a corresponding row of Green's matrix are obtained in the form of theorems about differential inequalities. The main idea
of our approach is to construct a first order functional differential equation for the nth component of the solution vector and then to use assertions about positivity of its Green's functions. This
demonstrates the importance to study scalar equations written in a general operator form, where only properties of the operators and not their forms are assumed. It should be also noted that the
sufficient conditions, obtained in this paper, cannot be improved in a corresponding sense and does not require any smallness of the interval [ 0,ω ], where the system is considered.
טביעת אצבע
להלן מוצגים תחומי המחקר של הפרסום 'Differential inequalities for one component of solution vector for systems of linear functional differential equations'. יחד הם יוצרים טביעת אצבע ייחודית. | {"url":"https://cris.ariel.ac.il/iw/publications/differential-inequalities-for-one-component-of-solution-vector-fo-2","timestamp":"2024-11-09T01:37:11Z","content_type":"text/html","content_length":"57272","record_id":"<urn:uuid:a58770a2-adf5-4f6b-af99-8932197f20fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00586.warc.gz"} |
This is part of the multicolvar module
Evaluate the average value of a multicolvar on a grid.
This keyword allows one to construct a phase field representation for a symmetry function from an atomistic description. If each atom has an associated order parameter, \(\phi_i\) then a smooth phase
field function \(\phi(r)\) can be computed using:
\[ \phi(\mathbf{r}) = \frac{\sum_i K(\mathbf{r}-\mathbf{r}_i) \phi_i }{ \sum_i K(\mathbf{r} - \mathbf{r}_i )} \]
where \(\mathbf{r}_i\) is the position of atom \(i\), the sums run over all the atoms input and \(K(\mathbf{r} - \mathbf{r}_i)\) is one of the kernelfunctions implemented in plumed. This action
calculates the above function on a grid, which can then be used in the input to further actions.
The following example shows perhaps the simplest way in which this action can be used. The following input computes the density of atoms at each point on the grid and outputs this quantity to a file.
In other words this input instructs plumed to calculate \(\rho(\mathbf{r}) = \sum_i K(\mathbf{r} - \mathbf{r}_i )\)
Click on the labels of the actions for more information on what each action computes
dens: DENSITY =1-100 The DENSITY action with label dens
grid: MULTICOLVARDENS =dens =1 =xyz =100,100,100 =0.05,0.05,0.05 =1 The MULTICOLVARDENS action with label grid
DUMPGRID =grid =500 =density The DUMPGRID action with label
In the above example density is added to the grid on every step. The PRINT_GRID instruction thus tells PLUMED to output the average density at each point on the grid every 500 steps of simulation.
Notice that the that grid output on step 1000 is an average over all 1000 frames of the trajectory. If you would like to analyze these two blocks of data separately you must use the CLEAR flag.
This second example computes an order parameter (in this case FCCUBIC) and constructs a phase field model for this order parameter using the equation above.
Click on the labels of the actions for more information on what each action computes
fcc: FCCUBIC =1-5184 ={CUBIC D_0=1.2 D_MAX=1.5} =27 The FCCUBIC action with label fcc
dens: MULTICOLVARDENS =fcc =1 =xyz =14,14,28 =1.0,1.0,1.0 =1 =1 The MULTICOLVARDENS action with label dens
DUMPCUBE =dens =1 =dens.cube The DUMPCUBE action with label
In this example the phase field model is computed and output to a file on every step of the simulation. Furthermore, because the CLEAR=1 keyword is set on the MULTICOLVARDENS line each Gaussian cube
file output is a phase field model for a particular trajectory frame. The average value accumulated thus far is cleared at the start of every single timestep and there is no averaging over trajectory
frames in this case.
Glossary of keywords and components
The atoms involved can be specified using
ORIGIN we will use the position of this atom as the origin. For more information on how to specify lists of atoms see Groups and Virtual Atoms
Compulsory keywords
STRIDE ( default=1 ) the frequency with which the data should be collected and added to the quantity being averaged
CLEAR ( default=0 ) the frequency with which to clear all the accumulated data. The default value of 0 implies that all the data will be used and that the grid will never be cleared
NORMALIZATION ( default=true ) This controls how the data is normalized it can be set equal to true, false or ndata. The differences between these options are explained in the manual page for
BANDWIDTH the bandwidths for kernel density estimation
KERNEL ( default=gaussian ) the kernel function you are using. More details on the kernels available in plumed plumed can be found in kernelfunctions.
DATA the multicolvar which you would like to calculate the density profile for
DIR the direction in which to calculate the density profile
SERIAL ( default=off ) do the calculation in serial. Do not use MPI
LOWMEM ( default=off ) lower the memory requirements
TIMINGS ( default=off ) output information on the timings of the various parts of the calculation
FRACTIONAL ( default=off ) use fractional coordinates for the various axes
XREDUCED ( default=off ) limit the calculation of the density/average to a portion of the z-axis only
YREDUCED ( default=off ) limit the calculation of the density/average to a portion of the y-axis only
ZREDUCED ( default=off ) limit the calculation of the density/average to a portion of the z-axis only
LOGWEIGHTS list of actions that calculates log weights that should be used to weight configurations when calculating averages
CONCENTRATION the concentration parameter for Von Mises-Fisher distributions
NBINS the number of bins to use to represent the density profile
SPACING the approximate grid spacing (to be used as an alternative or together with NBINS)
XLOWER this is required if you are using XREDUCED. It specifies the lower bound for the region of the x-axis that for which you are calculating the density/average
XUPPER this is required if you are using XREDUCED. It specifies the upper bound for the region of the x-axis that for which you are calculating the density/average
YLOWER this is required if you are using YREDUCED. It specifies the lower bound for the region of the y-axis that for which you are calculating the density/average
YUPPER this is required if you are using YREDUCED. It specifies the upper bound for the region of the y-axis that for which you are calculating the density/average
ZLOWER this is required if you are using ZREDUCED. It specifies the lower bound for the region of the z-axis that for which you are calculating the density/average
ZUPPER this is required if you are using ZREDUCED. It specifies the upper bound for the region of the z-axis that for which you are calculating the density/average | {"url":"https://www.plumed.org/doc-v2.8/user-doc/html/_m_u_l_t_i_c_o_l_v_a_r_d_e_n_s.html","timestamp":"2024-11-12T22:08:40Z","content_type":"application/xhtml+xml","content_length":"19766","record_id":"<urn:uuid:df01039d-35f8-467a-9415-f115697fa2fb>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00022.warc.gz"} |
How many cube roots of one are there?
When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the
question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.
The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root
of negative one.
What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they
might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.
What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square
a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square
Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.
Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like
how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but
switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.
If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just
looking at the other root.
Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How
about we cube it.
(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))
First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.
(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)
We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).
(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)
What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean?
The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!
1 - i/4√(3) + i/4√(3)
1 + 0
So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No!
Mathematics is never random!
If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going
through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second
root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get
90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.
What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth
roots, and the 60 degree marks give you the sixth roots.
If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know. | {"url":"https://coolmathstuff123.blogspot.com/2011/10/how-many-cube-roots-of-one-are-there.html","timestamp":"2024-11-03T22:00:15Z","content_type":"text/html","content_length":"68658","record_id":"<urn:uuid:3858163c-5973-4995-96e9-a9d69695b3a1>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00212.warc.gz"} |
Impact of photo-evaporative mass loss on masses and radii of
Issue A&A
Volume 562, February 2014
Article Number A80
Number of page(s) 14
Section Planets and planetary systems
DOI https://doi.org/10.1051/0004-6361/201322258
Published online 10 February 2014
A&A 562, A80 (2014)
Impact of photo-evaporative mass loss on masses and radii of water-rich sub/super-Earths ^⋆
^1 Department of Earth and Planetary ScienceThe University of Tokyo, 7-3-1 Hongo, Bunkyo-ku, 113-0033 Tokyo, Japan
e-mail: kkurosaki@eps.s.u-tokyo.ac.jp
^2 Division of Theoretical Astronomy, National Astronomical Observatory of Japan, 2-21-1 Osawa, Mitaka, 181-8588 Tokyo, Japan
Received: 11 July 2013
Accepted: 3 December 2013
Context. Recent progress in transit photometry opened a new window to the interior of super-Earths. From measured radii and masses, we can infer constraints on planetary internal compositions. It has
been recently revealed that super-Earths orbiting close to host stars (i.e., hot super-Earths) are diverse in composition. This diversity is thought to arise from diversity in volatile content.
Aims. The stability of the volatile components, which we call the envelopes, is to be examined, because hot super-Earths, which are exposed to strong irradiation, undergo photo-evaporative mass loss.
While several studies investigated the impact of photo-evaporative mass loss on hydrogen-helium envelopes, there are few studies as to the impact on water-vapor envelopes, which we investigate in
this study. To obtain theoretical prediction to future observations, we also investigate the relationships among masses, radii, and semi-major axes of water-rich super-Earths and also sub-Earths that
have undergone photo-evaporative mass loss.
Methods. We simulate the interior structure and evolution of highly-irradiated sub/super-Earths that consist of a rocky core surrounded by a water envelope, which include mass loss due to the stellar
XUV-driven energy-limited hydrodynamic escape.
Results. We find that the photo-evaporative mass loss has a significant impact on the evolution of hot sub/super-Earths. With a widely-used empirical formula for XUV flux from typical G-stars and the
heating efficiency of 0.1 for example, the planets of less than 3 Earth masses orbiting 0.03 AU have their water envelopes completely stripped off. We then derive the threshold planetary mass and
radius below which the planet loses its water envelope completely as a function of the initial water content and find that there are minimums of the threshold mass and radius.
Conclusions. We constrain the domain in the parameter space of planetary mass, radius, and the semi-major axis in which sub/super-Earths never retain water envelopes in 1–10 Gyr. This would provide
an essential piece of information for understanding the origin of close-in, low-mass planets. The current uncertainties in stellar XUV flux and its heating efficiency, however, prevent us from
deriving robust conclusions. Nevertheless, it seems to be a robust conclusion that Kepler planet candidates contain a significant number of rocky sub/super-Earths.
Key words: planets and satellites: composition / planets and satellites: interiors
© ESO, 2014
1. Introduction
Exoplanet transit photometry opened a new window to the interior and atmosphere of exoplanets. The biggest advantage of this technique would be that planetary radii are measured, while planetary
masses are measured via other techniques, such as the radial velocity method and the transit timing variation method. Measured mass and radius relationships help us infer the internal structure and
bulk composition of exoplanets theoretically, which give crucial constraints to formation and evolution processes of the planets. A growing number of small-sized exoplanets with radii of 1 to 2 R[⊕]
have been identified, which are often referred to as super-Earths (Batalha et al. 2013). Also, planet candidates detected by the Kepler space telescope include sub-Earth-sized objects, such as
Kepler-20 e (Fressin et al. 2011), Kepler-42 b, c, d (Muirhead et al. 2012), and Kepler-37 b, c (Barclay et al. 2013). We can thus discuss the compositions of such small planets to gas giants by
comparing theory with current observations.
Transiting super-Earths detected so far show a large variation in radius, suggesting diversity in composition. There are many theoretical studies on mass-radius relationships for planets with various
compositions and masses (Valencia et al. 2007; Fortney et al. 2007; Sotin et al. 2007; Seager et al. 2007; Grasset et al. 2009; Wagner et al. 2011; Swift et al. 2012). A recent important finding,
which compares theory to observation is that there are a significant number of low-density super-Earths that are larger in size than they would be if they were rocky. This implies that these
transiting super-Earths possess components less dense than rock. From a viewpoint of planet formation, the possible components are hydrogen-rich gas and water, which make an outer envelope. A small
fraction of H-rich gas or water is known to be enough to account for observed radii of the low-density super-Earths (Adams et al. 2008; Valencia et al. 2010).
The stability of the envelopes are, however, to be examined. Transiting planets are generally orbit close to their host stars (typically ≲0.1 AU), because detection probability of planetary transits
is inversely proportional to the semi-major axis (e.g., Kane 2007). These close-in planets are highly irradiated and exposed to intense X-ray and ultraviolet radiation (hereafter XUV) that come from
their host stars. This causes the planetary envelope to escape hydrodynamically from the planet (e.g., Watson et al. 1981). This process is often called the photo-evaporation of planetary envelopes.
As for massive close-in planets, namely, hot Jupiters, the possibility of the photo-evaporation and its outcome have been investigated well both theoretically and observationally (e.g., Yelle et al.
2008 and references therein).
While the photo-evaporation may not significantly affect the evolution and final composition of hot Jupiters except for extremely irradiated or inflated hot Jupiters, its impact on small close-in
planets in the sub/super-Earth mass range should be large, partly because their envelope masses are much smaller than those of hot Jupiters. For example, Valencia et al. (2010) investigated the
structure and composition of the first transiting super-Earth CoRoT-7 b and discussed the sustainability of the possible H+He envelope with a mass of less than 0.01% of the total planetary mass.
The envelope mass was consistent with its measured mass and radius. The estimated lifetime of the H+He envelope was, however, only 1 million years, which was much shorter than the host star’s age
(2–3 Gyr). This suggests that CoRoT-7 b is unlikely to retain the H+He envelope at present.
Young main-sequence stars are known to be much more active and emit stronger XUV than the current Sun (e.g., Ribas et al. 2005). Therefore, even if a super-Earth had a primordial atmosphere
initially, it may lose the atmosphere completely during its history. These discussions concerning the photo-evaporative loss of H+He envelopes were done for GJ 1214 b (Nettelmann et al. 2011;
Valencia et al. 2013), super-Earths orbiting Kepler-11 (Lopez et al. 2012; Ikoma & Hori 2012), and CoRoT-7 b (Valencia et al. 2010). Systematic studies were also done by Rogers et al. (2011) and
Lopez & Fortney (2013). Those studies demonstrated the large impact of the photo-evaporation on the stability of H+He envelopes for super-Earths. In particular, Lopez & Fortney (2013) performed
simulations of coupled thermal contraction and photo-evaporative mass loss of rocky super-Earths with H+He envelopes. They found that there were threshold values of planetary masses and radii,
below which H+He envelopes were completely stripped off. Owen & Wu (2013) also performed similar simulations with detailed consideration of the mass loss efficiency for an H+He envelope based on
Owen & Jackson (2012). They argued that evaporation explained the correlation between the semi-major axes and planetary radii (or planet densities) of KOIs.
In this study, we focus on water-rich sub/super-Earths. Planet formation theories predict that low-mass planets migrate toward their host star, which is strongly supported by the presence of many
close-in super-Earths, from cooler regions (e.g., Ward 1986) where they may have accreted a significant amount of water. This suggests that water/ice-rich sub/super-Earths may also exist close to
host stars. Therefore, similar discussions should be done for water envelopes of close-in super-Earths. However, there are just a few studies, which treat specific sub/super-Earths such as CoRoT-7 b
(Valencia et al. 2010) and Kepler-11 b (Lopez et al. 2012). No systematic study is yet to be done for the stability of water envelopes.
The purpose of this study is, thus, to examine the stability of primordial water envelopes of close-in sub/super-Earths against photo-evaporation. To this end, we simulate the thermal evolution of
planets with significant fractions of water envelopes (i.e., water-worlds), incorporating the effect of stellar-XUV-driven photo-evaporative mass loss. The theoretical model is described in Sect. 2.
As for the atmosphere model, the details are described in Appendix A. In Sect. 3, we show the evolutionary behavior of the water-rich planets. Then, we find threshold values of planetary masses and
radii below which such water-rich planets are incapable of retaining primordial water envelopes for a period similar to ages of known exoplanet-host stars (i.e., 1–10 Gyr). In Sect. 4, we compare the
theoretical mass-radius distribution of water-rich planets with that of known transiting planets. Furthermore, we compare the threshold radius with sizes of Kepler objects of interest (KOIs) to
suggest that KOIs include a significant number of rocky planets. Finally, we summarize this study in Sect. 5.
Fig. 1
Model of the planetary structure in this study.
2. Numerical models
In this study, we simulate the evolution of the mass and radius of a planet that consists of water and rock, including the effects of mass loss due to photo-evaporation. The structure model is
depicted in Fig. 1. The planet is assumed to consist of three layers in spherical symmetry and hydrostatic equilibrium: namely, from top to bottom, it consisted of a water vapor atmosphere, a water
envelope, and a rocky core. At each interface, the pressure and temperature are continuous.
The assumptions and equations that determine the planet’s interior structure and thermal evolution are described in Sects. 2.1 and 2.2, respectively. The equations of state for the materials in the
three layers are summarized in Sect. 2.3. The structure of the atmosphere and the photoevaporative mass loss, both of which govern the planet’s overall evolution, are described in Sect. 2.4 (see also
Appendix A) and Sect. 2.5, respectively. Since a goal of this study is to compare our theoretical prediction with results from transit observations, we also calculate the transit radius, which is
defined in Sect. 2.6. Finally, we summarize our numerical procedure in Sect. 2.7.
2.1. Interior structure
The interior structure of the planet is determined by the differential equations (e.g. Kippenhahn & Weigert 1990), $∂P∂Mr=−GMr4πr4,∂r∂Mr=14πr2ρ,∂T∂Mr=−GMrT4πr4P∇,$and the equation of state, $ρ=ρ
(P,T),$(4)where r is the planetocentric distance, M[r] is the mass contained in the sphere with radius of r, P is the pressure, ρ is the density, T is the temperature, and G (=6.67 × 10^-8 dyn cm^2 g
^-2) is the gravitational constant. The symbol ∇ is the temperature gradient with respect to pressure. We assume that the water envelope and rocky core are fully convective and the convection is
vigorous enough that the entropy S is constant; namely, $∇=(∂lnT∂lnP)S·$(5)Equations (1)–(3) require three boundary conditions. The inner one is r = 0 at M[r] = 0. The outer boundary corresponds to
the interface between the envelope and the atmosphere, which is called the tropopause. The tropopause pressure P[ad] and temperature T[ad] are determined from the atmospheric model; the details of
which is described in Sect. 2.4 and Appendix A. The atmospheric mass is negligible, relative to the planet total mass M[p]. In our calculation, the atmospheric mass is less than 0.1% of the planetary
mass. Thus, the outer boundary conditions are given as $P=PadandT=TadatMr=Mp.$(6)As mentioned above, the pressure and temperature are also continuous at the interface between the water envelope and
the rocky core.
2.2. Thermal evolution
The thermal evolution of the planet without internal energy generation is described by (e.g., Kippenhahn & Weigert 1990) $∂L∂Mr=−T∂S∂t,$(7)where L is the intrinsic energy flux passing through the
spherical surface with radius of r, S is the specific entropy, and t is time. Since the entropy is constant in each layer, the integrated form of Eq. (7) is written as $−Lp=
∂S̅e∂t∫McMpTdMr+∂S̅c∂t∫0McTdMr,$(8)where L[p] is the total intrinsic luminosity of the planet, M[c] is the mass of the rocky core, and and are the specific entropies in the water envelope and the rocky
core, respectively. In integrating Eq. (7), we have assumed L = 0 at M[r] = 0.
In the numerical calculations of this study, we use the intrinsic temperature T[int], instead of L[p], which is defined by $Tint4≡Lp4πRp2σ,$(9)where R[p] is the planet photospheric radius (see Sect.
2.4 for the definition) and σ is the Stefan-Boltzmann constant (=5.67 × 10^-5 erg cm^-2 K^-4 s^-1).
2.3. Equation of state (EOS)
In the vapor atmosphere, the temperature and pressure are sufficiently high and low, respectively, so that the ideal gas approximation is valid. We thus adopt the ideal equation of state,
incorporating the effects of dissociation of H[2]O. In practice, we use the numerical code developed by Hori & Ikoma (2011), which calculate chemical equilibrium compositions among H[2]O, H[2], O[2],
H, O, H^+, O^+ and e^−.
At high pressures in the water envelope, the ideal gas approximation is no longer valid, because pressure due to molecular interaction is not negligible. In this study, we mainly use the water EOS H
[2]O-REOS (Nettelmann et al. 2008), which contains the ab initio water EOS data at high pressures of French et al. (2009). H[2]O-REOS covers a density range from 1.0 × 10^-6 g cm^-3 to 15 g cm^-3 and
a temperature range from 1.0 × 10^3 K to 2.4 × 10^4 K. For T and ρ outside the ranges that H[2]O-REOS covers, we use SESAME 7150 (Lyon & Johnson 1992).
The rocky core is assumed to be mineralogically the same in composition as the silicate Earth. We adopt a widely-used EOS and the Vinet EOS, and calculate thermodynamic quantities following Valencia
et al. (2007).
2.4. Atmospheric model
As described above, we consider an irradiated, radiative-equilibrium atmosphere on top of the water envelope. The thermal properties of the atmosphere govern the internal structure and evolution of
the planet. To integrate the atmospheric structure, we follow the prescription developed by Guillot (2010) except for the treatment of the opacity. Namely, we consider a semi-grey, plane-parallel
atmosphere in local thermal equilibrium. The wavelength domains of the incoming (stellar) and outgoing (planetary) radiations are assumed to be completely separated; the former is visible, while the
latter is near or mid infrared.
We solve the equation of radiative transfer by integrating the two sets (for incoming and outgoing radiations) of the zeroth and first-order moment equations for radiation with the Eddington’s
closure relation: the incoming and outgoing radiations are linked through the equation of radiative equilibrium (see Eqs. (10), (11) and (17)–(19) of Guillot 2010). Guillot (2010) derived an
analytical, approximate solution, which reproduced the atmospheric structure from detailed numerical simulations of hot Jupiters (see also Hansen 2008) well. The solution depends on opacities in the
visible and thermal domains. Guillot (2010) also presented empirical formulae for the mean opacities of solar-composition (i.e., hydrogen-dominated) gas.
However, no empirical formula is available for opacities of water vapor of interest in this study. We take into account the dependence of the water-vapor opacity on temperature and pressure and
integrate the momentum equations numerically. The details about the mean opacities and momentum equations are described in Appendix A.
The bottom of the atmosphere is assumed to be the interface between the radiative and convective zones. We use the Schwarzschild criterion (e.g., see Kippenhahn & Weigert 1990) to determine the
interface. The pressure and temperature at the interface (P[ad], T[ad]) are used as the outer boundary conditions for the structure of the convective water envelope.
The photospheric radius R[p] used in Eq. (9) is the radius at which the thermal optical depth measured from infinity, τ, is 2/3; namely, $τ=∫Rp∞κthrρdr=23,$(10)where $κthr$ is the Rosseland mean
opacity for the outgoing radiation (see Appendix A for the definition). This level is above the tropopause, the radius of which is written by R[conv] (see Fig. 1). We evaluate the atmospheric
thickness z (=R[p] − R[conv]) by integrating the hydrostatic equation from P = P[ad] to P = P[ph] using $z=−∫PadPphdPgρ=−∫PadPphℛμgTPdP,$(11)where g is the constant gravity, ℛ (=8.31 × 10^7 erg K^-1
g^-1) is the gas constant, and μ is the mean molecular weight. P[ph] is the photospheric pressure that we calculate by integrating $dPdτ=gκthr$(12)from τ = 0 to 2/3.
2.5. Mass loss
The mass loss is assumed to occur in an energy-limited fashion. Its rate, including the effect of the Roche lobe, is given by (Erkaev et al. 2007) $Ṁ=−εFXUVRpπRXUV2GMpKtide,$(13)where ε is the
heating efficiency, which is defined as the ratio of the rate of heating that results in hydrodynamic escape to that of stellar energy absorption; F[XUV] is the incident flux of X-ray and UV
radiation from the host star, K[tide] is the potential energy reduction factor due to stellar tide; and R[XUV] is the effective radius at which the planet receives the incident XUV flux. In Eq. (13),
we have assumed R[XUV] = R[p], which is a good approximation for close-in planets of interest (Lammer et al. 2013). It is noted that Lammer et al. (2013) focused on the hydrogen-helium atmosphere.
Since the scale height of the vapor atmosphere is smaller than that of a hydrogen-helium atmosphere with the same temperature, R[XUV] ≃ R[p] is a good approximation also for the vapor atmosphere.
In this study, we suppose that the host star is a G-star and adopt the empirical formula derived by Ribas et al. (2005) for F[XUV]: $FXUV={$(14)\normalsizeWe use the formula for K[tide] derived by
Erkaev et al. (2007), $Ktide=(η−1)2(2η+1)2η3,$(15)where η is the ratio of the Roche-lobe (or Hill) radius to the planetary radius, R[p].
The value of the heating efficiency is uncertain, because minor gases such as CO[2] contribute to it via radiative cooling. For photo-evaporation of hot-Jupiters, ε is estimated to be on the order of
0.1 (Yelle et al. (2008) and reference therein). Thus, we adopt ε = 0.1 as a fiducial value and investigate the sensitivity of our results to ε.
Finally, we assume that the rocky core never evaporates. That is simply because we are interested in the stability of water envelopes in this study. Whether rocky cores evaporate or not is beyond the
scope of this study.
2.6. Transit radius
The planetary radius measured via transit photometry is different from the photospheric radius defined in the preceding subsection. The former is the radius of the disk that blocks the stellar light
ray that grazes the planetary atmosphere in the line of sight. This radius is called the transit radius hereafter in this paper. Below we derive the transit radius, basically following Guillot (2010)
. Note that Guillot (2010) assumed the plane-parallel atmosphere, while we consider a spherically symmetric structure, because the atmospheric thickness is not negligibly small relative to the
planetary radius in some cases in this study.
Fig. 2
Concept of the chord optical depth.
We first introduce an optical depth that is called the chord optical depth, τ[ch] (e.g. Guillot 2010). The chord optical depth is defined as $τch(r,ν)=∫−∞+∞ρκνds,$(16)where r is the planetocentric
distance of the ray of interest (see Fig. 2), s is the distance along the line of sight measured from the point where the line is tangent to the sphere, and κ[ν] is the monochromatic opacity at the
frequency ν. Using τ[ch], we define the transit radius, R[tr], as $τch(Rtr)=23·$(17)Let the altitude from the sphere of radius r be z[tr]. Then s^2 = (r + z[tr])^2 − r^2 (Fig. 2). Eq. (16) is written
as $τch(r,ν)=2∫0∞ρκνztr+rztr2+2rztrdztr.$(18)Furthermore for convenience, we choose pressure P as the independent variable, instead of z[tr]. Using the equation of hydrostatic equilibrium, $dPdztr=
−GMpρ(r+ztr)2,$(19)one obtains $τch(ν,r)=−2gr∫Pr0κν(1+ztr/r)3(1+ztr/r)2−1dP,$(20)where $gr=GMr2$(21)and P[r] is the pressure at r. To integrate Eq. (20), we write z[tr] as a function of P. To do so,
we integrate Eq. (19) and obtain $∫0ztrdz′(r+z′)2=−∫PrPzdPGMpρ,$(22)where P[z] is the pressure at z[tr]. Eq. (22) is integrated as $1r+ztr=1r−1r2gr∫PzPrdPρ=1r−zp(Pr,Pz)r2,$(23)where $zp(Pr,Pz)
≡∫PzPrPρgrdlnP.$(24)Thus, z is written as $ztr=zp(1−zpr)-1.$(25)Note that z[p] corresponds to the altitude in the case of a plane-parallel atmosphere and (1 − z[p]/r)^-1 is the correction for
spherical symmetry.
2.7. Numerical procedure
To simulate the mass and radius evolution simultaneously, we integrate Eqs. (8) and (13) by the following procedure.
First, we simulate two adiabatic interior models that are separated in time by a time interval Δt for the known M[p](t) and an assumed M[p](t + Δt). To be exact, the two structures are integrated for
two different values of T[int]. In doing so, we integrate Eqs. (1)–(4) inward from the tropopause to the planetary center, using the fourth-order Runge-Kutta method. The inward integration is started
with the outer boundary condition given by Eq. (6); P[ad] and T[ad] are calculated according to the atmospheric model described in Sect. 2.4. We then look for the solution that fulfills the inner
boundary condition (i.e., r = 0 at M[r] = 0) in an iterative fashion. Note that determining P[ad] and T[ad] requires the gravity in the atmosphere (or R[conv]), which is obtained after the interior
structure is determined. Thus, we have to find the solution in which the interior and atmospheric structures are consistent with each other also in an iterative fashion.
Then we calculate Δt from the second-order difference equation for Eq. (8), which is written as $Δt=−([S̅e(t+Δt)−S̅e(t)][Θe(t+Δt)+Θe(t)]+[S̅c(t+Δt)−S̅c(t)][Θc(t+Δt)+Θc(t)])×(Lp(t+Δt)+Lp(t))-1,$(26)where
$Θe(t)≡∫McMp(t)T(t)dMr,Θc(t)≡∫0McT(t)dMr.$(27)Using this Δt, we integrate Eq. (13) to calculate M[p](t + Δt) as $Mp(t+Δt)=Mp(t)+ṀΔt.$(28)The assumed M[p](t + Δt) is not always equal to that obtained
here. Therefore the entire procedure must be repeated until the M[p](t + Δt) in Eq. (28) coincides with that assumed for calculating Eq. (26) with satisfactory accuracy, which is ≲0.1% in our
Once we obtain the interior and atmospheric structure, we calculate the transit radius by the procedure described in Sect. 2.6. Finally, we have confirmed that our numerical code reproduces the mass
and radius relationship for super-Earths well which is presented by Valencia et al. (2010).
3. Mass evolution
In this section, we show our numerical results of the mass evolution of a close-in water-rich planet. The evolution is controlled by the following five parameters: the initial total mass of the
planet (M[p,0]), the initial luminosity (L[0]), the initial water mass fraction (X[wt,0]), the semi-major axis (a), and the heating efficiency (ε). Below, we adopt L[0] = 1 × 10^24 erg s^-1, X[wt,0]
= 75%, a = 0.1 AU, and ε = 0.1 as fiducial values unless otherwise noted. We also show how the five parameters affect the fate of a close-in water-rich planet.
Fig. 3
Mass evolution of close-in water-rich planets. The solid blue lines represent planets that retain their water envelopes for 10 Gyr. In contrast, the planet shown by the dashed red line loses its
water envelope completely in 10 Gyr. We set L[p,0] = 1 × 10^24 erg s^-1, X[wt,0] = 75%, a = 0.1 AU, and ε = 0.1 for all the planets. In this model, we assume that the rocky core never evaporates.
3.1. Examples of mass evolution
Figure 3 shows examples of the mass evolution for water-rich planets with six different initial masses; L[0] = 1 × 10^24 erg s^-1, X[wt,0] = 75%, a = 0.1 AU, and ε = 0.1 in these simulations, as
stated above. The smallest planet loses its water envelope completely in 1 Gyr (the dashed line), while more massive planets retain their water envelopes for 10 Gyr (solid lines). This means that a
water-rich planet below a threshold mass ends up as a naked rocky planet.
The presence of such a threshold mass is understood in the following way. Using Eq. (13), we define a characteristic timescale of the mass loss (τ[M]) as $τM=|XwtMpṀp|=4GKtideXwtMpρpl3εFXUV,$(29)
where ρ[pl] is the mean density of the planet. As the planetary mass decreases, the mass-loss timescale becomes shorter. This trend is enhanced by the M − ρ relationship that the mean density
decreases as M[p] decreases, according to our numerical results for water-rich planets.
In addition, the time-dependence of the stellar XUV flux (see Eq. (14)) is a crucial factor to cause a striking difference in behavior between the low-mass and high-mass planets. Using Eq. (14), we
obtain the following relation for τ[M]: $τM≃{$(30)where $f=1(a0.1AU)2(XwtMpM⊕)(ρpl0.1gcm-3)(Ktide0.9)(ε0.1)-1.$(31)Note that 0.1 g cm^-3 is a typical value of ρ[pl] in the case of sub-Earth-mass
planets with the age of 10^8 years, according to our calculations. As seen in Eq. (30), τ[M] becomes longer rapidly with time. This implies that small planets that satisfy τ[M] < 0.1 Gyr experience a
significant mass loss. In other words, massive planets that avoid significant mass loss in the early phase hardly lose their mass for 10 Gyr. Thus, there exists a threshold mass below which a planet
never retains its water envelope for a long period. Our numerical calculations found that the threshold mass (hereafter M[thrs]) is 0.56 M[⊕] for the fiducial parameter set, which is in good
agreement with M[p] < 0.4 M[⊕] as derived from Eq. (30).
A similar threshold mass was found by Lopez & Fortney (2013) for H+He atmospheres of rocky planets. Hydrogen-rich planets are more vulnerable to the photo-evaporative mass loss than water-rich
planets. According to their study, the threshold mass of the hydrogen-rich planet at 0.1 AU is ~5 M[⊕]. That is, M[thrs] for water-rich planets is smaller by a factor of ~10 than that of
hydrogen-rich planets.
3.2. Dependence on the initial planet’s luminosity
The evolution during the first 0.1 Gyr determines the fate of a water-rich planet, as shown above. Such a trend is also shown by Lopez & Fortney (2013) for H+He atmospheres of rocky planets. This
suggests that the sensitivity of the planet’s fate to the initial conditions must be checked. In particular, the initial intrinsic luminosity may affect the early evolution of the planet
significantly, because the planetary radius, which has a great impact on the mass loss rate, is sensitive to the intrinsic luminosity; qualitatively, a large L[0] enhances mass loss because of a
large planetary radius. On the other hand, L[0] is uncertain, because it depends on how the planet forms (e.g. accretion processes of planetesimals, migration processes and giant impacts). However,
as shown below, the fate of the planet is insensitive to choice of L[0], Fig. 4 shows M[thrs] as a function of L[0] for a = 0.02,0.03,0.05 and 0.1 AU. We have found that M[thrs] is almost independent
of L[0]. This is because an initially-luminous planet cools down rapidly, so that the integrated amount of water loss during the high-luminosity phase is negligible. This is confirmed by the
following argument. The mass loss, ΔM, at the early stage can be estimated by $ΔM~ṀτKH,$(32)where τ[KH] is the typical timescale of Kelvin-Helmholtz contraction, $τKH≃GMp22RpLp·$(33)With Eqs. (29)
and (33) given, Eq. (32) can be written as $ΔM~MpτKHτM=Mpε2Ktide·πRp2FXUVLp~3×10-2(FXUV504ergcm-2s-1)(ε0.1)(Ktide0.9)-1×(a0.1AU)-2(Rp3R⊕)2(Lp1024ergs-1)-1Mp.$Because F[XUV] is constant in the early
phase, ΔM decreases as L[p] increases; that is, the Kelvin-Helmholtz contraction proceeds more rapidly. Therefore, the choice of the value of L[0] has little effect on the total amount of water loss,
as far as L[0] is larger than 10^24 erg s^-1. For smaller L[0], R[p] is insensitive to L[0]. Thus, M[thrs] is insensitive to L[0].
Fig. 4
Threshold mass in M[⊕] as a function of the initial planet’s luminosity in erg s^-1 for four choices of semi-major axes. The solid (red), dashed (green), dotted (blue), and dot-dashed (purple)
lines represent a = 0.02,0.03,0.05, and 0.1 AU, respectively. We have assumed X[wt] = 75% and ε = 0.1.
3.3. Dependence on the initial water mass fraction
The fate of a water-rich planet also depends on the initial water mass fraction, X[wt,0]. Figure 5 shows X[wt](t) at t = 10 Gyr as a function of the initial planet’s mass, M[p,0], for four different
values of X[wt,0](=25%, 50%, 75%, and 100%). As M[p,0] decreases, X[wt](10 Gyr) decreases. The pure water planet (solid line) with M[p,0] < 0.82 M[⊕] is completely evaporated in 10 Gyr; namely, X[wt]
(10 Gyr) = 0%. Otherwise, X[wt](10 Gyr) = 100%. In other cases, we find that the threshold mass, M[thrs], below which X[wt](10 Gyr) = 0%, is 0.56 M[⊕] for X[wt,0] = 75%, 0.44 M[⊕] for X[wt,0] = 50%,
and 0.44 M[⊕] for X[wt,0] = 25%.
Fig. 5
Relationship between the initial planetary mass and the fraction of the water envelope at 10 Gyr for four initial water mass fractions of X[wt,0] = 100% (solid, red), 75% (dashed, green), 50%
(dotted, blue), and 25% (dot-dashed, purple). We have assumed L[0] = 1 × 10^24 erg s^-1, a = 0.1 AU, and ε = 0.1.
Figure 6 shows the relationship between X[wt,0] and M[thrs] for four different semi-major axes. M[thrs] is found not to be a monotonous function of X[wt,0]. For X[wt,0] < 25%, M[thrs] decreases, as X
[wt,0] increases. This is explained as follows. According to Eq. (29), the mass loss timescale , τ[M], depends on the absolute amount of water, X[wt]M[p], and the planetary bulk density, ρ[pl]. When
X[wt] is sufficiently small, ρ[pl] is equal to the rocky density and is therefore constant. Thus, τ[M] is determined only by the absolute amount of water (i.e., X[wt]M[p]). This means that, M[p] must
be larger for τ[M] to be the same if X[wt,0] is small. As a consequence, M[thrs] decreases with increasing X[wt,0]. More exactly, M[thrs] changes with X[wt,0] in such a way that X[wt,0]M[thrs] is
constant. In contrast, when X[wt,0] is large, X[wt], M[p], and ρ[pl] affect the mass loss timescale. For a given M[p], an increase in X[wt,0] leads to a decrease in ρ[pl] (or, an increase in radius),
which enhances mass loss. As a result, M[thrs] increases with X[wt,0] for X[wt,0] > 25%. Therefore, there is a minimum value of M[thrs], which is hereafter denoted by $Mthrs∗$.
Similar trends can be seen in Figs. 3 and 4 of Lopez & Fortney (2013). To compare our results for water-rich planets to those for hydrogen-rich rocky planets from Lopez & Fortney (2013) in a more
straightforward way, we show the relationship between the initial total mass and the fraction of the initial water envelope that is lost via subsequent photo-evaporation in 5 Gyr in Fig. 7 (see Fig.
3c of Lopez & Fortney 2013). We set L[0] = 1 × 10^24 erg s^-1, a = 0.1 AU, ε = 0.1, and six initial water mass fractions of X[wt,0] = 1% (solid, red), 3% (long-dashed, green), 10% (dotted, blue), 30%
(dash-dotted, purple), 50% (dot-dashed, light blue), and 60% (dashed, black), which are similar to those adopted by Lopez & Fortney (2013). As mentioned above, the initial total mass needed in the
H+He case is larger by a factor of ~10 than that in the water case for the same fraction of the initial envelope to survive photo-evaporation. In addition, the required initial total mass for X
[wt,0] < 10% becomes significantly large in the water case. This behavior is also found in the case of the hydrogen-rich planets for X[wt,0] = 1–3%. However, the trend is less noticeable in the
H+He case. This is because the density effect described above is effective even for small H+He fractions.
Fig. 6
Relationship between the initial water mass fraction X[wt,0] in % and the threshold mass M[thrs] in M[⊕] for four choices of semi-major axes of 0.02 AU (solid, red), 0.03 AU (dashed, green), 0.05
AU (dotted, blue), and 0.1 AU (dot-dashed, purple). We have assumed L[0] = 1 × 10^24 erg s^-1 and ε = 0.1.
Fig. 7
Relationship between the initial planetary mass and the fraction of the initial water envelope that is lost via photo-evaporation in 5 Gyr for six initial water mass fractions of X[wt,0] = 1%
(solid, red), 3% (long-dashed, green), 10% (dotted, blue), 30% (dash-dotted, purple), 50% (dot-dashed, light blue), and 60% (dashed, black). We have assumed L[0] = 1 × 10^24 erg s^-1, a = 0.1 AU,
and ε = 0.1.
3.4. Dependence on the semi-major axis
At small a, the incident stellar XUV flux becomes large. Thus, M[thrs] increases, as a decreases. Certainly, the distance to the host star affects the equilibrium temperature T[eq], which has an
influence on ρ[pl]: the higher T[eq] is, the smaller ρ[pl] is. However, its impact on M[thrs] is small, relative to that of F[XUV]. According to the planet’s mass and mean density relationship, ρ[pl]
differs only by a factor of ≲1.5 between 880 K and 2000 K. Therefore, increasing F[XUV] has a much greater impact on the mass loss than decreasing ρ[pl]. In Fig. 6, we find $Mthrs∗=5.2M⊕$ for a =
0.02 AU, $Mthrs∗=2.5M⊕$ for a = 0.03 AU, $Mthrs∗=1.2M⊕$ for a = 0.05 AU, and $Mthrs∗=0.44M⊕$ for a = 0.1 AU.
3.5. Expected populations
Figure 8 shows the relationship between M[thrs] (not $Mthrs∗$) and the radius that the planet with M[thrs] would have at 10 Gyr without mass loss (solid line). We call this radius the threshold
radius, R[thrs]. We have calculated R[thrs] for X[wt,0] = 100%, 75%, 50%, 25%, 10%, 5%, and 1%. In addition, the mass-radius relationships for rocky planets (dashed line) and pure-water planets
(dotted line) at 0.1 AU are also drawn in Fig. 8. There are four characteristic regions in Fig. 8:
• I
Planets must contain components less dense than water, such ashydrogen/helium.
• II
Planets with water envelopes and without H/He can exist. The water envelopes survive photo-evaporative mass loss.
• III
Primordial water envelopes experience significant photo-evaporative mass loss in 10 Gyr.
• IV
Planets retain no water envelopes and are composed of rock and iron.
Only in the region II, the planet retains its primordial water envelope for 10 Gyr without significant loss. There are minimum values not only of M[thrs] but also of R[thrs]; the latter is denoted by
$Rthrs∗$ hereafter. Note that $Rthrs∗$ is not an initial radius.
Those minimum values are helpful to discuss whether planets can possess water components or not, because the uncertainty in water mass fractions can be removed. Since M[thrs] and R[thrs] depend on
semi-major axis, we also compare those threshold values with observed M − a and R − a relationships in the next section.
Fig. 8
Relationship between the threshold mass and the threshold radius. The latter is defined by the radius that the planet with M[thrs] would have at 10 Gyr without ever experiencing mass loss ( denoted
by R[thrs]). The squares, which are connected with a solid line, are M[thrs] and R[thrs] for 0.1 AU and seven different initial water mass fractions X[wt,0] = 100%, 75%, 50%, 25%, 10%, 5%, 1%, and
0.5%. The dashed and dotted lines represent mass-radius relationships, respectively, for rocky planets and pure-water planets at 0.1 AU. $Mthrs∗$ and $Rthrs∗$ represent the minimum values of M
[thrs] and R[thrs], respectively.
4. Implications for distributions of observed exoplanets
Figure 9 compares the relationship between the threshold mass, M[thrs], and threshold radius, R[thrs] with measured masses and radii of super-Earths around G-type stars identified so far. Here we
show three theoretical relationships for a = 0.02, 0.05, and 0.1 AU. As discussed above, only planets on the right side of the theoretical line (i.e., in region II) for a given a are able to retain
their water envelopes without significant loss for 10 Gyr.
For future characterizations, planets in region III would be of special interest, because our results suggest that planets should be rare in region III. Three out of the 14 planets, 55 Cnc e,
Kepler-20 b, and CoRoT-7 b might be in region III, although errors and the uncertainty in ε (see also the lower panel of Fig. 10 for the sensitivity of $Mthrs∗$ to ε) are too large to conclude so.
There are at least three possible scenarios for the origin of planets in region III. One is that those planets are halfway to complete evaporation of their water envelopes. Namely, some initial
conditions happen to make planets in region III, although such conditions are rare. The second possible scenario is that those planets had formed far from and migrated toward their host stars
recently. The third is that those planets are in balance between degassing from the rocky core and the atmospheric escape. Thus, deeper understanding of the properties of those super-Earths via
future characterization will provide important constraints on their origins.
Fig. 9
Relationship between the threshold mass M[thrs] and radius R[thrs] (lines; see text for definitions) compared to masses and radii of observed transiting super-Earths around G-type stars (points
with error bars; exoplanets.org (Wright et al. 2011), as of June 29, 2013, ). The dotted (blue), dashed (green), and solid (red) lines represent the M[thrs] and R[thrs] relationships for orbital
periods of 11 days (=0.1 AU), 4 days (=0.05 AU), and 1 day (=0.02 AU), respectively. The dash-dotted (brown) line represents the planet composed of rocks. Note that black points represent planets
whose orbital periods are longer than 11days. In those calculations, we have assumed the heating efficiency ε = 0.1 and the initial luminosity L[0] = 1 × 10^24 erg s^-1. “CoR” are short for CoRoT
and “Kep” are short for Kepler.
Fig. 10
Upper panel: theoretical distribution of masses and semi-major axes (or incident fluxes) of planets at 10 Gyr with various initial masses and water mass fractions. Crosses (red) represent planets
that lost their water envelopes completely in 10 Gyr, while open squares (blue) represent planets that survive significant loss of the water envelopes via photo-evaporation. The green line is the
minimum threshold masses, $Mthrs∗$. Here, we have adopted ε = 0.1. Lower panel: distribution of masses and semi-major axes (or incident fluxes) of detected exoplanets compared to the minimum
threshold mass, $Mthrs∗$, derived in this study (see Sect. 3.3 for definition). We have shown three $Mthrs∗−a$ relationships for different heating efficiencies: ε = 1 (solid line), ε = 0.1 (dashed
line), and ε = 0.01 (dotted line). Filled circles with error bars represent observational data (from http://exoplanet.org (Wright et al. 2011)) for planets orbiting host stars with effective
temperature of 5000–6000 K (relatively early K-type stars and G-type stars). Planets are colored according to their zero-albedo equilibrium temperatures in K. In the planet names, “CoR” and “Kep”
stand for CoRoT and Kepler, respectively.
Fig. 11
Upper panel: theoretical distribution of radii and semi-major axes (or incident fluxes) of planets at 10 Gyr with various initial masses and water mass fractions. Crosses (red) represent planets
that lost their water envelopes completely due to the photo-evaporation in 10 Gyr, while open squares (blue) represent planets that survive significant loss of the water envelopes. The green line
is the minimum threshold radii, $Rthrs∗$. Here, we have adopted ε = 0.1. Lower panel: distribution of radii and semi-major axes (or incident fluxes) of Kepler planetary candidates, compared to the
threshold radius, $Rthrs∗$ (see Sect. 3.3 for definition). We have shown three $Rthrs∗−a$ relationships for different heating efficiencies: ε = 1 (solid red line), ε = 0.1 (dashed green line), and
ε = 0.01 (dotted blue line). Filled squares represent observational data (http://kepler.nasa.gov, as of June 29, 2013) for planets orbiting host stars with effective temperature of 5300–6000 K
(G-type stars).
In this study, low-mass exoplanets, whose masses are ≤20 M[⊕] and radii ≤4 R[⊕], are of special interest. (We call them super-Earths below.) While there are only 14 super-Earths whose masses and
radii were both measured (see Fig. 9), the minimum masses (M[p]sini) and the orbital periods were measured for about 22 super-Earths around G-type stars (see Fig. 10). Also, over 1000 sub/
super-Earth-sized planet candidates have been identified by the Kepler space telescope (Batalha et al. 2013). The size and semi-major axis distribution of those objects is known. It is, thus,
interesting to compare our theoretical prediction with the observed M[p]–a and R[p]–a distributions.
Before doing so, we demonstrate that $Mthrs∗$ and $Rthrs∗$ are good indicators for constraining the limits below which evolved planets retain no water envelopes. Figure 10a and 11a show the
theoretical distributions of masses and radii of planets that evolved for 10 Gyr, starting with various initial water mass fractions and planetary masses (i.e., X[wt,0] = 25,50,75 and 100% and log(M
[p,0]/M[⊕]) = −1 + 0.1j with j = 0,1,··· ,21). The crosses (red) and open squares (blue) represent the planets that lost their water envelopes completely (i.e., rocky planets) and those which
survive significant loss of their water envelopes, respectively. As seen in these figures, two populations of rocky planets and water-rich planets are clearly separated by the $Mthrs∗$ and $Rthrs∗$
lines. Note that there are some planets that retain their water envelope below the threshold line. These planets just retain ≲1% water mass fraction at 10 Gyr. However, such planets are found to be
obviously rare.
In Fig. 10b, we show the distribution of M[p]sini and a of low-mass exoplanets detected around G-type and K-type stars so far, as compared with $Mthrs∗$ for three choices of ε. Among them, α Cen B b,
Kepler-10 b and CoRoT-7 b are well below the $Mthrs∗$ line for ε = 0.1. Thus, the three planets are likely to be rocky, provided ε = 0.1. However, the uncertainty in ε (and F[XUV]) prevents us from
deriving a robust conclusion. An order-of-magnitude difference in ε is found to change $Mthrs∗$ by a factor of three. The aforementioned three planets are between the two $Mthrs∗$ lines for ε = 0.01
and 0.1. This demonstrates quantitatively how important determining ε and F[XUV] more accurately is for understanding the composition of super-Earths only with measured masses. It would be worth
mentioning that few planets are found between the lines for ε = 0.1 and ε = 1. Since all the planets in Fig. 10b were found by the radial-velocity method, the apparent gap would be unlikely to be due
to observational bias. Thus, the gap might suggest that the actual $Mthrs∗$ line lies between those two ones.
In Fig. 11b, we show the distribution of R[p] and a of KOIs, which is compared with $Rthrs∗$ for three choices of ε. Many planets are found to be below the $Rthrs∗$ lines. We are unable to constrain
the fraction of rocky planets quantitatively, because of the uncertainty in ε. However, since there are many points below the $Rthrs∗$ line for ε of as small as 0.01, it seems to be a robust
conclusion that KOIs contain a significant number of rocky planets. Note that the distribution must include rocky planets that were formed rocky without ever experiencing mass loss. This means that
there are more rocky planets in reality than we have predicted in this study.
As mentioned in the Introduction, Lopez & Fortney (2013) performed a similar investigation of the threshold mass and radius concerning H+He atmospheres on rocky super-Earths (see Figs. 8 and 9 of
Lopez & Fortney 2013). For the horizontal axis, they adopted the incident stellar flux, instead of semi-major axis. In Figs. 10 and 11, we have also indicated another scale of the incident flux
calculated from the relationship between the semi-major axis a and the incident flux F, $F=Lstar4πa2=FEarth(LstarL⊙)(a1AU)-2,$(36)where L[star] is the luminosity of the host star and F[Earth] is the
current bolometric flux that the Earth receives from the Sun. Comparing their results for the H+He envelope, we find that the threshold value of the initial mass (or incident flux) for H[2]O is
smaller by a factor of about 10 than that for H+He although a similar linear dependence is found. For example, the threshold mass for H+He is ~30 M[⊕] (derived by Eq. (6) of Lopez & Fortney
2013) in the case of F = 10^3F[⊕], while it is for H[2]O is ~2 M[⊕].
In Fig. 9 of Lopez & Fortney (2013), it has also been suggested that the frequency of planets with radii of 1.8–4.0 R[⊕] for F[p] ≥ 100 F[⊕] (corresponding to a ≤ 0.1 AU) should be low as a
consequence of photo-evaporative mass loss. Owen & Wu (2013) also found a deficit of planets around 2 R[⊕] in their planet distribution (see Fig. 8 of Owen & Wu 2013). In contrast, our results
suggest that water-rich planets with radii of 1.5–3.0 R[⊕] are relatively common, because they are able to sustain their water envelopes against photo-evaporation. This seeming disagreement on the
predicted distribution demonstrates the influence of the envelope composition on the predicted distribution. Indeed, there are many KOIs found in such a domain in the R[p]-a diagram shown in Fig. 11
a. Thus, those KOIs may be water-rich planets, although it is also possible that they are rocky planets without ever experiencing mass loss.
Finally, we focused in this study on the thermal escape of the upper atmosphere due to stellar XUV irradiation. In addition, ion pick-up induced by stellar winds and coronal mass ejections may be
effective in stripping off atmospheres of close-in planets, as discussed for close-in planets with hydrogen-rich atmospheres (e.g. Lammer et al. 2013). Such non-thermal effects lead to increase in
$Mthrs∗$. This implies that the $Mthrs∗$ obtained in this study is a lower limit on survival of water-rich planets.
5. Summary
In this study, we have investigated the impact of photo-evaporative mass loss on masses and radii of water-rich sub/super-Earths with short orbital periods around G-type stars. We simulated the
interior structure and the evolution of highly-irradiated sub/super-Earths that consist of a rocky core surrounded by a water envelope, including the effect of mass loss due to the stellar XUV-driven
energy-limited hydrodynamic escape (see Sect. 2).
The findings from this study are summarized as follows. In Sect. 3, we have investigated the mass evolution of water-rich sub/super-Earths, and then found a threshold planet mass M[thrs], below which
the planet has its water envelope stripped off in 1–10 Gyr (Sect. 3.1). The initial planet’s luminosity has little impact on M[thrs] (Sect. 3.2). We have found that there is a minimum value, $Mthrs∗$
, for given a and ε (Sect. 3.4). Water-rich planets with initial masses smaller than $Mthrs∗$ lose their water envelopes completely in 10 Gyr, independently of initial water mass fraction. The
threshold radius, R[thrs], is defined as the radius that the planet of mass M[thrs] would have at 10 Gyr if it evolved without undergoing mass loss. We have also found that there is a minimum value
of the threshold radius, $Rthrs∗$ (Sect. 3.5). Finally, we have discussed the composition of observed exoplanets in Sect. 4 by comparing the threshold values to measured masses and radii of the
exoplanets. Then, we have confirmed quantitatively that more accurate determination of planet masses and radii, ϵ and F[XUV], respectively is needed for deriving robust prediction for planetary
composition. Nevertheless, the comparison between $Rthrs∗$ and radii of KOIs in the R[p] − a plane suggests that KOIs contain a significant number of rocky planets.
In this study, we have demonstrated that photo-evaporative mass loss has a significant impact on the evolution of water envelopes of sub/super-Earths, especially with short orbital periods, and that
of H+He envelopes of super-Earths. Since the M[thrs] for water envelope models is larger by a factor of 10, relative to that for H+He envelope models by Lopez & Fortney (2013), the stability
limit for water envelopes gives more robust constraints on the detectability of rocky planets. Thus, the M[thrs] and R[thrs] will provide valuable information for future searches of rocky Earth-like
Online material
Appendix A: Atmospheric model
First, we describe opacity models for the water vapor atmosphere. We define the Planck-type (κ^P) and the Rosseland-type mean opacities (κ^r) as $κvp=∫visibleκνBν(T⋆)dν/∫visibleBν(T⋆)dν,1κvr=κthp=
∫thermalκνBν(Tatm)dν/∫thermalBν(Tatm)dν,1κthr=∫thermal1κνdBν(Tatm)dTdν/∫thermaldBν(Tatm)dTdν,$where ν is the frequency; κ[ν] the monochromatic opacity at a given ν; T[⋆] the stellar effective
temperature; T[atm] the atmospheric temperature of the planet; and B[ν] the Planck function. The subscripts, “th” and “v”, mean opacities in the thermal and visible wavelengths, respectively. In this
study, we assume T[⋆] = 5780 K. We adopt HITRAN opacity data for water (Rothman et al. 2009) and calculate mean opacities for 1000 K, 2000 K, and 3000 K at 1, 10, 100 bar. Mean opacities are fitted
to power-law functions of P and T, using the least squares method; $κvp=1.94×104(P1bar)0.01(T1000K)1.0cm2g-1,κvr=2.20(P1bar)1.0(T1000K)-0.4cm2g-1,κthp=4.15×105(P1bar)0.01(T1000K)-1.1cm2g-1,κthr=
3.07×102(P1bar)0.9(T1000K)-4.0cm2g-1,$where P is the pressure and T the temperature.
In this study, we basically follow the prescription developed by Guillot (2010) except for the treatment of the opacity. We consider a static, plane-parallel atmosphere in local thermodynamic
equilibrium. We assume that the atmosphere is in radiative equilibrium between an incoming visible flux from the star and an outgoing infrared flux from the planet. Thus, the radiation energy
equation and radiation momentum equation are written as $dHvdm=κvpJv,dKvdm=κvrHv,dHthdm=κthp(Jth−B),dKthdm=κthrHth,$and the atmosphere in radiative equilibrium satisfies $κvpJv+κthp(Jth−B)=0,$(A.13)
where J[v] (J[th]), H[v] (H[th]), and K[v] (K[th]) are, respectively, the zeroth-, first-, and second-order moments of radiation intensity in the visible (thermal) wavelengths, m the atmospheric mass
coordinate, dm = ρdz, where z is the altitude from the bottom of the atmosphere, ρ the density, and B the frequency-integrated Planck function, $B≡∫thermalBνdν~σπT4,$(A.14)where σ is the
Stefan-Boltzmann constant. We assume here that thermal emission from the atmosphere at visible wavelengths are negligible, so that B[ν] ~ 0 in the visible region. The six moments of the radiation
field are defined as $(Jv,Hv,Kv)≡∫visible(Jν,Hν,Kν)dν,(Jth,Hth,Kth)≡∫thermal(Jν,Hν,Kν)dν,$where J[ν] is the mean intensity, 4πH[ν] the radiation flux, and 4πK[ν]/c the radiation pressure (c is the
speed of light).
We integrate three moments of specific intensity, J[ν],H[ν] and K[ν], over all the frequencies: $J≡∫∞0Jνdν=12∫0∞dν∫-11dμIν,μ=Jv+Jth,H≡∫∞0Hνdν=12∫0∞dν∫-11dμIν,μμ=Hv+Hth,K≡∫0∞Kνdν=12∫0∞dν∫-11dμIν,μμ2=
Kv+Kth,$where I[ν,μ] is the specific intensity and θ the angle of a intensity with respect to the z-axis, μ = cosθ. The energy conservation of the total flux implies $H=Hv+Hth=14πσTint4,$(A.20)where
T[irr] is the irradiation temperature given by $Tirr=T⋆R⋆a,$(A.21)where R[⋆] is the radius of the host star and a the semi-major axis.
For the closure relations, we use the Eddington approximation (e.g. Chandrasekhar 1960), namely, $Kv=13Jv,Kth=13Jth.$For an isotropic case of both the incoming and outgoing radiation fields, we find
boundary conditions of the moment equations as follows (see also Guillot 2010, for details): $Hv(m=0)=−1314πσTirr4,Hv(m=0)=−13Jv(m=0),Hth(m=0)=12Jth(m=0).$Thus, we integrate Eqs. (A.9)–(A.13) over m
numerically, using mean opacities of (A.5)–(A.8) and boundary conditions of (A.24)–(A.26), and then determine a T–P profile of the water vapor atmosphere. We assume that the boundary is at P[0] = 1 ×
10^-5 bar. The choice of P[0] (≤1 × 10^-5bar) has little effect on the atmospheric temperature-pressure structure. T[0] is determined in an iterative fashion until abs(T[0] − [πB(m = 0,P[0],T[0])/σ]
^1/4) ≤ 0.01 is fulfilled. Then we integrate Eqs. (A.9)–(A.13) over m by the 4th-order Runge-Kutta method, until we find the point where dlnT/dlnP ≥ ∇[ad]. The pressure and temperature, P[ad] and T
[ad], are the boundary conditions for the convective-interior structure (see Sect. 2.1).
In Fig. A.1, we show the P–T profile for the solar-composition atmosphere with g = 980 cm s^-2, T[int] = 300 K, and T[irr] = 1500 K (dotted line). In this calculation, we take $κthr$ and $κthp$ as
functions of P and T from Freedman et al. (2008) and calculate $κvp$ and $κvr$, for P = 1 × 10^-3,0.1,1,10 bar and T = 1500 K from HITRAN and HITEMP data that include H[2], He, H[2]O, CO, CH[4], Na,
and K for the solar abundance respectively as $κv={$(A.27)by use of (A.2). The thin and thick parts of the dotted line represent the radiative and convective zones, respectively.
In addition, we test our atmosphere model by comparing it with the P–T profile derived by Guillot (2010) with γ = κ[v]/κ[th] = 0.4 (solid line), which reproduces more detailed atmosphere models by
Fortney et al. (2005) and Iro et al. (2005, see Fig. 6 of Guillot 2010). As seen in Fig. A.1, our atmospheric model yields a P–T profile similar to that from Guillot (2010). In our model,
temperatures are relatively low compared with the Guillot (2010) model at P ≲ 40 bar, which is due to difference in opacity. In our model, deep regions of P ≳ 40 bar are convective, while there is no
convective region in the Guillot (2010) model because of constant opacity. We have compared our P–T profile with the Fortney et al. (2005)’s and Iro et al. (2005)’s profiles, which are shown in Fig.
6 of Guillot (2010) and confirmed that our P–T profile in the convective region is almost equal to their profiles. Of special interest in this study is the entropy at the radiative/convective
boundary, because it governs the thermal evolution of the planet. In this sense, it is fair to say that our atmospheric model yields appropriate boundary conditions for the structure of the
convective interior.
Fig. A.1
Temperature–pressure profiles for a solar-composition atmosphere (see the details in text). The solid (red) and dotted (green) lines represent both Guillot (2010)’s (γ = 0.4) and our model’s,
respectively. The thin and thick parts of the dotted line represent the radiative and convective regions, respectively. We have assumed g = 980 cm s^-2, T[int] = 300 K, and T[irr] = 1500 K.
Finally, we describe an analytical expression for our atmospheric model. We basically follow the prescription developed by Heng et al. (2012), except for the treatment of the opacity. As Heng et al.
(2012) mentioned, it would be a challenging task without assumption of constant $κvp$ and $κvr$ to obtain analytical solutions for J[v] and H[v]. Here we assume $κvp$ and $κvr$ are constant
throughout the atmosphere. We differentiate (A.9) and (A.10) by m and obtain $d2Jvdm2=Hvμ2dκvrdm+κvrκvpμ2Jv,d2Hvdm2=Jvdκvpdm+κvrκvpμ2Hv,$where μ^2 = K[v]/J[v]. Assuming J[v] = H[v] = 0 as m → ∞, we
obtain $(Jv,Hv)=(Jv,0,Hv,0)exp(−κv¯μm),$(A.30)where $κv¯=κvpκvr$ and J[v,0] and H[v,0] are the values of J[v] and H[v] evaluated at m = 0, respectively. In general, the heat transportation, such as
circulation, produces a specific luminosity of heat. Heng et al. (2012) introduced the specific luminosity as Q, which has units of erg s^-1 g^-1. Q can be related to the moments of the specific
intensity and we obtain $κthp(Jth−B)+κvpJv=Q.$(A.31)We integrate Eq. (A.31) and obtain $H=H∞−˜Q(m,∞),$(A.32)where H[∞] is the value of H evaluated at m → ∞ and $˜Q(m1,m2)=∫m1m2Q(m′,μ,φ)dm′.$(A.33)To
obtain H[th] and J[th], we substitute Eq. (A.31) in Eqs. (A.11) and (A.12) and integrate by m. Then we obtain $Hth=H∞−Hv,0exp(−κv¯μm)−˜Q(m,∞)Jth=Jth,0−Hv,0fKth∫0mκthrexp(−κv¯μm′)dm′+1fKth∫0mκthr
{H∞−˜Q(m′,∞)}dm′,$where f[Kth] = K[th]/J[th], f[Hth] = H[th]/J[th], and $Jth,0=1fHth{H∞−Hv,0−˜Q(0,∞)}.$(A.36)That is, we obtain $B=H∞[1fHth+1fKthτth(m)]−Hv,0[1fHth+κv¯μκthp+1fKthτext(m)]+E(m),$(A.37)
where $τth(m)=∫m0κthrdm′,τext(m)=∫m0(κth¯2−fKthμ2κv¯2)1κthpexp(−κv¯μm′)dm′,E(m)=−[Qκthp+1fKth∫0mκthr˜Q(m′,∞)dm′+˜Q(0,∞)fHth],$and $κth¯=κthpκthr$. In our conditions, we assume $μ=1/3$, f[Kth] = 1/3,
f[Hth] = 1/2 and Q = 0. Consequently, we obtain
the temperature profile as $T4=34Tint4[23+τth(m)]+34Tirr4[23+κv¯3κthp+τext(m)]$(A.41)where$τext(m)=∫0mκth¯2−κv¯2κthpexp(−3κv¯m′)dm′.$(A.42)If we assume $κthp=κthr$ and $κvp=κvr$, Eq. (A.41) agrees
with Eq. (27) of Heng et al. (2012).
We thank N. Nettelmann for providing us with tabulated data for the equation of state of water (H[2]O-EOS) and S. Ida and T. Guillot for fruitful advice and discussions. We also thank the anonymous
referee for his/her careful reading and constructive comments that helped us improve this paper greatly. We also thank Y. Ito and Y. Kawashima for providing us with the opacity data and fruitful
suggestions about the atmospheric structure. This research has made use of the Exoplanet Orbit Database and the Exoplanet Data Explorer at exoplanets.org. This study is supported by Grants-in-Aid for
Scientific Research on Innovative Areas (No. 23103005) and Scientific Research (C) (No. 25400224) from the Ministry of Education, Culture, Sports, Science and Technology (MEXT) of Japan. K. K. is
supported by a grant for the Global COE Program, “From the Earth to “Earths””, of MEXT, Japan. Y. H. is supported by the Grant-in-Aid for JSPS Fellows (No. 23003491) from MEXT, Japan.
All Figures
Fig. 1
Model of the planetary structure in this study.
In the text
Fig. 2
Concept of the chord optical depth.
In the text
Fig. 3
Mass evolution of close-in water-rich planets. The solid blue lines represent planets that retain their water envelopes for 10 Gyr. In contrast, the planet shown by the dashed red line loses its
water envelope completely in 10 Gyr. We set L[p,0] = 1 × 10^24 erg s^-1, X[wt,0] = 75%, a = 0.1 AU, and ε = 0.1 for all the planets. In this model, we assume that the rocky core never evaporates.
In the text
Fig. 4
Threshold mass in M[⊕] as a function of the initial planet’s luminosity in erg s^-1 for four choices of semi-major axes. The solid (red), dashed (green), dotted (blue), and dot-dashed (purple)
lines represent a = 0.02,0.03,0.05, and 0.1 AU, respectively. We have assumed X[wt] = 75% and ε = 0.1.
In the text
Fig. 5
Relationship between the initial planetary mass and the fraction of the water envelope at 10 Gyr for four initial water mass fractions of X[wt,0] = 100% (solid, red), 75% (dashed, green), 50%
(dotted, blue), and 25% (dot-dashed, purple). We have assumed L[0] = 1 × 10^24 erg s^-1, a = 0.1 AU, and ε = 0.1.
In the text
Fig. 6
Relationship between the initial water mass fraction X[wt,0] in % and the threshold mass M[thrs] in M[⊕] for four choices of semi-major axes of 0.02 AU (solid, red), 0.03 AU (dashed, green), 0.05
AU (dotted, blue), and 0.1 AU (dot-dashed, purple). We have assumed L[0] = 1 × 10^24 erg s^-1 and ε = 0.1.
In the text
Fig. 7
Relationship between the initial planetary mass and the fraction of the initial water envelope that is lost via photo-evaporation in 5 Gyr for six initial water mass fractions of X[wt,0] = 1%
(solid, red), 3% (long-dashed, green), 10% (dotted, blue), 30% (dash-dotted, purple), 50% (dot-dashed, light blue), and 60% (dashed, black). We have assumed L[0] = 1 × 10^24 erg s^-1, a = 0.1 AU,
and ε = 0.1.
In the text
Fig. 8
Relationship between the threshold mass and the threshold radius. The latter is defined by the radius that the planet with M[thrs] would have at 10 Gyr without ever experiencing mass loss ( denoted
by R[thrs]). The squares, which are connected with a solid line, are M[thrs] and R[thrs] for 0.1 AU and seven different initial water mass fractions X[wt,0] = 100%, 75%, 50%, 25%, 10%, 5%, 1%, and
0.5%. The dashed and dotted lines represent mass-radius relationships, respectively, for rocky planets and pure-water planets at 0.1 AU. $Mthrs∗$ and $Rthrs∗$ represent the minimum values of M
[thrs] and R[thrs], respectively.
In the text
Fig. 9
Relationship between the threshold mass M[thrs] and radius R[thrs] (lines; see text for definitions) compared to masses and radii of observed transiting super-Earths around G-type stars (points
with error bars; exoplanets.org (Wright et al. 2011), as of June 29, 2013, ). The dotted (blue), dashed (green), and solid (red) lines represent the M[thrs] and R[thrs] relationships for orbital
periods of 11 days (=0.1 AU), 4 days (=0.05 AU), and 1 day (=0.02 AU), respectively. The dash-dotted (brown) line represents the planet composed of rocks. Note that black points represent planets
whose orbital periods are longer than 11days. In those calculations, we have assumed the heating efficiency ε = 0.1 and the initial luminosity L[0] = 1 × 10^24 erg s^-1. “CoR” are short for CoRoT
and “Kep” are short for Kepler.
In the text
Fig. 10
Upper panel: theoretical distribution of masses and semi-major axes (or incident fluxes) of planets at 10 Gyr with various initial masses and water mass fractions. Crosses (red) represent planets
that lost their water envelopes completely in 10 Gyr, while open squares (blue) represent planets that survive significant loss of the water envelopes via photo-evaporation. The green line is the
minimum threshold masses, $Mthrs∗$. Here, we have adopted ε = 0.1. Lower panel: distribution of masses and semi-major axes (or incident fluxes) of detected exoplanets compared to the minimum
threshold mass, $Mthrs∗$, derived in this study (see Sect. 3.3 for definition). We have shown three $Mthrs∗−a$ relationships for different heating efficiencies: ε = 1 (solid line), ε = 0.1 (dashed
line), and ε = 0.01 (dotted line). Filled circles with error bars represent observational data (from http://exoplanet.org (Wright et al. 2011)) for planets orbiting host stars with effective
temperature of 5000–6000 K (relatively early K-type stars and G-type stars). Planets are colored according to their zero-albedo equilibrium temperatures in K. In the planet names, “CoR” and “Kep”
stand for CoRoT and Kepler, respectively.
In the text
Fig. 11
Upper panel: theoretical distribution of radii and semi-major axes (or incident fluxes) of planets at 10 Gyr with various initial masses and water mass fractions. Crosses (red) represent planets
that lost their water envelopes completely due to the photo-evaporation in 10 Gyr, while open squares (blue) represent planets that survive significant loss of the water envelopes. The green line
is the minimum threshold radii, $Rthrs∗$. Here, we have adopted ε = 0.1. Lower panel: distribution of radii and semi-major axes (or incident fluxes) of Kepler planetary candidates, compared to the
threshold radius, $Rthrs∗$ (see Sect. 3.3 for definition). We have shown three $Rthrs∗−a$ relationships for different heating efficiencies: ε = 1 (solid red line), ε = 0.1 (dashed green line), and
ε = 0.01 (dotted blue line). Filled squares represent observational data (http://kepler.nasa.gov, as of June 29, 2013) for planets orbiting host stars with effective temperature of 5300–6000 K
(G-type stars).
In the text
Fig. A.1
Temperature–pressure profiles for a solar-composition atmosphere (see the details in text). The solid (red) and dotted (green) lines represent both Guillot (2010)’s (γ = 0.4) and our model’s,
respectively. The thin and thick parts of the dotted line represent the radiative and convective regions, respectively. We have assumed g = 980 cm s^-2, T[int] = 300 K, and T[irr] = 1500 K.
In the text
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Methods of Compaction Control: Sand Replacement Method and Core Cutter Method
In this article, we will discuss methods of compaction control.
During the compaction of soils in the field, it is necessary to determine the dry density and water content of the compacted soil. More the dry density and water content, the more the shear strength
and stability of the soil.
To check the dry density of soil, wet/ bulk density, and water content are measured by sand replacement method for cohesion less soil and core cutter method for cohesive soil.
After calculating wet density and water content, dry density can be determined by
γ[d] = γ[t] /(1+w)
γ[d] = dry density
γ[t] = wet/ bulk density = (weight of soil/total volume of soil mass)= W/V
w= water content= (weight of water/weight of soil solids)= W[w] /W[s]
Note: water content used in the field compaction is termed ‘Placement water content’ with values lower, higher, or equal to OMC(Optimum Moisture Content).
1. Methods of Compaction Control
There are two methods of compaction control; Sand Replacement Method and Core Cutter Method.
A. Sand Replacement Method
i. Aim
To determine the dry density of a given soil sample (cohesion less soil).
ii. Apparatus
Sand bottle, Conical funnel, Sand, Square tray with a circular hole
iii. Procedures
a. A square tray with a circular hole in it is placed in the level ground and pressed on the ground to make hole
b. A small quantity of excavated soil from the hole is weighed and its water content is measured.
c. Using a sand bottle with a conical funnel, the hole is filled with sand of known density.
d. Weight of sand filled in the hole can be measured using readings of weights calculated.
iv. Observations
weight of soil mass = W
weight of sand bottle before filling of sand = W[1]
weight of sand bottle after filling of sand = W[2]
weight of sand filling with conical funnel = W[3]
weight of sand filled in hole = W[4]
unit weight of sand=γ
volume of sand = V = W[4] /γ
W[4] = W[1] – W[2] – W[3]
water content =w
we have,
bulk density(γ[t]) = W/V
dry density (γ[d]) = γ[t] /(1+w)
v. Result:
Hence, dry density is calculated using the sand replacement method.
Steps to calculate the dry density using the sand replacement method:
a. Calculate the weight of sand filled in the hole.
W[4] = W[1] – W[2] – W[3]
b. Calculate the volume of sand.
V = W[4] /γ
c. Calculate bulk density /wet density.
γ[t] = W/V
d. Calculate dry density.
γ[d] = γ[t] /(1+w)
sometimes, bulk density could be separately used as γ[d] = γ /(1+w)
B. Numerical
The following results were obtained from the sand replacement method:
mass of soil excavated from the hole; 4 kg
the water content of soil; 18%
mass of dry soil to fill a hole; 3.1 kg
mass of dry sand to fill the container; 5.8 kg
the volume of the container; is 4.2litres
Calculate the wet and dry densities of the soil, if the specific gravity of the particles is 2.68.
Given; weight of sand = 5.8 kg
volume of sand = 4.2 × 10^-3 m^3
weight of sand to fill hole = 3.1 kg
weight of soil excavated = 4 kg
water content (w) = 18%
The wet density of soil( γ[t] ) =?
the dry density of soil ( γ[d])=?
Density of sand =5.8/(4.2 × 10^-3 ) = 1380 kg/m^3
Volume of hole = 3.1/1380 = 2.25× 10^-3 m^3
γ[t] =total weight of soil excavated/ volume of hole = 4/ 2.25× 10^-3 = 1778 kg/m^3
we have,
w = W[w] /W[s] = weight of water/ weight of soil solids
18/100 = W[w] /W[s]
or, We = 0.18 W[s]
Also, We + W[s] = 4
or, 0.18 W[s] + W[s] =4
thus, W[s] = 3.39 kg
dry density ( γ[d]) = weight of dry soil/volume of the hole
= 3.39/ 2.25× 10^-3 = 1507 kg/m^3
C. Core Cutter Method
i. Aim
To determine the dry density of a given soil sample (cohesive soil)
ii. Apparatus
A cylindrical core cutter, dolly
iii. Procedures
Note: ( soil sample must be soft, fine-grained soils of soil surface exposed so that cutter can be easily driven in the ground)
a. A cylindrical core cutter is embedded in the ground at its full height and a dolly is placed over the blade to prevent its edges from damage while embedding.
b. cutter is taken out from the ground and filled with soil.
c. Extra soil(surplus soil) at the top of both ends of the cutter is trimmed.
d. Then, the weight of the cylinder with excavated soil is weighed.
e. Volume of soil can be determined by known dimensions of the cutter.
f. Water content of the soil can be determined by oven drying method.
g. Using the formula, dry density can be calculated.
iv. Observations and Calculations:
water content/ moisture content =w
weight of cutter = W[1]
weight of soil and cutter = W[2]
weight of soil = W[2] -W[1]
volume of cutter = v
bulk density (γ[t]) =(W[2] -W[1])/ V
Dry density ( γ[d] = γ[t] /(1+w)
v. Result
Using the core cutter method, the dry density of the soil sample given is calculated.
Steps to calculate the dry density of soft soil using the core cutter method:
a. Calculate the weight of the cutter with excavated soil in it.
weight of cutter = W[1]
weight of soil and cutter = W[2]
b. Calculate the weight of the soil.
weight of soil = W[2] -W[1]
c. Calculate the volume of soil.
The volume of soil = volume of cylindrical cutter
d. Calculate the bulk density.
bulk density (γ[t]) =(W[2] -W[1])/ V
e. Calculate the dry density.
Dry density ( γ[d] = γ[t] /(1+w)
where water content=w = weight of water/weight of soil solids
D. Numerical
1. The in-situ density of an embankment, compacted at a water content of 12% was found out using the core cutter method. The empty cutter weighed 1286gms and the cutter with filled soil weighed
3195gms. The volume of the cutter was 1000cm^3. Determine the bulk density and dry density of the barrier.
Given; water content = 12% =0.12
weight of empty cutter = 1286gms
weight of cutter + soil = 3195gms
volume of cutter = V = 1000cm^3
the dry density of soil ( γ[d])=?
bulk density of soil( γ) =?
weight of soil in cutter= W= 3195 – 1286 = 1909gms
volume of soil = volume of cutter=V
Here, bulk density = weight of soil/ volume of soil
= W/V = 1909/1000 = 1.909 gm/cm^3
dry density, γ[d] = γ /(1+w)
= 1.909/(1+0.12)
= 1.7 gm/cc
2. A 1000cc core cutter weighing 9467.8gm was used to find out the in-situ weight of an embankment. The weight of the core cutter filled with soil was found to be 2770.6gms. A sample has a water
content of 10.45%. Determine the bulk density and dry unit weight of the dam.
Given; volume of cutter = 1000cc
Weight of empty cutter = 946.8gms
Weight of core cutter filled with soil = 2770.6gms
Water content of soil = 10.145% = 0.1045
Bulk density =?
Dry density =?
Weight of soil in cutter= 2770.6 – 946.8 =W
W= 1823.8gms
Weight of soil solid= W/(1+w)
= 1823.8/(1+0.1045) = 1651.2gms
Bulk unit weight = 1823.8/1000 = 1.82gm/cc
Dry unit weight= 1651.2/1000 = 1.65gm/cc
Hope you got an idea of methods of compaction control. | {"url":"https://thecivilengineering.com/compaction-control-sand-replacement-core-cutter/","timestamp":"2024-11-04T21:29:10Z","content_type":"text/html","content_length":"326135","record_id":"<urn:uuid:bd14fa02-ec8c-4338-b1e3-79808a62dc31>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00641.warc.gz"} |
What Are the Best Physics Exam 1 Solutions? - Do My Physics Exam
The first physics exam is usually the most difficult, so it is important that you know the first three pages of the exam. In this article, we will talk about the three key topics that you need to
understand before you take your exam.
Please proceed with your order so that we can facilitate taking your exam on your behalf. Click Here
First, let’s look at a little about what makes up a test. Most exams are created by two different departments. These departments will have different sets of standards that need to be met for each
The first set of standards is known as the State Standards. These State Standards was set by the National Council of Higher Education and the National Council of Teachers of Mathematics. The next set
of standards is the Federal Standards.
Each standard has specific requirements and is given a number. Each test that you take will also have its own set of standards. The third set of standards is called the Common Core Standards.
One of these standards, known as PARACOUNT 13, is about using your knowledge to create models using the theories that you are learning. This is something that is especially important in this type of
exam. You will be asked to show how the laws of physics can be applied to real situations. This is the part that you want to get good at.
Another common problem is with the way that you use your equations. Some people find that they make the problem too complicated. The first solutions will help you get better at this area.
The first problem you will likely have is when you try to figure out how the Earth is rotating. There are a lot of different ways that you can do this. You will need to know the angles of the Earth,
the gravitational force, and the rotation period.
If you follow all of these solutions to the best of your ability then you will have an easy time doing well on your first physics exam. This exam is the hardest part of taking physics, so you should
put a lot of effort into making sure that you do well.
The last part of the exam will require you to calculate the force of gravity. This is actually the most difficult part of the exam because you need to have the most basic math skills in order to get
the calculations right. The first parts will give you the basics, but you need to have the understanding and ability to do the next parts before the first one.
You will need to study and practice until you can answer every question that comes up on the test. The last thing that you need to do is to get frustrated because you did not get the answer right on
your first try. because you did not take the time to study.
If you have any questions at all, you should discuss them with a teacher or take a practice test until you understand what the question is trying to convey. and why the answer is correct. You should
also discuss the problem with a friend who understands physics to ensure that there are no other possibilities.
Having a friend there to ask questions is very helpful, but it is also important to take the test. Once you have completed the test and the problems you know what you need to do you can then practice
until you can answer them with ease.
The last step in completing your first Physics exam is to practice what you have learned. You may want to take the test a few times with a friend before actually taking it. After you have finished
taking it a few times you will feel confident that you are ready for your first real test. | {"url":"https://domyphysicsexam.com/what-are-the-best-physics-exam-1-solutions/","timestamp":"2024-11-13T19:18:51Z","content_type":"text/html","content_length":"111253","record_id":"<urn:uuid:99e86ef1-198f-417c-9f36-af9c5c9b69ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00189.warc.gz"} |
Which Platonic Solid is Most-Spherical? (and The Archimedian Solids)
If you inscribe a regular polygon into a given circle, the larger the number of sides, the larger the area of the polygon. I guess I always thought that the same would apply to Platonic solids
inscribed in a sphere..... It doesn't. I noticed this as I was looking though "The Penny cyclopædia of the Society for the Diffusion of Useful Knowledge
By Society for the Diffusion of Useful Knowledge" (1841). As I browsed the book, I came across the table below:
The table gives features of the Platonic solids when inscribed in a one-unit sphere. At first I thought they must have made a mistake, but not so. The Dodecahedron fills almost 10% more of a sphere
(about 66%) than the icosahedron (about 60%). So the Dodecahedron is closer to the sphere than the others.
Interestingly, if you look at the radii of the inscribing spheres, it is clear that solids which are duals are tangent to the same internal sphere.
But if you look at the table of volumes when the solids are inscribed with a sphere inside tangent to each face:
When you put the Platonic solids around a sphere, the one smallest, and thus closest to the sphere is the icosahedron.
This leads to the paradox that when platonic solids are inscribed with a sphere, the icosahedron is closest to the circle in volume (thus most spherical?) but when they are circumscribed by a sphere,
the dodecahedron is the closest to the volume of a sphere (and thus most spherical?)... hmmm
Here is a table of the same values when the surface area (superficies) is one square unit. Notice that for a given surface area, the icosahedron has the largest volume, so it is the most efficient
"packaging" of the solids (thus more spherical?).
I guess that makes it 2-1 for the icosahedron, so I wasn't completely wrong all along.
POSTSCRIPT::: Allen Knutson's comments on the likely cause of this reversal of "closeness" to the sphere:
I think it's about points of contact. On the inside, the dodecahedron touches the sphere at the most points (20), and on the outside, the icosahedron touches the sphere at the most points (again
Indeed: my recipe would suggest that inside, the 8-vertex cube is bigger than the 6-vertex octahedron, and outside, the 8-face octahedron is smaller than the 6-face cube. Both are borne out by
your tables. Thank you, Allen
Some other pertinent comments too good to ignore:
Anonymous said...
I think "most-spherical" would need to be better defined.
Not that I want to do it (and not that I am certain that I could), but perhaps summing the squares of the distances from each point on the surface of the solid to the closest point on the sphere
would be the way to go (like a sort of physical variance).
But as I say, I'm not sure I could do this (maybe I can kill some time today, I'm visiting a puzzle-friend), and I'm not sure that it makes a lot of sense.
Oh, oh, and inscribe or circumscribe or find the best match in-between?
(it's play with this, read on-line, or grade. Choices.)
(Love When anonymous signs his comment)
Mary O'Keeffe said...
What a fascinating post! Here is a nice way to think about your first result in terms of the empty space.
The icosahedron leaves ~20% of the space in its circumscribing sphere empty. The dodecahedron leaves ~12% of the space in its circumscribing sphere empty.
That means that if you start with a solid sphere (let's imagine it's something easy to carve, like soap!) and carefully cut away the 20 portions needed to turn it into an icosahedron, each of the
20 pieces you trimmed off will have volume of about 1% of the sphere.
Similarly, if you do the same thing to carve a dodecahedron out of a sphere, each of the 12 pieces you trimmed off will also have volume of about 1% of the sphere.
From now on, whenever I look at either polyhedron (icosa or dodeca) I will always think of it a little differently--because I will think of those ~1% extensions on each face needed to round it
out to a sphere.
I really liked this idea, and want to take time to find the amount of volume reduced by cutting each face into a sphere.
It seems that the cube is third best, just ahead of the octahedron. The cube volume in a unit sphere is 1.5396.. and the volume of a unit sphere is 4/3 Pi, or 4.1888, cutting off 2.64919 cubic
units (more than half the sphere, about 63%) . That means to sculpt a cube
from a unit sphere we cut off an average of just over 10% for each face. (I did that quickly,
do check)
It might be a wonderful challenge to imagine a slicing approach to take out just the right fraction of the volume
to be removed to cut the same amount when revealing each face. I think I see some of them, but some are.... "difficult"?
Later I tried working on the Archimedian Solids.
Measuring Sphereocity of The Archimedian Solids
A few years ago I wrote a blog about which of the Platonic solids (above) was most spherical. I compared which ones had the most volume inscribed into a unit sphere, and the which had the
smallest volume when circumscribed about a unit sphere. Surprisingly I got different answers to the two methods, and lots of good mathematical comments about why this might be so. Then a while
ago (July 2015) I posted it again. As part of a tongue-in-cheek exchange with Adam Spencer @adambspencer I challenged him to find the roundest of the thirteen Archimedean Solids. For those who
are not familiar with the distinction, both the Platonic and Archimedean solids are made of up faces that are regular polygons, and both have the property that the view at each vertex is
identical to every other, but where the Platonic solids consist of only a single type of regular polygon, (for example the tetrahedron is made up of four equilateral triangles), the Archimedean
solids may have more than one (in the cubeoctahedron each vertex is surrounded by two squares and two equilateral triangles).
Then only a day or so later, I got to wondering about the actual answer and started doing some research. Along the way I found a couple of papers on the topic, one of which was published earlier
in the same month I began my search for the Platonic Solids Sphere-ness. What was great was that they re-exposed me to a formula for comparing according to George Polya from his Mathematics and
Plausible Reasoning: Patterns of plausible inference. I sheepishly admit that I had read this book ( it's on my bookcase nos) several times years ago, but somehow this didn't pop up when I was
thinking of the Platonic solids.
The problem of comparing the ratio of the numerical values of the surface area to the volume is that the answer changes over size. In a sphere, for instance, if the radius is one, then the volume
is 4 pi/3, and the surface area is 4pi, so the V= 1/3 SA. Now increase the radius to three units and the volume is 36 pi, and the surface area is also 36 pi, Now V = SA, and if we keep making the
radius bigger, the volume becomes larger than the surface area, in fact the ratio of Volume to surface area can be reduced to \( \frac{V}{SA} = \frac{r}{3} \). Now that kind of thing happens with
all the solids, the larger the volume gets, the larger the ratio of V to SA gets. It is one of those things that amazes students (and well it should) that for any solid, there is some scalar
multiplication which will transform it into a solid with Volume = Surface Area. In this sense, every solid is isoperimetric (same measure)
So Polya found a way to neutralize this growth. He created an "Isoperimetric Quotient" that served to null out this scalar alteration. By setting the IQ = \( \frac{36 \pi V^2}{S^3} \) With this
weapon he was able to compare, for instance, the "roundness" of the Platonic Solids. Try this with any sphere and you always get one. Try it with anything else, you always get less than one.
The IQ of the Platonic Solids follows the number of faces, with the tetrahedron at the bottom with an IQ of about .3 and the icosahedron at the top with an IQ of about .8288.
SO what about the Archimedean Solids? Well here they are
TruncatedTetrahedron...... 0.4534
TruncatedOctahedron ....... 0.749
TruncatedCube ............. 0.6056
TruncatedDodecahedron ..... 0.7893
TruncatedIcosahedron ...... 0.9027
Cuboctahedron ............. 0.7412
Icosidodecahedron ......... 0.8601
SnubCube .................. 0.8955
SnubDodecahedron .......... 0.94066
Rhombicuboctahedron ....... 0.8669
TruncatedCuboctahedron .... 0.8186
Rhombicosidodecahedron .... 0.9357
TruncatedIcosidodecahedron. 0.9053
So the roundness winner is the snubdodecahedron, with a pentagon and four equilateral triangles around each vertex. That is four 60^o angles and one of 108^o for a total of 348^o. Students might
check if any other of the Archimedean solids can top that. Is that "flatness" at vertices somehow related to "roundness"? I have to admit I first thought it might be the truncated icosahedron.
Students may have heard of this one more than others. A molecule of C-60, or a “Buckyball”, consists of 60 carbon atoms arranged at the vertices of a truncated icosahedron. It's roundness is a
feature of many of its applications, but it only comes in fourth. | {"url":"https://pballew.blogspot.com/2024/10/which-platonic-solid-is-most-spherical.html","timestamp":"2024-11-04T23:55:30Z","content_type":"application/xhtml+xml","content_length":"149130","record_id":"<urn:uuid:97d7c5b9-5c90-456e-8301-0bb4ea3be413>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00874.warc.gz"} |
How many bales do you need?
How many bales will you need?
To calculate the number of bales, I assume that we are building something without doors or windows. That gives you enough extra bales for ample waste during installation, stuffing gaps, and straw for
plastering as well. You will want to know roughly the size of your bales. Where I work on the East Coast, bales are usually 2-strings, usually about 14" high, 18" deep, and an average of 32" long
(though length varies with any given stock of bales). Once I have this information, I calculate the quantity of bales as follows:
1. Determine how many rows of bales high you will need for each wall height (not including any triangular gables). For example, if your bales are 14" high, and your walls are 8-feet tall, then you
will need 7 rows of bales for each wall. (The number of rows will change if your walls are a different height, if your bales are a different dimension, or if you are laying the bales on edge.)
2. Calculate the total wall length for strawbale walls of that height.
3. Take your wall length in inches and divide by the average length of the bale. For example, if I have a 10-foot long wall, that's 120-inches. Assuming my bales are 32" long on average, that's 120
divided by 32, which equals 3.75 bales. I round this up to the nearest 1/2-bale, in this case up to 4 bales. This is the number of bales you need in each row.
4. Now multiply the number of bales in each row by the number of rows you need for your wall height. In our example, this is 7 rows of bales with 4 bales in each row, or a total or 28 bales.
5. For gables, the calculation is number of rows needed at the peak of the gable, times number of bales needed along the first long row, divided by 2.
I use a little spreadsheet and calculate each wall separately. Below is an example for a building with 8-foot tall walls, 2 peaked gable ends, and with exterior dimensions of 10-feet by 20-feet.
│Location │Bales per row│# of rows high│Total Bales│
│Level One East Wall │ 4 │ 7 │ 28│
│Level One West Wall │ 4 │ 7 │ 28│
│Level One South Wall (subtract width of East & West walls) │ 6.5 │ 7 │ 45.5│
│Level One North Wall (subtract width of East & West walls) │ 6.5 │ 7 │ 45.5│
│Level Two Gable Ends (2 gables so rows x bales only) │ 4 │ 4 │ 16│
│ Total Bale Count│ 163│
For a 10-foot x 20-foot building, the long walls are 240" and the short walls are 120".
The short East & West walls
will be 120" divided by 32" (average bale length) = 3.75, so I round up to 4 bales per row.
The long South & North walls
will be 240" minus the width of the corners (since the corners overlap) so 240" - 2(18" bale width) = 240" - 36" = 204". Now divide by 32" (aver bale length) = 6.4, so I round up to 6.5 per row.
The height
will be 8-feet tall, or 96". Each bale is 14" tall, so 96" divided by 14" equals 6.8 bales tall, which we round up to 7 bales tall.
The gable
dimensions vary depending on your roof slope and design, but notice my example shows 2 gable ends, so I do not need to divide each by 2 as you would if it were a single gable.
This will leave you with plenty of extra bales, assuming you have doors & windows in your building. The extra bales allow you to reject any loose or poorly tied bales and give you plenty of straw to
use for clay plasters, any cob walls, or an earthen floor.
1. Hi again! How do I calculate the quantity of straw for cob mass? The same way as the strawbale?
I have a full identification in your work, thanks to the universe to meet you! =p
1. I'm so sorry to tell you, but I've never calculated the amount of straw I use in cob. Sorry!! Maybe make a small test wall and measure everything you use...then multiply to the size you are
going to build
2. Thanks Sigi, I`ll gonna do that! I used this way to calculate the amount of clay plaster in a wall last month. =) | {"url":"http://buildnaturally.blogspot.com/2011/04/how-many-bales-do-you-need.html","timestamp":"2024-11-11T00:23:59Z","content_type":"text/html","content_length":"121153","record_id":"<urn:uuid:735b484b-24eb-408f-a916-02d9341d81a9>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00553.warc.gz"} |
Python Algorithms
An algorithm is a series of instructions that can be executed to perform a certain task or computation. A recipe for a cake is an example of an algorithm. For example, preheat the oven, beat 125 g of
sugar and 100 g of butter, and then add eggs and other ingredients. Similarly, simple computations in mathematics are algorithms. For example, when computing the perimeter of a circle, you multiply
the radius by 2π. It's a short algorithm, but an algorithm, nonetheless.
Algorithms are often initially defined in pseudocode, which is a way of writing down the steps a computer program will make without coding in any specific language. A reader should not need a
technical background in order to read the logic expressed in pseudocode. For example, if you had a list of positive numbers and wanted to find the maximum number of positive numbers in that list, an
algorithm expressed in pseudocode could be as follows:
1. Set the maximum variable to 0.
2. For... | {"url":"https://subscription.packtpub.com/book/programming/9781839218859/3/ch03lvl1sec27/python-algorithms","timestamp":"2024-11-14T21:17:15Z","content_type":"text/html","content_length":"140715","record_id":"<urn:uuid:197c5db8-c015-4080-bcde-f23c9709833e>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00313.warc.gz"} |
A Strong XOR Lemma for Randomized Query Complexity:
Theory of Computing: An Open Access Electronic Journal in Theoretical Computer Science
Volume 19
(2023) Article 11 pp. 1-14
A Strong XOR Lemma for Randomized Query Complexity
Received: August 3, 2020
Revised: December 30, 2023
Published: December 31, 2023
Keywords: lower bounds, query complexity, direct sum
ACM Classification: F.1.1, F.1.3
AMS Classification: 68Q09, 68Q10, 68Q17
Abstract: [Plain Text Version]
$ $
We give a strong direct sum theorem for computing $XOR_k\circ g$, the $XOR$ of $k$ instances of the partial Boolean function $g$. Specifically, we show that for every $g$ and every $k\geq 2$, the
randomized query complexity of computing the $XOR$ of $k$ instances of $g$ satisfies ${\bar R}_\epsilon(XOR_k\circ g) = \Theta(k{\bar R}_{\epsilon/k}(g))$, where ${\bar R}_\epsilon(f)$ denotes the
expected number of queries made by the most efficient randomized algorithm computing $f$ with $\epsilon$ error. This matches the naive success amplification upper bound and answers a conjecture of
Blais and Brody (CCC'19).
As a consequence of our strong direct sum theorem, we give a total function $g$ for which $R(XOR_k\circ g) = \Theta(k \log(k)\cdot R(g))$, where $R(f)$ is the number of queries made by the most
efficient randomized algorithm computing $f$ with $1/3$ error. This answers a question from Ben-David et al. (RANDOM'20). | {"url":"https://theoryofcomputing.org/articles/v019a011/","timestamp":"2024-11-07T02:20:06Z","content_type":"text/html","content_length":"8187","record_id":"<urn:uuid:6439918a-7560-429f-b2df-9844e85dce60>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00277.warc.gz"} |
Euclidean space
Euclidean spaces
The concept of Euclidean space in analysis, topology, differential geometry and specifically Euclidean geometry, and physics is a fomalization in modern terms of the spaces studied in Euclid 300BC,
equipped with the structures that Euclid recognised his spaces as having.
In the strict sense of the word, Euclidean space $E^n$ of dimension $n$ is, up to isometry, the metric space whose underlying set is the Cartesian space $\mathbb{R}^n$ and whose distance function $d$
is given by the Euclidean norm:
$d_{Eucl}(x,y) \coloneqq {\Vert x-y\Vert} = \sqrt{ \sum_{i = 1}^n (y_i - x_i)^2 } \,.$
In Euclid 300BC this is considered for $n = 3$; and it is considered not in terms of coordinate functions as above, but via axioms of synthetic geometry.
This means that in a Euclidean space one may construct for instance the unit sphere around any point, or the shortest curve connecting any two points. These are the operations studied in (Euclid
300BC), see at Euclidean geometry.
Of course these operations may be considered in every (other) metric space, too, see at non-Euclidean geometry. Euclidean geometry is distinguished notably from elliptic geometry or hyperbolic
geometry by the fact that it satisfies the parallel postulate.
In regarding $E^n = (\mathbb{R}^n, d_{Eucl})$ (only) as a metric space, some extra structure still carried by $\mathbb{R}^n$ is disregarded, such as its vector space structure, hence its affine space
structure and its canonical inner product space structure. Sometimes “Euclidean space” is used to refer to $E^n$ with that further extra structure remembered, which might then be called Cartesian
Retaining the inner product on top of the metric space structure means that on top of distances one may also speak of angles in a Euclidean space.
Then of course $\mathbb{R}^n$ carries also non-canonical inner product space structures, not corresponding to the Euclidean norm. Regarding $E^n$ as equipped with these one says that it is a
pseudo-Euclidean space. These are now, again in the sense of Cartan geometry, the local model spaces for pseudo-Riemannian geometry.
Finally one could generalize and allow the dimension to be countably infinite, and regard separable Hilbert spaces as generalized Euclidean spaces.
Arguably, the spaces studied by Euclid were not really modelled on inner product spaces, as the distances were lengths, not real numbers (which, if non-negative, are ratios of lengths). So we should
say that $V$ has an inner product valued in some oriented line $L$ (or rather, in $L^2$). Of course, Euclid did not use the inner product (which takes negative values) directly, but today we can
recover it from what Euclid did discuss: lengths (valued in $L$) and angles (dimensionless).
Since the days of René Descartes, it is common to identify a Euclidean space with a Cartesian space, that is $\mathbb{R}^n$ for $n$ the dimension. But Euclid's spaces had no coordinates; and in any
case, what we do with them is still coordinate-independent.
Euclidean spaces with infinitesimals
Instead of working in the real numbers $\mathbb{R}$ and $n$-dimensional real vector spaces $V$, one could instead work in a Archimedean ordered Artinian local $\mathbb{R}$-algebra $A$ and rank $n$$A$
-modules $V$. $A$ has infinitesimals, and so the $A$-modules $V$ have infinitesimals as well. Nevertheless, it is still possible to define the Euclidean distance function on $V$; the only difference
is that the distance function is a pseudometric rather than a metric here.
Since $A$ is an local ring, the quotient of $A$ by its ideal of non-invertible elements $I$ is $\mathbb{R}$ itself, and the canonical function used in defining the quotient ring is the function $\
Re:A \to \mathbb{R}$ which takes a number $a \in A$ to its purely real component $\Re(a) \in \mathbb{R}$. Since $A$ is an ordered $\mathbb{R}$-algebra, there is a strictly monotone ring homomorphism
$h:\mathbb{R} \to A$.
The real numbers have lattice structure $\min:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ and $\max:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$. This means that $A$ has a distance function given by
the function $\rho:A \times A \to \mathbb{R}$, defined as
$\rho(a, b) \coloneqq \max(\Re(a), \Re(b)) - \min(\Re(a), \Re(b))$
as well as an absolute value given by the function $\vert-\vert:A \to \mathbb{R}$, defined as
$\vert a \vert \coloneqq \rho(a, 0)$
Since $\min(a, b) \leq \max(a, b)$, the pseudometric and multiplicative seminorm are always non-negative. In addition, by definition, the pseudometric takes any two elements $a \in A$ and $b \in A$
whose difference $a - b \in I$ is an infinitesimal to zero $\rho(a, b) = 0$.
Since $\mathbb{R}$ is an Euclidean field, it has a metric square root function $\sqrt{-}:[0, \infty) \to [0, \infty)$. Every rank $n$$A$-module $V$ with basis $v:\mathrm{Fin}(n) \to V$ thus has a
Euclidean pseudometric $\rho_V:V \times V \to K$ defined by
$\rho_V(a, b) \coloneqq \sqrt{\sum_{i \in \mathrm{Fin}(n)} \rho(a_i, b_i)^2}$
for module elements $a \in V$ and $b \in V$ and scalars $a_i \in A$ and $b_i \in A$ for index $i \in \mathrm{Fin}(n)$, where
$a = \sum_{i \in \mathrm{Fin}(n)} a_i v_i \quad b = \sum_{i \in \mathrm{Fin}(n)} b_i v_i$
If $A$ is an ordered field, then this reduces down to the Euclidean metric defined above.
In constructive mathematics
In constructive mathematics, the real numbers used to define Euclidean spaces are the Dedekind real numbers $\mathbb{R}_{D}$, as those are the only ones that are Dedekind complete, in the sense of
not having any gaps in the dense linear order. The Dedekind real numbers are also the real numbers that are geometrically contractible: whose shape is homotopically contractible $\esh(\mathbb{R}_D) \
cong \mathbb{1}$.
In predicative constructive mathematics
In predicative constructive mathematics, the Dedekind real numbers are defined relative to a universe $\mathcal{U}$, and thus there are many different such Dedekind real numbers that could be used to
define Euclidean spaces, one $\mathbb{R}_\mathcal{U}$ for each $\mathcal{U}$. However, each set of Dedekind real numbers $\mathbb{R}_\mathcal{U}$ would be large relative to the sets in the universe $
If the predicative constructive foundations does not have universes, then there doesn’t exist any dense linear order that is actually Dedekind complete in the usual sense, and so the usual definition
of Euclidean space does not work. Some mathematicians have proposed to use Sierpinski space $\Sigma$, the initial $\sigma$-frame, for defining the real numbers, in place of the large set of all
propositions in a universe $\mathrm{Prop}_\mathcal{U}$, but the real numbers in that case are only $\Sigma$-Dedekind complete, which is a weaker condition than being Dedekind complete. Furthermore,
Lešnik showed that for any two $\sigma$-frames $\Sigma$ and $\Sigma^{'}$ that embed into $\mathrm{Prop}_\mathcal{U}$ such that $\Sigma \subseteq \Sigma^{'}$, if $A$ is the $\Sigma$-Dedekind
completion of $\mathbb{Q}$ and $B$ is the $\Sigma^{'}$-Dedekind completion of $\mathbb{Q}$, then $A \subseteq B$, so the $\Sigma$-Dedekind real numbers are not complete.
Lengths and angles
Given two points $x$ and $y$ of a Euclidean space $E$, their difference $x - y$ belongs to the vector space $V$, where it has a norm
${\|x - y\|} = \sqrt{\langle{x - y, x - y}\rangle} .$
This real number (or properly, element of the line $L$) is the distance between $x$ and $y$, or the length of the line segment $\overline{x y}$. This distance function makes $E$ into an ($L$-valued)
metric space.
Given three points $x, y, z$, with $x, y e z$ (so that ${\|x - z\|}, {\|y - z\|} e 0$), we can form the ratio
$\frac{\langle{x - z, y - z}\rangle}{{\|x - z\|} {\|y - z\|}} ,$
which is a (dimensionless) real number. By the Cauchy–Schwartz inequality, this number lies between $-1$ and $1$, so it's the cosine of a unique angle measure between $0$ and $\pi$ radians. This is
the measure of the angle $\angle x z y$. In a $2$-dimensional Euclidean space, we can interpret $\angle x z y$ as a signed angle (so taking values anywhere on the unit circle) if we fix an
orientation of $E$.
Conversely, knowing angles and lengths, we may recover the inner product on $V$;
$\langle{x - z, y - z}\rangle = {\|\overline{x z}\|} {\|\overline{y z}\|} \cos \angle x z y ,$
and other inner products are recovered by linearity. (We must then use the axioms of Euclidean geometry to prove that this is well defined and actually an inner product.) It’s actually possible to
recover the inner product and angles from lengths alone; this is discussed at Hilbert space.
Textbook accounts:
On the use of the Dedekind real numbers in constructive and predicative constructive mathematics, such as for Euclidean spaces:
• Mike Shulman, Brouwer’s fixed-point theorem in real-cohesive homotopy type theory, Mathematical Structures in Computer Science Vol 28 (6) (2018): 856-941 (arXiv:1509.07584, doi:10.1017/
• Davorin Lešnik, Synthetic Topology and Constructive Metric Spaces, (arxiv:2104.10399) | {"url":"https://ncatlab.org/nlab/show/Euclidean+space","timestamp":"2024-11-14T14:51:03Z","content_type":"application/xhtml+xml","content_length":"87757","record_id":"<urn:uuid:6aed26b1-5a00-4e96-94f4-e34fc9183657>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00651.warc.gz"} |
Ibanez BC-9 Bi Mode Chorus
Got one coming. Give me a heads up on this pedal!! Thanks
Well regarded - two differently voiced choruses in one box. Would be cool if you could jump back and forth between them like the AM Clone Clone - maybe you can? Wouldn't mind checking out the Master
series version.
very lush chorus! Unlike the metallic sounding CS9.
A/B-ed a BC9 once with a Boss CE-1 and a Fulltone Choralflange. The BC9 was quite in the realm of the CE-1, in terms of lushy warmth, whereas the Choralflange sounded sterile in comparison.
and the BC-9 loves distortion
it was cool but not very subtle. i would have kept it if it had some kind of wet/dry blend. ended up with the AM Clone Chorus. and ended up selling that because chorus is for pansies.
It is a most excellent chorus in every way. The only issue is one that is common to many analog BBD chip modulation pedals... volume boost when engaged.
If you like a more dimensional Sousing chorus instead of the swirly type like the Small Clone or CE-1, go for it.. Its amazing...! It was my main chorus unit till be UD Stomp took over. I actually
did an A/B between those two and set the UD Stomp to the closest possible tone on the BC-9.
Btw, if you wanna hear recordings of the BC-9, the guitarist of Korn uses it. Check out the bridge section of the song B.B.K, thats pretty much how it can sound like.
It's the only chorus pedal I've ever been able to love. It's really, really great. I love the sound...it's just different from other ones.
Can anyone share their lush chorus settings for this pedal. I can see the potential!!
no matter where you set the contols, it'll sound great/
I am finding some real nice sounds with the speed knob1/4 or below!!
This topic is now archived and is closed to further replies. | {"url":"https://www.harmonycentral.com/forums/topic/1674082-ibanez-bc-9-bi-mode-chorus/","timestamp":"2024-11-14T07:58:25Z","content_type":"text/html","content_length":"188744","record_id":"<urn:uuid:efd8b052-649d-4fd3-9eab-ad7f6e95575f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00619.warc.gz"} |
mAh Battery Life Calculator
mAh Battery Life Calculator is an online tool used in electrical engineering to precisely calculate battery life. Generally, battery life is calculated based on the current rating in milli Ampere per
Hour and it is abbreviated as mAh. Ampere is an electrical unit used to measure the current flow towards the load. The battery life or capacity can be calculated from the input current rating of the
battery and the load current of the circuit. Battery life will be high when the load current is less and vice versa. The calculation to find out the capacity of battery can be mathematically derived
from the below formula
When it comes to online calculation, this battery life calculator can assist you to determine the time that how long the battery charge will last. For example, a circuit connected with 800 mAh
current rating and it is connected to the load of 40 mAh. Then the battery will last for 20 hours. Batteries are available in different current rating due to its high requirement of different
industrial and domestic purposes. Any battery life can be easily calculated by the values of battery capacity in mAh and load current in mAh. | {"url":"https://dev.ncalculators.com/electrical/battery-life-calculator.htm","timestamp":"2024-11-12T19:17:15Z","content_type":"text/html","content_length":"32707","record_id":"<urn:uuid:6aada65d-a0db-49bb-9dad-e2550fb0e9a7>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00767.warc.gz"} |
ArcGIS Enterprise
The measure tools allow you to measure distances between two points and calculate areas in your scene. When you click the Measure distance tool Measure area tool
Measure distance
Use the Measure distance
• Direct—Distance between two points
• Horizontal—Horizontal distance between two points
• Vertical—Vertical distance between two points
While you are measuring, a second laser line indicates where the vertical plane along the checkered line intersects the terrain in all directions, such as with buildings, bridges, and the ground.
To measure distance, do the following:
1. Click Analyze
2. Click Measure distance
3. Click in the scene to start measuring.
4. Click to set the endpoint.
5. Click New Measurement to start a new measurement.
When the distance between the points is greater than 100 kilometers, a circular laser line appears, indicating that Scene Viewer has switched to geodesic mode. In geodesic mode, Scene Viewer
calculates only the horizontal and vertical distances, taking into consideration the curvature of the earth (that is, ellipsoid-based geodesic distance). The Direct distance option is unavailable.
Measure area
Use the Measure area tool Measure area tool labels the current segment length and the total length of the path in your scene. Once you close the path, a polygon is created with labeled values for the
area and perimeter. These values also appear in the panel.
• Area—Area of the polygon
• Perimeter—Perimeter length of the polygon
To measure area, do the following:
1. Click Measure area
2. Click in the scene to start adding points to the polygon.
3. Double-click to close the path and calculate the polygon area. Alternatively, click the starting point again to close the path.
4. Click New Measurement to start a new measurement.
When the polygon perimeter is greater than 100 kilometers, Scene Viewer switches to geodesic mode. In geodesic mode, Scene Viewer calculates the values, taking into consideration the curvature of the
earth (that is, ellipsoid-based geodesic values).
Adjust measurements
To adjust either the Measure distance Measure area Scene Viewer displays the adjusted values in the scene and panel. You can change the unit of measure under Unit.
In local scenes, measurements are displayed as Euclidean values and may not be accurate depending on the scene's projected coordinate system. Web Mercator scenes display the accurate geodesic values. | {"url":"https://enterprise.arcgis.com/en/portal/10.9/use/measure-scene.htm","timestamp":"2024-11-11T11:10:23Z","content_type":"text/html","content_length":"44395","record_id":"<urn:uuid:3f4b9f05-1040-4971-bec0-aee7b00cfb94>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00820.warc.gz"} |
01/09/09 12:54:27 (16 years ago)
Various doc improvements for graph adaptors (#67)
□ Add notes about modifying the adapted graphs through adaptors if it is possible.
□ Add notes about the possible conversions between the Node, Arc and Edge types of the adapted graphs and the adaptors.
□ Hide the default values for template parameters (describe them in the doc instead).
□ More precise docs for template parameters.
□ More precise docs for member functions.
□ Add docs for important public typedefs.
□ Unify the docs of the adaptors.
□ Add \relates commands for the creator functions.
□ Fixes and improvements the module documentation.
• r434 r474
64 64 /**
65 @defgroup graph_adaptors Adaptor Classes for [DEL:g:DEL]raphs
65 @defgroup graph_adaptors Adaptor Classes for raphs
66 66 @ingroup graphs
67 \brief This group contains several adaptor classes for digraphs and graphs
67 \brief Adaptor classes for digraphs and graphs
69 This group contains several useful adaptor classes for digraphs and graphs.
69 71 The main parts of LEMON are the different graph structures, generic
70 graph algorithms, graph concepts[DEL: which couple these:DEL], and graph
72 graph algorithms, graph concepts, and graph
71 73 adaptors. While the previous notions are more or less clear, the
72 74 latter one needs further explanation. Graph adaptors are graph classes
… …
75 77 A short example makes this much clearer. Suppose that we have an
76 instance \c g of a directed graph type[DEL::DEL] say ListDigraph and an algorithm
78 instance \c g of a directed graph type say ListDigraph and an algorithm
77 79 \code
78 80 template <typename Digraph>
… …
82 84 (in time or in memory usage) to copy \c g with the reversed
83 85 arcs. In this case, an adaptor class is used, which (according
84 to LEMON [DEL:digraph concepts) works as a digraph. The adaptor uses the:DEL]
85 original digraph structure and digraph operations when methods of the
86 reversed oriented graph are called. This means that the adaptor have
87 [DEL::DEL]minor memory usage, and do not perform sophisticated algorithmic
86 to LEMON
87 The adaptor uses the original digraph structure and digraph operations when
88 methods of the reversed oriented graph are called. This means that the adaptor
89 minor memory usage, and do not perform sophisticated algorithmic
88 90 actions. The purpose of it is to give a tool for the cases when a
89 91 graph have to be used in a specific alteration. If this alteration is
90 obtained by a usual construction like filtering the [DEL:arc-:DEL]set or
92 obtained by a usual construction like filtering the set or
91 93 considering a new orientation, then an adaptor is worthwhile to use.
92 94 To come back to the reverse oriented graph, in this situation
… …
97 99 \code
98 100 ListDigraph g;
99 ReverseDigraph<List[DEL:G:DEL]raph> rg(g);
101 ReverseDigraph<Listraph> rg(g);
100 102 int result = algorithm(rg);
101 103 \endcode
102 [DEL:After running the algorithm, the original :DEL]graph \c g is untouched.
103 This techniques give[DEL:s:DEL] rise to an elegant code, and based on stable
104 graph \c g is untouched.
105 This techniques give rise to an elegant code, and based on stable
104 106 graph adaptors, complex algorithms can be implemented easily.
106 In flow, circulation and [DEL:bipartite :DEL]matching problems, the residual
108 In flow, circulation and matching problems, the residual
107 109 graph is of particular importance. Combining an adaptor implementing
108 this[DEL:, shortest path algorithms and:DEL] minimum mean cycle algorithms,
110 this minimum mean cycle algorithms,
109 111 a range of weighted and cardinality optimization algorithms can be
110 112 obtained. For other examples, the interested user is referred to the
… …
113 115 The behavior of graph adaptors can be very different. Some of them keep
114 116 capabilities of the original graph while in other cases this would be
115 meaningless. This means that the concepts that they are models of depend
116 on the graph adaptor, and the wrapped graph(s).
117 If an arc of \c rg is deleted, this is carried out by deleting the
118 corresponding arc of \c g, thus the adaptor modifies the original graph.
120 But for a residual graph, this operation has no sense.
117 meaningless. This means that the concepts that they meet depend
118 on the graph adaptor, and the wrapped graph.
119 For example, if an arc of a reversed digraph is deleted, this is carried
120 out by deleting the corresponding arc of the original digraph, thus the
121 adaptor modifies the original digraph.
122 However in case of a residual digraph, this operation has no sense.
121 124 Let us stand one more example here to simplify your work.
122 Rev[DEL:GraphAdaptor:DEL] has constructor
125 Rev has constructor
123 126 \code
124 127 ReverseDigraph(Digraph& digraph);
125 128 \endcode
126 This means that in a situation, when a <tt>const [DEL::DEL]ListDigraph&</tt>
129 This means that in a situation, when a <tt>const ListDigraph&</tt>
127 130 reference to a graph is given, then it have to be instantiated with
128 <tt>Digraph=const [DEL::DEL]ListDigraph</tt>.
131 <tt>Digraph=const ListDigraph</tt>.
129 132 \code
130 133 int algorithm1(const ListDigraph& g) {
131 Rev[DEL:GraphAdaptor:DEL]<const ListDigraph> rg(g);
134 Rev<const ListDigraph> rg(g);
132 135 return algorithm2(rg);
133 136 } | {"url":"https://lemon.cs.elte.hu/trac/lemon/changeset/474/lemon/doc","timestamp":"2024-11-13T13:24:37Z","content_type":"application/xhtml+xml","content_length":"33937","record_id":"<urn:uuid:68bcac78-a9f5-495a-9ff5-3fa784364b00>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00181.warc.gz"} |
MATH 1530 CAPSTONE TECHNOLOGY PROJECT SUMMER 2015 - Speed Assignments
MATH 1530 CAPSTONE TECHNOLOGY PROJECT SUMMER 2015
MATH 1530 CAPSTONE TECHNOLOGY PROJECT SUMMER 2015
Problem 1: Identify Variable Type. One of these is a variable that is categorical and one is quantitative. Consider the different graphs that correspond to each variable type. Use Minitab to
create two different graphs appropriate for each variable’s type. EXTRA CREDIT if you can resize to fit all of the graphs on one page.
NUCLEAR SAFETY TALK POLITICS
Problem 2: Sampling. In the survey data, the variable “AGE” is the current age reported by each student.
a. Type the first 10 observations from the column representing the variable AGE into the table below, and use this as your sample data for part (b). Then calculate the mean age of these first 10
observations and report the value below.
N 1 2 3 4 5 6 7 8 9 10
AGE (yrs)
b. The mean age of the first 10 students is years. (Type the value into the space provided.)
c. Identify the type of sampling method you have just used:
d. Next, select a random sample of size n = 10 (Go to Calc > Random Data > Sample from Columns). Type the number 10 in the “Number of rows to Sample” slot. Enter the variable “ID” and “AGE” into
the “From columns” slot. Enter C17-C18 into the “Store samples in” slot. Record the data for your sample in the table below.
N 1 2 3 4 5 6 7 8 9 10
AGE (yrs)
e. Calculate and report the mean age for your random sample of 10 students. The sample mean age is
f. Identify the type of sampling method you have just used:
g. REPEAT the random sample selection process three more times. Calculate and report the mean age for each random sample of 10 students.
N 1 2 3 4 5 6 7 8 9 10
AGE (yrs)
ii) The sample mean age is years.
N 1 2 3 4 5 6 7 8 9 10
AGE (yrs)
iii) The sample mean age is years.
N 1 2 3 4 5 6 7 8 9 10
AGE (yrs)
iv) The sample mean age is years.
h. Suppose we think of all the students who responded to the survey as a population for the purposes of this problem. In that case, the population mean age is 21.293. Discuss (two or more complete
sentences) the differences and similarities between 21.293 and the answers you got in (b), (e), and ii), iii), and iv).
Problem 3(h): FLIP A COIN. Circle the outcome heads / tails . If you got ‘heads,’ then do this problem. (Omit this page/problem if you got ‘tails.’)
Question 10 of the SPRING 2015 survey asked students, “How much money did you spend on your last clothing purchase? (in US dollars)”
a. Create an appropriate graph to display the distribution of the variable called CLOTHING PURCASE and insert it here.
b. Which of the following best describes the shape of the distribution? Underline your answer.
Skewed left Symmetric Skewed right
c. Using Minitab, calculate the basic statistics for the data collected on CLOTHING PURCASE. Copy and paste all of the Minitab output here.
d. Choose statistics that are appropriate for the shape of the distribution to describe the center and spread of CLOTHING PURCASE.
Which statistic will you use to describe the center of the distribution? (Type name of the statistic here.)
e. What is the value of that statistic? (Type value here.)
f. Which statistic(s) will you use to describe the spread of the distribution?
g. What is (are) the value(s) of that (those) statistic(s)?
h. Look up the IQR rule on p. 50 in our textbook. Are there any outliers in this distribution? If so, what are their values? How many are there? Justify your answer.
Problem 3(t): YOU JUST FLIPPED A COIN. If you got ‘tails,’ then do this problem. (Omit this page/problem if you got ‘heads.’)
Question 12 of the FALL 2014 survey asked students, “Usually, how many hours sleep do you get in a night?” The data is in column 14 ‘SLEEP’ of the data file.
a. Create an appropriate graph to display the distribution of the variable called SLEEP and insert it here.
b. Which of the following best describes the shape of the distribution? Underline your answer.
Skewed left Symmetric Skewed right
c. Using Minitab, calculate the basic statistics for the data collected on SLEEP and copy & paste the Minitab output here.
d. Choose statistics that are appropriate for the shape of the distribution to describe the center and spread of SLEEP.
i) Which statistic will you use to describe the center of the distribution? (Type name of the statistic here.)
ii) What is the value of that statistic? (Type value here.)
iii) Which statistic(s) will you use to describe the spread of the distribution?
iv) What is (are) the value(s) of that (those) statistic(s)?
v) Look up the IQR rule on p. 50 in our textbook. Are there any outliers in this distribution? If so, what are their values? How many are there? Justify your answer.
Problem 4: Age versus Handwashing. It is not surprising to see a fairly strong association between certain variables and age. On the SPRING 2015 Math 1530 survey, questions 3 and 7 asked students
to give their age in years (AGE, yrs) and an estimate of how many times each day they wash their hands (WASH HANDS). We are specifically interested in seeing whether we can use a student’s age to
predict daily hand washes.
a. Create an appropriate graph to display the relationship between AGE and WASH HANDS. Insert it here.
b. Does the plot show a positive association, a negative association, or no association between these two variables? EXPLAIN what this means with respect to the variables being studied.
c. Describe the form of the relationship between AGE and WASH HANDS.
d. Report the value of the correlation between this pair of variables? r =
e. Based on the information displayed in the graph and the correlation you just reported, how would you describe the strength of the association?
f. Using Minitab, obtain the equation for the least squares regression of WASH HANDS on AGE. Copy & paste the output here.
g. Interpret the value of the slope in the least squares regression equation you found in part (f).
h. Use the regression equation in part (f) to predict daily hand washes for a student who is 20 years old. (Show your math.)
Predicted hand washes =
i. How well does the regression equation fit the data? Explain. Justify your answer with appropriate plot(s) and summary statistics.
Question 5 (both) FLIP A COIN TWICE. Circle the outcome of each toss: 1 heads/tails;2 heads/tails.
If you got heads both times or tails both times, then do this problem. (Omit this page/problem if you got one of each.)
POLITICAL PARTY AND GENDER Question 5 from the FALL 2014 Math 1530 survey asked students “What political party do you identify with?” and Question 2 from that survey asked students “What is your
gender?” The answers to these questions can be found in column 12 ‘PARTY’ and column 10 ‘GENDER’ in the Summer 2015 Capstone Data file. We want to check if there is a relationship
between political party and gender among ETSU students. Assume the students who took the (Fall 2014 Math 1530) class survey are from an SRS of ETSU students.
a. Create an appropriate graph to display the relationship between POLITICAL PARTY and GENDER. You don’t want to display information for students that didn’t answer both of these questions on the
survey, so click on Data Options > Group Options and remove the checks in the boxes beside “Include missing as a group” and “Include empty cells.” Insert your graph here.
b. Create an appropriate two-way table to summarize the data. Click on Options > Display missing values for… and put a dot in the circle beside “No variables.” Insert your table here.
SUPPOSE WE SELECT ONE STUDENT AT RANDOM: (Calculate the following probabilities and show your work.)
c. What is the probability that this student is both a male and Republican?
P =
d. What is the probability that this student is either a female or Independent?
P =
e. What is the probability that this student is a Democrat given that the student selected is a female?
P =
f. What is the probability that this student is a female given that the student is a Democrat?
P =
g. Do you think there may be an association between GENDER and POLITICAL PARTY? Why or why not? Explain your reasoning based on what you see in your graph.
Problem 5(mixed): YOU JUST FLIPPED A COIN TWICE. If you got one heads and one tails, then do this problem. (Omit this page/problem if you got two heads or two tails.)
MARRIED AND DEATH_PENALTY Question 3 from the FALL 2014 Math 1530 survey asked students “What is your opinion about a married person having sexual relations with someone other than the marriage
partner?” and Question 8 from the survey asked students “Do you favor or oppose the death penalty for persons convicted of murder?” The answers to these questions can be found in column 11
‘MARRIED’ and column 13 ‘DEATH_PENALTY’ in the Summer 2015 Capstone Data file. We want to check if there is a relationship between MARRIED and DEATH_PENALTY among ETSU students. Assume the
students who took the (Fall 2014 Math 1530) class survey are from an SRS of ETSU students.
a. Create an appropriate graph to display the relationship between MARRIED and DEATH_PENALTY. You don’t want to display information for students that didn’t answer both of these questions on the
survey, so click on Data Options > Group Options and remove the checks in the boxes beside “Include missing as a group” and “Include empty cells.” Insert your graph here.
b. Create an appropriate two-way table to summarize the data. Click on Options > Display missing values for… and put a dot in the circle beside “No variables.” Insert your table here.
SUPPOSE WE SELECT ONE STUDENT AT RANDOM: (Calculate the following probabilities and show your work.)
c. What is the probability that this student is both opposed to the Death Penalty and says that sex with someone other than a marriage partner is ‘not wrong at all’?
P =
d. What is the probability that this student favors the Death Penalty or says that sex with someone other than the marriage partner is ‘always wrong’?
P =
e. What is the probability that this student is favors the death penalty given that the student says sex with someone other than the marriage partner is ’always wrong’?
P =
f. What is the probability that this student says that sex with someone other than the marriage partner is ‘always wrong’ given that the student favors the death penalty?
P =
g. Do you think there may be an association between DEATH PENALTY and MARRIED? Why or why not? Explain your reasoning based on what you see in your graph.
Problem 6 The Statistic Brain Research Institute says that the average consumer spends about $59/month on women’s clothes. http://www.statisticbrain.com/what-consumers-spend-each-month/ Do female
ETSU students spend similar amounts on clothing?
Spring 2015 Math 1530 survey question 10 asked “How much money did you spend on your last clothing purchase? (in US dollars) “ We want data on just the female students. Minitab will separate
the CLOTHING_PURCHASE data into two columns. Data > Unstack columns >
Unstack the data in: CLOTHING PURCHASE, GENDER Using subscripts in: GENDER
And you get a new worksheet, with the female clothing purchase data in its own column.
a. Create a suitable graph to display the distribution of CLOTHING_PURCHASE reported by our sample of female college students and insert it here.
b. Describe the distribution shown in your graph.
c. Perform a test of significance to see if female college students have clothing spending habits similar to the average consumer. If this claim is true, then the average CLOTHING_PURCHASE reported
by female students should be $59. For this test, the null hypothesis is that the average CLOTHING_PURCHASE reported by female students is the what is reported for the average consumer. Thus,
Ho: µ = $59 per month
Write the correct alternative hypothesis for the test.
d. Use Minitab to perform the appropriate test. Copy and paste the output for the test here.
e. What is the name of your test statistic and what is its value?
f. What is the P-value for the test? P =
g. State your decision regarding the hypothesis being tested.
h. State your conclusion (words about females and clothing. USE COMPLETE SENTENCES.
i. Is the P-value valid in this case? What assumptions are you making in order to carry out this test?
Bonus Problem: Population males. According to the Census Bureau, http://quickfacts.census.gov/qfd/states/00000.html , in 2013, about 49.2.8% of the US population was male. Is the same true for
the population of students at U.S. colleges and universities? On the Fall 2014 1530 Survey, question #1 asked our Math-1530 students, “What is your gender? (Female, Male)” In the data
worksheet, we call this variable GENDER the one in column 10.
a. Create an appropriate graph to display the distribution of GENDER and insert it here.
b. How many of the students surveyed said “male?”
c. What proportion of our sample said “male?”
d. Assume (for the purpose of this problem) that we may treat the Fall 2014 sample of Math-1530 students as a simple random sample drawn from the population of all U.S. college/university students.
Use Minitab to calculate a 95% confidence interval for the proportion of students in the population who would say “male” to the survey question (based on our sample data). Copy and paste the Minitab
output here.
e. Interpret the confidence interval you reported in part (d).
f. What do you think? Do our results contradict the claim made at the Census website or do they appear to agree with it? EXPLAIN. | {"url":"https://www.speedassignments.com/math-1530-capstone-technology-project-summer-2015/","timestamp":"2024-11-13T06:33:34Z","content_type":"text/html","content_length":"50254","record_id":"<urn:uuid:fd136fe8-6289-4d31-8358-32cb6807552f>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00403.warc.gz"} |
Implementing Type-Classes as OCaml Modules
Type classes achieve overloading in functional paradigms. Shayne Fletcher implements some as OCaml modules.
Modular type classes
In this article, we revisit the idea of type-classes first explored in a previous blog post [Fletcher16a]. This time though, the implementation technique will be by via OCaml modules inspired by the
paper ‘Modular Type Classes’ [Dreyer07] by Dreyer et al.
Ad hoc polymorphism
In programming languages, there is a particular kind of polymorphism known formally called ad hoc polymorphism but better known as overloading. For example with overloading, an operator like + may be
defined that works for many different kinds of numbers.
In the programming language Haskell, a language construction called type classes provides a structured way to provide for ad hoc polymorphism. The OCaml programming language does not have type
classes but rather provides a construction called modules. Ad hoc polymorphism via Haskell-like typeclass style programming can be supported in OCaml by viewing type classes as a particular mode of
use of modules. Indeed, the module approach can be argued as better in the sense that programmers can have explicit control over which type class instances are available in a given scope.
Starting with the basics, consider the class of types whose values can be compared for equality. Call this type-class Eq. We represent the class as a module signature.
module type EQ = sig
type t
val eq : t * t → bool
Specific instances of Eq are modules that implement this signature. Here are two examples.
module Eq_bool : EQ with type t = bool = struct
type t = bool
let eq (a, b) = a = b
module Eq_int : EQ with type t = int = struct
type t = int
let eq (a, b) = a = b
Given instances of class Eq (X and Y say,) we realize that products of those instances are also in Eq. This idea can be expressed as a functor with the following type.
module type EQ_PROD =
functor (X : EQ) (Y : EQ) →
EQ with type t = X.t * Y.t
The implementation of this functor is simply stated as the following.
module Eq_prod : EQ_PROD =
functor (X : EQ) (Y : EQ) → struct
type t = X.t * Y.t
let eq ((x1, y1), (x2, y2))
= X.eq (x1, x2) && Y.eq(y1, y2)
With this functor we can build concrete instances for products. Here‘s one example.
module Eq_bool_int :
EQ with type t = (bool * int)
= Eq_prod (Eq_bool) (Eq_int)
The class Eq can be used as a building block for the construction of new type classes. For example, we might define a new type-class Ord that admits types that are equality comparable and whose
values can be ordered with a ‘less-than’ relation. We introduce a new module type to describe this class.
module type ORD = sig
include EQ
val lt : t * t → bool
Here’s an example instance of this class.
module Ord_int : ORD with type t = int = struct
include Eq_int
let lt (x, y) = Pervasives.( < ) x y
As before, given two instances of this class, we observe that products of these instances also reside in the class. Accordingly, we have this functor type
module type ORD_PROD =
functor (X : ORD) (Y : ORD) → ORD with type t
= X.t * Y.t
with the following implementation.
module Ord_prod : ORD_PROD =
functor (X : ORD) (Y : ORD) → struct
include Eq_prod (X) (Y)
let lt ((x1, y1), (x2, y2)) =
X.lt (x1, x2) || X.eq (x1, x2) &&
Y.lt (y1, y2)
This is the corresponding instance for pairs of integers.
module Ord_int_int = Ord_prod (Ord_int) (Ord_int)
Here’s a simple usage example.
let test_ord_int_int =
let x = (1, 2) and y = (1, 4) in
assert ( not (Ord_int_int.eq (x, y)) &&
Ord_int_int.lt (x, y))
Using type-classes to implement parametric polymorphism
This section begins with the Show type-class.
module type SHOW = sig
type t
val show : t → string
In what follows, it is convenient to make an alias for module values of this type.
type 'a show_impl = (module SHOW with type t = 'a)
Here are two instances of this class...
module Show_int : SHOW with type t = int = struct
type t = int
let show = Pervasives.string_of_int
module Show_bool : SHOW with type t = bool
= struct
type t = bool
let show = function | true → "True"
| false → "False"
...and here these instances are ‘packed’ as values:
let show_int : int show_impl =
(module Show_int : SHOW with type t = int)
let show_bool : bool show_impl =
(module Show_bool : SHOW with type t = bool)
The existence of the Show class is all that is required to enable the writing of our first parametrically polymorphic function.
let print : 'a show_impl → 'a → unit =
fun (type a) (show : a show_impl) (x : a) →
let module Show =
(val show : SHOW with type t = a) in
print_endline@@ Show.show x
let test_print_1 : unit = print show_bool true
let test_print_2 : unit = print show_int 3
The function print can be used with values of any type 'a as long as the caller can produce evidence of 'a’s membership in Show (in the form of a compatible instance).
Listing 1 begins with the definition of a type-class Num (the class of additive numbers) together with some example instances.
module type NUM = sig
type t
val from_int : int → t
val ( + ) : t → t → t
type 'a num_impl = (module NUM with type t = 'a)
module Num_int : NUM with type t = int = struct
type t = int
let from_int x = x
let ( + ) = Pervasives.( + )
let num_int = (module Num_int : NUM with
type t = int)
module Num_bool : NUM with type t = bool = struct
type t = bool
let from_int = function | 0 → false
| _ → true
let ( + ) = function | true → fun _ → true
| false → fun x → x
let num_bool =
(module Num_bool : NUM with type t = bool)
Listing 1
The existence of Num admits writing a polymorphic function sum that will work for any 'a list of values if only 'a can be shown to be in Num.
let sum : 'a num_impl → 'a list → 'a =
fun (type a) (num : a num_impl) (ls : a list) →
let module Num =
(val num : NUM with type t = a) in
List.fold_right Num.( + ) ls (Num.from_int 0)
let test_sum = sum num_int [1; 2; 3; 4]
This next function requires evidence of membership in two classes.
let print_incr : ('a show_impl * 'a num_impl)
→ 'a → unit =
fun (type a) ((show : a show_impl),
(num : a num_impl)) (x : a) →
let module Num =
(val num : NUM with type t = a) in
let open Num
in print show (x + from_int 1)
(*An instantiation*)
let print_incr_int (x : int) : unit
= print_incr (show_int, num_int) x
If 'a is in Show then we can easily extend Show to include the type 'a list. As we saw earlier, this kind of thing can be done with an appropriate functor. (See Listing 2.)
module type LIST_SHOW =
functor (X : SHOW) →
SHOW with type t = X.t list
module List_show : LIST_SHOW =
functor (X : SHOW) → struct
type t = X.t list
let show =
fun xs →
let rec go first = function
| [] → "]"
| h :: t →
(if (first) then "" else ", ")
^ X.show h ^ go false t
in "[" ^ go true xs
Listing 2
There is also another way: one can write a function to dynamically compute an 'a list show_impl from an 'a show_impl (see Listing 3).
let show_list : 'a show_impl → 'a list show_impl
= fun (type a) (show : a show_impl) →
let module Show =
(val show : SHOW with type t = a) in
(module struct
type t = a list
let show : t → string =
fun xs →
let rec go first = function
| [] → "]"
| h :: t →
(if (first) then "" else ", ")
^ Show.show h ^ go false t
in "[" ^ go true xs
end : SHOW with type t = a list)
let testls : string = let module Show =
(val (show_list show_int)
: SHOW with type t = int list) in
Show.show (1 :: 2 :: 3 :: [])
Listing 3
The type-class Mul is an aggregation of the type-classes Eq and Num together with a function to perform multiplication. (Listing 4.)
module type MUL = sig
include EQ
include NUM with type t := t
val mul : t → t → t
type 'a mul_impl = (module MUL with type t = 'a)
module type MUL_F =
functor (E : EQ) (N : NUM with type t = E.t)
→ MUL with type t = E.t
Listing 4
A default instance of Mul can be provided given compatible instances of Eq and Num. (See Listing 5.)
module Mul_default : MUL_F =
functor (E : EQ) (N : NUM with type t = E.t)
→ struct
include (E : EQ with type t = E.t)
include (N : NUM with type t := E.t)
let mul : t → t → t =
let rec loop x y = begin match () with
| () when eq (x, (from_int 0))
→ from_int 0
| () when eq (x, (from_int 1)) → y
| () → y + loop (x + (from_int (-1))) y
end in loop
module Mul_bool : MUL with type t = bool =
Mul_default (Eq_bool) (Num_bool)
Listing 5
Specific instances can be constructed as needs demand (Listing 6).
module Mul_int : MUL with type t = int = struct
include (Eq_int : EQ with type t = int)
include (Num_int : NUM with type t := int)
let mul = Pervasives.( * )
let dot : 'a mul_impl → 'a list → 'a list → 'a
= fun (type a) (mul : a mul_impl) →
fun xs ys →
let module M =
(val mul : MUL with type t = a) in
sum (module M : NUM with type t = a)
@@ List.map2 M.mul xs ys
let test_dot =
dot (module Mul_int : MUL with type t = int)
[1; 2; 3] [4; 5; 6]
Listing 6
Note that in this definition of dot, coercion of the provided Mul instance to its base Num instance is performed.
Listing 7 provides an example of polymorphic recursion utilizing the dynamic production of evidence by way of the show_list function presented earlier.
let rec replicate : int → 'a → 'a list =
fun n x → if n <= 0
then [] else x :: replicate (n - 1) x
let rec print_nested : 'a. 'a show_impl →
int → 'a → unit = fun show_mod → function
| 0 → fun x → print show_mod x
| n → fun x → print_nested
(show_list show_mod) (n - 1) (replicate n x)
let test_nested =
let n = read_int () in
print_nested (module Show_int : SHOW
with type t = int) n 5
Listing 7
This article was previously published as a blog post in 2016. [Fletcher16b] and the source is available at: https://github.com/shayne-fletcher/overload-2017/blob/master/mod.ml
[Dreyer07] Derek Dreyer, Robert Harper and Manuel M. T. Chakravarty, ‘Modular Type Classes’, 2007, available online at http://www.cse.unsw.edu.au/~chak/papers/mtc-popl.pdf
[Fletcher16a] Shayne Fletcher, ‘Haskell type-classes in OCaml and C++’, available at http://blog.shaynefletcher.org/2016/10/haskell-type-classes-in-ocaml-and-c.html
[Fletcher16b] Shayne Fletcher, ‘Implementing type-classes as OCaml modules’, available at http://blog.shaynefletcher.org/2016/10/implementing-type-classes-as-ocaml.html
[Kiselyov14] Oleg Kiselyov, ‘Implementing, and Understanding Type Classes’, updated November 2014, available at http://okmij.org/ftp/Computation/typeclass.html
Overload Journal #142 - December 2017 + Programming Topics
Browse in : All > Journals > Overload > o142 (7)
All > Topics > Programming (877)
Any of these categories - All of these categories | {"url":"https://members.accu.org/index.php/journals/2445","timestamp":"2024-11-12T13:25:52Z","content_type":"application/xhtml+xml","content_length":"34775","record_id":"<urn:uuid:e9ae0f5c-698e-45a9-8be0-b7517495538a>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00751.warc.gz"} |
Alist vs. hash-table
An alist is a simple data structure that holds key-value pairs in a linked list. When a key is looked up, the list is searched to find it. The time it takes is proportional to the length of the list,
or the number of entries.
A hash-table is a more complex data structure that holds key-value pairs in a set of "hash buckets". When a key is looked up, it is first "hashed" to find the correct bucket, then that bucket is
searched for the entry. The time it takes depends on a number of things, the hash algorithm, the number of buckets, the number of entries in the bucket, etc. A hash-table can be faster than an alist
because the hashing step is quick and the subsequent search step will have very few entries to search.
In theory, an alist takes time proportional to the number of entries, but a hash-table takes constant time independent of the number of entries. Let's find out if this is true for MIT/GNU Scheme.
I wrote a little program that measures how long it takes to look things up in an alist vs. a hash table. Here's what I measured:It does indeed seem that alists are linear and hash tables are constant
in lookup time. But the hashing step of a hash table does take time, so short alists end up being faster that hash tables. The breakeven point looks like a tad over 25 elements. So if you expect 25
or fewer entries, an alist will perform better than a hash table. (Of course different implementations will have different break even points.)
A tree data structure is slightly more complex than an alist, but simpler than a hash table. Looking up an entry in a tree takes time proportional to the logarithm of the number of entries. The
logarithm function grows quite slowly, so a tree performs pretty well over a very large range of entries. A tree is slower than an alist until you have about 15 entries. At this point, the linear
search of an alist cannot compete with the logarithmic search of a tree. The time it takes to search a tree grows, but quite slowly. It takes more than 100 entries before a tree becomes as slow as a
hash table.
With a big enough tree, the growth is so slow that you can pretend it is constant.
1 comment:
Paul F. Dietz said...
I ran into the slowness of hashing in another context: the Common Lisp function INTERN. I was using this to convert strings to keywords when parsing json. I could speed up the conversion by
special casing its for the common strings occurring as json field names, using someone's highly optimized STRING-CASE macro to build a fast decision tree. | {"url":"http://funcall.blogspot.com/2016/01/alist-vs-hash-table.html","timestamp":"2024-11-05T08:53:18Z","content_type":"text/html","content_length":"93302","record_id":"<urn:uuid:cc81adda-1e50-4f13-8033-9d55986ac778>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00676.warc.gz"} |
605 decimeters per square second to hectometers per square second
5,605 Decimeters per square second = 5.61 Hectometers per square second
Acceleration Converter - Decimeters per square second to hectometers per square second - 5,605 hectometers per square second to decimeters per square second
This conversion of 5,605 decimeters per square second to hectometers per square second has been calculated by multiplying 5,605 decimeters per square second by 0.001 and the result is 5.605
hectometers per square second. | {"url":"https://unitconverter.io/decimeters-per-square-second/hectometers-per-square-second/5605","timestamp":"2024-11-14T20:41:10Z","content_type":"text/html","content_length":"27118","record_id":"<urn:uuid:cd9338d7-4fcd-40c0-9a48-d20527045a85>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00043.warc.gz"} |
rocket nozzle calculator
12 Oct 2024
Title: Development of a Rocket Nozzle Calculator: A Comprehensive Tool for Design Optimization
This paper presents the development of a rocket nozzle calculator, a comprehensive tool designed to optimize the performance of rocket nozzles. The calculator takes into account various design
parameters, including nozzle shape, throat diameter, and exit velocity, to predict the optimal nozzle configuration for a given set of mission requirements. The calculator is based on established
theoretical frameworks, including the conservation of mass and momentum principles, and utilizes numerical methods to solve the governing equations.
Rocket nozzles play a crucial role in the performance of rocket propulsion systems, as they determine the exhaust velocity and thrust of the rocket. Optimal nozzle design is essential for achieving
efficient combustion, minimizing heat transfer, and maximizing specific impulse. However, designing an optimal nozzle configuration can be a complex task, requiring extensive knowledge of fluid
dynamics, thermodynamics, and numerical methods.
The calculator is based on the following theoretical framework:
1. Conservation of Mass: The mass flow rate of the exhaust gases is conserved throughout the nozzle, given by:
Q = ρA * V (1)
where Q is the mass flow rate, ρ is the density of the exhaust gases, A is the cross-sectional area of the nozzle, and V is the velocity of the exhaust gases.
1. Conservation of Momentum: The momentum of the exhaust gases is conserved throughout the nozzle, given by:
F = ρA * V^2 (2)
where F is the thrust force, ρ is the density of the exhaust gases, A is the cross-sectional area of the nozzle, and V is the velocity of the exhaust gases.
1. Thermodynamic Properties: The thermodynamic properties of the exhaust gases, such as temperature and pressure, are related to the specific impulse (Isp) and the nozzle efficiency (η), given by:
Isp = ∫(V * dM) / M (3)
where Isp is the specific impulse, V is the velocity of the exhaust gases, and M is the mass flow rate.
Calculator Development:
The calculator was developed using a combination of analytical and numerical methods. The following steps were taken:
1. Formulation: The governing equations (1-3) were formulated in ASCII format:
Q = ρA * V
F = ρA * V^2
Isp = ∫(V * dM) / M
2. Numerical Solution: A numerical solution was developed using a finite difference method to solve the governing equations.
3. User Interface: A user-friendly interface was designed to input design parameters, such as nozzle shape, throat diameter, and exit velocity, and output the optimal nozzle configuration.
The calculator was tested for various design scenarios, including different nozzle shapes (convergent, divergent, and converging-diverging) and throat diameters. The results showed that the
calculator accurately predicted the optimal nozzle configuration for each scenario, with a maximum error of 5%.
This paper presents the development of a rocket nozzle calculator, a comprehensive tool designed to optimize the performance of rocket nozzles. The calculator takes into account various design
parameters and utilizes numerical methods to solve the governing equations. The results show that the calculator accurately predicts the optimal nozzle configuration for various design scenarios.
The following formulae were used in the development of the calculator:
• Q = ρA * V (1)
• F = ρA * V^2 (2)
• Isp = ∫(V * dM) / M (3)
These formulae are presented in ASCII format for ease of use.
[1] Hall, J. M., & Knight, R. R. (2014). Fundamentals of Aerospace Engineering. Cambridge University Press.
[2] Sutton, G. P., & Biblarz, O. (2001). Rocket Propulsion Elements. John Wiley & Sons.
[3] Anderson, J. D. (2016). Computational Fluid Mechanics and Heat Transfer. McGraw-Hill Education.
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188,015 research outputs found
The state of cold quark matter really challenges both astrophysicists and particle physicists, even many-body physicists. It is conventionally suggested that BCS-like color superconductivity occurs
in cold quark matter; however, other scenarios with a ground state rather than of Fermi gas could still be possible. It is addressed that quarks are dressed and clustering in cold quark matter at
realistic baryon densities of compact stars, since a weakly coupling treatment of the interaction between constituent quarks would not be reliable. Cold quark matter is conjectured to be in a solid
state if thermal kinematic energy is much lower than the interaction energy of quark clusters, and such a state could be relevant to different manifestations of pulsar-like compact stars.Comment:
Proceedings of IWARA2009 (IJMP D
It is conjectured that cold quark matter with much high baryon density could be in a solid state, and strange stars with low temperatures should thus be solid stars. The speculation could be close to
the truth if no peculiar polarization of thermal X-ray emission (in, e.g., RXJ1856), or no gravitational wave in post-glitch phases, is detected in future advanced facilities, or if spin frequencies
beyond the critical ones limited by r-mode instability are discovered. The shear modulus of solid quark matter could be ~ 10^{32} erg/cm^3 if the kHz QPOs observed are relevant to the eigenvalues of
the center star oscillations.Comment: Revised significantly, ApJL accepted, or at http://vega.bac.pku.edu.cn/~rxxu/publications/index_P.ht
We study the instability development during a viscous liquid drop impacting a smooth substrate, using high speed photography. The onset time of the instability highly depends on the surrounding air
pressure and the liquid viscosity: it decreases with air pressure with the power of minus two, and increases linearly with the liquid viscosity. From the real-time dynamics measurements, we construct
a model which compares the destabilizing stress from air with the stabilizing stress from liquid viscosity. Under this model, our experimental results indicate that at the instability onset time, the
two stresses balance each other. This model also illustrates the different mechanisms for the inviscid and viscous regimes previously observed: the inviscid regime is stabilized by the surface
tension and the viscous regime is stabilized by the liquid viscosity.Comment: 4 pages, 5 figure
Let $u$ be a smooth convex function in $\mathbb{R}^{n}$ and the graph $M_{abla u}$ of $abla u$ be a space-like translating soliton in pseudo-Euclidean space $\mathbb{R}^{2n}_{n}$ with a translating
vector $\frac{1}{n}(a_{1}, a_{2}, \cdots, a_{n}; b_{1}, b_{2}, \cdots, b_{n})$, then the function $u$ satisfies $\det D^{2}u=\exp \left\{ \sum_{i=1}^n- a_i\frac{\partial u}{\partial x_{i}} +\sum_{i=
1}^n b_ix_i+c\right\} \qquad \hbox{on}\qquad\mathbb R^n$ where $a_i$, $b_i$ and $c$ are constants. The Bernstein type results are obtained in the course of the arguments.Comment: 9 page
The spectrum of the MeV tail detected in the black-hole candidate Cygnus X-1 remains controversial as it appeared much harder when observed with the INTEGRAL Imager IBIS than with the INTEGRAL
spectrometer SPI or CGRO. We present an independent analysis of the spectra of Cygnus X-1 observed by IBIS in the hard and soft states. We developed a new analysis software for the PICsIT detector
layer and for the Compton mode data of the IBIS instrument and calibrated the idiosyncrasies of the PICsIT front-end electronics. The spectra of Cygnus X-1 obtained for the hard and soft states with
the INTEGRAL imager IBIS are compatible with those obtained with the INTEGRAL spectrometer SPI, with CGRO, and with the models that attribute the MeV hard tail either to hybrid thermal/non-thermal
Comptonisation or to synchrotron emission.Comment: 6 pages, 7 figure
It is found that 1E 1207.4-5209 could be a low-mass bare strange star if its small radius or low altitude cyclotron formation can be identified. The age problems of five sources could be solved by a
fossil-disk-assisted torque. The magnetic dipole radiation dominates the evolution of PSR B1757-24 at present, and the others are in propeller (or tracking) phases.Comment: ApJL accepted, or at http: | {"url":"https://core.ac.uk/search/?q=authors%3A(Xu%2C%20R)","timestamp":"2024-11-08T08:38:45Z","content_type":"text/html","content_length":"121975","record_id":"<urn:uuid:7db4ef8d-5e58-4591-a914-8db1c233c576>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00314.warc.gz"} |
Problem Solving Paradigm
Problem Description - Given a weighted (family) tree of up to $N ≤ 80K $vertices with a special trait: Vertex values are increasing from root to leaves. Find the ancestor vertex closest to the root
from a starting vertex $v$ that has weight at least $ P$. There are up to $Q ≤ 20K$ such offline queries. Examine Figure 3.3 (left). If $P = 4$, then the answer is the vertex labeled with $B$ with
value $5$ as it is the ancestor of vertex v that is closest to root ‘$A$’ and has a value of $≥ 4$. If $P = 7$, then the answer is ‘$C$’, with value $7$.If $ P ≥ 9$, there is no answer.
One way to solve is do linear search $ O (QN) $ and since $Q$ is quite large this approach will give TLE.
Better solution is to traverse once starting at the node using the $ O(N) $ preorder tree traversal algorithm.
So we get these partial root-to-current-vertex sorted array:
{{3}, {3, 5}, {3, 5, 7},{3, 5, 7, 8}, backtrack, {3, 5, 7, 9}, backtrack,backtrack, backtrack, {3, 8}, backtrack,{3, 6}, {3, 6, 20}, backtrack, {3, 6, 10}, and finally {3, 6, 10, 20}, backtrack,
backtrack, backtrack (done)}.
Now we can use Binary search on these sorted arrays when we are queried. | {"url":"https://algo.minetest.in/CP3_Book/3_Problem_Solving_Paradigms/","timestamp":"2024-11-10T02:40:03Z","content_type":"text/html","content_length":"203943","record_id":"<urn:uuid:2db35451-e90d-4b27-bfef-645b103a8caa>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00143.warc.gz"} |
New algorithms for structure informed genome rearrangement
We define two new computational problems in the domain of perfect genome rearrangements, and propose three algorithms to solve them. The rearrangement scenarios modeled by the problems consider
Reversal and Block Interchange operations, and a PQ-tree is utilized to guide the allowed operations and to compute their weights. In the first problem, ConstrainedTreeToStringDivergence (CTTSD), we
define the basic structure-informed rearrangement measure. Here, we assume that the gene order members of the gene cluster from which the PQ-tree is constructed are permutations. The PQ-tree
representing the gene cluster is ordered such that the series of gene IDs spelled by its leaves is equivalent to that of the reference gene order. Then, a structure-informed genome rearrangement
distance is computed between the ordered PQ-tree and the target gene order. The second problem, TreeToStringDivergence (TTSD), generalizes CTTSD , where the gene order members are not necessarily
permutations and the structure informed rearrangement measure is extended to also consider up to d[S] and d[T] gene insertion and deletion operations, respectively, when modelling the PQ-tree
informed divergence process from the reference gene order to the target gene order. The first algorithm solves CTTSD in O(nγ^2· (m[p]· 1. 381 ^γ+ m[q])) time and O(n^2) space, where γ is the maximum
number of children of a node, n is the length of the string and the number of leaves in the tree, and m[p] and m[q] are the number of P-nodes and Q-nodes in the tree, respectively. If one of the
penalties of CTTSD is 0, then the algorithm runs in O(nmγ^2) time and O(n^2) space. The second algorithm solves TTSD in O(n2γ2dT2dS2m2(mp·5γγ+mq)) time and O(d[T]d[S]m(mn+ 5 ^γ)) space, where γ is
the maximum number of children of a node, n is the length of the string, m is the number of leaves in the tree, m[p] and m[q] are the number of P-nodes and Q-nodes in the tree, respectively, and
allowing up to d[T] deletions from the tree and up to d[S] deletions from the string. The third algorithm is intended to reduce the space complexity of the second algorithm. It solves a variant of
the problem (where one of the penalties of TTSD is 0) in O(nγ2dT2dS2m2(mp·4γγ2n(dT+dS+m+n)+mq)) time and O(γ^2nm^2d[T]d[S](d[T]+ d[S]+ m+ n)) space. The algorithm is implemented as a software tool,
denoted MEM-Rearrange, and applied to the comparative and evolutionary analysis of 59 chromosomal gene clusters extracted from a dataset of 1487 prokaryotic genomes.
• Breakpoint distance
• Gene cluster
• PQ-tree
ASJC Scopus subject areas
• Structural Biology
• Molecular Biology
• Computational Theory and Mathematics
• Applied Mathematics
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Algorithms Weekly by Petr Mitrichev
EGOI 2024 in Veldhoven, the Netherlands was the main event of last week (
, top 5 on the left,
). Eliška has won for the
second time in a row
, this time with an even more commanding margin — huge congratulations! She was one of the four girls (
with Lara, Vivienne and Anja) who has participated in all four EGOIs, but it seems that this was the last year for all of them. Nevertheless, the new stars are already there as well, with so many
medalists having several EGOI years ahead of them. It is great to see that this wonderful new community has been built and thrives!
Just like last year, I was onsite as a task author, but my task did not end up being used in the contest. I really need to up my game and submit more and better problems next year :) Of the problems
that did end up in the contest, I'd like to highlight
problem D
from the first day which had the unusual multirun format. There is a sequence of
bits (each 0 or 1)
that is unknown to you. In addition, there is a permutation
of size
that is known to you. Your goal is to apply this permutation to this sequence, in other words to construct the new sequence
. Your solution will be invoked as follows. On the first run, it will read the run number (0) and the permutation
, print an integer
, and exit. It will then be executed
more times. On the first of those runs, it will read the run number (1) and the permutation
again, and then it will read the sequence
from left to right, and after reading the
-th bit, it has to write the new value for the
-th bit before reading the next one. After processing the
bits, your program must exit. On the second of those runs, it will read the run number (2) and
again, and then read
(potentially modified during the first run) from right to left, and it has to write the new value for the i-th bit before reading the previous one. On the third of those runs, it goes left to right
again, and so on. After the
-th run is over, we must have
, where
is the final state of the sequence, and
is the original state of the sequence.
To get the full score on this problem you must always have w<=3, but I find that solving for w<=5 (which would give 95 points out of 100) is more approachable and potentially more fun. As another
hint, two of the subtasks in the problem were as follows:
• n=2
• the given permutation is a reverse permutation
Can you see how to move from those subtasks to a general solution with w<=5?
Codeforces ran Pinely Round 4 on Sunday (
, top 5 on the left,
). As Um_nik rightly
points out
, tourist has continued his amazing run of form, and is potentially one more win away from crossing the magical boundary of 4000. Very well done!
I also share Yui's sentiment: it is very cool and quite surprising that problems H and I can in fact be solved. During the contest, I did not manage to make any significant progress in both of them
in about 1 hour and 15 minutes I had left after solving the first 7. On the other hand, the (nicely written!) editorial almost makes them look easy :)
If you have not checked out the editorial yet, you can try crack problem I yourself. The statement is quite simple and beautiful: you are given an empty n times m grid (4 <= n, m <= 10). You need to
write all integers from 1 to nm exactly once in the grid. You will then play the following game on this grid as the second player and need to always win. The first player takes any cell of the grid
for themselves. Then the second player takes any cell that is not taken yet, but is adjacent (shares a side) to a taken cell, for themselves. Then the first player takes any cell that is not taken
yet, but is adjacent (shares a side) to a taken cell, for themselves, and so on until all cells are taken. The second player wins if the sum of the numbers of the cells they have in the end is
smaller than the sum of the numbers of the cells of the first player.
Given that the first player can choose the starting cell arbitrarily, it seems quite hard to believe that the second player can have any advantage. Can you see the way?
Codeforces Round 959 sponsored by NEAR was the main event of last week (problems, results, top 5 on the left, analysis). AWTF participants have probably already returned home and were ready to fill
the top five places in this round, but Egor has managed to solve everything and squeeze in the middle of their party. Congratulations to Egor on doing this and to tourist on winning the round!In my
previous summary, I have mentioned two problems. The first one came from AWTF: you are given n<=250000 slimes on a number line, each with weight 1. You need to choose k of them, all others will be
removed. Then, they will start moving: at each moment in time, every slime moves with velocity r-l, where r is the total weight of all slimes to the right of it, and l is the total weight of all
slimes to the left of it. When r-l is positive, it moves to the right, and when it is negative, it moves to the left. Since this rule pushes the slimes towards each other, sometimes they will meet.
When two or more slimes meet, they merge into one slime with weight equal to their total weight, which continues to move according to the above rule. Eventually, all slimes will merge into one
stationary slime, suppose this happens at time t. What is the maximum value of t, given that you can choose the k slimes to use freely?
Even though the problem does not ask for it, it is actually super helpful to think where will the final big slime be located. For me, this would have probably been the hardest part of solving this
problem. Why should one think about the final position if the problem only asks about the final time?..
After studying a few examples, one can notice that the final position is always equal to the average of the starting positions of the slimes. Having noticed this, it is relatively easy to prove:
consider the function sum(a[i]*w[i]) where a[i] is the position of the i-th slime, and w[i] is its weight, and consider two slimes (a[i],w[i]) and (a[j],w[j]). The first one contributes -w[i] to the
velocity of the second one, while the second one contributes w[j] to the velocity of the first one. Therefore together they contribute -w[i]*wj+w[j]*w[i]=0 to the velocity of value sum(a[i]*w[i]),
therefore that sum stays constant. And it also does not change when two slimes merge, therefore it is always constant and has the same value for the final big slime.
Now is the time for the next cool observation, this one is a bit more logical though. Consider the last two slimes that merge. They split the original slimes into two parts. What happens to the
weighted averages of the positions of those two parts sum(a[i]*w[i])/sum(w[i])? By the same argument as above, influences within each of the parts on that average cancel out. The influences of the
parts on each other do not cancel out though, but we can easily compute them and find out that those two averages are moving towards each other with constant velocity equal to the total weight, in
other words k.
Therefore if we know which two parts form the two final slimes we can find the answer using a simple formula: (sum(a[i]*w[i])/sum(w[i])-sum(a[j]*w[j])/sum(w[j]))/k, where j iterates over the slimes
in the first part, and i iterates over the slimes in the second part.
And here comes the final leap of faith, also quite logical: the answer can actually be found by finding the maximum of that amount over all ways to choose a prefix and a suffix as the two parts. This
can be proven by noticing that the difference between averages starts decreasing slower than k if some slimes merge across the part boundary, therefore the number we compute is both a lower bound and
an upper bound on the actual answer.
We have not yet dealt to the fact that we have to choose k slimes out of n, but seeing the above formula it is pretty clear that we should simply take a prefix of all available slimes for the sum
over j, and a suffix of all available slimes for the sum over i. Now all pieces of the puzzle fit together very well, and we have a full solution.
The second problem I mentioned came from Universal Cup: you are given a string of length n<=500000. We choose one its arbitrary non-empty substring and erase it from this string, in other words we
concatenate a prefix and a suffix of this string with total length less than n. There are n*(n+1)/2 ways to do it, but some of them may lead to equal strings. How many distinct strings can we get?
A natural question to ask is: how can two such strings be the same? If we align two different ways to erase a substring of the same length that lead to the same result, we get the following two
representations of the same string:
where in one case we delete the substring c, and in the other case we delete the substring e to obtain the result abd. We can notice that such pairs of equal prefix+suffix concatenations are in a 1:1
correspondence with pairs of equal substrings within the string (two occurrences of the substring b in the above example). It is not true though that our answer is simply the number of distinct
substrings, as we have merely proven that the sum of c*(c-1)/2 over all repeat quantities c of equal substrings is the same as the sum of d*(d-1)/2 over all repeat quantities d of equal
prefix+suffix, but that does not mean that the sum of c is the same as the sum of d.
However, this makes one think that probably the suffix data structures, such as the suffix tree, array or automaton, will help solve this problem in the same way they help count the number of
distinct substrings. This turns out to be a false path, as the actual solution is much simpler!
Consider the middle part of the above equation (bc=eb), and let us write out an example with longer strings for clarity:
We can notice that the following are also valid ways to obtain the same string:
So the structure here is simpler than in counting distinct substrings. In order to count only distinct prefix+suffix strings, let us count only canonical respresentations, for example those where the
prefix is the longest. The criteria for when we cannot make the prefix even longer is evident from the above example: the next character after the prefix (the one that is removed) must be different
from the first character of the suffix, if any. Therefore the answer is simply equal to the number of pairs of characters in the given string that differ, plus n to account for the representations
where the suffix is completely empty.
Thanks for reading, and check back next week!
AWTF24 was the main event of this week. I have mentioned its results in the previous post, so I want to use this one to discuss its problems briefly.
Problems A and B both had short solutions and were quick to implement, but required one to come up with a beautiful idea somehow. When Riku was explaining their solutions during the broadcast, I was
amazed but could not understand how to come up with them :) One way to do it, at least for problem A, was to actually start by assuming the problem is beautiful, and to try coming up with some
beautiful lower or upper bound for the answer, which can turn out to be the answer.
To test if you can walk this path, here is the problem statement: you are given n<=250000 slimes on a number line, each with weight 1. You need to choose k of them, all others will be removed. Then,
they will start moving: at each moment in time, every slime moves with velocity r-l, where r is the total weight of all slimes to the right of it, and l is the total weight of all slimes to the left
of it. When r-l is positive, it moves to the right, and when it is negative, it moves to the left. Since this rule pushes the slimes towards each other, sometimes they will meet. When two or more
slimes meet, they merge into one slime with weight equal to their total weight, which continues to move according to the above rule. Eventually, all slimes will merge into one stationary slime,
suppose this happens at time t. What is the maximum value of t, given that you can choose the k slimes to use freely?
Problem D turned out to be equivalent to an old Codeforces problem applied to the inverse permutation. Most of this week's finalists have participated in that round or upsolved it, so it was not too
unfair. The top two contestants ecnerwala and zhoukangyang did solve the Codeforces problem back in 2022, but did not remember it, and implemented the solution to D from scratch (even though of
course having solved the old problem might have helped come up with the correct idea here). ksun48 and heno239 in places 3 and 4 did copy-paste their code from 2022.
Problems C and E involved a bit more code and effort to figure out all details, but one could make gradual progress towards a solution when solving them, instead of having to pull a beautiful idea
out of thin air. Were I actually participating in this round, I would most likely spend the most time on, and maybe even solve those two problems.
Overall, this was a very nice round, and I'm looking forward to more AGCs in 2024 to try my hand at more amazing problems!
On the next day after the AWTF, the 3rd Universal Cup Stage 4: Hongō took place (problems, results, top 5 on the left). 8 out of 9 participants from the first three teams (congrats!), which
coincidentally are also the first three teams in the season ranking, were in Tokyo, so Riku and Makoto have organized an onsite version of this stage at the AtCoder office. Solving a 5-hour contest
with your team in person instead of online is already much more fun, but having other teams in the room and discussing the problems with them right after the round is even better. I guess I'm still
yearning for the Open Cup days :)
I was solving this round together with Mikhail and Makoto. Thanks a lot Makoto for briefly coming out of retirement to participate, it was great fun solving this round together, and I can confirm
that you're still very strong! Maybe we can have another go next year.
Problem N was not very difficult (I spent at least half an hour without any success, explained the problem to Makoto and he solved it almost immediately), but still enjoyable: you are given a string
of length n<=500000. We choose one its arbitrary non-empty substring and erase it from this string, in other words we concatenate a prefix and a suffix of this string with total length less than n.
There are n*(n+1)/2 ways to do it, but some of them may lead to equal strings. How many distinct strings can we get?
Thanks for reading, and check back next week.
AtCoder World Tour Finals 2024 took place today (problems, results, top 5 on the left, broadcast recording, analysis). This was my first 5-hour commentating experience, and I enjoyed it a lot! How
did it look from your side? Please share your improvement suggestions, just for the remote chance that I do not qualify again :)
This time the contestants had an opportunity to share their thoughts on stream (similar to the "confession booth" concept from some chess tournaments recently), and while not everybody used it, it
was great fun and great insight to listen to those who did (for example; did somebody maybe gather all timestamps?). I hope this practice gets expanded and improved at future contests!
ecnerwala also tried to share his thoughts before the contest even started, but unfortunately the stream had not started as well at that point and therefore his words were not recorded. Nevertheless,
maybe this helped him get into the right mood and solve problem E quickly, which was key to his win. Congratulations to him and to zhoukangyang and ksun48 who also won prizes!
Thanks for reading, and check back tomorrow for this week's summary.
On Wednesday, the contestants were gathering in the hotel. The contestants from Europe and America hat some very long flights behind them, so there was not much appetite for activities. Therefore we
played some board games in the hotel lobby in between short excursions to get some Japanese food. We did not actually meet most of the contestants from Asia — maybe the reason was that they actually
had more energy for exploring Tokyo and did not hang around in the hotel :)
The games of choice (well, those were the only ones I brought so there was not that much choice...) were Cascadia and (Level 1 H-Group) Hanabi. It turns out that the synergies of the H-Group
conventions are not so obvious at level 1, so probably next time we introduce somebody to them we should start at least with level 3.
We also got to witness the AtCoder admins printing the logos on the official t-shirts, as it turned out that the shop where one can print arbitrary content on a t-shirt in a self-service manner
happened to be on the lower floors of the hotel building. Even though this is not much different from a normal printer, seeing how one can slightly adjust the image and then get an actual t-shirt
with this image in a couple of minutes was quite impressive.
Today was a free day for the contestants, who have ventured a bit more into the city having rested from their travels. It was still funny with the timezone differences and jetlag, as the same meal
was breakfast for me, lunch for the locals, and dinner for the contestants from America. Some contestants warmed up their problem solving capabilities by doing escape rooms, while others opted for
actually solving old competitive programming problems for some last-ditch practice.
Tomorrow is the big day! The overall setup is similar to the last year, but with just one contest: 5 problems for 5 hours, the penalty time is computed as the time of the last successful submission (
not the usual ICPC sum) plus 5 minutes for each incorrect submission. You can find more details in Riku's post. And of course tune in to see my and Riku's commentary on the live broadcast which will
start at the same time as the contest, 12:30 Tokyo time, and last for 5 hours.
All 12 finalists are very strong, so it is hard to predict who will come out on top. zhoukangyang won 4 out of the last 6 AGCs, tourist has a lot of experience winning those contests, and jiangly has
won the AWTF last year — I guess we can keep an eye for those three, but anything can happen really.
Thanks for reading, and tune in tomorrow!
UPD: the live broadcast link has been updated.
There were no contests that I'd like to mention last week, so I can get straight to the new format of this blog for the coming week: a travel blog! I am going to the AtCoder World Tour Finals 2024 in
Tokyo. I did not manage to qualify this time, placing 14th when I needed to be in top 12, so I am going as a spectator and as a co-commentator for the stream, together with Riku, the main admin of
AtCoder.For a second year running, Tokyo welcomes the participants with an extreme weather warning in Japanese, this time for extreme heat. Please all take care, and focus on playing board games in
the air conditioned hotel!
Speaking for 5 hours while 12 contestants are facing, to put it mildly, challenging problems is also not a walk in the park. Please help us by suggesting topics that we should discuss or things we
should do on stream in comments!
In my previous summary, I have mentioned a Universal Cup problem: first, we draw the vertices of a regular n-gon (n<=10000) on the plane. Then we repeat the following process m times (m<=30000): take
any 3 already drawn points (either the original n-gon vertices, or ones drawn during the previous steps) A[i], B[i], C[i], and draw the point A[i]+B[i]-C[i] (the fourth vertex of a parallelogram).
Finally, we need to handle multiple queries of the form: given k drawn points (the sum of k over all queries <=30000), do they form the vertices of a regular k-gon in some order?
We can get points that are really close to a regular k-gon but not exactly one in this problem, and no feasible floating point precision is enough. Therefore we need to solve it using only integer
computations. Nevertheless, let us first consider how a floating-point solution might work.
We can imagine that the points lie on the complex plane, and the initial n points are e^2πi/n. Drawing a new point corresponds to computing A[i]+B[i]-C[i] using the complex number arithmetic. There
are many ways to check if k computed points form a regular k-gon, here is one: we need to check that the k points, in some order, can be represented as x, xy, xy^2, ..., xy^k-1, where y is such that
y^k=1 and no smaller power of y is equal to 1. Note that this order does not have to be the clockwise/counter-clockwise order of the vertices: multiplying by y can also represent a jump by any
coprime number of edges, and this criterion will still be valid. Also note that we can actually pick any of the vertices as x and such y will exist, moreover there will be φ(k) different values of y
that work for each vertex. So one way to find y is to just try a few other vertices z of the polygon, let y=z/x, and check if the criterion is satisfied. Since φ(k) is not too small compared to k, we
should find a valid y after a few attempts, let's say at most 50. Of course, we could've just said that y=e^2πi/k, but you will see below that the y=z/x approach leads to an interesting question that
I want to discuss.
If we denote the "base" point as u=e^2π/n, then all other initial points are powers of u, all computed points are polynomials of u, and the checks we are making boil down to whether a certain
polynomial of u with integer coefficients is equal to 0 or not (even though we also use division, checking if poly1(u)/poly2(u)=poly3(u)/poly4(u) is the same as checking if poly1(u)*poly4(u)-poly2(u)
*poly3(u)=0). We could try to maintain said polynomials with integer coefficients, but since the degrees would be on the order of n, and the coefficients themselves could be huge, this is not really
feasible within the time limit.
Here comes the key idea: instead of complex numbers and u=e^2π/n, let us do the same computations using integers modulo a prime p and using such u that the order of u modulo p is equal to n. Such u
exists if and only if p=sn+1 for some s, so we can search for a random big prime number of this form, which can be found quickly since all arithmetic progressions with coprime first element and
difference have a lot of prime numbers.
This actually works very nicely and allowed us to get this problem accepted. However, why does this actually work? The order of u is the same in our two models (complex numbers and modulo p), so
every polynomial of the form u^t-1 is equal to 0 in both or in neither model. However, this does not guarantee the same for an arbitrary polynomial with integer coefficients, or does it?
Of course it is not true for an arbitrary polynomial. For example, the polynomial p is equal to 0 modulo p, but not equal to 0 in complex numbers. However, we can deal with this by picking the modulo
p at random, and potentially also checking several moduli in case the judges actually create testcases against many potential moduli. So the real question is: are there polynomials for which being
equal to 0 differs between the complex numbers and computations modulo p for all or a significant fraction of all possible values of p?
Here we need to bring in some maths. When two polynomials with rational coefficients are equal to 0 for the given u their greatest common divisor also has rational coefficients and is also equal to 0
for the given u, which means that there must exist a minimal polynomial such that a polynomial with rational coefficients is equal to 0 for the given u if an only if the minimal polynomial divides
it. Such minimal polynomial for our u is called the n-th cyclotomic polynomial Φ[n](u).
Now, consider the equality u^n-1=Φ[n](u)*g[n](u) (where g[n](u) is just the result of dividing u^n-1 by Φ[n](u)). This equality is true in rational numbers, so it is also true modulo p where there is
no division by p in it, so for almost all p. The left-hand side is 0 modulo p because of our choice of u, so either Φ[n](u) or g[n](u) must be 0. However, from the structure of the cyclotomic
polynomials we know that g[n](u) is a product of cyclotomic polynomials of smaller order, so if it was equal to 0, it would mean that the identity u^t-1=0 would hold for some t<n, which contradicts
our choice of u. So we know that Φ[n](u)=0 modulo p, which means that every polynomial with integer coefficients that is equal to 0 for the given complex u will also be equal to 0 for the given u
modulo p. So we have proven one of the two required implications.
Now let us tackle the the opposite implication. Consider a polynomial h(u) with integer coefficients that is equal to 0 for all or a significant fraction of all possible values of p (with the
corresponding u). If Φ[n](u) divides this polynomial (as a polynomial with rational coefficients), then it is also equal to 0 for the given complex u, as needed. If Φ[n](u) does not divide it, then
we can find the greatest common divisor of Φ[n](u) and h(u), again doing computations using polynomials with rational coefficients. Since Φ[n](u) is irreducible over polynomials with rational
coefficients, this greatest common divisor will be 1, so we have 1=Φ[n](u)*a(u)+h(u)*b(u). The right side involves a finite number of different integers in denominators, so this equality will also
hold modulo p for all p except those dividing one of the denominators, in other words for almost all p. But since both Φ[n](u) and h(u) are equal to 0 for all or a significant fraction of all
possible values of p, this means that 1 is equal to 0 for all or a significant fraction of all possible values of p, which is a contradiction. Therefore we have also proven the opposite implication
and this solution does in fact work.
There are still a few things I do not fully understand about this setup. One is the following: it turns out that when n is odd, we can actually construct a regular 2n-gon (roughly speaking using the
fact that -1 helps generate the other n points; there was such example in the samples for n=3, k=6), so k does not have to divide n. In this case, the number y that we find as part of solving the
problem must have order 2n modulo p. However, note that in general it is not even guaranteed that there is any number with order 2n modulo p, as we only choose p in such a way that there is a number
with order n. Since we compute y=z/x, we can do this computation for any p where we can compute z and x. So it seems that the above also proves that for almost all primes p if there is a number of
odd order n modulo p, there is also a number of order 2n modulo p. This observation is in fact true for a straightforward reason: almost all primes are odd, so there is an even number p-1 of nonzero
remainders, therefore there is a number of order 2, and we can multiply the number of odd order n by the number of order 2 to get a number of order 2n. Still, I can't get rid of the feeling that I
might be missing something here. Any comments?
The second thing I don't fully understand is whether we truly need a full understanding of the structure of cyclotomic polynomials to prove that Φ[n](u)=0 modulo p. It feels that maybe there is an
easier way to explain this that does not require so much knowledge?
Thanks for reading, and check back for more AWTF updates! | {"url":"https://blog.mitrichev.ch/2024/07/","timestamp":"2024-11-02T19:00:52Z","content_type":"text/html","content_length":"141656","record_id":"<urn:uuid:c22c1558-c6b1-41a5-b4ef-f672f96f1e04>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00194.warc.gz"} |
Uniform Distribution
A distribution which has constant probability is called a uniform distribution, sometimes also called a Rectangular Distribution. The probability density function and cumulative distribution function
for a continuous uniform distribution are
With and , these can be written
The Characteristic Function is
The Moment-Generating Function is
The function is not differentiable at zero, so the Moments cannot be found using the standard technique. They can, however, be found by direct integration. The Moments about 0 are
The Moments about the Mean are
so the Mean, Variance, Skewness, and Kurtosis are
The probability distribution function and cumulative distributions function for a discrete uniform distribution are
for , ..., . The Moment-Generating Function is
The Moments about 0 are
and the Moments about the Mean are
The Mean, Variance, Skewness, and Kurtosis are
Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 531 and 533, 1987.
© 1996-9 Eric W. Weisstein | {"url":"http://drhuang.com/science/mathematics/math%20word/math/u/u033.htm","timestamp":"2024-11-08T18:25:23Z","content_type":"text/html","content_length":"29231","record_id":"<urn:uuid:54a7ef4d-1f85-4801-abbd-a15e420628c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00531.warc.gz"} |
800 micrograms to milligrams (800 mcg to mg)
This is where you learn how to convert 800 micrograms to milligrams. Before we continue, note that micrograms can be shortened to mcg, and milligrams can be shortened to mg. Therefore, 800 micrograms
to milligrams is the same as 800 micrograms to mg, 800 mcg to milligrams, and 800 mcg to mg.
There are 1000 micrograms per milligram. The illustration below shows you how one microgram fits into one milligram, to put it in perspective. The area in dark blue is one milligram, and the tiny
square in the top left corner is one microgram.
Since there are 1000 micrograms per milligram, you divide micrograms by 1000 to convert it to milligrams. Here is the Micrograms to Milligrams Formula (mcg to mg formula):
micrograms ÷ 1000 = milligrams
mcg ÷ 1000 = mg
To convert 800 micrograms to milligrams, we enter 800 into our formula to get the answer as follows:
mcg ÷ 1000 = mg
800 ÷ 1000 = 0.8
800 mcg = 0.8 mg Microgram to Milligram Converter
Now you know that 800 micrograms equals 0.8 milligrams. Here you can convert another amount of micrograms to milligrams.
801 mcg to mg
Here is the next weight in micrograms on our list that we have converted to milligrams for you!
Privacy Policy | {"url":"https://convertermaniacs.com/microgram-to-milligram/convert-800-mcg-to-mg.html","timestamp":"2024-11-10T16:00:06Z","content_type":"text/html","content_length":"6830","record_id":"<urn:uuid:a4d3aa70-0309-401e-b94c-d55b08083e01>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00881.warc.gz"} |
Evolution of the Relative Price of Goods and Services in a Neoclassical Model of Capital Accumulation
This paper provides an explanation for the secular increase in the price of services relative to that of manufactured goods that relies on capital accumulation rather than on an exogenous total
factor productivity growth differential. The key assumptions of the two-sector, intertemporal optimizing model are relatively high capital intensity in the production of goods and limited
cross-border capital mobility, allowing the interest rate to vary. With plausible parameterization, the model also predicts a decline in the employment share of the goods sector over time.
I. Introduction
An increase in the price of services relative to that of manufactured goods is a well-documented feature of economic development (Baumol and Bowen, 1966; Obstfeld and Rogoff, 1996). Because
manufactured goods are tradable across borders while services are largely not, one may observe a secular increase in the price of nontradbles relative to that of tradables in an open economy – the
celebrated Balassa-Samuelson effect.^2
The explanation for this phenomenon relies largely on the difference in the growth of labor productivity over time between the two sectors.^3 As labor productivity goes up in manufacturing, wages
will increase in that sector. In the presence of intersectoral labor mobility, upward pressure will be put on the wages in services. If labor productivity growth lags behind in the services sector,
unit labor cost will rise in that sector, and the price of services will have to go up for their provision to remain profitable.
Labor productivity growth can be decomposed into contributions from capital deepening (increase in the capital-labor ratio) and total factor productivity (TFP) growth. While early models were
agnostic about the source of labor productivity differential (e.g., Baumol, 1967), lately the emphasis has been on TFP. This is particularly true of open-economy models (e.g., Obstfeld and Rogoff,
1996), where it is customary to assume the domestic interest rate to equal a given world interest rate, the parity being maintained by perfect capital mobility. As technological parameters and the
rate of interest completely determine capital-labor ratios and relative prices, TFP growth becomes the only possible source of a change in the relative price of the two sectors’ outputs.
This approach rests on three postulates: (1) labor productivity growth is driven primarily by TFP growth; (2) the rate of TFP growth is faster in manufacturing than in services; and (3) the domestic
interest rate equals the world interest rate when expressed in terms of tradable (manufactured) goods. Definitely, none of these propositions are universal truths, and their applicability should be
assessed on a case-by-case basis. These presumptions have been challenged even in some cases where strong priors existed in their favor. One of the most vivid examples is the controversy generated by
Young’s (1995) accounting exercise on the sources of growth in East Asia, where he downplayed the role of TFP growth. Regarding proposition (2), Triplett and Bosworth (2003) have established that TFP
in the services sector has been growing as fast as that in manufacturing in the United States since 1995.^4 And, of course, evidence abounds that even nowadays and even among the most open economies
capital mobility is far from perfect.
Increases in the relative price of services may occur, and have occurred, in circumstances where propositions (1)–(3) do not apply. While TFP growth drives economic progress in the long run,
developments in the medium term may be determined by capital accumulation. This is particularly true of economies that have high level of human capital and access to advanced technology, but
relatively low physical capital stock. This description fits well post-war economies, like western European countries or Japan after World War II. Other examples could include the Asian “tigers” at
the beginning of their takeoff and, arguably, transitional economies. One can also note that fast growth and change in relative prices occurred in many of these economies when cross-border capital
movements were quite restricted, so that the assumption of an exogenously given interest rate was not applicable.
This paper proposes an explanation for the change over time of the relative price of services and manufactures, or nontradables and tradables, that does not rely on TFP differential. The driving
force in the model is capital accumulation, which leads to an increase in the relative price of services under the assumption that this sector is relatively less capital intensive than manufacturing.
According to Obstfeld and Rogoff (1996), this assumption reflects reality.
While our result seems intuitive and has a familiar counterpart in trade theory, we have not been able to trace this particular application in the literature. Brock (1994) provides a rare example of
an intertemporal optimizing model that avoids the use of exogenous technological change as an explanation of changes in relative prices. Brock’s objective is to emphasize the importance of
investment, but as he assumes perfect capital mobility, he has to introduce a very complicated production structure (three factors of production and three goods, with two alternative technologies
with different capital intensities available for the production of one of them) to generate the effect. Elsewhere in the literature the possibility that uneven capital accumulation may be responsible
for changes in relative prices has been mentioned (e.g., Lipschitz et al., 2002), but no formal treatment has been provided.
Our model we dismisses the assumption of perfect capital mobility to allow interest rate changes and gradual capital accumulation. To highlight the contrast with perfect capital mobility models,
capital cannot move across borders in our model. However, one can readily see that the dynamics will be essentially similar in a system with imperfect capital mobility, where the differential between
the domestic and the world interest rate declines over time as the economy develops.
The model also sheds additional light on the issue of “deindustrialization” – a decline in the share of labor employed in manufacturing that accompanies high labor productivity growth in that sector.
While this phenomenon has been well documented (Baumol et al., 1989; Rowthorn and Ramaswamy, 1999), a model that assumes faster TFP growth in manufacturing than in services and a constant interest
rate would predict an increase in manufacturing employment share for any reasonable parameter values (Obstfeld and Rogoff, 1996). In contrast, our model predicts a shift of labor from manufacturing
into services over the course of development for the benchmark case of Cobb-Douglas preferences and production functions.
II. Model
The economy produces goods G and services S using capital K and labor L by means of neoclassical production functions F and H. We assume uniform labor-augmenting technological progress in both
$\begin{array}{ccc}{Q}_{G}=F\left({K}_{G},{e}^{xt}{L}_{G}\right)\equiv {e}^{xt}{L}_{G}f\left({\stackrel{^}{k}}_{G}\right),& {\stackrel{^}{k}}_{G}\equiv \frac{{K}_{G}}{{e}^{xt}{L}_{G}}& \\ {Q}_{S}=H\
left({K}_{S},{e}^{xt}{L}_{S}\right)\equiv {e}^{xt}{L}_{S}h\left({\stackrel{^}{k}}_{S}\right),& {\stackrel{^}{k}}_{S}\equiv \frac{{K}_{S}}{{e}^{xt}{L}_{S}}\end{array}\begin{array}{cc}\phantom{\rule
{7.0em}{0ex}}& \left(1\right)\end{array}$
In our notation, capital letters will be used to denote variables in levels (i.e., economy-wide aggregates), small letters will be used for per worker/per capita quantities (as well as prices, wages
and interest rates), and “hats” represent variables per unit of effective labor (i.e. per worker variables divided by the efficiency factor e^xt). The technology for producing goods is more capital
intensive than that for services, so that we always have k[G] > k[S], where
${k}_{G}\equiv \frac{{K}_{G}}{{L}_{G}}\text{ }\text{ }\text{and}\text{ }\text{ }{k}_{S}\equiv \frac{{K}_{S}}{{L}_{S}}{.}^{5}$
Factors of production are freely mobile between the two sectors, but cannot move across the border. Neither can the residents of the country borrow abroad – the capital account of the balance of
payments is closed. The current account could be closed as well, or we could allow trade in goods, but not in services, across the border. In the latter case, the ratio of services to goods prices
can also be interpreted as the relative price of nontradables and tradables, or the real exchange rate. Of course, the financing constraint would impose balanced trade.
The infinitely lived households maximize the present discounted value of a logarithmic Cobb-Douglas utility scaled at each moment of time by the number of household members. Population is initially
normalized to one and is assumed to grow exponentially at a rate n.
$\begin{array}{ccc}U=\underset{0}{\overset{\infty }{\int }}\left(\alpha \text{ }\text{ }{log}\text{ }{c}_{G}+\left(1-{\alpha }\right)\text{ }{log}\text{ }{c}_{S}\right){e}^{-\left(\rho -n\right)t}dt.
& \phantom{\rule{7.0em}{0ex}}& \left(2\right)\end{array}$
To guarantee that attainable utility is bounded, the rate of discount is assumed to exceed the rate of population growth:
$\begin{array}{ccc}\rho >n.& \phantom{\rule{7.0em}{0ex}}& \left(3\right)\end{array}$
We will normalize the price of the consumption good to unity, so that all the prices and wages are measured in units of consumption good.
The household supplies inelastically e^nt units of labor to the market. Each worker receives wage w regardless of the sector in which they are employed. The household can also hold assets of two
kinds – physical capital, which is convertible into the consumption good and back, so its unit price is one, and a riskless bond, denominated in the units of consumption good. The agents view both
types of assets as perfect substitutes, so the interest rate on the bond r equals the rental rate of capital.^6 As all agents are identical and the economy is closed, no bonds will be issued in
equilibrium, and household assets will consist only of capital. The household will maximize its utility (2) subject to an asset accumulation equation:
$\begin{array}{ccc}\stackrel{˙}{K}=rK+w{e}^{nt}-{C}_{G}-p{C}_{S},& \phantom{\rule{7.0em}{0ex}}& \left(4\right)\end{array}$
where p is the price of services.
It is straightforward to rewrite the objective function and the accumulation equation in “efficiency units.” The household will maximize^7
$\begin{array}{ccc}\stackrel{˜}{U}=\underset{0}{\overset{\infty }{\int }}\left(\alpha \text{ }\text{ }{log}\text{ }{\stackrel{^}{c}}_{G}+\left(1-{\alpha }\right)\text{ }{log}\text{ }{\stackrel{^}{c}}
_{S}\right){e}^{-\left(\rho -n\right)t}dt& \phantom{\rule{7.0em}{0ex}}& \left(5\right)\end{array}$
subject to the constraint
$\begin{array}{ccc}\stackrel{˙}{\stackrel{^}{k}}=\left(r-n-x\right)\stackrel{^}{k}+\stackrel{^}{w}-{\stackrel{^}{c}}_{G}-p{\stackrel{^}{c}}_{S}.& \phantom{\rule{7.0em}{0ex}}& \left(6\right)\end
From the household optimization problem we can derive an intratemporal condition
$\begin{array}{ccc}\frac{{\stackrel{^}{c}}_{G}}{p{\stackrel{^}{c}}_{S}}=\frac{\alpha }{1-\alpha }& \phantom{\rule{7.0em}{0ex}}& \left(7\right)\end{array}$
and an intertemporal Euler equation
$\begin{array}{ccc}\frac{{\stackrel{˙}{\stackrel{^}{c}}}_{G}}{{\stackrel{^}{c}}_{G}}=r-x-\rho .& \phantom{\rule{7.0em}{0ex}}& \left(8\right)\end{array}$
We will find it convenient to introduce total household expenditure in terms of goods:
$\begin{array}{ccc}C\equiv {C}_{G}+p{C}_{S}.& \phantom{\rule{7.0em}{0ex}}& \left(9\right)\end{array}$
With this notation, the Euler equation and the capital accumulation equation take the following form:
$\begin{array}{ccc}\frac{\stackrel{˙}{\stackrel{^}{c}}}{\stackrel{^}{c}}=r-x-\rho & \phantom{\rule{7.0em}{0ex}}& \left(10\right)\end{array}$
$\begin{array}{ccc}\stackrel{˙}{\stackrel{^}{k}}=\left(r-n-x\right)\stackrel{^}{k}+\stackrel{^}{w}-\stackrel{^}{c}.& \phantom{\rule{7.0em}{0ex}}& \left(11\right)\end{array}$
Profit maximization implies equality between the value marginal products of capital and labor in both sectors and the prices of these factors – the rental rate of capital and the wage rate. These
four marginal conditions determine four variables, ${\stackrel{^}{k}}_{S},r,\stackrel{^}{w}$ and p, as functions of $\text{ }{\stackrel{^}{k}}_{G}$ . If we fully differentiate the system, we obtain
the following responses of these variables to changes in ${\stackrel{^}{k}}_{G}$:
({\stackrel{^}{k}}_{S}\right)}d{\stackrel{^}{k}}_{G}& \phantom{\rule{7.0em}{0ex}}& \left(12\right)\end{array}$
$\begin{array}{ccc}dr={f}^{″}\left({\stackrel{^}{k}}_{G}\right)d{\stackrel{^}{k}}_{G}& \phantom{\rule{7.0em}{0ex}}& \left(13\right)\end{array}$
$\begin{array}{ccc}d\stackrel{^}{w}=-{\stackrel{^}{k}}_{G}{f}^{″}\left({\stackrel{^}{k}}_{G}\right)d{\stackrel{^}{k}}_{G}& \phantom{\rule{7.0em}{0ex}}& \left(14\right)\end{array}$
$\begin{array}{ccc}dp=\frac{{\stackrel{^}{k}}_{S}-{\stackrel{^}{k}}_{G}}{h\left({\stackrel{^}{k}}_{S}\right)}{f}^{″}\left({\stackrel{^}{k}}_{G}\right)d{\stackrel{^}{k}}_{G}& \phantom{\rule{7.0em}
{0ex}}& \left(15\right)\end{array}$
Next we combine market clearing conditions for capital, labor, and services and obtain an equation that links $\stackrel{^}{k},\stackrel{^}{c}$ and ${\stackrel{^}{k}}_{G}$:
$\begin{array}{ccc}\stackrel{^}{k}={\stackrel{^}{k}}_{G}-\left({\stackrel{^}{k}}_{G}-{\stackrel{^}{k}}_{S}\left({\stackrel{^}{k}}_{G}\right)\right)\frac{\left(1-\alpha \right)\stackrel{^}{c}}{\
stackrel{^}{w}\left({\stackrel{^}{k}}_{G}\right)+r\left({\stackrel{^}{k}}_{G}\right){\stackrel{^}{k}}_{S}\left({\stackrel{^}{k}}_{G}\right)}.& \phantom{\rule{7.0em}{0ex}}& \left(16\right)\end{array}$
This equation implicitly determines the capital intensity in the goods sector as an increasing function of consumption expenditure and the stock of capital in efficiency units:
$\begin{array}{ccc}{\stackrel{^}{k}}_{G}={\stackrel{^}{k}}_{G}\left(\underset{+}{\stackrel{^}{k}},\underset{+}{\stackrel{^}{c}}\right).& \phantom{\rule{7.0em}{0ex}}& \left(17\right)\end{array}$
$\begin{array}{cc}\frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{k}}=\frac{1}{{l}_{G}-{l}_{S}\frac{{\left({\stackrel{^}{k}}_{G}-{\stackrel{^}{k}}_{S}\right)}^{2}{f}^{″}\left({\stackrel
{G}}}>0& \\ \frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{c}}=\frac{\left(1-\alpha \right)\left({\stackrel{^}{k}}_{G}-{\stackrel{^}{k}}_{S}\right)}{ph\left({\stackrel{^}{k}}_{S}\right)}
\frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{k}}>0\end{array}\begin{array}{cc}\phantom{\rule{7.0em}{0ex}}& \left(18\right)\end{array}$
With this, we obtain a system of two differential equations in two unknowns:
$\begin{array}{ccc}\frac{\stackrel{˙}{\stackrel{^}{c}}}{\stackrel{^}{c}}=r\left({\stackrel{^}{k}}_{G}\left(\stackrel{^}{k},\stackrel{^}{c}\right)\right)-x-\rho & \phantom{\rule{7.0em}{0ex}}& \left(19
_{G}\left(\stackrel{^}{k},\stackrel{^}{c}\right)\right)-\stackrel{^}{c}.& \phantom{\rule{7.0em}{0ex}}& \left(20\right)\end{array}$
We can apply standard techniques to solve the dynamic system. The steady state (which corresponds to the balanced growth path of the economy) will be found at the intersection of the loci satisfying
the conditions $\stackrel{˙}{\stackrel{^}{c}}=0$ and $\stackrel{˙}{\stackrel{^}{k}}=0$. The former is given by the equation
$\begin{array}{ccc}f\prime \left({\stackrel{^}{k}}_{G}^{*}\right)=x+\rho .& \phantom{\rule{7.0em}{0ex}}& \left(21\right)\end{array}$
The locus of points ${\stackrel{^}{k}}_{G}\left(\stackrel{^}{k},\stackrel{^}{c}\right)={\stackrel{^}{k}}_{G}^{*}$ is a downward sloping straight line with the slope equal
$\begin{array}{ccc}\frac{d\stackrel{^}{c}}{d\stackrel{^}{k}}{׀}_{\stackrel{˙}{\stackrel{^}{c}}=0}=-\frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{k}}/\frac{\partial {\stackrel{^}{k}}_
{G}}{\partial \stackrel{^}{c}}=\frac{\stackrel{^}{w}*+\left(p+x\right){\stackrel{^}{k}}_{S}^{*}}{\left(1-\alpha \right)\left({\stackrel{^}{k}}_{G}^{*}-{\stackrel{^}{k}}_{S}^{*}\right)}.& \phantom{\
rule{7.0em}{0ex}}& \left(22\right)\end{array}$
To the right of that line ${\stackrel{^}{k}}_{G}>{\stackrel{^}{k}}_{G}^{*}$, so r is smaller than ρ+x, and consumption per unit of effective labor is declining. To the left of that line $\stackrel{˙}
{\stackrel{^}{c}}$ is positive.
The constancy of capital stock per unit of effective labor requires
$\begin{array}{ccc}\left[r\left({\stackrel{^}{k}}_{G}\right)-n-x\right]\stackrel{^}{k}+\stackrel{^}{w}\left({\stackrel{^}{k}}_{G}\right)-\stackrel{^}{c}=0.& \phantom{\rule{7.0em}{0ex}}& \left(23\
The points satisfying this equation are located on an upward sloping line^8 with a slope
$\begin{array}{ccc}\frac{d\stackrel{^}{c}}{d\stackrel{^}{k}}{׀}_{\stackrel{˙}{\stackrel{^}{k}}=0}=\frac{r-n-x-\left({\stackrel{^}{k}}_{G}-\stackrel{^}{k}\right){f}^{″}\frac{\partial {\stackrel{^}{k}}
_{G}}{\partial \stackrel{^}{k}}}{1+\left({\stackrel{^}{k}}_{G}-\stackrel{^}{k}\right){f}^{″}\frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{c}}}=\frac{r-n-x-\left({\stackrel{^}{k}}_{G}-{\
stackrel{^}{k}}_{S}\right){f}^{″}{l}_{S}\frac{\partial {\stackrel{^}{k}}_{G}}{\partial \stackrel{^}{k}}}{1+\left({\stackrel{^}{k}}_{G}-{\stackrel{^}{k}}_{S}\right){f}^{″}{l}_{S}\frac{\partial {\
stackrel{^}{k}}_{S}}{\partial \stackrel{^}{c}}}.& \phantom{\rule{7.0em}{0ex}}& \left(24\right)\end{array}$
The stock of capital is increasing to the right of this line and decreasing to its left.
These findings are presented in the graphical form in the phase diagram in Figure 1. The system will evolve along the stable arm of the saddle path, starting from the point where $\stackrel{^}{k}$
equals the initial stock of capital and converging over time to the balanced growth path.
If initially the capital stock is below its equilibrium value, as one would expect for an emerging market, over time $\stackrel{^}{k}$ and $\stackrel{^}{c}$ will rise, as can be seen from the phase
diagram. Being an increasing function of $\stackrel{^}{k}$ and $\stackrel{^}{c}$ (see equation (17)), the capital intensity in the goods sector ${\stackrel{^}{k}}_{G}$ will rise as well. This means
that the capital intensity in the services sector ${\stackrel{^}{k}}_{S}$ will also go up (according to equation (12)), and so will the relative price of services p (equation (15)). This establishes
the main result of the paper. The wage rate per unit of effective labor $\stackrel{^}{w}$ will increase in terms of both goods and services. The interest rate, on the other hand, will go down, as the
marginal product of capital declines.
These simple considerations do not establish the direction of the evolution of several other macroeconomic variables. In particular, the way in which employment will shift between the two sectors
cannot be established in the general case. The Appendix looks further into this issue and derives more definite results for certain special cases. We establish that labor will shift from the goods
sector into the services sector in the course of development if the production technology in both sectors is Cobb-Douglas. This fits the pattern observed in many countries across the world. In
contrast, an open economy model with perfect capital mobility, which assumes a constant interest rate and relies on TFP growth differential to generate change in relative prices, predicts an increase
in manufacturing employment as a result of an increase in manufacturing productivity if consumer preferences are Cobb-Douglas (which is arguably a reasonable baseline) and would require a very low
elasticity of substitution between goods and services in combination with other extreme parameter values to reverse the result (Obstfeld and Rogoff, 1996, p. 224).
III. Conclusion
This paper has presented a model in which an increase in the price of services relative to goods is generated by capital accumulation under the assumption that goods production technology is
relatively more capital intensive. While the idea that changes in relative price can be driven by capital accumulation has been mentioned in the literature, no formal optimizing models of that
process have been developed. Traditionally, the secular decline in the price of goods relative to that of services has been accounted for by a differential in total factor productivity growth. In
reality, these two mechanisms complement each other, but the capital channel has been given short shrift.
The relative importance of TFP and capital accumulation channels depends on the magnitude of TFP growth differential, the gap in capital intensity (as reflected in capital income shares), and the
rate of capital deepening. The capital accumulation channel will be particularly important for economies whose stock of capital has been depleted by wars or natural disasters, or was far below
potential because of government policies, which start growing fast, primarily through high rates of investment, once impediments have been removed.
The pattern of relative price evolution derived in the model is quite general. We did not specialize the production functions other than imposing the standard neoclassical properties. The choice of
the utility function was more restrictive, but it is easy to see that it is not critical for obtaining the secular increase in the relative price of services. Indeed, the relative price will go up
over time as capital is accumulated as long as the goods sector is relatively more capital intensive. The relative capital intensity condition is therefore critical for the result.^9
The evolution of sectoral output and employment does depend on the choice of the utility function. We have shown that in the baseline case of Cobb-Douglas preferences and production functions, labor
will shift from manufacturing into services as the country develops. This prediction accords with the pattern observed across the world. It is worth noting that models that assume perfect
international capital mobility and inter-industry TFP differential make the opposite prediction for this baseline case.
In the modern world, the assumption of a completely closed capital account of the balance of payments is arguably as unrealistic as the opposite assumption of perfect capital mobility. We certainly
agree with that. We would emphasize, however, that as long as imperfect capital mobility creates room for domestic interest rate movements, capital accumulation will have an effect on the evolution
of the relative price of goods and services along the lines drawn in our model.
Evolution of Sectoral Output and Employment over Time
As indicated in the main text, the direction of the evolution on the convergence path of the output of goods and services and the labor employment in the two sectors is in the general case ambiguous.
We will explore the evolution of these variables in the vicinity of the steady state using the following approach. For notational simplicity, we will assume away population growth and technological
We know that if the initial capital endowment is smaller than the steady-state value, consumption expenditure increases monotonically over time. Hence, a variable will increase over time if and only
if its full derivative with respect to C is positive. For example, the production of services will increase if and only if $\frac{d{Q}_{S}}{dC}>0$.
Now, as
${Q}_{S}={C}_{S}=\frac{\left(1-\alpha \right)C}{p},$
we can write
$\frac{d{Q}_{S}}{dC}=\frac{\left(1-\alpha \right)}{p}\left[1-\frac{C}{p}\frac{dp}{dC}\right]=\frac{\left(1-\alpha \right)}{p}\left[1+\frac{C\left({k}_{G}-{k}_{S}\right){f}^{″}}{ph}\frac{d{k}_{G}}{dC}
According to ^equation (17), k[G] is a function of K and C. In addition, along the convergence path K and C are related one to one. Hence,
$\frac{d{k}_{G}}{dC}=\frac{\partial {k}_{G}}{\partial C}+\frac{\partial {k}_{G}}{\partial K}\frac{dK}{dC}.$
The partial derivatives are given by ^equations (18). The inverse of the slope of the convergence line, $\frac{dK}{dC}$, cannot be found in the general case, as it would require solving the dynamic
system. However, we can express this derivative in the vicinity of the steady state via the parameters of the model. To do that, we linearize the model around the steady state.
$\frac{d}{dt}\left[\begin{array}{c}C-{C}^{*}\\ K-{K}^{*}\end{array}\right]\approx \left[\begin{array}{cc}{C}^{*}\frac{\partial {k}_{G}}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)& {C}^{*}\frac{\
partial {k}_{G}}{\partial K}{f}^{″}\left({k}_{G}^{*}\right)\\ -\left[1+{L}_{S}^{*}\left({k}_{G}^{*}-{k}_{S}^{*}\right)\frac{\partial {K}_{G}}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)\right]& \left
[\rho -{L}_{S}^{*}\left({k}_{G}^{*}-{k}_{S}^{*}\right)\frac{\partial {k}_{G}}{\partial K}{f}^{″}\left({k}_{G}^{*}\right)\right]\end{array}\right]\text{ }\left[\begin{array}{c}C-{C}^{*}\\ K-{K}^{*}\
The determinant of the matrix equals ${C}^{*}{f}^{″}\left({k}_{G}^{*}\right)\left[\frac{\partial {k}_{G}}{\partial K}+\rho \frac{\partial {k}_{G}}{\partial C}\right]$. Since the partial derivatives
of k[G] with respect to K and C are positive, the determinant has a negative sign, which confirms the saddle point stability in the vicinity of the steady state.
The trace equals simply ρ, as
$\begin{array}{c}{C}^{*}\frac{\partial k}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)-{L}_{S}^{*}\left({k}_{G}^{*}-{k}_{S}^{*}\right)\frac{\partial k}{\partial K}{f}^{″}\left({k}_{G}^{*}\right)={C}^
{*}\frac{\partial k}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)\left[1-\frac{{L}_{S}^{*}}{C*}\frac{\frac{\partial k}{\partial K}}{\frac{\partial k}{\partial c}}\left({k}_{G}^{*}-{k}_{S}^{*}\right)\
right]=\\ ={C}^{*}\frac{\partial k}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)\left[1-\frac{\left(1-\gamma \right)}{{p}^{*}h\left({k}_{s}^{*}\right)}\frac{p*h\left({k}_{S}^{*}\right)}{\left(1-\gamma
The rate of convergence, equal to the modulus of the negative eigenvalue of the above matrix, can be expressed as $\lambda =\frac{1}{2}\left[\sqrt{{\text{trace}}^{2}-4×\text{det}}-\text{trace}\right]
The derivative $\frac{dK}{dC}$ at the steady state can be expressed through the coefficients in the top row of the matrix of the linearized system and the convergence rate in the following fashion:
$\frac{dK}{dC}=-\frac{C*\frac{\partial {k}_{G}}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)+\lambda }{C*\frac{\partial {k}_{G}}{\partial K}{f}^{″}\left({k}_{G}^{*}\right)}.$
This implies
$\frac{d{k}_{G}}{\mathit{dC}}=\frac{\partial {k}_{G}}{\partial C}+\frac{\partial {k}_{G}}{\partial K}\frac{dK}{dC}=\frac{\partial {k}_{G}}{\partial C}-\frac{\partial {k}_{G}}{\partial K}\frac{{C}^{*}
\frac{\partial {k}_{G}}{\partial c}{f}^{″}\left({k}_{G}^{*}\right)+\lambda }{{C}^{*}\frac{\partial {k}_{G}}{\partial K}{f}^{″}\left({k}_{G}^{*}\right)}=-\frac{\lambda }{{C}^{*}{f}^{″}\left({k}_{G}^
Combining all these results, we obtain in the vicinity of the steady state
$\frac{d{Q}_{S}}{\mathit{dC}}=\frac{\left(1-\alpha \right)}{{p}^{*}}\left[1+\frac{\lambda \left({k}_{G}^{*}-{k}_{S}^{*}\right)}{p*h\left({k}_{S}^{*}\right)}\right]$
Hence, the condition for the production of services to increase over time is
$\lambda <\frac{{p}^{*}h\left({k}_{S}^{*}\right)}{{k}_{G}^{*}-{k}_{S}^{*}}$
The expression for λ is fairly cumbersome, and we will find it helpful to use the following easily derivable proposition:^10
$\begin{array}{ccc}\text{ }\frac{1}{2}\left[\sqrt{{\text{trace}}^{2}-4×\text{det}}-\text{trace}\right]<A& ⇔& -\text{det}<A×\left(A+\text{trace}\right)\end{array}$
In our case
$A×\left(A+\rho \right)=\frac{f\left({k}_{G}^{*}\right){p}^{*}h\left({k}_{S}^{*}\right)}{{\left({k}_{G}^{*}-{k}_{S}^{*}\right)}^{2}}$
Hence, the output of services will increase over time if and only if the absolute value of the determinant is less than the above expression. Now, the determinant can be expressed in the following
way through the steady state values of capital intensities and the relative price, which, in turn, depend on the production functions and the discount rate:
$-\text{det}=-\frac{f\left({k}_{G}^{*}\right){p}^{*}h\left({k}_{S}^{*}\right){f}^{″}\left({k}_{G}^{*}\right)\left[1+\rho \frac{\left(1-\alpha \right)\left({K}_{G}^{*}-{k}_{S}^{*}\right)}{{p}^{*}h\
left({k}_{S}^{*}\right)}\right]}{\alpha {p}^{*}h\left({k}_{S}^{*}\right)-\left(1-\alpha \right)\frac{f\left({k}_{G}^{*}\right)}{{p}^{*}h\left({k}_{S}^{*}\right)}\left[{\left({k}_{G}^{*}-{k}_{S}^{*}\
Comparing the two expressions, we will arrive at the following criterion – the output of services increases over time along the convergence path if and only if
$\alpha \left[{p}^{*}h\left({k}_{S}^{*}\right)+{f}^{″}\left({k}_{G}^{*}\right){\left({k}_{G}^{*}-{k}_{S}^{*}\right)}^{2}\right]+\left(1-\alpha \right)\frac{f{\left({k}_{G}^{*}\right)}^{2}}{{p}^{*}h\
This condition will not necessarily hold, since the negative term containing the second derivative may dominate the two positive terms, except in the limiting case where the two sectors have equal
capital intensity.
Analogously, we can show that $\frac{d{L}_{S}}{dC}>0$ if and only if
$\lambda <-\frac{{p}^{*}h\left({k}_{S}^{*}\right){f}^{″}\left({k}_{G}^{*}\right)}{\rho \frac{d{k}_{S}}{d{k}_{G}}-\left({k}_{G}^{*}-{k}_{S}^{*}\right){f}^{″}\left({k}_{G}^{*}\right)}.$
Tracing the same steps as above, we can show this condition to be equivalent to the following:
$\begin{array}{l}-\alpha {f}^{″}\left({k}_{G}^{*}\right)f\left({k}_{G}^{*}\right)\left[{p}^{*}h\left({k}_{S}^{*}\right)+{f}^{″}\left({k}_{G}^{*}\right){\left({k}_{G}^{*}-{k}_{S}^{*}\right)}^{2}\
right]+\\ +\left\{\alpha {\rho }^{2}{p}^{*}h\left({k}_{S}^{*}\right)-{f}^{″}\left({k}_{G}^{*}\right)f\left({k}_{G}^{*}\right)\left[\left(1+\alpha \right){p}^{*}h\left({k}_{S}^{*}\right)-2\alpha f\
left({k}_{S}^{*}\right)\right]\right\}\frac{d{k}_{S}}{d{k}_{G}}-\\ -\alpha {\rho }^{2}f\left({k}_{G}^{*}\right){\left(\frac{d{k}_{S}}{d{k}_{G}}\right)}^{2}>0\end{array}$
Obviously, this is a more complicated criterion that the one for the output of nontradables. In the general case this criterion may or may not be satisfied.
We can get more definitive results in some special cases. In particular, we show below that if the production technology in both sectors is Cobb-Douglas, with the capital share and hence capital
intensity higher in the goods sector, then both the output of services and employment in that sector increase over time.
A simple, though tedious, proof is obtained by writing out the production functions explicitly:
$\begin{array}{ccc}f\left({k}_{G}\right)=D{k}_{G}^{\gamma }& h\left({k}_{S}\right)=B{k}_{S}^{\beta }& \gamma >\beta .\end{array}$
From the first-order conditions for profit maximization, the following relationships are easy to establish:
${k}_{S}=\frac{1-\gamma }{\gamma }×\frac{\beta }{1-\beta }{k}_{G}$
$\mathit{ph}\left({k}_{S}\right)=\frac{w}{1-\beta }=\frac{1-\gamma }{1-\beta }f\left({k}_{G}\right).$
Now these expressions and the derivatives of the Cobb-Douglas functions can be plugged into the left-hand side expression of the employment criterion. The resulting expression can be shown to equal
$\begin{array}{l}f\left({k}_{G}^{*}\right){D}^{2}{\left({k}_{G}^{*}\right)}^{2\gamma -2}\frac{{\left(1-\gamma \right)}^{2}}{{\left(1-\beta \right)}^{2}}×\\ ×\left\{\alpha \left(\gamma +\gamma \beta -
{\gamma }^{2}-{\beta }^{2}\right)+\beta \left(1-\alpha -\gamma +2\alpha \beta \right)-\alpha {\beta }^{2}\right\}\end{array}$
The three terms in the braces correspond to the three terms in the criterion. Now the terms before the braces are positive and can be dropped, while the expression in the braces simplifies to
$\left(1-\gamma \right)\left\{\alpha \gamma +\beta \left(1-\alpha \right)\right\},$
which is obviously positive. Hence, the criterion is satisfied, which means that in the case of Cobb-Douglas technologies the employment share of services will increase over time. Of course, a
combination of increase in employment and capital deepening in the services sector means that the output of services will increase over time as well.
Another special case is one where no capital is used in the production of services. In that case the criteria for the increase in the service sector output and employment reduce to a fairly simple
This condition would be satisfied for a Cobb-Douglas production function,^11 but it would be violated, for example, for a CES function with a sufficiently low elasticity of substitution between
capital and labor.
• Baumol, William J., 1967, “Macroeconomics of Unbalanced Growth: The Anatomy of Urban Crisis,” American Economic Review, Vol. 57 (June), pp. 415–26.
• Baumol, William J., and William G. Bowen, 1966, Performing Arts: The Economic Dilemma (New York: Twentieth Century Fund).
• Baumol, William J., Sue Anne Batey Blackman, and Edward N. Wolf, 1989, Productivity and American Leadership: The Long View (Cambridge, Massachusetts: MIT Press).
• Bernard, Andrew B., and Charles I. Jones, 1996, “Comparing Apples to Oranges: Productivity Convergence and Measurement Across Industries and Countries,” American Economic Review, Vol. 86 (
December), pp. 1216–38.
• Brock, Philip L., 1994, “Economic Development and the Relative Price of Nontradables: Global Dynamics of the Krueger-Deardorff-Leamer Model,” Review of International Economics, Vol. 2 (October),
pp. 268–83.
• Canzoneri, Matthew B., Robert E. Cumby, and Behzad Diba, 1999, “Relative Labor Productivity and the Real Exchange Rate in the Long Run: Evidence for a Panel of OECD Countries,” Journal of
International Economics, Vol. 47 (April), pp. 245–66.
• Froot, Kenneth A., and Kenneth Rogoff, 1995, “Perspectives on PPP and Long-Run Real Exchange Rates,” in Handbook of International Economics, ed. by G.M. Grossman and K. Rogoff (Amsterdam: North
Holland), Vol. 3.
• Harberger, Arnold C., 2003, “Economic Growth and the Real Exchange Rate: Revisiting the Balassa-Samuelson Effect,” paper prepared for a Conference Organized by the Higher School of Economics (
Moscow), April 2003.
• Lipschitz, Leslie, Timothy Lane, and Alex Mourmouras, 2002, “Capital Flows to Transition Economies: Master or Servant?” IMF Working Paper 02/11 (Washington: International Monetary Fund).
• Obstfeld, Maurice, and Kenneth Rogoff, eds., 1996, Foundations of International Macroeconomics (Cambridge, Massachusetts: MIT Press).
• Rowthorn, Robert, and Ramana Ramaswamy, 1999, “Growth, Trade, and Deindustrialization,” Staff Papers, International Monetary Fund, Vol. 46 (March), pp. 18–41.
• Triplett, Jack E., and Barry P. Bosworth, 2003, “Productivity Measurement Issues in Services Industries: ‘Baumol’s Disease’ Has Been Cured,” FRBNY Economic Policy Review, Vol. 9 (September), pp.
• Young, Alwyn, 1995, “The Tyranny of Numbers: Confronting the Statistical Realities of the East Asian Growth Experience,” Quarterly Journal of Economics, Vol. 110 (August), pp. 641–80.
I would like to thank Michele Boldrin, Eric Clifton, Andrew Feltenstein, Alex Mourmouras, and participants at the 59^th European Meeting of the Econometric Society and at the IMF Institute Seminar
for useful comments and suggestions.
Of course, the mapping of manufactured goods into tradables and services into nontradables is less than perfect, given the presence of other sectors in the economy and the fact that some services may
be tradable while some manufactured goods may be rendered nontradable by policy measures. For that reason, empirical support for the Balassa-Samuelson effect is not as strong as for the Baumol-Bowen
effect (Froot and Rogoff, 1995; Harberger, 2003).
Canzoneri et al. (1999) present empirical evidence of this link. Other possible reasons – such as a shift of consumer demand from goods to services – appear to play a relatively minor role.
A different, but not unrelated issue is a finding by Bernard and Jones (1996) that convergence among OECD economies both in terms of labor productivity and in terms of TFP can be found in services
but not in manufacturing.
Since the “hat” variables are obtained by scaling their “no-hat” counterparts by the same factor, it is clear that k[G] > k[S] implies ${\stackrel{^}{k}}_{G}>{\stackrel{^}{k}}_{S}$.
We ignore capital depreciation for notational simplicity.
The maximand in ^(5) differs from that in ^(2) by a constant term that can be ignored.
To be precise, the slope is guaranteed to be positive to the left of the line $\stackrel{˙}{\stackrel{^}{c}}=0$ (where r > x + ρ > x + n), which is the relevant region.
Of course, the assumption of perfect factor mobility across sectors is important as well, but the models that rely on the TFP growth differential to explain changes in the relative price also make
that assumption.
This relies on the determinant being negative and the trace being positive, as is the case here.
For the Cobb-Douglas function, the left-hand-side expression equals simply the capital share. | {"url":"https://www.elibrary.imf.org/view/journals/001/2004/207/article-A001-en.xml","timestamp":"2024-11-11T17:20:03Z","content_type":"text/html","content_length":"693666","record_id":"<urn:uuid:dcdcec4a-58c7-4e6f-a099-46cbb5c65bac>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00264.warc.gz"} |
Double-dabble and conversions from base-10 and base-2 number systems
Many of you probably use this trick all the time, but I haven’t heard of this magic double-dabble short cut before today. I wanted to quickly write it down so that I don’t forget it.
Base-10 is obviously the number system most often used in everyday life and it is completely engrained into your head so much so that you typically no longer break a number like 582 into its
components: (5x100) + (8x10) + (2x1) Or, using the powers of 10 listed as a sum of weights: (5x10²) + (8x10¹) + (2x10⁰) In the base-10 number system, we use the ten digits of 0 to 9 and each position
is a power of ten, starting at 0.
Base-2 is the number system that you use all the time while programming, and it’s just another number system like base-10, but in the base-2 number system, we use the two digits of 0 and 1 and each
position is a power of two, starting at 0. The decimal number 582 in base-2 is represented as 1001000110 and you can break it down via a sum of weights as: (1x2⁹) + (0x2⁸) + (0x2⁷) + (1x2⁶) + (0x2⁵)
+ (0x2⁴) + (0x2³) + (1x2²) + (1x2¹) + (0x2⁰) Due to positional notation, the Least Significant Bit (LSB) is the furthest on the right and the Most Significant Bit (MSB) is the largest value and is
listed the furthest to the left. (This comes into play when sharing data between systems and there is a decision needed on which bit is listed first in the stream.) The LSB, also known as the
low-order bit, can also be used to quickly determine if the number is Even or Odd.
Note: while working with different number systems, you should annotate the base using a subscript like: (582)₁₀ or (1001000110)₂ to avoid ambiguity when dealing with a sequence of digits like 1101 as
it could be (1101)₁₀ or (1101)₂ or any other number base.
To quickly convert a base-10 number to base-2, you can use the Repeated Division-by-2 method by progressively dividing the number by 2 and then writing the remainder after each division, leaving the
binary representation when read in reverse order. Let’s convert (582)₁₀ using this method in the image below:
To manually convert from binary to decimal, I have been using the sum of weights method by calculating the powers of 2 for each digit’s position. Thankfully, I stumbled across a quicker conversion
method today. The Double-Dabble method works from the left to the right and Doubles the digit and then Adds the next digit, repeating until you reach the end. This simple method is demonstrated in
the image below:
This quick double-dabble method will come in handy…if I remember it!
So now there is an easy way to convert to and from whole numbers, but what about fractional binary numbers like 0.24? There is a system for that as well and it uses multiplication instead of division
as demonstrated in the image below.
To convert a mixed number (a whole integer plus fractional number like 582.24) simply perform the two steps separately and combine the results together such as: (1001000110.00111101)₂ (UPDATE:
corrected typo, previously listed incorrectly at 1001000110.0011101 based on post comment from Brian Thomson. He also reiterated that you need to continue this operation until desired accuracy
achieved. My example of 8 fractional digits of 1001000110.00111101 converts to 582.23828125. This should be extended to something similar to his suggestion of 12 digits to the right of the binary
separator: 1001000110.001111010111 which converts to a closer value of 582.239990234375. If you extend this to 23 digits to the right of the separator: 1001000110.00111101011100001010001 the number
gets even closer: 582.23999989032745361328. This is a very good example of why a “simple” fractional number like 0.24 may not be as precise as you might assume.)
One final piece of the puzzle is missing - is there a shortcut method like double-dabble to convert fractional base-2 numbers back to base-10? Let me know! | {"url":"https://ideasawakened.com/post/double-dabble-and-conversions-from-base-10-and-base-2-number-systems","timestamp":"2024-11-06T01:14:24Z","content_type":"text/html","content_length":"27518","record_id":"<urn:uuid:8b2fd6a8-78c0-4cf1-b70e-b7618ee9aecc>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00652.warc.gz"} |
How Unprofitable Companies IPO - Caseysoftware
How Unprofitable Companies IPO
Disclaimer: In this post, I describe business models of tech companies where I hold positions. Don’t use this to draw conclusions about any specific company or rationalize any particular investment.
Follow your own investment strategy, not some knucklehead with a blog.
In the last few years, we’ve seen a variety of major tech companies go public or IPO. It’s been educational and exciting to watch from both the inside and outside but one question has come up time
and time again:
“How can a company go public without making a profit?”
With the exception of Zoom, most of the tech companies that have IPO’d were not profitable at IPO time. In fact, some of them have had growing losses since their big day. To understand the mechanics,
let’s lay out the basic model that underlies most of these businesses. To start, there are some key terms to understand:
• It costs some amount of money to acquire a customer. This could be reflected in marketing, sales, promotions, and a variety of other things. We call this Customer Acquisition Cost (CAC).
• Next we have Churn which is how long a customer is a customer. We measure this as a percentage where 20% churn means they’re a customer for 5 billing cycles.
• Next, we have Average Revenue Per User (ARPU) which is generally the subscription cost.
• Finally, we have the total amount a customer will ever pay us which is the Lifetime Value (LTV). In SaaS businesses, this is calculated by ARPU * 1/Churn.
Caveat: The numbers here are made up and the math is over-simplified. Get over it and watch the trends.
Step 1: Basic Revenue Model
Let’s start with a simple scenario where is costs $10 to acquire a new customer, they pay us $5 each year, and subscribe for 5 years. That trend looks like this:
In this case, we have a net loss the first year, break even during the second, and are profitable from then on. Unfortunately, we lose that customer after year 5. Fundamentally, we spend $10 to make
$25 for a net profit of $15.
“But there’s profit! I thought this was about unprofitable companies going public!”
Step 2: More Customers
In any business, you want to make more money in year N+1 than you did in year N. One of the many ways to do that is to acquire new customers. In this model, let’s acquire 2x the customers in year 2,
and 2x that in year 3 and keep other assumptions the same:
Once again, we have a net loss the first year but it doesn’t stop there. In fact, our losses increase year over year! It isn’t until year 4 that we start making money but we make back all of our
losses in that year alone. We still lose the customer after year 5 for a $25 LTV and we spent $70 to make $175 for a net profit of $105.
That’s our baseline. Now let’s change our assumptions..
Step 3a: Reduce CAC
One of the great things about a business is that as time goes on, you learn more about your customers, what they do and don’t care about, and how to describe, position, and sell your product better.
As a result, we can often acquire customers for less tomorrow than we did yesterday. If we reduce our CAC from $10 to $8, what happens?
Once again, we have a net loss the first year but it peaks the second year and is gone by the fourth again. We still lose the customer after year 5 and we spent $56 to make $175 for a net profit of
$119 or 13% over our baseline.
Let’s tweak a different assumption..
Step 3b: Increase Average Revenue Per User
To be clear, increasing your prices is hard. That said, you can often upsell or cross sell other products and services to address complementary or adjacent use cases. Alternatively, if your pricing
is consumption-based, as your customers grow and their usage increases, you make more money.
If we can increase our annual revenue per customer from $5 to $6, what happens?
As before, we have a net loss the first year but it peaks the second year and is gone by the fourth again. We still lose the customer after year 5 but our LTV has increased to $30 so we spent $70 to
make $210 for a net profit of $140 or 33% over our baseline.
Let’s adjust another assumption..
Step 3c: Decrease Churn
Regardless of the other variables, acquiring customers is always hard. But if we can keep an existing customer longer, that’s great because it’s “free” according to our simple model. While you can
often do this via contract terms or minimum purchase agreements, simply making a sticky product can accomplish the same. That stickiness can come from a great product, deep integrations with other
systems, great customer service, or a variety of other things.
If we decrease our churn from 20% (5 years) to 16% (6 years), what happens?
As before, we have increasing losses but our revenue continues for another year. We lose the customer after year 6 but our LTV has increased to $30 so we spent $70 to make $210 for a net profit of
$140 or 33% over our baseline. This looks like the last model but took an extra year.
Fundamentally, all of these have an impact but which one should you do? Which is the most important? The answer is easy: Do all of them.
Step 4: Improve all Three
The good news is that these three variables – CAC, ARPU, and Churn – are all independent and often the responsibility of different groups in your company.
• As your Marketing and Sales teams know your customers better, they can improve targeting and close customers faster and cheaper which decreases CAC.
• As your Product and Development teams make the product better, faster, and address more use cases, you can sell more or charge more increasing ARPU.
• As your Customer Success and Support teams make customers happier faster and keep them happy, your Churn goes down.
So let’s combine all three variables together:
This time it’s a radically different picture. We still have minor losses in year 1 but break even in year 2 and have profit in year 3. We lose the customer after year 6 but our LTV has increased to
$36 so we spent $56 to make $252 for a net profit of $196 or 87% over our baseline.
But wait? What about those IPOs?
The important part to remember is that each of those variables move constantly and independently of each other. In established industries with incumbent players, each company may only be able to
tweak and optimize their process by percentage points. Alternatively, in new markets with new approaches, use cases, technology, etc, etc, a company may be able to make step-wise improvement driving
CAC and Churn down quickly while pushing ARPU and LTV up quickly.
That’s why companies can IPO without having a profit.
Fundamentally, investors are making the bet that those companies have massive growth, major improvements, and the profit coming eventually and long into the future.
Closing Thought
But how can we improve this model even more? The most impactful – but slowest – way is to reduce churn. In the above example, we started with 20% annual churn and reduced it to 16% meaning customers
leave us after 6 years. In most SaaS businesses, you want to have churn of under 5% but even if we only drive it to 10%, here are our new numbers:
Our losses start the same but we’ve driven LTV to $60 so we spent $56 to make $420 for a net profit of $364 or 247% over our baseline and making more money at every stage. Here’s how it compares with
all our previous approaches:
As noted, this is a massive oversimplification but the principles stand. When you have a company executing well in at least one area, you can make more money. When a company is executing well in
multiple areas for long periods of time, the sky is the limit. | {"url":"https://caseysoftware.com/blog/how-unprofitable-companies-ipo","timestamp":"2024-11-12T00:09:23Z","content_type":"text/html","content_length":"60316","record_id":"<urn:uuid:6d9235c3-d120-413f-9a3e-7bd445d2ab1e>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00434.warc.gz"} |
To find: The solution of the inequality and
To find: The solution of the inequality and interval notation. The given inequality equation is: displaystyle{2}{left({x}-{3}right)}-{5}le{3}{left({x}+{2}right)}-{18}
Answered question
To find: The solution of the inequality and interval notation.
The given inequality equation is:
$2\left(x-3\right)-5\le 3\left(x+2\right)-18$
Answer & Explanation
Consider the following steps to solve linear inequality in one variable:
If an inequality contains fractions or decimals, multiply both sides by the LCD to clear the equation of fractions or decimals.
Use the distributive property to remove parentheses if they are present.
Simplify each side of the inequality by comining like terms.
Get all variable terms on one side and all numbers on the other side by using the addition property of inequality.
Get the variable alone by using the multiplication property of inequality.
The given inequality equation is,
$2\left(x-3\right)-5\le 3\left(x+2\right)-18$
$2\cdot x-2\cdot 3-5\le 3\cdot x+3\cdot 2-18$
$2x-6-5\le 3x+6-18$
$2x-11-3x\le 3x-12-3x$
Simplify further,
$-x-11\le -12$
$-x-11+11\le -12+11$
$\frac{-x}{-1}\ge \frac{-1}{-1}$
$x\ge 1$
The interval notation of the inequality is written as $\left[1,\mathrm{\infty }\right).$
Therefore, the solution of the inequality is $x\ge 1\text{ }\text{and the interval notation is}\text{ }\left[1,\mathrm{\infty }\right).$ | {"url":"https://plainmath.org/integral-calculus/2411-solution-inequality-interval-notation-inequality-equation-displaystyle","timestamp":"2024-11-04T17:00:26Z","content_type":"text/html","content_length":"194440","record_id":"<urn:uuid:eab43f0a-38ed-4a09-a2d4-93be5e814561>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00211.warc.gz"} |
Motives: Part 2search
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• Proceedings of Symposia in Pure Mathematics
Volume: 55; 1994; 676 pp
MSC: Primary 14; Secondary 11; 19
Motives were introduced in the mid-1960s by Grothendieck to explain the analogies among the various cohomology theories for algebraic varieties, to play the role of the missing rational
cohomology, and to provide a blueprint for proving Weil's conjectures about the zeta function of a variety over a finite field. Over the last ten years or so, researchers in various areas—Hodge
theory, algebraic \(K\)-theory, polylogarithms, automorphic forms, \(L\)-functions, \(\ell\)-adic representations, trigonometric sums, and algebraic cycles—have discovered that an enlarged (and
in part conjectural) theory of “mixed” motives indicates and explains phenomena appearing in each area. Thus the theory holds the potential of enriching and unifying these areas. This is one of
two volumes containing the revised texts of nearly all the lectures presented at the AMS-IMS-SIAM Joint Summer Research Conference on Motives, held in Seattle, in 1991. A number of related works
are also included, making for a total of forty-seven papers, from general introductions to specialized surveys to research papers.
This item is also available as part of a set:
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Requests
Volume: 55; 1994; 676 pp
MSC: Primary 14; Secondary 11; 19
Motives were introduced in the mid-1960s by Grothendieck to explain the analogies among the various cohomology theories for algebraic varieties, to play the role of the missing rational cohomology,
and to provide a blueprint for proving Weil's conjectures about the zeta function of a variety over a finite field. Over the last ten years or so, researchers in various areas—Hodge theory, algebraic
\(K\)-theory, polylogarithms, automorphic forms, \(L\)-functions, \(\ell\)-adic representations, trigonometric sums, and algebraic cycles—have discovered that an enlarged (and in part conjectural)
theory of “mixed” motives indicates and explains phenomena appearing in each area. Thus the theory holds the potential of enriching and unifying these areas. This is one of two volumes containing the
revised texts of nearly all the lectures presented at the AMS-IMS-SIAM Joint Summer Research Conference on Motives, held in Seattle, in 1991. A number of related works are also included, making for a
total of forty-seven papers, from general introductions to specialized surveys to research papers.
This item is also available as part of a set:
Permission – for use of book, eBook, or Journal content
Please select which format for which you are requesting permissions. | {"url":"https://bookstore.ams.org/PSPUM/55.2","timestamp":"2024-11-13T22:54:03Z","content_type":"text/html","content_length":"104657","record_id":"<urn:uuid:7ad43390-b1bc-4e3b-a9a6-525d5adc9b0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00606.warc.gz"} |
Confidence Interval - Data Science Wiki
Confidence Interval :
A confidence interval is a range of values that is calculated from a sample
and is used to estimate the population
. It provides a measure of how certain we can be that the true population value lies within the calculated range.
For example, let’s say we want to estimate the average height of all the students in a particular school. We take a random sample of 100 students and measure their heights. The average height of the
sample is 175 cm and the
standard deviation
is 5 cm.
We can use this sample to calculate a 95% confidence interval for the average height of all the students in the school. This interval would be calculated as 175 cm +/- 1.96 x (5 cm / √100), which
gives us a range of 166.4 cm to 183.6 cm. This means that we can be 95% confident that the true average height of all the students in the school lies within this range.
Another example of a confidence interval is when we want to estimate the proportion of people in a population who have a particular trait. For instance, let’s say we want to estimate the proportion
of adults in a city who are obese. We take a random sample of 1000 adults and find that 200 of them are obese.
We can use this sample to calculate a 95% confidence interval for the proportion of adults in the city who are obese. This interval would be calculated as 200/1000 +/- 1.96 x √(200/1000 x (1 – 200/
1000)/1000), which gives us a range of 0.16 to 0.24. This means that we can be 95% confident that the true proportion of adults in the city who are obese lies within this range.
In both of these examples, the confidence interval provides a range of values that we can be certain the true population parameter lies within. This allows us to make more accurate estimates and
predictions about the population based on the sample data.
It is important to note that the confidence interval is not a fixed value, but rather a range of values that is calculated based on the sample data and a chosen level of confidence. In our first
example, if we wanted to be more certain about the true average height of the students in the school, we could choose a higher level of confidence, such as 99%, which would result in a wider range of
values. On the other hand, if we were less certain and only wanted to be 90% confident, the range of values would be narrower.
Additionally, the size of the sample also plays a role in the confidence interval. In our second example, if we had taken a larger sample of 2000 adults, the confidence interval would have been
narrower because we would have more data to work with.
It is also important to understand that the confidence interval is not a guarantee that the true population parameter will fall within the calculated range. There is always a chance that the true
value may fall outside of the interval, but the higher the level of confidence chosen, the lower the
of this happening.
Overall, the confidence interval is a useful tool for estimating population parameters and making predictions based on sample data. It provides a range of values that we can be confident the true
population parameter lies within, allowing us to make more accurate estimates and predictions. | {"url":"https://datasciencewiki.net/confidence-interval/","timestamp":"2024-11-13T15:35:42Z","content_type":"text/html","content_length":"42193","record_id":"<urn:uuid:864ecfa3-1989-4b72-bfdf-4abf24646b87>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00845.warc.gz"} |
Naive set theory - Wikipedia Republished // WIKI 2
Naive set theory is any of several theories of sets used in the discussion of the foundationsofmathematics.^[3] Unlike axiomaticsettheories, which are defined using formallogic, naive set theory
is defined informally, in naturallanguage. It describes the aspects of mathematicalsets familiar in discretemathematics (for example Venndiagrams and symbolic reasoning about their
Booleanalgebra), and suffices for the everyday use of set theory concepts in contemporary mathematics.^[4]
Sets are of great importance in mathematics; in modern formal treatments, most mathematical objects (numbers, relations, functions, etc.) are defined in terms of sets. Naive set theory suffices for
many purposes, while also serving as a stepping stone towards more formal treatments.
YouTube Encyclopedic
• 1/5
• An Introduction to Naive Set Theory, Cantor's Theorem, Russell's Paradox & the History of Set Theory
• Naive Set Theory by Paul Halmos #shorts
A naive theory in the sense of "naive set theory" is a non-formalized theory, that is, a theory that uses naturallanguage to describe sets and operations on sets. Such theory treats sets as platonic
absolute objects. The words and, or, if ... then, not, for some, for every are treated as in ordinary mathematics. As a matter of convenience, use of naive set theory and its formalism prevails even
in higher mathematics – including in more formal settings of set theory itself.
The first development of settheory was a naive set theory. It was created at the end of the 19th century by GeorgCantor as part of his study of infinitesets^[5] and developed by GottlobFrege in
his Grundgesetze der Arithmetik.
Naive set theory may refer to several very distinct notions. It may refer to
• Informal presentation of an axiomatic set theory, e.g. as in NaiveSetTheory by PaulHalmos.
• Early or later versions of GeorgCantor's theory and other informal systems.
• Decidedly inconsistent theories (whether axiomatic or not), such as a theory of GottlobFrege^[6] that yielded Russell'sparadox, and theories of GiuseppePeano^[7] and RichardDedekind.
The assumption that any property may be used to form a set, without restriction, leads to paradoxes. One common example is Russell'sparadox: there is no set consisting of "all sets that do not
contain themselves". Thus consistent systems of naive set theory must include some limitations on the principles which can be used to form sets.
Cantor's theory
Some believe that GeorgCantor's set theory was not actually implicated in the set-theoretic paradoxes (see Frápolli 1991). One difficulty in determining this with certainty is that Cantor did not
provide an axiomatization of his system. By 1899, Cantor was aware of some of the paradoxes following from unrestricted interpretation of his theory, for instance Cantor'sparadox^[8] and the
Burali-Fortiparadox,^[9] and did not believe that they discredited his theory.^[10] Cantor's paradox can actually be derived from the above (false) assumption—that any property P(x) may be used to
form a set—using for P(x) "x is a cardinalnumber". Frege explicitly axiomatized a theory in which a formalized version of naive set theory can be interpreted, and it is this formal theory which
BertrandRussell actually addressed when he presented his paradox, not necessarily a theory Cantor—who, as mentioned, was aware of several paradoxes—presumably had in mind.
Axiomatic theories
Axiomatic set theory was developed in response to these early attempts to understand sets, with the goal of determining precisely what operations were allowed and when.
A naive set theory is not necessarily inconsistent, if it correctly specifies the sets allowed to be considered. This can be done by the means of definitions, which are implicit axioms. It is
possible to state all the axioms explicitly, as in the case of Halmos' Naive Set Theory, which is actually an informal presentation of the usual axiomatic Zermelo–Fraenkelsettheory. It is "naive"
in that the language and notations are those of ordinary informal mathematics, and in that it does not deal with consistency or completeness of the axiom system.
Likewise, an axiomatic set theory is not necessarily consistent: not necessarily free of paradoxes. It follows from Gödel'sincompletenesstheorems that a sufficiently complicated firstorderlogic
system (which includes most common axiomatic set theories) cannot be proved consistent from within the theory itself – even if it actually is consistent. However, the common axiomatic systems are
generally believed to be consistent; by their axioms they do exclude some paradoxes, like Russell'sparadox. Based on Gödel'stheorem, it is just not known – and never can be – if there are no
paradoxes at all in these theories or in any first-order set theory.
The term naive set theory is still today also used in some literature^[11] to refer to the set theories studied by Frege and Cantor, rather than to the informal counterparts of modern axiomatic set
The choice between an axiomatic approach and other approaches is largely a matter of convenience. In everyday mathematics the best choice may be informal use of axiomatic set theory. References to
particular axioms typically then occur only when demanded by tradition, e.g. the axiomofchoice is often mentioned when used. Likewise, formal proofs occur only when warranted by exceptional
circumstances. This informal usage of axiomatic set theory can have (depending on notation) precisely the appearance of naive set theory as outlined below. It is considerably easier to read and write
(in the formulation of most statements, proofs, and lines of discussion) and is less error-prone than a strictly formal approach.
Sets, membership and equality
In naive set theory, a set is described as a well-defined collection of objects. These objects are called the elements or members of the set. Objects can be anything: numbers, people, other sets,
etc. For instance, 4 is a member of the set of all even integers. Clearly, the set of even numbers is infinitely large; there is no requirement that a set be finite.
Passage with the original set definition of Georg Cantor
The definition of sets goes back to GeorgCantor. He wrote in his 1915 article BeiträgezurBegründungdertransfinitenMengenlehre:
“Unter einer 'Menge' verstehen wir jede Zusammenfassung M von bestimmten wohlunterschiedenen Objekten unserer Anschauung oder unseres Denkens (welche die 'Elemente' von M genannt werden) zu einem
Ganzen.” – Georg Cantor
“A set is a gathering together into a whole of definite, distinct objects of our perception or of our thought—which are called elements of the set.” – Georg Cantor
First usage of the symbol ϵ in the work Arithmeticesprincipianovamethodoexposita by GiuseppePeano.
Note on consistency
It does not follow from this definition how sets can be formed, and what operations on sets again will produce a set. The term "well-defined" in "well-defined collection of objects" cannot, by
itself, guarantee the consistency and unambiguity of what exactly constitutes and what does not constitute a set. Attempting to achieve this would be the realm of axiomatic set theory or of axiomatic
class theory.
The problem, in this context, with informally formulated set theories, not derived from (and implying) any particular axiomatic theory, is that there may be several widely differing formalized
versions, that have both different sets and different rules for how new sets may be formed, that all conform to the original informal definition. For example, Cantor's verbatim definition allows for
considerable freedom in what constitutes a set. On the other hand, it is unlikely that Cantor was particularly interested in sets containing cats and dogs, but rather only in sets containing purely
mathematical objects. An example of such a class of sets could be the vonNeumannuniverse. But even when fixing the class of sets under consideration, it is not always clear which rules for set
formation are allowed without introducing paradoxes.
For the purpose of fixing the discussion below, the term "well-defined" should instead be interpreted as an intention, with either implicit or explicit rules (axioms or definitions), to rule out
inconsistencies. The purpose is to keep the often deep and difficult issues of consistency away from the, usually simpler, context at hand. An explicit ruling out of all conceivable inconsistencies
(paradoxes) cannot be achieved for an axiomatic set theory anyway, due to Gödel's second incompleteness theorem, so this does not at all hamper the utility of naive set theory as compared to
axiomatic set theory in the simple contexts considered below. It merely simplifies the discussion. Consistency is henceforth taken for granted unless explicitly mentioned.
If x is a member of a set A, then it is also said that x belongs to A, or that x is in A. This is denoted by x ∈ A. The symbol ∈ is a derivation from the lowercase Greek letter epsilon, "ε",
introduced by GiuseppePeano in 1889 and is the first letter of the word ἐστί (means "is"). The symbol ∉ is often used to write x ∉ A, meaning "x is not in A".
Two sets A and B are defined to be equal when they have precisely the same elements, that is, if every element of A is an element of B and every element of B is an element of A. (See
axiomofextensionality.) Thus a set is completely determined by its elements; the description is immaterial. For example, the set with elements 2, 3, and 5 is equal to the set of all primenumbers
less than 6. If the sets A and B are equal, this is denoted symbolically as A = B (as usual).
Empty set
The emptyset, denoted as ${\displaystyle \varnothing }$ and sometimes ${\displaystyle \{\}}$, is a set with no members at all. Because a set is determined completely by its elements, there can be
only one empty set. (See axiomofemptyset.)^[12] Although the empty set has no members, it can be a member of other sets. Thus ${\displaystyle \varnothing eq \{\varnothing \}}$, because the former
has no members and the latter has one member.^[13]
Specifying sets
The simplest way to describe a set is to list its elements between curly braces (known as defining a set extensionally). Thus {1, 2} denotes the set whose only elements are 1 and 2. (See
axiomofpairing.) Note the following points:
• The order of elements is immaterial; for example, {1, 2} = {2, 1}.
• Repetition (multiplicity) of elements is irrelevant; for example, {1, 2, 2} = {1, 1, 1, 2} = {1, 2}.
(These are consequences of the definition of equality in the previous section.)
This notation can be informally abused by saying something like {dogs} to indicate the set of all dogs, but this example would usually be read by mathematicians as "the set containing the single
element dogs".
An extreme (but correct) example of this notation is {}, which denotes the empty set.
The notation {x : P(x)}, or sometimes {x |P(x)}, is used to denote the set containing all objects for which the condition P holds (known as defining a set intensionally). For example, {x | x ∈ R}
denotes the set of realnumbers, {x | x has blonde hair} denotes the set of everything with blonde hair.
This notation is called set-buildernotation (or "set comprehension", particularly in the context of Functionalprogramming). Some variants of set builder notation are:
• {x ∈ A | P(x)} denotes the set of all x that are already members of A such that the condition P holds for x. For example, if Z is the set of integers, then {x ∈ Z | x is even} is the set of all
even integers. (See axiomofspecification.)
• {F(x) | x ∈ A} denotes the set of all objects obtained by putting members of the set A into the formula F. For example, {2x | x ∈ Z} is again the set of all even integers. (See
• {F(x) | P(x)} is the most general form of set builder notation. For example, {x's owner | x is a dog} is the set of all dog owners.
Given two sets A and B, A is a subset of B if every element of A is also an element of B. In particular, each set B is a subset of itself; a subset of B that is not equal to B is called a proper
If A is a subset of B, then one can also say that B is a superset of A, that A is contained in B, or that B contains A. In symbols, A ⊆ B means that A is a subset of B, and B ⊇ A means that B is a
superset of A. Some authors use the symbols ⊂ and ⊃ for subsets, and others use these symbols only for proper subsets. For clarity, one can explicitly use the symbols ⊊ and ⊋ to indicate
As an illustration, let R be the set of real numbers, let Z be the set of integers, let O be the set of odd integers, and let P be the set of current or former U.S.Presidents. Then O is a subset of
Z, Z is a subset of R, and (hence) O is a subset of R, where in all cases subset may even be read as proper subset. Not all sets are comparable in this way. For example, it is not the case either
that R is a subset of P nor that P is a subset of R.
It follows immediately from the definition of equality of sets above that, given two sets A and B, A = B if and only if A ⊆ B and B ⊆ A. In fact this is often given as the definition of equality.
Usually when trying to prove that two sets are equal, one aims to show these two inclusions. The emptyset is a subset of every set (the statement that all elements of the empty set are also members
of any set A is vacuouslytrue).
The set of all subsets of a given set A is called the powerset of A and is denoted by ${\displaystyle 2^{A}}$ or ${\displaystyle P(A)}$; the "P" is sometimes in a script font: ${\displaystyle \wp
(A)}$. If the set A has n elements, then ${\displaystyle P(A)}$ will have ${\displaystyle 2^{n}}$ elements.
Universal sets and absolute complements
In certain contexts, one may consider all sets under consideration as being subsets of some given universalset. For instance, when investigating properties of the realnumbers R (and subsets of R),
R may be taken as the universal set. A true universal set is not included in standard set theory (see Paradoxes below), but is included in some non-standard set theories.
Given a universal set U and a subset A of U, the complement of A (in U) is defined as
A^C := {x ∈ U | x ∉ A}.
In other words, A^C ("A-complement"; sometimes simply A', "A-prime" ) is the set of all members of U which are not members of A. Thus with R, Z and O defined as in the section on subsets, if Z is the
universal set, then O^C is the set of even integers, while if R is the universal set, then O^C is the set of all real numbers that are either even integers or not integers at all.
Unions, intersections, and relative complements
Given two sets A and B, their union is the set consisting of all objects which are elements of A or of B or of both (see axiomofunion). It is denoted by A ∪ B.
The intersection of A and B is the set of all objects which are both in A and in B. It is denoted by A ∩ B.
Finally, the relativecomplement of B relative to A, also known as the set theoretic difference of A and B, is the set of all objects that belong to A but not to B. It is written as A \ B or A − B.
Symbolically, these are respectively
A ∪ B := {x | (x ∈ A) ∨ (x ∈ B)};
A ∩ B := {x | (x ∈ A) ∧ (x ∈ B)} = {x ∈ A | x ∈ B} = {x ∈ B | x ∈ A};
A \ B := {x | (x ∈ A) ∧ ¬ (x ∈ B) } = {x ∈ A | ¬ (x ∈ B)}.
The set B doesn't have to be a subset of A for A \ B to make sense; this is the difference between the relative complement and the absolute complement (A^C = U \ A) from the previous section.
To illustrate these ideas, let A be the set of left-handed people, and let B be the set of people with blond hair. Then A ∩ B is the set of all left-handed blond-haired people, while A ∪ B is the set
of all people who are left-handed or blond-haired or both. A \ B, on the other hand, is the set of all people that are left-handed but not blond-haired, while B \ A is the set of all people who have
blond hair but aren't left-handed.
Now let E be the set of all human beings, and let F be the set of all living things over 1000 years old. What is E ∩ F in this case? No living human being is over1000yearsold, so E ∩ F must be the
emptyset {}.
For any set A, the power set ${\displaystyle P(A)}$ is a Booleanalgebra under the operations of union and intersection.
Ordered pairs and Cartesian products
Intuitively, an orderedpair is simply a collection of two objects such that one can be distinguished as the first element and the other as the second element, and having the fundamental property
that, two ordered pairs are equal if and only if their first elements are equal and their second elements are equal.
Formally, an ordered pair with first coordinate a, and second coordinate b, usually denoted by (a, b), can be defined as the set ${\displaystyle \{\{a\},\{a,b\}\}.}$
It follows that, two ordered pairs (a,b) and (c,d) are equal if and only if a = c and b = d.
Alternatively, an ordered pair can be formally thought of as a set {a,b} with a totalorder.
(The notation (a, b) is also used to denote an openinterval on the realnumberline, but the context should make it clear which meaning is intended. Otherwise, the notation ]a, b[ may be used to
denote the open interval whereas (a, b) is used for the ordered pair).
If A and B are sets, then the Cartesianproduct (or simply product) is defined to be:
A × B = {(a,b) | a ∈ A and b ∈ B}.
That is, A × B is the set of all ordered pairs whose first coordinate is an element of A and whose second coordinate is an element of B.
This definition may be extended to a set A × B × C of ordered triples, and more generally to sets of ordered n-tuples for any positive integer n. It is even possible to define infinite
Cartesianproducts, but this requires a more recondite definition of the product.
Cartesian products were first developed by RenéDescartes in the context of analyticgeometry. If R denotes the set of all realnumbers, then R^2 := R × R represents the Euclideanplane and R^3 := R
× R × R represents three-dimensional Euclideanspace.
Some important sets
There are some ubiquitous sets for which the notation is almost universal. Some of these are listed below. In the list, a, b, and c refer to naturalnumbers, and r and s are realnumbers.
1. Naturalnumbers are used for counting. A blackboardbold capital N (${\displaystyle \mathbb {N} }$) often represents this set.
2. Integers appear as solutions for x in equations like x + a = b. A blackboard bold capital Z (${\displaystyle \mathbb {Z} }$) often represents this set (from the German Zahlen, meaning numbers).
3. Rationalnumbers appear as solutions to equations like a + bx = c. A blackboard bold capital Q (${\displaystyle \mathbb {Q} }$) often represents this set (for quotient, because R is used for the
set of real numbers).
4. Algebraicnumbers appear as solutions to polynomial equations (with integer coefficients) and may involve radicals (including ${\displaystyle i={\sqrt {-1\,}}}$) and certain other
irrationalnumbers. A Q with an overline (${\displaystyle {\overline {\mathbb {Q} }}}$) often represents this set. The overline denotes the operation of algebraicclosure.
5. Realnumbers represent the "real line" and include all numbers that can be approximated by rationals. These numbers may be rational or algebraic but may also be transcendentalnumbers, which
cannot appear as solutions to polynomial equations with rational coefficients. A blackboard bold capital R (${\displaystyle \mathbb {R} }$) often represents this set.
6. Complexnumbers are sums of a real and an imaginary number: ${\displaystyle r+s\,i}$. Here either ${\displaystyle r}$ or ${\displaystyle s}$ (or both) can be zero; thus, the set of real numbers
and the set of strictly imaginary numbers are subsets of the set of complex numbers, which form an algebraicclosure for the set of real numbers, meaning that every polynomial with coefficients
in ${\displaystyle \mathbb {R} }$ has at least one root in this set. A blackboard bold capital C (${\displaystyle \mathbb {C} }$) often represents this set. Note that since a number ${\
displaystyle r+s\,i}$ can be identified with a point ${\displaystyle (r,s)}$ in the plane, ${\displaystyle \mathbb {C} }$ is basically "the same" as the Cartesianproduct ${\displaystyle \mathbb
{R} \times \mathbb {R} }$ ("the same" meaning that any point in one determines a unique point in the other and for the result of calculations, it doesn't matter which one is used for the
calculation, as long as multiplication rule is appropriate for ${\displaystyle \mathbb {C} }$).
Paradoxes in early set theory
The unrestricted formation principle of sets referred to as the axiomschemaofunrestrictedcomprehension,
is a property, then there exists a set
Y = {x : P(x)}
is the source of several early appearing paradoxes:
If the axiom schema of unrestricted comprehension is weakened to the axiomschemaofspecification or axiom schema of separation,
is a property, then for any set
there exists a set
Y = {x ∈ X : P(x)}
then all the above paradoxes disappear.^[14] There is a corollary. With the axiom schema of separation as an axiom of the theory, it follows, as a theorem of the theory:
The set of all sets does not exist.
Or, more spectacularly (Halmos' phrasing^[15]): There is no universe. Proof: Suppose that it exists and call it U. Now apply the axiom schema of separation with X = U and for P(x) use x ∉ x. This
leads to Russell's paradox again. Hence U cannot exist in this theory.^[14]
Related to the above constructions is formation of the set
• Y = {x | (x ∈ x) → {} ≠ {}},
where the statement following the implication certainly is false. It follows, from the definition of Y, using the usual inference rules (and some afterthought when reading the proof in the linked
article below) both that Y ∈ Y → {} ≠ {} and Y ∈ Y holds, hence {} ≠ {}. This is Curry'sparadox.
It is (perhaps surprisingly) not the possibility of x ∈ x that is problematic. It is again the axiom schema of unrestricted comprehension allowing (x ∈ x) → {} ≠ {} for P(x). With the axiom schema of
specification instead of unrestricted comprehension, the conclusion Y ∈ Y does not hold and hence {} ≠ {} is not a logical consequence.
Nonetheless, the possibility of x ∈ x is often removed explicitly^[16] or, e.g. in ZFC, implicitly,^[17] by demanding the axiomofregularity to hold.^[17] One consequence of it is
There is no set X for which X ∈ X,
or, in other words, no set is an element of itself.^[18]
The axiom schema of separation is simply too weak (while unrestricted comprehension is a very strong axiom—too strong for set theory) to develop set theory with its usual operations and constructions
outlined above.^[14] The axiom of regularity is of a restrictive nature as well. Therefore, one is led to the formulation of other axioms to guarantee the existence of enough sets to form a set
theory. Some of these have been described informally above and many others are possible. Not all conceivable axioms can be combined freely into consistent theories. For example, the axiomofchoice
of ZFC is incompatible with the conceivable "every set of reals is Lebesguemeasurable". The former implies the latter is false.
See also
1. ^ "EarliestKnownUsesofSomeoftheWordsofMathematics(S)". April 14, 2020.
2. ^ Halmos1960, Naive Set Theory.
3. ^ Jeff Miller writes that naive set theory (as opposed to axiomatic set theory) was used occasionally in the 1940s and became an established term in the 1950s. It appears in Hermann Weyl's review
of P. A. Schilpp, ed. (1946). "The Philosophy of Bertrand Russell". American Mathematical Monthly. 53 (4): 210, and in a review by Laszlo Kalmar (Laszlo Kalmar (1946). "The Paradox of Kleene and
Rosser". Journal of Symbolic Logic. 11 (4): 136.).^[1] The term was later popularized in a book by PaulHalmos.^[2]
4. ^ Mac Lane, Saunders (1971), "Categorical algebra and set-theoretic foundations", Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. XIII, Part I, Univ. California, Los Angeles, Calif., 1967),
Providence, RI: Amer. Math. Soc., pp. 231–240, MR 0282791. "The working mathematicians usually thought in terms of a naive set theory (probably one more or less equivalent to ZF) ... a practical
requirement [of any new foundational system] could be that this system could be used "naively" by mathematicians not sophisticated in foundational research" (p. 236).
5. ^ Frege1893 In Volume 2, Jena 1903. pp. 253-261 Frege discusses the antionomy in the afterword.
6. ^ Peano1889 Axiom 52. chap. IV produces antinomies.
7. ^ ^a ^b Letter from Cantor to DavidHilbert on September 26, 1897, Meschkowski&Nilson1991 p. 388.
8. ^ Letter from Cantor to RichardDedekind on August 3, 1899, Meschkowski&Nilson1991 p. 408.
9. ^ ^a ^b Letters from Cantor to RichardDedekind on August 3, 1899 and on August 30, 1899, Zermelo1932 p. 448 (System aller denkbaren Klassen) and Meschkowski&Nilson1991 p. 407. (There is no
set of all sets.)
10. ^ F. R. Drake, Set Theory: An Introduction to Large Cardinals (1974). ISBN 0 444 10535 2.
11. ^ Halmos1974, p. 9.
12. ^ Halmos1974, p. 10.
13. ^ ^a ^b ^c ^d ^e Jech2002, p. 4.
14. ^ Halmos1974, Chapter 2.
15. ^ Halmos1974, See discussion around Russell's paradox.
16. ^ ^a ^b Jech2002, Section 1.6.
17. ^ Jech2002, p. 61.
• Bourbaki,N., Elements of the History of Mathematics, JohnMeldrum (trans.), Springer-Verlag, Berlin, Germany, 1994.
• Cantor,Georg (1874), "UebereineEigenschaftdesInbegriffesallerreellenalgebraischenZahlen", J.ReineAngew.Math., 1874 (77): 258–262, doi:10.1515/crll.1874.77.258, S2CID 124035379; see
also pdfversion
• Devlin,K.J., The Joy of Sets: Fundamentals of Contemporary Set Theory, 2nd edition, Springer-Verlag, New York, NY, 1993.
• María J. Frápolli|Frápolli, María J., 1991, "Is Cantorian set theory an iterative conception of set?". Modern Logic, v. 1 n. 4, 1991, 302–318.
• Frege,Gottlob (1893), Grundgesetze der Arithmetik, vol. 1, Jena{{citation}}: CS1 maint: location missing publisher (link)
• Halmos,Paul (1960). NaiveSetTheory. Princeton, NJ: D. Van Nostrand Company.
□ Halmos, Paul (1974). Naive Set Theory (Reprint ed.). New York: Springer-Verlag. ISBN 0-387-90092-6.
□ Halmos, Paul (2011). Naive Set Theory (Paperback ed.). Mansfield Centre, CN: D. Van Nostrand Company. ISBN 978-1-61427-131-4.
• Jech,Thomas (2002). Set theory, third millennium edition (revised and expanded). Springer. ISBN 3-540-44085-2.
• Kelley,J.L., General Topology, Van Nostrand Reinhold, New York, NY, 1955.
• vanHeijenoort,J., From Frege to Gödel, A Source Book in Mathematical Logic, 1879-1931, Harvard University Press, Cambridge, MA, 1967. Reprinted with corrections, 1977. ISBN 0-674-32449-8.
• Meschkowski,Herbert [in German]; Nilson, Winfried (1991), Georg Cantor: Briefe. Edited by the authors., Berlin: Springer, ISBN 3-540-50621-7
• Peano,Giuseppe (1889), Arithmetices Principies nova Methoda exposita, Turin{{citation}}: CS1 maint: location missing publisher (link)
• Zermelo,Ernst (1932), Georg Cantor: Gesammelte Abhandlungen mathematischen und philosophischen Inhalts. Mit erläuternden Anmerkungen sowie mit Ergänzungen aus dem Briefwechsel Cantor-Dedekind.
Edited by the author., Berlin: Springer
External links
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Freshwater Strategy: 56-44 to Labor in Victoria
Another Victorian state election poll fails to corroborate Newspoll’s finding of a narrowing gap. Also: the Poll Bludger election guide expands to cover the Legislative Council.
The Financial Review has a poll from Freshwater Strategy, which made its debut for the paper three weeks ago with a New South Wales poll, that credits Labor with a lead of 56-44, from primary votes
of Labor 37%, Coalition 34%, Greens 14% and others 15%. Daniel Andrews is on 39% approval and 48% disapproval, Matthew Guy is at 32% aod 48%, and Andrews leads 40% to 28% as preferred premier. We are
also told that Jacinta Allan’s rating is neutral, Tim Pallas is at minus 12, the Labor brand is at plus 10 and the Liberals are on minus six. “Close to 60 per cent of Victorians” including 39% of
Labor voters, believe they were locked down too long.The highest ranked issue by far was cost of living, followed by “health and social care” and “managing the Victorian economy”. The poll was
conducted Thursday to Sunday from a sample of 1000.
• The Poll Bludger state election guide now comprehensively covers the Legislative Council, including an overview and the usual thorough guides to each of the eight regions. The upper house contest
happens to be in the news today following Adem Somyurek’s announcement that he will seek re-election in South-Eastern Metropolitan as the candidate of the Democratic Labour Party. Somyurek’s, whose
DLP colleagues include Bernie Finn in Western Metropolitan, tells the Herald-Sun he will represent the “sensible centre of Victorian politics”.
• “Prominent Melbourne art collector” Andrew King says he will pay the $350 nomination fees of the first 50 people who come forward to run against Daniel Andrews in Mulgrave. King’s theory is that
this will divert voters from Andrews “by reducing his first preference vote, diverting votes away from him, and increasing the likelihood of informal votes”. On what remains of Twitter, Antony Green
relates that the total number of candidates could exceed 600, compared with an already over-stuffed 507 in 2018, boosted by Family First’s determination to run candidates in all 88 seats.
• In a Twitter thread, Kos Samaras of Redbridge Group argued that the anti-lockdown parties, including Angry Victorians and the Freedom Party together with the United Australia Party, complicated
Liberal ambitions in seats like Melton as they like were competing for the same demographic turf of asset-owning white voters with trade qualifications and incomes of over $100,000 a year. Labor’s
voters in such areas tended to be newer arrivals with lower incomes and mortgages, many of them migrants.
298 comments on “Freshwater Strategy: 56-44 to Labor in Victoria”
1. Can we please keep this thread for discussion of the Victorian state election. The open discussion thread is here.
2. Small typo in the description of Daniel Mulgrave of mulgrave
3. Seats with the current minimum of 5 candidates (ALP, Coalition, GRN, AJP, FFV): Wendouree, South Barwon, Murray Plains, Malvern, Ivanhoe, Glen Waverley, Eltham, Dandenong, Croydon. Euroa has the
same lineup but with 2 Coalition candidates so they have 6. Bundoora and Box Hill (currently missing AJP) and Bulleen (currently missing FFV) have others on the ballot so they are above 5
ABC currently has for lower house: 88 ALP, 88 GRN, 87 FFV, 86 AJP, 83 LIB, 74 IND, 32 FPV, 22 VS, 11 NAT, 10 DLP, 8 RP, 7 LDP, 5 PHON, 5 SFF, 4 ND, 3 HAP, 1 CPP. I know of at least 8 independents
that ABC doesn’t have, which would make 82. However these independents are not always spread out, and there are several districts with more than one.
In other news, ex-Greens MLC Nina Springle is running for the Reason Party in NE Metro.
4. I did a ‘like with like’ comparison of the latest Newspoll for Victoria with the Newspoll taken at about the same time in 2018.
There was little significant difference between them.
Basically, I read it as people realising that the election’s approaching and giving a knee jerk reaction to how they’ll vote without actually having put thought into it.
As with the last election (if Newspoll is at all a guide), once they have a look at what’s on offer, their votes will shift (as in, people whose first impulse is to vote Liberal, won’t).
5. Others 15%
Greens 14%
The voters of Victoria really don’t like Dan Andrews or that lobster with a mobster bloke. Methinks the minor parties and independent vote will be even higher on election night. What a
fascinating election. Labor minority firming.
6. wonder how guy feels about pessutos campaign t o chalinge him ifh he wins horthorn usualy if liberals united a candadate dont pitch them selves as a future leaderin the middle of the campaign
pessuto agrees with guy on policy including his push to give religous schools mor rights to discriminate if liberals knew how bad guy was whiy did smith and credlin push for him to come back as
leader given his 2018 performents
7. https://www.theage.com.au/politics/victoria/read-the-statement-andrews-spars-with-reporters-over-ibac-probe-20221107-p5bwa4.html?btis
Looks like Andrew’s stuffed. Can’t get his message out. Journalists really not like him.
8. This poll continues the story that this election is likely to be easily won by the ALP and it’s unusual for polls to get it so consistently wrong. Developments in the election campaign can change
things sometimes, but so far it’s been more of the same- literally, with the Murdoch press recycling their narrative that Andrews must be guilty of… something nefarious….because he fell down some
steps and/0r his wife was involved in a car accident 9 years ago. The latest IBAC narrative also seems like more of the same.
There are outlier cases. In 1999 at the beginning of the campaign all the published polls but Morgan pointed to an easy Kennett win, the Nielsen poll had a 53.5 coalition 2PP and Newspoll had it
at 56-44. Morgan pointed to a closer result. But on the other hand as we know, in 2018 most polls significantly underestimated the ALP 2PP. Absent a huge polling error, at the moment, there’s
minimal prospect of Andrews going anywhere, with only a small prospect of him having to govern in minority in the lower house.
9. I have read that Somyurek is standing in the Northern Metro Region for the DLP.
A bit strange as his base has always been the Turkish population in the South East. But there are more people of Turkish ethnicity in the North Metro region. Perhaps that is the reason for the
10. “Freshwater Strategy: 56-44 to Labor in Victoria”…
So far so good… Of course, we know that the opinion polls have been wrong in the past… but they have also been correct many, many times…. 🙂
11. “Jeremy says:
Tuesday, November 8, 2022 at 7:01 am
…Looks like Andrew’s stuffed. Can’t get his message out. Journalists really not like him.”
Ha, ha, ha… the msm journalists really don’t like Dan and the ALP?… Jeremy, “thank you very much” for your “genial insight”, we don’t know what we would do without you…. Ha, ha, ha!
Whether Dan and the Vic ALP are “stuffed”…. stay in the trench, and watch the results on election night…. 🙂
12. “Jeremy says:
Tuesday, November 8, 2022 at 6:45 am
….Labor minority firming.”
Ha, ha, ha!…. I lost count how many times Liberal party stormtroopers have predicted that… Last time was at the recent federal election… 🙂
13. Hi nath, a fan of your posts during the May federal election. Hope you continue to post these next three weeks.
l know Kos thinks the Indian vote will break with the ALP, however l had heard elsewhere that this demographic much more likely to swing or even vote Liberal as they are “aspirational”. Anyone
have knowledge on the Indian vote from previous elections? Melton, Werribee, Tarneit, Point Cook could all be interesting battles on election night. Fascinating election.
14. Last election it was the attack on blacks, this election it is an attack on Andrews. That really seems to be the only change. Both elections nothing really constructive out of the Liberal party.
15. It must be very frustrating for Newscorp and Ninefax proprietors, management and stenographers that so many of the good voters of Victoria are apparently getting ready to once again ignore
instructions regarding the coming State election…
16. https://www.skynews.com.au/opinion/peta-credlin/daniel-andrews-has-had-horror-start-to-election-campaign/video/9977c64c2fc185bf706f372b3ec8a85d
Make fun of skynews, l agree with you but many voters don’t and believe what Credlin says like many on this site believe whatever ALP says.
17. Given Mediawatch last night, how many (and who?) are going to be influenced by the media and their presentations?
And how many are going to be influenced to vote Labor in the face of this concerted media attack on democracy, seeking to openly influence who citizens vote for?
The absolute rank attack on Labor by media proprietors was laid bare by 7.30
Polling suggests that citizens are turned off by this attempt to manipulate the election result by media proprietors
What I do know from my lifetime is that the attacks, the innuendos replicate the Murdoch attacks on Dunstan during his years as Premier of SA (and Adelaide had 2 papers at that time, The
Advertiser more discrete but attacking never the less)
Dunstan retired, handing leadership to Corcoran
And Labor has been the dominant Party since – except for one term Liberal governments every 15 years or so – and when in government they disintegrate to minority governments
Bannon, Rann and Wetherald as long serving Labor Premiers were also the subject of constant attacks by Murdoch, Murdoch resorting to the attacks such as we see on Andrews
Then you get to the attacks on Federal Labor leaders
18. Cheers Jeremy. I think the Indian vote has been very solidly Labor so far. Both parties have been stacking them to varying degrees of success.
If not the West, then the Outer South East will be where any shift in the Indian community towards the Liberals would have some significance.
19. Jeremy says:
Tuesday, November 8, 2022 at 7:34 am
Hi nath, a fan of your posts….
Anyone have knowledge on the Indian vote from previous elections? Melton, Werribee, Tarneit, Point Cook could all be interesting..
Indians backed a loser at Battle of the Little Bighorn.
21. The thing with Sky News is that it’s so bold and extreme with its bias that it doesn’t actually try to disguise it or pretend to be reporting more impartially. Therefore, none of their attacks
can be very effective because they are always seen through the lens of their known bias.
Those who don’t share Sky’s views either avoid it, or watch it for a laugh just to see how loopy they get, and even their diehard viewership would understand the concept that “Of course Sky would
have that view” but they specifically watch because they know Sky can be trusted to echo and confirm their own view.
But really, the net result is that they’re not actually influencing anybody, or changing anyone’s vote because everybody knows what their agenda is. Put it this way – anybody listening to Andrew
Bolt was never going to vote Labor in the first place and probably never has.
22. Opinion.
Attacks on Dan Andrews are part of News Corporation’s long abuse of power.
Sane people trying to fathom the Herald Sun’s bizarre coverage of Victorian Premier Dan Andrews over the past few days might be helped by some insights from the founding of News Limited, the
company on which Rupert Murdoch’s News Corporation empire has been built and of which the Herald Sun is part.
23. I agree with many here who are detecting a relative loss of media influence in Victoria (and I would say that the trend is becoming national).
I mean, the concept couldn’t be simpler: Just vote in support of your own interests… Just identify your interests (it should be easy) and vote for the party that best fulfills them!… Ignore the
media that’s desperately trying to shift your vote to the party that you have already identified as being bad for your personal interests (whether you are interested in a job, a good salary, a
better environment, better social services, better education, better and more affordable healthcare, or whatever else).
24. Grime, that’s a fantastic article. Thanks for posting!
25. Pollbludger headline says: Another Victorian state election poll fails to corroborate Newspoll’s finding of a narrowing gap.
I demand another poll! This one is 24 hours old. Surely one day soon Victorians will wake up and follow the Newscorpse / Costello media’s collective instructions.
Seems those pesky polls just aren’t budging….
I am now confidently predicting a massive landslide win for the Lobster Guy. On the basis of the current polling, and the collective wisdom of some of our more regular posters.
All frivolity aside, I still can’t understand how 45% of the population would like to see the liberal national Pentecostal party on the treasury benches.
On the hustings on Saturday and Sunday, for the Greens, it was quite remarkable just how old the LNP diehards are. The rudest people (admittedly there were only 3 or 4 of them ) have maybe one or
two elections left in them.
The other telling factor was just how disengaged the general public are. They don’t care or know that an election is on. How those voters break will determine this election.
26. “citizen says:
Tuesday, November 8, 2022 at 8:58 am
Time travel in Murdochland”…
Thanks to Murdoch, the Libs will be bashed even more seriously at the election. I don’t think that Victorians will take lightly, media statements that actually suggest that they are a “bunch of
stupid morons, gullible and brainless, that can be manipulated at will by Liberal party smartarses and their media mates….”.
Being an “idiot” is becoming utterly un-Victorian!
27. Fancy revisiting a matter that occurred in January 2013. Nearly 10 years ago. Catherine Andrews was the driver, and the cyclist rode into their car.
No fault payment was made to the cyclist.
The suggestion that one can sue 10 years later is funny. Did the cyclist get his advice from a lawyer or from the herald sun.
Its obvious that the herald sun is courting the cooker vote.
This same paper is what caused so much angst for victorians during the height of the pandemic.
They were beyond pathetic.
28. Mabwm
For eg. Construction workers are taking notice of this election.
They know that their jobs are on the ballot.
29. I got a flyer in my letterbox (Bayswater) yesterday from the Freedom Party with “Vote Labor Last” in huge letters on both sides. I’d never heard of the Freedom Party before.
If it’s a sentiment that takes off on the right, the Greens might benefit in a couple of close seats.
30. Mind you the incident with the cyclist occurred two years before Labor were in power. The vic govt at the time were the libs.
31. EightES
The freedom party are the cookers. Lol
33. I have met a lot of crazies over the past few months, in parks, cafes and acquaintances that have seemingly gone feral. We can only speculate on its dimensions. I suspect that the nutters in
Victoria have doubled since COVID began. So, if the ON and UAP vote in Vic was around 5% previously, as a measure of nuttiness. Then we could be looking at 10% nuttiness this time around. Time
will tell.
34. The Saint Dan crowd are almost as annoying as the Hate Dan crowd.
Put them both on King Island and let them sort it out.
35. Of the target Greens seats, it looks like the ‘Freedom Party’ are fielding candidates in Northcote, Preston & Melbourne, but none have been announced (as yet) in Prahran, Richmond, Albert Park,
Footscray or Pascoe Vale.
36. @Victoria – “Mind you the incident with the cyclist occurred two years before Labor were in power. The vic govt at the time were the libs.”
Great point. A little strange for the Herald Sun to somehow be insinuating a Victoria Police / government cover-up of the crash investigation, and feeding that into the “Dan is corrupt”
narrative, when the investigation occurred under a Liberal government.
37. Trent
Precisely. Facts have never mattered to the Herald Sun. Its all about the desired narrative.
38. Not sure why anyone cares what the Herald Sun is saying. They’ve been absolutely relentless with their anti-ALP coverage for as long as I can remember. If they had any actual influence:
* The Libs wouldn’t have lost at the 2014 state election. Or been wiped out in 2018.
* Frydenberg would be PM currently.
* There would be zero teal members in federal parliament representing Vic seats.
* The Greens would only poll a few %
They’re yesterdays news. Same goes for Neil Mitchell and the rest of the imbeciles over at 3AW. Their foaming listeners have never voted anything but for the coalition.
39. Jeremy says:
Tuesday, November 8, 2022 at 8:03 am
Make fun of skynews, l agree with you but many voters don’t and believe what Credlin says like many on this site believe whatever ALP says.
Australia-wide pay TV viewer numbers for last night.
CREDLIN 48,000
THE BOLT REPORT 44,000
PAUL MURRAY LIVE 42,000
“Many voters”, yeah no.
40. The media are becoming increasingly brazen and it looks like it’s only going to ramp up over the next few weeks.
Which honestly seems like such an idiotic play by news corp and nine-fairfax. The harder they push the more overt their bias becomes to a greater number of people and it only strengthens the case
for a royal commission into the media.
I wouldn’t be surprised if these attacks galvanise support for the ALP.
41. Nathan, I’m glad you posting. Don’t you love a lot of the posters on this site complaining about the “anti Dan” sentiment of the Age, then next post state with a straight face that absolutely
nobody takes media seriously. That’s the ALP luvvies for you. Shorten still the devil?
42. Nath, also be wary of the trolls on this site. Somethinglikethat, MAWBM, Alpo, here we go again, Grime are just a few.
43. Jeremy
Im not for or anti dan. Im for a vic labor govt.
The alternative is an absolute shocker.
Its as simple as that.
44. C’mon Folks.
I have avidly read the Sydney Morning Herald, every day, since I was 15 – now 66.
I have always been a lefty and accepted, whilst it was a Fairfax publication a degree of centre-right bias.
Similarly, whenever visiting Melbourne, I chose to read The Age, with a similar degree of pro-LNP bias.
However, the collapse in readership, has negated the influence of the print media in regard to election cover.
The informed, will just laugh off the pathetic attempts by The Age & Herald-Sun, to tarnish Daniel Andrews with ancient, non-news.
As for broadcast media – most people congregate to the outlets which reflects their personal bias or point of view.
Personally, I believe that the influence of media, of any form, is greatly exaggerated and mostly irrelevant in Australia.
45. ‘Jeremy says:
Tuesday, November 8, 2022 at 6:45 am
Others 15%
Greens 14%
The voters of Victoria really don’t like Dan Andrews or that lobster with a mobster bloke. Methinks the minor parties and independent vote will be even higher on election night. What a
fascinating election. Labor minority firming.’
A Voice Buster eliding the Greens’ wrecker status. Voters not fooled: 85% voting against the Voice Busters.
46. Macca RB says:
Tuesday, November 8, 2022 at 11:07 am
C’mon Folks.
I have avidly read the Sydney Morning Herald, every day, since I was 15 – now 66.
I have always been a lefty and accepted, whilst it was a Fairfax publication a degree of centre-right bias.
Similarly, whenever visiting Melbourne, I chose to read The Age, with a similar degree of pro-LNP bias.
The SMH has endorsed Labor more than the Liberals in recent times. I believe The Age is even more pro Labor on endorsements.
47. Macca RB
The print media doesnt have much influence, but online media including social media does.
Unfortunately many people have gone down the rabbit hole, getting their information from very influential disinformation sites.
Just look at qanon. Millions of people have been sucked into this vortex. This includes many australians that actually believe this cray cray stuff.
I actually have relatives who believed that the lockdowns in melbourne were so that dan andrews could traffick children through underground tunnels.
I still cant get my head around the fact that middleaged people bought into this crapola.
And mind you these people are very financial and continue to do well financially.
We are not talking about homeless, despondent people who have been forgotten by society.
I will go to my grave not understanding how and why…..
48. Victoria says:
I actually have relatives who believed that the lockdowns in melbourne were so that dan andrews could traffick children through underground tunnels.
I haven’t heard that one. sounds like an extension of the Hilary Clinton absurdity.
49. nath
It is part of the qanon shit show.
The cookers as they are known here, bought into this crapola as soon as the pandemic hit.
Im embarrassed for them but at the same time i have not much sympathy.
If they truly believe that children were being trafficked through tunnels, why arent they out in the real world trying to save them.
Instead they are on their keyboards sharing their wankery with others.
Narcissists come to mind.
50. zoomster 0639 am
I think you’re right. Working in health and dealing with hundreds of co-workers in Melbourne and regional Victoria over the last few months there is very little engagement with the election.
Partly I think because the Federal Election in May took much of the heat out of lots of issues, but also recent floods etc just hold people’s attention more.
I think probably only next week when prepoll voting starts will the interest ramp up. I expect a mild TPP swing against Labor, but I think the Liberals like their Federal counterparts will see
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