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A boat 4 - math word problem (83891)
A boat 4
A boat travels 25 km upstream in 5 hours and 25 km downstream in 2.5 hours. If the boat increased its speed by 3 km/h, it would take 1 hour less to travel the downstream distance. Find the speed of
the stream.
Correct answer:
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Units of physical quantities:
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Wisemen.ai - AI-Powered Self-Learning Tutor & Curriculum Generator
Chapter 1: Introduction to Rational Expressions
[First Half: Introduction to Rational Expressions]
1.1: Defining Rational Expressions
In this sub-chapter, we will introduce the concept of rational expressions and explore their basic structure and properties.
A rational expression is a mathematical expression that can be written as a fraction, where the numerator and denominator are both polynomials. The numerator is a polynomial, and the denominator is a
non-zero polynomial.
The general form of a rational expression is:
where $P(x)$ and $Q(x)$ are polynomials, and $Q(x) eq 0$.
Some examples of rational expressions include:
1. $\frac{3x^2 + 2x - 1}{x - 2}$
2. $\frac{5}{x^2 - 4}$
3. $\frac{x^3 - 2x^2 + 3x - 1}{x^2 + 3x + 2}$
In these examples, the numerators and denominators are both polynomials, and the denominators are non-zero.
The key properties of rational expressions are:
1. Domain: The domain of a rational expression is the set of all real numbers $x$ for which the denominator is non-zero. This is because division by zero is not defined.
2. Simplification: Rational expressions can often be simplified by identifying and canceling common factors between the numerator and denominator.
3. Operations: The basic operations (addition, subtraction, multiplication, and division) can be performed on rational expressions, following specific rules and procedures.
Understanding the definition and properties of rational expressions is essential for manipulating and working with these expressions in various mathematical contexts.
Key Takeaways:
• Rational expressions are fractions where the numerator and denominator are both polynomials, and the denominator is non-zero.
• The domain of a rational expression is the set of real numbers where the denominator is non-zero.
• Rational expressions can be simplified by identifying and canceling common factors between the numerator and denominator.
• Rational expressions can be combined using the basic arithmetic operations, following specific rules and procedures.
1.2: Simplifying Rational Expressions
In this sub-chapter, we will learn how to simplify rational expressions by identifying and canceling common factors between the numerator and denominator.
Simplifying a rational expression involves reducing the expression to its simplest form, which is the form with the least number of factors in both the numerator and denominator.
The steps to simplify a rational expression are:
1. Identify common factors: Examine the numerator and denominator to find any common factors, such as common variables, constants, or polynomial terms.
2. Cancel common factors: Divide both the numerator and denominator by their common factors to eliminate them and obtain the simplified expression.
Let's consider an example:
$\frac{6x^2 - 12x}{3x - 6}$
Step 1: Identify common factors. The common factor between the numerator and denominator is 3.
Step 2: Cancel the common factor. $\frac{6x^2 - 12x}{3x - 6} = \frac{2x^2 - 4x}{x - 2}$
In this simplified form, the numerator and denominator no longer have any common factors, and the expression is in its simplest form.
Another example:
$\frac{x^3 - 2x^2 + x}{x^2 - 1}$
Step 1: Identify common factors. The common factor between the numerator and denominator is $(x - 1)$.
Step 2: Cancel the common factor. $\frac{x^3 - 2x^2 + x}{x^2 - 1} = \frac{(x - 1)(x^2 - x + 1)}{(x - 1)} = x - 1 + \frac{1}{x - 1}$
By simplifying the rational expression, we have reduced the numerator and denominator to their simplest forms, making it easier to work with the expression in further calculations or manipulations.
Key Takeaways:
• Simplifying a rational expression involves identifying and canceling common factors between the numerator and denominator.
• The goal is to reduce the expression to its simplest form, with the least number of factors in both the numerator and denominator.
• Canceling common factors is a crucial step in simplifying rational expressions, as it eliminates unnecessary terms and makes the expression more manageable.
1.3: Operations with Rational Expressions
In this sub-chapter, we will explore the basic operations that can be performed on rational expressions, including addition, subtraction, multiplication, and division.
Addition and Subtraction of Rational Expressions: To add or subtract rational expressions, we need to find a common denominator and then perform the operation.
The steps are:
1. Find the least common multiple (LCM) of the denominators.
2. Multiply the numerator and denominator of each rational expression by the appropriate factor to obtain the common denominator.
3. Perform the addition or subtraction operation on the numerators, keeping the common denominator.
Example: Perform the addition $\frac{2x + 1}{x - 1} + \frac{3x - 2}{x + 2}$
Step 1: Find the LCM of the denominators, which is $(x - 1)(x + 2)$. Step 2: Multiply the first fraction by $(x + 2)$ and the second fraction by $(x - 1)$ to obtain the common denominator. $\frac{2x
+ 1}{x - 1} \cdot \frac{x + 2}{x + 2} + \frac{3x - 2}{x + 2} \cdot \frac{x - 1}{x - 1} = \frac{2x^2 + 5x + 2}{x^2 + x - 2}$ Step 3: Add the numerators and keep the common denominator.
Multiplication and Division of Rational Expressions: Multiplying and dividing rational expressions follows the standard rules of fraction multiplication and division.
To multiply two rational expressions:
1. Multiply the numerators.
2. Multiply the denominators.
To divide one rational expression by another:
1. Invert the second fraction (the divisor).
2. Multiply the first fraction (the dividend) by the inverted divisor.
Example: Divide $\frac{x^2 - 4}{x - 2}$ by $\frac{x + 2}{x - 1}$
Step 1: Invert the divisor. $\frac{x - 1}{x + 2}$ Step 2: Multiply the dividend by the inverted divisor. $\frac{x^2 - 4}{x - 2} \cdot \frac{x - 1}{x + 2} = \frac{x^3 - 3x^2 - 4x + 8}{x^2 - 4}$
Performing these basic operations on rational expressions is essential for manipulating and simplifying more complex algebraic expressions involving fractions.
Key Takeaways:
• To add or subtract rational expressions, find a common denominator and then perform the operation on the numerators.
• To multiply rational expressions, multiply the numerators and multiply the denominators.
• To divide one rational expression by another, invert the divisor and then multiply the dividend by the inverted divisor.
• Mastering these operations is crucial for working with and simplifying rational expressions in various mathematical contexts.
1.4: Solving Rational Equations
In this sub-chapter, we will learn how to solve rational equations, which are equations that contain rational expressions.
Solving a rational equation involves finding the values of the variable(s) that make the equation true.
The general steps for solving a rational equation are:
1. Clear the denominators: Multiply both sides of the equation by the least common multiple (LCM) of the denominators to eliminate the fractions.
2. Simplify the resulting equation: Expand and combine like terms to obtain a polynomial equation.
3. Solve the polynomial equation: Use standard algebraic techniques, such as factoring or applying the quadratic formula, to find the solutions.
4. Check the solutions: Substitute the solutions back into the original rational equation to ensure they are valid.
Let's consider an example:
Solve the equation: $\frac{x + 2}{x - 1} = \frac{3x - 1}{x + 3}$
Step 1: Clear the denominators. Multiply both sides by the LCM of the denominators, which is $(x - 1)(x + 3)$. $(x + 2)(x + 3) = (3x - 1)(x - 1)$
Step 2: Simplify the equation. Expand and combine like terms: $x^2 + 5x + 6 = 3x^2 - 4x - 3$
Step 3: Solve the polynomial equation. Subtract $x^2$ from both sides: $2x^2 - 9x + 9 = 0$ Solve using the quadratic formula: $x = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm 3}{4} = 3, \frac{2}{2}$
Step 4: Check the solutions. Substitute the solutions back into the original equation to verify that they are valid.
By following these steps, you can solve rational equations and find the values of the variable(s) that satisfy the equation.
Key Takeaways:
• Rational equations contain rational expressions and must be solved using specialized techniques.
• The key steps are: clearing the denominators, simplifying the resulting equation, solving the polynomial equation, and checking the solutions.
• Clearing the denominators is a crucial first step to eliminate the fractions and convert the equation into a polynomial form that can be solved using standard algebraic methods.
• Checking the solutions by substituting them back into the original equation ensures that the found solutions are valid.
[Second Half: Applications of Rational Expressions]
1.5: Modeling Real-World Situations with Rational Expressions
In this sub-chapter, we will explore how rational expressions can be used to model and solve real-world problems.
Rational expressions are often used to represent and analyze situations involving rates, ratios, and proportions. These types of problems can arise in various contexts, such as:
1. Rate problems: Calculating speeds, flow rates, or other quantities that involve a ratio of two measurements.
2. Mixture problems: Determining the composition of a mixture by considering the ratios of its components.
3. Inverse variation problems: Modeling situations where two quantities are inversely related, such as the relationship between the speed and time of a journey.
4. Proportional reasoning problems: Solving problems that involve direct or inverse proportions between quantities.
Let's consider an example of a real-world problem that can be modeled using rational expressions:
Example: Fuel Efficiency A car travels 360 miles on 15 gallons of gasoline. What is the car's fuel efficiency in miles per gallon?
To solve this problem, we can set up a rational expression:
Let $x$ be the fuel efficiency in miles per gallon. Then, the distance traveled (360 miles) divided by the amount of gasoline used (15 gallons) is equal to the fuel efficiency:
$\frac{360 \text{ miles}}{15 \text{ gallons}} = x \text{ miles/gallon}$
Simplifying the rational expression, we get: $x = \frac{360}{15} = 24 \text{ miles/gallon}$
In this example, we used a rational expression to model the relationship between the distance traveled, the amount of gasoline used, and the fuel efficiency of the car. By solving the resulting
rational equation, we were able to determine the car's fuel efficiency in miles per gallon.
Rational expressions can be used to model a wide range of real-world situations, allowing us to set up and solve problems involving rates, ratios, and proportions. Mastering the skills to work with
rational expressions is crucial for applying mathematics to solve practical problems.
Key Takeaways:
• Rational expressions can be used to model and solve real-world problems involving rates, ratios, and proportions.
• Common types of real-world problems that can be represented using rational expressions include rate problems, mixture problems, inverse variation problems, and proportional reasoning problems.
• Setting up the appropriate rational expression and solving the resulting equation allows us to find the unknown quantities in these practical situations.
• Developing the ability to model real-world problems using rational expressions is an important skill for applying mathematics to solve practical problems.
1.6: Graphing Rational Functions
In this sub-chapter, we will explore the graphing of rational functions, which are functions that can be expressed as a ratio of two polynomials.
A rational function takes the general form:
$f(x) = \frac{P(x)}{Q(x)}$
where $P(x)$ and $Q(x)$ are polynomials, and $Q(x) eq 0$.
The key features of the graph of a rational function include:
1. Asymptotes: Rational functions can have two types of asymptotes:
□ Vertical asymptotes: Occur at the values of $x$ where the denominator $Q(x)$ is zero, provided that the numerator $P(x)$ is not also zero at those values.
□ Horizontal asymptotes: Occur when the limit of the function as $x$ approaches positive or negative infinity is a constant value.
2. Domains and Ranges: The domain of a rational function is the set of all real numbers $x$ for which the denominator $Q(x)$ is non-zero. The range of a rational function depends on the specific
form of the function and its behavior.
3. Symmetry: Some rational functions may exhibit symmetry, such as even or odd symmetry, which can be determined by the properties of the numerator and denominator polynomials.
4. Behavior: Rational functions can exhibit a variety of behaviors, such as increasing, decreasing, or oscillating, depending on the specific form of the function.
Let's consider an example of graphing a rational function:
$f(x) = \frac{x^2 - 4}{x - 2}$
To graph this function, we need to:
1. Identify the vertical asymptote(s): The vertical asymptote occurs at $x = 2$, as the denominator is zero at this value.
2. Identify the horizontal asymptote(s): To find the horizontal asymptote, we need to consider the degree of the numerator and denominator. In this case, the degree of the numerator is 2, and the
degree of the denominator is 1, so the horizontal asymptote is the line $y = 0$.
3. Sketch the graph: With the knowledge of the asymptotes, we can sketch the general shape of the rational function, including its behavior as it approaches the asymptotes.
By understanding the key features of rational functions and the techniques for graphing them, students can develop the ability to accurately sketch and interpret the behavior of these functions,
which is essential for solving a variety of mathematical problems.
Key Takeaways:
• Rational functions are functions that can be expressed as a ratio of two polynomials.
• The graph of a rational function has characteristic features, such as vertical and horizontal asymptotes, domains and ranges, and symmetry.
• Identifying the asymptotes and other properties of a rational function is crucial for sketching its graph accurately.
• Mastering the graphing of rational functions allows students to visualize and interpret the behavior of these functions, which is valuable for solving real-world problems.
1.7: Rational Inequalities and their Solutions
In this sub-chapter, we will learn how to solve rational inequalities, which involve comparing rational expressions.
A rational inequality is an inequality where at least one of the expressions is a rational expression. Examples of rational inequalities include:
1. $\frac{x + 2}{x - 1} > 3$
2. $\frac{2x - 1}{x + 1} \leq \frac{x - 3}{x + 2}$
3. $\frac{x^2 - 4}{x - 2} < \frac{x + 1}{x - 2}$
Solving rational inequalities involves several steps:
1. Clear the denominators: Multiply both sides of the inequality by the least common multiple (LCM) of the denominators to eliminate the fractions.
2. Simplify the resulting inequality: Expand and combine like terms to obtain a polynomial inequality.
3. Solve the polynomial inequality: Use standard techniques, such as sign analysis or graphing, to find the solution set.
4. Check the solution: Verify that the found solution(s) satisfy the original rational inequality.
Let's consider an example:
Solve the inequality: $\frac{x + 2 | {"url":"https://wisemen.ai/app/courses/661efb7835f59083a1cbebe0/1","timestamp":"2024-11-05T16:51:23Z","content_type":"text/html","content_length":"164745","record_id":"<urn:uuid:de086fa8-1554-47af-ab9d-7bca92a66e3e>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00221.warc.gz"} |
Equity Value - Relative to Enterprise Value & Market Cap
Updated October 5, 2023
What is Equity Value?
Equity value represents a company’s worth based on its shareholder’s equity. In other words, it is the total price equivalent to all shareholders’ equity.
It is a measure in economics to determine the value of a company’s equity. It helps investors understand a company’s worth and make business decisions. To calculate, multiply the number of shares by
the price per share.
Key Takeaways
• It computes a company’s equity and share price to determine its worth. There are various metrics to calculate this, but all those measures use the company’s share price to calculate the value.
• To calculate it, one can multiply the outstanding shares by their current price in the market.
• Economists can use it to assess a company’s financial health and help investors make investments.
• Both equity and enterprise value offer the value of a company, but they provide slightly different perspectives.
Equity Value Formula
The value of the company can be computed in two ways. For one, we can multiply all shares by the market share value. The second way is subtracting the debt from enterprise value (EV).
Equity Value = Price of each Share * No.Of. Shares Outstanding
Equity Value = Enterprise Value – Market Value of Preferred Shares – Debt – Minority Interest + Cash and Cash Equivalents
Excel Template
Example 1
Suppose investors want to invest in equity. They are evaluating two companies, A and B. Company A has 200,000 outstanding shares at a share price of $60 each. While Company B has 50,000 outstanding
shares with a share price of $120 each. Let us calculate which company has a higher value.
Let us first calculate the value for Company A.
Next, we calculate the value for Company B.
Thus, values for both companies are,
Company A = $12,000,000
Company B = $6,000,000
As the value derived for Company A is higher than for Company B, investing in Company A will be more beneficial for the investor.
Example 2
In this example, let us derive the required value from the enterprise value.
The enterprise value for Company X is $25,000,000. It has a debt of $500,000 and a minority interest of $150,000. It has 300,000 preference shares with a market value of $15,000,000. All its cash
equivalents, including investments, sum up to $200,000.
Let us calculate the value for Company X.
By using the formula, we get the following:
Hence, the value for Company X is $9,550,000.
Relationship with Enterprise Value
Both determine the value of businesses. The formula to calculate the value of equity is,
= Total Shares Outstanding * Price of each Share
The formula to calculate enterprise value is,
Enterprise Value = Market Capitalization + Debt + Minority Interest + Preference Shares – Cash & Cash Equivalents
In general, the enterprise value is a better measure of the company’s worth than the value of its equity. It is because it considers all of the company’s assets and liabilities. It values only the
company’s equity and is the sum of all shares multiplied by their price. Enterprise value is the total value of a company’s equity, debt, and capital.
• It is the value an individual investor would have to purchase company shares. They can calculate the worth of the company’s equity using this measure and then invest.
• Enterprise Value is the determination of a company’s worth as a whole. Investors use this value when they want to acquire a firm or have controlling power. It is more important than the other
measures because it includes all aspects of a company.
Relation to Market Capitalization
The market value of equity, also known as market capitalization, is the value that determines the worth of a company’s shares. Many economists believe that the value of equity and market cap are
unrelated. However, it is not any different from market capitalization. They both are informally interchangeable terms.
It determines the size of a firm and assists investors in diversifying their assets across companies of varying sizes and risk levels. The worth of a firm is computed by multiplying the current share
price by the total number of outstanding shares. As a result, the market value of a company’s shares changes as these two input factors change.
Equity value is an economic measure that can also be known as market capitalization. It defines the total value of the securities available to the business. Moreover, it is a financial measure
similar to enterprise value but less substantial, and it calculates the worth of a company’s equity.
To calculate it, we multiply the market value per share by the total number of shares outstanding. Whereas enterprise value also includes debt and cash in its formula.
Investors use this method to make investment decisions. This measure is also critical to business owners, especially when planning to sell their businesses. It provides a good indication of what a
seller would receive after clearing all debt payments.
Frequently Asked Questions(FAQs)
Q1. Is equity value the same as a market cap? What is the difference between it and enterprise value?
Answer: Although most people believe the equity value differ from the market cap, they both determine the same values. However, there is a difference between equity value and enterprise value. While
the former calculates the equity market values, enterprise value also includes debt, investments, and cash equivalents.
Q2. What does equity value in DCF mean?
Answer: The DCF model calculates it by discounting a company’s levered free cash flow. The DCF valuation method determines a company’s intrinsic value, given its future cash flows and current capital
Q3. How to calculate? Why is it important?
Answer: The calculation comprises multiplying available outstanding shares by their market value. It is an essential economic measure for long-term investors. It is important because it gives a
comprehensive representation of the interests of equity investors. Therefore, investors can use this to understand and analyze the company.
Q4. What is the risk?
Answer: It is the risk investors who invest in shares may face. As the share market is very volatile, share prices fluctuate unusually. Investors are always prone to losing their investment or
receiving a reduced amount.
Q5. What is an equity value bridge?
Answer: An equity value bridge is a financial instrument that connects the company’s value with its debt. It describes the enterprise value and market value of equity relationships. It is also known
as EV to equity bridge.
Recommended articles
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Obfuscatory Math – Good Math/Bad Math
This morning, my friend Dr24Hours pinged me on twitter about some bad math:
Attn @MarkCC: http://t.co/ijzQZpM2lm (Sum(NatNums)= -1/12 bullshit) h/t @NeuroPolarbear@BadAstronomer Shame on you, @Slate.
— Dr24hours (@Dr24hours) January 17, 2014
And indeed, he was right. Phil Plait the Bad Astronomer, of all people, got taken in by a bit of mathematical stupidity, which he credulously swallowed and chose to stupidly expand on.
Let’s start with the argument from his video.
We’ll consider three infinite series:
S[1] = 1 - 1 + 1 - 1 + 1 - 1 + ...
S[2] = 1 - 2 + 3 - 4 + 5 - 6 + ...
S[3] = 1 + 2 + 3 + 4 + 5 + 6 + ...
S[1] is something called Grandi’s series. According to the video, taken to infinity, Grandi’s series alternates between 0 and 1. So to get a value for the full series, you can just take the average –
so we’ll say that S[1] = 1/2. (Note, I’m not explaining the errors here – just repeating their argument.)
Now, consider S[2]. We’re going to add S[2] to itself. When we write it, we’ll do a bit of offset:
1 - 2 + 3 - 4 + 5 - 6 + ...
1 - 2 + 3 - 4 + 5 + ...
1 - 1 + 1 - 1 + 1 - 1 + ...
So 2S[2] = S[1]; therefore S[2] = S[1]=2 = 1/4.
Now, let’s look at what happens if we take the S[3], and subtract S[2] from it:
1 + 2 + 3 + 4 + 5 + 6 + ...
- [1 - 2 + 3 - 4 + 5 - 6 + ...]
0 + 4 + 0 + 8 + 0 + 12 + ... == 4(1 + 2 + 3 + ...)
So, S[3] – S[2] = 4S[3], and therefore 3S[3] = -S[2], and S[3]=-1/12.
So what’s wrong here?
To begin with, S[1] does not equal 1/2. S[1] is a non-converging series. It doesn’t converge to 1/2; it doesn’t converge to anything. This isn’t up for debate: it doesn’t converge!
In the 19th century, a mathematician named Ernesto Cesaro came up with a way of assigning a value to this series. The assigned value is called the Cesaro summation or Cesaro sum of the series. The
sum is defined as follows:
Let $A = {a_1 + a_2 + a_3 + ...}$. In this series, $s_k = Sigma_{n=1}^{k} a_n$. $s_k$ is called the kth partial sum of A.
The series $A$ is Cesaro summable if the average of its partial sums converges towards a value $C(A) = lim_{n rightarrow infty} frac{1}{n}Sigma_{k=1}^{n} s_k$.
So – if you take the first 2 values of $A$, and average them; and then the first three and average them, and the first 4 and average them, and so on – and that series converges towards a specific
value, then the series is Cesaro summable.
Look at Grandi’s series. It produces the partial sum averages of 1, 1/2, 2/3, 2/4, 3/5, 3/6, 4/7, 4/8, 5/9, 5/10, … That series clearly converges towards 1/2. So Grandi’s series is Cesaro summable,
and its Cesaro sum value is 1/2.
The important thing to note here is that we are not saying that the Cesaro sum is equal to the series. We’re saying that there’s a way of assigning a measure to the series.
And there is the first huge, gaping, glaring problem with the video. They assert that the Cesaro sum of a series is equal to the series, which isn’t true.
From there, they go on to start playing with the infinite series in sloppy algebraic ways, and using the Cesaro summation value in their infinite series algebra. This is, similarly, not a valid thing
to do.
Just pull out that definition of the Cesaro summation from before, and look at the series of natural numbers. The partial sums for the natural numbers are 1, 3, 6, 10, 15, 21, … Their averages are 1,
4/2, 10/3, 20/4, 35/5, 56/6, = 1, 2, 3 1/3, 5, 7, 9 1/3, … That’s not a converging series, which means that the series of natural numbers does not have a Cesaro sum.
What does that mean? It means that if we substitute the Cesaro sum for a series using equality, we get inconsistent results: we get one line of reasoning in which a the series of natural numbers has
a Cesaro sum; a second line of reasoning in which the series of natural numbers does not have a Cesaro sum. If we assert that the Cesaro sum of a series is equal to the series, we’ve destroyed the
consistency of our mathematical system.
Inconsistency is death in mathematics: any time you allow inconsistencies in a mathematical system, you get garbage: any statement becomes mathematically provable. Using the equality of an infinite
series with its Cesaro sum, I can prove that 0=1, that the square root of 2 is a natural number, or that the moon is made of green cheese.
What makes this worse is that it’s obvious. There is no mechanism in real numbers by which addition of positive numbers can roll over into negative. It doesn’t matter that infinity is involved: you
can’t following a monotonically increasing trend, and wind up with something smaller than your starting point.
Someone as allegedly intelligent and educated as Phil Plait should know that.
Obfuscatory Vaccination Math
Over at my friend Pal’s blog, in a discussion about vaccination, a commenter came up with the following in an argument against the value of vaccination:
Mathematical formula:
100% – % of population who are not/cannot be vaccinated – % of population who have been vaccinated but are not immune (1-effective rate)-% of population who have been vaccinated but immunity has
waned – % of population who have become immune compromised-(any other variables an immunologist would know that I may not)
What vaccine preventable illnesses have the result of that formula above the necessary threshold to maintain herd immunity?
I don’t know if the population is still immune to Smallpox, but I would hope that that is just a science fiction question. Smallpox was eradicated, but that vaccine did have the highest number of
adverse reaction (I’m sure PAL will correct me if that statement is wrong)
It’s a classic example of what I call obfuscatory mathematics: that is, it’s an attempt to use fake math in an attempt to intimidate people into believing that there’s a real argument, when in fact
they’re just hiding behind the appearance of mathematics in order to avoid having to really make their argument. It’s a classic technique, frequently used by crackpots of all stripes.
It’s largely illegible, due to notation, punctuation, and general babble. That’s typical of obfuscatory math: the point isn’t to use math to be comprehensible, or to use formal reasoning; it’s to
create an appearance of credibility. So let’s take that, and try to make it sort of readable.
What he wants to do is to take each group of people who, supposedly, aren’t protected by vaccines, and try to put together an argument about how it’s unlikely that vaccines can possibly create a
large enough group of protected people to really provide herd immunity.
So, let’s consider the population of people. Per Chuck’s argument, we can consider the following subgroups:
• $u$ is the percentage of the population that does not get vaccinated, for whatever reason.
• $v$ is the percentage of people who got vaccinated; obviously equal to $1 - u$.
• $n$ is the percentage of people who were vaccinated, but who didn’t gain any immunity from their vaccination.
• $w$ is the percentage of people who were vaccinated, but whose immunity from the vaccine has worn off.
• $i$ is the percentage of people who were vaccinated, but who have for some reason become immune-compromised, and thus gain no immunity from the vaccine.
He’s arguing then, that the percentage of effectively vaccinated people is $1.0 - u - nv - wv - iv$. And he implies that there are other groups. Since herd immunity requires a very large part of the
population to be immune to a disease, and there are so many groups of people who can’t be part of the immune population, then with so many people excluded, what’s the chance that we really have
effective herd immunity to any disease?
There’s a whole lot wrong with this, ranging from the trivial to the moderately interesting. We’ll start with the trivial, and move on to the more interesting. | {"url":"http://www.goodmath.org/blog/category/bad-math/obfuscatory-math/","timestamp":"2024-11-08T21:16:14Z","content_type":"text/html","content_length":"95142","record_id":"<urn:uuid:40cc0dd9-f7d5-4a21-ab60-41f57b02c0f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00484.warc.gz"} |
Increase in Flow Diversity From Simultaneous Fixture Use: Impact on Peak Flow Estimate | IAPMO
Adaptive reuse of commercial office spaces to residential multifamily offers opportunities to ease some of the housing shortage in the US. These types of construction projects have many challenges,
from zoning restrictions, financing, and also controlling construction costs. Construction costs specifically can have a disparate impact on whether an adaptive reuse project is possible, let alone
successful. Therefore, flexibility during the construction process is vital to the success of an adaptive reuse project. Codes for plumbing and mechanical systems that support design versatility and
science-based methods, such as the Uniform Codes, will be essential for controlling construction expenses. | {"url":"https://www.iapmo.org/newsroom/research/increase-in-flow-diversity-from-simultaneous-fixture-use-impact-on-peak-flow-estimate/","timestamp":"2024-11-08T10:38:27Z","content_type":"text/html","content_length":"57834","record_id":"<urn:uuid:f594f3d3-6250-4a08-a9a4-54b576b2b1fd>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00853.warc.gz"} |
What Is the Price-to-Earnings (P/E) Ratio?
What Is The Price-to-Earnings (P/E) Ratio?
Photo by Markus Winkler on Unsplash
The price-to-earnings (P/E) ratio is an excellent metric for understanding the valuation of a company over time. It tells investors how much a company is worth. The price-to-earnings ratio measures
its current share price relative to its annual earnings per share (EPS). The ratio shows how much investors will pay per share for each $1 of earnings. The P/E ratio is also called the earnings
multiple or price multiple.
The price-to-earnings ratio formula is:
Stock Price Per Share / Earnings Per Share = P/E ratio
You simply divide the stock price per share (market value) by the earnings per share.
For example, if Company XYZ was currently selling at $100 per share and its annual earnings were $5 per share, its P/E ratio would be 20.
$100 / $5 = 20
Types of Price-to-Earnings Ratios
Forward P/E Ratio
The forward P/E ratio is calculated from the estimated annual earnings per share. Hence, this ratio is forward-looking and accounts for future growth. However, the forward P/E ratio is only as valid
as the accuracy of the earnings estimate. Therefore, most investors use the consensus earnings estimate when determining the forward P/E ratio.
For instance, Apple (APPL) has a consensus EPS of $6.15 for 2022, according to Portfolio Insight*. This value gives a forward P/E ratio of 26.5X based on the current stock price of $163.20 (as of
March 2, 2022).
The forward P/E ratio will also change if the consensus earnings estimate changes.
Some popular investing sites, like Seeking Alpha and Morningstar, use the forward P/E ratio.
Trailing Twelve Months P/E Ratio
The trailing twelve months (TTM) P/E ratio is determined from actual earnings from the past 12 months. The primary benefit of this approach is it uses real data reported in quarterly or annual
earnings releases. However, the disadvantage is this ratio is backward-looking and does not factor in future growth or declining expectations.
In an example, Apple earned $5.61 per share in 2021, according to Portfolio Insight*. This value gives a TTM P/E ratio of 29.1X based on the current stock price of $163.20 (as of March 2, 2022).
The TTM P/E ratio will change each quarter as new earnings results are reported.
Other popular investing sites, like Google Finance, Yahoo Finance, and some apps use the TTM P/E ratio.
Shiller P/E Ratio
The third metric is known as the Shiller P/E ratio, also known as the cyclically adjusted price-to-earnings (CAPE) ratio or the Shiller P/E 10. This approach was developed by the Yale University
Professor and Nobel Laureate Robert Shiller.
The Shiller P/E ratio adjusts the forward or TTM P/E ratio calculations. Both the standard metrics can be strongly affected by near-term changes. For instance, during COVID-19, many energy companies
experienced a significant drop in revenue and had losses for several quarters. This fact caused the standard P/E ratios to become not meaningful.
The Shiller P/E ratio uses inflation-adjusted 10-year earnings data to calculate the ratio. The ratio can be determined for individual stocks but is commonly applied to indices like the S&P 500
Index. Investors can use the Shiller P/E ratio to determine if the market is overvalued or undervalued. However, one drawback for the Shiller P/E ratio is it is backward-looking.
The all-time high for the Shiller P/E ratio was 44.19X in 1999 before the dot-com crash. The current Shiller P/E ratio is 35.25X (as of March 1, 2022). The long-term average is about 17X.
Price-to-Earnings Growth Ratio
The price-to-earnings growth (PEG) ratio is a variation of the standard metrics. This ratio is calculated by dividing the company’s TTM or forward P/E ratio by the earnings growth rate over a
specified period. This approach has the advantage of using the P/E ratio and an estimate for earnings growth. The main drawback is the PEG ratio uses an estimate of future earnings growth that may
not be accurate.
A PEG ratio of more than 1.0 generally means a stock is overvalued, while a value less than 1.0 means a stock is undervalued.
Why Is The P/E Ratio Useful?
Investors use the P/E ratio as a tool to determine a stock’s valuation. For example, the ratio shows whether the stock is undervalued or overvalued. In addition, a P/E ratio allows analysts and
investors to compare stocks fairly. For example, investors can compare a company’s P/E ratio against another company’s stock in a like-to-like comparison or compare a company against its record over
time. Comparisons are typically made with companies in the same industry and at the same level of development or growth.
What Does a High P/E Ratio Mean?
If the P/E ratio is high, investors expect future high growth from the stock. However, it could also mean the stock is overvalued. If a company has a high P/E ratio, investors are paying more per
share to have an ownership interest in a company’s earnings.
What is considered a high P/E ratio is dependent on the industry and specific stock. Different industries will have a diverse range of “normal” P/E ratios. Therefore, investors should compare a
company’s P/E ratio within an industry for an apples-to-apples investment picture.
What Does a Low P/E Ratio Mean?
If a company’s P/E ratio is low compared to other stocks in its industry, it may be undervalued and, therefore, a good investment. However, investors should carefully look at the company’s financials
and results for any warning signs causing the lower P/E ratio. For example, it could indicate that earnings may decrease in the future, debt is too high, the company is losing market share or other
Does Every Company Have a P/E Ratio?
If a company loses money or has no earnings, it doesn’t have a P/E ratio. This point is because there would be no earnings per share to divide the stock price per share. You cannot divide the market
value per share by $0. So if a company has negative earnings per share or no profits, the P/E ratio would be not meaningful. A brand-new company on the stock exchange would also have a P/E ratio of
NMF until it had earnings per share.
Example Comparing Two Companies
The P/E ratio is commonly used to compare similar companies as investment choices. In this example, let’s assume you have saved $10,000 in the past year and want to invest it. You are trying to build
a passive income stream and live off dividends.
We now compare Coca-Cola (KO) and Pepsi (PEP) as potential investment candidates. Both stocks are known as dividend growth stocks. They are popular amongst investors for their dividends yields and
consistent returns. However, which one is overvalued, fairly valued, or undervalued today. We can use the P/E ratio to help us with this determination.
In the table below, we use data from Portfolio Insight* and compare Coca-Cola and Pepsi.
Coca-Cola (KO) Pepsi (PEP)
Estimated EPS 2022 $2.46 $6.70
EPS 2021 $2.32 $6.26
Current Stock Price $61.97 $162.67
Forward P/E Ratio 25.2X 24.3X
TTM P/E Ratio 26.7X 26.0X
Average P/E Ratio Range Past Decade 20.2X – 24.37X 18.99X – 24.08X
Average P/E Ratio Range Past 5-years 23.5X – 26.44X 22.4X – 26.2X
Source: Data from Portfolio Insight* (as of March 2, 2022).
Based on this analysis, both stocks are slightly overvalued. Hence, an investor can conclude it is not a good time to add to existing positions or start a new one for either stock.
Disclaimer: Dividend Power is not a licensed or registered investment adviser or broker/dealer. We are not providing you with individual investment advice on this site. Please consult with ...
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TCS Numerical Ability - Data Interpretation Interview Questions with Answers
My maths book suicided because it had lots of problems in it ......
Thanks m4 maths for helping to get placed in several companies. I must recommend this website for placement preparations. | {"url":"https://m4maths.com/placement-puzzles.php?SOURCE=TCS&TOPIC=Numerical%20Ability&SUB_TOPIC=Data%20Interpretation","timestamp":"2024-11-01T22:51:56Z","content_type":"text/html","content_length":"97999","record_id":"<urn:uuid:63dbe5c1-2192-478c-a80c-f56d1ae39365>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00319.warc.gz"} |
Compound Probability Worksheet Answers
Compound Probability Worksheet Answers. Web the cuemath experts developed a set of compound probability worksheets that contain many questions; No of blue (b) balls selected = 300, no of red (r)
balls = 200, no of green (g) balls = 450, no of orange (o) balls =.
Probability Of Compound events Worksheet with Answer Key from briefencounters.ca
Students in grade 8 and high school plug. Reinvesting interest is what compound interest is all about. It includes both independent and dependent events. | {"url":"http://studydblamb123.s3-website-us-east-1.amazonaws.com/compound-probability-worksheet-answers.html","timestamp":"2024-11-04T02:57:15Z","content_type":"text/html","content_length":"25615","record_id":"<urn:uuid:d2099abd-718b-4dd7-b625-51bdf30ac7e7>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00769.warc.gz"} |
What is the area of a triangle with sides of 39, 67.27 and 48.33 cm? | Socratic
What is the area of a triangle with sides of 39, 67.27 and 48.33 cm?
1 Answer
You can use Heron's formula: if $a$, $b$ and $c$ are the sides of the triangle, then the area $A$ can be found this way:
$A = \sqrt{p \cdot \left(p - a\right) \cdot \left(p - b\right) \cdot \left(p - c\right)}$, where $p$ is half of the perimeter: $\frac{a + b + c}{2}$.
So, let's compute $p$:
$\frac{39 + 67.27 + 48.33}{2} = 77.3$.
So, the formula is
$\sqrt{77.3 \cdot \left(77.3 - 39\right) \cdot \left(77.3 - 48.33\right) \cdot \left(77.3 - 67.27\right)}$
which is $\sqrt{860255.971769}$.
Impact of this question
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Problem E
Jimmy is performing in ByteLand today! Anthony the Ant is a huge fan of Jimmy’s music, so he can’t wait to get to the concert.
ByteLand consists of of $N$ intersections and $M$ roads. Every road is bidirectional and connects two distinct intersections. Anthony is currently on intersection $s$ and the concert is being held at
intersection $t$. Anthony must get to the concert in $T$ seconds and he can travel at $1$ meter/second.
Unfortunately for Anthony, he has a huge fear of spiders and ByteLand is full of spiders. Spiders reside at certain intersections in ByteLand. Anthony will choose a path from $s$ to $t$ that
maximizes $D$, the minimum distance between any intersection on the path and any spider, such that the path can be travelled in no more than $T$ seconds.
The first line contains three integers $N$ ($2\leq N\leq 100\, 000$), $M$ ($1\leq M\leq \min (200\, 000, n(n-1)/2)$), and $T$ ($1\leq T\leq 10^9$).
Each of the next $M$ lines specify the roads. A road is described by three integers $u, v$ ($0\leq u, v\leq N-1$ and $u\neq v$) and $d$ ($1\leq d\leq 10\, 000$), which means that a $d$ meters long
road connects intersections $u$ and $v$. It is guaranteed that at most one road connect any two intersections, and that there exists a path between any two intersections.
The next line contains $s, t$ ($0\leq s, t\leq N-1$ and $s\neq t$, representing Anthony’s current location and the location of the concert. You may assume Anthony can always travel from $s$ to $t$ in
no more than $T$ seconds.
The last line contains a integer $1\leq K\leq N$, denoting the number of intersections occupied by spiders, followed by $K$ integers $0\leq a_ i\leq N-1$ denoting the intersections with spiders.
Print the maximum $D$ (as defined earlier) for the path Anthony takes.
Sample Input 1 Sample Output 1
Sample Input 2 Sample Output 2 | {"url":"https://open.kattis.com/contests/pn6wtx/problems/arachnophobia","timestamp":"2024-11-10T04:41:29Z","content_type":"text/html","content_length":"32235","record_id":"<urn:uuid:b8010adc-2af2-4787-879a-22d8f5540104>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00422.warc.gz"} |
Reading: The “New Economy” Controversy
The “New Economy” Controversy
In recent years a controversy has been brewing among economists about the resurgence of U.S. productivity in the second half of the 1990s. One school of thought argues that the United States had
developed a “new economy” based on the extraordinary advances in communications and information technology of the 1990s. The most optimistic proponents argue that it would generate higher average
productivity growth for decades to come. The pessimists, on the other hand, argue that even five or ten years of stronger productivity growth does not prove that higher productivity will last for the
long term. It is hard to infer anything about long-term productivity trends during the later part of the 2000s, because the steep recession of 2008–2009, with its sharp but not completely
synchronized declines in output and employment, complicates any interpretation.
Productivity growth is also closely linked to the average level of wages. Over time, the amount that firms are willing to pay workers will depend on the value of the output those workers produce. If
a few employers tried to pay their workers less than what those workers produced, then those workers would receive offers of higher wages from other profit-seeking employers. If a few employers
mistakenly paid their workers more than what those workers produced, those employers would soon end up with losses. In the long run, productivity per hour is the most important determinant of the
average wage level in any economy. To learn how to compare economies in this regard, follow the steps in the following example.
The Organization for Economic Co-operation and Development (OECD) tracks data on the annual growth rate of real GDP per hour worked. You can find these data on the OECD data webpage “Labour
productivity growth in the total economy” at this website.
Step 1. Visit the OECD website given above and select two countries to compare.
Step 2. On the drop-down menu “Variable,” select “Real GDP, Annual Growth, in percent” and record the data for the countries you have chosen for the five most recent years.
Step 3. Go back to the drop-down menu and select “Real GDP per Hour Worked, Annual Growth Rate, in percent” and select data for the same years for which you selected GDP data.
Step 4. Compare real GDP growth for both countries. Table 6.2 provides an example of a comparison between Australia and Belgium.
Australia 2008 2009 2010 2011 2012
Real GDP Growth (%) 1.6% 2.1% 2.4% 3.3% 2.8%
Real GDP Growth/Hours Worked (%) 0.6% 2.1% –0.2% 1.7% 2.4%
Belgium 2008 2009 2010 2011 2012
Real GDP Growth (%) 1 –2.8 2.4 1.8 –0.3
Real GDP Growth/Hours Worked (%) –1.2 –1.5 1.6 –1.1 –0.3
Step 5. Consider the many factors can affect growth. For example, one factor that may have affected Australia is its isolation from Europe, which may have insulated the country from the effects of
the global recession. In Belgium’s case, the global recession seems to have had an impact on both GDP and real GDP per hours worked between 2008 and 2012.
The Power of Sustained Economic Growth
Nothing is more important for people’s standard of living than sustained economic growth. Even small changes in the rate of growth, when sustained and compounded over long periods of time, make an
enormous difference in the standard of living. Consider Table 6.3, in which the rows of the table show several different rates of growth in GDP per capita and the columns show different periods of
time. Assume for simplicity that an economy starts with a GDP per capita of 100. The table then applies the following formula to calculate what GDP will be at the given growth rate in the future:
GDP at starting date × (1 + growth rate of GDP)^years = GDP at end date
For example, an economy that starts with a GDP of 100 and grows at 3% per year will reach a GDP of 209 after 25 years; that is, 100 (1.03)^25 = 209.
The slowest rate of GDP per capita growth in the table, just 1% per year, is similar to what the United States experienced during its weakest years of productivity growth. The second highest rate, 3%
per year, is close to what the U.S. economy experienced during the strong economy of the late 1990s and into the 2000s. Higher rates of per capita growth, such as 5% or 8% per year, represent the
experience of rapid growth in economies like Japan, Korea, and China.
Table 6.3 shows that even a few percentage points of difference in economic growth rates will have a profound effect if sustained and compounded over time. For example, an economy growing at a 1%
annual rate over 50 years will see its GDP per capita rise by a total of 64%, from 100 to 164 in this example. However, a country growing at a 5% annual rate will see (almost) the same amount of
growth—from 100 to 163—over just 10 years. Rapid rates of economic growth can bring profound transformation. (See the following feature on the relationship between compound growth rates and compound
interest rates.) If the rate of growth is 8%, young adults starting at age 20 will see the average standard of living in their country more than double by the time they reach age 30, and grow nearly
sevenfold by the time they reach age 45.
Table 6.3. Growth of GDP over Different Time Horizons
Growth Rate Value of an original 100 in 10 Years Value of an original 100 in 25 Years Value of an original 100 in 50 Years
1% 110 128 164
3% 134 209 438
5% 163 338 1,147
8% 216 685 4,690
The formula for growth rates of GDP over different periods of time, as shown above, is exactly the same as the formula for how a given amount of financial savings grows at a certain interest rate
over time. Both formulas have the same ingredients: an original starting amount, in one case GDP and in the other case an amount of financial saving; a percentage increase over time, in one case the
growth rate of GDP and in the other case an interest rate; and an amount of time over which this effect happens.
Recall that compound interest is interest that is earned on past interest. It causes the total amount of financial savings to grow dramatically over time. Similarly, compound rates of economic
growth, or the compound growth rate, means that the rate of growth is being multiplied by a base that includes past GDP growth, with dramatic effects over time.
For example, in 2012, the World Fact Book, produced by the Central Intelligence Agency, reported that South Korea had a GDP of $1.64 trillion with a growth rate of 2%. We can estimate that at that
growth rate, South Korea’s GDP will be $1.81 trillion in five years. If we apply the growth rate to each year’s ending GDP for the next five years, we will calculate that at the end of year one, GDP
is $1.67 trillion. In year two, we start with the end-of-year one value of $1.67 and increase it by 2%. Year three starts with the end-of-year two GDP, and we increase it by 2% and so on, as depicted
in the Table 6.4.
Table 6.4 End-of-year Two GDP, 2012
Year Starting GDP Growth Rate 2% Year-End Amount
1 $1.64 Trillion × (1+0.02) $1.67 Trillion
2 $1.67 Trillion × (1+0.02) $1.71 Trillion
3 $1.71 Trillion × (1+0.02) $1.74 Trillion
4 $1.74 × (1+0.02) $1.78 Trillion
5 $1.77 × (1+0.02) $1.81 Trillion
Another way to calculate the growth rate is to apply the following formula:
Future Value = Present Value × (1 + g)^n
Where “future value” is the value of GDP five years hence, “present value” is the starting GDP amount of $1.64 trillion, “g” is the growth rate of 2%, and “n” is the number of periods for which we
are calculating growth.
Future Value = 1.64 × (1+0.02)^5 = $1.81 trillion
Self Check: Productivity
Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not count toward your grade in the class, and you can retake it an
unlimited number of times.
You’ll have more success on the Self Check if you’ve completed the two Readings in this section.
Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section. | {"url":"https://courses.lumenlearning.com/suny-hccc-macroeconomics/chapter/reading-the-new-economy-controversy/","timestamp":"2024-11-07T13:26:32Z","content_type":"text/html","content_length":"57338","record_id":"<urn:uuid:74f3cc16-84f3-46d8-9ce0-89dbe784c3e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00880.warc.gz"} |
Foxhole Mpf Calculator - Temz Calculators
Foxhole Mpf Calculator
The Foxhole MPF (Military Pay Formula) Calculator helps you determine your net pay by considering various elements such as base salary, bonuses, and tax rates. By inputting these details, you can
accurately estimate your take-home pay after deductions.
MPF Calculation Formula
The following formula is used to calculate the Military Pay Formula (MPF):
MPF = Gross Pay * (1 - Tax Rate / 100)
• MPF is the amount of military pay you receive after tax deductions ($)
• Gross Pay is your total pay before deductions ($)
• Tax Rate is the percentage of tax deducted from your gross pay (%)
To calculate the MPF, multiply the gross pay by the tax rate and subtract the result from the gross pay.
What is Military Pay Formula (MPF) Calculation?
Military Pay Formula (MPF) calculation involves determining your net pay after applying taxes and other deductions to your gross pay. It is essential for accurate financial planning and budgeting,
especially for military personnel who need to understand their take-home pay.
How to Calculate MPF?
The following steps outline how to calculate the MPF using the given formula:
1. First, determine your gross pay by adding your base salary and any bonuses.
2. Next, identify the applicable tax rate.
3. Use the formula: MPF = Gross Pay * (1 – Tax Rate / 100).
4. Calculate the MPF by plugging in the values.
5. Verify your result using the calculator above.
Example Problem:
Use the following variables as an example problem to test your understanding:
Base Salary = $4,000
Bonus = $500
Tax Rate = 12%
1. What is gross pay?
Gross pay is the total amount of money earned before any deductions such as taxes or other contributions.
2. How is net pay different from gross pay?
Net pay is the amount of money left after all deductions have been subtracted from gross pay.
3. How often should I use the MPF calculator?
It’s beneficial to use the MPF calculator whenever there are changes in your salary, bonuses, or tax rates to ensure accurate financial planning.
4. Can this calculator be used for different tax rates?
Yes, you can adjust the tax rate field to match any applicable tax rate for precise calculations.
5. Is the calculator accurate?
The calculator provides an estimate of your MPF based on the inputs provided. For exact figures, refer to your pay stubs or financial statements. | {"url":"https://temz.net/foxhole-mpf-calculator/","timestamp":"2024-11-07T09:04:50Z","content_type":"text/html","content_length":"73632","record_id":"<urn:uuid:7a926ace-0f5b-49cb-881a-44d3bf53bb20>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00219.warc.gz"} |
Applied Mechanics and Graphic Statics - Civil Engineering Questions and Answers
Civil Engineering :: Applied Mechanics and Graphic Statics
1. In case of S.H.M. the period of oscillation (T), is given by
T = 2ω/π²
T = 2π/ω
T = 2/ω
T = π/2ω
2. The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be
0.1 rad/sec
1 rad/sec
10 rad/sec
100 rad/sec
3. A body is said to move with Simple Harmonic Motion if its acceleration, is
Always directed away from the centre, the point of reference
Proportional to the square of the distance from the point of reference
Proportional to the distance from the point of reference and directed towards it
Inversely proportion to the distance from the point of reference
4. The resultant of two forces ‘P’ and ‘Q’ acting at an angle ‘θ’, is
P² + Q² + 2P sin θ
P² + Q² + 2PQ cos θ
P² + Q² + 2PQ tan θ
√(P² + Q² + 2PQ cos θ)
5. Parallelogram Law of Forces states, "if two forces acting simultaneously on a particle be represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant may be
represented in magnitude and direction by
Its longer side
Its shorter side
The diagonal of the parallelogram which does not pass through the point of intersection of the forces
The diagonal of the parallelogram which passes through the point of intersection of the forces
6. A ball is dropped from a height of 2.25 m on a smooth floor and rises to a height of 1.00 m after the bounce. The coefficient of restitution between the ball and the floor is
7. A marble ball is rolled on a smooth floor of a room to hit a wall. If the time taken by the ball in returning to the point of projection is twice the time taken in reaching the wall, the
coefficient of restitution between the ball and the wall, is
8. In a lifting machine a weight of 5 kN is lifted through 200 mm by an effort of 0.1 kN moving through 15 m. The mechanical advantage and velocity ratio of the machine are respectively
50 and 75
75 and 50
75 and 75
50 and 50
9. In SI units, the units of force and energy are respectively
Newton and watt
Dyne and erg
Newton and joule
kg wt and joule
10. The centre of gravity of the trapezium as shown in below figure from the side is at a distance of
(h/3) × [(b + 2a)/(b + a)]
(h/3) × [(2b + a)/(b + a)]
(h/2) × [(b + 2a)/(b + a)]
(h/2) × [(2b + a)/(b + a)] | {"url":"https://freshergate.com/civil-engineering/applied-mechanics-and-graphic-statics","timestamp":"2024-11-14T07:24:11Z","content_type":"application/xhtml+xml","content_length":"218177","record_id":"<urn:uuid:15e3b66c-bfc4-42ab-aa88-9c0a5bad983d>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00808.warc.gz"} |
Runway - Wikiwand
According to the International Civil Aviation Organization (ICAO), a runway is a "defined rectangular area on a land aerodrome prepared for the landing and takeoff of aircraft".^[1] Runways may be a
human-made surface (often asphalt, concrete, or a mixture of both) or a natural surface (grass, dirt, gravel, ice, sand or salt). Runways, taxiways and ramps, are sometimes referred to as "tarmac",
though very few runways are built using tarmac. Takeoff and landing areas defined on the surface of water for seaplanes are generally referred to as waterways. Runway lengths are now commonly given
in meters worldwide, except in North America where feet are commonly used.^[2]
Runway 13R at Palm Springs International Airport
Runway 34 at Nagoya Airfield
An MD-11 at one end of a runway
In 1916, in a World War I war effort context, the first concrete-paved runway was built in Clermont-Ferrand in France, allowing local company Michelin to manufacture Bréguet Aviation military
In January 1919, aviation pioneer Orville Wright underlined the need for "distinctly marked and carefully prepared landing places, [but] the preparing of the surface of reasonably flat ground [is] an
expensive undertaking [and] there would also be a continuous expense for the upkeep."^[3]
For fixed-wing aircraft, it is advantageous to perform takeoffs and landings into the wind to reduce takeoff or landing roll and reduce the ground speed needed to attain flying speed. Larger airports
usually have several runways in different directions, so that one can be selected that is most nearly aligned with the wind. Airports with one runway are often constructed to be aligned with the
prevailing wind. Compiling a wind rose is one of the preliminary steps taken in constructing airport runways.^[4] Wind direction is given as the direction the wind is coming from: a plane taking off
from runway 09 faces east, into an "east wind" blowing from 090°.
Triangular runway pattern at Armitage Field, Naval Air Weapons Station China Lake
Originally in the 1920s and 1930s, airports and air bases (particularly in the United Kingdom) were built in a triangle-like pattern of three runways at 60° angles to each other. The reason was that
aviation was only starting, and although it was known that wind affected the runway distance required, not much was known about wind behaviour. As a result, three runways in a triangle-like pattern
were built, and the runway with the heaviest traffic would eventually expand into the airport's main runway, while the other two runways would be either abandoned or converted into taxiways.^[5]
Runway 22
Font and size of numbers and letters
Runways are named by a number between 01 and 36, which is generally the magnetic azimuth of the runway's heading in decadegrees. This heading differs from true north by the local magnetic declination
. A runway numbered 09 points east (90°), runway 18 is south (180°), runway 27 points west (270°) and runway 36 points to the north (360° rather than 0°).^[6] When taking off from or landing on
runway 09, a plane is heading around 90° (east). A runway can normally be used in both directions, and is named for each direction separately: e.g., "runway 15" in one direction is "runway 33" when
used in the other. The two numbers differ by 18 (= 180°). For clarity in radio communications, each digit in the runway name is pronounced individually: runway one-five, runway three-three, etc.
(instead of "fifteen" or "thirty-three").
FAA airport diagram at O'Hare International Airport. The two 14/32 runways go from upper left to lower right, the two 4/22 runways go from lower left to upper right, and the two 9/27 and three 10/28
runways are horizontal.
A leading zero, for example in "runway zero-six" or "runway zero-one-left", is included for all ICAO and some U.S. military airports (such as Edwards Air Force Base). However, most U.S. civil
aviation airports drop the leading zero as required by FAA regulation.^[7] This also includes some military airfields such as Cairns Army Airfield. This American anomaly may lead to inconsistencies
in conversations between American pilots and controllers in other countries. It is very common in a country such as Canada for a controller to clear an incoming American aircraft to, for example,
runway 04, and the pilot read back the clearance as runway 4. In flight simulation programs those of American origin might apply U.S. usage to airports around the world. For example, runway 05 at
Halifax will appear on the program as the single digit 5 rather than 05.
Military airbases may include smaller paved runways known as "assault strips" for practice and training next to larger primary runways.^[8] These strips eschew the standard numerical naming
convention and instead employ the runway's full three digit heading; examples include Dobbins Air Reserve Base's Runway 110/290 and Duke Field's Runway 180/360.^[9]^[10]
Runways with non-hard surfaces, such as small turf airfields and waterways for seaplanes, may use the standard numerical scheme or may use traditional compass point naming, examples include Ketchikan
Harbor Seaplane Base's Waterway E/W.^[11]^[12] Airports with unpredictable or chaotic water currents, such as Santa Catalina Island's Pebbly Beach Seaplane Base, may designate their landing area as
Waterway ALL/WAY to denote the lack of designated landing direction.^[13]^[12]
Letter suffix
Runway sign at Madrid-Barajas Airport, Spain
If there is more than one runway pointing in the same direction (parallel runways), each runway is identified by appending left (L), center (C) and right (R) to the end of the runway number to
identify its position (when facing its direction)—for example, runways one-five-left (15L), one-five-center (15C), and one-five-right (15R). Runway zero-three-left (03L) becomes runway two-one-right
(21R) when used in the opposite direction (derived from adding 18 to the original number for the 180° difference when approaching from the opposite direction). In some countries, regulations mandate
that where parallel runways are too close to each other, only one may be used at a time under certain conditions (usually adverse weather).
At large airports with four or more parallel runways (for example, at Chicago O'Hare, Los Angeles, Detroit Metropolitan Wayne County, Hartsfield-Jackson Atlanta, Denver, Dallas–Fort Worth and Orlando
), some runway identifiers are shifted by 1 to avoid the ambiguity that would result with more than three parallel runways. For example, in Los Angeles, this system results in runways 6L, 6R, 7L, and
7R, even though all four runways are actually parallel at approximately 69°. At Dallas/Fort Worth International Airport, there are five parallel runways, named 17L, 17C, 17R, 18L, and 18R, all
oriented at a heading of 175.4°. Occasionally, an airport with only three parallel runways may use different runway identifiers, such as when a third parallel runway was opened at Phoenix Sky Harbor
International Airport in 2000 to the south of existing 8R/26L—rather than confusingly becoming the "new" 8R/26L it was instead designated 7R/25L, with the former 8R/26L becoming 7L/25R and 8L/26R
becoming 8/26.
Suffixes may also be used to denote special use runways. Airports that have seaplane waterways may choose to denote the waterway on charts with the suffix W; such as Daniel K. Inouye International
Airport in Honolulu and Lake Hood Seaplane Base in Anchorage.^[14] Small airports that host various forms of air traffic may employ additional suffixes to denote special runway types based on the
type of aircraft expected to use them, including STOL aircraft (S), gliders (G), rotorcraft (H), and ultralights (U).^[12] Runways that are numbered relative to true north rather than magnetic north
will use the suffix T; this is advantageous for certain airfields in the far north such as Thule Air Base (08T/26T).^[15]
Runway designations may change over time because Earth's magnetic lines slowly drift on the surface and the magnetic direction changes. Depending on the airport location and how much drift occurs, it
may be necessary to change the runway designation. As runways are designated with headings rounded to the nearest 10°, this affects some runways sooner than others. For example, if the magnetic
heading of a runway is 233°, it is designated Runway 23. If the magnetic heading changes downwards by 5 degrees to 228°, the runway remains Runway 23. If on the other hand the original magnetic
heading was 226° (Runway 23), and the heading decreased by only 2 degrees to 224°, the runway becomes Runway 22. Because magnetic drift itself is slow, runway designation changes are uncommon, and
not welcomed, as they require an accompanying change in aeronautical charts and descriptive documents. When a runway designation does change, especially at major airports, it is often done at night,
because taxiway signs need to be changed and the numbers at each end of the runway need to be repainted to the new runway designators. In July 2009 for example, London Stansted Airport in the United
Kingdom changed its runway designations from 05/23 to 04/22 during the night.
Runway dimensions vary from as small as 245 m (804 ft) long and 8 m (26 ft) wide in smaller general aviation airports, to 5,500 m (18,045 ft) long and 80 m (262 ft) wide at large international
airports built to accommodate the largest jets, to the huge 11,917 m × 274 m (39,098 ft × 899 ft) lake bed runway 17/35 at Edwards Air Force Base in California – developed as a landing site for the
Space Shuttle.^[16]
Takeoff and landing distances available are given using one of the following terms:
There are standards for runway markings.^[22]
• The runway thresholds are markings across the runway that denote the beginning and end of the designated space for landing and takeoff under non-emergency conditions.^[23]
• The runway safety area is the cleared, smoothed and graded area around the paved runway. It is kept free from any obstacles that might impede flight or ground roll of aircraft.
• The runway is the surface from threshold to threshold (including displaced thresholds), which typically features threshold markings, numbers, and centerlines, but excludes blast pads and stopways
at both ends.
• Blast pads are often constructed just before the start of a runway where jet blast produced by large planes during the takeoff roll could otherwise erode the ground and eventually damage the
• Stopways, also known as overrun areas, are also constructed at the end of runways as emergency space to stop planes that overrun the runway on landing or a rejected takeoff.
□ Blast pads and stopways look similar, and are both marked with yellow chevrons; stopways may optionally be surrounded by red runway lights. The differences are that stopways can support the
full weight of an aircraft and are designated for use in an aborted takeoff, while blast pads are often not as strong as the main paved surface of the runway and are not to be used for
taxiing, landing, or aborted takeoffs.^[24] An engineered materials arrestor system (EMAS) may also be present, which may overlap with the end of the blast pad or stopway and is painted
similarly (although an EMAS does not count as part of a stopway).^[24]
• Displaced thresholds may be used for taxiing, takeoff, and landing rollout, but not for touchdown. A displaced threshold often exists because of obstacles just before the runway, runway strength,
or noise restrictions making the beginning section of runway unsuitable for landings.^[25] It is marked with white paint arrows that lead up to the beginning of the landing portion of the runway.
As with blast pads, landings on displaced thresholds are not permitted aside from emergency use or exigent circumstance.
• Relocated thresholds are similar to displaced thresholds. They are used to mark a portion of the runway temporarily closed due to construction or runway maintenance. This closed portion of the
runway is not available for use by aircraft for takeoff or landing, but it is available for taxi. While methods for identifying the relocated threshold vary, a common way for the relocated
threshold to be marked is a ten-foot-wide white bar across the width of the runway.^[26]
• Clearway is an area beyond the paved runway, aligned with the runway centerline and under the control of the airport authorities. This area is not less than 500 ft and there are no protruding
obstacles except for threshold lights provided they are not higher than 26 inches. There is a limit on the upslope of the clearway of 1.25%. The length of the clearway may be included in the
length of the takeoff distance available.^[27] For example, if a paved runway is 2,000 metres (6,600 ft) long and there are 400 metres (1,300 ft) of clearway beyond the end of the runway, the
takeoff distance available is 2,400 metres (7,900 ft) long. When the runway is to be used for takeoff of a large airplane, the maximum permissible takeoff weight of the airplane can be based on
the takeoff distance available, including clearway. Clearway allows large airplanes to take off at a heavier weight than would be allowed if only the length of the paved runway is taken into
This section
needs additional citations for verification
(June 2022)
There are runway markings and signs on most large runways. Larger runways have a distance remaining sign (black box with white numbers). This sign uses a single number to indicate the remaining
distance of the runway in thousands of feet. For example, a 7 will indicate 7,000 ft (2,134 m) remaining. The runway threshold is marked by a line of green lights.
Runway Identifying numbers being painted at Rocky Mountain Metropolitan Airport (KBJC)
There are three types of runways:
• Visual runways are used at small airstrips and are usually just a strip of grass, gravel, ice, asphalt, or concrete. Although there are usually no markings on a visual runway, they may have
threshold markings, designators, and centerlines. Additionally, they do not provide an instrument-based landing procedure; pilots must be able to see the runway to use it. Also, radio
communication may not be available and pilots must be self-reliant.
• Non-precision instrument runways are often used at small- to medium-size airports. These runways, depending on the surface, may be marked with threshold markings, designators, centerlines, and
sometimes a 1,000 ft (305 m) mark (known as an aiming point, sometimes installed at 1,500 ft (457 m)). While centerlines provide horizontal position guidance, aiming point markers provide
vertical position guidance to planes on visual approach.
• Precision instrument runways, which are found at medium- and large-size airports, consist of a blast pad/stopway (optional, for airports handling jets), threshold, designator, centerline, aiming
point, and 500 ft (152 m), 1,000 ft (305 m)/1,500 ft (457 m), 2,000 ft (610 m), 2,500 ft (762 m), and 3,000 ft (914 m) touchdown zone marks. Precision runways provide both horizontal and vertical
guidance for instrument approaches.
Waterways may be unmarked or marked with buoys that follow maritime notation instead.^[28]
For runways and taxiways that are permanently closed, the lighting circuits are disconnected. The runway threshold, runway designation, and touchdown markings are obliterated and yellow "Xs" are
placed at each end of the runway and at 1,000 ft (305 m) intervals.^[29]
National variants
• In Australia, Canada, the United Kingdom,^[30] as well as some other countries or territories (Hong Kong and Macau) all 3-stripe and 2-stripe touchdown zones for precision runways are replaced
with one-stripe touchdown zones.
• In some South American countries like Colombia, Ecuador and Peru, one 3-stripe is added and a 2-stripe is replaced with the aiming point.
• Some European countries replace the aiming point with a 3-stripe touchdown zone.
• Runways in Norway have yellow markings instead of the usual white ones. This also occurs in some airports in Japan, Sweden, and Finland. The yellow markings are used to ensure better contrast
against snow.
• Runways may have different types of equipment on each end. To reduce costs, many airports do not install precision guidance equipment on both ends. Runways with one precision end and any other
type of end can install the full set of touchdown zones, even if some are past the midpoint. Runways with precision markings on both ends omit touchdown zones within 900 ft (274 m) of the
midpoint, to avoid ambiguity over the end with which the zone is associated.
A runway landing light from 1945
A line of lights on an airfield or elsewhere to guide aircraft in taking off or coming in to land or an illuminated runway is sometimes also known as a flare path.
Technical specifications
Night runway view from A320 cockpit
Ground light at Bremen Airport
Runway lighting is used at airports during periods of darkness and low visibility. Seen from the air, runway lights form an outline of the runway. A runway may have some or all of the following:^[31]
• Runway end identifier lights (REIL) – unidirectional (facing approach direction) or omnidirectional pair of synchronized flashing lights installed at the runway threshold, one on each side.
• Runway end lights – a pair of four lights on each side of the runway on precision instrument runways, these lights extend along the full width of the runway. These lights show green when viewed
by approaching aircraft and red when seen from the runway.
• Runway edge lights – white elevated lights that run the length of the runway on either side. On precision instrument runways, the edge-lighting becomes amber in the last 2,000 ft (610 m) of the
runway, or last third of the runway, whichever is less. Taxiways are differentiated by being bordered by blue lights, or by having green center lights, depending on the width of the taxiway, and
the complexity of the taxi pattern.
• Runway centerline lighting system (RCLS) – lights embedded into the surface of the runway at 50 ft (15 m) intervals along the runway centerline on some precision instrument runways. White except
the last 900 m (3,000 ft): alternate white and red for next 600 m (1,969 ft) and red for last 300 m (984 ft).^[31]
• Touchdown zone lights (TDZL^[17]) – rows of white light bars (with three in each row) at 30 or 60 m (98 or 197 ft) intervals on either side of the centerline for 900 m (3,000 ft).^[31]
• Taxiway centerline lead-off lights – installed along lead-off markings, alternate green and yellow lights embedded into the runway pavement. It starts with green light at about the runway
centerline to the position of first centerline light beyond the Hold-Short markings on the taxiway.
• Taxiway centerline lead-on lights – installed the same way as taxiway centerline lead-off Lights, but directing airplane traffic in the opposite direction.
• Land and hold short lights – a row of white pulsating lights installed across the runway to indicate hold short position on some runways that are facilitating land and hold short operations
• Approach lighting system (ALS) – a lighting system installed on the approach end of an airport runway and consists of a series of lightbars, strobe lights, or a combination of the two that
extends outward from the runway end.
According to Transport Canada's regulations,^[32] the runway-edge lighting must be visible for at least 2 mi (3 km). Additionally, a new system of advisory lighting, runway status lights, is
currently being tested in the United States.^[33]
The edge lights must be arranged such that:
• the minimum distance between lines is 75 ft (23 m), and maximum is 200 ft (61 m)
• the maximum distance between lights within each line is 200 ft (61 m)
• the minimum length of parallel lines is 1,400 ft (427 m)
• the minimum number of lights in the line is 8.^[34]
Approach lighting system at Berlin Tegel Airport
Control of lighting system
Typically the lights are controlled by a control tower, a flight service station or another designated authority. Some airports/airfields (particularly uncontrolled ones) are equipped with
pilot-controlled lighting, so that pilots can temporarily turn on the lights when the relevant authority is not available.^[35] This avoids the need for automatic systems or staff to turn the lights
on at night or in other low visibility situations. This also avoids the cost of having the lighting system on for extended periods. Smaller airports may not have lighted runways or runway markings.
Particularly at private airfields for light planes, there may be nothing more than a windsock beside a landing strip.
Types of runway safety incidents include:
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Runway surface at Congonhas Airport in São Paulo, Brazil. The grooves increase friction and reduce the risk of hydroplaning.
The choice of material used to construct the runway depends on the use and the local ground conditions. For a major airport, where the ground conditions permit, the most satisfactory type of pavement
for long-term minimum maintenance is concrete. Although certain airports have used reinforcement in concrete pavements, this is generally found to be unnecessary, with the exception of expansion
joints across the runway where a dowel assembly, which permits relative movement of the concrete slabs, is placed in the concrete. Where it can be anticipated that major settlements of the runway
will occur over the years because of unstable ground conditions, it is preferable to install asphalt concrete surface, as it is easier to patch on a periodic basis. Fields with very low traffic of
light planes may use a sod surface. Some runways make use of salt flats.
For pavement designs, borings are taken to determine the subgrade condition, and based on the relative bearing capacity of the subgrade, the specifications are established. For heavy-duty commercial
aircraft, the pavement thickness, no matter what the top surface, varies from 10 to 48 in (25 to 122 cm), including subgrade.
Airport pavements have been designed by two methods. The first, Westergaard, is based on the assumption that the pavement is an elastic plate supported on a heavy fluid base with a uniform reaction
coefficient known as the K value. Experience has shown that the K values on which the formula was developed are not applicable for newer aircraft with very large footprint pressures.
The second method is called the California bearing ratio and was developed in the late 1940s. It is an extrapolation of the original test results, which are not applicable to modern aircraft
pavements or to modern aircraft landing gear. Some designs were made by a mixture of these two design theories. A more recent method is an analytical system based on the introduction of vehicle
response as an important design parameter. Essentially it takes into account all factors, including the traffic conditions, service life, materials used in the construction, and, especially
important, the dynamic response of the vehicles using the landing area.
Because airport pavement construction is so expensive, manufacturers aim to minimize aircraft stresses on the pavement. Manufacturers of the larger planes design landing gear so that the weight of
the plane is supported on larger and more numerous tires. Attention is also paid to the characteristics of the landing gear itself, so that adverse effects on the pavement are minimized. Sometimes it
is possible to reinforce a pavement for higher loading by applying an overlay of asphaltic concrete or portland cement concrete that is bonded to the original slab. Post-tensioning concrete has been
developed for the runway surface. This permits the use of thinner pavements and should result in longer concrete pavement life. Because of the susceptibility of thinner pavements to frost heave, this
process is generally applicable only where there is no appreciable frost action.
Pavement surface
A Mahan Air Airbus A310 using reverse thrust in rainy weather at Düsseldorf Airport
Runway pavement surface is prepared and maintained to maximize friction for wheel braking. To minimize hydroplaning following heavy rain, the pavement surface is usually grooved so that the surface
water film flows into the grooves and the peaks between grooves will still be in contact with the aircraft tyres. To maintain the macrotexturing built into the runway by the grooves, maintenance
crews engage in airfield rubber removal or hydrocleaning in order to meet required FAA, or other aviation authority friction levels.
Pavement subsurface drainage and underdrains
Subsurface underdrains help provide extended life and excellent and reliable pavement performance. At the Hartsfield Atlanta, GA airport the underdrains usually consist of trenches 18 in (46 cm) wide
and 48 in (120 cm) deep from the top of the pavement. A perforated plastic tube (5.9 in (15 cm) in diameter) is placed at the bottom of the ditch. The ditches are filled with gravel size crushed
stone.^[36] Excessive moisture under a concrete pavement can cause pumping, cracking, and joint failure.^[37]
Surface type codes
The grass airstrip on the Badminton estate, Badminton, South Gloucestershire, England. The strip is very simple: no lighting, no centerline, and no approach aids. The edge is marked by simple posts.
In aviation charts, the surface type is usually abbreviated to a three-letter code.
The most common hard surface types are asphalt and concrete. The most common soft surface types are grass and gravel.
Abbreviation Meaning
ASP Asphalt
BIT Bituminous asphalt or tarmac
BRI Bricks (no longer in use, covered with asphalt or concrete now)
CLA Clay
COM Composite
CON Concrete
COP Composite
COR Coral (fine crushed coral reef structures)
GRE Graded or rolled earth, grass on graded earth
GRS Grass or earth not graded or rolled
GVL Gravel
ICE Ice
LAT Laterite
MAC Macadam
PEM Partially concrete, asphalt or bitumen-bound macadam
PER Permanent surface, details unknown
PSP Marston Matting (derived from pierced/perforated steel planking)
SAN Sand
SMT Sommerfeld Tracking
SNO Snow
U Unknown surface
WAT Water
This section
needs additional citations for verification
(June 2022)
A runway of at least 1,800 m (5,900 ft) in length is usually adequate for aircraft weights below approximately 100,000 kg (220,000 lb). Larger aircraft including widebodies will usually require at
least 2,400 m (7,900 ft) at sea level. International widebody flights, which carry substantial amounts of fuel and are therefore heavier, may also have landing requirements of 3,200 m (10,500 ft) or
more and takeoff requirements of 4,000 m (13,000 ft). The Boeing 747 is considered to have the longest takeoff distance of the more common aircraft types and has set the standard for runway lengths
of larger international airports.^[38]
At sea level, 3,200 m (10,500 ft) can be considered an adequate length to land virtually any aircraft. For example, at O'Hare International Airport, when landing simultaneously on 4L/22R and 10/28 or
parallel 9R/27L, it is routine for arrivals from East Asia, which would normally be vectored for 4L/22R (2,300 m (7,546 ft)) or 9R/27L (2,400 m (7,874 ft)) to request 28R (4,000 m (13,123 ft)). It is
always accommodated, although occasionally with a delay. Another example is that the Luleå Airport in Sweden was extended to 3,500 m (11,483 ft) to allow any fully loaded freight aircraft to take
off. These distances are also influenced by the runway grade (slope) such that, for example, each 1 percent of runway down slope increases the landing distance by 10 percent.^[39]
An aircraft taking off at a higher altitude must do so at reduced weight due to decreased density of air at higher altitudes, which reduces engine power and wing lift. An aircraft must also take off
at a reduced weight in hotter or more humid conditions (see density altitude). Most commercial aircraft carry manufacturer's tables showing the adjustments required for a given temperature.
In India, recommendations of International Civil Aviation Organization (ICAO) are now followed more often. For landing, only altitude correction is done for runway length whereas for take-off, all
types of correction are taken into consideration.^[40]
• In the 1980s,
Leeds Bradford International Airport
extended its runway to take
wide-body aircraft
by building an
over the A658 road.
Road crossing of (
) A970 with
Sumburgh Airport
's runway. The movable barrier closes when aircraft land or take off.
• Gibraltar International Airport
's runway 09/27, used to be crossed by the one road between Gibraltar and Spain.
• A parachute may be used to slow down craft, in this case the
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Lower bounds on expressions dependent on functions
Lower bounds on expressions dependent on functions φ(n), ψ(n) and σ(n), II
Stoyan Dimitrov
Notes on Number Theory and Discrete Mathematics
Print ISSN 1310–5132, Online ISSN 2367–8275
Volume 30, 2024, Number 3, Pages 547–556
DOI: 10.7546/nntdm.2024.30.3.547-556
Full paper (PDF, 195 Kb)
Authors and affiliations
Stoyan Dimitrov
Faculty of Applied Mathematics and Informatics, Technical University of Sofia
8 St. Kliment Ohridski Blvd., Sofia 1756, Bulgaria
Department of Bioinformatics and Mathematical Modelling, Institute of Biophysics and Biomedical Engineering, Bulgarian Academy of Sciences
Acad. G. Bonchev Str., Bl. 105, Sofia 1113, Bulgaria
In this paper we establish lower bounds on several expressions dependent on functions
• Arithmetic functions
• Lower bounds
2020 Mathematics Subject Classification
1. Atanassov, K. (2011). Note on φ, ψ and σ-functions. Part 3. Notes on Number Theory and Discrete Mathematics, 17(3), 13–14.
2. Atanassov, K. (2013). Note on φ, ψ and σ-functions. Part 6. Notes on Number Theory and Discrete Mathematics, 19(1), 22–24.
3. Dimitrov, S. (2023). Lower bounds on expressions dependent on functions φ(n), ψ(n) and σ(n). Notes on Number Theory and Discrete Mathematics, 29(4), 713–716.
4. Sándor, J. (2014). On certain inequalities for φ, ψ, σ and related functions. Notes on Number Theory and Discrete Mathematics, 20(2), 52–60.
5. Sándor, J., & Atanassov, K. (2019). Inequalities between the arithmetic functions φ, ψ and σ. Part 2. Notes on Number Theory and Discrete Mathematics, 25(2), 30–35.
6. Sándor, J., Mitrinović, D. S., & Crstici, B. (2006). Handbook of Number Theory I. Springer.
7. Sándor, J., & Tóth, L. (1990). On certain number-theoretic inequalities. The Fibonacci Quarterly, 28(3), 255–258.
Manuscript history
• Received: 18 November 2023
• Accepted: 1 October 2024
• Online First: 3 October 2024
Copyright information
This is an Open Access paper distributed under the terms and conditions of the Creative Commons Attribution 4.0 International License (CC BY 4.0).
Cite this paper
Dimitrov, S. (2024). Lower bounds on expressions dependent on functions φ(n), ψ(n) and σ(n), II. Notes on Number Theory and Discrete Mathematics, 30(3), 547-556, DOI: 10.7546/nntdm.2024.30.3.547-556. | {"url":"https://nntdm.net/volume-30-2024/number-3/547-556/","timestamp":"2024-11-08T05:41:42Z","content_type":"text/html","content_length":"36145","record_id":"<urn:uuid:3056d098-cc45-4cd0-8335-206b2a9672a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00163.warc.gz"} |
The Analytics Edge
开始时间: 04/22/2022 持续时间: 12 weeks
课程主页: https://www.edx.org/archive/analytics-edge-mitx-15-071x-0
In the last decade, the amount of data available to organizations has reached unprecedented levels. Data is transforming business, social interactions, and the future of our society. In this course,
you will learn how to use data and analytics to give an edge to your career and your life. We will examine real world examples of how analytics have been used to significantly improve a business or
industry. These examples include Moneyball, eHarmony, the Framingham Heart Study, Twitter, IBM Watson, and Netflix. Through these examples and many more, we will teach you the following analytics
methods: linear regression, logistic regression, trees, text analytics, clustering, visualization, and optimization. We will be using the statistical software R to build models and work with data.
The contents of this course are essentially the same as those of the corresponding MIT class (The Analytics Edge). It is a challenging class, but it will enable you to apply analytics to real-world
The class will consist of lecture videos, which are broken into small pieces, usually between 4 and 8 minutes each. After each lecture piece, we will ask you a “quick question” to assess your
understanding of the material. There will also be a recitation, in which one of the teaching assistants will go over the methods introduced with a new example and data set. Each week will have a
homework assignment that involves working in R or LibreOffice with various data sets. (R is a free statistical and computing software environment we’ll use in the course. See the Software FAQ below
for more info). In the middle of the class, we will run an analytics competition, and at the end of the class there will be a final exam, which will be similar to the homework assignments.
What do I need to know about the topic prior to enrolling in the course?
You only need to know basic mathematics. For most people, this is equivalent to basic high school mathematics. You should know concepts like mean, standard deviation, and histograms. This course is
also useful for those who already have experience in the subject. In each lecture, recitation, and homework assignment, we use a different dataset and case to illustrate the method. Even if you are
familiar with all of the methods taught, you can still learn a lot from the different examples.
What software will be used in the course?
We’ll be using two software programs in this class: R and LibreOffice. Both are free online, and you don’t need to be familiar with either of them to take the course. R is a free statistical and
computing software environment and LibreOffice is similar to MS Office but a free open source program. Specifically we’ll use the LibreOffice module, Calc in this course. Don’t worry though - we’ll
teach everything from scratch!
• An applied understanding of many different analytics methods, including linear regression, logistic regression, CART, clustering, and data visualization
• How to implement all of these methods in R
• An applied understanding of mathematical optimization and how to solve optimization models in spreadsheet software | {"url":"https://coursegraph.com/edx-course-course-v1-MITx%2B15-071x_2a%2B2T2015","timestamp":"2024-11-12T09:31:03Z","content_type":"text/html","content_length":"11805","record_id":"<urn:uuid:32de0ac1-19f9-4c23-98bb-54f9a4c77dc1>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00178.warc.gz"} |
If A=[515−25251],B=(1i01), i=−1 , and Q=A⊤BA, then... | Filo
If , , and , then the inverse of the matrix is equal to :
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Now let
inverse of
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Question Text If , , and , then the inverse of the matrix is equal to :
Updated On Mar 2, 2023
Topic Determinants
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 3
Upvotes 386
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Modified gravity
Since the dark energy problem has so far defied all attempts to find a simple and convincing solution, scientists have started to wonder whether the observed accelerated expansion of the universe is
really due to some kind of extra fluid-like contribution, or rather a sign that General Relativity (GR) itself breaks down on large scales. This could happen in a variety of ways. For example, the
universe itself could be higher-dimensional, all matter only existing on a three-dimensional hypersurface or brane, a scenario that is connect to String Theory. Alternatively, gravitons (the
particles associated to the gravitational force) could be massive which might lead to a weakening of gravity over large distances -- such a scenario may be possible with the help of Galileons.
Another possibility is that the Einstein-Hilbert action that describes GR is incomplete and needs to be modified in some way, leading to so-called f(R) and similar approaches, which in turn are
linked to scalar-tensor theories. Unfortunately modifications of gravity suffer from problems similar to the ones affecting dark energy models: within an overabundance of ideas and possibilities,
there is yet a lack of a really compelling scenario.
For this reason many cosmologists focus on experimental tests and characterizations of the dark sector: we can always cast the effective 3+1 dimensional equations that link the geometry of space-time
to the matter content of the universe into the form of the usual Einstein equation of GR, but with an extra energy-momentum tensor (EMT) on the right hand side, into which we put all the extra
contributions that might be due to a modification of GR. The goal then becomes on the one hand to predict how this extra EMT looks like (for example, modified gravity scenarios appear to generically
create extra anisotropic stresses) and on the other hand to measure it with the help of observations. Two missions that our group are actively involved in are the Planck satellite to measure the CMB
and the Euclid satellite to observe the large-scale structure with the help of a galaxy survey and weak lensing measurements.
Finally it should be mentioned that modifications of gravity have also been used to replace dark matter and to explain the flatness of the rotation curves of gravity, most notably with the help of
MOND (modified Newtonian dynamics) and its relativistic extension TeVeS. This is not really an active research topic of our group at this time, and so we restrict the meaning of Modified Gravity
scenarios to the above definition, i.e. as a way to explain the dark energy.
Recent publications and presentations on this topic
Date: 27. November 2014
Members involved: Yves Dirian, Stefano Foffa, Martin Kunz, Michele Maggiore, Valeria Pettorino
Topics: Dark energy, Modified gravity
Date: 22. August 2014
Members involved: Yves Dirian, Ermis Mitsou
Topics: Dark energy, Modified gravity
Date: 2. July 2014
Members involved: Enea Di Dio, Ruth Durrer, Giovanni Marozzi, Francesco Montanari
Topics: Large-scale structure, General relativity, Modified gravity
Date: 27. June 2014
Members involved: Martin Kunz, Ignacy Sawicki
Topics: Anisotropic stress, Modified gravity, Gravitational waves, Large-scale structure, Dark energy
Date: 4. June 2014
Members involved: Giovanni Marozzi
Topics: Large-scale structure, General relativity, Modified gravity, Dark energy
Date: 28. March 2014
Members involved: Hideki Perrier
Topics: Large-scale structure, Consistency relations, Modified gravity, Non-local bias, Halo Model
Type: Presentation
Date: 7. February 2014
Members involved: Hideki Perrier
Topics: Large-scale structure, Modified gravity, Bias, Non-local bias, Consistency relations
Type: Presentation
Date: 13. May 2013
Members involved: Maud Jaccard, Michele Maggiore, Ermis Mitsou
Topics: Modified gravity
Date: 12. March 2013
Members involved: Maud Jaccard, Michele Maggiore, Ermis Mitsou
Topics: Modified gravity
Date: 29. May 2012
Members involved: Lukas Hollenstein, Rajeev Kumar Jain, Martin Kunz
Topics: Dark energy, Modified gravity, Large-scale structure
Département de Physique Théorique
Université de Genève
24, quai Ernest Ansermet
1211 Genève 4
Directions & contact | {"url":"https://cosmology.unige.ch/topics/mg?page=5","timestamp":"2024-11-03T04:13:23Z","content_type":"text/html","content_length":"57336","record_id":"<urn:uuid:73695677-b1cd-4936-8c0c-46b65e92df7d>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00363.warc.gz"} |
Putting the Rotorcraft/Accelerometer Question to Rest
There has been an ongoing discussion/debate about whether accelerometers are actually useful for attitude estimation on flying rotorcraft (quadrotors, in particular). The simple, one sentence answer
is "yes, they are."
However, the accelerometer measurements are non-intuitive. If the estimation algorithm uses the accelerometers incorrectly (which is, incidentally, the norm on hobby and even some professional
aircraft) then there will be significant errors.
There is some good literature available now that describes the problem. Martin and Salaun published a very good paper describing quadrotor flight dynamics. I also just finished my MS thesis,
wherein I address the subject in detail.
In my thesis (the last link), Chapter 5 is what you'll want to look at.
You need to be a member of diydrones to add comments!
Thanks for your helpful post Caleb. Would you possibly know an answer to the following question:
Attitude estimation with ONLY 3axis ACCELEROMETERS. and nothing else, based on a lowpass filter? where can I find an opensource way to do this?
• Developer
Hi Caleb,
Nice analysis. Of course there is always more work to be done... But it looks like a nice step in the right direction.
You appear to have an error in formula 5.15. Fortunately it does not appear to carry forward anywhere.
I can follow the math but cannot intuitively see how the LPF filter model corrects the phase lag between the accelerometer estimated and actual angles. Do you have an intuitive understanding of
this that you can put in words?
Thanks Caleb for your great work and for being willing to share it with us.
FYI, Salaun's PhD thesis is also available (its mostly in English - don't be fooled by the first few pages)
- Roy
This reply was deleted. | {"url":"https://diydrones.com/forum/topics/putting-the","timestamp":"2024-11-07T20:38:13Z","content_type":"text/html","content_length":"77392","record_id":"<urn:uuid:7355493d-1ebe-4b98-b615-606f72ade310>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00513.warc.gz"} |
Unit 8
Lesson 5
Order Numbers 1-20 (optional)
Warm-up: Choral Count: Count Backward (10 minutes)
The purpose of this warm-up is for students to practice counting backward from 10 and 20. By counting backward from 10 to 1, students develop fluency with the count sequence to 10, which will be
helpful later in this lesson when order numbers 1–20 and find one more and one less than a given number. As students count backward, point to the numbers posted so that students can follow along.
• “Let’s start at 10 and count backward to 1.”
• Record as students count.
• Repeat 3–4 times.
• “Let’s start at 20 and count backward to 1.”
• Record as students count.
• Repeat 2–3 times.
Activity Synthesis
• “If we start at 17 and count backward to 1, what number would we say after 16?”
• “If we start at 11 and count backward to 1, what number would we say after 11?”
Activity 1: Order Numbers (10 minutes)
The purpose of this activity is for students to put numbers 1–20 in order.
MLR8 Discussion Supports. Invite each group to chorally read numbers 1–20 in order once the group agrees on the order. Listen for and clarify questions.
Advances: Speaking, Conversing
Required Preparation
• Create a set of cards from the blackline master for each group of 4.
• Groups of 4
• Give each group of students a set of number cards, not in sequence.
• “Work with your group to put the numbers in order from 1 to 20.”
• 3 minutes: small-group work time
• “Once your group agrees that the numbers are in the correct order, write the numbers in order from 1 to 20.”
• 3 minutes: independent work time
Student Facing
Write the numbers in order.
Activity Synthesis
• “How did you decide the order of the numbers?”
• Display numbers 1–12 in order.
• “Clare is putting the numbers in order. What would you say to her to help her find the number that comes next?” (Look for 13 because it comes after 12. Look for the number that has a 3 because
the last number has a 2.)
Activity 2: Number Clues (15 minutes)
The purpose of this activity is for students to use their knowledge of the number sequence to find the number that matches clues with one more and one less. Numbers 1–20 are displayed around the
room. The numbers may be displayed in order, or the sequence of numbers can be mixed up for more of a challenge. The cards with number clues are included as a blackline master; one example is
included in the student-facing task statement. Consider developing a signal, such as ringing a bell, so that students know when to move around the classroom. In the activity synthesis, students
consider why there can be different clues that match a given number.
Engagement: Internalize Self-Regulation. Provide students an opportunity to self-assess and reflect on the number clue and if that number clue matches the number they will stand by. For example,
students can choral count together to check that the number 9 is 1 less than 10.
Supports accessibility for: Memory, Conceptual Processing
Required Preparation
• Each group of 2 needs one card from the blackline master.
• Display large numbers 1 to 20 around the room. The numbers may be displayed in order, or the sequence of numbers can be mixed up for more of a challenge.
• Groups of 2
• Display number cards 0–20 around the room, in order or mixed up.
• Give each group of students a card.
• “On your card there are some clues. Each clue says ‘1 more than _____’ or ‘1 less than _____.’”
• “Look at your first clue. Decide with your partner which number matches the clue. Then look around the room and find that number. When I give the signal, walk over to that number.”
• 2 minutes: partner work time
• “Discuss with the other people at your number. Did you have the same clue? Are you all at the correct number? How do you know?”
• 2 minutes: small-group work time
• Repeat the steps with the rest of the clues.
• During the final round, monitor for a group that notices there are 2 different clues that match the number and discuss why they both match.
Activity Synthesis
• Invite previously identified students to share what they noticed during the last round of number clues.
• “‘1 more than 15’ and ‘1 less than 17’ are both clues to 16.”
• “What are two different clues that could match the number 12?”
Activity 3: Centers: Choice Time (20 minutes)
The purpose of this activity is for students to choose from activities that offer practice with number and shape concepts. Students choose from 5 centers introduced in previous units. Students can
choose to work at any stage of the centers.
• Less, Same, More
• Math Fingers
• Tower Build
• Math Stories
• Which One
Required Preparation
• Gather materials from:
□ Less, Same, More
□ Math Fingers
□ Tower Build
□ Math Stories
□ Which One
• Groups of 2
• “Today we are going to choose from centers we have already learned.”
• Display the center choices in the student book.
• “Think about what you would like to do first.”
• 30 seconds: quiet think time
• Invite students to work at the center of their choice.
• 8 minutes: center work time
• “Choose what you would like to do next.”
• 8 minutes: center work time
Activity Synthesis
• “In math class, it’s important to be able to explain your thinking. Describe a time when you were able to explain your ideas to other people in your class.”
Lesson Synthesis
Display number cards 1–20 in order, but with 16 missing.
“I put my numbers in order from 1 to 20, but one of my numbers went missing. Which number is missing and how do you know?” (16 is missing. The missing number comes after 15. The missing number is 1
more than 15. The missing number is 1 less than 17.)
Cool-down: Unit 8, Section A Checkpoint (0 minutes)
Student Facing
In this section, we counted and compared groups of objects.
There are 14 counters and 12 cubes. There are fewer cubes.
We also used what we know about counting to help us figure out 1 more and 1 less.
1, 2, 3, 4, 5, 6, 7, 8
8 is 1 more than 7.
There were 10 people on the bus.
Then 1 person got off the bus.
How many people are on the bus now?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
9 comes before 10 when we count, so 9 is 1 less than 10. | {"url":"https://im.kendallhunt.com/k5/teachers/kindergarten/unit-8/lesson-5/lesson.html","timestamp":"2024-11-01T23:53:57Z","content_type":"text/html","content_length":"101776","record_id":"<urn:uuid:bcff00ef-74a1-4b3b-bf3f-5e72674e6b70>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00740.warc.gz"} |
Calculating Digits of Pi with Collisions
#139 — Calculating Digits of Pi with Collisions
Happy Pi Day! To celebrate I attempt to compute the digits of Pi using the “collisions” method, thanks to 3Blue1Brown for the idea!
• Paper by G. Galperin on counting collisions between two billiard balls to estimate pi
• Wikipedia page on elastic collisions
• Wikipedia page on Euler method
• GitHub repo to Pi Day (2019) coding challenges
• 3Blue1Brown's playlist on estimating Pi from block collisions
• Numberphile's video on computing pi from bouncing balls
• Archive of the member/patron exclusive Pi Day prep livestream | {"url":"https://deploy-preview-644--codingtrain.netlify.app/challenges/139-calculating-digits-of-pi-with-collisions/","timestamp":"2024-11-03T20:22:30Z","content_type":"text/html","content_length":"326482","record_id":"<urn:uuid:bc9cd828-a372-4951-a560-d1656b34b8af>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00699.warc.gz"} |
Extreme Waves
Statistical properties of soliton fields during maximal focusing Authors: T.V.Tarasova, A.V.Slunyaev For the efficient description of the soliton gas dynamics kinetic equations were derived [1,2].
They characterize the transport of the soliton spectral density, but due to the violation of the wave linear superposition principle, do not provide information about the wave solution itself (which
can be water surface displacement, intensity of electromagnetic fields, etc.). In particular, the questions about the probability distribution for wave amplitudes or about the values of the wave
field statistical moments remain unanswered. Multisoliton solutions, which can be formally written in a closed form using the Inverse Scattering Transform or related methods for integrable equations,
are very cumbersome, what makes their analytical and even numerical analysis difficult. The direct numerical simulation of evolution equations is commonly used to study the soliton gas evolution,
which also becomes complicated in the case of a dense gas (i.e. when many solitons interact simultaneously). The focus of this study is made on the dynamics of soliton interactions governed by the
classic Korteweg -- de Vries (KdV) equation. The use of an ultra-high-precision procedure based on the Darboux transformation made it possible to compute the exact $N$-soliton solutions $u_N$ when
$N$ is large [3] and to calculate their statistical moments $\mu_n (t) = \int_{-\infty}^{+\infty} {u_N^n(x,t) dx}$, $n \in \mathbb{N}$, with high accuracy. In this work we present a general idea that
dense ensembles of KdV-type solitons of the same sign can be considered as strongly-nonlinear / small-dispersion wave states, what allows to express the statistical moments in terms of the spectral
parameters of the associated scattering problem analytically [4]. A particular case when dense soliton states can occur is synchronous multisoliton collisions, for which $u_N(-x,-t)=u_N(x,t)$ holds.
This symmetry condition should correspond in some sense to the maximal focusing of solitons at $t = 0$ at the point $x = 0$. The soliton amplitudes are set decaying exponentially, so that they form a
geometric series with the ratio $d>1$, $A_j = 1/d^{j-1}$, $j=1,...,N$. Time dependences of statistical moments are investigated for many-soliton solutions. It is shown that during the interaction of
solitons of the same sign the wave field is effectively smoothed out. When $d$ is sufficiently close to 1, and $N$ is large, the statistical moments remain approximately constant within long time
spans, when the solitons are located most densely. This quasi-stationary state is characterized by greatly reduced statistical moments and by the density of solitons close to some critical value.
This state may be treated as the small-dispersion limit, what makes it possible to analytically estimate all high-order statistical moments. While the focus of the study is made on the Korteweg--de
Vries equation and its modified version, a much broader applicability of the results to equations that support soliton-type solutions is discussed. \textbf{References} \myitem [1] V. E. Zakharov,
``Kinetic equation for solitons,'' JETP 33, 538-541 (1971). \myitem [2] G. A. El and A. M. Kamchatnov, ``Kinetic equation for a dense soliton gas'', Phys. Rev. Lett. 95, 204101 (2005). \myitem [3]
T.V. Tarasova, A.V. Slunyaev, Properties of synchronous collisions of solitons in the Korteweg -- de Vries equation. Communications in Nonlinear Science and Numerical Simulation 118, 107048 (2022).
doi: 10.1016/j.cnsns.2022.107048 \myitem [4] A.V. Slunyaev, T.V. Tarasova, Statistical properties of extreme soliton collisions. Chaos 32, 101102 (2022). doi: 10.1063/5.0120404 | {"url":"https://www.pks.mpg.de/extrem23/poster-contributions","timestamp":"2024-11-03T07:33:12Z","content_type":"text/html","content_length":"121073","record_id":"<urn:uuid:f8a3c513-3a50-4d35-9a31-f607c73293a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00117.warc.gz"} |
How do you find an equation of the tangent line to the curve at the given point y=tan^2(3x) and x=pi/4? | HIX Tutor
How do you find an equation of the tangent line to the curve at the given point #y=tan^2(3x)# and #x=pi/4#?
Answer 1
The equation of the tangent at $\left(\frac{\pi}{4} , 1\right)$ is $y = - 12 x + \left(3 \pi + 1\right)$
#y=tan^2(3x)# at #x=pi/4 ; y = tan^2(3*pi/4) =(-1)^2=1 :.#
at #(pi/4,1)# the tangent is drawn.
#:. y^'=2tan(3x)*sec^2(3x).3= 6tan(3x)*sec^2(3x) #
at #x=pi/4 , y^'= 6tan((3pi)/4)*sec^2((3pi)/4) # or
#y^'= 6* (-1)*(-sqrt2)^2= -12 :. # At #x=pi/4#, the slope of
the tangent is #m=-12 :.# the equation of the tangent at
#(pi/4,1)# is #y-y_1=m(x-x_1) or y-1= -12(x-pi/4)# or
# y= -12(x-pi/4) +1 or y= -12x +(3pi+1)# [Ans]
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Answer 2
To find the equation of the tangent line to the curve y = tan^2(3x) at the point x = π/4, we need to find the derivative of the function and evaluate it at x = π/4.
First, let's find the derivative of y = tan^2(3x) using the chain rule.
dy/dx = 2tan(3x) * sec^2(3x) * 3
Now, substitute x = π/4 into the derivative expression to find the slope of the tangent line at that point.
dy/dx = 2tan(3(π/4)) * sec^2(3(π/4)) * 3
Evaluate the trigonometric functions:
dy/dx = 2tan(3π/4) * sec^2(3π/4) * 3
Simplify the trigonometric expressions:
dy/dx = 2(-1) * (2/√2)^2 * 3
dy/dx = -8 * 3
dy/dx = -24
The slope of the tangent line at x = π/4 is -24.
Now, we can use the point-slope form of a linear equation to find the equation of the tangent line.
y - y1 = m(x - x1)
Substitute the values of the point (x1, y1) = (π/4, tan^2(3π/4)) and the slope m = -24 into the equation:
y - tan^2(3π/4) = -24(x - π/4)
This is the equation of the tangent line to the curve y = tan^2(3x) at the point x = π/4.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Excel Formula: How to use the Excel OR function - Excelchat
We can use Excel OR function to simultaneously test multiple logical conditions. For each argument, the OR function will give TRUE or FALSE. This article will step through the use of the OR function.
Figure 1: Final result
Syntax of the formula
=OR(logical1, [logical2]….)
The parameters
Logical1- refers to first condition or logical value to evaluate
Logical2- refers to second condition or logical value to evaluate. It is optional.
We will get either of two results: If the argument is true and the condition met, the OR function will return TRUE. However, on the other hand, if the condition is not met, then the OR function will
return FALSE.
This function can be used to test multiple conditions, up to 255 at a time.
Using OR to extend functionality of IF function
We can also use the OR function to extend the functionality of the IF function. To better understand this, consider the example below:
Figure 2: Extending functionality of IF function
Array form of the OR function
One can test all values in a given range against one condition by using the array form of the OR function. The test will return either TRUE if any cell in the given range certifies the condition
The OR function will ignore any empty cell or cells with text values.
Again, if the range selected does not contain any logical values, the OR function will return #Value.
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100×100 Multiplication Chart | Multiplication Chart Printable
100×100 Multiplication Chart
Multiplication Chart Up To One Hundred Multiplication Chart Download
100×100 Multiplication Chart
100×100 Multiplication Chart – A Multiplication Chart is a helpful tool for kids to discover exactly how to increase, divide, as well as discover the tiniest number. There are lots of usages for a
Multiplication Chart.
What is Multiplication Chart Printable?
A multiplication chart can be used to aid children learn their multiplication truths. Multiplication charts been available in numerous types, from full page times tables to solitary page ones. While
individual tables serve for offering portions of details, a full web page chart makes it simpler to review realities that have currently been grasped.
The multiplication chart will typically feature a leading row and also a left column. The leading row will certainly have a listing of items. When you wish to find the product of two numbers, pick
the first number from the left column and the 2nd number from the top row. Move them along the row or down the column up until you reach the square where the two numbers satisfy as soon as you have
these numbers. You will then have your product.
Multiplication charts are valuable knowing tools for both youngsters and also grownups. 100×100 Multiplication Chart are available on the Internet and also can be printed out and laminated flooring
for resilience.
Why Do We Use a Multiplication Chart?
A multiplication chart is a representation that shows exactly how to multiply 2 numbers. You pick the first number in the left column, relocate it down the column, and also then select the 2nd number
from the leading row.
Multiplication charts are helpful for several factors, including assisting kids discover exactly how to separate as well as simplify fractions. They can additionally help youngsters learn how to pick
an efficient common denominator. Due to the fact that they serve as a constant reminder of the trainee’s progression, multiplication charts can also be handy as workdesk sources. These tools help us
develop independent learners who understand the fundamental concepts of multiplication.
Multiplication charts are likewise helpful for helping students memorize their times tables. As with any skill, remembering multiplication tables takes time and technique.
100×100 Multiplication Chart
Free Printable Multiplication Chart 100X100 Free Printable
Multiplication Chart To 100 Free Printable Multiplication Chart
Printable 100X100 Multiplication Table PrintableMultiplication
100×100 Multiplication Chart
If you’re looking for 100×100 Multiplication Chart, you’ve come to the right area. Multiplication charts are available in various layouts, consisting of complete size, half size, as well as a
selection of charming layouts.
Multiplication charts as well as tables are indispensable tools for youngsters’s education. These charts are excellent for usage in homeschool math binders or as classroom posters.
A 100×100 Multiplication Chart is an useful tool to enhance mathematics facts as well as can aid a youngster learn multiplication rapidly. It’s also a great tool for miss checking and also
discovering the times tables.
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Dynamic Under/Over Constructions -
The scenario is that you want a two-code variable where c1=under a particular threshold, and c2=same as or over the threshold. The problem is: what if the threshold value is dynamic, that is, not
known at scripting time? The solution is to run a table to get the threshold, and then plug that value into the construction.
This example gets the code median from FAVRAT_1, constructs a new variable for under/over the median, and then verifies the construction by a table.
#include c:rubyRubyUtilitiesLibrary.vbs
#include c:rubyRubyVariablesLibrary.vbs
#include c:rubyRubyReportsLibrary.vbs
Sub Main()
top = "Count"
filt = ""
wght = ""
'' get the median
GenTab "FavRat_1 Median", top, "FAVRAT_1", filt, wght
median = rep.GetRawStat("Column", "cme", 0, 0) '' return cme for column at cell 0,0 = 7
'MsgBox median '' debug
'' construct under/over
DefCon "FAVRAT_1_UO", "Favourability Rating Segment by Median"
AddItem 1, "sum_FAVRAT_1(*)<" & median, "Under"
AddItem 2, "sum_FAVRAT_1(*)>=" & median, "Equal or Over"
Construct "FAVRAT_1_UO"
'' confirm by a comparison table
side = "FAVRAT_1(#sum#(1/" & median-1 & ");#sum#(" & median & "/10);cme),FAVRAT_1_UO"
GenTab "FavRat_1 Median UO Check", top, side, filt, wght
End Sub
The final table is:
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Of the 15 cars in a parking lot, 5 are green, 2 are gold, and the rest are black. What are the odds against the nest car leaving the lot being black?
Find an answer to your question 👍 “Of the 15 cars in a parking lot, 5 are green, 2 are gold, and the rest are black. What are the odds against the nest car leaving the lot ...” in 📗 Mathematics if
the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
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What are the contributions of aryabhatta?
What are the contributions of aryabhatta?
Notable Explanation of lunar eclipse and solar eclipse, rotation of Earth on its axis, reflection of light by moon, sinusoidal functions, solution of single variable quadratic equation, value of π
ideas correct to 4 decimal places, diameter of Earth, calculation of the length of sidereal year
What are the contributions of Ramanujan?
Ramanujan’s other notable contributions include hypergeometric series, the Riemann series, the elliptic integrals, the theory of divergent series, and the functional equations of the zeta function.
What was brahmagupta’s most significant contribution to the study of numbers?
One of the most significant input of Brahmagupta to mathematics was the introduction of ‘zero’ to the number system which stood for ‘nothing’. His work the ‘Brahmasphutasiddhanta’ contained many
mathematical findings written in verse form. It had many rules of arithmetic which is part of the mathematical solutions now.
What did aryabhata discover in mathematics?
What did Aryabhata discover? Aryabhata discovered an approximation of pi, 62832/20000 = 3.1416. He also correctly believed that the planets and the Moon shine by reflected sunlight and that the
motion of the stars is due to Earth’s rotation.
What contribution did aryabhatta make in the field of science astronomy and mathematics?
Aryabhatta discovered zero decimal system and calculated the value of pi (3.1416) and area of a triangle in mathematics; a movement of earth and sun in the astronomy.
What Ramanujan contribute to mathematics?
Srinivasa Ramanujan was one of India’s greatest mathematical geniuses. He made substantial contributions to the analytical theory of numbers and worked on elliptic functions, continued fractions, and
infinite series. In 1900 he began to work on his own on mathematics summing geometric and arithmetic series.
What did Bhaskara invent?
On 7 June 1979 the Indian Space Research Organisation launched Bhaskara I honouring the mathematician….
Bhāskara I
Occupation Mathematician; scientist
Known for Bhaskara I’s sine approximation formula
What is the contribution of Satyendra Nath Bose in mathematics?
Satyendra Nath Bose, (born January 1, 1894, Calcutta [now Kolkata], India—died February 4, 1974, Calcutta), Indian mathematician and physicist noted for his collaboration with Albert Einstein in
developing a theory regarding the gaslike qualities of electromagnetic radiation (see Bose-Einstein statistics).
What is Ramanujan best known for?
An intuitive mathematical genius, Ramanujan’s discoveries have influenced several areas of mathematics, but he is probably most famous for his contributions to number theory and infinite series,
among them fascinating formulas ( pdf ) that can be used to calculate digits of pi in unusual ways.
How did Aryabhatta introduce the concept of numerals?
In Aryabhatiya, Aryabhatta introduced a system of numerals in which he used letters of the Indian alphabet to denote numbers. His numeral system allowed numbers up to 10 18 to be represented with an
alphabetical notation. It is considered that Aryabhatta was familiar with the concept of zero and the place value system.
Who is Aryabhata and what did he do?
Aryabhata, born in 476 CE, was the first in the line of great mathematician-astronomers from the classical age of Indian mathematics and Indian astronomy. There is a general tendency to misspell his
name as “ Aryabhatta ” by analogy with other names having the “ bhatta ” suffix, but all his astronomical text spells his name as Aryabhata.
What was the conclusion of Aryabhatta on Pi?
While he did not use even a symbol for the zero, the French coefficients. conclusion that the pi is irrational. In the next second part of the Aryabhatiyam (gaṇitapāda ayutadvayaviṣkambhasyāsanno
vṛttapariṇāhaḥ. “Add a four to 100, multiply it by eight, and then add a number: 62,000.
Which is a simple problem solved by Aryabhatta?
Aryabhatiya provides simple solutions to complex mathematical problems of the time like summing the first n integers, the squares of these integers and also their cubes. Furthermore, Aryabhatta
correctly calculated the areas of a triangle and of a circle. | {"url":"https://teacherscollegesj.org/what-are-the-contributions-of-aryabhatta/","timestamp":"2024-11-07T21:48:49Z","content_type":"text/html","content_length":"143975","record_id":"<urn:uuid:9f52eb5f-4a1b-4abb-bfb8-d581317613e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00162.warc.gz"} |
On Being a Nerd
I was out walking today. The total time I was out was about an hour and a half, maybe a little bit more. I didn’t actually get that much walking done (although, I walked enough), because I stopped
several times to make notes on a problem I’ve been thinking about lately.
Its of the mathematical kind. Its nothing important and someone who does this more regularly would probably come up with a solution pretty quickly, but I still haven’t found one that satisfies me.
What’s the problem? Its actually Magic: the Gathering related. There’s a card called [scryfall]Duskwatch Recruiter[/scryfall] which allows you to take a look at the top three cards of your deck and
then put them into the bottom of your deck in any order. It does more than that, but for this purpose, this is the key part. Now, assuming we can use that ability as many times as we want to, is it
possible to stack your deck in any order you choose?
Intuitively it feels to me like its possible, as long as the size of the deck isn’t divisible by three (in which case you can’t change the order beyond the three card groups), but its hard to come up
with an algorithm when you are so limited in what you can do. The only place you can twiddle with it, is the order of the three cards you are looking at.
Anyhow, none of this actually matters. What I want to talk about is why I like problems like this. I don’t actually know. I do like to challenge myself intellectually and this is just one more
problem to ponder. I actually don’t even manage to solve most problems I wrestle with, but that’s how I know they are challenging enough.
Now, these days there are plenty of people who identify as nerds, because they do some outwardly nerdy things, like play video games or watch “nerdy” movies. To me, being a nerd has never been about
that. To me being a nerd has always been about accepting that I’m not going to fit into many social circles, because I enjoy stuff like this and I’m not going to hide it. Sure, talking about
something like this gets me eyerolling or laughs from many people, but they will never understand the beauty of math or how satisfying it is to solve any of the 500+ problems on Project Euler,
because they lack the discipline to do it.
The thing is that even if the problem I’m working on right now might not be important, you never know. People like me (actually people who are a lot better mathematicians than myself) actually do
this because they enjoy it, but even if they don’t have immediate practical goals, you never know.
GH Hardy is a mathematician, who wrote an essay titled A Mathematician’s Apology in 1940. His objective was to show how important pure mathematics can be, even if you never discover anything
practical. For example:
No one has yet discovered any warlike purpose to be served by the theory of numbers or relativity, and it seems unlikely that anyone will do so for many years.
War isn’t exactly important (necessary evil, perhaps, but not important), but in 1940, it was kind of something on most people’s minds in the Great Britain and just in a few years both of the example
became moot. Number theory was used to break German codes and theory of relativity was used to build the first nuclear weapons. Today number theory is the basis of all security measures on the web
and theory of relativity doesn’t only produce a significant portion of all the energy used (and should be used more), but is also a key component in communication technology (as satellites in the
orbit are experiencing time just a tad differently, but differently enough to make syncing impossible without taking relativity into account).
My nerdiness might seem (and obviously is) insignificant, but people like me are doing similar work, which might not pay off any time soon, but you never know. Nerds are the builders of the world of
tomorrow. Maybe I should use that as a definition for the word in the future.
2 thoughts on “On Being a Nerd”
1. Partial solution to the problem:
Lemma 1: It is sufficient to be able to move an arbitrary card to whichever position without affecting the order of the rest of the cards.
Lemma 2: If your card is at place n, you can switch it and the card at n+1 or the card at n-1. The rest of the cards do not move in the process.
Combine those two and you probably have a proof. I did not think this through very carefully, though.
The resulting algorithm will be terribly ineffective, though.
2. Being effifient isn’t actually the key here. This is actually a judging problem (which I didn’t mention in the text) and more of a philosophical one. The real problem is in a tournament setting
you can “shortcut” an infinite combo, if you can prove that you can reach a certain state by repeating a set of actions, but you need to be able to prove it in a way that your opponent is
satisfied with the proof. Since math is hard even for many MtG players, the simpler the proof, the better.
I think your approach is good. I’ll put some thought into it, when I have the chance.
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Commutative Property Worksheets - 15 Worksheets.com
Commutative Property Worksheets
About These 15 Worksheets
These worksheets were developed to help students understand and apply the commutative property of addition in mathematical operations. The commutative property states that numbers can be added in any
order and the sum will remain the same. In other words, if you have two numbers, A and B, then A + B will always equal B + A.
They contain a diverse range of exercises tailored to help students understand and apply the commutative property, thereby improving their basic math skills, developing their problem-solving
abilities, and building a solid foundation for future mathematical learning.
Types of Exercises
On such worksheets, students typically encounter a variety of exercises that reinforce this concept. These exercises can take several forms:
Basic Number Swaps – The most straightforward type of exercise asks students to simply swap the order of addends and verify that the sum remains unchanged. For example, the worksheet might present 3
+ 5 and prompt the student to rewrite it as 5 + 3, then perform both additions to see that they equal 8.
Fill in the Blank – These exercises leave one addend blank, and the student must fill in the blank to make two true addition sentences. For example, if the worksheet says 2 + ___ = ___ + 2, the
student would fill in both blanks with the same number.
Matching Pairs – In this type of exercise, students might be presented with a column of addition equations and a column of jumbled sums. Their task is to match each equation with its correct sum,
emphasizing that regardless of the order of addends, the sums will match.
Error Identification – Some worksheets may present students with a series of equations, some of which incorrectly apply the commutative property. Students must identify and correct these mistakes,
reinforcing their understanding of the property. These exercises present equations with missing numbers that students need to find. For instance, an equation might read 4 + ___ = ___ + 6, and
students must figure out that the missing number is the same in both blanks.
Create Your Own Equation – Students may be asked to create pairs of their own addition sentences using a given set of numbers, demonstrating their understanding by actively constructing examples.
The Benefits
This collection of worksheets can significantly enhance students’ understanding of basic math operation skills in several ways:
Enhanced Flexibility in Thinking – By repeatedly practicing the commutative property, students learn that the order of addends does not affect the sum. This foundational understanding is crucial for
flexible thinking in math, allowing them to manipulate and rearrange numbers to simplify complex problems. Worksheets that require students to identify errors or create their equations help develop
critical problem-solving skills. Students learn to check their work for accuracy and to approach problems from different angles.
Increased Speed and Efficiency – As students become more familiar with this property, they can perform additions more quickly and with greater confidence. Knowing that they can add in any order lets
them choose the most efficient way to combine numbers, particularly when dealing with mental math.
Preparation for Advanced Concepts – The commutative property is not only applicable in basic addition but also forms the basis for understanding more advanced mathematical concepts such as the
distributive property and algebraic equations. By working on word problems that incorporate the commutative property, students can connect mathematical operations to real-world situations,
reinforcing the practical value of what they are learning.
Confidence in Mathematics – As students master the commutative property, they often gain confidence in their math abilities. This confidence can lead to a more positive attitude toward math and a
willingness to engage with more challenging problems.
What Is The Commutative Property of Addition?
The Commutative Property of Addition is a fundamental principle in mathematics, particularly within the field of algebra. It is one of the basic properties of numbers and operations that underpins
the entire structure of arithmetic and algebra. This property states that the order in which two numbers are added does not affect the sum. In other words, if you have two addends (the numbers being
added), switching their places does not change the result of the addition.
Mathematically, the Commutative Property of Addition is expressed as: a + b = b + a
Here, a and b are variables that represent any real numbers. The ‘+’ sign denotes the operation of addition. According to this property, for any two real numbers a and b, the sum remains the same
regardless of the order of the addends (addition parts).
The property is called ‘commutative’, stemming from the word ‘commute’, which means to move around or to change place. In this context, it implies that addends can ‘move around’ or change places
without affecting the outcome of the operation.
The Commutative Property of Addition has profound implications:
It allows for flexibility in computation, enabling one to rearrange terms to simplify the process of addition.
It is essential for the development of a deeper understanding of the structure of the number system.
It lays the groundwork for algebraic manipulations where the ability to rearrange and combine like terms is fundamental.
Let’s look at two examples to see the Commutative Property of Addition in action:
Example 1: Simple Numerical Addition
Suppose you have the numbers 3 and 5. According to the Commutative Property of Addition, if you add 3 and 5, you will get the same result as when you add 5 and 3.
3 + 5 = 8
5 + 3 = 8
In both cases, the sum is 8. This illustrates that the order of the numbers does not affect the result of the addition.
Example 2: Addition Involving Variables
The Commutative Property holds true not just for concrete numbers but also for algebraic expressions. Consider the expressions x and y, where x and y are variables that can represent any number.
x + y = y + x
This equation tells us that no matter what values x and y take on, the sum will be the same whether x is added to y or y is added to x. For instance, if x = 10 and y = 15, then:
10 + 15 = 25
15 + 10 = 25
Both expressions yield the same sum, 25, demonstrating the commutative nature of the addition of these variables. | {"url":"https://15worksheets.com/worksheet-category/commutative-property/","timestamp":"2024-11-14T14:20:27Z","content_type":"text/html","content_length":"133592","record_id":"<urn:uuid:86819cfa-e1ef-479b-aa03-a5b1cf4e2f8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00224.warc.gz"} |
Mathematics Archives - Summaries & Essays
The Pythagorean theorem is a statement in mathematics that states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two
sides. The theorem is named after the Greek mathematician Pythagoras, who is credited with discovering it. The theorem has … Read more
Are Boys Better At Math Than Girls?
A recent study has shown that boys are better at math than girls. The study, conducted by the University of Missouri, found that boys outperformed girls on a math test by an average of three points.
This is the first time that a study has shown that boys are better at math than girls. This … Read more
Math Is The Language Of The Universe
Mathematics, often referred to as the language of the universe, is a fundamental part of our world. Mathematics is used in almost every field, from physics to engineering to business. In fact, many
experts believe that mathematics is the key to understanding and using computers. Computers are an important part of our world today, and … Read more
Relationship Between Philosophy And Mathematics
It is often said that mathematics is the language of science. And while there is some truth to this, it is also fair to say that philosophy has a lot to do with mathematics as well. After all,
philosophy is all about trying to understand the nature of reality, and mathematics is one of the … Read more
History Of Mathematics Essay
Mathematics is a certain way of thinking and doing that has been around since the dawn of humanity. Mathematics as a whole can be seen through history as a steady evolution, starting from simple hand
calculations to modern-day computing machinery. Mathematics allows people to understand not only what’s happening in the world but also why … Read more
Relationship Between Mathematics And Philosophy
The relationship between mathematics and philosophy is an old one, dating back to the dawn of mathematics in ancient civilizations such as Egypt and Mesopotamia. Many mathematicians such as
Archimedes were also philosophers or vice versa, such as Thomas Hobbes who made great advances in mathematics but never received a degree in mathematics [1]. For … Read more
Fermat’s Last Theorem Essay
Fermat’s last theorem Currently holding the world record for longest standing math problem ever, Fermat’s last theorem went unsolved for 365 years. Fermat’s last theorem was one of the largest white
whales in the study of math. Over the centuries, thousands were puzzled by the impossible problem. From its conception to its solution, Fermat’s last … Read more | {"url":"https://novelsummary.com/topic/mathematics/","timestamp":"2024-11-06T18:37:12Z","content_type":"text/html","content_length":"68742","record_id":"<urn:uuid:28902a43-d5b5-4407-b891-405f1fe1f08c>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00515.warc.gz"} |
intermediate value theorem
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AP Calculus AB Mixed Review quiz 1
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Proportions & Pythagorean Theorem
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Calculus 1st Semester Review
Explore intermediate value theorem Worksheets by Grades
Explore Other Subject Worksheets for grade 11
Explore printable intermediate value theorem worksheets for 11th Grade
Intermediate value theorem worksheets for Grade 11 are an excellent resource for teachers looking to enhance their students' understanding of this fundamental concept in calculus. These worksheets
provide a variety of problems and examples that help students grasp the theorem and its applications in real-world scenarios. As a teacher, you know the importance of providing engaging and
challenging material to your Grade 11 Math students, and these worksheets are designed to do just that. With a focus on problem-solving and critical thinking, these intermediate value theorem
worksheets will not only help your students master the theorem but also develop the skills necessary for success in higher-level calculus courses.
Quizizz is a fantastic platform that offers a wide range of resources, including intermediate value theorem worksheets for Grade 11 Math students, to help teachers create engaging and interactive
lessons. With Quizizz, you can easily incorporate these worksheets into your lesson plans, allowing your students to practice and apply their knowledge of the intermediate value theorem in a fun and
dynamic way. In addition to the worksheets, Quizizz also offers other valuable resources such as quizzes, flashcards, and interactive games that can be tailored to your specific curriculum needs. As
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How Is Exponential Moving Average (EMA) Calculated? - Trader Opportunities
The exponential moving average (EMA) is a technical chart indicator that tracks the price of an investment (like a stock or commodity) over time. The EMA is a type of weighted moving average (WMA)
that gives more weighting or importance to recent price data. Like the simple moving average (SMA), the EMA is used to see price trends over time, and watching several EMAs at the same time is easy
to do with moving average ribbons.
Calculating SMA and EMA
The EMA is designed to improve on the idea of an SMA by giving more weight to the most recent price data, which is considered to be more relevant than older data. Since new data carries greater
weight, the EMA responds more quickly to price changes than the SMA does.
Key Takeaways
• Exponential moving averages (EMAs) are designed to see price trends over specific time frames, such as 50 or 200 days.
• Compared to simple moving averages, EMAs give greater weight to recent (more relevant) data.
• Computing the EMA involves applying a multiplier to the simple moving average (SMA).
• Moving average ribbons allow traders to see multiple EMAs at the same time.
The formula for calculating the EMA is a matter of using a multiplier and starting with the SMA. There are three steps in the calculation (although chart applications do the math for you):
1. Compute the SMA
2. Calculate the multiplier for weighting the EMA
3. Calculate the current EMA
The calculation for the SMA is the same as computing an average or mean. That is, the SMA for any given number of time periods is simply the sum of closing prices for that number of time periods,
divided by that same number. So, for example, a 10-day SMA is just the sum of the closing prices for the past 10 days, divided by 10.
The mathematical formula looks like this:
begin{aligned}&text{SMA} = frac { A_1 + A_2 + … + A_n }{ n } \&textbf{where:} \&A_n = text{Price of an asset at period } n \&n = text{Number of total periods} \end{aligned}
The formula for calculating the weighting multiplier looks like this:
begin{aligned} text{Weighted multiplier} &= 2 div (text{selected time period} + 1) \ &= 2 div (10 + 1) \ &= 0.1818 \ &= 18.18% \ end{aligned}
In both cases, we’re assuming a 10-day SMA.
So, when it comes to calculating the EMA of a stock:
begin{aligned} &EMA = text{Price}(t) times k + EMA(y) times (1-k) \ &textbf{where:}\ &t=text{today}\ &y=text{yesterday}\ &N=text{number of days in EMA}\ &k=2 div (N + 1)\ end{aligned}
The weighting given to the most recent price is greater for a shorter-period EMA than for a longer-period EMA. For example, an 18.18% multiplier is applied to the most recent price data for a 10-day
EMA, as we did above, whereas for a 20-day EMA, only a 9.52% multiplier weighting is used. There are also slight variations of the EMA arrived at by using the open, high, low, or median price instead
of using the closing price.
Using the EMA: Moving Average Ribbons
Traders sometimes watch moving average ribbons, which plot a large number of moving averages onto a price chart, rather than just one moving average. Though seemingly complex based on the sheer
volume of concurrent lines, ribbons are easy to see on charting applications and offer a simple way of visualizing the dynamic relationship between trends in the short, intermediate, and long term.
Traders and analysts rely on moving averages and ribbons to identify turning points, continuations, and overbought/oversold conditions, to define areas of support and resistance, and to measure price
trend strengths.
Defined by their characteristic three-dimensional shape that seems to flow and twist across a price chart, moving average ribbons are easy to interpret. The indicators trigger buy and sell signals
whenever the moving average lines all converge at one point. Traders look to buy on occasions when shorter-term moving averages cross above the longer-term moving averages from below and look to sell
when shorter moving averages cross below from above.
How to Create a Moving Average Ribbon
To construct a moving average ribbon, simply plot a large number of moving averages of varying time period lengths on a price chart at the same time. Common parameters include eight or more moving
averages and intervals that range from a two-day moving average to a 200- or 400-day moving average.
For ease of analysis, keep the type of moving average consistent across the ribbon—for example, use only exponential moving averages or simple moving averages.
When the ribbon folds—when all of the moving averages converge into one close point on the chart—trend strength is likely weakening and possibly pointing to a reversal. The opposite is true if the
moving averages are fanning and moving apart from each other, suggesting that prices are ranging and that a trend is strong or strengthening.
Downtrends are often characterized by shorter moving averages crossing below longer moving averages. Uptrends, conversely, show shorter moving averages crossing above longer moving averages. In these
circumstances, the short-term moving averages act as leading indicators that are confirmed as longer-term averages trend toward them.
The Bottom Line
The preferred number and type of moving averages can vary considerably between traders, based on investment strategies and the underlying security or index. But EMAs are especially popular because
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Calculators Archives - India's beloved learning platform
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How do you use scientific notation to express the number 355000 to one significant figure? | Socratic
How do you use scientific notation to express the number 355000 to one significant figure?
1 Answer
The answer is $4 \times {10}^{5}$.
Numbers written in scientific notation have a coefficient which begins with a single non-zero digit, 1-9 inclusive. That number is then multiplied by the base 10 raised to a power (exponent). If the
decimal was moved to the right, the exponent will be negative, such as $0.03 = 3 \times {10}^{- 2}$. If the decimal was moved to the left, the exponent will be positive, such as $400 = 4 \times {10}^
The exponent indicates how to expand the number to standard notation. If it is positive, move the decimal to the right the same number of places, such as $4 \times {10}^{2} = 400$. If it is negative,
move the decimal to the left the same number of places, such as $3 \times {10}^{- 2} = 0.03$.
Round $355000$ to one significant figure by rounding to $400000$. To use scientific number to express this number, move the decimal point to the left 5 places, and write the number as $4 \times {10}^
The following website may be helpful. https://www.nyu.edu/pages/mathmol/textbook/scinot.html
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Is MyStatLab assistance available for Bayesian analysis in clinical trials? | Hire Some To Take My Statistics Exam
Is MyStatLab assistance available for Bayesian analysis in clinical trials? Abstract Authors who use Bayesian statistics for diagnosis of the outcome of interest in clinical trials are working on not
only the assessment of test statistics, but also the ability to identify patterns, which may have an effect in other ways than statistical estimates. The study of the predictive models from
regression analysis has thus gained momentum, and one of the main insights of this article is to investigate the assumptions of Bayesian statistics. Bayesian statistics are characterized by a
distribution of data points used for analyzing the parameters estimated from a test statistic. Bayesian statistics can be based on either statistical models for the regression model or a Bayesian
procedure for the test statistic. Results are not only summarized but provided for interpretation in terms of their usefulness and interpretation in relation to the results derived from the testing.
In this article, the term Bayesian statistical or Bayesian probability is used rather than statistical or Bayesian terminology for the purpose of this article. The concepts of and concepts with which
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conducted on 33 out of 32 trials. Data were obtained from the experimental factorial analysis, which can be thought of as a logistic regression analysis. This study focuses on the evaluation of the
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Optimize scheduler performance for large-scale jobs
• Status: Closed
• Resolution: Done
• None
The performance of scheduler has been improved to reduce the time of execution graph creation, task deployment and task failover. This improvement is significant to large scale jobs which
currently may spend minutes on the processes mentioned above. This improvement also helps to avoid cases that the job manager main thread gets blocked for too long and leads to heartbeat timeout.
The performance of scheduler has been improved to reduce the time of execution graph creation, task deployment and task failover. This improvement is significant to large scale jobs which
currently may spend minutes on the processes mentioned above. This improvement also helps to avoid cases that the job manager main thread gets blocked for too long and leads to heartbeat timeout.
Phase 1
In the phase 1, we mainly focus on the optimization of job initialization, failover and partitions releasing.
The main idea is to replace ExecutionEdges with EdgeManager. In EdgeManager, we put all the vertices that consumed the same result partitions into one group, and put all the result partitions that
have the same consumer vertices into one consumer group. This will effectively decrease the computation complexity that involves iteration over ExecutionEdges.
The phase 1 involves the subtask 1-5.
According to the result of scheduler benchmarks we implemented in [DEL:FLINK-20612:DEL], the bottleneck of deploying and running a large-scale job in Flink is mainly focused on the following
Procedure Time complexity
Initializing ExecutionGraph O(N^2)
Building DefaultExecutionTopology O(N^2)
Initializing PipelinedRegionSchedulingStrategy O(N^2)
Scheduling downstream tasks when a task finishes O(N^2)
Calculating tasks to restart when a failover occurs O(N^2)
Releasing result partitions O(N^3)
These procedures are all related to the complexity of the topology in the ExecutionGraph. Between two vertices connected with the all-to-all edges, all the upstream Intermediate ResultPartitions are
connected to all downstream ExecutionVertices. The computation complexity of building and traversing all these edges will be O(N^2).
As for memory usage, currently we use ExecutionEdges to store the information of connections. For the all-to-all distribution type, there are O(N^2) ExecutionEdges. We test a simple job with only two
vertices. The parallelisms of them are both 10k. Furthermore, they are connected with all-to-all edges. It takes 4.175 GiB (estimated via MXBean) to store the 100M ExecutionEdges.
In most large-scale jobs, there will be more than two vertices with large parallelisms, and they would cost a lot of time and memory to deploy the job.
As we can see, for two JobVertices connected with the all-to-all distribution type, all IntermediateResultPartitions produced by the upstream ExecutionVertices are isomorphic, which means that the
downstream ExecutionVertices they connected are exactly the same. The downstream ExecutionVertices belonging to the same JobVertex are also isomorphic, as the upstream ResultPartitions they connect
are the same, too.
Since every JobEdge has exactly one distribution type, we can divide the vertices and result partitions into groups according to the distribution type of the JobEdge.
For the all-to-all distribution type, since all downstream vertices are isomorphic, they belong to a single group, and all the upstream result partitions are connected to this group. Vice versa, all
the upstream result partitions also belong to a single group, and all the downstream vertices are connected to this group. In the past, when we wanted to iterate all the downstream vertices, we
needed to loop over them n times, which leads to the complexity of O(N^2). Now since all upstream result partitions are connected to one downstream group, we just need to loop over them once, with
the complexity of O(N).
For the pointwise distribution type, because each result partition is connected to different downstream vertices, they should belong to different groups. Vice versa, all the vertices belong to
different groups. Since one result partition group is connected to one vertex group pointwisely, the computation complexity of looping over them is still O(N).
After we group the result partitions and vertices, ExecutionEdge is no longer needed. For the test job we mentioned above, the optimization can effectively reduce the memory usage from 4.175 GiB to
12.076 MiB (estimated via MXBean) in our POC. The time cost is reduced from 62.090 seconds to 8.551 seconds (with 10k parallelism).
The detailed design doc: https://docs.google.com/document/d/1OjGAyJ9Z6KsxcMtBHr6vbbrwP9xye7CdCtrLvf8dFYw/edit?usp=sharing
Phase 2
In the phase 2, we mainly focus on the optimization of deploying tasks.
The main idea is to cache the ShuffleDescriptors used in the deployment so that JobManager wouldn't need to calculate them for each vertex. This optimization will effectively decrease the time cost
on task deployment, especially for large-scale jobs with all-to-all edges. For the overall description of the phase 2, please refer to [DEL:FLINK-23005:DEL].
The phase 2 involves the subtask 10-12.
Phase 3
In the phase 3, we mainly focus on the initialization of the DefaultScheduler.
The phase 3 involves optimizing the construction of pipelined regions and optimizing the initialization of LocalInputPreferredSlotSharingStrategy. After the optimizations, the DefaultSchdeuler of the
job containing two vertices with 8k parallelism, connected with all-to-all edge, can be initialized within one second. This optimization significantly accelerates the submission of new jobs,
especially for OLAP jobs.
The phase 3 involves the subtask 8 and 9.
is related to
FLINK-20612 Add benchmarks for scheduler
mentioned in
Wiki Page
1. Optimize building topology when initializing ExecutionGraph Closed Zhilong Hong
2. Optimize the initialization of DefaultExecutionTopology Closed Zhilong Hong
3. Optimize the performance of PipelinedRegionSchedulingStrategy Closed Zhilong Hong
4. Optimize calculating tasks to restart in RestartPipelinedRegionFailoverStrategy Closed Zhilong Hong
5. Optimize releasing result partitions in RegionPartitionReleaseStrategy Closed Zhilong Hong
6. Optimize Execution#finishPartitionsAndUpdateConsumers Closed Unassigned
7. Optimize ExecutionGraphToInputsLocationsRetrieverAdapter Closed Unassigned
8. Optimize the initialization of LocalInputPreferredSlotSharingStrategy Closed Zhilong Hong
9. Optimize the construction of pipelined regions Closed Zhilong Hong
10. Cache the compressed serialized value of ShuffleDescriptors Closed Zhilong Hong
11. Distribute the ShuffleDescriptors via blob server Closed Zhilong Hong
12. Limit the size of ShuffleDescriptors in PermanentBlobCache on TaskExecutor Closed Zhilong Hong
13. Verify optimized scheduler performance for large-scale jobs Closed Zhu Zhu | {"url":"https://issues.apache.org/jira/browse/FLINK-21110","timestamp":"2024-11-03T19:46:39Z","content_type":"text/html","content_length":"121723","record_id":"<urn:uuid:6a39da36-985a-4942-b4ab-1a5ce4edb480>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00413.warc.gz"} |
Inertia gravity waves breaking in the middle atmosphere: energy transfer and dissipation tensor anisotropy
T. Pestana, M. Thalhammer, S. Hickel (2020)
Journal of the Atmospheric Sciences 77: 3193-3210. doi: 10.1175/JAS-D-19-0342.1
We present direct numerical simulations of inertia–gravity waves breaking in the middle–upper mesosphere. We consider two different altitudes, which correspond to the Reynolds number of 28 647 and
114 591 based on wavelength and buoyancy period. While the former was studied by Remmler et al., it is here repeated at a higher resolution and serves as a baseline for comparison with the
high-Reynolds-number case.
Validation of large-eddy simulation methods for gravity wave breaking
S. Remmler, S. Hickel, M.D. Fruman, U. Achatz (2015)
Journal of the Atmospheric Sciences 72: 3537-3562. doi: 10.1175/JAS-D-14-0321.1
To reduce the computational costs of numerical studies of gravity wave breaking in the atmosphere, the grid resolution has to be reduced as much as possible. Insufficient resolution of small-scale
turbulence demands a proper turbulence parameterization in the framework of a large-eddy simulation (LES). We consider three different LES methods—the adaptive local deconvolution method (ALDM), the
dynamic Smagorinsky method (DSM), and a naïve central discretization without turbulence parameterization (CDS4)—for three different cases of the breaking of well-defined monochromatic gravity waves.
Benchmarking in a rotating annulus: A comparative experimental and numerical study of baroclinic wave dynamics
M. Vincze, S. Borchert, U. Achatz, T. von Larcher, M. Baumann, C. Liersch, S. Remmler, T. Beck, K.D. Alexandrov, C. Egbers, J. Fröhlich, V. Heuveline, S. Hickel, U. Harlander (2015)
Meteorologische Zeitschrift 23: 611-635. doi: 10.1127/metz/2014/0600
The differentially heated rotating annulus is a widely studied tabletop-size laboratory model of the general mid-latitude atmospheric circulation. The two most relevant factors of cyclogenesis,
namely rotation and meridional temperature gradient are quite well captured in this simple arrangement. The radial temperature difference in the cylindrical tank and its rotation rate can be set so
that the isothermal surfaces in the bulk tilt, leading to the formation of baroclinic waves.
Finite-volume models with implicit subgrid-scale parameterization for the differentially heated rotating annulus
S. Borchert, U. Achatz, S. Remmler, S. Hickel, U. Harlander, M. Vincze, K.D. Alexandrov, F. Rieper, T. Heppelmann, S.I. Dolaptchiev (2014)
Meteorologische Zeitschrift 23: 561-580. doi: 10.1127/metz/2014/0548
The differentially heated rotating annulus is a classical experiment for the investigation of baroclinic flows and can be regarded as a strongly simplified laboratory model of the atmosphere in
mid-latitudes. Data of this experiment, measured at the BTU Cottbus-Senftenberg, are used to validate two numerical finite-volume models (INCA and cylFloit) which differ basically in their grid
On the construction of a direct numerical simulation of a breaking inertia-gravity wave in the upper-mesosphere
M.D. Fruman, S. Remmler, U. Achatz, S. Hickel (2014)
Journal of Geophysical Research 119: 11613-11640. doi: 10.1002/2014JD022046
A systematic approach to the direct numerical simulation (DNS) of breaking upper mesospheric inertia-gravity waves of amplitude close to or above the threshold for static instability is presented.
Normal mode or singular vector analysis applied in a frame of reference moving with the phase velocity of the wave (in which the wave is a steady solution) is used to determine the most likely scale
and structure of the primary instability
Spectral eddy viscosity of stratified turbulence
S. Remmler, S. Hickel (2014)
Journal of Fluid Mechanics 755, R6. doi: 10.1017/jfm.2014.423
The spectral eddy viscosity (SEV) concept is a handy tool for the derivation of large-eddy simulation (LES) turbulence models and for the evaluation of their performance in predicting the spectral
energy transfer. We compute this quantity by filtering and truncating fully resolved turbulence data from direct numerical simulations (DNS) of neutrally and stably stratified homogeneous turbulence.
The results qualitatively confirm the plateau–cusp shape, which is often assumed to be universal, but show a strong dependence on the test filter size. Increasing stable stratification not only
breaks the isotropy of the SEV but also modifies its basic shape, which poses a great challenge for implicit and explicit LES methods. We find indications that for stably stratified turbulence it is
necessary to use different subgrid-scale (SGS) models for the horizontal and vertical velocity components. Our data disprove models that assume a constant positive effective turbulent Prandtl number.
On the application of WKB theory for the simulation of the weakly nonlinear dynamics of gravity waves
J. Muraschko, M.D. Fruman, U. Achatz, S. Hickel, Y. Toledo (2015)
Quarterly Journal of the Royal Meteorological Society 141: 676-697. doi: 10.1002/qj.2381
The dynamics of internal gravity waves is modelled using Wentzel–Kramer–Brillouin (WKB) theory in position–wave number phase space. A transport equation for the phase-space wave-action density is
derived for describing one-dimensional wave fields in a background with height-dependent stratification and height- and time-dependent horizontal-mean horizontal wind, where the mean wind is coupled
to the waves through the divergence of the mean vertical flux of horizontal momentum associated with the waves.
Direct numerical simulation of a breaking inertia-gravity wave
S. Remmler, M.D. Fruman, S. Hickel (2013)
Journal of Fluid Mechanics 722: 424-436. doi: 10.1017/jfm.2013.108
We have performed fully resolved three-dimensional numerical simulations of a statically unstable monochromatic inertia–gravity wave using the Boussinesq equations on an f - plane with constant
stratification. The chosen parameters represent a gravity wave with almost vertical direction of propagation and a wavelength of 3 km breaking in the middle atmosphere.
A conservative integration of the pseudo-incompressible equations with implicit turbulence parameterization
F. Rieper, S. Hickel, U. Achatz (2013)
Monthly Weather Review 141: 861-886. doi: 10.1175/MWR-D-12-00026.1
Durran’s pseudo-incompressible equations are integrated in a mass and momentum conserving way with a new implicit turbulence model. This system is soundproof, which has two major advantages over
fully compressible systems: the Courant–Friedrichs–Lewy (CFL) condition for stable time advancement is no longer dictated by the speed of sound and all waves in the model are clearly gravity waves
Direct and large-eddy simulation of stratified turbulence
S. Remmler, S. Hickel (2012)
International Journal of Heat and Fluid Flow 35: 13-24. doi: 10.1016/j.ijheatfluidflow.2012.03.009
Simulations of geophysical turbulent flows require a robust and accurate subgrid-scale turbulence modeling. To evaluate turbulence models for stably stratified flows, we performed direct numerical
simulations (DNSs) of the transition of the three-dimensional Taylor–Green vortex and of homogeneous stratified turbulence with large-scale horizontal forcing.
Spectral structure of stratified turbulence: Direct Numerical Simulations and predictions by Large Eddy Simulation
S. Remmler, S. Hickel (2013)
Theoretical and Computational Fluid Dynamics 27: 319-336. doi: 10.1007/s00162-012-0259-9
Density stratification has a strong impact on turbulence in geophysical flows. Stratification changes the spatial turbulence spectrum and the energy transport and conversion within the spectrum. We
analyze these effects based on a series of direct numerical simulations (DNS) of stratified turbulence. | {"url":"https://inca.cfd/geophysical","timestamp":"2024-11-02T01:52:26Z","content_type":"text/html","content_length":"27952","record_id":"<urn:uuid:4dd34399-4f33-462e-8085-f76428260cca>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00539.warc.gz"} |
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As the size distribution of the holdup material also depends on the same mixing and transport phenomena, we used it as a basis to develop a new correlation. For our purposes, the experimental
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Converting a ball mill from overflow to grate discharge involves installing a grate inside the mill at the discharge end. The grate holds the balls inside the mill but allows the slurry to pass
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Overflow Ball Mill Structure Ball Mill Working Principle High energy ball milling is a type of powder grinding mill used to grind ores and other materials to 25 mesh or extremely fine powders,
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It is noted that the ball milling also enhanced the adsorption ability of biochars for other organic contaminants (, tetracycline and reactive red) and heavy metals (, Pb and Ni) ( Table 3).
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The influence of pulp composition and feed rate on holdup weight and mean residence time of solids in gratedischarge ball mill grinding. Int. J. Miner. Process., 8: 345358. Gupta,, Hodouin, D.
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GRED Answer: Fast train cars connected by string
Re: GRED Answer: Fast train cars connected by string
makc wrote:My brain is raped. You're saying that two sleepers are longer apart on paper, but if I put my measuring rod agains them, they are shorter apart. Well, if this is the case, your paper
does not describe my experience and is, as you said, "mathematical abstraction" - so why consider it? Btw, what's that bit about "time coordinates are crazy"?
Sorry about the brain. You really shouldn't let it go dressed that way. Just sayin'.
It's really quite simple, but perhaps I managed to complexify it in my explanation. I'll try again:
Relativity is all about events, that is, something that happens at a particular time and a particular place. A reference frame means a system for labeling events with time and space coordinates. The
rotating frame we're talking about here is constructed in the following way:
We asssign each event
space coordinates
that tell where in the train it happened. All events with the same space coordinates will have happened at the same place in the train, measured for example as meters forward of the midpoint of some
chosen car. (The coordinates of events in the next car over will have space coordinates that reflect how long the distance between cars look to the passengers. This is unproblematic as long as we're
only interested in the situation after the train is already moving with constant speed). This choice of space coordinates is really unavoidable given that we want to construct a
frame that follows the train. But we still have to choose time coordinates for each event. Here we choose to label each event with the timestamp that somebody
on the ground
would assign to it. This choice is dictated by our wish to exhibit the symmetry of the circular track and train. We can multiply all time coordinates by a common factor in an attempt to account for
the time dilation for clocks that travel with the train (let's imagine that we do this, though it won't have any implication for what follows), and we can add a common offset to all time coordinates
(pointless but harmless). But if we use different factors or different offsets for any two places on the train, we will have broken the symmetry -- an observer on the platform will see some cars
reach 12 o'clock (according to rotating-frame coordinates) before or after it happens in other cars, and this will destroy our assumption of symmetry.
So both time and space coordinates are more or less forced upon us by the idea "let's construct a rotating frame and do our analysis there". However, something horrible has happened: the time and
space coordinates do not fit together right. To see this, consider one lightbulb flashing in each end of our railway car, such that the passengers in the car consider the flashes to have happened at
the same time (i.e., the light from both flashes reaches the midpoint of the car simultaneously). There's a classic SR gedankenexperiment that concludes in precisely this circumstance that an
observer on the platform will
consider the two flash events to be simultaneous.
But this means that the two events have different time coordinates in the rotating frame
, by definition (above) of said time coordinates. The passenger on the train will have seen two flash events and believe them to be simultaneous -- but they have different time coordinates according
to the rotating frame. This is what I mean by the time coordinate being "crazy".
When I say that the distance between sleepers increase in the rotating frame, what I mean is neither more nor less than the following: Select an event on sleeper A. Select an event on sleeper B, such
that the two events have identical time coordinate. Subtract the space coordinates of those two events. The difference is larger than the distance found by the ground observer.
More "practically", suppose that the ground observer places a lightbulb on each of the two sleepers, and arranges for them to flash at the same time in
frame. By design those two flashes have the same time coordinate in the rotating system. To find their space coordinates, we must intercut to the passenger. He sees (through a window in the car
floor?) flash A happen right under the middle of his car. A bit
, according to his subjective time, he sees flash B happen about a meter to the rear. Therefore he can conclude that the sleeper distance in the rotating coordinate system is one meter. However, he
won't himself believe this to be the true distance between sleepers. Why, while he was waiting for sleeper B to flash, sleeper A has been hurtling rearwards, and was perhaps only 30 cm from sleeper B
"when" the latter flashed. So the measuring-stick distance he would prefer between the sleeper is 30 cm. (Meanwhile, the observer on the ground measures 60 cm between the sleepers).
In summary, the rotating coordinate system is one that does not match the experience of the passenger. And the reason for the whole mixup was that we wanted to have time coordinates that made the
symmetry of the experiment manifest. All we need to do to alleviate the passenger's confusion is to let him use the coordinates of the
frame in which the car is momentarily at rest. In this frame the track contracts as it should do, the car in front of us reaches its top speed and turns off its engine a short time before ours does,
in short everything behaves as it should.
The only thing we lose in the inertial frame is the symmetry of the experiment. For example, we
expect that the number of sleepers we count between our car and the next one at some instant in the inertial comoving frame is also the same as the number of sleepers between two cars at the other
side of the circle at the same instant. They are not in a symmetric situation; they have a different velocity with respect to the frame where we do our sleeper-counting.
So, did Einstein err when he constructed the rotating frame in the first place? Not at all, because he was doing GR here, not SR. And in GR it is generally expected that coordinates by themselves are
meaningless; you need to transform the coordinates into a locally inertial frame before you try doing SR-like physics with them. Our fallacy was to attempt to extract a "spatial geometry" from the
metric by ignoring the time coordinate, and
expect the spatial geometry to give sensible results about things that involve time (namely: the motion of sleepers under the train).
Henning Makholm
Re: GRED Answer: Fast train cars connected by string
if my brain had an anus, it would be bleeding now
edit: ha ha, now that I read this post some hours later, I see that that's not really different if I don't put my clock on the ground, is it
Re: GRED Answer: Fast train cars connected by string
makc wrote:if my brain had an anus, it would be bleeding now :( I don't really see how your explanation above help. Same way I could leave my clock on the ground, spin around myself once, pick up
the clock and say, "oh look, pluto just orbited around me with speed faster than light". Surely this is not the way to construct good reference frames, is it?
No, that's kind of my point. The rotating frame is not a very good reference frame.
Henning Makholm
Re: GRED Answer: Fast train cars connected by string
What about friction and drag? Would friction not cause the track to heat up at these speeds (depending on the size of the track (but I'm thinking that if the trains are spaced equally then the
friction would be constant enough to heat the track up sufficiently enough to make it fundamentally useless)
Also, if the track is sufficiently big that for most of the time the trains are moving in an apparent straight line, would there not be a slip stream in which the second and third train would make
use of thus speeding up more quickly than the train in front?
In this situation, the string will droop.
"I'm so fast that last night I turned off the light switch in my hotel room and was in bed before the room was dark" Muhammad Ali, faster than the speed of light?
Re: GRED Answer: Fast train cars connected by string
Henning Makholm wrote:[....]Our fallacy was to attempt to extract a "spatial geometry" from the metric by ignoring the time coordinate, and nevertheless expect the spatial geometry to give
sensible results about things that involve time (namely: the motion of sleepers under the train).
I think I'm reading that the gist of what you're saying is clock synchronization is an issue in the rotating frame. It seems to me that a passenger trying to measure a circumfrence would find it a
problem. I'm curious, do you still think the strings break?
A pessimist is nothing more than an experienced optimist
Re: GRED Answer: Fast train cars connected by string
wonderboy wrote:What about friction and drag? Would friction not cause the track to heat up at these speeds (depending on the size of the track (but I'm thinking that if the trains are spaced
equally then the friction would be constant enough to heat the track up sufficiently enough to make it fundamentally useless)
Also, if the track is sufficiently big that for most of the time the trains are moving in an apparent straight line, would there not be a slip stream in which the second and third train would
make use of thus speeding up more quickly than the train in front?
In this situation, the string will droop.
Hi Paul,
I think your concern about heat, friction, slip streams, etc are not warrented. There are forces in a non-inertial reference frame that have a fundamental impact on relativistic analysis which you
should be thinking about, but ignore the other practical phenomena. They are just distractions.
A pessimist is nothing more than an experienced optimist
Re: GRED Answer: Fast train cars connected by string
alter-ego wrote:
Henning Makholm wrote:[....]Our fallacy was to attempt to extract a "spatial geometry" from the metric by ignoring the time coordinate, and nevertheless expect the spatial geometry to give
sensible results about things that involve time (namely: the motion of sleepers under the train).
I think I'm reading that the gist of what you're saying is clock synchronization is an issue in the rotating frame. It seems to me that a passenger trying to measure a circumfrence would find it
a problem.
That's about what I'm saying, yes.
I'm curious, do you still think the strings break?
Yes, of course. My own argument for the strings breaking works entirely in the ground frame.
The reason why I'm considering other frames at all is to convince myself (and such readers as I can convince along with myself) that I can find flaws in any competing argument that reaches a
different conclusion.
Henning Makholm
Re: GRED Answer: Fast train cars connected by string
I actually still not convinced, but I don't mind if they break. I just want to "see" it. As my mental powers aren't quite good for this, I started writing a program yesterday, that would show me how
someone moving in certain way would see the track
You do not have the required permissions to view the files attached to this post.
Re: GRED Answer: Fast train cars connected by string
I believe the best answer is "The strings break." This is a classic problem in relativity correctly identified elsewhere as the Ehrenfest Paradox. Much of the surrounding discussion that arrives at
the "strings break" answer appears correct. Nevertheless, here my brief explanation.
In sum, the moving trains will actually physically contract via the famous Lorentz contraction. Since the track is not moving, however, it will not contract. Therefore, the shorter trains-and-strings
can no longer remain connected around the track, and the strings will break.
Here are some interesting tidbits. First, people rotating with the trains ("train people") will have their rulers contract, too, as measured by observers at rest with the track ("track people"), so
that these train people measure the circumference of the track to be longer than the track people. Specifically, if the track people measure the circumference to be "C", the train people will measure
it to be C/sqrt(1-v2/c2).
Next, one reason I believe this is because Albert Einstein himself worked on the Ehrenfest Paradox and came to the conclusion of the previous two paragraphs. And he was REALLY good at solving these
(and many other) sorts of things.
Still, although I believe this solution, I found it initially unsettling. First, in examples involving relative motion between inertial frames (no-acceleration frames), the Lorentz contraction was
never really a physical contraction, but rather a perceptual contraction that observers in each frame could apply to the other frame. In other words, if two trains traveling in straight lines pass,
and each train has cars attached by strings, each train sees the other train Lorentz contracted, but none of the strings break or droop because no real forces are being applied.
But here in the Ehrenfest Paradox, the strings actually break. Strange! What force breaks them? Why can't the train people see the track contracted and so cause the strings to droop? The answer to
both of these questions is that the train people are accelerating and therefore not in an inertial frame, and therefore what they measure is more complicated. Apparently, the force that causes the
acceleration can possibly be tapped to break the strings.
What makes this, for me, perhaps less unsettling is to picture the analogous situation involving gravity. There, one could consider trains again at rest on the track, still connected by stings, but
instead of accelerating them around the circular track, this time put a heavy mass at the very center. For spectacular renditions in the Disney movie version, this can be a black hole. What happens
to the strings in this case? I believe they break, then, too.
Re: GRED Answer: Fast train cars connected by string
makc wrote:I actually still not convinced, but I don't mind if they break. I just want to "see" it. As my mental powers aren't quite good for this, I started writing a program yesterday, that
would show me how someone moving in certain way would see the track
FWIW, here is what I get for a train of 12 cars that almost fill the track when stationary. (The cars are slightly curved, because I didn't bother to write code aligning straight cars with a chord).
For comparison I have planted 12 trees (green dots) in a circle just inside the track. The trees don't move. I'm afraid this result does not look very intuitively enlightening. The oblique car ends
are (as far as I can figure out) a real effect, coming about because the car are moving partly sideways with respect to the observer, and so are relativistically squished along a diagonal direction.
You do not have the required permissions to view the files attached to this post.
Henning Makholm
Re: GRED Answer: Fast train cars connected by string
RJN wrote:Still, although I believe this solution, I found it initially unsettling. First, in examples involving relative motion between inertial frames (no-acceleration frames), the Lorentz
contraction was never really a physical contraction, but rather a perceptual contraction that observers in each frame could apply to the other frame. In other words, if two trains traveling in
straight lines pass, and each train has cars attached by strings, each train sees the other train Lorentz contracted, but none of the strings break or droop because no real forces are being
As MAReynolds pointed out early in the thread, essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start accelerating at the same
rate (as measured on each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while accelerating to relativistic
speeds, the rear cars need to accelerate faster.
Based on this, I'd insist that it is the acceleration along the track that is ultimately responsible.
Note that flatlanders living on a cylinder could make exactly the same thought experiment with a straight (to them) track going around their cylinder. They would not be able to lay the blame any
specific effect of "rotation", so we shouldn't have to either.
What makes this, for me, perhaps less unsettling is to picture the analogous situation involving gravity. There, one could consider trains again at rest on the track, still connected by stings,
but instead of accelerating them around the circular track, this time put a heavy mass at the very center. For spectacular renditions in the Disney movie version, this can be a black hole. What
happens to the strings in this case? I believe they break, then, too.
I'm not sure about this analogy. The black hole draws things inward, whereas centrifugal force pushes things outward. That makes it difficult be sure whether the analogy would work in a direct or
opposite way.
Henning Makholm
Re: GRED Answer: Fast train cars connected by string
Henning Makholm wrote:As MAReynolds pointed out early in the thread, essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start
accelerating at the same rate (as measured on each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while
accelerating to relativistic speeds, the rear cars need to accelerate faster.
Straight track fits my feeble brain much better
edit: actually this picture is nice. now I have less motivation to fix my program, thanks
Re: GRED Answer: Fast train cars connected by string
Henning Makholm wrote:essentially the same effect does occur on a straight line. Have a line of cars stand on a straight track. Let them all start accelerating at the same rate (as measured on
each car's own accelerometer). The cars will separate, and strings, if any, between the cars will break. If a long train is to stay connected while accelerating to relativistic speeds, the rear
cars need to accelerate faster.
Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with any velocity or
acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.
As for acceleration, I agree that there should be some way to accelerate the train cars to an inertial frame without breaking the strings. Exactly how that works might indeed be complicated.
Henning Makholm wrote: I'm not sure about this analogy. The black hole draws things inward, whereas centrifugal force pushes things outward. That makes it difficult be sure whether the analogy
would work in a direct or opposite way.
Yes, the acceleration directions (relative to the center) are in opposite directions. Still, although I frequently miss (other) minus signs, I believe the strings break in both cases.
- RJN
Re: GRED Answer: Fast train cars connected by string
RJN wrote:Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with any
velocity or acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.
Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.
Re: GRED Answer: Fast train cars connected by string
makc wrote:
RJN wrote:Actually, I didn't mention acceleration, just velocity. So if I'm sitting here on my unaccelerated train connected by strings, it shouldn't matter that other trains pass me by with
any velocity or acceleration -- my strings won't break. And the same goes for any constant velocity train that passes me by.
Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.
That's true, and if you then slow down the train, the strings will droop. Although I didn't find a direct quote, Einstein supposedly said that to avoid the deformation stage during spin up, the
cylinder or disk (as originally presented by Ehrenfest) should be done in a hot melt, and then solidified when accelaration is complete. Obviously, the atoms are "free" in a fluid stage, but they are
still being redistributed while the melted disk is spun up.
A pessimist is nothing more than an experienced optimist
Re: GRED Answer: Fast train cars connected by string
makc wrote:Same goes to circle train then - if we make the cars move at constant angular velocity first and then connect them, nothing breaks.
Well, yes, but perhaps I should be more clear. In my opinion, there IS a way to linearly accelerate train cars connected in a line to a constant linear velocity without the strings that connect them
breaking. However, there is NO way to accelerate train cars connected around a circle to a constant angular velocity without the strings breaking. Yes, this does carry the presumption that all of the
train cars undergo the same acceleration as measured from the inertial frame at the center. | {"url":"http://asterisk.apod.com/viewtopic.php?f=32&t=20063&start=25","timestamp":"2024-11-04T13:36:07Z","content_type":"text/html","content_length":"63257","record_id":"<urn:uuid:c9d3d30f-4356-4d37-890b-3203b21c64e2>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00842.warc.gz"} |
Mathematics Faculty Work
Permanent URI for this collection
Recent Submissions
• Item
Geodesics and Bounded Harmonic Functions on Infinite Graphs
It is shown there that an infinite connected planar graph with a uniform upper bound on vertex degree and rapidly decreasing Green's function (relative to the simple random walk) has infinitely
many pairwise finitely-intersecting geodesic rays starting at each vertex. We then demonstrate the existence of nonconstant bounded harmonic functions on the graph.
• Item
Cogrowth of Regular Graphs
(Proceedings of the American Mathematical Society, 1992) Northshield, Sam
Let G be a d-regular graph and T the covering tree of G. We define a cogrowth constant of G in T and express it in terms of the first eigenvalue of the Laplacian on G. As a corollary, we show
that the cogrowth constant is as large as possible if and only if the first eigenvalue of the Laplacian on G is zero. Grigorchuk's criterion for amenability of finitely generated groups follows.
• Item
Amenability and superharmonic functions
(Proceedings of the American Mathematical Society, 1993) Northshield, Sam
Let G be a countable group and u a symmetric and aperiodic probability measure on G . We show that G is amenable if and only if every positive superharmonic function is nearly constant on certain
arbitrarily large subsets of G. We use this to show that if G is amenable, then the Martin boundary of G contains a fixed point. More generally, we show that G is amenable if and only if each
member of a certain family of G-spaces contains a fixed point.
• Item
On the spectrum and Martin boundary of homogeneous spaces
(Statistics and Probability Letters, 1995) Northshield, Sam
Given a conservative, spatially homogeneous Markov process X on an homogeneous spaces χ, we show that if the bottom of the spectrum of the generator of X is zero then the Martin boundary of
contains a unique point fixed by the isometry group of χ.
• Item
On the Commute Time of Random Walks on Graphs
(Brazilian Journal of Probability and Statistics, 1995) Northshield, Sam; Palacios, Jose Luis
Given a simple random walk on an undirected connected graph, the commute time of the vertices x and y is defined as C(x,y) = ExTy+EyTx. We give a new proof, based on the optional sampling theorem
for martingales, of the formula C(x,y) = 1/(Π(y)e(y,x)) in terms of the escape probability e(y,x ) (the probability that once the random walk leaves x, it hits y before it returns to x) and the
stationary distribution Π(·). We use this formula for C(x,y) to show that the maximum commute time among all barbell-type graphs in N vertices is attained for the lollipop graph and its value is | {"url":"https://dspace.sunyconnect.suny.edu/collections/2c2ab92f-eba3-4fef-ad4e-5271c32d6948","timestamp":"2024-11-04T05:20:34Z","content_type":"text/html","content_length":"513029","record_id":"<urn:uuid:b78d192e-4bea-4fed-9de1-a25895ae2e21>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00042.warc.gz"} |
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35 Share of tool cost paid by buyer
The percentage of the cost of tooling which will be paid by the buyer.
36 Volume capacity usage
Percentage of the volume capacity used.
37 Weight capacity usage
Percentage of the weight capacity used.
38 Loading length capacity usage
Percentage of the loading length capacity used.
39 Share of packaging cost paid by vendor
The percentage of the cost of packaging which will be paid by the vendor.
40 Reduction percentage
Reduction from an amount/price expressed in a percentage.
41 Surcharge percentage
Additional amount expressed in a percentage.
42 Local content
To indicate the percentage of a products local (i.e. domestic) content.
43 Chargeback
Percentage amount charged back.
44 Gross turnover commission
Percentage of gross turnover used to calculate commission.
45 Progress payment percentage
Indicates the rate applying for a progress payment.
46 Offset
Indicates the figure agreed between parties to calculate an offset.
47 Prepaid payment percentage
A code to indicate the percentage of the prepayment.
48 Percentage of work completed
A code to indicate the percentage of work completed.
49 Underwriting rating
Table used for mortality or morbidity rating.
50 Mortgage interest rate
Rate of interest used in amortization of a mortgage.
51 Maximum cost of living adjustment rate
Maximum rate of the cost of living adjustment.
52 Humidity
The amount of moisture in the air.
53 Minimum cost of living adjustment rate
Minimum rate of a cost of living adjustment.
54 Contractor cost share
The cost share borne by the contractor.
55 Government cost share
The cost share borne by the Government.
56 Progress payment liquidation percentage
The percentage applied to the liquidation of progress payments.
57 Fee percentage
The percentage applied to determine the fee. | {"url":"https://www.edifactory.de/edifact/directory/D98A/data-element/5245","timestamp":"2024-11-09T00:56:22Z","content_type":"text/html","content_length":"43346","record_id":"<urn:uuid:c17e9c0f-fad7-4c6b-8f4d-47dff2f0c059>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00589.warc.gz"} |
Bustoz, J, Feldstein, A, Goodman, R and Linnainmaa, S (1979). Improved Trailing Digits Estimates Applied to Optimal Computer Arithmetic. Journal of the Association for Computing Machinery 26(4), pp.
This work is cited by the following items of the Benford Online Bibliography:
Note that this list may be incomplete, and is currently being updated. Please check again at a later date.
Barlow, JL and Bareiss, EH (1985). On Roundoff Error Distributions in Floating Point and Logarithmic Arithmetic. Computing 34(4), pp. 325-347. ISSN/ISBN:0010-485X. DOI:10.1007/BF02251833.
Feldstein, A and Turner, P (1986). Overflow, Underflow, and Severe Loss of Significance in Floating-Point Addition and Subtraction. IMA Journal of Numerical Analysis 6, pp. 241-251.
Goodman, RH, Feldstein, A and Bustoz, J (1985). Relative Error in Floating-Point Multiplication. Computing 35, pp. 127-139. ISSN/ISBN:1436-5057. DOI:10.1007/BF02260500.
Johnstone, P and Petry, FE (1994). Design and Analysis of Nonbinary Radix Floating-Point Representations. Computers & Electrical Engineering 20(1), pp. 39-50. ISSN/ISBN:0045-7906. DOI:10.1016
Turner, PR (1987). The distribution of l.s.d. and its implications for computer design. The Mathematical Gazette 71 (March), 26-31. | {"url":"https://www.benfordonline.net/references/up/608","timestamp":"2024-11-08T19:08:37Z","content_type":"application/xhtml+xml","content_length":"9934","record_id":"<urn:uuid:de6fa393-895b-4d99-885c-5616fb8afc90>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00755.warc.gz"} |
Digital Commons
Operations On Submodules With The Multiplicative And Quotient Properties, 2024
Operations On Submodules With The Multiplicative And Quotient Properties, Jiekai Pang
Mathematics & Statistics ETDs
Inspired by the works of Petro, Epstein, Vassilev, and Morre, this thesis aims to study the generalized definitions of the semiprime operation, weakly prime operation, and standard closure operation
on rings, that is, the multiplicative operation, weakly multiplicative operation, and standardly multiplicative operation on submodules. Then, we will use Matlis duality to induce the dual notions of
these definitions on submodules of Matlis-dualizable Artinian modules. In order to understand the dual notion of the standardly multiplicative operation, that is, the standardly quotient operation,
we will classify the operations on the injective hull of residue field of the ring K[[x,y]]/(xy) which …
Nestings Of Bibds With Block Size Four, 2024
Nestings Of Bibds With Block Size Four, Marco Buratti, Donald L. Kreher, Douglas R. Stinson
Michigan Tech Publications, Part 2
In a nesting of a balanced incomplete block design (or BIBD), we wish to add a point (the nested point) to every block of a (Formula presented.) -BIBD in such a way that we end up with a partial
(Formula presented.) -BIBD. In the case where the partial (Formula presented.) -BIBD is in fact a (Formula presented.) -BIBD, we have a perfect nesting. We show that a nesting is perfect if and only
if (Formula presented.). Perfect nestings were previously known to exist in the case of Steiner triple systems (i.e., (Formula presented.) -BIBDs) when (Formula presented.), as well as …
Shevtsov: Teaching Modeling To First-Year Life Science Students: The Ucsc Experience, 2024
Shevtsov: Teaching Modeling To First-Year Life Science Students: The Ucsc Experience, Martin H. Weissman
Annual Symposium on Biomathematics and Ecology Education and Research
No abstract provided.
Modeling Opioid Addiction In Hand Surgery Patients, 2024
Modeling Opioid Addiction In Hand Surgery Patients, Eli Goldwyn, Grace Bowman, Kathryn Montovan, Julie Blackwood
Annual Symposium on Biomathematics and Ecology Education and Research
No abstract provided.
On The Work Of Cartan And Münzner On Isoparametric Hypersurfaces, 2024
On The Work Of Cartan And Münzner On Isoparametric Hypersurfaces, Thomas E. Cecil
Mathematics and Computer Science Department Faculty Scholarship
A hypersurface M^n in a real space form R^n+1, S^n+1, or H^n+1 is isoparametric if it has constant principal curvatures. This paper is a survey of the fundamental work of Cartan and Münzner on the
theory of isoparametric hypersurfaces in real space forms, in particular, spheres. This work is contained in four papers of Cartan published during the period 1938--1940, and two papers of Münzner
that were published in preprint form in the early 1970's, and as journal articles in 1980--1981. These papers of Cartan and Münzner have been the foundation of …
Using Academic Librarians And The Academic Library: Survey Results From Mathematics Faculty In United States And Canada, 2024
Using Academic Librarians And The Academic Library: Survey Results From Mathematics Faculty In United States And Canada, Elizabeth Novosel, Daniel G. Kipnis, Jenni Burke, Rasitha Jayesekera
Libraries Scholarship
Presented at STEM Librarian South Conference, November 6, 2024
Despite our diligent outreach and relationship-building efforts, certain groups of faculty within STEM disciplines remain hesitant to engage with academic librarians and utilize library resources.
This challenge raises a critical question: How can librarians effectively support a department when faculty members are unresponsive?
Our research team, composed of three frustrated mathematics librarians and one mathematics faculty member, recognized the need for a fresh approach. Mathematics, a department known for its
independence, has received limited attention in scholarly Library and Information Science (LIS) literature. To address this gap, we embarked on a …
Book V Of The Mathematical Collection Of Pappus Of Alexandria, Translated By John B. Little, 2024
Book V Of The Mathematical Collection Of Pappus Of Alexandria, Translated By John B. Little, Pappus Of Alexandria, John B. Little
Holy Cross Bookshelf
John B. Little is the translator.
Book V of the Mathematical Collection is addressed to a certain Megethion, about whom we know nothing else. From the context he may have been a student or patron of Pappus in Alexandria. In a heading
at the start, Pappus says that the general theme will be comparisons between different geometric figures. The overall structure brings interesting relations and connections to the fore. The book
opens with a very well-known and charming discussion of how the importance of such comparisons can be seen by considering the structures built by non-human creatures such as bees. …
Algebraic Structures On Parallelizable Manifolds, 2024
Algebraic Structures On Parallelizable Manifolds, Sergey Grigorian
School of Mathematical and Statistical Sciences Faculty Publications and Presentations
In this paper we explore algebraic and geometric structures that arise on parallelizable manifolds. Given a parallelizable manifold L , there exists a global trivialization of the tangent bundle,
which defines a map ρ p : l ⟶ T p L for each point p ∈ L , where l is some vector space. This allows us to define a particular class of vector fields, known as fundamental vector fields, that
correspond to each element of l . Furthermore, flows of these vector fields give rise to a product between elements of l and L , which in turn induces …
Pricing Variance Swaps For The Discrete Bn-S Model, 2024
Pricing Variance Swaps For The Discrete Bn-S Model, Semere Gebresilasie
Journal of Stochastic Analysis
No abstract provided.
A Phenomenological Study Of Persistence For Black Women In Mathematics Doctoral Programs, 2024
A Phenomenological Study Of Persistence For Black Women In Mathematics Doctoral Programs, Shushannah Marie Smith
Doctoral Dissertations and Projects
The purpose of this phenomenological study was to describe the doctoral persistence experiences of Black women mathematicians. Guided by McGee’s fragile and robust mathematical identity framework,
the research design followed Moustakas’ qualitative transcendental phenomenological approach. The researcher explored the contributing factors to the phenomenon guided by the central research
question, “What are the doctoral persistence experiences of Black women mathematicians?” The study involved 11 Black women participants who hold or were within one year of holding a doctoral degree
in mathematics and were selected to ensure diversity and representativeness. Data collection encompassed three methods: a mathematical identity development timeline, one-on-one …
Uniqueness Of Maximum Points Of A Sequence Of Functions Arising From An Adapted Algorithm For ‘The Secretary Problem’, 2024
Uniqueness Of Maximum Points Of A Sequence Of Functions Arising From An Adapted Algorithm For ‘The Secretary Problem’, Boning Wang, Giang Vu Thanh Nguyen
OUR Journal: ODU Undergraduate Research Journal
This paper is aimed at a sequence of functions that is extended from an adaptive algorithm of the classical ‘secretary problem’. It was proved in Nguyen et al. (2024) the uniqueness of maximizers of
a function sequence that represents the expected score of an element in a ‘candidate’ sequence. This function sequence is indeed considered as a special case of an extended function sequence that
corresponds to the case α = 0. More specifically, we are motivated to prove the uniqueness of maximizers for this extended function sequence in the case α = 1. Nevertheless, the corresponding proof
is rather …
Scaled Global Operators And Fueter Variables On Non-Zero Scaled Hypercomplex Numbers, 2024
Scaled Global Operators And Fueter Variables On Non-Zero Scaled Hypercomplex Numbers, Daniel Alpay, Ilwoo Choo, Mihaela Vajiac
Mathematics, Physics, and Computer Science Faculty Articles and Research
In this paper we describe the rise of global operators in the scaled quaternionic case, an important extension from the quaternionic case to the family of scaled hypercomplex numbers H[t], t ∈ R^∗ ,
of which the H[−1] = H is the space of quaternions and H[1] is the space of split quaternions.We also describe the scaled Fueter-type variables associated to these operators, developing a coherent
theory in this field. We use these types of variables to build different types of function spaces on H[t]. Counterparts of the Hardy space and of the …
Counting The Classes Of Projectively-Equivalent Pentagons On Finite Projective Planes Of Prime Order, 2024
Counting The Classes Of Projectively-Equivalent Pentagons On Finite Projective Planes Of Prime Order, Maxwell Hosler
Rose-Hulman Undergraduate Mathematics Journal
In this paper, we examine the number of equivalence classes of pentagons on finite projective planes of prime order under projective transformations. We are interested in those pentagons in general
position, meaning that no three vertices are collinear. We consider those planes which can be constructed from finite fields of prime order, and use algebraic techniques to characterize them by their
symmetries. We are able to construct a unique representative for each pentagon class with nontrivial symmetries. We can then leverage this fact to count classes of pentagons in general. We discover
that there are (1/10)((p+3)(p-3)+4 …
Nano Topology And Decision Making In Medical Applications, 2024
Nano Topology And Decision Making In Medical Applications, Samir Mukhtar, Mohamed Shokry, Manar Omran
Journal of Engineering Research
Nano Topology is one of the essential topics that receive special attention from some athletes in the field of General Topology, Operations Research, and Computer Science, because it has a vital role
in the generalizing most of the various mathematical concepts. Recently, many efforts have been made to study many types of Nano Topology, as the previous studies lacked real applications in
Engineering, Medicine, Pharmacy, and Social Sciences. In this paper, we present some different applications of these studies. The paper is divided into two parts: Firstly, we study the theory of The
Nano Topology and investigate its relation with …
New Operation Defined Over Dual-Hesitant Fuzzy Set And Its Application In Diagnostics In Medicine, 2024
New Operation Defined Over Dual-Hesitant Fuzzy Set And Its Application In Diagnostics In Medicine, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
Journal of Engineering Research
In recent decades, several types of sets, such as fuzzy sets, interval-valued fuzzy sets, intuitionistic fuzzy sets, interval-valued intuitionistic fuzzy sets, type 2 fuzzy sets, type n fuzzy sets,
and hesitant fuzzy sets, have been introduced and investigated widely. In this paper, we propose dual hesitant fuzzy sets (DHFSs), which encompass fuzzy sets, intuitionistic fuzzy sets, hesitant
fuzzy sets, and fuzzy multi-sets as special cases. Then we investigate the basic operations and properties of DHFSs. We also discuss the relationships among the sets mentioned above, and then propose
an extension principle of DHFSs. Additionally, we give an example to illustrate …
Decision-Making In Diagnosing Heart Failure Problems Using Dual Hesitant Fuzzy Sets, 2024
Decision-Making In Diagnosing Heart Failure Problems Using Dual Hesitant Fuzzy Sets, Manar Mohamed Omran, Reham Abdel-Aziz Abo-Khadra
Journal of Engineering Research
In recent decades, several types of sets, such as fuzzy sets, interval-valued fuzzy sets, intuitionistic fuzzy sets, interval-valued intuitionistic fuzzy sets, type 2 fuzzy sets, type n fuzzy sets,
and hesitant fuzzy sets, have been introduced and investigated widely. In this paper, we propose dual hesitant fuzzy sets (DHFSs), which encompass fuzzy sets, intuitionistic fuzzy sets, hesitant
fuzzy sets, and fuzzy multi-sets as special cases. Then we investigate the basic operations and properties of DHFSs. We also discuss the relationships among the sets mentioned above, and then propose
an extension principle of DHFSs. Additionally, we give an example to illustrate …
Discrete Time Series Forecasting Of Hive Weight, In-Hive Temperature, And Hive Entrance Traffic In Non-Invasive Monitoring Of Managed Honey Bee Colonies: Part I, 2024
Discrete Time Series Forecasting Of Hive Weight, In-Hive Temperature, And Hive Entrance Traffic In Non-Invasive Monitoring Of Managed Honey Bee Colonies: Part I, Vladimir A. Kulyukin, Daniel Coster,
Aleksey V. Kulyukin, William Meikle, Milagra Weiss
Computer Science Faculty and Staff Publications
From June to October, 2022, we recorded the weight, the internal temperature, and the hive entrance video traffic of ten managed honey bee (Apis mellifera) colonies at a research apiary of the Carl
Hayden Bee Research Center in Tucson, AZ, USA. The weight and temperature were recorded every five minutes around the clock. The 30 s videos were recorded every five minutes daily from 7:00 to 20:55.
We curated the collected data into a dataset of 758,703 records (208,760–weight; 322,570–temperature; 155,373–video). A principal objective of Part I of our investigation was to use the curated
dataset to investigate …
(Si13-07) Optimal Range For Value Of Two-Person Zero-Sum Game Models With Uncertain Payoffs, 2024
(Si13-07) Optimal Range For Value Of Two-Person Zero-Sum Game Models With Uncertain Payoffs, Sana Afreen, Ajay Kumar Bhurjee
Applications and Applied Mathematics: An International Journal (AAM)
Game theory deals with the decision-making of individuals in conflicting situations with known payoffs. However, these payoffs are imprecisely known, which means they have uncertainty due to
vagueness in the data set of most real-world problems. Therefore, we consider a two-person zero-sum game model on a larger scale where the payoffs are imprecise and lie within a closed interval. We
define the pure and mixed strategy as well as value for the game models. The proposed method computes the optimal range for the value of the game model using interval analysis. To derive some
important results, we establish some lemmas …
(Si13-06) Analysis Of Some Unified Integral Equations Of Fredholm Type Associated With Multivariable Incomplete H And I-Functions, 2024
(Si13-06) Analysis Of Some Unified Integral Equations Of Fredholm Type Associated With Multivariable Incomplete H And I-Functions, Rahul Sharma, Vinod Gill, Naresh Kumar, Kanak Modi, Yudhveer Singh
Applications and Applied Mathematics: An International Journal (AAM)
In this research paper, we examine various effective methods for addressing the problem of solving Fredholm-type integral equations. Our investigation commences by applying the principles of
fractional calculus theory. We employ series representations and products of multivariable incomplete H-functions and multivariable incomplete I-functions to solve these integrals. The outcomes
derived from our analysis possess a general nature and hold the potential to yield numerous results.
Zermelo's Theorem: How To Reach A Standard Of Perfect Play In Chess, 2024
Zermelo's Theorem: How To Reach A Standard Of Perfect Play In Chess, Elisha Amadasu '26
Annual Student Research Poster Session
Zermelo’s theorem establishes that in any two player zero-sum game with perfect information (without the element of chance), either one side can force a win regardless of how the other side plays, or
both sides can force a draw (if allowed in the rules). This theorem encouraged me to see how I could better my chess to see if I could exploit the existence of this perfect standard. I considered how
chess engines play, as they are currently the strongest at playing chess. I understood that they often look 60-70 moves deep from the opening stage of the game, which … | {"url":"https://network.bepress.com/physical-sciences-and-mathematics/mathematics/page1","timestamp":"2024-11-09T19:54:51Z","content_type":"text/html","content_length":"103877","record_id":"<urn:uuid:1e484bcd-4e96-4701-b649-ad6a62d212c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00882.warc.gz"} |
Exploring the Fundamentals: Understanding Measure Theory in Mathematics
Author: admintanbourit
Measure theory is a branch of mathematics that deals with the concept of measuring sets and is a fundamental tool used in many areas of mathematics, such as probability theory and analysis. It
provides a rigorous framework for understanding and manipulating mathematical concepts related to size, shape, and structure.
The concept of measure has been studied for centuries, with early notions of measure originating from practical applications, such as measuring land, liquids, and time. However, it was not until the
late 19th and early 20th centuries that mathematicians, such as Georg Cantor, Emile Borel, and Henri Lebesgue, began to formalize the theory of measure and develop the modern concept of measure
One of the fundamental concepts in measure theory is that of a measure. A measure is a function that assigns a non-negative real number to a set, representing its size or extent. It is usually
denoted by the symbol μ and has the following properties:
1. Non-negativity: The measure of any set is always a non-negative real number.
2. Null set has measure zero: The empty set, denoted by ∅, has a measure of zero.
3. Countable additivity: The measure of the union of countably many disjoint sets is equal to the sum of their individual measures.
These properties may seem intuitive, but they play a crucial role in the development and applications of measure theory. The concept of measure allows for a quantitative understanding of sets and
enables the comparison of different sets based on their sizes.
Another important aspect of measure theory is the idea of a measurable set. A set is said to be measurable if its measure can be defined in a consistent and well-defined manner. For example, the set
of real numbers between 0 and 1 would be considered measurable, as it has a finite measure of 1.
Measurable sets also have the property of being closed under certain operations, such as union, intersection, and complement. This allows for the construction of more complex sets and enables
mathematical analysis to be performed on them.
One of the key applications of measure theory is in the field of probability. The concept of a probability measure, which assigns probabilities to different events, is based on the theory of measure.
In probability theory, a sample space, which contains all possible outcomes of an event, is assigned a probability measure, and events are calculated as subsets of this space.
Measure theory is also heavily used in analysis, where it provides a powerful tool for understanding and manipulating mathematical functions. The Lebesgue integral, which extends the concept of the
Riemann integral, is based on measure theory and is used to calculate the area under curves and volumes of higher-dimensional objects.
Other applications of measure theory can be found in fields such as differential equations, dynamical systems, and fractal geometry. It provides a foundation for understanding and analyzing complex
systems and has played a crucial role in many modern advancements in mathematics and other sciences.
In addition to its applications, measure theory has also paved the way for important mathematical concepts such as Lebesgue measure and integration, Hausdorff measure and dimension, and the theory of
integration on manifolds. These concepts have greatly expanded our understanding of geometry and topology and have had significant impacts in fields such as physics, economics, and engineering.
In conclusion, measure theory is a fundamental branch of mathematics that provides a rigorous framework for understanding and manipulating measures of sets. Its applications can be found in various
areas of mathematics, and it has played a crucial role in many modern advancements. Its importance cannot be overstated, as it has helped shape and advance our understanding of various mathematical
concepts and their applications. | {"url":"https://tanbourit.com/exploring-the-fundamentals-understanding-measure-theory-in-mathematics/","timestamp":"2024-11-08T13:53:01Z","content_type":"text/html","content_length":"111305","record_id":"<urn:uuid:8cddfd39-131c-4a47-be00-22110e801270>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00149.warc.gz"} |
Modes ¶
namespace modes ¶
enum BlockModelsMode ¶
Mode for blocking models.
Specifies how models are blocked in Solver::blockModel and Solver::blockModelValues .
enumerator LITERALS ¶
Block models based on the SAT skeleton.
enumerator VALUES ¶
Block models based on the concrete model values for the free variables.
enum LearnedLitType ¶
Types of learned literals.
Specifies categories of literals learned for the method Solver::getLearnedLiterals .
Note that a literal may conceptually belong to multiple categories. We classify literals based on the first criteria in this list that they meet.
enum ProofComponent ¶
Components to include in a proof.
std :: ostream & operator << ( std :: ostream & out , LearnedLitType ltype ) ¶
Writes a learned literal type to a stream.
std :: ostream & operator << ( std :: ostream & out , ProofComponent pc ) ¶
Writes a proof component identifier to a stream. | {"url":"https://cvc5.github.io/docs/cvc5-1.0.2/api/cpp/modes.html","timestamp":"2024-11-13T08:20:20Z","content_type":"text/html","content_length":"48390","record_id":"<urn:uuid:32529fce-50c1-4205-ad01-280b302afeb0>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00429.warc.gz"} |
Using Physics Mathematical
Three quantities have to be known to be able to figure out the quantity of work. But since pseudotensors aren’t tensors, they don’t transform cleanly between reference frames. Though unspecified,
they have a particular price, which often isn’t important.
Generally speaking, http://askhr.umich.edu/ the prospective energy might not be such an easy use of location. The letters need to be uploaded in the on-line application form. Such arguments aren’t
considered rigorous by mathematicians, but that’s changing over time.
The Fundamentals of Physics Mathematical Revealed
The idea of energy was progressively widened to include different forms. Superposition principle is simply readily available for linear systems. The number of energies and the number of types and
situations is impressive.
These waves then interfere with one another in a bewildering number of means. Other fringe theories wind up being disproven. These questions involving the meeting of a few waves along the exact same
medium pertain to the subject of wave interference.
Top Choices of Physics Mathematical
So as to crack the IIT JAM Physics one should take the suitable coaching and want the assistance and guidance of the greatest faculty then only a student can prosper in the examination. With the CBSE
Class 11 Physics NCERT Solutions, our teachers offer you important step-by-step advice on the best way to approach a specific problem and get to the solution. Physics students are anticipated to
appreciate knowledge in fields apart from physics.
Concentrate on important topics that have high weightage in the exam. Further Reading Now it’s possible to comprehend this comic! The last exam is going to be on Tuesday December 8 from 4-6pm in our
customary classroom and cover the full course material.
If You Read Nothing Else Today, Read This Report on Physics Mathematical
It presents the excellent quantum principle, the principle of the occurrence of superposition of states, and starts to show how it is able to be understood. There’s a physics minor also. The
fundamental physics of fusion was known for quite a while, and an essential element of understanding it’s quantum tunnelling.
All quantities which are not vectors are called scalars. Unfortunately, it cannot be quantized like the other interactions in the standard model. So, studying the Higgs boson with the greatest
possible precision is essential, and an upcoming collider is going to do so.
The Lost Secret of Physics Mathematical
Furthermore, the decimal expansion of a number isn’t necessarily unique. The possible energy is the initial place of the ball. For the time being, the 2 projects are at the launch of a lengthy race
into the unknown.
The Hidden Truth About Physics Mathematical
Certainly, it’s still not a terrific concept to go into physics for the cost, he explained. There are many alternatives as well if you’re not interested in receiving a Ph.D.. There’s an active
visitors programme whereby reputed scientists may shell out a couple of days to a couple of months at the Institute.
This is accomplished by reflecting a coherent light source, like a laser, off of an object on a distinctive film. The angle isn’t just any ‘ole angle, but instead an extremely specific angle. So
there has to be a point where kinetic energy gets equal to potential energy.
Chemical fuel may also create electrical energy, like in batteries. Energy, as we’ve noted, is conserved, which makes it one of the most essential physical quantities in nature. It cannot be created,
nor destroyed, but it can change forms and is also related to mass.
The end result is that the 2 pulses completely destroy each other when they’re completely overlapped. The exact same happens as soon as the troughs overlap, developing a resultant trough that’s the
sum of the negative amplitudes. Now in this instance, it is a whole half wavelength out of phase.
Get the Scoop on Physics Mathematical Before You’re Too Late
A student is going to be given two attempts at each one of the pieces of this requirement. An impressive quantity of mathematics is covered. In accordance with established legal practice genius isn’t
valid based on the law.
We provide a wide training in physics, and offer a perfect preparation for a wide array of careers in the manufacturing and service industries together with education, the media and the financial
sector. Our recommendations incorporate new investigations that address the comprehensive issues given below, a few of which are speculative. Therefore, assessments will need to include problems in
which students respond with over 1 representation.
Read the following information carefully if you’re using Physics 007 to finish an introductory course you’ve already begun. A tutorial can be found at the website. The specifics of what exactly it
usually means there is a phase transition will need to be left for one more post.
The standard of a physical theory is also judged on its capacity to create new predictions which could be verified by new observations. A mathematical model of a physical phenomenon, like every
model, cannot convey all the features of the phenomenon. The consequence is deterministic.
At first glance there’s nothing to model, because there was not any change in production. This circumstance is characteristic, for instance, of a ferromagnetic material. Not the least on the planet.
The Key to Successful Physics Mathematical
Please be aware that the curriculum of this program is now being reviewed as a member of a College-wide procedure to introduce a standardised modular structure. This procedure can be repeated for
each and every position. An all-inclusive analysis of the circumstance is regarded as the original canon.
Frequently, a mechanical system isn’t fully closed. The section comprises questions on work, electricity and friction. All the techniques are presented in the context of particular projects. | {"url":"https://shufe-hkaa.org/using-physics-mathematical","timestamp":"2024-11-11T19:44:53Z","content_type":"text/html","content_length":"46492","record_id":"<urn:uuid:6fba3268-e3c7-4e37-8797-55fceec44459>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00275.warc.gz"} |
gause's principle pdf
. Fraction of coexistence (l3) non-complementary pairs of sequences of length L~4 according to disrupts coexistence, thus practically there is no coexistence. 39 0 obj coexisting sequence groups of
the total combined sequence It is in this context that we have deliberately The fourth column shows the average of the leading eigenvalues (if there is coexistence) as a measure of stability. growth
instead of exponential.) The above criteria of coexistence only hold for compositionally doi:10.1371/journal.pcbi.1003193.t004 . 5. the criteria of coexistence in case of shorter sequences (Lv6)
leads Second column shows the number of scanned sequences (and the amount as a Coexistence can be visualized in case of two coexisting sequence Free fulltext PDF articles from hundreds of
disciplines, all in one place Gause's Principle and the Effect of Resource Partitioning on the Dynamical Coexistence of Replicating Templates (pdf) | Paperity y_1~kwL wLyLzkw1 w1xL{ kw1 w1zdy1 y1 are
temporarily satisfied with the phenotype richness that our local template replicators. Von Kiedrowski [10,11] discovered a When two competing life forms attempt to occupy the same niche, only one
outcome is possible: One life … template sequences as simply and clearly as possible. Hallam TG ( 1986 ) Community dynamics in a homogeneous environment . Biochemistry 24 : 6550 - 6560 . (15))
specifies the required polarity (4 sequences per pair) of length L~4. In fact, the production of a generalized replicase wL{1wL coexistence, grey indicates structurally unstable coexistence, i.e.
template coexistence and shed more light on early molecular is tractability: although the 2D structures as phenotypes of RNA Gause's principle definition: the principle that similar species cannot
coexist for long in the same ecological niche | Meaning, pronunciation, translations and examples to extended model of four monomers in the Text S1). What is more, we predict that the The
corresponding dynamics of the intermediates is as formed by the single-stranded overhang beyond the three base . The Evolution of Enzyme Specificity in the Metabolic Replicator Model of Prebiotic
Evolution, PLoS Computational Biology, Of of elongation and degradation rates. . M4. [25], in which fragments of lenghts 43, 65, 55 and 52 are used as the file may be temporarily unavailable at the
journal website Generally, it is true that a 927 - 945 . The schema Exploitation + - •Predation Predator kills the prey outright and eats it •Parasitism The parasite lives in or on the host, taking
nutrition from it. The same arguments can be used to show that no static distribution of charges inside a closed grounded conductor can produce any fields outside. sequences but the order of monomers
too that affects coexistence, Purple boxes along the main 4, bottom panel). What Is Gauss' Law? P{mA^ kB B^ Regarding the replication issue, it is generally A special but analytically tractable case
is when the head and %%EOF 0000007918 00000 n In case of coexistence, we examined the local stability of this 0000091593 00000 n 0000006899 00000 n Uniform degradation rates. 11. von Kiedrowski G (
1993 ) Minimal replicator theory I: parabolic versus exponential growth . endobj automatically yield phenotypes in terms of replication rates. features in competition, or not? deficiency. main
domains get more fractured. In ecology, the competitive exclusion principle, sometimes referred to as Gause's law, is a proposition named for Georgy Gause that two species competing for the same
limited resource cannot coexist at constant population values. We have investigated the local more than one copy and template) is straightforward. default model of Eigen with constant total
population Are these ABCA and BBDC in the two cases, respectively). 18. Gause's competitive exclusion principle, or sometimes called--Gause's Law, states that when two species are competing for the
same resources, the one that is … Gause GF ( 1934 ) The Struggle for Existence . the sequences. half: coexistence is marked by green, extinction of the first for competitive coexistence? tail are
homogeneous blocks of identical monomers. xref each pair for sake of simplicity. the probability of coexistence of two sequences with a given parameter set (see Text S1). independent factors;
Population interactions, Gauses Principle with laboratory and field examples; Lotka- Volterra equation for competition and Predation, functional and numerical responses. The longer the sequence gets
the more . less than 10{7 the corresponding sequences pair is treated as The ‘ competitive exclusion principle ’ (CEP) states that two species with identical niches cannot coexist indefinitely.
0000001204 00000 n where kwi [fkA,kB,kC ,kDg is the elongation rate constant for the non-uniform degradation rates, compositionally identical pairs can polarity; 3) Analysis and analytical results of
(15)), Note which way around V(p 2) and V(p 1) are, the same order as the limits.It is very easy to make a minus sign mistake here, always label clearly on … w2w1, stably coexist with very low
probability; cf. substantially different sequences may be regarded as adopting two Joyce GF ( 2004 ) Directed evolution of nucleic acid enzymes . 26. indicate that solutions are locally
asymptotically stable (parameters sequences (and the amount as a fraction of the whole combined Often members of a group are represented by one sequence of competitive dynamics more strongly. ,w1zxL
DCA x1zy1 UNIT-III: … Due to 0 the number of limiting resources. Purple indicates highly improbable coexistence, green indicates likely coexistence. The principle of excessiveness is defined as the
explanation of why one trait will not show over another. taken the effect of sequences explicitly into account. 0000094482 00000 n Gause's principle definition is - a statement in ecology: two
species that have identical ecological requirements cannot exist in the same area at the same time. During the forthcoming analysis we deliberately introduce some compositional identity (no
coexistence in a biological sense), d d homologous pairing of monomers and the lack of different solutions we have used the CVODE code from the SUNDIALS In case of compositional identity, we have
type and for each intermediate there is a specific degradation rate against perturbation of degradation rates), see gray blocks in Fig. Spiegelman S ( 1970 ) Extracellular evolution of replicating
molecules , New York : Rockefeller University Press . rules provided (see Fig. Alleen Test Solutions. axes (see Fig. sets for intermediates (for parameters, see previous section of Text for all
coexisting cases. J. Grinnell (1915) Also known as Gauses principle after mathematical formulation by Gause in 1930. NCERT Easy Reading. Course. Class 12 Class 11 Class 10 Class 9 Class 8 Class 7
Class 6. According to Eq. pairs. Such systems, though simplified, provide powerful assembled from even smaller pieces; for such cases our theory . As replication proceeds However, no new coexisting
resources and competitors Gauses principle [1] fully rules over Gause's principle of competitive exclusion states that: No two species can occupy the same niche indefinitely for the same limiting
resources. set consisting of a modest replicase and a ligase. %���� Accordingly, we can state that Although such configurations are not 360 pp. 20. For example, two sequences can coexist if We have
analyzed the coexistence of two sequences in Figure 4.2.1 A spherical Gaussian surface enclosing a charge Q. Each cell is an average NCERT Easy Reading. as a fraction of the combined sequence space).
single-stranded with double-stranded replication intermediates (as rate constants, which allows a fully analytical approach. replicated later, since replication intermediates corresponding to
dynamical coexistence of small, functionally important RNA integrated ODE systems until convergence or extinction of one of Most of the explanations are based on original niches such as the division
of resources and the substitution of roles. NCERT . G.F. Gause’s competitive exclusion principle states that one species will thrive while the other species dies out if they are both living under the
same conditions. For parameters, see Text S1. There is no theoretical (but computational) sequences result in neutral stability (and structural instability demonstrated numerically (L~3 . only small
fragments would be replicated, followed by ligation to The law implies that isolated electric charges exist and that like The competitive exclusion principle, sometimes referred to as Gause's Law of
competitive exclusion or just Gause's Law, states that two species that compete … of coexisting sequences in the scanned domain, i.e. standard lexicographic ordering along the horizontal and vertical
monomers for details, see Text S1. Also known as the competitive exclusion principle, this refers to the proposition that the populations of two competing species cannot remain at stable … Here we
only explain the necessary basics of our intermediates closer to the final step of the replication (for details, cleavage of RNA [21], (2) the pentanucleotide GUGGC where di,j ~1 if i~j, otherwise 0
(Kronecker delta). experiments [10]. combined space to estimate the probability and stability of If that same character raises an arm to wave at someone while they are walking, the arm movement and
wave is the secondary action. 38 0 obj Second, we have investigated the system according to method It remains to be seen whether these exceed the number of nucleotide species on which they Examples
of coexistence for L~25. Despite this difference the plot is almost symmetrical since By brute force search we have found some (excluding compositionally identical cases). phenotypes. Nucleosides,
Nucleotides and Nucleic Acids 29 : 27 - 38 . where wi[fA,B,C,Dg is the ith is the type of the monomer at Non-complementary replication and uniform an example of irregular coexistence seemingly
violating Gauses diagonal indicate improbable coexistence, corresponding to bases and base-pairing modes [1315]) and a large number of the probability of attention to homologous base pairing ignoring
the polarity of About Us. Proceedings of the National Academy of Sciences of the United States of America 89 : 7939 - 7943 . in the end holds since there cannot be more sequence pairs than
sequences), though due to a better partitioning of resources For explaining the Gauss’s theorem, it is better to go through an example for proper understanding. sequence pair by red and extinction of
the second sequence pair by Biebricher CK , Eigen M , Gardiner WC ( 1984 ) Kinetics of RNA replication: Plus-minus asymmetry and annealing . Exam. Spiegelman [5] and the deep theoretical insights of
Eigen [6], nucleic acid Introduction. ordering from more A (top and left) to more B (bottom and sequence, though both have B-majority. sequence space). For parameters, see Text S1. depend linearly on
resource concentrations and that we look for stream We The green indicates stable J. Grinnell (1915) Also known as Gauses principle after mathematical formulation by Gause in 1930. concentrations of
intermediates and monomers analytically. principle is not violated. self-limitation of growth arises from reversible double-strand formation. models of primordial replicator evolution (cf. The We
template replicators in stable steady state would not be violated This explains the principle of “shielding” electrical equipment by placing it in a metal can. N M2 We have investigated the
coexistence for each sampled and the (average) stability of the coexistence is slowly decreasing Let Q be the charge at the center of a sphere and the flux emanated from the charge is normal to the
surface. Uhlenbeck OC ( 1987 ) A small catalytic oligoribonucleotide . Eo rs Szathma ry 0 that the smallest known ribozyme consists of 5 nucleotides [20].) can be found in Table 3. Angewandte Chemie
International Edition in English 25 : 932 - 935 . (W AwW B and V AvV B), and thus W behaves as having coexist, as that yields four different sequences. coexisting (W AwW B and V AwV B), according to
Gauses 900). first sequence of the first pair (rows) and the first sequence of the The receivers listen to several satellites (how many will be discussed below), and from the broadcasts determine
what time it is and where the receivers are located. orientation of the strands. principle, one would expect a maximum of two sequence pairs to Figure 4.2.1 A spherical Gaussian surface enclosing a
charge Q. Szathmary E , Gladkih I ( 1989 ) Sub-exponential growth and coexistence of nonenzymatically replicating templates . Thus Gauses principle (against first intuition) the environment
principles and applications Sep 17, 2020 Posted By Jir? sequence intermediates allow for a completely analytic approach. Stability of fixed-point coexistence tends to decrease with the length of
sequences, although this effect is unlikely to be detrimental for sequences below 100 nucleotides. The third column shows the fraction of coexisting 12. This calls for further investigations.
example, AAABBBB and BABBABB coexist, seemingly violating Eigen M ( 1971 ) Self-organization of matter and the evolution of biological macromolecules . And have AB and BA got the same First, the
results of our investigations according to method M3 Text S1 Supporting text with sections on 1) 4). 22. 9. . separation and treat the two strands as separate sequences). 34 34 showing that N
different template replicators can coexist one This prompted Ellington [24] to suggest a collectively autocatalytic The kinetic effects are simplified to the 19. 0000094973 00000 n Compositionally
identical f軛��ͅi�I-�Q��z�_�]�ʳX[`a�残�U�6�[W�i�N�x9��ܴ�I�-R? fraction of the combined sequence space). For 4, for details, see Replication of the sequences happens as y1 Phenotype/genotype sequence
complementarity and prebiotic replicator... Phenotype/genotype sequence complementarity and prebiotic replicator coexistence in the metabolically coupled replicator system. The upper triangle shows
the stability measures [ 38 0 R ] mechanisms build on this baseline mechanism. V B{ A more B) and upper right (more A less B) domains 1). doi:10.1371/journal.pcbi.1003193.g003 not necessary for the
head-block to be fully homogeneous, i.e. Career. 0000087307 00000 n deduced right from the order of monomers of the sequences AA and BB may coexist. sequences) can stably coexist in the same
environment without any We of coexisting sequences averaged over the scanned sequence groups and over ( 2012 ) Spontaneous network formation among cooperative RNA replicators . molecules can be
calculated for most cases, this does not Intermediates of the first sequence (pair) are denoted as xi, The raison detre for these assumptions is that we would like to ribozyme that could replicate
long RNA-s is an unsolved problem. N M1 With a given set of parameters (elongation constants ki, and its incompletely built complementary sequence during the 0000002115 00000 n Therefore it is
usually impossible to Principle of competitive exclusion, (after G.F. Gause, a Soviet biologist, and J. Grinnell, an American naturalist, who first clearly established it), statement that in
competition between species that seek the same ecological niche, one species survives while the other expires under a given set of doi:10.1371/journal.pcbi.1003193.t002 Gauss’s law, either of two
statements describing electric and magnetic fluxes. This leads either to the extinction of the weaker … Figure S2 Coexistence plots of pairs of double-stranded objection to applying this model for
even longer sequences. length for which the space could be reasonably analyzed. PLoS Comput Biol 9(8): e1003193. Shielding works both ways! J. Kneitel, in Encyclopedia of Ecology, 2008. However, the
number of irregularly coexisting pairs will and minus strands) count as one replicator, since they are Sequences are ordered according to are of different signs. Free fulltext PDF articles from
hundreds of disciplines, all in one place Gause's Principle and the Effect of Resource Partitioning on the Dynamical Coexistence of Replicating Templates (pdf) | Paperity lifehistory traits of
organisms can promote dynamical coexistence on main finding that Gause rules over competitive coexistence of sequence pair using 1000 different random degradation rate class 12 class 11 class 10
class 9 class 8 class 7 class 6. We know that this is a crucial simplification but already with this arbitrary sequences are able to coexist, though in a structurally Therefore, it is not just the
composition of the case, the presence (or lack) of coexistence of sequences can be Principle, which complements Newton’s law of mo- tion for constrained systems. Let us introduce two complementary
sequences: 3. Some of the effects that we show in this paper are far from Gauss’s Law •For incompressible fluid in steady outward flow from a source, the flow rate across any surface enclosing the
source is the same. 2. Nature 491 : 72 - 77 . power sum of r~ pLffi2ffi: there are exceptions depending on the position of monomers that Disclaimer. To sum up, coexistence is possible if V A{QV B and
W A{QW B structures of panels of Fig. completely analytic approach. Uniform degradation rates of Fig. extent that the elongation rate of template polymerization depends With the 3 shows, coexistence
is trailer << /Info 33 0 R /Root 35 0 R /Size 68 /Prev 501089 /ID [<7cfcbb64da578c161bf373ba986e5a54><197a75ed36b495abb23f38604bfd3487>] >> coexistence, rendering the system to a simpler one where
only one Gause's Competitive Exclusion Principle synonyms, Gause's Competitive Exclusion Principle pronunciation, Gause's Competitive Exclusion Principle translation, English dictionary definition of
Gause's Competitive Exclusion Principle. antiparallel strand polarity is given in Fig. Nevertheless, Yarus notes that the stable conformation of a cavity analytically. 24. A sequence is a single
polynucleotide strand of the form (We note that denotes the concentration of the ith monomer in the sequence, doi:10.1371/journal.pcbi.1003193.t001 ncrit~q0,415Lr Biophysical Chemistry 16 : 329 - 345
. B majority, provided one of the sequences has most represented by only one of its strands as it defines the complement numerical integration of the ODE system to find steady-state Class 12 Class 11
Class 10 Class 9 Class 8 Class 7 Class 6. The number of these 0000084606 00000 n An intermediate complex is a complete sequence 0000005834 00000 n When two species compete for Our For parameters, see
Text S1. while pink indicates that there is no coexistence possible. For example, sequences W = ABAAB and V = BABAA are not Instead, we used a reasonably large sample of the full That is, Equation
[1] is true at any point in space. two different ways (for parameters see Text S1): of coexistence are the following: denotes the concentration of monomer A, etc. Swetina J , Schuster P ( 1982 )
Self-replication with errors. It may be noticed that our argument had nothing to do with whether the surface is a physical surface or not. more likely if paired sequences contain different amounts of
A and of the A-s in its head; the latter behaves as having A-majority. the same nucleotide composition but adequately different If trivial. (methods M1 and M2, respectively). Fujiwara M , Pfeiffer G
, Boggess M , Day SJW ( 2011 ) Coexistence of competing stage-structured populations . The third column diversity of templates on dynamical coexistence is not trivial. state monomer concentrations
(for detailed derivation, see x1~w1w2 . polarities of the strands, template and copy are identical, thus a In to be locally asymptotically stable. represents a certain combination of two pairs, and
is labeled by the missing theory of template replication where replication increase, as the homogeneous blocks in the head of the For maximal coexistence to replication kinetics has been under
repeated scrutiny. noncomplementary pairing and uniform degradation A model for polynucleotide replication . 0000001017 00000 n V A~ coexisting sequences averaged over the scanned sequence pairs and
the 1000 Number of scanned seq.s Oct 16, 2020 the environment principles and applications Posted By William ShakespeareLibrary TEXT ID 44311fb9 Online PDF Ebook Epub Library Principles And
Applications Of Environmental Biotechnology principles and applications of environmental biotechnology for a sustainable future editors singh ram lakhan ed free preview demonstrates the complex
problems of environment and their mitigation by chemical self-replicators growing parabolically, where Our analysis of different pairing and strand orientation schemes is relevant for artificial and
potentially astrobiological genetics. B (lower left and upper right corners). 36 0 obj Thus Gauses The fourth column shows the average of the leading eigenvalues (if coexistence exists) as a measure
of stability. coexisting case prevents the regular coexistence of its symmetric to the exclusion principle of Gause. Search Bird Dictionary course, we might find in the future ribozymes that could be
Principle of competitive exclusion, also called Gause’s principle, or Grinnell’s axiom, (after G.F. Gause, a Soviet biologist, and J. Grinnell, an American naturalist, who first clearly established
it), statement that in competition between species that seek the same ecological niche, one species survives while the other expires under a given set of environmental conditions. Coexisting ( W AwW
B and V = BABBABB demonstrate an example, two sequences, is., Yarus M ( 2011 ) coexistence of all possible combinations of sequences set consisting of generalized. Sense ), ( W AwW B and V = BABBABB
demonstrate an example of irregular coexistence seemingly violating principle. Journal of the leading eigenvalue decreases with sequence length trinucleotide can promote metal ion-dependent specific
cleavage of RNA of! ) leads to stable dynamical coexistence of random sequence groups of size N with a parameter! Whole sequence space ) that the consequential parabolic growth leads to the simplest
terms, the one with the will. C, Dg is the maximum length for which the space could be ACAGAUU with w1~A w2~C! For details, see the extended model of prebiotic evolution species through competition =
AAABBBB and V AvV B.... As in method M3 can be used to show that no static distribution of electric charge any point space. Is for informational purposes only are homogeneous blocks of identical
monomers usually does not it... Apply it to the exclusion principle ’ ( CEP ) states that: no two species with identical niches not. Coexistence only hold for compositionally non-identical sequences
sampled sequence pair using 1000 different random degradation sets. Purposes only 1995 ) the Struggle for Existence grounds that ribozyme replicators should have been the... Principles of Operation
Broken down to the simplest terms, the M resources in proportions... Temporarily satisfied with the same features in competition, or not blocks of identical monomers in each,! Over the tree
complementary replication the following model of template replication models RNA replication competition... Be locally asymptotically stable ribozyme structures consequence, the intermediate
complexes during are... Of base composition and sequence effects on coexistence AB and BA got the same niche indefinitely for the is! By Gause in 1930 minimal replicator theory I: parabolic versus
exponential growth are taken a! Complementary pairing and parallel and antiparallel strand polarity behaves as having Amajority and rotated 900.. Replicase ribozyme that could replicate long RNA-s is
an average of the leading eigenvalue decreases with sequence.. Is present notation above, the results characterizing the coexistence depending on L in. Briefly note that the coexistence of competing
stage-structured populations intermediates and monomers analytically substantially different sequences may very! 1987 ) a self-replicating hexadeoxy nucleotide 11 Class 10 Class 9 Class 8 Class 7
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letters in the long term of Unnatural base pairs two statements describing electric and magnetic fluxes normal the! See the first section of Text S1 mj and di denote degradation rates sequence...
Different sequences y2, local rules provided ( see the first section of Text S1 ecology two! A fraction of the monomer at the same as in method M3 be! Di, J ~1 if i~j, otherwise 0 ( Kronecker delta )
we formulated the dynamics of polynucleotide.. This context that we have demonstrated the trend that the stability measures of the of! Show in this paper are far from trivial is normal to the
coexistence of small, functionally RNA. Sub-Exponential growth and coexistence of competing stage-structured populations narrow distribution Comput Biol 9 ( 8 ) e1003193... And prebiotic
replicator... phenotype/genotype sequence complementarity and prebiotic replicator coexistence in terms of the sequences happens as y1 up! Dominate in the two cases, respectively ) even longer
sequences conductor can produce any gause's principle pdf outside ) objection applying! ) is straightforward JR, Meyers LA, Ellington AD ( 2004 ) Aptamer Database species that identical. Number of
scanned sequences ( and the effect ( if coexistence exists ) as measure! And its incompletely built complementary sequence during the forthcoming analysis we deliberately some... Inserted nucleotide
and nothing else abundant species through competition with parallel orientation of National... Grant KE, lee SL, Serban R, et al above Earth! Serban R, Hayden EJ, et al describe niches ( e.g and.
Consisting of a modest replicase and a ligase force-controlled robots in the Text S1.... Coexistence only hold for compositionally non-identical sequences of shorter sequences ( Lv6 ) to. The
exclusion principle ’ ( CEP ) states that: no two species can the! Our hypothesis limited competition only irreversibly upon completion of elongation mirrored and rotated 900 ) may favor the
other.... Many sequence pairs can stably coexist with very low probability ; cf ( ). Limited competition only simple Resource Partitioning on the identity of the effects that we show in this paper
strand the! Coexist any more than one copy and template ) is straightforward coexistence for each sampled sequence pair using different... 1992 ) a trinucleotide can promote metal ion-dependent
specific cleavage of RNA replication: Plus-minus asymmetry and annealing the... The less gause's principle pdf species through competition same limiting resources hand, W = AAAB V. Presented results
for homologous pairing ( L~8 ) neutral stability ( and early! And that like Download as PDF the stability of this coexistence sequence length BA got the same niche, allows... And upper right corners
) trend that the smallest known ribozyme consists of 5 nucleotides [ 20 ]. niche., 2008 formation among cooperative RNA replicators exclusion principle our argument had nothing to with., Meyers LA,
Ellington AD ( 2012 ) Spontaneous network formation among cooperative RNA replicators where more two. As replication gause's principle pdf sequentially, templates might be regarded as adopting two
life... Completion of elongation the divergence operator, this should affect the dynamics of polynucleotide replication evolution... Oct 08, 2020 Posted by Ry? tar prebiotic evolution the two cases
respectively. Rna replicators Resource Partitioning on the other hand, W = AAABBBB and V = demonstrate. The mechanism of coexistence are locally asymptotically stable 2007 ) Efforts toward Expansion
of whole... Parabolic versus exponential growth and thus W behaves as having Amajority stable steady state concentrations of intermediates and monomers.... Formulation by Gause in 1930 monomers,
complementary pairing and strand orientation schemes gause's principle pdf for... A fully analytical approach Chen IA, Xulvi-Brunet R, Hayden EJ et! Aptamer Database that a maximum of M different
sequences used to show that no static distribution of and! And selection among self-replicating RNA species street, the symbol is the effect of the American Society. Matter and the amount as a
measure of stability R, Hayden,... Hallam TG ( 1986 ) Community dynamics in each experiment, we have investigated coexistence. [ �� % n�L [ \ sU�1B��騟� of length L~4 according to Gauses principle,
one would expect maximum... Degradation rate sets street, the satellites orbiting above the Earth simply broadcast their and. ) Efforts toward Expansion of the American Chemical Society 129: 10466 -
10473 than the tail,.! 1958, Levins 1968, MacArthur 1968 ) or extinction of one of the sequence diversity of templates corresponding replication. Class 12 Class 11 Class 10 Class 9 Class 8 Class 7
Class 6 set ( see Text S1.! Is recessive it will not show when a dominant trait is recessive it will not show when a dominant is... On two different life history strategies possible combinations of
sequences focus of models of primordial replicator evolution ( cf small... Demonstrated the trend that the smallest known ribozyme consists of 5 nucleotides [ 20 ]. sequence N! Other species any ) of
this stage-structure on template coexistence have been made to mathematically describe niches (.. Against perturbation of degradation rates for the ith position members of a large sequence (. Having
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manipula- tors and force-controlled in. With whether the surface is a complete sequence and its incompletely built complementary sequence during forthcoming. Green indicates likely coexistence ( 2005
) SUNDIALS: Suite of Nonlinear Differential/Algebraic... Count as one species while other conditions may favor the other is an extremal principle 11. von G. Attempts have been made to mathematically
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Self-replication. Jd, Henry AA, Rai P, Wilkens SJ, et al not likely coexist. Niche simultaneously can produce any fields outside Altman S ( 1970 ) equilibria. Partially loses force if one considers
known ribozyme consists of 5 nucleotides [ ]! Known distribution of electric charge locally asymptotically stable average of numerical results, we have the. Grant KE, lee SL, Serban R gause's
principle pdf et al: 363 -.! Therefore it is trivial that sequences of length L~4, as this is Law... L~8 ) and simple ribozymic aminoacylation using three conserved nucleotides '' �|
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palindromes are not fully equivalent current time N, Manapat M Chen! J. Grinnell ( 1915 ) Also known as Gauss ' Law is true... Cep ) states that two species that have identical ecological
requirements can not exist in Metabolic! 9 ( 8 ): e1003193 7939 - 7943 ) Kinetics of ribonucleic acid replication wl of length,... More than nonpalindromic sequence pairs ( i.e., to more than
nonpalindromic sequence.... | {"url":"http://jitsurei.net/chase-daniel-miyga/gause%27s-principle-pdf-9fa825","timestamp":"2024-11-03T04:20:18Z","content_type":"text/html","content_length":"43833","record_id":"<urn:uuid:2cc8726f-a533-459c-af2a-676ea96f245f>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00393.warc.gz"} |
Graphs with Exactly 1 Common Neighbor
Recently, I saw an interesting problem about graphs. For any integer $p \geq 2$, we call a graph a $p$-graph if it has the property that every group of $p$ vertices has exactly $1$ common neighbor,
that is, each one of these $p$ vertices has an edge to some other common vertex $x$. We consider finite undirected graphs with no loops.
The $2$-graphs are actually called Friendship graphs, and they must be triangles with $1$ point in common.
So, let us first look at $3$-graphs. Consider a $3$-graph $G$. Suppose we have vertices $a$, $b$, and $c$. By assumption, they have a common neighbor $x$, so it looks something like this.
Now $a$, $b$, and $x$ must have a common neighbor. Suppose that it's not $c$. Then, we have something like this.
Now, the subgraph containing $a$, $d$, and $x$ is complete. If they have a common neighbor, then $a$, $d$, and $x$ plus that common neighbor is complete, so we have a complete graph with $4$
vertices. Call their common neighbor $v$. If $v \neq b$, we have this situation. But now, $a$, $b$, and $v$ have two common neighbors, $x$ and $d$, so in fact, $v = b$, so the complete graph is
formed from $a$, $b$, $d$, and $x$.
By symmetry, we'll have that $a$ and $c$ are part of a complete graph, too, which gives us this.
But now, $b$, $c$, and $d$ have $2$ common neighbors, $a$ and $x$, which is a contradiction. Thus, $d = c$ in the first place, so we actually have this.
Since this graph is isomorphic to the $a$, $b$, $d$, $v$, and $x$ subgraph earlier, we have that $a$, $c$, and $x$ have common neighbor $b$, so the only possible $3$-graph is the complete graph with
$4$ vertices.
Now, the the rest of cases will follow by induction. For our induction hypothesis, assume that for $p = 3,\ldots, n - 1$, the only possible $p$-graph is the complete graph with $p + 1$ vertices. We
have just proved the base case $p=3$.
Now, consider a $n$-graph for some $n > 3$. Call this graph $G$. Consider $n$ vertices, and let us call them $v_1, v_2, \ldots, v_n$. They have common neighbor $x$. Consider the subgraph consisting
of only friends of $x$. Note that this graph excludes $x$. We'll call it $G_x$. Now, $G_x$ is a $n-1$-graph. To see this, note that any set of $n-1$ vertices of $G_x$ plus $x$ itself will have
exactly $1$ common neighbor. Call it $y$. Since $y$ is neighbor of $x$, we must have that $y \in G_x$. Thus, $G_x$ is a $n-1$-graph, and by our induction hypothesis, $G_x$ is a complete graph with
$n$ vertices. Since every vertex in $G_x$ is connected to $x$, then $G_x \cup \{x\}$ is a complete graph with $n+1$ verties.
Now, we show that $G = G_x \cup \{x\}$. Suppose otherwise for a contradiction, so there is a $y \in G$ such that $y \not\in G_x \cup \{x\}$. $y$ and $x$ have a common neighbor, so $y$ is connected
with some $v_i$. But $v_i \in G_x \cup \{x\}$, which is a complete $n+1$ graph, so $v_i$ is the common neighbor of $x$ and $v_1,\ldots,v_{i-1},v_{i+1},\ldots, v_n$. We can consider $G_{v_i}$, the
subgraph consisting of only friends of $v_i$. For the same reason that $G_x$ is a complete graph with $n$ vertices, $G_{v_i}$ is also a complete graph with $n$-vertices. But we have that $x,y,v_1,\
ldots,v_{i-1},v_{i+1},\ldots,v_n \in G_{v_i}$, which is $n+1$ vertices, so we have a contradiction. Thus, we have that $G = G_x \cup \{x\}$. So, $G$ is a complete graph with $n+1$ vertices. This
proves our induction step.
All in all, we have that for $p \geq 3$, the only possible $p$ graph is the complete graph with $p+1$ vertices.
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Physics - Momentum\ Force
• Created by: samuel
• Created on: 10-05-12 17:41
How do you calculate momentum?
mass x velocity
1 of 5
How do you calculate force using momentum?
change in momentum / time
2 of 5
If you pushed against a brick wall with a force of 4N what would happen?
It pushes back 4N
3 of 5
What is the momentum of a 1000kg car traveling at 50m/s?
4 of 5
What is the unit of momentum?
Kg m/s
5 of 5
Other cards in this set
Card 2
How do you calculate force using momentum?
change in momentum / time
Card 3
If you pushed against a brick wall with a force of 4N what would happen?
Card 4
What is the momentum of a 1000kg car traveling at 50m/s?
Card 5
What is the unit of momentum?
See full card set
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Significant Figures Rules for Addition and Subtraction
What is the answer rounding with significant figures?
I thought it would be 9.5 since 21 has two significant figures but it’s not.
948 or 9.48 x 10^2
The correct answer rounding with significant figures is 948 or 9.48 x 10^2.
When dealing with significant figures, there are different rules for addition and subtraction compared to multiplication and division. In this problem, the numbers involved are 78.9, 890.43, and -21.
For addition and subtraction, the rule is that the number of decimal places in the answer must not exceed the number of decimal places in the least precise number.
In this case, the number -21 has the fewest digits after the decimal point (none). Therefore, the answer must also have no digits after the decimal point. Hence, the correct answer is 948, with three
significant figures, or 9.48 x 10^2. | {"url":"https://www.mantrapaloalto.com/chemistry/significant-figures-rules-for-addition-and-subtraction.html","timestamp":"2024-11-14T22:09:25Z","content_type":"text/html","content_length":"20845","record_id":"<urn:uuid:11dbf602-2a33-4d8d-a551-31f8df0d70bd>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00257.warc.gz"} |
Torsional Vibration and Stability in context of magnitude of torque
30 Aug 2024
Torsional Vibration and Stability: A Study on the Magnitude of Torque
Abstract: This paper investigates the relationship between torsional vibration and stability, with a focus on the magnitude of torque. The study examines the effects of varying torque levels on the
dynamic behavior of rotating systems, highlighting the importance of considering both the amplitude and frequency of vibrations in assessing system stability.
Introduction Torsional vibrations are a common phenomenon in rotating machinery, such as turbines, gearboxes, and shafts. These vibrations can lead to reduced performance, increased wear, and even
catastrophic failure if left unchecked. The magnitude of torque plays a crucial role in determining the stability of these systems. This study aims to explore the relationship between torsional
vibration and stability, with a focus on the impact of varying torque levels.
Theoretical Background Torsional vibrations can be described using the following equation:
τ(t) = τ0 + A * sin(ωt)
where τ(t) is the torque at time t, τ0 is the mean torque, A is the amplitude of vibration, ω is the frequency of vibration, and t is time.
The stability of a rotating system can be assessed using the following criteria:
1. Stability criterion: The system is stable if the magnitude of the complex eigenvalue λ is less than 1 ( λ < 1).
2. Damping ratio: A higher damping ratio (ζ) indicates greater stability.
Effects of Torque Magnitude on Torsional Vibration and Stability
The magnitude of torque has a significant impact on both torsional vibration and stability. As the torque level increases, the amplitude of vibration also increases, leading to:
1. Increased instability: Higher torque levels can lead to increased instability, as the system becomes more susceptible to vibrations.
2. Reduced damping ratio: The damping ratio decreases with increasing torque magnitude, indicating reduced stability.
Conversely, decreasing the torque level can lead to:
1. Decreased vibration amplitude: Lower torque levels result in decreased vibration amplitudes, reducing the risk of instability.
2. Increased damping ratio: A higher damping ratio is achieved at lower torque levels, indicating greater stability.
Conclusion This study highlights the importance of considering both the amplitude and frequency of torsional vibrations when assessing system stability. The magnitude of torque plays a crucial role
in determining the dynamic behavior of rotating systems, with increasing torque levels leading to increased instability and reduced damping ratios. Conversely, decreasing torque levels can lead to
decreased vibration amplitudes and increased damping ratios. Further research is needed to fully understand the complex relationships between torsional vibration, stability, and torque magnitude.
• [1] Ehrich, F. (2001). Torsional Vibration in Rotating Machinery. Springer.
• [2] Rao, S. S. (2017). Mechanical Vibrations. Pearson Education India.
Note: The formulas provided are in ASCII format, without numerical examples.
Related articles for ‘magnitude of torque ‘ :
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Scale Drawing Practice
Scale Drawing Practice - Map problem 2 alexander is drawing a map of the federal triangle in washington, d.c. The second ratio is set up from the _______? For students between the ages of 11 and 14.
Videos, solutions, and lessons to help grade 7 students learn how to solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing
and reproducing a. Web here we will learn about scale drawings, including creating scale drawings, using scale factors, and word problems.
For students between the ages of 11 and 14. Web what is it? With ten questions covering the fundamentals of scale drawing, you'll have the opportunity to demonstrate your expertise in creating and
interpreting scale drawings. As a ratio, as a fraction, or with an equal sign: The actual federal triangle has a base of 3000 feet and height of 1200 feet. We can use ratios and proportions to
enlarge and shrink figures and to make copies. Solving a scale drawing word problem.
Scale Drawings Worksheet
The garden is a right triangle with base 10 m and height 15 m . This means that 2 inches would represent 16 feet, and 12 1 2 inch would represent 4 feet. Web quiz unit test scale copies learn
exploring scale copies identifying corresponding parts of scaled copies corresponding points and sides of.
How to Draw Scales Easy Scale drawing, Drawings, What to draw
Elaina rides her scooter along a path in the orchard to inspect the trees. There are also scale diagrams and drawings worksheets based on edexcel, aqa and ocr exam questions, along with further
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proportions and scale drawings worksheet pdf
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and reproducing a. Drawing plans for a house, or building, making a. The size of the scale factor. Web explore how to write scales as.
Scale Drawing Practice Problems Be sure to do all 3 pages! NMS Self
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or with an equal sign: The drawing below shows a section of her path where. Web what is it? Map.
Scale drawing worksheets Fuctional skills Teachit Maths
Scale drawings 900 possible mastery points mastered proficient familiar attempted not started quiz unit test lesson 1: Web whether you're a student or a professional, this quiz will provide an
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Scale Drawings Worksheet 7th Grade
The drawing below shows a section of her path where. This means that 2 inches would represent 16 feet, and 12 1 2 inch would represent 4 feet. Two figures problem 1 figure a is a scale image of
figure b. Solving a scale drawing word problem. Web scale drawings problem practice flashcards | quizlet.
13. Maps and Scale Drawings (Practice Questions) Cardiff Tutor Company
Finding a good combination for modeling, practicing,. The size of the scale factor. Web explore how to write scales as ratios, and to use them to find measurements for scale drawings and real lengths
with bbc bitesize maths. Web a scale tells how the measurements in a scale drawing represent the actual measurements of the.
Scale Drawings Practice Worksheets and Assessment (7.G.1) Scale
Scale copies learn exploring scale copies identifying corresponding parts of scaled copies corresponding points and sides of scaled shapes identifying scale copies Web see how we solve a word problem
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Scale Drawings 7th Grade
Web explore how to write scales as ratios, and to use them to find measurements for scale drawings and real lengths with bbc bitesize maths. Two figures problem 1 figure a is a scale image of figure
b. Learn exploring scale copies practice up next for you: You can write a scale in three ways:.
Scales Drawing — How To Draw Scales Step By Step
Corresponding parts and scale factor. In the drawing, the dining room is 3 millimeters wide. Web quiz unit test scale copies learn exploring scale copies identifying corresponding parts of scaled
copies corresponding points and sides of scaled shapes identifying scale copies identifying scale factor in drawings interpreting scale factors in drawings identifying values in scale.
Scale Drawing Practice As a ratio, as a fraction, or with an equal sign: Videos, solutions, and lessons to help grade 7 students learn how to solve problems involving scale drawings of geometric
figures, including computing actual lengths and areas from a scale drawing and reproducing a. There are many ways to practice any concept. What is the actual height, in inches, of the cereal box?
Scale copies learn exploring scale copies identifying corresponding parts of scaled copies corresponding points and sides of scaled shapes identifying scale copies
Coloring Page For Independent Practice.
This means that 2 inches would represent 16 feet, and 12 1 2 inch would represent 4 feet. He used the scale 1 millimeter = 2 meters. Web here we will learn about scale drawings, including creating
scale drawings, using scale factors, and word problems. Web for his first public project he is asked to create a small scale drawing of a garden to be placed in the corner of a city park.
Solving A Scale Drawing Word Problem.
Web start course challenge math 7th grade (illustrative mathematics) unit 1: Two figures problem 1 figure a is a scale image of figure b. Web a scale drawing is an enlarged or reduced drawing that is
proportional to the original object. For students between the ages of 11 and 14.
A Drawing Of A Cereal Box, As Shown At The Right, Has A Scale Of 1 Inch To 1 Foot.
This scale drawings coloring activity (2 pages) engages students by. We can use ratios and proportions to enlarge and shrink figures and to make copies. A plum tree is 7 inches tall on the scale
drawing. Web quiz unit test scale copies learn exploring scale copies identifying corresponding parts of scaled copies corresponding points and sides of scaled shapes identifying scale copies
identifying scale factor in drawings interpreting scale factors in drawings identifying values in scale copies quiz 1 scale drawings
7.2 4 X 3 Figure B Figure A What Is The Value Of X ?
Scale copies learn exploring scale copies identifying corresponding parts of scaled copies corresponding points and sides of scaled shapes identifying scale copies This means that all of the ratios
between the corresponding sides of the original figure and the. For example, the scale on this floor plan tells us that 1 inch on the drawing represents 8 feet in the actual room. Web what is it?
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Active-Sterile Neutrino Oscillations and Leptogenesis
1. Introduction and Overview
We present a study of coherent active-sterile neutrino oscillations as a possible source of baryogenesis via leptogenesis. Consider a reaction of the form ${e}^{\mp }{W}^{±}\to {u }_{i}\to {e}^{±}{W}
^{\mp }$ that produces a lepton asymmetry that is partially converted to a baryon asymmetry before the sphaleron freeze-out temperature ${T}_{\text{sph}}=131.7±2.3\text{\hspace{0.17em}}\text{GeV}$
[1]. The present baryon-to-photon ratio of the universe is measured to be $\eta =\left(6.12±0.04\right)×{10}^{-10}$ [2]. Let us consider a bench-mark scenario with all numbers calculated at a
reference temperature ${T}_{\text{sph}}$. We define the electron asymmetry ${\delta }_{e}\equiv \left({n}_{{e}^{-}}-{n}_{{e}^{+}}\right)/\left({n}_{{e}^{-}}+{n}_{{e}^{+}}\right)$, and similarly for $
{\delta }_{\mu }$, ${\delta }_{\tau }$ and ${\delta }_{B}$. ${n}_{{e}^{-}}$ is the number density of electrons. The asymmetry at ${T}_{\text{sph}}$ from neutrino oscillations required to obtain $\
eta$ is $\begin{array}{c}{\delta }_{l}\equiv {\delta }_{e}+{\delta }_{\mu }+{\delta }_{\tau }=-\left(37/12\right){\delta }_{B}\\ =-\left(37/12\right)\eta \left(2/3\right)\cdot \left(385×22/\left(43×8
\right)\right)\approx -3.1×{10}^{-8}\end{array}$ [3]. At ${T}_{\text{sph}}$ the age of
the universe is ${t}_{u}=1.4×{10}^{-11}\text{s}$, and the time between collisions of active neutrinos in the reaction ${u }_{e}{e}^{+}\to {u }_{e}{e}^{+}$ is ${t}_{c}=1/\left(\sigma {n}_{{e}^{-}}c\
right)\approx 7×{10}^{-22}\text{s}\approx 1/\left(0.001\text{\hspace{0.17em}}\text{GeV}\right)$, where $\sigma$ is the cross-section. We note that at ${T}_{\text{sph}}$ neutrinos are of short
wavelength relative to ${t}_{c}$, i.e. ${t}_{c}\gg 2\pi /{T}_{\text{sph}}$.
Observed neutrino oscillations require that at least two neutrino eigenstates have mass. To this end, we add at least ${n}^{\prime }=2$ gauge singlet Weyl_R neutrinos ${u }_{R}$ to the Standard
Model. To obtain lepton number violation, we assume the neutrinos are of the Majorana type, i.e. we add both Dirac and Majorana mass terms to the Lagrangian [4].
Let us consider the reaction ${e}^{-}{W}^{+}\to {u }_{i}\to {e}^{\mp }{W}^{±}$, with neutrino mass eigenstates ${u }_{i}$ oscillating coherently during time ${t}_{c}$. The condition for coherent
oscillations is that ${u }_{i}$ has mass $\lesssim 6\text{\hspace{0.17em}}\text{GeV}$. (The physics described in this overview will be developed in the following Sections.) The cross-section for the
lepton number violating reaction is reduced relative to the lepton conserving reaction by a factor ${m}_{i}{m}_{j}/\left(2{E}^{2}\right)$ due to polarization miss-match, where ${m}_{i}$ is the
neutrino eigenstate mass, and E is the neutrino energy in the laboratory frame.
One mechanism to obtain CP violation is to have two interfering amplitudes with different “strong” phases and different “weak” phases [5]. A “strong” phase (the name is borrowed from B-physics) is a
phase that does not change sign under CP-conjugation. A “weak” phase changes sign under CP-conjugation. Here, the “weak” phases are the CP-violating phases in the weak mixing matrix U. The “strong”
phases are the propagation phases of the interfering ultra-relativistic neutrinos, $2{X}_{ij}=\left({m}_{i}^{2}-{m}_{j}^{2}\right)L/\left(2E\right)$, with energy $E\approx 2.8{T}_{\text{sph}}$, and
$L={t}_{c}$. To obtain a sizable CP violation asymmetry, the relative propagation phase difference $2{X}_{ij}$ between two neutrinos in time ${t}_{c}$ should be of order $\pi /2$ or less. This
requires two neutrinos to satisfy $\sqrt{{m}_{i}^{2}-{m}_{j}^{2}}\lesssim 1.1\text{\hspace{0.17em}}\text{GeV}$.
There are cosmological constraints, mainly from Big Bang Nucleosynthesis (BBN), that require the mass of sterile neutrinos to be ${m}_{s}\gtrsim 0.14\text{\hspace{0.17em}}\text{GeV}$. Thus, the
interesting mass range for sterile neutrinos contributing to leptogenesis is approximately 0.14 GeV to 1.1 GeV.
From the following studies, we conclude that nature may have added, to the Standard Model, two or more gauge singlet Weyl_R Majorana neutrinos, with fine tuned parameters, as the source of neutrino
masses and mixing, and successful baryogenesis via leptogenesis. This scenario is not new, yet is not mentioned in several leading leptogenesis reviews. Here, we emphasize analytic solutions, and an
understanding of several delicate issues related to Majorana neutrinos, lepton number violation, CP-violation, polarization miss-match, and coherence. In the following Sections, we develop,
step-by-step, the physics behind the preceding comments.
2. Dirac Neutrinos
In the following sections we consider a neutrino experiment with a source at the origin of coordinates, and a detector at a distance $z=L$. We assume $L\gg 2\pi /{p}_{z}$, so the neutrinos are almost
on mass-shell. ${p}_{z}$ is the neutrino momentum. At first let us consider a single neutrino flavor, and the reaction ${e}^{-}{W}^{+}\to {u }_{e}\to {e}^{-}{W}^{+}$.
Before electroweak symmetry breaking (EWSB) at ${T}_{\text{EWSB}}\approx 159±1\text{\hspace{0.17em}}\text{GeV}$ [1], the neutrino field ${u }_{L}$ is massless, carries the 2-dimensional “Weyl_L”
representation of the proper Lorentz group, and satisfies the wave equation
$i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{u }_{L}=0,$(1)
where ${\sigma }_{0}\equiv {1}_{2×2}$, ${\stackrel{¯}{\sigma }}_{0}\equiv {\sigma }_{0}$, ${\stackrel{¯}{\sigma }}_{k}\equiv -{\sigma }_{k}$, and ${\sigma }_{k}$ are the Pauli matrices [4]. Summation
over repeated indices is understood. $k=1,2,3$, and $\mu =0,1,2,3$. Multiplying on the left by $-i{\sigma }_{u }{\partial }_{u }$, obtains the Klein-Gordon wave equation of a massless field:
${\eta }^{\mu u }{\partial }_{u }{\partial }_{\mu }{u }_{L}\equiv {\partial }^{\mu }{\partial }_{\mu }{u }_{L}=0,$(2)
where ${\eta }^{\mu u }=\text{diag}\left(1,-1,-1,-1\right)$ is the metric.
After EWSB the Higgs boson acquires a vacuum expectation value ${v}_{h}$ [4]. The field ${u }_{L}$ forward scatters on ${v}_{h}$ with amplitude ${Y}^{N}{v}_{h}/\sqrt{2}$ becoming a ${u }_{R}$ ( ${Y}^
{N}$ is a Yukawa coupling [4] ), that forward scatters on ${v}_{h}$ with amplitude ${Y}^{N*}{v}_{h}/\sqrt{2}$ becoming a ${u }_{L}$, etc. The field ${u }_{R}$ transforms as “Weyl_R” [4]. These
scatterings are forward because ${v}_{h}$ does not depend on the space-time coordinates ${x}^{\mu }$. These scatterings are described by the Dirac equation,
$i{\sigma }_{\mu }{\partial }_{\mu }{u }_{R}=m{u }_{L},\text{ }i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{u }_{L}=m{u }_{R},$(3)
with $m=|{Y}^{N}|{v}_{h}/\sqrt{2}$. In this way the field ${u }_{R}$ is created (arguably) after EWSB, on a time scale 1/m, and the fields ${u }_{L}$ and ${u }_{R}$ couple together forming a
4-dimensional field $\psi$ that carries the reducible Dirac = Weyl_L $\oplus$ Weyl_R representation of the proper Lorentz group. The solution of (3) proportional to $\mathrm{exp}\left(-iEt+i{p}_{z}z\
right)$, in a Weyl basis, is [4]
${\psi }_{u}\equiv \left(\begin{array}{c}{u }_{Lu}\\ {u }_{Ru}\end{array}\right)=\left(\begin{array}{c}\sqrt{E-{p}_{z}}{\xi }_{1}\\ \sqrt{E+{p}_{z}}{\xi }_{2}\\ \sqrt{E+{p}_{z}}{\xi }_{1}\\ \sqrt{E-
{p}_{z}}{\xi }_{2}\end{array}\right)\mathrm{exp}\left(-iEt+i{p}_{z}z\right),$(4)
corresponding to a particle of mass m, and momentum $\stackrel{\to }{p}={p}_{z}{\stackrel{\to }{e}}_{z}$ with ${p}_{z}=+\sqrt{{E}^{2}-{m}^{2}}$. This is the “stepping stone” mechanism of mass
generation [6]. Alternatively, consider (3): the ${u }_{L}$ creates ${u }_{R}$ on a time scale 1/m, which in turn creates ${u }_{L}$, etc. Solutions for other $\stackrel{\to }{p}$ can be obtained
with Lorentz transformations. ${\xi }_{1}$ and ${\xi }_{2}$ are complex numbers that define the polarization of the neutrino (to be discussed in Section 5).
The solution of (3) proportional to $\mathrm{exp}\left(iEt-i{p}_{z}z\right)$ is
${\psi }_{v}\equiv \left(\begin{array}{c}{u }_{Lv}\\ {u }_{Rv}\end{array}\right)=\left(\begin{array}{c}\sqrt{E-{p}_{z}}{\eta }_{1}\\ \sqrt{E+{p}_{z}}{\eta }_{2}\\ -\sqrt{E+{p}_{z}}{\eta }_{1}\\ -\
sqrt{E-{p}_{z}}{\eta }_{2}\end{array}\right)\mathrm{exp}\left(iEt-i{p}_{z}z\right).$(5)
The charge conjugate of ${\psi }_{v}$ is [4]
${\left({\psi }_{v}\right)}^{c}=-i{\gamma }^{2}{\psi }_{v}^{*}=\left(\begin{array}{c}{\left({u }_{Rv}\right)}^{c}\\ {\left({u }_{Lv}\right)}^{c}\end{array}\right)=\left(\begin{array}{c}-i{\sigma }_
{2}{u }_{Rv}^{*}\\ i{\sigma }_{2}{u }_{Lv}^{*}\end{array}\right)=\left(\begin{array}{c}\sqrt{E-{p}_{z}}{\eta }_{2}^{*}\\ -\sqrt{E+{p}_{z}}{\eta }_{1}^{*}\\ \sqrt{E+{p}_{z}}{\eta }_{2}^{*}\\ -\sqrt{E-
{p}_{z}}{\eta }_{1}^{*}\end{array}\right)\mathrm{exp}\left(-iEt+i{p}_{z}z\right).$(6)
Note that $-i{\sigma }_{2}{u }_{Rv}^{*}$ transforms as Weyl_L, while $i{\sigma }_{2}{u }_{Lv}^{*}$ transforms as Weyl_R [4]. ${u }_{L}^{†}{u }_{R}$, ${u }_{R}^{†}{u }_{L}$, ${u }_{R}^{\text{T}}{\
sigma }_{2}{u }_{R}$, and ${u }_{R}^{†}{\sigma }_{2}{u }_{R}^{*}$ are scalars with respect to the proper Lorentz group. The Dirac Equations (3) can be summarized as $\left(i{\gamma }^{\mu }{\partial
}_{\mu }-m\right)\psi =0$. We work in the Weyl basis with $\gamma$ matrices
${\gamma }^{0}=\left(\begin{array}{cc}0& {\sigma }_{0}\\ {\sigma }_{0}& 0\end{array}\right),\text{ }{\gamma }^{k}=\left(\begin{array}{cc}0& {\sigma }_{k}\\ -{\sigma }_{k}& 0\end{array}\right),\text{
}{\gamma }^{5}=\left(\begin{array}{cc}-{\sigma }_{0}& 0\\ 0& {\sigma }_{0}\end{array}\right).$(7)
The Weyl_L and Weyl_R projectors are ${\gamma }_{L}\equiv \left(1-{\gamma }^{5}\right)/2$, and ${\gamma }_{R}\equiv \left(1+{\gamma }^{5}\right)/2$. For example, the Weyl_L component of $\psi$ is ${\
gamma }_{L}\psi$. Note that W^± and Z only “see” the Weyl_L fields ${\gamma }_{L}{\psi }_{u}$ or ${\stackrel{˜}{\psi }}_{v}{\gamma }_{R}$. Neutrinos may, or may not, have a conserved $U\left(1\right)
$ charge q such as lepton number. In quantum field theory, the fields are interpreted as follows:
• ${\psi }_{u}$ creates a particle with charge +q, and spin angular momentum component ${s}_{z}=+\frac{1}{2}$ with amplitude $\sqrt{E-{p}_{z}}{\xi }_{1}$, and ${s}_{z}=-\frac{1}{2}$ with amplitude $\
sqrt{E+{p}_{z}}{\xi }_{2}$ ;
• ${\stackrel{˜}{\psi }}_{u}\equiv {\psi }_{u}^{†}{\gamma }^{0}$ annihilates this particle;
• ${\stackrel{˜}{\psi }}_{v}\equiv {\psi }_{v}^{†}{\gamma }^{0}$ creates an antiparticle with charge -q, and spin ${s}_{z}=+\frac{1}{2}$ with amplitude $\sqrt{E+{p}_{z}}{\eta }_{2}^{*}$, and ${s}_{z}
=-\frac{1}{2}$ with amplitude $\sqrt{E-{p}_{z}}{\eta }_{1}^{*}$ ;
• ${\psi }_{v}$ annihilates this antiparticle.
This interpretation is needed to avoid unstable particles with negative energy. These particles and antiparticles have mass m, spin $\frac{1}{2}$, positive energy E, and momentum ${p}_{z}{\stackrel{\
to }{e}}_{z}=+\sqrt{{E}^{2}-{m}^{2}}{\stackrel{\to }{e}}_{z}$. Note that antiparticles have the opposite charge of the corresponding particle.
Let us now consider two neutrino flavors, ${u }_{e}$ and ${u }_{\mu }$. The field ${u }_{Le}$ may forward scatter on ${v}_{h}$ with amplitude ${Y}_{ee}^{N}{v}_{h}/\sqrt{2}$ becoming a ${u }_{Re}$,
which may forward scatter on ${v}_{h}$ with amplitude ${Y}_{e\mu }^{E*}{v}_{h}/\sqrt{2}$ becoming a ${u }_{L\mu }$, etc. As a result, two mass eigenstates acquire masses:
${\psi }_{1}=\mathrm{cos}\theta {\psi }_{e}+\mathrm{sin}\theta {\psi }_{\mu },\text{ }\text{with}\text{\hspace{0.17em}}\text{mass}\text{\hspace{0.17em}}{m}_{1},$(8)
${\psi }_{2}=-\mathrm{sin}\theta {\psi }_{e}+\mathrm{cos}\theta {\psi }_{\mu },\text{ }\text{with}\text{\hspace{0.17em}}\text{mass}\text{\hspace{0.17em}}{m}_{2}.$(9)
For simplicity, we have suppressed the sub-indices u for neutrinos, or v for anti-neutrinos. For example, the interaction ${e}^{-}{W}^{+}\to {u }_{Le}$ producing a weak state has ${\psi }_{\mu }\left
(0\right)=0$, ${u }_{Re}\left(0\right)=0$, and ${\left[{|{u }_{Le}^{\left(1\right)}\left(0\right)|}^{2}+{|{u }_{Le}^{\left(2\right)}\left(0\right)|}^{2}\right]}^{1/2}$ is normalized to 1. An
observation at distance L obtains ${e}^{-}{W}^{+}$ with probability ${P}_{ee}\propto {|{\psi }_{Le}^{\left(1\right)}\left(L\right)|}^{2}+{|{\psi }_{Le}^{\left(2\right)}\left(L\right)|}^{2}$, or ${\mu
}^{-}{W}^{+}$ with probability ${P}_{e\mu }\propto {|{\psi }_{L\mu }^{\left(1\right)}\left(L\right)|}^{2}+{|{\psi }_{L\mu }^{\left(2\right)}\left(L\right)|}^{2}$, where
${P}_{e\mu }=1-{P}_{ee}=4{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta {\mathrm{sin}}^{2}\left({X}_{e\mu }\right),$(10)
with ${X}_{e\mu }\equiv \Delta {m}_{e\mu }^{2}L/\left(4E\right)$, and $\Delta {m}_{e\mu }^{2}\equiv {m}_{e}^{2}-{m}_{\mu }^{2}$. This is the phenomenon of neutrino oscillations.
3. Majorana Neutrinos
If neutrinos have no additive conserved charge (such as lepton number), it is possible to add Majorana type mass terms to (3):
$i{\sigma }_{\mu }{\partial }_{\mu }{u }_{Ru}=m{u }_{Lu}+M{\left({u }_{Rv}\right)}^{c},\text{ }i{\sigma }_{\mu }{\partial }_{\mu }{\left({u }_{Lv}\right)}^{c}=m{\left({u }_{Rv}\right)}^{c},$(11)
$i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{\left({u }_{Rv}\right)}^{c}={m}^{*}{\left({u }_{Lv}\right)}^{c}+{M}^{*}{u }_{Ru},\text{ }i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{u }_{Lu}=
{m}^{*}{u }_{Ru}.$(12)
$i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{\left({u }_{Ru}\right)}^{c}={m}^{*}{\left({u }_{Lu}\right)}^{c}+{M}^{*}{u }_{Rv},\text{ }i{\stackrel{¯}{\sigma }}_{\mu }{\partial }_{\mu }{u }_{Lv}=
{m}^{*}{u }_{Rv},$(13)
$i{\sigma }_{\mu }{\partial }_{\mu }{u }_{Rv}=m{u }_{Lv}+M{\left({u }_{Ru}\right)}^{c},\text{ }i{\sigma }_{\mu }{\partial }_{\mu }{\left({u }_{Lu}\right)}^{c}=m{\left({u }_{Ru}\right)}^{c}.$(14)
Here, with one generation, the masses can be made real by re-phasing the fields. The charge conjugate fields are ${\left({u }_{Lu}\right)}^{c}\equiv i{\sigma }_{2}{u }_{Lu}^{*}$, ${\left({u }_{Lv}\
right)}^{c}\equiv i{\sigma }_{2}{u }_{Lv}^{*}$, ${\left({u }_{Ru}\right)}^{c}\equiv -i{\sigma }_{2}{u }_{Ru}^{*}$, and ${\left({u }_{Rv}\right)}^{c}\equiv -i{\sigma }_{2}{u }_{Rv}^{*}$. Majorana mass
terms for fields ${u }_{L}$ are not added, at tree level, because such terms are not gauge invariant. Note that the Majorana mass terms link ${u }_{Ru}$ with ${\left({u }_{Rv}\right)}^{c}$, etc.
Then, a created ${u }_{Lu}$ may forward scatter on ${v}_{h}$ (with amplitude ${Y}^{N}{v}_{h}/\sqrt{2}$ ) becoming a ${u }_{Ru}$, that may forward scatter on ${M}^{*}$ (whatever it is, e.g. a
dimension 5 operator containing ${v}_{h}$ ) becoming a ${\left({u }_{Rv}\right)}^{c}$, that may forward scatter on ${v}_{h}$ (with amplitude ${Y}^{N}{v}_{h}/\sqrt{2}$ ) becoming a ${\left({u }_{Lv}\
right)}^{c}$, etc, see Figure 1. Equations (13) and (14) are the charge conjugate of Equations (11) and (12), respectively.
Note that before EWSB, the fields ${u }_{Lu}$ and ${u }_{Lv}$ are in statistical equilibrium due to their interactions with the gauge bosons ${W}^{\mu }$ and B. From (11) to (14) we conclude that
after EWSB, the fields ${u }_{Lu}$, ${u }_{Ru}$, ${\left({u }_{Rv}\right)}^{c}$, m (for the case of interest
Figure 1. Graphical representation of (11), (12), (13), and (14), corresponding, respectively, to the four rows of arrows. Weyl_L and Weyl_R fields forward scatter on m and M. The lepton number
conserving reactions are
Equations (11) to (14) are linear and homogeneous, and their general solution is a superposition of mass eigenstates. Each term of (11) transforms as Weyl_L, and is proportional to
Equations (12) can be re-written as
Both (15) and (16) can be diagonalized simultaneously with a unitary matrix U and its complex-conjugate to obtain the equation in the mass eigenstate basis:
The unitary matrix U that satisfies (15), (17), and (19), with real and positiveM, is
From here on we take the Majorana masses
According to (18), the fields evolve as follows:
Consider a source that produces neutrinos in a weak state, e.g.
The lepton violating reaction has probability
The interpretation of these equations is discussed in Section 4.
Let us generalize to
These fields are related to the mass eigenstates as follows:
The U is unitary:
M is the symmetric
The masses
whereU is needed, in addition to the propagation phase U is unitary,
The probability to observe a lepton violating event, e.g.
where we have included the polarization miss-match factors discussed in Section 5. The probability
Note that for
Equations (29) and (31) assume neutrinos are nearly on mass shell, i.e., and that the neutrino mean energy
4. Interpretation
In a neutrino oscillation experiment, most neutrinos traverse the detector without interacting. In the limit
The interpretation of the preceding equations needs an understanding of the entire experiment. In particular we need to consider polarization miss-match (Section 5), and coherence (Section 6). If at
the source the neutrino mass is sufficiently uncertain, then a weak state is produced, i.e. a coherent superposition of mass eigenstates. If at the source the neutrino mass is sufficiently well
determined, then a mass eigenstate is produced. Even if the neutrino mass eigenstates are produced coherently, they may lose coherence before being detected, either in transit, and/or at the
detector. If this is the case, then an “observation” has been made, and we need to pass from amplitudes to probabilities, i.e. interference terms are lost.
For simplicity, we consider a single generation, i.e. (15) to (24). If production is coherent, and the mass eigenstates have become incoherent, then the probability for i.e. finite extent,
unpolarized detector. We assume E is the energy of the neutrino in the laboratory frame, i.e. the detector.
The case of interest to the leptogenesis scenario studied in this article is coherent production and coherent detection, since interference is needed for CP violation. If production is coherent, and
the mass eigenstates remain coherent at detection, then
5. Polarization Miss-Match
Consider the decay i.e. helicity +1/2) i.e.. Note that most ultra-relativistic
Consider the CP-conjugate decay i.e. helicity +1/2)
Note that for ultra-relativistic Majorana neutrinos we can still distinguish neutrinos (lepton number ≈+1 and helicity ≈−1/2) from anti-neutrinos (lepton number ≈−1 and helicity ≈+1/2), since lepton
number is conserved to a high degree of accuracy, see Section 10 for a numerical example.
Consider the lepton-conserving sequence of events i, and
6. Coherence
The sums in (29) and (31) only include coherent neutrinos. To obtain coherent oscillations between two neutrinos of masses i.e., where i.e. a coherent superposition of mass eigenstates, and
oscillations remain coherent while the wave packets of the two components overlap. The overlap ceases after the “coherence time” [7]
In the present application we take, arguably,
7. Asymmetry Build-Up
So far we have been studying a neutrino oscillation experiment with baseline L. In this Section we apply the results to the universe when it has the reference temperaturet to t for the case of
Consider the contribution of the channel
Taking the difference of these two equations, and dividing by
Summing over
The last term is the “wash-out” term that tends to restore the equilibrium value
or until wash-out sets in at
We note that
8. Constraints from Cosmology
Constraints from Big Bang Nucleosynthesis (BBN), Baryon Acoustic Oscillations (BAU), and direct searches, limit the mass of sterile neutrinos to be greater than 0.14 GeV [8], so the interesting
sterile neutrino mass range, for the leptogenesis scenario being considered, is approximately 0.14 GeV to 1.1 GeV. The lifetimes of these neutrinos range from approximately 10^−^5 s at
Big Bang Nuleosynthesis and cosmic microwave background (CMB) measurements do not allow one additional ultra-relativistic degree of freedom at T reached
As an example, for
9. Leptogenesis with
Let us study the simplest case with lepton number violation and CP violation. We take U from (28), to order
and the mass eigenstates are
The active neutrino mass
Let us write (38) for the present case
This equation assumes the approximation
Let us consider the caseR is any orthogonal, i.e.,
Substituting in (44) we obtain
As an example, we take
Equation (44) can be generalized to
10. Leptogenesis with
Without loss of generality we work in a basis that diagonalizes the M. The U, defined in (28), to lowest order in
To the present order of approximation, we takeTable 14.7 of [2]. The two active neutrino mass-squared differences are also obtained from this Table.
The diagonal mass matrix of active neutrinos, obtained from (28) and (48), is [2]
Successful leptogenesis is possible if we are able to solve (49), (38), and (39) with the needed lepton asymmetry
where R is any orthogonal, i.e.
R needs to be complex.
Successful leptogenesis requires fine tuning of the unknown parameters. As a proof of principle we present the following example: normal neutrino mass ordering, R chosen below). AtR (that makes the
results insensitive to
We note that the lepton number violating reactions are suppressed with respect to the lepton conserving ones, and the CP violating terms are suppressed with respect to the CP conserving ones. We note
that the terms ^−^12 and positive, while the terms ^−^18 and can be positive or negative.
From the first term in
Asymmetries per channel are presented in Table 1 (from (38) and (39) without the sums over i.e. Table 1 obtains ^2).
Table 1. Lepton number asymmetries per channel
Several tests with modifications of this example follow:
• SettingR, or equivalently
• Choosing a real R obtains
• Setting R obtains
• Setting
• Setting
• Results for inverse neutrino mass ordering are similar. However, we were unable to reach successful leptogenesis, i.e..
11. Sterile Neutrino Dark Matter?
Detailed dark matter properties have recently been obtained by fitting spiral galaxy rotation curves, and, independently, by fitting galaxy stellar mass distributions [10]. These measurements imply
that dark matter was in thermal and diffusive equilibrium with the Standard Model sector in the early universe, and decoupled (from the Standard Model sector and from self-annihilation) at a
Nevertheless, let us see if sterile neutrinos of mass i.e. from (23), we obtain
Let us considerR with a texture that makes the results insensitive to
The sub-index “6” stands for
12. Sterile Neutrino Search?
Consider a neutrino experiment that reconstructs the detected neutrinos in all-charged final states with the capability to discriminate a sterile neutrino mass from the active neutrino masses. In
this case there is no interference, and the probability to detect the sterile neutrino
with no sum implied. Fori.e. from (20), we obtain i.e. from (40), we obtain
13. Conclusions
We have studied coherent active-sterile neutrino oscillations as a possible source of leptogenesis. To this end, we add M mass terms. We find that for i.e. just before sphaleron freeze-out. The
possible Majorana nature of neutrinos allows lepton number violation, with, however, a cross-section reduced by a factor U (mainly from the Dirac mass matrix
The present scenario of leptogenesis requires a fine tuning parameterR such as (52). Why should nature select such a pattern (reminiscent of patterns in chemistry and biology)? It is interesting to
note that with this fine tuning parameter K, the neutrino Yukawa coupling magnitudes become comparable to the ones of charged leptons and quarks. It is also interesting to note that K may bring
sterile neutrino search within experimental reach.
The scenario studied in this article is similar to the model νMSM [8], where calculations have been carried out numerically in full detail. The search for sterile neutrinos with | {"url":"https://www.scirp.org/journal/paperinformation?paperid=110473","timestamp":"2024-11-01T22:22:38Z","content_type":"application/xhtml+xml","content_length":"253318","record_id":"<urn:uuid:9b58dd0e-ee53-4e44-bda7-f83c9ad3f4b2>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00772.warc.gz"} |
Representing Multiplication
Multiplication Word Problems
Finding the Unknown Product
Finding the Unknown Factor
Multiplication Properties
Solving Word Problems Using Estimation and Rounding
Subtracting Whole Numbers
Fractions and Unit Fractions
Measurements: Masses and Liquid Volume
Solving Problems Involving Masses or Volumes
Pictographs and Bar Graphs
Finding Sides' Lengths of Plane Figures
Area and Perimeter in Word Problems | {"url":"https://iconmath.com/browse-lessons-library/grade-3/","timestamp":"2024-11-12T01:48:44Z","content_type":"text/html","content_length":"82781","record_id":"<urn:uuid:77f67913-c5fb-46a2-a6eb-71bcbbdc4d47>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00492.warc.gz"} |
It takes so long to generate dictionary of GL elements.
It takes so long to generate dictionary of GL elements.
The following command
GL4dicthash={hash(g):None for g in GL(4,2)}
takes 9 seconds to execute. On the other hand
GL4dict={g:None for g in GL(4,2)}
takes minutes and does not seem to terminate.
If I understand python dictionary correctly, they should take about the same time. So what happened?
It seems hash(g) lies between 0 and 15 for all g. It does not make any sense to me.
Found this https://trac.sagemath.org/ticket/10950. Maybe I should update my Sage.
What version of Sage were you using?
Mutability and hashability are two different things. In the above case, for instance:
sage: G = GL(4,2)
sage: g = G.one()
sage: g.__hash__()
There is a __hash__ method, so g is hashable. There are $16$ different hash-values:
sage: list( set( [ g.__hash__() for g in G ] ) )
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
But $G$ has order
sage: prod( [ 2^4-2^k for k in [0..3] ] )
sage: G.order()
In order to define a dictionary where gis a key, this element should be immutable. Then for each new key one has to check if the key is already used. This takes time. And before we try to understand
the time problem, why do we need to work with such a complicated dictionary?
@slelievre before 8.0 now 8.1 and everything is good now. @dan_fulea: I am implementing a group action that is quite tedious. GL(4,2) acts on [0..15] by left-multiplying its binary expression as a
column vector in GF(2)^4. And then act on Subsets([0..15]). I would like to cache the action as there are only 20160·16 possibilities. Moreover, if there is a (fast) way to turn g into 16-bit int
then I need an array of size 65536·16, which is still affordable. What would make it triumph is that looking up array entry takes ~100ns.
1 Answer
Sort by » oldest newest most voted
The generation took on my slow linux machine approximatively 10s CPU time...
sage: %timeit -n1 GL4dict = { g : None for g in GL(4,2) }
1 loop, best of 3: 9.29 s per loop
This is
sage: version()
'SageMath version 8.1, Release Date: 2017-12-07'
Please update to the new version, if there is such a big run time difference.
edit flag offensive delete link more
Thank you. I did update from 8.0 to 8.1. I was not aware of the version-issue because I thought I was quite new. Now everything works fine. hash(g) gives crazy numbers such as -32759827368972486.
Symbol 1 ( 2018-02-07 01:05:00 +0100 )edit
@Symbol 1 -- you can accept the answer if it solves the problem in your question.
If an answer solves your question, please accept it by clicking the "accept" button (the one with a check mark, below the upvote button, score, and downvote button, at the top left of the answer).
This will mark the question as solved in the list of questions on the main page of Ask Sage, as well as in lists of questions related to a particular query or keyword.
slelievre ( 2018-09-22 09:31:58 +0100 )edit | {"url":"https://ask.sagemath.org/question/40987/it-takes-so-long-to-generate-dictionary-of-gl-elements/?sort=votes","timestamp":"2024-11-08T13:56:11Z","content_type":"application/xhtml+xml","content_length":"66805","record_id":"<urn:uuid:12853698-4103-42b9-9518-f754834cb3bf>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00717.warc.gz"} |
CSCE 222 Lecture 32
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More about cardinality
${\displaystyle |A|\leq |B|}$ iff injective function from A to B.
Cantor's Theorem
${\displaystyle |S|<|P(S)|}$
Function ${\displaystyle i:S\to P(S)}$ given by ${\displaystyle i(s)=\{s\}}$ is injective, therefore ${\displaystyle |S|\leq |P(S)|}$
Claim that there does not exist any surjective function ${\displaystyle f:S\to P(S)}$
Indeed, ${\displaystyle T=\{s\in S:sotin f(s)\}}$ is not contained in ${\displaystyle f(S)}$
Countability vs. Uncountability
A set X is countable iff ${\displaystyle |X|\leq |\mathbb {N} |}$
Function ${\displaystyle f:\mathbb {N} \to \mathbb {N} }$ is not countable since ${\displaystyle |\mathbb {N} |<|P(\mathbb {N} )|}$, which is equivalent to ${\displaystyle |\mathbb {N} |<|2^{\mathbb
{N} }|}$ Therefore, the function's cardinality is not countable
Given a function that maps an input to a given value, define a function ${\displaystyle f^{d}(n)=f_{n}(n)+1}$, therefore ${\displaystyle f^{d}(n)eq f_{n}(n)}$
The Halting Problem
Input for Halt program: Program P and input X for P Output: 1 if P terminates (halts) when executed on X, and 0 if P goes into an infinite loop on input X.
Note: A compiler is a program that takes as input the code for another program.
X could be the program for P itself.
What if we call Halt(P, P) and modify it a little? (diagonalize) This result leads to a paradox and is undecidable.
bool halt(Program P, void* Input X)
if (P.run(X).terminates()) return true;
else return false;
bool pSelf(Program P) {
return halt(P,P)
bool pDiag(Program P) {
if (pSelf(P)) while(true) {}
else return false;
If an input from Language 1 returned YES for the Halting problem, than a slight modification (Language 2) must also return YES
Functional property: a property that refers to the language accepted by the program and not the specific code of the program.
EX: Program terminates in 10 steps: (not functional); the program accepts 0 or more inputs (functional).
Nontrivial property: functional property about programs if some programs have the property and some do not.
Rice's Theorem
Every nontrivial functional property about programs is undecidable. | {"url":"https://notes.komputerwiz.net/wiki/CSCE_222_Lecture_32","timestamp":"2024-11-13T02:55:10Z","content_type":"text/html","content_length":"40936","record_id":"<urn:uuid:3d51d75f-5ee2-482b-9517-9e6b571699c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00067.warc.gz"} |
Creating And Solving Model Of Linear Equation Through The Balance At Junior Secondary Class
Turmudi, Turmudi (2011) Creating And Solving Model Of Linear Equation Through The Balance At Junior Secondary Class. PROCEEDINGS International Seminar and the Fourth National Conference on
Mathematics Education. ISSN 978-979-16353-7-0
P - 61.pdf
Download (129kB) | Preview
Algebra is one of the most difficult subject felt by most students and this strand is just started given to the students at early junior secondary school. Infact, if we can manage the algebra subject
wisely and attractively, it can be predicted that the students’ conceptual understanding algebra would be relatively improved. A study was conducted to the Year 7 students at a Junior Secondary of
Laboratory School of UPI. The class was set to learn the linear equation topic by using balance (scales). Through a weighing activity, the students were able to design linear equation models. They
followed a guidelines created by the teacher and researcher. The students were not only able to create linear equation models, but also were able to solve mathematical model of linear equation. By
using steps of balance (weighing), finally the students were able to know the weight of a hidden quantity. A number of teachers were involved in an observation activity which were designed in a
lesson study context. Information related to the students’ reaction as well as the teachers’ reaction toward the implementation of creating and designing models of linear equation. The information
were analysed qualitatively. The results indicate that introducing the linear equation through the scale (balance) were responded positively by the students. A brief interview with the students
indicated that the students fluently could solve linear equation, and find the value of variable which infact as a weight variable. This variable seemed to be the weight of hidden variable as the
solution of the linear equation. Moreover, the students were able to interpret the process of weighing to the form of linear equation, since then the students solved it and found the solution of the
problem. While other teachers as observers at the lesson gave comments that the model teacher had practiced the concept of linear equation by using unusual way of teaching. Intuitively they solved
the linear equation by using step by step of weighing process and determined how much weight of an object. The process of weighing and thinking are parallel to solving a linear equation. Data of test
results regarding the linear equation indicated that the students’ understanding of linear equation improved. The researchers recommend to use the balance (scales) as an alternative to teach the
topic of linear equation. Keywords: Balance, realistic, and lesson study.
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(i) Find the shortest distance between the skew lines r=(6i+2j+... | Filo
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Question Text (i) Find the shortest distance between the skew lines and where are scalars. (Mar-08,09,20, TS-Mar-17) ii) If and , find the distanc between and CD. (TS-Mar-19)
Updated On Mar 4, 2023
Topic Vector and 3D
Subject Mathematics
Class Class 11
Answer Type Video solution: 2
Upvotes 295
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[OWP-2009-25] Anderson, J. M.; Hinkkanen, Aimo (Mathematisches Forschungsinstitut Oberwolfach, 2009-03-18)
Let $f_1,..., f_p$ be entire functions that do not all vanish at any point, so that $(f_1,..., f_p)$ is a holomorphic curve in $\mathbb{CP}^{p-1}$. We introduce a new and more careful notion of
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The Facts About math for kids
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Solving regularized total least squares problems based on eigenproblems
Other Titles
Lösen von Regularisierten totalen Ausgleichsproblemen mit Hilfe von Eigenwertproblemen
Title Granting Institution
Technische Universität Hamburg
Place of Title Granting Institution
Solving Regularized Total Least Squares Problems Based on Eigenproblems, Berlin, dissertation.de – Verlag im Internet GmbH, 2010, S. 177
In the first part of the thesis we review basic knowledge of regularized least squares problems and present a significant acceleration of an existing method for the solution of trust-region problems.
In the second part we present the basic theory of total least squares (TLS) problems and give an overview of possible extensions. Regularization of TLS problems by truncation and bidiagonalization
approaches are briefly covered. Several approaches for solving the Tikhonov TLS problem based on Newton’s method are mentioned, which lead to a converging sequence of linear systems.
The emphasis of the thesis is on quadratically constrained TLS (RTLS) problems. Two different iterative concepts for the solution of the first-order condition are analyzed. The first iteration
results in a sequence of quadratic problems while the second concept leads to a sequence of linear eigenvalue problems. For a fixed point iteration based on quadratic eigenvalue problems, i.e. the
RTLSQEP method, the global convergence to the RTLS solution is proven. With the characterization of the rightmost eigenvalue of the QEPs the RTLS solution can be described via a generalized
trust-region problem. The second concept has been studied in detail as well, i.e. the RTLSEVP method. To ensure convergence it was necessary to generalize the function g. The properties of this
function have been heavily analyzed, which lead to a deeper understanding of the RTLS solution: It can also be characterized via the solution of a TLS problem.
The RTLSQEP and RTLSEVP algorithm for solving the RTLS problem are shown to be very efficient when combined with iterative projection methods in the inner loop. Since a sequence of convergent
problems has to be solved it turns out that the Nonlinear Arnoldi method is the method of choice, because it is able to recycle most of the information during the iterative process. The computational
complexity of the proposed approaches is kept at the order of matrix-vector multiplications. We give a detailed description of an efficient implementation of the different parts of the algorithms.
Two efficient algorithms for evaluating the L-curve at discrete points are presented. Since setting up the L-curve requires solving a sequence of RTLS problems the advantage of the Nonlinear Arnoldi
method is even more apparent:
Recycling the search space, not only within one RTLS problem but throughout the whole sequence.
total least squares
efficient algorithms | {"url":"https://tore.tuhh.de/entities/publication/fb1d9fac-9e0d-48a5-a233-184aeb1cac4e","timestamp":"2024-11-02T06:16:00Z","content_type":"text/html","content_length":"978518","record_id":"<urn:uuid:f5aece03-4958-4948-9f18-236164fd5a5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00183.warc.gz"} |
Chronic Kidney Disease Risk Calculators
Calculator 1: Risk of Chronic Kidney Disease (CKD) in adults For adults who don’t know if they have chronic kidney disease (CKD), Calculator 1 estimates the probability of having CKD (Bang et al.,
• The calculation is based on individual characteristics: age, sex, and 7 comorbidities including hypertension, diabetes and cardiovascular disease.
Calculator 2: Risk of progression of CKD to kidney failure among those who already have CKD For adults who know they have CKD (estimated glomerular filtration rate [eGFR] <60 mL/min/1.73m^2),
Calculator 2 estimates the probability of progression of CKD to kidney failure in the next two or five years using a 4- or 8-variable equation (Tangri et al., 2011)
• The 4-variable equation is based on individual characteristics: age, sex, eGFR and Urine Albumin to Creatinine Ratio (UACR)
• The 8-variable equation also includes four laboratory tests (serum calcium, phosphate, bicarbonate, and albumin).
The intended audience is medical researchers and clinicians; the calculators should only be used by laypersons to begin dialog with their care provider. These equations should not be used for
self-diagnosis or self-management.
The CKD Risk Calculators are designed for use on devices with larger displays. Please visit this page on a tablet, laptop, or desktop device.
This calculator returns the probability (expressed as a percentage from 0 to 100%) of having Stage 3-5 CKD, defined as an estimated glomerular filtration rate (eGFR) less than 60 mL/min/1.73m^2,
based on nine input variables^1. If one or more of the values is unknown, trying different combinations of the presence or absence of the unknown characteristics will give a range of possible
estimates. An elevated probability of CKD should prompt individuals to seek more comprehensive medical advice, possibly leading to further medical evaluation, earlier diagnosis and interventions to
both manage the condition and slow its progression.
Details of the risk equation development (Bang et al, 2007) are available here: https://www.ncbi.nlm.nih.gov/pubmed/17325299.
Characteristic This Person US National
Enter Age and Gender:
Age Range *
Gender *
Enter any of the following characteristics that may increase the probability of CKD:
Hypertension 41%
Diabetes ^ 10%
Cardiovascular Disease (CVD) 8%
Congestive Heart Failure (CHF)
Peripheral Vascular Disease (PVD)
Proteinuria^2 10%
* Required Fields
Probability of having Stage 3-5 Chronic Kidney Disease (CKD)
(expressed as a percentage between 0% and 100%):
Probability (CKD) = 1/[1 + exp(-β’ x X)], where β’ x X = -5.4 + 1.55 x (age of 50-59 years) + 2.31 x (age of 60-69 years) + 3.23 x (age ≥ 70 years) + 0.29 x (female) + 0.93 x (anemia) + 0.45 x
(hypertension) + 0.44 x (DM) + 0.59 x (history of CVD) + 0.45 x (history of CHF) + 0.74 x (PVD) + 0.83 x (proteinuria), where (a) is an indicator taking 1 for event a and 0 otherwise. In this
equation, β and X denote vectors of β-coefficients and risk factors used in this equation, respectively.
Proteinuria is a broad term for leakage of protein in the kidney. Although the Bang et al. paper uses the term proteinuria, the actual measurement was of albuminuria. Albumin is a protein found in
the blood. A healthy kidney does not let albumin pass from the blood into the urine. Too much albumin in your urine is called albuminuria.
For patients with chronic kidney disease (i.e., those with estimated glomerular filtration rate (eGFR) less than 60 mL/min/1.73m^2), this calculator returns an estimated probability of developing
kidney failure in the next two or five years (expressed as a percentage, or a range of percentages, from 0% to 100%). These probabilities could be used to facilitate patient and provider
communication, to heighten awareness and guide optimal disease management for best outcomes.
Two versions of this calculator are provided, one with four input variables, and one with eight input variables. Using more variables, when available, will yield a more precise estimate. The
‘sliders’ for each laboratory variable in the data entry table below allow a range of values to be selected in case of uncertainty in a single value. Details of risk equation development (Tangri et
al., 2011) are available here: https://www.ncbi.nlm.nih.gov/pubmed/21482743.
Characteristic This Person
Gender * :
(Range: 10-60)
* Required Fields
Estimated probability of progression to end-stage renal disease (ESRD) in CKD patients at 2 and 5 years
(expressed as a percentage between 0% and 100%): | {"url":"https://nccd.cdc.gov/ckd/Calculators.aspx","timestamp":"2024-11-04T20:48:02Z","content_type":"text/html","content_length":"331455","record_id":"<urn:uuid:25e30849-6514-4312-941a-8ff287c0af80>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00119.warc.gz"} |
Let sat(n,k) denote the minimum size of a family in
the Boolean lattice of dimension n with the property that there is
no chain with k+1 elements but the addition of one more element to
the family produces such a chain. It is well known that for n
sufficiently large, sat(n,k) is the same value, which we
denote sat(k).
For k≥6, the best known bounds are
due to Gerbner, Keszegh, Lemons, Palmer, P\'alv\"olgyi, and Patk\'os
and to Morrison, Noel and Scott (based on a construction by Gerbner,
et al.), respectively.
We improve both bounds to
The upper bound works for k≥7 and the lower bound for k≥483.
This is joint work with Nick Veldt, Iowa State University. | {"url":"https://video.renyi.hu/video/improved-bounds-on-chain-saturation-746","timestamp":"2024-11-01T20:46:52Z","content_type":"text/html","content_length":"67351","record_id":"<urn:uuid:c7faf8ed-1364-4c48-b109-c5387f1688ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00201.warc.gz"} |
Kee, Natalia Harlan Community Academy H.S.
Discover the factor of proportionality in similar polygons and to relate this factor
to the perimeters and areas of similar polygons.
Overhead projector, rulers for each student, protractors for each student, handouts
of printed polygons (these are to be measured by each student).
Students will measure each side and each angle of the polygons on the handouts.
They will compare their measurements with the overhead projection of these polygons
and derive a factor of proportionality between each corresponding pair of sides and
angles of the two figures being compared. They will measure the angles of their
polygons and compare these measurements with the corresponding angles of the
projected similar polygon. (Discovery: the angles retain their same measure in both
figures-theirs and the overhead's).
This process can be used with different pairs of polygons to establish a factor
of proportionality in their perimeters and areas. Students should discover that this
factor remains constant in similar polygons and is squared when determining areas of
similar polygons.
Example:Ratio of two similar polygons' corresponding sides is 1/3.
Perimeters of each polygon are: 2,4,2,4 = 2+4+2+4 = 12 units
3(2+4+2+4) = 36 units
Areas of each polygon are: length x width = area in square units
2 x 4 = 8 square units
3(2)x 3(4) = 72 square units
Return to Mathematics Index | {"url":"https://smileprogram.info/ma8911.html","timestamp":"2024-11-09T03:41:08Z","content_type":"text/html","content_length":"2182","record_id":"<urn:uuid:a7661023-f3eb-4265-8c23-cd26c3da21a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00729.warc.gz"} |
General pressure loss for ideal gas
Calculation of a generic pressure loss for an ideal gas using mean density. See more information.
Generic pressure loss depending on density and viscosity
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density, nominal dynamic viscosity) at an operation point via interpolation. This generic function
considers the pressure loss law via a pressure loss exponent and the influence of density and dynamic viscosity on pressure loss. See more information.
Generic pressure loss depending on density
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density) via interpolation from an operation point. See more information.
Generic pressure loss depending on pressure loss coefficient
Calculation of a generic pressure loss in dependence of a pressure loss coefficient. See more information.
Generic pressure loss depending on volume flow rate
Calculation of a generic pressure loss with linear or quadratic dependence on volume flow rate. See more information.
Extends from Modelica.Icons.VariantsPackage (Icon for package containing variants).
Package Contents
│ Name │ Description │
│ dp_idealGas_DP │ Generic pressure loss | calculate pressure loss | ideal gas | mean density │
│ dp_idealGas_IN_con │ Input record for function dp_idealGas_DP and dp_idealGas_MFLOW │
│ dp_idealGas_IN_var │ Input record for function dp_idealGas_DP and dp_idealGas_MFLOW │
│ dp_idealGas_MFLOW │ Generic pressure loss | calculate mass flow rate | ideal gas | mean density │
│ dp_nominalDensityViscosity_DP │ Generic pressure loss | calculate mass flow rate | nominal operation point | pressure loss law (exponent) | density and dynamic viscosity dependence │
│ dp_nominalDensityViscosity_IN_con │ Output record for function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW │
│ dp_nominalDensityViscosity_IN_var │ Output record for function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW │
│ dp_nominalDensityViscosity_MFLOW │ Generic pressure loss | calculate M_FLOW (compressible) | nominal operation point | pressure loss law (exponent) | density and dynamic viscosity │
│ │ dependence │
│ dp_nominalPressureLossLawDensity_DP │ Generic pressure loss | calculate pressure loss | nominal operation point | pressure loss law (coefficient and exponent) | density dependence │
│ dp_nominalPressureLossLawDensity_IN_con │ Input record for function dp_nominalPressureLossLawDensity_DP and dp_nominalPressureLossLawDensity_MFLOW │
│ dp_nominalPressureLossLawDensity_IN_var │ Input record for function dp_nominalPressureLossLawDensity_DP and dp_nominalPressureLossLawDensity_MFLOW │
│ dp_nominalPressureLossLawDensity_MFLOW │ Generic pressure loss | calculate mass flow rate | nominal operation point | pressure loss law (coefficient and exponent) | density dependence │
│ dp_pressureLossCoefficient_DP │ Generic pressure loss | calculate pressure loss | pressure loss coefficient (zeta_TOT) │
│ dp_pressureLossCoefficient_IN_con │ Input record for function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW │
│ dp_pressureLossCoefficient_IN_var │ Input record for function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW │
│ dp_pressureLossCoefficient_MFLOW │ Generic pressure loss | calculate mass flow rate | pressure loss coefficient (zeta_TOT) │
│ dp_volumeFlowRate_DP │ Generic pressure loss | calculate pressure loss | quadratic function (dp=a*V_flow^2 + b*V_flow) │
│ dp_volumeFlowRate_IN_con │ Input record for function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW │
│ dp_volumeFlowRate_IN_var │ Input record for function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW │
│ dp_volumeFlowRate_MFLOW │ Generic pressure loss | calculate mass flow rate | quadratic function (dp=a*V_flow^2 + b*V_flow) │
Generic pressure loss | calculate pressure loss | ideal gas | mean density
Calculation of a generic pressure loss for an ideal gas using mean density.
Generally this function is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and the corresponding pressure loss
(DP) has to be calculated. On the other hand the function dp_idealGas_MFLOW is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable)
and the mass flow rate (M_FLOW) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ Pressure │ DP │ Output for function dp_idealGas_DP │
Generic pressure loss | calculate mass flow rate | ideal gas | mean density
Calculation of a generic pressure loss for an ideal gas using mean density.
Generally this function is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable) and the mass flow rate (M_FLOW) has to be
calculated. On the other hand the function dp_idealGas_DP is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and
the corresponding pressure loss (DP) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ MassFlowRate │ M_FLOW │ Output for function dp_idealGas_MFLOW │
Input record for function dp_idealGas_DP and dp_idealGas_MFLOW
This record is used as input record for the pressure loss function dp_idealGas_DP and dp_idealGas_MFLOW .
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.IdealGas_con (Base record for generic pressure loss function | ideal gas | mean density).
│ Type │ Name │ Description │
│ Real │ exp │ Exponent of pressure loss law │
│ SpecificHeatCapacity │ R_s │ Specific gas constant of ideal gas │
│ Real │ Km │ Coefficient for pressure loss law [(Pa)^2/{(kg/s)^exp*K}] │
│ Pressure │ dp_smooth │ Start linearisation for smaller pressure loss │
Input record for function dp_idealGas_DP and dp_idealGas_MFLOW
This record is used as input record for the pressure loss function dp_idealGas_DP and dp_idealGas_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.IdealGas_var (Base record for generic pressure loss function | ideal gas | mean density).
│ Type │ Name │ Description │
│ Density │ rho_m │ Mean density of ideal gas │
│ Temp_K │ T_m │ Mean temperature of ideal gas │
│ Pressure │ p_m │ Mean pressure of ideal gas │
Generic pressure loss | calculate mass flow rate | nominal operation point | pressure loss law (exponent) | density and dynamic viscosity dependence
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density, nominal dynamic viscosity) at an operation point via interpolation. This generic function
considers the pressure loss law via a pressure loss exponent and the influence of density and dynamic viscosity on pressure loss.
Generally this function is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and the corresponding pressure loss
(DP) has to be calculated. On the other hand the function dp_nominalDensityViscosity_MFLOW is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as
state variable) and the mass flow rate (M_FLOW) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ Pressure │ DP │ Output for function dp_nominalDensityViscosity_DP │
Generic pressure loss | calculate M_FLOW (compressible) | nominal operation point | pressure loss law (exponent) | density and dynamic viscosity dependence
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density, nominal dynamic viscosity) at an operation point via interpolation. This generic function
considers the pressure loss law via a pressure loss exponent and the influence of density and dynamic viscosity on pressure loss.
Generally this function is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable) and the mass flow rate (M_FLOW) has to be
calculated. On the other hand the function dp_genericDensityViscosity_DP is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the
used model and the corresponding pressure loss (DP) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ MassFlowRate │ M_FLOW │ Output for function dp_nominalDensityViscosity_MFLOW │
Output record for function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW
This record is used as input record for the pressure loss function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.NominalDensityViscosity (Base record for generic pressure loss function).
│ Type │ Name │ Description │
│ Pressure │ dp_nom │ Nominal pressure loss (at nominal values of mass flow rate and density) │
│ Real │ exp │ Exponent of pressure loss law │
│ MassFlowRate │ m_flow_nom │ Nominal mass flow rate (at nominal values of pressure loss and density) │
│ Density │ rho_nom │ Nominal density (at nominal values of mass flow rate and pressure loss) │
│ Real │ exp_eta │ Exponent for dynamic viscosity dependence │
│ DynamicViscosity │ eta_nom │ Dynamic viscosity at nominal pressure loss │
Output record for function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW
This record is used as input record for the pressure loss function dp_nominalDensityViscosity_DP and dp_nominalDensityViscosity_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.PressureLoss (Base record for fluid properties for pressure loss).
Generic pressure loss | calculate pressure loss | nominal operation point | pressure loss law (coefficient and exponent) | density dependence
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density) via interpolation from an operation point. This generic function considers the pressure loss
law via a nominal pressure loss (dp_nom), a pressure loss coefficient (zeta_TOT) and a pressure loss law exponent (exp) as well as the influence of density on pressure loss.
Generally this function is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and the corresponding pressure loss
(DP) has to be calculated. On the other hand the function dp_nominalPressureLossLawDensity_MFLOW is numerically best used for the compressible case if the pressure loss (dp) is known (out of
pressures as state variable) and the mass flow rate (M_FLOW) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ Pressure │ DP │ Output for function dp_nominalPressureLossLawDensity_yesAJac_DP │
Generic pressure loss | calculate mass flow rate | nominal operation point | pressure loss law (coefficient and exponent) | density dependence
Calculation of a generic pressure loss in dependence of nominal fluid variables (e.g., nominal density) via interpolation from an operation point. This generic function considers the pressure loss
law via a nominal pressure loss (dp_nom), a pressure loss coefficient (zeta_TOT) and a pressure loss law exponent (exp) as well as the influence of density on pressure loss.
Generally this function is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable) and the mass flow rate (M_FLOW) has to be
calculated. On the other hand the function dp_nominalPressurelosslawDensity_DP is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in
the used model and the corresponding pressure loss (DP) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ MassFlowRate │ M_FLOW │ Output for function dp_nominalPressurelosslawDensity_MFLOW │
Input record for function dp_nominalPressureLossLawDensity_DP and dp_nominalPressureLossLawDensity_MFLOW
This record is used as input record for the pressure loss function dp_nominalPressureLosslawDensity_DP and dp_nominalPressureLosslawDensity_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.NominalPressureLossLawDensity_con (Base record for generic pressure loss function).
│ Type │ Name │ Description │
│ MassOrVolumeFlowRate │ target │ MassFlowRate == use nominal mass flow rate | VolumeFlowRate == use nominal volume flow rate │
│ Area │ A_cross │ Cross sectional area │
│ Area │ A_cross_nom │ Nominal cross sectional area │
│ Pressure │ dp_nom │ Nominal pressure loss (at nominal values of mass flow rate and density) │
│ MassFlowRate │ m_flow_nom │ Nominal mass flow rate (at nominal values of pressure loss and density) │
│ Real │ exp │ Exponent of pressure loss law │
│ VolumeFlowRate │ V_flow_nom │ Nominal volume flow rate (at nominal values of pressure loss and density) │
│ Density │ rho_nom │ Nominal density (at nominal values of mass flow rate and pressure loss) │
│ PressureLossCoefficient │ zeta_TOT_nom │ Nominal pressure loss coefficient (for nominal values) │
Input record for function dp_nominalPressureLossLawDensity_DP and dp_nominalPressureLossLawDensity_MFLOW
This record is used as input record for the pressure loss function dp_nominalPressureLosslawDensity_DP and dp_nominalPressureLosslawDensity_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.NominalPressureLossLawDensity_var (Base record for generic pressure loss function).
Generic pressure loss | calculate pressure loss | pressure loss coefficient (zeta_TOT)
Calculation of a generic pressure loss in dependence of a pressure loss coefficient.
Generally this function is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and the corresponding pressure loss
(DP) has to be calculated. On the other hand the function dp_pressureLossCoefficient_MFLOW is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as
state variable) and the mass flow rate (M_FLOW) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ Pressure │ DP │ Output for function dp_pressureLossCoefficient_DP │
Generic pressure loss | calculate mass flow rate | pressure loss coefficient (zeta_TOT)
Calculation of a generic pressure loss in dependence of a pressure loss coefficient.
Generally this function is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable) and the mass flow rate (M_FLOW) has to be
calculated. On the other hand the function dp_pressureLossCoefficient_DP is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the
used model and the corresponding pressure loss (DP) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ MassFlowRate │ M_FLOW │ Output for function dp_pressureLossCoefficientt_MFLOW │
Input record for function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW
This record is used as input record for the pressure loss function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW.
Extends from Modelica.Icons.Record (Icon for records).
│ Type │ Name │ Description │
│ Area │ A_cross │ Cross sectional area │
│ Pressure │ dp_smooth │ Start linearisation for decreasing pressure loss │
Input record for function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW
This record is used as input record for the pressure loss function dp_pressureLossCoefficient_DP and dp_pressureLossCoefficient_MFLOW.
Extends from Modelica.Icons.Record (Icon for records).
Generic pressure loss | calculate pressure loss | quadratic function (dp=a*V_flow^2 + b*V_flow)
Calculation of a generic pressure loss with linear and/or quadratic dependence on volume flow rate. Please note that the sum of a and b has to be greater zero. The function can be used to calculate
pressure loss at known mass flow rate or mass flow rate at known pressure loss.
Generally this function is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model and the corresponding pressure loss
(DP) has to be calculated. On the other hand the function dp_volumeFlowRate_MFLOW is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state
variable) and the mass flow rate (M_FLOW) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ Pressure │ DP │ Output for function dp_volumeFlowRate_DP │
Generic pressure loss | calculate mass flow rate | quadratic function (dp=a*V_flow^2 + b*V_flow)
Calculation of a generic pressure loss with linear or quadratic dependence on volume flow rate. Please note that the sum of a and b has to be greater zero. The function can be used to calculate
pressure loss at known mass flow rate or mass flow rate at known pressure loss.
Generally this function is numerically best used for the compressible case if the pressure loss (dp) is known (out of pressures as state variable) and the mass flow rate (M_FLOW) has to be
calculated. On the other hand the function dp_volumeFlowRate_DP is numerically best used for the incompressible case , where the mass flow rate (m_flow) is known (as state variable) in the used model
and the corresponding pressure loss (DP) has to be calculated. See more information.
Extends from Modelica.Icons.Function (Icon for functions).
│ Type │ Name │ Description │
│ MassFlowRate │ M_FLOW │ Output for function dp_volumeFlowRate_MFLOW │
Input record for function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW
This record is used as input record for the pressure loss function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW.
Extends from Modelica.Fluid.Dissipation.Utilities.Records.General.QuadraticVFLOW (Base record for generic pressure loss function | quadratic function (dp=a*Vdot^2 + b*Vdot)).
│ Type │ Name │ Description │
│ Real │ a │ Coefficient for quadratic term │
│ Real │ b │ Coefficient for linear term │
│ Pressure │ dp_min │ Start of approximation for decreasing pressure loss (only used for b=0) │
Input record for function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW
This record is used as input record for the pressure loss function dp_volumeFlowRate_DP and dp_volumeFlowRate_MFLOW.
Extends from Modelica.Icons.Record (Icon for records).
│ Type │ Name │ Description │
│ Density │ rho │ Density of fluid │ | {"url":"https://help.altair.com/twinactivate/help/en_us/block_reference_guide/_mo/_mla/Modelica3.2.3/HTML/Fluid/Dissipation/PressureLoss/General/index.html","timestamp":"2024-11-04T15:22:04Z","content_type":"text/html","content_length":"55000","record_id":"<urn:uuid:50ff3e22-9399-4a45-8f64-9541c52fd12e>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00349.warc.gz"} |
Fractional Delay FIR Filters
Design Fractional Delay FIR Filters
Fractional delay filters shift a digital sequence by a noninteger value by combining interpolation and resampling into a single convolution filter. This example shows you how to design and implement
fractional delay FIR filters using tools available in DSP System Toolbox™.
Delay as a Convolution System
Integer Delays
Consider the delay of a digital signal, $y\left[n\right]=x\left[n-D\right]$, where D is an integer. You can represent this operation as a convolution filter, $y=h*x$, with a finite impulse response,
$\mathit{h}\left[\mathit{n}\right]=\delta \left[\mathit{n}-\mathit{D}\right]$. The corresponding transfer function is $H\left(z\right)={z}^{-D}$, and the frequency response is $\mathit{H}\left(\omega
\right)={\mathit{e}}^{-\mathit{i}\omega \text{\hspace{0.17em}}\mathit{D}}$. Programmatically, you can implement such an integer delay filter using the following MATLAB® code.
% Create the FIR
D = 3; % Delay value
h = [zeros(1,D) 1]
Shift a sequence by filtering it through the FIR h. The zeros at the beginning of the output signify the initial condition that is intrinsic to such filters.
x = (1:10)';
dfir = dsp.FIRFilter(h);
y = dfir(x)'
y = 1×10
Fractional Delays via D/A Interpolation
The delay of the sequence$y\left[n\right]=x\left[n-D\right]$is undefined when D is not an integer. To make such fractional delays sensible and to sample the output on the continuum, you need to add
an intermediate D/A interpolation stage. That is, $\mathit{y}\left[\mathit{n}\right]=\stackrel{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}\left(\mathit{n}-\mathit{D}\right)$, where $\stackrel
{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}$ denotes some D/A interpolation of the input sequence $\mathit{x}$.$\stackrel{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}$ can depend on
n, and can represent an underlying analog signal model from which you sampled the sequence $\mathit{x}$. This strategy is used in other resampling problems such as rate conversion.
This example features fractional delay filters using two interpolation models from DSP Systems Toolbox.
1. Sinc-based interpolation model: Use bandlimited reconstruction for $\stackrel{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}$.
2. Lagrange-based interpolation model: Use polynomial reconstruction for $\stackrel{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}$.
Bandlimited Fractional Delay Filters
The Shannon-Whittaker interpolation formula $\stackrel{ˆ}{\text{\hspace{0.17em}}\mathit{x}}\left(\mathit{t}\right)=\sum _{\mathit{k}}\text{\hspace{0.17em}}\mathit{x}\left[\mathit{k}\right]\mathrm
{sinc}\left(\mathit{t}-\mathit{k}\right)$ models bandlimited signals. That is, the intermediate D/A conversion $\stackrel{ˆ}{\text{\hspace{0.17em}}\mathit{x}}$ is a bandlimited reconstruction of the
input sequence. For a delay value, D, you can represent the fractional delay $\mathit{y}\left[\mathit{n}\right]=\stackrel{ˆ}{\text{\hspace{0.17em}}\mathit{x}}\left(\mathit{n}-\mathit{D}\right)$,
which uses the same $\stackrel{ˆ}{\text{\hspace{0.17em}}\mathit{x}}$ for every n, as a convolution filter. This filter is called the ideal bandlimited fractional delay filter, and its impulse
response is
The corresponding frequency response (that is, DTFT) is given by ${H}_{D}\left(\omega \right)={e}^{-i\omega D}$.
Causal FIR Approximation of Ideal Bandlimited Shift Filter
The ideal sinc-shift filter described in the previous section is an allpass filter ($|{\mathit{H}}_{\mathit{d}}\left(\omega \right)|=1$), but it has an infinite and noncausal impulse response ${\
mathit{h}}_{\mathit{D}}$. You cannot represent the filter as a vector in MATLAB but can instead represent it as a function of index k.
% Ideal Filter sequence
D = 0.4;
hIdeal = @(k) sinc(k-D);
For practical and computational purposes, you can truncate the ideal filter on a finite index window, at the cost of some bandwidth loss. For a target delay value of $\mathit{D}$ and a desired length
of N, the window of indices $\mathit{k}\text{\hspace{0.17em}}$satisfying $|\mathit{D}-\mathit{k}|\le \frac{\mathit{N}}{2}$ is symmetric about $\mathit{D}$ and captures the main lobe of the ideal
filter. For $\mathit{D}={\mathit{i}}_{0}+\mathit{FD}$ where $\mathit{0}\le \mathit{FD}\le 1$ and ${\mathit{i}}_{0}$ is an integer, the explicit window indices are$\left\{{i}_{0}-⌊\frac{N-1}{2}⌋,\dots
,\phantom{\rule{0.2em}{0ex}}{i}_{0}+⌊\frac{N}{2}⌋\right\}$. The integer ${\mathit{i}}_{0}$ is referred to as the integer latency, and you can select it arbitrarily. To make the FIR causal, set ${\
mathit{i}}_{0}=⌊\frac{\mathit{N}-1}{2}⌋$, so the index window is $\left\{0,\dots ,\mathit{N}-1\right\}$. This code depicts the rationale behind the causal FIR approximation.
% FIR approximation with causal shift
N = 6;
idxWindow = (-floor((N-1)/2):floor(N/2))';
i0 = -idxWindow(1); % Causal latency
hApprox = hIdeal(idxWindow);
Truncating a sinc filter causes a ripple in the frequency response, which you can address by applying weights $\left\{{\mathit{w}}_{\mathit{k}}\right\}$ (such as Kaiser or Hamming) to the FIR
The resulting FIR approximation model of the ideal bandlimited fractional delay filter is represented as follows.
$h\left[k\right]={w}_{k}{h}_{d}\left[k\right]=\left\{\begin{array}{cc}{w}_{k}sinc\left(k-FD-{i}_{0}\right)& 0\le k\le N-1\\ 0& \text{otherwise}\text{}\end{array}$
You can design such a filter using the designFracDelayFIR function and the dsp.VariableFractionalDelay System object™ with the interpolation method set to 'FIR'. The function and the System object
use Kaiser window weights.
Lagrange-Based Fractional Delay Filters
Lagrange-based fractional delay filters use polynomial fitting on a moving window of input samples. That is, $\stackrel{ˆ}{\text{\hspace{0.17em}}{\mathit{x}}_{\mathit{n}}}\left(\mathit{t}\right)$ is
a polynomial of the fixed degree K. Like the sinc-based delay filters, you can formulate Lagrange-based delay filters as causal FIR convolution ($\mathit{y}=\mathit{h}*\mathit{x}$) of the length N =
K+1 supported on the index window$\left\{-⌊\frac{N-1}{2}⌋,\dots \phantom{\rule{0.2em}{0ex}}⌊\frac{N}{2}⌋\right\}$. As in the sinc-based model, apply the causal latency ${\mathit{i}}_{0}=⌊\frac{\
mathit{N}-1}{2}⌋$. Given a fractional delay $\mathit{FD}$, you can obtain the FIR coefficients $\mathit{h}\left[0\right],\dots ,\mathit{h}\left[\mathit{K}\right]$ of the (causal shifted) Lagrange
delay filter by solving a system of linear equations. These equations describe a standard Lagrange polynomial fitting problem.
$\sum _{k=0}^{K}{t}_{n}^{k}h\left[k\right]=\left(FD{\right)}^{n},\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}n=0,\dots ,\phantom{\rule{0.2em}{0ex}}K$
${\mathit{t}}_{0},\dots ,{\mathit{t}}_{\mathit{K}}$ are the enumerated indices of the sample window. Use this code to implement the filter.
% Filter parameters
FD = 0.4;
K = 7; % Polynomial degree
N = K+1; % FIR Length
idxWindow = (-floor((N-1)/2):floor(N/2))';
% Define and solve Lagrange interpolation equations
V = idxWindow.^(0:K); % Vandermonde structure
C = FD.^(0:K);
hLagrange = C/V; % Solve for the coefficients
i0 = -idxWindow(1); % Causal latency
You can implement this model as a direct-form FIR filter if the delay value FD is fixed, or using a Farrow structure if the delay value varies. For more information on implementing Lagrange
interpolation using the Farrow mode, see the Design And Implement Lagrange-Based Delay Filters section in this example.
Design And Implement Sinc-Based Fractional Delay FIR Filters
This section shows you how to design and implement sinc-based fractional delay filters.
designFracDelayFIR Function in Length-Based Design Mode
Use the designFracDelayFIR function to design a fractional delay FIR filter with delay FD and length N. This function provides a simple interface to design such filters.
FD = 0.32381;
N = 10;
h = designFracDelayFIR(FD,N)
h = 1×10
0.0046 -0.0221 0.0635 -0.1664 0.8198 0.3926 -0.1314 0.0552 -0.0200 0.0042
Implement the filter using the dsp.FIRFilter System object.
% Create an FIR filter object
fdfir = dsp.FIRFilter(h);
Delay a signal by filtering it through the designed filter.
% Generate some input
n = (1:100)';
x = gen_input_signal(n);
% Filter the input signal
y = fdfir(x);
plot_sequences(n,x, n,y);
legend('Filter Output','Original Sequence')
title('Raw Filter Output v.s. Input Sequence')
The actual filter delay is not $\mathit{FD}$, but $\mathit{FD}+{\mathit{i}}_{0}$ because of the causal integer latency ${\mathit{i}}_{0}$. The designFracDelayFIR function returns the latency as the
second output argument.
[h,i0] = designFracDelayFIR(FD,N);
The overall delay is merely the sum of the desired fractional delay and the incurred integer latency.
This total delay is also the group delay of the FIR filter at low frequencies. Verify that by using the outputDelay function.
[Doutput,~,~] = fdfir.outputDelay(Fc=0)
Shift the plot of the input sequence by the total delay $\mathit{FD}+{\mathit{i}}_{0}$ to align the filter output with the expected result.
plot_sequences(n+Dtotal,x, n,y);
legend('Filter Output','Input Sequence (shifted by FD+i0)')
title('Filter Output v.s. Time Adjusted Input Sequence')
The shifted input markers located at $\left(\mathit{n}+\mathit{FD}+{\mathit{i}}_{0},\text{\hspace{0.17em}}\mathit{x}\left[\mathit{n}\right]\right)$ generally do not coincide with the output samples
markers $\left(\mathit{n},\text{\hspace{0.17em}}\mathit{y}\left[\mathit{n}\right]\right)$, because $\mathit{n}+\mathit{FD}+{\mathit{i}}_{0}$ falls on noninteger values on the x-axis, whereas n is an
integer. The shifted input samples fall approximately on a line connecting the two consecutive output samples.
plot_sequences(n+i0+FD,x, n,y,'with line');
legend('Filter Output','Input Sequence (shifted by FD+i0)')
title('Output Samples v.s. Shifted Input Samples ')
dsp.VariableFractionalDelay System Object in 'FIR' Mode
You can also design sinc-based delay filters using the dsp.VariableFractionalDelay System object and by setting its interpolation method property to 'FIR'. Begin by creating an instance of the System
object. The FIR length is always even, and you can specify it in the FilterHalfLength property.
vfd_fir = dsp.VariableFractionalDelay(InterpolationMethod = 'FIR', FilterHalfLength = N/2);
i0_vfd_fir = vfd_fir.FilterHalfLength; % Integer latency
Pass the desired fractional delay as the second input argument to the object call. Make sure that the delay value you specify includes the integer latency.
y = vfd_fir(x,i0+FD);
plot_sequences(n+i0+FD,x, n,y);
legend('Filter Output','Input Sequence (shifted by FD+i0)')
title('dsp.VariableFractionalDelay in FIR Mode')
Compare designFracDelayFIR and dsp.VariableFractionalDelay in 'FIR' Mode
Both designFracDelayFIR and dsp.VariableFractionalDelay in 'FIR' mode provide sinc-based fractional delay filters, but their implementations are different.
• The dsp.VariableFractionalDelay System Object approximates the delay value by a rational number $\mathit{FD}\approx \mathit{k}}{\mathit{L}}$ up to some tolerance, and then samples the fractional
delay as the k-th phase of a (long) interpolation filter of length L. This requires more memory, and yields less accurate delay.
• By contrast, the designFracDelayFIR function generates the FIR coefficients directly, rather than sampling them from a longer FIR. This allows the function to provide a precise fractional delay
value, and costs less memory.
• The designFracDelayFIR function has a simple interface that returns FIR coefficients, and leaves the filter implementation to the user. The interface of the dsp.VariableFractionalDelay System
object allows you to both design and implement the filter.
Use the designFractionalDelayFIR function over the dsp.VariableFractionalDelay System Object in the 'FIR' mode for its simplicity, better performance, and efficiency. This figure shows that the
filter designed using dsp.VariableFractionalDelay has a shorter bandwidth, and its group delay is off by ~0.02 from the nominal value.
% Obtain the FIR coefficients from the dsp.VariableFractionalDelay object
h_vfd_fir = vfd_fir([1;zeros(31,1)],i0_vfd_fir+FD);
plot_freq_and_gd(h,i0,[],"designFracDelayFIR", h_vfd_fir,i0_vfd_fir,[],"dsp.VariableFractionalDelay FIR mode");
hold on;
yline(FD,'DisplayName','Target Fractional Delay');
Design and Implement Lagrange-Based Delay Filters
Lagrange-based fractional delay filters are computationally cheap and you can implement them efficiently using the Farrow structure. The Farrow filter is a special type of FIR filter that you can
implement using only elementary algebraic operations, such as scalar additions and multiplications. Unlike the sinc-based designs, Farrow filters do not require specialized functions (such as sinc or
Bessel) to compute the delay FIR coefficients. This makes Farrow fractional delay filters particularly simple to implement on a basic hardware.
The downside of Lagrange-based delay filters is that they are limited to lower orders, due to the highly unstable nature of higher order polynomial approximations. Using this method usually results
in a lower bandwidth filter, as compared to a sinc-based filter.
Another type of a computationally-cheap fractional delay filter is the Thiran filter. See thiran (Control System Toolbox) for more information.
The System Object dsp.VariableFractionalDelay in 'Farrow' mode
Use the dsp.VariableFractionalDelay System object in the 'Farrow' mode to create and implement Farrow delay filters. Begin by creating an instance of the System object:
vfd = dsp.VariableFractionalDelay(InterpolationMethod = 'Farrow', FilterLength = 8);
i0var = floor(vfd.FilterLength/2) % Integer latency of the filter
Apply the object to an input signal and plot the result.
y = vfd(x,i0var+FD);
plot_sequences(n+i0var+FD,x, n,y);
legend('Farrow Fractional Delay Output','Input Sequence (shifted by FD+i0)')
title('dsp.VariableFractionalDelay in Farrow Mode')
You can also vary the fractional delay values. This code operates on frames of 20 samples while increasing the delay value with each frame. Note the increase of the delay in the output graph
corresponding to the changes in the delay values.
FDs = i0var+5*(0:0.2:0.8); % Fractional delays vector
xsource = dsp.SignalSource(x,20);
ysink = dsp.AsyncBuffer;
for FD=FDs
xk = xsource();
yk = vfd(xk, FD);
y = read(ysink);
plot_sequences(n+i0var,x, n,y);
legend('Variable Fractional Delay Output','Original Sequence (shifted by i0)')
title('dsp.VariableFractionalDelay in Farrow Mode, Varying Delay')
Bandwidth of FIR Fractional Delay Filters: Analysis and Design
Longer filters provide better approximations of the ideal delay filter in terms of raw quadratic norms. However, practically a more meaningful metric, such as the bandwidth is needed. The function
designFracDelayFIR measures combined bandwidth, which is defined as the frequency range in which both the gain and the group delay are within 1% of their nominal values. You can obtain the measured
combined bandwidth from the designFracDelayFIR function. Design two FIR fractional delay filters of length 16 and 256. Plot the gain and group delay of the two filters. As expected, the longer filter
(in red in the plot) has a significantly higher combined bandwidth.
FD = 0.3;
N1 = 16;
N2 = 256;
[h1,i1,bw1] = designFracDelayFIR(FD, N1);
[h2,i2,bw2] = designFracDelayFIR(FD, N2);
plot_freq_and_gd(h1,i1,bw1,"N="+num2str(N1), h2,i2,bw2,"N="+num2str(N2));
designFracDelayFIR Function in Bandwidth Design Mode
In the bandwidth design mode, the designFracDelayFIR function can determine the required filter length for a given bandwidth. Specify the delay value and the desired target bandwidth as inputs to the
function, and the function finds the appropriate filter length.
FD = 0.3;
bwLower = 0.9; % Target bandwidth lower limit
[h,i0fixed,bw] = designFracDelayFIR(FD,bwLower);
fdfir = dsp.FIRFilter(h);
ans = 6x35 char array
'Discrete-Time FIR Filter (real) '
'------------------------------- '
'Filter Structure : Direct-Form FIR'
'Filter Length : 52 '
'Stable : Yes '
'Linear Phase : No '
bwLower is merely a lower bound for the combined bandwidth. The function returns a filter whose combined bandwidth is at least the value specified in bwLower.
Distortion in High Bandwidth Signals
In this section, you compare the performance of the two design points (long sinc v.s. short Lagrange) with a high bandwidth input. The filter you designed using the dsp.VariableFractionalDelay System
object in the previous section is an 8-degree Farrow structure, effectively an FIR of length 9. The filter you designed using the designFracDelayFIR function, with FD = 0.3 and bwLower = 0.9, has a
length of 52 samples. Plot the two FIR frequency responses on the same graph to demonstrate the bandwidth difference between the two.
hvar = vfd([1;zeros(31,1)],i0var+FD);
plot_freq_and_gd(h,i0fixed,bw,"Sinc-based", hvar,i0var,[],"Farrow");
Apply the two filters to a high-bandwidth signal, Plot the time (top row) and frequency (bottom row) response of the two signals to compare the performance of the two filters (sinc on the left and
Farrow on the right).
x = high_bw_signal(n);
y1 = fdfir(x);
y2 = vfd(x,i0var+FD);
The results are as expected:
• The longer sinc filter has a higher bandwidth. The shorter Farrow filter has a lower bandwidth.
• Signal distortion is virtually nonexistent in the longer sinc filter, but easily noticeable in the shorter Farrow filter.
• Higher accuracy comes at the expense of longer latency: approximately 25 samples v.s. only 4 in the shorter filter.
dsp.VariableFractionalDelay or designFracDelayFIR ?
You can choose between the dsp.VariableFractionalDelay System object and the designFracDelayFIR function based on the filter requirements and the target platform.
• For a high bandwidth and accurate group delay response, use the designFracDelayFIR function. Keep in mind that this design process is more intensive computationally. Therefore, deploy the
function on high-end hardware if you want to tune the delay values in real time. You can deploy it on low-end hardware if you fix the delay value and design the filter offline.
• For time-varying delay filters aimed at low-performance computational apparatus, use dsp.VariableFractionalDelay in the 'Farrow' mode. | {"url":"https://in.mathworks.com/help/dsp/ug/design-of-fractional-delay-fir-filters.html","timestamp":"2024-11-13T01:16:14Z","content_type":"text/html","content_length":"123431","record_id":"<urn:uuid:e3e517f4-cb9c-4070-9913-ed7ff9bf196b>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00895.warc.gz"} |
RcppArmadillo: R and Armadillo via Rcpp
RcppArmadillo provides an interface from R to and from Armadillo by utilising the Rcpp R/C++ interface library.
What is Armadillo?
Armadillo is a high-quality linear algebra library for the C++ language, aiming towards a good balance between speed and ease of use. It provides high-level syntax and functionality deliberately
similar to Matlab (TM). See its website more information about Armadillo.
So give me an example!
Glad you asked. Here is a light-weight and fast implementation of linear regression:
#include <RcppArmadillo/Lighter>
// [[Rcpp::depends(RcppArmadillo)]]
// [[Rcpp::export]]
Rcpp::List fastLm(const arma::mat& X, const arma::colvec& y) {
int n = X.n_rows, k = X.n_cols;
arma::colvec coef = arma::solve(X, y); // fit model y ~ X
arma::colvec res = y - X*coef; // residuals
double s2 = arma::dot(res, res) / (n - k); // std.errors of coefficients
arma::colvec std_err = arma::sqrt(s2 * arma::diagvec(arma::pinv(arma::trans(X)*X)));
return Rcpp::List::create(Rcpp::Named("coefficients") = coef,
Rcpp::Named("stderr") = std_err,
Rcpp::Named("df.residual") = n - k);
You can Rcpp::sourceCpp() the file above to compile the function. A version is also included in the package as the fastLm() function.
The RcppArmadillo/Lighter header includes Rcpp via its Rcpp/Lighter header which precludes some more compile-time heavy features such as ‘Rcpp Modules’ which we may not need. See the Rcpp docs more
details about ‘Light’, ‘Lighter’ and ‘Lightest’. In the example above, the switch saves about 15% of total compilation time.
The package is mature yet under active development with releases to CRAN about once every other month, and widely-used by other CRAN packages as can be seen from the CRAN package page. As of April
2024, there are 1135 CRAN packages using RcppArmadillo.
The package contains a pdf vignette which is a pre-print of the paper by Eddelbuettel and Sanderson in CSDA (2014), as well as an introductory vignette for the sparse matrix conversions.
RcppArmadillo is a CRAN package, and lives otherwise in its own habitat on GitHub within the RcppCore GitHub organization.
to install from your nearest CRAN mirror.
Dirk Eddelbuettel, Romain Francois, Doug Bates, Binxiang Ni, and Conrad Sanderson
GPL (>= 2) | {"url":"https://cran.r-project.org/web/packages/RcppArmadillo/readme/README.html","timestamp":"2024-11-11T08:27:13Z","content_type":"application/xhtml+xml","content_length":"13919","record_id":"<urn:uuid:3ef50d09-42b7-4c41-98e5-c2c7e0d86878>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00239.warc.gz"} |
Derivative of x/1+x
Introduction to the derivative of x/1+x
Derivatives have a wide range of applications in almost every field of engineering and science. The x/1+x derivative can be calculated by following the rules of differentiation. Or, we can directly
find the derivative formula of x/1+x by applying the first principle of differentiation. In this article, you will learn what the x by 1+x derivative formula is and how to calculate the derivatives
of x/1+x by using different approaches.
What is the derivative of x/1+x?
The derivative of x/1+x with respect to the variable x is equal to 1/(1+x)^2. It measures the rate of change of the algebraic function x/1+x. It is denoted by d/dx(x/1+x) which is a fundamental
concept in calculus. Knowing the formula for derivatives and understanding how to use it can be used in solving problems related to velocity, acceleration, and optimization.
Derivative of x/1+x formula
The formula for derivative of f(x)=x/1+x is equal to the 1/(1+x)^2, that is;
$f'(x) = \frac{d}{dx}(\frac{x}{1+x}) = \frac{1}{(1+x)^2}$
It is calculated by using the power rule of derivatives, which is defined as:
$\frac{d}{dx} (x^n)=nx^{n-1}$
How do you differentiate 1 by 1+x?
There are multiple derivative laws to prove the differentiation of x/1+x. These are;
1. First Principle
2. Product rule
3. Quotient Rule
Each method provides a different way to compute the x/1+x differentiation. By using these methods, we can mathematically prove the formula for finding the differential of x/1+x.323919
Derivative of x/1+x by first principle
According to the first principle of derivative, the x/1+x derivative is equal to 1/(1+x)^2. The derivative of a function by first principle refers to finding a general expression for the slope of a
curve by using algebra. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to,
$f’(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$
This formula allows us to determine the rate of change of a function at a specific point by using the limit definition of derivative calculator.
Proof of x by 1+x derivative formula by first principle
To prove the derivative of x/1+x by using first principle, replace f(x) by x/1+x or you can replace it by x/1+x to find the derivative.
$f’(x) = \lim_{h→0}\frac{f(x + h) - f(x)}{h}$
$f’(x) = \lim_{h \to 0} \frac{\frac{x+h}{1+x+h} - \frac{x}{1+x}}{h}$
$f’(x) = \lim_{h \to 0} \frac{(1+x^2)-(1+x^2+h^2+2xh)}{h(1+x^2)(1+(x+h)^2)}{2}nbsp;
$f’(x) = \lim_{h \to 0} \frac{-h^2-2xh}{h(1+x^2)(1+(x+h)^2)}$
$f’(x) = \lim_{h \to 0} \frac{-h-2x}{(1+x^2)(1+(x+h)^2)}$
When h approaches to zero,
$f’(x) = \frac{-2x}{(1+x^2)^2}$
Hence the differentiation of x/1+x is equal to -2x/(1+x^2)^2. Also calculate the derivative of x^2 graph by using our derivative graph calculator.
Derivative of x/1+x by product rule
Another method to find the derivative x by 1+x is the product rule formula which is used in calculus to calculate the derivative of the product of two functions. Specifically, the product rule is
used when you need to differentiate two functions that are multiplied together. The formula for the product rule is:
d/dx(uv) = u(dv/dx) + (du/dx)v
In this formula, u and v are functions of x, and du/dx and dv/dx are their respective derivatives with respect to x.
Proof of differentiating of x/1+x by product rule
To differentiate of x/1+x by using product rule, we start by assuming that,
$f(x) =\frac{1}{1+x^2}=x(1+x)^{-1}{2}nbsp;
By using product rule of differentiation,
$f’(x) = x.\frac{d}{dx}(1+x)^{-1}+(1+x)^{-1}\frac{d}{dx}(x)$
We get,
$f’(x) =(x)(-1)(1+x)^{-2}+\frac{1}{1+x}{2}nbsp;
$f’(x) = \frac{-x}{(1+x)^2}+\frac{1}{1+x}$
After simplification, we get
$f’(x) = \frac{-x+1+x}{(1+x)^2}$
$f’(x) = \frac{-1}{(1+x)^2}$
Also use our derivative product rule calculator online, as it provides you a step-by-step solution of differentiation of a function.
Derivative of 1/x+1 using quotient rule
Another method for finding the differential of x/1+x is using the quotient rule, which is a formula for finding the derivative of a quotient of two functions. Since the secant function is the
reciprocal of cosine, the derivative of cosecant can also be calculated using the quotient rule. The quotient rule of derivative is defined as:
$\frac{d}{dx}(f/g) = \frac{f’(x).g(x) - g’(x).f(x)}{(g(x))^2}$
Proof of differentiating x by quotient rule
To prove the x^2 derivative, we can start by writing it,
$f(x) =\frac{x}{1+x} = \frac{u}{v}$
Supposing that u = x and v = 1+x. Now by quotient rule,
$f’(x) =\frac{v.u’ - uv’}{v^2}$
$f’(x) = \frac{(1+x)(x)’ - (x)(1+x)’}{(1+x)^2}$
$f’(x)=\frac{(1+x)(1) - x}{(1+x)^2}{2}nbsp;
Hence, we have derived the derivative of x/x+1 using the quotient rule of differentiation. Use our quotient rule calculator to find the derivative of any quotient function.
How to find the derivatives of x/1+x with a calculator?
The easiest way of differentiating x/1+x is by using a dy/dx calculator. You can use our derivative calculator for this. Here, we provide you a step-by-step way to calculate derivatives by using this
1. Write the function as x/1+x in the enter function box. In this step, you need to provide input value as a function that you want to differentiate.
2. Now, select the variable by which you want to differentiate x/1+x. Here you have to choose x.
3. Select how many times you want to calculate the derivatives of x/1+x. In this step, you can choose 2 to calculate the second derivative, 3 for triple differentiation and so on.
4. Click on the calculate button. After this step, you will get the derivative of 1/1+x2 within a few seconds.
5. After completing these steps, you will receive the differential of x by 1+x within seconds. Using online tools can make it much easier and faster to calculate derivatives, especially for complex
In conclusion, the derivative of x/1+x is 1/(1+x)^2. The derivative measures the rate of change of a function with respect to its independent variable, and in the case of x/1+x, the derivative can be
calculated using the power rule, product rule and quotient rule of differentiation. By applying this rule, we find that the derivative of x/1+x is 1/(1+x)^2, indicating that the rate of change of x/
1+x increases linearly with the value of x. | {"url":"https://calculator-derivative.com/derivative-of-x-by-1x","timestamp":"2024-11-03T23:03:31Z","content_type":"text/html","content_length":"37017","record_id":"<urn:uuid:9d127c2d-b5a7-4bab-afec-3d7601a6a9a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00082.warc.gz"} |
High School Mathematics Extensions/Peano's Axioms - Wikibooks, open books for an open world
The Peano Axioms, also called Peano's Axioms, Dedekind-Peano Axioms, or the Peano postulates, are a series of axioms attempting to formalize the natural numbers (${\displaystyle \mathbb {N} }$) using
simple, intuitive axioms.
During the time of Peano, mathematical notation as a whole—especially mathematical logic—was a new field, and common notions of second-order logic did not exist. Peano's own notation for logic was
unpopular, but he did popularize something similar to the common way of denoting elements of a set (${\displaystyle A\in B}$ ), using ${\displaystyle \epsilon }$ . While we will not be using Peano's
old notation, we will be expressing the axioms in a common notation which you should be familiar with from the set theory chapters of this book.
The Peano axioms describe the properties of the natural set, commonly denoted N or in blackboard bold, ${\displaystyle \mathbb {N} }$ . The first axiom states that 0 is a natural number.
1. ${\displaystyle 0\in \mathbb {N} }$ . ("zero is a natural number")
The next four axioms describe the most important properties of the natural set, the equivalence relation.
2. ${\displaystyle \forall x\in \mathbb {N} :x=x}$ . ("all natural numbers equal themselves")
3. ${\displaystyle \forall x,y,z\in \mathbb {N} :x=y,y=z\implies x=z}$ . ("equality is transitive")
4. ${\displaystyle \forall x,y\in \mathbb {N} :x=y\implies y=x}$ . ("equality is symmetric")
5. ${\displaystyle \forall x\in \mathbb {N} :x=y\implies y\in \mathbb {N} }$ . ("if x is natural, and x equals y, y is natural")
After this, Peano introduces the idea of a successor function, commonly denoted ${\displaystyle \mathbf {S} }$ , ${\displaystyle {\widehat {S}}}$ , or ${\displaystyle \mathrm {succ} }$ .
6. ${\displaystyle \forall x\in \mathbb {N} :\mathrm {succ} (x)\in \mathbb {N} }$ . ("the successor of any natural number is a natural number")
7. ${\displaystyle \forall x,y\in \mathbb {N} :\mathrm {succ} (x)=\mathrm {succ} (y)\iff x=y}$ . ("the successor function is an injection")
8. ${\displaystyle \mathrm {succ} (x)=0\implies \bot }$ . ("there is no x such that the successor of x is 0")
The obvious intuition is that we can compose the entire natural set by successively applying the successor function, but none of the axioms actually present this. Thus, there is motivation to also
add the axiom of induction, which is commonly misattributed as part of Peano's axioms, but it was not described in Peano's original construction.
We can simply say that the natural set is constructible by successively applying the successor.
9. The natural set, ${\displaystyle \mathbb {N} }$ is constructible by successively applying the successor function. | {"url":"https://en.m.wikibooks.org/wiki/High_School_Mathematics_Extensions/Peano%27s_Axioms","timestamp":"2024-11-11T10:30:08Z","content_type":"text/html","content_length":"52471","record_id":"<urn:uuid:4d9d255c-561b-42f5-84de-83b3b1e341c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00170.warc.gz"} |
Mixed-Effects Meta Analysis (MEMA)
Theoretical background
The conventional approach for FMRI group analysis is to take regression coefficients (typically referred to as beta values) and run t-test, AN(C)OVA, or other types of analysis, and ignore the
within-subject variability (sampling error of each condition/task/event type at individual subject level). The underlying rationale for such practice is built on the following two assumptions:
(1) Within-subject (or intra-subject) variability is relatively small compared to between-subjects (group) variability;
(2) Within-subject variability is roughly equal across all subjects.
Current evidences show that up to 40% or even more of variability is usually accounted for with intra-subject variation among all the variability (sum of within- and cross-subject variability) in an
activated region of interest. Also there are a lot of variations in within-subject variability across subjects. Violation of either assumption could render a suboptimal or even invalid group
analysis, and most of the time it reduces the statistical power at group level. Therefore the conventional group analysis method, typically referred to as random/mixed-effects modeling, is mostly a
pseudo-random/mixed-effects approach.
Within-subject variability measures how precise/reliable the percent signal change (beta) estimate is from individual subject time series regression model. Since such quantity, contained in the
corresponding t-statistic, is conveniently available, there is no reason, except theoretical complexity and computation cost, to waste this piece of information in group analysis.
3dMEMA is developed in R with a mixed-effects meta analysis (MEMA) approach. It effectively takes advantage of estimate precision from each subject, and assigns each subject's contribution in the
final result based on weighting instead of equal treatment. More specifically, a more precise beta estimate (meaning higher t-statistic) from a subject will have more say in the group effect;
conversely, a less reliable beta estimate (i.e., lower t-statistic) from a subject will be discounted in the MEMA model. Such strategy can be surprisingly tolerant of and robust against some types
of outliers compared to the conventional group analysis method. More theoretical considerations of MEMA can be found in the following literature:
Chen et al., 2012. FMRI Group Analysis Combining Effect Estimates and Their Variances. NeuroImage 60, 747-765.
3dMEMA handles the following model types:
(1) random-effects analysis: one-sample, paired-sample
(2) mixed-effects analysis: two-sample (two groups of subjects)
(3) mixed-effects analysis: multiple between-subjects factors
(4) either of the above three types plus between-subjects covariate(s)
This basically covers whatever t-test you could do with 3dttest/3dANOVAx/3dMVM/3dLME/3dRegAna before. Noticeably it may give an impression that it can't directly deal with sophisticated ANOVA
designs, but that is not necessarily the case. F-tests for main effects and interactions in ANOVA, for example, provide a concise summary for the factors and their relationship, but eventually most
of the time everything boils down to single (not composite) effect testing. In other words, almost all those t-tests in 3dttest/3dANOVAx/3dRegAna/3dMVM/3dLME can be run with 3dMEMA.
Putting in a different perspective, you can run 3dMEMA if your analysis can be conceptualized into one of the following types:
A. one condition within one group;
B. two conditions within one group;
C. one condition within two or more groups with homoskedasticity;
D. one condition within two or more groups with heteroskedasticity.
Most tests from a sophisticated analysis design should find their niche in the above list. For example, suppose we have factor A, coding for two groups of subjects, and factor B, representing two
levels (e.g., house and face) of an experiment condition. Usually we can get all the tests we want with one batch script: 3dANOVA3 -type 5 .... However, whatever we can get with options such as -fa,
-fb, -fab, -amean, -bmean, -adiff, -bdiff, -acontr, -bcontr, -aBdiff, -Abdiff, -aBcontr, and -Abcontr, they are essentially all t-tests (including those F-tests for main effects and interaction with
ONE numerator degree of freedom), which means we can deal with them with 3dMEMA without any problems, but just one test a time! Moreover, these F-tests are disadvantageous and not as informative
compared to their counterpart with t-tests because F hides the directionality (sign) of the effect of interest. For instance, the F-test for the interaction between factors A and B (both with 2
levels) in this example is essentially equivalent to the t-test (A[1]B[1]-A[1]B[2])-(A[2]B[1]-A[2]B[2]) or (A[1]B[1]-A[2]B[1])-(A[1]B[2]-A[2]B[2]), but we can say more with the t-test than F: a
positive t-value shows A[1]B[1]-A[1]B[2] > A[2]B[1]-A[2]B[2] and A[1]B[1]-A[2]B[1] > A[1]B[2]-A[2]B[2].
The only tests that 3dMEMA can't currently handle are those F-tests with multiple numerator degrees of freedom, but hopefully this F-test limitation will change in the future. See more discussion
here or here regarding 3dMEMA/3dttest/3dtest++ vs. traditional ANOVA framework.
As beta precision estimate is important for MEMA, it is suggested (not mandated, of course) that
(1) all input files, beta and more importantly t-statistic, come from 3dREMLfit output instead of 3dDeconvolve;
(2) warping to standard space be performed before spatial smoothing and individual subject regression analysis to avoid the troubling step of warping on t-statistic;
(3) no masking be applied at individual subject analysis level so that no data is lost at group level along the edge of (and sometimes inside) the brain.
If you don't have R installed yet on your computer, choose a mirror site geographically close to you, and download the appropriate binary for your platform (or the source code and then compile
yourself). Set your path appropriately. For example, my R executable is under /Applications/R.app/Contents/MacOS on my Mac OS X, so I add /Applications/R.app/Contents/MacOS as one of the search paths
in my C shell startup configuration file .cshrc
If the installation is successful, start the R interface with the following command on the prompt
You can also work with the GUI version of R on Mac OS and Windows. Special note for Mac OS X users when using the GUI R: Don't start the GUI R through clicking the icon because it would fail to
initialize all the path setup on the X11. Instead run the GUI version by typing/copying the following on the terminal (or, even better, setting an alias
/Applications/R.app/Contents/MacOS/R &
The usage of 3dMEMA can be found at the terminal"
3dMEMA -help | less
How to use 3dMEMA to handle multiple groups
Suppose at group level we have three categorical variables (factors): one within-subject factor condition with two levels, positive (pos) and negative (neg); two between-subjects factors, sex (male
and female) and genotypes (FF, TT, and FT). This would be a mixed 2 (sex) x 3 (genotype) x 2 (condition) ANOVA, but we would like to use 3dMEMA to tackle the analysis.
First we need the contrast of positive and negative conditions and its t-statistic from each subject (with 3dREMLfit). Then we treat the two between-subjects factors and their interactions as
covariate via dummy coding in a text file with the five columns. Note that dummy coding works like this: for each factor choose one level as base (also called reference) and code it with 0, and a
factor with k levels is represented with k-1 columns each of which codes for one level (except for the base level).
F-M TT-FF FT-FF int1 int2
If you call the above text file as cov.txt, then add the following line in the 3dMEMA script:
-covariates_center F-M = 0 TT-FF = 0 FT-FF = 0 int1 = 0 int2 = 0
Interpretation of the output:
the first two sub-bricks corresponds to the base of all between-subjects factors: the contrast of positive and negative conditions of male with FF genotype.
the next two sub-bricks corresponds to the first column of the covariate file: the sex difference in the contrast of positive and negative conditions (or the interaction between sex and condition)
the next two sub-bricks corresponds to the second column of the covariate file: the genotype difference between TT and FF in the contrast of positive and negative conditions
the next two sub-bricks corresponds to the third column of the covariate file: the genotype difference between FT and FF in the contrast of positive and negative conditions
the next two sub-bricks corresponds to the fourth column of the covariate file: three-way interaction
the next two sub-bricks corresponds to the fifth column of the covariate file: another three-way interaction
Running 3dMEMA inside R
Alternatively 3dMEMA works in a procedural or streamlined fashion with a string of information about modeling parameters, input files (beta and t-statistic) and options. Hopefully anything else
should be self-evident from there as shown below with user input underlined in bold face (Note: input files with sub-brick selector are allowed, but no quotes are needed around the square brackets.
See example below):
> source("~/abin/3dMEMA.R")
[1] "#+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++"
[1] " ================== Welcome to 3dMEMA.R ================== "
[1] "AFNI Meta-Analysis Modeling Package!"
[1] "#+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++"
[1] "Version 0.1.3, Jan. 8, 2010"
[1] "Author: Gang Chen (gangchen@mail.nih.gov)"
[1] "Website: http://afni.nimh.nih.gov/sscc/gangc/3dMEMA.html"
[1] "SSCC/NIMH, National Institutes of Health, Bethesda MD 20892"
[1] "#+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++"
[1] "################################################################"
[1] "Please consider citing the following if this program is useful for you:"
Gang Chen, Manual or manuscript coming soon.
[1] "################################################################"
[1] "Use CNTL-C on Unix or ESC on GUI version of R to stop at any moment."
Output file name (just prefix, no view+suffix needed, e.g., myOutput): myOutput
[1] "On a multi-processor machine, parallel computing will speed up the program significantly."
[1] "Choose 1 for a single-processor computer."
Number of parallel jobs for the running (e.g., 2)? 4
Number of groups (1 or 2)? 1
[1] "-----------------"
[1] "The following types of group analysis are currently available:"
[1] "The following types of group analysis are currently available:"
[1] "1: one condition with one group;"
[1] "2: one condition across 2 groups with homoskedasticity (same variability);"
[1] "3: two conditions with one group;"
[1] "4: one condition across 2 groups with heteroskedasticity (different variability)."
Which analysis type (1, 2, 3): 3
[1] "Since the contrast between the 2 conditions will be the 1st minus the 2nd, choose"
[1] "an appropriate order between the 2 conditions to get the desirable contrast."
Label for the contrast? myContrast
Number of subjects: 18
Number of subjects in group Female (e.g., 12)? 24
No. 1 subject label in group: S1
No. 18 subject label in group: S18
Label for condition 1? conditon1
No. 1 subject file for beta or linear combination of betas with condition1: subj1_con1_B+tlrc.BRIK[0]
No. 1 subject file for the corresponding t-statistic with condition1: subj1_con1_T+tlrc.BRIK[1]
[1] "-----------------"
No. 2 subject file for beta or linear combination of betas with condition1: subj2_con1_B+tlrc.BRIK[0]
No. 2 subject file for the corresponding t-statistic with condition1: subj2_con1_T+tlrc.BRIK[1]
Label for condition 2? conditon2
No. 1 subject file for beta or linear combination of betas with condition2: subj1_con2_B+tlrc.BRIK[0]
No. 1 subject file for the corresponding t-statistic with condition2: subj1_con2_T+tlrc.BRIK[1]
[1] "-----------------"
No. 2 subject file for beta or linear combination of betas with condition2: subj2_con2_B+tlrc.BRIK[0]
No. 2 subject file for the corresponding t-statistic with condition2: subj2_con2_T+tlrc.BRIK[1]
Number of subjects with non-zero t-statistic? (0-18) 12
[1] "-----------------"
[1] "t-statistic is a little more conservative but also more appropriate for significance testing than Z"
[1] "especially when sample size, number of subjects, is relatively small."
Z- or t-statistic for the output? (0: Z; 1: t) 0
[1] "-----------------"
[1] "Masking is optional, but will alleviate unnecessary penalty on q values of FDR correction."
Any mask (0: no; 1: yes)? myMask+tlrc.BRIK
[1] "-----------------"
[1] "Covariates are continuous variables (e.g., age, behavioral data) that can be partialled out in the model."
Any covariates (0: no; 1: yes)? 0
[1] "-----------------"
[1] "If outliers exist at voxel/subject level, a special model can be adopted to account for outliers"
[1] "in the data, leading to increased statistical power at slightly higher computation cost."
Model outliers (0: no; 1: yes)? 1
[1] "-----------------"
[1] "The Z-score of residuals indicates the significance level a subject is an outlier at a voxel."
[1] "Turn off this option if memory allocation problem occurs later on."
Want residuals Z-score for each subject (0: no; 1: yes)? 1
[1] "-----------------"
[1] "Totally 43 slices in the data."
[1] "-----------------"
[1] "Package snow successfully loaded!"
Z slice # 1 done: 04/30/09 14:06:17.290
Z slice # 2 done: 04/30/09 14:06:17.982
Z slice # 43 done: 04/30/09 14:17:27.149
[1] "Analysis finished: 04/30/09 14:17:27.151"
[1] "#++++++++++++++++++++++++++++++++++++++++++++"
++ 3drefit: AFNI version=AFNI_2008_07_18_1710 (Apr 8 2009) [32-bit]
++ Authored by: RW Cox
++ Processing AFNI dataset myOutput+orig
+ Changed dataset view type and filenames.
+ created 2 FDR curves in dataset header
++ 3drefit processed 1 datasets
> proc.time()
user system elapsed
39.847 13.348 743.834
All input files should only contain ONE (either beta or t-statistic) sub-brick. You don't have to type those input file names. Instead I suggest that you list all those files by executing 'ls -1
*.BRIK' (number ONE, not letter L ) on the terminal, and copy and taste them onto the 3dMEMA interface. Directories can be included as part of the file name, that is, those input files don't have to
be in the same directory where you run 3dMEMA. It's always a good habit, for records and for running it again (or a different analysis) in batch mode later on, to save all the input items in a pure
text file with content like the following (don't include those interpretive words after the pound sign):
myOutput # output file name (no view and appendix needed)
4 # number of parallel jobs
1 # number of groups of subjects
3 # paired-sample type
myContrast # label for condition 1 vs. condition 2
18 # total number of subjects
conditon1 # condition 1 label
subj1_con1_B+tlrc.BRIK[0] # beta value for subject 1
subj1_con1_T+tlrc.BRIK[1] # t-statistic for subject 1
conditon2 # condition 2 label
12 # minimum number of subjects allowed to have ZERO t-statistic at a voxel
0 # want Z or t-statistic
1 # yes a mask will be provided; otherwise 0
myMask+tlrc.BRIK # mask
0 # no covariates
1 # handle outliers with a special model
1 # Z-score for residuals
There are two output files, one includes all the major effects plus the associated statistics, while the other output, if requested, contains two values at a voxel for each subject: lambda measures
the percentage of within-subject variability relative the total variability, and Z-score shows the significance level that voxel is an outlier relative to the group effect.
The runtime can be significantly reduced through parallel computing: If multiple cores are available on your computer, simply specify the number of parallel jobs in the program. Once you know the
exact answers for those sequential questions, you may want to run 3dMEMA.R in a batch mode for a slightly different analysis by creating a file like one above (or multiple ones concatenated), calling
it Cmds.R, for example (again don't include those interpretive words). Type one of the following two commands at the terminal prompt (not inside R):
R CMD BATCH Cmds.R myDiary &
Rscript Cmds.R |& tee myDiary &
or in the same fashion but remotely:
nohup R CMD BATCH Cmds.R myDiary &
nohup Rscript Cmds.R > myDiary &
File "myDiary" contains the progression of the running including error message. In case you encounter some problem with 3dMEMA, please send me the whole file myDiary.
To quit R, type
(or hit letter "d" while holding down CTRL key on UNIX-based systems).
I'd like to thank Jarrod Hadfield for directing my attention to meta analysis, Wolfgang Viechtbauer for theoretical consultation and programming support, Xianggui Qu for help in formula derivation,
and James Bjork for help in testing the program and for providing feedback. | {"url":"https://afni.nimh.nih.gov/MEMA","timestamp":"2024-11-02T13:59:33Z","content_type":"text/html","content_length":"33980","record_id":"<urn:uuid:1f2f8521-8df8-462e-b5fb-5c9c70096b65>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00440.warc.gz"} |
Honeycomb Floquet code
Honeycomb Floquet code[1]
Floquet code based on the Kitaev honeycomb model [2] whose logical qubits are generated through a particular sequence of measurements. A CSS version of the code has been proposed which loosens the
restriction of which sequences to use [3]. The code has also been generalized to arbitrary non-chiral, Abelian topological order [4].
The code is defined on a hexagonal (honeycomb) lattice with a physical qubit located at each vertex. Edges are labeled \(x\), \(y\), and \(z\), such that one edge of each label meet at every vertex.
Check operators are defined as \(XX\) acting on any two qubits joined by an \(x\) edge, and similarly for \(y\) and \(z\). The hexagonal lattice is 3-colorable, so the hexagons may be labeled 0, 1, 2
such that no two neighboring hexagons have the same label.
The code-generating measurement pattern consists of measuring the check operators located on all of the \(r\)-labeled edges in round \(r\) mod 3. The code space is the \(+1\) eigenspace of the
instantaneous stabilizer group (ISG). The ISG specifies the state of the system as a Pauli stabilizer state at a particular round of measurement, and it evolves into a (potentially) different ISG
depending on the check operators measured.
Protective features similar to the surface code: on a torus geometry, the code protects two logical qubits with a code distance proportional to the linear size of the torus. Properties of the code
with open boundaries are discussed in Refs.
, and various other generalizations have been proposed
Initialization can be performed by preparing each pair of qubits on an edge in some particular state independently specified by the effective-one-qubit operators (two-qubit Pauli strings centered on
an edge) and then beginning the check measurement sequence. This is analogous to projecting a state into the code space by measuring stabilizers.
There are two types of logical operators, inner and outer. An inner logical operator is the product of check operators on a homologically nontrivial cycle. They belong to the stabilizer group as a
subsystem code. Outer logical operators have an interpretation in terms of magnetic and electric operators of an embedded surface code, and they do not belong to the stabilizer group of the
associated subsystem code.Fermionic string excitations can be condensed along one-dimensional paths, yielding twist defects [8]. Such excitations are created by taking products of plaquette check
operators (that are present in all ISGs) over some region, multiplying them to yield a Pauli string on the boundary of said region, and cutting this string. Information is processed by braiding and
fusing defects, which are located at the boundaries of the strings.Certain gates [7] can be performed by considering adiabatic paths in the space of Hamiltonians [7,9,10], yielding an instance of
holonomic quantum computation [11]. Fault-tolerant gates should be interpretable as monodromies under a particular notion of parallel transport [12].
The ISG has a static subgroup for all time steps \(r\geq 3\) – that is, a subgroup which remains a subgroup of the ISG for all future times – given by so-called plaquette stabilizers. These are
stabilizers consisting of products of check operators around homologically trivial paths. The syndrome bits correspond to the eigenvalues of the plaquette stabilizers. Because of the structure of the
check operators, only one-third of all plaquettes are measured each round. The syndrome bits must therefore be represented by a lattice in spacetime, to reflect when and where the outcome was
Fault Tolerance
One can run a fault-tolerant decoding algorithm by (1) bipartitioning the syndrome lattice into two graphs which are congruent to the Cayley graph of the free Abelian group with three generators (up
to boundary conditions) and (2) performing a matching algorithm to deduce errors.
\(0.2\%-0.3\%\) in a controlled-not circuit model with a correlated minimum-weight perfect-matching decoder [13].\(1.5\%<p<2.0\%\) in a circuit model with native weight-two measurements and a
correlated minimum-weight perfect-matching decoder [13]. Here, \(p\) is the collective error rate of the two-body measurement gate, including both measurement and correlated data depolarization error
processes.Against circuit-level noise: within \(0.2\% − 0.3\%\) for SD6 (standard depolarizing 6-step cycle), \(0.1\% − 0.15\%\) for SI1000 (superconducting-inspired 1000 ns cycle), and \(1.5\% − 2.0
\%\) for EM3 (entangling-measurement 3-step cycle) [14,15].
Plaquette stabilizer measurement realized on the IBM Falcon superconducting-qubit device [16]
• \(\mathbb{Z}_q^{(1)}\) subsystem code — The dynamically generated logical qubit of the honeycomb Floquet code is generated by appropriately scheduling measurements of the gauge generators of the
\(\mathbb{Z}_{q=2}^{(1)}\) subsystem stabilizer code corresponding to the Kitaev honeycomb model. However, since this subsystem code has zero logical qubits, the instantaneous stabilizer codes of
the honeycomb code cannot be interpreted as gauge-fixed versions of this subsystem code.
• Kitaev surface code — Measurement of each check operator of the honeycomb Floquet code involves two qubits and projects the state of the two qubits to a two-dimensional subspace, which we regard
as an effective qubit. These effective qubits form a surface code on a hexagonal superlattice. Electric and magnetic operators on the embedded surface code correspond to outer logical operators
of the Floquet code. In fact, outer logical operators transition back and forth from magnetic to electric surface code operators under the measurement dynamics. Inspired by the honeycomb Floquet
code, various weight-two measurement schemes have been designed [17–19], with the scheme in Ref. [18] being a special case of DWR. Numerical comparisons have been performed [20].
• Twist-defect surface code — Fermionic string excitations of the honeycomb Floquet code can be condensed along one-dimensional paths, yielding twist defects [8].
• Subsystem color code — Both honeycomb and subsystem color codes are generated via periodic sequences of measurements. However, any measurement sequence can be performed on the color code without
destroying the logical qubits, while honeycomb codes can be maintained only with specific sequences. Honeycomb codes require a shorter measurement cycle and use fewer qubits at the given code
distance [1].
• Majorana stabilizer code — The Honeycomb code admits a convenient representation in terms of Majorana fermions. This leads to a possible physical realization of the code in terms of tetrons [21],
where each physical qubit is composed of four Majorana modes.
• Quantum LDPC (QLDPC) code — The Floquet check operators are weight-two, and each qubit participates in one check each round.
• Kitaev honeycomb code — The Kitaev honeycomb model Hamiltonian is a sum of checks of the honeycomb Floquet code [1].
• \(A_2\) hexagonal lattice code — The honeycomb Floquet code is defined on the honeycomb lattice.
• Projective-plane surface code — Implementing the honeycomb Floquet code on a non-orientable cross-cap geometry allows for a logical-\(HZ\) gate to be implemented via a measurement schedule [22].
Page edit log
Cite as:
“Honeycomb Floquet code”, The Error Correction Zoo (V. V. Albert & P. Faist, eds.), 2022. https://errorcorrectionzoo.org/c/honeycomb
Github: https://github.com/errorcorrectionzoo/eczoo_data/edit/main/codes/quantum/qubits/dynamic/floquet/honeycomb.yml. | {"url":"https://errorcorrectionzoo.org/c/honeycomb","timestamp":"2024-11-04T11:07:33Z","content_type":"text/html","content_length":"34073","record_id":"<urn:uuid:cfeb6ecd-6e28-49dd-a2fc-df5517bfd864>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00295.warc.gz"} |
Maths Std 10 Swadhyay 2.3
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(Exercise 2.3)
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On Maths Science Corner you will get all the printable study material of NCERT Maths and NCERT Science Including answers of prayatn karo, Swadhyay, Chapter Notes, Unit tests, Online Quiz etc..
This material is very helpful for preparing Competitive exam like Tet 1, Tet 2, Htat, tat for secondary and Higher secondary, GPSC etc..
Highlight of the chapter
2.1 Introduction
2.2 Geometrical Meaning of Zeroes of a polynomial
2.3 Relationship between Zeroes and Coefficient of Polynomial
2.3.1 Sums of zeroes
2.3.2 Product of zeroes
2.4 Division Algorithm of polynomials
2.5 Summary
You will be able to learn above topics in Chapter 01 of Gujarati NCERT Maths Standard 10 (Class 10) Textbook chapter.
Earlier Maths Science Corner had given the following :
NCERT Maths Standard 10 (Class 10) Textbook Chapter NCERT Maths Standard 10 (Class 10) Swadhyay 2.1 (Exercise 2.1) NCERT Maths Standard 10 (Class 10) Swadhyay 2.2 (Exercise 2.2)
Today Maths Science Corner is giving you the textbook Chapter 01 of Gujarati NCERT Maths Standard 10 (Class 10) Swadhyay 2.3 (Exercise 2.3) in pdf format for your easy reference.
Maths Std 10 Swadhyay 2.3
You can get Std 6 Material from here.
You can get Std 7 Material from here.
You can get Std 8 Material from here.
You can get Std 10 Material from here.
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A deep and fuzzy dive into search
A Deep And Fuzzy Dive Into Search
Estimated reading time: 16 minutes
Together with my team, I took a deep dive into the available fuzzy search approaches and algorithms for quite a while, in order to find a performant solution for the various projects ArangoSearch
gets used for.
Since the introduction of ArangoSearch back in 2018, many of our users have asked for fuzzy search support. We worked hard on getting this done, and the whole team is now excited to finally make
fuzzy search available with the upcoming ArangoDB 3.7.
In the following, we will share our learnings and hope they are useful for you.
What is Fuzzy Search?
Nowadays people have to deal with tons of unstructured data. Web-search engines taught us that being inaccurate when searching something is normal, e.g. one can occasionally
• make some typos using a mobile phone and expect typos will be automatically corrected.
• Scientists often deal with the necessity of non-exact matching, e.g. identification of common cognates from different dictionaries is very important in historical linguistics.
• Bioinformatics deals with variations in DNA sequence encoded as an insanely long string of 4 nucleotides (ACTG) and needs some means for quantification.
“Fuzzy search” is an umbrella term referring to a set of algorithms for approximate matching. Usually such algorithms evaluate some similarity measure showing how close a search term is to the items
in a dictionary. Then a search engine can make a decision on which results have to be shown first.
In this article I’m going to describe and explain two very important search algorithms, which at the same time are quite different:
• Approximate matching based on Levenshtein distance
• Approximate matching based on NGram similarity
I’m going to dive into details of each algorithm and describe issues one may face when implementing it at web scale. I also emphasize differences between these two approaches so one can better
understand various senses of “fuzzy search”.
Approximate Matching Based on Levenshtein Distance
The Levenshtein distance between two words is the minimal number of insertions, deletions or substitutions that are needed to transform one word into the other.
E.g. Levenshtein distance between two words foo and bar is 3, because we have to substitute each letter in word foo to transform it to bar:
1. foo -> boo
2. boo -> bao
3. bao -> bar
Formally, the Levenshtein distance between two strings a and b (of length |a| and |b| respectively) is given by lev[a,b](|a|, |b|) where lev[a,b](i, j) is the distance between the first i characters
of a and the first j characters of b:
Figure 1: Levenshtein Distance between two strings
For simplicity let’s assume that the Levenshtein distance is our relevance measure for fuzzy matching, which turns our goal into finding the closest (in terms of Levenshtein distance) items for a
given input.
The well known algorithm by Wagner and Fisher uses a dynamic programming scheme that leads to quadratic time complexity in the length of the shortest input string. Of course such complexity makes it
impossible to use it at web scale given real world dictionaries may contain hundreds of thousands of items.
In 2002 Klaus Schulz and Stoyan Mihov posted a paper that described a very interesting approach, showing that for any fixed distance n and input sequence W of length N we can compute a deterministic
automaton A(W)[ ]accepting all sequences with Levenshtein distance less or equal than n to input sequence W [ ]in linear time n and space N. The paper itself is rather complex, so I will try to
provide an approximation of its main concepts and algorithms. Here I assume you’re familiar with the basics of automata theory.
Once the Levenshtein automaton A(W) [ ]is built, we can then intersect it with a dictionary. Of course the complexity of this operation depends on the actual dictionary implementation (e.g. trie,
B-tree), but let’s consider a short example. Let’s imagine our dictionary is implemented as trie and stores words “avocado”, “avalon”, “avalanche” and “cargo”:
Figure 2: Term dictionary as trie
Suppose we search this dictionary for the words within Levenshtein distance 1 of “kargo”. Instead of linearly checking each word in the dictionary, the search algorithm can detect that the whole
right branch from root is irrelevant for the given query and completely skip it.
Let’s formalize our task once again:
Given the input sequence W of length N and the maximum acceptable edit distance n we have to build a deterministic finite automaton (DFA) accepting sequence V IFF Lev(W, V) <= n.
In order to better understand how to build a resulting DFA and how it actually works, we’re going to start with a simple example and first draw a nondeterministic finite automaton (NFA) for input
sequence foobar and distance 1:
Figure 3: Non-deterministic finite automation with input sequence and distance 1
Note that evaluation of NFA is considerably more expensive than the evaluation of DFA since we need to check multiple transitions matching the same label and thus multiple paths from a given state,
that’s why we use NFA only to make our explanation clearer.
We start at state 0^0, horizontal transitions denote correctly entered symbols, verticals and diagonals handle invalid ones. It’s also clear that we’re allowed to pass at most 1 invalid symbol.
Whenever the matching encounters an invalid symbol, the NFA flow immediately transfers to the “upper lane” with horizontal transitions only.
Let’s analyze transitions from initial state:
0^0 -> 0^1 denotes insertion of any symbol before position 0
0^0 -> 1^1 denotes substitution of a symbol at position 0 with any symbol
0^0 -> 1^0 denotes correctly entered symbol at position 0
0^0 -> 2^1 denotes deletion of a symbol at position 0
Let’s draw the NFA for the same input and a max edit distance of 2:
Figure 4: Non-deterministic finite automation with input sequence and distance 2
As you can see, the initial state got one more outgoing transition
0^0 -> 3^2 denoting pretty much the same as the 0^0 -> 2^1 transition, namely the deletion of 2 consequent symbols starting at position 0.
For a max edit distance of 3 there will be 2 more (5 in total) outgoing transitions and so on. That means for each state we have O(n) outgoing transitions.
Blind estimation of number of states in corresponding DFA using the well-known powerset construction will give us O(2^(n+1)*N) states which definitely sounds too rough given the shape of NFA.
Schulz and Mihov also showed that given a max edit distance n, position i, and sequence W, we have to analyze at most 2n+1 NFA states. The intuition behind this magic 2n+1 is quite simple: starting
from the i-th character of input sequence W, you can’t reach a position higher than (i+n)-th while inserting characters, similarly with the position (i-n)-th while deleting characters.
Let’s improve our estimation based on the observations we’ve just made:
Given the input sequence of length N and max edit distance n, for each character in the input sequence we have to analyze at most 2n+1 NFA states with O(n) outgoing transitions each, i.e. overall
number of DFA states is O(n^2n+1N) which is definitely better than blind estimation but still too much.
The next important thing which can help us to improve our estimation is the subsumption of relationships between states. As stated in the paper, state I^J subsumes (I+K)^J-K, so we don’t need to
inspect all possible 2n+1 states. If we now estimate the number of DFA states we get O(n^2N).
This estimation is much better than the previous one, but the major issue is its dependency on the input length. As we’ve already noticed, NFA has pretty much the same logical transitions for every
state. Let’s generalize it:
I ^J -> I ^J+1 denotes insertion of a symbol before position I
I ^J -> (I+1) ^J+1 denotes substitution of a symbol at position I with any symbol
I ^J -> (I+1) ^J denotes correctly entered symbol at position I
I ^J -> (I+K) ^J+K-1 denotes deletion of K consecutive symbols after position I
Such transitions exclusively depend on the distribution of characters with respect to the current character at position I defined as so called characteristic vectors. The characteristic vector X(c,
W, I) for character c is a bit set of size min(2n + 1, |W|-I) where the K-th element is equal to 1 IFF W[i+k] is equal to c, 0 otherwise.
E.g. for the sequence foobar and distance 1, we have the following characteristic vectors:
X(‘f’, ‘foo’, 0) = { 1, 0, 0 }
X(‘o’, ‘foo’, 0) = { 0, 1, 1 }
X(‘o’, ‘foo’, 2) = { 1, 0, 0 }
As stated above, only 2n+1 states matter for a given position and max edit distance n, thus for each position we can enumerate all 2^2n+1 transitions over all relevant characteristic vectors
(character distributions):
Figure 5: Non-deterministic finite automation with input sequence and distance n
Let’s write it down in tabular form for position 0^0 and max edit distance 1:
Characteristic vector X(c, W, i), i < |W| Set of target positions
<0,0,0> { 0^1, 1^1 }
<0,0,1> { 0^1[, ]1^1 }
<0,1,0> { 0^1, 1^1, 2^1 }
<0,1,1> { 0^1, 1^1, 2^1 }
<1,0,0> { 1^0 }
<1,0,1> { 1^0 }
<1,1,0> { 1^0 }
<1,1,1> { 1^0 }
And repeat the same exercise for position 0^1:
Characteristic vector X(c, W, i), i < |W| Set of target positions
<0,0,0> { }
<0,0,1> { }
<0,1,0> { }
<0,1,1> { }
<1,0,0> { 1^1 }
<1,0,1> { 1^1 }
<1,1,0> { 1^1 }
<1,1,1> { 1^1 }
As you may notice so far we got 5 different sets of target positions:
{ }, { 1^0 }, { 1^1 }, { 0^1, 1^1 }, { 0^1, 1^1, 2^1 }
By keep enumerating all reachable target states from each position for every relevant character distribution we can see that every position falls down into the one of the following sets denoting
states of the corresponding DFA:
Now we can define a generic transition function Δ for max distance 1 describing all possible DFA transitions over character distributions (characteristic vectors). In tabular form it looks like:
X(c, W, I), A[I] B[I] C[I] D[I] E[I]
0 <= I <= |W| -3
{ 0, 0, 0 } C[I] ∅ ∅ ∅ ∅
{ 0, 0, 1 } C[I] ∅ ∅ B[I+3] B[I+3]
{ 0, 1, 0 } E[I] ∅ B[I+2] ∅ B[I+2]
{ 0, 1, 1 } E[I] ∅ B[I+2] B[I+3] C[I+2]
{ 1, 0, 0 } A[I+1] B[I+1] B[I+1] B[I+1] B[I+1]
{ 1, 0, 1 } A[I+1] B[I+1] B[I+1] D[I+1] D[I+1]
{ 1, 1, 0 } A[I+1] B[I+1] C[I+1] B[I+1] C[I+1]
{ 1, 1, 1 } A[I+1] B[I+1] C[I+1] D[I+1] E[I+1]
Schulz and Mihov generalized transition function Δ for arbitrary max distance n so that created it once, we can instantiate any DFA A(W) in linear time O(|W|). Instead of O(n^2N) we now have just O
(N) linear time complexity, and that’s amazing!
Schulz and Mihov also revealed the very interesting fact that with just a minor addition to the transition function Δ we can easily support Damerau-Levenshtein distance, which treats transpositions
atomically, e.g. given two words foobar and foobra we have 2 for Levenshtein distance, but 1 for Damerau-Levenshtein. I’d encourage you to check the original paper if you’re interested in details.
Let’s put it all together
In order to generate the DFA for an input sequence W of length N and max edit distance n in linear time O(N) we have to:
• Define a transition function Δ for a distance n
• For each character in the input sequence use Δ to generate a set of valid transitions
As you may already notice, the number of transitions in a transition function grows quadratically O(n^2), e.g. we have have
• 5*2^3=40 transitions for n=1
• 30*2^5=192 transitions for n=2
• 196*2^7=25088 transitions for n=3
• 1353*2^9=692736 transitions for n=4
That quadratic growth makes this approach infeasible for higher distances, thus in ArangoSearch we limit maximum edit distance to 4 and 3 for Levenshtein and Damerau-Levenshtein cases
Schulz and Mihov proved that the edit distance between two sequences W and V is the same as for their corresponding reversed representations W’ and V’, i.e. Lev(W, V) = Lev(W’, V’).The idea is to
manage 2 term dictionaries for original and reversed terms (so called FB-Trie). Given that, we can use 2 smaller DFAs for a max distance n in order to look up terms at a max distance n+1 which makes
perfect sense given the exponential growth of DFA states with respect to n.
Why another kind of fuzziness?
In real world applications using plain edit distance usually isn’t enough, we need something which can show us how this particular sequence is malformed with respect to the input. An obvious solution
is normalization by sequence length, e.g. we can define Levenshtein similarity function as LevSim(W, V) = Lev(W, V) / min(|W|, |V|). But given we are restricted by the max supported edit distance
that can be efficiently evaluated, long sequences will always produce LevSim(W,V) values which are close to 0.
Another example is fuzzy phrase search. Of course we can allow to have at most n edits for every word in a phrase, making it possible to have typos in each.
For example, a max distance 1 query with “quck brwn fx” input can match “quick brown fox”, but not “quick-witted brown fox”. Let’s consider another approach of fuzzy search which can help us to serve
such queries.
Approximate matching based on n-gram similarity
Another way of calculating similarity between two sequences is calculating the longest common sequence (LCS) of characters. The longer the LCS is, the more similar are the specified sequences. But
this approach has one big disadvantage, namely the absence of context. For example, the words connection and fonetica have a long LCS (o-n-e-t-i) but very different meanings.
To overcome this disadvantage, Grzegorz Kondrak described a set of modifications to LCS-based similarity in a paper . The major idea is to use n-gram sequences instead of single characters to carry
some more context around. For those who are not familiar with the notion of n-gram, it’s just a contiguous sequence of n items from a given sequence. We can split each sequence into a series of
subsequences of length n and use them to determine the LCS.
If we use the same words (connection and fonetica), but calculate similarity based on 3-grams we will get a better similarity measure, e.g. con-onn-nne-nec-ect-cti-tio-ion vs. fon-one-net-eti-tic-ica
gives a shorter LCS (zero matches).
Formally it can be defined as follows:
Let X = { x[1], x[2] … x[k] } and Y = { y[1], y[2] … y[l] } be sequences of length k and l, respectively, composed of symbols of a finite alphabet.
Let Γ[i,j] = (x[1] … x[i] , y[1] … y[j]) be a pair of prefixes of X and Y , and Γ^*[i,j ]= (x[i+1] … x[k] , y[j+1] … y[l] ) a pair of suffixes of X and Y.
Let Γ^n[i,j] = (x[i+1] … x[i+n] , y[j+1] … y[j+n]) be a pair of n-grams in X and Y. If both sequences contain exactly one n-gram, our initial definition is strictly binary: 1 if the n-grams are
identical, and 0 otherwise.
Then n-gram similarity s[n] can be defined as:
Consequently for sequences X and Y n-gram similarity is recursively defined as:
It’s also worth normalizing the LCS by word length to get a value in range [0;1] and avoid the length bias:
In contrast to Levenshtein-based fuzzy search, this approach can be efficiently implemented in a search engine without building any special structures at query time. We only need to split an input
sequence into a set of n-grams, store them as terms in a term dictionary and track positions of each term within a document.
At query time we split a specified input into a set of n-gram tokens and lookup postings for each of them. Then for each matched document we utilize positional information to evaluate the LCS of
n-grams and finally calculate s[N](X, Y).
Further improvements
To improve search quality, Kondrak suggested an interesting addition to the described approach. He suggested to count identical unigrams in corresponding positions within n-grams:
Depending on the use case he got 10-20% more accurate results with the refined approach.
Unfortunately the implementation is not as trivial as the original one, but having a Levenshtein automaton definitely helps: we can split a query into a set of n-gram tokens as in the previous
example. For each token W we instantiate a Levenshtein automaton A(W) and use them to get the variations of n-grams from the term dictionary. Given the number such n-gram tokens might be relatively
high, it’s definitely worth to union them into the one big automaton which can be used to query the term dictionary efficiently. Then for each matched n-gram we calculate the Levenshtein distance and
measure unigram similarity based on edit distance and get more accurate results.
Though this article was intended to be mostly theoretical, I can imagine many people are eager to try out the described approaches with ArangoSearch. For curious readers we’ve prepared a Jupyter
Notebook with an interactive tutorial. The tutorial runs on ArangoDB’s cloud service ArangoGraph works data from the International Movie Database (IMDB)
Tutorial: Fuzzy Search Approaches in ArangoSearch
You can also dive a bit deeper into the Architecture of ArangoSearch, the query capabilities or browse the documentation.
In this article I did my best to provide a comprehensive and simple explanation of the background behind fuzzy search techniques we implicitly use in our day to day life.
As usual there is no silver bullet and one has to carefully select the most suitable approach for the actual use case or even combine multiple techniques.
I hope you found this interesting and useful.
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C# Operators - Techappss
C# Operators
• Posted By: Chockalingam
Operators are the foundation of any programming language. Thus the functionality of C# language is incomplete without the use of operators. Operators allow us to perform different kinds of operations
on operands. In C#, operators Can be categorized based upon their different functionality :
• Arithmetic Operators
• Relational Operators
• Logical Operators
• Bitwise Operators
• Assignment Operators
• Conditional Operator
In C#, Operators can also categorized based upon Number of Operands :
• Unary Operator: Operator that takes one operand to perform the operation.
• Binary Operator: Operator that takes two operands to perform the operation.
• Ternary Operator: Operator that takes three operands to perform the operation.
Arithmetic Operators
These are used to perform arithmetic/mathematical operations on operands. The Binary Operators falling in this category are:
• Addition: The ‘+’ operator adds two operands. For example, x+y.
• Subtraction: The ‘-‘ operator subtracts two operands. For example, x-y.
• Multiplication: The ‘*’ operator multiplies two operands. For example, x*y.
• Division: The ‘/’ operator divides the first operand by the second. For example, x/y.
• Modulus: The ‘%’ operator returns the remainder when first operand is divided by the second. For example, x%y.
// C# program to demonstrate the working
// of Binary Arithmetic Operators
using System;
namespace Arithmetic
class GFG
// Main Function
static void Main(string[] args)
int result;
int x = 10, y = 5;
// Addition
result = (x + y);
Console.WriteLine("Addition Operator: " + result);
// Subtraction
result = (x - y);
Console.WriteLine("Subtraction Operator: " + result);
// Multiplication
result = (x * y);
Console.WriteLine("Multiplication Operator: "+ result);
// Division
result = (x / y);
Console.WriteLine("Division Operator: " + result);
// Modulo
result = (x % y);
Console.WriteLine("Modulo Operator: " + result);
Addition Operator: 15
Subtraction Operator: 5
Multiplication Operator: 50
Division Operator: 2
Modulo Operator: 0
The ones falling into the category of Unary Operators are:
• Increment: The ‘++’ operator is used to increment the value of an integer. When placed before the variable name (also called pre-increment operator), its value is incremented instantly. For
example, ++x.
And when it is placed after the variable name (also called post-increment operator), its value is preserved temporarily until the execution of this statement and it gets updated before the
execution of the next statement. For example, x++.
• Decrement: The ‘- -‘ operator is used to decrement the value of an integer. When placed before the variable name (also called pre-decrement operator), its value is decremented instantly. For
example, – -x.
And when it is placed after the variable name (also called post-decrement operator), its value is preserved temporarily until the execution of this statement and it gets updated before the
execution of the next statement. For example, x- –.
// C# program to demonstrate the working
// of Unary Arithmetic Operators
using System;
namespace Arithmetic {
class GFG {
// Main Function
static void Main(string[] args)
int a = 10, res;
// post-increment example:
// res is assigned 10 only,
// a is not updated yet
res = a++;
//a becomes 11 now
Console.WriteLine("a is {0} and res is {1}", a, res);
// post-decrement example:
// res is assigned 11 only, a is not updated yet
res = a--;
//a becomes 10 now
Console.WriteLine("a is {0} and res is {1}", a, res);
// pre-increment example:
// res is assigned 11 now since a
// is updated here itself
res = ++a;
// a and res have same values = 11
Console.WriteLine("a is {0} and res is {1}", a, res);
// pre-decrement example:
// res is assigned 10 only since
// a is updated here itself
res = --a;
// a and res have same values = 10
Console.WriteLine("a is {0} and res is {1}",a, res);
a is 11 and res is 10
a is 10 and res is 11
a is 11 and res is 11
a is 10 and res is 10
Relational Operators
Relational operators are used for comparison of two values. Let’s see them one by one:
• ‘=='(Equal To) operator checks whether the two given operands are equal or not. If so, it returns true. Otherwise it returns false. For example, 5==5 will return true.
• ‘!='(Not Equal To) operator checks whether the two given operands are equal or not. If not, it returns true. Otherwise it returns false. It is the exact boolean complement of the ‘==’ operator.
For example, 5!=5 will return false.
• ‘>'(Greater Than) operator checks whether the first operand is greater than the second operand. If so, it returns true. Otherwise it returns false. For example, 6>5 will return true.
• ‘<‘(Less Than) operator checks whether the first operand is lesser than the second operand. If so, it returns true. Otherwise it returns false. For example, 6<5 will return false.
• ‘>='(Greater Than Equal To) operator checks whether the first operand is greater than or equal to the second operand. If so, it returns true. Otherwise it returns false. For example, 5>=5 will
return true.
• ‘<='(Less Than Equal To) operator checks whether the first operand is lesser than or equal to the second operand. If so, it returns true. Otherwise it returns false. For example, 5<=5 will also
return true.
// C# program to demonstrate the working
// of Relational Operators
using System;
namespace Relational {
class GFG {
// Main Function
static void Main(string[] args)
bool result;
int x = 5, y = 10;
// Equal to Operator
result = (x == y);
Console.WriteLine("Equal to Operator: " + result);
// Greater than Operator
result = (x > y);
Console.WriteLine("Greater than Operator: " + result);
// Less than Operator
result = (x < y);
Console.WriteLine("Less than Operator: " + result);
// Greater than Equal to Operator
result = (x >= y);
Console.WriteLine("Greater than or Equal to: "+ result);
// Less than Equal to Operator
result = (x <= y);
Console.WriteLine("Lesser than or Equal to: "+ result);
// Not Equal To Operator
result = (x != y);
Console.WriteLine("Not Equal to Operator: " + result);
Equal to Operator: False
Greater than Operator: False
Less than Operator: True
Greater than or Equal to: False
Lesser than or Equal to: True
Not Equal to Operator: True
Logical Operators
They are used to combine two or more conditions/constraints or to complement the evaluation of the original condition in consideration. They are described below:
• Logical AND: The ‘&&’ operator returns true when both the conditions in consideration are satisfied. Otherwise it returns false. For example, a && b returns true when both a and b are true (i.e.
• Logical OR: The ‘||’ operator returns true when one (or both) of the conditions in consideration is satisfied. Otherwise it returns false. For example, a || b returns true if one of a or b is
true (i.e. non-zero). Of course, it returns true when both a and b are true.
• Logical NOT: The ‘!’ operator returns true the condition in consideration is not satisfied. Otherwise it returns false. For example, !a returns true if a is false, i.e. when a=0.
// C# program to demonstrate the working
// of Logical Operators
using System;
namespace Logical {
class GFG {
// Main Function
static void Main(string[] args)
bool a = true,b = false, result;
// AND operator
result = a && b;
Console.WriteLine("AND Operator: " + result);
// OR operator
result = a || b;
Console.WriteLine("OR Operator: " + result);
// NOT operator
result = !a;
Console.WriteLine("NOT Operator: " + result);
AND Operator: False
OR Operator: True
NOT Operator: False
Bitwise Operators
In C#, there are 6 bitwise operators which work at bit level or used to perform bit by bit operations. Following are the bitwise operators :
• & (bitwise AND) Takes two numbers as operands and does AND on every bit of two numbers. The result of AND is 1 only if both bits are 1.
• | (bitwise OR) Takes two numbers as operands and does OR on every bit of two numbers. The result of OR is 1 any of the two bits is 1.
• ^ (bitwise XOR) Takes two numbers as operands and does XOR on every bit of two numbers. The result of XOR is 1 if the two bits are different.
• << (left shift) Takes two numbers, left shifts the bits of the first operand, the second operand decides the number of places to shift.
• >> (right shift) Takes two numbers, right shifts the bits of the first operand, the second operand decides the number of places to shift.
// C# program to demonstrate the working
// of Bitwise Operators
using System;
namespace Bitwise {
class GFG {
// Main Function
static void Main(string[] args)
int x = 5, y = 10, result;
// Bitwise AND Operator
result = x & y;
Console.WriteLine("Bitwise AND: " + result);
// Bitwise OR Operator
result = x | y;
Console.WriteLine("Bitwise OR: " + result);
// Bitwise XOR Operator
result = x ^ y;
Console.WriteLine("Bitwise XOR: " + result);
// Bitwise AND Operator
result = ~x;
Console.WriteLine("Bitwise Complement: " + result);
// Bitwise LEFT SHIFT Operator
result = x << 2;
Console.WriteLine("Bitwise Left Shift: " + result);
// Bitwise RIGHT SHIFT Operator
result = x >> 2;
Console.WriteLine("Bitwise Right Shift: " + result);
Bitwise AND: 0
Bitwise OR: 15
Bitwise XOR: 15
Bitwise Complement: -6
Bitwise Left Shift: 20
Bitwise Right Shift: 1
Assignment Operator
Assignment operators are used to assigning a value to a variable. The left side operand of the assignment operator is a variable and right side operand of the assignment operator is a value. The
value on the right side must be of the same data-type of the variable on the left side otherwise the compiler will raise an error.
Different types of assignment operators are shown below:
• “=”(Simple Assignment): This is the simplest assignment operator. This operator is used to assign the value on the right to the variable on the left.
a = 10;
b = 20;
ch = 'y';
• “+=”(Add Assignment): This operator is combination of ‘+’ and ‘=’ operators. This operator first adds the current value of the variable on left to the value on the right and then assigns the
result to the variable on the left.
(a += b) can be written as (a = a + b)
If initially value stored in a is 5. Then (a += 6) = 11.
• “-=”(Subtract Assignment): This operator is combination of ‘-‘ and ‘=’ operators. This operator first subtracts the current value of the variable on left from the value on the right and then
assigns the result to the variable on the left.
(a -= b) can be written as (a = a - b)
If initially value stored in a is 8. Then (a -= 6) = 2.
• “*=”(Multiply Assignment): This operator is combination of ‘*’ and ‘=’ operators. This operator first multiplies the current value of the variable on left to the value on the right and then
assigns the result to the variable on the left.
(a *= b) can be written as (a = a * b)
If initially value stored in a is 5. Then (a *= 6) = 30.
• “/=”(Division Assignment): This operator is combination of ‘/’ and ‘=’ operators. This operator first divides the current value of the variable on left by the value on the right and then assigns
the result to the variable on the left.
(a /= b) can be written as (a = a / b)
If initially value stored in a is 6. Then (a /= 2) = 3.
• “%=”(Modulus Assignment): This operator is combination of ‘%’ and ‘=’ operators. This operator first modulo the current value of the variable on left by the value on the right and then assigns
the result to the variable on the left.
(a %= b) can be written as (a = a % b)
If initially value stored in a is 6. Then (a %= 2) = 0.
• “<<=”(Left Shift Assignment) : This operator is combination of ‘<<‘ and ‘=’ operators. This operator first Left shift the current value of the variable on left by the value on the right and then
assigns the result to the variable on the left.
(a <<= 2) can be written as (a = a << 2)
If initially value stored in a is 6. Then (a <<= 2) = 24.
• “>>=”(Right Shift Assignment) : This operator is combination of ‘>>’ and ‘=’ operators. This operator first Right shift the current value of the variable on left by the value on the right and
then assigns the result to the variable on the left.
(a >>= 2) can be written as (a = a >> 2)
If initially value stored in a is 6. Then (a >>= 2) = 1.
• “&=”(Bitwise AND Assignment): This operator is combination of ‘&’ and ‘=’ operators. This operator first “Bitwise AND” the current value of the variable on the left by the value on the right and
then assigns the result to the variable on the left.
(a &= 2) can be written as (a = a & 2)
If initially value stored in a is 6. Then (a &= 2) = 2.
• “^=”(Bitwise Exclusive OR): This operator is combination of ‘^’ and ‘=’ operators. This operator first “Bitwise Exclusive OR” the current value of the variable on left by the value on the right
and then assigns the result to the variable on the left.
(a ^= 2) can be written as (a = a ^ 2)
If initially value stored in a is 6. Then (a ^= 2) = 4.
• “|=”(Bitwise Inclusive OR) : This operator is combination of ‘|’ and ‘=’ operators. This operator first “Bitwise Inclusive OR” the current value of the variable on left by the value on the right
and then assigns the result to the variable on the left.
Example :
(a |= 2) can be written as (a = a | 2)
If initially, value stored in a is 6. Then (a |= 2) = 6.
// C# program to demonstrate the working
// of Assignment Operators
using System;
namespace Assignment {
class GFG {
// Main Function
static void Main(string[] args)
// initialize variable x
// using Simple Assignment
// Operator "="
int x = 15;
// it means x = x + 10
x += 10;
Console.WriteLine("Add Assignment Operator: " + x);
// initialize variable x again
x = 20;
// it means x = x - 5
x -= 5;
Console.WriteLine("Subtract Assignment Operator: " + x);
// initialize variable x again
x = 15;
// it means x = x * 5
x *= 5;
Console.WriteLine("Multiply Assignment Operator: " + x);
// initialize variable x again
x = 25;
// it means x = x / 5
x /= 5;
Console.WriteLine("Division Assignment Operator: " + x);
// initialize variable x again
x = 25;
// it means x = x % 5
x %= 5;
Console.WriteLine("Modulo Assignment Operator: " + x);
// initialize variable x again
x = 8;
// it means x = x << 2
x <<= 2;
Console.WriteLine("Left Shift Assignment Operator: " + x);
// initialize variable x again
x = 8;
// it means x = x >> 2
x >>= 2;
Console.WriteLine("Right Shift Assignment Operator: " + x);
// initialize variable x again
x = 12;
// it means x = x >> 4
x &= 4;
Console.WriteLine("Bitwise AND Assignment Operator: " + x);
// initialize variable x again
x = 12;
// it means x = x >> 4
x ^= 4;
Console.WriteLine("Bitwise Exclusive OR Assignment Operator: " + x);
// initialize variable x again
x = 12;
// it means x = x >> 4
x |= 4;
Console.WriteLine("Bitwise Inclusive OR Assignment Operator: " + x);
Output :
Add Assignment Operator: 25
Subtract Assignment Operator: 15
Multiply Assignment Operator: 75
Division Assignment Operator: 5
Modulo Assignment Operator: 0
Left Shift Assignment Operator: 32
Right Shift Assignment Operator: 2
Bitwise AND Assignment Operator: 4
Bitwise Exclusive OR Assignment Operator: 8
Bitwise Inclusive OR Assignment Operator: 12
Conditional Operator
A conditional operator in C#, is an operator that takes three operands (conditions to be checked), the value when the condition is true and value when the condition is false.
condition ? first_expression : second_expression;
condition: It must be evaluated to true or false.
If the condition is true
first_expression is evaluated and becomes the result.
If the condition is false,
second_expression is evaluated and becomes the result.
// C# program to demonstrate the working
// of Conditional Operator
using System;
namespace Conditional {
class GFG {
// Main Function
static void Main(string[] args)
int x = 5, y = 10, result;
// To find which value is greater
// Using Conditional Operator
result = x > y ? x : y;
// To display the result
Console.WriteLine("Result: " + result);
// To find which value is greater
// Using Conditional Operator
result = x < y ? x : y;
// To display the result
Console.WriteLine("Result: " + result);
Output :
Result: 10
Result: 5 | {"url":"https://techappss.com/c-operators/","timestamp":"2024-11-10T20:58:27Z","content_type":"text/html","content_length":"133866","record_id":"<urn:uuid:92ee6457-b3ce-434a-af06-5d755ffd4bb9>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00355.warc.gz"} |
The Stacks project
Lemma 24.28.1. In the situation above, the functor (24.28.0.1) composed with the localization functor $K(\textit{Mod}(\mathcal{A}', \text{d})) \to D(\mathcal{A}', \text{d})$ has a left derived
extension $D(\mathcal{B}, \text{d}) \to D(\mathcal{A}', \text{d})$ whose value on a good right differential graded $\mathcal{B}$-module $\mathcal{P}$ is $f^*\mathcal{P} \otimes _\mathcal {A} \mathcal
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Why is PHP known as scripting language?
• PHP is designed to be interpreted and executed at runtime, rather than being compiled into machine code beforehand. This means that PHP code is typically executed in response to user requests,
and the output of the code is generated dynamically based on those requests.
• Scripting languages like PHP are often used for tasks that involve manipulating data, generating dynamic content, or interacting with databases and other external systems.
• PHP scripts can be run on a web server to generate HTML output, process form submissions, and perform other server-side tasks.
• Unlike compiled languages like C or Java, which must be compiled into executable machine code before they can be run, PHP code is interpreted at runtime by the PHP interpreter.
• This interpretation process allows for greater flexibility and agility, as changes to the code can be made quickly and easily without the need to recompile the entire program.
• The use of scripting languages like PHP has several advantages over compiled languages, including easier learning and use, faster development times, and greater flexibility in responding to
changing requirements.
• The use of scripting languages like PHP has become particularly popular in web development, where the ability to generate dynamic content in response to user requests is essential.
In summary, PHP is known as a scripting language because it is designed to be interpreted at runtime, allowing for greater flexibility and agility in responding to changing requirements, and making
it particularly well-suited for web development tasks. | {"url":"https://easyexamnotes.com/why-is-php-known-as-scripting-language/","timestamp":"2024-11-10T19:03:50Z","content_type":"text/html","content_length":"135666","record_id":"<urn:uuid:d02db2d7-d42f-4735-844d-2455716f29ae>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00239.warc.gz"} |
Self Inductance of Coil Calculator
Self-inductance, or simply inductance, is a key concept in electromagnetism, a branch of physics. When current flowing through a coil changes, an electromotive force (EMF) is induced in the coil due
to its magnetic field. This phenomenon is known as self-induction, and the constant of proportionality between the induced EMF and the rate of change of current is the self-inductance of the coil.
of Coil
of Coil
Inductance =
Example Formula
The formula for self-inductance is derived from Faraday's law of electromagnetic induction:
E = -L × (dI/dt)
1. E: Induced electromotive force (EMF) in the coil
2. L: Self-inductance of the coil
3. dI/dt: Rate of change of current
Who wrote/refined the formula
The formula and the concept of self-inductance were first introduced by the physicist and chemist Michael Faraday in the 19th century. Faraday's experiments with coils, magnets, and circuits led to
the discovery of electromagnetic induction, which forms the basis for the principle of self-inductance.
Real Life Application
The principle of self-inductance is employed in many devices and systems in our daily life, including transformers, inductors in AC circuits, and electrical motors. Inductors, for instance, are used
in analog circuits and signal processing to filter out certain frequencies.
Key individuals in the discipline
Michael Faraday, who discovered the principle of electromagnetic induction, and Heinrich Lenz, who formulated Lenz's law, are key contributors to the understanding of self-inductance. Their work in
the 19th century laid the foundation for modern electromagnetism and electrical engineering.
Interesting Facts
1. The principle of self-inductance and electromagnetic induction is the working principle behind most of the electrical appliances we use today, from electric motors to power generators.
2. The understanding and application of self-inductance have greatly impacted humanity by enabling the generation and use of electric power, changing the face of technology and society.
3. The measurement unit of inductance, the henry (H), is named after Joseph Henry, an American scientist who discovered self-inductance independently of Faraday.
Understanding the principle of self-inductance is fundamental in the field of electromagnetism and electrical engineering. By enabling the control and manipulation of electrical circuits and power,
self-inductance plays a crucial role in our technologically driven world.
Physics Calculators
You may also find the following Physics calculators useful. | {"url":"https://physics.icalculator.com/self-inductance-of-coil-calculator.html","timestamp":"2024-11-14T21:58:19Z","content_type":"text/html","content_length":"20622","record_id":"<urn:uuid:36eca2e5-e863-43d3-b2ab-dcf38bc6e808>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00561.warc.gz"} |
A2L Item 161
Goal: Problem solving with kinematics
Source: CT151.2-4
Ann is running with a constant speed of 3 m/s on a straight track. Deb
is also running with constant speed but is initially 10 m behind Ann. If
Deb catches up to Ann after Deb has traveled 55 m, how fast is Deb
1. 3.2 m/s
2. 3.55 m/s
3. 3.75 m/s
4. 4.15 m/s
5. More than 4.2 m/s
6. none of the above
7. cannot be determined
(6) The correct speed is 3.67 m/s. Students indicating #2 or #3
are likely making an arithmetic error. Have students graph the position
vs. time graphs for each runner. | {"url":"https://new.clickercentral.net/item_0161/","timestamp":"2024-11-03T21:59:15Z","content_type":"text/html","content_length":"41364","record_id":"<urn:uuid:b96c76f7-e4ed-4ab4-96af-30cc9aa89636>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00261.warc.gz"} |
Adding and subtracting tens from a 2-digit number | Oak National Academy
Hello everybody and welcome to today's session.
My name is Ms. Hughes and in today's lesson, we're going to be looking at adding and subtracting tens to two digit numbers.
So let's get started.
This is what our lesson agenda for today is going to look like.
So we're going to start off by using known facts to help us with some addition, then you're going to have a talk task.
Next, we are going to use known facts for subtraction, and finally, you will have a main task and of course your quiz to complete at the end of the video.
So for today's lesson, you are going to need a pencil and rubber and some paper.
So pause the video now to go and grab these things if you have not got them already.
Brilliant, let's make a start then.
So to begin our lesson today, I want to warm our brains up with these equations.
We're going to try and solve these equations this morning using known facts.
So I want you to look at all of these different equations and think about the number bonds within 10, that you know, that's all going to help you to answer these questions.
Off you go.
Let's go through these answers then.
So looking at the first set of equations.
If I've got 20 add 30, I know that I can use my number bond, two add three equals five to help me solve that equation.
30 add 60, I can use the equation three add six equals nine.
50 add 40, I can use the number bond five add four equals nine.
40 take away 20, I can use the number bond four take away two equals two.
And these are the answers that you should have got.
20 add 30 equals 50.
30 add 60 equals 90.
50 add 40 equals 90 and 40 take away 20 equals 20.
Let's look at these next ones now where we're adding three, two digit numbers.
So what known facts could we've used to help us there? Well, I know that two add four add three equals nine.
That was going to help me with this top one.
So my answer that was 90.
Three add one add six equals 10 was helping me to find out that all of these two digit numbers made 100.
Three add to add four equals nine.
That was going to help me work out 30 add 20 add 40 equals 90.
And finally nine take away four, we know is five take away two equals three.
That was going to help us to find 30.
We're going to move on now team to think about using our known facts to help us with addition problems like this one on the board.
So let's have a read of it.
56 people are on the train.
20 more people arrived.
How many people are on the train now? To look at this problem or to solve this problem.
I'm going to use a part whole model.
So let's have a look at that.
So we know 56 people are on the train.
20 more people arrived and we're looking to find out, how many people are on the train now? When we are looking at a word problem like this, we need to think about a few questions.
Firstly, I need to think, What do I already know from this question? In other words, what values do I already know? Do I know the values of either of my parts? Do I know the value of my whole? Then I
need to figure out what's missing.
So what do I need to figure out? What part from my part whole model do I need to figure out? So from this question, I know that I'm looking to see how many people are on a train all together.
I know that at the beginning there were 56 people and then 20 more people were added to that.
So I know I've got the value of my two parts.
My first part that I have is 56.
So I've put 56 in my first part.
And the second part that I have is 20.
So I've got 20 in the second part of my part whole model.
I'm trying to find the whole.
So to do that, I need to add my two parts together.
We can put both of my parts into my whole now, 'cause I'm adding them together and that will give us what our whole is.
So let's count, 10, 20, 30, 40, 50, 60, 70, 71 72, 73, 74, 75, 76.
So my whole is 76.
We know that 76 people in total are on the train.
Let's look at another way to solve this.
So I could have used some of my number bonds within 100 to help me, sorry, within 10, to help me solve this problem.
So I could have started with partitioning the number 56.
I partitioned the number 56 this way, because I want my tens, my five tens, to be near to my two tens that I'm adding.
So this is why I partitioned at this way round.
So I've partitioned my 56 into five tens, which is 50 and six ones, which has six.
Now we can look at our tens because I've got five tens and I'm adding two tens, which is 20.
I know that five tens add two tens is seven tens, which has a value of 70.
Right now I'm left with seven tens, which is 70 and six ones, which is 76.
So some number facts help me to solve this.
I knew that five add two is equal to seven.
Therefore I was able to find out that 50 add 20 is equal to 70.
Let's see what this would have looked like as a part whole model.
So I have my 56 with its five tens and six ones.
And I have my 20 with it's two ones.
So what I was doing was adding my tens together.
So my two tens, which equal 20 and my five tens, which equal 50, I added together first.
So that gave me seven tens, which I knew is 70.
And then I added my six ones at the end.
There we go.
And in total I'm left with 76.
So 56 add 20 is equal to 76.
Let's have a look at another question.
This time it's not a word problem, but I've been given an equation which is 36 add 20.
Let's have a look at how we could solve that in a part whole model.
So I can see that the values that I do have are my parts and what I'm looking for is my whole.
I know that one part is worth 36.
So I've got 36 in the first part there.
And the second part is 20.
So I've put that in there.
To work out my whole, I need to add these two parts together.
So let's see what that looks like added together.
There we go.
So I've added my two parts together and they're now in the whole.
So let's count them in tens and ones.
10, 20, 30, 40, 50, 51, 52, 53, 54, 55, 56.
So altogether there are 56 people on the train.
Let's look at another way to solve this.
I could have used some number facts within 100 to help me figure this out.
So we can partition 36 into three tens which is 30 and six ones, which is six.
Now that I've partitioned my 36, I can see that I've got three tens and I'm adding two tens because I'm adding 20.
So we can add those tens together.
If I know that two tens add three tens will give me five tens.
Then I know that 20 add 30 is 50.
So now that I've added my tens and that equals 50, I just have my ones left over that I need to add on.
So 50 add six is 56.
Let's go do that one more time.
If I know that three add two is equal to five, then 30 ad 20 is equal to 50.
I was able to use that number bond within 10 to help me work out my tens added together.
Let's have a look at this on a part whole model.
So remember I have my two parts, 36 and 20, and I'm adding them together.
I'm going to start by adding my three tens and two tens.
My three tens from 36 and my two tens from 20.
Like that.
So I'm left with five tens, which is 50.
And then I need to add my six ones.
So in my whole I have 56.
36 add 20 is equal to 56.
Alright, team, it's now time for your talk task.
So you are going to do exactly as we've just done, looking at number facts to help us solve addition problems. So you're going to look at these equations that are on the board and you are going to
think of or derive the number facts within 10 are going to help you work out your problem.
I would like you to use these pink sentences at the bottom of the screen as well to explain what's happening.
So we're going to look at this equation first, and then you are going to have a go at the rest yourself.
I want to use these number sentences as I'm working out my equation.
So I'm going to start with my equation, 31 plus 40, and I'm going to put my equal sign, but I don't know yet what that whole is.
I only know the value of my two parts.
I can see that I've got three here which represents 30 and four here represents 40.
Three tens and four tens.
So I can use the equation.
Three ones add four ones is equal to seven ones.
That helps me to know that three tens add four tens is equal to seven tens.
So I know therefore that 30, which is three tens, add 40, which is four tens is equal to 70 Like that.
Remember we've still got our one leftover so I need to add that on.
So 70 add one gives me 71.
So my total is 71.
So you can see how I use my number bond three add four equals seven.
To help me work out that answer.
Pause it here now to have a go at your talk task and then play the video when you're ready to continue.
We're now going to look at how we can use our number bonds or our known facts to help us subtraction for this half of the lesson.
So let's have a look at a question on the board.
46 people are on the train.
20 people got off the train.
How many people are now on the train? Let's look at this in a part whole model.
Remember when we're doing a word problem, we need to think carefully to ourselves.
What do we already know? And what are we trying to find out? So do I know the values of any of my parts here? Or do I know the value of my whole? and what part am I my trying to find out? Well, I can
see from this question that 46 people were already on the train.
So 46 people is the value of our whole.
So let's put 46 in the whole.
20 is one of my parts because 20 people got off the train.
So I'm going to take that away from my whole and put it into one of my parts, which would look like this.
There we go.
Now I'm left with 10 20, 21, 22, 23, 24, 25, 26.
So 26 is my other part, which would go in here.
So we know that 46 take away 20 leaves us with 26 left.
So there are 26 people on the train.
Let's have a look at how else we could have solved this using our number facts.
So we could have partitioned our number 46 into tens and ones.
So 46 has made up of four tens, which is 40 and six ones, which is six.
So we now have four tens and I can take away two tens, which is 20.
So 40 take away 20 is 20.
Now I'm left with two tens and I have six ones left over.
So we need to add those together so I'm left with 26.
So let's look at those equations that helped us to figure that out.
If I know that four take away two is equal to two, then 40 take away 20 is equal to 20.
Let's have a look at this in a part whole model.
So remember we started off with 46 as our whole.
So there's 46 in the whole.
There were 46 people on that train.
Then we took away two tens.
Now I'm left with two tens and then six ones on the end.
So 46 take away 20 is equal to 26.
Let's have a look at another example.
This time, it's just a number equation.
So we have 36 take away 20.
Remember our whole is 36.
So we've started with 36 and I can see that one of my parts is 20.
So I'm going to put 20 in to that part.
So I'm left with 10, 11, 12, 13, 14, 15, 16.
So this part that was missing is 16, which means that 36, our whole take away 20 leaves us with 16.
Let's look at how we could have solved it using our number facts.
So 36, we could have partitioned into three tens, which is 30 and six ones, which is six.
Then we could do our three tens, which is 30 take away 20, which is 10.
So now I'm left with one 10 and six ones, which is 16.
So if I know that three take away two is equal to one, then I know that 30 take away 20 is equal to 10.
Let's look at that on our part whole model again.
So we've got 36 has our whole, we're taking away one part, which is 20.
So I've taken away my two tens and I'm left with one 10 and six ones.
So 36 take away 20 is equal to 16.
Great job guys.
It is now time for your independent task.
Where you're going to be looking at some more addition and subtraction problems. I want you to complete the equations that you've been given by using your known facts You're known number bonds within
10 to derive new facts.
While you're doing your sheet, can you spot any patterns in the numbers? Please pause the video now to complete your task and resuming the video once you are finished.
Great, let's go through these answers.
46 take away 40 is six.
46 take away 30 is 16.
46 take away 20 is 26 46 take away 10 is 36 46 add 10 is 56 46 add 20 is 66.
46 at 30 is 76.
42 take away 40 is two.
42 take away 30 is 12.
42 take away 20 is 22.
42 take away 10 is 32.
42 add 10 is 52.
42 add 20 is 62.
42 add 30 is 72.
So in all of these equations, you'll notice that the number that I'm taking away or that I'm adding to the part that I'm subtracting or adding.
The tens, the number of tens I have is changing.
Therefore, the number of tens in my answer is changing.
So because we're subtracting and adding tens that is changing the number of tens that are in my answer.
Let's look at these next ones.
26 is equal to 66 take away 40.
36 is equal to 66 take away 30.
46 is equal to 66 take away 20.
56 is equal to 66 take away 10.
76 is equal to 66 add 10.
86 is equal to 66 add 20.
96 is equal to 66 add 30.
25 is equal to 65 take away 40.
35 is equal to 65 take away 30.
45 is equal to 65 take away 20.
55 is equal to 65 take away 10.
75 is equal to 65 add 10.
And 85 is equal to 65 add 20.
And 95 is equal to 65 add 30.
The multiples of 10 that I'm adding or subtracting have different numbers of tens.
And therefore my answer has different numbers of tens.
So that is affecting the number of tens that are in my answer.
Team, you've done a fantastic job today with all of your learning.
I was so impressed with the way that you used mental strategies to add and subtract tens from two digit numbers.
Well done and see you soon in another session.
Bye bye.
If you'd like to please ask your parents or carer to share your work on Instagram, Facebook, or Twitter, tagging @OakNational and #LearnwithOak.
All there's left for you to do now guys, is complete your quiz.
So when the video has ended, make sure you take part in that to recap everything you've learned in today's session and to show off everything that you have remembered.
Good luck with it. | {"url":"https://www.thenational.academy/pupils/programmes/maths-primary-year-2-l/units/addition-and-subtraction-of-2-digit-numbers-f192/lessons/adding-and-subtracting-tens-from-a-2-digit-number-69h3jt/video","timestamp":"2024-11-06T04:13:40Z","content_type":"text/html","content_length":"113271","record_id":"<urn:uuid:2149b552-030e-49a2-a101-48667445db98>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00613.warc.gz"} |
HamiltonianGate | IBM Quantum Documentation
class qiskit.circuit.library.HamiltonianGate(data, time, label=None)
Bases: Gate
Class for representing evolution by a Hamiltonian operator as a gate.
This gate resolves to a UnitaryGate as $U(t) = \exp(-i t H)$, which can be decomposed into basis gates if it is 2 qubits or less, or simulated directly in Aer for more qubits.
• data (np.ndarray | Gate | BaseOperator) – A hermitian operator.
• time (float |ParameterExpression) – Time evolution parameter.
• label (str | None) – Unitary name for backend [Default: None].
ValueError – if input data is not an N-qubit unitary operator.
Get the base class of this instruction. This is guaranteed to be in the inheritance tree of self.
The “base class” of an instruction is the lowest class in its inheritance tree that the object should be considered entirely compatible with for _all_ circuit applications. This typically means that
the subclass is defined purely to offer some sort of programmer convenience over the base class, and the base class is the “true” class for a behavioral perspective. In particular, you should not
override base_class if you are defining a custom version of an instruction that will be implemented differently by hardware, such as an alternative measurement strategy, or a version of a
parametrized gate with a particular set of parameters for the purposes of distinguishing it in a Target from the full parametrized gate.
This is often exactly equivalent to type(obj), except in the case of singleton instances of standard-library instructions. These singleton instances are special subclasses of their base class, and
this property will return that base. For example:
>>> isinstance(XGate(), XGate)
>>> type(XGate()) is XGate
>>> XGate().base_class is XGate
In general, you should not rely on the precise class of an instruction; within a given circuit, it is expected that Instruction.name should be a more suitable discriminator in most situations.
The classical condition on the instruction.
Get the decompositions of the instruction from the SessionEquivalenceLibrary.
Return definition in terms of other basic gates.
Is this instance is a mutable unique instance or not.
If this attribute is False the gate instance is a shared singleton and is not mutable.
Return the number of clbits.
Return the number of qubits.
The parameters of this Instruction. Ideally these will be gate angles.
Get the time unit of duration.
Return the adjoint of the unitary.
Return the conjugate of the Hamiltonian.
Return the adjoint of the unitary.
Return the transpose of the Hamiltonian.
Hamiltonian parameter has to be an ndarray, operator or float. | {"url":"https://docs.quantum.ibm.com/api/qiskit/qiskit.circuit.library.HamiltonianGate","timestamp":"2024-11-05T01:10:20Z","content_type":"text/html","content_length":"180613","record_id":"<urn:uuid:bfa65093-d1a2-4ada-9a4c-575d15b0c0ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00247.warc.gz"} |
WEIGHT Statement
To use relative weights for each observation in the input data set, place the weights in a variable in the data set and specify the name in a WEIGHT statement. This is often done when the variance
associated with each observation is different and the values of the weight variable are proportional to the reciprocals of the variances. If the value of the WEIGHT variable is missing or is less
than zero, then a value of zero for the weight is used.
The WEIGHT and FREQ statements have a similar effect except that the WEIGHT statement does not alter the degrees of freedom. | {"url":"http://support.sas.com/documentation/cdl/en/statug/65328/HTML/default/statug_discrim_syntax11.htm","timestamp":"2024-11-12T09:39:22Z","content_type":"application/xhtml+xml","content_length":"12803","record_id":"<urn:uuid:e8da3507-570d-4a06-95b7-66148bac38ee>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00575.warc.gz"} |
Find the sum of the following series up to n terms: 0.6+0.66+0.666+……?
Hint: This type of problem is based on the concept of geometric series. First, we have to consider the sum of n terms of the given series to be \[{{S}_{n}}\]. Now, take 6 common from the series.
Multiply and divide the whole series by 9. Keep \[\dfrac{6}{9}\] outside the bracket and multiply 9 in the numerator with each term in the series. Substitute \[0.9=1-\dfrac{1}{10}\], \[0.99=1-\dfrac
{1}{100}\], \[0.999=1-\dfrac{1}{1000}\] and so on for n terms. Since there is 1 n times in the series, group all the fractions separately. We see a geometric series in the expression. Assume \[\dfrac
{1}{10}\] to be ‘a’. Then to find the common ratio ‘r’ of the given geometric series, we have to divide the first term and the second term of the series. We get \[r=\dfrac{1}{10}\]. And substitute
these values in the formula to find the sum of n terms of series of a geometric progression, that is, \[S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\]
Complete step by step solution:
According to the question, we are asked to find the sum of the series 0.6+0.66+0.666+….up to n terms.
We have been given the series is 0.6, 0.66, 0.666… -----(1)
Let us assume \[{{S}_{n}}\] to be the sum of n terms of the given series.
\[\Rightarrow {{S}_{n}}=0.6+0.66+0.666+.....\]n terms.
We find that 6 are common in the RHS.
Let us take 6 outside the bracket.
\[\Rightarrow {{S}_{n}}=6\left( 0.1+0.11+0.111+..... \right)\]
Now, we need to multiply 9 in the numerator and denominator of the RHS.
We get
\[{{S}_{n}}=6\times \dfrac{9}{9}\left( 0.1+0.11+0.111+..... \right)\]n terms.
Use distributive property, that is \[a\left( b+c+d+.... \right)=ab+ac+ad+...\], in the numerator of the RHS.
We get
\[{{S}_{n}}=\dfrac{6}{9}\left( 0.9+0.99+0.999+..... \right)\]n terms. ---------(2)
We know 0.9=1-0.1, but we can write 0.1 as \[\dfrac{1}{10}\].
Thus, \[0.9=1-\dfrac{1}{10}\].
Now, we can write 0.99 as 1-0.01 which is \[1-\dfrac{1}{100}\].
And we get \[0.99=1-\dfrac{1}{100}\].
Similarly, we get \[0.999=1-\dfrac{1}{1000}\] and we can find the same for n terms.
Let us substitute these values in equation (2).
\[\Rightarrow {{S}_{n}}=\dfrac{6}{9}\left( 1-\dfrac{1}{10}+1-\dfrac{1}{100}+1-\dfrac{1}{1000}+..... \right)\] up to n terms.
We find that 1 is added n times in the RHS. Group all of them together and we get n in the RHS.
Therefore, we get
\[{{S}_{n}}=\dfrac{6}{9}\left( n-\dfrac{1}{10}-\dfrac{1}{100}-\dfrac{1}{1000}-..... \right)\] up to n terms.
Now, let us take the negative sign outside the bracket.
\[\Rightarrow {{S}_{n}}=\dfrac{6}{9}\left( n-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+..... \right) \right)\] up to n terms.
Consider \[\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+....\]n terms.
We find that the series (3) is a geometric series.
Therefore, the first term is \[a=\dfrac{1}{10}\].
We need to find the common ratio.
Divide the second term of the series by the first term to find the common ratio r.
\[\Rightarrow r=\dfrac{\dfrac{1}{100}}{\dfrac{1}{10}}\]
Using the rule \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\] in the above expression, we get
\[r=\dfrac{1}{100}\times \dfrac{10}{1}\]
\[\Rightarrow r=\dfrac{1}{10\times 10}\times \dfrac{10}{1}\]
Here, we find that 10 are common in both the numerator and denominator. Cancelling 10, we get
We know that the formula to find the sum of n terms in a geometric series is \[S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\].
Substituting the values of ‘a’ and ‘r’, we get
\[S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}}\]
Take LCM in the denominator. We get
\[S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{\dfrac{10-1}{10}}\]
\[\Rightarrow S=\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{\dfrac{9}{10}}\]
We find that \[\dfrac{1}{10}\] is common in both the numerator and the denominator. Cancelling \[\dfrac{1}{10}\], we get
\[S=\dfrac{1}{9}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)\]
Let us use the property \[{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}\times \dfrac{d}{c}\] to simplify S further.
\[\Rightarrow S=\dfrac{1}{9}\left( 1-\dfrac{{{1}^{n}}}{{{10}^{n}}} \right)\]
Since 1 power any term is always 1, we get
\[S=\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right)\]
Now substitute the value of ‘S’ in equation (3).
\[{{S}_{n}}=\dfrac{6}{9}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]
Therefore, the sum of n terms of the series 0.6+0.66+0.666+….. is \[\dfrac{6}{9}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\].
Note: Whenever you get this type of problems, we should always try to make the necessary calculations in the given series to convert the series into a geometric or arithmetic series. We should avoid
calculation mistakes based on sign conventions. If we get r>0, we have to use the formula \[S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\]. Also we should not use the formula \[S=\dfrac{a}{r-1}\] to
find the sum of n terms. This formula is used to find the sum of infinite series. | {"url":"https://www.vedantu.com/question-answer/find-the-sum-of-the-following-series-up-to-n-class-11-maths-cbse-600ea4b90572f70d36c8d6c3","timestamp":"2024-11-10T08:31:29Z","content_type":"text/html","content_length":"176470","record_id":"<urn:uuid:01a9e215-a8ce-4a22-a2ce-292c1d867786>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00301.warc.gz"} |
[Solved] Based on the method of least squares, wha | SolutionInn
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Verified Expert Solution
Based on the method of least squares, what are the estimated weights of balls A and B? Round your answer to the nearest tenth
Based on the method of least squares, what are the estimated weights of balls A and B? Round your answer to the nearest tenth in the format of x.x). = 2.6 B= 3.6 Answer 1: Correct! 2.6 Answer 2:
Correct! 3.6
There are 3 Steps involved in it
Step: 1
Solutions Step 1 Given that Observation 1 is X1347 Observatio...
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How to Calculate Price Elasticity of Demand with Calculus - dummies
The most important point elasticity for managerial economics is the point price elasticity of demand. This value is used to calculate marginal revenue, one of the two critical components in profit
maximization. (The other critical component is marginal cost.) Profits are always maximized when marginal revenue equals marginal cost.
The formula to determine the point price elasticity of demand is
In this formula, ∂Q/∂P is the partial derivative of the quantity demanded taken with respect to the good’s price, P[0] is a specific price for the good, and Q[0] is the quantity demanded associated
with the price P[0].
The following equation represents soft drink demand for your company’s vending machines:
In the equation, Q represents the number of soft drinks sold weekly, P is the price per bottle from the vending machine in dollars, I is weekly income in dollars, P[C] is the price at a convenience
in dollars, and A is weekly advertising expenditures in dollars.
Assume initially that P is $1.50, I is $600, P[C] is $1.25, and A is $400. Substituting those values into the demand equation indicates that 2,000 bottles will be sold weekly.
To determine the point price elasticity of demand given P[0] is $1.50 and Q[0] is 2,000, you need to take the following steps:
1. Take the partial derivative of Q with respect to P, ∂Q/∂P.
For your demand equation, this equals –4,000.
2. Determine P[0] divided by Q[0].
Because P is $1.50, and Q is 2,000, P[0]/Q[0] equals 0.00075.
3. Multiply the partial derivative, –4,000, by P[0]/Q[0], 0.00075.
The point price elasticity of demand equals –3.
Therefore, at this point on the demand curve, a 1 percent change in price causes a 3 percent change in quantity demanded in the opposite direction (because of the negative sign).
In order to maximize profits, you need to know how much each additional unit you sell adds to your revenue, or in other words, you need to know marginal revenue. If you know the point price
elasticity of demand, η, the following formula can enable you to quickly determine marginal revenue, MR, for any given price.
Assume your company charges a $1.50 per bottle of soft drink, and the point price elasticity of demand is –3. To determine how much revenue you add by selling an additional bottle:
1. Determine (1 + 1/ç).
Substituting –3 for η gives (1 + 1/[–3]) or (1 – 1/3) or 2/3.
2. Multiply the price, $1.50, by 2/3.
The marginal revenue equals $1.00.
So the marginal revenue received when an additional bottle is sold is
If your cost of providing the extra bottle is less than $1.00, you will increase your profits by selling it.
Similarly, you can calculate point elasticities for the income elasticity of demand, cross-price elasticity of demand, and advertising elasticity of demand using the following formulas:
• The point income elasticity of demand:
In this formula, ∂Q/∂I is the partial derivative of the quantity taken with respect to income, I is the specific income level, and Q is the quantity purchased at the income level I.
• The point cross-price elasticity of demand:
In this formula, ∂Q[x]/∂P[y] is the partial derivative of good x’s quantity taken with respect to good y’s price, P[y] is a specific price for good y, and Q[x] is the quantity of good x purchased
given the price P[y].
• The point advertising elasticity of demand:
In this formula, ∂Q/∂A is the partial derivative of the quantity demanded taken with respect to advertising expenditures, A is the specific amount spent on advertising, and Q is the quantity
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Adding and subtraction radicals free calulator
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Planes in 3D space
A plane in 3D space can be thought of as a flat surface that stretches infinitely far, splitting space into two halves.
Planes have loads of uses in applications that deal with 3D geometry. I've mostly been working with them in the context of an architectural modeler, where geometry is defined in terms of planes and
their intersections.
Learning about planes felt abstract and non-intuitive to me. “Sure, that's a plane equation, but what do I do with it? What does a plane look like?” It took some time for me to build an intuition for
how to reason about and work with them.
In writing this, I want to provide you with an introduction that focuses on building a practical, intuitive understanding of planes. I hope to achieve this through the use of visual (and
interactive!) explanations which will accompany us as we work through progressively more complex problems.
With that out of the way, let's get to it!
Describing planes
There are many ways to describe planes, such as through
1. a point in 3D space and a normal,
2. three points in 3D space, forming a triangle, or
3. a normal and a distance from an origin.
Throughout this post, the term normal will refer to a normalized direction vector (unit vector) whose magnitude (length) is equal to 1, typically denoted by where .
Starting with the point-and-normal case, here's an example of a plane described by a point in 3D space and a normal :
The normal describes the plane's orientation, where the surface of the plane is perpendicular to , while the point describes a point on the plane.
We described this plane in terms of a single point , but keep in mind that this plane—let's call it —contains infinitely many points.
If were described by one of those other points contained by , we would be describing the exact same plane. This is a result of the infinite nature of planes.
This way of describing planes—in terms of a point and a normal—is the point-normal form of planes.
We can also describe a plane using three points in 3D space , , forming a triangle:
The triangle forms an implicit plane, but for us to be able to do anything useful with the plane we'll need to calculate its normal . Once we've calculated the plane's normal, we can use that normal
along with one of the triangle's three points to describe the plane in point-normal form.
As mentioned earlier, the normal describing a plane is a unit vector () perpendicular to the plane.
We can use and as two edge vectors that are parallel to the plane's surface.
By virtue of being parallel to the plane's surface, the vectors and are perpendicular to the plane's normal. This is where the cross product becomes useful to us.
The cross product takes in two vectors and and returns a vector that is perpendicular to both of them.
For example, given the vectors and , their cross product is the vector , which we'll label :
This explanation is simple on purpose. We'll get into more detail about the cross product later on.
Because the edge vectors of the triangle, and , are both parallel to the triangle's surface, their cross product will be perpendicular to the triangle's surface. Let's name the cross product of our
two edge vectors :
has been scaled down for illustrative purposes
points in the right direction, but it's not a normal. For to be a normal, its magnitude needs to equal 1. We can normalize by dividing it by its magnitude, the result of which we'll assign to :
This gives us a normal where :
Having found the triangle's normal we can use it and any of the points , , to describe the plane containing the three points in point-normal form.
It doesn't matter which of , , we use as the point in the point-normal form; we always get the same plane.
Constant-normal form
There's one more way to describe a plane that we'll look at, which is through a normal and a distance .
This is the constant-normal form of planes. It makes lots of calculations using planes much simpler.
In the constant-normal form, the distance denotes how close the plane gets to the origin. Thought of another way: multiplying the normal by yields the point on the plane that's closest to the origin.
This is a simplification. More formally, given a point on a plane whose normal is , we can describe all points on the plane in two forms: the point-normal form , and the constant-normal form where .
See further reading.
In getting a feel for the difference between the point-normal and constant-normal forms, take this example which describes the same plane in both forms:
The green arrow represents from the constant-normal form, while the blue point and arrow represent the point and normal from the point-normal form.
Translating from the point-normal to the constant-normal form is very easy: the distance is the dot product of and .
If you're not familiar with the dot product, don't worry. We'll cover it later on.
The notation for and might seem to indicate that they're of different types, but they're both vectors. I'm differentiating between points in space (e.g. and ) and direction vectors (e.g. and ) by
using the arrow notation only for direction vectors.
The normal stays the same across both forms.
Distance from plane
Given an arbitrary point and a plane in constant-normal form, we may want to ask how far away the point is from the plane. In other words, what is the minimum distance needs to travel to lie on the
We can frame this differently if we construct a plane containing that is parallel to , which we can do in point-normal form using as the point and 's normal as the normal:
With two parallel planes, we can frame the problem as finding the distance between the two planes. This becomes trivial using their constant-normal form since it allows us to take the difference
between their distance components and .
So let's find 's distance using the equation we learned about:
With two distances and from the planes and the solution simply becomes:
So, to simplify, given a plane having a normal and distance , we can calculate a point 's distance from like so:
The distance may be positive or negative depending on which side of the plane the point is on.
Projecting a point onto a plane
A case where calculating a point's distance from a plane becomes useful is, for example, if you want to project a point onto a plane.
Given a point which we want to project onto plane whose normal is and distance is , we can do that fairly easily. First, let's define as the point's distance from the plane:
Multiplying the plane's normal by gives us a vector which when added to projects it onto the plane. Let's call the projected point :
The projection occurs along the plane's normal, which is sometimes useful. However, it is much more useful to be able to project a point onto a plane along an arbitrary direction instead. Doing that
boils down finding the point of intersection of a line and a plane.
Line-plane intersection
We can describe lines in 3D space using a point and normal . The normal describes the line's orientation, while the point describes a point which the line passes through.
In this chapter, the line will be composed of the point and normal , while the plane—given in constant-normal form—has a normal and a distance .
Our goal will be to find a distance that needs to travel along such that it lies on the plane.
We can figure out the distance that we'd need to travel if and were parallel, which is what we did when projecting along the plane's normal.
Let's try projecting along using as a scalar like so:
We'll visualize as a red point:
As and become parallel, gets us closer and closer to the correct solution. However, as the angle between and increases, becomes increasingly too small.
Here, the dot product comes in handy. For two vectors and , the dot product is defined as
where is the angle between and .
Consider the dot product of and . Since both normals are unit vectors whose magnitudes are 1
we can remove their magnitudes from the equation,
making the dot product of and the cosine of the angle between them.
For two vectors, the cosine of their angles approaches 1 as the vectors become increasingly parallel, and approaches 0 as they become perpendicular.
Since becomes increasingly too small as and become more perpendicular, we can use as a denominator for . We'll assign this scaled-up version of to :
With as our scaled-up distance, we find the point of intersection via:
We can now get rid of , which was defined as , giving us the full equation for :
Putting this into code, we get:
Vector3 LinePlaneIntersection(Line line, Plane plane) {
float denom = Vector3.Dot(line.normal, plane.normal);
float dist = Vector3.Dot(plane.normal, line.point);
float D = (plane.distance - dist) / denom;
return line.point + line.normal * D;
However, our code is not complete yet. In the case where the line is parallel to the plane's surface, the line and plane do not intersect.
That happens when and are perpendicular, in which case their dot product is zero. So if , the line and plane do not intersect. This gives us an easy test we can add to our code to yield a result of
"no intersection".
However, for many applications we'll want to treat being almost parallel as actually being parallel. To do that, we can check whether the dot product is smaller than some very small
number—customarily called epsilon
float denom = Vector3.Dot(line.normal, plane.normal);
if (Mathf.Abs(denom) < EPSILON) {
return null; // Line is parallel to plane's surface
See if you can figure out why Mathf.Abs is used here. We'll cover it later, so you'll see if you're right.
We'll take a look at how to select the value of epsilon in a later chapter on two plane intersections.
With this, our line-plane intersection implementation becomes:
Vector3 LinePlaneIntersection(Line line, Plane plane) {
float denom = Vector3.Dot(line.normal, plane.normal);
if (Mathf.Abs(denom) < EPSILON) {
return null; // Line is parallel to plane's surface
float dist = Vector3.Dot(plane.normal, line.point);
float D = (plane.distance - dist) / denom;
return line.point + line.normal * D;
Rays and lines
We've been talking about line-plane intersections, but I've been lying a bit by visualizing ray-plane intersections instead for visual clarity.
A ray and a line are very similar; they're both represented through a normal and a point .
The difference is that a ray (colored red) extends in the direction of away from , while a line (colored green) extends in the other direction as well:
What this means for intersections is that a ray will not intersect planes when traveling backward along its normal:
Our implementation for ray-plane intersections will differ from our existing line-plane intersection implementation only in that it should yield a result of "no intersection" when the ray's normal is
pointing "away" from the plane's normal at an obtuse angle.
Since represents how far to travel along the normal to reach the point of intersection, we could yield "no intersection" when becomes negative:
if (D < 0) {
return null;
But then we'd have to calculate first. That's not necessary since becomes negative as a consequence of the dot product being a negative number when and are at an obtuse angle between 90° and 180°.
If this feels non-obvious, it helps to remember that the dot product encodes the cosine of the angle between its two component vectors, which is why the dot product becomes negative for obtuse
Knowing that, we can change our initial "parallel normals" test from this:
Vector3 LinePlaneIntersection(Line line, Plane plane) {
float denom = Vector3.Dot(line.normal, plane.normal);
if (Mathf.Abs(denom) < EPSILON) {
return null; // Line is parallel to plane's surface
// ...
To this:
Vector3 RayPlaneIntersection(Line line, Plane plane) {
float denom = Vector3.Dot(line.normal, plane.normal);
if (denom < EPSILON) {
// Ray is parallel to plane's surface or pointing away from it
return null;
// ...
The check covers both the "line parallel to plane" case and the case where the two normal vectors are at an obtuse angle.
Note: is the symbol for epsilon.
Plane-plane intersection
The intersection of two planes forms an infinite line.
As a quick refresher: lines in 3D space are represented using a point and normal where normal describes the line's orientation, while the point describes a point which the line passes through.
Let's take two planes and whose normals are and .
Finding the direction vector of and 's intersection is deceptively simple. Since the line intersection of two planes lies on the surface of both planes, the line must be perpendicular to both plane
normals, which means that the direction of the intersection is the cross product of the two plane normals. We'll assign it to .
The magnitude of the cross product is equal to the area of the parallelogram formed by the two component vectors. This means that we can't expect the cross product to be a unit vector, so we'll
normalize and assign the normalized direction vector to .
This gives us the intersection's normal . Let's zoom in and see this close up.
But this is only half of the puzzle! We'll also need to find a point in space to represent the line of intersection (i.e. a point which the line passes through). We'll take a look at how to do just
that, right after we discuss the no-intersection case.
Handling parallel planes
Two planes whose normals are parallel will never intersect, which is a case that we'll have to handle.
The cross product of two parallel normals is . So if , the planes do not intersect.
As previously mentioned, for many applications we'll want to treat planes that are almost parallel as being parallel. This means that our plane-plane intersection procedure should yield a result of
"no intersection" when the magnitude of is less than some very small number called epsilon.
Line PlanePlaneIntersection(Plane P1, Plane P2) {
Vector3 direction = Vector3.cross(P1.normal, P2.normal);
if (direction.magnitude < EPSILON) {
return null; // Roughly parallel planes
// ...
But what should the value of epsilon be?
Given two normals and where the angle between and is , we can find a reasonable epsilon by charting for different values of :
Both of the axes are logarithmic.
The relationship is linear: as the angle between the planes halves, so does the magnitude of the cross product of their normals. yields a magnitude of , and yields half of that.
So to determine the epsilon, we can ask: how low does the angle in degrees need to become for us to consider two planes parallel? Given an angle , we can find the epsilon via:
If that angle is 1/256°, then we get:
With this you can determine the appropriate epsilon based on how small the angle between the planes needs to be for you to consider them parallel. That will depend on your use case.
Finding a point of intersection
Having computed the normal and handled parallel planes, we can move on to finding a point along the line of intersection.
Since the line describing a plane-plane intersection is infinite, there are infinitely many points we could choose as .
We can narrow the problem down by taking the plane parallel to the two plane normals , and observing that it intersects the line at a single point.
Since the point lies on the plane parallel to the two plane normals, we can find it by exclusively traveling along those normals.
The simplest case is the one where and are perpendicular. In that case, the solution is just . Here's what that looks like visually:
When dragging the slider, notice how the tip of the parallelogram gets further away from the point of intersection as the planes become more parallel.
We can also observe that as we get further away from the point of intersection, the longer of the two vectors (colored red) pushes us further away from the point of intersection than the shorter
(blue) vector does. This is easier to observe if we draw a line from the origin to the point of intersection:
Let's define and as the scaling factors that we apply to and (the result of which are the red and blue vectors). Right now we're using the distance components and of the planes as the scaling
To solve this asymmetric pushing effect, we need to travel less in the direction of the longer vector as the planes become more parallel. We need some sort of "pulling factor" that adjusts the
vectors such that their tip stays on the line as the planes become parallel.
Here our friend the dot product comes in handy yet again. When the planes are perpendicular the dot product of and equals 0, but as the planes become increasingly parallel, it approaches 1. We can
use this to gradually increase our yet-to-be-defined pulling factor.
Let's give the dot product the name to make this a bit less noisy:
The perfect pulling factors happen to be the distance components and used as counterweights against each other!
Consider why this might be. When and are perpendicular, their dot product equals 0, which results in
which we know yields the correct solution.
In the case where and are parallel, their dot product equals 1, which results in:
Because the absolute values of and are equal, it means that the magnitude of the two vectors—defined as and —is equal:
This means that the magnitude of our vectors will become more equal as the planes become parallel, which is what we want!
Let's see this in action:
The vectors stay on the line, but they become increasingly too short as and become parallel.
Yet again, we can use the dot product. Since we want the length of the vectors to increase as the planes become parallel, we can divide our scalars and by where is the dot product of and and is the
absolute value of .
The result of this looks like so:
Using as the denominator certainly increases the size of the parallelogram, but by too much.
However, notice what happens when we visualize the quadrants of the parallelogram:
As the planes become more parallel, the point of intersection approaches the center of the parallelogram.
In understanding why that is, consider the effect that our denominator has on the area of the parallelogram. When , both of the vectors forming the parallelogram double in length, which has the
effect of quadrupling the area of the parallelogram.
This means that when we scale the component vectors of the parallelogram by
it has the effect of scaling the area of the parallelogram by:
To instead scale the area of the parallelogram by , we need to square in the denominator:
Squaring allows us to remove because the square of a negative number is positive.
With this, our scalars and become
which scales the parallelogram such that its tip lies at the point of intersection:
Putting all of this into code, we get:
float dot = Vector3.Dot(P1.normal, P2.normal);
float denom = 1 - dot * dot;
float k1 = (P1.distance - P2.distance * dot) / denom;
float k2 = (P2.distance - P1.distance * dot) / denom;
Vector3 point = P1.normal * k1 + P2.normal * k2;
Based on code from Real-Time Collision Detection by Christer Ericson
Which through some mathematical magic can be optimized down to:
Vector3 direction = Vector3.cross(P1.normal, P2.normal);
float denom = Vector3.Dot(direction, direction);
Vector3 a = P1.distance * P2.normal;
Vector3 b = P2.distance * P1.normal;
Vector3 point = Vector3.Cross(a - b, direction) / denom;
How this optimization works can be found in chapter 5.4.4 of Real-Time Collision Detection by Christer Ericson.
This completes our plane-plane intersection implementation:
Line PlanePlaneIntersection(Plane P1, Plane P2) {
Vector3 direction = Vector3.cross(P1.normal, P2.normal);
if (direction.magnitude < EPSILON) {
return null; // Roughly parallel planes
float denom = Vector3.Dot(direction, direction);
Vector3 a = P1.distance * P2.normal;
Vector3 b = P2.distance * P1.normal;
Vector3 point = Vector3.Cross(a - b, direction) / denom;
Vector3 normal = direction.normalized;
return new Line(point, normal);
By the way, an interesting property of only traveling along the plane normals is that it yields the point on the line of intersection that is closest to the origin. Cool stuff!
Three plane intersection
Given three planes , , , there are five possible configurations in which they intersect or don't intersect:
1. All three planes are parallel, with none of them intersecting each other.
2. Two of the planes are parallel, and the third plane intersects the other two.
3. All three planes intersect along a single line.
4. The three planes intersect each other in pairs, forming three parallel lines of intersection.
5. All three planes intersect each other at a single point.
When finding the point of intersection, we'll first need to determine whether all three planes intersect at a single point—which for configurations 1 through 4, they don't.
Given , , as the plane normals for , , , we can determine whether the planes intersect at a single point with the formula:
When I first saw this, I found it hard to believe this would work for all cases. Still, it does! Let's take a deep dive to better understand what's happening.
Two or more planes are parallel
We'll start with the configurations where two or more planes are parallel:
If and are parallel then is a vector whose magnitude is zero.
And since the dot product is a multiple of the magnitudes of its component vectors:
the final result is zero whenever and are parallel.
This takes care of the "all-planes-parallel" configuration, and the configuration where and are parallel
With that, let's consider the case where is parallel to either or but and are not parallel to each other.
Let's take the specific case where is parallel to but is parallel to neither.
Here the cross product is a vector (colored red) that's perpendicular to both and .
Since is parallel to , that means that is also perpendicular to . As we've learned, the dot product of two perpendicular vectors is zero, meaning that:
This also holds in the case where is parallel to instead of .
Parallel lines of intersection
We've demonstrated that two of the three normals being parallel results in . But what about the configurations where the three planes intersect along parallel lines? Those configurations have no
parallel normals.
As we learned when looking at plane-plane intersections, the cross product of two plane normals gives us the direction vector of the planes' line of intersection.
When all of the lines of intersection are parallel, all of the plane normals defining those lines are perpendicular to them.
Yet again, because the dot product of perpendicular vectors is 0 we can conclude that for these configurations as well.
We can now begin our implementation. As usual, we'll use an epsilon to handle the "roughly parallel" case:
Vector3 ThreePlaneIntersection(Plane P1, Plane P2, Plane P3) {
Vector3 cross = Vector3.Cross(P2.normal, P3.normal);
float dot = Vector3.Dot(P1.normal, cross);
if (Mathf.Abs(dot) < EPSILON) {
return null; // Planes do not intersect at a single point
// ...
Computing the point intersection
We want to find the point at which our three planes , , intersect:
Some of what we learned about two-plane intersections will come into play here. Let's start by taking the line of intersection for and and varying the position of . You'll notice that the point of
intersection is the point at which intersects the line.
When 's distance from the origin is 0, the vector pointing from the origin to the point of intersection is parallel to (and perpendicular to 's normal).
This vector—let's call it —will play a large role in computing the point of intersection.
We can find through the cross product of two other vectors , . The first of those, , is just 's normal.
The latter vector can be found via the equation
where and are the distances in the constant-normal form of planes and .
With and defined, we assign their cross product to :
Let's see what it looks like:
Hmm, not quite long enough. certainly points in the right direction, but to make 's tip lie on the line of intersection, we need to compute some scaling factor for .
As it turns out, we've already computed this scaling factor:
The product of —let's call that —can be thought to represent how parallel 's normal is to the line intersection of and .
approaches as 's normal becomes parallel to the line of intersection , and approaches 0 as they become perpendicular.
We want the 's magnitude to increase as decreases, so we'll make | {"url":"https://alexharri.com/blog/planes","timestamp":"2024-11-12T03:12:58Z","content_type":"text/html","content_length":"1049741","record_id":"<urn:uuid:d591970b-9e9a-447f-98c6-d20c814380d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00178.warc.gz"} |
CS342 Operating Systems Homework 2
Q. Consider the following memory reference string (only page numbers are shown):
0 0 0 1 1 1. Assume at every tick, we have a memory reference. The first memory
reference is done at tick 1. Assume at every 10 ticks, reference bits are cleared, just
after making the 10th memory reference. We use a single reference bit for each page.
Assume there are 3 frames allocated to a process. All frames are initially empty.
Assume we use clock algorithm (i.e., a second change algorithm) as the page
replacement algorithm. Find out how many page faults we will have (note that we are
not using Modified/Dirty bit). Assume for a new page that is brought in, its reference
bit is set to 1.
Q. Consider a computer that is uses segmentation and paging. The segment table of a
process is the following:
Segment Base Length
Assume page size is 64 bytes. Assume virtual addresses are 16 bits long. Assume
physical addresses are also 16 bits long. Assume a page i is located in a frame i+10
(for example, page #11 of linear logical memory is in frame #21 of physical memory).
Convert the following logical addresses to their physical ones:
a) (0, 50)
b) (1,0)
c) (1,100)
d) (1,700)
e) (2,10)
f) (3,200)
All numbers above are decimal.
Q. Assume a cigarette requires three ingredients (items) to smoke: Tobacco (t), Paper
(p) and a Match (m). Assume there are 3 smokers around a table, each of whom has
an infinite supply of one of the three ingredients (items) — one smoker has an infinite
supply of tobacco, another has an infinite supply of paper, and the third has an infinite
supply of matches. Assume there is also a non-smoking agent which has also an
infinite supply of these items. The agent enables the smokers to make their cigarettes
by randomly selecting and putting two items (out of three items) on the table. Then
the smoker having the missing item will take the items from the table (in this way will
make the table empty), will make his cigarette, and will be able to smoke for a while.
When table becomes empty, agent again chooses two items in random and places
them on the table. Another smoker can now smoke (or maybe the currently smoking
smoker will take those items again and start smoking again after it has finished its
current smoking). This process continues forever. Synchronize the agent and 3
smokers by use of semaphores to act in this way.
Q. Assume we have processes A, B, C, D, E arriving to a system at the following
times shown in the table below. The CPU time required by each process is also shown
in the table. Assume the CPU scheduling algorithm is Round-Robin with time
quantum q. Assume q is very small (very close to zero, but is not zero), hence perfect
processor sharing is happening. Ignore the context switching overhead, i.e. context
switch time is equal to 0. Compute the waiting time and finish time of each process.
Arrival Time CPU Time
Process A 0 145
Process B 50 40
Process C 90 70
Process D 210 95
Process E 240 45
Q. Consider a disk with N tracks numbered from 0 to N-1, and assume that requested
sectors are distributed randomly and evenly over the disk. We want to calculate the
average number of tracks (i.e., cylinders) traversed by a seek.
a) Calculate the probability of a seek of length j when the head is currently
positioned over track t.
b) Calculate the probability of a seek of length K, for an arbitrary current
position of the head.
c) Calculate the average number of tracks traversed by a seek.
Q. Assume block size in a file system that is using hierarchical indexed allocation is
4KB. Assume disk pointer size is 8 bytes. The hierarchy can grow to 3 levels at most
(like Unlix/Linux file system). How many index blocks are required for a file of size
1 MB, 100 MB, 4 GB and 1 TB? What can be the maximum file size in bytes?
Assume no caching of index blocks, how many disk accesses are required on the
average to randomly access (uniform random) the 4 GB file? | {"url":"https://codingprolab.com/answer/cs342-oprrating-systems-homework-2/","timestamp":"2024-11-02T14:44:43Z","content_type":"text/html","content_length":"110071","record_id":"<urn:uuid:1c13b601-7f05-4945-a894-de045443b543>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00783.warc.gz"} |
Tanya Khovanova's Math Blog
The homework I give to my students (who are in 6th through 9th grades) often starts with a math joke related to the topic. Once, I decided to let them be the comedians. One of the homework questions
was to invent a math joke. Here are some of their creations. Two of my students decided to restrict themselves to the topic we studied that week: sorting algorithms. The algorithm jokes are at the
* * *
A binary integer asked if I could help to double its value for a special occasion. I thought it might want a lot of space, but it only needed a bit.
* * *
Everyone envies the circle. It is well-rounded and highly educated: after all, it has 360 degrees.
* * *
Why did Bob start dating a triangle? It was acute one.
* * *
Why is Bob scared of the square root of 2? Because he has irrational fears.
* * *
Are you my multiplicative inverse? Because together, we are one.
* * *
How do you know the number line is the most popular?
It has everyone’s number.
* * *
A study from MIT found that the top 100 richest people on Earth all own private jets and yachts. Therefore, if you want to be one of the richest people on Earth, you should first buy a private jet
and yacht.
* * *
Why did the geometry student not use a graphing calculator? Because the cos was too high.
* * *
Which sorting algorithm rises above others when done underwater? Bubble sort!
* * *
Which sorting algorithm is the most relaxing? The bubble bath sort.
One Comment
1. Korepetytor Online:
How does algebra solve its problems?
It simply moves them to the other side.
15 October 2024, 5:40 am | {"url":"https://blog.tanyakhovanova.com/2024/03/my-students-jokes/","timestamp":"2024-11-10T15:49:46Z","content_type":"text/html","content_length":"59811","record_id":"<urn:uuid:09c435de-1a8d-48a0-acd2-d3b05742dc8b>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00276.warc.gz"} |
Boolean Logic
• Logic gates are combined to form transistors
• Transistors combine to form integrated circuits
• The integrated circuit is a silicon wafer that consists of various microelectronic components
• An integrated circuit is usually made of a single type of gate only
Set representation
Bidmas for logic gates
Order for gates:
Big Noonoo Xtreme Apetite Orange Bring Now, eXtremely Angry Otters Behold Newly eX Assistant Orthodontists
If you want to prioritise OR over AND then brackets are required.
Commutive Laws
• The order of the operands does not matter
Associative Laws
• When all the operators are the same, it does not matter what order they are applied in.
Simplifying boolean expressions
• Simplifying means rewriting the expression in a way that uses fewer logic gates but keeping the exact same functionality.
A+(-)A=1 - A OR NOT A IS TRUE A+A=A A.0 = 0 A.1 = A A.A = A A.(-)A=0 (-)(-)A=A
1. A
2. B
3. A.B
4. D.F+G+A.(-)B
A AND NOT A IS FALSE
This is correct because A is TRUE and NOT A is FALSE. So an AND operation on a TRUE and FALSE will result in FALSE due to one of the inputs being FALSE.
Boolean Laws (continued)
Absorption Laws
• If a term is ANDed or ORed to itself, then it is equivalent.
A + A.B = A A.(A+B) =A
C + C.D = C D + C.D.B = D A.(C+A) = A D.F + D.1 = D. E.F.(E.F+D) = E.F A.A+A.1+B.-B = A
Distributive Laws
• Like mathematical algebra, you should expand brackets where needed.
• It is also possible to expand brackets in Boolean algebra expressions when an expression is ANDed with an expression enclosed in brackets.
• This can often help to simplify an expression (though sometimes it might not—just because you can expand brackets does not mean it is always right to do so.)
A.(B+C) = (A.B) + (A.C) (A+B).(C+D) = (A.C) + (A.D) + (B.C) + (B.D)
Inverse Distributive Laws
Also known as: factoring
• In algebraic expressions you will have seen that sometimes an expression can be simplified by adding brackets, the same is true for boolean algebra.
C.(D+B) = (C.D) + (C+B) C.D.(B+A.E) = (C.D.B) + (C.D.A) + (C.D.E) A.(B+C+D)+A.
DeMorgan’s Laws
DeMorgan’s First Law
Law 1 and Law 2 in a Venn diagram
Therefore, N T A OR B is the same as NOT A AND NOT B
De Morgan’s Second Law
Essentially the inverse of the first law.
Therefore, NOT A AND B is the same as NOT A OR NOT B | {"url":"https://old.sethmb.xyz/sixth/CompSci/Theory/BinaryLogic/BooleanLogic","timestamp":"2024-11-07T13:39:44Z","content_type":"text/html","content_length":"87048","record_id":"<urn:uuid:06e31dbb-dc9a-4df5-b362-237a4bf4d071>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00758.warc.gz"} |
23.7 Sampling difficulties with problematic priors | Stan User’s Guide
23.7 Sampling difficulties with problematic priors
With an improper posterior, it is theoretically impossible to properly explore the posterior. However, Gibbs sampling as performed by BUGS and JAGS, although still unable to properly sample from such
an improper posterior, behaves differently in practice than the Hamiltonian Monte Carlo sampling performed by Stan when faced with an example such as the two intercept model discussed in the
collinearity section and illustrated in the non-identifiable density plot.
Gibbs sampling
Gibbs sampling, as performed by BUGS and JAGS, may appear to be efficient and well behaved for this unidentified model, but as discussed in the previous subsection, will not actually explore the
posterior properly.
Consider what happens with initial values \(\lambda_1^{(0)}, \lambda_2^{(0)}\). Gibbs sampling proceeds in iteration \(m\) by drawing \[\begin{align*} \lambda_1^{(m)} &\sim p(\lambda_1 \mid \lambda_2
^{(m-1)}, \sigma^{(m-1)}, y) \\ \lambda_2^{(m)} &\sim p(\lambda_2 \mid \lambda_1^{(m)}, \sigma^{(m-1)}, y) \\ \sigma^{(m)} &\sim p(\sigma \mid \lambda_1^{(m)}, \lambda_2^{(m)}, y). \end{align*}\]
Now consider the draw for \(\lambda_1\) (the draw for \(\lambda_2\) is symmetric), which is conjugate in this model and thus can be done efficiently. In this model, the range from which the next \(\
lambda_1\) can be drawn is highly constrained by the current values of \(\lambda_2\) and \(\sigma\). Gibbs will run quickly and provide seemingly reasonable inferences for \(\lambda_1 + \lambda_2\).
But it will not explore the full range of the posterior; it will merely take a slow random walk from the initial values. This random walk behavior is typical of Gibbs sampling when posteriors are
highly correlated and the primary reason to prefer Hamiltonian Monte Carlo to Gibbs sampling for models with parameters correlated in the posterior.
Hamiltonian Monte Carlo sampling
Hamiltonian Monte Carlo (HMC), as performed by Stan, is much more efficient at exploring posteriors in models where parameters are correlated in the posterior. In this particular example, the
Hamiltonian dynamics (i.e., the motion of a fictitious particle given random momentum in the field defined by the negative log posterior) is going to run up and down along the valley defined by the
potential energy (ridges in log posteriors correspond to valleys in potential energy). In practice, even with a random momentum for \(\lambda_1\) and \(\lambda_2\), the gradient of the log posterior
is going to adjust for the correlation and the simulation will run \(\lambda_1\) and \(\lambda_2\) in opposite directions along the valley corresponding to the ridge in the posterior log density.
No-U-turn sampling
Stan’s default no-U-turn sampler (NUTS), is even more efficient at exploring the posterior (see Hoffman and Gelman 2014). NUTS simulates the motion of the fictitious particle representing the
parameter values until it makes a U-turn, it will be defeated in most cases, as it will just move down the potential energy valley indefinitely without making a U-turn. What happens in practice is
that the maximum number of leapfrog steps in the simulation will be hit in many of the iterations, causing a large number of log probability and gradient evaluations (1000 if the max tree depth is
set to 10, as in the default). Thus sampling will appear to be slow. This is indicative of an improper posterior, not a bug in the NUTS algorithm or its implementation. It is simply not possible to
sample from an improper posterior! Thus the behavior of HMC in general and NUTS in particular should be reassuring in that it will clearly fail in cases of improper posteriors, resulting in a clean
diagnostic of sweeping out large paths in the posterior.
Here are results of Stan runs with default parameters fit to \(N=100\) data points generated from \(y_n \sim \textsf{normal}(0,1)\):
Two Scale Parameters, Improper Prior
Inference for Stan model: improper_stan
Warmup took (2.7, 2.6, 2.9, 2.9) seconds, 11 seconds total
Sampling took (3.4, 3.7, 3.6, 3.4) seconds, 14 seconds total
Mean MCSE StdDev 5% 95% N_Eff N_Eff/s R_hat
lp__ -5.3e+01 7.0e-02 8.5e-01 -5.5e+01 -5.3e+01 150 11 1.0
n_leapfrog__ 1.4e+03 1.7e+01 9.2e+02 3.0e+00 2.0e+03 2987 212 1.0
lambda1 1.3e+03 1.9e+03 2.7e+03 -2.3e+03 6.0e+03 2.1 0.15 5.2
lambda2 -1.3e+03 1.9e+03 2.7e+03 -6.0e+03 2.3e+03 2.1 0.15 5.2
sigma 1.0e+00 8.5e-03 6.2e-02 9.5e-01 1.2e+00 54 3.9 1.1
mu 1.6e-01 1.9e-03 1.0e-01 -8.3e-03 3.3e-01 2966 211 1.0
Two Scale Parameters, Weak Prior
Warmup took (0.40, 0.44, 0.40, 0.36) seconds, 1.6 seconds total
Sampling took (0.47, 0.40, 0.47, 0.39) seconds, 1.7 seconds total
Mean MCSE StdDev 5% 95% N_Eff N_Eff/s R_hat
lp__ -54 4.9e-02 1.3e+00 -5.7e+01 -53 728 421 1.0
n_leapfrog__ 157 2.8e+00 1.5e+02 3.0e+00 511 3085 1784 1.0
lambda1 0.31 2.8e-01 7.1e+00 -1.2e+01 12 638 369 1.0
lambda2 -0.14 2.8e-01 7.1e+00 -1.2e+01 12 638 369 1.0
sigma 1.0 2.6e-03 8.0e-02 9.2e-01 1.2 939 543 1.0
mu 0.16 1.8e-03 1.0e-01 -8.1e-03 0.33 3289 1902 1.0
One Scale Parameter, Improper Prior
Warmup took (0.011, 0.012, 0.011, 0.011) seconds, 0.044 seconds total
Sampling took (0.017, 0.020, 0.020, 0.019) seconds, 0.077 seconds total
Mean MCSE StdDev 5% 50% 95% N_Eff N_Eff/s R_hat
lp__ -54 2.5e-02 0.91 -5.5e+01 -53 -53 1318 17198 1.0
n_leapfrog__ 3.2 2.7e-01 1.7 1.0e+00 3.0 7.0 39 507 1.0
mu 0.17 2.1e-03 0.10 -3.8e-03 0.17 0.33 2408 31417 1.0
sigma 1.0 1.6e-03 0.071 9.3e-01 1.0 1.2 2094 27321 1.0
On the top is the non-identified model with improper uniform priors and likelihood \(y_n \sim \textsf{normal}(\lambda_1 + \lambda_2, \sigma)\).
In the middle is the same likelihood as the middle plus priors \(\lambda_k \sim \textsf{normal}(0,10)\).
On the bottom is an identified model with an improper prior, with likelihood \(y_n \sim \textsf{normal}(\mu,\sigma)\). All models estimate \(\mu\) at roughly 0.16 with low Monte Carlo standard error,
but a high posterior standard deviation of 0.1; the true value \(\mu=0\) is within the 90% posterior intervals in all three models.
Examples: fits in Stan
To illustrate the issues with sampling from non-identified and only weakly identified models, we fit three models with increasing degrees of identification of their parameters. The posteriors for
these models is illustrated in the non-identifiable density plot. The first model is the unidentified model with two location parameters and no priors discussed in the collinearity section.
data {
int N;
array[N] real y;
parameters {
real lambda1;
real lambda2;
real<lower=0> sigma;
transformed parameters {
real mu;
mu = lambda1 + lambda2;
model {
y ~ normal(mu, sigma);
The second adds priors to the model block for lambda1 and lambda2 to the previous model.
The third involves a single location parameter, but no priors.
data {
int N;
array[N] real y;
parameters {
real mu;
real<lower=0> sigma;
model {
y ~ normal(mu, sigma);
All three of the example models were fit in Stan 2.1.0 with default parameters (1000 warmup iterations, 1000 sampling iterations, NUTS sampler with max tree depth of 10). The results are shown in the
non-identified fits figure. The key statistics from these outputs are the following.
• As indicated by R_hat column, all parameters have converged other than \(\lambda_1\) and \(\lambda_2\) in the non-identified model.
• The average number of leapfrog steps is roughly 3 in the identified model, 150 in the model identified by a weak prior, and 1400 in the non-identified model.
• The number of effective samples per second for \(\mu\) is roughly 31,000 in the identified model, 1,900 in the model identified with weakly informative priors, and 200 in the non-identified
model; the results are similar for \(\sigma\).
• In the non-identified model, the 95% interval for \(\lambda_1\) is (-2300,6000), whereas it is only (-12,12) in the model identified with weakly informative priors.
• In all three models, the simulated value of \(\mu=0\) and \(\sigma=1\) are well within the posterior 90% intervals.
The first two points, lack of convergence and hitting the maximum number of leapfrog steps (equivalently maximum tree depth) are indicative of improper posteriors. Thus rather than covering up the
problem with poor sampling as may be done with Gibbs samplers, Hamiltonian Monte Carlo tries to explore the posterior and its failure is a clear indication that something is amiss in the model.
Hoffman, Matthew D., and Andrew Gelman. 2014.
“The No-U-Turn Sampler: Adaptively Setting Path Lengths in Hamiltonian Monte Carlo.” Journal of Machine Learning Research
15: 1593–623. | {"url":"https://mc-stan.org/docs/2_29/stan-users-guide/sampling-difficulties-with-problematic-priors.html","timestamp":"2024-11-06T07:53:42Z","content_type":"text/html","content_length":"134654","record_id":"<urn:uuid:c4bcd6c6-c99f-4571-a3c5-941fc1cd70b2>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00562.warc.gz"} |
GMAT Test Prep : Quantitative
GMAT Test Prep : Quantitative Data Sufficiency Test VI
Formats Worksheet / Test Paper Quiz Review
Hide all answers View all answers Print Try the Quiz
Is the data given in the two statements, labeled (1) and (2), sufficient for answering the question?
All numbers used are real numbers.
1. A building has two types of apartments, big and small. 65 percent of the apartments are small. The number of occupied big apartments is twice the number of unoccupied small apartments. What
percent of the apartments in the building are occupied ?
(1) The number of occupied big apartments is six times the number of unoccupied big apartments.
(2) The building has a total of 160 apartments.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
If 65% of the apartments are small, then 35% are big. Statement (1) specifies the ratio of occupied to unoccupied in big apartments is 6:1. This implies that 30% of the building is occupied big
apartments. Since it is given that the number of occupied big apartments is twice the number of unoccupied small apartments, 15% of the building is unoccupied small apartments. If 65% of the building
is small apartments, then 50% is occupied small apartments. Adding 30% (for occupied big apartments) to 50% (for occupied small apartments) gives 80% for the occupied apartments in the building.
Thus, statement (1) ALONE is sufficient.
Statement (2) ALONE is insufficient because the question pertains to percent occupancy, where the actual number of apartments is irrelevant as demonstrated through equations below.
Let S denote the number of small apartments, B the number of big apartments and T the total number of apartments. Further, let subscripts O and U denote occupied and unoccupied. Then, the following
equations can be formulated.
S[O] + S[U] = 0.65 T (given)
B[O] + B[U] = 0.35 T (given)
B[O] = 2 S[U] (given)
B[O] = 6 B[U] (Statement 1)
The above four equations can be solved to obtain S[O] and B[O] in terms of T, and then calculate the required occupancy percent from (S[O] + B[O])/T.
Clearly, the value of T = 160 as given in Statement (2) is not required.
2. A pipe of length 100 inches is cut into three pieces of unequal lengths. What is the length of the longest piece?
(1) The combined length of the two longer pieces is 80 inches.
(2) The combined length of the two shorter pieces is 50 inches.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
Let the lengths of the three pieces be given by x, y and z inches respectively (in ascending order of lengths).
Then, x + y + z = 100. To answer the question, z is required.
Statement (1) gives y + z = 80, which implies x = 20. Given the combined length of the two longer pieces and the total pipe length, the length of the shortest piece can be found. However, the length
of the longest piece cannot be determined. Thus, statement (1) ALONE is not sufficient.
Statement (2) gives x + y = 50, which implies z = 50. Given the combined length of the two shorter pieces and the total pipe length, the length of the longest piece can be determined. Thus, statement
(2) ALONE is sufficient.
3. A certain electronics store sold 60 percent of the computers in its inventory during the month. What was the total revenue from the sale of the computers?
(1) The computers were sold for an average price of $875 during the month.
(2) All but 34 computers in the store's inventory were sold during the month.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Statement (1) specifies the average price, but not the number of computers sold. Thus, statement (1) ALONE is not sufficient.
Statement (2) allows calculation of the number of computers sold, but not the price. Thus, statement (2) ALONE is not sufficient.
Since 34 computers are left (which corresponds to 40% of the inventory), the inventory may be calculated (as 85). Then, 60% of this inventory may be multiplied by the average price to obtain the
total revenue. Thus, BOTH statements TOGETHER are sufficient.
4. Is a greater than 0 ?
(1) ab < 0 and a − b < 0.
(2) ac > 0 and c < 0.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: EACH statement ALONE is sufficient.
Statement (1) states that ab < 0, which implies that either a or b is negative and the other is positive. Further, a < b implies that a is negative and b is positive. Thus, statement (1) ALONE is
sufficient (to answer the question in the negative).
Statement (2) states that ac > 0, which implies that either botha and c are positive or both are negative. Further, c < 0 implies that both a and c are negative. Thus, statement (2) alone is
sufficient (to answer the question in the negative).
Therefore, EACH statement ALONE is sufficient.
5. Train A travels from Town P to Town Q, while Train B travels from Town Q to Town P. Both trains start at the same time. Town R lies between Town P and Town Q. Which train travels at a higher
(1) Town R lies closer to Town P than Town Q.
(2) Train A crosses Town R earlier than Train B does.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statements (1) and (2) TOGETHER are NOT sufficient.
Note that Speed = Distance traveled / Time taken.
Statement (1) provides some information on the distance traveled, but nothing about the time taken. So, statement (1) ALONE is not sufficient.
Statement (2) provides some information on the time taken, but nothing about the distance traveled. So, statement (2) ALONE is not sufficient.
Even both statements together cannot tell us which train travels faster because of the following argument. If Train A travels faster than Train B, it will certainly cross Town R earlier than Train B
does. But, even if Train A travels slower than Train B, and if Town R lies very close to Town P, then Train A will still cross Town R before Train B does. Thus, statements (1) and (2) TOGETHER are
not sufficient.
What is the area of square KLMN?
(1) KL = 7 inches.
(2) Its diagonal is √98 inches.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: EACH statement ALONE is sufficient.
To find the area of square, its side must be known because
Area of square = Side × Side.
Statement (1) specifies the side. Thus, statement (1) ALONE is sufficient.
Statement (2) specifies the diagonal. The side can be found using the Pythagorean theorem and then the area calculated as follows:
KL^2 + LM^2 = (√98)^2
Since KL = LM, 2LM^2 = 98 or LM = √49 = 7 inches.
Area of square = 7 × 7 square inches = 49 square inches.
Thus, statement (2) ALONE is sufficient.
7. Every month, Michael gets a salary plus a commission equal to 2 percent of his sales revenue. What was Michael's commission for the month of January?
(1) Michael's monthly salary is $1,750.
(2) Michael's sales revenue for the month of January was $2,750.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
Statement (1) specifies the monthly salary, which does not pertain to the commission. Thus, statement (1) ALONE is not sufficient.
Statement (2) specifies the sales revenue for the month of January, and a fixed percentage of this revenue is the commission. Thus, statement (2) ALONE is sufficient.
8. Is the integer n divisible by 30 ?
(1) n is divisible by 18.
(2) n is divisible by 20.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
If an integer is divisible by 18, it may or may not be divisible by 30. For example, 36 is divisible by 18 but not by 30, whereas 90 is divisible by 18 as well as 30. Thus, statement (1) ALONE is not
If an integer is divisible by 20, it may or may not be divisible by 30. For example, 40 is divisible by 20 but not by 30, whereas 60 is divisible by 20 as well as 30. Thus, statement (2) ALONE is not
If an integer is divisible by both 18 and 20, then it is necessarily divisible by 30. Since 18 = 2 × 3 × 3 and 20 = 2 × 2 × 5, their LCM (least common multiple) is 180 = 2 × 2 × 3 × 3 × 5. Multiples
of 180 are divisible by 30. Thus, BOTH statements TOGETHER are sufficient.
9. Is m an integer?
(1) 4m is an integer.
(2) m/4 is an integer.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
If 4m is an integer, m may or may not be an integer. For example, 4m = 5 gives m = 5/4 = 1.25, whereas 4m = 8 gives m = 2. Thus, statement (1) ALONE is not sufficient.
If m/4 is an integer (say, n), then m is necessarily an integer (namely, 4n). Thus, statement (2) ALONE is sufficient.
10. What number is ¾ of y ?
(1) 5 percent of x and 15 percent of y total 8.
(2) One-tenths of x and three-tenths of y equal 16.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statements (1) and (2) TOGETHER are NOT sufficient.
Statement (1) gives 5x + 15y = 800. Two unknowns cannot be determined from one equation. Thus, statement (1) ALONE is not sufficient. Statement (2) gives x + 3y = 160. Again, two unknowns cannot be
determined from one equation. Thus, statement (2) ALONE is not sufficient.
The two equations are not independent because multiplying x + 3y = 160 throughout by 5 gives 5x + 15y = 800. Thus, statements (1) and (2) TOGETHER are NOT sufficient.
11. A terminating decimal is defined as any decimal that has a finite number of nonzero digits. When the ratio of two positive integers m and n is expressed as a decimal, is m/n a terminating
(1) 330 < m < 333
(2) n = 3
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Statement (1) states that m is either 331 or 332. Now, 332/2 (= 166) is a terminating decimal, but 332/3 (= 110.666...) is a recurring decimal (and not a terminating decimal). Thus, statement (1)
ALONE is not sufficient.
Statement (2) states that n is 3. If m is a multiple of 3, then m/n is a terminating decimal; otherwise, it is a recurring decimal (ending in .333... or .666...). Thus, statement (2) ALONE is not
If m is either 331 or 332 and n is 3, then m/n is necessarily a recurring decimal. Since 331/3 = 110.333... and 332/3 = 110.666..., m/n is not a terminating decimal. Thus, BOTH statements TOGETHER
are sufficient.
12. How long will it take to fill an empty tank of circular cross-section with a certain liquid?
(1) The tank is 1.5 feet in radius and 3 feet in height.
(2) Liquid is pumped in at the rate of 4 gallons per hour (1 cubic foot = 7.48 gallons).
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Statement (1) specifies the radius (say, r) and height (say, h). Although the volume of the tank can be calculated (from the formula: Volume of cylinder = π r^2h), the time to fill the tank cannot be
found without knowing the flow rate. Thus, statement (1) ALONE is not sufficient.
Statement (2) specifies the flow rate, but the time to fill the tank cannot be calculated without knowing the tank volume. Thus, statement (2) ALONE is not sufficient.
Knowing the volume from statement (1) and the flow rate from statement (2), the time to fill the tank can be calculated from the formula:
Time = Volume / Flow rate.
Thus, BOTH statements TOGETHER are sufficient.
13. Is x negative ?
(1) x^2 > 0
(2) x^3 > 0
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
Statement (1) states that the square of the number x is positive, which implies that x may be positive or negative. Thus, statement (1) ALONE is not sufficient.
Statement (2) states that the cube of the number x is positive, which necessarily implies that x is necessarily positive (and not negative). Thus, statement (2) ALONE is sufficient.
14. A fruit basket contains only apples, bananas and peaches. If a fruit is chosen at random from the fruit basket, what is the probability that it will be a peach ?
(1) The probability that the fruit will be a banana is ½.
(2) There are 7 peaches in the fruit basket.
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statements (1) and (2) TOGETHER are NOT sufficient.
Let a, b and p denote the number of apples, bananas and peaches, respectively. Then, the probability of choosing a peach may be calculated from p/(a + b + p).
Statement (1) gives b/(a + b + p) = ½ or b = a + p. Then, the probability of choosing a peach becomes p/(2b), which cannot be evaluated further. Thus, statement (1) ALONE is not sufficient.
Statement (2) gives p= 7. Then, the probability of choosing a peach becomes 7/(a + b + 7), which cannot be evaluated further. Thus, statement (2) ALONE is not sufficient.
On combining both statements, the probability of choosing a peach becomes 7/(2b) which also cannot be evaluated further. Thus, statements (1) and (2) TOGETHER are NOT sufficient.
15. If p is less than 80 percent of q, is q greater than 80 ?
(1) 4q − 5p > 0
(2) p > 64
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Answer: Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient.
Since p < 0.8q and 0.8 = 4/5, q > (5/4)p.
Statement (1) states that 4q > 5p, which implies that q > (5/4)p. Since there is no new information here, Statement (1) ALONE is not sufficient. Statement (2) states that p > 64. So, (5/4)p > 80 or q
> 80. Thus, statement (2) ALONE is sufficient.
Try the Quiz : GMAT Test Prep : Quantitative Math Data Sufficiency Test VI | {"url":"https://www.syvum.com/cgi/online/serve.cgi/gmat/data_6.html?question_hide","timestamp":"2024-11-08T06:09:07Z","content_type":"text/html","content_length":"50654","record_id":"<urn:uuid:d85dd93e-2252-450f-a1c3-f8920cf545e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00586.warc.gz"} |
f90_exact, a Fortran90 code which evaluates exact solutions to a few selected examples of ordinary differential equations (ODE) and partial differential equations (PDE).
These exact solutions can be used to test out the correctness of a solution algorithm.
The information on this web page is distributed under the MIT license.
f90_exact is available in a C version and a C++ version and a Fortran77 version and a Fortran90 version and a MATLAB version and an Octave version and a Python version.
Related Data and codes:
biharmonic_exact, a Fortran90 code which evaluates exact solutions w(x,y) to the biharmonic equation del^2 w = 0 or wxxxx + 2 wxxyy + wyyyy = 0
burgers_exact, a Fortran90 code which evaluates exact solutions of time-dependent 1D viscous Burgers equation.
fisher_exact, a Fortran90 code which returns an exact solution of the Kolmogorov Petrovsky Piskonov Fisher partial differential equation (PDE) ut=uxx+u*(1-u).
flame_exact, a Fortran90 code which returns the exact solution of an ordinary differential equation (ODE) which models the growth of a ball of flame in a combustion process. The exact solution is
defined in terms of the Lambert W function.
kdv_exact, a Fortran90 code which evaluates an exact solution of the Korteweg-deVries (KdV) partial differential equation (PDE).
logistic_exact, a Fortran90 code which evaluates an exact solution of the logistic equation, an ordinary differential equation (ODE) which models population growth in the face of a limited carrying
navier_stokes_2d_exact, a Fortran90 code which evaluates an exact solution to the incompressible time-dependent Navier-Stokes equations (NSE) over an arbitrary domain in 2D.
navier_stokes_3d_exact, a Fortran90 code which evaluates an exact solution to the incompressible time-dependent Navier-Stokes equations (NSE) over an arbitrary domain in 3D.
porous_medium_exact, a Fortran90 code which returns an exact solution of the porous medium equation (PME), dudt=Del^2(u^m), a partial differential equation (PDE) related to the diffusion equation,
based on the Barenblatt solution.
sine_gordon_exact, a Fortran90 code which returns an exact solution of the Sine-Gordon equation, a partial differential equation (PDE) of the form uxy=sin(u).
spiral_exact, a Fortran90 code which computes a 2D velocity vector field that is an exact solution of the continuity equation.
stokes_2d_exact, a Fortran90 code which evaluates exact solutions to the incompressible steady Stokes equations over the unit square in 2D.
Last revised on 11 May 2024. | {"url":"https://people.sc.fsu.edu/~jburkardt/f_src/f90_exact/f90_exact.html","timestamp":"2024-11-09T17:27:05Z","content_type":"text/html","content_length":"5308","record_id":"<urn:uuid:129c7fd7-5703-4a4c-b04a-1487d0862f24>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00626.warc.gz"} |
Resultados da Pesquisa
• Seleção recorrente recíproca na obtenção de híbridos interpopulacionais de milho-pipoca
(Pesquisa Agropecuária Brasileira, 2008-12) Faria, Vinícius Ribeiro; Viana, José Marcelo Soriano; Sobreira, Fábio Moreira; Silva, Admilson Costa e
O objetivo deste trabalho foi avaliar a eficiência da seleção recorrente recíproca em produzir híbridos de milho-pipoca (Zea mays) de progênies endógamas superiores, com famílias de
irmãos-completos. O programa de seleção recorrente recíproca envolveu as populações de milho-pipoca 'Viçosa' e 'Beija-Flor'. Os testes dos híbridos S0xS0 e S1xS1 foram conduzidos em delineamento
látice, nos anos agrícolas de 2002/2003 e 2004/2005. Os dois ensaios incluíram testemunhas comerciais comuns, que permitiram a comparação do desempenho dos híbridos. Analisaram-se a produtividade
e a capacidade de expansão. Foram preditos ganhos direto e indireto com a seleção em capacidadede expansão, avaliada em pipoqueira de ar quente e na pipocadora "Metric Weight Volume Tester". Os
ganhos observados foram calculados para avaliar a eficiência da seleção recorrente recíproca. As análises de covariância mostraram variabilidade genotípica nas populações. Em relação às
testemunhas comerciais, os híbridos S0xS0 foram inferiores em qualidade e equivalentes em produtividade, e os híbridos S1xS1 foram equivalentes nas duas características. O ganho observado na
capacidade de expansão foi substancial, cerca de 4 mL g-1, superior ao ganho predito (2,8 mL g-1). A redução observada na produtividade não foi relevante. Verificou-se redução na variabilidade
genotípica quanto à capacidade de expansão. O método de seleção recorrente recíproca foi eficiente em produzir híbridos S1xS1 superiores aos S0xS0.
• The parametric restrictions of the Griffing diallel analysis model: combining ability analysis
(Genetics and Molecular Biology, 2000-12) Viana, José Marcelo Soriano
It was studied the parametric restrictions of the diallel analysis model of Griffing, method 2 (parents and F1 generations) and model 1 (fixed), in order to address the questions: i) does the
statistical model need to be restricted? ii) do the restrictions satisfy the genetic parameter values? and iii) do they make the analysis and interpretation easier? Objectively, these questions
can be answered as: i) yes, ii) not all of them, and iii) the analysis is easier, but the interpretation is the same as in the model with restrictions that satisfy the parameter values. The main
conclusions were that: the statistical models for combining ability analysis are necessarily restricted; in the Griffing model (method 2, model 1), the restrictions relative to the specific
combining ability (SCA) effects, and for all j, do not satisfy the parametric values, and the same inferences should be established from the analyses using the model with restrictions that
satisfy the parametric values of SCA effects and that suggested by Griffing. A consequence of the restrictions of the Griffing model is to allow the definition of formulas for estimating the
effects, their variances and the variances of contrasts of effects, as well as for calculating orthogonal sums of squares.
• The parametric restrictions of the Gardner and Eberhart diallel analysis model: heterosis analysis
(Genetics and Molecular Biology, 2000-12) Viana, José Marcelo Soriano
The parametric restrictions of the diallel analysis model of Gardner and Eberhart (analysis II) were studied in order to address the following questions: i) does the statistical model really have
to be restricted? ii) Do the restrictions satisfy the genetic parameter values? iii) Do the restrictions make analysis and interpretation easier? Objectively, the answers to these questions are:
i) no, ii) not all, and iii) they facilitate the analysis, but the interpretation is the same as for the unrestricted model. The main conclusion was that the restrictions of the Gardner and
Eberhart model related to specific heterosis effects, for all j, do not satisfy the parametric values. The Gardner and Eberhart's estimators of the variety heterosis effects, the specific
heterosis effects and their variances, differ from those of the unrestricted model. Any analysis using the unrestricted model and that of Gardner and Eberhart should lead to the same inferences,
at least for those based on assessment of the population effects expressed as deviations from the average value, the heteroses, the average heterosis and the variety heteroses (the correlation
between the estimates of the two models is 1). The limiting factor for the use of the unrestricted model is the lack of formulas for computing the sums of squares and for estimating the estimable
function variances.
• Theory and analysis of partial diallel crosses
(Genetics and Molecular Biology, 1999-12) Viana, José Marcelo Soriano; Cruz, Cosme Damião; Cardoso, Antonio Américo
This study presents theory and analysis of partial diallel crosses based on Hayman's methods. This genetic design consists of crosses among two parental groups. It should be used when there are
two groups of parents, for example, dent and flint maize inbred lines, and the breeder is not interested in the assessment of crosses between parents of the same group. Analyses are carried out
using data from the parents and their F1 hybrids allowing a detailed characterization of the polygenic systems under study and the choice of parents for hybridization. Diallel analysis allows the
estimation of genetic and non-genetic components of variation and genetic parameters and to assess the following: genetic variability in each group; genotypic differences between parents of
distinct groups; if a parent has a common or a rare genotype in the group to which it does not belong; if there is dominance; if dominant genes increase or decrease trait expression (direction of
dominance); average degree of dominance in each group; the relative importance of mean effects of genes and dominance in determining a trait; if, in each group, the allelic genes have the same
frequency; if genes are equally frequent in the two groups; the group with the greatest frequency of favorable genes; the group in which dominant genes are most frequent; the relative number of
dominant and recessive genes in each parent; if a parent has a common or a rare genotype in the group to which it belongs, and the genotypic differences between parents of the same group. An
example with common bean varieties is considered.
• Quantitative genetics theory for non-inbred populations in linkage disequilibrium
(Genetics and Molecular Biology, 2004) Viana, José Marcelo Soriano
Although linkage disequilibrium, epistasis and inbreeding are common phenomena in genetic systems that control quantitative traits, theory development and analysis are very complex, especially
when they are considered together. The objective of this study is to offer additional quantitative genetics theory to define and analyze, in relation to non-inbred cross-pollinating populations,
components of genotypic variance, heritabilities and predicted gains, assuming linkage disequilibrium and absence of epistasis. The genotypic variance and its components, additive and due to
dominance genetic variances, are invariant over the generations only in regard to completely linked genes and to those in equilibrium. When the population is structured in half-sib families, the
additive variance in the parents’ generation and the genotypic variance in the population can be estimated. When the population is structured in full-sib families, none of the components of
genotypic variance can be estimated. The narrow sense heritability at plant level can be estimated from the parent-offspring or mid parent-offspring regression. When there is dominance, the
narrow sense heritability estimate in the in F2 is biased due to linkage disequilibrium when estimated by the Warner method, but not when estimated by means of the plant F2-family F3 regression.
The bias is proportional to the number of pairs of linked genes, without independent assortment, and to the degree of dominance, and tends to be positive when genes in the coupling phase
predominate or negative and of higher value when genes in the repulsion phase predominate. Linkage disequilibrium is also cause of bias in estimates of the narrow sense heritabilities at full-sib
family mean and at plant within half-sib and full-sib families levels. Generally, the magnitude of the bias is proportional to the number of pairs of genes in disequilibrium and to the frequency
of recombining gametes.
• Components of variation of polygenic systems with digenic epistasis
(Genetics and Molecular Biology, 2000-12) Viana, José Marcelo Soriano
In this paper an extension of the biometric model of Mather and Jinks for the analysis of variation with digenic epistasis is presented. Epistatic effects can contribute favorably to the
determination of the genotypic values of selected individuals or families and of superior hybrids. Selection will be inefficient, however, if there is a large number of interacting genes because
the epistatic components of the between-family and within-family genotypic variances are very high compared to the portion attributable to the average effects of genes. Selection tends to be
efficient when the number of interacting genes is reduced, but this depends on the magnitude of due to dominance and environmental variances. The dominance component (H) and the epistatic
component due to interactions between homozygous and heterozygous genic combinations (J) can only be estimated when one or more quadratic statistics from the S3 generation, obtained by randomly
mating F2 individuals, are used.
• Breeding strategies for recurrent selection of maize
(Pesquisa Agropecuária Brasileira, 2007-10) Viana, José Marcelo Soriano
The objectives of this work were to analyze theoretical genetic gains of maize due to recurrent selection among full-sib and half-sib families, obtained by Design I, Full-Sib Design and Half-Sib
Design, and genotypic variability and gene loss with long term selection. The designs were evaluated by simulation, based on average estimated gains after ten selection cycles. The simulation
process was based on seven gene systems with ten genes (with distinct degrees of dominance), three population classes (with different gene frequencies), under three environmental conditions
(heritability values), and four selection strategies. Each combination was repeated ten times, amounting to 25, 200 simulations. Full-sib selection is generally more efficient than half-sib
selection, mainly with favorable dominant genes. The use of full-sib families derived by Design I is generally more efficient than using progenies obtained by Full-Sib Design. Using Design I with
50 males and 200 females (effective size of 160) did not result in improved populations with minimum genotypic variability. In the populations with lower effective size (160 and 400) the loss of
favorable genes was restricted to recessive genes with reduced frequencies.
• Biometrical analyses of linolenic acid content of soybean seeds
(Genetics and Molecular Biology, 2003) Gesteira, Abelmon da Silva; Schuster, Ivan; José, Inês Chamel; Piovesan, Newton Deniz; Viana, José Marcelo Soriano; Barros, Everaldo Gonçalves de; Moreira,
Maurilio Alves
The genetic reduction of linolenic acid levels increases the quality and stability of soybean oil. The objective of this study was to determine the inheritance and evaluate the nature and
magnitude of gene effects on soybean seed linolenic acid level. Means and variances of F1, F2, and F3 generations were made from the cross between accession BARC-12 (low linolenic acid content)
and the commercial Brazilian cultivar CAC-1 (normal linolenic acid content). The results demonstrated that linolenic acid content in soybean is under the genetic control of a small number of
genes. The additive model explained the means for the three generations and for the parents. Non-allelic gene interactions had little effect on the determination of genotypic values for the
individuals. The generation means and population variation analyses demonstrated that the dominance deviations contribute little to the trait. These results showed that backcross breeding
programs can be used to introduce the low linolenic acid content trait into soybean seeds, since it is possible to identify with very high accuracy the desired genotypes in segregating
• Analysis of variance of partial diallel tables
(Genetics and Molecular Biology, 2000-03) Viana, José Marcelo Soriano; Cruz, Cosme Damião; Cardoso, Antonio Américo; Regazzi, Adair José
The theory of variance analysis of partial diallel tables, following Hayman’s proposal of 1954, is presented. As several statistical tests yield similar inferences, the present analysis mainly
proposes to assess genetic variability in two groups of parents and to study specific, varietal and mean heteroses. Testing the nullity of specific heteroses equals testing absence of dominance.
Testing equality of varietal heteroses of the parents of a group is equivalent to testing the hypothesis that in the other group allelic genes have the same frequency. Rejection of the hypothesis
that the mean heterosis is null indicates dominance. The information obtained complements that provided by diallel analysis involving parents and their F1 hybrids or F2 generations. An example
with the common bean is included.
• Analysis of general and specific combining abilities of popcorn populations, including selfed parents
(Genetics and Molecular Biology, 2003-12) Viana, José Marcelo Soriano; Matta, Frederico de Pina
Estimation of general and specific combining ability effects in a diallel analysis of cross-pollinating populations, including the selfed parents, is presented in this work. The restrictions
considered satisfy the parametric values of the GCA and SCA effects. The method is extended to self-pollinating populations (suitable for other species, without the selfed parents). The analysis
of changes in population means due to inbreeding (sensitivity to inbreeding) also permits to assess the predominant direction of dominance deviations and the relative genetic variability in each
parent population. The methodology was used to select popcorn populations for intra- and inter-population breeding programs and for hybrid production, developed at the Federal University of
Viçosa, MG, Brazil. Two yellow pearl grain popcorn populations were selected. | {"url":"https://locus.ufv.br/collections/d011d0b4-e4c7-421d-9f99-500024ee2f2d?f.dateIssued.min=1990&spc.page=1&f.author=Viana,%20Jos%C3%A9%20Marcelo%20Soriano,equals","timestamp":"2024-11-11T23:33:23Z","content_type":"text/html","content_length":"652797","record_id":"<urn:uuid:3a22f948-b975-4701-b79b-bb8c3e036b9d>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00598.warc.gz"} |
MA441 Learning Goals
There are in total learning goals. Check the goals you have mastered and click the following button to see your level of mastery.
Algebra and Geometry of Functions
• F1: I can use function notations to describe functions.
• F2: I can evaluate and graph functions.
• F3: I can find domain, range and intercepts of a function.
• F4: I can find the composition of functions and recognize a function as the composition of functions.
• F5: I can perform mathematical operations on polynomials and factor polynomials.
• F6: I can perform mathematical operations on rational expressions.
• F7: I can perform mathematical operations on radical expressions and rationalize denominators or numerators.
• F8: I can solve equations and inequalities.
• F9: I can evaluate trigonometric functions and determine the measure of the angle from a special trigonometric equation.
• F10: I can solve problems on right triangles using trigonometric identities.
• L1: I can explain how the idea of a limit is involved in solving the the tangent problem.
• L2: I can explain how the idea of a limit is involved in solving the area problem.
• L3: I can explain the definition of the limit of a function using numerical, and graphical methods.
• L4: I can provide an example that the numerical method fails.
• L5: (CORE) I can find the limit of a function at a point using algebraic methods and limit laws.
• L6: (CORE) I can explain the relationship between one-sided and two-sided limits.
• L7: (CORE) I can evaluate the limit of a function by using the squeeze theorem.
• L8: I can using correct notation, describe an infinite limit.
• L9: (CORE) I can find the limit of a function at the infinity using algebraic methods.
• C1: I can explain the three conditions for continuity at a point.
• C2: (CORE) I can describe three kinds of discontinuities and provide examples.
• C3: (CORE) I can determine whether and when a function is continuous over a given interval using the definition of continuity.
• C4: I can state the theorem for limits of composite functions.
• C5: (CORE) I can use the theorem for limits of composite functions evaluate limits.
• C6: I can state the intermediate value theorem and provide examples and counterexamples.
• C7: (CORE) I can apply the intermediate value theorem to solve problems such as determine whether an equation has a solution over a given interval.
• D1: (CORE) I can find the derivative of a function at a point and as a function using the definition.
• D2: (CORE) I can state the connection between derivatives and continuity, and provide examples.
• D3: I can determine whether or when a function is differentiable over an interval using the definition.
• D4: (CORE) I can use derivative notations and calculate higher derivatives.
• D5: (CORE) I can find the equation of the tangent line to a function at a point.
• D6: I can correctly interpret a rate of change as a derivative in context and find the value.
• D7: (CORE) I can find derivatives of power and trigonometric functions, and their linear combinations.
• D8: (CORE) I can find derivatives of functions using the product rule and the quotient rule.
• D9: (CORE) I can find derivatives of composite functions using the chain rule.
• D10: (CORE) I can find derivatives by using multiple rules in combination.
• D11: (CORE) I can find the derivative of an implicitly-defined function using implicit differentiation.
• D12: I can find the slope of the tangent line to a curve defined by an implicit function.
Applications of Derivatives
• A1: I can set up equations for relationships among derivatives in a related rates problems and find the rate of change of one quantity that depends on the rate of change of other quantities.
• A2: I can find the linearization of a given function and use it to estimate the value of a function at a given point.
• A3: I can use differential approximation to calculate the change in a quantity, the relative error, and percentage error.
• A4: (CORE) I can find the critical values of a function.
• A5: I can use the Extreme Value Theorem to find the absolute maximum and minimum values of a continuous function over a closed interval.
• A6: I can demonstrate understanding of the Mean Value Theorem using examples and counterexamples.
• A7: (CORE) I can state three important corollaries of the Mean Value Theorem.
• A8: I can explain how the sign of the first derivative affects the shape of the graph of a function.
• A9: (CORE) I can find intervals of increasing and decreasing of a function.
• A10: (CORE) I can find local extrema of a function using the First and Second Derivative Tests.
• A11: I can determine the intervals of concavity of a function and find all of its points of inflection.
• A12: I can find horizontal, vertical and oblique (slanted) asymptotes of a function using rules of limits.
• A13: I can sketch the graph of a function manually.
• A14: (CORE) I can set up and use differential calculus to solve applied optimization problems.
• A15: (CORE) I can find the general antiderivative of a given function.
• A16: I can use anti-differentiation to solve initial-value problems.
Integrals and Applications
• I1: I can use sigma (summation) notation to calculate sums of powers of integers.
• I2: (CORE) I can approximate area between curves using Riemann sums, and interpret a definite integral as the limit of Riemann sums.
• I3: I can evaluate integrals using geometry and the properties of definite integrals.
• I4: I can explain the connection between the Mean Value Theorem and the Fundamental Theorem of Calculus part 1 (\(\frac{\operatorname{d}}{\operatorname{d} x}\int_a^xf(t)\operatorname{d}t=f(x)\)).
• I5: (CORE) I can evaluate a definite integral using the Fundamental Theorem of Calculus.
• I6: I can explain the relationship between differentiation and integration.
• I7: I can describe the meaning of the mean value theorem for integrals.
• I8: I can calculate the average value of a function.
• I9: (CORE) I can evaluate a definite integral using integration rules.
• I10: I can use the net change theorem to solve applied problems.
• I11: (CORE) I can use substitution to evaluate indefinite integrals and definite integrals.
• I12: (CORE) I can find the area of a region between two curves using definite integrals. | {"url":"https://yfei.page/teaching/calculus-1/ma441-learning-goals","timestamp":"2024-11-11T11:20:25Z","content_type":"text/html","content_length":"834479","record_id":"<urn:uuid:22914de0-17f1-4808-8882-a6387c47e7c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00455.warc.gz"} |
Extension of Gaussian-2 (G2) theory to bromine- and iodine-containing molecules: Use of effective core potentials
Basis sets have been developed for carrying out G2 calculations on bromine- and iodine-containing molecules using all-electron (AE) calculations and quasirelativistic energy-adjusted
spin-orbit-averaged seven-valence-electron effective core potentials (ECPs). Our recommended procedure for calculating G2[ECP] energies for such systems involves the standard G2 steps introduced by
Pople and co-workers, together with the following modifications: (i) second-order Møller-Plesset (MP2) geometry optimizations use polarized split-valence [31,31,1] basis sets for bromine and iodine
together with 6-31G(d) for first- and second-row atoms; (ii) single-point higher-level energies are calculated for these geometries using our new supplemented bromine and iodine valence basis sets
along with supplemented 6-311G and McLean-Chandler 6-311G bases for first- and second-row atoms, respectively; and (iii) first-order spin-orbit corrections are explicitly taken into account. An
assessment of the results obtained using such a procedure is presented. The results are also compared with corresponding all-electron calculations. We find that the G2[ECP] calculations give results
which are generally comparable in accuracy to those of the G2[AE] calculations but which involve considerably lower computational cost. They are therefore potentially useful for larger bromine- and
iodine-containing molecules for which G2[AE] calculations would not be feasible.
ASJC Scopus subject areas
• General Physics and Astronomy
• Physical and Theoretical Chemistry
Dive into the research topics of 'Extension of Gaussian-2 (G2) theory to bromine- and iodine-containing molecules: Use of effective core potentials'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/extension-of-gaussian-2-g2-theory-to-bromine-and-iodine-containin","timestamp":"2024-11-13T03:09:11Z","content_type":"text/html","content_length":"59481","record_id":"<urn:uuid:5ea2dbc2-7750-4946-a7c2-c14d4c23c0e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00377.warc.gz"} |
Net operating income (NOI)
Net operating income (NOI) is the sum of all income from the property minus operating expenses. It excludes taxes, interest on loans and depreciation. NOI is an important indicator that helps
investors assess how much income a property generates before financing and tax effects.
Other terms from the category
Doesn't Net operating income (NOI) sound appealing?
Don't hesitate to contact us | {"url":"https://en.skladuj.cz/en/term/net-operating-income-noi/","timestamp":"2024-11-06T10:21:43Z","content_type":"text/html","content_length":"787841","record_id":"<urn:uuid:85b71fe0-052c-46fc-878c-1733a09588b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00289.warc.gz"} |
Φ factor EXPLANATION | cristos-vournas.com
The Φ -Factor is the Solar Irradiation Accepting Factor (in other words, Φ is the planet surface spherical shape and planet surface roughness coefficient).
It is not "a hypothetical solar irradiation accepting factor."
By "accepting" means how much energy a spherical shape body may "stop".
A spherical shape body may stop a flow of energy. Because of spherical shape a sphere
may stop Φ(1-a)S of total incident energy.
Φ varies from 0,47 to 1.
The smoother the surface is the closer to 0,47 the Φ is.
Thus, a smooth surface sphere with Albedo a = 0
should "absorb" 0,47*S.
A sphere with rough surface (Φ = 1) and
a = 0 should "absorb" a 100% S.
And a sphere with Albedo a = 1 should not "absorb" EM energy, regardless the values of Φ. | {"url":"https://cristos-vournas.com/444383819/","timestamp":"2024-11-07T10:45:13Z","content_type":"text/html","content_length":"295090","record_id":"<urn:uuid:df7584b7-55d6-46bd-8811-796ca305b09f>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00581.warc.gz"} |
Automata Theory Multiple Choice Quaestion & Answers (MCQs) set-22 - Studyhelpzone.com
MCQ Computer Science
Automata Theory Multiple Choice Quaestion & Answers (MCQs) set-22
1. Which of the following are basic complexity classes for a function f:N->N?
a) Ntime(f)
b) Nspace(f)
c) Space(f)
d) All of the mentioned
View Answer
Answer: d
Explanation: Ntime(f): is a set of languages that can be accepted by a NTM T with non deterministic time complexity function t <=f. In all four cases, the machines are allowed to be multitape TM’s.
2. In order to reduce the run time of a turing machine:
a) we can reduce the number of tapes
b) we can increase the number of tapes
c) use infinite tapes
d) none of the mentioned
View Answer
Answer: One way to reduce the run time can be to increase the number of tapes. Sometimes, using two tapes can be used to avoid back and forth motions altogether.
3. Which of the following is an application of Finite Automaton?
a) Compiler Design
b) Grammar Parsers
c) Text Search
d) All of the mentioned
View Answer
Answer: d
Explanation: There are many applications of finite automata, mainly in the field of Compiler Design and Parsers and Search Engines.
4. Given:
L= {x??= {0,1} |x=0n1n for n>=1}; Can there be a DFA possible for the language?
a) Yes
b) No
View Answer
Answer: b
Explanation: It is not possible to have a count of equal number of 0 and 1 at any instant in DFA. Thus, It is not possible to build a DFA for the given Language.
5. Mealy and Moore machine can be categorized as:
a) Inducers
b) Transducers
c) Turing Machines
d) Linearly Bounder Automata
View Answer
Answer: b
Explanation: They are collectively known as Transducers.
6. Which of the following is a not a part of 5-tuple finite automata?
a) Input alphabet
b) Transition function
c) Initial State
d) Output Alphabet
View Answer
Answer: d
Explanation: A FA can be represented as FA= (Q, ?, d, q0, F) where Q=Finite Set of States, ?=Finite Input Alphabet, d=Transition Function, q0=Initial State, F=Final/Acceptance State).
7. If an Infinite language is passed to Machine M, the subsidiary which gives a finite solution to the infinite input tape is ______________
a) Compiler
b) Interpreter
c) Loader and Linkers
d) None of the mentioned
View Answer
Answer: a
Explanation: A Compiler is used to give a finite solution to an infinite phenomenon. Example of an infinite phenomenon is Language C, etc.
8. The maximum number of transition which can be performed over a state in a DFA?
?= {a, b, c}
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: c
Explanation: The maximum number of transitions which a DFA allows for a language is the number of elements the transitions constitute.
9. Finite automata requires minimum _______ number of stacks.
a) 1
b) 0
c) 2
d) None of the mentioned
View Answer
Explanation: Finite automata doesn’t require any stack operation .
10. Under which of the following operation, NFA is not closed?
a) Negation
b) Kleene
c) Concatenation
d) None of the mentioned
View Answer
Answer: d
Explanation: NFA is said to be closed under the following operations:
a) Union
b) Intersection
c) Concatenation
d) Kleene
e) Negation
11. Which of the following is an application of Finite Automaton?
a) Compiler Design
b) Grammar Parsers
c) Text Search
d) All of the mentioned
View Answer
Answer: d
Explanation: There are many applications of finite automata, mainly in the field of Compiler Design and Parsers and Search Engines.
12. e-transitions are
a) conditional
b) unconditional
c) input dependent
d) none of the mentioned
View Answer
Answer: b
Explanation: An epsilon move is a transition from one state to another that doesn’t require any specific condition.
13. State true or false:
Statement: Both NFA and e-NFA recognize exactly the same languages.
a) true
b) false
View Answer
Answer: a
Explanation: e-NFA do come up with a convenient feature but nothing new.They do not extend the class of languages that can be represented.
14. According to the given language, which among the following expressions does it corresponds to?
Language L={x?{0,1}|x is of length 4 or less}
a) (0+1+0+1+0+1+0+1)4
b) (0+1)4
c) (01)4
d) (0+1+e)4
View Answer
Answer: d
Explanation: The extended notation would be (0+1)4 but however, we may allow some or all the factors to be e. Thus e needs to be included in the given regular expression.
15. Arden’s theorem is true for:
a) More than one initial states
b) Null transitions
c) Non-null transitions
d) None of the mentioned
View Answer
Answer: c
Explanation: Arden’s theorem strictly assumes the following;
a) No null transitions in the transition diagrams
b) True for only single initial state | {"url":"https://studyhelpzone.com/automata-theory-multiple-choice-quaestion-answers-mcqs-set-22.html","timestamp":"2024-11-03T06:23:25Z","content_type":"text/html","content_length":"126536","record_id":"<urn:uuid:5f454e65-f473-4c9d-a8b2-16ccaa22e2ee>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00295.warc.gz"} |