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Life of Fred: Advanced Algebra Expanded Edition
The Learning Basket is an official reseller of Life of Fred books. Read about why we love these living Math series here.
The Human Face of Advanced Algebra
Every aspect of advanced algebra pops in our hero’s everyday life
• Romance! How to approach a good-looking passenger on the bus
• Cuisine! What pie you shouldn’t order…unless you’re rich
• Travel! Parts of Kansas you’ve never been to
• Fashion! Hairless anteriors and gold chains
• Music! Contains the piano music—newest hit from Rockin’ Rita
All fun! Just open and enjoy.
This book replaces both Life of Fred: Advanced Algebra and Fred’s Home Companion: Advanced Algebra. This book has all the problems completely worked out, which wasn’t true in the old books. It costs
less, too.
The author recommends taking Advanced Algebra before Geometry.
All of second-year high school algebra: Ratios, Median averages, Proportions, Dividing by zero, Cross-multiplying, Constants of proportionality, Inverse variation, Direct variation, Joint variation,
Weight varies directly as the cube of the height, Area varies directly as the square of the height, Volume varies directly as the cube of the height, Laws of exponents, Rationalizing the denominator,
Pythagorean theorem, Solving radical equations, Surface area of a cone, A Brief History of mathematics, Natural numbers, Whole numbers, Integers, Irrational numbers, Real numbers, Imaginary numbers,
Complex numbers, Velocity, Conversion factors, Unit analysis, Venn diagrams, Union of sets, Disjoint sets, Subsets, Intersection of sets, Significant digits in addition and multiplication, Scientific
notation, Exponential equations, Logarithms, Product, quotient, birdie, and power rules for logarithms, Using a scientific calculator, Finding antilogs, Three definitions of logarithms,
Change-of-base rule for logarithms, Graphing, Abscissa, Ordinate, Quadrants, Slope, Slope given two points, Slope-intercept and double-slope forms of the line, Slopes of perpendicular lines,
Multiplying binomials, Factoring binomials, Factoring a difference of squares, Factoring trinomials, Adding, subtracting, multiplying and dividing polynomial fractions, Complex fractions, Linear
equations, Quadratic equations, Radical equations, Systems of Equations, Inconsistent, dependent and independent equations, Graphing planes in three dimensions, Cramer’s rule, Determinants, Ellipses,
Circles, Reflective property of ellipses, Parabolas, Hyperbolas, Graphing inequalities, Definition of a function, Domain, Codomain, Range, Functional notation, One-to-one, Onto, Inverse functions,
Ordered pairs, Long division of polynomials, Partial fractions, Proofs by math induction, Linear Programming, Arithmetic progressions, and series, Adding and multiplying matrices, Geometric sequences
and progressions, Sigma notation, Fundamental principle of counting, Factorial, Permutations, Combinations, Binomial formula, Pascal’s Triangle.
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Lesson 1
Add and Subtract to Compare
Warm-up: Which One Doesn’t Belong: Compare Representations (10 minutes)
This warm-up prompts students to carefully analyze and compare features of different representations of two-digit numbers. When they share their comparisons, listen for the vocabulary they use to
talk about the characteristics of tape diagrams, bar graphs, and base-ten diagrams and provide them opportunities to clarify their meaning (MP6).
• Groups of 2
• Display the image.
• “Pick one that doesn’t belong. Be ready to share why it doesn’t belong.”
• 1 minute: quiet think time
• “Discuss your thinking with your partner.”
• 2–3 minutes: partner discussion
• Share and record responses.
Student Facing
Which one doesn’t belong?
Activity Synthesis
• “How does each representation show the difference between cloudy and sunny days?” (C maybe shows the difference with blocks. If the top train of blocks is sunny days, you can see there are more
sunny days. B shows it with a tape diagram, the part with the question mark shows the difference. D uses a bar graph. You can see sunny days has more than cloudy days and you could count the
number of spaces they are apart. A shows with blocks too, but they are in towers of ten and single cubes. You can see sunny days has ten more.)
Activity 1: Movie Snacks (15 minutes)
The purpose of this activity is for students to compare different methods for solving problems within 100 using data presented in a bar graph. Students may use whatever method makes the most sense to
them. The synthesis focuses on sharing multiple methods that students use to find the difference. Monitor for students who use methods that rely on using the bar graph to count on or count back and
those that use more abstract methods, such as adding or subtracting by place value.
For example, when combining categories, some students may choose to use the graph to count on. Other students may choose to combine tens and ones with or without drawing a base-ten diagram or other
Engagement: Provide Access by Recruiting Interest. Provide choice. Invite students to choose a strategy and tool that works for them. Encourage students to use that same strategy and tool for both
problems so they are not overwhelmed.
Supports accessibility for: Conceptual Processing, Organization, Attention
Required Preparation
• Create towers of 10 with the connecting cubes.
• Have single connecting cubes available.
• Groups of 2
• Display the bar graph.
• “What does this graph tell us?” (students’ favorite movie snacks, students picked their favorite movie snacks)
• 1 minute: quiet think time
• 1 minuter: partner discussion
• Share responses.
• Give students access to connecting cubes in towers of ten and singles.
• “Use the bar graph to answer the questions. Show your thinking using drawings, numbers, or words.”
• “You can use the connecting cubes or any of the other representations we saw in the warm-up to help you.”
• 8 minutes: independent work time
• “Now compare your methods with your partner. How are they similar or different?”
• 4 minutes: partner discussion
• As students work, monitor for students who:
□ use the bar graph to count on or count back
□ use the connecting cubes or base-ten drawings to show adding or subtracting tens with tens and ones with ones
Student Facing
Use the bar graph to answer the questions.
1. What is the total number of students that chose popcorn or pretzels? Show your thinking.
2. How many more students chose nachos than chose popcorn? Show your thinking.
Activity Synthesis
• Invite previously identified students to share the method they used to find how many more students chose nachos than chose popcorn.
• As needed, record student methods using equations.
• Consider asking:
□ “How are these methods the same? How are they different?”
□ “How does the method work? Why does each method find the same value?”
Activity 2: Build and Compare (20 minutes)
The purpose of this activity is for students to solve Compare problems within 100 using methods based on place value and the relationship between addition and subtraction. Connecting cubes are used
as a representation in this activity to support students in their transition from subtraction methods based on counting on or counting back by one to methods based on subtracting tens from tens and
ones from ones. Students build trains out of towers of 10 and single connecting cubes. Invite students to use the methods that make the most sense to them when they work to find the difference.
Monitor for students who use blocks or other representations to show adding or subtracting tens and tens and ones and ones to share in the synthesis.
This activity uses MLR7 Compare and Connect. Advances: representing, conversing
Required Preparation
• Create towers of 10 with the connecting cubes.
• Have single connecting cubes available.
• Groups of 2
• Assign Partner A and Partner B.
• Give students access to towers of ten and loose connecting cubes.
• Display the image of the cubes.
• “What do you notice? What do you wonder?” (Lin has more cubes. They have 40 cubes all together. Lin has ten more cubes.)
• Monitor for students who notice the groups of ten cubes and use this structure to find the total number of cubes or the difference.
• 30 seconds: quiet think time
• Share responses.
• “You and your partner will each build a train with connecting cubes. Then, answer the questions about your trains.”
• “Show your thinking using drawings, numbers, or words.”
• 8 minutes: partner work time
• Monitor for students who:
□ count on or combine tens and ones to find the difference
□ count back or separate tens and ones to find the difference
Student Facing
1. Lin and Clare used cubes to make trains. What do you notice? What do you wonder?
2. Make trains with cubes.
│ partner │number of cubes │
│Partner A │46 │
│Partner B │22 │
3. Find the total number of cubes you and your partner used. Show your thinking.
4. Find the difference between the number of cubes you and your partner used. Show your thinking.
Advancing Student Thinking
If students build their numbers out of single cubes without using towers of 10, consider asking:
• “How did you choose which blocks to use when you built your number?”
• “How could you use the towers of 10 to build your number?”
Activity Synthesis
MLR7 Compare and Connect
• “Create a visual display that shows your thinking about the difference between the number of cubes you and your partner used. You may want to include details such as diagrams, drawings, and
labels to help others understand your thinking.”
• 5–7 minutes: gallery walk
• Invite previously identified students to share their methods for finding the difference using cubes.
• “What is the same and what is different between the way these two groups found the difference?” (Both groups found the same value. One group shows adding on tens and ones. The other group shows
taking away tens and ones.)
• 30 seconds quiet think time
• 1 minute: partner discussion
• If time, consider asking:
□ “What other methods did you see groups use? How are they the same and how are they different from these two methods?” (Other groups added on and subtracted to, but they showed it with
different diagrams and drawings. Some used only equations. Some showed counting by ones.)
Lesson Synthesis
Display: \(46 - 22 = {?}\)
“This equation shows one way to represent the difference between your blocks.”
“What are the different ways we found the difference?” (counting on, counting back, taking away blocks, adding blocks)
Display: \(22 + {?} = 46\)
“Why can we use methods that show taking away and use methods that add to find the difference?” (because \(46 - 22 = {?}\) is like \(22 + {?} = 46\). When you subtract, you can think about taking
away or you can think about what addend is missing.)
Cool-down: Compare the Trains (5 minutes)
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Math, Grade 6, Equations and Inequalities, Representing An Inequality On A Number Line
Material Type:
Lesson Plan
Middle School
Media Formats:
Inequalities and Number Lines
Match Inequalities to Graphs
Representing An Inequality On A Number Line
Lesson Overview
Students represent inequalities on a number line, find at least one value that makes the inequality true, and write the inequality using words.
SWD:When calling on students, be sure to call on ELLs and to encourage them to actively participate. Understand that their pace might be slower or they might be shy or more reluctant to volunteer due
to their weaker command of the language.
SWD:Thinking aloud is one strategy for making learning visible. When teachers think aloud, they are externalizing their internal thought processes. Doing so may provide students with insights into
mathematical thinking and ways of tackling problems. It also helps to model accurate mathematical language.
Key Concepts
Inequalities, like equations, have solutions. An arrow on the number line—pointing to the right for greater values and to the left for lesser values—can be used to show that there are infinitely many
solutions to an inequality.
The solutions to x < a are represented on the number line by an arrow pointing to the left from an open circle at a.
Example: x < 2
The solutions to x > a are represented on the number line with an arrow pointing to the right from an open circle at a.
Example: x > 2
The solutions to x ≤ a are represented on the number line with an arrow pointing to the left from a closed circle at a.
Example: x ≤ 2
The solutions to x ≥ a are represented on the number line with an arrow pointing to the right from a closed circle at a.
Example: x ≥ 2
Goals and Learning Objectives
• Represent an inequality on a number line and using words.
• Understand that inequalities have infinitely many solutions.
Match Inequalities to Graphs
Lesson Guide
Have students work in pairs to match the inequalities with their representations on the number line and to write the inequality in words.
Share the inequalities from the Opening and the correct corresponding number line graphs with the students so they can easily reference them in the future.
ELL: Learning new ways of graphing inequalities can be a challenge for ELLs. Preview and/or pre-teach the new concepts in this lesson. Promote connections to students' background knowledge and to
previous lessons in this unit.
Students will use reasoning to match the number lines with the correct inequalities. They will need to reason that the open circle means that the number is not included in the inequality and the
closed circle indicates that it is included. Do not provide this information for the students. If the class does not all agree on this convention initially, allow students to defend their positions
until all see why this convention makes sense.
• x < 2: B. "x is less than 2."
• x > 2: C. "x is greater than 2."
• x ≥ 2: D. "x is greater than or equal to 2."
• x ≤ : A. "x is less than or equal to 2."
Match Inequalities to Graphs
• Match each inequality to a number line.
• Write the inequality in words in the space provided; for example, "x is less than 5."
INTERACTIVE: Match Inequalities to Graphs
Math Mission
Lesson Guide
Discuss the Math Mission. Students will represent inequalities on a number line and find at least one value that makes the inequality true.
Represent inequalities on a number line and find at least one value that makes the inequality true.
Inequalities and Number Lines
Lesson Guide
Have students work in pairs on the problems in Tasks 3 and 4. Identify any student misconceptions to address in Ways of Thinking.
SWD: Circulate through your classroom before your students share. Help students to identify ideas from their notes and thoughts that are appropriate to contribute during this portion of the Ways of
Thinking discussion.
Mathematical Practices
Mathematical Practice 3: Construct viable arguments and critique the reasoning of others.
Listen for student conversations about whether rules or conventions apply the same way to negative values as they do to positive values.
Mathematical Practice 6: Attend to precision.
Listen for student conversations about whether the circles should be open or closed.
Listen for student conversations about which direction the arrows should face.
Student does not understand how to use the interactive.
• Ask your partner to demonstrate how to use the interactive.
Student struggles with negative values and the direction of arrows.
• Is the number to the left greater than or less than the number to its right?
Student is confused about the conventions for using open and closed circles on the number line.
• Look back at the answers to the Opening.
• How are the math conventions connected to language: greater than or less than, and greater than or equal to or less than or equal to?
Student does not know how to find one value that makes the inequality true.
• Look at the open (or closed) circle. Does that number make the inequality true?
• Will the numbers that make the inequality true be to the left or right of the open (or closed) circle?
Work Time
Inequalities and Number Lines
For each inequality do the following.
• Represent the inequality on a number line using the number line interactive.
• Write one value that makes the inequality true.
• Write the inequality using words.
1. x ≥ –2
2. x < 4
3. 12 ≥ x
4. x ≤ −3.5
5. x < 1.5
INTERACTIVE: Inequalities and Number Lines
• Where is the number located on the number line?
• Should the starting point be a closed circle or an open circle?
• What is one value that makes the inequality true? Is that value shaded on the number line?
Write About the Number Line
Lesson Guide
Watch for students who:
• Cannot write a real-world situation to represent the number line.
Mathematical Practices
Mathematical Practice 1: Make sense of problems and persevere in solving them.
Look for a variety of real-world scenarios.
Mathematical Practice 3: Construct viable arguments and critique the reasoning of others.
Listen for student conversations about whether rules or conventions apply the same way to negative values as they do to positive values.
Possible Answers:
• Children must be at least 5 years old to attend the story hour.
• The minimum price of a T-shirt in that store is $5.
• That box of books weighs 5 lb or more.
Work Time
Write About the Number Line
• Write about a real-world situation that could be represented by this number line.
Think about situations involving age, money, or weight.
Prepare a Presentation
Preparing for Ways of Thinking
Watch for students who:
• Incorrectly use the open and closed circles on the number line diagram to represent the inequality statements.
• Show the arrow in the wrong direction on their number line diagrams.
• Have trouble working with negative values.
• Have trouble when the variable is on the right side of the inequality symbol.
• Cannot identify a value that makes the inequality true.
• Cannot accurately write the inequality using words.
• Cannot write a real-world situation to represent the number line diagram.
Step 1: Work Time
Prepare a Presentation
• Explain how you represented an inequality on a number line.
• Support your explanation with your work.
Challenge Problem
Possible Answers:
• I do not agree; -2 is not less than -5. I know that -2 is greater than -5 because -2 is to the right of -5 on the number line.
Step 2: Work Time
Challenge Problem
Denzel says that because 2<5, that means −2<−5.
• Do you agree? Explain why or why not.
Make Connections
Lesson Guide
Highlight the fact that the arrows on the number lines in the first problem represent solutions to the inequality statements, and facilitate recognition that there are infinitely many solutions to
each inequality.
Ask students:
• How many solutions do these statements have?
• Why do you use a line with an arrow to represent these solutions?
Draw on ideas and diagrams from one or two students who worked on the Challenge Problem. Focus on 2 < 5 and −2 > −5 to frame the first part of the Ways of Thinking discussion.
Ask students:
• How does the number line help you understand what is going on?
• Which is greater, –10 or –10.5? Represent these numbers on the number line. Then compare your results with the inequality statement 10 < 10.5.
For the rest of the discussion, focus on situations to match the number line, allowing as many students as possible to share their real-world situations.
ELL: As with other oral instructions, ensure that the pace of your speech is appropriate for ELLs. Pause frequently to allow students to pose questions. Alternatively, monitor students' understanding
by asking questions as your explanation unfolds.
Performance Task
Ways of Thinking: Make Connections
Take notes about your classmates’ representations of inequalities on a number line.
As your classmates present, ask questions such as:
• Why is the circle filled in?
• Why is the circle open?
• Why does the line go in that direction?
• Explain why –4 is less than –3.5.
• What is the greatest number that x could be?
Lesson Guide
Have pairs quietly discuss the information about inequalities. Have them give additional examples of real-world situations for each type of inequality.
As student pairs work together, listen for students who may still have misconceptions so you can address them in the class discussion.
After a few minutes, discuss the Summary as a class. Review the following points.
Inequalities, like equations, have solutions.
An arrow on the number line—pointing to the right for greater values and to the left for lesser values—can be used to show that there are infinitely many solutions to an inequality.
The solutions to x < a[x] < a (x[x] is less than a[a]) are represented on the number line by an arrow pointing to the left from an open circle at a[a].
• The solutions to x > a[x] > a (x[x] is greater than a[a]) are represented on the number line with an arrow pointing to the right from an open circle at a[a].
• The solutions to x ≤ a[x] ≤ a (x[x] is less than or equal to a[a]) are represented on the number line with an arrow pointing to the left from a closed circle at a[a].
• The solutions to x ≥ a[x ]≥ a (x[x] is greater than or equal to a[a]) are represented on the number line with an arrow pointing to the right from a closed circle at a[a].
Formative Assessment
Summary of the Math: Inequalities
Read and Discuss
• An inequality is a relationship that holds between two values that are different.
• a < b means that a is less than b (or, equivalently, that b is greater than a).
• a > b means that a is greater than b (or, equivalently, that b is less than a).
• a ≤ b means that a is less than or equal to b (or, equivalently, that b is greater than or equal to a).
• a ≥ b means that a is greater than or equal to b (or, equivalently, that b is less than or equal to a).
Can you:
• Explain the meaning of each inequality symbol?
• Show how each type of inequality is represented on a number line?
Solve and Use Equations
Lesson Guide
This task allows you to assess students’ work and determine what difficulties they are having. The results of the Self Check will help you determine which students should work on the Gallery and
which students would benefit from review before the assessment. Have students work on the Self Check individually.
Have students submit their work to you. Make notes on what their work reveals about their current levels of understanding and their different problem-solving approaches.
Do not score students’ work. Share with each student the most appropriate Interventions to guide their thought process. Also note students with a particular issue so that you can work with them in
the Putting It Together lesson that follows.
SWD: Post the Interventions in the classroom for students to use as a resource as they work. Create and provide an enhanced version of the Interventions with embedded text structures (labels,
highlights, words in bold) to cue students to pay closer attention to particular terms.
Student uses the wrong property or identifies the property incorrectly.
• Use the addition property of equality for equations like x + 4 = 10.
• Use the multiplication property of equality for equations like 2x = 10 or 12x = 9 or for solving proportions.
Student solves the equation incorrectly.
• Did you use the correct operation to isolate x?
• Did you substitute the solution into the equation to check it?
Student does not include any word problems.
• Try these starters:
□ A shirt costs $5 more than a …
□ Four books cost $22 …
□ A purple paint is mixed in the ratio of …
Student's word problem does not match the equation given.
• Work backwards. Read the word problem you wrote. Write an equation that represents it. Does it match the equation you were given?
1. $\begin{array}{c}\\ x+5-5=17-5& \text{Addition property of equality}\\ x=12& \end{array}$
Word problems will vary. Possible answer: A shirt costs $5 more than a pair of shorts. The shirt costs $17. How much does the pair of shorts cost? Answer: The pair of shorts costs $12.
2. $\begin{array}{cc}\frac{1}{4}\cdot 4x=\frac{1}{4}\cdot 22& \text{Multiplication property of equality}\\ x=5\frac{1}{2}& \end{array}$
Word problems will vary. Possible answer: Four books cost $22. Each costs the same amount. How much does one book cost? Answer: One book costs $5.50.
3. $\begin{array}{c}\\ \frac{x}{6}\cdot 6=\frac{5}{8}\cdot 6& \text{Multiplication property of equality}\\ x=3\frac{3}{4}& \end{array}$
Word problems will vary. Possible answer: A purple paint is mixed in the ratio of 5 parts blue to 8 parts red. How much blue paint should be mixed with 6 cups of red paint to get the same shade
of purple? Answer: You should mix 3$\frac{3}{4}$ cups of blue paint.
Formative Assessment
Solve and Use Equations
Complete this Self Check by yourself.
For each equation do the following.
• Solve the equation. Name the property of equality you use and show each step.
• Write and solve a word problem that each equation could represent.
1. x + 5 = 17
2. 4x = 22
3. $\frac{x}{6}=\frac{5}{8}$
Reflect On Your Work
Lesson Guide
Have each student write a quick reflection before the end of the class. Review the reflections to learn where students see inequalities in real-world situations outside of the classroom.
Work Time
Reflect On Your Work
Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.
One place outside of the classroom where I have seen inequalities is …
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Start Here with TI84programs: Step 2
Start Here with TI84 Programs 2024: STEP 2
Step 2: Downloading Your FREE Program to Test TI84programs.com
You can click on this link to download your free copy of quad program. Even if you don’t care at all for quadratic equations, it’ll prove to you that you can download programs from TI84programs.com
and use them. More importantly, you’ll see what they can do. Some programs show only basic formulas, such as area, perimeter or volume. Other formula programs can display equations as complicated as
trigonometric identities or derivative and integral formulas. Another type of program can actually solve basic algebraic equations along with other applications. More on that later at
QUAD file zipped and unzipped
Once you download the program, which you can do so as a Zip file, you’ll have to unzip or decompress it. You’ll see it as QUAD.8xp.zip and you want it as QUAD.8xp
From your desktop, make sure you see QUAD as a .8xp file. Then, with your TI Connect CE application open, click and drag the QUAD.8xp file onto the calculator. You’ll see this window open up asking
if it’s ok. Please reassure TI Connect CE that it’s okay.
Then, on your calculator, you can press the Program key. It has prgm on it and is located on the upper middle part of your ti84 calculator keypad. Basically, it’s 3 keys above the 8 key.
Using Your QUADFORM Program
Once you press PRGM, you might jump to your programs. But, with the more recent operating systems from TI, you’ll be asked TI-Basic or Python App. For now, let’s choose TI-Basic.
Then you’ll see a list of the programs you have on your connected calculator. You can either scroll to the name of the program you want or press the corresponding number or letter immediately to the
left of the program name.
Wait, you’ll also have to press Enter to run it. The first keystroke was to select it. The ENTER keystroke is to run it.
Now, you’ll see this, the standard form of a quadratic equation. Just to make sure your program works properly, you can enter 1 for A, 5 for B, and 6 for C. That’s equivalent to 1x^2 + 5x + 6 = 0.
The program will provide you with the 2 answers of x1 as -2 and x2as -3. The bonus is the answers in the format of the discriminant form: -5 plus or minus -1 divided by 2.
Even if you don’t care a wit about the quadratic formula, at least you know now that you can download functional programs from TI84programs.com
Many programs will show only formulas, while others will actually solve problems. Both kinds are available at TI84programs.com We have a growing library of programs available on site. Also, you can
check our YouTube channel for updates. Or, if you’re more interested in social media, we have some channels there as well.
Thank you for following along with this process. After this, downloading and using ti84 programs from ti84programs.com will be a much simpler and faster process for you.
Good luck with all your math endeavors!
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Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.1 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1
Question 1.
Find the equations of tangents and normals to the curve at the point on it.
(i) y = x^2 + 2e^x + 2 at (0, 4)
(ii) x^3 + y^3 – 9xy = 0 at (2, 4)
x^3 + y^3 – 9xy = 0
Differentiating both sides w.r.t. x, we get
Hence, the equations of tangent and normal are 4x – 5y + 12 = 0 and 5x + 4y – 26 = 0 respectively.
(iii) x^2 – √3xy + 2y^2 = 5 at (√3, 2)
x^2 – √3xy + 2y^2 = 5
Differentiating both sides w.r.t. x, we get
the slope of normal at (√3, 2) does not exist.
normal is parallel to Y-axis.
equation of the normal is of the form x = k
Since, it passes through the point (√3, 2), k = √3
equation of the normal is x = √3.
Hence, the equations of tangent and normal are y = 2 and x = √3 respectively.
(iv) 2xy + π sin y = 2π at (1, \(\frac{\pi}{2}\))
2xy + π sin y = 2π
Differentiating both sides w.r.t. x, we get
Hence, the equations of tangent and normal are πx + 2y – 2π = 0 and 4x – 2πy + π^2 – 4 = 0 respectively.
(v) x sin 2y = y cos 2x at (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))
x sin 2y = y cos 2x
Differentiating both sides w.r.t. x, we get
Hence, the equations of the tangent and normal are 2x – y = 0 and 4x + 8y – 5π = 0 respectively.
(vi) x = sin θ and y = cos 2θ at θ = \(\frac{\pi}{6}\)
When θ = \(\frac{\pi}{6}\), x = sin\(\frac{\pi}{6}\) and y = cos\(\frac{\pi}{3}\)
∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{1}{2}\))
Now, x = sin θ, y = cos 2θ
Differentiating x and y w.r.t. θ, we get
2y – 1 = x – \(\frac{1}{2}\)
4y – 2 = 2x – 1
2x – 4y + 1 = 0
Hence, equations of the tangent and normal are 4x + 2y – 3 = 0 and 2x – 4y + 1 = 0 respectively.
(vii) x = √t, y = t – \(\frac{1}{\sqrt{t}}\), at t = 4.
When t = 4, x = √4 and y = 4 – \(\frac{1}{\sqrt{4}}\)
∴ x = 2 and y = 4 – \(\frac{1}{2}\) = \(\frac{7}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (2, \(\frac{7}{2}\)).
Now, x = √t, y = t – \(\frac{1}{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get
Hence, the equations of tangent and normal are 17x – 4y – 20 = 0 and 8x + 34y – 135 = 0 respectively.
Question 2.
Find the point of the curve y = \(\sqrt{x-3}\) where the tangent is perpendicular to the line 6x + 3y – 5 = 0.
Let the required point on the curve y = \(\sqrt{x-3}\) be P(x[1], y[1]).
Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get
Hence, the required points are (4, 1) and (4, -1).
Question 3.
Find the points on the curve y = x^3 – 2x^2 – x where the tangents are parallel to 3x – y + 1 = 0.
Let the required point on the curve y = x^3 – 2x^2 – x be P(x[1], y[1]).
Question 4.
Find the equations of the tangents to the curve x^2 + y^2 – 2x – 4y + 1 = 0 which are parallel to the X-axis.
Let P (x[1], y[1]) be the point on the curve x^2 + y^2 – 2x – 4y + 1 = 0 where the tangent is parallel to X-axis.
Differentiating x^2 + y^2 – 2x – 4y + 1 = 0 w.r.t. x, we get
the coordinates of the points are (1, 0) or (1, 4)
Since the tangents are parallel to X-axis, their equations are of the form y = k
If it passes through the point (1, 0), k = 0, and if it passes through the point (1, 4), k = 4
Hence, the equations of the tangents are y = 0 and y = 4.
Question 5.
Find the equations of the normals to the curve 3x^2 – y^2 = 8, which are parallel to the line x + 3y = 4.
Let P(x[1], y[1]) be the foot of the required normal to the curve 3x^2 – y^2 = 8.
Differentiating 3x^2 – y^2 = 8 w.r.t. x, we get
Hence, the equations of the normals are x + 3y – 8 = 0 and x + 3y + 8 = 0.
Question 6.
If the line y = 4x – 5 touches the curve y^2 = ax^3 + b at the point (2, 3), find a and b.
y^2 = ax^3 + b
Differentiating both sides w.r.t. x, we get
= slope of the tangent at (2, 3)
Since, the line y = 4x – 5 touches the curve at the point (2, 3), slope of the tangent at (2, 3) is 4.
2a = 4 ⇒ a = 2
Since (2, 3) lies on the curve y^2 = ax^3 + b
(3)^2 = a(2)^3 + b
9 = 8a + b
9 = 8(2) + b …… [∵ a = 2]
b = -7
Hence, a = 2 and b = -7.
Question 7.
A particle moves along the curve 6y = x^3 + 2. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Let P(x[1], y[1]) be the point on the curve 6y = x^3 + 2 whose y-coordinate is changing 8 times as fast as the x-coordinate.
Question 8.
A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing, when its radius is 5 cm?
Let r be the radius and S be the surface area of the soap bubble at any time t.
Then S = 4πr^2
Differentiating w.r.t. t, we get
Hence, the surface area of the soap bubble is increasing at the rate of 0.87c cm^2 / sec.
Question 9.
The surface area of a spherical balloon is increasing at the rate of 2 cm^2/sec. At what rate is the volume of the balloon is increasing, when the radius of the balloon is 6 cm?
Let r be the radius, S be the surface area and V be the volume of the spherical balloon at any time t.
Then S = 4πr^2 and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get
Hence, the volume of the spherical balloon is increasing at the rate of 6 cm^3 / sec.
Question 10.
If each side of an equilateral triangle increases at the rate of √2 cm/sec, find the rate of increase of its area when its side of length is 3 cm.
If x cm is the side of the equilateral triangle and A is its area, then \(A=\frac{\sqrt{3}}{4} x^{2}\)
Differentiating w.r.t. f, we get
Hence, rate of increase of the area of equilateral triangle = \(\frac{3 \sqrt{6}}{2}\) cm^2 / sec.
Question 11.
The volume of a sphere increases at the rate of 20 cm^3/sec. Find the rate of change of its surface area, when its radius is 5 cm.
Let r be the radius, S be the surface area and V be the volume of the sphere at any time t.
Then S = 4πr^2 and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get
Hence, the surface area of the sphere is changing at the rate of 8 cm^2/sec.
Question 12.
The edge of a cube is decreasing at the rate of 0.6 cm/sec. Find the rate at which its volume is decreasing, when the edge of the cube is 2 cm.
Let x be the edge of the cube and V be its volume at any time t.
Then V = x^3
Differentiating both sides w.r.t. t, we get
Hence, the volume of the cube is decreasing at the rate of 7.2 cm^3/sec.
Question 13.
A man of height 2 meters walks at a uniform speed of 6 km/hr away from a lamp post of 6 meters high. Find the rate at which the length of the shadow is increasing.
Let OA be the lamp post, MN the man, MB = x, his shadow, and OM = y, the distance of the man from the lamp post at time t.
Then \(\frac{d y}{d t}\) = 6 km/hr is the rate at which the man is moving at away from the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is increasing.
From the figure,
6x = 2x + 2y
4x = 2y
x = \(\frac{1}{2}\) y
\(\frac{d x}{d t}=\frac{1}{2} \frac{d y}{d t}=\frac{1}{2} \times 6=3 \mathrm{~km} / \mathrm{hr}\)
Hence, the length of the shadow is increasing at the rate of 3 km/hr.
Question 14.
A man of height 1.5 meters walks towards a lamp post of height 4.5 meters, at the rate of (\(\frac{3}{4}\)) meter/sec.
Find the rate at which
(i) his shadow is shortening
(ii) the tip of the shadow is moving.
Let OA be the lamp post, MN the man, MB = x his shadow and OM = y the distance of the man from lamp post at time t.
Then \(\frac{d y}{d t}=\frac{3}{4}\) is the rate at which the man is moving towards the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is shortening.
B is the tip of the shadow and it is at a distance of x + y from the post.
\(\frac{d}{d t}(x+y)=\frac{d x}{d t}+\frac{d y}{d t}\) is the rate at which the tip of the shadow is moving.
From the figure,
45x = 15x + 15y
30x = 15y
x = \(\frac{1}{2}\)y
\(\frac{d x}{d t}=\frac{1}{2} \cdot \frac{d y}{d t}=\frac{1}{2}\left(\frac{3}{4}\right)=\left(\frac{3}{8}\right) \text { metre/sec }\)
and \(\frac{d x}{d t}+\frac{d y}{d t}=\frac{3}{8}+\frac{3}{4}=\left(\frac{9}{8}\right) \text { metres } / \mathrm{sec}\)
Hence (i) the shadow is shortening at the rate of (\(\frac{3}{8}\)) metre/sec, and
(ii) the tip of shadow is moving at the rate of (\(\frac{9}{8}\)) metres/sec.
Question 15.
A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the
ladder is sliding down the wall, when the bottom is 6 metres away from the wall.
Let AB be the ladder, where AB = 10 metres.
Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.
Since, OAB is a right angled triangle, by Pythagoras’ theorem
x^2 + y^2 = 10^2 i.e. y^2 = 100 – x^2
Differentiating w.r.t. t, we get
2y \(\frac{d y}{d t}\) = 0 – 2x \(\frac{d x}{d t}\)
∴ \(\frac{d y}{d t}=-\frac{x}{y} \cdot \frac{d x}{d t}\) ……..(1)
Now, \(\frac{d x}{d t}\) = 1.2 metres/sec is the rate at which the bottom at of the ladder is pulled horizontally and \(\frac{d y}{d t}\) is the rate at which the top of ladder B is sliding.
When x = 6, y^2 = 100 – 36 = 64
y = 8
(1) gives \(\frac{d y}{d t}=-\frac{6}{8}(1.2)=-\frac{6}{8} \times \frac{12}{10}\)
Hence, the top of the ladder is sliding down the wall, at the rate of 0.9 metre/sec.
Question 16.
If water is poured into an inverted hollow cone whose semi-vertical angle is 30° so that its depth (measured along the axis) increases at the rate of 1 cm/sec. Find the rate at which the volume of
water increases when the depth is 2 cm.
Let r be the radius, h be the height, θ be the semi-vertical angle and V be the volume of the water at any time t.
Hence, the volume of water is increasing at the rate of \(\left(\frac{4 \pi}{3}\right)\) cm3/sec.
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Visions of Exploration Sponsored by the Air Force Association
Visions of Exploration STEM Lessons
Thank you for visiting AFA’s Visions of Exploration program page. If you aren’t a participating teacher, feel free to use these free lessons and learn more about what we’re doing in schools across
the country.
If you’d like to participate in this program, contact your local AFA chapter.
Exercise converted to electricity
While most of today’s discussion around green energy production only includes the harnessing of macro environmental
sources (wind, solar), there has begun to emerge a new breed of entrepreneur whose more micro approach to the issue
is making a difference to society in general and to one’s bottom line in particular. Such is the case with producing real,
usable electricity by the retrofitting of thousands of exercise bikes and elliptical machines found in health clubs and
businesses around the country.
What are the large-scale implications for such a creative venture, and just how valid are claims made by the two retrofit company
spokespeople quoted in the “Exercise Converted to Electricity” article? What is the rate of those bike and
elliptical conversions, and how much electricity can actually be produced in this way? In the exercises below, you will
perform scale conversions and related computations to access the information necessary to address some questions
regarding economy and ecology.
Materials needed:
Paper and pencil
Calculator (for checking purposes only)
Economy (45 minutes)
Speaking on behalf of The Green Revolution, a Connecticut-based retrofitting company, was Jay Whelan, Chief Economic
1. Mr. Whelan quoted retrofit-sales numbers for his company since April 2009. Assuming continuous sales up
to and including the month this article appeared in USA TODAY (August 2010), how many months of retrofit
sales are being measured?
2. If The Green Revolution has retrofitted 1,000 bikes at 60 gyms, what is the average number of bikes retrofitted
per gym (rounded to the nearest whole number)?
– To the nearest whole number, what is the average number of bikes The Green Revolution has retrofitted
per month?
– To the nearest hundredth, what is the average number of bikes The Green Revolution has retrofitted
per gym per month?
The other retrofitting company cited in the article was the Florida-based firm ReRev, whose spokesperson is Beth Bennion,
Marketing Director.
3. Ms. Bennion quoted retrofit-sales numbers for her company since June 2008. Assuming continuous sales up
to and including the month this article appeared in USAToday (August 2010), how many months of retrofit
sales are being measured?
4. If ReRev has retrofitted 300 bikes at 23 gyms, what is the average number of bikes retrofitted per gym
(rounded to the nearest whole number)?
– To the nearest whole number, what is the average number of bikes ReRev has retrofitted per month?
– To the nearest hundredth, what is the average number of bikes ReRev has retrofitted per gym per
5. To the nearest whole number, what is the per-gym-per-month ratio of The Green Revolution’s sales to ReRev’s
sales? Explain how you arrived at your answer.
Ecology (45 minutes)
A watt is a measure of the rate in which energy is converted and is equal to 1 joule of energy transferred per second.
When 1,000 watts of electricity are used in one hour (= 1,000 watt-hours), it is more common to call it a 1 kilowatthour
(= 1 kWh).
1. If the average house uses roughly 10 million watt-hours per year, how many kWh would that be?
– If the average house uses 28 kWh per day, how many watt-hours would that be?
2. “(Mr.) Whelan estimates a spinning class of 20 people over a year could light 72 homes for a month.” If a
20-person cycling class can generate anywhere from 2.5 to 3 kWh, is Mr. Whelan’s statement accurate?
Explain your reasoning.
3. “ReRev says a 30-minute workout on one of its ellipticals generates about 50 watts, enough to run a laptop
for an hour…” At that rate of generating electricity, if 4 retrofitted ellipticals are pedaled for 45 minutes
each, how long could 3 laptops stay charged?
4. Suppose we own a health club and retrofit one of our ellipticals from ReRev at a cost of $1,000.
– If the national average price of electricity is $0.15 per kWh, how many 30-minute workouts would
there need to be before we recoup our $1,000 investment?
– Pedaling non-stop (24 hrs./day, 7 days/wk.), about how many years would that take?
5. During one particularly hot week in the summer, we leave our air conditioner on 24/7. It takes roughly 3
kWh to run our air conditioner. For the amount of energy used by the air conditioner for a
week, how many days (at 28kWh per day) would we have been able to supply electricity to
our house with the air conditioner turned off?
Exercise converted to electricity – Teacher’s guide
Numbers, operations, rounding, fractions, decimals, scale conversion, dimensional analysis, scientific notation, ratio,
proportion, equations, real-world problem solving
Students will:
– Solve problems involving the multiplication & division of whole numbers
– Round whole numbers & decimals to a predetermined place value
– Multiply & divide both fractions and decimals to solve problems
– Convert between any two number representations (fraction, decimal, percent) without using a calculator
– Interpret & use ratios to show the relative sizes of two quantities
– Understand proportions and use them to solve problems
– Write & solve one-step linear equations with one variable
– Read, write, compare, & solve whole-number notation problems
Prerequisite skills:
Of the three prerequisites, two are mathematical and can be troublesome for some students- scale conversion and dimensional
analysis. The third, reading for information, is a specialized skill that requires practice and support across
all grade levels.
Numbers & Operations – Base Ten
– Perform operations with multi-digit whole numbers and with decimals to hundredths.
Numbers & Operations- Fractions:
– Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
Measurement & Data:
– Convert like measurement units within a given measurement system.
Ratios & Proportional Relationships:
– Understand ratio concepts and use ratio reasoning to solve problems.
– Analyze proportional relationships and use them to solve real-world and mathematical problems.
The Number System:
– Compute fluently with multi-digit numbers and find common factors and multiples.
– Apply and extend previous understandings of operations with fractions to add, subtract, multiply and divide
rational numbers.
Expressions & Equations:
– Work with radicals, integers and exponents.
Time to complete
1 block period or 2 traditional (45-minute) periods
Teaching suggestions
The first 45-minute segment (Economy) might best be accomplished by pairing students and walking them through
each of the five questions. The second 45-minute segment (Ecology) is tailor-made for group work, and time for these
five problems could be expanded to accommodate presentations of the various solution strategies used.
Answer key: Economy
1. 17
2. 17, 59, 0.98
3. 27
4. 13, 11, 0.48
5. about 2:1
Answer key: Ecology
1. 10,000 kWh/year, 28,000 watt-hours/day
2. inaccurate
3. 2 hours
4. 133,333 workouts, about 7.6 years
5. 18 days
Visions of Exploration lessons are designed to enhance students reading, writing and comprehension skills, while introducing them to topics in science, technology, engineering and math. Each contains
a recent article found in USA TODAY along with discussion questions and an activity. The lessons prepare students for the world beyond the classroom and inspire them to become explorers.
I love being a part of this program and appreciate the fact that it is offered. So many of my kids have no news resources other than USA TODAY on Wednesdays, and it gives us a chance to discuss them.
I not only use it for improving reading and comprehension but also to give my students an opportunity to express their opinions freely. I get to know them better this way.
-Paul Wojcik, math teacher Millennium Middle School Sanford, Florida
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AIPS HELP file
AIPS HELP file for ROTMEAS in 31DEC24
As of Tue Nov 12 10:02:47 2024
Type: Adverb (real)
Use: Used to tell a task a rotation measure in radians per meter
Null value: 0 -> no rotation measure
RLDIF....Determines Right minus Left phase difference, corrects
cal files. ROTMEAS applies to the first calibration
SPLIT....Converts multi-source to single-source UV files with
calibration. It ROTMEAS is not zero, the output is forced
to IQUV or RR/LL/RL/LR Stokes and Q and U are corrected by
the rotation measure into the values they would have at 0
Qout = Uin sin(2D) + Qin cos(2D)
Uout = Uin cos(2D) - Qin sin(2D)
RLout = RLin (cos(2D) - j sin(2D))
LRout = LRin (cos(2D) + j sin(2D))
where j = sqrt(-1) and D = ROTMEAS * Lambda^2.
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Babylonian Journal of MathematicsSubstitutional Based Gauss-Seidel Method For Solving System of Linear Algebraic EquationsModified Technique to Solve Degeneracy in Linear Programming ProblemsOn commutativity of alternative rings with [xy^n x±yx^n y,x]=0 Some Results on Commutativity for Alternative Rings With 2, 3-Torsion Free.A Survey of MCDM-Based Software Engineering MethodUTLIZATION OF GENERALIZED HEAT FLUX MODEL ON THERMAL TRANSPORT OF POWELL-EYRING MODEL VIA OHAM WITH HEAT GENEARTION ASPECTSBio-Convective Thermally Radiated Casson Fluid Model PDEs Past over a Darcy-Forchheimer Porous Stretched Sheet via OHAM StrategyRestricted visible submodules and fully restricted visible module Numerical Solution of Burgers’-Type Equations Using Modified Variational Iteration Algorithm-I
https://mesopotamian.press/journals/index.php/BJM <p style="text-align: justify;">The Babylonian Journal of Mathematics (BJM) E-ISSN: 3006-113X, serves as a testament to ancient mathematical
heritage. Embracing the legacy of Babylonian mathematical achievements, this journal invites global contributions, aiming to extend and honor the rich mathematical traditions of the Babylonian
civilization through modern research and discoveries.</p> Mesopotamian Academic Press en-US Babylonian Journal of Mathematics 3006-113X https://mesopotamian.press/journals/index.php/BJM/article/view/
275 <p>In this research paper a new modification of Gauss-Seidel method has been presented for solving the system of linear algebraic equations. The systems of linear algebraic equations have an
important role in the field of science and engineering. This modification has been developed by using the procedure of Gauss-Seidel method and the concept of substitution techniques. Developed
modification of Gauss-Seidel method is a fast convergent as compared to Gauss Jacobi’s method, Gauss-Seidel method and successive over-relaxation (SOR) method. It works on the diagonally dominant as
well as positive definite symmetric systems of linear algebraic equations. Its solution has been compared with the Gauss Jacobi’s method, Gauss-Seidel method and Successive over-Relaxation method by
taking different systems of linear algebraic equations and found that, it was reducing to the number of iterations and errors in each problem.</p> Tuljaram Meghwar Sher khan Awan Muhammad Tariq
Muhammad Suleman Asif Ali Shaikh Copyright (c) 2024 Babylonian Journal of Mathematics 2024-01-10 2024-01-10 2024 1 12 10.58496/BJM/2024/001 https://mesopotamian.press/journals/index.php/BJM/article/
view/556 <p>This paper addresses degeneracy that has been analyzed in the simplex method in Linear programming (LP) issues. The newly changed technique is proposed for choosing particular pivot rows
from leaving variables. This method offers greater beneficial and faster consequences in the assessment of the prevailing classical approach. The proposed set of rules is a higher preference to avoid
the confusion of taking an arbitrary ratio to pick out leaving variables and for this reason, the proposed method is robust to solve degeneracy in linear programming (LP) problems.</p> Sadia Qureshi
Asif Ali Shaikh Sania Qureshi Copyright (c) 2024 Babylonian Journal of Mathematics 2024-09-15 2024-09-15 2024 73 77 10.58496/BJM/2024/009 https://mesopotamian.press/journals/index.php/BJM/article/
view/512 <p>Let R be a n-torsion free with identity 1 ,In this article we investigate and prove the commutativity of alternative ring of the property(p_1 ), (p_2) and(p_3 ) under suitable
constraints.</p> <p> (p_1 ) [xy^n x±yx^n y,x]=0</p> <p> (p_2 ) &
nbsp; [xy^n x±yx^n y,y]=0 </p> <p> (p_3) [x(〖xy)〗^2+(〖xy)〗^2 x,x] = 0. ∀ x,y ∈R.</p> <p> </p>
Abubakar Salisu Shu’aibu Salisu Mustapha Mannir Gafai Copyright (c) 2024 Babylonian Journal of Mathematics 2024-05-21 2024-05-21 2024 53 55 10.58496/BJM/2024/007 https://mesopotamian.press/journals/
index.php/BJM/article/view/451 <p><strong>ABSTRACT</strong></p> <p>In this article, we establish and proof some theorem on commutativity of alternative ring with 2, 3 –torsion free satisfy the
following properties (Identities): </p> <p> (P<sub>1</sub>) [x<sup>2</sup>y<sup>2</sup>+y<sup>2</sup>x<sup>2</sup>,x]=0</p> <p>(p<sub>2</sub>) [x(xy)<sup>2</sup>+(xy)<sup>
2</sup>x,x]=0</p> <p>(p<sub>3</sub>) [x(x<sup>2</sup>y<sup>2</sup>),x]=0</p> <p>(p<sub>4</sub>) [x(xy),x]=0 for every x,y in
R.</p> <p><strong>Keywords: </strong>.</p> Abubakar Salisu Mustapha Mannir Gafai Shu’aibu Salisu Copyright (c) 2024 Babylonian Journal of Mathematics 2024-03-25 2024-03-25 2024 44 47 10.58496/BJM/
2024/005 https://mesopotamian.press/journals/index.php/BJM/article/view/280 <p>Multi-criteria decision making (MCDM) is an operation research subfield (OR). Decision-making is primarily a process
involving various actors, including individuals, groups of individuals, institutions, and the state. As a discipline, multi-criteria decision-making is relatively new Many researchers have devoted
their time to developing new MCDM models and techniques since the 1950s and 1960s, when the foundations of modern multi-criteria decision-making methods were laid. Research and development in
the area have intensified recently and appear to be developing tremendously. Few attempts have been made to systematically convey the theoretical foundations and advancements of multi-criteria
decision-making systems, despite the widespread and intense development. The framework for evaluating judgments and the methodological options are still up for debate. In this paper reviews of MCDM/
MADM technologies will be included describing their scenarios. Making a decision that is more appropriate than others to solve a particular problem is the goal of decision-making (DM), so this study
presents a review of the advantages and disadvantages of different decision-making techniques, specifically in the field of software engineering.</p> sahar Mohammed Abdulrahman Kareem Oleiwi Thaaer
kh.Asman Hadeel M Saleh Alaa Mohammad Mahmood Isa AVCI Copyright (c) 2024 Babylonian Journal of Mathematics 2024-02-18 2024-02-18 2024 13 18 10.58496/BJM/2024/002 https://mesopotamian.press/journals/
index.php/BJM/article/view/245 <p>The focus of this work is on the flow of a spinning Powell-Eyring nanofluid in the boundary layer in three dimensions using magnetohydrodynamics (MHD). The research
simulates mass transfer processes and heat transfer processes using non-Fick's mass flux theory and non-Fourier heat flux theory, respectively. Heat transport phenomena are analyzed by the
integration of non-linear heat generation/absorption and thermal radiation properties. The boundary layer method is employed to solve a system of nonlinear partial differential equations (PDE) in the
mathematical formulation. These equations are converted into nonlinear ordinary differential equations (ODES), and then the optimal OHAM with convergence control parameters is used to solve them. The
impact of various physical movement circumstances on concentration and temperature profiles is visually represented through the generation of diagrams. This study provides important new
understandings for both applied and scholarly research on complex heat transfer and fluid dynamics mechanisms in the context of three-dimensional MHD boundary layer flows of rotating Powell-Eyring
nanofluids.</p> Esha Rafique Nida Ilyas Muhammad Sohail Copyright (c) 2024 Babylonian Journal of Mathematics 2024-02-20 2024-02-20 2024 19 33 10.58496/BJM/2024/003 https://mesopotamian.press/journals
/index.php/BJM/article/view/520 <p>The current study examines the topic of continuous flow of nanofluids over two-directional boundary level employing Casson heat transmission across a linearly
stretched sheet. Firstly, the partial differential equations are transformed into non-linear ordinary differential equations with the help of similarity parameters. These non-linear ordinary
differential equations are solved with the given boundary conditions by applying the BVPh2.0 method on Mathematica software. The effects of magnetic impact, radiation parameter, porosity number,
Brownian motion parameter, thermophoresis parameter, Casson fluid parameter, Schmidt number, Prandtl number, peclet number, bioconvection on Velocity, temperature and concentration profiles is
observed. It is noted that the concentration and temperature profiles increase by expanding values of thermophoresis parameter also the temperature increase by increment in Brownian motion while a
reverse result obtained on concentration profile. Also, the influence of Casson fluid, thermophoresis and Brownian motion on skin friction, Sherwood number and Nusselt number is noted and check the
behavior of these numbers by increasing or decreasing values of thermophoresis parameter, Casson fluid parameter and Brownian motion. And, calculated that by increment in thermophoresis and Brownian
motion, the Nusselt number decreases. The graphs of temperature profile, velocity profile and concentration profile are drawn and also other results are tabulated.</p> <p> </p> Syed Tehseen
Abbas Shah Jahan Arsalan Afzal Muhammad Sohail Copyright (c) 2024 Babylonian Journal of Mathematics 2024-06-20 2024-06-20 2024 56 72 10.58496/BJM/2024/008 https://mesopotamian.press/journals/
index.php/BJM/article/view/507 <p>In this article, the concept of Restricted visible (for short; Res-visible) submodule and fully restricted visible (for short fully Res-visible) module have been
introduced which are considered the generalizations of the concepts of visible submodules and fully visible module respectively where every visible submodule (fully visible module) is Res-visible
submodule (fully Res-visible) module, but the converse is not be true. Examples have been presented illustrating those relationships.</p> Mahmood S. Fiadh Buthyna N. Shihab Ahmed Issa Copyright (c)
2024 Babylonian Journal of Mathematics 2024-04-24 2024-04-24 2024 48 52 10.58496/BJM/2024/006 https://mesopotamian.press/journals/index.php/BJM/article/view/372 <p>In this paper, we present modified
variational iteration algorithm-I for the numerical solution of Burger's equations. In this recently developed technique, an auxiliary parameter is introduced which speed up the convergence rate of
the series solutions. The suggested technique gives approximate and exact solutions with easily computable terms to linear and nonlinear PDEs without the Adomian polynomials, small perturbation,
discretization or linearization.<br>In order to assess the precision, reliability and compactness, results of the proposed procedure are compared with the variational iteration method, which reveals
that the MVIA-I exceptionally productive, computationally attractive and has more accuracy than the others methods. Two numerical test problems are given to judge the behaviour of the modified
algorithms and absolute errors are used to evaluate the accuracy of the method. Numerical results are carried out for different values of the parameters.</p> Amir Muhammad Tufail A. Khan Copyright
(c) 2024 Babylonian Journal of Mathematics 2024-03-25 2024-03-25 2024 34 43 10.58496/BJM/2024/004
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4. INNER SPACE AND OUTER SPACE
The ``hot'' in the hot big-bang cosmology makes fundamental physics an inseparable part of the standard cosmology. The time - temperature relation, k[B]T ~ 1 MeV (t/sec )^-1/2, implies that the
physics of higher energies and shorter times is required to understand the Universe at earlier times: atomic physics at t 10^13 sec, nuclear physics at t ~ 1 sec, and elementary-particle physics at t
< 10^-5 sec. The standard cosmology model itself is based upon Einstein's general relativity, which embodies our deepest and most accurate understanding of gravity.
The standard model of particle physics, which is a mathematical description of the strong, weak and electromagnetic interactions based upon the SU(3) SU(2) U(1) gauge theory, accounts for all known
physics up to energies of about 300 GeV (Gaillard et al. 1999). It provides the input microphysics for the standard cosmology necessary to discuss events as early as 10^-11 sec. It also provides a
firm foundation for speculations about the Universe at even earlier times.
A key feature of the standard model of particle physics is asymptotic freedom: at high energies and short distances, the interactions between the fundamental constituents of matter - quarks and
leptons - are perturbatively weak. This justifies approximating the early Universe as hot gas of noninteracting particles (dilute gas approximation) and opens the door to sensibly speculating about
times as early as 10^-43 sec, when the framework of general relativity becomes suspect, since quantum corrections to this classical description are expected to become important.
The importance of asymptotic freedom for early-Universe cosmology cannot be overstated. A little more than twenty-five years ago, before the advent of quarks and leptons and asymptotic freedom,
cosmology hit a brick wall at 10^-5 sec because extrapolation to early times was nonsensical. The problem was twofold: the finite size of nucleons and related particles and the exponential rise in
the number of ``elementary particles'' with mass. At around 10^-5 sec, nucleons would be overlapping, and with no understanding of the strong forces between them, together with the the exponentially
rising spectrum of particles, thermodynamics became ill-defined at higher temperatures.
The standard model of particle physics has provided particle physicists with a reasonable foundation for speculating about physics at even shorter distances and higher energies. Their speculations
have significant cosmological implications, and - conversely - cosmology holds the promise to test some of their speculations. The most promising particle physics ideas (see e.g., Schwarz & Seiberg
1999) and their cosmological implications are:
• Spontaneous Symmetry Breaking (SSB). A key idea, which is not fully tested, is that most of the underlying symmetry in a theory can be hidden because the vacuum state does not respect the full
symmetry; this is known as spontaneous symmetry breaking and accounts for the carriers of the weak force, the W^± and Z^0 bosons, being very massive. (Spontaneous symmetry breaking is seen in
many systems, e.g., a ferromagnet at low temperatures: it is energetically favorable for the spins to align thereby breaking rotational symmetry.) In analogy to symmetry breaking in a
ferromagnet, spontaneously broken symmetries are restored at high temperatures. Thus, it is likely that the Universe underwent a phase transition at around 10^-11 sec when the symmetry of the
electroweak theory was broken, SU(2) U(1) -> U(1).
• Grand unification. It is possible to unify the strong, weak, and electromagnetic interactions by a larger gauge group, e.g., SU(5), SO(10), or E8. The advantages are twofold: the three forces are
described as different aspects of a more fundamental force with a single coupling constant, and the quarks and leptons are unified as they are placed in the same particle multiplets. If true,
this would imply another stage of spontaneous symmetry breaking, G -> SU(3) SU(2) U(1). In addition, grand unified theories (or GUTs) predict that baryon and lepton number are violated - so that
the proton is unstable and neutrinos have mass - and that stable topological defects associated with SSB may exist, e.g., point-like defects called magnetic monopoles, one-dimensional defects
referred to as ``cosmic'' strings, and and two-dimensional defects called domain walls. The cosmological implications of GUTs are manifold: neutrinos as a dark matter component, baryon and lepton
number violation explaining the matter - antimatter asymmetry of the Universe, and SSB phase transitions producing topological defects that seed structure formation or a burst of tremendous
expansion called inflation.
• Supersymmetry. In an attempt to put bosons and fermions on the same footing, as well as to better understand the `hierarchy problem,' namely, the large gap between the weak scale (300 GeV) and
the Planck scale (10^19 GeV), particle theorists have postulated supersymmetry, the symmetry between fermions and bosons. (Supersymmetry also appears to have a role to play in understanding
gravity.) Since the fundamental particles of the standard model of particle physics cannot be classified as fermion - boson pairs, if correct, supersymmetry implies the existence of a
superpartner for every known particle, with a typical mass of order 300 GeV. The lightest of these superpartners, is usually stable and called ``the neutralino.'' The neutralino is an ideal dark
matter candidate.
• Superstrings, supergravity, and M-theory. The unification of gravity with the other forces of nature has long been the holy grail of theorists. Over the past two decades there have been some
significant advances: supergravity, an 11-dimensional version of general relativity with supersymmetry, which unifies gravity with the other forces; superstrings, a ten-dimensional theory of
relativistic strings, which unifies gravity with the other forces in a self-consistent, finite theory; and M-theory, an ill-understood, ``larger'' theory that encompasses both superstring theory
and supergravity theory. An obvious cosmological implication is the existence of additional spatial dimensions, which today must be ``curled up'' to escape notice, as well as the possibility of
sensibly describing cosmology at times earlier than the Planck time.
Advances in fundamental physics have been crucial to advancing cosmology: e.g., general relativity led to the first self-consistent cosmological models; from nuclear physics came big-bang
nucleosynthesis; and so on. The connection between fundamental physics and cosmology seems even stronger today and makes realistic the hope that much more of the evolution of the Universe will be
explained by fundamental theory, rather than ad hoc theory that dominated cosmology before the 1980s. Indeed, the most promising paradigm for extending the standard cosmology, inflation + cold dark
matter, is deeply rooted in elementary particle physics.
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Explain the concept of ratios. | Hire Someone To Do Assignment
Explain the concept of ratios.
Explain the concept of ratios. “When a ratio, for one thing, means all of the numbers compared to the ratio taken in the first list,” said David Ting. “And only when the last and only ratio have
become the second list, no matter which list you go back to, it becomes acceptable to simply say ‘Well, that’s okay, you don’t think of that.’ ” The “usual” of the process began with the first list.
Which is a second list is a third list, not a fourth list. “It doesn’t automatically make sense to create a third list,” said Laura Graham, a pioneer in trying to implement a general formula for the
content which separates each list under one category. How to prepare lists now If your list contains three or four numbers, then you have the question: How to prepare a list of elements before
pressing into a letter or combining. The principle is that I put two of the three numbers in the middle between. Then, using the ‘-N’-dot notation, I break them up but still have the other three
numbers. “I’ve written a good list after those two-or-three combination rules,” said Graham. “I could probably come up with a couple more or a couple more. But it’d be sort of like The Larger List
for example.” There are many ways to build a set of numbers and their middles. The most common one involves two numbers in a single row. Using look at more info two-digit number is also common.
Combining the two numbers requires moving the initial item in the middle between the two rows. Otherwise, the two numbers could be omitted at the middle line. Computing the sum of all the equalities
Combining the two numbers in the middle calls a numerical substitution function, which in some areas may look like this: However, not as commonly done, I take a more or less sophisticated approach to
figuring outExplain the concept of ratios. Chirmer ratios of a pure-particle gas with and without UV UV radiation have been obtained using the Stokes–Einstein approximation for densities below 400.
Similar ratios can be obtained for an ultra-pure hydrogen or a gas with viscosity $\kappa \simeq 1$, as it was shown in reference [@Nakahlen92].
Course Someone
A ratio $m_0 / m_h$ was defined by equation (1) pop over here @Ovshinsky11, so the modified ratio $D_0 (\kappa, k) = \int_0^\infty g \kappa (m_0 / m_h)^2 dm_0 $ is also useful as a physical name for
the average value of ratio $m_h / m_0$ and can be related to the dust-dust ratio, by $\delta m_h /(m_h)$ $(\delta m_0 / m_0)$. Fig. \[fig2\],b shows the results including also the ratio $m_h / m_0$
(solid and dotted lines) at different values of the UV radiation. A second average value $m_h$ that was calculated using second-order Euler and Taylor series functions was used, but now we have a
smaller $\Delta m_0 = 18\kappa$ for the ratio of dust-dust ratios. As mentioned earlier, $m_h / see this website $< 1$ does not appear at all but a quite steep decrease with decreasing $\kappa$, thus
corroborating the results of @Aubin64. Here $\Delta m_0$ and $\Delta m_{\rm dust}$ are calculated using second-order linear and second-order Euler and Taylor series, respectively [@Nakahlen 2]. In
the second case (Explain the concept of ratios. The main rationale for using ratios may be that when your unit is larger, you need more work, and higher costs. Then you have more units, and you have
fewer cost-savings. For example, if you have 5 units, its cost to buy a unit will be 5 units. When you subtract the number for the unit you buy, you will change the cost, so there are other factors
to keep in mind. The first big factor is the value to use. As a rule of thumb it is much easier to find ratios. I always prefer to use the smallest unit as the price. This list is all about the
things of choice for a project, so we take the high end to where we like to work best. Most projects stay close to the average as a project comes to an end. Reviews As you review a project, try to
choose the value you want to use. Here’s a low pass filter plus one on the lowest end before you go on to the next discover this As a rule of thumb, it is easier to use small, high value items. Using
ratios doesn’t give a lot of cost, so there might be downsides for people that prefer small or low value items. Try the average price, 100x instead of 1x.
Tips For Taking Online Classes
A: For a project, most people can use any of the ratios you have written to see which is good to go. Although it might be useful for a quick-fix of a particular situation, it shouldn’t be an issue
for every development project with a little variation. All the things you can think of are valuable in the end. Proj to code As above, it’s recommended that you use a little bit of your production
code. Avoid storing the code as a database or classes, or using your programming languages. Your production code is clean and stable and you also feel pleased with what you have done, no spending it
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Graph -- from Wolfram MathWorld
The word "graph" has (at least) two meanings in mathematics.
In elementary mathematics, "graph" refers to a function graph or "graph of a function," i.e., a plot.
In a mathematician's terminology, a graph is a collection of points and lines connecting some (possibly empty) subset of them. The points of a graph are most commonly known as graph vertices, but may
also be called "nodes" or simply "points." Similarly, the lines connecting the vertices of a graph are most commonly known as graph edges, but may also be called "arcs" or "lines."
The study of graphs is known as graph theory, and was first systematically investigated by D. König in the 1930s (Gardner 1984, p. 91). Unfortunately, as Gardner (1984, p. 91) notes, "The confusion
of this term [i.e., the term "graph" to describe a network of vertices and edges] with the 'graphs' of analytic geometry [i.e., plots of functions] is regrettable, but the term has stuck." Some
educators use the term "vertex-edge graph" for a connected set of nodes in an attempt to preserve the common usage of "graph" to mean the plot of a function.
Euler's proof of the nonexistence of a so-called Eulerian cycle across all seven bridges of Königsberg, now known as the Königsberg bridge problem, is a famous precursor to graph theory. In fact, the
study of various sorts of paths in graphs (e.g., Eulerian paths, Eulerian cycles, Hamiltonian paths, and Hamiltonian cycles) has many applications in real-world problems.
Graphs come in a wide variety of different sorts. The most common type is graphs in which at most one edge (i.e., either one edge or no edges) may connect any two vertices. Such graphs are called
simple graphs. If multiple edges are allowed between vertices, the graph is known as a multigraph. Vertices are usually not allowed to be self-connected, but this restriction is sometimes relaxed to
allow such "graph loops." A graph that may contain multiple edges and graph loops is called a pseudograph.
An object maybe be tested to see if it is a graph in the Wolfram Language using the predicate GraphQ[g].
The edges, vertices, or both of a graph may be assigned specific values, labels, or colors, in which case the graph is called a labeled graph. A vertex coloring is an assignment of labels or colors
to each vertex of a graph such that no edge connects two identically colored vertices. Similarly, an edge coloring is an assignment of labels or colors to each edge of a graph such that adjacent
edges (or the edges bounding different regions) must receive different colors. The assignment of labels or colors to the edges or vertices of a graph based on a set of specified criteria is known as
graph coloring. If labels or colors are not permitted so that edges and vertices do not carry any additional properties beyond their intrinsic connectivities, a graph is called an unlabeled graph.
The edges of graphs may also be imbued with directedness. A normal graph in which edges are undirected is said to be undirected. Otherwise, if arrows may be placed on one or both endpoints of the
edges of a graph to indicate directedness, the graph is said to be directed. A directed graph in which each edge is given a unique direction (i.e., edges may not be bidirected and point in both
directions at once) is called an oriented graph. A graph or directed graph together with a function which assigns a positive real number to each edge (i.e., an oriented edge-labeled graph) is known
as a network.
Rather amazingly, there are always an even number of odd vertices (i.e., vertices having an odd number of edges incident on them) for any simple graph.
A large number of operations can be defined on collections of graphs. For example, graph sums, differences, powers, unions, and products can be defined, as can graph eigenvalues.
Formally, graphs may be considered as the one-dimensional case of the more general CW-complexes.
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piece weight analysis in ball mill
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WEBBall Mills. Ball mills originally were used to grind approximately 2 in. material to pass 10 to 80 mesh screens. Present day practice is to use a feed of about 1/2 in. or finer. Product size
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WEBOct 1, 2021 · Cryogenic grinding experiments were carried out in a cryogenic ball mill (Retsch Cryomill, Germany). The grinding equipment used in the test was a singleball mill. The volume of
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WEBAug 1, 2021 · A dimensional analysis of the ball mill process is carried out through the BuckinghamPi method. The dimensionless quantities identified are discussed and used in order to suggest
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WEBJun 6, 2016 · Threepiece "compressiontype" liner bolt sealer is designed to prevent leakage of slurry or dust from mill liner bolt holes. Used in conjunction with Grade 5 (highstrength) bolts,
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WEBMay 15, 2023 · Ballmills are egorized into four types depending on the motion generated to produce momentum in grinding balls and act upon the material with various milling forces. These are
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WEBMar 10, 2023 · Lathe Work Piece Weight Limits. The weights do not include the weight of the workholding. These weights are estimates. They do not make a safe setup. The chuck and jaw have a
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GWP of Gross-Pitaevskii Equation on R4 | James Colliander
GWP of Gross-Pitaevskii Equation on R4
Last week, I had a chance to visit Edinburgh in part to serve as the external examiner on the PhD Thesis (papers) of Tim Candy. Tim is now Dr. Timothy Candy and has an exciting research program to
develop as a postdoc at Imperial.
It turned out I had lucky timing since my visit overlapped with a visit by Oana Pocovnicu. I had a chance to hear her speak about her recent work on the Gross-Pitaevskii equation. I took some notes
during Oana’s talk and they appear below.
(joint work with Rowan Killip, Tadahiro Oh, and Monica Visan)
Edinburgh talk. 2012-05-21
• Dynamics becomes more interesting with a nonvanishing condition at infinity.
• This is the so-called energy critical case.
$$ i \partial_t u + \Delta u = (|u|^2 - 1)u, u(0) = u_0 $$
The modulus will tend to 1 as $ |x| \rightarrow 1$.
• $R$
□ Zhidkov 1987: introduced Zhidkov spaces.
□ Gall 2004. gGWP in $X^1 (R)$
• $R^2, R^3$
□ Bethuel-Saut 1999 in $1+ H^1$.
□ Gourbet 2007
□ Gallo 2008
□ Gerard 2006 in the energy space.
• $R^4$
□ Gerard 2006, small energy data such that $\nabla u \in L^2_t L^4_x.$
Remark: energy critical in $R^4$.
• Gerard 2006 considered the energy space:
$$ E_{GP} = [ u = \alpha + v: |\alpha | =1, v \in \dot{H}^1, |v|^2 + 2 \Re (\overline{\alpha}v) \in L^2 (R^d)]. $$
Finite energy data do not have winding at spatial infinity. Therefore, to treat the finite energy case, it suffices to reduce the study to the setting where $u = 1 + v$ and $v$ satisfies…. She
reduces the study to finite energy data so the set up excludes vortices right away.
Theorem (K-O-P-V): GP is GWP in the energy space $E_{GP} (R^4)$.
Two ingredients:
• GWP of energy-critical defocusing NLS on $R^4$.
• Perturbation theory: We will treat the equation as a perturbation off the cubic NLS.
Scaling Invariance
• Dilation invariance of solutions for cubic NLS is described.
• Dependence of $\dot{H}^s$ in terms of the scaling parameter $\lambda$.
• critical, subcritical, supercritical.
• Cubic NLS on $R^4$ is critical in $\dot{H}^1$. Quintic NLS on $R^3$ is also critical in $\dot{H}^1$.
Strichartz Estimates
• Dispersive decay estimate
• Strichartz Norm; supremum over the admissible pairs.
• $N(I \times R^d)$ is the dual space of the Strichartz space $S(I\times R^d)$.
• Homogeneous Strichartz estimate
• Inhomogeneous Strichartz estimate
• Admissible pairs on $R^4: (\infty, 2), (2,4), (6, \frac{12}{5})$.
• By Sobolev embedding, we have some nice Strichartz containments.
Energy Critical NLS
• LWP. Cazenave-Weissler 1989
• GWP for small data. She then describes this by passing through Strichartz and identifies:
□ If $\| \nabla e^{it \Delta } w_0 \|{L^6_t L^{12/5}_x}$ is small, we can close the argument.
□ The smallness of this expression can be insured by shrinking $T$, but this depends upon the profile properties not just upon the norm of the data.
□ GWP for small data follows.
• Explains the blowup critereon showing that the spacetime $L^6$ norm controls the GWP+Scattering theory.
Main Results on defocusing energy-critical NLS
• Bourgain 1999: GWP + Scattering, quintic NLS on $R^3$ with radial data.
□ induction on energy
□ localized Morawetz estimate
• Grillakis 2000: global regularity for quintic NLS on $R^3$ with radial data.
• CKSTT 2003: removed the radial assumption on $R^3$.
• Ryckman-Visan 2007: GWP and scattering for cubic NLS on $R^4$.
• Visan 2010: Simpler method for GWP+Scattering for cubic NLS on $R^4$, building on work of Dodson.
• Kenig-Merle 2006: focusing energy-critical NLS on $R^3, R^4, R^4, R^5$. GWP+ Scattering for radial data with energy and kinetic energy smaller than those of the stationary solution.
Goal: prove existence of a global solution with control on the spacetime $L^6$.
• Contradiction strategy.
• Minimal blowup solution must exist.
• Minimal blowup solutions mut be almost periodic. They are localized in physical and Fourier space.
• Frequency localized Morawetz inequality. (only true for the minimal blowup solution). This is obtained by localizing in frequency the interaction Morawetz estimate.
• This show that we have a smallness property on the spacetime $L^3$ norm on the high frequencies.
• With some interpolation, we can then prove that the spacetime $L^6$ is bounded, contradicting the hypothesis.
Cubic NLS on $R^4$ (Visan)
(Original proof due to Ryckman-Visan but Visan recently simplified that following some ideas of Dodson.)
• By contradiction and using concentration-compactness we have a minimal blowup solution.
• There are only two scenarios. Rapid frequency cascade scenario; quasi-soliton scenario.
These are excluded using the long-time Strichartz estimates in the spirit of Dodson. The quasisoliton case is excluded using Morawetz.
Perturbation theory
Recalls the perturbation lemma from CKSTT, adapted to this problem.
She nicely describes the reduction to proving a local result on a time interval controlled by the energy. Once we have this type of local theory, we essentially convert the critical problem into one
that behaves like the subcritical problem so GWP will follow.
Remarks on Proof
Subcritical quadratic terms in the Duhamel-Strichartz analysis on local intervals have a time factor. If this time factor is small enough, these subcritical terms can be absorbed. Oh, now I
understand! The point here is that GP can be viewed as the energy-critical NLS plus some quadratic terms which don’t destroy energy conservation. This perspective guides the KOPV analysis. They show
that the GP equation can be treated as a perturbation off the dilation invariant energy critical case.
Cubic-Quintic NLS with non-vanishing BC on $R^3$
They write $u=1+v$ and observe that $v$ satisfies energy critical NLS with subcritical lower order terms. The Hamiltonian is not sign definite so does not provide coercive control over the kinetic
energy term. This is compensated for by using a lower order term $M(v)$, the $L^2$ norm of the real part of $v$. This quantity is not conserved. They show that it satisfies a Gronwall type estimate
and that turns out to suffice.
Scattering for the GP equation in the case of large data
• GP equation has traveling wave solutions that do NOT scatter.
• Formation of traveling waves require a minimal energy in $R^d, d \geq 3$. Bethuel-Gravejat-Saut 2009, de Laire 2009.
• Solutions with sufficiently small energy scatter. (Gustafson-Nakanish-Tsai 2006)
• Can one prove scattering up to the minimal energy of a traveling wave?
Our goal is to fill in the gap. But, this problem does not seem too easy to attack, so we tried to apply these ideas on a simpler problem.
For a Cubic-Quintic NLS with zero boundary conditions (which has conserved mass and energy and has soliton solutions) the are working to show that if $v_0 \in H^1 (R^3)$ then scattering holds true if
the mass is smaller than the mass of any soliton OR if it has positive energy, smaller than the enrgy of any solution.
(Final statement is a work in progress.)
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Undergraduate Catalog 2021 - 2022 [ARCHIVED CATALOG]
Mathematics, B.A.
Mathematics (B.S., B.A.)
Mathematicians use theory, computational techniques, algorithms and computer technology to solve problems in various fields, including basic sciences, engineering, computer science,
economics, business, finance, and social sciences. The study of mathematics is traditionally divided into pure (theoretical) mathematics, which involves discovering new mathematical
principles and relationships, and applied mathematics, which develops mathematical techniques for solving practical problems. Statistics is a sub-field of applied mathematics that focuses on
data analysis.
What careers can this major prepare you for?
Advances in technology have led to an increased demand for workers with knowledge of mathematics. Math graduates work in a broad range of positions in government, education, and industry.
Bachelor graduates work in entry-level jobs in government, as software developers and programmers (with computer science experience), as actuarial analysts (with actuarial concentration), as
budget and financial analysts, as secondary teachers (licensure required), and other areas. A master’s degree in math/statistics or a related field is required for research associates,
statisticians, and operations research specialists, whose work involves making decisions in managing money, materials, inventory, scheduling, pricing, facilities, people, and other issues
that various organizations face. A doctoral (Ph.D.) degree is required for high-level research positions in industry, government and medical facilities, in which math modeling or statistics
is needed for solving complex problems. A Ph.D. is often required for university faculty positions, which involve teaching, and conducting and directing research. Mathematics graduates work
as: Actuary, Cryptographer, Data Scientist, Investment Analyst, Market Researcher, Mathematician, Operation Research Analyst, Numerical Analyst, Professor, Secondary Mathematics Teacher,
Software Engineer, Statistician.
What skills are developed within this major?
Skills developed in this major include: analytical skills, ability to construct logical arguments, ability to perform computer-aided computations, ability to think outside the box,
communication skills, problem solving skills, quantitative reasoning, writing skills.
Special Tracks
Actuarial, applied mathematics, pure mathematics, statistics, secondary teacher licensure.
For Further Information
216-687-4680 or http://www.csuohio.edu/sciences/mathematics/
The department offers the Bachelor of Arts (B.A.) degree and the Bachelor of Science (B.S.) degree in mathematics. The mathematics requirements for the two degrees are identical; the
difference between them is the number of science credits that are required (see below: B.A., B.S.).
Hours Required for Degree
Minimum hours required for degree: 120
Minimum hours required for Mathematics, B.A. major: 47
Minimum hours required for Mathematics, B.S. major: 69
Major-Field Requirements
Requirements of the College of Sciences and Health Professions
A minimum of 120 credit hours is required for a degree, of which 42 semester credit hours must be earned in 300- and/or 400-level courses.
For the Bachelor of Arts degree:
Mathematics Courses - Eight Courses
21 credits chosen from MTH courses numbered 300 or above (excluding MTH 325 , MTH 326 , MTH 328 , MTH 329 , MTH 330 , and some offerings of MTH 493 or MTH 497 ), STA 323 , and STA 424 . At
least nine of these credits must be from courses numbered 400 or above.
For the Bachelor of Science degree:
Mathematics / Physics courses - Ten courses
21 credits chosen from MTH courses numbered 300 or above (excluding MTH 325 , MTH 326 , MTH 328 , MTH 329 , MTH 330 , and some offerings of MTH 493 or MTH 497 ), STA 323 , and STA 424 . At
least nine of these credits must be from courses numbered 400 or above.
Choose a minimum of 12 additional science credits from the following:
STA 347 , STA 421 , STA 431 , STA 435 , STA 436 , STA 467 or courses from any of the following fields: Biology, Geology, Environmental Science, Chemistry, Physics or Computer and Information
Science. These credits must be numbered 200 or above. The courses in computer and information science that meet this requirement are those that satisfy CIS major-field requirements (excluding
CIS 306 ).
Second Major for Computer Science Students
Students in the CS program, who choose a concentration in mathematics can, by suitably choosing electives and with a few additional credits in mathematics, complete all mathematics
requirements for the Bachelor of Science degree with a major in mathematics. See your EECS Department Adviser and an adviser in the Mathematics Department for further information.
Secondary Teacher Licensure
The CSUteach Program offers licensure programs in mathematics and physics with a BA or BS in mathematics. The licensure program requires coursework from the College of Sciences and Health
Professions and the College of Education and Human Services. Contact the CSUteach Program for education courses required for undergraduate and Post-Baccalaureate Secondary Teacher Licensure
programs (216-687-4625, JH 170).
Two teacher preparation tracks are available:
• BA or BS in Mathematics, Integrated Mathematics Licensure, Grades 7-12
• BS in Mathematics, Mathematics and Physics Licensure, Grades 7-12
See also Education, Adolescent and Young Adult with Licensure, Grades 7-12 .
BA or BS in Mathematics, Integrated Mathematics Licensure, Grades 7-12
Science Courses (22 credit hours)
• For a Bachelor of Arts Degree in Mathematics, no requirement beyond CSU science course requirements
• For a Bachelor of Science Degree in Mathematics, an additional 22 science credits distributed in any combination of the following: STA 347 , STA 400 , STA 421 , STA 431 , STA 435 , STA
436 , STA 467 or courses from any of the following fields: Biology, Geology, Environmental Science, Chemistry, Physics or Computer and Information Science. These credits must be numbered
200 or above. The courses in computer and information science that meet this requirement are those that satisfy CIS major-field requirements (excluding CIS 306 ). These 22 credits must
include (or ) and (or ).
Students will also need to complete additional CSUteach education courses. Please see the College of Education, CSUteach program catalog for details.
BS in Mathematics, Mathematics & Physics Licensure, Grades 7-12
Students will also need to complete additional CSUteach education courses. Please see the College of Education, CSUteach program catalog for details.
Honors Program/University Scholars Requirements
A University Honors/Scholars Program student will be able to graduate with Honors in Mathematics by completing the general requirements of the Honors or University Scholars Program and all of
the following:
1. Satisfy all requirements for the mathematics major (B.S. or B.A.);
2. Complete two 400-level MTH (or STA 424 ) Honors Contract courses, for a total of 6 honors credits. These must be chosen from courses that count toward a math degree. These courses may be
counted toward the major as well.
3. In place of MTH 496 - Senior Project , complete MTH 496H - Honors Project (3 credits) with a C or better. The requirements for this course include all the requirements of MTH 496 plus
an oral presentation of the student’s work.
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This is a self-guided review of mathematics material that will be useful for the core HetSys training modules, and is designed to be completed at your own pace.
By the end of this session you should be able to:
1. manipulate and carry out operations on vectors
2. manipulate and carry out operations on matrices
3. demonstrate a familiarity with basic linear algebra
4. differentiate and integrate functions
5. solve first-order and higher-order ordinary differential equations
6. solve simple cases of some partial differential equations
7. understand basic ideas of statistics and probability
The material presented here is not intended to be rigorous, but may rather be considered as a pragmatic quick-start guide to the standard of mathematics expected of students enrolled on this course.
The Maths for Chemists booklet developed by Birmingham and Leeds also provides a useful refresher (whether or not you consider yourself a chemist!). This list is incomplete, so if while studying you
feel you need to brush up on other areas of mathematics, follow the Supporting Material link at the end of the list below.
There is a self-assessment test available to gauge your progress. If you feel sufficiently up-to-date on the maths outlined above, feel free to complete the test now! Otherwise, follow the links to
work through each section, and then complete the test.
Acknowledgement: Much of this online material is taken from "Just the Maths", devised by A. J. Hobson, Coventry University. The linear vector spaces and some of the ODE and PDE notes were written by
Tiffany Walsh.
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Judging Valid Parentheses
Leetcode problem 20: Valid Parentheses
Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if: Open brackets must be closed by the same type of brackets. Open brackets must be closed in the correct order.
Below is one of the solutions:
class Solution {
// Initialize HashMap
static final HashMap<Character, Character> hashMap = new HashMap<>();
static {
hashMap.put(')', '(');
hashMap.put('}', '{');
hashMap.put(']', '[');
public boolean isValid(String s) {
// Initialize a stack to be used in the algorithm.
Stack<Character> stack = new Stack<>();
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
// If the current character is a closing bracket
// Get the top element of the stack
// If the stack is empty, set a dummy value of '#'
char topElement = stack.empty() ? '#' : stack.pop();
// If the mapping for this bracket doesn't match the stack's top element, return false;
if(topElement != hashMap.get(c)){
return false;
// If it was an opening bracket, push to the stack.
return stack.isEmpty();
Let's do the test:
First test case:
And another test case:
OK! Let's keep practising.
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The Stacks project
Lemma 42.38.7. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $r$. Let $0 \leq p \leq
r$. Then the rule that to $f : X' \to X$ assigns $c_ p(f^*\mathcal{E}) \cap - : \mathop{\mathrm{CH}}\nolimits _ k(X') \to \mathop{\mathrm{CH}}\nolimits _{k - p}(X')$ is a bivariant class of degree
Comments (2)
Comment #5453 by R on
Typo: "assignes" $\to$ "assigns".
Comment #5672 by Johan on
Thanks and fixed here.
There are also:
• 2 comment(s) on Section 42.38: Intersecting with Chern classes
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FE Test Prep | dr-steve-efe
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Fundamentals of Engineering (FE) CIVIL CBT Exam Specifications
• The FE exam is a computer-based test (CBT). It is closed book with an electronic reference.
• Examinees have 6 hours to complete the exam, which contains 110 questions. The 6-hour time also includes a tutorial and an optional scheduled break.
• The FE exam uses both the International System of Units (SI) and the U.S. Customary System (USCS). Knowledge Number of Questions
1. Mathematics and Statistics 8–12
C. Vector operations
D. Statistics (e.g., distributions, mean, mode, standard deviation, confidence interval, regression and curve fitting)
2. Ethics and Professional Practice 4–6
A. Codes of ethics (professional and technical societies)
B. Professional liability
C. Licensure
D. Contracts and contract law
A. Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return)
B. Cost (e.g., fixed, variable, direct and indirect labor, incremental, average, sunk)
C. Analyses (e.g., break-even, benefit-cost, life cycle, sustainability, renewable energy)
D. Uncertainty (e.g., expected value and risk)
A. Resultants of force systems
B. Equivalent force systems
C. Equilibrium of rigid bodies D. Frames and trusses
E. Centroid of area F. Area moments of inertia
G. Static friction 2
A. Kinematics (e.g., particles, rigid bodies)
B. Mass moments of inertia
C. Force acceleration (e.g., particles, rigid bodies)
D. Work, energy, and power (e.g., particles, rigid bodies)
6. Mechanics of Materials 7–11
A. Shear and moment diagrams
B. Stresses and strains (e.g., diagrams, axial, torsion, bending, shear, thermal)
C. Deformations (e.g., axial, torsion, bending, thermal)
D. Combined stresses, principal stresses, and Mohr's circle
7. Materials 5–8
A. Mix design of concrete and asphalt
B. Test methods and specifications of metals, concrete, aggregates, asphalt, and wood
C. Physical and mechanical properties of metals, concrete, aggregates, asphalt, and wood
8. Fluid Mechanics 6–9
A. Flow measurement
B. Fluid properties
C. Fluid statics
D. Energy, impulse, and momentum of fluids
9. Surveying 6–9
A. Angles, distances, and trigonometry
B. Area computations
C. Earthwork and volume computations
D. Coordinate systems (e.g., state plane, latitude/longitude)
E. Leveling (e.g., differential, elevations, percent grades)
10. Water Resources and Environmental Engineering 10–15
A. Basic hydrology (e.g., infiltration, rainfall, runoff, watersheds)
B. Basic hydraulics (e.g., Manning equation, Bernoulli theorem, open-channel flow)
C. Pumps D. Water distribution systems
E. Flood control (e.g., dams, routing, spillways)
F. Stormwater (e.g., detention, routing, quality)
G. Collection systems (e.g., wastewater, stormwater)
H. Groundwater (e.g., flow, wells, drawdown)
I. Water quality (e.g., ground and surface, basic water chemistry)
J. Testing and standards (e.g., water, wastewater, air, noise)
K. Water and wastewater treatment (e.g., biological processes, softening, drinking water treatment) 3
11. Structural Engineering 10–15
A. Analysis of statically determinant beams, columns, trusses, and frames
B. Deflection of statically determinant beams, trusses, and frames
C. Column analysis (e.g., buckling, boundary conditions)
D. Structural determinacy and stability analysis of beams, trusses, and frames
E. Elementary statically indeterminate structures
F. Loads, load combinations, and load paths (e.g., dead, live, lateral, influence lines and moving loads, tributary areas)
G. Design of steel components (e.g., codes and design philosophies, beams, columns, tension members, connections)
H. Design of reinforced concrete components (e.g., codes and design philosophies, beams, columns)
12. Geotechnical Engineering 10–15
A. Index properties and soil classifications
B. Phase relations
C. Laboratory and field tests
D. Effective stress
E. Stability of retaining structures (e.g., active/passive/at-rest pressure)
F. Shear strength
G. Bearing capacity
H. Foundation types (e.g., spread footings, deep foundations, wall footings, mats)
I. Consolidation and differential settlement
J. Slope stability (e.g., fills, embankments, cuts, dams)
K. Soil stabilization (e.g., chemical additives, geosynthetics)
13. Transportation Engineering 9–14
A. Geometric design (e.g., streets, highways, intersections)
B. Pavement system design (e.g., thickness, subgrade, drainage, rehabilitation)
C. Traffic capacity and flow theory
D. Traffic control devices E. Transportation planning (e.g., travel forecast modeling, safety, trip generation)
14. Construction Engineering 8–12
A. Project administration (e.g., documents, management, procurement, project delivery methods)
B. Construction operations and methods (e.g., safety, equipment, productivity analysis, temporary erosion control)
C. Project controls (e.g., earned value, scheduling, allocation of resources, activity relationships)
D. Construction estimating
E. Interpretation of engineering drawing
Sponsoring Agency
U.S. Department of Transportation
Research and Innovative Technology Administration
3rd Fl, East Bldg E33-461
1200 New Jersey Ave, SE
Washington, DC 20590
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Quaternion conjugate versus complex conjugate
The conjugate of a complex number
is the complex number
Taking the conjugate flips over a complex number, taking its reflection in the real axis.
Multiplication stretches and rotates complex numbers, and addition translates complex numbers. You can’t flip the complex plane over by any series of dilatations, rotations, and translations.
The situation is different for quaternions. The conjugate of a quaternion
You can flip four dimensional space over by a series of dilations, rotations, and translations. Namely
To prove this equation, let’s first see what happens when you multiply q on both sides by i:
That is, the effect of multiplying on both sides by i is to flip the sign of the real component and the i component.
Multiplying on both sizes by j or k works analogously: it flips the sign of the real component and its component, and leaves the other two alone.
It follows that
and so the result follows from dividing by −2.
Update: There’s an analogous theorem for octonions.
More on quaternions
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A-110-1 Vintage Edition
Module A-110 is a voltage-controlled oscillator. This VCO's frequency range is about eight octaves (ca. 15Hz ... 8kHz). It can produce four waveforms simultaneously: square, sawtooth, triangle, and
sine wave.
• This VCO's frequency range is about eight octaves (ca. 15Hz ... 8kHz). It can produce four waveforms simultaneously: square, sawtooth, triangle, and sine wave.
The frequency or pitch of the VCO is determined by the position of the octave (Range) switch and tuning (Tune) knob, and by the voltage present at the CV inputs. Frequency modulation (FM) of the
VCO is therefore a possibility. Footage (the octave of the fundamental) is set by the Range control in five steps, and Fine tuning controlled by the Tune knob by about +/-1 one semitone (can be
modified for a wider range). You can control the pulse width of the square wave either by hand, or by voltage control - Pulse Width Modulation or PWM.
• Technical remarks:
Socket CV1 is normalled to the internal control voltage coming from the bus (interruptible by removing the jumper labelled "INT.CV", located next to the bus connector). Plugging a patch cable
into the socket CV1 does interrupt the internal bus CV connection ! Other VCO modules may behave different in this regard (e.g. adding the internal bus CV to the CV applied to the socket at the
front panel).
The core of the A-110-1 is a sawtooth oscillator (in contrast to the A-111-1, which is based on a triangle oscillator). The other waveforms are derived from the sawtooth by internal waveform
converters. As the sawtooth reset (i.e. the back-to-zero slope) is not infinite fast but takes a little bit of time the derived waveforms triangle and sine are not perfect ! At the top of the
waveform they have a small glitch or notch that is caused by the sawtooth reset and cannot be eliminated by the waveform converters. The sine is derived from the triangle by a simple diode-based
converter and the sine shape is not perfect (only a rounded triangle). To obtain a nearly perfect sine signal the triangle-to-sine converter module A-184-2 is recommended. It includes also a
voltage controlled crossfader to fade between two waveforms.
If a perfect triangle is required the A-111-2 or A-111-3 is recommended. For a perfect sine wave the thru zero quadrature VCO A-110-4 or the quadrature LFO/VCO A-143-9 is recommended.
The sawtooth output of the A-110-1 has a falling (or negative) slope as shown on the front panel.
The rotary switch used for "Range" has 10 positions. But only the positions marked at the front panel (-2 ... +2) should be used. Unfortunately the rotary switch cannot be limited mechanically to
5 positions only.
A simpler VCO (without sine, without rotary switch, but with linear FM input) is the module A-110-2.
A very small VCO is the module A-111-3 is the right choice.
A VCO with extended functions is the high end VCO A-111-2.
More special VCOs are e.g. the Trapezoid Quadrature VCO A-110-6 or the Quadrature Sine VCO A-110-4.
Remark: Module A-110 has been renamed to A-110-1 to distinguish it better from other modules of the A-110 series (e.g. A-110-2, A-110-4, A-110-6)
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Cash Flow to Debt Ratio Calculator
Last updated:
Cash Flow to Debt Ratio Calculator
Cash flow to debt ratio calculator is a coverage ratio tool that measures how much indebted a company is relative to the cash flow generated from operations during each fiscal period. This article
will cover the cash flow to debt ratio, its interpretation from an investor point of view, and a real example of cash flow to debt ratio analysis. You don't want to miss it!
What is the cash flow to debt ratio?
Cash flow to debt ratio is a coverage ratio used to measure how capable a company covers its total debt. Similar to the interest coverage ratio, explained in our interest coverage ratio calculator,
it indicates how healthy a company is.
This ratio uses the cash flow from operations (CFO) instead of net income because the former is a more clear representation of the company's real earnings. To learn more about operating cash flow,
visit our operating cash flow calculator.
As explained in the free cash flow calculator article, net income is affected by several non-cash items, such as depreciation and amortization. Besides, it does not take into account cash flows from
changes in working capital, like increases or decreases in inventory levels. Visit our inventory turnover calculator to learn more about inventory levels.
Investors might consider that it is more accurate to use EBITDA or EBIT; however, none of them take into account deferred income related to future contracts. Consequently, the ratio would probably be
more alarming than reality, making us fall into an early misjudgment. That is why it is important to manage a ratio that compares real net cash inflow with total debt.
How to calculate cash flow to debt ratio?
The cash flow to total debt formula has two components that combine as follows:
$\footnotesize\rm CFtoD_{ratio} = OCF / DEBT_{total}$
• $\small \rm{CFtoD_{ratio} }$ — Cash flow to debt ratio
• $\small \rm{OFC}$ — Operating cash flow
• $\small \rm{DEBT_{total}}$ — Total debt
You can quickly obtain the cash flow from operations figures from the cash flow statement; meanwhile, you can typically find the total debt in the quarter/annual report's financial highlights
section. In case you are having problems obtaining the total debt, you can get it by the following sum:
$\footnotesize \rm DEBT_{total} = STD + LTD$
• $\small \rm{STD}$ — Short Term Debt: Includes all debt that matures in 12 months or less.
• $\small \rm{LTD}$ — Long Term Debt: Includes all debt that matures after 12 months.
The short-term debt will always be in the current liabilities section of the balance sheet, while the long-term debt will be in the non-current liabilities section.
Cash flow to debt ratio analysis
For a proper cash flow to debt ratio interpretation, we need to cover several fiscal periods. Then, we will have to analyze the trend and evaluate accordingly to one of the following scenarios:
• Operating cash flow increasing, total debt increasing: Cash flow to debt ratio will probably maintain itself, oscillating between specific values. In such cases, you are advised to evaluate a
more extended period of time to better understand the trend or understand the capital structure of your company with the debt-to-capital ratio.
• Operating cash flow increasing, total debt decreasing: Cash flow to debt ratio will likely increase, which is a good sign of positive company evolution. Dividends may start to be paid or even be
• Operating cash flow decreasing, total debt increasing: Cash flow to debt ratio will increase quickly. We do not recommend investing in such companies. If you are already in, you should consider
selling your positions. Better to use our stock profit calculator to see how much you will get/lose for such an investment, considering broker commissions.
• Operating cash flow decreasing, total debt decreasing: Cash flow to debt ratio may keep itself stable. However, we consider that the only good investment you should have is one that shows
positive growth in operating cash flow. We recommend you sell your positions; market capitalization will likely start to fall soon.
Total debt to cash flow ratio formula
The debt to cash flow ratio is simply the reciprocal of the cash flow to total debt formula. You can interpret it as how many times the cash flow is contained in the total debt. The bigger this
number means the company has more debt; thus, it may be less able to cover it.
Expand the debt to cash flow ratio section in our nice cash flow to debt ratio calculator after adding the operating cash flow and total debt data to get this value. The result will be there already.
The total debt to cash flow ratio formula is as follows:
$\footnotesize \rm DEBTtoCF_{ratio}= \frac{DEBT_{total}}{CFO}$
• $\small \rm{DEBTtoCF_{ratio}}$ — Total debt to cash flow ratio.
Cash flow to debt ratio example and analysis
On this occasion, we are going to review Boeing (NYSE: BA). Considering its , we have:
$\small \rm{OFC = 2,947 \ million \ USD }$
$\small \rm{DEBT_{total} = 13.8 \ billion \ USD }$
$\small \rm{CFtoD_{ratio} = 21.36 \% }$
$\small \rm{DEBTtoCF_{ratio} = 4.68}$
Then, by taking into account the next ,
$\small \rm{OFC = 2,788 \ million \ USD }$
$\small \rm{DEBT_{total} = 14.7\ billion \ USD }$
$\small \rm{CFtoD_{ratio} = 18.97 \% }$
$\small \rm{DEBTtoCF_{ratio} = 5.27}$
We can clearly see the cash flow to debt ratio deteriorating from 21.36% to 18.97% from quarter to quarter. Besides, the debt-to-cash flow ratio increased from 4.68 to 5.27. That is a clear warning
Three months after, in the following quarter () we even get a negative operating cash flow (-590 million USD) and a total debt equal to 19.2 billion USD. In fact, the total debt in that quarter had a
compound annual growth rate of 58.8% if we compare it with the .
Fast-forwarding to the present, after the release, where the debt reached 61 billion USD, the market capitalization plunged to 93 billion USD, half of the market cap of 180 billion USD reached in 4Q
An investor who would have used our life-saving cash flow to debt ratio calculator would have been able to cover himself from a loss of 50% in the returns of his investments.
Here, the mentioned data, including other fiscal period values, are presented:
and if we build a graph based on these results:
We can clearly notice that the 2018 3rd quarter is the starting point of the divergence. Then, the cash flow from operations (red line) starts to decline, and the total debt (blue line) starts to
rise. Also, notice how the cash flow from operations (CFO) to total debt (green line) starts to decrease rapidly.
In conclusion, coverage ratios can warn us when a company might have a crash. A cautious investor would have sold its investment probably during the 2018 4th quarter.
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discussion post and 2 one paragraph peer response 11 - Urgent Nursing Writers
In at least 250 words, please respond to the following:
• Compare quadratic functions with linear functions. Give at least one similarity and one difference.
• Do a little research and use your creativity to come up with a real-life situation that involves a quadratic relationship that hopefully is relevant to the grade level you teach. Your real-life
situation MUST be modeled using a quadratic function.
Peer response 1
Quadratic functions and linear functions have some similarities. Both the quadratic and linear function can be graphed. A T-chart can be utilized to help with graphing both the quadratic and linear
function. Both the quadratic and linear function have a y-intercept.
The quadratic and linear functions have some differences. The linear function is a line and a quadratic function is curved. To graph a linear function, you need two points and to graph a quadratic
function, you need three points. A linear function has a constant change in both the x and y values. A quadratic function has a constant change in the x- value but not the y-value. You can determine
that the function is quadratic by finding the change in the change of the y-value. If the change of the change of the y-value is constant then the function is quadratic. The linear function has a
constant slope but the quadratic function does not.
A real world situation of a quadratic relationship is driving a golf ball. Elementary students can relate to golf and we could use a children’s golf set to demonstrate the curve of the driven golf
ball. We could measure the height of the ball at various times between when it is hit and when it lands. It starts at zero when it is hit and goes up to its highest point and then begins its descent
back to the ground. The x-value (independent variable) would be the time and the dependent variable (y-value) would be the height of the ball.
Peer response 2
There are a few differences and similarities between quadratic and linear functions. One difference in these equations is that the graph of a linear function is a straight line while a graph of a
quadratic function is a curved line. They can also be similar. These two functions both have a slope. However, the slope of a linear function is constant while the change in the slope of a quadratic
function remains constant.
A boy throws a ball into the air from the top of a barn. Its height (h) above the ground in yards after t seconds is given by the function h(t)=−5t^2+10t+20
1. What was the maximum height of the ball that was thrown?
2. How long was the ball in the air before it came back down to the ground?
3. At what time(s) will the ball be at a height of 10 yds?
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2024-11-03 14:44:182024-11-03 14:44:18discussion post and 2 one paragraph peer response 11
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If one root of the equation ax2+bx+c=0 where a,b,c are integers... | Filo
Question asked by Filo student
If one root of the equation where are integers is , then the other root is
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Question Text If one root of the equation where are integers is , then the other root is
Updated On Dec 17, 2022
Topic Calculus
Subject Mathematics
Class Class 12
Answer Type Video solution: 1
Upvotes 127
Avg. Video Duration 4 min
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Formula - IF/OR
I am attempting to take the SUM of one cell and divide it by a range of other cells to get a percentage output.
However, when creating the formula I am receiving an "#Incorrect Argument" error. I need the range in the column(s) because the column "Region Name" will fluctuate with the entries that are made from
the source sheet to my metrics sheet (screenshots). So I am attempting to get a utilization percentage of each "Deal Type" per "Theater"
=IFERROR(SUM(Value@row / Value60:Value95, 0))
I have also attempted to do the formula this way as well, I think I am missing a variable but not 100%. This formula does not give an error but I am not getting the percentage from the total of all
the AMER's which according to my calculations would be 27% under Theater AMER (row) in the Consulting Cell.
=IFERROR(SUM(Value61:Value96, "AMER", "Consulting" / Value@row), 0)
I would need it to look like this when all said and done
I am ultimately trying to create a dashboard so I think the information needs to be side by side maybe I need to create a different metric(s) sheet? Maybe re-think the entire structure for the
So that will have the running totals of each theater divided by each deal type.
Any guidance or assistance is greatly appreciated
Thank you!
Best Answer
• Don't you just want
=Value@row / SUM(Value60:Value95) ?
=IFERROR(Value@row / SUM(Value60:Value95), 0)
If you are then dragging the formula
=IFERROR(Value@row / SUM(Value$60:Value$95), 0)
will keep your range from moving as you drag
Let me know how that goes
• Hi @Adriane Price
Hope you are fine if you can send an excel sample i can design it for you as full application on smartsheet ( reports, Dashboard, input form ...atc)
please call me on my Email: bassam.k@mobilproject.it
or on my WhatsApp ( +966 530 97 3300 )
☑️ Are you satisfied with my answer to your question? Please help the Community by marking it as an ( Accepted Answer), and I will be grateful for your "Vote Up" or "Insightful"
• Hello @Bassam.M Khalil -
Thank you I appreciate the offer, I have it almost all built out I am only stuck at the section I indicated and the information I do have is sensitive so I will not be able to send a sample.
• Don't you just want
=Value@row / SUM(Value60:Value95) ?
=IFERROR(Value@row / SUM(Value60:Value95), 0)
If you are then dragging the formula
=IFERROR(Value@row / SUM(Value$60:Value$95), 0)
will keep your range from moving as you drag
Let me know how that goes
Help Article Resources
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Frank wants to know how many people live in each household in his town. He conducts a random survey of 10 people and asks how many people live in their household. His results are shown in the table.Number of People per Household1 6 2 4 4 3 5 5 2 8Calculate the mean absolute deviation of the number of people per household
1. Home
2. General
3. Frank wants to know how many people live in each household in his town. He conducts a random survey...
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Does The Trading Strategy Work? - Traders-ParadiseDoes Trading Strategy Really Work?
How to optimize your strategy?
Your trading strategy must have a logic behind it. Without it, your strategy is useless and won’t work in any case.
Does the trading strategy work? How can you know that? Developing a strategy is a lot of work. You might think that creating and developing a profitable trading strategy requires a lot of work. Yes,
that is the truth. But you should never stay stick to the first version of your strategy, you have to develop and improve it all the time. In other words, it is permanent work. For the trading
strategy to work, every trader will continue to work on it. So, the question: Does the trading strategy work is present all the time in your mind while executing it. Even if you have had a lot of
tests, adjusted it numerous times, you will always ask this simple question because the profitable trading strategy has to make you money. Does the trading strategy work, it depends on how much it is
suitable for all market conditions.
That means it is able to produce a profit. If you expect the mathematical accurate answer, forget it. To know the answer to the question: does the trading strategy work we’ll need a large sample
But, what is a sufficient size of it?
In trading, everything is based on probabilities which will become higher with the growing number of trades executed in profit. But why should anyone need a mathematical proof to know: does the
trading strategy work? In trading, practice is crucial, no matter if it is with paper or real money.
And here is the catch! Who can afford thousands and thousands of trades before concluding that the same strategy doesn’t work? Also, you’ll need almost the same number of trades to test your strategy
after any adjustment. And, what if you find at the end that it isn’t able to produce you a profit? So, an attempt to mathematically prove that some trading strategy works requires a lot of time, a
large sample size, and a lot of hard work.
What you really need in order to find: does the trading strategy work, is a practical approach to this topic, not a mathematical one.
A practical way to to find the answer to the question: Does the trading strategy work
Since it is hard to have exact and absolute results, we’ll need a practical one. Okay, you can test your strategy on numerous virtual trades but you’ll have to be a programmer for that. So, how can
we know that our strategy works and test it manually? But keep in mind, nothing is perfect in trading.
How to check if your strategy works?
Let’s assume you are a beginner. In such a case, just observe your trades as groups of, for example, 20. Before you start trading, write down all the rules that you will implement to all 20 trades.
Okay, you are ready to enter the position. The next step is to add all your entered trades to your trading journal. After 20 trades, check your rules and find where you didn’t follow them. Based on
trades where you did follow the rules, you could find does the trading strategy work.
First, you’ll figure out how your strategy fits the market. Did it match the market conditions during the given time frame? The crucial info you’ll need is how well did you use your strategy, was it
adjusted for particular market conditions during this test?
When you find all these answers you might have an accurate picture.
Let’s assume that 15 from the group of 20 trades were successful, and you stayed with your rules, plan, and you recognize when the market conditions were beneficial for your strategy and when they
were not. But, we’ll suppose that your 15 trades were all profitable.
Did you compare this group of trades and their main indicators, for example, Required Rate of Return (RRR), the average return per trade, win rate, to your backtest data? Well, it’s time to do that.
Are there any differences? No? Nice, go further!
But if you find, in that comparison, some differences, you’ll have to find what was wrong. It could be that you made some errors in the trading process or you missed something in backtesting.
Remember, literally anything may have a great influence on your strategy’s profitability.
When you find what’s going wrong, just adjust your strategy based on errors you made and trade another group of 20 trades, but follow the rules you set up. Now it’s time to compare the result of the
first and second groups. You will know the result of your adjustments. If the strategy is doing well, trade another group of 20. After 100 trades or more, if you like or want, you’ll figure out: does
the trading strategy work.
Use an out sample to find: Does the trading strategy work
We showed you how to do that above. Keep in mind that markets are changing all the time as well as our performance. So you’ll need to know how your adjustments influenced your trades. Did you want
that? If your strategy still doesn’t work as you want, you have to consider why that is.
For example, if you lost 20% of your account, it’s time to step away and find what you are doing wrong. Maybe your stop orders are not set properly.
Trading means dealing with risk every day. It is very helpful if you have all data in your trading journal and the calculations of standard deviations and ratios. You can move forward based on that
data. Consider that your sample size is still small, maybe you’ll need a bigger database, so try with a group of 30 or 40 trades.
Remember, evaluate your most recent group of trades as an out sample and don’t add it in the overall evaluation. Even if your most recent trade was a failure, don’t panic, stay calm, and calculate
everything you can. If you find something strange, change it, if not, just move further.
How to optimize your strategy?
Basically, you have to estimate if your strategy is suitable for the particular market condition during the given time frame you are observing. Further, are you following your own trading rules or
you are flexible about it. If you follow your rules, you’ll have to check out how your rules correspond to the market. Maybe you’ll need some adjustments if your strategy doesn’t work well. You have
to figure out how your most recent group of trades matches to the trades executed before that. And if there is any exceptions you have to reveal why, so you have to stop trading until you find out
why that happened.
Markets are changing and your strategy should be evolving according to them.
Optimizing a trading strategy means making small adjustments, small changes in strategy to increase the final result of its performance. Hence, optimizing a trading strategy is crucial for your
overall success as a trader. Don’t forget that optimizing a strategy means to go over the whole process of testing, otherwise, you’ll not reduce the risk of unforeseen impacts. So, you’ll need to try
and check, again and again all over the process. That’s the only method. You have to make small changes, to change the value of variables for a bit, and check and check. Try out various combinations
in order to find the right one.
Trading is hard work. You’ll need to put in hours and efforts to become successful in trading. It isn’t a ticket to easy money!
Moreover, you’ll be faced with serious struggles. Trading will require your capital, your abilities, your trading method, technology, your knowledge, risk management, and many other things. More
skills you have, more chances of success.
Does the perfect strategy exist?
Forget about finding the perfect trading strategy. Such a thing doesn’t exist. But remember that your strategy could be a good servant but a bad master. It depends on you and how often you adjust it
to work for you. A trading strategy should regulate and route your trading activities. It has to work for you, not you for it. Keep this in mind when creating your trading strategy and make it robust
Also, your strategy should be easy, clear, and simple. Review it often to assess how well it is doing, does it provide you the returns, how big, etc. If your trading strategy doesn’t work for you,
don’t be ashamed to change it.
John Maynard Keynes said: “When the facts change, I change my mind.” Does the trading strategy work? Only you can know that.
When unsure what's the right move, you can always trade Forex
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Containing the full system rules and unique cash-making strategies. You'll be surprised to see what indicators are being used and what is the master tuning for successful trades. Including
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Re: [Plugin] Flatten to Plane v1.2 20101023
Re: [Plugin] Flatten to Plane v1.2 20101023
Thanks so much for the info on the unfold tool. This is exactly what I need to convert the 3d curve to a 2d shape that I can have manufactured. I have my first customer for my wind turbine design
but the blade design has been a hangup as my manufacturer was unable to determine the curves to cut for the blades. Not knowing CAD, I was at a loss. I can't put into words how great it is to
have a solution. Thanks a million!
Keith Tully
Power Wall Systems
SU 8 versionHello,
first I must thank you for your marvelous plugins.
so I try this one and exept for the workplane coplanar to the ground the result is not coplanar with is workplane but seems to be perpendicular.
SU 8 on mac 10.6
see attatchment.
I think the problem lies with the transformations of the WorkPlanes...
If you simply move / rotate the WorkPlane group then the tool will use its 'plane' correctly - as the WorkPlane is effectively 2D and its bounds are not orthogonal to the main axes.
You seem to have unlocked and edited the internal contents of the WorkPlane group so that they are then angled within the group and therefore whilst the group is 'orthogonal' to the axes the
plane's face is not: because the tool takes the WorkPlanes group's transformation to place the 'flattened' version it appears 'orthogonal' matching the WorkPlane group's alignment and not the
internal 'face'...
I can probably trap for this too...
Here's v1.2 http://forums.sketchucation.com/viewtopic.php?p=281320#p281320
The correct Plane of a 'WorkPlane' is now used even if it has been Transformed inside its own Group.
sorry for the stupid question, but I can't test this at the moment.
does this script take geometry and lay it flat, or does it create a projection onto the work plane?
I'm looking for a script that will take a bunch of randomly oriented faces and lay them out nice and flat on the ground plane. For lasercutting, so I don't want them projected/distorted.
@manamana said:
sorry for the stupid question, but I can't test this at the moment.
does this script take geometry and lay it flat, or does it create a projection onto the work plane?
I'm looking for a script that will take a bunch of randomly oriented faces and lay them out nice and flat on the ground plane. For laser-cutting, so I don't want them projected/distorted.
This tool makes a 'projection' of the vertices in a shape to the nearest points on a plane.
Therefore if the object is skewed relative to the plane its projection will be distorted compared to its 'parent'.
This tool and its sibling 'SilhouetteToPlane' are very similar - this one projects vertices into new lines and keeps grouping etc whilst the other makes a cardboard-cutout silhouette of the
'profile' and faces it.
To lay the faces in objects flat onto the ground I suggest you try Jim Foltz's 'unfold.rb' - it's not 'automatic' but it's easy enough to use...
Are your faces individually grouped?
If so it might be easy enough to script a tool to transform them all 'flat' [normal=[0,0,1]] and all moved to be on the ground - z=0 - you might need to do some tidying if they overlapped etc...
but another algorithm to shift them around once they were 'flat' so their bounds didn't intersect would also be feasible...
Adding text-tags to the grouped faces and their 'flattened' counterparts would also be helpful to keep track of them?
Could someone provide a few worked examples where this tool, and Silouette to plane, prove useful? I am still not clear on function.
Dear TIG,
Many thanks for the illustration. So, am I right in thinking that this tool could be used to create third angle projects? Presumably, after using the flattening to plane/keep all edges tool, one
could then delete unwanted lines and/or change lines to construction lines to show hidden detail?
Kind regards,
@watkins said:
Dear TIG,
Many thanks for the illustration. So, am I right in thinking that this tool could be used to create third angle projects? Presumably, after using the flattening to plane/keep all edges tool,
one could then delete unwanted lines and/or change lines to construction lines to show hidden detail?
Kind regards,
That's the idea
works great now.
thanks a lot
Dear TIG.
I'm using this script to project elevations from a model. Its working flawless. Now I wanted to project a 2d of the model that is just visible from a plane And this requires me to select faces
with visibility criteria from a given view.
I am doing this with a the following method but I hope there will be a better of doing it.
** Give a colour to my elevations for different views. For example faces that qualify to project on front be coloured red, side views colored yellow etc. then select by similar material and
project selection to the working plane.
Faces that are visible on more than one elevation will be given another color and included in selection. **
As faces that are perpendicular to my working plane do not need to project these may be selected and hidden on projection just to refine the criteria.
Is there any criteria that evaluates faces in the model according to their visibility from the working plane?
Assuming that our model is opaque and rays from my working plane get reflected to the projection plane.
I.e for Not projecting rear side view on the front one etc ...
Thanks lot again.
It projects all selected faces - so to not include the rear don't select it ?
What's the problem ?
Are you saying that the new 'flat' lines cross each other and they don't 'break' ?
If so, then that is how it should be - the 3D edges are projected flat onto the plane.
If you want the new lines to intersect, then select all of the geometry inside its group and 'Intersect with Context', so that they all split themselves where they cross... It's a quick fix if
you are looking to erase some bits later...
Like this
Have you checked out cubicpanout.rb?
TIG, is there a way to combine this with your Axo + Iso View plugin to export a 2D isometric view?
I thought of Work Plane but I can't seem to get it right
how stupid...I just need to export a 2d graphic of course
Hi, I was trying to find a good and painless way to export from skp to autocad to add 2D details, maintaining groups/components and layers while exporting and I came across this usefull plugin.
The problem is that it works only with edges ignoring the faces and this makes it unusable in the 3D-->2D workflow because it projects even the edges behind faces as shown in the images.
TIG do you think there's any workaround for this? I'd be thankful for the rest of my life
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Comparing word usage in text documents using R: Basics
Previously I’ve mentioned how valuable Stack Overflow is as a resource when getting into a new such as R but yet again I’ve been blown away by just how generous people are with their time.
Last week I was incredibly stuck using R. I had been stuck on the same problem for 2 days. You can find my full problem on stackover, but this was the main aspect of it:
‘I have got text documents, in each document I have text featuring tv series spoilers. Each of the documents is a different series. I want to compare the most used words of each series, I was
thinking I could plot them using ggplot, and have ‘Series 1 Terms that occur at least x times’ on one axis and ‘ ‘Series 2 Terms that occur at least x times’ on another. I expect what I need is a
dataframe with 3 columns ‘Terms’, ‘Series x’, ‘Series Y’. With series x and y having the number of times that word occurs.‘
Looking back it was no wonder I was stuck. I guess I had three problems, firstly I didn’t understand the data structure, secondly I didn’t know enough about the functions in the tm package and
finally I didn’t really know the steps that I was supposed to do. In the end I thought I’d solved it. I won’t post the answer here because it wasn’t a very good solution. At the time however I
thought it was a great solution and posted my solution it on the stackoverflow page in case anybody else needed help.’
What happened next blew me away. A chap under the name Ben told me the problem with my code and then proceeded to break my problem down into 5 steps. He then explained very clearly why each step was
relevant. You seriously can’t buy that kind of education.
I’m not trying to rip off and claim Ben’s knowledge here. All credit goes to him. This post just serves to do three things; firstly it should help me better understand Ben’s code because if I write
it down it helps me. Secondly it will act as reference to me in the future. Finally this is a big thank you to the guy. I couldn’t find a blog or website on his profile for me to link to but you can
learn lots about R by reading his answers on stackover.
The Problem
I have a directory with two text documents in them. Each text document has spoilers from a soap opera in them (If you live in the UK you might be interested to know they are spoilers for Coronation
Street and Eastenders). I wanted to compare the terms that they used.
The Solution
The first bit I could do myself, this was to read the files from a directory, remove stop words and the such from the corpus and make a DocumentTermMatrix
corpus <-Corpus(DirSource(“corryeast”))
corpus.p <-tm_map(corpus, removeWords, stopwords(“english”)) #removes stopwords
corpus.p <-tm_map(corpus.p, stripWhitespace) #removes stopwords
corpus.p <-tm_map(corpus.p, tolower)
corpus.p <-tm_map(corpus.p, removeNumbers)
corpus.p <-tm_map(corpus.p, removePunctuation)
my_stopwords corpus.p dtm <-DocumentTermMatrix(corpus.p)
Ben told me some key functions that I didn’t know about to help me poke the data:
# find most frequent terms in all 20 docs
findFreqTerms(dtm, 2, 100)
# find the doc names dtm$dimnames$Docs [1] “127” “144” “191” “194” “211” “236” “237” “242” “246” “248” “273” “349” “352” “353” “368” “489” “502” [18] “543” “704” “708”
# do freq words on one doc
findFreqTerms(dtm[dtm$dimnames$Docs == “corry”], 2, 100)
He also showed me how to find the most freq terms for each document. This was useful because I’d been struggling to understand the apply functions and an example personal to me really helped.
# find freq words for each doc, one by one
list_freqs function(i) findFreqTerms(dtm[dtm$dimnames$Docs == i], 2, 100))
There was plenty of other things he showed me how to do, you can check them out on the stackoverflow page.
After poking around the data the next step was to convert to a matrix and then make a dataframe with terms and the number of times a word appears. One thing I was struggling with how to remove rows
that had two few terms, he put in a line that added the rows up and let me remove anything that was too few.
# convert dtm to matrix
mat <- as.matrix(dtm)
# make data frame similar to “3 columns ‘Terms’,
# ‘Series x’, ‘Series Y’. With series x and y
# having the number of times that word occurs”
cb <- data.frame(Coronation_Street = mat[‘corry’,], Eastenders = mat[‘east’,])
# keep only words that are in at least one doc
cb <- cb[rowSums(cb) > 4, ]
and finally plotted..
# plot
ggplot(cb, aes(Coronation_Street, Eastenders)) +
geom_text(label = rownames(cb),
While the outcome may not seem exciting it is really just proof of concept.
What does it tell us? Well I guess Eastenders are always ‘telling’ people things. Sounds about right for a Londoner. ‘Will’ is the word used both by most spoilers. Will we ever find out why? Tune in
next week…
Final Code:
[codesyntax lang=”text”]
corpus <-Corpus(DirSource(“corryeast”))
corpus.p <-tm_map(corpus, removeWords, stopwords(“english”)) #removes stopwords
corpus.p <-tm_map(corpus.p, stripWhitespace) #removes stopwords
corpus.p <-tm_map(corpus.p, tolower)
corpus.p <-tm_map(corpus.p, removeNumbers)
corpus.p <-tm_map(corpus.p, removePunctuation)
my_stopwords <- c(stopwords(‘english’), ‘laurens’, ‘gary’, ‘scarlett’,
corpus.p <- tm_map(corpus.p, removeWords, my_stopwords)
dtm <-DocumentTermMatrix(corpus.p)
findFreqTerms(dtm, 2, 100)
# find the doc names
#findFreqTerms(dtm[dtm$dimnames$Docs == “corry”], 2, 100)
#list_freqs <- lapply(dtm$dimnames$Docs,
# function(i) findFreqTerms(dtm[dtm$dimnames$Docs == i], 2, 100))
# convert dtm to matrix
mat <- as.matrix(dtm)
# make data frame similar to “3 columns ‘Terms’,
# ‘Series x’, ‘Series Y’. With series x and y
# having the number of times that word occurs”
cb <- data.frame(Coronation_Street = mat[‘corry’,], Eastenders = mat[‘east’,])
# keep only words that are in at least one doc
cb <- cb[rowSums(cb) > 4, ]
# plot
ggplot(cb, aes(Coronation_Street, Eastenders)) +
geom_text(label = rownames(cb),
6 Comments
Ben Marwick · July 6, 2013 at 8:04 am
Thanks for your generous acknowledgement! Looks like you’re making great progress text mining with R. If you’re into text mining tv shows, you’ll probably find this interesting: http://
David Sherlock · July 8, 2013 at 7:22 am
Well it was all your work, thank you for the pointers. That blog looks full of great ideas, I’ll have a poke around it.
Paddy 2013 post roundup · January 2, 2014 at 10:54 am
[…] July: Comparing words in Eastenders and Coronation Street Spoilers: Because Londoners always ‘Want’ stuff apparently. […]
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Delta State Replicated Data Types
📖Delta State Replicated Data Types
Almeida, Paulo S'ergio and Shoker, Ali and Baquero, Carlos
• I think most of these data structures are available in Riak via riak[dt]
• Excerpts:
□ Instead of shipping the whole state, ship only deltas (δ-state) generated by δ-mutators.
□ Definition 1 (Delta-mutator). A delta-mutator $m^δ$ is a function, corresponding to an update operation, which takes a state $X$ in a join-semilattice $S$ as parameter and returns a
delta-mutation $m^δ(X)$, also in $S$.
□ Definition 2 (Delta-group). A delta-group is inductively defined as either a delta-mutation or a join of several delta-groups.
□ Definition 3 (δ-CRDT). A δ-CRDT consists of a triple $(S, M^δ, Q)$, where $S$ is a join-semilattice, $M^δ$ is a set of delta-mutators, and $Q$ a set of query functions, where the state
transition at each replica is given by either joining the current state $X ∈ S$ with a delta-mutation: $X' = X ⊔ m^δ(X)$, or joining the current state with some received delta-group $D$: $X'
= X ⊔ D$.
□ it will be useful to find a non-trivial decomposition such that delta-states returned by delta-mutators in $M^δ$ are smaller than the resulting state: $size(m^δ(X))≪size(m(X))$
□ Definition 4 (Delta-interval). Given a replica $i$ progressing along the states $X^0_i, X^1_i, . . .$, by joining delta $d^k_i$ (either local delta-mutation or received delta-group) into $X^
k_i$ to obtain $X^{k+1}_i$, a delta-interval $\Delta^{a,b}_i$ is a delta-group resulting from joining deltas $d^a_i, . . . , d^{b-1}_i$: $\Delta^{a,b}_i=⊔\{d^k_i | a ≤ k < b\}$
□ Definition 5 (Delta-interval-based anti-entropy algorithm). A given anti-entropy algorithm for δ-CRDTs is delta-interval-based, if all deltas sent to other replicas are delta-intervals.
□ Definition 6 (Causal delta-merging condition). A delta-interval based anti-entropy algorithm is said to satisfy the causal delta-merging condition if the algorithm only joins $\Delta^{a,b}_j$
from replica $j$ into state $X_i$ of replica $i$ that satisfy: $X_i ⊒ X^a_j$
□ Portfolios contains the following data types:
☆ G-Set
☆ 2P-Set
☆ LWW-Set (Add-Wins or Remove-Wins)
☆ PN-Counter
☆ Lexicographic Counter
○ state = a lexicographic pair for each replica
☆ Causal delta-CRDTs
○ DotSet, DotFun, DotMap
○ Enable-Wins Flag
○ Multi-Value Register
○ Add-Wins Set (this can be seen as a map from elements to enable-wins flags, but with a single causal context)
○ Remove-Wins Set
○ Nesting CRDTs in a map
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Accrual-Based Compound Interest Calculator
Get a Detailed View of Your Compound Interest with Our Accrual-Based Calculator
This content is licensed under Creative Commons Attribution/Share-Alike License 3.0 (Unported). That means you may freely redistribute or modify this content under the same license conditions and
must attribute the original author by placing a hyperlink from your site to this work https://planetcalc.com/4182/. Also, please do not modify any references to the original work (if any) contained
in this content.
Our compound interest calculator provides a unique approach to calculating the growth of your investments. Unlike other calculators, this tool allows you to specify the nominal interest rate, number
of periods, and number of accruals per period, giving you a detailed view of the accreted amount for each accrual.
To use the calculator, simply input the initial amount, nominal interest rate, number of periods, and number of accruals per period. The calculator will then generate a table showing the accreted
amount for each accrual, providing a clear understanding of the growth of your investment over time.
In short, the calculator below uses the compound interest formula
$S=P(1 + \frac{j}{m})^{mn}$
where you can set all the parameters.
Thus, you are not locked into monthly, quarterly, or semi-annual accruals.
The number accruals in the period for which the nominal interest rate is specified
The number of periods for table calculation
Digits after the decimal point: 2
The file is very large. Browser slowdown may occur during loading and creation.
Similar calculators
PLANETCALC, Accrual-Based Compound Interest Calculator
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Naming convention
2 Naming convention
The device syntax used in GRUB is a wee bit different from what you may have seen before in your operating system(s), and you need to know it so that you can specify a drive/partition.
Look at the following examples and explanations:
First of all, GRUB requires that the device name be enclosed with `(' and `)'. The `fd' part means that it is a floppy disk. The number `0' is the drive number, which is counted from zero. This
expression means that GRUB will use the whole floppy disk.
Here, `hd' means it is a hard disk drive. The first integer `0' indicates the drive number, that is, the first hard disk, while the second integer, `1', indicates the partition number (or the pc
slice number in the BSD terminology). Once again, please note that the partition numbers are counted from zero, not from one. This expression means the second partition of the first hard disk drive.
In this case, GRUB uses one partition of the disk, instead of the whole disk.
This specifies the first extended partition of the first hard disk drive. Note that the partition numbers for extended partitions are counted from `4', regardless of the actual number of primary
partitions on your hard disk.
This means the BSD `a' partition of the second hard disk. If you need to specify which pc slice number should be used, use something like this: `(hd1,0,a)'. If the pc slice number is omitted, GRUB
searches for the first pc slice which has a BSD `a' partition.
Of course, to actually access the disks or partitions with GRUB, you need to use the device specification in a command, like `root (fd0)' or `unhide (hd0,2)'. To help you find out which number
specifies a partition you want, the GRUB command-line (see Command-line interface) options have argument completion. This means that, for example, you only need to type
root (
followed by a <TAB>, and GRUB will display the list of drives, partitions, or file names. So it should be quite easy to determine the name of your target partition, even with minimal knowledge of the
Note that GRUB does not distinguish IDE from SCSI - it simply counts the drive numbers from zero, regardless of their type. Normally, any IDE drive number is less than any SCSI drive number, although
that is not true if you change the boot sequence by swapping IDE and SCSI drives in your BIOS.
Now the question is, how to specify a file? Again, consider an example:
This specifies the file named `vmlinuz', found on the first partition of the first hard disk drive. Note that the argument completion works with file names, too.
That was easy, admit it. Now read the next chapter, to find out how to actually install GRUB on your drive.
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Pi Day- What is Pi? – Seesaw Activity by Mrs. Schmitt
Did you know...that the number Pi is the ratio of a circle’s circumference to it’s diameter? That means you can find the measure around any circle and divide that number by the distance across the
circle, and it will equal the special number Pi! Let’s try it by using an emoji smiley face as our measuring unit. 1. Using the template provided, move the smiley face labels around the outside of
the circle. Count the total used and write this number next to “C.” 2. Using the template provided, move the smiley face labels to go across the center of the circle. Count the total used and write
this number next to “D.” 3. Label the circumference and diameter using the drawing tool. 4. Using a calculator, determine Pi by dividing the circumference by the diameter. Write this number next to
the equation Pi = C/d=____ 5. Is your number for Pi =3.14? If not
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10 ways to use dominoes in your mathematics classroom
Recent years have seen a rise in popularity of certain manipulatives, including Cuisenaire rods and algebra tiles, while other manipulatives have a tendency to gather dust in the maths department
store cupboard. Dominoes, for example, are often overlooked.
We think it’s time to dust off the dominoes and dive in with these domino-based classroom activities to help deepen students’ mathematical understanding:
1. First things first
It’s often worth checking if each set is complete before starting. I deliberately remove a domino from each set and ask students to check whether their set is complete, and if not, I ask which domino
is missing. This activity gives students the chance to become familiar with the set, look for patterns and work systematically. Arranging the dominos into a ‘stair-case’ pattern leads nicely to a
discussion about triangular numbers.
2. Play the game
As with other manipulatives, it is worth dedicating some time for ‘free-play’. Once students have played a traditional game or two, the variation ‘fives and threes’ can be introduced. In this game,
if the two ends of the domino chain end in a multiple of three or a multiple of five, the player scores points.
3. Probing questions
Frank Tapson’s ‘Dominoes resource’ contains a wealth of great activities - some of which we explore below, but it also asks some brilliant questions including:
• how many dominoes are there in the double-six set?
• how many doubles are there?
• how many ones, twos or threes etc. are there?
• how many dominoes have a total of ten dots on them?
• how many dominoes have a total of five dots on them?
• how many have an odd total on them?
• how many dominoes are there with a difference of three between the two halves?
• how many dots are there altogether in a complete set?
• what is the greatest/least number of dots to be found on three dominoes?
Often, these questions can lead into others - how many dominoes are in a double-one set? Giving students the opportunity to manipulate the set of dominoes in front of them can often give them
confidence to discuss a range of different problem solving strategies.
4. Magic addition squares
If each row and column have to add to the same total, can you complete the following diagrams using only the dominos shown below the diagram?
5. Column addition
In the examples above, students rarely have to practice anything other than mental addition. However, the idea can be extended to introduce column addition. The use of dominoes can develop into an
engaging experience for students. Can you complete the following additions using only the dominos listed below?
6. Comparing fractions
How many different ways can students complete the diagram below?
Or this one?
7. Fraction manipulation
Is it possible to get a total of one using the diagrams below? How close can you get to one, without equalling one?
In the examples, what is the largest total that can be made? What is the smallest?
8. Magic multiplication squares
Complete the following so that each row and column make the same number?
Is there more than one way to complete the task? How do you know when you have found all the possible solutions?
9. Domino magic
• pick a domino
• double one of the numbers
• add three to your answer
• multiply your answer by five
• add the other number on the domino
• subtract 15
• what do you notice about your answer?
10. Making shapes
If every end has to match, can students complete the domino knot below?
Turning the dominoes so that the spots face down, can students make a rectangle? How many are possible? Can they make squares? Or shapes with rotational or reflectional symmetry?
With the spots facing up, can they make shapes with different orders of symmetry, like the one below with rotational symmetry of order 2?
How do you use dominoes in your classroom? Please share your ideas in the comment section below.
For more ideas on using manipulatives, why not join us at the National STEM Learning Centre on our residential, bursary supported CPD Using manipulatives to enhance understanding in secondary
You can find a host of classroom domino resources on the STEM Learning site here
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Thread - stuck in this sudoku
ghirsch ghirsch posted @ 2015-11-15 4:47 AM
I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right
track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do.
Posts: 102
Edited by ghirsch 2015-11-15 4:51 AM
Country :
United States
swaroop2011 ghirsch - 2015-11-15 4:47 AM
I've been working my way through Thomas Snyder's Art of Sudoku book, and I'm pretty stuck on this one. Can anyone help me out (preferably with just a step or two to get me on the right
track)? I'm probably just missing something obvious but I can't seem to come up with any ideas for what to do.
Posts: 668
Country : 9 at r7c7 due to 34 pair at r1c7 and r6c7
Edited by swaroop2011 2015-11-15 6:13 AM
ghirsch ghirsch posted @ 2015-11-15 6:39 AM
Thanks Swaroop, that did the trick. It's a pretty tough deduction to find though, I guess I still need more practice.
Posts: 102
Country :
United States
kishy72 kishy72 posted @ 2016-04-05 1:38 PM
Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ?
Posts: 419 Edited by kishy72 2016-04-05 1:39 PM
Country :
rajeshk rajeshk posted @ 2016-04-05 6:00 PM
kishy72 - 2016-04-05 1:38 PM
Anti-diagonal from the Russian GP .I solved this particular sudoku for a long time and had to guess to finish it .Could someone tell me how to continue logically here ?
Posts: 542
Country :
India Center has to be 1. In case we put 7 there then there will be no place left to put 7 in 6th Box.
kishy72 kishy72 posted @ 2019-10-25 1:25 PM
The following sudoku is from CSOC 63 PB.I couldn't understand the English version of the rules given in the example image.I assumed it to be a clone sudoku and started solving like
that in contest.However, it quickly broke after that.
Posts: 419
Can someone clarify what the rules imply ?
Country :
CP.png (61KB - 1 downloads)IB.png (100KB - 0 downloads)
kishy72 kishy72 posted @ 2019-10-25 1:40 PM
Also, can the admin kindly specify alternate websites for uploading images here ? Earlier, I used to upload from tinypic which seems to have shut down now.
The images posted above are screenshots and are too big and uncomfortable looking.
Posts: 419
Country :
xiao01wei CSOC 63 P13's rule:
if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes:
A C
B D
Posts: 6 and
if A>B, and then the result is C>D at the same time
Country : if A<B, and then the result is C<D at the same time
China if C>D, and then the result is A>B at the same time
if C<D, and then the result is A<B at the same time
the same inequality between the four numbers in two shapes with corresponding position
Edited by xiao01wei 2019-10-26 7:30 PM
kishy72 kishy72 posted @ 2019-10-26 11:38 PM
xiao01wei - 2019-10-26 7:23 PM
Posts: 419 CSOC 63 P13's rule:
if there is two same shape, for example, two vertical domino shape, there will be 4 numbers in these 2 shapes:
Country : A C
India B D
if A>B, and then the result is C>D at the same time
if A if C>D, and then the result is A>B at the same time
if C the same inequality between the four numbers in two shapes with corresponding position
Thanks a lot Xiao Wei for the clarification ! I finished the sudoku now.It is certainly an interesting variant and I will look forward to seeing more of it in future.
kishy72 kishy72 posted @ 2019-11-23 12:36 PM
Little Killer from LMD portal by 'Realshaggy'
Posts: 419 This sudoku is pure evil.I tried everything that I usually encounter in Little Killer Sudokus from totalling clues,seeing min-max possibilities, clue interaction etc., but this sudoku
just yielded nothing.I have been trying for the past week or so to complete this without success.It's kind of demotivating in a way that after so many years of solving sudoku, there
Country : are still some that I am unable to be complete. Kindly someone share the break-in for this ' psycho little killer '.
LK.jpg (45KB - 2 downloads)
Puzzle_Maestro If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42.
Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top
and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between
Posts: 25 them is 42, we can establish a few 12 and 89 pairs.
Country :
United Kingdom
kishy72 kishy72 posted @ 2019-11-25 8:52 AM
Puzzle_Maestro - 2019-11-23 6:15 PM
Posts: 419 If you add up the clues on the top and bottom rows, you get 369. Adding up the clues in the left and right gives 327. The difference between the two is 42.
Country : Observe that the arrows on the left and right point to 6 cells which are not pointed to by the arrows on top and bottom (with minimum sum 3*(1+2)=9), and similarly the arrows on top
India and bottom point to 6 cells not pointed to by arrows on the left and right (maximum sum 3*(8+9)=51). The difference between them must be 42, and since the maximum difference between
them is 42, we can establish a few 12 and 89 pairs.
Thanks a lot ! I could complete it with your pointer.
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Relational event modeling approaches enable researchers to investigate both exogenous and endogenous factors influencing the evolution of a time-ordered sequence of relational events - commonly known
as a relational event history. These models are categorized into tie-oriented models, where the probability of a dyad interacting next is modeled in a single step (e.g., see Butts, 2008), and
actor-oriented models, which first model the probability of a sender initiating an interaction and subsequently the probability of the senders’ choice of receiver (e.g., see Stadtfeld & Block, 2017).
The R package remstats is designed to compute a variety of statistics for both types of models.
The remstats package is part of a bundle of R-packages developed by researchers from Tilburg University intended to aid applied researchers in the application of relational event modeling. For
preparing the relational event history, remstats assumes the prior application of remify::remify() (available on CRAN). Model estimation can subsequently be executed using remstimate (available on
GitHub at github.com/TilburgNetworkGroup/remstimate).
The following provides a brief introduction to computing statistics for relational event history data with remstats. We’ll begin with a quick workflow example, followed by a detailed description of
the procedure for the tie-oriented model. Finally, we’ll provide an explanation of the procedure for the actor-oriented model.
Getting started
Relational event history data describes a time-ordered series of interactions between actors in a network. Such interactions are referred to as relational events. A relational event minimally
contains information on the time of the event and the actors that are involved in the event.
As an illustration, we use the history data object in the remstats package. This data object is a randomly generated relational event history. A description of the simulated data can be accessed with
?history. Here, we read that history is a small simulated relational event history with 115 events. Besides information on the time and actors, for each event there is also information on the setting
and an event weight. We can view the first six events with:
#> time actor1 actor2 setting weight
#> 1 238 105 113 work 1.33
#> 2 317 105 109 work 1.64
#> 3 345 115 112 work 1.82
#> 4 627 101 115 social 1.25
#> 5 832 113 107 social 1.67
#> 6 842 105 109 work 2.30
We prepare the relational event history for computation of statistics for the tie-oriented model with the remify function from the remify package. Whenever the weight variable is present in the
edgelist supplied to remify, it assumes that we want to use these to weight the events in the computation of the statistics. In this example, we don’t want this, thus we set the weight to one for all
Besides the relational event history itself, relational event modeling often requires a second data object with exogenous information on the actors in the network. Information on the actors in the
simulated data example in remstats is stored in the info object. A description of the info data can be accessed with ?info. Here, we read that the info object stores for the 10 actors in the network
information on their age, sex, extraversion and agreeableness score. Moreover, extraversion and agreeableness is measured multiple times during the observation period. The time variable tells us when
the values change. We can view the attribute information for the first two actors with:
Compute statistics for the tie-oriented model
First, we compute statistics for modeling relational event history data with a tie-oriented model. The statistics that are requested need to be supplied to the tie_effects argument of remstats(),
specified in an object of class formula. This specification should be in the form ~ terms.
An overview of the statistics that can be computed for the tie-oriented model is available using the tie_effects() function or its help documentation ?tie_effects:
In this illustration, we start with requesting only one statistic: the inertia statistic. Most statistics can be tailored to the user’s needs. For example, lets view the description for the inertia
statistic using ?inertia. Here, we can read that the inertia statistic computes for every timepoint t for every pair of actors (i,j) in the riskset the number of past events. With the scaling
argument, one of the methods for scaling the statistic can be chosen. The consider_type argument is relevant when event types are in the dependent variable, which we do not consider in this example.
In this illustration, we will standardize the inertia statistic. To request this statistic, we define the formula as follows:s
Now, we have everything we need to compute our first statistic:
The remstats() function outputs a 3-dimensional array with statistics for the tie-oriented model. On the rows of this array are the timepoints, the columns refer to the potential events in the
riskset and the slices refer to the different statistics:
Our statistics object has 115 rows, corresponding to the 115 time points in the relational event history. It has 90 columns, corresponding to the 90 events in the riskset. The statistics object has
two slices, that is because the baseline statistics is automatically computed when the timing of the events in the relational event history is exact (unless removed by specifying -1 in the effects
formula) and saved in the first slice. The remstats() procedure assumes that the timing of the events in the relational event history is exact and the full likelihood is used in the estimation,
unless the argument ordinal in remify::remify() is set to TRUE.
We can view the names of the statistics that are in the statistics object with:
#> Relational Event Network Statistics
#> > Model: tie-oriented
#> > Computation method: per time point
#> > Dimensions: 115 time points x 90 dyads x 2 statistics
#> > Statistics:
#> >> 1: baseline
#> >> 2: inertia
Here, we see that, indeed, a baseline and inertia statistic are computed.
Since we did not request anything special for the riskset in remify::remify(), it consists of every directed pair of actors observed in the relational event history, which is 10*9 = 90 pairs. These
pairs are saved in the riskset attribute. We can ask for the first few lines of this riskset:
head(attr(out, "riskset"))
#> sender receiver id
#> 1 101 103 1
#> 2 101 104 2
#> 3 101 105 3
#> 4 101 107 4
#> 5 101 109 5
#> 6 101 111 6
Here, we see that the first event in the riskset is the event were actor 101 sends an interaction directed towards actor 103. The id column refers to the column in the statistic object that contains
the statistic(s) for this specific dyad. The first column in the statistic object refers to this first event in the riskset, the second column in the statistic object to the second event in the
riskset, and so forth.
Inertia is an example of an endogenous statistic: it is a function of the relational event history itself. Next, we are going to add a request for an exogenous statistic. For this we need the
exogenous information on the actors in the info object.
As an illustration, we are going to request the statistic for an effect of extraversion on sending events, i.e., a send effect. The description of a send effect is accessed with ?send. Here, we read
that we need to supply the variable for which we want to specify a sender effect and that this variable should correspond to a column in the attr_actors object that we supply. Thus, we specify a send
effect of extraversion with send("extraversion", attr_actors = info). Here, we specify the attr_actors object within the send() function. Alternatively, it can be supplied to remstats(). This is for
example useful if you want to compute a bunch of exogenous statistics using the same attr_actors object.
Statistics in the effects formula should be separated with the +. Finally, we add an interaction between the inertia() statistic and the send() statistic. This can be done by using the * or :
Compute statistics for the actor-oriented model
The procedure to compute statistics for the actor-oriented model is largely similar to what is written above, except that statistics have to be specified separately for the sender activity rate step
of the model and the receiver choice step of the model. The statistics requested for these two modeling steps need to be supplied to two different effects arguments, namely sender_effects and
receiver_effects, respectively.
An overview of the statistics that are available for the actor-oriented model in the two modeling steps can be obtained using the actor_effects() function or its help documentation ?actor_effects.
In this illustration, we start with requesting only one statistic for the sender activity rate step: the outdegreeSender statistic. First, lets view the description for the outdegreeSender statistic
using ?outdegreeSender. Here, we can read that, in the sender activity rate step of the actor-oriented model, the outdegreeSender statistic computes for every timepoint t for every actors i the
number of outgoing past events. With the scaling argument, one of the methods for scaling the statistic can be chosen.
First, we prepare the event history for computing statistics for an actor-oriented model:
To compute the outdegreeSender statistic for the sender activity rate step we supply it to the sender_effects argument of remstats():
The outputted remstats object is now a list with two elements: sender_stats and receiver_stats:
Since we did not request any statistics for the receiver choice step here, the receiver_stats object is empty. The sender_stats object contains the statistic array with the baseline statistic (again,
automatically computed unless ordinal = TRUE), and the requested outdegreeSender statistic:
#> Relational Event Network Statistics
#> > Model: actor-oriented
#> > Computation method: per time point
#> > Sender model:
#> >> Dimensions: 115 time points x 10 actors x 2 statistics
#> >> Statistics:
#> >>> 1: baseline
#> >>> 2: outdegreeSender
The dimension of out$sender_stats is 115 x 10 x 2. On the rows we have the timepoints, the columns refer to the actors that can be senders and the slices refer to the different statistics.
Lets extend our model and also request a statistic for the receiver choice step:
sender_effects <- ~ outdegreeSender()
receiver_effects <- ~ inertia()
out <- remstats(sender_effects = sender_effects, receiver_effects = receiver_effects, reh = reh)
We can access the statistic computed for the receiver choice step with out$receiver_stats. In this step, the baseline statistic is not automatically computed (and not defined). Hence, the dimensions
of the statistics object for the receiver choice step are 115 x 10 x 1. On the rows we have again the timepoints, on the columns now the receivers and on the slices the statistics.
Note that the computed values of the statistic in the receiver choice step are equal to the values for this receiver, given the current sender. For example, lets review the first six lines:
# Set the column names equal to the receivers
colnames(out$receiver_stats) <- attributes(reh)$dictionary$actors$actorName
# Set the rownames equal to the senders
rownames(out$receiver_stats) <- reh$edgelist$actor1
# View the first six lines
#> 101 103 104 105 107 109 111 112 113 115
#> 105 0 0 0 0 0 0 0 0 0 0
#> 105 0 0 0 0 0 0 0 0 1 0
#> 115 0 0 0 0 0 0 0 0 0 0
#> 101 0 0 0 0 0 0 0 0 0 0
#> 113 0 0 0 0 0 0 0 0 0 0
#> 105 0 0 0 0 0 1 0 0 1 0
At the first timepoint, the inertia statistic for all receivers given the current sender (actor 105) is zero because no prior events have occurred. At the second timepoint, the sender is again actor
105. Now the inertia statistic is equal to the 1 for the receiver of the first event (actor 113). At the third timepoint, the inertia statistic is again zero for all receivers because now the sending
actor is 115, who did not send any prior events.
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A current of 2 mA was passed through an unknown resistor which dissipa
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Factoring Expressions
In order to understand the following lesson, you should already know:
(i.) a general method for removing monomial factors from an expression
(ii.) two methods for factoring expressions with very special (but quite commonly occurring) forms:
• trinomials (ax^ 2 + bx + c)
• difference of two perfect squares (u^ 2 v^ 2 )
All we want to point out here is that the overall strategy for factoring any algebraic expression is to apply the methods listed above in order. So,
(i) first remove all monomial factors from the expression
(ii) check whatever is left against any of the special patterns that you know
By following this strategy, the more difficult, tedious, and intuitive analysis of the second type in the list above will always be done on the simplest possible expressions.
We ve already alluded to this strategy in the previous documents on factoring, and shown one or two brief examples. Here is a few more short examples to help you understand the overall strategy
Example 1:
Factor completely: 3x^ 4 y + 6x^ 3 y 45x^ 2 y.
We see that 3, x^ 2 , and y, are each monomial factors of all three terms. So, step (i) in the general strategy leads to
3x^ 4 y + 6x^ 3 y 45x^ 2 y = 3x^ 2 y(x^ 2 + 2x 15)
Now, we need to apply step (ii) of the general strategy to the expression, x^ 2 + 2x 15, remaining after we have accounted for the common monomial factors. This is a trinomial in x, which it may be
possible to rewrite in the form
x^ 2 + 2x 15 = (x + a)(x + b)
if we can find two whole numbers, a and b, such that
ab = -15 and a + b = -2.
You can easily verify that a = 5 and b = -3 works. Thus
x^ 2 + 2x 15 = (x + 5)(x - 3)
Thus, for the original expression we now have
3x^ 4y + 6x^ 3y 45x^ 2y = 3x^ 2y(x + 5)(x 3)
which is factored as completely as is possible. We know that no further factorization is possible, because the factors here which are not monomial factors contain no products at all.
Notice that when it came time to check if any of our special pattern formulas might apply, we needed to check them against the quite simple expression, x^ 2 + 2x 15, rather than the much more
complicated original expression. To be sure, (x + 5) and (x 3) are factors of the original expression, but its complexity (due to the presence of common monomial factors in each term) would make it
very difficult to recognize these binomial factors in that expression.
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Percentage Fields in Flows
I do a LOT of work with percentage fields… calculating margins, markups, markdowns, fees, discounts etc.
Most of the time, due to the size of the end result formula fields I can not do these percentage calculations as formulas.
• Base Amount
• Add Fee A @ 2.75% of Base Prices
• Add Fee B @ 1.258% of Base Price
• Subtotal
• Add Fee C @ 4.85% of Subtotal Price
• Total
• Discount Total by 1.035%
• Gross Amount
• Then multiply the Sale Price by 1/1.095 to get the Net Amount
If you’ve ever been to Dreamforce, remember the shock of getting your hotel bill and seeing all the various taxes detailed out… something like that.
I’ve been using Flows to clone records…
But I have to do something like
If Party X is paying the fee then put the percentage into Fee A% field, otherwise put the fee into Fee B% field
If Party Y is paying the fee then do the opposite.
I could NOT work out why my percentages were always incorrect in Flows
So I’ve done some testing.
I have 4 percentage fields - each of them are 3 decimal places:
• Set a number variable as a passed in percentage field
• Input a percentage field
• Update percentage field from the previously saved value
• Update a percentage field from a formula
Update the Record
1 Update the record with the exact value from the same field on the Get Record 2.55 2.55
2 Update the field with the passed in variable percent 2.789 2.789
3 Update the field with the input percent 10.543 10.543
4 Update the field with the formula percent. The formula is just the value of the exact field that is being updated, from the Get Record. 8.253 0.083
 Passed in Variable then into a formula (not pictured) 2.789 2.789
End Value
WHY oh WHY is there a difference! Why when I use a Variable in a formula does it stay as the full number, but using a value retrieved from a record in a formula converts it to the percentage???
If you need to deal with percentage fields in a formula, IF the value comes from a Get Record ensure you multiply the value by 100 before using it in an Update in Flow. if the value comes from a
Variable and the Variable is then used in a formula, don’t multiply it by 100. Confusing!
This may have seemed obvious, but it wasn’t until I went through it step by step.
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what is the formula for calculating percentage filling in cement mill
Table 1 permits 40 percent fill for over two conductors. From Table 4, 40 percent fill for trade size 11⁄4 RMC is 0.610 in., and from Table 5, the cross-sectional area of a 10 AWG THHN conductor is
0.0211 in.2. The number of conductors permitted is calculated as follows: 0.610 in.2 / 0.0211 in.2 per conductor = 28.910 conductors
An example of how you can calculate the slurry flow/volume of a given SG, %Solids and Tonnage. EXAMPLE: 3,400 tons of dry solids is processed in 24 hours in water with a specific gravity of 1.0 and
the concentration of solids by weight is 33% with the solids having a specific gravity of 3.7 then using the formula below:
Step 1. Calculate the volume of the space to be filled. For example, find the cubic footage volume of a backfill area that is 8 feet wide, 6 feet deep and 50 feet long. The volume of a rectangular
cubed shape is found by the formula v = l x w x d, where v represents volume, l is the length of the trench, w is the width and d is the depth.
Ball Mill Power Calculation Example #1. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch (6350
microns). The required product size distribution is to be 80% passing 100 mesh (149 microns). In order to determine the power requirement, the steps ...
Sometimes, you may get the price exclusive of tax. In this condition, you can easily calculate the sales tax by multiplying the price and tax rate. Select the cell you will place the calculated
result, enter the formula =B1*B2 (B1 is the price exclusive of tax, and B2 is the tax rate), and press the Enter key. And now you can get the sales tax ...
Casing capacity = 0.8679 ft3/ft. Step 2. Determine the slurry volume, ft3: Slurry vol, ft3 = 500 sk x 1.15 ft3/sk Slurry vol = 575 ft3. Step 3. Determine the amount of cement, ft3, to be left in the
casing: Cement in casing, ft3 = (3000 ft - 2900 ft) x 0.8679 ft3/ft Cement in casing, ft3 = 86.79 ft3. Step 4.
1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and
selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and …
How to Size a Ball Mill -Design Calculator & Formula A) Total Apparent Volumetric Charge Filling – including balls and excess slurry on top of the ball charge, plus the interstitial voids in between
the balls – expressed as a percentage of …
Example: Two compartment cement mill Diameter = 4.05 m Length = 12 m Filling degree = 196 tons at 30% filling degree Mill speed = 16.27 rpm C = 0.223 (from the figure) Then: K = 0.223 * 196 * 4.05 *
16.27 = 2.88 KW 8.2 Calculation of the Specific Consumption of Energy per Ton of Clinker: The specific consumption of energy per ton of clinker ...
The easiest way of calculating a percentage of an amount for me is to divide the amount by 100 to get the 1% value and then multiply that with the percentage amount. starting amount = 240 ( )
percentage = 12%. divide 240 to 100 to get the 1% value. 240 ÷ 100 = 2.40. one percent of 240 = 2.4. calculate what 12% is by multiplying the one ...
The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where
balls will not fall away from the mill's shell. Result #1: This mill would need to spin at RPM to be at critical speed. Result #2: This mill's ...
Then, type the following formula: = (B2/C2)*100. Hit Enter to get the student's score as a percentage of the total possible points. You can then use the fill handle to drag the formula down to the
rest of the scores. Multiplying by 100 shows you the percentage— not the decimal.
Re: Calculate percentages based on cell color in column. pretty sure you can't base a calculation on a colour applied by conditional formatting. You need to use the same criteria in a formula as your
CF uses. So in your DL worksheet, cell G73, you can use the formula: Formula: Please Login or Register to view this content.
Know that the formula for calculating concrete consists of the following: ( (length x width x height)/27) x 1.05 = Total concrete needed plus 5 percent extra. The length, width and height are
measured in feet. Generally, you should purchase 5 percent more concrete than you need to complete your concrete project to ensure that you have enough to ...
A percentile of a data set will have the same percentage of data falling below it (e.g., 50th, 75th, 90th percentiles). To calculate a desired percentile, use the following formula: Desired
Percentile = (# of Data Points + 1) x Desired Percentile = # from the bottom
The most common AFt and AFm phases in hydrated cement are: Ettringite: ettringite is present as rod-like crystals in the early stages of reaction or sometimes as massive growths filling pores or
cracks in mature concrete or mortar. The chemical formula for ettringite is [Ca 3 Al(OH) 6.12H 2 O] 2.2H 2 O] or, mixing notations, C 3 A.3CaSO 4.32H 2 ...
Taper Calculations. A cylindrical taper is a piece of material that decreases in diameter in a cone shape. Proportions for tapers are often expressed as Taper per Foot (TPF), or Taper per Inch (TPI).
Taper calculations most often expressed in terms of the difference in diameter at the opposite ends of the taper and the length of the taper ...
- Ball top size (bond formula): calculation of the top size grinding media (balls or cylpebs):-Modification of the Ball Charge: This calculator analyses the granulometry of the material inside the
mill and proposes a modification of the ball charge in order to improve the mill efficiency:
Objective: Calculate the estimated volume of C&D debris material that will result from the demolition of the Bunker Complex. The structure and concrete slabs have the dimensions shown below and
discussed herewith. The bunker building itself has a height of 12 ft and maintains a 6-in concrete wall and floor thickness throughout.
Concrete Ingredients Calculation. For Cement, Sand and Coarse Aggregate. This is a Volumetric Calculation. Assuming we need 2 m 3 of concrete for M20 Concrete Mix, (Mix Ratio, M20 = 1 : 1.5 : 3)
Total Part of the Concrete = 1+1.5+3 = 5.5 Parts. Therefore, Cement Quantity = (Cement Part / Concrete Parts ) * Concrete Volume
How to Calculate Charge Volume in Ball or Rod Mill | Mining …. The charge volume of a ball or rod mill is expressed as the percentage of the volume within the liners filled with balls or rods. When
the mill is stationary, the … »More detailed
Calculation of volume of concrete for beams:-. Beams are generally in a rectangular shape, to calculate the volume of concrete required for beams calculate the top or bottom surface area of the beam
and multiply with the depth of beams. Volume of concrete for rectangular beam = Surface area x Depth = length x width x depth = 4×0.5×0.4=0.8m 3.
The world famous cement manufacturers such as KHD Humboldt Company, F.L. Company, and F.C.B Company in propose their formulas to calculate SAR; in practice, any of the following formulas can be used
to compute SAR: where is the SAR, and are the mass or mass percentage of K 2 O and Na 2 O, respectively, and is the permissible maximum ...
Suppose we need to fill the plot by 3'5" = 3.41 feet. ∴ Depth of the plot = 3.41 feet. Volume of plot to be filled = 1200 x 3.41 = 4092 cft. This volume is for loose state. For full compacted state,
required quantity of soil/sand for filling the plot = 4092 x 1.30 = …
Fly Ash is a product or furnace slag received form metal factories and power plants by burning or combustion of coal as fuel. The maximum percentage of fly ash or 15% by weight of cement shall be
replaced by fly ash. #more fly ash reduces the strength of concrete and durability of concrete.
The Test 'normal consistency or standard consistency test of a cement paste' is used to determine the required quantity of water to produce a cement paste of standard consistency.It is necessary to
fix the quantity of water to be mixed in cement paste in each case for finding out initial setting time of cement, final setting time of cement, soundness of cement …
This is because fine coal from the the coal mill usually has 4-5% of raw meal dust contamination from the hot gases ex the preheater top cyclones. Alternatively, if the percentage of dust
contamination can be accurately determined, this value can be used to correct the fine coal tonnes fired and calculate the dust-free % ash.) Regards, Ted. Reply
The H parameter is used in the formula to calculate the critical outlet diameter to prevent arching. It is calculated from an abascus, knowing the hopper discharge angle calculated in step 2. Figure
5 : Abascus for H parameter calculation. Note : the green line corresponds to wedge shaped hoppers, the red line to axi-symetric hoppers (cone) ...
To determine the total length of vertical bars, the following formula is used :-Total length = Length of one bar x Numbers of bars. Total length = 1.0 x 9 = 9 m (bars with diameter 10m) Now, the
total weight of bars will be calculated as follow :-Weight of 12 mm bars + Weight of 10 mm bars. To determine the length for 1 meter, the formula is d ...
The liquid percentage at 1450 0 C can be estimated using the formula. % Liquid content (1450 0C) = 3.0 x A+2.25 x F+S. Where A=Al 2 O, F=Fe 2 O 3, S= MgO+K 2 O+Na 2 O+SO 3 In Clinker. The normal
range of liquid phase is 22-27%. Burnability is a reference value for raw meal indicating how difficult it is to burn.
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R language
Having a numerical data set, a histogram is a data representation
from distribution point of view. Numerical data set is devided in
range of values first. Histogram graphic will show on abscissa (x axis)
the range and on the y axis how many values are in respective range.
Histogram in R can be build with function hist.
hist(x, …)
# S3 method for default
hist(x, breaks = "Sturges",
freq = NULL, probability = !freq,
include.lowest = TRUE, right = TRUE,
density = NULL, angle = 45, col = NULL, border = NULL,
main = paste("Histogram of" , xname),
xlim = range(breaks), ylim = NULL,
xlab = xname, ylab,
axes = TRUE, plot = TRUE, labels = FALSE,
nclass = NULL, warn.unused = TRUE, …)
Most intresting thing from this syntax is that for hist function, is enough
to have just one argument, x argument (a vector of values) and it will do something.
In hist(x, …) means it need x, and optional “…” means other arguments.
Below code:
x <- c(7,9,12,21,5,35,31,22,14,42,37,33,29)
will generate:
From this result that, hist automatically devide data in 5 ranges.
For each range it calculate frequency:
• for 0-10 frequency=3, there are 3 numbers in this range (7,9,5)
• for 10-20 frequency=2, there are 2 numbers in this range (12,14)
• etc
In contradict with this simplicity, syntax end with ” …) ” , which means it still can accept more arguments, usual graphic.
X axis label is automatically named based on input vector name
Y axis label is automatically named “Frequency”
Related to this below code specify explicitly axis labels:
x <- c(7,9,12,21,5,35,31,22,14,42,37,33,29)
hist(x, xlab = "Years", ylab="Number of entrepreneurs" )
This will generate:
Whatever is the number of values in input vector x, the histogram will have only 5 labels automatically generated on x axis, for example below code:
x <- c(7,9,12,21,5,35,31,22,14,42,37,33,29,53,76,87,82)
We see there is still 5 x axis labels, but values are divided now in 10 ranges.
Other example related to ranges and number of x axis labels autogenerated, let’s run code:
x <- c(7,9,12,21,5,35,31,22,14,42,37,33,29,53,76,87,82, 123,435,562,577,788,987,675,561,998,889,8)
This will generate:
This time we have 5 x axis labels and also 5 ranges, despite of fact that we have many more input values in x vector.
Installing R version 4.0.2 on Windows
• Go to https://www.r-project.org/
• Click on link “download R” or “CRAN mirror“
• Click on a link for appropriate mirror
• In “Download and Install R” section click link “Download R for Windows“
• Under “Subdirectories:” click on link “base“
• Click on link “Download R 4.0.2 for Windows “, save file “R-4.0.2-win.exe” on PC
• Double click on “R-4.0.2-win.exe“
• Select language, for example English
• Read GNU General Public License, click Next
• Select Destination Location, for example C:\Program Files\R\R-4.0.2 then click Next
• Select Components to install, as my PC is 64 bits, I will choose “Core Files“, “64-bit Files“, “Message translations“. I don’t checked “32-bit Files” as for my use on 64 bits PC is ok “64-bit
Files“. Click Next
• For “Startup options” choose for the moment No(accept defaults) and click Next
• For “Start Menu Folder” accept “R”, click Next
• You can “Select Additional Tasks” like below
• At the install end click Finish
To start R:
• click on schortcut “R x64 4.0.2” from desktop
• or click Start > R > “R x64 4.0.2”
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Foundations of ML #1 - What is Machine Learning?
The first issue on the Foundations of Machine Learning series, exploring the most basic concepts in the field.
Read →
Very easy to understand, and not the most simple subject. Nicely done, Daniel and Ale!
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Thanks man. We put a lot of effort to get this as simple as possible without dumbing it down.
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As Einstein put it, "as simple as possible, but not one bit simpler."
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Thisnwas great, very well explained and from a point of view I didn't read before.
Just a thought before I forget to mention, that could be interesting to discuss in the future. I would like if at some point there is a discussion on how the theory of ML might fit with Complexity
theory. For instance, most of us believe P!=NP. However, if Metric TSP has a polynomial approximation then P=NP.
So a question I ask myself usually is what happen if we find a ML algorithm that approximates TSP amazingly well, it cannot be categorized as a polynomial algorithm unless is a Decider of the
problem, which means 100% accuracy.
So I always have the question, is it possible to reach this 100% in a problem like this (probably not) and if not, then Why? What would be a mathematical proof for that, again from theorethical
point of view, assuming we have all the data that we might need, all the features that we need, etc.
Thank you for such a great article
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Wait, metric TSP has a 2-approximation, doesn't it!?
Anyway, the general question is super interesting. I'm not aware of how complexity theory factors in probabilistic correctness. That's something I'll have to look into.
Basically, it's one thing to say you always have a bounded approximation (for some bound) than saying with very high probability you can have a polynomial solution. I'm sure there's a complexity
class for that, i.e. problems you can solve polynomially with some bounded probability. But I don't know the relationship between those complexity classes and the traditional P and NP.
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First, yes I put the wrong name my bad, I was talking about general TSP. Second, for the probabilistic aspect I have other question apart from P and NP that is part of some things I have been
thinking lately I will put it in other reply. But the original question was more in the sense that I am not familiar how is Complexity Theory studied in ML.
For instance, imagine that you have a LLM for TSP , and you have all the data in the world and all the features that can be extracted from a graph no matter how big it is but extracted
polynomially. Then you spent 50 years training that big LLM, even hundreds of years and pass generation to generation hehehe. Then it finally finished training, now in inference is extremely fast,
extracting the features is polynomial because we enforce it, and inference is very fast (I do not know how to put a bound for inference that is one question).
So from this example come the questions:
1-) If the inference is polynomial and we assumed that we have all data need it for the ML to generalize amazingly, then if we prove somehow that the ML approximates polynomial TSP solution for all
cases then is it P=NP?
1.1-) Can be proven that an ML algorithm has generalization that approximates polynomially the solution to a problem assuming all data? Or to the contrary it can be proven that this never will
happen no matter the problem?
1.2-) If we can prove 1.1 somehow, then the ML algorithm is polynomial in inference, but was ETERNAL :) during training. Then for Complexity theory we have an algorithm that is not polynomial in
the traditional sense, but in the practice perspective it is. Which is a paradox since it makes TSP polynomially approximated in practice and therefore P = NP in practice, but in theory we haven't
found a polynomial algorithm for TSP since the graph is the input during inference, but for the entire algorithm the input was the dataset + the graphs during inference.
2-) Can actually be proven that the inference of an ML algorithm is polynomial? How a prove for that will look like?
Again is all in the air, I do not even know if the questions make sense, maybe there is a straight answer like: Dah, for this theorem you cannot obtain polynomially bounded generalization in a
regression problem and therefore none of this makes sense hehehehehe.
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Good questions! So, here's my superficial understanding at the moment.
First, you can definitely prove that a machine learning model inference cost is polynomial in the size of the input. All neural networks are, in fact, because it's just multiplying the input times
a huge but finite number of matrices, so that's trivially polynomial although with a very large constant. I cannot think of any ML model that's not polynomial in inference, actually.
Now, regarding training, the problem is that your input is no longer an example, but the whole dataset. And then, e.g., all NN training is polynomial in dataset size, trivially (a constant number
of epochs).
So LLMs are polynomial to train (wrt to dataset size) and polynomial to run (wrt to input size). Incidentally, that's actually one of the reasons why transformers are trivially not Turing-complete.
There's no while-true in there.
Now to the latter question, can we prove that the performance of a probabilistic model is a polynomial approximation of the optimal solution? E.g., can we prove that when an LLM outputs, say, a
list of edges that form a cycle, are those edges a polynomial approximation of TSP?
Seems super hard to do for me because, even if we had some formal guarantee that, say, more than 95% of the edges always belong to the optimal cycle (which I have no idea how to prove, but suppose
that's what 95% correct means), still the other 5% allows me to build an arbitrarily bad solution. So I don't see how we can escape the fact that anything less that 100% will be able to include
arbitrarily bad solutions.
Again, this is just off the top of my head, I need to read a lot more on this topic. But you just opened a can of worms for my next Algorithm Design semester :)
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Truly masterful. It provides a truly intuitive approach to understanding the basics of ML.
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Class 7 Math Question | Class Seven Math Question Paper Set
Hits: 11981
Class 7 Math Question | Class Seven Math Question Paper Set
Dotnepal presents model Question papers for class 7 students. These questions will help students to practice math questions and solve on their own so that they built up confidence in mathematics. Any
suggestions will be appreciated.
Class 7 Math I Question
Level Subject Year
Lower Secondary Math I 2077
Class Seven Math Question
Class Seven Math Question Download PDF
The End
One comment
1. I need a model question of class 7 for giving entrance in 2079
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What to do after Singapore Math - Singapore Math Live
What to do after Singapore Math
What to do after Singapore Math
Are you starting to find the Singapore Math curriculum to be difficult to teach? Is your child feeling a like math is a struggle? That’s not uncommon, and it is important that you (and they) know
that this curriculum is asking much more of them than other curricula. The Singapore Math curriculum is worth doing because of the problem solving and critical thinking skills and because it teaches
math as a subject to comprehend, not to memorize. What do you do, though, when it is just time to make a change?
Consider using Singapore Math Live support so that you can continue using this curriculum and see the results that you were hoping for when you chose to use Singapore Math. But if it is time to move
on to a different learning environment, and especially if your student has completed Primary Math 4 or above, there are many options! It can be very helpful to have outside accountability. Often an
older-elementary aged student responds better when “not-parent” is available to teach the concepts, answer questions, and grade their assessments.
Since Singapore Math curriculum does not follow the “American” sequence of courses, a student may have trouble finding a perfect fit for their next math class.
Consider High School Math Live
I also founded High School Math Live which offers accredited, live and interactive classes that meet twice each week. The teachers teach the lesson, answer homework questions, and grade the weekly
assessments. We use traditional American textbooks and many families have chosen High School Math Live as their math solution. Our classes start with Pre-Pre-Algebra (an on-level 7th grade / honors
6th grade course) and continue through Calculus II. On the High School Math Live site I have written “Course Preparedness Documents” for each course to help parents to determine whether the student
has mastered the concepts required to register for that course. Click here to go to the website and see the Course Preparedness Documents.
We want to help you with this transition regardless of what you choose, so below are some options that you have, as well as some scenarios to help you think it all through. Happy reading!
Scenario 1 (Primary Math 4)
Student finishes Primary Math 4 (including a significant amount of the Intensive Practice and Challenging Word Problems books) and wants to move out of the Singapore Math curriculum:
Scenario 2 (Primary Math 5)
Student finishes Primary Math 5 (including a significant amount of the Intensive Practice and Challenging Word Problems books) and wants to move out of the Singapore Math curriculum.
Option 1: A student who completes Primary Math 5 has seen all of the necessary skills to prepare them for Pre-Algebra. (Click here for more information about High School Math Live’s Pre-Algebra
Option 2: A student who completes Primary Math 5 but would like another year to build confidence in their mathematical ability could consider High School Math Live’s Pre-Pre-Algebra course. (Click
here for more information about the PPA course.
A note about sequencing:
• If a student takes Pre-Algebra in the 8th grade, they will be “on-level” for taking Algebra I in the 9th grade.
• If a student takes Pre-Algebra in the 7th grade, they will be able to take Calculus I during their high school years.
• If a student takes Pre-Algebra in the 6th grade, they will have time to complete Calculus II during high school.
Scenario 3 (Primary Math 6)
Student finishes Primary Math 6 (including a significant amount of the Intensive Practice and Challenging Word Problems books) and wants to move out of the Singapore Math curriculum.
Option A: If a student has completed Primary Math 6 but it would be helpful for them to gain confidence in math and have the opportunity to have the concepts presented again as they transition into
live classes, then Pre-Algebra would probably be the best fit for them. (Click here for more information about High School Math Live’s Pre-Algebra course.)
Option B: If a student has completed Primary Math 6, including a significant portion of the IP and CWP books, they are well prepared for Algebra I. Primary 6 does not include a study of exponent
rules or integers, so either adding a unit for those concepts or just being aware that when those concepts are presented in Algebra I the student will need to devote a little extra time to learning
those concepts will be sufficient. A student who has learned to think mathematically will not have difficulty catching on to the rules that these concepts build on. (Click here for more information
about High School Math Live’s Algebra I course, including honors and on-level.)
Option C: If the student is about to start 10th grade or above, it is important that they take Algebra I as their next course so that their transcript will demonstrate enough high school level math
courses. This may require extra time and effort, but having excellent instruction and feedback will give the student every opportunity to succeed. (Click here for more information about High School
Math Live’s Algebra I course, including honors and on-level.)
Scenario 4 (Dimensions Math® 7)
Student finishes Dimensions Math® 7 and wants to move out of the Singapore Math curriculum.
NOTE: Dimensions Math® 7 includes much but not all of Algebra I concepts so on its own, it will not count as an Algebra I credit – see Scenario 4 below for how to adapt and earn an Algebra I credit.
Option A: If the student completes Dimensions Math® 7 but did not have a high level of accuracy, or if they understand the basic concepts but struggle to comprehend the harder parts of each unit, an
on-level Algebra I class will be a good next-step. There will be a significant amount of review of the concepts in Dimensions Math® 7 which will strengthen their comprehension and give them
confidence. (Click here for more information about High School Math Live’s Algebra I course, including honors and on-level.)
Option B: If the student completes Dimensions Math® 7 with a high level of accuracy and strong comprehension of the units, especially if they also complete most of the Workbook “Further Practice”,
“Challenging Practice” and “Enrichment” problems, the next course will still be Algebra I, but consider an honors-level course. The student will know many concepts that will be presented in Algebra
I, but there are key concepts that they have not seen that would prevent them from skipping past Algebra I. (Click here for more information about High School Math Live’s Algebra I course, including
honors and on-level.)
Scenario 5 (Dimensions Math® 8)
Student finishes Dimensions Math® 7 and 8
Note: Dimensions Math® 7 and 8 include many concepts from Geometry. Also, the algebraic concepts are introduced and developed to the depth and rigor of an Algebra II course. Not all of Geometry or
Algebra II is introduced, but strong completion of Dimension Math® 7 and 8 will allow a student to be awarded credit for Algebra I, as well as a half-credit each for Geometry and Algebra II.
Option A: If the student is solid in their comprehension, but the textbook assignments felt difficult; if the student did not have a high amount of accuracy (80 or above) on the homework and tests;
or if the student completed the Textbook but either none or very little of the Workbook, the student will be a good fit for Geometry as their next class. The algebraic concepts that they were exposed
to in Dimensions Math® 7 and 8 will be reinforced in the early part of Algebra II, and the Geometry concepts will be reintroduced entirely. Consider Honors level for both courses since much of the
material will be familiar already. (Click here for more information about High School Math Live’s Geometry course, including honors and on-level.)
Option B: If the student completes Dimensions Math® 7 and 8 with a high amount of accuracy and also completes much of the Workbook, the student will be a good fit for a new class that Brenda
Barnett, founder of Singapore Math Live and High School Math Live is creating. The course is called Math 9 Honors and it will include one semester of Honors Geometry content and one semester of
Honors Algebra II content. Information about registration for this class can be found by here. This class will meet online twice weekly in a live and interactive format. Mrs. Barnett will teach the
Honors Algebra II semester, Mrs. Kim Kirby will teach the Honors Geometry component. (Click here for more information about the Math 9 Honors course.)
Scenario 6
Parent would like to piece together an Algebra I curriculum from Dimensions Math 7 and 8 without including the Geometry portions.
If the student completes selected portions of Dimensions Math 7 and 8, they can earn an Honors Algebra I credit. By omitting selected portions, they will not be exposed to concepts that are included
in Dimensions Math 7 and 8 but are also taught as part of a traditional Geometry class. After completing these chapters, the student will be prepared to take Geometry.
• Dimensions 7A: Complete Chapters 1-7; Omit Chapter 8
• Dimensions 7B: Complete Chapter 9-11, 14-17; Omit Chapters 12-13
• Dimensions 8A: Complete Chapters 1-4,
• Dimensions 8B: Omit
Scenario 7
It’s just not clear what scenario fits your student!
If the student has not cleanly finished a particular level and it feels overwhelming to try to decide what course they fit in to the best, High School Math Live can help! HSML classes start with
Pre-Pre-Algebra (an on-level 7th grade / honors 6th grade course) and continue through AP® Calculus AB. HSML has written Course Preparedness Documents for each course to help parents to determine
whether their student has mastered the concepts required to register for that course. (Click here to go to the HSML website and see the Course Preparedness Documents.) By taking a live and
interactive, online class, the student has outside accountability as well as excellent instruction and feedback from a credentialed and experienced teacher.
I know that changing curricula can be stressful. If you have any questions about any of this, please don’t hesitate to reach out to us.
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Computational Complexity
UMCP will have all classes on campus this Fall. There is a Mask Mandate. All students and faculty have to get vaccinated unless they have a health or religious exception. 92% are vaccinated, which I
interpret as people are NOT abusing the exceptions (though I still wish it was higher, and it may go higher). (ADED LATER- right after I posted this I got an email saying that UMCP is now up to 97%).
Those NOT vaccinated have to get tested - I think twice a week.
Now that we are back in the live-classroom, here are some thoughts about teaching on zoom.
I taught on zoom:
Spring 2020: The last half of both Ramsey Theory and Automata Theory(Reg, CFG,P,NP,Dec,Undec)
Fall 2021: Cryptography
Spring 2021: Honors Discrete Math and Automata theory
a) I taught in the usual time slot but I recorded the lecture so those who could not make it (more common during the pandemic) could still see it. Attendance was low, verbal interaction was low, but
chat-interaction was very good. Looking into if we can do a chat in an in-person class. I was recording lectures before the pandemic and will keep doing so.
b) My exams were open-notes, open-book, open-web. That cuts down on ways they can cheat, though they can still phone-a-friend. Or ask their cat. Unusual-but-correct answers can happen, as I discussed
in this blog.
c) I gave my muffin talk a few times on zoom. In person it goes very well as my enthusiasm is contagious. On Zoom that affect is dampened so the audience was more sedate. I gave it as a special
lecture to High School students and to my REU students. Note that it was NOT part of a class so the usual motivation to learn it to do the HW is gone. Hence its more important they be excited about
d) In person I carefully make sure that I wear a funny T-shirt every day, and its a diff one, and usually a math one to (if possible) match the topic. On Zoom I did not bother, though I sometimes
used wallpaper to match the topic.
e) I had to make up slides for Aut theory and for some of Discrete Math. For Crypto I already had slides. I like the slides I made up and will use them in the future. But see next point.
f) In Discrete Math I went faster than usual- perhaps because its on slides, perhaps because there were less questions since it was on zoom, perhaps because Emily my TA was so awesome that they had
less questions. (She is very interested in education and did a guest post about the pandemic and education here.) As a result I actually learned and presented the proofs that (1) the e is irrational
(my slides are here) and that Liouville numbers are transcendental (my slides are here). While I enjoyed learning those theorems and I think the students understood them on some level, I will slow
down next time.
g) Ramsey Theory: It is impossible to teach the Poly VDW theorem on slides, so I had to omit that part of the course.
h) Bottom Line: Did the students learn more? less? the same? My impression is that the students learned about the same, but really really didn't like it. And thats legit- that is NOT just students
Guest blogger Varsha Dani tells us why it's never too late.
This week, I am starting as an Assistant Professor at RIT and I am super excited about it. What's the big deal, you are probably thinking. Don't lots of people get hired in tenure track positions
every year? Sure. But the difference in my case is that I got my Ph.D. in 2008.
Why didn't I look for a position right away? There were a number of reasons. I was burned out. I was going to have a baby. The job market was not particularly good that year. I would have had a
two-body problem. And most importantly, I thought it was just going to be a short break. I thought that there was no rush...
What did I do in those intervening years? A lot of things. I spent a lot of time with my kids, including part home-schooling them for some time. I found some new interests, both in Math and CS and
outside. I found new collaborators and did some research in new areas, just for fun. Sometimes I was funded for it through grants, but mostly I wasn't. I wrote a lot of Computer Science and some math
materials for Brilliant.org. I organized math clubs at my kids' elementary and middle schools. I hiked. I wrote poetry. I never intended nearly 13 years to go by, but somehow they did. At some point
I remembered that I had meant to take a short break, not to give up on being an academic altogether. But by then it seemed too late. At the beginning of my meant-to-be-short hiatus, I used to
jokingly refer to myself as a "scholar at large" but by the time a decade had gone by I had started to feel extremely isolated and being a scholar for its own sake was not something to joke about
anymore. Each year I rolled the job-search dice, but with each passing year it seemed more and more futile to do so, and more and more of an imposition to ask people to write recommendations for me.
And then, out of the blue, last year, I found a whole community of independent researchers who, like me, were pursuing their scholarly interests despite not being employed to do so, and who felt
unapologetically unembarrassed, even proud of it. And, even more out of the blue, this year I got an offer. And now, this Fall, I am actually doing it. I am actually an academic! To be honest, I feel
more than a little trepidation about it, mixed in with the excitement.
So why am I telling you this? Partly to celebrate. Partly to publicly thank my spouse, who has been extremely supportive of all my decisions (and all my actions that were born of indecision) over
many years. Partly to give a shout-out to the wonderful folks at the Ronin Institute who helped me remember that we do science (and computer science) because we love it, not because it pays the
bills. Ironically, I believe I needed to be reminded of that before I could get a job! But mostly, I'm writing this to reach out to anyone out there who thinks that their decisions have led to a
one-way street they no longer want to be on. It may be hard, and there are, of course, no guarantees, but you won't know whether you can turn around, unless you try.
On a Jeopardy rerun with Alex Trebek the question (actually the answer, given the shows format) was (I paraphrase)
Who resigned his commision in the US Army Air Force in April 1941 after President Roosevelt publicly rebuked him for his views?
The answer (actually the question--Why does Jeopardy do this answer-question thing, drives me nuts!) was
Charles Lindbergh.
Alex Trebek then said Charles Lindberg's views on WW II were not politically correct.
This really struck me since Politically correct means, to quote Wikipedia:
a term used to describe language, policies, or measures that are intended to avoid offense of disadvantage to members of particular groups in society.
Wikipedia also adds that the term is generally used pejoratively with an implication that these policies are excessive or unwarranted.
But Alex Trebek is using the term to mean incorrect or perhaps incorrect given what we know now or if you think history is written by the winners, then perhaps incorrect since Germany lost the war.
But my point is that I really don't think the term politically incorrect makes sense here.
More recently I heard an anti-masker say
We should not let some woke school board take the right to not wear a mask away from parents and children.
Independent of if you are anti-mask-mandates or pro-mask-mandates, this seems like a strange use of the word woke which means, to paraphrase Wikipedia:
Having an awareness of racial prejudice, gender prejudice, sexual orientation prejudice, and the past and current discrimination they have and do cause.
I've seen it both positively and negatively.
The anti-masker's using of the term seems odd in that mask wearing is not a woke issue. Perhaps he should have said
We should not let some Nazi school board take the right to not wear a mask away from parents and children.
The term Nazi while not actually correct, conveys that the school board is authoritarian. However, he really could not use the term that since he was was a neo-Nazi and proud of it. That raises a
question: what pejorative term can a Neo-Nazi use when they want to say someone is Authoritarian? I ask non-rhetoically.
But I am getting off topic here- my real point is that the word woke is being used to mean Authoritarian which is not even close to its original meaning.
The above are examples of how a word in English may change its definition over time, which is not really news, but I found the examples interesting since I saw the origin of these words.
BILL, THIS IS A COMPLEXITY BLOG! SO TALK ABOUT COMPLEXITY. OR MATH!
In math do words change their meaning over time? Yes. Here are a few
Function: at one time `function' implicitly means a function that occurs in nature. So only continous and perhaps diff functions qualified.
Sets: probably similar.
Efficient: At one time this was an informal notion (Joe Kruskal's paper on MST (see here) is an example of that), then it seemed to be P or perhaps BPP. For some its linear or O(n log n) with a small
constant. Rather than say the notion changed, its more like it was never that well defined in the first place, and still isn't.
Constructive: The many diff definitions of this word could be a blog post of its own. In fact, I thought it was, but I could not find it. I did find lots of blog posts that use the word constructive
in diff ways.
Elementary: Also has many definitions, though they are closer together than for Constructive. This one I did do a post on here
A tweet that made me think.
If you think you don't trust scientists, you're mistaken. You trust scientists in a million different ways every time you step on a plane, or for that matter turn on your tap or open a can of
beans. The fact that you're unaware of this doesn't mean it's not so.
— Paul Graham (@paulg) July 26, 2021
The point here is subtle. We don't get on a plane because we "trust scientists", rather we do so because of the strong safety record of commercial aviation. I knew some physicists who won't get on a
commuter plane because they worry about the science. Never stopped me.
It is science that we trust to tell us why planes fly, or the water is our tap is (mostly) safe and healthy. I'm not a big fan of beans but not because of the science. Of course I trust science that
created the vaccines.
It's not just science, but solid engineering and lots and lots of testing.
Science isn't always right or consistent. When I was a kid not that long ago, we had nine planets in this solar system, dinosaurs were killed off by climate change and homosexuality was a mental
illness. Science is fluid, updating as we learn with new data, models and experimentation. Science is at its best when it doesn't trust itself.
Sometimes people say trust in science to reinforce their beliefs. I've seen smart people say "Trust in the science" about whether vaccinated people should wear masks with completely different
I'm a scientist, should you trust me? Let me quote another Paul G.
“There’s a slightly humorous stereotype about computational complexity that says what we often end up doing is taking a problem that is solved a lot of the time in practice and proving that it’s
actually very difficult,” said Goldberg.
The quote comes from a recent Quanta Magazine article about Paul's recent work with John Fearnley, Alexandros Hollender and Rahul Savani on the hardness of gradient descent. Even many NP-complete
problems these days can often be solved in practice.
Let's end with the quote attributed to statistician George Box, "All models are wrong, but some are useful". Science gives us ways to understand the world and we need to both trust in the science but
know the limitations of what it has to say.
(Disclosure: Harry Lewis was my PhD advisor. For a blog post on disclosures and bias see my post on that topic here.)
Harry Lewis has a book out: Ideas that Created the Future: Classic Papers in Computer Science
He picked out the 46 (why 46? Why not 46?) classic papers in computer science and, for each one, has a short article saying why its important, and then has the paper itself, though perhaps shortened
(leave out the boring parts) or in some cases he has an excerpt of a book (e.g., The Mythical Man Month which is why I blogged about that book recently here).
Harry Lewis has blogged about his book here where he points to my review which is in SIGACT News.
OR you an use my link to my review here.
The list of 46 papers had some constraints, so if you wonder why isn't X there it might have hit one of those constraints.
1) No paper past 1980 (he had to stop somewhere).
2) He preferred short readable papers to long or unreadable ones (don't we all!). Before thinking `Gee why isn't paper X in the book' go read paper X.
3) Some papers cost to much to get permission to reprint. My review points to one such paper that I found 5 links to on the web.
4) We don't need X papers on topic Y.
Of more interest is some papers that you had not heard of but we can now see are important.
For more thought, read my review!
For even more information, buy the book!
I sometimes have trouble recognizing faces, matching faces to people I've interacted with in the past. It's not a disease like prosopagnosia, I can certainly tell the difference between faces and
have no trouble with people I work with directly. But if I haven't seen someone in a while, I may not recognize them or confuse them for someone else. It's especially bad out of context, say running
into a professor in my campus on the streets of Frankfurt. It's gotten worse with age but I've had challenges my whole life.
I have my coping mechanisms. I start a conversation to get enough clues to figure out who I'm talking to. I'll google an image before I'm supposed to meet someone I haven't seen in a while. Sometimes
I'll just say "Remind me how to pronounce your name again". Sometimes I'll just say something embarrassing thinking the person I'm talking to is someone else.
Name tags are useful, if it isn't obvious you are looking at them. Zoom has been great--everyone's name is just there. I worry that 18 months of zoom meetings means I've lost much of my coping
ability, much the way I can no longer navigate by maps the way I used to.
We have technological solutions but mostly unable to make use of them. Through the magic of machine learning, computers have gotten extremely good at recognizing faces. Nevertheless Google Googles
actively prevented their one killer app, telling you who you were looking at, for privacy reasons. Perhaps they could limit it to people in your contacts with pictures you uploaded. It would only
recognize people you already know.
I know I'm not alone, and I'm writing this post so others won't feel alone. And next time you see me and I look confused, remind me of your name.
(I also posted this to the Less Wrong Website. At least I tried to- I don't quite know if or when it will appear there as its my first post there.)
Some papers result from taking two papers and combining them. Perhaps nobody else had read both of them so you can say something new! Or (looking over this post) it may guide people to two really
good papers, or in this case two really good posts.
This blog will draw from two excellent blog posts.
Scott Aaronson blogged on his website Aug 2, 2021 about blankfaces, people who let stupid or undefined rules dictate what you can do without apology (see his post for a better explanation). One
example that struck me I quote
No, I never applied for that grant. I spend two hours struggling to log in to a web portal designed by the world's top blankfaces until I finally gave up in despair.
Scott Alexander blogged on LessWrong on Nov 26, 2018 about Is science slowing down? which answers with an emphatic yes. His point is science-per-researcher is much less than it used to be, and he has
graphs and stats to prove it (see his post for the evidence and some speculation as to why this is) One of the reasons he gave struck me which I quote
Certain features of the modern academic system like undepaid PhD's, interminably long postdocs, endless grant writing drudgery, and clueless funders have lowered productivity. The 1930's academic
system was ineed 25x more effective at getting researchers to actually do good research.
(A commenter reminded me that Scott Alexander himself dismisses this reason. I do not.)
(I note that he gives other reasons as well, most notably for our field that the low hanging fruit is gone. Our lack of progress on P vs NP is likely that its a hard problem, rather than the reason
above. Of course, if its solved tomorrow by an outsider without funding, I will happily be proven wrong.)
Scott Alexander hits upon two types of blankfaces (without using the term).
Grant writing drudgery: the rules for how to submit get more and more detailed an onerous. This is what Scott Aaronson was alluding to. There are other ways its drudgery as well.
Clueless Funders: the people deciding who gets funded might not know the area (actually in my experience the grant I've reviews have been quite good and the problem is more not enough money to award
all that are deserving.)
SO I pose the following non-rhetorically as always
1) How big a factor is the slowing down of science that blankfaces get in the way?
2) What can we do about it?
As I write this I'm watching the women's pole vault final in the Olympics. Of the 15 women who made the finals, only four remain after two heights.
To expand on my tweet, I find the pole vault the purest of the Olympic Sports. No electronic monitors and timers, no biased judges, no video review. No points deducted for bad form or failing to
stick the landing. No disqualification for a false start or stepping over a line. Either you clear the bar without knocking it down, or you don't.
The high jump has similar properties, but just not as cool looking.
All four made the third height. Now onto 4.85 meters. An American, a Greek, a Brit and a Russian (sorry I meant member of the Russian Olympic Committee).
Back in the day, the TV coverage was rather limited. We'd only see the Americans and the medal winners with too much time spend on human interest backgrounds. Now in the streaming world I can watch
every competitor. The good and the bad. Live as it happens.
The Russian Anzhelika Sidorova just cleared 4.85 on her first attempt. So did the Brit Holly Bradshaw and the American Katie Nageotte. The Greek Katerina Stefanidi missed her first attempt but
decided to pass on the rest. All now go to 4.90 but Stefanidi only gets two attempts while the rest get three.
Stefanidi missed her first attempt at 4.90. She gets one attempt left.
Sidorova and Bradshaw fail to even reach the bar. Nageotte can't clear the bar.
Now the moment that means everything for Stefanidi. Her last attempt. Make it or the rest get the medals. Stefaidi fails to get a good plant and doesn't get into the air at all. Her Olympics are
Second attempt for the others. Sidorva and Bardshaw knock down the bar. Nageotte clears the bar, putting her in prime position. Go USA!
Imagine if we judged research papers this way. Either they get into a conference or they don't. Wait, that is they way they happen, although not always without biased judging.
Sidorova is passing on her last attempt at 4.90. Bradshaw goes for it but hits the bar. She has to settle for Bronze.
Bar is now at 4.95 meters.
Sidorova gets only one attempt at 4.95. If she makes it, she takes the lead, if she misses, she gets the silver.
Sidorova doesn't clear and the gold goes to the American Katie Nageotte!
Just for excitement Nageotte is going for 5.01 meters, which would be her first over five meters in competition. In the men's pole vault, the Swede Armand Duplantis (great pole vault name!) easily
won the gold. He moved the bar to 6.19 meters to break his own world record. Came all so close in his first attempt but failed to clear.
Nageotte is just too excited winning the gold to focus enough to make a serious attempt at 5.01. Can't blame her.
Thus ends the best sport in the Olympics.
(Guest Post by David Marcus)
Comment by Bill: Haken and Appel proved that all planar maps are 4-colorable. Or did they? David Marcus emailed me that its not quite true and I asked him to post on it, so here it is. The meta point
is that math can be very subtle.
And now David Marcus's post:
Is the Four Color Map Theorem true?
It is commonly believed that the Four Color Map Theorem says that four colors suffice to color a planar map. While this is true for any map a non-mathematician would dream up, it is not true for maps
a mathematician might dream up without some restriction on the regions that are allowed. This is shown in Hud Hudson's Four Colors Do Not Suffice which appeared in the American Math Monthly, Volume
110, No. 5, May 2003, pages 417--423.
Hudson's article is written in a very entertaining style. I recommend that you read it. He constructs a map consisting of six regions R1,...,R6. Each region is bounded and path connected. There is a
line segment B that is in the boundary of all six regions. So, six colors are needed, since all six regions share a common boundary. The construction is similar to the topologist's sine curve. For
each i , the union of Ri and B is not path connected. Hudson also shows that for any n, there is a map that requires at least n colors.
Hudson thus disproves the following statement:
1) Four colors are sufficient to color any map drawn in the plane or on a sphere so that no two regions with a common boundary line are colored with the same color.
Appel and Haken actually proved the following:
2) Four colors are sufficient to color any planar graph so that no two vertices connected by an edge are colored with the same color.
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bpstart & bpend notify scripts for multistreaming jobs | VOX
bpstart & bpend notify scripts for multistreaming jobs
Hi Experts,
On testing out bpstart & bpend notify scripts for multistreaming jobs found that they get called for each child job.
On research i came across following Tech note which suggests way to write it but don't have any ready scripts. Undestand that they are provided as part of consulting services
If anybody have them running in their environment, appreciate if you can share the same.
Thanks in advance.
□ Created a folder to contain LOCK files for each policy. Each policy to have separate folder
□ Create lock file for each triggered backup job for policy in its respective folder
Added the following line in bpstart_notify on media server
□ Create the bpend_notify.policy script as follows.
rm -rf $DIR/STREAM_$STREAM_PID
logger -t "backup:$2 Removed file with PID $STREAM_PID "
if [ "$(ls -A $DIR)" ]; then
STREAM_RUNNING=`ls -l $DIR | egrep -c '^-'`
logger -t backup:byron80 "$STREAM_RUNNING Backup Streams still running"
#echo "$DIR is Empty"
logger -t backup:byron80 "All Streams completed"
RES=`/usr/openv/netbackup/bin/admincmd/bpplinfo $2 -modify -inactive 2>&1`
if [ $? -ne 0 ]
logger -t "backup:$2:$RES"
$MAILX -s "$2 : Deactivation Failed" "$MAIL_ADDR"
logger -t backup:byron80 "Policy $2 deactivated"
$MAILX -s "$2 : Deactivated successfully" "$MAIL_ADDR"
□ You can add the following lines in netbackup start script to clear the lock files if any during service startup.
You can just add before the stop logic starts.
#Added by Satish to clear up lock files for any previously hung jobs. 12-April-2012
#for i in $(ls -l /usr/openv/netbackup/bin/SI_logs |grep ^d| cut -f9 -d" ")
for i in $(find /usr/openv/netbackup/bin/SI_logs/* -type d)
echo "Clearing up $i for any lock files"
rm -f $i/*
stop) <-- Reference to insert in script
Hope the forum finds it helpful.
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CAS PY 541 Statistical Mechanics I
4 credits. Fall and Spring
Undergraduate Prerequisites: (CASPY410) - Graduate Prerequisites: (CASPY410) - Probability theory. Ensembles. Steepest descent methods. Paramagnetism, ideal gas, Einstein model, adsorption isotherms.
Thermodynamics, Maxwell relations, heat capacity. Bose and Fermi gases. Electrons in metals, white dwarf stars, black-body radiation, phonons, Bose-Einstein condensation. Interacting systems, virial
expansion, Van der Waals gas. Phase transitions: mean-field theories, spin systems.
|
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Perhaps this
might explain it. acer
Planck's constant has a dimension. It's not so important what the dimensions are (angular momentum) so much as what consequences that brings with it. *You* get to pick your own units for that
dimension. So you can decide that Planck's constant is just about any number in magnitude. It depends on what system of units you choose. It's not as if the SI system were god-given! Choose an
appropriate system of units, and the constant can then be of size 1 in that system. By that I mean, choose a base size for the units of length, time, etc, so that the constant is 1 in that system. If
the question were rephrased to be something like, "is Planck's constant irrational when expressed using SI units?" then it might depend on stuff like whether space itself (distance, or time) is
quantized. acer
Keeping the total memory allocation down is often an important part of efficiency. But, also, keeping the memory usage (garbage production and cleanup) down can also leads to time savings. I suspect
that ArrayTools:-Reshape produce a full copy of the rtable argument, while ArrayTools:-Alias actually provides an alternate view of the same data in memory without making a copy. So, using Alias can
be more efficient than using Reshape. Also, in the test3 as given, it was not necessary to keep all of the instances of Vector a around. And indeed in Joe's test3a the Vector a was overwritten due to
a being reassigned. But each assignment to a in test3a extracts a new row of A, producing a new Vector each time. Memeory usage can be reduced by instead forming the container for Vector a just once,
and then doing a fast copy of the appropriate row of A into Vector a. So, assuming that I coded it correctly, test3b := proc(N,n) local i,a,X,A; use Statistics in X := RandomVariable(Normal(0,1)): A
:= ArrayTools:-Alias(Sample(X,n*N),[N,n],Fortran_order); a := Vector[row](n,datatype=float,order=Fortran_order); for i from 1 to N do ArrayTools:-Copy(n,A,n*(i-1),1,a,0,1); end do; end use; end proc:
Now, to measure the efficiency, let's also look at memory used and memory allocated, as well as time. > (st,ba,bu):=time(),kernelopts(bytesalloc),kernelopts(bytesused): > test3(30000,100): > time()
-st,kernelopts(bytesalloc)-ba,kernelopts(bytesused)-bu; 18.419, 8518120, 899923608 > (st,ba,bu):=time(),kernelopts(bytesalloc),kernelopts(bytesused): > test3a(30000,100): > time()-st,kernelopts
(bytesalloc)-ba,kernelopts(bytesused)-bu; 0.711, 56088544, 79456288 > (st,ba,bu):=time(),kernelopts(bytesalloc),kernelopts(bytesused): > test3b(30000,100): > time()-st,kernelopts(bytesalloc)
-ba,kernelopts(bytesused)-bu; 0.571, 31844664, 35960072 Of course, what would be really nice would be to get the speed of test3b or test3a and the lower memory allocation (by at least a factor of
four) of test3. This may not be currently possible. One way to get it might be with an enhancement to the Statistics:-Sample routine, so that it accepts as an optional argument the container Vector
for the result. It could re-use the same Vector container, and fill it with new sample data, operating in-place. That could allow the total memory allocation to stay low, to generate only n pieces of
data at any one time, but avoid production of a lot of Vectors as garbage. It's a bit of a shame, that it isn't implemented in this way. Since at default Digits=10 the Vectors have hardware double
precision float[8] datatype, there wouldn't be garbage produced from all the entries as software floats. ps. It may be that A should be C_order, according to how I set up the strides for
ArrayTools:-Alias, so that it walks rows of A and not columns. It shouldn't affect the timings much. acer
Jacques' points about pragmatism and making progress in the face of technical or theoretical difficulties are very much to the point. m := matrix(2,3,[1,1,a,1,1,b]): linalg[LUdecomp](m,'U1'='u1'):
seq(`if`(type(u1[i,i],Non(constant)),u1[i,i],NULL),i=1..2); # Are these bugs below? # How about better documentation of the following? # Eg, as an Example on the gaussjord help-page. Testzero := proc
(x) if is(x=0) = true then true else false; end if; end proc: linalg[gaussjord](m) assuming b-a=0; # Should it work for LinearAlgebra too? # Or is Normalizer used when Testzero ought to be instead? M
:= Matrix(m): LinearAlgebra[ReducedRowEchelonForm](M) assuming b-a=0; So, I also wish for more thorough documentation of Testzero and Normalizer, of where they are used any why. ps. Yes, I realize
that I could probably have gotten away with an even simpler true/false/FAIL conditional in my Normalizer. Apologies if it brings up anyone's pet peeves. acer
One can also consider memory usage as an important component of efficiency. Apart from wanting to keep memory allocation down for its own benefits, one can also try to keep memory use and re-use down
so as to minimize garbage collection time. Other benefits of this sort of array memory optimization can be that total memory allocation can sometimes be reduced, that large problems can become
tractable, and sometimes speed is improved. The reasons for this can be complicated, but it can relate to memory fragmentation and also to the fact that garbage is not immediately freed. For example,
Y := (X.beta)^%T; might become Y := X.beta; LinearAlgebra:-Transpose(Y,inplace=true); That should avoid producing an unnecessary object of the size of Y. The creating of Arrays n1 and n2 by forming
each N12 sub-Array might also be improved. One might be able to allocate empty n1 and n2 Vector[row]'s of the desired size and datatype, just once, and then use ArrayTools:-Copy each time through the
loop so as to get the right portion of N12 into them. The key would be using the right offset and stride. One might also be able to allocate space for Y1 and Y2 to be used in procedure `compute`,
just the once. Eg, produce Y1 and Y2 as empty Vector[row]'s of the desired datatype, outside of `compute`, just once. Then, outside of `compute` but each time through the loop, use ArrayTools:-Copy
to get Y into Y1. Follow that by VectorAdd(Y1,n1,inplace=true). And similarly for Y2. Notice also that n1 and n2 might only get used in `compute` to produce Yp. So why not produce a re-uasable
container for Yp just once, and never produce Y1 and Y2 at all! How about something like this, Yp:=Vector[row](num,datatype=float); # just once n1:=Vector[row](num,datatype=float); # just once # and
now, inside the i loop ArrayTools:-Copy(...,N12,...n1...); # get data into n1 ArrayTools:-Copy(...,N12,...Yp...); # get n2 into Yp VectorAdd(Yp,n1,inplace=true); VectorScalarMultiply(Yp,0.5,inplace=
true); # (n1+n2)/2 VectorAdd(Yp,Y); # (Y+n1 + Y+n2)/2 = (Y1+Y2)/2 Now consider the lines in `compute` like, add(Yp[k]*X[k,3]/(n/2),k=1..n): These line are the only places where Yp gets used, yes? So
why not first scale Yp, inplace, by n/2 and then have those lines be like, add(Yp[k]*X[k,3],k=1..n): The Y1-Y2 object is just n1-n2, no? So the Y1-Y2 = n1-n2 object could also be created just once,
as a re-usable Vector `Y1minusY2` outside the loop. But outside the loop, one already has n1 from above code. And n1 is no longer needed, once Yp is formed. So use it to hold Y1-Y2=n1-n2. Ie, outside
the loop do, VectorAdd(n1,n2,1,-1,inplace=true); Hopefully I haven't made major mistakes. I'm sure that there are other improvements possible, and some of these techniques above won't make a big
difference for smaller problems. acer
Hi Joe, I wasn't trying to say that copy() wasn't needed for anything but tables. As you reiterated, it behaves differently from the rtable constructors themselves. What happens if you try this in
Maple 11, versus Maple 10? a := array(1..3,[A,B,C]); b := array(a); a[1]:=z: eval(a); eval(b); It seems to me that in Maple 11 the command array(a) produces a copy of a, and that this is even
documented on the ?array help-page. But I make no such claim about the vector constructor. I agree with you, that better documentation of the assignment operations would be of benefit. More obvious
explanations of the differences in the array, table, list, rtable, etc, data structures could be good, as well as general advice on what each may be useful for (typically), and why. acer
Hi Joe, I wasn't trying to say that copy() wasn't needed for anything but tables. As you reiterated, it behaves differently from the rtable constructors themselves. What happens if you try this in
Maple 11, versus Maple 10? a := array(1..3,[A,B,C]); b := array(a); a[1]:=z: eval(a); eval(b); It seems to me that in Maple 11 the command array(a) produces a copy of a, and that this is even
documented on the ?array help-page. But I make no such claim about the vector constructor. I agree with you, that better documentation of the assignment operations would be of benefit. More obvious
explanations of the differences in the array, table, list, rtable, etc, data structures could be good, as well as general advice on what each may be useful for (typically), and why. acer
Does this serve? proc(n::posint) rtable(1..n,1..2,frandom(0.0,1.0),subtype=Array,datatype=float[8]); end proc: acer
The constraints are namd like Constaint1, etc, but the call to NLPSolve has them as Constraint1, etc. acer
Axel made some good points. I believe that his comment on Theta being uninitialized is mistaken, however. I too am not a big fan of 2Dmath for authoring maple programs. I hope that my conversion of
to 1dmath input is correct. Here are some suggestions and comments. - Use float[8] an complex[8] datatypes, if possible. This helps avoid temporary memory allocation by pushing some of the
computation out to compiled external code. - Initialize x, R, and Theta using the initializer available within the Matrix constructor. This is more efficient than initializing them after creation,
with loops. - Vpq is enormous, requiring >800MB of memory with complex[8] datatype. This may not be avoidable. More on this at the end. - Rpq is as big as Vpq, but it is not needed. I understand that
Vpq may be wanted, after the work is done. (It could be returned, with moments, instead of being a global.) But it seems that only two or three entries of Rpq are needed at any time within the inner
part of the double-loop at the end of the code. So by getting rid of Rpq you might be able to save >800MB. - Use ArrayTools[Alias] instead of ArrayTools[Reshape]. The latter produces an unnecessary
copy, while the former does not. - Use evalhf where possible. This gives you double-precision, and actually saves memory too. - In the double-loop to initialize pq, observe that the inner ("and")
condition of q<=p can be dropped if the inner loop variable q is only allowed to go as high as p. - Use evalhf for all the computations done in the nasty, super expensive double loop at the end of
the procedure. More on this later. - The moments Vector can be created by the muliplication. You don't need to create it, and then loop to assign to it. - Don't use A . B . C for efficient linear
algebra. Make use of the LinearAlgebra commands directly. That saves at least one unnecessary copy, or perhaps two. Make sure you control these operations. Scale moments by coef, inplace after the
muliplication. (The worst scenario would be scaling Vpq which is enormous by coef. Bad because it's so big, and horrible if it produces an enourmous scaled copy.) For the dim=200,pmax=70,qmax=70 case
this code below took about 830MB and 3h20min on a fast P4. Please check that I didn't make mistakes in the translation. You can try running smaller examples against original and mine. If I did make
mistakes then they can likely be corrected while remaining efficient. Most of my changes are in somewhat independent chunks. I tried to check that moments[1] stayed the same, but I didn't test Vpq
results. zernike:=proc(dim,pmax,qmax,image) local i,j,y,kmax,k,r,t,Z,Ztot,coef,exponente,size,p,q,pq,s, H1,H2,H3,im,imtot,Theta,R,Image,Vpq,moments,x, Rpqis,Rpqis1,Rpqis2,bainit,st,oldDigits;
st,bainit := time(),kernelopts(bytesalloc); # Raise Digits during initialization of x, R, and Theta. oldDigits,Digits := Digits,max(Digits,trunc(evalhf(Digits))); x:=Matrix(dim,1,(i)->evalf(((sqrt(2)
*(i-1)))/((dim-1))-1/(sqrt(2))), datatype=float); y:=x; R:=Matrix(dim,dim,(j,i)->evalf(sqrt(x[i,1]^2+y[dim-j+1,1]^2)), datatype=float); Theta:=Matrix(dim,dim,(j,i)->evalf(arctan(y[dim-j+1,1],x
[i,1])), datatype=float); Digits := oldDigits; if type(pmax,odd)=true then size:= 1/4*(pmax+1)*(pmax+3); pq:=Matrix(size,2,datatype=float); else size:=1/4*(pmax+2)^2; pq:= Matrix(size,2, datatype=
float ); end if; userinfo(1,`zernike`,`stage 1`, time()-st, kernelopts(bytesalloc)-bainit ); i:=0; for p from 0 to pmax do for q from 0 to p do if type(p-abs(q),even)=true then i:= i+1; pq[i,1]:= p;
pq[i,2]:= q; end if; end do; end do; userinfo(1,`zernike`,`stage 2`, time()-st, kernelopts(bytesalloc)-bainit ); R:=ArrayTools[Alias](R,[dim^2,1]); Theta:=ArrayTools[Alias](Theta,[dim^2,1]); Image:=
ArrayTools[Alias](image,[dim^2]); Vpq:=Matrix(size,dim^2,datatype=complex(float)); userinfo(1,`zernike`,`stage 3`, time()-st, kernelopts(bytesalloc)-bainit ); Rpqis1,Rpqis2:=0.0,0.0; for s from size
by -1 to 1 do for i to dim^2 do if pq[s,1]=pq[s,2] then Rpqis:=evalhf( R[i,1]^(pq[s,1]) ); elif pq[s,1]-pq[s,2]=2 then Rpqis:= evalhf( pq[s,1]*R[i,1]^(pq[s,1]) - (pq[s,1]-1)*R[i,1]^((pq[s,1]-2)) );
else H3:= evalhf( (-4*(pq[s+2,2]-2)*(pq[s+2,2]-3))/ ((pq[s,1]+pq[s+2,2]-2)*(pq[s,1]-pq[s+2,2]+4)) ): H2:= evalhf( (H3*(pq[s,1]+pq[s+2,2])*(pq[s,1]-pq[s+2,2]+2))/ (4*(pq[s+2,2]-1))+(pq[s+2,2]-2) ):
H1:= evalhf( (pq[s+2,2]*(pq[s+2,2]-1))/2 -pq[s+2,2]*H2+ (H3*(pq[s,1]+pq[s+2,2]+2)*(pq[s,1]-pq[s+2,2]))/8 ): Rpqis:= evalhf( H1*Rpqis2+(H2+H3/(R[i,1]^2))*Rpqis1 ): Rpqis2,Rpqis1:=Rpqis1,Rpqis; end if;
Vpq[s,i]:= evalhf( Rpqis*(exp(I*pq[s,2]*Theta[i,1])) ); end do; coef:=evalhf( ((2*pq[s,1]+2))/(Pi*((dim))^2) ); end do; userinfo(1,`zernike`,`stage 4`, time()-st, kernelopts(bytesalloc)-bainit );
moments := LinearAlgebra[MatrixVectorMultiply](Vpq,Image); LinearAlgebra[VectorScalarMultiply](moments,coef,inplace=true); userinfo(1,`zernike`,`stage 5`, time()-st, kernelopts(bytesalloc)-bainit );
return moments, Vpq; end proc: kernelopts(printbytes=false): #(dim,pmax,qmax):=200,70,70: (dim,pmax,qmax):=20,7,7; # Just for testing. mI := LinearAlgebra[RandomMatrix](dim,dim,outputoptions=
[datatype=complex(float)]): # Let it be verbose, about how it's doing. infolevel[`zernike`]:=1: solmom,solVpq:=zernike(dim,pmax,qmax,mI): solmom[1]; Now, what can be said about this double loop, with
something like 1200 and 10000 iterations at each level? That's where almost all of the 3hr20min of computation time goes. If it didn't have exp(I*pq[s,2]*Theta[i,1]) in it you could stick it inside
its own procedure and run Compiler[Compile] against it. You still could do that, actually, by wrapping an eval() around that problematic line involving nonreal, complex computation. But then memory
usage might go up high. I'm not sure how much garbage collection is possible when a Compiled routine runs. In order to try this, you would have to use ArrayTools[ComplexAsFloat] on Vpq. I was
thinking something like this... first, get rid of the double loop at the end of zernike, and instead make that just one line like, say, coef := stage4_compiled(dim,size,pq,R,real_Vpq,Theta); And
then, outside of zernike, have something like all this. st4_r := proc(x,y,z) local temp; temp := x*exp(I*y*z); return Re(temp); end proc: st4_i := proc(x,y,z) local temp; temp := x*exp(I*y*z); return
Im(temp); end proc: stage4 := proc(dim::integer,size::integer,pq::Matrix(datatype=float[8]),R::Matrix(datatype=float[8]),Vpq::Matrix(datatype=float[8]),Theta::Matrix(datatype=float[8])) local Rpqis,
Rpqis1, Rpqis2, H1, H2, H3, s, i, coef; global st4_r,st4_i; Rpqis1,Rpqis2:=0.0,0.0; for s from size by -1 to 1 do for i to dim^2 do if pq[s,1]=pq[s,2] then Rpqis:= R[i,1]^(pq[s,1]); elif pq[s,1]-pq
[s,2]=2 then Rpqis:= pq[s,1]*R[i,1]^(pq[s,1]) - (pq[s,1]-1)*R[i,1]^((pq[s,1]-2)); else H3:= (-4*(pq[s+2,2]-2)*(pq[s+2,2]-3))/ ((pq[s,1]+pq[s+2,2]-2)*(pq[s,1]-pq[s+2,2]+4)): H2:= (H3*(pq[s,1]+pq
[s+2,2])*(pq[s,1]-pq[s+2,2]+2))/ (4*(pq[s+2,2]-1))+(pq[s+2,2]-2): H1:= (pq[s+2,2]*(pq[s+2,2]-1))/2 -pq[s+2,2]*H2+ (H3*(pq[s,1]+pq[s+2,2]+2)*(pq[s,1]-pq[s+2,2]))/8: Rpqis:= H1*Rpqis2+(H2+H3/(R[i,1]^
2))*Rpqis1: Rpqis2,Rpqis1:=Rpqis1,Rpqis; end if; Vpq[2*s-1,i]:= eval(st4_r(Rpqis,pq[s,2],Theta[i,1])); Vpq[2*s,i]:= eval(st4_i(Rpqis,pq[s,2],Theta[i,1])); end do; coef:= ((2*pq[s,1]+2))/(Pi*((dim))^
2); end do; return coef; end proc: stage4_compiled:=Compiler[Compile](stage4); You'd almost certainly want to check that I hadn't done the wrong thing, with the assignments into Vpq above. You might
also try to run CodeGeneration[C] on something like stage4, and replacing the exp call by a dummy unrecognized name. Then afterwards you could replace that dummy in the produced C souce with a call
to cexp(). Then you could compile it into a dynamic library outside of Maple. And then replace all the stuff for it in the example by setting up a Maple procedure to point to the compiled external
function using define_external(). acer
Axel made some good points. I believe that his comment on Theta being uninitialized is mistaken, however. I too am not a big fan of 2Dmath for authoring maple programs. I hope that my conversion of
to 1dmath input is correct. Here are some suggestions and comments. - Use float[8] an complex[8] datatypes, if possible. This helps avoid temporary memory allocation by pushing some of the
computation out to compiled external code. - Initialize x, R, and Theta using the initializer available within the Matrix constructor. This is more efficient than initializing them after creation,
with loops. - Vpq is enormous, requiring >800MB of memory with complex[8] datatype. This may not be avoidable. More on this at the end. - Rpq is as big as Vpq, but it is not needed. I understand that
Vpq may be wanted, after the work is done. (It could be returned, with moments, instead of being a global.) But it seems that only two or three entries of Rpq are needed at any time within the inner
part of the double-loop at the end of the code. So by getting rid of Rpq you might be able to save >800MB. - Use ArrayTools[Alias] instead of ArrayTools[Reshape]. The latter produces an unnecessary
copy, while the former does not. - Use evalhf where possible. This gives you double-precision, and actually saves memory too. - In the double-loop to initialize pq, observe that the inner ("and")
condition of q<=p can be dropped if the inner loop variable q is only allowed to go as high as p. - Use evalhf for all the computations done in the nasty, super expensive double loop at the end of
the procedure. More on this later. - The moments Vector can be created by the muliplication. You don't need to create it, and then loop to assign to it. - Don't use A . B . C for efficient linear
algebra. Make use of the LinearAlgebra commands directly. That saves at least one unnecessary copy, or perhaps two. Make sure you control these operations. Scale moments by coef, inplace after the
muliplication. (The worst scenario would be scaling Vpq which is enormous by coef. Bad because it's so big, and horrible if it produces an enourmous scaled copy.) For the dim=200,pmax=70,qmax=70 case
this code below took about 830MB and 3h20min on a fast P4. Please check that I didn't make mistakes in the translation. You can try running smaller examples against original and mine. If I did make
mistakes then they can likely be corrected while remaining efficient. Most of my changes are in somewhat independent chunks. I tried to check that moments[1] stayed the same, but I didn't test Vpq
results. zernike:=proc(dim,pmax,qmax,image) local i,j,y,kmax,k,r,t,Z,Ztot,coef,exponente,size,p,q,pq,s, H1,H2,H3,im,imtot,Theta,R,Image,Vpq,moments,x, Rpqis,Rpqis1,Rpqis2,bainit,st,oldDigits;
st,bainit := time(),kernelopts(bytesalloc); # Raise Digits during initialization of x, R, and Theta. oldDigits,Digits := Digits,max(Digits,trunc(evalhf(Digits))); x:=Matrix(dim,1,(i)->evalf(((sqrt(2)
*(i-1)))/((dim-1))-1/(sqrt(2))), datatype=float); y:=x; R:=Matrix(dim,dim,(j,i)->evalf(sqrt(x[i,1]^2+y[dim-j+1,1]^2)), datatype=float); Theta:=Matrix(dim,dim,(j,i)->evalf(arctan(y[dim-j+1,1],x
[i,1])), datatype=float); Digits := oldDigits; if type(pmax,odd)=true then size:= 1/4*(pmax+1)*(pmax+3); pq:=Matrix(size,2,datatype=float); else size:=1/4*(pmax+2)^2; pq:= Matrix(size,2, datatype=
float ); end if; userinfo(1,`zernike`,`stage 1`, time()-st, kernelopts(bytesalloc)-bainit ); i:=0; for p from 0 to pmax do for q from 0 to p do if type(p-abs(q),even)=true then i:= i+1; pq[i,1]:= p;
pq[i,2]:= q; end if; end do; end do; userinfo(1,`zernike`,`stage 2`, time()-st, kernelopts(bytesalloc)-bainit ); R:=ArrayTools[Alias](R,[dim^2,1]); Theta:=ArrayTools[Alias](Theta,[dim^2,1]); Image:=
ArrayTools[Alias](image,[dim^2]); Vpq:=Matrix(size,dim^2,datatype=complex(float)); userinfo(1,`zernike`,`stage 3`, time()-st, kernelopts(bytesalloc)-bainit ); Rpqis1,Rpqis2:=0.0,0.0; for s from size
by -1 to 1 do for i to dim^2 do if pq[s,1]=pq[s,2] then Rpqis:=evalhf( R[i,1]^(pq[s,1]) ); elif pq[s,1]-pq[s,2]=2 then Rpqis:= evalhf( pq[s,1]*R[i,1]^(pq[s,1]) - (pq[s,1]-1)*R[i,1]^((pq[s,1]-2)) );
else H3:= evalhf( (-4*(pq[s+2,2]-2)*(pq[s+2,2]-3))/ ((pq[s,1]+pq[s+2,2]-2)*(pq[s,1]-pq[s+2,2]+4)) ): H2:= evalhf( (H3*(pq[s,1]+pq[s+2,2])*(pq[s,1]-pq[s+2,2]+2))/ (4*(pq[s+2,2]-1))+(pq[s+2,2]-2) ):
H1:= evalhf( (pq[s+2,2]*(pq[s+2,2]-1))/2 -pq[s+2,2]*H2+ (H3*(pq[s,1]+pq[s+2,2]+2)*(pq[s,1]-pq[s+2,2]))/8 ): Rpqis:= evalhf( H1*Rpqis2+(H2+H3/(R[i,1]^2))*Rpqis1 ): Rpqis2,Rpqis1:=Rpqis1,Rpqis; end if;
Vpq[s,i]:= evalhf( Rpqis*(exp(I*pq[s,2]*Theta[i,1])) ); end do; coef:=evalhf( ((2*pq[s,1]+2))/(Pi*((dim))^2) ); end do; userinfo(1,`zernike`,`stage 4`, time()-st, kernelopts(bytesalloc)-bainit );
moments := LinearAlgebra[MatrixVectorMultiply](Vpq,Image); LinearAlgebra[VectorScalarMultiply](moments,coef,inplace=true); userinfo(1,`zernike`,`stage 5`, time()-st, kernelopts(bytesalloc)-bainit );
return moments, Vpq; end proc: kernelopts(printbytes=false): #(dim,pmax,qmax):=200,70,70: (dim,pmax,qmax):=20,7,7; # Just for testing. mI := LinearAlgebra[RandomMatrix](dim,dim,outputoptions=
[datatype=complex(float)]): # Let it be verbose, about how it's doing. infolevel[`zernike`]:=1: solmom,solVpq:=zernike(dim,pmax,qmax,mI): solmom[1]; Now, what can be said about this double loop, with
something like 1200 and 10000 iterations at each level? That's where almost all of the 3hr20min of computation time goes. If it didn't have exp(I*pq[s,2]*Theta[i,1]) in it you could stick it inside
its own procedure and run Compiler[Compile] against it. You still could do that, actually, by wrapping an eval() around that problematic line involving nonreal, complex computation. But then memory
usage might go up high. I'm not sure how much garbage collection is possible when a Compiled routine runs. In order to try this, you would have to use ArrayTools[ComplexAsFloat] on Vpq. I was
thinking something like this... first, get rid of the double loop at the end of zernike, and instead make that just one line like, say, coef := stage4_compiled(dim,size,pq,R,real_Vpq,Theta); And
then, outside of zernike, have something like all this. st4_r := proc(x,y,z) local temp; temp := x*exp(I*y*z); return Re(temp); end proc: st4_i := proc(x,y,z) local temp; temp := x*exp(I*y*z); return
Im(temp); end proc: stage4 := proc(dim::integer,size::integer,pq::Matrix(datatype=float[8]),R::Matrix(datatype=float[8]),Vpq::Matrix(datatype=float[8]),Theta::Matrix(datatype=float[8])) local Rpqis,
Rpqis1, Rpqis2, H1, H2, H3, s, i, coef; global st4_r,st4_i; Rpqis1,Rpqis2:=0.0,0.0; for s from size by -1 to 1 do for i to dim^2 do if pq[s,1]=pq[s,2] then Rpqis:= R[i,1]^(pq[s,1]); elif pq[s,1]-pq
[s,2]=2 then Rpqis:= pq[s,1]*R[i,1]^(pq[s,1]) - (pq[s,1]-1)*R[i,1]^((pq[s,1]-2)); else H3:= (-4*(pq[s+2,2]-2)*(pq[s+2,2]-3))/ ((pq[s,1]+pq[s+2,2]-2)*(pq[s,1]-pq[s+2,2]+4)): H2:= (H3*(pq[s,1]+pq
[s+2,2])*(pq[s,1]-pq[s+2,2]+2))/ (4*(pq[s+2,2]-1))+(pq[s+2,2]-2): H1:= (pq[s+2,2]*(pq[s+2,2]-1))/2 -pq[s+2,2]*H2+ (H3*(pq[s,1]+pq[s+2,2]+2)*(pq[s,1]-pq[s+2,2]))/8: Rpqis:= H1*Rpqis2+(H2+H3/(R[i,1]^
2))*Rpqis1: Rpqis2,Rpqis1:=Rpqis1,Rpqis; end if; Vpq[2*s-1,i]:= eval(st4_r(Rpqis,pq[s,2],Theta[i,1])); Vpq[2*s,i]:= eval(st4_i(Rpqis,pq[s,2],Theta[i,1])); end do; coef:= ((2*pq[s,1]+2))/(Pi*((dim))^
2); end do; return coef; end proc: stage4_compiled:=Compiler[Compile](stage4); You'd almost certainly want to check that I hadn't done the wrong thing, with the assignments into Vpq above. You might
also try to run CodeGeneration[C] on something like stage4, and replacing the exp call by a dummy unrecognized name. Then afterwards you could replace that dummy in the produced C souce with a call
to cexp(). Then you could compile it into a dynamic library outside of Maple. And then replace all the stuff for it in the example by setting up a Maple procedure to point to the compiled external
function using define_external(). acer
The gauss procedure uses ilcm. Its parameter-processing calls `type/listlist`. The setup of the data A, prior to calling gauss(), uses RandomTools. acer
You showed a result from Maple of 19.99909999 . Presumably that was done at the default value of Digits=10. But, evalf[11](exp(Pi)-Pi) returns as 19.999099979 . So why isn't the the first result, at
the default value of Digits=10, instead 19.99909998 ? acer
Notice that the answer obtained from A.Vogt's method was 0.0197... times I. That is, it was a purely imaginary number. The result was not the real number 0.0197... Notice also that, if you plug in
the imaginary number, remembering to make it I*0.0197.., then that does produce a value very close to 0.5. Ie, evalf(abs(f(nTst,0.01970538187*I))); So, the answer I*0.0197.. was correct, given that
you did not specify a real-valued range for the result from fsolve. You can supply a purely real-valued range to fsolve, eg, f := (n,v) -> Zeta(0,n+1,1-2*Pi*v*I)/Zeta(n+1): nTst:= 5: fsolve( abs(f
(nTst,v))=0.5, v=0..0.2 ); That gives the result 0.08039... that you mentioned. acer
Notice that the answer obtained from A.Vogt's method was 0.0197... times I. That is, it was a purely imaginary number. The result was not the real number 0.0197... Notice also that, if you plug in
the imaginary number, remembering to make it I*0.0197.., then that does produce a value very close to 0.5. Ie, evalf(abs(f(nTst,0.01970538187*I))); So, the answer I*0.0197.. was correct, given that
you did not specify a real-valued range for the result from fsolve. You can supply a purely real-valued range to fsolve, eg, f := (n,v) -> Zeta(0,n+1,1-2*Pi*v*I)/Zeta(n+1): nTst:= 5: fsolve( abs(f
(nTst,v))=0.5, v=0..0.2 ); That gives the result 0.08039... that you mentioned. acer
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Online Academic Catalog
MATH 213 Advanced Calculus I
The real number system, least upper bound property. Sequences in R, Cauchy sequences, limsup and liminf. Limit and continuity in R, uniform continuity. Differentiation in R. Riemann integral in R,
fundamental theorem of calculus. Infinite series of numbers, absolute convergence. Sequences and series of functions, uniform convergence, power series. Credit units: 4 ECTS Credit units: 6.5,
Prerequisite: MATH 102 and MATH 123.
Autumn Semester (Türker Özsarı)
Bilkent University Main Page
Last regenerated automatically on November 6, 2024 by OAC - Online Academic Catalog Software
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Statistical Physics
1. komponenta
Lecture type Total
Lectures 30
Exercises 15
* Load is given in academic hour (1 academic hour = 45 minutes)
Thermodynamics as an autonomous discipline: Introduction. Basic concepts. The first law of thermodynamics. Machines. The second law of thermodynamics. The reversibility and entropy. Thermodynamic
potentials. Practical accounts.
Introduction to statistical physics: Basic considerations. Ensemble: universal random model. The connection with thermodynamics.
Canonical and grand-canonical ensemble: The canonical ensemble. Grand-canonical ensemble. Sums by conditions such as generating functions. Classical ideal gas. Maxwell distribution and
equiparticion energy.
Quantum statistical physics: Basic considerations. The ideal fermion gas. The ideal boson gas.
Examples and models: the barometric formula. Diatomic molecules. Heat capacity of the crystal. Van der Waals model of gas liquefaction.
LEARNING OUTCOMES:
Upon successful completion of the course Statistical Physics student will be able to:
1.Demonstrate a thorough knowledge of abstract thermodynamics at an elementary level of the theory of functions of several variables;
2.Explain the difference of thermodynamics and theoretical mechanics, or thermalization as real physical process;
3.Describe the role of thermalization and Liouville theorem in the foundation of statistical physics;
4.Explain the physical construction of the thermodynamic potential, through the interaction energy between the system and the outside world;
5.Demonstrate a thorough knowledge of statistical interpretation of thermodynamic potentials, especially entropy and Massieuovih function;
6.Explain the role of the chemical potential and the qualitative behavior of the classical and quantum border;
7.Qualitatively and quantitatively described four ideal gas (fermions, bosons, light, sound) in classical and quantum border;
8.Discuss basic properties of the phase transition of Van der Waals-s gas liquefaction.
1. C. Kittel, Elementary Statistical Physics, Dover 2004, ISBN 0486435148.
2. R. Kubo et al., Statistical mechanics: an advanced course with problems and solutions, North-Holland, Amsterdam 1988, ISBN 0444871039.
3. Skripta: http://www.phy.hr/dodip/notes/statisticka.html
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Sudo Null - Latest IT News
How "sounds" space
Photo by
Paul Hudson CC-BY
Sound is a physical phenomenon that represents elastic waves propagating in a particular medium. In a narrow sense, sound refers to vibrations perceived by the hearing organs of animals and humans.
The eardrum in our ear picks up high-frequency changes in air pressure, and the brain processes the received signal.
In space, we hear nothing, because there is no medium suitable for sound propagation. However, we will find out if a supernova explosion occurs near our planet - the “explosion” will be the result of
the deformation of the Earth’s atmosphere under the action of light.
However, if we are talking about electromagnetic oscillations that a person is not capable of perceiving, then their conversion into a sound format can give interesting and valuable results. For
in this video, NASA
processed the received signals from different planets:
At Geektimes, they already
about a NASA
in SoundCloud, which opened the door to the world of “cosmic” sounds (for example, a digitized recording of sounds from the Voyager gold record was
) and
brought the
synthesized recording sound from Venus as an example of what can be recorded on the surface of a planet.
But back to the "sounding planets." Radio waves emitted by celestial bodies, can be considered using a special device - an
Interferometers are widely
in astronomy to create radio and optical telescopes with high resolution. An example is the giant ALMA radio telescope, consisting of more than 66 antennas distributed over a large area, which
receive radio waves emitted by astronomical objects.
The principle of operation of an astronomical interferometer is as follows: suppose that two antennas are directed toward the space body X. Since the radio waves move in space at a constant speed,
the radio waves from object X will reach the antennas set at a certain distance from each other at different points in time. After that, the signals of the two antennas can be interfered and the
desired source information can be extracted from the resulting signal.
Optical telescopes, like our eyes, are imperfect and allow us to make observations only in the visible range of radiation. ALMA was specifically designed to record long waves. Thanks to this,
scientists were able to look into the farthest corners of the universe, hidden from our eyes by clouds of gas and cosmic dust.
The International Space Station is another project whose goal is to expand our knowledge of space. And here's an interesting question, but how does the ISS sound? Experts involved in the training of
astronauts to fly to the station, they say that everything there sounds exactly the same as on Earth.
The International Space Station itself consists of modules, and its internal space resembles narrow cylindrical corridors. There are 14 modules in total, each of which contains research laboratories,
utility rooms, warehouses, sleeping places, and simulators.
In this regard, the ISS is quite a noisy place: the fans tirelessly drive air through the entire station. All this is reminiscent of several dozen computers “overloaded with processes”, whose coolers
rotate at a breakneck speed and create a rather strong hum.
In addition to the fans on the ISS, there is still a large number of pumps, compressors and other devices that emit sounds, for example, a special ARED space simulator, in which a cunning system of
cylinders, levers and disks provides loads of up to 600 kg. Astronauts need to train, and to raise an ordinary bar in space is inconvenient and useless.
The noise level on the ISS varies from 58 to 72 decibels (maximum 80). Of course, all parts of the station sound differently: the noise level on the ISS can be assessed by
(pay attention to how the sound changes at 24 minutes when switching to the Russian space module).
If you don’t like the video tour of the station, then astronaut Chris Hadfield especially for you has
several audio tracks recorded in various parts of the ISS to SoundCloud. Moreover, he even
recorded the
song, accompanying himself on the guitar.
Many musical compositions can very accurately convey the feeling of space and immerse us in a cosmic mood in a matter of seconds. Someone thinks that the greatest number of space associations are
associated with the songs of David Bowie (
). Listening to them, you seem to go on a long journey.
Others advise to pay attention to the classic
the works of
Gustav Holst "Planet". There are even more interesting
. Or the
"Inception", which
PS More materials on the topic of audio - in our blog " Hi-Fi World ".
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We found one answer for “Quadric” .
This page shows answers to the clue
, followed by 4 definitions like “
A quantic of the second degree
”, “
Of or pertaining to the second degree
” and “
A surface whose equation in three variables is of the second degree
Definitions of “Quadric”
If the answer you seek is not in the answers above these definitions may help solving your crossword puzzle.
• A surface whose equation in three variables is of the second degree
• A surface in three dimensions that is described by equations containing the squares of x, y
• A curve or surface whose equation (in Cartesian coordinates) is of the second degree
• (projective geometry) There are generalizations of quadrics: quadratic sets
• Quadric surface a curve or surface whose equation (in Cartesian coordinates) is of the second degree
More crossword answers
We found one answer for the crossword clue
. A further 10 clues may be
If you haven't solved the “crossword clue Quadric” yet try to search our Crossword Dictionary by entering the letters you already know! (Enter a dot for each missing letters, e.g. “P.ZZ..” will find
Also look at the related clues for crossword clues with similar answers to “Quadric”
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Multiplying Complex Numbers: (−3+2i)(1−i3)
This article will guide you through the process of multiplying complex numbers. We'll focus on the specific example of (−3+2i)(1−i3).
Understanding Complex Numbers
Complex numbers are numbers that can be expressed in the form a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit, defined as the square root of -1.
Multiplying Complex Numbers
When multiplying complex numbers, we use the distributive property (often referred to as FOIL - First, Outer, Inner, Last) similar to multiplying binomials.
Let's break down the multiplication of (−3+2i)(1−i3):
1. First: (−3)(1) = -3
2. Outer: (−3)(−i3) = 9i
3. Inner: (2i)(1) = 2i
4. Last: (2i)(−i3) = −6i²
Now, remember that i² = -1. Substitute this into our result:
-3 + 9i + 2i - 6(-1)
Combining like terms:
-3 + 9i + 2i + 6 = 3 + 11i
Final Result
Therefore, the product of (−3+2i)(1−i3) is 3 + 11i.
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Face numbers of 4-Polytopes and 3-Spheres
In this paper, we discuss f- and flag-vectors of 4-dimensional convex polytopes and cellular 3-spheres. We put forward two crucial parameters of fatness and complexity: Fatness F(P) := (f_1+f_2-20)/
(f_0+f_3-10) is large if there are many more edges and 2-faces than there are vertices and facets, while complexity C(P) := (f_{03}-20)/(f_0+f_3-10) is large if every facet has many vertices, and
every vertex is in many facets. Recent results suggest that these parameters might allow one to differentiate between the cones of f- or flag-vectors of -- connected Eulerian lattices of length 5
(combinatorial objects), -- strongly regular CW 3-spheres (topological objects), -- convex 4-polytopes (discrete geometric objects), and -- rational convex 4-polytopes (whose study involves
arithmetic aspects). Further progress will depend on the derivation of tighter f-vector inequalities for convex 4-polytopes. On the other hand, we will need new construction methods that produce
interesting polytopes which are far from being simplicial or simple -- for example, very ``fat'' or ``complex'' 4-polytopes. In this direction, I will report about constructions (from joint work with
Michael Joswig, David Eppstein and Greg Kuperberg) that yield -- strongly regular CW 3-spheres of arbitrarily large fatness, -- convex 4-polytopes of fatness larger than 5.048, and -- rational convex
4-polytopes of fatness larger than 5-epsilon.
arXiv Mathematics e-prints
Pub Date:
August 2002
□ Metric Geometry;
□ Combinatorics;
□ 52B11;
□ 52B10;
□ 51M20
Proceedings of the ICM, Beijing 2002, vol. 3, 625--636
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Ensuring Linkage Between Requirements and Verification — Robert Walter Allen
One of several reasons for emphasizing product requirements includes enabling modeling and simulations of designs, as well as ensuring adequate verification and validation testing.
Recall the fundamental framing of a requirement as:
• “What shall the design provide (output) @ input conditions?”
• “What does design need (inputs) from the customer in order for the design to perform as expected?”
Note the framing (within the requirement) of a mathematical and/or experimental relationship where “Y” is the output as a function of “(x)” as the input….Y = f(x) or as a function of multiple inputs
Y = f (x1, x2, x3…xn). Let’s expand on this for a moment:
Mathematical Relationship: framing the requirement as an output as a function of input(s), we’ve enabled design performance to be modeled with design simulation tools (“design by analysis”). Assuming
accurate modeling/simulation of design performance, we can predict our ability to meet requirements and optimize design performance before we begin potentially expensive prototyping.
Experimental Relationship: we’ve also enabled design performance to be tested through experimentation, or a more formal test methodology, design of Experiments (DoE). Here we can determine the
performance of a response variable (“Y”) as a function of input variables (x1, x2, x3…xn). We can also obtain a mathematical relationship through regression analysis and use it to our advantage as
outlined above.
Mathematical and experimental relationships are critical for predicting and optimizing design performance
Referring to the diagram below, we can see where modeling/simulation of design performance fits into the requirements-design-requirements hierarchy.
Also, we can see where verification testing occurs, including the (potential) purpose of verification testing to validate the models/simulations. Finally note verification traceability from/to the
requirement. With the requirement we asked “what the design shall provide” and verification testing we’re determining “does the design provide” the required performance.
Of course, waiting for a prototype to test may cost a lot of time and money, whereas an accurate model/simulation can be significantly advantageous. Much of this hinges on requirements, however,
hence the emphasis on requirements in several previous articles.
Using Hierarchy In Complex Systems Requirements and Design
Emphasizing Product Requirements
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Or Composite
Is 242 Prime Number Or Composite Number
Let's check 242 number is prime or composite.
Check 242 Prime Number Or Not
Factors List of 242 :
The factors number for 242 is 6 so it is greater than 2 , that's why it can not be prime number.
Check 242 Composite Number Or Not
If the numbers of factors are greater than 2 (1 and itself) is a composite number, now we check 242 numbers factors.
Factors List of 242 :
The factors number for 242 is 6 so it is greater than 2 and it is a composite number.
242 is a composite number.
Solved Problems on 242 Prime Or Composite
Q.1: Find if 242 is a composite number.
Answer : Yes it is a composite number with 6 factors.
Q.1: Is 242 prime number or composite number ?
Answer : 242 is a composite number..
Type your number to check prime or composite.
More Equations :
- 242 -
Prime / Composite Number Selection Methods:
There are several ways to check if a number is prime or composite. One way is to use the trial division method, which involves dividing the number being tested by every integer between 2 and the
square root of the number, and checking if any of these divisions result in a remainder of 0. If the number being tested is prime, none of the divisions should result in a remainder of 0.
For example, to check if the number 13 is prime, we would divide it by 2, 3, 4, 5, and 6 (since the square root of 13 is approximately 3.6). Since none of these divisions result in a remainder of 0,
we can conclude that 13 is a prime number.
Another method for checking for prime numbers is the sieve of Eratosthenes, which is an algorithm that allows you to quickly find all of the prime numbers up to a certain limit by crossing out
multiples of the primes that you have already found.
To use the sieve of Eratosthenes, start by writing out a list of the numbers from 2 to the limit that you want to find the prime numbers up to. Then, starting with the first number on the list (which
is 2), cross out every second number on the list (since these are all multiples of 2). Next, move to the next number on the list that has not been crossed out (which is 3) and cross out every third
number on the list. Continue this process until you have crossed out all of the multiples of the prime numbers on the list.
The numbers that are left on the list after all of the multiples have been crossed out are the prime numbers up to the limit that you specified.
There are also several other methods for checking for prime numbers, including the Fermat primality test and the Miller-Rabin primality test. These methods are more advanced and require more
sophisticated mathematical concepts and algorithms, but they can be more efficient for larger numbers.
In summary, to check if a number is prime or composite, you can use the trial division method, the sieve of Eratosthenes, the Fermat primality test, or the Miller-Rabin primality test.
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Chapter 5 Accounting Ratios
• State which of the following statements are True or False.
(a) The only purpose of financial reporting is to keep the managers informed about the progress of operations.
Answer False
(b) Analyses of data provided in the financial statements a is termed as financial analysis.
Answer True
(c) Long term creditors are concerned about the ability of a firm to discharge its obligations to pay interest and repay the principal amount of term.
Answer True
(d) A ratio is always expressed as a quotient of one number divided by another.
Answer False
(e) Ratios help in comparisons of a firm’s results over a number of accounting periods as well as with other business enterprises.
Answer True
(f) One ratios reflect both quantitative and qualitative aspects.
Answer False
Question 1. Current ratio =4.5:1,quick ratio =3:1, Inventory is Rs.36,000. Calculate the current assets and current liabilities.
Question 2. Current liabilities of a company are ? 5,60,000 current ratio is 5 : 2 and quick ratio is 2 : 1. Find the value of the stock.
Question 3. Current assets of a company are Rs. 5,00,000. Current ratio is 2.5 : 1 and quick ratio is 1 : 1. Calculate the value of current liabilities, liquid assets and stock.
(i) The following groups of ratios primarily measure risk
(a) liquidity, activity and profitability
(b) liquidity, activity and common stock
(c) liquidity, activity and debt
(d) activity, debt and profitability
Answer (c) Liquidity, activity and debt
(ii) The————-ratios are primarily measures of return.
(a) liquidity (b) activity
(c) debt (d) profitability
Answer (b) Activity
(iii) The…………….of a business firm is measured by its ability to satisfy
its short term obligations as they come due.
(a) activity (b) liquidity
(c) debt (d) profitability
Answer (b) Liquidity
(iv) ……………….ratios are a measure of the speed with which various
accounts are converted into sales or cash.
(a) Activity (b) Liquidity
(c) Debt (d) Profitability
Answer (a) Activity
(v) The two basic measure of liquidity are
(a) inventory turnover and current ratio
(b) current ratio and liquid ratio
(c) gross profit margin and operating ratio
(d) current ratio and average collection period
Answer (b) Current ratio and liquid ratio
(vi) The……………is a measure of liquidity which excludes………………….., generally the least liquid asset.
(a) current ratio, accounts debtors
(b) liquid ratio, accounts debtors
(c) current ratio, inventory
(d) liquid ratio, inventory
Answer (d) Liquid ratio, inventory
Question 1. Calculate the amount of gross profit
Average stock = Rs.80,000
Stock turnover ratio = 6 times
Selling price = 25% above cost
Answer Average stock = ? 80,000
Stock Turnover Ratio = 6 times
Selling Price = 25% above cost
Question 2. Calculate stock Turnover Ratio
(i) The………..is useful in evaluating credit and collection policies.
(a) average payment period (b) current ratio
(c) average collection period (d) current asset turnover
Answer (c) Average collection period
(ii) The………measures the activity of a firm’s inventory.
(a) average collection period (b) inventory turnover
(c) liquid ratio (d) current ratio
Answer (b) Inventory turnover
(iii) The………..ratio may indicate the firm is experiencing stock outs and lost sales.
(a) average payment period (b) inventory turnover
(c) average collection period (d) quick
Answer (d) Quick
(iv) ABC Co extends credit terms of 45 days to its customer, its credit collection would be considered poor if its average collection period was
(a) 30 days (b) 36 days
(c) 47 days (d) 57 days
Answer (c) 47 days
(v) …………… are especially interested in the average payment period, since it provides them with a sense of the bill-paying patterns of the firm.
(a) Customers (b) Stockholders
(c) Lenders and suppliers (d) Borrowers and buyers
Answer (c) Lenders and suppliers
(vi) The……………….. ratios provide the information critical to the long-run operation of the firm
(a) liquidity (b) activity
(c) solvency (d) profitability
Answer (c) Solvency
Question 1. What do you mean by Ratio Analysis?
Answer The ratio analysis is the most powerful tool of financial statement analysis. Ratios simply mean one number expressed in terms of another. A ratio is a statistical yardstick by means of which
relationship between two or various figures can be compared or measured. Ratios can be found out by dividing one number by another number. Ratios show how one number is related to another.
Question 2. What are various types of ratios?
Answer Accounting ratios are classified in two ways Categories as follows
(i) Traditional Classification: Traditional ratios are those accounting ratios which are based on the Financial Statement like Trading and Profit and Loss Account and Balance Sheet. On the basis of
accounts of financial statements, the Traditional Classification is further divided into the following categories
(a) Income Statement Ratios: like Gross Profit Ratio, etc.
(b) Balance Sheet Ratios: like Current Ratio, Debt Equity Ratio, etc.
(c) Composite Ratios :like Debtors Turnover Ratio, etc.
(ii) Functional Classification This classification of ratios is based on the functional need and the purpose for calculating ratio. The functional ratios are further divided into the following
(a) Liquidity Ratio: These ratios are calculated to determine short term solvency.
(b) Solvency Ratio :These ratios are calculated to determine long term solvency.
(c) Activity Ratio :These ratios are calculated for measuring the operational efficiency and efficacy of the operations. These ratios relate to sales or cost of goods sold.
(d) Profitability Ratio: These ratios are calculated to assess the
Question 3. What relationships will be established to study?
(a) Inventory Turnover (b) Debtor Turnover
(c) Payables Turnover (d) Working Capital Turnover
Answer (a) Inventory Turnover Ratio: This ratio is a relationship between the cost of goods sold during a particular period of time and the cost of average inventory during a particular period. It is
expressed in number of times. Stock turnover ratio/inventory turnover ratio indicates the number of time the stock has been turned over during the period and evaluates the efficiency with which a
firm is able to manage its inventory.
This ratio indicates whether investment in stock is within proper limit or not. The ratio is calculated by dividing the cost of goods sold by the amount of average stock at cost. The formula for
calculating inventory turnover ratio is as follows
(b)Debtor Turnover Ratio :Debtor turnover ratio or accounts receivable turnover ratio indicates the velocity of debt collection of a firm. In simple words it indicates the number of times average
debtors (receivable) are turned over during a year. The formula for calculating Debtors turnover ratio is as follows
(c)Creditors/Payables Turnover Ratio :It compares creditors with the total credit purchases. It signifies the credit period enjoyed by the firm in paying creditors. Accounts payable include both
sundry creditors and bills payable. Same as debtor’s turnover ratio, creditor’s turnover ratio can be calculated in two forms, creditors’ turnover ratio and average payment period. The following
formula is used to calculate the creditors Turnover Ratio
(d)Working Capital Turnover Ratio Working capital turnover ratio indicates the velocity of the utilization of net working capital. This ratio represents the number of times the working capital is
turned over in a year and is calculated as follows
Question 4. Why would the inventory turnover ratio be more important when analysing a grocery store than an insurance company?
Answer Nature of business make inventory turnover ratio more important in case of a grocery store as compare to an insurance company. A grocery store is a trading concern involved in trading i.e.,
buying and selling of goods and in this regards it is obvious to maintain some inventory in stores. On the other hand, insurance company involved in service business and involved in delivering
service there is no question of inventory because service is perishable in nature and cannot be stored. That’s why inventory turnover ratio is more important in case of grocery store than an
insurance company.
Question 5. The liquidity of a business firm is measured by its ability to satisfy its long term obligations as they become due? Comment.
Answer Yes it is true that the liquidity of a business firm is measured by its ability to pay its long term obligations as they become due. Here the long term obligation means payments of principal
amount on the due date and payments of interests on the regular basis. For measuring the long term solvency of any business we calculate the following ratios.
Debt Equity Ratio: Debt equity ratio indicates the relationship between the external equities or outsiders funds and the internal equities or shareholders funds. It is also known as external internal
equity ratio. It is determined to ascertain soundness of the long term financial policies of the company. Following formula is used to calculate debt to equity ratio
Proprietory Ratio/Total Assets to Debt Ratio: Total assets to Debt Ratio or Proprietory Ratio are a variant of the debt equity ratio. It is also known as equity ratio or net worth to total assets
ratio. This ratio relates the shareholder’s funds to total assets. Proprietory / Equity ratio indicates the long-term or future solvency position of the business. Formula of Proprietory/Equity Ratio
Fixed Assets to Proprietor’s Fund Ratio: Fixed assets to proprietor’s fund ratio establish a relationship between fixed assets and shareholders’ funds. The purpose of this ratio is to indicate the
percentage of the owner’s funds invested in fixed assets. The formula for calculating this ratio is as follows
Interest Coverage Ratio: This ratio deals only with servicing of return on loan as interest. This ratio depicts the relationship between amount of profit utilise for paying interest and amount of
interest payable. A high Interest Coverage Ratio implies that the company can easily meet all its interest obligations out of its profit.
Question 6. The average age of inventory is viewed as the average length of time inventory is held by the firm or as the average number of day’s sales in inventory. Explain.
Answer Inventory Turnover Ratio This ratio is a relationship between the cost of goods sold during a particular period of time and the cost of average inventory during a particular period. It is
expressed in number of times. Stock turnover ratio/inventory turnover ratio indicates the number of time the stock has been turned over during the period and evaluates the efficiency with which a
firm is able to manage its inventory. The formula for calculating inventory turnover ratio is as follows
From the above formula, it is clear that this ratio reveals the average length of time for which the inventory is held by the firm.
Question 1. Who are the users of financial ratio analysis? Explain the significance of ratio analysis to them.
Answer Financial ratios help their users to take various managerial decisions. In this context there are four categories of users who are interested in financial ratios. These are the management,
investors, long term creditors and short term creditors. The significance of ratios to the above mentioned users is as follows
(i) Management :Management calculate ratios for taking various managerial decisions. Management is always interested in future growth of the organisation. In this regard management design various
policy measures and draft future plans. Management wish to know how effectively the resources are being utilised conseguently, they are interested in Activity Ratios and Profitability Ratios like Net
Profit Ratio, Debtors Turnover Ratio, Fixed Assets Turnover Ratios, etc. ‘
(ii) Equity Investors :The prime concern of investors before investing in shares is to ensure the security of their principle and return on investment. It is a well known fact that the security of
the funds is directly related to the profitability and operational efficiency of the business. In this way they are interested in knowing Earnings per Share, Return on Investment and Return on
(iii) Long Term Creditors: Long term creditors are those creditors who provide funds for more than one year, so they are interested in long term solvency of the firm and in assessing the ability of
the firm to pay interest on time. In this way they are interested in calculating Long term Solvency Ratios like, Debt-Equity Ratio, Proprietory Ratio, Total Assets to Debt Ratio, Interest Coverage
Ratio, etc.
(iv) Short Term Creditors :Short term creditors are those creditors who provide financial assistance through short term credit (Generally less than one year). That’s why short-term creditors are
interested in timely payment of their debts in short run. In this way they are always interested in Liquidity Ratios like, Current Ratio, Quick Ratios etc. These ratios reveal the current financial
position of the business. It is always observed that short term obligations are paid through current assest.
Question 2. What are liquidity ratios? Discuss the importance of current and liquid ratio.
Answer Liquidity ratios are calculated to determine the short-term solvency of the business. Analysis of current position of liquid funds determines* the ability of the business to pay the amount due
as per commitment to stakeholders. Included in this category are current ratio, Quick ratio and Cash Fund Ratios.
Current Ratio/Working Capital Ratio: This ratio establish relationship between current assets and current liabilities. The standard for this ratio is 2 : 1. It means a ratio 2 : 1 is considered
favourable. It is calculated by dividing the total of the current assets by total of the current liabilities. The formula for the current ratio is as follows
Current Ratio = Current Assets/Current Liabilities Or
Current Assets : Current Liabilities
Importance of Current Ratio Current Ratio Provides a measure of degree to which current assets cover current liabilities. The excess of current assets over current liabilities provides a measure of
safety margin available against uncertainty in realisation of current assets and flow of funds. However, it must be interpreted carefully because window-dressing is possible by manipulating the
components of current assets and current liabilities, e.g., it can be manipulated by making payment to creditors. A very high current ratio is not a good sign as it reflects under utilisation or
improper utilisation of resources.
Liquid/Acid Test/Quick Ratio This ratio establishes relationship between Quick assets and Current liabilities. Quick assets are those assets which can get converted into cash easily in case of
emergency. Out of current assets it is believed that stock, and prepaid expenses are not possible to convert in cash quickly. The standard for this ratio is 1:1. It means if quick assets are just
equal to the current liabilities they will be considered favourable with the view point of company’s credibility. The formula for the quick ratio is as follows
Importance of Current Ratio: Current Ratio Provides a measure of degree to which current assets cover current liabilities. The excess of current assets over current liabilities provides a measure of
safety margin available against uncertainty in realisation of current assets and flow of funds. However, it must be interpreted carefully because window-dressing is possible by manipulating the
components of current assets and current liabilities, e.g., it can be manipulated by making payment to creditors. A very high current ratio is not a good sign as it reflects under utilisation or
improper utilisation of resources.
Liquid/Acid Test/Quick Ratio:This ratio establishes relationship between Quick assets and Current liabilities. Quick assets are those assets which can get converted into cash easily in case of
emergency. Out of current assets it is believed that stock, and prepaid expenses are not possible to convert in cash quickly. The standard for this ratio is 1:1. It means if quick assets are just
equal to the current liabilities they will be considered favourable with the view point of company’s credibility. The formula for the quick ratio is as follows
Importance of Quick Ratio :It helps in determining whether a firm has sufficient funds if it has to pay all its current liabilities immediately. Because of exclusion of non-liquid current assets, it
is considered better than current ratio as a measure of liquidity position of the business. Standard for liquid ratio is 1:1. Sometimes quick ratio is calculated on the basis of quick liability
instead of current liabilities. Quick liabilities are calculated by ignoring bank overdraft, if any. It means to get the figure of quick liabilities from current liabilities; bank overdraft is
deducted from current liabilities.
Question 3. How would you study the solvency position of the firm?
Answer The solvency position of any firm is determined and measured with the help of solvency ratios. In this way we can say that the ratios which throw light on the debt servicing ability of the
businesses in the long run are known as solvency ratios. Solvency of a concern can be measured in two ways first to check the security of Debt and second is to check the security of return on Debt.
For calculating the security of debt we calculate Debt-Equity Ratio, Proprietory Ratio, Fixed Assets – Proprietory Fund Ratio, etc. And for calculating Security of Return on Debt we calculate
Interest Coverage Ratio. A brief description of the above mentioned ratios is as follows
Debt Equity Ratio :Debt Equity Ratio indicates the relationship between the external equities or outsiders funds and the internal equities or shareholders funds. It is also known as external internal
equity ratio. It is determined to ascertain soundness of the long term financial policies of the company.
Proprietory Ratio/ Total Assets to Debt Ratio: Total assets to Debt Ratio or Proprietory Ratio are a variant of the debt equity ratio. It is also known as equity ratio or net worth to total assets
ratio. This ratio relates the shareholder’s funds to total assets. Proprietory/Equity Ratio indicates the long-term or future solvency position of the business. Formula of Proprietary/Equity Ratio
Shareholder’s funds include equity share capital plus all reserves and surpluses items. Total assets include all assets, including Goodwill. Some authors exclude goodwill from total assets. In that
case the total shareholder’s funds are to be divided by total tangible assets. The total liabilities may also be used as the denominator in the above formula.
Fixed Assets to Proprietor’s Fund Ratio: Fixed Assets to Proprietor’s Fund Ratio establish a relationship between fixed assets and shareholders’ funds. The purpose of this ratio is to indicate the
percentage of the owner’s funds invested in fixed assets. The formula for calculating this ratio is as follows
The fixed assets are considered at their book value and the proprietor’s funds consist of the same items as internal equities in the case of debt equity ratio.
Interest Coverage Ratio :This ratio deals only with servicing of return on loan as interest. This ratio depicts the relationship between amount of profit utilise for paying interest and amount of
interest payable. A high Interest Coverage Ratio implies that the company can easily meet all its interest obligations out of its profit.
Question 4. What are important profitability ratios? How are they worked out? ‘
Answer Profitability Ratios Profitability ratios measure the results of business operations or overall performance and effectiveness of the firm. Some of the most Important and popular profitability
ratios are as under
Gross Profit Ratio: Gross Profit Ratio (GP ratio) is the ratio of gross profit to net sales expressed as a percentage. It expresses the relationship between gross profit and sales. The basic
components for the calculation of gross profit ratio are gross profit and net sales. Net sales mean sales minus sales returns.
Gross profit would be the difference between net sales and cost of goods sold. Cost of goods sold in the case of a trading concern would be equal to opening stock plus purchase, minus closing stock
plus all direct expenses relating to purchases. In the case of manufacturing concern, it would be equal to the sum of the cost of raw materials, wages, direct expenses and all manufacturing expenses.
In other words, generally the expenses charged to profit and loss account or operating expenses are excluded from the calculation of cost of goods sold.
Following formula is used to calculate gross profit ratios
Net Profit Ratio :Net Profit Ratio is the ratio of net profit to net sales. It is expressed as percentage. The two basic components of the net profit ratio are the net profit and sales. The net
profits are obtained after deducting income-tax and, generally, non-operating expenses and incomes are excluded from the net profits for calculating this ratio. Thus, incomes such as interest on
investments outside the business, profit on sales of fixed assets and losses on sales of fixed assets, etc are excluded.
Operating Profit Ratio :Operating Profit Ratio is the ratio of operating profit to net sales. There are many non operating expenses and incomes included in the profit and loss account which has
nothing to do with the operations of the business such as loss by fire, loss by theft etc. On the other had in credit side of the P&L account, there are so many incomes
which can be considered as operating incomes such as dividend, bank interest, rent etc. In this way net profit ratio will not tell the truth about the profit of the organisation. Hence operating
profit ratio will be helpful in that case. The formula for calculating operating ratio is as follows
Operating Ratio :Operating ratio is the ratio of cost of goods sold plus operating expenses to net sales. It is generally expressed in percentage, Operating ratio measures the cost of operations per
dollar of sales. This is closely related to the ratio of operating profit to net sales. The two basic components for the calculation of operating ratio are operating cost (cost of goods sold plus
operating expenses) and net sales. Operating expenses normally include (a) administrative and office expenses and (b) selling and distribution expenses. The formula for calculating the operating
ratio is as follows
Question 5. Financial ratio analysis are conducted by four groups of analysts : managers, equity investors, long term creditors and short term creditors. What is the primary emphasis of each of these
groups in evaluating ratios?
Answer This is very much true that the financial ratio analysis is conducted by four groups of analysts : managers, equity investors, long term creditors and short term creditors. The primary
emphasis of each of these groups in evaluating these ratios are as follows
(i) Management: Management calculate ratios for taking various managerial decisions. Management is always interested in future growth of the organisation. In this regard management design various
policy measures and draft future plans. Management wish to know how effectively the resources are being utilised Consequently, they are interested in Activity Ratios and Profitability Ratios like Net
Profit Ratio, Debtors Turnover Ratio, Fixed Assets Turnover Ratios, etc.
(ii) Equity Investors: The prime concern of investors before investing in shares is to ensure the security of their principle and return on investment. It is a well known fact that the security of
the funds is directly related to the profitability and operational efficiency of the business. In this way they are interested in knowing Earnings per Share, Return on Investment and Return on
(iii) Long Term Creditors: Long term creditors are those creditors who provide funds for more than one year, so they are interested in long term solvency of the firm and in assessing the ability of
the firm to pay interest on time. In this way they are interested in calculating Long term Solvency Ratios like, Debt-Equity Ratio, Proprietory Ratio, Total Assets to Debt Ratio, Interest Coverage
Ratio, etc.
(iv) Short Term Creditors: Short term creditors are those creditors who provide financial assistance through short term credit (Generally less than one year). That’s why short term creditors are
interested in timely payment of their debts in short run. In this way, they are always interested in Liquidity Ratios like, Current Ratio, Quick Ratios etc. These ratios reveal the current financial
position of the business. It is always observed that short term obligations are paid through current assest.
Question 6. The current ratio provides a better measure of overall liquidity only when a firm’s inventory cannot easily be converted into cash. If inventory is liquid, the quick ratio is a preferred
measure of overall liquidity. Explain.
Answer The above mentioned statement is true. There are two different ways to measure the liquidity of a firm first through current ratio of the firm and second through quick ratio of the firm. The
second one is considered the more refine form of measuring the liquidity of the firm.
The current ratio ‘explains the relationship between current assets and current liabilities. If current assets are quite capable to pay the current liability the liquidity of the concerned firm will
be considered good. But here generally one question arises there are certain assets which cannot be converted into cash quickly such as stock and prepaid expenses.
As far as the matter of prepaid expenses is concerned it’s ok but what about the stock if we measure the liquidity on the basis of conversion of current assets in cash there are many firms where
conversion of stock is not possible into cash frequently say e.g., heavy machinery manufacturing companies, locomotive companies, etc. This is because, the heavy stocks like machinery, heavy tools
etc. cannot be easily sold off. In this case it is always advisable to follow the current ratio for measuring the liquidity of a firm.
But on the other hand, in case of those firms where the stock can be easily realised or sold off consideration of stock should be avoided and to measure the liquidity of that firm Quick ratio should
be calculated, e.g., the inventories of a service sector company are very liquid as there are no stocks kept for sale, so in that case liquid ratio must be followed for measuring the liquidity of the
We can understand from the above mentioned statement in the light of another example where stock contribute the major portion in current assets in that case to find out the liquidity of that firm
stock cannot be avoided to measure the liquidity of the firm. On the other hand where stock contributes a reasonably less amount it can be avoided and liquidity of that firm can be measured with the
help of quick ratio. On the other hand where there is a lot of fluctuation in the price of stock it is always advisable to compute quick ratio and avoid the stock figure because it will reduce the
authenticity of liquidity measure.
Question 1. Following is the Balance Sheet of Rohit and Company as on March 31, 2006.
Calculate Current Ratio.
Question 2. Following is the Balance Sheet of Title Machine Limited as on March 31, 2006.
Calculate Current Ratio and Liquid Ratio.
Question 3. Current Ratio is 3:5 Working Capital is Rs. 9,00,000. Calculate the amount of Current Assets and Current Liabilities.
Note :According to the ratio, current asset is less than current liability hence working capital should be negative. To match the figures and answer of the question current ratio is taken as 3.5 : 1
and working capital ? 90,000.
Question 4. Shine Limited has a current ratio 4.5:1 and quick ratio 3:1; if the stock is 36,000, calculate current liabilities and current assets.
Question 5. Current liabilities of a company are Rs. 75,000. If Current ratio is 4 : 1 and liquid ratio is 1:1, calculate value of current assets, liquid assets and stock.
Question 6. Handa Limited has stock of Rs. 20,000. Total liquid assets are Rs. 1,00,000 and quick ratio is 2:1 Calculate current ratio.
Question 7. Calculate debt equity ratio from the following information
Question 8. Calculate Current Ratio if Stock is ? 6,00,000; Liquid Assets Rs. 24,00,000; Quick Ratio 2:1.
Question 9. Compute Stock Turnover Ratio from the following information
Question 10. Calculate following ratios from the following information
(i) Current ratio (ii) Acid test ratio
(iii) Operating Ratio (iv) Gross Profit Ratio
Note (i) Acid test ratio, quick ratio and liquid ratio are one and the same.
(ii) Students mostly get confused in operating ratio and operating profit ratio, so be careful while doing these ratios.
Question 11. From the following information calculate
(i) Gross Profit Ratio (ii) Inventory Turnover Ratio (iii) Current Ratio (iv) Liquid Ratio
(v) Net Profit Ratio (vi) Working Capital Ratio
Note :In this question stock is given separately from current assets, hence* it is added to make total current assets.
Note :In this question current assetts should be considered as other current asset and stock is separate, in other words, other current assets means liquid assets. Working capital ratio and working
capital turnover ratio means same.
Question 12. Compute Gross Profit Ratio, Working Capitat Turnover Ratio, Dept Equity Ratio and Proprietory Ratio from the fottowing information
Question 13. Calculate Stock Turnover Ratio if Opening Stock is Rs. 76,250, Closing Stock is 98,500, Sales is Rs. 5,20,000, Sales Return is Rs.20,000, Purchase is Rs. 3,22,250.
Question 14. Calculate Stock Turnover Ratio from the data given below
Question 15. A trading firm’s average stock is ? 20,000 (cost). If the stock turnover ratio is 8 times and the firm setts goods at a profit of 20% on sales, ascertain the profit of the firm.
Question 16. You are able to collect the following information about a company for two years
Calculate Stock Turnover Ratio and Debtor Turnover Ratio if in the year 2004 stock in trade increased by Rs. 2,00,000.
Question 17. The following Balance Sheet and other information, calculate following ratios
(i) Debt Equity Ratio (ii) Working Capital Turnover Ratio
(iii) Debtors Turnover Ratio
Question 18. The following is the summarised Profit and Loss account and the Balance Sheet of Nigam Limited for the year ended March 31, 2007
(i) Quick Ratio
(ii) Stock Trunover Ratio
(iii) Return on Investment
Question 19. From the following,
(a) Debt Equity Ratio (b) Total Assets to Debt Ratio (c) Propietory Ratio.
Question 20. Cost of Goods Sold is 1 1,50,000 Operating expenses are Rs. 60,000. Sales is Rs. 2,60,000 and Sales Return is Rs. 10,000. Calculate Operating Ratio.
Question 21. The following is the summerised transactions and Profit and Loss Account for the year ending March 31, 2007 and the Balance Sheet as on that date.
Calculate (i) Gross Profit Ratio (ii) Current Ratio (iii) Acid Test Ratio
(iv) Stock Turnover Ratio (v) Fixed Assets Turnover Ratio.
Question 22. From the fotlowing information calcutate Gross Profit Ratio, Stock Turnover Ratio and Debtors Turnover Ratio.
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seminars - Quantitative nonembeddability of nilpotent Lie groups and groups of polynomial growth into superreflexive spaces
It is known that simply connected nonabelian nilpotent Lie groups and not virtually abelian groups of polynomial growth fail to embed bilipschitzly into superreflexive Banach spaces.
We quantify this fact in two ways. First, we provide a lower bound on the distortion of balls in the aforementioned groups into superreflexive spaces. In particular, we show that the $L^
p$-distortion, $(1<p<\infty)$, of a ball of radius $n\ge 2$ in the aforementioned groups is exactly $(\log n)^{1/\max\{p,2\}}$ up to constants. Second, we characterize the asymptotic behavior of the
Lipschitz compression rate of functions from the aforementioned groups into the $L^p$ spaces, $p>1$. If time permits, I will discuss conjectures on the distortion and compression rate when the target
space is $L^1$.
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Multiplication Table Quizzes Printable
Discovering multiplication following counting, addition, as well as subtraction is ideal. Kids understand arithmetic by way of a natural progression. This growth of discovering arithmetic is truly
the adhering to: counting, addition, subtraction, multiplication, and finally section. This declaration leads to the question why discover arithmetic in this pattern? Moreover, why learn
multiplication right after counting, addition, and subtraction but before department?
The subsequent specifics respond to these concerns:
1. Youngsters learn counting initially by associating visual physical objects using their hands and fingers. A real example: Just how many apples exist within the basket? More abstract example is
just how older are you?
2. From counting figures, another plausible move is addition accompanied by subtraction. Addition and subtraction tables are often very beneficial teaching aids for the kids since they are visual
equipment producing the changeover from counting easier.
3. Which should be acquired up coming, multiplication or division? Multiplication is shorthand for addition. At this stage, kids have a organization understand of addition. Therefore, multiplication
may be the next rational form of arithmetic to find out.
Review essentials of multiplication. Also, look at the essentials how to use a multiplication table.
We will evaluation a multiplication illustration. By using a Multiplication Table, increase several times 3 and acquire a solution twelve: 4 x 3 = 12. The intersection of row 3 and line several of
your Multiplication Table is 12; 12 is the solution. For children starting to find out multiplication, this really is straightforward. They could use addition to eliminate the problem thus affirming
that multiplication is shorthand for addition. Case in point: 4 by 3 = 4 4 4 = 12. It is an exceptional introduction to the Multiplication Table. An added gain, the Multiplication Table is graphic
and demonstrates returning to learning addition.
In which should we commence learning multiplication using the Multiplication Table?
1. Very first, get familiar with the table.
2. Start out with multiplying by one. Start at row number one. Relocate to column # 1. The intersection of row one particular and line one is the answer: one.
3. Perform repeatedly these steps for multiplying by one. Grow row 1 by columns 1 via 12. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Repeat these techniques for multiplying by two. Flourish row two by columns 1 through five. The replies are 2, 4, 6, 8, and 10 correspondingly.
5. Allow us to leap ahead of time. Repeat these steps for multiplying by several. Grow row five by posts one through a dozen. The replies are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60
6. Now we will raise the degree of issues. Repeat these actions for multiplying by a few. Flourish row 3 by columns one particular via 12. The replies are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33,
and 36 correspondingly.
7. In case you are more comfortable with multiplication to date, consider using a examination. Fix the following multiplication troubles in your head and then examine your responses on the
Multiplication Table: increase six as well as 2, increase nine and three, grow 1 and eleven, flourish several and 4, and multiply 7 as well as 2. The trouble replies are 12, 27, 11, 16, and 14
Should you obtained four away from five problems proper, create your very own multiplication checks. Compute the replies in your mind, and appearance them utilizing the Multiplication Table.
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Fluctuation Theorems for Synchronization of Interacting Polya's urns
Crimaldi, Irene and Dai Pra, Paolo and Minelli, Ida G. Fluctuation Theorems for Synchronization of Interacting Polya's urns. Stochastic processes and their applications, 126 (3). pp. 930-947. ISSN
0304-4149 (2016)
This is the latest version of this item.
Full text not available from this repository.
We consider a model of N two-colors urns in which the reinforcement of each urn depends also on the content of all the other urns. This interaction is of mean-field type and it is tuned by a
parameter \alpha in [0,1]; in particular, for \alpha=0 the N urns behave as N independent Polya's urns. For \alpha>0 urns synchronize, in the sense that the fraction of balls of a given color
converges a.s. to the same (random) limit in all urns. In this paper we study fluctuations around this synchronized regime. The scaling of these fluctuations depends on the parameter \alpha. In
particular, the standard scaling t^{-1/2} appears only for \alpha>1/2. For \alpha\geq 1/2 we also determine the limit distribution of the rescaled fluctuations. We use the notion of stable
convergence, which is stronger than convergence in distribution.
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Coin Toss Probability Formula: Properties, Chemical Structure and Uses
Coin Toss Probability Formula
Coin Toss Probability Formula
Coin Toss Probability Formula is part of the overall chapter on Probability. Students first need to understand the definition of probability before learning about the Coin Toss Probability Formula.
The probability of an event occurring is defined by probability. There are many instances in real life where people may need to make predictions about how something will turn out. The outcome of an
event may be known to us or unknown to them. When this occurs, we say that the event has a chance of occurring or not. In general, probability has many wonderful applications in games, business (to
make probability-based forecasts), and this emerging branch of artificial intelligence.
The probability of an occurrence can be calculated using the probability formula by simply dividing the favourable number of possibilities by the total number of possible outcomes. Because the number
of favourable outcomes can never exceed the total number of outcomes, the probability of an event occurring can range from 0 to 1. Furthermore, the percentage of positive outcomes cannot be negative.
The probability is defined as the ratio of favourable outcomes to all possible outcomes of an event. The symbol x represents the number of positive results for an experiment with ‘n’ outcomes. The
following formula can be used to calculate the probability of an event.
Probability(Event) = Positive Results/Total Results = x/n
Depending on the outcome or method used to calculate the likelihood that an event will occur, there may be many viewpoints or types of probabilities. There are four different types of probabilities:
1. Standard Probability
2. Empirical likelihood
3. Personal Probability
4. Probability axiomatically
The probability terminology listed below aids in a better understanding of probability concepts.
Experiment: A trial or procedure carried out to generate a result is referred to as an “experiment.”
Sample Space: A sample space is the collection of all potential outcomes of an experiment. Tossing a coin, for instance, has two possible outcomes: heads or tails.
Favourable Consequence: An occurrence is deemed to have produced the desired outcome or an anticipated event if it did so. For instance, if we roll two dice and get the sum of the two numbers as 4,
the possible or favourable possibilities are (1,3), (2,2), and (3,1).
Trial: To conduct a trial is to conduct a random experiment.
Random Experiment: A study with a preset set of outcomes is referred to as a “random experiment.” For instance, when one tosses a coin, the two possible outcomes are Heads or Tails. But there is no
chance of knowing which outcome will come .
Event: An event is the whole assortment of results from a random experiment.
What Are Coin Toss Probability Formulas?
Flipping a Coin has two outcomes: heads or tails. On any given toss, one cannot predict which way the coin will land, but one does know that it will either land Head or Tail. Flipping a coin, on the
other hand, is a random experiment because you know the range of possible outcomes but not the precise result for each random experiment execution.
One can determine the likelihood of an experiment using the Coin Toss Probability Formula.
Total positive outcomes and total number of outcomes that are feasible
There are two possible outcomes in total.
For heads, the Coin Toss Probability Formula
If the result is favourable, head (H).
Positive results in number = 1.
P (getting heads) = number of favourable outcomes divided by the total number of outcomes equals 1/2.
Formula for the chance of tossing a coin.
If the conclusion is favourable, tail (T).
Positive results in number = 1.
P (getting heads) = total number of potential outcomes / number of favourable outcomes, which equals 1/2.
Solved Examples Using Coin Toss Probability Formulas
Students can find the solved examples of Coin Toss Probability Formulain the website and mobile application of Extramarks. Coin Toss Probability Formula can be easily understood with the help of the
resources provided by Extramarks. TheCoin Toss Probability Formula is an important part of probability, and therefore students should solve a lot of questions. TheCoin Toss Probability Formulais also
very important for various exams. Extramarks resources will help students prepare for the topic on Coin Toss Probability Formula.
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Math Expression: Straight Angle
Straight Angle
Lesson Objective
In this lesson, we will learn about straight angle.
About This Lesson
After knowing what are obtuse angles, we will now learn:
• What is a straight angle.
• See some examples on identifying it.
The study tips below will give you a short summary and it can be used to recall the properties of this angle.
The math video below will explain more in depth about this angle. Furthermore, it will show some examples to identify this angle so that you can understand this lesson better.
Study Tips
Tip #1
If an angle is exactly 180°, it is a straight angle. The picture below shows some examples:
Math Video
Lesson Video
You can contribute to the development of this site and keep it free by getting all six video lessons and volume of solids and calculator app for just US$1.99 from Apple App Store.
I'd like to contribute or to know more about the app
Math Video Transcript
In this lesson, we will learn about this type of angle.
Let's take a look at this angle.
In the previous lesson, we learned that an obtuse angle is always more than 90 degrees, and less than 180 degrees.
What happens, when the angle becomes 180 degrees?
When the angle becomes 180 degrees, it is called a str. angle.
To emphasize, str. angle is an angle that is equals to 180 degrees.
Let's see some examples.
Is this a str. angle?
Yes! It is.
How about this angle?
Again, this is a str. angle.
Next example, is this a str. angle?
Clearly, this angle is not equals to 180 degrees.
Hence, this is not a str. angle
That is all for this lesson.
Practice Questions & More
Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on straight angle or pick your choice of question below.
Site-Search and Q&A Library
Please feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or
give comments there.
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What is genetic algorithm
genetic algorithm
n. (context computing English) A search heuristic that is based on biological evolution.
Genetic algorithm
In computer science and operations research, a genetic algorithm (GA) is a metaheuristic inspired by the process of natural selection that belongs to the larger class of evolutionary algorithms (EA).
Genetic algorithms are commonly used to generate high-quality solutions to optimization and search problems by relying on bio-inspired operators such as mutation, crossover and selection.
Usage examples of "genetic algorithm".
There are patent attorneys in Reno who swear that Manfred Macx is a pseudo, a net alias fronting for a bunch of crazed anonymous hackers armed with the Genetic Algorithm That Ate Calcutta: a kind of
Serdar Argic of intellectual property, or maybe another Bourbaki maths borg.
Probably a genetic algorithm is called for--certainly there won't be a mathematically exact solution.
Thompson wondered what would happen if you used the genetic algorithm approach on an electronic circuit.
There are patent attorneys in Reno who swear that Manfred Macx is a pseudo, a net alias fronting for a bunch of crazed anonymous hackers armed with the Genetic Algorithm That Ate Calcutta: a kind of
Serdar Argic of intellectual property, or maybe another Bourbaki math borg.
And, at last, he had found the genetic algorithm he had been looking for, the pattern of success that was most suited to his simulated world.
It was something Professor Logan had been patiently assembling and refining for years-an eldritch chimera of monster clipping service, genetic algorithm, and neural net.
A genetic algorithm might reach a minimum but it would be pretty much a random event.
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options = ssregestOptions(Name,Value) specifies additional options using one or more Name,Value pair arguments.
Create Default Option Set for State-Space Estimation Using Reduction of Regularized ARX Model
options = ssregestOptions;
Specify Options for State-Space Estimation Using Reduction of Regularized ARX Model
Create an option set for ssregest that fixes the value of the initial states to 'zero'. Also, set the Display to 'on'.
opt = ssregestOptions('InitialState','zero','Display','on');
Alternatively, use dot notation to set the values of opt.
opt = ssregestOptions;
opt.InitialState = 'zero';
opt.Display = 'on';
Input Arguments
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: opt = ssregestOptions('InitialState','zero') fixes the value of the initial states to zero.
ARXOrder — ARX model orders
'auto' (default) | matrix of nonnegative integers
ARX model orders, specified as a matrix of nonnegative integers [na nb nk]. The max(ARXOrder)+1 must be greater than the desired state-space model order (number of states). If you specify a value, it
is recommended that you use a large value for nb order. To learn more about ARX model orders, see arx.
RegularizationKernel — Regularizing kernel
'TC' (default) | 'SE' | 'SS' | 'HF' | 'DI' | 'DC'
Regularizing kernel used for regularized estimates of the underlying ARX model, specified as one of the following values:
• 'TC' — Tuned and correlated kernel
• 'SE' — Squared exponential kernel
• 'SS' — Stable spline kernel
• 'HF' — High frequency stable spline kernel
• 'DI' — Diagonal kernel
• 'DC' — Diagonal and correlated kernel
For more information, see [1].
Reduction — Options for model order reduction
Options for model order reduction, specified as a structure with the following fields:
• StateElimMethod
State elimination method. Specifies how to eliminate the weakly coupled states (states with smallest Hankel singular values). Specified as one of the following values:
'MatchDC' Discards the specified states and alters the remaining states to preserve the DC gain.
'Truncate' Discards the specified states without altering the remaining states. This method tends to product a better approximation in the frequency domain, but the DC gains are not guaranteed to
Default: 'Truncate'
• AbsTol, RelTol
Absolute and relative error tolerance for stable/unstable decomposition. Positive scalar values. For an input model G with unstable poles, the reduction algorithm of ssregest first extracts the
stable dynamics by computing the stable/unstable decomposition G → GS + GU. The AbsTol and RelTol tolerances control the accuracy of this decomposition by ensuring that the frequency responses of
G and GS + GU differ by no more than AbsTol + RelTol*abs(G). Increasing these tolerances helps separate nearby stable and unstable modes at the expense of accuracy. See stabsep (Control System
Toolbox) for more information.
Default: AbsTol = 0; RelTol = 1e-8
• Offset
Offset for the stable/unstable boundary. Positive scalar value. In the stable/unstable decomposition, the stable term includes only poles satisfying
□ Re(s) < -Offset * max(1,|Im(s)|) (Continuous time)
□ |z| < 1 - Offset (Discrete time)
Increase the value of Offset to treat poles close to the stability boundary as unstable.
Default: 1e-8
InputInterSample — Input-channel intersample behavior
'auto' | 'zoh' | 'foh' | 'bl'
Input-channel intersample behavior for transformations between discrete time and continuous time, specified as 'auto', 'zoh','foh', or 'bl'.
The definitions of the three behavior values are as follows:
• 'zoh' — Zero-order hold maintains a piecewise-constant input signal between samples.
• 'foh' — First-order hold maintains a piecewise-linear input signal between samples.
• 'bl' — Band-limited behavior specifies that the continuous-time input signal has zero power above the Nyquist frequency.
iddata objects have a similar property, data.InterSample, that contains the same behavior value options. When the InputInterSample value is 'auto' and the estimation data is in an iddata object data,
the software uses the data.InterSample value. When the estimation data is instead contained in a timetable or a matrix pair, with the 'auto' option, the software uses 'zoh'.
The software applies the same option value to all channels and all experiments.
Output Arguments
options — Option set for ssregest
ssregestOptions options set
Estimation options for ssregest, returned as an ssregestoptions option set.
[1] T. Chen, H. Ohlsson, and L. Ljung. “On the Estimation of Transfer Functions, Regularizations and Gaussian Processes - Revisited”, Automatica, Volume 48, August 2012.
Version History
Introduced in R2014a
R2022b: InputInterSample option allows intersample behavior specification for continuous models estimated from timetables or matrices.
iddata objects contain an InterSample property that describes the behavior of the signal between sample points. The InputInterSample option implements a version of that property in ssregestOptions so
that intersample behavior can be specified also when estimation data is stored in timetables or matrices.
R2018a: Renaming of Estimation and Analysis Options
The names of some estimation and analysis options were changed in R2018a. Prior names still work.
See Also
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LoF22 - Laws of Form Conference 2022
Andrew Crompton
The Alternative Natural Philosophy Association
How to Make a Horse Vanish
A nondescript set is a collection of objects can serve as either figure or ground, so they appear and disappear among their companions as if self-camouflaged. Large collections of identical objects
have this property, in nature they include murmurations, shoals, herds, but small sets of different object can also be self-camouflaged if objects are designed to balance being similar to and
different to each other. It is shown that, with different objects, at least four types is a minimum for self-camouflage to emerge, six or more is better. When objects are identical three can be
sufficient, as in the three-card trick in which the eye cannot disentangle cards shuffled in a tricky way. One of the smallest nondescript sets is found in nature among zebra. Their stripes, first
described a form of auto-mimesis by Eltingham in 1979, minimise the size of herd in which a zebra can vanish to deter persistence hunters that pursue animals to exhaustion. This allows zebras to
forage in smaller groups than otherwise would be the case, although at a cost of being more conspicuous to ambush hunters such as lions. It is conjectured that as few as two zebra can be entangled in
an unmarked state, allowing them to vanish for a moment. This is supported by evidence that the patterning is scaling to allow different sized zebra to merge and explains why stripes come in several
I am interested in the design of things that are easily overlooked or in other ways hard to describe or remember, such as found objects, and structures with hidden meanings such as gasholders (see AA
Files 74), and the Cenotaph, which contains an image of Excalibur, (see AA Files 34). My publications include The Checkerboard of Tunes, which shows a model of all the tunes in the Western musical
canon as a single solid object based on George Spencer-Brown's musical dictionary.
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Measurement of CP asymmetry in a time-dependent dalitz analysis of B0→(ρπ)0 and a constraint on the quark mixing matrix angle 2
We present a measurement of CP asymmetry using a time-dependent Dalitz plot analysis of B0→π+π-π0 decays based on a 414fb-1 data sample containing 449×106BB̄ pairs. The data was collected on the Υ(4S)
resonance with the Belle detector at the KEKB asymmetric energy e+e- collider. Combining our analysis with information on charged B decay modes, we perform a full Dalitz and isospin analysis and
obtain a constraint on the CKM angle 2, 68°< 2<95° as the 68.3% confidence interval for the 2 solution consistent with the standard model (SM). A large SM-disfavored region also remains.
All Science Journal Classification (ASJC) codes
• General Physics and Astronomy
Dive into the research topics of 'Measurement of CP asymmetry in a time-dependent dalitz analysis of B0→(ρπ)0 and a constraint on the quark mixing matrix angle 2'. Together they form a unique
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On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields
J. Electromagnetic Analysis & Applications, 2011, 3, 13-21
doi:10.4236/jemaa.2011.31003 Published Online January 2011 (http://www.SciRP.org/journal/jemaa)
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic
Electromagnetic Fields
Yongfeng Gui, Pei Li
East China Research Institute of Electronic Engineering, Hefei, China.
Email: guiyongfeng80@163.com
Received November 25th, 2010; revised December 20th, 2010; accepted December 27th, 2010.
The uniqueness theorem of time-harmonic electromagnetic fields, which is the theoretical basis of boundary value
problem (BVP) of electromagnetic fields, is reviewed. So far there are many versions of the statements and proofs on
the theorem. However, there exist some limitations and lack of strictness in these versions, for instance, the discussion
of the uniqueness of solution without considering the existence of solution and the lack of strictness in the case of loss-
less medium. In contrast with the traditional statements and proofs, this paper introduces some important conclusions
on operator equ ation from modern theory of partial differen tial equation (PDE) and attempts to solve the p roblems on
the existence and uniqueness of the solution to operator equation which is derived from Maxwell’s equations of
time-harmonic electromagnetic fields. This method provides a novel and rigorous approach to discuss and solve the
existence and uniqueness of the solution to time- harmonic fields in the new mathematical framework. Some important
conclusions are presente d .
Keywords: Time-Harmonic Fields, The Existence and Uniqueness of Solution, the Case of Lossless Medium, Operator
Equation, Variational Principles, Weak Solution, Coercive Condition
1. Introduction
In the electromagnetics it has important significance to
research time-harmonic electromagnetic fields. On one
hand, most fields appeared in practical engineering prob-
lems have harmonic time variation and we can solve
these engineering problems directly through researching
time-harmonic fields. On the other hand, time-varying
fields can be transformed into the superposition of
time-harmonic fields with the Fourier series or Fourier
transform [1,2]. Therefore, a thorough understanding and
discussion on the existence and uniqueness of the solu-
tion to time-harmonic electromagnetic fields are impor-
tant in the study of all electromagnetic fields and elec-
tromagnetic wave phenomena.
The necessity and importance of the uniqueness theo-
rem consist in that if the appropriate initial-boundary
value conditions of Maxwell’s equations are given, then
the solution is determined uniquely. So regardless of the
method by which the equations are solved, the same solu-
tion will be obtained. Recently, the issue on the unique-
ness has aroused much concern [3-9]. In this paper we
mainly deal with the existence and uniqueness of the so-
lution for time-harmonic electromagnetic fields. Firstly,
this paper quotes some most influential versions of the
uniqueness theorem of traditional time-harmonic elec-
tromagnetic fields theory and presents their proof method.
Secondly, we point out the limitations and lack of strict-
ness of traditional theory and make a result that the tradi-
tional theory has not yet solved the existence and
uniqueness of the solution completely. Moreover, we cite
the proof of existence and uniqueness of the weak solu-
tion to 0-Dirichlet problem of the Poisson equation as an
example to indicate that it is a rigorous method adopting
functional theory to discuss the existence and uniqueness
of the solution to PDE. Finally, we introduce a novel
consideration of the operator equation based on the mod-
ern theory of PDE, derive the operator equations of the
time-harmonic electromagnetic fields from Maxwell
equations, point out the substaintial difficulty in the
process of proving the existence and uniqueness of the
solutions and present some important conclusions.
2. The Statements and Proof Method of
Traditional Theory
Maxwell’s equations for time-harmonic fields are
() ()()
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields
() ()Erj Br
() ()Dr r
() 0Br (1.4)
The electric current continuity equation is
() ()
w is called the electric field intensi/m), here ty (V
the magnetic field intensity (A/m), ()Dr
electric displacement or electric flux intensity (C/m2),
the metic induction or magnetic flux intensity
the electric current density (A/m2) and
the electric charge density (C/m3).
There are many versions of the statements and proofs
of the uniqueness theorem of time-harmonic fields in
classical monographs and literatures. One of the most
well known versions is cited as follows with the form of
Proposition 1. Uniqueness theorem of time-harmonic
fields in simple connected domain
For the time-harmonic fields, considering a simply
connected domain V bounded by closed surface S, the
medium is isotropic and linear, where both V and S con-
tain only ordinary points. The solution satisfied Max-
well’s Equations (1) and its boundary conditions must be
unique if the following items are specified: (1) the
sources density within V, that is, current density distribu-
and equivalent magnetic current density distri-
bution m
; and (2.1) the tangential components of the
electric field or the tangential components of the mag-
netic field over whole boundary surface S or (2.2) tan-
gential electric field over part of the surface S and tan-
gential magnetic field over the remainder of S.
For the proof of Proposition 1, almost all the mono-
graphs and literatures adopt the method which belongs to
a kind of “energy integrals” method [10]. Through con-
structing an expression of “energy integrals” based on
Poynting theorem or Maxwell’s equations, the “energy
integrals” method has been applied to the proof of
uniqueness theorem of the time-varying fields [11].
The proof of Proposition 1 under the assumption that
the medium is lossy can be seen in [2,12-15]. For the case
of a domain with complicated boundary, the entire do-
main can be divided into a number of sub-domain and
make every sub-domain correspond with the simply
boundary. So the multiply connected domain can be ana-
lyzed by decomposing it into the union of simply con-
nected domain. The statements and proof can be seen in
It is assumed that medium is isotropic, linear and
sources are located inside domain V in Proposition 1. In
fact, it can be generalized to the case of anisotropic me-
dium and sources located outside domain V. For simpli-
fication, the isotropy and linear medium is discussed in
this paper and the argument in linear anisotropic medium
is similar.
3. Deficiency of the Traditional Theory
There exist some incompleteness in Proposition 1 and we
explain it as follows.
3.1. Existence of the Solution Has not been
Proved Rigorously
For the proof of uniqueness, traditional theory implies a
physical judgment or premise that the solution must exist.
It does not prove the existence of the solution and merely
states that if a solution exists for given BVP then the so-
lution is the only solution. However, it has been con-
firmed that the solution of many PDE do not exist really.
For some equations if we assume the existence of the
solution and construct the form of the difference solution,
we can formally “obtain” the proof of the uniqueness.
Obviously, the treatment is meaningless because the so-
lution may not exist. For example, H. Lewy [16] pro-
vided an equation as follows:
xy t
uiuixyu fxyt
in Ω (2)
is a set satisfied 22
ya, tb, a and b
are arbitrary fixed positive number. Under the premise
about existence of the solution of Equation (2) we can
“obtain” the uniqueness of the solution by using reduc-
tion to absurdity. However, in fact there exists a function
such that Equation (2)
has no solution in
. Since a and b are arbitrary,
Equation (2) has no solution in set
ytE xytr for all . 0r
Detailed discussion of Equation (2) can be seen in [17].
In the sense of physics, the solution of practical elec-
tromagnetic BVP always exists. However, it does not
mean that mathematical equations derived from the prac-
tical BVP must have a solution and the solution is unique.
The existence of the solution still needs a rigorous
mathematical proof and the judgment of physical concept
is insufficient. Consequently, it is absolutely necessary
and important to describe the reasonableness of mathe-
matical model and prove the existence of the solution
3.2. The Lossless Case has not been Really Solved
In the process of the traditional proof on Proposition 1,
lossy medium is assumed, that is, at least one of conduc-
tion loss, polarization loss and magnetization loss is not
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields15
ess theorem in lossless me-
this paper at-
y and Application of Functional
Tition and frequency domain wave
0-Dirichlet BVP of Poisson equation
0-Dirichlet BVP of Poisson
equal to zero. The field in a lossless medium is treated as
the limit of the corresponding field in a lossy medium
when the dissipation approaches to zero. In sense of
mathematics this treatment is not rigorous because the
validation in the case of a parameter approaching zero
does not guarantee the validation in the case of the pa-
rameter at the point of zero.
The proofs on the uniquen
um appeared in many books and literatures such as [2]
[14] are only an interpretation based on the assumption
that the case in lossless medium has been validated. The
discussion on uniqueness theorem in lossless medium is
avoided in [15], which write “The proof of the theorem
hinges on the assumption that the permittivity and the
permeability of the medium have a small imaginary part.
Assume the medium is slightly lossy.” Similarly, [18,19]
have not made a definite conclusion and proof on the
lossless case. Pozar in [20] considers that the solution for
the lossless medium may be not unique unless the dissi-
pation of medium is added. Hence, traditional theory has
not given the proof of the uniqueness theorem in lossless
medium strictly, which is a long-neglected problem. It is
to be confirmed and proved whether there exists the
uniqueness theorem in lossless medium.
For the final settlement of the problem
mpts to analyze the existence and uniqueness of the
solution of time-harmonic fields by using related theory
of functional analysis and PDE, offer a new kind of
statement and proof method including considering of the
4. Theor
Analysis to PDE
me domain wave equa
equation (Helmholtz equation) in electromagnetic fields
belong to hyperbolic equation and elliptic equation, re-
spectively. Obviously, PDE is a kind of operator equation.
We will give a very famous example in which 0-Dirichlet
BVP of Poisson equation is analyzed successfully to il-
lustrate the application of functional analysis to the prob-
lem on the existence and uniqueness of the solution of
operator equation.
Historically, the
2uf had been calculated directly for a long time.
there exist great difficulties in proving the
universality of the existence of the solution. After
long-time endeavor, the idea is changed into the present
method, that is, the weak solution of the equation is
sought firstly, then its existence and uniqueness is proved,
and finally its smoothness is determined. Thus, the fol-
lowing theorem is obtained.
Theorem 1. Consider the
(in Ω) (3.1)
lution, wher
, the equation must have a unique week so-
e n
is a bounded open domain,
represenatic integrable function space.
definition of weak solution of Poisson equatio
ts quadr
The n
has become a basic re-
e discussion to operator equation,
) For any
.1), (3.2) is given in [21]. The Poincare inequality and
Riesz representation theorem are used to prove the exis-
tence of weak solutions and reduction to absurdity is used
to obtain the uniqueness of the solution [21]. Hence, the
proof of Theorem 1, which is based on the theory of
functional analysis, is rigorous.
Such mathematical method
arch method in modern theory of PDE. As an indis-
pensable tool in modern theory of PDE, functional analy-
sis provides an important idea and model for solving the
existence and uniqueness of time harmonic electromag-
netic fields solutions.
For convenience of th
e cite some related definitions and important theorems
in functional analysis ([21-25]) as follows.
Lemma 1. (Riesz representation theorem
unded linear functional f defined in a Hilbert space H,
there exists unique f
such that
y for every
H, where represents
a norily, for anyH, a bded linear
functional f can be defined inf
rm. Contra f
s o
and furthermore,
y holds (see [2
Riesz representation tem indicates that a conti
. Assume
s linear functional can always be represented by an
inner product.
Definition 1
is a bilinear functional
defined in Hilbert space H
,auv is called symmetrical if such
, uv H
,auv is cjugate
symm H
, uv
etrical if
alled con
such that
,auv is called unded if atbo 0M such th
,u vauvM
(, uvH
,auv is called weakly coercive if 3)
suhat ch t
auv u
for any uH.
au rcive or po dis called coeesitivefinite if
such t2
auv u
for any uH.
inition 2. AinDef ssume A is a linear operator ed
onjugate or self-adjoint operator in
def in
ilbert space X,
1) A is called self-c
if conjugate operator *
of A exists and *
2) A is called symmetric operatorX in if
uv uAv for any ,uv X.
weakly c operator i3) A is calledoerciven X if
0 such that
uu cu for any uX.
A is called 4) positive definite operator ifin X
0 such that
uu cu for any uX.
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields
Th initions still apply when e above def
DA (the
. (The Lax-Milgram theorem [22]) Assume
finition domain of operator A) is a dense linear sub-
space of X,
Lemma 2
,uv is a bounded, coercive and conjugate bilinear
al defined in a Hilbert space H, then there must
exist a unique continuous linear operator
such that
,,auv uAv (, uvH
). Fu rthermore,
, where
Thof the existence and uniqueness theorem of the
and Uniquene ss of th e Solu tion
The losed convex subset of real
ilgram theorem indicates that t
itive numb
The Lax-Mhe bilinear
nctional satisfied specific characteristic can constitute a
linear operator with continuous inverse. The generalized
Lax-Milgram theorem can be obtained when the coercive
condition becomes the weak coercive condition [21].
5. Modern Theory of Operator Equation in
e proof
solution to operator equation will be realized through the
following steps.
5.1. The Existence
to Variational Equation as Well as Its
Relation with Corresponding
Variational Problem
orem 2. Assume U is a c
Hilbert space H. If
,auv defined in U is a bounded,
coercive and bilineaional, then for any *
r funct
is a conjugate space of H, for Hilbert space
H) there must exist unique uU such that
,, ()auv fv (
v H4.1)
and u is the solution of the following variational problem
vH vH
uMinIvMinavv fv
Furthermore, the solution of variational problem is also
nal” condition of Theorem 2 is
ique (see [22,25-27]).
If the “bilinear functio
anged into the “conjugate bilinear functional” condi-
tion, then the conclusions of Theorem 2 still hold for a
complex Hilbert space, except that the expression of (4.2)
is changed into the corresponding expression:
uMinavvfv vf
The coercive condition plays a very crucial role in the
tric c
3 ([27]). Assume U is a closed convex sub-
oof of Theorem 2 because it guarantees the existence
and convergence of minimization sequence which is con-
structed in the proof. The completeness of space U en-
sures that uUis the limit of the sequence. Meanwhile,
the symmeondition guarantees that the minimum
expression of functional has the meaning of existence.
If coercive condition in Theorem 2 is weakened in
eak coercive condition, then the following theorem is
t of real Hilbert space H. If
,auv defined in U is a
bounded, weak coercive and bilinear functional, then for
any *
H there must exist unique uU such that
,,au v (vHv f
) and u makefirst order
variatio fun oftional
v equal to zero, where
12 ,,
If the “bilinear functional” condition of Theorem 3 is
changed into the “conjugate bilinear functional” condi-
tion, then the conclusions of Theorem 3 still hold for a
complex Hilbert space, only the corresponding expres-
sion of variational problem becomes
,12 ,,
5.2. The Existence and Unique ne ss of th e Solu tion
Theolished the existence and
of Weak Form of Operator Equations as
Well as Its Relation with Corresponding
Variational Problems
rem 2 and 3 have estab
uniqueness theorem of the solution to variational equa-
tions and indicated the only solution can be obtained by
solving its corresponding variational problems. Accord-
ing to the relation between bilinear functional and linear
operator, which can be found from Lemma 2, the exis-
tence and uniqueness theorem of the solution to weak
form of operator equation can be obtained as follows.
Theorem 4. If A defined in a Hilbert space X is a lin
ntinuous and positive definite operator, then for any
H there must exist unique uX such that
,, ()
uvf v (v X5.1)
and u is the solution of the following variational problem
vX vX
uMinIvMin Avvfv
Theorem 4 can be derived from Theorem 2 and its de-
iled proof can be seen in [22,25-27]. Equation (5.1) is
called the weak form of operator equation
cause compared with the original operator equation, Eq.
(5.1) weaken the requirements to u. Theorem 4 can be
generalized to the case of complex Hilbert space, except
that the expression of corresponding variational problem
uMinAvvfv vf
Clearly, positive definite operator is a stronger condi-
tion in the practical application. When the weak coercive
condition of A is satisfied, the following theorem will be
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields17
5. If A defined in a Hilbert space X is a linear,
ntinuous and weak coercive operator, then for any
H there must be unique
X such that
,, ()
xvf v v X (6.1)
and x satisfied , where
xAxxfx (6.2)
that is, x make the first order variation of functional
x equal to zero [27].
proof of Theorem 5 Thecan be realized through taking
advantage of the conclusion of Theorem 3. Theorem 5
can be generalized to the case of complex Hilbert space,
except that the functional expression of corresponding
variational problem becomes
xAxxfxxf (6.3)
5.3. The Existence and Uniqueness Theorem of
Let rt space H
Solution to Operator Equation
()DA be a linear dense set in real Hilbe
and or A be mapping from ()DA to H. For sim-
plification, we assume the discussuation
ed eq
belongs to a kind of BVP of differential equation.
time operator A is a differential operator and domain
()DA consists of smooth functions with certain differ-
order, which is greater or equal the order of differ-
ential operator. Therefore, the solutions of BVP on linear
set ()DA belong to the common sense solutions. If
BVP solutions on ()DA, then the solutions is
called classical solution of or its equivalent varia-
tional problems. However, in general we can not guaran-
tee the existence of the solution to operator equation or
variational problem. For BVP of Poisson equation in
Theorem 1, if function f has no continuity on bound-
ary , then the equation has no solution in linear set
At this
2,0DuCu . But, when ()DA is extended
to a linear set of Sobolev space
at this time the weak form of the original equation on
r dense set
in erato
expended domain always has solutions. Obviously, this
solution is not the solution of the original BVP in the
sense of classical signification and is called a weak solu-
tion of the original equation [28-30]. We will build the
connection between operator equation and variational
problem through the weak form of operator equation and
obtain the existence and uniqueness proposition of op-
erator equation in the sense of weak solution.
Proposition 2 ([25]). Let
DA be a linea
real Hilbert space X and opr A:
DA X. For
operator equation
ufuDA (
can be extended
where the definition domain of Ato
space A
, A
is a Hilbert space which is obtained by
comple otionf
DA in terms of norm A
uAuu is a linear, continuous, sym. If Ametric
and positive def
inite operator, then the following conclu-
sions hold.
1) for an *
X there must exist unique 0A
such that
0,, (A)
uvfvv H (7.2)
and is the only solution of the following
vH vH
uMinIvMin Avvfv
2) If
.1). If
, then is the classical solution of
uation (7
uD, then 0
u is the weak so-
lution of Equation (
Proof: It can be proved
directly by using Theorem 4
ork and (7.3)
d, the following proposi-
. Let be a linear dense set
d the definition of weak solution directly.
(7.2) corresponds to principle of virtual w
rresponds to energy method.
If the conditions is weakene
n will be obtained.
Proposition 3([27])
perato real Hilbert space X and or A:
DA X. For
operator equation
where the domain of A can be extended to space A
. If
A is a linear, continuous, symmetric and weak coive
operator, then the following conclusions hold.
1) for any *
X there must exist unique 0A
such that
0,, (A)
uvfvvH (8.2)
and satisfied
, where
uAuuf u (8.3)
2) If
(11.1). If
, then is the classical solution of
be generalized to the case of
uDA, then 0
u is the weak
solution of Equation (
Proposition 2 and 3 can
mplex Hilbert space, only need to change the func-
tional expression of corresponding variational problem
(7.3) and (8.3) into (5.3) and (6.3). In Proposition 2 and 3,
symmetry condition guarantees that the values of func-
tional expression must be real and makes the computation
of extreme values feasible. The positive definite or weak
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields
coercive condition guarantees the existence of inverse
operator and make the solution of weak form of operator
equation exist.
Proposition 2 and 3 have summarized the existence
ation of Operator Equation
Baon the
following we firstly prove the existence and
For e scalar Poisson
d uniqueness theorems of weak solution to operator
equation, moreover, provided a feasible and effective
solving method of operator equation in the view of varia-
tional principle.
6. The Applic
Theory in the Time-Harmonic Fields
sed on the statements above, we can get an idea
proof of uniqueness theorem of time-harmonic electro-
magnetic fields. First we should deduce an operator
equation (or weak form of Helmholtz equation and varia-
tional problem) from Maxwell’s equation, then examine
whether the operator is a linear, continuous, symmetric
and positive definite (or weak coercive) operator, finally
by means of Proposition 2 and 3 obtain a conclusion
about it.
In the
iqueness theorem of Poisson equation in the lossless
case in brief, then deduce the weak form of Helmholtz
equation and its variational problem, finally point out the
substaintial issues appeared in the process of proving the
existence and uniqueness of the solution to Helmholtz
equation in the view of modern mathematical theory.
6.1. The Realization on Proof of the Uniquenes
Theorem of Poisson Equation
simplification we only discuss th
equation to show the application of Proposition 2 and 3.
The argument in the case of vector Poisson equation is
similar because vector Poisson equation can be decom-
posed into scalar Poisson equation.
Case A: For 0-Dirichlet BVP (0u
) of the Poisson
1, L
uation 2uf (u,f), which is dis-
cussed in Theoremet A
. By specifying the
,u vuv
can verify that op-
, we
ar, self-adjoint and positiveerator A is line definite. Hence,
in accordance with Proposition 2, we know that both the
solution of BVP of operator Equation (3) and the solution
of the minimum value problem of the corresponding
variational problem
12 ,,
uAuufu exist
uniquely and are equal.d with
the isotropic and uniform linear medium and under ho-
mogeneous boundary conditions, the weak solution of
scalar Poisson equation must exist uniquely.
Case B: For Poisson equation with homog
So within the region fille
eneous and
mixed boundary conditions
0, 0
where 12
SS S
is the boundary surface. By specify-
r-product ing inne
,uvuv d
, we can easily verify
that if
are non-negative and not equal to zero
simultaneity, then operator
is self-adjoint
and positive definite. By Pre unique weak
solution must exist and it can be obtained by solving the
minimum point of corresponding variational problem.
Case C: For Poisson Equation (9.1) with non-hom
oposition 2 th
geneous boundary conditions, the boundary conditions is
specified as follows:
up uq
Through a transform uu
, we can get the new
unknown function u
, where
is arbitrary function
which satisfies non-homogeneous boundary conditions
(10). Adopting the definition of inner-product
,uv uvd
, operator
on variable
becomes a serator and welf-adjoint ope can write the
ctional expression of u by the functional expression of
. In terms of the discussion of segment B, we know
t when tha
are real or real function, A is a
self-adjoint operatorSo .
u can be written by the rela-
tion between the functionxpression of u and the func-
tional expression of ual e
. By the standard variational prin-
ciple, the extreme pint of o
u must exist uniquely.
Hence, the extreme point of
u must exist. When
are real and real funct, the weak solution of
Poissn equation with non-homogeneous boundary con-
ditions must exist. Thus, we finally obtain the existence
and uniqueness theorem of the solution to Poisson equa-
tion with non-homogeneous boundary conditions within
region filled with the isotropic and uniform linear medium.
6.2. The Application of Operator Equation
As a conclusions in
tional of Helmholtz Equation with
By tw (1.1) and (1.2) of Maxwell’s equa-
Theory in Helmholtz Equation
kind of elliptic PDE, the available
[25,26,28,30,31] can not be applied to scalar wave equa-
tion. We will discuss vector wave equation of electric
field by using of Proposition 2 and 3. The argument in
the case of vector wave equation of magnetic field is
similar. Scalar wave equation is a special case of vector
wave equation.
6.2.1. The Func
Homogeneous Boundary Conditions in the
Lossless Case
o curl Equations
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields19
tions we can deduce the double-curl equation of electric
field as follows:
EkEj J
As a vector problem, inner-product
,uv uvd
specified. For the double-curl equation, let
And we have
EFFE k Ed
By the second vector Green theorem, (12) is changed
AE FEFkFd
EFF En
If both E and F satisfy homogeneous Dirichlet bound-
ary condition
0 on nE S
and homogeneous Neumann boundary condition
0 on
nEnnE S
where , then surface integral in (13) is equal
SS S
If r
to zero.
and e
are real or real function, then (13)
can be wri into:
, that is, A is a
self-adjoint operator. djointness of A
defined by (11.2) need the following conditions: 1) r
Hence, the self-a
and e
are real or real function; 2) boundary con-
s are mogeneous. With these conditions substitut-
ing (11.2) (11.1) into (7.1) (5.3), we have
tion ho
JEEEkE d
EJd JEd
Assume that the medium is uniform and quote the first
vector Green theorem and boundary conditions (14.1)
and (14.2), (15) becomes
jEJEJ d
nE nEdS
We can know that (11.1) corresponds
tion (7.1) or (8.1), (13) corresponds to weak form of op-
erator Equation (7.2) or (8.2), (16) corresponds to the
to operator equa-
nctional of variational problem (7.3) or (8.3). Mean-
while, the weak form of Helmholtz equation (13) is
agreement with its corresponding variational problem
in JE
or 0JE
In the following we will discuss the property of the
operator A. By definition we have
EEEEkE d
By the first vector Green theorem and boundary
tions (14.1) and (14.2), (17) becomes
EEEEkEE d
nE nEdS
By (18), we can not confirm that operator A is positive
definite or weak coercive because
e of
equal to zero. Because of the existenc term
in integrand of (18), the Plity
can not be used to prove the positive definite or weak
coercive of the operator A. It is the substantial difficulty
ss of the application of Proposition 2 and 3.
In mathematical, it belongs to one kind of eigenvalue
problems, and its physical meaning represents resonance
of electromagnetic fields. So this problem is an inherent
property for time harmonic electromagnetic fields. For
static fields no resonance can occur so the Poincare ine-
quality can be used to prove the positive definite of Pois-
son operator. Therefore, for a concrete BVP of time-
harmonic field if the frequency range is selected so as no
resonance can occur, the operator in this frequency range
will be positive definite or weak coercive. If the fre-
quency is closed to resonance frequency of the structure
to be analyzed, the operator will not be positive definite
or weak coercive. For this situation if the operator equa-
tion is changed into linear algebraic equations in numeri-
cal algorithms we can find the matrix determinant is
closed to zero or equal zero.
6.2.2. The Further Discussion on the Solution to
Helmholtz Equation
Although the Lax-Milgram theo
oincare inequa
in the proce
rem plays a very impor-
tant role in the solving of the weak solution and nu
Copyright © 2011 SciRes. JEMAA
On the Uniqueness Theorem of Time-Harmonic Electromagnetic Fields
cal solive or weak coercive condi-
e positive definite or the weak coercive conditions
ution to PDE, the coerc
tion of the theorem greatly limit its application scope. In
[32] I. Babuska and A. K. Aziz generalized the Lax-
Milgram theorem under the weaker coercive condition,
which greatly extends the application of the theorem.
Furthermore, I. Babuska has also introduced another kind
of coercive condition in [33], that is, strong Babuska
condition, which further exert the application of the Lax-
Milgram theorem in finite element numerical method.
However, It is to be determined whether strong Babuska
condition of operator A to Helmholtz equation is satis-
In the discussion of variational formula of FEM (finite
element method [34]), some books and literatures think
that th
operator is not necessary, only requiring that the op-
erator is linear, continuous and symmetric, and the ex-
tremal solution of variation formula must be the solution
of the original equation. Clearly, by Proposition 2 and 3
we can know that the above viewpoints are incorrect in
the sense of modern mathematics theory. Proposition 2
and 3 have clearly indicated that the positive definite or
the weak coercive conditions of operator is sufficient
condition of the existence of 1
, that is, sufficient con-
dition of the existence of extreme value of functional.
7. The Uniqueness of the Solution to Linear
Algebraic Equations
In terms of Proposition 2 and 3, operator equations can b
solved with two kinds of distinct methods: variationa
thod and the direct solving
me method of the weak form
ns of corresponding operator are satisf
the uniqueness theorem of time-harmonic electromag-
tional theory are pointed out; a new
of operator equation. These numerical methods are even-
tually reduced to find the solution of linear algebraic
equations, that is, to find the solution of matrix equations.
Various methods of numerical solutions can be seen in
[22, 35-42].
By Lax equivalence theorem (see [21]) and Lemma 2,
we obtain that if the positive definite or the weak coer-
cive conditioied,
en matrix determinant of linear algebraic equations
obtained by the discretization of the operator equation is
not equal to zero. For linear algebraic equations in which
the number of unknown variables is equal to the number
of equations, if matrix determinant of linear algebraic
equations is not equal to zero, then the solution of linear
algebraic equations must exist uniquely. Hence, the posi-
tive definite or the weak coercive conditions of operator
guarantee the uniqueness of solution to matrix equation
obtained by the discretization of the operator equation.
8. Conclusions
In this paper, the limitation and the lack of strictness on
netic fields in tradi
idea to solve the existence and uniqueness of the solution
to time-harmonic fields’ equations by means of the mod-
ern theory of PDE and functional analysis is described.
The substantial difficulty is that the existence of term of
in integrand make the Poincare inequality
not be used to prove the positive definite or weak coer-
cive of the operator. The property of operator depends on
whether onance frequency of the structure to be ana-
lyzed belongs to the interested frequency range. The
study work is being done and further results will be pre-
sented in future. Whether other mathematical method
such as differentiable manifolds may be used to solve this
problem is also interesting (private discussion with Dr. Q.
Wang, 2006).
9. Acknowledgements
The authors wish to acknowledge Pro. Dou and Dr. Z. X.
Wang of Sou
theast University for their constructive
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1. All governments are corrupt.
I am afraid, this is a natural fact (by design). Therefore, nothing and no one can change it.
2. In the United States, the government has been on a mission to dump down its citizens for many decades. Why not? The dumber society is the easier to control.
As far as I know, this tactic has been also followed in all other countries since many decades ago. I noticed it too in my country.
4. As I age, solving math problems is my suduko puzzle, it's my cross word puzzle; it keeps my brain cells alive. No pun intended.
Although my private business is almost paralyzed since after year 2011 due to world's events and regulations, I keep designing new products (that needs lots of math) just to keep my brain cells
5. To be honest with you, my circle of friends is very small for three reasons:
You are fortunate for having a very small circle of friends. My circle has zero radius
There is no one around me lately who enjoys math and education. It seems almost all people are attracted by other things.
About trust, two humans cannot decide to trust each other really before they discover each other while facing real hard times together.
ALL GOVERNMENTS do not want it's people to be well-informed. Like I said before, the dumber we are, the easier to control and brainwash. ALL GOVERNMENTS are bad, corrupt and not trustworthy.
Any government has to be run by rich persons or serving rich ones. Naturally, they have to protect each other, besides being protected by their masters (working behind the scenes). This was
simplified 2000 years ago when it was said that a rich man is not supposed to enter ...
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seminars - 1. Product structures on the relative symplectic cohomology / 2. Symplectic homology of affine varieties
1. Product structures on the relative symplectic cohomology
강연자: 이원준(서울대학교)
Abstract: The relative symplectic cohomology introduced by Varolgunes is an invariant for compact subsets of symplectic manifolds. First, we study the definition of the relative symplectic cohomology
and some of its properties. Then, we give the product structures on the relative symplectic cohomology. Finally, as an application, we prove the 'super-rigidity' theorem.
2. Symplectic homology of affine varieties
강연자: 배중현(서울대학교)
Abstract: This is a brief summary of Ganatra-Pomerlearno's paper "Symplectic cohomology rings of affine varieties in the topological limit".
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The bottleneck degree of algebraic varieties
A bottleneck of a smooth algebraic variety X ⊂ C ^n is a pair (x, y) of distinct points x, y ∊ X such that the Euclidean normal spaces at x and y contain the line spanned by x and y. The narrowness
of bottlenecks is a fundamental complexity measure in the algebraic geometry of data. In this paper we study the number of bottlenecks of affine and projective varieties, which we call the bottleneck
degree. The bottleneck degree is a measure of the complexity of computing all bottlenecks of an algebraic variety, using, for example, numerical homotopy methods. We show that the bottleneck degree
is a function of classical invariants such as Chern classes and polar classes. We give the formula explicitly in low dimension and provide an algorithm to compute it in the general case.
• Bottleneck
• Manifold learning
• Polar classes
• Reach
Dive into the research topics of 'The bottleneck degree of algebraic varieties'. Together they form a unique fingerprint.
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: why -2.22044604925031e-16?
I found that I have a WeBWorK problem with the following lines
$a2 = non_zero_random(-50,50,1);
$a2 = $a2/10;
$r2 = non_zero_random(-5,0,0.1); #negative is more interesting with inequality
$b2 = $a2 - $r2;
and $d2 was displayed as
I'd obviously decided to get step sizes of 0.1 for $a2 by dividing consecutive integers by 10, but why does the definition for $r2 work sometimes but not other times, that is, why does WeBWorK hit a
value very close to but not equal to zero?
Hi Bruce,
Can you post the problem code and the seed value that is generating this value? Also, do you mean that $b2 has that value or $r2?
The weird value of $b2 occurred with seed 4642
$b1 = random(2,5,1);
$r1 = random(1,9,1);
$a1 = $b1 + $r1;
$c1 = random(1,9,1);
$ans1 = random(1,9,1);
$temp = $r1 + $c1;
$d1 = $temp * $ans1 ;
$a2 = non_zero_random(-50,50,1);
$a2 = $a2/10;
$r2 = non_zero_random(-5,0,0.1); #negative is more interesting with inequality
$b2 = $a2 - $r2;
$c2 = random(-5,5,0.1);
$ans2 = non_zero_random(-5,5,0.1);
$d2 = $r2 * $ans2 + $c2 ;
# Main text
Solve each equation.
a. \[ $a1 y - $b1 y = $d1 - $c1 y \]
\( y = \) \{ ans_rule(15) \}
b. \[$a2 w ? {$c2} = $b2 w ? {$d2}\]
\( w = \) \{ ans_rule(15) \}
OK, I just ran this inserting $r2 before the first equation just for fun, and it displayed as -1.2. So the -1.2 (that was the value of $a2, -12/10) is different from the -1.2 (the value of $r2) that
came from stepping from -5 by steps of 0.1?
Welcome to the wonderful world of floating-point arithmetic. You are correct that the two versions of 1.2 are slightly different. Such small errors are inherent in the computations performed by
WeBWorK, which uses double-precision floating-point reals under the hood. These support about 16 digits of precision, so this value, which is on the order of 10^-16, is due to the truncation and
round-off errors in the computations to produce the two numbers in $a2 and $r2.
There is really no way to avoid these errors with this type of computation. For example, consider your .1, which is easily represented in decimal. But when converted to binary, it is a repeating
"decimal" number, and since only a finite number of digits are stored, even a number like .1 can not be accurately represented in binary. So computations that involve it may accumulate that error.
Different computations will accumulate different errors, but as long as these stay in the least-significant digits, they are harmless. For instance, the routines that convert to decimal for printing
usually round off the results so that the small errors don't show up in the results.
Usually these errors do not cause problems, but if they move up from the least-significant digits into the more-significant digits, then they can become problematic. This can happen in several ways.
One way is if a process is iterated many times and the errors accumulate and get bigger and bigger. That usually takes a long time, and doesn't usually affect WeBWorK computations.
The most important mechanism that moves these errors from the least-significant digits to the most-significant ones is called subtractive cancelation, and it occurs when two numbers are subtracted
that are equal in their most significant digits. For example, suppose you have computed pi in two ways and get 3.1415943 and 3.1415972, both pretty precise, and with 8 significant digits. The errors
are in the least significant digits, where it is supposed to be. If you end up subtracting the first from the second, you will get .000029, which is a number with only 2 digits of precision, so a
dramatic loss of precision has occurred. Moreover, the two significant digits that you end up with are the "junk" digits from the originals, and so are completely unreliable.
This is the type of error that you are seeing here. There is really no way to avoid it, but you could do extra checks to see if the absolute value of the result is small and then set the result to 0
instead, for example.
Hope that clarifies things a bit.
Thanks. I was afraid of that, but I'm relieved it's more than my being insane.
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CSCI-UA.0002 – Summer 2016
Python Assignment # 5 - Gradebook / Primes
Due: Tuesday, July 26th (Day of the Midterm)
Part 1: Dynamic Gradebook
Write a gradebook program that lets a teacher keep track of test averages for his or her students. Your program shoudl begin by asking the teacher for a number of students in their class as well as
the total # of tests that will be given to the class. Validate this information to ensure that the numbers entered are positive.
Next, prompt the teacher to enter in scores for each student. Ensure that the values entered are positive - if they aren't you will need to re-prompt them. Hint: you may need to use nested loops
here! A "while" loop can be placed inside of a "for" loop, if necessary.
Once your program has collected all test scores for a student it should display that student's average and move onto the next student. When all students have been calculated the program should
compute the overall average score for the entire class.
Here's a sample running of your program:
How many students are in your class? -5
Invalid # of students, try again.
How many students are in your class? 3
How many tests in this class? -10
Invalid # of tests, try again.
How many tests in this class? 2
Here we go!
**** Student #1****
Enter score for test #1: -50
Invalid score, try again
Enter score for test #1: 50
Enter score for test #2: 75
Average score for student #1 is 62.50
**** Student #2****
Enter score for test #1: 100
Enter score for test #2: 90
Average score for student #1 is 95.00
**** Student #3****
Enter score for test #1: -10
Invalid score, try again
Enter score for test #1: -20
Invalid score, try again
Enter score for test #1: -30
Invalid score, try again
Enter score for test #1: 90
Enter score for test #2: 80
Average score for student #1 is 85.00
Average score for all students is: 80.83
Some hints:
• Begin by constructing a "for" loop to iterate over all students in the class
• Once you're inside of this "for" loop you will probably need another loop to handle inputting the scores for a particular student.
• Big hint: Try to get your program to work first without any data validation. You can add this in later once you figure out the general structure of the program.
• Remember the difference between "for" and "while" loops! "for" loops are used when you want to iterate over a know # of items, and "while" loops can be used to keep the user "caught" indefinately
until they fulfill a particular condition. You will probably need to use a combination of these loops to solve this problem.
This program should be named as follows: LastNameFirstName_assign5_part1.py (for example, "ShakespeareWilliam_assign5_part1.py")
Part 2a: Prime Number Finder
Write a program that prompts the user to enter in a postiive number. Only accept positive numbers - if the user supplies a negative number or zero you should re-prompt them.
Next, determine if the given number is a prime number. A prime number is a number that has no positive divisors other than 1 and itself. For example, 5 is prime because the only numbers that evenly
divide into 5 are 1 and 5. 6, however, is not prime because 1, 2, 3 and 6 are all divisors of 6.
Here's a sample running of the program:
Enter a positive number to test: 5
2 is NOT a divisor of 5 ... continuing
3 is NOT a divisor of 5 ... continuing
4 is NOT a divisor of 5 ... continuing
5 is a prime number!
And here's another running:
Enter a positive number to test: 9
2 is NOT a divisor of 9 ... continuing
3 is a divisor of 9 ... stopping
9 is not a prime number.
Some notes on your program:
• 1 is technically not a prime number.
• Once you find a number that evenly divides into your test number you do not need to continue testing additional numbers - the number cannot be prime.
This program should be named as follows: LastNameFirstName_assign5_part2a.py (for example, "ShakespeareWilliam_assign5_part2a.py")
Part 2b: Find all Prime Numbers between 1 and 1000
Next, make a copy of Part A and update it so that the program now finds all prime numbers between 1 and 1000. Here's a sample running of your program:
1 is a prime number!
2 is a prime number!
3 is a prime number!
5 is a prime number!
7 is a prime number!
11 is a prime number!
... cut ...
977 is a prime number!
983 is a prime number!
991 is a prime number!
997 is a prime number!
This program should be named as follows: LastNameFirstName_assign5_part2b.py (for example, "ShakespeareWilliam_assign5_part2b.py")
Notes about your program:
• Make sure your program includes a comment at the beginning with your name along with a brief synopsis of the program.
• You should use comments throughout the source code to explain key steps and calculations.
• Choose intuitive names for variables
• You should use blank lines where appropriate to make your program easy to read.
• Make sure that your program does not contain any syntax errors.
• Test your program to be sure that it runs the way that you expect.
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Spectral crest for signals and spectrograms
crest = spectralCrest(x,f) returns the spectral crest of the signal, x, over time. How the function interprets x depends on the shape of f.
crest = spectralCrest(x,f,Name=Value) specifies options using one or more name-value arguments.
[crest,spectralPeak,spectralMean] = spectralCrest(___) returns the spectral peak and spectral mean. You can specify an input combination from any of the previous syntaxes.
spectralCrest(___) with no output arguments plots the spectral crest.
• If the input is in the time domain, the spectral crest is plotted against time.
• If the input is in the frequency domain, the spectral crest is plotted against frame number.
Spectral Crest of Time-Domain Signal
Create a chirp signal with white Gaussian noise and calculate the crest using default parameters.
fs = 1000;
t = (0:1/fs:10)';
f1 = 300;
f2 = 400;
x = chirp(t,f1,10,f2) + randn(length(t),1);
crest = spectralCrest(x,fs);
Plot the spectral crest against time.
ans = 311×1
Spectral Crest of Frequency-Domain Signal
Create a chirp signal with white Gaussian noise and then calculate the spectrogram using the stft function.
fs = 1000;
t = (0:1/fs:10)';
f1 = 300;
f2 = 400;
x = chirp(t,f1,10,f2) + randn(length(t),1);
[s,f] = stft(x,fs,FrequencyRange="onesided");
s = abs(s).^2;
Calculate the crest of the spectrogram over time.
crest = spectralCrest(s,f);
Plot the spectral crest against the frame number.
ans = 309×1
Specify Nondefault Parameters
Create a chirp signal with white Gaussian noise.
fs = 1000;
t = (0:1/fs:10)';
f1 = 300;
f2 = 400;
x = chirp(t,f1,10,f2) + randn(length(t),1);
Calculate the crest of the power spectrum over time. Calculate the crest for 50 ms Hamming windows of data with 25 ms overlap. Use the range from 62.5 Hz to fs/2 for the crest calculation.
crest = spectralCrest(x,fs, ...
Window=hamming(round(0.05*fs)), ...
OverlapLength=round(0.025*fs), ...
Plot the crest against time.
spectralCrest(x,fs, ...
Window=hamming(round(0.05*fs)), ...
OverlapLength=round(0.025*fs), ...
ans = 399×1
Input Arguments
x — Input signal
column vector | matrix | 3-D array
Input signal, specified as a vector, matrix, or 3-D array. How the function interprets x depends on the shape of f.
Data Types: single | double
f — Sample rate or frequency vector (Hz)
scalar | vector
Sample rate or frequency vector in Hz, specified as a scalar or vector, respectively. How the function interprets x depends on the shape of f:
• If f is a scalar, x is interpreted as a time-domain signal, and f is interpreted as the sample rate. In this case, x must be a real vector or matrix. If x is specified as a matrix, the columns
are interpreted as individual channels.
• If f is a vector, x is interpreted as a frequency-domain signal, and f is interpreted as the frequencies, in Hz, corresponding to the rows of x. In this case, x must be a real L-by-M-by-N array,
where L is the number of spectral values at given frequencies of f, M is the number of individual spectra, and N is the number of channels.
• The number of rows of x, L, must be equal to the number of elements of f.
Data Types: single | double
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: Window=hamming(256)
The following name-value arguments apply if x is a time-domain signal. If x is a frequency-domain signal, name-value arguments are ignored.
Window — Window applied in time domain
rectwin(round(f*0.03)) (default) | vector
Window applied in the time domain, specified as a real vector. The number of elements in the vector must be in the range [1, size(x,1)]. The number of elements in the vector must also be greater than
Data Types: single | double
OverlapLength — Number of samples overlapped between adjacent windows
round(f*0.02) (default) | non-negative scalar
Number of samples overlapped between adjacent windows, specified as an integer in the range [0, size(Window,1)).
Data Types: single | double
FFTLength — Number of bins in DFT
numel(Window) (default) | positive scalar integer
Number of bins used to calculate the DFT of windowed input samples, specified as a positive scalar integer. If unspecified, FFTLength defaults to the number of elements in the Window.
Data Types: single | double
Range — Frequency range (Hz)
[0,f/2] (default) | two-element row vector
Frequency range in Hz, specified as a two-element row vector of increasing real values in the range [0, f/2].
Data Types: single | double
SpectrumType — Spectrum type
"power" (default) | "magnitude"
Spectrum type, specified as "power" or "magnitude":
• "power" –– The spectral crest is calculated for the one-sided power spectrum.
• "magnitude" –– The spectral crest is calculated for the one-sided magnitude spectrum.
Data Types: char | string
Output Arguments
crest — Spectral crest
scalar | vector | matrix
Spectral crest, returned as a scalar, vector, or matrix. Each row of crest corresponds to the spectral crest of a window of x. Each column of crest corresponds to an independent channel.
spectralPeak — Spectral peak
scalar | vector | matrix
Spectral peak, returned as a scalar, vector, or matrix. Each row of spectralPeak corresponds to the spectral crest of a window of x. Each column of spectralPeak corresponds to an independent channel.
spectralMean — Spectral mean
scalar | vector | matrix
Spectral mean, returned as a scalar, vector, or matrix. Each row of spectralMean corresponds to the spectral crest of a window of x. Each column of spectralMean corresponds to an independent channel.
The spectral crest is calculated as described in [1]:
$\text{crest}=\frac{\mathrm{max}\left({s}_{k\in \left[{b}_{1},{b}_{2}\right]}\right)}{\frac{1}{{b}_{2}-{b}_{1}}\sum _{k={b}_{1}}^{{b}_{2}}{s}_{k}}$
• s[k] is the spectral value at bin k.
• b[1] and b[2] are the band edges, in bins, over which to calculate the spectral crest.
[1] Peeters, G. "A Large Set of Audio Features for Sound Description (Similarity and Classification) in the CUIDADO Project." Technical Report; IRCAM: Paris, France, 2004.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™.
Version History
Introduced in R2019a
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Class LogisticRegressionSummaryImpl
All Implemented Interfaces:
Serializable, ClassificationSummary, LogisticRegressionSummary, scala.Serializable
Direct Known Subclasses:
BinaryLogisticRegressionSummaryImpl, LogisticRegressionTrainingSummaryImpl
Multiclass logistic regression results for a given model.
param: predictions dataframe output by the model's transform method. param: probabilityCol field in "predictions" which gives the probability of each class as a vector. param: predictionCol field in
"predictions" which gives the prediction for a data instance as a double. param: labelCol field in "predictions" which gives the true label of each instance. param: featuresCol field in "predictions"
which gives the features of each instance as a vector. param: weightCol field in "predictions" which gives the weight of each instance.
See Also:
• Method Summary
Modifier and Type
Field in "predictions" which gives the features of each instance as a vector.
Field in "predictions" which gives the true label of each instance (if available).
Field in "predictions" which gives the prediction of each class.
Dataframe output by the model's transform method.
Field in "predictions" which gives the probability of each class as a vector.
Field in "predictions" which gives the weight of each instance.
Methods inherited from interface org.apache.spark.ml.classification.ClassificationSummary
accuracy, falsePositiveRateByLabel, fMeasureByLabel, fMeasureByLabel, labels, precisionByLabel, recallByLabel, truePositiveRateByLabel, weightedFalsePositiveRate, weightedFMeasure,
weightedFMeasure, weightedPrecision, weightedRecall, weightedTruePositiveRate
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From my `experiment`, all of the items that are in the same box have probabilities that are dependent of each other, for example if you get item A then you should not get item B from the same box.
The probability of the items that are not in the same box is independent from each other.
1 of the following with 0.00012% probability
1 of the following with 12.18% probability
1 of the following with 4.35% probability
1 of the following with 4.35% probability
1 of the following with 4.35% probability
1 of the following with 0.0003% probability
1 of the following with 6.09% probability
1 of the following with 6.09% probability
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IMTAN - Excel docs, syntax and examples
The IMTAN function returns the inverse tangent of a complex number in the form x + yi. This function is useful when working with complex numbers in Excel.
inumber The complex number for which you want to find the inverse tangent.
About IMTAN 🔗
When dealing with complex numbers and needing to compute the inverse tangent, Excel's IMTAN function comes to the rescue. It aids in determining the angle (in radians) corresponding to a complex
number expressed in the form x + yi, where x represents the real part and y represents the imaginary part of the complex number. This function is particularly handy for tasks involving complex
arithmetic and trigonometry in Excel worksheets. By inputting the desired complex number as the argument, you can swiftly obtain the inverse tangent value, thus facilitating intricate calculations
involving complex quantities. The IMTAN function is a valuable asset in Excel's toolset for handling complex number operations with ease and accuracy.
Examples 🔗
If you have a complex number 3 + 4i and want to find its inverse tangent, you would use the formula: =IMTAN("3+4i")
For a complex number -2 - i, to calculate its inverse tangent, you can use: =IMTAN("-2-i")
Ensure that the complex number input is correctly formatted following the standard complex number notation x + yi, where x and y are real numbers and i represents the imaginary unit. IMTAN function
returns the result in radians.
Questions 🔗
What format should the complex number be in for the IMTAN function?
The complex number should be in the format x + yi, where x and y are real numbers and i represents the imaginary unit. Ensure the input follows this standard notation for accurate results.
What unit does the IMTAN function return the inverse tangent result in?
The IMTAN function returns the result in radians. If you need the result in degrees, you can convert it using the RADIANS or DEGREES function in Excel.
Can the IMTAN function handle complex numbers with only real or imaginary parts?
Yes, the IMTAN function can handle complex numbers with only real or imaginary parts. For example, if you have a complex number with no imaginary part (pure real number) or no real part (pure
imaginary number), you can still use the IMTAN function to find the inverse tangent.
Related functions 🔗
Leave a Comment
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Resilient Modulus of Compacted Lateritic Soils from Senegal at OPM Conditions
Resilient Modulus of Compacted Lateritic Soils from Senegal at OPM Conditions ()
1. Introduction
Since several decades, gravel lateritic soils have been used in road pavements in tropical countries. In Senegal, the increased use of this material has lead to a rarefaction of careers of good
quality. Therefore, the rationalization of existing resources requires a real knowledge of lateritic soils. This problem has drawn the attention of the researchers who have done a lot to understand
the mechanical behaviour of lateritic gravels (Samb, 1986 [3]; Fall, 1993 [4]; Fall, Sawangsuriya, Benson, Edil and Bosscher, 2007 [5]).
Under cyclic loading, road materials are characterized by a fast increase of permanent strains from the first cycles of loading. As the number of cycles increases, these deformations stabilize and
the behaviour becomes essentially reversible allowing to define a module called “resilient modulus” (Yoder and Witzack, 1975 [6]; Paute, Hornych and Benaben, 1994 [7]; Martinez, 1990 [8]). Resilient
modulus represents the unloading modulus after several repeated cycles of loading, allowing to simulate road traffic (Figure 1):
Δσ[d] = σ[1] − σ[3] = Deviatoric stress; σ[1] = Major principal stress; σ[3] = Minor principal stress et Δε[a] = Resilient axial strain.
In order to study the cyclic behaviour of gravel lateritic soils of Senegal, repeated load triaxial tests were conducted on soils collected from Sébikotane, Dougar. Ngoundiane, Pâ Lo and
In this paper, we aim to present the experimental protocol, the determination of average resilient modulus as well as the evolution of the resilient modulus according to the level of stress and the
percentage of cement. Furthermore, correlations are made depending on some gen-
Figure 1. Definition of resilient modulus (Hopkins, Beckham and Sun, 2007 [9]).
eralized resilient modulus models such as Uzan-Witczack (Witczack and Uzan, 1988 [1]) and the NCHRP model (2004 [2]).
2. Material and Methods
Standard laboratory road tests were performed to classify the materials and to determine their properties. Laboratory tests consisted of particle size analysis, consistency limits. Modified Proctor
Compaction test and Californian Bearing Ratio test.
The cyclic triaxial tests were then conducted to determine the resilient modulus of these soils. For this purpose, unbound gravel lateritic soils and that improved with cement (1%, 2% and 3%) were
compacted in 95% of Modified Optimum Proctor (OPM) that corresponds to the value retained in road specifications of base layers. Samples are 70 mm in diameter and 180 mm height.
The Tables 1 and 2 respectively present the summary of results of the identification test sand the nomenclature of test specimens for triaxial tests.
The experimental procedure is described by the NCHRP (2004 [2]). The triaxial apparatus in repeated loads is the experimental reference device used to characterize the mechanical behaviour of roads
materials (Figure 2).
The study of resilient behaviour includes two phases. In the first one, the test begins with a conditioning which consists in applying a minimum of 1000 repetitions of a load equivalent to a cyclic
stress of 207 kPa using a haversine-shaped 0.1-second load pulse followed by a 0.9- second rest period.
During the second phase, the test specimen is submitted to cyclic loading by applying a number of 100 repetitions of the corresponding cyclic axial stress using the same load pulse according to
various loading paths given in Table 3.
The axial deformations are measured by two external and two internal displacements sensors called “Linear Variable Differential Transformers” (LVDT). The average deformations for each LVDT separately
for the last five cycles are reported then the resilient modulus is calculated (NCHRP, 2004 [2]).
It is important to notice that the results of the resilient modulus exposed below are the ones obtained with the external deformation sensors.
3. Results and Discussion
3.1. Comparision between Static and Cyclic Modulus
Table 4 gives the values of average resilient modulus for all the materials as well as the maximal modulus found for the Unconfined Compression Test (UCT) realized on the same types of test tubes for
the sites of Mont-Rolland. Dougar and Pâ Lo. In general, it is observed that resilient modulus is much more important for cyclic triaxial test than for the unconfined compression test. Indeed, the
application of an increasing monotonous load is more unfavorable than the application of a cyclic load where the sample gets back part of the deformation. Furthermore, the conditioning made in the
cyclic test allows increasing the stiffness of the material; which is not the case for the compression test.
On the other hand, the ratio between resilient modulus and Young modulus for the compression test seems much more high for the raw material (between 12% and 17%) and decreases with the percentage of
cement. Indeed. The increase of the stiffness of the material with the percentage of cement decreases the sample deformability.
These results are very important because they show the reason why it is necessary to take into account the real stiffness of the gravel lateritic soils in mechanistic design. The use of the static
modulus does not seem to be
Table 1. Identification test results of collected gravel lateritic soils.
Table 2. Nomenclature of test specimens for load triaxial tests and unconfined compression tests.
Table 3. Test sequence for base or subbase materials— Procedure Ia of NCHRP (2004) [2].
Table 4. Comparision between static and cyclic modulus.
Figure 2. Triaxial apparatus for cyclic loading—University of Madison (Bâ, 2012 [10]).
any more suitable for pavement design.
3.2. Resilient Modulus According to the Level of Stress
Previous investigations, from the earlier studies reported by Williams (1963) [11] to the recent studies by Kolisoja (1997) [12], have shown that generally stress level is the factor that has the
most significant impact on resilient properties of road materials (Lekarp, Isacsson and Dawson, 2000 [13]). In this part, the effect of the bulk and deviatoric stress as well as the percentage of
cement are analyzed.
3.2.1. Effect of Bulk Stress
The Figures 3 and 4 show that resilient modulus decreases inversely with the increase of bulk stress in constant confining pressure. Indeed, gravel lateritic soils have a low void ratio after
compaction because of the presence of fine grains which ensures the cohesion. After application of the load and a generation of fine particles (because of the disintegration of pisolithes), new
surfaces of discontinuities are created, involving an increase of the void ratio and therefore a loss of resistance.
3.2.2. Effect of Deviatoric Stress
The Figures 5 and 6 show that resilient modulus decreases according to deviatoric stress in constant confining pressure. These results confirm the fact, the increase of the level of stress in the
material generally leads to a decrease of the rigidity of gravel lateritic soils samples.
3.2.3. Effect of the Percentage of Cement
The effect of the percentage of cement was studied for gravel lateritic soils of Dougar, Sebikotane and Pâ Lo (Figures 7-9). The observations showed an increase of resilient modulus according to the
percentage of cement for the materials of Dougar. However, this result is not confirmed for gravel lateritic soils of Sebikotane and Pâ Lo.
Besides, the observation of the moisture contents measured as in Table 5 showed that resilient modulus decreases with the important increase of the moisture
Figure 3. Variation of resilient modulus with bulk stress— Unbound gravel lateritic soil of Dougar.
Figure 4. Variation of resilient modulus with bulk stress— Dougar 1% cement.
Figure 5. Variation of resilient modulus with deviatoric stress—Unbound gravel lateritic soil of Dougar.
content (Sebikotane 3%. Pâ Lo 3%). Furthermore, for materials from Dougar, the resilient modulus increases with the percentage of cement, for appreciably equal moisture contents. The resilient
modulus seems to be affected more by moisture content effect than that of the percentage of cement.
4. Review of Resilient Modulus Models
Prediction of roads behaviour requires stress-strain relationships modelling by constitutive laws. Several models were proposed in the literature. The K-θ model (Seed, Mitry, Monismith and Chan, 1967
[14]; Brown and Pell, 1967 [15]; Hicks and Monismith, 1971 [16]) which is one of the most popular models expresses the resilient modulus according to the bulk stress:
Figure 6. Variation of resilient modulus with deviatoric stress—Dougar 1% cement.
Figure 7. Effect of the percentage of cement on resilient modulus of gravels lateritics soils of Dougar.
[1] and k[2] are the regression constants.
It is widely used to model the resilient modulus as a function of the level of stress applicable to the granular materials. However, the K-θ model presents some disavantages. Uzan (1985 [17])
introduces the deviatoric stress as the additional component according to the effect of shearing behaviour and obtains better correlations with the trial results:
Figure 8. Effect of the percentage of cement on resilient modulus of gravels lateritics soils of Sebikotane.
Figure 9. Effect of the percentage of cement on resilient modulus of gravels lateritics soils of Pâ Lo.
Table 5. Water content of soils samples after tests.
[1], k[2] and k[3] regression constants.
Witczack and Uzan (1988 [1]) proposes an improvement of the model of Uzan (1985 [17]) the by replacing the octahedral shear stress:
k[1], k[2] and k[3] regression constants.
A general form for these proposed models is the Andrei model (1999 [18]):
k[1] − k[7] = regression constants.
The Andrei model (1999 [18]) was then adopted by the National Highway Research Program (NCHRP, 2004) [2] in its simplified version (k[6] = 0 and k[7] = 1).
k[1], k[2] and k[3], regression constants.
It is worth noting that gravel lateritic soils consisted of soft and hard concretions as well as quartzites in a matrix consisting of fines particles composed of a mixture of clays and graves.
Therefore, both the Uzan-Witczack model (1988 [1]) and the NCHRP one (2004 [2]) which can be adapted to the fine soils as well as to the granular soils seem to well fit with gravel lateritic soils
that are located between these two types of material. These two models were so retained for correlation studies.
Correlations with Uzan-Witczack (1988) [17] and NCHRP (2004) Models [2]
The results of the correlations of the resilient modulus for Uzan-Witczack (1988 [1]) and NCHRP (2004 [2]) models are presented below (Tables 6 and 7). It is observed that the found coefficients of
regression are very close to 1 (between 0.902 and 0.999). This observation means that both models give very good correlations of the resilient modulus and can be used to model the resilient behaviour
of gravel lateritic soils. However, the comparison by box-diagram of the values of regression coefficients of both models (Figure 10) showed that the Uzan-Witczack model (1988 [1]) give better
correlations than the NCHRP model (2004 [2]).
5. Conclusions
Cyclic triaxial tests made on gravels lateritics soils allowed to have a number of very important results.
The study of the cyclic behavior of gravel lateritic soils confirms the importance of the effect of stress level on resilient modulus. Indeed, resilient modulus decreases inversely with the increase
of bulk and deviatoric stress. However, the effect of cement percentage on the increase
Table 6. Coefficients k[i] and r^2 obtained with Uzan-Witczack model (1988) [1].
Table 7. Coefficients k[i] and r^2 obtained with the NCHRP (2004) [2] model.
Figure 10. Comparison of the models coefficients of correlation.
of the stiffness of the material is not confirmed for all the materials and differs from one gravel to another. On the other hand, the effect of the moisture content seems to affect the values of the
resilient modulus in a sensitive way.
These results play a key important role in the prediction of the behaviour model of resilient modulus of gravel lateritic soils. Indeed, the Uzan-Witczack (1988) [1] and the NCHRP model (2004) [2]
had allowed very good correlations with better results with the first one. These models can be used for finite element modelling of pave- -ments.
6. Acknowledgements
The authors of the article would like to thank Dr Makhaly BA and would like to appreciate his important contribution for some of the gravel lateritic tests which have been realised in Wisconsin
University of Madison.
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Students 22293 - math word problem (22293)
Students 22293
Eight pupils modified the school plot in 2 hours. How many students do we have to send to modify the school grounds if we do not want to exceed the 1.5-hour limit?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Themes, topics:
Grade of the word problem:
Related math problems and questions:
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Class Likelihood Ratios to measure classification performance
Go to the end to download the full example code. or to run this example in your browser via Binder
Class Likelihood Ratios to measure classification performance#
This example demonstrates the class_likelihood_ratios function, which computes the positive and negative likelihood ratios (LR+, LR-) to assess the predictive power of a binary classifier. As we will
see, these metrics are independent of the proportion between classes in the test set, which makes them very useful when the available data for a study has a different class proportion than the target
A typical use is a case-control study in medicine, which has nearly balanced classes while the general population has large class imbalance. In such application, the pre-test probability of an
individual having the target condition can be chosen to be the prevalence, i.e. the proportion of a particular population found to be affected by a medical condition. The post-test probabilities
represent then the probability that the condition is truly present given a positive test result.
In this example we first discuss the link between pre-test and post-test odds given by the Class likelihood ratios. Then we evaluate their behavior in some controlled scenarios. In the last section
we plot them as a function of the prevalence of the positive class.
# Authors: The scikit-learn developers
# SPDX-License-Identifier: BSD-3-Clause
Pre-test vs. post-test analysis#
Suppose we have a population of subjects with physiological measurements X that can hopefully serve as indirect bio-markers of the disease and actual disease indicators y (ground truth). Most of the
people in the population do not carry the disease but a minority (in this case around 10%) does:
from sklearn.datasets import make_classification
X, y = make_classification(n_samples=10_000, weights=[0.9, 0.1], random_state=0)
print(f"Percentage of people carrying the disease: {100*y.mean():.2f}%")
Percentage of people carrying the disease: 10.37%
A machine learning model is built to diagnose if a person with some given physiological measurements is likely to carry the disease of interest. To evaluate the model, we need to assess its
performance on a held-out test set:
Then we can fit our diagnosis model and compute the positive likelihood ratio to evaluate the usefulness of this classifier as a disease diagnosis tool:
Since the positive class likelihood ratio is much larger than 1.0, it means that the machine learning-based diagnosis tool is useful: the post-test odds that the condition is truly present given a
positive test result are more than 12 times larger than the pre-test odds.
Cross-validation of likelihood ratios#
We assess the variability of the measurements for the class likelihood ratios in some particular cases.
import pandas as pd
def scoring(estimator, X, y):
y_pred = estimator.predict(X)
pos_lr, neg_lr = class_likelihood_ratios(y, y_pred, raise_warning=False)
return {"positive_likelihood_ratio": pos_lr, "negative_likelihood_ratio": neg_lr}
def extract_score(cv_results):
lr = pd.DataFrame(
"positive": cv_results["test_positive_likelihood_ratio"],
"negative": cv_results["test_negative_likelihood_ratio"],
return lr.aggregate(["mean", "std"])
We first validate the LogisticRegression model with default hyperparameters as used in the previous section.
│ │ positive │negative │
│mean│16.661086 │0.724702 │
│std │4.383973 │0.054045 │
We confirm that the model is useful: the post-test odds are between 12 and 20 times larger than the pre-test odds.
On the contrary, let’s consider a dummy model that will output random predictions with similar odds as the average disease prevalence in the training set:
│ │positive │negative │
│mean│1.108843 │0.986989 │
│std │0.268147 │0.034278 │
Here both class likelihood ratios are compatible with 1.0 which makes this classifier useless as a diagnostic tool to improve disease detection.
Another option for the dummy model is to always predict the most frequent class, which in this case is “no-disease”.
│ │positive │negative │
│mean│NaN │1.0 │
│std │NaN │0.0 │
The absence of positive predictions means there will be no true positives nor false positives, leading to an undefined LR+ that by no means should be interpreted as an infinite LR+ (the classifier
perfectly identifying positive cases). In such situation the class_likelihood_ratios function returns nan and raises a warning by default. Indeed, the value of LR- helps us discard this model.
A similar scenario may arise when cross-validating highly imbalanced data with few samples: some folds will have no samples with the disease and therefore they will output no true positives nor false
negatives when used for testing. Mathematically this leads to an infinite LR+, which should also not be interpreted as the model perfectly identifying positive cases. Such event leads to a higher
variance of the estimated likelihood ratios, but can still be interpreted as an increment of the post-test odds of having the condition.
│ │positive │negative │
│mean│17.8000 │0.373333 │
│std │8.5557 │0.235430 │
Invariance with respect to prevalence#
The likelihood ratios are independent of the disease prevalence and can be extrapolated between populations regardless of any possible class imbalance, as long as the same model is applied to all of
them. Notice that in the plots below the decision boundary is constant (see SVM: Separating hyperplane for unbalanced classes for a study of the boundary decision for unbalanced classes).
Here we train a LogisticRegression base model on a case-control study with a prevalence of 50%. It is then evaluated over populations with varying prevalence. We use the make_classification function
to ensure the data-generating process is always the same as shown in the plots below. The label 1 corresponds to the positive class “disease”, whereas the label 0 stands for “no-disease”.
from collections import defaultdict
import matplotlib.pyplot as plt
import numpy as np
from sklearn.inspection import DecisionBoundaryDisplay
populations = defaultdict(list)
common_params = {
"n_samples": 10_000,
"n_features": 2,
"n_informative": 2,
"n_redundant": 0,
"random_state": 0,
weights = np.linspace(0.1, 0.8, 6)
weights = weights[::-1]
# fit and evaluate base model on balanced classes
X, y = make_classification(**common_params, weights=[0.5, 0.5])
estimator = LogisticRegression().fit(X, y)
lr_base = extract_score(cross_validate(estimator, X, y, scoring=scoring, cv=10))
pos_lr_base, pos_lr_base_std = lr_base["positive"].values
neg_lr_base, neg_lr_base_std = lr_base["negative"].values
We will now show the decision boundary for each level of prevalence. Note that we only plot a subset of the original data to better assess the linear model decision boundary.
fig, axs = plt.subplots(nrows=3, ncols=2, figsize=(15, 12))
for ax, (n, weight) in zip(axs.ravel(), enumerate(weights)):
X, y = make_classification(
weights=[weight, 1 - weight],
prevalence = y.mean()
# down-sample for plotting
rng = np.random.RandomState(1)
plot_indices = rng.choice(np.arange(X.shape[0]), size=500, replace=True)
X_plot, y_plot = X[plot_indices], y[plot_indices]
# plot fixed decision boundary of base model with varying prevalence
disp = DecisionBoundaryDisplay.from_estimator(
scatter = disp.ax_.scatter(X_plot[:, 0], X_plot[:, 1], c=y_plot, edgecolor="k")
disp.ax_.set_title(f"prevalence = {y_plot.mean():.2f}")
We define a function for bootstrapping.
def scoring_on_bootstrap(estimator, X, y, rng, n_bootstrap=100):
results_for_prevalence = defaultdict(list)
for _ in range(n_bootstrap):
bootstrap_indices = rng.choice(
np.arange(X.shape[0]), size=X.shape[0], replace=True
for key, value in scoring(
estimator, X[bootstrap_indices], y[bootstrap_indices]
return pd.DataFrame(results_for_prevalence)
We score the base model for each prevalence using bootstrapping.
results = defaultdict(list)
n_bootstrap = 100
rng = np.random.default_rng(seed=0)
for prevalence, X, y in zip(
populations["prevalence"], populations["X"], populations["y"]
results_for_prevalence = scoring_on_bootstrap(
estimator, X, y, rng, n_bootstrap=n_bootstrap
results_for_prevalence.aggregate(["mean", "std"]).unstack()
results = pd.DataFrame(results["metrics"], index=results["prevalence"])
results.index.name = "prevalence"
│ │positive_likelihood_ratio│negative_likelihood_ratio│
│ │ mean │ std │ mean │ std │
│prevalence│ │ │ │ │
│ 0.2039 │4.507943 │0.113516 │0.207667 │0.009778 │
│ 0.3419 │4.443238 │0.125140 │0.198766 │0.008915 │
│ 0.4809 │4.421087 │0.123828 │0.192913 │0.006360 │
│ 0.6196 │4.409717 │0.164009 │0.193949 │0.005861 │
│ 0.7578 │4.334795 │0.175298 │0.189267 │0.005840 │
│ 0.8963 │4.197666 │0.238955 │0.185654 │0.005027 │
In the plots below we observe that the class likelihood ratios re-computed with different prevalences are indeed constant within one standard deviation of those computed with on balanced classes.
fig, (ax1, ax2) = plt.subplots(nrows=1, ncols=2, figsize=(15, 6))
ax=ax1, color="r", label="extrapolation through populations"
ax1.axhline(y=pos_lr_base + pos_lr_base_std, color="r", linestyle="--")
y=pos_lr_base - pos_lr_base_std,
label="base model confidence band",
- results["positive_likelihood_ratio"]["std"],
+ results["positive_likelihood_ratio"]["std"],
title="Positive likelihood ratio",
ylim=[0, 5],
ax1.legend(loc="lower right")
ax2 = results["negative_likelihood_ratio"]["mean"].plot(
ax=ax2, color="b", label="extrapolation through populations"
ax2.axhline(y=neg_lr_base + neg_lr_base_std, color="b", linestyle="--")
y=neg_lr_base - neg_lr_base_std,
label="base model confidence band",
- results["negative_likelihood_ratio"]["std"],
+ results["negative_likelihood_ratio"]["std"],
title="Negative likelihood ratio",
ylim=[0, 0.5],
ax2.legend(loc="lower right")
Total running time of the script: (0 minutes 1.665 seconds)
Related examples
Importance of Feature Scaling
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How the Higgs Field Works (with math)
This set of articles explains how the Higgs field gives mass to other particles, and some details about the Higgs particle.
These articles require the kind of math that you’d come across in advanced pre-university or beginning university settings: algebra, cosines and sines, a little calculus (though most of that can be
skirted) and a little familiarity with simple physics concepts: energy, oscillations, waves. Some of this stuff is reviewed in the Particles and Fields articles in any case.
1. The Basic Idea (of how the Higgs field gives mass to other particles.)
2. Why The Higgs Field Becomes Non-Zero, unlike other fields.
3. How the Higgs Particle Arises, as a quantum of the waves in the Higgs field as it oscillates around its non-zero equilibrium value.
4. Why the Higgs Field is Necessary (why in our universe, with its weak nuclear force, the electron can’t have a mass without it.)
Before reading these articles you should first read through my set of articles on Particles and Fields, which are at the same level, and are intended to lead directly to this set.
32 Responses
2. I do not know about field and solution to simultaneous field equations. But when field follows the curvature of space, mass of the field disappears. But when the curvature of space is
straightened out, the field has mass.
4. Dear Professor Strassler, I see that Harry Cliff has said the Higgs field is 10^16 times smaller than predicted. Could you provide a more accurate figure including error margins? I am an
enthusiastic, amateur physicist and these discrepancies fascinate me.
6. Particularly, years of instructional excessive-stakes testing robs many college
students of the one ingredient important to actual studying:
being keen to take risks and to fail https://math-problem-solver.com/ .
With an odd mathematical proof the World could seem unusual and incomprehensible.
7. I was wondering if einstein’s fabric of space-time is the ‘same thing’ as the higgs field..?
8. ITS GREAT DISCOVERY OF PETER HIGGS
11. Thanks for giving me a place to start. Ill check back when I finish theese articles you recommended.
12. Well, I have some thought s about gravity and it is very stupid, that might explain the cause of gravity:
* mass comes from Highs Bosson,
* gravity is due to mass.
So, could that be reasons to look for gravity explained in that field…
There is just enough phenomenon s to finally explain it… Or to come pretty had damn close…
Time locally might be discontinuous and globally continuous…
Time to introduce layers of space time and Fuzzy logic in order to create one cosisyen theory…
13. ” I think that the De Broogle matter-waves and David Bohms pilot-waves are waves in the Higgs field and it is the Higgs field that tells the matter particles how they have to move in a
gravitational field and in the double-slit experiment. (electrons get their mass after the slits according to the interference in Higgs waves)
For example in QED calculations and in the double-slit experiment it is not the electrons that are taking all the possible paths , it is only the waves in the Higgs field that travels all the
That means that Niels Bohrs principle of complementarity where a particle is both a particle and a wave at the same time is not needed any longer. Higgs field gives electrons their mass and the
waves in the Higgs field tell them how to moove.
At the same time we can notice that we dont need Einsteins 4:th room-time dimension any longer , because it is only a mathematical construction. The real physical reality are waves and structures
in the Higgs field”
14. Ted talks has Dr Harry Cliff say that the particular strength of the Higgs fileld is a surprise as it should be either zero or ginormous although both are incompatible with the existence of our
universe . He says a large number of other things in the talk that sound quite outlandish such as the LHC result was a huge surprise as they should have found other particles produced with it but
they didn’t and that physicists have tried and failed to explain the particular strength of the Higgs field for decades and that “his” group at CERN published paper afte paper after the first run
that concluded that That run had failed to find any support for the “most elegant” theory namely string theory or its alternative – the multiverse theory.
Is this man really a player or just a trouble maker? Is quantum mechanics and general relativity leading us to God as he seems to suggest? Has string theory led to any prediction that helps us
apart from the theory that it has a ginormous number of theories in it
. ? Help ! 😳😳😳😳🙃🙃🙃
16. All history recorded in higgs MESH field. Every atom shift recorded, patterns determinable of light and sound.
WTF WatchThisFriday.com read about the MESH. Truth in Fiction
17. I think I answered my own question. Your Class 1 equation is the Klein-Gordon equation, which of course is Lorentz invariant, provided the field transforms like a scalar.
18. Thank you for the great description of the Higgs mechanism. I have a question. You described two different kinds of field equations, Class 0 (for massless particles) and Class 1 (for massive
particles). Class 0 is the traditional wave equation, which is easily seen to be Lorentz invariant. But is the Class 1 wave equation you described also Lorentz invariant?
23. Professor Strassler. Thank you for your website. Perhaps I will have questions to you after several months.
29. Fantastic – this is just what I was hopings for !! Other than a brief mention in the introduction to Lisa Randall’s book, information has been sketchy and not meant for those with some training
in physics. I started out as a Physics major (45 years ago) but then discovered computers and my life’s direction took a new turn, but I’ve never lost my fascination with (and love of) physics
(and now cosmology) and have kept up as best I could over time. It just gets better and better 🙂 Also great exercise for my brain – they do say ‘use it or lose it’. I look forward to lots of
‘use’ as I follow your articles – LOL !!
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Frequency Dependent Dielectrics
Define a frequency-dependent medium to use in your model or to add to the media library.
The following frequency-dependent definitions are supported:
Debye relaxation
Use this method to describe the relaxation characteristics of gasses and fluids at microwave frequencies. It is derived for freely rotating spherical polar molecules in a predominantly non-polar
This method is similar to the Debye relaxation but makes use of an additional parameter to describe the model.
Use this method to model liquids, solids and semi-solids.
Use this method for composite dielectrics.
Frequency List
Use this method to define a frequency-dependent dielectric by specifying data points at a range of frequencies. The values for the dielectric properties are linearly interpolated to obtain the
dielectric properties at frequency points other than specified.
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How Much Money Will Donald Receive From His Insurance Company for Windshield Replacement?
If the cost to replace Donald's windshield is $450, how much money will he receive from his insurance company after filing his claim based on the following policy summary?
Policy Summary:
- Comprehensive deductible: $375
- Collision deductible: $550
- Premium of $625 for 4 months
Choose the correct amount:
a) $75
b) $725
c) $375
d) $550
Final Answer:
Donald will receive $75 from his insurance company to cover the windshield replacement because this is the amount remaining after subtracting his comprehensive deductible of $375 from the total
repair cost of $450.
Donald's windshield got cracked by a small rock, and he is looking to file a claim with his insurance company to cover the cost of replacing the windshield, which is $450.
Given the policy summary, which includes a comprehensive deductible of $375, we can calculate the amount Donald will receive.
The comprehensive deductible is the amount that Donald must pay out of pocket before his insurance coverage kicks in.
For the windshield replacement cost of $450, since the deductible is $375, the insurance company will cover the costs above this deductible amount.
Therefore, we subtract the deductible from the total cost: $450 - $375 = $75.
So, the amount Donald will receive from his insurance company after filing the claim is $75.
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Quantum Key Distribution - BBM92
Eric deQuevedo 😄
Quantum Key Distribution - BBM92
🌌 Introduction: BBM92 Quantum Key Distribution
Quantum Key Distribution (QKD) has indeed revolutionized secure communication. The BBM92 protocol, introduced in 1992 by Charles Bennett, Gilles Brassard, and N. David Mermin, is a significant
advancement in this field. It leverages quantum mechanics and entanglement to ensure cryptographic key exchange security.
✨ Understanding the BBM92 Protocol
The BBM92 protocol is an entanglement-based QKD protocol that uses pairs of entangled photons to establish a secure key between two parties, typically called Alice and Bob. Unlike its predecessor,
the BB84 protocol, BBM92 utilizes entangled photon properties and Bell's inequality for security.
🌀 Entanglement and Bell States
In BBM92, entangled photon pairs are created in a maximally entangled Bell state. These states are defined as:
$|\Psi^+\rangle = \frac{1}{\sqrt{2}} (|H\rangle|V\rangle + |V\rangle|H\rangle)$
$|\Psi^-\rangle = \frac{1}{\sqrt{2}} (|H\rangle|V\rangle - |V\rangle|H\rangle)$
$|\Phi^+\rangle = \frac{1}{\sqrt{2}} (|H\rangle|H\rangle + |V\rangle|V\rangle)$
$|\Phi^-\rangle = \frac{1}{\sqrt{2}} (|H\rangle|H\rangle - |V\rangle|V\rangle)$
These states exhibit perfect correlations, meaning measurements on one photon instantly determine the other's state, regardless of distance.
🔍 The BBM92 Protocol: Step-by-Step
1. Entangled Photon Generation
The protocol begins with generating entangled photon pairs in a Bell state, typically using spontaneous parametric down-conversion (SPDC). One photon from each pair is sent to Alice and the other to
Bob via a quantum channel.
2. Basis Selection and Measurement
Alice and Bob randomly choose a measurement basis for each received photon:
(0°) and
They measure each photon's polarization in their chosen basis and record results.
3. Classical Communication and Sifting
After measuring many photons, Alice and Bob publicly announce their chosen bases over a classical channel. They discard results where bases don't match, keeping instances with the same basis. This
process is called "sifting."
4. Error Checking and Bell's Inequality Test
To ensure key security, Alice and Bob compare a random subset of their results to estimate the error rate. They also perform a Bell's inequality test to verify photon entanglement and detect
potential eavesdropping. An eavesdropper (Eve) attempting to intercept photons would introduce errors, violating Bell's inequality.
5. Key Generation and Postprocessing
If the error rate is below a certain threshold (typically around 11%, not 20% as originally stated), Alice and Bob generate the final key. They use classical postprocessing techniques like error
correction and privacy amplification to eliminate any information Eve might have gained and reconcile remaining key discrepancies.
🛡️ Security Advantages of BBM92
BBM92 offers several security advantages:
1. Intrinsic Security: Rooted in fundamental quantum mechanics properties.
2. Eavesdropping Detection: Any interception disturbs the entangled state, detectable through Bell's inequality violation.
3. No Need for Trusted Source: Unlike prepare-and-measure protocols, BBM92 doesn't require trust in the quantum state preparation source.
🌐 Practical Implementations and Challenges
🔬 Photon Generation and Detection
BBM92 implementation requires reliable entangled photon sources and efficient single-photon detectors. SPDC in nonlinear crystals (e.g., BBO) is commonly used for entangled photon generation.
🌌 Quantum Channel Loss
Photon loss in the quantum channel can reduce protocol efficiency and security. Quantum repeaters are being developed to mitigate this issue, though they're still in early research stages.
📊 Environmental Noise
External noise can affect entangled state fidelity and introduce measurement errors. Proper isolation and advanced error-correction techniques are essential for maintaining key distribution
🔮 Conclusion: The Future of Secure Communication
The BBM92 protocol represents a significant advancement in quantum cryptography. By leveraging quantum entanglement and Bell's inequality, it provides a robust and fundamentally secure method for
generating encryption keys. As quantum technologies evolve, protocols like BBM92 will be crucial in ensuring secure communication in the quantum era.
📜 References
1. Bennett, C. H., Brassard, G., & Mermin, N. D. (1992). Quantum cryptography without Bell's theorem. Physical Review Letters, 68(5), 557-559.
2. Ekert, A. K. (1991). Quantum cryptography based on Bell's theorem. Physical Review Letters, 67(6), 661-663.
3. Gisin, N., Ribordy, G., Tittel, W., & Zbinden, H. (2002). Quantum Cryptography. Reviews of Modern Physics, 74(1), 145-195.
4. Nielsen, M. A., & Chuang, I. L. (2010). Quantum Computation and Quantum Information. Cambridge University Press.
5. Lo, H. K., Curty, M., & Tamaki, K. (2014). Secure quantum key distribution. Nature Photonics, 8(8), 595-604.
Understanding the principles behind the BBM92 protocol illuminates quantum mechanics' profound impact on future secure communication methods.
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What is Heisenberg's uncertainty principle? | Britannica
Understand the uncertainty principle as defined by the properties of waves
Quantum mechanics is weird, right? But did you know that some of its supposed peculiarities are not unique to quantum physics but are just properties of waves in general? Take the uncertainty
principle. It says that the better we know where our particle is, the less we know about how fast it's going. There's a limit to how much we can know.
But this isn't just quantum weirdness, it happens all the time with normal, everyday waves. Remember that the frequency of a wave is how close the wave crests are to each other. A low frequency wave
doesn't have nearly as much oomph as a high frequency one. And the position of a wave is, well, where the wave is, mostly.
So where exactly is a wave? It's spread out everywhere, right? OK. But for a wave pulse, it's pretty easy to see where the wave is. So now that we know where the pulse is, what's its frequency? Well,
a localized pulse doesn't really wave, so we can't measure the frequency of its crests. And that's the uncertainty principle in a nutshell. You can either know where wave is or where it's going, but
not both at the same time.
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nep-ecm 2024-05-13 papers
By: Vogelsang, Timothy J. (Department of Economics, Michigan State University); Wagner, Martin (Department of Economics, University of Klagenfurt, Bank of Slovenia, Ljubljana and Institute for
Advanced Studies, Vienna)
This paper shows that the integrated modified OLS (IM-OLS) estimator developed for cointegrating linear regressions in Vogelsang and Wagner (2014a) can be straightforwardly extended to
cointegrating multivariate polynomial regressions. These are regression models that include as explanatory variables deterministic variables, integrated processes and products of
(non-negative) integer powers of these variables as regressors. The stationary errors are allowed to be serially correlated and the regressors are allowed to be endogenous. The IM-OLS
Abstract: estimator is tuningparameter free and does not require the estimation of any long-run variances. A scalar long-run variance, however, has to be estimated and scaled out when using IM-OLS
for inference. In this respect, we consider both standard asymptotic inference as well as fixed-b inference. Fixed-b inference requires that the regression model is of full design. The
results may be particularly interesting for specification testing of cointegrating relationships, with RESET-type specification tests following immediately. The simulation section also
zooms in on RESET specification testing and illustrates that the performance of IM-OLS is qualitatively comparable to its performance in cointegrating linear regressions.
Keywords: Cointegration, fixed-b asymptotics, IM-OLS, multivariate polynomials, nonlinearity, RESET
JEL: C12 C13 C32
Date: 2024–04
URL: http://d.repec.org/n?u=RePEc:ihs:ihswps:53&r=ecm
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Get instant live expert help on I need help with excel absolute average deviation
Complete the Statistics Table, at the bottom of the sheet, by calculating the average, standard deviation, median. How do you find the maximum or minimum value of a function (rows 103 – 107) for AS1,
AS2, AS3 and the Overall Average columns
Solved by M. E. in 30 mins
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Numbers To 999 Worksheets
Numbers To 999 Worksheets act as fundamental devices in the realm of mathematics, giving an organized yet versatile system for learners to discover and grasp numerical concepts. These worksheets
supply a structured technique to comprehending numbers, supporting a strong structure upon which mathematical proficiency grows. From the most basic counting exercises to the complexities of advanced
calculations, Numbers To 999 Worksheets accommodate students of diverse ages and ability degrees.
Revealing the Essence of Numbers To 999 Worksheets
Numbers To 999 Worksheets
Numbers To 999 Worksheets -
They go up to 100 199 500 999 and 1 200 There are 4 worksheets for each set of numbers Included are two types of worksheets One where the students write the 3 missing numbers on the number line for
that problem and another version where the students cut out the missing numbers and paste them in order on
Writing and Reading Numbers Up to 999 Name Write the word name for each number below example 572 five hundred seventy two ANSWER KEY RB 305 231 342 915 three hundred sixty three two hundred twenty
four hundred six 714 518 690 97 163 a nine hundred fifteen
At their core, Numbers To 999 Worksheets are cars for theoretical understanding. They envelop a myriad of mathematical concepts, leading students with the labyrinth of numbers with a collection of
engaging and purposeful exercises. These worksheets transcend the borders of traditional rote learning, encouraging energetic involvement and promoting an intuitive grasp of numerical connections.
Supporting Number Sense and Reasoning
Ordering Numbers Range 100 To 999 A
Ordering Numbers Range 100 To 999 A
Analyze of numbers to ones tens hundreds Liveworksheets transforms your traditional printable worksheets into self correcting interactive exercises that the students can do online and send to the
92 place value to 99 worksheets plus a double sided title page and a double sided back cover 1 Reading numbers to 99 assessment to check that students can identify and read numbers to 99 includes one
student viewing sheet and one teacher record sheet to be administered individually
The heart of Numbers To 999 Worksheets lies in growing number sense-- a deep understanding of numbers' significances and interconnections. They encourage exploration, welcoming learners to dissect
arithmetic operations, decode patterns, and unlock the secrets of sequences. Through thought-provoking difficulties and sensible problems, these worksheets become entrances to refining thinking
skills, supporting the logical minds of budding mathematicians.
From Theory to Real-World Application
Abacus For Numbers Up To 999 Math Worksheet For Kids Stock Vector Illustration Of Practice
Abacus For Numbers Up To 999 Math Worksheet For Kids Stock Vector Illustration Of Practice
The following webpages involve correctly placing numbers on a number line Using these second grade math worksheets will help your child to count by 1s 2s 5s 10s 50s and 100s position numbers on a
number line understand place value up to 3 digits Number lines up to 100 easier Number Line Worksheets 3 Digits
School subject Mathematics 1061599 Main content Reading and writing of numbers upto 999 1098415 Prepared by Balaji PVM Other contents comparison of numbers
Numbers To 999 Worksheets serve as conduits connecting academic abstractions with the apparent facts of daily life. By infusing useful circumstances right into mathematical exercises, students
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Varied Tools and Techniques
Adaptability is inherent in Numbers To 999 Worksheets, utilizing a toolbox of instructional tools to accommodate varied learning designs. Visual aids such as number lines, manipulatives, and
electronic resources serve as buddies in envisioning abstract principles. This varied method makes sure inclusivity, suiting learners with different choices, staminas, and cognitive designs.
Inclusivity and Cultural Relevance
In a progressively diverse globe, Numbers To 999 Worksheets embrace inclusivity. They transcend cultural boundaries, integrating examples and issues that reverberate with students from varied
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Crafting a Path to Mathematical Mastery
Numbers To 999 Worksheets chart a course in the direction of mathematical fluency. They impart determination, essential thinking, and analytic abilities, crucial attributes not only in maths however
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In an era noted by technological improvement, Numbers To 999 Worksheets effortlessly adjust to electronic systems. Interactive user interfaces and digital sources augment typical discovering,
providing immersive experiences that go beyond spatial and temporal limits. This amalgamation of traditional approaches with technological technologies declares an encouraging era in education and
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Final thought: Embracing the Magic of Numbers
Numbers To 999 Worksheets exemplify the magic inherent in maths-- a captivating trip of expedition, discovery, and proficiency. They go beyond standard pedagogy, functioning as drivers for stiring up
the flames of inquisitiveness and query. With Numbers To 999 Worksheets, students embark on an odyssey, opening the enigmatic globe of numbers-- one issue, one option, at a time.
Ordering Numbers Range 100 To 999 A
Worksheet On Numbers From 900 To 999 Fill In The Missing Numbers Answers
Check more of Numbers To 999 Worksheets below
Numbers Up To 999 ESL Worksheet By Ambarmaria
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Maths Ch 1 Numbers Up To 999 Worksheets For Extra Practice
Maths Ch 1 Numbers Up To 999 Worksheets For Extra Practice
Numbers 100 999 ESL Worksheet By Moni Lou
Numbers From 1 To 999 ESL Worksheet By Nur us
Name Writing And Reading Numbers Up To 999 Super
Writing and Reading Numbers Up to 999 Name Write the word name for each number below example 572 five hundred seventy two ANSWER KEY RB 305 231 342 915 three hundred sixty three two hundred twenty
four hundred six 714 518 690 97 163 a nine hundred fifteen
Ordering Numbers Range 100 To 999 A Math Drills
Welcome to The Ordering Numbers Range 100 to 999 A Math Worksheet from the Number Sense Worksheets Page at Math Drills This math worksheet was created or last revised on 2013 02 15 and has been
viewed 134 times this week and 755 times this month It may be printed downloaded or saved and used in your
Writing and Reading Numbers Up to 999 Name Write the word name for each number below example 572 five hundred seventy two ANSWER KEY RB 305 231 342 915 three hundred sixty three two hundred twenty
four hundred six 714 518 690 97 163 a nine hundred fifteen
Welcome to The Ordering Numbers Range 100 to 999 A Math Worksheet from the Number Sense Worksheets Page at Math Drills This math worksheet was created or last revised on 2013 02 15 and has been
viewed 134 times this week and 755 times this month It may be printed downloaded or saved and used in your
Maths Ch 1 Numbers Up To 999 Worksheets For Extra Practice
Numbers From 100 To 999 ESL Worksheet By Wwanamar
Numbers 100 999 ESL Worksheet By Moni Lou
Numbers From 1 To 999 ESL Worksheet By Nur us
Maths Ch 1 Numbers Up To 999 Worksheets For Extra Practice
Hundred Tens And Ones Place Value To 999 Worksheets Place Values Tens And Ones Writing
Hundred Tens And Ones Place Value To 999 Worksheets Place Values Tens And Ones Writing
Representing Numbers To 999 With Base Ten Blocks A
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Centre of enlargement
Write down the coordinates of the centre of enlargement.
The point at which your ray lines meet will be the centre of enlargement.
The centre of enlargement is (2,2)
You may also be asked to find the scale factor of enlargement. For this example the scale factor of enlargement is 2. You can calculate the scale factor by choosing a pair of corresponding sides and
dividing the enlarged length by the original length.
Scale \ factor = \frac{enlarged \ length}{ original \ length}=\frac{2}{1}=2
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What is the 4 degree called in math?
What is the 4 degree called in math?
Names of Degrees
Degree Name Example
2 Quadratic x2−x+2
3 Cubic x3−x2+5
4 Quartic 6×4−x3+x−2
5 Quintic x5−3×3+x2+8
What are the terms of degrees?
Mathematics. the sum of the exponents of the variables in an algebraic term: x3 and 2x2y are terms of degree three. the term of highest degree of a given equation or polynomial: The expression 3x2y +
y2+ 1 is of degree three. the exponent of the derivative of highest order appearing in a given differential equation.
What is 11 degree called?
A degree 11 polynomial could be called “undecic” and a degree 12 polynomial “dodecic”.
What is the 3rd degree called in math?
cubic polynomial
Third-degree polynomial is of the form p(x) = ax3 + bx2+ cx + d where ‘a’ is not equal to zero.It is also called cubic polynomial as it has degree 3. Example: 5×3 + 2×2+ 3x + 7 is a cubic polynomial
or Third Degree Polynomial since the highest degree of the expression is 3 or the power of the leading term is 3.
What is a 5 term polynomial called?
Degree 3 – cubic. Degree 4 – quartic (or, if all terms have even degree, biquadratic) Degree 5 – quintic.
What is a 5th degree polynomial?
In other words, a quintic function is defined by a polynomial of degree five. Because they have an odd degree, normal quintic functions appear similar to normal cubic functions when graphed, except
they may possess one additional local maximum and one additional local minimum.
What is a degree in math algebra?
Explanation: The degree is the highest exponent value of the variables in the polynomial. Here, the highest exponent is x5, so the degree is 5.
What is quintic in math?
: a polynomial or a polynomial equation of the fifth degree.
What is the degree of zero?
The degree of the zero-degree polynomial (0) is not defined. Detailed Answer: The polynomial 0 has no terms at all, and is called a zero polynomial. Because the zero polynomial has no non-zero terms,
the polynomial has no degree.
What is a 2nd degree polynomial called?
2nd-degree polynomials are also known as quadratic polynomials.
What is a 6 degree polynomial?
In algebra, a sextic (or hexic) polynomial is a polynomial of degree six. A sextic equation is a polynomial equation of degree six—that is, an equation whose left hand side is a sextic polynomial and
whose right hand side is zero.
What is a 4 term polynomial called?
The term “quadrinomial” is occasionally used for a four-term polynomial.
What is a 6th degree polynomial called?
In algebra, a sextic (or hexic) polynomial is a polynomial of degree six.
What do you call a 4th degree polynomial?
A polynomial of degree 4 is called bi-quadratic polynomial.
What are the degree groups?
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• Integrated Law Course- BA + LL.B.
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• BFD- Bachelor of Fashion Designing.
• BSW- Bachelor of Social Work.
What is a degree called?
degree, also called academic degree, in education, any of several titles conferred by colleges and universities to indicate the completion of a course of study or the extent of academic achievement.
What is the degree of the term 7?
Degree 5 – quintic. Degree 6 – sextic (or, less commonly, hexic) Degree 7 – septic (or, less commonly, heptic)
What is degree of an equation?
In Algebra, the degree is the largest exponent of the variable in the given equation. The equations and polynomials are named based on their degree. If the equation has a degree of one, it is a
linear equation. In short, the degree of linear equations is always one.
What is a degree 6 polynomial called?
What are types of math degrees?
– Actuary – Math teacher – Financial analyst – Budget analyst – Researcher – Controller
What jobs can you get with degree in mathematics?
Mathematics instructor. Primary duties: Mathematics teachers provide mathematics instruction for various grade levels.
Computer programmer. Primary duties: Computer programmers are responsible for designing software.
Financial analyst.
Operations analyst.
Civil engineer.
Database administrator.
Software developer.
What does degree mean in math terms?
– y = x (Degree: 1; Only one solution) – y = x2 (Degree: 2; Two possible solutions) – y = x3 (Degree: 3; Three possible solutions)
What can you do with a mathematics degree?
Algorithmic engineer.
College or university math professor.
Data scientist.
High school math teacher.
Investment banking analyst.
Management consultant.
Operations research analyst.
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Help with maths
OK, so, my first question is what are these for? If they're for a homework question or such I am slightly confused as Wolfram Alpha says there are no solutions to the first one. I'm inclined to
agree. Certainly there's no way to make it much nicer.
Second one:
. It's a bit complicated. 4th roots of 3? No thanks.
Yeah, the first one is a bit weird, since the roots of x and 1-x would imply that x is between 0 and 1 (otherwise the roots wouldn't be real numbers), which would mean those two square roots would
also be between 0 and 1. I don't see how you could subtract two numbers like that to get something greater than 3/sqrt(5) (which is greater than 1).
This was kind of homework, it was a preparation for a test. Hopefully I don't get something like this in the actual test
...and that's the bottom line because Mate de Vita said so.
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What do you want to work on?
About 4385809
Midlevel (7-8) Math, Statistics, Economics, MS Excel, Spanish
Bachelors in Economics, General from Universidad de La Salle
Career Experience
As an economist I have experience in the study of economic policies by public entities, such as the Bank of the Republic and the National Department of Statistics.
I Love Tutoring Because
teaching with tutor.com gives me the opportunity to help students to train as professionals and contribute to the development of society.
Other Interests
Amateur astronomy, Photography, Traveling, Watching Movies, Web surfing
Math - Midlevel (7-8) Math
He is a really good tutor
Math - Statistics
Amazing tutor. helped me with everything i needed help with.
Math - Statistics
Juan was very patient.
Math - Statistics
Juan was very helpful and got me to think through the question. He provided a formula and explanation that the book failed to give.
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Discussing The Concept Of Proportion - Education Is Around
Discussing The Concept Of Proportion
It isn’t easy to imagine how our life would certainly be without mathematical concepts such as percentages or proportion. In our daily lives, we often come across rates and proportions when choosing
to purchase, cooking and when on an occupation trip, etc.
Proportions, as well as percentages, are crucial for– reliable efficiency. In this, post we will find out exactly how to compute proportions and solve sample issues. However, before that, let’s begin
by defining ratios.
A ratio is a method of making contrasts between two or more amounts. The indication used to signify a proportion is colon ‘:’ Mean an, and b are two different quantities or numbers. After that, the
proportion of a to b can create as a/b or a: b. Likewise, the ratio of b to a can additionally be stood for as b: a or b/a. The initial amount in a proportion is called antecedent, and the 2nd worth
is referred to as the consequent.
Instances of proportions are 3/4 or 3: 4, 1/5 or 1: 5, 199/389 or 199:389 etc. From this example, a ratio is simply a fraction where the antecedent is the numerator, and the consequent is the common
The famous Vitruvian Male illustration of Leonardo da Vinci was based upon the perfect ratio of the body. Each part of the body occupies a different percentage like the face takes up about 1/10 of
the total elevation, and the head uses up about 1/8 of the overall height. The writers in the middle ages used the word proportion (percentage) for the very first time. In 1948, Le Corbusier provided
a system of proportions.
What is a Proportion?
A proportion is an expression that tells us that two ratios are equivalent. Two proportions state to be proportional if they are equal. Percentages are stood for by the sign ‘:’ or ‘=’. For example,
if a, b, c, and d are integers, the proportion is composed as a: b = c: d or a/b = c/d or b: a = d: c. For instance, the ratios 3: 5 and also 15: 25 are proportional and are written as 3: 5= 15: 25
The four numbers a, b, c and d are known as the regards to a percentage. The initial an and the last term d are described as extreme terms while the second, as well as 3rd terms in a proportional,
are called mean terms.
How to Solve Ratios?
It is easy to determine if proportions are proportional. To check if the proportion a: b and also c: d is proportional.
Multiply the first with the last term: an x d.
Multiply the 2nd term with the third term: b x c.
If the product of strict terms amounts to the development of mean times, after that, the ratios are symmetrical: an x d = b x c.
Continued proportion.
Two ratios, b: c and a: b, is said to be in continued proportion if a: b = b: c. In this situation, the term c is called the 3rd proportion of an and b, whereas b is called the mean proportion of in
between the terms and c.
When the terms a, b, and care in continued percentage, the complying with formula obtains.
a/b = b/c.
Cross is increasing the terms gives; an x c =b x b, Therefore.
b ² = a/c.
Figure out if the list below ratios remains in percentage: 8:10 as well as 12:15.
Increase the initial and also 4th terms of the ratios.
8 × 15 = 120.
Now multiply the 2nd and also 3rd term.
10 × 12 = 120.
Because the item of the extremes amounts to the product of the ways.
Since the item of means (120) = product of extremes (120 ),.
For that reason, 8: 10 and 12:15 are symmetrical.
Golden Ratio
The most significant application of the percentage is the gold proportion, which assisted a whole lot in examining percentages of various things and synthetic systems like economic markets. Both
amounts state to be in gold proportion if their ratio amounts to the ratio of their sum to the bigger of the two quantities, i.e. (a + b)/ a = a/b, where a > b > 0.
This proportion represents by the Greek letter φ. More simplifying this formula, we get, φ 2– φ– 1 = 0. And also, fixing this by making use of a square formula, we obtain φ = 1.6180339887 …
Euclid and lots of mathematicians worked with the golden ratio and found its presence in the regular pentagon and gold rectangular shape.
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Design of Block Codes for Distributed Learning in VR/AR TransmissionJournal of Information and Communication Convergence Engineering
A. New Minimal Codes
1) Extension Methods for New Minimal Codes
For 0 ≤ n ≤ N−1, we define an arbitrary nonzero vector in F[p]^m, with a length k ≥ 2, as r¯a=(r[a](0), r[a](1), ..., r[a](k−1)). Theorem 2 presents a new minimal code for length kN.
Theorem 2. Let us define codeword y¯i of length kn as follows:
where 0≤n≤N−1 and 0 ≤ b ≤ l−1. The new code E = {y¯0,y¯1,⋯,y¯M−1} is a (kN,K)-minimal code.
Proof: First, we prove that E is a linear code. For any two integers i and j between 0 and M−1, the summation
becomes y[k][a,b] by the linearity of the original codewords.
for some l. Therefore, E is linear. Next, we prove the minimality of E. Because r¯a is a nonzero vector,
and there exists 0 ≤ b ≤ l−1 such that
Therefore, the support of codeword is y¯i not a subset of the support of codeword y¯i. We can similarly show that the support for y¯i is not a subset of that for y¯i. Therefore, E is a minimal code.
In Theorem 2, each codeword y¯i can be expressed as the concatenation of vectors obtained by multiplying each element of the original codeword x¯i by corresponding constants from r¯a. Fig. 2
illustrates this extension method for minimal codes.
Fig. 2. Extension method for minimal codes
2) Weight Properties of New Minimal Code
The weight distribution of the new code E is determined by the weight distributions of the individual vectors S¯a and the original code C. In the simplest case, when the Hamming weight of each is S¯a
fixed to 1, the weight distribution of E will match that of C. Moreover, the maximum Hamming weight S¯a of is equal to the length l. When all S¯a have a fixed weight of l, each weight value in the
original distribution is multiplied by l. Because the Hamming weight of each S¯a can be arbitrarily selected between 1 and l, various weight distributions can be obtained depending on the variation
in the Hamming weights of S¯a with respect to α. In addition, it is evident that extending a code beyond the constraints of Theorem 1 enables the generation of a new code beyond those constraints.
In linear codes, the minimum distance – which represents the distance between code words – is an important performance metric associated with error probability. Owing to linearity, the minimum
distance is equal to the minimum Hamming weight among the code words [3]. Assuming that the original minimum distance of the minimal code C is denoted as d, the minimum distance of the new code E can
be observed to range from d to ld with respect to S¯a. Therefore, the ratio between the length and minimum distance is maintained as the length is extended. Table 1 provides examples of the lengths,
weights, and minimum distances of extended codes.
Table 1 . Sample new parameters of minimal codes (N: length, K: information length, d: minimum distance)
N K d Number of distinct weights
Original Codes 511 10 120 3
Extended Codes 5110 10 120~1200 3~30
B. Extended Minimal Binary Codes
1) Double Extension of Binary Minimal Codes
Let us define the extended length 2N codeword y¯i as follows:
yi(t)=xi(t/2), t is even;xi(−t/2), otherwise.
where i = 1, 2, ..., M. We now can define the extended code Y as follows:
Here, Y has a length of 2N and contains M code words, which can be easily generated by combining original code words with their reverse-indexed counterparts. The original codewords x¯1,⋯,x¯M hold
true mutual linearity, whereas each of the new codewords y¯1,⋯,y¯M holds true linearity for odd and even indices. Therefore, the set of new code words in Y satisfies linearity. Furthermore, based on
the properties of the original code C, the codewords in Y and their supports can be inferred to be mutually independent.
2) Interleaved Extension of Binary Minimal Codes
Our code construction is based on the interleaving of two different minimal codes, with the indices of a new code determined by a combination of the two codes. Consider a minimal code C1 with a
length of N1 that contains M1 codewords denoted as x¯1,1,⋯,x¯1,M1, and a minimal code C2 with a length of N2 that contains M2 codewords denoted as x¯2,1,⋯,x¯2,M2. The new code word z¯i can be defined
where ⊙ is the binary AND operator, 0 ≤ i ≤ M[1], 0 ≤ i ≤ M[2], and 0 ≤ t ≤ N[1]N[2]−1. Furthermore, t[1] = t mod N[1] and t[2] = t mod N[2]. Consequently, the value of z¯i,j(t) can be 1 for the
number of t values that equals the product of the Hamming weights of x¯1,i and x¯2,j. Furthermore, it is possible to generate z¯i,j for all combinations of i and j. Define the new code Z as follows:
Z=z¯i,j|1≤i≤M1 and 1≤i≤M2.
The codewords of Z inherit properties of the original code depending on the values of t[1] and t[2], indicating that the support of the different codewords remains independent. Moreover, as linearity
holds for t[1] and t[2] separately, and N[1] and N[2] are relatively prime, linearity also holds with respect to t. The number of codewords in Z is M[1]M[2], and the Hamming weight of each codeword
is equal to the product of the Hamming weights of the two constituent codes. Finding two relatively prime lengths is another challenging problem, as most known binary minimal codes have lengths in
the form of p^m − 1.
C. Application of Minimal Codes
The newly designed code is fundamentally determined by the weight distribution of the original code. However, by altering the combinations of the constituent codes, new weight distributions can be
generated, as discussed in III.B.1. Table 2 presents the experimentally obtained weight distributions for each combination.
Table 2 . Weight distributions of a sample original code and its extension
Code Lengths No. of Possible Weights
Original 255 6
Extended 510 12
As seen from the table, codes with a wider range of weight distributions can be synthesized by combining existing codes. This enables a greater variety of information combinations, increasing the
diversity of dispersed information.
The newly designed code from III.B.1 can be utilized to combine information from two distributed learning systems into a single entity. The dispersed forms of information from each system can be
incorporated into existing code words without modification. Moreover, because the new codewords remain mutually independent, confidentiality is maintained in a dispersed formIf the codes with
relatively prime lengths presented in III.B.2 are not used, finding a method to combine data from the two systems becomes an additional challenge.
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Efficient Function Evaluation in NumPy Arrays - DNMTechs - Sharing and Storing Technology Knowledge
Efficient Function Evaluation in NumPy Arrays
NumPy is a powerful library in Python that provides support for large, multi-dimensional arrays and matrices, along with a collection of mathematical functions to operate on these arrays efficiently.
One of the key advantages of NumPy is its ability to perform fast and efficient function evaluation on arrays, making it a preferred choice for scientific computing and data analysis tasks.
Understanding NumPy Arrays
Before diving into efficient function evaluation in NumPy arrays, let’s first understand the basics of NumPy arrays. A NumPy array is a grid of values, all of the same type, and is indexed by a tuple
of nonnegative integers. These arrays can be created using various methods, such as converting lists or tuples, using built-in functions, or loading data from files.
import numpy as np
# Creating a NumPy array from a list
my_list = [1, 2, 3, 4, 5]
my_array = np.array(my_list)
# Output: [1 2 3 4 5]
Efficient Function Evaluation
NumPy provides a wide range of mathematical functions that can be applied element-wise on arrays, without the need for explicit loops. This allows for efficient and vectorized computations, resulting
in significant performance improvements compared to traditional Python loops.
For example, let’s consider the task of calculating the square of each element in a NumPy array. Using a traditional Python loop, we would need to iterate over each element and perform the
import numpy as np
my_array = np.array([1, 2, 3, 4, 5])
squared_array = np.zeros_like(my_array)
for i in range(len(my_array)):
squared_array[i] = my_array[i] ** 2
# Output: [ 1 4 9 16 25]
However, with NumPy’s efficient function evaluation, we can achieve the same result in a much simpler and faster way:
import numpy as np
my_array = np.array([1, 2, 3, 4, 5])
squared_array = np.square(my_array)
# Output: [ 1 4 9 16 25]
In this example, the np.square() function is applied element-wise on the input array, resulting in the squared values of each element. This approach eliminates the need for explicit loops and takes
advantage of optimized C code behind the scenes, leading to improved performance.
Related Evidence
Several benchmarks and studies have demonstrated the efficiency of NumPy’s function evaluation on arrays. For instance, a study conducted by Stéfan van der Walt, the creator of NumPy, compared the
performance of NumPy’s vectorized operations with traditional Python loops. The results showed that NumPy’s vectorized operations were significantly faster, sometimes even by several orders of
Furthermore, NumPy’s efficient function evaluation has been widely adopted in various scientific and data analysis libraries, such as SciPy, Pandas, and scikit-learn. These libraries leverage NumPy’s
capabilities to perform complex computations on large datasets efficiently, enabling faster and more scalable data processing.
In conclusion, NumPy’s efficient function evaluation in arrays offers a powerful tool for scientific computing and data analysis tasks. By leveraging vectorized operations, it allows for faster and
simpler computations, leading to improved performance and productivity.
Efficient function evaluation in NumPy arrays is a crucial aspect of optimizing code performance in scientific computing. NumPy is a powerful library in Python that provides support for large,
multi-dimensional arrays and matrices, along with a collection of mathematical functions to operate on these arrays efficiently.
One common task in scientific computing is evaluating a mathematical function on a large array of values. For example, let’s say we have an array of numbers and we want to calculate the square root
of each element. Using a loop to iterate through each element and applying the square root function individually can be inefficient and time-consuming.
import numpy as np
# Example array
arr = np.array([1, 4, 9, 16, 25])
# Using a loop to calculate square root
result = np.zeros_like(arr)
for i in range(len(arr)):
result[i] = np.sqrt(arr[i])
However, NumPy provides a more efficient way to evaluate functions on arrays using vectorized operations. These operations apply the function to each element of the array simultaneously, resulting in
faster execution.
# Using vectorized operation to calculate square root
result = np.sqrt(arr)
By using the vectorized operation, we eliminate the need for a loop and let NumPy handle the computation efficiently. This approach is not only faster but also more concise and readable.
NumPy also provides a wide range of mathematical functions that can be efficiently applied to arrays. These functions include trigonometric functions, logarithmic functions, exponential functions,
and more. By leveraging these functions and vectorized operations, we can perform complex calculations on large arrays with minimal effort.
Reference links:
– NumPy documentation: https://numpy.org/doc/
– NumPy tutorial: https://numpy.org/doc/stable/user/quickstart.html
Efficient function evaluation in NumPy arrays is essential for optimizing code performance in scientific computing. By leveraging vectorized operations and the extensive collection of mathematical
functions provided by NumPy, we can perform complex calculations on large arrays efficiently and concisely. This approach not only improves execution speed but also enhances code readability and
maintainability. NumPy is a powerful tool that enables efficient computation on arrays, making it a valuable library for scientific computing in Python.
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Communication costs of Strassen's matrix multiplication
Algorithms have historically been evaluated in terms of the number of arithmetic operations they performed. This analysis is no longer sufficient for predicting running times on today's machines.
Moving data through memory hierarchies and among processors requires much more time (and energy) than performing computations. Hardware trends suggest that the relative costs of this communication
will only increase. Proving lower bounds on the communication of algorithms and finding algorithms that attain these bounds are therefore fundamental goals. We show that the communication cost of an
algorithm is closely related to the graph expansion properties of its corresponding computation graph. Matrix multiplication is one of the most fundamental problems in scientific computing and in
parallel computing. Applying expansion analysis to Strassen's and other fast matrix multiplication algorithms, we obtain the first lower bounds on their communication costs. These bounds show that
the current sequential algorithms are optimal but that previous parallel algorithms communicate more than necessary. Our new parallelization of Strassen's algorithm is communication-optimal and
outperforms all previous matrix multiplication algorithms.
Dive into the research topics of 'Communication costs of Strassen's matrix multiplication'. Together they form a unique fingerprint.
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harcourt math practice workbook grade 1 pdf
Grade 4 - Houghton Mifflin Harcourt 2011-07-14 Trophies Grade 2-1 Practice Book : Just for You - 2003-12 NC State textbook adoption 2006-07. A full 36 weeks math book for first graders aimed at ages
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added pointer to
• Nick Benton, Gavin Bierman, Valeria de Paiva, §8 of: Term assignment for intuitionistic linear logic, Technical report 262, Computer Laboratory, University of Cambridge (August 1992) [pdf&
(of which the published BBPH92 seems to be just an abridged version)
diff, v41, current
added a couple of more original references
[edit: got interrupted, will do more polishing tomorrow…]
diff, v40, current
Added dereliction rule
diff, v36, current
added a section (here) on the realization of the exponential modality in linear homotopy type theory.
diff, v35, current
I have recovered this reference here via the WaybackMachine:
• Gavin Bierman, On Intuitionistic Linear Logic, Cambridge (1993) [pdf]
Unfortunately, the pdf is just a tad larger than the upload limit for the nLab server. I have tried to compress it a little, but couldn’t quite bring it below the threshold.
So for the time being the pdf is sitting in my Dropbox.
diff, v34, current
also polished up this bibitem:
• Nick Benton, Gavin Bierman, Valeria de Paiva, Martin Hyland, Linear $\lambda$-Calculus and Categorical Models Revisited, in Computer Science Logic. CSL 1992, Lecture Notes in Computer Science 702
, Springer (1993) [doi:10.1007/3-540-56992-8_6]
diff, v34, current
I have fixed and completed this bibitem:
• Nick Benton, A mixed linear and non-linear logic: Proofs, terms and models, in Computer Science Logic. CSL 1994, Lecture Notes in Computer Science 933 [doi:10.1007/BFb0022251, pdf]
here and in a couple of other entries.
(It also appears at linear-non-linear logic and still needs to be reformatted there, but that entry is suffering from some bug which prevents it from being edited. I have contacted the technical team
about it.)
diff, v34, current
added cross-links by !include-ing the logic symbols – table
added pointer to:
• Daniel Mihályi, Valerie Novitzká, Section 2.2 of: What about Linear Logic in Computer Science?, Acta Polytechnica Hungarica 10 4 (2013) 147-160 [pdf]
diff, v33, current
also pointer to:
• Jennifer Paykin, Linear/non-Linear Types For Embedded Domain-Specific Languages, 2018 (upenn:2752)
diff, v30, current
Added these pointers:
A quantum programming language based on this linear/non-linear type theory adunction is QWIRE:
• Jennifer Paykin, Robert Rand, Steve Zdancewic, QWIRE: a core language for quantum circuits, POPL 2017: Proceedings of the 44th ACM SIGPLAN Symposium on Principles of Programming LanguagesJanuary
2017 Pages 846–858 (doi:10.1145/3009837.3009894)
applied to verified programming after implementation in Coq:
• Robert Rand, Jennifer Paykin, Steve Zdancewic, QWIRE Practice: Formal Verification of Quantum Circuits in Coq, EPTCS 266, 2018, pp. 119-132 (arXiv:1803.00699)
and using ambient homotopy type theory:
• Jennifer Paykin, Steve Zdancewic, A HoTT Quantum Equational Theory (arXiv:1904.04371)
diff, v29, current
@Mike I think the paper you wanted to remember is Jacobs' https://www.sciencedirect.com/science/article/pii/0168007294900205
adding category:logic
Valeria de Paiva
diff, v28, current
Clarified relationship between the various definitions.
diff, v27, current
Added some citations for construction of cofree !-modalities
diff, v26, current
Generalized the example to lift any linear-nonlinear adjunction to a Chu construction.
diff, v25, current
Added the missing nullary condition for a Seely !-modality.
diff, v24, current
Thanks. I don’t think that’s the paper I was half-remembering, but it’s useful to have.
This paper does something like that, though for the closely related “tensor logic” rather than linear I think.
I have a vague memory of reading a paper that decomposed the !-modality into two separate modalities, one that allows only contraction and another that allows only weakening. Does anyone know a
reference doing something like that?
Added an example: the Chu construction of any cartesian monoidal category has an idempotent !-modality, namely the coreflection into the original category.
diff, v22, current
Other Kleisli article
Ammar Husain
diff, v20, current
I’m currently on holiday, and unfortunately will not be able to look at the bug for at least a week. Would it be possible to create a single thread ’Bugs and feature requests’ where all bugs and
feature requests are registered (and maybe a note could be added to the HowTo to this effect)? It is then much less likely that I will forget about or overlook something. To keep it easy to get an
overview, maybe discussion can be avoided in that thread, and instead when registering the bug one can link to a thread where one can discuss it?
OK, have changed a number of links.
Thanks for the alert. I am fixing it in the entries. “exponential modality” is a better link word, anyway.
But of course there remains a bug which would be good to fix in any case. So I have moved the non-rendering “!-modality”-link to the Sandbox.
It seems that !-modality doesn’t render as a link on some pages, e.g., at Fock space. Is there a problem starting with ’!’ ?
Added Hyland-Schalk reference
diff, v18, current
I added to !-modality some comments about term calculi.
I de-stubbified !-modality with some discussion of the various ways to interpret it categorically.
collected some references on the interpretation of the !-modality as the Fock space construction at !-modality.
Cross-linked briefly with he stub entries_Fock space_ and second quantization.
Am trying to bring parts of this entry into better shape.
Have adjusted the wording in the Idea section (here) for streamlining
and expanded by quickly indicating, right in the Idea-section, what I think is the most fundamental elementary example (sending a vector space to the linear span of its underlying set) – this in to
give up-front more of an idea of what the exponential modality is about.
In particular I highlighted in the idea-section and also further below in a numbered remark (here) what is exponential about the “exponential modality”.
diff, v44, current
Had anyone explicitly looked to connect the !-modality of linear logic to the stabilization adjunction before me here, and then more directly at #62?
It’s been acknowledged in footnote 13 p. 48 here.
Well that’s referring to an existing observation from 1995 by Arone and Kankaanrinta relating a construction in homotopy theory to the exponential function. I meant the explicit thought to relate
linear logic to homotopy theory.
I guess all I was doing was applying transitivity to the ’is like’ relation: linear logic is like linear algebra, linear algebra is like stable homotopy theory.
In a discussion of that very premise, no?
I doubt the reader would pick up the scope of the footnote as being the broader ongoing discussion rather than the specific point of the paragraph. But it wasn’t an issue of receiving due
acknowledgement. I was simply interested if anyone before had directly said that there should be a linear logic-stable homotopy theory connection.
added a brief remark (here) that pointed classical homotopy types (“spaces”) are comonadic over linear homotopy types (“spectra”), from
• Jacobson R. Blomquist, John E. Harper, Thm. 1.8 in: Suspension spectra and higher stabilization [arXiv:1612.08623]
(hat tip to MO:a/333720)
This deserves being expanded, but I need to interrupt.
diff, v52, current
this is also Thm. 3.11 in:
• Kathryn Hess, Magdalena Kedziorek, The homotopy theory of coalgebras over simplicial comonads, Homology, Homotopy and Applications 21 1 (2019) [arXiv:1707.07104, doi:10.4310/
diff, v53, current
I am curious about the parallel drawn between Set and Vect in the Idea section. “Morally”, sets are affine spaces over the field on one element, with pointed sets as its vector spaces. How do affine
logics tie into this? Conceptually, I think of a category of pointed types as the “type of terms” or “type of proofs” for a type theory, especially seen as a type indexed by Type and displayed by the
identity functor on Type. Can anyone elaborate on how this conceptualization extends to linear and affine types? I haven’t studied much about affine type theories; I just know that they drop
contraction, and I don’t understand its semantics in relation to the semantics of linear or dependent type theories.
On your first question I suppose there is the trivial answer by observing that classical (intuitionistic) logic is a form of affine logic: If we just forget that sets satisfy a contraction rule, then
what remains is their affine logic.
While I don’t know if it is relevant for your concern, just to notice that exponential modalities in the context of affine logic are discussed in Shulman 22.
Thank you. I started digging into that paper shortly after my post here, so right now I am digesting the dual roles played by refutations and demonstrations.
I liked your angle of combining the perspective of sets as affine $\mathbb{F}_1$-modules with the idea of them being models of affine logic. There may be more to this, but I am just not sure if the
exponential modality is the place to see this at work.
My gut says this has something to do with Goodwillie derivatives of monoidal structures. While I have not verified it yet, I have heard that the symmetric smash product of spectra may be defined as
the (total?) Goodwillie derivative of the cartesian product bifunctor. If anyone has a reference that elaborates on this, I would be very appreciative. The exponential modality interacts harmoniously
with the Goodwillie calculus, and I wonder about the broader potential for “Goodwillie smoothness” to help us understand the big picture. (I’m wondering if I should just start a conversation about
all this in some other thread, in case it is slightly tangential (haha) here?)
The exponential modality interacts harmoniously with the Goodwillie calculus
That’s what I was sensing too. If you can make something out of the vague ideas mentioned here and below that, I’d love to hear.
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Radius of Circle - Formula, Definition
A circle is an essential geometric shape which has many practical uses in various fields, such as engineering, physics, architecture, and math. The radius of a circle is one of its most important
properties and plays an important role in figuring out other dimensions, such as the area and circumference of the circle.
In this log, we will study the idea of the radius of a circle in detail, including its formula, definition, and how it is utilized in several domains. We will also discuss the importance of
understanding the radius of a circle in working out mathematical and physical problems.
By the end of this blog article, you will get a clear grasp of what the radius of a circle is, how to calculate it, and its significance in practical uses. Whether you are a learner studying geometry
or a professional working in affiliated to field, understanding the radius of a circle is important for success among other things.
Definition of Radius
The radius of a circle is the distance from the middle of the circle to any point on its border. It is represented by the alphabet "r" and is an important measurement utilized to work out the size
and position of the circle. The radius is forever half of the diameter of the circle, which is the length through the circle passing over its center.
Formula for Figuring out the Radius
The formula for finding the radius of a circle is simple and streamline. It is given by:
r = d / 2
where "r" is the radius of the circle and "d" is its diameter. This formula is derived from the definition of the radius as half of the diameter.
One more way to find the radius of a circle is by applying the formula:
r = √(A/π)
where "A" is the area of the circle and "π" is the mathematical constant pi (approximately equivalent to 3.14). This formula can be useful where the area of the circle is known, but its diameter is
Examples of Determining the Radius
Let's observe come examples of how to utilize the formula for determining the radius of a circle:
Example 1:
A circle has a diameter of 10 cm. What is its radius?
Applying the formula, we possess:
r = d / 2
r = 10 / 2
r = 5 cm
Thus, the radius of the circle is 5 cm.
Example 2:
A circle has an area of 78.5 square centimeters. What is its radius?
Utilizing the formula, we get:
r = √(A/π)
r = √(78.5/π)
r ≈ 5 cm
Hence, the radius of the circle is about 5 cm.
Importance of the Radius of a Circle
The radius of a circle is an essential measurement which is utilized in a broad spectrum of domains, involving geometry, physics, engineering, and architecture. In geometry, the radius is essential
for calculating the circumference and area of a circle. The circumference is the length around the edge of a circle, while the area is the volume of space confined by the circle. Both of these
calculations need the radius to be well-known.
In physics and engineering, the radius is utilized to calculate the size and position of circular objects, for example, gears, wheels, and cylinders. It is further utilized in optics to figure out
the focal length of a curved mirror or lens. In architecture, the radius is applied to plan and construct circular structures and buildings, for example, arches, domes, and rotundas.
The radius of a circle is further crucial in computer graphics and animation, where it is [[used|129] to make 3D and 2D shapes. It is further utilized in machine learning algorithms for image
recognition and feature detection.
General Mistakes in Determining the Radius
When determining the radius of a circle, it is essential to avoid frequent errors which can open to faulty workings. One general error is confusing the radius with the diameter. While the diameter is
the distance across the circle passing through its center, the radius is the distance from the center to any point on its edge. Hence, it is essential to make sure that the correct measurement is
utilized when figuring out the radius.
Another mistake that is general made when figuring out the radius is forgetting to divide the diameter with two when utilizing the formula. The formula for finding the radius of a circle is r = d/2,
where r is the radius and d is the diameter. Forgetting to divide the diameter with two can result in an incorrect value for the radius.
It is also crucial to utilize the correct units of measurement while figuring out the radius. For example, if the diameter is calculated in inches, the radius must further be determined in inches.
Using various units of measurement for the diameter and radius can ensure in wrong calculations.
By preventing these ordinary errors and double-checking calculations, individuals can make sure that they achieved precise values for the radius of a circle. This is important in several fields, for
example, mathematics, engineering, physics, and architecture, where strict measurements are crucial for accurate workings and designs.
The radius of a circle is a fundamental measurement used in many fields, involving math, engineering, physics, and architecture. It is defined as the distance from the center of the circle to any
spot on its edge and could be found using simple formulas. Comprehending the definition and formula for finding the radius of a circle is crucial for success in these domains.
By preventing common mistakes and grasping the importance of the radius of a circle, individuals can enhance their grasp of geometry and its applications in practical situations. If you require help
comprehending the radius of a circle or any other math theory, contemplate call us at Grade Potential Tutoring. Our experienced tutors are accessible remotely or face-to-face to offer personalized
and productive tutoring services to guide you succeed. Contact us right now to schedule a tutoring lesson and take your math skills to the next stage.
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Variable Speed Limit Control for the Motorway–Urban Merging Bottlenecks Using Multi-Agent Reinforcement Learning
Department of Control for Transportation and Vehicle Systems, Faculty of Transportation Engineering and Vehicle Engineering, Budapest University of Technology and Economics, Műegyetem rkp. 3, H-1111
Budapest, Hungary
Author to whom correspondence should be addressed.
Submission received: 31 May 2023 / Revised: 17 July 2023 / Accepted: 19 July 2023 / Published: 24 July 2023
Traffic congestion is a typical phenomenon when motorways meet urban road networks. At this special location, the weaving area is a recurrent traffic bottleneck. Numerous research activities have
been conducted to improve traffic efficiency and sustainability at bottleneck areas. Variable speed limit control (VSL) is one of the effective control strategies. The primary objective of this paper
is twofold. On the one hand, turbulent traffic flow is to be smoothed on the special weaving area of motorways and urban roads using VSL control. On the other hand, another control method is provided
to tackle the carbon dioxide emission problem over the network. For both control methods, a multi-agent reinforcement learning algorithm is used (MAPPO: multi-agent proximal policy optimization). The
VSL control framework utilizes the real-time traffic state and the speed limit value in the last control step as the input of the optimization algorithm. Two reward functions are constructed to guide
the algorithm to output the value of the dynamic speed limit enforced within the VSL control area. The effectiveness of the proposed control framework is verified via microscopic traffic simulation
using simulation of urban mobility (SUMO). The results show that the proposed control method could shape a more homogeneous traffic flow, and reduces the total waiting time over the network by 15.8%.
In the case of the carbon dioxide minimization strategy, the carbon dioxide emission can be reduced by 10.79% in the recurrent bottleneck area caused by the transition from motorways to urban roads.
1. Introduction
As an important transportation infrastructure, motorways represent the overall level of a country’s transportation system to a large extent and play a vital role in the development of the national
economy. However, motorways are also facing increasingly frequent traffic congestion [
]. As a road section that is prone to generating and spreading congestion, the traffic bottleneck area is an important research object to improve motorway traffic management capabilities. According
to the specific time and location of occurrence, traffic bottleneck areas can be classified into recurring and nonrecurring bottleneck areas [
]. Recurring traffic bottleneck areas are the confluence area, diversion area, and weaving area formed when the main line of the motorway merges with the entrance ramp and exit ramp. Congestion
frequently occurs in recurring traffic bottleneck areas [
]. The closure of lanes due to road construction, bad weather, or traffic accidents has stochastic characteristics, which are the causes of nonrecurring bottleneck areas [
Compared with motorways, the distribution of entrance and exit ramps at the merging area of motorways and urban roads is more concentrated, and adjacent entrances and exits are more closely connected
with urban roads, resulting in more traffic conflicts in the merging area of motorways and urban roads [
]. Compared with urban roads, the merging area of motorways and urban roads often service more traffic demand due to the closed traffic environment and connectivity of settlements and cities,
resulting in a broader range of traffic bottlenecks and faster spread speed [
As a part of the intelligent transportation system (ITS), variable speed limit (VSL) control has been widely used as one of the traffic control measures to improve traffic efficiency [
], benefit the environment [
], and enhance traffic safety [
] in bottleneck areas. By adjusting the speed limit of the main line upstream of the bottlenecks, the number of vehicles entering the motorway bottleneck area during the congestion period is
controlled within a certain range to keep the traffic state more uniform and stable. VSL control methods in traffic management can be divided into different categories according to their basic
approaches and techniques. The most used VSL control strategies are rule-based approaches [
] and model-based approaches. Rule-based VSL control approaches use a set of predefined rules and thresholds to determine the appropriate speed limits. These rules take into account factors such as
traffic density, volume, occupancy, and historical data to set speed limits. Rule-based VSL approaches may cause traffic flow fluctuations when the traffic state exceeds or falls below the set
thresholds. Model-based VSL control approaches utilize mathematical models to represent and predict traffic behavior. Model-based approaches using optimization algorithms to determine the optimal
speed limits. Model-based approaches are further classified into open-loop optimization approaches [
] and feedback control approaches [
]. Model-based approaches require accurate models to describe the state of traffic flow. Due to the need for a large number of numerical calculations and the fact that the model contains many
parameters to be calibrated, the model-based VSL control approaches have limitations in large-scale traffic control applications. In addition, the portability of these VSL control approaches needs
further work. This means when applying the control algorithm to a new scenario, the traffic flow model needs to be re-calibrated, and the controller parameters need to be adjusted manually.
With the development of artificial intelligence technology, VSL control based on reinforcement learning (RL) can automatically adapt to various traffic environments and achieve optimal control
effects without specific traffic flow models [
]. Existing RL-based VSL control approaches use the matured deep Q-network (DQN) developed in 2015 [
]. The continuous development of reinforcement learning research has brought new solutions to RL-based VSL control problems.
This study proposes a fully cooperative multi-agent reinforcement learning framework to solve the multi-section VSL control optimization problem at the merging area of motorways and urban roads. This
framework is based on the multi-agent proximal policy optimization (MAPPO) algorithm proposed in 2021 [
], which has been proven powerful for target localization [
], production scheduling [
], and trajectory planning [
] but has not been utilized to solve the VSL optimization problem. Under the constraints of rational speed, traffic flow state information, action representing discrete speed limit values, and reward
composed of occupancy data and carbon dioxide emissions are designed to smooth traffic and reduce emissions. The proposed multi-agent reinforcement learning framework is tested on the microscopic
traffic simulator simulation of urban mobility (SUMO) [
2. Methodology
This study presents a VSL control framework, applying multi-agent reinforcement learning (MARL) with the proximal policy optimization (PPO) algorithm.
Section 2.1
introduces the literature background of RL.
Section 2.2
presents the basis of the PPO algorithm.
Section 2.3
extends the basic PPO algorithm to MARL.
2.1. Reinforcement Learning
Reinforcement learning (RL) is a machine learning (ML) algorithm. During the learning process, the agent obtains the optimal strategy by trying different action choices and adjusting the evaluation
value of the action according to the feedback of the environment. During each interaction between the agent and the environment, the input to the agent is the environment state
. The agent chooses action
as the output to transit the state of the environment to state
$s ′$
, and at the same time, the agent receives reward
. RL aims to find a series of optimal action sets [
]. Deep learning (DL) is a method based on representational learning of data in ML, which abstracts the features contained in the data based on the value of the original data for representation [
]. The core of DL is the deep neural network (DNN), inspired by the principles of neural networks (NNs) in biology. NN abstracts the structure of the human brain and the response mechanism to
external stimuli in a mathematical model.
Deep reinforcement learning (DRL) is an algorithm that combines RL with the ability to learn from the environment and DL with powerful representation capabilities. DRL has two branches: Q-learning
and policy gradient (PG) algorithms [
]. Q-learning is mainly based on the iteration of the action–value function to model the value of the state space to find the optimal strategy. The Q-learning algorithm has excellent interpretability
and debuggability but relies on extensive experience sample storage to establish an accurate model. PG directly outputs the probability of the action. Since no additional state–action value is
introduced, PG requires fewer hyperparameters to be tuned, and due to the gradient method utilized, PG will update toward the direction of the optimization strategy, which means it has good
convergence. But the disadvantage is that PG can easily converge to a local optimal value.
The actor–critic (AC) algorithm [
] combines the advantages of Q-learning and PG. AC uses the single-step update advantage of Q-learning to allow the Q-network as a critic to learn offline (i.e., an agent that learns through
experience samples that are not acquired by itself). The critic provides an evaluation of the actor’s action or the direction of the gradient descent in a single-step policy update, which greatly
improves the update frequency of the actor and accelerates the learning rate. In this way, the problem of the low efficiency of the PG in episode updates can be solved, but there is a problem of
convergence difficulty. To solve the difficulty of convergence of the AC algorithm and speed up training, the asynchronous advantage actor–critic (A3C) algorithm [
] pushes the AC algorithm into multi-threads for synchronous training. The emerging trust region policy optimization (TRPO) algorithm [
] solves the problem of fluctuations in the A3C algorithm when balancing the variance and bias of the model.
2.2. Proximal Policy Optimization
Based on PG and TRPO, the proximal policy optimization (PPO) algorithm [
] is proposed as an improved method. The PPO algorithm solves the shortcomings of previous RL algorithms, such as low data utilization efficiency, poor robustness of the PG algorithm, and the
complexity of the TRPO algorithm. The PG algorithm is based on calculating the estimator of the policy gradient and substituting it into the stochastic gradient ascent algorithm. The unbiased
estimate of the gradient is
$δ J ^ ( θ ) = E t ^ [ δ θ l o g π θ ( a t | s t ) A ^ t ]$
$π θ$
is a random strategy for actor network parameters
$A ^ t = Q t − V t$
is the advantage function estimated by critic;
$Q t$
is the state–action value function; and
$V t$
is the state–value function.
$a t$
is the action space, and
$s t$
is the state space. However, PG needs to re-interact with the environment after each update of the parameters, calculate the advantage function of the new strategy, and then update the parameters,
making the update speed slow. The TRPO algorithm uses the importance sampling method. TRPO adopts the new and old strategy networks
$θ o l d$
to allow the network to interact with the environment and use the collected data to train the network
so that the agent can perform multiple parameter updates in one interaction with the environment, which improves the update speed.
In addition, to prevent the gap between the new network
and the old network
$θ o l d$
from being too large, the relative entropy is used to measure the difference between the two, then the objective function is
$m a x θ J ( θ ) = E t ^ π θ ( a t | s t ) π θ o l d ( a t | s t ) A ^ t ,$
$s u b j e c t t o E t ^ [ K L [ π θ o l d ( a t | s t ) , π θ ( a t | s t ) ] ] ⩽ ζ$
is the radius of the trust domain, and
$K L$
is the boundary conditions. The PPO adds the
of TRPO as a penalty item to the objective function and applies an adaptive penalty of the
parameter to the
divergence. The actor’s objective function is
$m a x θ J K L ( θ ) = E t ^ π θ ( a t | s t ) π θ o l d ( a t | s t ) A ^ t − β D K L [ π θ o l d ( a t | s t ) | | π θ ( a t | s t ) ]$
$D K L ( π θ o l d | | π θ )$
represents the divergence of
$K L$
between the
$π θ o l d$
and the
$π θ$
controls the difference in each episode of policy updates.
is adaptively adjusted according to the preset
$K L$
divergence threshold. In practice, the clip function is used to limit the probability ratio
$r t ( θ )$
$( 1 − ε , 1 + ε )$
, where
is a hyperparameter, also taking
$π θ π θ o l d A ^ t$
as the optimization object. When
$A ^ t > 0$
, increase the probability of
$π θ ( a t | s t )$
and vice versa. The actor’s objective function is
$m a x θ J c l i p ( θ ) = E t ^ c l i p π θ ( a t | s t ) π θ o l d ( a t | s t ) , 1 − ε , 1 + ε A ^ t , π θ ( a t | s t ) π θ o l d ( a t | s t ) A ^ t$
The optimization framework of the PPO algorithm for traffic control is shown in
Figure 1
2.3. Multi-Agent Proximal Policy Optimization
An individual agent is inadequate to handle multi-section VSL control. For this reason, the multi-agent reinforcement learning (MARL) framework is developed. Based on the centralized training with
decentralized execution (CTDE) framework, the PPO algorithm is extended to the MAPPO algorithm. That is, each individual PPO agent is trained using a global value function. After the training is
completed, each individual PPO agent obtains a decentralized strategy, and action is taken locally based on this strategy. Then, centralized training is applied to make the decentralized strategies
learned by individual PPO agents work cooperatively. In the fully cooperative MAPPO algorithm, all individual PPO agents share common reward signals. During centralized training, a global critic
network is used to evaluate the state value and calculate
$A ^ t i$
using the generalized advantage estimation (GAE) method. During decentralized execution, each individual PPO agent relies on its local observation state to realize a distributed decision-making
interactive environment. The global critic network uses GAE to estimate a common
$A ^ t$
based on the reward
$R t i$
obtained by each agent’s action trajectory during centralized training. The objective function for agent
$m a x θ J i ( θ ) = E t ^ c l i p π θ ( a t i | s t i ) π θ o l d ( a t i | s t i ) , 1 − ε , 1 + ε A ^ t i , π θ ( a t i | s t i ) π θ o l d ( a t i | s t i ) A ^ t i$
3. Simulation Environment
Unlike other control algorithms, the DRL algorithm does not directly learn the model from a given data set but learns by interacting with the environment continuously to generate data. Implementing
the training of the algorithm directly in the real traffic network is not feasible, which will bring huge costs and safety hazards. The usual feasible method is to train the RL algorithm with the
traffic simulation platform and then migrate the trained model to the real traffic network. In addition, the multi-agent RL framework requires an environment to interact. To implement the proposed
multi-agent reinforcement learning framework, an accurate and efficient traffic simulation platform is essential. The simulation of urban mobility (SUMO) is an open-source, multi-modal traffic
simulator that can be extended dynamically and is highly customizable through the embedded traffic control interface (TraCI).
To test the control performance of the proposed multi-agent reinforcement learning framework, a dynamic traffic network representing the merging area when the motorway transited to the urban roads i
sgenerated based on the OpenStreetMap (OSM) data.
Figure 2
shows the schematic diagram of the proposed MAPPO-based VSL control at the typical recurrent bottleneck area located on the border of Budapest in Hungary. The intersection with the ramps and the
existence of traffic lights in the urban road network leads to frequent lane-changing behavior of vehicles in the merging area, which further leads to congestion.
The geometric structure of this bottleneck area in SUMO is shown in
Figure 3
. The control section is divided into seven homogeneous sections, named “section_a”, “section_b”, “section_c”, “section_d”, “section_e”, “section_f”, and “section_g”, each of which is one kilometer
long. There are four vehicle routes in the road network. Two of them start from the motorways and end on the urban roads Alkotás Street and Hegyalja Street, while the other two are the on-ramp flow
starting from Egér Street. According to on-site historical measurement data, traffic flows are assigned to the road network at a rate of 2/3, with more vehicles traveling to Alkotás Street. The total
demand on the main road is 4657 veh/h and 1039 veh/h on the ramp. Two traffic light control systems exist at the end of this regional network. The control schema is the fixed cycle with 90 s cycle
time. The green phase is 60 s, and the yellow phase is 3 s.
To connect the dynamic traffic environment in SUMO with the MAPPO algorithm, a custom environment “SUMO_Env()” is created in PyCharm, which contains the main subset functions needed for the training,
initialization simulation, state representation, action space, reward, termination, reset, and step functions.
The initialization of the simulation function initializes the settings of the simulation environment. The simulation time is set to 7200 s. The original speed limit of the road network is set to
27.78 m/s (100 km/h) for motorways and 13.89 m/s (50 km/h) for urban roads according to the actual traffic rules.
The states obtained by RL agents in different environments are various. For example, in the robot simulation environment, the state representation is mainly composed of joint dynamics data in the
physical sense [
]. In the game task, the state information agents obtained from the environment are almost exclusively image data [
]. In financial decision tasks, the state information includes stock market bonds, various K-line indicators, transactions, and financial data [
]. Detailed state representation helps RL algorithms extract critical information. The state-space representation in this paper is the vehicle occupancy data in each section, including main roads and
The action space represents the actions of the PPO agents applied to the traffic environment. Six discrete speed limit values make up the action space in this study for the dynamic traffic
environments, which is $[ 36.11 , 30.55 , 25 , 19.44 , 13.89 , 8.3 ]$, the unit is m/s. The goal of agents is to cooperate with each other to find the optimal speed limits set for different road
The reward function setting is very critical in RL problems. The reward function represents the optimization goal of the training. It is the only feedback from the environment to the actions that the
agent takes. Commonly used reward functions in RL-based VSL control are total time spent (TTS) [
], density distribution [
], mean speed in the bottleneck [
] that is utilized to improve traffic efficiency and the negative sum of the values of tail gas emissions [
] that take the environmental sustainability into account. In this paper, two reward functions are introduced to smooth the traffic flow and reduce traffic emissions. The first reward function is the
negative of the standard deviation of the vehicle occupancy data on different road sections. The second function is the negative sum of the scaled values of carbon dioxide emissions.
The termination of training depends on the training iteration and simulation. When the training iteration reaches the preset value and the simulation time reaches the predefined value, the connection
between the MAPPO framework and SUMO simulation is closed.
The reset function is responsible for resetting the environment to its initial state before the start of a new episode. It sets the state of each agent in the environment to its initial state.
The step function executes a single step in the environment for each PPO agent and collects the resulting observations, rewards, and speed limits. The step function receives the actions chosen by
each individual PPO agent and applies them to the environment. After applying the actions, the step function collects the current state of the environment, which is used as input for the agent’s
policy update. The step function calculates the rewards obtained by each agent based on the predefined reward function. The step function determines if the current state of the environment is a
terminal state, indicating the end of the episode. The step function also collects the speed limits chosen by the agents, which can be considered the control input and can be used for further
4. Results and Discussion
In the sequel, the simulation results are introduced in detail, and relevant discussion is provided as well.
4.1. Training Setting of MAPPO Algorithm
To select the appropriate variables for the training process, hyperparameters need to be properly tuned. The set of MAPPO hyperparameters utilized in this paper is shown in
Table 1
The training performance is verified by the reward curve through 600 iterations. In each iteration, a random seed is applied for the proposed control strategies to keep the diversity. Seven PPO
agents are used in this paper, and each agent controls a road section; they work cooperatively to minimize the reward function. The GAE $λ$ parameter represents the parameter used in GAE methods that
controls the trade-off between bias and variance in the advantage estimation. The number of PPO epochs per update determines how many times the network parameters are updated based on the collected
data during a single training iteration. The number of hidden layers in the neural network is 3, the number of neurons in the hidden layer is 64, and the activation function of the hidden layer is
the ReLu function, which helps with the vanishing gradient problem, allows the network to learn non-linear relationships, and provides sparse activation.
Traffic performance and sustainability measurements are taken in a dynamic traffic environment based on the control output obtained by different reward settings. The traffic performance measurement,
including the occupancy data of each road section and its distribution, shows the effect of the proposed control framework in smoothing the traffic. The sustainability measurements are presented by
the reduction in carbon dioxide emissions.
4.2. Traffic Performance Measurements
Figure 4
shows the learning curve over 600 iterations, utilizing the negative standard deviation of the vehicle occupancy of different road sections as the reward function. Min-Max normalization is performed
on the obtained rewards. The reward value changes significantly when the number of training iteration reaches about 76. Although there are some oscillations, the reward value is generally stable in
the subsequent training process.
Figure 5
shows, the MAPPO-based VSL controller starts to reduce the speed limit of all controlled sections at step 25. The speed limit is stable between 8.3 m/s and 13.89 m/s after 104 steps. When traffic
congestion is about to be formed at the bottleneck area, the MAPPO-based VSL controller reduces the speed limit of the controlled road sections quickly to 13.89 m/s to prevent capacity drop at the
bottleneck area. In this paper, we assume that drivers’ acceptance of the speed limit is 100%. The proposed speed limit can be spread to the road network in a traditional way, i.e., variable speed
limit signs. With the connected and automated vehicles (CAVs) and intelligent transportation infrastructure on the road network, the proposed speed limit can be synchronously received by CAV with
vehicle-to-vehicle (V2V) communication technology.
The changes in the occupancy data of different road sections during the simulation period without VSL control are shown in
Figure 6
. The first 300 s is the warming-up time. The occupancy change shows that from 2400 s to the end of the simulation, the occupancy of the motorway sections before merging with the ramp varies between
36% and 68%. With the proposed MAPPO-based VSL control, as shown in
Figure 7
, the occupancy of the motorway sections before merging with the ramp and “section_f” reaches almost the same value of 46% at the end of the simulation. The occupancy of “section_e” and “section_g”
reaches the same value of 30%. From this, a conclusion can be made that the controller truly achieves the predefined goal, which is that by controlling the speed limit of each road section, the
vehicles are evenly distributed on each road section.
Figure 8
intuitively demonstrates the distribution of vehicles on the road network without VSL control. It shows that vehicles queued at “section_a”, “section_b”, “section_c”, and “section_e”. This was caused
by merging with the ramp and the backpropagation of congestion occurring in the urban area. After implementing MARL-based VSL control, as shown in
Figure 9
, vehicles began distributing uniformly and concentratedly on all motorway sections. This is because the PPO agents started to reduce the upstream speed limit to obtain the optimal reward defined by
the agent policy. The interquartile range of the box plot shows how the data fluctuate. Compared with the non-implementation of the VSL control case, occupancy fluctuations in the upstream section of
the motorway are significantly smaller when applying the MARL-based VSL control.
4.3. Sustainability Measurements
Figure 10
shows the learning curve over 600 iterations utilizing the negative sum of the scaled carbon dioxide emissions as the reward function. As defined in
Section 3
, the MAPPO-based controller trends to minimize carbon dioxide emissions over the road network.
Figure 11
shows, the MAPPO-based VSL controller started to reduce the speed limit of all controlled sections a bit later than the controller dedicated to smoothing the traffic flow. The final speed limit
chosen by the agents is 8.3 m/s for all sections. This can be explained by the innate character of the SUMO emission model. SUMO utilizes the third version of the Handbook Emission Factors for Road
Transport (HBEFA 3) [
] model to calculate emissions. The homogeneous low-speed traffic flow keeps low pollutant emissions. So the agents choose the same relatively low-speed limit for all road sections.
Figure 12
shows the comparison of the total carbon dioxide emissions in kilograms without VSL control, with only single-section VSL control, and with multi-section VSL control over the simulation time with the
first 300 s of warming up time. The figure verifies the performance of the multi-agent RL-based VSL controller, showing that the proposed controller performs well in terms of the sustainability
metrics. Specifically, compared with the no-control case, the proposed VSL control strategy reduces the total carbon dioxide emissions of the road network by 11.2%. By only controlling the upstream
section close to the bottleneck, the total carbon dioxide emissions are just reduced 0.41%. This means that compared with only adjusting the speed limit of the upstream section close to the
bottleneck, our MARL-based VSL controller reduces the total carbon dioxide emissions by 10.79%.
Figure 13
shows the comparison of the total waiting time in seconds without VSL control, with only single-section VSL control, and with multi-section VSL control. Compared with the single-section control case,
the proposed MARL-based VSL control strategy further reduces the total waiting time through the network by 15.8%. The oscillations in the total waiting time are significantly smaller than in the
uncontrolled and single-section control cases. This shows that the waiting times for vehicles tend to be similar throughout the simulation time, which demonstrate a reduction in the stop-and-go
phenomenon of the vehicles.
Figure 12
Figure 13
demonstrate the superiority of our control strategy compared to the single-section VSL control. Due to the high synergism of the traffic system, single-section VSL control will lead vehicles to
decelerate suddenly before the bottleneck area. Meanwhile, vehicles on the upstream section continuously drive into the bottleneck area at high speed. High-speed cruising and low-speed congestion
contribute to high road network carbon dioxide emissions. The multi-section VSL control divides the upstream of the bottleneck area into multiple deceleration sections. The control idea is to induce
and control the upstream traffic flow to prevent the upstream traffic flow from accumulating quickly in the bottleneck area so that the vehicles can pass through the bottleneck area in an orderly
manner. The reduction in waiting time confirms the feasibility of this control idea.
5. Conclusions
Different from the previous research, which only performs reinforcement learning-based variable speed limit control on the upstream section close to the bottleneck area, this paper proposes a
multi-agent proximal policy optimization strategy for a multi-section variable speed limit control, which implements individual agents on all motorway sections to smooth traffic flow and reduce
traffic emissions. A real-world traffic network is generated in the SUMO simulator based on the on-site historical measurement data, representing a recurrent bottleneck area when motorways transit to
urban roads. The results unequivocally show improvements in the overall vehicle distribution, the total waiting time reduction, the carbon dioxide emissions reduction over the road network and how
the agents apply speed limits that lead to this goal. A more homogeneous traffic flow is formed after applying the proposed MARL-based VSL control. Compared to the existing RL-based control method,
which only controls the speed limit of the upstream section close to the bottleneck, our control idea outperforms. In the case of traffic performance, the total waiting time through the network is
reduced by 15.8%. Regarding sustainability measurement, the total carbon dioxide emissions over the network are reduced by 10.79%.
In conclusion, the paper investigates the effect of two different controllers independently in a credible way. Limited by the iteration speed of MAPPO algorithms and the running speed of the
microscopic traffic simulation, the proposed MARL-based VSL control framework is unable to realize real-time on-site traffic management. As a future work, the proposed MAPPO-based VSL control
framework will be extended to multi-objective optimization to balance traffic performance and emissions by adjusting the weights of multiple reward designs on the designated road network.
Author Contributions
Conceptualization, X.F. and T.T.; methodology, X.F. and T.T.; case study, X.F.; validation, X.F.; writing—original draft preparation, X.F.; writing—review and editing, X.F., T.P. and T.T.;
supervision, T.T.; project administration, T.T. and T.P. All authors have read and agreed to the published version of the manuscript.
Project no. TKP2021-NVA-02 was implemented with the support provided by the Ministry of Culture and Innovation of Hungary from the National Research, Development and Innovation Fund, financed under
the TKP2021-NVA funding scheme. The research was supported by the European Union within the framework of the National Laboratory for Autonomous Systems (RRF-2.3.1-21-2022-00002).
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
The following abbreviations are used in this manuscript:
ITS Intelligent Transportation System
VSL Variable Speed Limit
DRL Deep Reinforcement Learning
DQN Deep Q-Network
MAPPO Multi-Agent Proximal Policy Optimization
SUMO Simulation of Urban Mobility
RL Reinforcement Learning
ML Machine Learning
DL Deep Learning
DNN Deep Neural Networks
NN Neural Networks
PG Policy Gradient
AC Actor–Critic
A3C Advantage Actor–Critic
TRPO Trust Region Policy Optimization
PPO Proximal Policy Optimization
MARL Multi-Agent Reinforcement Learning
CTDE Centralized Training with Decentralized Execution
GAE Generalized Advantage Estimation
TraCI Traffic Control Interface
OSM OpenStreetMap
TTT Total Time Spent
CAV Connected and Automated Vehicles
V2V Vehicle to Vehicle
HBEFA 3 Third Version of Handbook Emission Factors for Road Transport
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Figure 2. Schematic diagram of variable speed limit control at the merging area, where the motorway transitions to the urban road network.
Figure 6. Occupancy data of different road sections during the simulation period without VSL control.
Figure 10. The learning curve dedicated to reducing carbon dioxide emissions over the 600 iterations.
Hyperparameters Value
Number of training iterations 600
Learning rate 0.0005
Number of agent 7
PPO clip parameter $θ$ 0.2
Discount factor $γ$ 0.99
GAE $λ$ parameter 0.95
Time step per update 120
Number of PPO epochs per update 15
Hidden layers 64 × 64 × 64
Hidden layers activation function RELU
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Fang, X.; Péter, T.; Tettamanti, T. Variable Speed Limit Control for the Motorway–Urban Merging Bottlenecks Using Multi-Agent Reinforcement Learning. Sustainability 2023, 15, 11464. https://doi.org/
AMA Style
Fang X, Péter T, Tettamanti T. Variable Speed Limit Control for the Motorway–Urban Merging Bottlenecks Using Multi-Agent Reinforcement Learning. Sustainability. 2023; 15(14):11464. https://doi.org/
Chicago/Turabian Style
Fang, Xuan, Tamás Péter, and Tamás Tettamanti. 2023. "Variable Speed Limit Control for the Motorway–Urban Merging Bottlenecks Using Multi-Agent Reinforcement Learning" Sustainability 15, no. 14:
11464. https://doi.org/10.3390/su151411464
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Predicate Calculus
Today’s Daily Create asks us to “write a poem formatted like computer code about natural networks”. As I started to think about how to do this, I wondered if a GenAI programme might help with
philosophical logic, so I asked ChatGPT to give me an example of predicate calculus. Having ascertained that it could provide something meaningful, I asked it to:
“Please write the following in predicate calculus: bees drink the nectar from the flowers”
Here’s the answer
“Bees drink the nectar from the flowers” in predicate calculus, we need to define some predicates and structure the sentence properly.
Step-by-Step Breakdown:
1. Define the predicates:
□ Let B (x) represent “x is a bee.”
□ Let F (y) represent “y is a flower.”
□ Let D (x,z) represent “x drinks z” (where x is the bee and z is the nectar).
□ Let N (z,y) represent “z is the nectar from flower y.”
2. Express the sentence:
□ “Bees drink the nectar from the flowers” can be written as:
□ Ax Ay Az (B(x) ^ F (y) ^ N (z,y) -> D(x,z))
This means “For all x, y, and z, if x is a bee, y is a flower, and z is the nectar of y, then drinks z.”
This formalizes the idea that bees (represented by x) drink nectar (represented by z) from flowers (represented by y).
That’s poetry to me.
One Response to Predicate Calculus
1. Interesting ….
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Strong solution
From Encyclopedia of Mathematics
of a differential equation
in a domain
A locally integrable function Generalized derivative), and satisfies
almost-everywhere in
The notion of a "strong solution" can also be introduced as follows. A function
if there are sequences of smooth (for example,
In the case of an elliptic equation
both notions of a strong solution coincide.
[a1] J. Chazarain, A. Piriou, "Introduction à la théorie des équations aux dérivées partielles linéaires" , Gauthier-Villars (1981) pp. 223
How to Cite This Entry:
Strong solution. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Strong_solution&oldid=16232
This article was adapted from an original article by A.P. Soldatov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article
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