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https://physics-network.org/how-can-i-make-physics-teaching-interesting/ | 1,695,307,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00425.warc.gz | 516,224,975 | 57,763 | # How can I make physics teaching interesting?
1. Have Confidence in Your Students’ Abilities.
2. Expressing Said Confidence to Students.
3. Allowing More Initiative.
4. Memorisation.
5. Enjoyability of Learning.
6. Set Reasonable Homework.
7. Engaging One-On-One.
8. Staying in Touch With Parents.
## What is the best physics based game?
• Grand Theft Auto V.
• Kerbal Space Program.
• Just Cause 4.
• Crayon Physics Deluxe.
• Totally Accurate Battle Simulator.
• Besiege.
• Line Rider.
• Human Fall Flat.
## How is physics used in games?
Modern video games use physics to achieve realistic behaviour and special effects. Everything from billiard balls, to flying debris, to tactical fighter jets is simulated in games using fundamental principles of dynamics.
## What was the first game with physics?
After a little debugging, the first video game was ready for its debut. They called the game Tennis for Two. Players could turn a knob to adjust the angle of the ball, and push a button to hit the ball towards the other player.
## What is a physics based puzzle game?
Physics puzzle games combine the basics of physics with challenging puzzle action. Use problem solving skills to solve each fun level.
## What is game physics and how it affects the game?
Computer animation physics or game physics are laws of physics as they are defined within a simulation or video game, and the programming logic used to implement these laws. Game physics vary greatly in their degree of similarity to real-world physics.
## How does gravity work in games?
Some video games may have a constant fall speed (which isn’t really gravity) but most have an actual approximation of gravity in them. The coding of gravity in a game is pretty straightforward. You apply a downward acceleration on any objects that are not currently touching (colliding with) another object beneath them.
## How is physics used in the real world?
Physics extends well into your everyday life, describing the motion, forces and energy of ordinary experience. In actions such as walking, driving a car or using a phone, physics is at work. For everyday living, all the technologies you might take for granted exploit the rules of physics.
## How do you introduce students in physics?
1. Engage your students by asking them how they think physics might be related to driving cars.
2. Divide your students into pairs and show your class the slow-motion videos of car wrecks.
3. Explain to your students that Newton’s laws deal with the physics of force and motion.
## How can I understand physics easily?
1. Mastering the basics is necessary: One of the easiest ways of learning physics is to master the basic theories.
2. Simplification is a Key Rule:
3. Create Flashcards:
5. A good tutor makes a BIG difference!
6. Use diagrammatic representations to learn the concepts:
## Is class 11 physics easy?
Studying for Class 11 Physics is not easy. Hence, we at Vedantu, have prepared the solutions for all the chapters of Class 11 accurately to help you during your exam times.
## How is AI used in games?
In video games, artificial intelligence (AI) is used to generate responsive, adaptive or intelligent behaviors primarily in non-player characters (NPCs) similar to human-like intelligence. Artificial intelligence has been an integral part of video games since their inception in the 1950s.
## Is physics important for game development?
yes absolutely physics plays an important role in game dev. because you often observe when you play any kind of game .
## What was the first ragdoll physics game?
Puzzle Games Stair Dismount (originally named Porrasturvat) is notable for being one of the first games to use ragdoll physics as a gameplay feature.
## Who owns Interactive Physics?
Interactive Physics was a physics simulation created in 1989 by Knowledge Revolution, which was founded by David Baszucki, the same man behind Roblox. Windows variant. Macintosh variant. Co-owned with Tencent.
## What physics does Unity use?
Unity comes with dedicated and optimized 2D physics, with many more features and optimizations to power your game. 2D Colliders enable accurate detection of your sprites’ shapes, from primitive to custom shapes. If they also include a Rigidbody 2D, the objects will react to gravity and behave as solid objects.
## How do physics simulations work?
A physics simulation starts with a mathematical model whose variables define the state of the system at a given time. Each variable represents the position or velocity of some part of the system. The heart of a physics simulation is the set of differential equations that describe how the variables evolve over time.
## What is the best physics simulation?
• Giant Army, Spiderling Studios, Brilliant Game Studios.
• Landfall.
• Brilliant Game Studios.
• Spiderling Studios.
• Flashbulb.
• Nolla Games.
• Dry Cactus.
• Giant Army.
## What is Python physics?
Python programming quantum mechanics (Schrödinger’s) and many others are used to model simple or complicated phenomena. By using Python, we’ll show you how to numerically solve these equations. – It gives you independence and self-reliance in analyzing any kind of experimental data.
## What is physical simulation?
Physical simulation is the reproduction, on. a laboratory scale and in real time, of the thermal and mechanical parameters of a real-world production process.
## What is gravity caused by?
Earth’s gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That’s what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.
## How do you add gravity in Java?
1. GRAVITY = 10;
2. TERMINAL_VELOCITY = 300;
3. vertical_speed = 0;
4. public void fall(){
5. this. vertical_speed = this. vertical_speed + GRAVITY;
6. if(this. vertical_speed > TERMINAL_VELOCITY){
7. this. vertical_speed = TERMINAL_VELOCITY; | 1,258 | 5,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.918421 |
https://www.freemathhelp.com/forum/threads/77402-Question-on-Parametric-Equations | 1,516,629,979,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00412.warc.gz | 911,408,775 | 11,536 | # Thread: Question on Parametric Equations
1. ## Question on Parametric Equations
Hey everyone, I am having a hard time understanding a question regarding parametric equations. Basically:
A ladder 16 feet in length slides down a wall as its bottom is pulled away from the wall, using the angle theta as parameter, find the parametric equations for the path followed by the point P located 5 feet from the top of the ladder.
I am unsure as to how to approach/solve this problem. Any feedback and help is appreciated.
Thanks
2. Where is angle theta located?
(By the way, only the top of the ladder is sliding down the wall.)
Introduce a coordinate system. Use Quadrant I.
The y-axis (wall), the x-axis (floor), and the 16-unit ladder form a bunch of right triangles, as the top of the ladder slides down the wall.
Draw a picture of one of these representative triangles. If you drop a vertical line segment from point P to the x-axis (or draw a horizontal line segment from point P to the y-axis), then you have similar triangles.
The hypotenuse of the bigger triangle is always 16 units, and you apparently know one of the acute angles (the parameter).
You know how to express the legs of those triangles in terms of theta, yes?
There is likely more than one approach to this exercise, but can you make a start now?
Please explain what you've thought about or tried, if you would like more guidance.
Cheers
3. ^^^
Angle theta is located at the base of the ladder sliding down the wall, across from the 90 degree angle in a right triangle.
Your method is a bit confusing. We just started learning about parametric equations and derivatives of parametric equations and I fail to see how this problem quite relates to the subject matter. I would appreciate a bit more guidance.
4. Originally Posted by Edder
the base of the ladder sliding down the wall
Perhaps, you mispoke? The base of the ladder is sliding along the floor -- away from the wall.
The top of the ladder is sliding down the wall.
Really, I think that we may freeze time, and just consider the triangle formed at that instant as representative of all the triangles.
Your method is a bit confusing.
I posted no method.
I posted some clues that I hoped would get you thinking.
I fail to see how this problem quite relates to the subject matter.
Ah, thank you for reminding me that you asked for an explanation of the exercise. You may always feel free to do that.
We assume that the bottom of the ladder does not sway to the left or right, but moves in a straight line away from the wall. Hence, point P moves within the plane that contains the ladder, and so we can describe the location of point P using an xy-coordinate system with a line segment representing the ladder. Point P will trace out a curved path, from start to finish.
For example, before the ladder moves, point P is located at (0,11). Do you agree? Likewise, when the ladder stops moving, the location of point P will be (5,0).
We may describe the path of point P with a system of two equations. Each equation will be a function definition, and these functions' input will be the measure of angle theta in radians (a Real number).
The output of the first function will be the x-coordinate of point P, for any given Real number theta within the domain.
The output of the second function will be the y-coordinate of point P, for the same Real number theta.
x(θ) = some expression containing θ
y(θ) = some expression containing θ
The two equations above are parametric equations. They give P's location (x(θ), y(θ)) in terms of the parameter theta. When plotted across the domain, the graph shows the path of P.
Do you remember Right-Triangle Trigonometry? Can you determine what the value of θ is before the ladder moves? When the ladder stops moving, it is obvious what θ equals, yes? (These beginning and ending values define the domain of θ. Stating the domain is a required part of the system of parametric equations, so you should report it.)
Have you drawn a diagram, yet? Do you understand how to draw a representative triangle in Quadrant I? Do you understand how to connect P to the x-axis with a vertical line segment?
In order to attempt clearing up confusion, I first need to know exactly where the confusion lies.
Cheers
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https://www.assignmentexpert.com/homework-answers/mathematics/trigonometry/question-43686 | 1,537,604,195,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158205.30/warc/CC-MAIN-20180922064457-20180922084857-00460.warc.gz | 682,590,172 | 11,960 | 64 635
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# Answer to Question #43686 in Trigonometry for yuba raj
Question #43686
prove that sina.cos2a = 1/4 sin4a.sec a
We can write 1/4 sin(4a)sec(a) as:
1/4 sin(4a)sec(a)=1/4*2sin(2a)*cos(2a)*1/(cos(a))=1/2*2sin(a)*cos(a)*cos(2a)*1/(cos(a))=sin(a)*cos(2a).
Here we use the next formulas: sin(2a)=2*sin(a)*cos(a);
sec(a)=1/(cos(a)).
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prove that sin^3a+ cos^3a/sin a + cos a = 1- 1/2 sin2a | 362 | 919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-39 | latest | en | 0.657305 |
https://countfast.com/product/countfast-9/ | 1,709,541,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00625.warc.gz | 187,643,267 | 58,566 | Math Flash Cards
# CountFast 9
\$5.00
This module introduces a learning tool or trick to help solve a number multiplied by 9 in a cool way. While the trick is simple, it is great for young learners and will get them excited to show it to other kids.
The cards contains visual pictures of the trick and also have tools to speed up the calculation.
### OVERVIEW & PURPOSE
Week 5 of the 3rd Grade CountFast program focuses on strategies to add, subtract, and multiply with the digit 9. Spend 15 minutes each day on one of the activities listed in this module. Card decks should go home with students each day for additional practice with a parent at home. Each week, a new deck is introduced, and the previous deck is for the student to keep at home for continued practice.
#### EDUCATION STANDARDS
1. NCTM Standard: develop a sense of whole numbers and represent and use them in flexible ways, including relating, composing, and decomposing numbers
2. Math.Content.3.OA.B.5
Apply properties of operations as strategies to multiply and divide.
#### OBJECTIVES
1. Mentally add and subtract with the number nine.
2. Mentally multiply one and two-digit numbers by nine.
#### MATERIALS NEEDED
1. One CountFast 9 card deck for each student. This deck is for school and home use. Discuss routine and expectations for taking home the deck and returning it to school each day.
2. One writing utensil per student, optional per teacher plans.
#### CountFast 9 Pack – Day 1
Teacher Model/Direct: Use the yellow cards from the deck. Teach students that the fastest way to add 9 to any number is to think of 9 as being one less than 10. Add ten to the original number, and then back up one spot (in a negative direction) on the number line. For example, to add 18 + 9, first realize that 18 + 10 is 28, then back up one spot on the number line, which would be 27 for a final answer. Practice with the class using the yellow cards.
Student Activity: Give each student a deck and ask them to take out the yellow cards. Working with a partner, take turns holding up the cards for the partner to add 9 using the strategy from the lesson.
Home Activity: Students will take home the deck and the “CountFast Home Connection” letter. Students will review the strategy for adding 9 to a number. Parents can record how quickly the child adds all of the yellow cards correctly.
#### CountFast 9 Pack – Day 2
Teacher Model/Direct: Using the same procedure as Day 1, introduce the blue cards in the deck. Today we will be subtracting 9 from a number using the same idea that 9 is the same as 10 – 1. Subtract 10 from the original number, and then move up one place on the number line. For example, solve 23 – 9 by first subtracting 10. This puts you at 13. Then move up (in a positive direction) on the number line one spot, which gives the final answer of 24. Practice with the class using the blue cards.
Student Activity: Students should work with a partner, using the blue cards, to practice subtracting 9 from any number. If time allows, review the yellow cards and adding 9 to a number.
Home Activity: Students will review the strategy for subtracting 9 from a number. Parents will use the blue cards with child and record how quickly the child answers all cards correctly each round.
#### CountFast 9 Pack – Day 3
Teacher Model/Direct: Use the pink cards from the deck. Today, show students the ‘finger trick’ for multiplying a single-digit number by nine. Model the technique several times with the class, then have students look at the pink cards to see that the pictures are the same as the examples you modeled for them. (Holding both hands up in front of you, think about ‘reading’ the fingers as 1 through 10, going from left to right. Hold down the finger that 9 is being multiplied by – for example, 9 X 3 – hold down the third finger. The fingers on the left of the one held down represent the ‘tens’ number of the answer. The fingers on the right represent the ‘ones’ digit of the answer. For 9 X 3, there are 2 fingers on the left, and 7 on the right, giving the answer of 27.)
Student Activity: With a partner, practice using the ‘finger trick’ method of multiplying a single-digit number by nine. If time allows, review the strategies for adding and subtracting 9 with the yellow and blue cards.
Home Activity: Students will share and practice the ‘finger trick’ method of multiplying by nine with parents.
#### CountFast 9 – Day 4
Teacher Model/Direct: Use the green cards in the deck for today’s lesson. Review how to quickly add, subtract, and multiply by nine. Then introduce the strategy to multiply two-digit numbers by nine. The quick way to mentally multiply by 9 is to multiply the original number by 10 (by simply adding a zero to the original number) and then subtract the original number to get the final answer. For example, 15 X 9 is solved quickly by first multiplying 15 X 10 to get 150, and then subtracting 15 from 150 to get a final answer of 135.
Student Activity: Students will work with partner using green cards to use the lesson strategy to multiply double-digit numbers by nine. If time allows, review the other cards from previous lessons this week.
Home Activity: Students will share and practice the strategy for multiplying a two-digit number by nine with parents. Parents can track the speed in which the child works through all of the green cards each round of practice.
#### CountFast 9 – Day 5
Teacher Model/Direct: Today, review with students the strategies for adding/subtracting 9. Also review the finger method of multiplying a single-digit number by 9, and the strategy for multiplying a two-digit number by nine. This can be made more challenging by shuffling all of the cards in the deck to that students have to really think about what operation is on the card.
Student Activity: With a partner, shuffle the deck and take turns answering the problems on the cards using the strategies for working with 9.
Home Activity: Students will review this week’s strategies for working with 9. | 1,459 | 6,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-10 | longest | en | 0.912875 |
https://naturallyvegetarianrecipes.com/qa/what-is-transit-duration.html | 1,579,401,168,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00541.warc.gz | 582,041,393 | 7,419 | # What Is Transit Duration?
Transit Duration:
The total transit duration, , defined as the time during which any part of the planet obscures the disc of the star, depends on how the planet transits the host star.
## What transit duration tells us?
The Transit Duration
The duration does not tell us anything physically about the planet. However, like the transit depth, the duration must by consistent for all the transits of a given planet-star combination.
## What is transit period?
The transit period is a period of fructification of events for a person.
## How do you determine transit depth?
Ratio of the size of the planet to the size of the star: The transit depth is the ratio of the surface area of the star’s disk blocked out by the planet’s disk. So the transit depth is the square of the planet radius divided by the star’s radius.
## How does the transit technique work?
Transit Photometry. This method detects distant planets by measuring the minute dimming of a star as an orbiting planet passes between it and the Earth. The passage of a planet between a star and the Earth is called a “transit.”
## What’s the transit?
Transit means “journey,” and saying you’re “in transit” means you’re on your way somewhere. While it sometimes seems like “mass transit” is synonymous with “unreliable, crowded and inefficient,” the phrase just means “public transportation.”
## What is the transit method of planet detection?
Transit Photometry
This method detects distant planets by measuring the minute dimming of a star as an orbiting planet passes between it and the Earth. The passage of a planet between a star and the Earth is called a “transit.”
## What is a transit person?
conveyance or transportation from one place to another, as of persons or goods, especially, local public transportation: city transit.Compare mass transit.
## How long does a Venus transit last?
During a transit, Venus can be seen from Earth as a small black dot moving across the face of the Sun. The duration of such transits is usually several hours (the transit of 2012 lasted 6 hours and 40 minutes).
## What is a transit chart?
Planets on the move are called transiting planets. They make aspects to the “moment in time” planets of a chart for an event, entity or person (the birth chart). When astrologers refer to a transit, they’re talking about a specific event, like a planet aligning with the natal chart.
## Which planet has the largest semimajor axis?
ch 3 quiz.docx – Which major planet has the largest semi-major axis Neptune Which major planet has the largest orbital eccentricity Mercury Which major.
## How many planets are in the Goldilocks zone?
Potential habitable zone status
In November 2013, astronomers reported, based on Kepler space mission data, that there could be as many as 40 billion Earth-sized planets orbiting in the habitable zones of Sun-like stars and red dwarfs in the Milky Way, 11 billion of which may be orbiting Sun-like stars.
## What is the astrometric method?
Astrometry is the science (and art!) of precision measurement of stars’ locations in the sky. If such a periodic shift is detected, it is almost certain that the star is being orbited by a companion planet. Astrometry is the oldest method used to detect extrasolar planets.
## What planets do we sometimes observe a transit?
Brilliantly, some planets can be indirectly discovered through the transit method as astronomers can measure the slight perturbations that other planets exert on those whose transits they are observing. Through the transit method, terrestrial planets have been found within the habitable zones of their parent stars.
## Can we see exoplanets?
There are many methods of detecting exoplanets. Transit photometry and Doppler spectroscopy have found the most, but these methods suffer from a clear observational bias favoring the detection of planets near the star; thus, 85% of the exoplanets detected are inside the tidal locking zone.
## Why is Pluto not considered a planet?
The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one- it “has not cleared its neighboring region of other objects.” | 914 | 4,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-05 | latest | en | 0.930139 |
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# 4-2 Some Ways To Prove Triangles Congruent Filled Out
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### Transcript
• 1. 4-2 some ways to prove triangles congruent
• 2. Think (Don’t write yet)
• If all three angles and all three sides of two triangles are congruent then we can conclude that the triangles are congruent.
• Is there a shortcut we can take????
• 3. SSS
• SSS Postulate
• If 3 sides of one ∆ are congruent to 3 sides of another ∆ then the ∆’s are congruent
• (Volunteer needed for demonstration)
• 4. SAS
• SAS postulate
• If 2 sides and the included angle of one ∆ are congruent to two sides and the included angle of another ∆ then the triangles are congruent.
• 5. ASA Postulate
• If 2 angles and the included side of one ∆ are congruent to two angles and the included side of another ∆ then the triangles are congruent.
• 6. Tricks to remember
• 7. STATEMENT REASON
• 8. Homework
• Page 124 #2-18 even | 344 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2014-52 | longest | en | 0.851995 |
https://meangreenmath.com/2017/04/29/my-favorite-one-liners-part-88/ | 1,685,681,353,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00532.warc.gz | 444,338,251 | 27,954 | # My Favorite One-Liners: Part 88
In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.
In the first few weeks of my calculus class, after introducing the definition of a derivative,
$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$,
I’ll use the following steps to guide my students to find the derivatives of polynomials.
1. If $f(x) = c$, a constant, then $\displaystyle \frac{d}{dx} (c) = 0$.
2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$.
3. If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$.
4. If $f(x) = x^n$, where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$.
5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$.
After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.
Example 1. Let $A(r) = \pi r^2$. Notice I’ve changed the variable from $x$ to $r$, but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)
What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$.
Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)
Generally, students start waking up even though it’s near the end of class. I continue:
Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$. Does this remind you of anything? (Students answer: the volume of a sphere.)
What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$.
Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)
By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:
Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)
This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.
If you’d like to see the answer, see my previous post on this topic.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 812 | 2,711 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-23 | latest | en | 0.822745 |
http://excel.bigresource.com/Formula-for-determining-if-two-date-columns-fall-within-specific-date-range-xtelDcTd.html | 1,579,471,191,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00471.warc.gz | 60,241,457 | 12,275 | # Formula For Determining If Two Date Columns Fall Within Specific Date Range
Apr 21, 2006
Let's say I have thousands of employees, but I need to determine who worked for me during a particular date range, and all I have to go on is their start date in one column and their end date in another column.
If:
A1 contains beginning date of employment
B1 contains ending date of employment
C1 contains specified beginning date (criteria)
D1 contains specified ending date (criteria)
## Formula That Will Determine The Number Of Days That Fall In A Specific Month Based On A Date Range
Jul 31, 2009
I'm needing a formula that will determine the number of days that fall in a specific month based on a date range. For example, if I have a date range of 10/15/2009 to 01/13/2009, I need the formula to determine the number of days in each month within the range (October has 15 days in the date range; November has 30, December has 31, and January has 13.) I have a large spreadsheet that would be so much easier to manage with such a formula. Currently, my spreadsheet is setup as follows. I need the forumla automatically fill in the number of days under each month.
Stard Date End Date Oct-09 Nov-09 Jan-10 Feb-10
10/15/2009 01/13/2009
I'm using Excel 2007.
## REcord That Fall Within Date Range, AND Meet Text Criteria
Dec 8, 2006
I have a large database that is updated daily. From within the database I need to:
1. Select all records where date field A is 14 days or less than date field B
AND
2. Where a field C matches a text criteria, i.e., =DOGGIE
AND
3. Append KITTY to field D of all records that match criteria 1 and 2
Finally
4. SAVE results as a text file.
## Determining If Date In Row Is Within 90 Days Of Date Accessed?
Jul 27, 2012
ive created an excel spreadsheet with expiration dates in one row and im trying to write a code which will pull out all information in the column if the expiration date in a certain row is within 90 days of the date the excel file is accessed
## Determining Date Based On Other Date
Mar 9, 2013
So i have have a two week period that is established by two dates (a start date "F3"... And an end date "F4")
I have a formula that determine if the "1st" of the month falls within the 13 days preceding the start day. This will notify me that I can the deposit my tenant first check on the start date.
Now that that is known, i would like to create a formula that will establish a second date in that month that will notify me when I can deposit my other renters payment. It will always be two after i deposit the first renters cheque.
## Fill Date Across Columns Based On User Date Range Input?
Mar 5, 2014
based on user date ranges entered on sheet1, I'm trying to write code that will write each month of the date range on other sheets across the 2nd row. at this point I'm getting "object required" error at "Set DateStart = Cells(2, 6)"
I also want the date format to be mmm-yy (Mar 14) on the sheets even if sheet1 has a different format. I tried using sourcerange instead of DateStart, but that didnt work either.
Code:
Dim projStartDate As Date
Dim projEndDate As Date
Dim DateStart As Date[code]....
## Only Countr Cell Values That Fall Within A Specific Range
Oct 24, 2007
I need to keep track of tardy occurences, but I only have to consider occurrences that have occurred in the past six months from the current day. Column A contains the names of the employees, Column C thru IV contains daily dates beginning with 1/1/2007. Column B contains a CountIf function to count the number of times T appears in columns c - iv.
## How To Calculate Date Fall In Which Week
Jul 28, 2009
Need a solution for calcuate a week in user enterd date?
Example
A1 A2
07/01/2009 1
07/01/2009 5
## Show / Sort Dates That Fall In Between Date (offset)
Jan 27, 2014
I have in column A the units, and in B and C the date changes. I want, when I pick a month from the box, that it only picks the IN and OUT date that applies to the month that I picked (per unit). For ex: if I choose Month march, it should show only for unit 19902506 for example: A2 and A5. And not A7 and A11. As March falls in between the Start and End Date. Is that passible? Something with OFFSET? I managed to find the latest known date with the formula =lookup (Column M:N) but Im not sure
## Determining Date Ranges
Mar 31, 2009
I have about 20 files with check posting dates stretched over several years. Maybe a million checks total. The company uses 13 four week periods each year, not a calendar month/year. I need to look up each posting date and attach the correct period to it.
I could do a VLOOKUP with one row for each day of the year, but that would require 2 x 365 row's for just two years. There must be a more elegant way, maybe using an array or some IF statement.
Attached is a sample file, xlsx, with the perior/year info and sample posting date info. Can someone point me in the right direction so that I end up with the correct period/year in column 'C' for the applicable posting date?
## Determining Date Of 2nd Or Last Friday Of Any Month?
Sep 14, 2008
Without VB, is there a way to write a formula which when given a certain month, will output the date of, let's say, 2nd or last friday of the month?
## Two Columns - Lookup Rate Corresponding To Specific Date
Jan 30, 2012
I have 2 cols of dates; col1 effective from, col2 effective to. In col c i have a list of corresponding rates. How can i look up the rate corresponding to a specific date. I've tried lookups and matching formulas but not working.
## Determining If Datetime Falls Within Start And End Date Times
Sep 25, 2013
I have been using this statement
if(and(c1>=a1:a144,c1<=b1:b144),"yes","no"))
And it just works for the first 2 values c1, c2 and doesn't fit for the others.
The case is i have more than one event at the same video and i need to confirm that no event was taken unless it is between start and end.
Here are some samples:
Start dtime End Dtime Event Dtime
16/09/2013 22:13:34 16/09/2013 22:14:18 16/09/2013 22:13:38
16/09/2013 22:15:57 16/09/2013 22:24:30 16/09/2013 22:16:02
16/09/2013 22:24:30 16/09/2013 22:33:49 16/09/2013 22:17:32
16/09/2013 22:33:53 16/09/2013 22:35:05 16/09/2013 22:19:02
16/09/2013 22:35:05 16/09/2013 22:39:57 16/09/2013 22:20:02
So as you can see there are more than one event between one start and end dtimes.
## Determining If Datetime Falls Within Start And End Date Times
Sep 25, 2013
I have been using this statement.
if(and(c1>=a1:a144,c1
## How To Auto-Populate Date Range (working Days Only) From Start Date And End Date
May 25, 2014
I am now trying to create a excel macro to auto populate all the dates with reference to a start date and end date. The catch is that only working days are required in the range. My reference cells (start and end date) are in Sheet 1 while the destination cell range are in Sheet 2. The reason for creating a macro instead of a function is that the intervals between the start date and end date changes frequently (annual, semi-annual and quarterly) Best case scenario would be a button which I can just press after i input the dates to generate the range of dates in another sheet.
## IF Date Formula: IF Formula, That States If A Date Is More Than A Month After The Date In The Cell Then It Is Timely
Jun 2, 2006
I need an IF formula, that states if a date is more than a month after the date in the cell then it is timely and if it isn't then it is not timely. For example the date in A1 is 12/11/05, if the cell is A2 is 13/12/05 (or any date after that) then A3 should read Not Timely, if A2 is 12/12/05 (or any date before that, including a date before the date in A1) then A3 should read Timely. Is there any way to do this?? At the moment I am having to go through manually and put in either Timely or Not Timely.
## Date Range: Reference A Single Date And Output Date Ranges
Oct 11, 2008
I need to create formulas that reference a single date and output date ranges. The objective is to have a person input a Monday date in any given month and receive a four weeks out worth of dates and ranges. For example: In a lone cell, the person inputs 10/13/08. Automatically, the sheet produces the next full week range: October 19 – October 25 in a single cell and also produces a cell for each date. Example: Sunday 19, Monday 20, Tuesday 21, etc…. It should look like:
Monday Date:
10/13/08
October 19 – October 25
Sunday 19
Monday 20
Tuesday 21
Wednesday 22
Thursday 23
Friday 24
Saturday 25
and then repeat for three more weeks. I thought I had it figured out until the month changed. The dates continued in October instead of adding a month. This report will be ran weekly, so simply adding a +1MONTH to some cells will not benefit me as I’ll have to change the formula every week. I want the formula to compute the data without any manipulation over the next several years. The only change will be the Monday date.
## Date Range Need To Have Specific Value Returned
Jan 7, 2014
Every week I have a set of open tickets with various dates. I need to break these Dates down to certain ranges.
Current DateAged Date Range Value to assign
10/20/2013 10/18/13< = 3 days A1
10/20/2013 10/7/13> 3 days & < 2 weeksA2
10/20/2013 9/23/132 weeks - 4 weeksA3
10/20/2013 9/22/13> 4weeks A4
I need in cell H:2 a value returned of with A1 thru A4 based on the results of the information in Columns A:D.
So basically it shoudl have in column H a bunch that fall in the reange of 3 day or less with "A1" then the dates that fall in the range of greater then 3 days but less then 2 weeks a value of "A2", etc. I hope i made this clear enuf to understand.
## Counting Occurences Within Specific Date Range
Sep 14, 2009
I’m looking for a formula that will count the number of occurrences of “YES” in column A between a specific date range (column B) BUT only if the it's categorized as "LAB" (column C).
Up until now I’ve jerry rigged the spread sheet to do it with various filters, if, and countif formulas but I’m looking for something that will fit in one cell and lower the overall size of the file.
To make it even more complicated I’d like to feed the formula the date range information from another editable cell so that it can basically be queried for whatever dates my boss is concerned about at that second.
## Calculate Specific Days Within Specified Date Range?
Jan 20, 2013
I have a form with grouped check boxes for each day of the week, and start date and end date from the Microsoft Date and Time picker control. I want to know the total days within the date range based on check boxes. For example, Tuesday and Thursday are checked off. The start date is December 1, 2012 and end date is January 31, 2013. What is the total number of Tuesdays and Thursays in that two month date range.
## Create Calculator To Show Start And End Date Based On Specific Date Provided
Jan 12, 2014
I am trying to create a calculator which will show start date and end date based on the specific date provided.
EX: column A has January 1, 2013, column B has Wednesday, Column C should have a start date which supposed to be 4 days ago (December 28).
## Lookup Date And Return Specific Cell Value From A Range
Jan 31, 2014
I have a table that has a number of new starters and corresponding appointments offered, what I originally required was to lookup the chronological date after the new start date.
However this has now been scuppered by my boss who has requested that not only lookup the date, but also add who the appointment is with, but if I do this the first array formula stops working and to tell the trust I'm not to sure how to do it anyway.
## Minimum And Maximum Values Within Specific Date Range
Sep 30, 2009
I have a table of dates and values and would very much like to know a formula for returning the min and mean values for each column within a specified time (date) period.
Table is as follows
Date value(a) Value (b) ....
1/1/09 10 8
2/1/09 8 6
3/1/09 5 2
## Automatically Date & Time Stamp Changes To Specific Range
Jun 4, 2008
i need to put a date stamp when a change is made in b3:b31 into e3:e31 for each row also i need to put a date stamp into g3:g31 when a change is made in F3:f31 i try to use 1 "worksheet-change" and it is fine once i use 2 i get Ambiguous Name Detected errors
Private Sub Worksheet_Change(ByVal Target As Excel.Range)
With Target
If .Count > 1 Then Exit Sub
If Not Intersect(Range("b3:b31"), .Cells) Is Nothing Then
Application.EnableEvents = False
If IsEmpty(.Value) Then
.Offset(0, 3).ClearContents
Else
With .Offset(0, 3) ............
## How To Create Formula To Add 30 Days Out From Specific Date
Sep 15, 2009
I am creating a tracking spreadsheet and I will need to f/u on a process 30/60/90 days out from a particular date. Example: CPAP Therapy began 1/22/09 and I need to follow-up by 2/22/09. How do I create formula to add 30/60 or 90 days out from the setup date?
## How To Keep Date Entry TODAY Frozen On That Specific Date
Mar 6, 2014
I am working on a spreadsheet to create a school 'tracking' system based on excel. What I need is that as soon as a box in column B is ticked, a date (TODAY() is entered in column A. That is not so difficult, and I have used the formula in cells in column A:
=IF(ISBLANK(Bx);"";TODAY()), where x is the specific row.
So, keep cell A clear untill something is written in cell B. This works fine.
But, for the purpose I need to keep the TODAY() date fixed after entry. And it need to be erased again if the tickbox in cell B is emptied again. It need then to re-enter a new date after cell B is re-used.
## Calculate Total Amount Of Kilos For A Specific Date With Given Time Range
Mar 26, 2014
I am trying to calculate the total amount of kilos for a specific date with a given time range.
As well as the average time they have been handled with in the same specifications.
Attached is a sample sheet of the info I am working with but I cant seem to get the formulas to work.
Book1.xlsx
## Automatic Current Date Formula Extracts From Specific Cell
Mar 6, 2007
formula that can automatically use the current month date to extract information from other cells on my spreadsheet.
For example: This formula should be located in cell C6. Cells D6, F6, H6, J6, L6, N6, P6, R6, T6, V6, X6, and Z6 (these cells correspond to the months of the year, January to December respectively) contain the values I need. The month names are labeled above in row five.
Let's say the current month is March, I need the formula in C6 to automatically know that it is the month of March and to pull the information from cell H6 (which is the March cell). Then when April comes along, it will know that it is April and to only pull the information from cell J6; and so on until the end of the year.
## Date/Month :: Specific Date Falls In?
Feb 26, 2008
I have a column with dates populated(examp.Fri, 15 Feb 08)in it. In the next column I need to return the month and year (month,year format) this specific date falls in?
## Formula To Count Dates Between 2 Date Periods If Specific Condition Exists
Apr 6, 2012
I am currently using this formula to count dates between 2 date periods If specific condition exists.
=COUNTIFS(
Data!S1:S100000,"*KP*",
Data!X1:X100000,">=10/1/2010",
Data!X1:X100000,"=10/1/2010",
Data!X1:X100000," | 4,057 | 15,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-05 | latest | en | 0.912861 |
https://www.numbersaplenty.com/90612 | 1,680,099,413,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00172.warc.gz | 995,256,853 | 3,175 | Search a number
90612 = 2233839
BaseRepresentation
bin10110000111110100
311121022000
4112013310
510344422
61535300
7525114
oct260764
9147260
1090612
1162095
1244530
1332322
1425044
151bcac
hex161f4
90612 has 24 divisors (see below), whose sum is σ = 235200. Its totient is φ = 30168.
The previous prime is 90599. The next prime is 90617. The reversal of 90612 is 21609.
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (90617) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 312 + ... + 527.
It is an arithmetic number, because the mean of its divisors is an integer number (9800).
290612 is an apocalyptic number.
It is an amenable number.
90612 is an abundant number, since it is smaller than the sum of its proper divisors (144588).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
It is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (117600).
90612 is a wasteful number, since it uses less digits than its factorization.
90612 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 852 (or 844 counting only the distinct ones).
The product of its (nonzero) digits is 108, while the sum is 18.
The square root of 90612 is about 301.0182718707. The cubic root of 90612 is about 44.9153965944.
The spelling of 90612 in words is "ninety thousand, six hundred twelve". | 462 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.902041 |
http://www.ima.umn.edu/~arnold/complex-maps/ | 1,438,445,500,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988840.31/warc/CC-MAIN-20150728002308-00242-ip-10-236-191-2.ec2.internal.warc.gz | 511,381,032 | 1,853 | # Visualization of complex functions by animation
### Douglas N. Arnold
This is a collection of flash animations for visualizing maps of one complex variable. Each animation shows the image of a domain in the complex plane as it morphed by a given function. In mathematical terms we display the image of the domain under a homotopy of maps ranging from the identity to the given function, so the first frame of the animation shows the domain and the last frame shows its image under the function. The domain is colored with a pattern to make it easier to follow the homotopy.
• function: f(z)=z2, domain: unit circle
• function: f(z)=z2, domain: two squares
• function: f(z)=z2, domain: unit square (colored with my picture)
• function: f(z)=ez, domain: [-2,2]x[-π,-π]
• function: f(z)=ez, domain: [-2,2]x[0,2π]
• function: f(z)=ez, domain: [-2,2]x[-2π,2π]
• function: a cubic polynomial, domain: a square
These animations were made with Mathematica. The Mathematica notebook is available. | 258 | 993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-32 | longest | en | 0.841067 |
https://instantnursingpapers.com/typical-reasoning-assignment/ | 1,679,522,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00746.warc.gz | 348,773,663 | 14,130 | # Typical Reasoning assignment
Typical Reasoning assignment
Typical Reasoning
All work must be COMPLETELY ORIGINAL AS IT GOES THROUGH A turnitin PROGRAM
Assignment 2: Discussion—Typical Reasoning
People often take shortcuts in problem solving and quickly arrive at answers. Known as heuristics, these shortcuts may increase the speed of decisions but may also decrease the accuracy of those decisions. The experiment used in this assignment deals with inaccurate decisions based on the conjunction fallacy, where people think the chance of two events happening at the same time is greater than just one event occurring. However, the chance of one event occurring is greater than two events occurring; hence, the fallacy.
Access the CogLab demonstration Typical Reasoning. Follow the instructions to complete the demonstration. Next, answer the following questions:
• For this demonstration, on average, do participants give higher ratings for single events or conjunctions of events? Based on the demonstration results, did you make your judgments by using objective probabilities? Why or why not?
• What is a stereotype? How do stereotypes relate to the findings of this demonstration?
• Respond to the following two situations:
• You and two of your coworkers have just interviewed a candidate for a job opening at your law firm. Your boss asks you what inferences you made about the candidate during the interview. What can you do to maximize your likelihood of making a correct inference?
• John is a young, energetic, muscular, and outgoing individual. Estimate the following for him:
• He is tall and likes sports
• He is tall, likes sports, and has lots of friends
Write your initial response in 4–5 paragraphs. Apply APA standards to citation of sources.
By Sunday, March 1, 2015, post your response to the appropriate Discussion Area. Through Wednesday, March 4, 2015, review and comment on at least two peers’ responses.
This discussion assignment is worth 40 points and will be graded using the discussion rubric.
Likewise, large type, large margins, large indentations, triple-spacing, increased leading (space between lines), increased kerning (space between letters), and any other such attempts at “padding” to increase the length of a paper are unacceptable, wasteful of trees, and will not fool your professor.
The paper must be neatly formatted, double-spaced with a one-inch margin on the top, bottom, and sides of each page. When submitting hard copy, be sure to use white paper and print out using dark ink. If it is hard to read your essay, it will also be hard to follow your argument.
Discussion Questions (DQ)
• Initial responses to the DQ should address all components of the questions asked, include a minimum of one scholarly source, and be at least 250 words.
• Successful responses are substantive (i.e., add something new to the discussion, engage others in the discussion, well-developed idea) and include at least one scholarly source.
• One or two sentence responses, simple statements of agreement or “good post,” and responses that are off-topic will not count as substantive. Substantive responses should be at least 150 words.
• I encourage you to incorporate the readings from the week (as applicable) into your responses.
Weekly Participation
• Your initial responses to the mandatory DQ do not count toward participation and are graded separately.
• In addition to the DQ responses, you must post at least one reply to peers (or me) on three separate days, for a total of three replies.
• Participation posts do not require a scholarly source/citation (unless you cite someone else’s work).
• Part of your weekly participation includes viewing the weekly announcement and attesting to watching it in the comments. These announcements are made to ensure you understand everything that is due during the week.
APA Format and Writing Quality
• Familiarize yourself with APA format and practice using it correctly. It is used for most writing assignments for your degree. Visit the Writing Center in the Student Success Center, under the Resources tab in LoudCloud for APA paper templates, citation examples, tips, etc. Points will be deducted for poor use of APA format or absence of APA format (if required).
• Cite all sources of information! When in doubt, cite the source. Paraphrasing also requires a citation.
• I highly recommend using the APA Publication Manual, 6th edition.
Use of Direct Quotes
• I discourage overutilization of direct quotes in DQs and assignments at the Masters’ level and deduct points accordingly.
• As Masters’ level students, it is important that you be able to critically analyze and interpret information from journal articles and other resources. Simply restating someone else’s words does not demonstrate an understanding of the content or critical analysis of the content.
• It is best to paraphrase content and cite your source.
LopesWrite Policy
• For assignments that need to be submitted to LopesWrite, please be sure you have received your report and Similarity Index (SI) percentage BEFORE you do a “final submit” to me.
• Once you have received your report, please review it. This report will show you grammatical, punctuation, and spelling errors that can easily be fixed. Take the extra few minutes to review instead of getting counted off for these mistakes.
• Review your similarities. Did you forget to cite something? Did you not paraphrase well enough? Is your paper made up of someone else’s thoughts more than your own?
• Visit the Writing Center in the Student Success Center, under the Resources tab in LoudCloud for tips on improving your paper and SI score.
Late Policy
• The university’s policy on late assignments is 10% penalty PER DAY LATE. This also applies to late DQ replies.
• Please communicate with me if you anticipate having to submit an assignment late. I am happy to be flexible, with advance notice. We may be able to work out an extension based on extenuating circumstances.
• If you do not communicate with me before submitting an assignment late, the GCU late policy will be in effect.
• I do not accept assignments that are two or more weeks late unless we have worked out an extension.
• As per policy, no assignments are accepted after the last day of class. Any assignment submitted after midnight on the last day of class will not be accepted for grading.
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• Questions to Instructor Forum: This is a great place to ask course content or assignment questions. If you have a question, there is a good chance one of your peers does as well. This is a public forum for the class.
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If you happen to hop into our site in search of nursing coursework help, look no further. We are the number one solution for the best nursing coursework assignments. Therefore, to complete the best APA nursing coursework, seek help from professional nursing coursework experts at instantnursingpapers.com | 2,065 | 10,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-14 | latest | en | 0.941071 |
latestdigital.info | 1,519,281,266,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814036.49/warc/CC-MAIN-20180222061730-20180222081730-00233.warc.gz | 711,532,110 | 7,548 | # Understanding the claims like”How I made \$100 in a Day”
This is the most important thing to be clearly understood. Let us analyze the claims like “How I made \$100 in a Day” Out of some 20 people who claims this, really 10 will have made it but the other 10, claims are false! So called “proofs” will be shown with important ID numbers and others masked.
Let us now think of the correct claims. The 10 people who say that they have made \$100 a day, WILL HAVE MADE IT ON A PARTICULAR DAY but not every day! What I am saying is that the average income of the month will not be \$100 X 30 = \$3000.
I will give the best example of a salaried person with a monthly salary of \$5000. He can claim that he earned a salary of \$5000 in one day that is the first of the month, when the salary is credited to his account! In a similar way, in some particular day, one can get \$100 from affiliate marketing, not because of the effort of that day but because of the accumulated efforts and timings of the whole month or even more! This may also be due to some affiliate networks or companies updating the income report once in a month. The correct index will be average yearly income.
Let us think of a person working in a company. He will never try to disclose his salary income to almost any one expect for family members! He will never write it in blogs and show the “proof” of his salary to claim that he is getting a good salary, even if he is really getting a fat salary! Then why are these affiliate marketing people declaring their income and even trying to give “proofs”?
I will give a logical answer for this in my next post. If you do not want to miss it, subscribe to this website. | 387 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-09 | longest | en | 0.971659 |
https://www.saintvespaluus.com/tag/algebra-linear-equation-worksheet/ | 1,638,534,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00078.warc.gz | 1,022,178,136 | 14,649 | # algebra linear equation worksheet
## 6+ Two Step Algebra Equations Worksheets
How Do You Solve Two-step Algebra Equations? https://www.youtube.com/watch?v=AP5MbH88cdo What is a 2 step equation example? Step 1: Subtract -6 from both sides of the equation to isolate the variable x. Step 2: Divide both sides of the equation by 2 to solve for x. Hence, we have solved the equation 2x + 3 = 12 …
## 5+ Pre Algebra Two Step Equations Worksheets
What Are The Steps To Solving 2 Step Equations? https://www.youtube.com/watch?v=MP5I9EO-PM0 What are two-step algebra equations? A two-step equation is an algebraic equation that takes you two steps to solve. You’ve solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign. via … | 202 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-49 | latest | en | 0.888109 |
https://wiki.opencog.org/w/MOSES_for_Noobs_-_Logic_Gates | 1,632,023,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00065.warc.gz | 651,059,761 | 6,758 | MOSES for Noobs - Logic Gates
Description
This page will attempt to explain how to use MOSES in an even simpler way than described on the MOSES Tutorial page. If you feel this tutorial is too simple for you, please check that page.
Before you start, you may want to take a look at the MOSES terminology, and maybe browse through the MOSES man page.
Contents
This tutorial expands on the previous MOSES for Noobs tutorial by letting MOSES learn logic gates (AND, OR, XOR, etc.).
Running MOSES
The files you can use are available below
To let MOSES process a file, put the file somewhere you can let MOSES find it, and then include it in the command line. Assuming you put it in a subfolder called Logic, it would be
```moses -i Logic/AND.txt -w1
```
The w1 parameter tells MOSES to use the column names found in the file in the generation of output, rather than just assigning indexes to all the input and output variables.
Understanding the output
The output should be something like
```0 and(\$foo \$bar)
-1 \$foo
-1 \$bar
-2 and(!\$foo !\$bar)
-2 and(!\$foo \$bar)
-2 and(\$foo !\$bar)
-2 or(!\$foo \$bar)
-2 or(\$foo !\$bar)
-2 or(\$foo \$bar)
-3 true
```
Each line first gives a score for how well the generated program approximates the target feature (0 is perfect, anything lower is worse). Behind the score it will output the COMBO program it used to achieve the score.
The highest scoring program is on the first line. The 0 indicates there is 0 distance between the target feature we wanted the program to approximate and the actual output of this program. As we can see it takes both input parameters, applies the AND operator, and achieves a perfect score for every case in our input file.
Next Steps
Feel free to run MOSES on all the files provided above to see how smart it is. When you're done with that, you could move on to MOSES for Noobs - Flowers.
Any questions? | 463 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | latest | en | 0.844109 |
http://passyworldofmathematics.com/similar-triangles/ | 1,495,770,302,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608633.44/warc/CC-MAIN-20170526032426-20170526052426-00510.warc.gz | 356,210,127 | 16,461 | # Similar Triangles
Image Source: www.ebay.com
Similar Triangles turn up in the strangest of places, even in Jewellery made from crystals of the gem stone “Tourmaline”.
In this lesson we look at the nature of similar figures, concentrating on Similar Triangles.
Similar objects have the exact same shape but are different in size.
We cover the methods and rules for establishing similarity.
We also work out how to find unknown sides using Similarity Ratios.
If you are not sure about Ratios, we suggest a review of these previous lessons:
http://passyworldofmathematics.com/introduction-to-ratios/
http://passyworldofmathematics.com/finding-ratio-amounts-of-proportions/
We highly recommend that you have also done our Congruent Triangles lesson, before doing Similar Triangles.
http://passyworldofmathematics.com/congruent-triangles/
In the Similar Triangles lesson here, we do not cover composite similar triangle questions, or applications of Similar Triangles, as these are covered in a separate lesson.
Similar Objects
Two or more items are similar to each other if they have the same shape, but are different sizes.
SIMILAR = SAME SHAPE, DIFFERENT SIZE
Image Copyright 2013 by Passy’s World of Mathematics
Scale Factor
The amount by which we increase, or decrease, the size of an object is called the “Scale Factor” or “S.F.”
The following examples of enlarging and reducing the size of a photo illustrate the concept of Scale Factor.
Image Copyright 2013 by Passy’s World of Mathematics
Image Copyright 2013 by Passy’s World of Mathematics
Calculating the Scale Factor
We calculate the SCALE FACTOR by comparing matching sides, using Ratios.
Image Copyright 2013 by Passy’s World of Mathematics
The following video gives a good introduction to Scale Factor, and also shows some real world applications that involve using Ratio Fractions and Cross Multiplying.
If you are not sure about Ratios, and the Cross Multiplication Method, then check out our previous lesson on this at the link below:
http://passyworldofmathematics.com/finding-ratio-amounts-of-proportions/
Checking for Similar Objects
For two objects to be similar, all their measurements must be changed in the exact same Ratio.
The two objects will then be proportional to each other.
In the example below we first enlarge a 4:3 aspect ratio photo to be double its original size.
The beginning small photo, and the ending larger photo are the exact same shape, and are similar objects.
In the second photo enlargement in our example, we have changed from Standard digital camera 4:3 ratio, to Wide Angle 16:9 aspect ratio.
When we do the ratio proportion mathematics, we find that the ratios of the corresponding sides are NOT the same.
In this 4:3 to 16:9 enlargement, we do not have the same shape, we now have a much wider photo, and so the two photos are NOT Similar.
Image Copyright 2013 by Passy’s World of Mathematics
Similar Triangles
Similar Triangles are the exact Same Shape, but are Different Sizes.
In the remainder of this lesson we will be looking at Similar Triangles.
The following example of two similar triangles involves one triangle, and then a second half size copy of the triangle.
For any two similar triangles their angles will be identical.
However, the sides of the second triangle will be either an Enlargement or a Reduction of the sides of the first triangle.
Image Copyright 2013 by Passy’s World of Mathematics
Introductory Video on Similar Triangles
The following video gives a good introduction to Similar Triangles, including some proofs and problem solving.
Determining Similar Triangles
Similar Triangles need to have the exact same shape, which will happen when their angles are all the same.
Similar Triangles are NOT the same size, which means that their matching sides are NOT equal.
Instead their matching sides are “In Proportion”, eg. either Enlarged or Reduced.
Image Copyright 2013 by Passy’s World of Mathematics
Similar Triangle Rules
Like was the case for Congruent Triangles, there are some “shortcut” rules we can use to prove that two triangles are similar.
Eg. We do not have to check that all three angles are equal, or that all three sides are in proportion.
The shortcut Rules for working out if two triangles are similar to each other are a lot like the rules we use for Congruent Triangles.
There are four Rules for Similar Triangles:
Angle Angle Angle or “AAA”, which turns out to really be just the Angle Angle or “AA” Rule
Proportional Side, Proportional Side, Proportional Side or “PPP” or “SSS” Rule
Proportional Sides, Equal Included Angle, Proportional Sides or “PAP” or “SAS” Rule
Right Angle, Proportional Hypotenuses, Proportional Sides or “RHS” Rule.
Here are some summaries of those four rules.
Angle Angle Angle AAA Rule
If all three Angles are the exact same sizes, (but sizes of triangles are different), then the triangles must be similar.
We actually only need two pairs of matching angles the same, because the third pair will automatically match, because the total angle size in any triangle adds up to 180 degrees.
Image Copyright 2013 by Passy’s World of Mathematics
Proportional Sides PPP or SSS Rule
Here at Passy’s World we like to call this the “PPP” rule, rather than the “SSS” Rule, and leave the “SSS” Rule as being for Congruent equal sized triangles only.
The “PPP” Rule is true if all three sides of the two triangles produce the same Scale Factor value.
If all three sides have the same S.F. then the sides are all in proportion and the two triangles are similar.
Image Copyright 2013 by Passy’s World of Mathematics
Proportional Sides with Angle PAP or SAS Rule
We prefer to call this the “PAP” rule, rather than the “SAS” Rule, and leave the “SAS” Rule as being for Congruent equal sized triangles only.
Image Copyright 2013 by Passy’s World of Mathematics
Right Angled Triangles RHS Rule
This rule is a lot like the RHS rule for congruent equal sized triangles.
However, in this similar triangles rule, the hypotenuses and either pair of the two sides are in Proportion to each other, rather than being equal to each other.
Image Copyright 2013 by Passy’s World of Mathematics
The following video covers the four Similar Triangles Rules
Similar Triangles in the Real World
Similar Triangles can be found in Nature, in Art and Craft, and in many structures we design and build.
Amazing Similar Triangles are found inside the crystals of the beautifully colored “Tourmaline” gemstone.
Image Copyright 2013 by Passy’s World of Mathematics
Image Copyright 2013 by Passy’s World of Mathematics
Similar Triangles can create striking effects when used in Art and Craft.
Image Copyright 2013 by Passy’s World of Mathematics
Image Copyright 2013 by Passy’s World of Mathematics
Similar Triangles provide reinforced strength and rigidity to structures, as well as greatly reducing the weight of the objects they are used in.
Image Copyright 2013 by Passy’s World of Mathematics
Examples of Using Similar Triangle Rules
In this first example, we are asked to use either AAA or PPP or PAP or RHS to prove that we have Similar Triangles.
It turns out we need to use the “AAA” Rule.
Image Copyright 2013 by Passy’s World of Mathematics
In this second example, we are asked to use either AAA or PPP or PAP or RHS to prove that we have Similar Triangles.
The two triangles have two sides and the included Angles inbetween these two sides.
Therefore we need to used the PAP / SAS Rule.
Image Copyright 2013 by Passy’s World of Mathematics
Examples of Solving Similar Triangles
In this first Example, we need to first prove that the Two Trianges are Similar.
After this we can find the Scale Factor that exists between the two tiangles.
Using the S.F. value, we then find the unknown side.
Image Copyright 2013 by Passy’s World of Mathematics
Using the Scale Factor works well for situation where we have the known side on a small triangle, and we need to find the unknown length of the matching side on a big triangle.
However, if the situation if the other way around, we get fraction Scale Factors such as one third 1/3 to deal with and this makes the mathematical working out a bit fiddly.
Another way to solve similar triangles is to write two rations and then use “cross multiplying”, (or “Cross Products”).
At Passy’s World we prefer to use this “Cross Products” method for solving triangle questions, because it helps avoid dealing with fractions.
In “Example 3B” below, we redo “Example 3A”, but this time use the Cross Products Method to solve our triangle question.
Image Copyright 2013 by Passy’s World of Mathematics
In this final “Example 3C” our question is the other way around, and we have to find an unknown side on the smaller triangle.
We could work out that the Scale Fator is 10/30 and reduce this down to 1/3. But then we get into fractions work to get to the final answer.
Here at Passy’s World, we have found that students have difficulty with fractions, and so we have worked out “Example 3C” using Cross Multiplyling as shown below.
Using Cross Multiplying avoids having to deal with fractions, and we believe that is a good thing.
Image Copyright 2013 by Passy’s World of Mathematics
Here are some videos where unknown sides are found for Similar Triangles
Similarity concepts can also be applied to Quadrilaterals as well as Trinagles.
This is demonstrated very well in the following video by Mr Bill Konst.
Here is a bit of a humorous rap song about Similar Triangles.
There are some extremely, but a bit way out, mathematical concepts related to triangles covered in this entertaining little story.
Interactive Similar Triangles
The above Online Manipulative, (which can go full screen), allows us to drag corner vertices around and make any kind of shaped pair of similar triangles.
Click the link below to try it out:
http://www.mathopenref.com/similartriangles.html
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Passy | 2,521 | 11,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-22 | longest | en | 0.88506 |
https://www.wyzant.com/resources/answers/3769/ab_9_ab_8 | 1,521,801,124,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00128.warc.gz | 917,126,944 | 16,570 | 0
(ab - 9)(ab + 8)
how do you foil this?
Nancy N. | Outstanding Elementary and Spanish Teacher working with grades K-12Outstanding Elementary and Spanish Teach...
4.9 4.9 (35 lesson ratings) (35)
0
ab and ab are put together creating "a" squared and "b "squared Then you multiply (8*ab) and (-9*ab) which gives you +8ab and -9ab -9*8 which gives you a squared*b squared - 1 ab-72 or a squared b squared - 1ab-72. Sorry, but I have no idea how to do square numbers on the computer.
Nancy n.
Andres F. | Great Mathematician & Best tutorGreat Mathematician & Best tutor
4.8 4.8 (88 lesson ratings) (88)
0
Step 1: (ab-9)(ab+8)
Step 2: ab x ab +8 x ab -9 x ab -9 x 8
Step 3: ab +8ab-9ab-72=
Step 4: ab2 -ab-72
solved!!! that how you would solve it everything will continue as normal
Kevin S. |
5.0 5.0 (4 lesson ratings) (4)
0
Same way you would it "a" was a number instead of a variable:
(ab)(ab) + (8)(ab) - (9)(ab) - (9)(8)
a2b2 - ab - 72 | 329 | 949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-13 | longest | en | 0.874158 |
http://interactive.onlinemathlearning.com/tri_pythagoras.php?action=generate&numProblems=10 | 1,508,265,470,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00679.warc.gz | 168,913,098 | 5,516 | OML-IZ Search
## Pythagorean Theorem
With reference to the diagram above, calculate and enter the answers. Round your answers to the nearest hundredth. You may use the TAB key to move to the next question. When you are done, click Submit.
1. a = 2.5, c = 6.2, b =
2. b = 3.2, c = 6, a =
3. a = 5.7, c = 9.6, b =
4. a = 7.6, b = 3.2, c =
5. a = 8.5, b = 1.8, c =
6. a = 4.5, b = 5.9, c =
7. a = 1, b = 3.8, c =
8. a = 5.6, b = 7.4, c =
9. a = 7.4, c = 7.8, b =
10. a = 1.3, b = 1.8, c =
We hope that the free math worksheets have been helpful. We encourage parents and teachers to adjust the worksheets according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles. | 300 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-43 | latest | en | 0.907379 |
https://www.physicsforums.com/threads/evaluate-the-divergence-and-curl-of-the-following-vector.345899/ | 1,603,979,643,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00524.warc.gz | 829,746,768 | 14,271 | # Evaluate the divergence and curl of the following vector
## Homework Statement
Evaluate the divergence and curl of the following vectors.
A(r) is everywhere parallel to the y-axis with a magnitude A = cx + A0 , where c and
A0 are constants.
## The Attempt at a Solution
I can evaluate the div and curl, but i dont know how to work out what the actual vector is, so can anyone help me work out what the vector is?
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
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$\vec{j}$ is "parallel to the y axis". $\vec{i}$ is parallel to the x axis. So any vector that is "always parallel to the y axis" must be of the form $A\vec{j}$. | 174 | 707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.837642 |
http://jean-pierre.moreau.pagesperso-orange.fr/Pascal/cdetmat_pas.txt | 1,606,151,173,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00313.warc.gz | 44,827,286 | 2,702 | {************************************************************ * Determinant of a complex square matrix * * By Gauss Method with full pivoting * * --------------------------------------------------------- * * SAMPLE RUN: * * Calculate the determinant of complex matrix: * * ( 47,-15) ( 62,5) ( 0,-72) (61, 20) * * ( 6, 14) (-17,3) (-102,91) ( 7,-12) * * ( 13, 55) ( 32,8) ( 41, 7) (25, 1) * * (111,25) ( 40,0) ( 12,-82) (58,-30) * * * * Det = 1.74165640000000E+0007 - 1.05983200000305E+0007 I * * * * --------------------------------------------------------- * * Ref.: "Algèbre, Algorithmes et programmes en Pascal * * By Jean-Louis Jardrin, DUNOD Paris, 1988". * ************************************************************} Program CDetMat; Uses WinCrt; Const NMAX = 20; Type Complex = Record R, I: Real End; Matc = Array[1..NMAX,1..NMAX] of Complex; Veci = Array[1..NMAX] of Integer; Var n: Integer; eps: Real; A: Matc; det:Complex; Function CABS(Z:Complex): Real; Begin CABS := sqrt(Z.R*Z.R+Z.I*Z.I) End; Procedure CADD(Z1,Z2:Complex; Var Z:Complex); Begin Z.R := Z1.R + Z2.R; Z.I := Z1.I + Z2.I End; Procedure CDIF(Z1,Z2:Complex; Var Z:Complex); Begin Z.R := Z1.R - Z2.R; Z.I := Z1.I - Z2.I End; Procedure CMUL(Z1,Z2:Complex; Var Z:Complex); Begin Z.R := Z1.R*Z2.R - Z1.I*Z2.I; Z.I := Z1.R*Z2.I + Z1.I*Z2.R End; Procedure CDIV(Z1,Z2:Complex; Var Z:Complex); Var d:Real; C:Complex; Begin d := Z2.R*Z2.R + Z2.I*Z2.I; if d<1E-10 then writeln(' Complex Divide by zero !') else begin C.R:=Z2.R; C.I:=-Z2.I; CMUL(Z1,C,Z); Z.R:=Z.R/d; Z.I:=Z.I/d end End; {**************************************************************** * TSCGT procedure implements the triangularization algorithm of * * Gauss with full pivoting at each step for a complex matrix, A * * and saves the made transformations in KP and LP. * * ------------------------------------------------------------- * * INPUTS: * * N: size of complex matrix A * * A: complex matrix of size N x N * * OUTPUTS; * * it: =0 if A is singular, else =1. * * C: contains the upper triangular matrix and the * * multipliers used during the process. * * KP: contains the column exchanges. * * LP: contains the line exchanges. * ****************************************************************} Procedure TSCGT(eps:Real; N:integer; A:Matc; Var it:integer; Var C:Matc; Var KP,LP:Veci); Var i,j,k,k0,l0:integer; C0,C1,P0,T0:Complex; Begin C:=A; it:=1; K:=1; While (it=1) and (k CABS(P0) then begin P0:=C[i,j]; l0:=i; k0:=j end; LP[k]:=l0; KP[k]:=k0; if CABS(P0) < eps then it:=0 else begin if l0<>k then For j:=k to N do begin T0:=C[k,j]; C[k,j]:=C[l0,j]; C[l0,j]:=T0 end; if k0<>k then For i:=1 to N do begin T0:=C[i,k]; C[i,k]:=C[i,k0]; C[i,k0]:=T0 end; For i:=k+1 to N do begin C0:=C[i,k]; CDIV(C0,P0,C[i,k]); For j:=k+1 to N do begin C0:=C[i,j]; CMUL(C[i,k],C[k,j],C1); CDIF(C0,C1,C[i,j]) end end; Inc(k) end end; if (it=1) and (CABS(C[N,N]) < eps) then it:=0 End; {TSCGT} {*************************************************************** * The DCGT procedure calculates the complex determinant of a * * complex square matrix by the Gauss method with full pivoting * * ------------------------------------------------------------ * * INPUTS: * * eps: required precision * * N : size of matrix A * * A : complex matrix of size N x N * * OUTPUT: * * det: complex determinant. * ***************************************************************} Procedure DCGT(eps:Real; N:integer; A:Matc; Var det:Complex); Var it,k,l: integer; C0,Z0,Z1: Complex; C:Matc; KP,LP:Veci; Begin Z0.R:=0.0; Z0.I:=0.0; Z1.R:=1.0; Z1.I:=0.0; TSCGT(eps,N,A,it,C,KP,LP); if it=0 then det:=Z0 else begin det:=Z1; For k:=1 to N do begin C0:=det; CMUL(C0,C[k,k],det) end; l:=0; For K:=1 to N-1 do begin if LP[k]<>k then Inc(l); if KP[k]<>k then Inc(l) end; if Odd(l) then begin C0:=det; det.R:=-C0.R; det.I:=-C0.I end end end; {DCGT} {main program} BEGIN (* Example #1 N:=3; {size of complex matrix A} A[1,1].R:=1.0; A[1,1].I:=0.0; A[1,2].R:=0.0; A[1,2].I:=1.0; A[1,3].R:=0.0; A[1,3].I:=1.0; A[2,1].R:=0.0; A[2,1].I:=1.0; A[2,2].R:=1.0; A[2,2].I:=0.0; A[2,3].R:=1.0; A[2,3].I:=0.0; A[3,1].R:= 0.0; A[3,1].I:=1.0; A[3,2].R:= 1.0; A[3,2].I:=0.0; A[3,3].R:=-1.0; A[3,3].I:=0.0; ( Det = -4.0 + 0 I ) *) { Example #2 } N:=4; A[1,1].R:=47.0; A[1,1].I:=-15.0; A[1,2].R:=62.0; A[1,2].I:= 5.0; A[1,3].R:= 0.0; A[1,3].I:=-72.0; A[1,4].R:=61.0; A[1,4].I:= 20.0; A[2,1].R:= 6.0; A[2,1].I:= 14.0; A[2,2].R:=- 17.0; A[2,2].I:= 3.0; A[2,3].R:=-102.0; A[2,3].I:= 91.0; A[2,4].R:= 7.0; A[2,4].I:=-12.0; A[3,1].R:= 13.0; A[3,1].I:=-55.0; A[3,2].R:= 32.0; A[3,2].I:= 8.0; A[3,3].R:= 41.0; A[3,3].I:= 7.0; A[3,4].R:= 25.0; A[3,4].I:= 1.0; A[4,1].R:=111.0; A[4,1].I:= 25.0; A[4,2].R:= 40.0; A[4,2].I:= 0.0; A[4,3].R:= 12.0; A[4,3].I:=-82.0; A[4,4].R:= 58.0; A[4,4].I:=-30.0; eps:=1E-10; DCGT(eps,N,A,det); writeln; write(' Det = ',det.R); if det.I>=0.0 then write(' + ') else write(' - '); writeln(abs(det.I),' I'); writeln; ReadKey; DoneWinCrt END. {end of file cdetmat.pas} | 2,031 | 4,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-50 | longest | en | 0.399255 |
https://amazingconverter.com/density-converter/density-of-water-g-ml | 1,669,724,882,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00319.warc.gz | 138,665,112 | 9,374 | lb/ft3
kg/m3
# Density of water g/ml
Published: 3/16/2021 Last updated: 3/16/2021
Density of water g/ml – do you know what its value is? Do you know why the water density varies at different temperatures? And how to find density of water? All these things you will learn here.
Table of contents:
Density of water g/ml – are you wondering what its value is? Do you want to know why the value of the density of water g/ml vary at different temperatures? We have the answers to both these questions. And, of course, we are going to share them with you, so keep reading.
The density of water g/ml is the density used mostly when it comes to smaller mass or volume. This unit of density, that is g/ml, will be perfect, for instance, to measure density of water in a glass.
Have you ever tried to calculate the density of water g/ml in a case like that mentioned above? Did you have some troubles then? But… Do you know that it is possible to know the topic of density of water g/ml clearly and finally can use this knowledge without any problems in practice?
And that’s what we want to do in this article – explain the topic of the density of water g/ml as easy as possible so all of you can finally understand it. Without any further ado, let’s start.
## Density of water grams per ml at 4 degrees Celsius
We will start the topic of the density of water grams per ml at 4 degrees Celsius. Why exactly 4, and not, for example, 0 degrees Celsius?
You have to know that the water density is the biggest at 4 degrees Celsius. It is caused by 2 physical effects acting together. They ‘fight’ each other, so that’s why the water density increases as much.
What are these effects? First, faster molecules moving. You need to know that the molecules move faster when the temperature increases. Second, forming clusters. When the temperature is coming to 0 degrees Celsius, the clusters start forming.
So after that explanation, let’s go back to the value of the density of water grams per ml at 4 degrees Celsius. The density of water g/ml at 4°C is equal to 1. See this result also down below:
• the density of water = 1 g/ml
Why is the density of water 1 g ml? We hope that all of you can answer this question now.
How do the values of density of water present at higher temperatures? Let’s check it in the next part of the article.
### Density of water g/ml at higher temperatures
So what are the values of the density of water g/ml at higher temperatures than 4 degrees Celsius? As you may know, these values at higher temperatures will be smaller, because the clusters stop forming when the temperature increases.
Do you know that even at 4.4 degrees Celsius the value of the density of water g/ml is not equal to 1? The density of water is equal to 0.9999 g/ml then. Physics is fascinating, don’t you think?
Of course, the higher temperature, the bigger difference. You can see it clearly even when it comes to not super high temperature, like 10 degrees Celsius. The density of water g/ml is then equal to 0.99975.
So what about the value of the density of water grams per ml at, for instance, 60 degrees Celsius? The difference is even bigger. The result is 0.98338. Now, even if you round off the result, for instance, to 2 decimal places, it will differ significantly from the previous results.
At the boiling point, that is at 100 degrees Celsius, the water density g/ml is even lower. The value is equal to 0.95865 only.
We talked about only a few examples, but we gathered more in the table. The density of water g/ml chart you can find down below:
Temperature (°C / °F) The density of water g/ml 4.0 / 39.2 1.00000 4.4 / 40 0.99999 10 / 50 0.99975 15.6 / 60 0.99907 21 / 70 0.99802 26.7 / 80 0.99669 32.2 / 90 0.99510 37.8 / 100 0.99318 48.9 / 120 0.98870 60 / 140 0.98338 71.1 / 160 0.97729 82.2 / 180 0.97056 93.3 / 200 0.96333 100 / 212 0.95865
### How to find density of water
At the end, we want to quickly mention how to find the density of water. Why quickly? Because to check the density of water g/ml in a way we want to show you, you need only one thing – the temperature.
So how can you find the density of water g/ml? You need to check the temperature of water and then find its value in the table above in the first column. Then you can easily check what is the value of the density of water at this temperature in the second column. That’s all.
As we promised, it’s a quick and easy way to find the density of water, isn’t it?
We hope that now the topic of the density of water g/ml is finally clear for you. We explained to you not only what the maximum density of water is, but also how the values of the water density look at other, higher temperatures. We showed you how to find the density of water too.
It’s a quite big dose of knowledge, don’t you think? So let’s use it in practice now!
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# 201-tutorial-4 - Econ 201 Tutorial#4 Date Week of Feb 8-14...
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Econ 201 Tutorial #4 Date: Week of Feb. 8-14, 2010 Coverage: Mainly Chapter 5 Welfare Economics I. Multiple Choice Questions: 1. The term "consumer surplus" refers to a. the diminishing marginal utility that consumers receive from additional consumption of a good. b. the total amount paid for the purchase. c. the total value that a consumer places upon a good. d. the area under the demand curve. e. the difference between the value of a purchase and the total amount paid for the purchase. 2. If the government imposes a price ceiling below the equilibrium price, then a. the quantity consumed will rise. b. consumer surplus will fall. c. consumer surplus may rise or fall. d. the price will rise. e. consumer surplus will rise. 3. If the government introduces a subsidy into a market, then a. consumer surplus will rise. b. the quantity consumed will rise. c. consumer surplus will fall. d. both a and b. e. both b and c. 4. Suppose demand is Qd=12-2P and supply is Qs=P. Producer surplus in equilibrium is a. \$4 b. \$8 c. \$12 d. \$16 e. \$18 5. Continue from #4: The value of consumer surplus is a. \$4 b. \$8 c. \$12 d. \$16 e. \$18 6. Currently Joe and Hana are consuming the same amount of strawberries, but Joe's demand curve is much more elastic than Hana's. Which statement is true? a. Hana's consumer surplus exceeds Joe's. b. Any comparison of consumer surplus depends on the price of strawberries. c. Hana's consumer surplus equals Joe's. d. No statement can be made comparing consumer surpluses. e. Joe's consumer surplus exceeds Hana's. 7. At the efficient quantity of a good, a. producer surplus exceeds consumer surplus by the greatest possible amount. b. total consumer surplus is zero. c. the sum of consumer surplus and producer surplus is maximized. d. total producer surplus is zero. e. consumer surplus exceeds producer surplus by the greatest possible amount. 1
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II. Short Questions 1. Consider the market for whale watching boat rides in Tofino, British Columbia. In this market, the whale watching boat drivers are producers and the whale watching riders are consumers. The demand curve and supply curves in the market for whale watching rides are given by Demand: P = 10 – 0.5Q Supply: P = 0.5Q (i) Find the equilibrium price and quantity in the market for whale watching rides. (ii) Calculate the consumer surplus and producer surplus in the market for whale watching rides. Sum consumer and producer surplus to get the total surplus in the market for whale watching rides. Assume now that the government in Tofino imposes a quota of 8 rides per year.
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twice as many people get on the bus
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twice as many as what?
2x<something>
## Similar Questions
1. ### psych
Given the following problem: Pretend that you are a bus driver. You begin the day with an empty bus. At the first stop, you let on 10 passengers. At your second stop, 4 people get off the bus, and 12 people on. At your third stop, …
2. ### Maths
There are x people in the bus. When the bus stops y people get off and 6 people get on. How many people are yhere now on the bus ?
3. ### Maths
60% of the people on a bus are male. After 1/3 of the males alighted from the bus, what percentage of the people left on the bus are females?
4. ### 8TH GRADE MATH EXPRESSIONS
THE TOTAL COST FOR 4 PEOPLE TO RIDE THE BUS IF THE FARE OF .75 IS INCREASED BY D DOLLARS.
5. ### elementry
translating words to expressions 3 times as old as 6 years old
6. ### elementary
translating words to expressions twice as many people get on the bus .there were 12 people on the bus
7. ### Math
* How many different odd 3-digit numbers can be formed using the digits 1,2,3,4,5 if none the numbes have repeating digits?
8. ### Math
A survey finds that 85% of the people asked use the local bus. 35.3% of those who use the bus use it at least once a week. 240 people took part in the survey. (a) Work out how many people say they use the bus once a week. (b) What …
9. ### algebra
There are n people on a bus . At the next stop y people get off and then t people get on. How many people are there now on the bus.
10. ### math
There were 24 people on the bus. At the bus stop 9 people got off the bus. How many people are on the bus now?
More Similar Questions | 470 | 1,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-34 | latest | en | 0.940703 |
https://functions.wolfram.com/Bessel-TypeFunctions/BesselY/21/01/02/02/01/02/0001/ | 1,722,858,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00747.warc.gz | 205,188,431 | 8,968 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
BesselY
http://functions.wolfram.com/03.03.21.0019.01
Input Form
Integrate[BesselY[\[Nu], a z^r]/E^(I a z^r), z] == (2^\[Nu] z Csc[Pi \[Nu]] HypergeometricPFQ[{1/2 - \[Nu], 1/r - \[Nu]}, {1 - 2 \[Nu], 1 + 1/r - \[Nu]}, -2 I a z^r])/(a z^r)^\[Nu]/ ((-1 + r \[Nu]) Gamma[1 - \[Nu]]) + (z (a z^r)^\[Nu] Cot[Pi \[Nu]] HypergeometricPFQ[ {1/2 + \[Nu], 1/r + \[Nu]}, {1 + 1/r + \[Nu], 1 + 2 \[Nu]}, -2 I a z^r])/2^\[Nu]/((1 + r \[Nu]) Gamma[1 + \[Nu]])
Standard Form
Cell[BoxData[RowBox[List[RowBox[List["\[Integral]", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "a", " ", SuperscriptBox["z", "r"]]]], RowBox[List["BesselY", "[", RowBox[List["\[Nu]", ",", RowBox[List["a", " ", SuperscriptBox["z", "r"]]]]], "]"]], RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", RowBox[List[RowBox[List[RowBox[List["(", RowBox[List[SuperscriptBox["2", "\[Nu]"], " ", "z", " ", SuperscriptBox[RowBox[List["(", RowBox[List["a", " ", SuperscriptBox["z", "r"]]], ")"]], RowBox[List["-", "\[Nu]"]]], " ", RowBox[List["Csc", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List[FractionBox["1", "2"], "-", "\[Nu]"]], ",", RowBox[List[FractionBox["1", "r"], "-", "\[Nu]"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "-", RowBox[List["2", " ", "\[Nu]"]]]], ",", RowBox[List["1", "+", FractionBox["1", "r"], "-", "\[Nu]"]]]], "}"]], ",", RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]]]], "]"]]]], ")"]], "/", RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", RowBox[List["r", " ", "\[Nu]"]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List["1", "-", "\[Nu]"]], "]"]]]], ")"]]]], "+", RowBox[List[RowBox[List["(", RowBox[List[SuperscriptBox["2", RowBox[List["-", "\[Nu]"]]], " ", "z", " ", SuperscriptBox[RowBox[List["(", RowBox[List["a", " ", SuperscriptBox["z", "r"]]], ")"]], "\[Nu]"], " ", RowBox[List["Cot", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List[FractionBox["1", "2"], "+", "\[Nu]"]], ",", RowBox[List[FractionBox["1", "r"], "+", "\[Nu]"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "+", FractionBox["1", "r"], "+", "\[Nu]"]], ",", RowBox[List["1", "+", RowBox[List["2", " ", "\[Nu]"]]]]]], "}"]], ",", RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]]]], "]"]]]], ")"]], "/", RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List["1", "+", RowBox[List["r", " ", "\[Nu]"]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List["1", "+", "\[Nu]"]], "]"]]]], ")"]]]]]]]]]]
MathML Form
- a z r Y ν ( a z r ) z 2 ν z csc ( π ν ) ( a z r ) - ν ( r ν - 1 ) Γ ( 1 - ν ) 2 F 2 ( 1 2 - ν , 1 r - ν ; 1 - 2 ν , - ν + 1 r + 1 ; - 2 a z r ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox["2", TraditionalForm]], SubscriptBox["F", FormBox["2", TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[RowBox[List[FractionBox["1", "2"], "-", "\[Nu]"]], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[RowBox[List[FractionBox["1", "r"], "-", "\[Nu]"]], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, False]], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List["1", "-", RowBox[List["2", " ", "\[Nu]"]]]], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[RowBox[List[RowBox[List["-", "\[Nu]"]], "+", FractionBox["1", "r"], "+", "1"]], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, False]], ";", TagBox[RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]], HypergeometricPFQ, Rule[Editable, True]]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]], Rule[Editable, False]], HypergeometricPFQ] + 2 - ν z cot ( π ν ) ( a z r ) ν ( r ν + 1 ) Γ ( ν + 1 ) 2 F 2 ( ν + 1 2 , ν + 1 r ; ν + 1 r + 1 , 2 ν + 1 ; - 2 a z r ) TagBox[TagBox[RowBox[List[RowBox[List[SubscriptBox["\[InvisiblePrefixScriptBase]", FormBox["2", TraditionalForm]], SubscriptBox["F", FormBox["2", TraditionalForm]]]], "\[InvisibleApplication]", RowBox[List["(", RowBox[List[TagBox[TagBox[RowBox[List[TagBox[RowBox[List["\[Nu]", "+", FractionBox["1", "2"]]], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[RowBox[List["\[Nu]", "+", FractionBox["1", "r"]]], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, False]], ";", TagBox[TagBox[RowBox[List[TagBox[RowBox[List["\[Nu]", "+", FractionBox["1", "r"], "+", "1"]], HypergeometricPFQ, Rule[Editable, True]], ",", TagBox[RowBox[List[RowBox[List["2", " ", "\[Nu]"]], "+", "1"]], HypergeometricPFQ, Rule[Editable, True]]]], InterpretTemplate[Function[List[SlotSequence[1]]]]], HypergeometricPFQ, Rule[Editable, False]], ";", TagBox[RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]], HypergeometricPFQ, Rule[Editable, True]]]], ")"]]]], InterpretTemplate[Function[HypergeometricPFQ[Slot[1], Slot[2], Slot[3]]]], Rule[Editable, False]], HypergeometricPFQ] z -1 a z r BesselY ν a z r 2 ν z ν a z r -1 ν r ν -1 Gamma 1 -1 ν -1 HypergeometricPFQ 1 2 -1 ν 1 r -1 -1 ν 1 -1 2 ν -1 ν 1 r -1 1 -2 a z r 2 -1 ν z ν a z r ν r ν 1 Gamma ν 1 -1 HypergeometricPFQ ν 1 2 ν 1 r -1 ν 1 r -1 1 2 ν 1 -2 a z r [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["\[Integral]", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["-", "\[ImaginaryI]"]], " ", "a_", " ", SuperscriptBox["z_", "r_"]]]], " ", RowBox[List["BesselY", "[", RowBox[List["\[Nu]_", ",", RowBox[List["a_", " ", SuperscriptBox["z_", "r_"]]]]], "]"]]]], RowBox[List["\[DifferentialD]", "z_"]]]]]], "]"]], "\[RuleDelayed]", RowBox[List[FractionBox[RowBox[List[SuperscriptBox["2", "\[Nu]"], " ", "z", " ", SuperscriptBox[RowBox[List["(", RowBox[List["a", " ", SuperscriptBox["z", "r"]]], ")"]], RowBox[List["-", "\[Nu]"]]], " ", RowBox[List["Csc", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List[FractionBox["1", "2"], "-", "\[Nu]"]], ",", RowBox[List[FractionBox["1", "r"], "-", "\[Nu]"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "-", RowBox[List["2", " ", "\[Nu]"]]]], ",", RowBox[List["1", "+", FractionBox["1", "r"], "-", "\[Nu]"]]]], "}"]], ",", RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]]]], "]"]]]], RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", RowBox[List["r", " ", "\[Nu]"]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List["1", "-", "\[Nu]"]], "]"]]]]], "+", FractionBox[RowBox[List[SuperscriptBox["2", RowBox[List["-", "\[Nu]"]]], " ", "z", " ", SuperscriptBox[RowBox[List["(", RowBox[List["a", " ", SuperscriptBox["z", "r"]]], ")"]], "\[Nu]"], " ", RowBox[List["Cot", "[", RowBox[List["\[Pi]", " ", "\[Nu]"]], "]"]], " ", RowBox[List["HypergeometricPFQ", "[", RowBox[List[RowBox[List["{", RowBox[List[RowBox[List[FractionBox["1", "2"], "+", "\[Nu]"]], ",", RowBox[List[FractionBox["1", "r"], "+", "\[Nu]"]]]], "}"]], ",", RowBox[List["{", RowBox[List[RowBox[List["1", "+", FractionBox["1", "r"], "+", "\[Nu]"]], ",", RowBox[List["1", "+", RowBox[List["2", " ", "\[Nu]"]]]]]], "}"]], ",", RowBox[List[RowBox[List["-", "2"]], " ", "\[ImaginaryI]", " ", "a", " ", SuperscriptBox["z", "r"]]]]], "]"]]]], RowBox[List[RowBox[List["(", RowBox[List["1", "+", RowBox[List["r", " ", "\[Nu]"]]]], ")"]], " ", RowBox[List["Gamma", "[", RowBox[List["1", "+", "\[Nu]"]], "]"]]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 3,090 | 8,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.194737 |
https://www.certicom.com/content/certicom/en/21-elliptic-curve-addition-a-geometric-approach.html | 1,553,390,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203123.91/warc/CC-MAIN-20190324002035-20190324024035-00501.warc.gz | 728,591,255 | 6,512 | P + Q = R is the additive property defined geometrically.
Elliptic curve groups are additive groups; that is, their basic function is addition. The addition of two points in an elliptic curve is defined geometrically.
The negative of a point P = (xP,yP) is its reflection in the x-axis: the point -P is (xP,-yP). Notice that for each point P on an elliptic curve, the point -P is also on the curve.
### 2.1.1
Adding distinct points P and Q
Suppose that P and Q are two distinct points on an elliptic curve, and the P is not -Q. To add the points P and Q, a line is drawn through the two points. This line will intersect the elliptic curve in exactly one more point, call -R. The point -R is reflected in the x-axis to the point R. The law for addition in an elliptic curve group is P + Q = R. For example: | 211 | 810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-13 | longest | en | 0.960067 |
https://www.sepsis-q.cz/2022-07-15+14625.html | 1,695,674,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00739.warc.gz | 1,079,083,568 | 8,812 | ## Convert Liter/minute to Cubic Meter/second - Cubic Yards to Cubic Meters conversion
Apr 30, 2021· Boy is our Air Bad! Image by Ralf Vetterle from Pixabay. We wanted the air to be monitored and it was not a good report for benzene. A new effort to measure the levels of benzene, a cancer-causing air pollutant, along the perimeters of U.S. refineries found that five of the 13 facilities with the highest levels are in Louisiana.Apr 21, 2021· The 72-million-cubic-meter wall of water behind the dam is drained—gradually to prevent flooding downstream. Steel sheet piling along the crest, along with fish …
## Cubic Meters Per Second to Cubic Feet Per Second | Kyle's - How the Largest Dam Removal Project in History Will ...
Cubic Meter to Cubic Feet Conversion Example. Task: Convert 50 cubic meters to cubic feet (show work) Formula: m 3 ÷ 0.0283168466 = ft 3 Calculations: 50 m 3 ÷ 0.0283168466 = 1,765.73333558 ft 3 Result: 50 m 3 is equal to 1,765.73333558 ft 3.ADVANCED METER DATA INTO THE CORE BALANCING PROCESS (A.17-10-002) (3RD DATA REQUEST FROM SOUTHERN CALIFORNIA GENERATION COALITION AND INDICATED SHIPPERS) _____ 6 • The second scheduled Data Warehouse Load process begins at 2:30 PM and transfers hourly interval data in cubic feet that has been through the VEE process.
## Governors' Biofuels Coalition - Gallon to Cubic Meter Conversion (gal to m³)
The Iowa Air Coalition. (515) 242-6131. EH/PH agencies can order Radon Test kits by contacting one of the following: Radon Hotline 1-800-383-5992. Order radon kits from Air Check ( ) Contact Makala McFee ( makala@radon ), customer service and Order Processing to set up an account ensure that you get discounted kits. Health House.Mar 25, 2018· Readers will be interested in the findings of a recent study completed by Cary Kopczynski & Company, that compares the cost of the structural frame of a hypothetical 10-story residential building located in the Pacific Northwest using cross laminated timber (CLT) …
## Convert billion cubic metre to cubic metre - Conversion of - National Ambient Air Quality Standards (NAAQS)
Cubic Meters : The cubic meter (also written "cubic meter", symbol: m3) is the SI derived unit of volume. It is defined as the volume of a cube with edges one meter in length. Another name, not widely used any more, is the kilolitre. It is sometimes abbreviated to cu m, m3, M3, m^3, m**3, CBM, cbm. Volume and Capacity Conversion CalculatorTONNE OF COAL EQUIVALENT TO TONNE/CUBIC METER (T coal TO T/m³) CHART 1 tonne of coal equivalent in tonne/cubic meter = 29307600.
## Rails-with-Trails: Lessons Learned - Gallons to Cubic Meters conversion
Cubic Meters. One cubic meter is equal to the volume of a cube with each edge measuring one meter. The cubic meter, or cubic metre, is the SI derived unit for volume in the metric system. Cubic meters can be abbreviated as m³, and are also sometimes abbreviated as cu m, CBM, cbm, or MTQ. For example, 1 cubic meter can be written as 1 m³, 1 cu ...100 Cubic Meters Per Second to Cubic Feet Per Second = 3531.4667. 5 Cubic Meters Per Second to Cubic Feet Per Second = 176.5733. 200 Cubic Meters Per Second to Cubic Feet Per Second = 7062.9333. 6 Cubic Meters Per Second to Cubic Feet Per Second = 211.888. 300 Cubic Meters Per Second to Cubic Feet Per Second = 10594.4.
## The true cost of water | Greenbiz - tonne of coal equivalent to tonne/cubic meter (T coal to T ...
Jul 22, 2018· Cubic Meters. A metric unit of volume, commonly used in expressing concentrations of a chemical in a volume of air. One cubic meter equals 35.3 cubic feet or 1.3 cubic yards. One cubic meter also equals 1000 liters or one million cubic centimeters.This is a very easy to use cubic yard to cubic meter converter.First of all just type the cubic yard (yd³) value in the text field of the conversion form to start converting yd³ to m³, then select the decimals value and finally hit convert button if auto calculation didn't work.Cubic Meter value will be converted automatically as you type.. The decimals value is the number of digits to be ...
## Governors' Wind Energy Coalition - Cubic Meters conversion calculators, tables and forumas
We assume you are converting between kilogram [water] and cubic metre. You can view more details on each measurement unit: kilogram [water] or cubic metre The SI derived unit for volume is the cubic meter. 1 kilogram [water] is equal to 0.001 cubic meter.B. It must have ventilation equipment installed capable of removing smoke particles as small as 0.01 microns and maintaining a particulate matter PM 2.5 level equal to or less than 35.4 micrograms per cubic meter, which matches the EPA standard for good to moderate air quality. C).
## Tighter silica rules needed to protect miners from black - Water & Sewer Service Charges - GCWW - Cincinnati
Mostly used in measuring natural gas reserves. 1 Trillion cubic feet is equal to 28 316 846 592 cubic meters (SI base unit). 1 TCF = 28316846592 m 3. 1 Cubic Meter: Volume made by a cube having one meter per side. 1 m * 1 m * 1 m. Link to Your Exact Conversion;In order to execute pallet shippings as efficiently as possible, various calculation units are frequently used in road freight transport. Cubic meter (m3) is an example of a calculation unit with which both shippers as well as freight forwarders, carriers and logistics companies are confronted with on a daily basis.. Computing the cubic meters of cargo can often be a hassle.
## COALITION FOR FAIR LUMBER IMPORTS - Comparing the Costs of Cross Laminated Timber and ...
The current annual standard — 12 micrograms per cubic meter of air — should be reduced to between 8 and 10 micrograms per cubic meter of air, the panel said; the yearly threshold of 35 micrograms per cubic meter of air should similarly be chopped to a number between 25 and 30 micrograms per cubic meter …Conversion between cubic meter and square meter. Cubic meter to Square meter Calculator
## Convert cubic meters to cubic feet - Volume Conversions - Cubic Millimeters to Cubic Meters conversion
This is a very easy to use gallon to cubic meter converter.First of all just type the gallon (gal) value in the text field of the conversion form to start converting gal to m³, then select the decimals value and finally hit convert button if auto calculation didn't work.Cubic Meter value will be converted automatically as you type.. The decimals value is the number of digits to be calculated ...1 Cubic meter (m3) is equal to 264.172052 gallons. To convert cubic meters to gallons, multiply the cubic meter value by 264.172052. For example, to find out how many gallons there are in 2 cubic meters, multiply 2 by 264.172052, that makes 528.3441 gallons in 2 m3.
## How many centimeters are in a standard ruler? - Quora - Metric ton per cubic meter (t/m3) Conversion - Density ...
parts per billion (ppb) by volume, and micrograms per cubic meter of air (µg/m3). (1) Final rule signed October 15, 2008. The 1978 lead standard (1.5 µg/m3 as a quarterly average) remains in effect until one year after an area is designated for the 2008 standard, except that in areas designated nonattainment for the 1978, the 19781.33 fl oz to cubic meter conversion result above is displayed in three different forms: as a decimal (which could be rounded), in scientific notation (scientific form, standard index form or standard form in the United Kingdom) and as a fraction (exact result).
## Calculate cubic meters for freight: use Quicargo's m3 - Convert million gallon per day to cubic metre/day ...
Apr 14, 2021· This growing coalition is calling on MSHA to lower the Permissible Exposure Limit for silica from 100 to 50 micrograms per cubic meter, which would grant coal miners the same protections as workers in every other industry. Miners and their advocates are also demanding that MSHA implement effective enforcement mechanisms targeting silica levels ...EPA's current annual standard, set in 2012, is 12 micrograms per cubic meter of air. But the report notes that U.S. studies "indicate a linear relationship at levels as low as" 5 micrograms per cubic meter. Particularly vulnerable are children, based on "strong evidence of impaired lung function growth," and minorities, the draft adds.
# LCDY
خلال 30 عامًا من العمل الشاق ، بنى موظفو LCDY تفوقًا في المصداقية والجودة الممتازة وخدمة العلامة التجارية "LCDY"
#### ابقى على تواصل
رقم 1688 ، طريق Gaoke شرق ، حي بودونغ الجديد ، شنغهاي ، الصين.
#### النشرة الإخبارية
تقوم الشركة بشكل أساسي بتصنيع الكسارات المتنقلة والكسارات الثابتة وآلات صنع الرمل
حقوق النشر © 2023.LCDY كل الحقوق محفوظة.خريطة الموقع | 2,178 | 8,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.851357 |
https://www.vuvickypro.com/2019/08/cs302-current-paper-final-spring-2019.html | 1,696,094,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00696.warc.gz | 1,175,944,168 | 51,219 | **Keep visiting this site and link. Same links will be updated with time**
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PAPER 1:
5 marks
Draw circuit diagram of binary weighted inputs convertor
Write test vector of 3 bit up down counter
Timing diagram of frequency dividor. We have to make the waveform of the final output
Other short questions:
Characteristics of serial in serial out shift register
3 guidelines of state assignment
2 applcations of ROM
Why certain counters like 74hc160 is called pre-set counters
How many bits a flip flop can save.
Purpose of stop bits and start bit
Which is the normal mode of operation of PROM (exactly yehi question statement tha confusing sa)
A circuit has 2 flip flops. What will be the power dissipation of the whole circuit is 5 volts and 5 mA current is applied to it. (I wrote 50mW. As for each flip flop its 25mW so 25+25 = 50..not sure)
PAPER 2:
QNO1: Analyze the following timing dagiram and tell which counter is being depicted by the diagram.
QNO2: give three guidelines for the selection of state assignment?
QNO3: what does the statement given below (using PLD programming) mean? “YPIN23 ISTYPE.’com’”
QNO4: draw a diagram of a 3 bit-up-down counter. Use an external inputX; when X sets to logic 1, the counter downward otherwise countup wards.
QNO5: what is meant by state assignment?
QNO6: on the fifth clock pluse a 4 bit jhonson sequence is Q0=0, Q1=1, Q2=1, Q3=1. What will be the sequence after sixth clock pluse.
Assume Q3 represent the least significant.
QNO7: what is the minimum sample frequency if the highest frequency component in analog signalis 20KHZ?
QNO8: what is meant by stop time and hold time of a flip flop?
Paper 3:
CS302 Final Term Paper
2 Marks Question:
1. In ABEL, an Input (source) file contains the module which has three sections. Name any two.
2. Write down two functions of registers.
3. Suppose a 2-bit up-counter, having states, “A, B, C, D”. Write down GOTO statements to show how present states change to next states.
4. Missing code is one of the Analog-to-Digital converters errors. Name other two types of errors.
3 Marks Question:
5. Convert the decimal 5846 number to octal using repeated division method, write down all steps.
6. Write down three characteristics of serial in/serial out 4-bit right shift register.
7. Draw a circuit diagram of flip-flop based static memory cell.
8. Draw the circuit diagram of operational amplifier used as an inverting amplifier.
5 Marks Question:
9. Convert the following numbers into its equivalent decimal number with complete procedure 1101101.1011.
10. Consider a state sequence a,f,d,c,a,b,c,f,d,c. Starting from initial state “a”. draw a table for the inputs and outputs for the state diagram given below.
11. Mealy Machine Per Tha. State Table
12. Draw the circuit diagram of binary-weighted-input digital to analog converter.
a | 778 | 2,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | longest | en | 0.890934 |
https://www.shaalaa.com/question-bank-solutions/resolving-power-microscope-telescope-the-numerical-aperture-objective-microscope-012-limit-resolution-when-light-wavelength-6000a-used-view-object_569 | 1,544,917,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00231.warc.gz | 1,045,853,310 | 13,987 | HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution - The Numerical Aperture of Objective of a Microscope is 0.12. the Limit of Resolution, When Light of Wavelength 6000A is Used to View an Object is - HSC Science (Electronics) 12th Board Exam - Physics
ConceptResolving Power of a Microscope and Telescope
#### Question
The numerical aperture of objective of a microscope is 0.12. The limit of resolution, when light of wavelength 6000A is used to view an object is.......................
1. 0.25 * 10-7m
2. 2.5 * 10-7m
3. 25 * 10-7m
4. 250 * 10-7m
#### Solution
(c) 25 * 10-7m
The minimum resolvable linear distance between two nearby objects is given by
"x"=lambda/(2"NA")=(6000xx10^-10)/(2xx0.12)=2.5xx10^-6"m"=25xx10^-7"m"
Is there an error in this question or solution?
#### APPEARS IN
Solution for question: The Numerical Aperture of Objective of a Microscope is 0.12. the Limit of Resolution, When Light of Wavelength 6000A is Used to View an Object is concept: Resolving Power of a Microscope and Telescope. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)
S | 351 | 1,194 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-51 | longest | en | 0.779101 |
https://de.mathworks.com/matlabcentral/answers/506950-getting-the-value-of-a-graph-at-a-certain-point | 1,639,060,150,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00491.warc.gz | 284,067,041 | 24,667 | # Getting the value of a graph at a certain point
110 views (last 30 days)
prateek bhadauria on 22 Feb 2020
i want to know that how can i find the green line graph values at equidistance at x axis interval (0-5) ,i want to know the exact values of x and y coordinates at a particular intervals and at equidistance ie -like this when x is 0 then y value is 0.35,then again same like this how could i find the x and y values of graph,i have to evaluate the points at equidistance .
##### 2 CommentsShowHide 1 older comment
prateek bhadauria on 2 Mar 2020
Thank you for your response darova and sorry for late reply also , i have attach the graph properly and i want to find out the MSE(Mean sqaure error ) so how could i proceed kindly suggest .
Priyanshu Mishra on 25 Feb 2020
Hi Prateek,
You can use interp1 function to know the y values for the corresponding x values. You may also refer to follwing example taken from the documentaion of interp1 function.
x = 0:pi/4:2*pi;
v = sin(x);
xq = 0:pi/16:2*pi;
figure
vq1 = interp1(x,v,xq);
plot(x,v,'o',xq,vq1,':.');
xlim([0 2*pi]);
title('(Default) Linear Interpolation');
prateek bhadauria on 2 Mar 2020
Thank you response priyanshu this is helpful for me to know the basics of interp 1 function but one step ahead i want to know the MSE(Mean Square Error) at the peaks of graphs which is like first peak at the (4, 0.61) so i want to find the mse at this interval when the graph rises from (0.35 to 0.61 on y axis when the value 4 at x axis) . | 431 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-49 | latest | en | 0.828315 |
https://nyx.meccahosting.com/~a0008617/ti_programs/tislvpoly2.htm | 1,642,874,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00511.warc.gz | 472,122,982 | 4,473 | E-mail Webmaster
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SOLVING POLYNOMIALS BY SUCCESSIVE SYNTHETIC DIVISION
TI-82 & TI-83/Plus
This is a menu driven program that solves polynomials by successively reducing the degree of the polynomial by synthetic division to a quadratic. Although the quadratic could be solved by further synthetic division if it has real zeros, the quadratic solution will provide a solution if the roots are either imaginary or real. Nevertheless, it is left to the student to choose the method from the menu. A quadratic with either real or imaginary roots may be solved by selecting item 6 from the menu after first entering the quadratic coefficients.
To conserve memory, and in deference to those who are going to enter this program by hand, I have written this program so that all coefficients are entered at the same time. As an example, suppose that you want to divide x3+4x2 +6x +3 by x+1. When the program prompts "ZERO=" respond with -1. When it prompts "COEFFS=", respond with {1,4,6,3}. Be sure to use braces rather than brackets or parentheses. Since you’re inputting to a list, L1, you’ll get an error if you don’t. This is a program using lists. So, if you get errors, first thing to look for is parentheses or brackets where there should be braces.
Memory Usage etc.: This version of the program uses 413 bytes of memory. I estimate it will take an inexperienced programmer about 20 minutes to enter by hand. If you don't know how to enter the coding, you can find it in Chapter 16 of the TI-83 User's Guide. I may also include some often-used entries on my web page. Look for it if you need it.
DISCLAIMER: This program is free, and, therefore, I make no claims about it's efficacy, efficiency, or proper operation. If you find a problem with this program, or can suggest an improvement, please e-mail me at knosummath@aol.com .
Use of this Program: You may use this program freely for your own personal use and for the use of other students, but use for publication or any means of profit requires my permission.
Revisions: Version V1.0, date 8/28/03. Version V1.1, date 10/2/04
SLVPOLY2 Disp "V1.1" Disp "FKizer" Lbl 0 Lbl 5 Input "COEFFS=",L1 2→I dim(L1)→N {0}→L2 N→dim(L2) Lbl 7 Menu ("SELECT","FIRST ZERO",1, "RETRY",2, "NEXT ZERO",3,"SLV RED QDRTC",4,"STRT OVR",5,"SLV NEW QDRTC",6 ) Lbl 1 Input "ZERO=", Z L1(1)→L2(1) For (I,2,N) Z*L2(I-1)+L1(I)→L2(I) End Disp "NEW EQUA",L2 Disp "REM=", L2(N) Pause Goto 7 Lbl 2 Goto 1 Lbl 3 N-1® N N® dim(L1) N® dim(L2) L2® L1 Goto 1 Lbl 4 L2(1) ® A L2(2) ® B L2(3) ® C a +bi Disp "X1 IS ",(-B+Ö (B2-4AC))/(2A) Disp "AND X2 IS ",(-B-Ö (B2-4AC))/(2A) Stop Lbl 6 L1→L2 Goto 4
Revised 10/1/04 | 1,099 | 3,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.672292 |
http://mathhelpforum.com/advanced-algebra/36045-linear-algebra.html | 1,481,111,839,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542060.60/warc/CC-MAIN-20161202170902-00349-ip-10-31-129-80.ec2.internal.warc.gz | 175,781,983 | 10,523 | 1. ## Linear Algebra
Is anyone able to help me do the following proof please:
Let f:R3 > R3 be given by
f(x1,x2,x3)=(x3,x1+x2,x1-x2)
Prove that f is a linear transformation.
Many thanks to anyone who can help.
2. Hi
You have to prove two things :
• For all $x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix},\,y=\b egin{pmatrix}y_1\\y_2\\y_3\end{pmatrix} \in \mathbb{R}^3$, $f(x+y)=f(x)+f(y)$
• For all $\lambda \in \mathbb{R}, \,x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\in\m athbb{R}^3$, $f(\lambda x)=\lambda f(x)$
3. Originally Posted by flyingsquirrel
Hi
You have to prove two things :
• For all $x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix},\,y=\b egin{pmatrix}y_1\\y_2\\y_3\end{pmatrix} \in \mathbb{R}^3$, $f(x+y)=f(x)+f(y)$
• For all $\lambda \in \mathbb{R}, \,x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\in\m athbb{R}^3$, $f(\lambda x)=\lambda f(x)$
Or obtain a matrix A such that f(x) = Ax(You should know the result that this is always a linear map).
If $x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\in\mat hbb{R}^3, \,\, A = \begin{pmatrix}0&0&1\\1&1&0\\1&-1&0\end{pmatrix}$ | 476 | 1,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2016-50 | longest | en | 0.502808 |
https://www.lancaster.ac.uk/stor-i-student-sites/ben-lowery/2022/04/the-monty-hall-problem-and-its-generalisations-part-2/ | 1,721,777,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00531.warc.gz | 747,211,323 | 22,652 | # The Monty Hall problem and its generalisations: Part 2
In the previous blog post we looked at the infamous Monty Hall problem and its controversial (but correct) solution. The main problem has been talk of countless blog posts, news articles, and youtube videos; providing a nice introduction to probability and Bayes theorem. And while it is fun to rehash the same story, it might be worth looking at how to broaden the problem and see how the same core principles can be applied to a more obtuse setting. With this we can explore some of these re-formulations, starting with expanding the number of doors in our game.
## Monty Hall: Live from the Hilbert Hotel!
Let’s envision the following fictitious scenario; in which after the success of his three door final showdown, Monty and his team have been gifted a bigger budget to make a more elaborate show, and a new, possibly infinitely big studio. Here, Monty utilises this increased budget to order his producers to purchase more doors. Now that he possesses a studio filled to the brim of disused doors, Monty again places one car behind a door and keeps count of where it is placed. While a flood of goats trundle in and hide behind the rest, he asks a contestant, now slightly more intimidated than their predecessors (see below gif), to pick a door. The contestant hesitantly chooses, giving way to our host opening every door but the contestants and one final door.
Now we again pose the question, stick or switch? Given the information we attained from the original Monty Hall situation, it makes sense to have an intuitive guess that switching will be in the contestants best interest. And we can test this again by using Bayes’ Theorem and some basic Probability theory.
Like in the original incarnation, we define events and variables for the new version. Instead of 3 doors, we now possess $d$ doors. Each of these doors are assigned the event it may possess the prize behind it, we can define these events as $D_1,...,D_d$. While also allowing $G$ to be the event we open all but doors 1 and $d$ to reveal a goat. So with this in mind, we can formulate the following Bayes equation for any door in particular, say $i$:
$$\mathbb{P}[D_i|G]=\frac{\mathbb{P}[D_i]\mathbb{P}[G|D_i]}{\mathbb{P}[G]}.$$
The individual probabilities for the right hand side are calculated as:
$$\mathbb{P}[D_1]=...=\mathbb{P}[D_d]=1/d \\ \mathbb{P}[G|D_1]=\frac{1}{d-1} \\ \textrm{(As we are just restricted to opening every other door if the prize is here)} \\ \mathbb{P}[G|D_d]=1 \\ \textrm{(As we are just restricted to opening every other door if the prize is here)} \\ \mathbb{P}[G|D_2]=...=\mathbb{P}[G|D_{d-1}]=0 \\ \textrm{(If the prize lies in all these doors we want to open, we clearly can’t open them).}$$
Since $D_1$ to $D_d$ cannot occur simultaneously, then we can use some more simple statistical properties, specifically mutually exclusivity, to express the probability of a goat behind all but doors 1 and $d$ as follows in this slightly long, but hopefully intuitive derivation:
$$\mathbb{P}[G]=\sum_{i=1}^d \mathbb{P}[D_i]\mathbb{P}[G|D_i]\\= \mathbb{P}[D_1]\mathbb{P}[G|D_1]+ \mathbb{P}[D_d]\mathbb{P}[G|D_d]+\sum_{i=2}^{d-1}\mathbb{P}[D_i]\mathbb{P}[G|D_i]\\ =\frac{1}{d}\cdot \frac{1}{d-1}+\frac{1}{d}\cdot 1=\frac{1}{d}\left(\frac{1}{d-1}+1\right)=\frac{1}{d}\left(\frac{d}{d-1}\right).$$
Remember we opened all but door 1 and $d$, so all doors in-between will have a probability 0 of having the car behind it. Thus, we substitute the above derivations back into Bayes Theorem equations, but only for doors 1 and $d$ are,
$$\mathbb{P}[D_1|G]=\frac{1/d\cdot (1/d-1)}{1/d\cdot d/(d-1)}=\frac{1}{d} \\ \mathbb{P}[D_d|G]=\frac{1/d\cdot 1}{1/d\cdot d/(d-1)}=\frac{d-1}{d}$$
While tedious, this derivation is pivotal in allowing a generalisation of the problem. Generalisations are crucial in mathematics, allowing us to expand our problem from an initial set of constrained numbers (like only 3 available doors) to as many doors as we want, and all we need to do is plug that number into $d$.
To test this – and provide a little sanity check – we can hark back to our original Monty Hall Problem and seeing that substituting 3 doors gives us probabilities of switching and staying as 2/3 and 1/3 respectively. So now we have a rather straightforward method to show, no matter how many doors we have, if we open all but one door and the original door, it is in our best interest to switch. In addition to this, although trivial to point out, we see that with more doors, our likelihood of winning when switching increases. For example, with $d=7$ doors, we should, by plugging our values in, attain the staying probability (i.e. door 1) as 1/7, and switching (to door $d$) as 6/7.
To see this in practice let’s run some simulations for 3, 5, 7 and 9 doors after carrying out Monty Hall’s deal 3000 times.
## Winning isn’t everything
Our second generalisation into Monty Hall’s problem is one in which we are looking to try flip the odds back into the favour of the host. Given the generosity of winning when we expand the problem to $d$ doors, it is now worth seeing if limiting the number of doors Monty opens can make it more – or less – likely for the contestant to win when switching. If we think about this in a logical manner, it should be the case that now we have the option of which door we can switch to, we are less likely to get the prize than in the previous scenario if we switch. But it is worth calculating how much of a detriment this new rule is to our contestant. And analysing how drastically our odds can change by opening less and less doors.
This time we will take a scheduled commercial break out of laziness to relieve ourselves of any more probability equations and focus solely on numerical computations. Consider an example where, given $d$ doors, we open $k$ of these. More specifically let’s look at the case of having 10 doors and we open a subset of these, analysing the number of times we win if we switch, we win if we stay, and the new third case that we don’t win if we stayed or switched. This is seen in the following graph.
As we can see in the choice between 10 doors, when opening 8 (which is the max number of doors we can open) and opening 6 (in which we have a choice as to what we can switch to), there exists a significant dip in the probability of winning when switching, with it then being more likely to not win the game whatever we do. This is due to the truly random choice we now have with the selection of the door we might want to switch to. Despite all these changes, the chance of switching consistently gives better odds than staying. We can see this in the generalisation of opening $k$ from a set of $d$ doors. This is given as the following equation:
$$\mathbb{P}[\textrm{Winning when switching}]=\left(\frac{d-1}{d}\right)\cdot\left(\frac{1}{d-k-1}\right)$$
For those in dire need of a Bayesian derivation (as I normally am), one can refer here. As a little test, we can take $d=10$ and $k=2$ and see that the probability of winning when switching as $\approx 0.129$, which leaves our simulation pretty damn close to what we want.
And with this we can conclude our investigation into suspected goat farmer Monty Hall and his mystery doors. But was this investigation as concrete as the numbers suggest?
## Statistical stage fright
In these two blogs we’ve seen how applying some mathematical rigour allows us to understand, dissect and create advantages in a game of seemingly random luck. With this being said, as often is the case of applying Mathematics to the real world, our logical reasoning may still not be perfect, nor reveal the true solution to the problem. since arguments could be made in that randomising the choices of contestants in the simulation and using conditional probability detracts from the human element in the game. That in which the host, the atmosphere and the audience play a crucial role in the dilemma posed to the contestant, perhaps leading to a bias in the options available. This is something that probability and random simulations simply cannot account for. Hence it could even be contested that in reality, based on the host’s hints and approach towards the contestant, the probability of finding the winning car can range from 1/2 to 1. A paper on this dilemma of the human element can be see here.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,129 | 8,466 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 26, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-30 | latest | en | 0.936466 |
https://www.questia.com/library/journal/1P3-2191380001/subjective-randomness-and-natural-scene-statistics | 1,586,522,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00356.warc.gz | 1,088,523,997 | 16,670 | Academic journal article Psychonomic Bulletin & Review
# Subjective Randomness and Natural Scene Statistics
Academic journal article Psychonomic Bulletin & Review
# Subjective Randomness and Natural Scene Statistics
## Article excerpt
Accounts of subjective randomness suggest that people consider a stimulus random when they cannot detect any regularities characterizing the structure of that stimulus. We explored the possibility that the regularities people detect are shaped by the statistics of their natural environment. We did this by testing the hypothesis that people's perception of randomness in two-dimensional binary arrays (images with two levels of intensity) is inversely related to the probability with which the array's pattern would be encountered in nature. We estimated natural scene probabilities for small binary arrays by tabulating the frequencies with which each pattern of cell values appears. We then conducted an experiment in which we collected human randomness judgments. The results show an inverse relationship between people's perceived randomness of an array pattern and the probability of the pattern appearing in nature.
(ProQuest: ... denotes formulae omitted.)
People are very sensitive to deviations from their expectations about randomness. For example, the game Yahtzee involves repeatedly rolling 5 six-sided dice. If you were to roll all sixes 6 times in a row, you would probably be quite surprised. The probability of such a sequence arising by chance is 1/630. However, the low probability of such an event is not sufficient to explain its apparent nonrandomness, since any other ordered sequence of the same number of dice rolls has the same probability. Consequently, recent accounts of human subjective randomness (our sense of the extent to which an event seems random) have focused on the regularities in an event. These regularities suggest that a process other than chance might be at work (Falk & Konold, 1997; Feldman, 1996, 1997; Griffiths & Tenenbaum, 2003, 2004). The basic idea behind these accounts is that stimuli will appear random when they do not express any regularities.
An important challenge for any account of subjective randomness based on the presence of regularities is to explain why people should be sensitive to a particular set of regularities. In the example given above, systematic runs of the same number may suggest loaded dice or some other nonrandom process influencing the outcomes. However, for other kinds of stimuli, such as the one- or twodimensional binary arrays used in many subjective randomness experiments, explanations are more difficult to come by. A common finding in these experiments is that people consider arrays in which cells take different values from their neighbors (such as the one-dimensional array 0010101101) more random than arrays in which cells take the same values as their neighbors (such as 0000011111) (Falk & Konold, 1997). This result makes it clear that people are sensitive to certain regularities, such as cells having the same values as their neighbors. However, it is difficult to explain why these regularities should be more important than others that seem a priori plausible, such as neighboring cells differing in their values.
In this article, we explore a possible explanation for the origins of the regularities that influence subjective randomness judgments for one class of stimuli: two-dimensional binary arrays. These stimuli are essentially images, with the cells in the array having the appearance of a grid of black and white pixels (see Figure 1). We might thus expect that the kinds of regularities detected by the visual system should play an important role in determining their perceived randomness. A great deal of recent research suggests that the human visual cortex efficiently codes for the structure of natural scenes-scenes containing natural elements, such as trees, flowers, and shrubs, that represent the visual environment in which humans evolved (Olshausen & Field, 2000; Simoncelli & Olshausen, 2001). We consider the possibility that the kinds of regularities that people detect in two-dimensional binary arrays are those that are characteristic of natural scenes. …
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# Lab One Linear Motion
Lab Assignment 1: Linear Motion
Instructor’s Overview
This lab experiment focuses on freely falling objects. We developed the kinematic equations for freely falling objects in Module 2. Now we will have an opportunity to directly experiment with free fall and apply our quantitative knowledge of kinematics.
This activity is based on Lab 6 of the eScience Lab kit. Although you should read all of the content in Lab 6, we will be performing a targeted subset of the eScience experiments.
Our lab consists of two components. These components are described in detail in the eScience manual (pages 71-73).
This document serves as your lab report. Please include detailed descriptions of your experimental methods and observations.
EXPERIMENT 1: Time vs Distance of a Dropped Object.
Reference: Section 2-7 in your text.
Theory: When a object is dropped the magnitude of its velocity (speed) increases as it is accelerated by the force of gravity. Assuming air friction is negligible all objects will have accelerated (increase their speed) by 9.8 m/s every second. At the end of 1 second the speed is 9.8 m/s, at the end of 2 seconds the speed is 19.6 m/s, etc. This acceleration, the acceleration due to gravity is named g. We write this as 9.8 m/s per second, which then becomes 9.8 m/s/s and then 9.8 m/s2. We say g is 9.8 meters per second squared.
For a falling object the following equations prevail:
y = ½ g t2 and t =
y is the distance dropped and t is the time from releasing the object until it hits the floor. Note other letters are sometimes used for the variable, d being a popular choice.
Procedure:
You will drop one of the hex nuts from a height of about 6 ft. Stand near a wall, door jam, post of some kind and place a tape or have some other way of marking the height you will drop the hex nut from. Record this height.
With a stop watch time the drop of the hex nut and record in the data table below. Repeat this for a total of 10 trials.
Height of Drop: ___________
Data table for hex nut drop experiment:
Drop Number Time (sec) 1 2 3 4 5 6 7 8 9 10 Average Standard Deviation
Calculations:
Using the average time from the table calculate the distance the hex nut dropped and compare to your recorded distance. Discuss sources of error. What does the Standard Deviation of your table imply? By-the-way, average reaction time for most people is about 0.2 seconds. That implies if you are driving and see a change in the traffic light it will take 0.2 seconds before your foot begins to move.
EXPERIMENT 2 – Time intervals of a series of dropped objects.
In this experiment you will make two strings of separated hex nuts and explore the interval of time between the impacts on the floor.
:
Construction:
String 1: Evenly spaced hex nuts:
d = 0.1 m (10 cm). Tie the first nut 10 cm from the end of the string and then continue tying hex nuts 10 cm above the previous until 6 hex nuts are tied to the string.
Record the distance the hex nut is from the end of the string; this will be the distance the hex nut falls to the floor.
Hex Nut Y (m) 1 0.1 2 3 4 5 6
Procedure:
Consider placing a pan lid, plate, or perhaps aluminum foil on the floor to make the impact sound louder.
With the end of the string just touching the floor and you holding up the extended string drop the string and listen to the impacts.
Results: Record your observation of the impacts. Do the succeeding impacts occur in equal time intervals or are the time intervals different and if so how?
String 2: Unevenly space hex nuts:
d = 0.1 m (10 cm). Tie the first nut 10 cm from the end of the string and then continue tying hex nuts with the spacing of: 30 cm, 50 cm, and 70 cm.
Again record the distance the nuts are from the end of the string. This will be the distance they fall to the floor.
Hex Nut Y (m) 1 0.1 2 3 4
Procedure:
With the end of the string just touching the floor and you holding up the extended string drop the string and listen to the impacts.
Results: Record your observation of the impacts. Do the succeeding impacts occur in equal time intervals or are the time intervals different and if so how?
What was the difference between the noise patterns for equally spaced nuts compared to the second spacing given to you?
Analysis and Discussion
ï What gives a falling object its acceleration?
ï At any given time, all of the nuts should have the same velocity. Why are the time intervals between the evenly spaced nuts different?
To understand this quantitatively, consider the following diagrams that model our string-hex nut systems:
Set d = 10 cm (0.1 m)
Using the kinematic equations, calculate the time it takes for each of the equally-spaced masses in the left diagram to hit the ground. Perform the same calculation for the staggered mass system in the right diagram. How does the time change between impacts of successive masses in both scenarios? Does this agree with your observations?
ï In the staggered hex nut system that you dropped, which hex nut had the highest velocity when hitting the ground? Calculate this velocity using the proper kinematic equations.
ï Say you have a very long string and want the hex nuts to hit the ground 1 second apart. Using the kinematics equations, determine the spacing for 5 nuts to hit with equal timing. How much string would you need? (Hint: plug in time values starting with t = 1 second into the kinematics equations to find each nut distance.)
Conclusions
References | 1,296 | 5,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.916577 |
https://laptrinhx.com/the-power-of-pca-228289885/ | 1,628,042,642,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00008.warc.gz | 357,110,721 | 13,269 | ### What is PCA — Principal Component Analysis?
Principal Component Analysis or PCA for short is a dimensionality reduction method, which is used to reduce the dimensions of large datasets by reducing the number of features within the dataset. But as you reduce the number of features, you will deal with some trade-off between accuracy and the number of features. It is obvious that as you will be reducing the dimension of the dataset, you will lose some “knowledge” about the dataset and thus lose some accuracy. PCA is very helpful because its result work, a dataset with reduced dimensions is easier to utilize, explore and visualize. Moreover, a subsequent benefit of PCA is that with this machine learning algorithms will perform faster and development speed gets faster. To make things clear let’s define some terminologies I used in this article:
• Feature
• Dimension
• Visualization of datasets
Imagine we have a tabular dataset with 100 columns. These columns form our features. And thus our dataset made up of 100 features. The number of rows on each column would define our dataset’s dimension. For instance, for each 100 hundred columns we have 50 different records(rows) of some data. These 50 rows of data would define our dataset’s dimension. In order to be able to visualize this dataset, you would normally plot it into some 2D or 3D space. This process is called visualization of the dataset.
### Where is it used?
PCA is widely used across different disciplines such as data science projects, face recognition, image compression. Let me explain the example of a data science project — imagine you have tabular data with hundreds of features(columns) and as we already know, in such kinds of projects, the visualization of such dataset is very important, so visualizing the dataset with such size becomes almost impossible. To be able to visualize your dataset on 2D/3D space you will need to reduce the number of features to some small number. This is where PCA comes to help. PCA reduces the number of features that best describe the dataset into a small number: e.g 2–3.
### When it should be used?
So, among the main scenarios I can highlight TOP-3 use cases where PCA can be helpful:
1. Better Perspective and less Complexity: PCA is useful when we need to get an intuitive understanding of a given dataset and having so many features is not necessary.
2. Better Visualization: When we cannot get a good visualization due to a high number of dimensions we use PCA to reduce it into a shadow of 2D or 3D features
3. Reduce size: When we have too much data and the result of applying PCA would give us only a reduced number of features (e.g 1%) that would explain the maximum variance in our dataset.
### Is PCA always recommended to use?
Spoiler — No!
After all these praises towards this algorithm, now it is time to look at the other side of the coin. Applying PCA randomly in any project has always a bad idea. PCA may be very helpful on one kind of ML application but not preferred on another. So before applying PCA to any of your projects you should be aware of why you need it and lastly consider its pros and cons. Also, I want to note one important detail — before applying PCA, you should know whether your dataset has characteristics that make PCA the right choice for your project.
#### Limitations:
One of the main limitations of PCA is its linearity. PCA is a linear model that tries to find linear relationships between variables but in reality, your variables may have non-linear relationships, and the latter fact makes PCA a bad choice for your project. It is recommended to apply PCA only when it is really needed. And so here is my recommendation: Do not blindly apply PCA on every ML you work on, always consider its pros and cons before starting your project.
### Thanks for reading! If you enjoyed this article, please hit the clap button 👏 as many times as you can. It would mean a lot and encourage me to keep writing stories like this. Let’s connect on Twitter!🐦 Cheers!
The power of PCA. was originally published in DataDrivenInvestor on Medium, where people are continuing the conversation by highlighting and responding to this story. | 885 | 4,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.918672 |
http://support.hp.com/us-en/document/bpia5140 | 1,490,738,888,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189903.83/warc/CC-MAIN-20170322212949-00040-ip-10-233-31-227.ec2.internal.warc.gz | 346,188,688 | 24,740 | Actions
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# HP 6s Calculator - Using Statistics
Introduction
The HP 6s scientific calculator can be put into the statistics mode by pressing MODE, then SD. With the statistics mode, data can be entered and analyzed.
Calculator symbol key
The procedures in this document use the following text to represent symbol keys:
Key Description Text representation Sum of x. SUMx Sum of x2 SUMx2 Sample standard deviation STDV Population standard deviation PSTDV
Entering a list of data items to analyze
Press DATA after each data element. When DATA is pressed, the calculator displays the number of data elements entered. For example to enter a list of data consisting of 5, 8, and -3, use the following keystrokes: Press 5, DATA, 8, DATA, 3, +/-, then DATA.
To enter the results of a calculation as a data item, perform the calculation normally, and then press DATA when the answer is displayed.
Fixing the data entered into the calculator
• To cancel the last entry made, press C/CE before pressing DATA.
##### note:
When C/CE is pressed, “0” is displayed.
• Press INV, then n to display the number of data elements stored in the calculator. To delete a data item that was previously entered, enter the value again, then press INV, then DEL.
Using statistical functions on the data
Once a list of data values is entered the following statistical functions can be used.
Keys Description Press INV, then n To display the number of data elements entered. Press INV, then STDV To display the sample standard deviation. Pressing INV, then PSTDV To display the population standard deviation. Pressing INV, then xy To display the arithmetic mean. Pressing INV, then SUMx To display the sum of each data element. Pressing INV, then SUMx2 To display the sum of the data element squared.
Example of finding the standard deviation
Find the standard deviation of the data 5, 9, 13, and 6.
Keys Display Press MODE, SD, 5, DATA, 9, DATA, 13, DATA, 6, DATA, INV, then STDV SD3.593976442
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Actions | 558 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-13 | latest | en | 0.75666 |
https://discourse.julialang.org/t/reversediff-for-loss-function-with-zygote-derivatives/94382 | 1,685,851,491,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00298.warc.gz | 242,962,422 | 6,128 | # ReverseDiff for loss function with Zygote derivatives
I’m trying to use Flux.jl to find the solution to a function equation with a form a bit like g(f(x)[1]) = h(f’(x)), where f(x) is an unkown function to be approximated by a neural network.
The loss function for this setup needs to include the derivative of the neural network (with respect to inputs).
From reading posts from others that have had difficulty with this, I understand that currently the most reliable way to make that work is to use Zygote to take the derivative internal to the loss function and then ReverseDiff for the derivative of the loss function with respect to the neural network parameters.
I’m having trouble figuring out the syntax to make that work though…I keep getting errors.
Here’s a small example that I’m working on:
``````using Flux, Zygote, ReverseDiff
mod1 = Chain(Dense(1 => 4, relu),
Dense(4 => 4, relu),
Dense(4 => 3)) #Full scale example needs 3 outputs so including here to make sure it works
dMdx3(m,x) = [Zygote.gradient( w -> m(w)[3], y)[1][1] for y in x]
``````
``````julia> dMdx3(mod1,rand(3)) #checking to make sure this works
3-element Vector{Float64}:
-0.35708116668549555
-0.35708116668549555
-0.35708116668549555
``````
``````loss(f,x) = sum(dMdx3(f,x))
``````
``````julia> loss(mod1,rand(3)) #again checking functionality
-1.0712435000564866
``````
``````julia> ReverseDiff.gradient(m -> loss(m,rand(3)), mod1)
ERROR: MethodError: no method matching ReverseDiff.GradientConfig(::Chain{Tuple{Dense{typeof(relu), Matrix{Float32}, Vector{Float32}}, Dense{typeof(relu), Matrix{Float32}, Vector{Float32}}, Dense{typeof(identity), Matrix{Float32}, Vector{Float32}}}})
Closest candidates are:
ReverseDiff.GradientConfig(::Tuple, ::Type{D}) where D at C:\Users\Patrick\.julia\packages\ReverseDiff\YkVxM\src\api\Config.jl:45
ReverseDiff.GradientConfig(::Tuple, ::Type{D}, ::Vector{ReverseDiff.AbstractInstruction}) where D at C:\Users\Patrick\.julia\packages\ReverseDiff\YkVxM\src\api\Config.jl:45
...
Stacktrace:
[1] gradient(f::Function, input::Chain{Tuple{Dense{typeof(relu), Matrix{Float32}, Vector{Float32}}, Dense{typeof(relu), Matrix{Float32}, Vector{Float32}}, Dense{typeof(identity), Matrix{Float32}, Vector{Float32}}}})
[2] top-level scope
@ REPL[19]:1
``````
I got the basic syntax for using ReverseDiff to get the gradient of a neural network from the DiffEqFlux.jl documentation here. But it was used in the context of a Hamiltonian neural network so not sure if it’s correct here.
Would greatly value any advice on how to make this work properly.
I think I’ve got it figured out now. Code below borrows heavily from the source code for DiffEqFlux.jl
Big thanks to @ChrisRackauckas for figuring this out in DiffEqFlux.jl
``````using Flux
using ReverseDiff
#Create modified Flux NN type that carries destructured parameters
struct ModdNN{M, R, P}
model::M
re::R
p::P
function ModNN(model; p = nothing)
_p, re = Flux.destructure(model)
if p === nothing
p = _p
end
return new{typeof(model), typeof(re), typeof(p)}(model, re, p)
end
end
#function to compute the derivative of third NN output with respect to input
function diff3destruct(re, p, x)
H = [Flux.gradient(w -> re(p)(w)[3], [y])[1][1] for y in x]
end
#Create an instance of modified type
mod1 = ModNN(Chain(Dense(1 => 32, relu),
Dense(32 => 32, relu),
Dense(32 => 3)))
#Record parameter values to check that they're updating correctly
checkP = copy(mod1.p)
#Loss function (simple aggregation of NN outputs)
loss(re, p, x) = sum(diff3destruct(re, p, x))
#Get the gradients using Reverse diff
gs = ReverseDiff.gradient(p -> loss(mod1.re, p, rand(3)), mod1.p)
``````julia> maximum(mod1.p .- checkP) | 1,062 | 3,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-23 | latest | en | 0.72859 |
https://www.teacherspayteachers.com/Product/Fractions-4th-Grade-2316754 | 1,526,896,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00286.warc.gz | 825,211,444 | 19,891 | # Fractions 4th Grade
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Fractions 4th Grade: This Fractions 4th Grade PowerPoint with Practice Pages and Flipbook is packed full of important concepts and skills to encourage a deeper understanding of fractions. Sit back and let this Fractions 4th Grade PowerPoint teach your students everything they need to know about 4th Grade Fractions.
The following standards are represented in this product:
TEK 4.3a | TEK 4.3b | TEK 4.3c | TEK 4.3d | TEK 4.3e | TEK 4.3f | TEK 4.3g| 4.NF.A1 | 4.NF.A2| 4.NF.B.3 | 4.NF.C.5 | 4.NF.C.6 |
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This Fractions 4th Grade resource is easy enough for a substitute to provide a week full of meaningful content and productive learning experiences. The ready-to-use instructional guide will help students apply key concepts as they work to construct a deeper understanding of fractions. The practice pages will secure a better understanding of concept attainment. This product is bundled together with a Fraction Flipbook that is perfect for the interactive notebook or study guide. It also includes 2 of my favorite fraction activities. Both products contain much of the same content in completely different formats.
Use Fractions 4th Grade PowerPoint as a multi-lesson introduction to fractions or use it as a review with practice pages as a test prep activity. But don’t stop there….You can continue to use it by breaking it up into sections and placing each section on class computers for students to rotate through stations while completing their practice pages.
All of my math PowerPoints are editable and can be easily customized for your students’ specific needs.
★★★★★★However, the fraction font is not compatible on an Apple Device★★★★★★
In the Fractions 4th Grade PowerPoint with Practice Pages and Flipbook the key concepts include:
* represent fractional unit
* represent fractions on a number line
* understanding unit fractions
* compose and decompose fractions
* partitioning equal parts or set of objects
* equivalent fractions
* comparing fractions
* use fractions with multiple denominators
* adding and subtraction fractions
* properties of operation – associative, commutative
* relating fractions to decimals
* mixed numbers
* improper fractions
* benchmark fractions
* represent fractions and decimals to tenth and hundredths
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 811 | 3,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-22 | latest | en | 0.912177 |
https://oeis.org/A117901 | 1,680,324,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00571.warc.gz | 479,350,303 | 3,804 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A117901 Inverse of number triangle A117898. 3
1, -1, 1, 0, -2, 1, 0, 3, -2, 1, 0, 0, 0, -1, 1, 0, -6, 3, 0, -2, 1, 0, 12, -6, 0, 3, -2, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, -24, 12, 0, -6, 3, 0, -2, 1, 0, 48, -24, 0, 12, -6, 0, 3, -2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, -96, 48, 0, -24, 12, 0, -6, 3, 0, -2, 1, 0, 192, -96, 0, 48, -24, 0, 12, -6, 0, 3, -2, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,5 COMMENTS Row sums are A117902. Diagonal sums are A117903. LINKS FORMULA G.f.:(2x^6*y^2+x^5*y(2-y)+4x^4(1-y)-x^3(2y^2-3y+4)+x^2*y(y-2)+x(y-1)+1)/((1-4x^3)(1-x^3*y^3)) EXAMPLE Triangle begins 1, -1, 1, 0, -2, 1, 0, 3, -2, 1, 0, 0, 0, -1, 1, 0, -6, 3, 0, -2, 1, 0, 12, -6, 0, 3, -2, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, -24, 12, 0, -6, 3, 0, -2, 1, 0, 48, -24, 0, 12, -6, 0, 3, -2, 1 CROSSREFS Sequence in context: A131358 A291895 A177351 * A074984 A112658 A190693 Adjacent sequences: A117898 A117899 A117900 * A117902 A117903 A117904 KEYWORD easy,sign,tabl AUTHOR Paul Barry, Apr 01 2006 STATUS approved
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https://turtletoy.net/turtle/36ead8f7d2 | 1,725,921,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00475.warc.gz | 561,103,016 | 10,842 | Colliding Flow 2
Snakes grow at their heads and tails following a vector field that changes over time. They stop when they hit another snake.
```const RAD = 80; // min=10, max=100, step=1
const N_PARTICLES = 100; // min=10, max=300, step=1
const ITERS = 200; // min=10, max=300, step=1
const STEPSIZE = 1; // min=0.25, max=5, step=0.25
const COLLIDE_DIST = 1.25; // min=0.5, max=5, step=0.5
const OUTER_BOUNDS = 92; // kill particles that exceed this
const GROW_TAILS = true;
// x and y in the range -100 to 100
// t in the range 0 to ITERS * STEPSIZE
const SWIRL = 25; // min=-50, max=50, step=1
let field = (x, y, t) => ({
x: -x/50 + Math.sin(y / 10 + t/50) / 2 - y * SWIRL/1000,
y: -y/50 + 1 - t/20 + Math.sin(x/10) / 10 + x* SWIRL/1000,
})
//=========================================================================
let rand = (min, max) =>
Math.random() * (max - min) + min;
let randInt = (min, max) =>
Math.floor(rand(min, max));
let remap = (x, oldmin, oldmax, newmin, newmax) => {
let t = (x - oldmin) / (oldmax - oldmin);
return t * (newmax - newmin) + newmin;
}
let choose = (arr) =>
arr[randInt(0, arr.length)];
let dist = (x1, y1, x2, y2) =>
Math.sqrt((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2));
let len = (p) =>
dist(0, 0, p.x, p.y);
let normalize = (p) => {
let length = len(p);
if (length === 0) { return {x: 0, y: 0}; }
return {
x: p.x / length,
y: p.y / length,
};
}
let makeParticle = (x, y, headAlive, tailAlive) => ({
points: [{x:x, y:y}], // [head, ... history ..., tail]
tailAlive: tailAlive,
})
particle.points[0];
let getTail = (particle) =>
particle.points[particle.points.length-1];
let moveTo = (particle, isHead, x, y) => {
let p2 = {x: x, y: y};
}
let moveBy = (particle, isHead, dx, dy) => {
let p2 = {x: p.x + dx, y: p.y + dy};
}
let FAR = 99999999;
let distToTrail = (part1, isHead, part2) => {
// closest distance from p1's current head or tail to p2's whole trail
if (part1 === part2) { return FAR; }
let lowestDist = FAR;
for (let p2 of part2.points) {
lowestDist = Math.min(lowestDist, dist(
p1.x, p1.y, p2.x, p2.y
));
}
return lowestDist;
}
let distToAny = (particle, isHead, particles) => {
let lowestDist = FAR;
for (let p2 of particles) {
lowestDist = Math.min(lowestDist, distToTrail(particle, isHead, p2));
}
return lowestDist
}
//=========================================================================
// make initial circle of particles
console.log('-------------------');
let PARTICLES = [];
for (let ii = 0; ii < N_PARTICLES; ii++) {
let theta = remap(ii, 0, N_PARTICLES, 0, 2 * Math.PI);
let particle = makeParticle(
);
PARTICLES.push(particle);
}
//=========================================================================
// iterate through vector field
// iterate!
for (let iter = 0; iter < ITERS; iter++) {
let t = iter * STEPSIZE;
let numAlive = 0;
//for (let particle of PARTICLES) {
for (let rr = 0; rr < N_PARTICLES; rr++) {
let particle = choose(PARTICLES);
// instead of looping through each particle in the same order each time,
// update N particles each iteration (with replacement)
// to allow some to get ahead of others
numAlive += 1;
if (Math.abs(p.x) > OUTER_BOUNDS || Math.abs(p.y) > OUTER_BOUNDS) {
continue;
}
let force = field(p.x, p.y, t);
moveBy(particle, true, force.x * STEPSIZE, force.y * STEPSIZE);
if (distToAny(particle, true, PARTICLES) < COLLIDE_DIST) {
}
}
if (particle.tailAlive) {
numAlive += 1;
let p = getTail(particle);
if (Math.abs(p.x) > OUTER_BOUNDS || Math.abs(p.y) > OUTER_BOUNDS) {
particle.tailAlive = false;
continue;
}
let force = field(p.x, p.y, t);
moveBy(particle, false, -force.x * STEPSIZE, -force.y * STEPSIZE);
if (distToAny(particle, false, PARTICLES) < COLLIDE_DIST) {
particle.tailAlive = false;
}
}
}
if (numAlive === 0) {
console.log('all particles stopped after ' + iter + ' iters');
break;
}
}
//=========================================================================
// draw
// The walk function will be called until it returns false.
let numSegs = 0;
for (let particle of PARTICLES) {
numSegs += particle.points.length - 1;
}
console.log('' + numSegs + ' line segments');
Canvas.setpenopacity(1);
const turtle = new Turtle();
turtle.pendown();
function walk(ii) {
let points = PARTICLES[ii].points;
turtle.jump(points[0].x, points[0].y);
for (let jj = 1; jj < points.length; jj++) {
turtle.goto(points[jj].x, points[jj].y);
numSegs += 1;
}
return ii < PARTICLES.length - 1;
}
``` | 1,380 | 4,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.50395 |
simulatedominion.wordpress.com | 1,369,482,126,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705939136/warc/CC-MAIN-20130516120539-00006-ip-10-60-113-184.ec2.internal.warc.gz | 244,612,438 | 20,684 | Home > Simulation > “Analyzing the Chapel” or Lies, Damned Lies and Statistics
## “Analyzing the Chapel” or Lies, Damned Lies and Statistics
This article explores ways to gather insight from strategies. Why does a particular strategy do well? I’m developing statistics analyzers for the Dominator, and want to put them in action in an actual case. To make it interesting, I examine a Chapel strategy competing with Big Money. To be precise: I run Single Chapel canonical 1 against Big Money canonical. Now I know both are not competitive strategies, but they are equal in the sense that they both don’t buy Duchies and Estates. Including these would complicate things too much for this early analysis. I will investigate real Chapel strategies later.
But first: pop quiz
If you play “Single Chapel canonical 1″ against “Big Money canonical”, what are your win chances?
A) Playing Chapel gives you roughly an 80% win chance.
B) Playing Chapel gives you roughly a 65% win chance.
C) Playing Chapel gives you roughly a 50% win chance.
The Chapel has been hyped a lot, even having been called the most powerful card in the game on more than one occasion. For instance on dominionstrategy.wordpress.com. My guess was that it is one of those cards that is very strong on its own, not needing any other kingdom cards to make it work.
So I ran the simulation and the results were not what I expected at all. Garion even didn’t believe them, so I wanted to fully understand them before writing about them. And in order to do this I added some victory points statistics logging. I’m recording the amount of victory points each player has in each turn, to get an average number of victory points for each strategy per turn. This resulted in the following graph:
What does this graph tell us? There is the expected drop in VP of the Chapel strategy in the early turns as the Estates get trashed. Then at turn 12 the Chapel overtakes the big money in average VP. Since only 1 game in my 10,000 runs ended in turn 12, this graph tells you that the Chapel strategy must win most of the games.
What about the actual results?
Chapel wins: 4886 (49%)
BigMoneyCanonical wins: 5103 (51%)
ties: 11
First of all I’ll bet good money that very few people actually choose answer C off the bat. Playing a single Chapel does not improve your odds?
And if the simulation is right -which I believe it is, because I checked against using another simulator- then my initial average VP graph must be flawed. Well indeed I did not tell the whole story about the way the average is calculated. Calculating that average the big question is: how do you combine the data from games that do not last the same number of turns. Say that you have a normal game that lasts 16 turns and one that lasts 17 turns. Everything is easy for calculating the average number of points for turn 16, but what to do with the data for turn 17. Consider two games where game 1 ends in turn 16, and game 2 last until turn 17:
turn chapel strategy score game 1 chapel strategy score game 2 16 30 26 17 n/a 26
I thought it would be acceptable to extrapolate that the first game has 30 points in turn 17 as well.
Average points in turn 16: 30 + 26 / 2 = 28 points
Average points in turn 17: 30 + 26 / 2 = 28 points.
But as we have seen the resulting averages are misleading and of no use. So I tried again, this time not extrapolating scores from games. I needed to keep track of how many games ended in each turn, and this allowed me to calculate the average scores using only the games that provided data for it.
Average points in turn 16: 30 + 26 / 2 = 28 points
Average points in turn 17: 0 + 26 / 1 = 26 points.
You immediately see that this can have strange consequences: the average number of points drops. To me this was counter intuitive because (apart from trashing Estates) the actual number of points only rises during games. Still, this method of averaging gives a much less misleading graph:
Here we see that the Chapel strategy is only superior in games that end between turn 12 and 16. If the Chapel strategy take longer than that, then -on average- the big money strategy is superior. It may be that in those cases the Chapel strategy is hurt by the extra Provinces.
Anyway if we look at the histogram of the number of turns played we see that the whole story starts to look plausible:
These totals also add up nicely to the win percentages of the strategies:
The total numbers for games ending up to turn 16 is: 4943
The total numbers for games ending after to turn 16 is: 5057
OK, some more graphs to see if we really understand why the Chapel does so poorly compared to expectations. Let’s look at the average value of each card. The basic idea of this version of the Chapel strategy is to raise the average value by trashing low value cards.
It is nice to see that the graph for Chapel drops flat after turn 10. From that point onwards the influx of Provinces balances the influx of money. I mentioned previously that I expected that once the Chapel deck start buying Provinces it would starts degrading. This does not seem to happen, so there must be another reason that the Chapel strategy is better in shorter games.
Let’s take a quick side step. One theory suggests that you need to have at least one Gold before buying your first Province. This should be very visible in the average card value graph, and so it is:
Now we do see a drop in average value after turn 9. Unfortunately the extra Gold does not actually improve the win-rate of this strategy, which stays at roughly 50%. So trading a Province for a Gold does as much damage as it does good.
But I’m still a bit at a loss. We saw the average VP of the Chapel strategy decline for longer games, but we also see the buying power steadily increase, especially for longer games. Maybe that average buying power does not translate into turn where you can actually buy a Province. One turn with 12 money plus one with 6 money only gives you one Province, even though the average is a whopping 9 money. The graph below shows the percentage of the time that a player had 8 or more money to spend:
This graph is final proof that the Chapel strategy is capable of outbuying Big Money when it comes to buying Provinces. So how on earth can it be that the Big Money strategy still is just as good?
My last stab at solving the mystery is by thinking about the victory conditions. A Chapel deck will always have less Estates than Big Money. So once Big Money buys its 4th Province, the Chapel deck can no longer win. Nonetheless my Chapel strategy implementation will never buy a card that will cost it the game. And it is easy to count how many times that happens for the 10k games simulated. The numbers:
Games where the Chapel deck did not buy a Province: 1801
Total number of Provinces not bought: 3498
Since there is no other way to gain victory points, in those games the Chapel deck will still lose, just a couple of turns later, judging by the numbers by about 2 turns. Now back to the win chance. Given that the Chapel deck will need 5 Provinces to win, and the big money only needs 4, we can relate the actual number of Provinces bought by these targets:
Chapel buys 42973 Provinces for an adjusted number of 42973 / 5 = 8594
Big Money buys 37027 Provinces for an adjusted number of 37027 / 4 = 9256
(Sanity check: 42973 + 37027 = 80,000 Provinces sold in total.)
In this respect Big Money is actually doing better than the Chapel strategy. So although the Chapel is generating more buying power than Big Money, it just is not enough…
Conclusion
The Chapel will increase the buying power, but in this simple form it also increases the number Provinces it needs to win. These effects compensate for each other, making this simple single Chapel strategy no better than canonical Big Money.
Next up is adding Duchies and Estates into the mix. And to give you a sneak preview: my best Chapel strategy so far (that does not use any other kingdom cards) scores 54% against Bmu. More on the esteemed Chapel later…
[Edit: I decided that card names should be capitalized. And thx Willie for pointing out a very basic calculation error.]
1. January 11, 2011 at 01:20 | #1
You misinterpreted the dominion strategy post. Chapel does not require any specific card, it doesn’t have a strong bias toward any one action in particular. It helps very many them all a lot. If your simulation isn’t showing that a chapel based strategy is significantly outperforming a straight big money strategy, you probably need to do a lot of work on making your simulations more realistic. Otherwise, your experiments do not match the view of many of the players who do very well on the isotropic dominion server.
Take a look here, for example. http://boardgamegeek.com/thread/586435/i-still-love-dominion-but-im-getting-annoyed-by-th/page/4
Wodan held your thesis, that chapel is not much better than big money because of the results of simulations. theory, the author of the claim that chapel is best card in the game, played 3 games against wodan. Wodan bought never bought a chapel and lost all 3 games.
2. January 11, 2011 at 23:10 | #2
I was quoting the nice “This is the best card in the game.” phrase from dominionstrategy, so I cannot see how I am misinterpreting it. I do not entirely agree with it, but that is another matter.
The BGG topic is great by the way, thx for that. Shame I did not read it earlier.
But I stand by my simulations. Especially in this simple case where no other cards, no duchies or estates are bought. Allow me to recap the essence of this article in fewer words.
Since Chapel trashes estates, it needs 5 provinces to win. Big Money only needs 4 province.
Big Money (no duchies/estates) needs 16.8 turns to get 4 provinces.
Chapel (no other cards, no duchies/estates) needs 16.93 turns to get 5 provinces.
Therefore, in this simple case: big money is on par with chapel. Other simulations agree with these numbers. Try: http://www.westpark-gamers.de/index.html?/Reviews/bericht291.php&wpgref=none
3. January 12, 2011 at 06:59 | #3
The problem is that chapel + straight big money is not why chapel is so great.
Chapel is so great because it lets you consistently hit card combos.
I am not saying your numbers are wrong, just that your are spending way too much effort analyzing an overly simple model. I think that less analysis on a more realistic strategy is a lot more fruitful in terms of relevance to actual reasonable high level Dominion play.
• January 12, 2011 at 12:53 | #4
Ghe ghe, as a matter of fact you did. But indeed this post is just about base stuff. I needed to do this to develop analyzers that allow me to understand more complex situations. I am still very curious how the numbers will stack up for realistic chapel combo’s.
But let’s turn things around: give me a combo (preferably one that requires few decisions to play) and I will see if can give you some numbers on a realistic chapel case. To allow comparison with Big Money you cannot include attack cards in the combo.
4. January 12, 2011 at 22:10 | #5
Here are first two solitaire strategy games that I played with one nice combo available, chapel/quarry/grand market. I hit 8 provinces in 14 turns and 13 turns respectively, with King’s Court/Grand Market in the first, and City/Upgrade/Grand Market in the second.
I did branch out a little and not use exclusively the named cards, but this is just me playing quickly and without much thought as a fairly skilled human.
5. January 14, 2011 at 19:40 | #6
Choke! Those hardly represent easy to simulate strategies, I counted 7 kingdom cards in one game. I’m sure chapel contributes to a strategy that already uses King’s Court, Grand Market and Quarry (not to mention City, Laboratory and Remodel) but I think those cards represent a power combo all by themselves.
6. January 17, 2011 at 06:30 | #7
Thank you for this. Very interesting presentation. This indicates how important unwanted estates can turn out. Keep it up!
7. January 19, 2011 at 01:05 | #8
I think you have a math error or typo here:
“One turn with 12 money plus one with 6 money only gives you one province, even though the average is a whopping 10 money.”
• January 19, 2011 at 06:38 | #9
Doh! That should have read 9 money obviously! Thx for pointing that out.
8. December 8, 2011 at 19:10 | #10
i’m curious did you simulate trashing provinces with chapel. Against straight big money Chapel should have an early lead on the province race. Trashing provinces keeps your deck from over greening while keeping big money from winning. just a thought
• December 8, 2011 at 20:05 | #11
Hi Seamus,
thx for the suggestion. Personally I don’t think this is a good idea for I don’t think the single Chapel will build up a large enough lead to justify this. And since I’m not very active with simulations at the moment, I don’t think I will simulate this. However: if you want to get your hands dirty, I can provide you with a copy of the code.
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Semantic Web App (iOS & Android) | 663 | 2,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.792834 |
https://www.cfd-online.com/Forums/main/1638-3d-2d-les-print.html | 1,495,545,154,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00110.warc.gz | 876,941,765 | 5,756 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- Main CFD Forum (https://www.cfd-online.com/Forums/main/)
- - 3D-2D LES (https://www.cfd-online.com/Forums/main/1638-3d-2d-les.html)
Jan Ramboer December 10, 1999 08:09
3D-2D LES
Hello,
I know that 2D LES is not a good thing to do. But since I was not interested in absolute values and wanted to do some fast calculations I took a turbulent 3D channel flow with LES and calculated it in 2D. Now it seems that by reducing the length of the spanwise direction my flow became laminar. However if I kept the same length in the spanwise direction than the ratio's of the dimensions of my cells (very big length in the spanwise direction) seemed to spoil my convergence. Is there anybody out there who has some experience in 2D LES and who could give me some advice?
Jan Ramboer
Hongyi Xu December 11, 1999 22:32
Re: 3D-2D LES
Hi Jan,
It seems I can share some relevant experiences with you about performing 2-D LES. First, I did LES in a square duct. The initial velocity profile was set as a laminar flow (parabolic distribution on the cross-section of square duct) superimposed by some perturbation from the solution of Orr-Sommerfeld equation. Periodic condition was used in streamwise direction. In the flow evolution, the velocity profile was found to approach to a turbulent profile. The results can be found in my home-page. Then I reduced my 3-D program to 2-D and repeated the same calculation, I found the 2-D laminar velocity profile was not approaching to the expected turbulent distribution, instead, it stayed around the laminar one. It seems that 3-D calculation is a necessary condition for the flow to transit from laminar to turbulent. If you really would like to do 2-D LES, you have to extract the instantaneous velocity profile from a 3-D channel LES and use this as inlet condition to perform 2-D calculation. However, I am still doubting about how much turbulence physical information will be contained in such kind of simulation, or in another word, to what extent this simulation can reflect the relevant turbulence physics. That is just my opinion and experiences Thanks for your attention.
Ben R. Hodges December 13, 1999 01:33
Re: 3D-2D LES
There is a reason that 2D LES is not a good idea for a duct flow: the initiation of turbulence from the boundary layer begins with the smallest scales of motion and requires the 3D bursts of horshoe vorticies (related to the slow-speed streaks) to get the correct qualitative and quantitative picture. Furthermore, the use of LES in the viscous sublayer region is problematic even in 3D - essentially, you must get DNS level resolution to get the correct spacing of the slow-speed streaks in the boundary layer. If you don't get the correct spacing of the slow speed streaks you will not get the correct turbulent bursts out of the boundary layer - from which it follows that you won't produce the large scale eddies that are necessary for the LES to work. Remember LES assumes that you have the large scale eddies correct. In a turbulent boundary layer, creation of the large scale eddies required that you get the physics of the burst/sweep mechanisms correct (i.e you must capture the small scales). There is no free lunch.
John C. Chien December 13, 1999 02:34
Re: 3D-2D LES
(1). I like your answer. (2). The other day, when I was running a 3-D flow through turbine passage with a low Reynolds number two-layer model, I decided to keep the mesh fixed and increase the Reynolds number of the flow. (3). The results was the resultant Y+ next to the wall became much larger than one. (4). And the previously captured secondary flow suddenly disappeared from the picture. (5).It was just amazing that it is hard to fool the mother nature.
Jan Ramboer December 13, 1999 10:51
Re: 3D-2D LES
Hello,
First thanks for answering. Another questions: did you try to investigate at which limits of the spanwise dimension your flow became laminar? I mean more or less which dimensions of the spanwise direction should you use to keep it turbulent?
Greetings,
Jan
Hongyi Xu December 13, 1999 12:12
Re: 3D-2D LES
No, because my calculation is on square duct flow instead of Channel flow, the dimension sizes in the two cross-streamwise directions are same, as can be seen at http://mach.me.queensu.ca/~hongyi/. But it would be interesting to purposely make it a rectangular duct and to increase the dimension size in one direction. Probably that will answer your question.
To make some comments on Ben's response:
Based on my experience, I agree with Ben's first comment, that is the 3D bursts of horshoe vorticies play an important part in the transition from laminar to turbulent since when I started the calculation from perturbed laminar flow, I did observe a kind of ordered structure near wall and the corners of square duct and the corner regions was the most turbulence active region. Something I'd like to correct is that I tried to do both DNS and LES on 2-d channel, cause in 2-D case, the current capability of SGI workstation allows me to comfortably distribute enough grid points (500*256) into the turbulent sublayer region, the y+ of first grid point away from wall was about 0.1. Also, I tried to switch on and off the subgrid model (Smg SGS). The simulation did not show much difference, which means the grid was fine enough so that LES was basically approaching to DNS. Still I did not see any sign that the velocity profile evolve towards a turbulent one. The point is that turbulence burst is a 3-D phenomena which can not be well presented in 2-D. The grid resolution is not a critical issue. For instance, in the 3-D LES with grid (130*34*34), the transition phenomena can be clearly observed which may not be 100% digitally correct. However, in the 2-D DNS with grid 500*256, the transition was not found after 10,000 time step calculation.
John C. Chien December 13, 1999 14:03
Re: 3D-2D LES
(1). If we let u=U+u', v=V+v', and w=W+w', and substitute it into the continuity equation, (du/dx)+(dv/dy)+(dw/dz)=0, then we have, { (dU/dx) + (dV/dy) +(dW/dz) } + { (du'/dx) + (dv'/dy) + (dw'/dz) } =0. where U,V,and W are time averaged mean values, and u',v' and w' are fluctuating components. (2). We can assume that the mean flow also satisfy the equation, { (dU/dx) + (dV/dy) + (dW/dz) }=0. From here, for 2-D channel flow, one can assume that (dW/dz)=0, and the 2-D mean flow will satisfy the mean flow continuity equation of { (dU/dx) + (dV/dy) }=0. This is possible, because it is possible to make (dW/dz)=0, where W is a mean quantity only. (3). This leaves the fluctuating components to satisfy the equation of { (du'/dx) + (dv'/dy) + (dw'/dz) }=0. Where u'=u'(x,y,z,t), v'=v'(x,y,z,t), and w'=w'(x,y,z,t). It is difficult to make u'=u'(x,y,t), v'=v'(x,y,t), and w'=w'(x,y,t).(or (dw'/dz)=0 ) For this to happen, organized motion has to happen at any instant so that we can use { (du'/dx) + (dv'/dy) }=0 for 2-D simulation. This organized instantaneous motion, was not observed in the random turbulent motion of fluid. (4). As a result, the continuity equation can only be { (dU/dx) + (dV/dy) }=0 and { (du'/dx) + (dv'/dy) + (dw'/dz) }=0 for 2-D channel flow. That means the instantaneous components must always be 3-D transient. And the 2-D simulation will require the assumption of organized instantaneous fluid motion, which can not be classified as turbulent flow as is observed in nature.
Ben R. Hodges December 15, 1999 23:26
Re: 3D-2D LES
The spanwise grid resolution is critical in obtaining the correct turbulent boundary layer. The slow speed streaks in a boundary layer generally are spaced at ~100 y+ units (observed in both laboratory and well-resolved DNS). These streaks are the initiation of the coherent motions and need on the order of 10 grid cells between the centers of the streaks to resolve the motion. Thus, if your spanwise resolution is ~20 y+ units, the streaks in a 3D LES will end up spaced about 200 y+ units. My experience shows the coarser spanwise resolution gives larger scale (and fewer) burst-sweep events. This produces a catch 22: if an LES model accurately captures the dissipation caused by the large scale motion, then the larger (and fewer) eddies in an under-resolved flow will not create enough global dissipation and the mean flow will be to fast. If an LES captures the correct mean flow then the dissipation computation by the LES must be incorrect: that is, since it has fewer eddies to work with, the LES must be computing an amplified dissipation for each eddy. Again, there is no free lunch. The fundamental premise of LES is that the larger scale motions are captured so that the smaller scales can be computed. However, when you don't create enough large scales you don't get the right smaller scales unless the model "fudges" the physics. This is why some 3D LES produce the correct mean flows and poor representations of turbulent statistics while other methods produce good turbulent statistics and poor mean flows. Your use of 2D LES has infinite grid spacing in the spanwise and cannot ever resolve the spanwise streaks. Thus you cannot produce the burst-sweep events that characterise boundary layer turbulence. 2D LES should only be used in flows that are actually 2D - i.e. it has successfully been used in atmospheric flows that model the atmosphere as a 2D space. Boundary layers have been and always will be 3D, thus, they require DNS level accuracy in the near wall region to capture turbulence. If you are trying to capture the effects of turbulence rather than the turbulence itself, then you should proceed with some kind of RANS modeling that accounts for the 3D physics that you are missing or underresolving. Alternatively, one could parameterize the 3D physics of boundary layers into a 2D LES (as far as I know, it hasn't been done). Simple application of any existing LES to a boundary layer in 2D cannot produce a good representation of the physics.
Ben
Jan Xu January 4, 2000 14:31
Re: 3D-2D LES
Hi, dear Jan,
Happy new year!
I just wonder whether you could kindly tell me where I can download a basic and, of course, free code for 2D-LES? My supervisor want me to dive into this subject, I am also very interseted, but I am afraid it will be too much for me to build up a palace from plain. Any hints would be highly appreciated. Thank you in advance!
Jan Xu
All times are GMT -4. The time now is 09:12. | 2,615 | 10,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-22 | longest | en | 0.958088 |
https://www.nagwa.com/en/videos/183185256107/ | 1,586,480,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00258.warc.gz | 1,036,653,918 | 5,394 | # Video: Solving Simple Quadratic Equations Using Square Roots
Solve π₯Β² = 16/169.
00:45
### Video Transcript
Solve π₯ squared equals 16 over 169.
π₯ squared equals 16 over 169. If we take the square root of π₯ squared, we get π₯. If we take the square root of 16 over 169, we get four over 13. Itβs not just the positive four over 13, but also the negative value of four over 13.
π₯ could be equal to four over 13 or π₯ is equal to negative four over 13. | 160 | 478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-16 | latest | en | 0.775058 |
zilb.pinkland.ru | 1,524,323,434,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945232.48/warc/CC-MAIN-20180421145218-20180421165218-00584.warc.gz | 565,682,661 | 3,685 | im not the greatest archer but found your formula works for me. Apostolos 03/10/07 : Archimedes: you'd better use one of the other calculator that work on principle of draw weight and arrow weight.Charles: it could be tweaked but it would require two more input variables. Roy 14/11/07 : an excellent tool but for one thing! would it be possible to produce a version with imperial input values for those of us who still live in the stone age...;-) Apostolos 21/05/08 : The calculator is back in order! It mesures it very i have some the calculation inculdes the air friction,arrow weight ext?
It's very helpfull and so helped me to verifie my estimation of the speed of my arrows. But then thinking about the speed of most compounds (300 fps) this would make sense.
I'm not a ballistic engineer, but very curious about how you find the formula. Just to test the calculator I entered a 10cm difference between to distances - then you can see how the decceleration, gravitational and other forces start to show a non-linear relationship. Mark 25/09/09 : Robin or Locksley : The things you mention are constants for the two shots from you bow setup and your arrows. Decceleration, gravitational forces and other forces are taken into account by the formular - see my earlier comment.
Most (if not all) archery clubs have a radar device in which you place it near the target and shoot the arrow through it (from about 10 feet away / the arrow will have maximum velocity shortly after it's release). Ludek 13/08/09 : Hello, it is also good for bare bow, i tried it.
This would be the most accurate way to determine the speed for the particular setup you're using. Patrick Andresz 17/07/09 : Thank you very much for this nice programm ! Don Walker 22/09/09 : an excellent tool but for one thing! would it be possible to produce a version in Yards and diference in site marks settings in mm for old field archers Mark 25/09/09 : I didn't believe it when I looked at the numbers, as the difference between each 5M distance for time and sight markings seemed to be almost constant.
This program was meant mostly as an easy way to estimate your arrow speed.
However if you give the program first your 20m/30m sight marks and then your 70m/90m, you can find out how much speed your arrow loses at longer distances. If you fire two shots using the same pin at different distances and measure the arrow drop, you should be able to determine the speed without knowing distance to the eye, non? Dan 21/10/08 : Good calculator however it does not seem to account for decceleration and would only be the average arrow speed from distance one to distance 2, beyond distance 2 your arrow would be constantly deccelerating and thus your extrapolated sight pin settings are slightly off.
The Muricay Beach is located further away, at 4 kilometers, but the white sands and clear seas more than make up for the distance.
There are also mangrove tracts and seaweed plantations nearby, for those who are interested in such plant specimens.
A nice place to start would be the Dao-Dao Islands some seven to ten minutes away from the seaport, when riding a motorboat. | 723 | 3,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-17 | latest | en | 0.936425 |
http://www.chegg.com/homework-help/questions-and-answers/f-vc-t-2s-10v-circuit-shown-find-energy-wc-stored-capacitor-power-p-supplied-source-t-6s-t-q1038619 | 1,469,316,590,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00257-ip-10-185-27-174.ec2.internal.warc.gz | 376,344,593 | 14,209 | f Vc(t=2s)=10V in the circuit shown, find the energy, wc stored in the capacitor and the power, P, supplied by the source at t=6s
Take the values of the components to be C1=05FI1=2AR1=3 and R2=6.
wc= J
P= W | 74 | 206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.857152 |
https://ctftime.org/writeup/14723 | 1,618,838,300,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00601.warc.gz | 286,076,095 | 7,546 | Rating:
# i can count
Description:
Let's do this together. You do know how to count, don't you?
"i can count" is a 32bit Linux binary with symbols that requests the user to count up until a function identifies the entered number as the flag.
## Solution
The verification process consists out of a function that "encodes" each ASCII digit that was entered and checks if they match with a fixed array.
If 19 correct digits are entered the number is treated as correct and is outputted as the flag.
By mapping each digit to the encoded value and then mapping the compare array back to the digits the values originate from results in the flag PCTF{2052419606511006177}.
I automated this task with a small gdb script:
python
gdb.execute("b *main") # break somewhere
gdb.execute("b *check_flag+35") # beginning of rol code
gdb.execute("b *check_flag+1330") # compare at the end
gdb.execute("run") # start the process
d = {} # dictionary of "encoded" values
for c in "0123456789": # iterate over all number character
gdb.execute("jump *check_flag") # start the function
gdb.execute("set *((int*)($ebp-0x1C)) = 0") # reset the index just in case gdb.execute("p flag_buf@1 = 0x%02X" % ord(c)) # set the first buffer character gdb.execute("continue") # encode the character v1 = int(gdb.parse_and_eval("$al"))&0xFF # read the encoded value
d[v1] = c # save input and output character in a dictionary
res = [] # array containing finished number
for i in range(0x13):
cmp = int(gdb.parse_and_eval("*((unsigned char*)(check_buf + "+str(i)+"))")) # check buffer value for the given index
res.append(d[cmp]) # map the "encoded" value to a number
print("PCTF{%s}" % ''.join(res))
# Space Saver
Description:
we couldn't think of anything cute so here you go
Space Saver is a 100MB file starting with a DOS boot sector and some data within surrounded by zero bytes.
## Solution
Running "binwalk -e <binary>" extracts a RAR archive with a password protected picture called "final.png" from the image and also hints at the existence of PNG files in the file system.
Examining the image in a hex editor manually and searching for the PNG Header shows multiple files, but 3 of them contain a short string at the end which didn't belong to the file itself ("Spac", "3ei2", "herE").
Using the password "Spac3ei2herE" to extract the final.png file works and reveals the flag:
# can you guess me
"can you guess me" consists out of a service that serves a python script that allows 10 unique characters to be entered and evaluated.
The challenge services source file:
python
#! /usr/bin/env python3
from sys import exit
from secret import secret_value_for_password, flag, exec
print(r"")
print(r"")
print(r" ____ __ __ ____ __ __ ")
print(r" / ___|__ _ _ _\ \ / /__ _ _ / ___|_ _ ___ ___ ___| \/ | ___ ")
print(r"| | / _ | '_ \ V / _ \| | | | | _| | | |/ _ \/ __/ __| |\/| |/ _ \ ")
print(r"| |__| (_| | | | | | (_) | |_| | |_| | |_| | __/\__ \__ \ | | | __/ ")
print(r" \____\__,_|_| |_|_|\___/ \__,_|\____|\__,_|\___||___/___/_| |_|\___| ")
print(r" ")
print(r"")
print(r"")
try:
val = 0
inp = input("Input value: ")
count_digits = len(set(inp))
if count_digits <= 10: # Make sure it is a number
val = eval(inp)
else:
raise
if val == secret_value_for_password:
print(flag)
else:
print("Nope. Better luck next time.")
except:
print("Nope. No hacking.")
exit(1)
## Solution
Using the build-in help command on the flag results in the service giving it out:
____ __ __ ____ __ __
/ ___|__ _ _ _\ \ / /__ _ _ / ___|_ _ ___ ___ ___| \/ | ___
| | / _ | '_ \ V / _ \| | | | | _| | | |/ _ \/ __/ __| |\/| |/ _ \
| |__| (_| | | | | | (_) | |_| | |_| | |_| | __/\__ \__ \ | | | __/
\____\__,_|_| |_|_|\___/ \__,_|\____|\__,_|\___||___/___/_| |_|\___|
Input value: help(flag)
No Python documentation found for 'PCTF{hmm_so_you_were_Able_2_g0lf_it_down?_Here_have_a_flag}'.
Use help() to get the interactive help utility.
Use help(str) for help on the str class.
Nope. Better luck next time.
Opening the interactive help() console and reading out the secret module also reveals that there is no correct number input:
DATA
flag = 'PCTF{hmm_so_you_were_Able_2_g0lf_it_down?_Here_have_a_flag}'
secret_value_for_password = 'not even a number; this is a damn string;...
trollface = '\n@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@...@@@@@...
Original writeup (https://github.com/Pusty/writeups/tree/master/PlaidCTF2019#space-saver). | 1,255 | 4,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-17 | latest | en | 0.785042 |
https://www.jobduniya.com/Aptitude-Resources/Numerical-Aptitude/Numerical-Aptitude-Questions-Part-7.html | 1,542,471,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117170031-00415.warc.gz | 931,339,830 | 11,413 | # Aptitude Papers: Numerical Aptitude Questions (Part 7 of 12)
Examrace Placement Series prepares you for the toughest placement exams to top companies.
1. Vivek purchased 120 tables at a price of Rs. 110 per table. He sold 30 tables at a profit of Rs. 12 per table and 75 tables at a profit of Rs. 14 per table. The remaining tables were sold at a loss of Rs. 7 per table. What is the average profit per table?
1. Rs. 12.875
2. Rs. 10.04
3. 10.875
4. Rs. 12.80
5. Rs. 13.80
Answer: c
2. A book dealer allows 16 per cent discount to a retailer and the retailer allows 10 per cent discount to a customer. These both discounts are given on the printed price of the book. If the customer pays Rs. 270 for a book and the dealer makes a profit of 5 per cent what is the cost price of the book for the dealer?
1. Rs. 252
2. Rs. 243
3. Rs. 240
4. Rs. 250
5. Rs. 350
Answer: c
3. The difference between the greatest and the least numbers of eight digits which begin with 8 and end with 6 is
1. 99999999
2. 10000000
3. 80000006
4. 9999996
5. None of these
Answer: d
4. If a * b = a2 + b2 then 3 * 5 is equal to
1. 16
2. 34
3. 8
4. 15
5. 18
Answer: b
5. The Simple interest on a sum of money at 8 per cent per annum for 6 years is half the sum. The sum is
1. Rs. 4800
2. Rs. 6000
3. Rs. 8000
4. Rs. 7000
5. Data inadequate
Answer: d
6. How long will it take a sum of money invested at 5 per cent per annum simple interest to increase its value by 40 per cent?
1. 5 years
2. 6 years
3. 7 years
4. 8 years
5. 10 years
Answer: d
7. A number, when successively divided by 3 and 5, leaves remainder of 2 and 1, when the same number is divided by 15, the remainder is
1. 1
2. 2
3. 5
4. 7
5. 10
Answer: c
8. The numbers 1, 3, 5, … 25 are multiplied together. The numbers of zeros at the right end of the product is
1. 1
2. 0
3. 2
4. 3
5. 8
Answer: b
9. What will be the compound interest on Rs. 240 for 2 years at 4 per cent per annum?
1. Rs. 19.20
2. Rs. 9.60
3. Rs. 19.18
4. Rs. 19.58
5. Rs. 20.58
Answer: d
10. Find the compound interest on Rs. 15000 at 8 per cent per annum payable half-yearly for 1 year.
1. Rs. 1500
2. Rs. 1432
3. Rs. 1200
4. Rs. 1224
5. Rs. 1324
Answer: d | 793 | 2,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-47 | longest | en | 0.858908 |
http://list.seqfan.eu/pipermail/seqfan/2010-March/004213.html | 1,586,393,161,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371826355.84/warc/CC-MAIN-20200408233313-20200409023813-00032.warc.gz | 103,133,642 | 2,249 | # [seqfan] Re: Zeros in A172390 and A172391
Paul D Hanna pauldhanna at juno.com
Mon Mar 29 23:20:44 CEST 2010
```SeqFans,
Please forgive the typos - I should have written:
"then the following functions equal the sum of the respective quadrasections defined by:
sqrt( sqrt(A(x^2)) + 4*x ) = QA0(x^4) + x*QA1(x^4)
sqrt( sqrt(B(x^2)) + 4*x ) = QB0(x^4) + x*QB1(x^4)"
as the examples demonstrated.
It is significant that in both of these functions defined above, 2 of the 4 quadrasections are zero,
and quite unexpected that the products of the non-zero quadrasections would be constant.
Paul
---------- Original Message ----------
From: "Paul D Hanna" <pauldhanna at juno.com>
To: seqfan at list.seqfan.eu
Subject: [seqfan] Re: Zeros in A172390 and A172391
Date: Mon, 29 Mar 2010 21:06:44 GMT
SeqFans,
To expand on Joerg's observation,
here is a relation that both sequences A172390 and A172391 seem to have in common.
No doubt the relation is due to their elliptic connections.
Let
A(x) = g.f. of A172390, and
B(x) = g.f. of A172391;
then the following functions equal the sum of the respective quadrasections defined by:
sqrt( sqrt(A(x^2)) + 4*x ) = QA0(x^4) + x^2*QA1(x^4)
sqrt( sqrt(B(x^2)) + 4*x ) = QB0(x^4) + x^2*QB1(x^4)
such that the quadrasections satisfy the product:
QA0(x^4)*QA1(x^4) = QB0(x^4)*QB1(x^4) = 2
and also
QA0(x^2)^2 + x*QA1(x^2)^2 = sqrt(A(x)) ;
QB0(x^2)^2 + x*QB1(x^2)^2 = sqrt(B(x)). | 492 | 1,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-16 | latest | en | 0.833382 |
https://www.nagwa.com/en/videos/973125484080/ | 1,540,226,487,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00434.warc.gz | 1,016,793,482 | 11,610 | # Video: Interior Angles of a Regular Polygon
Lauren McNaughten
Use the fact that the sum of measures of interior angles (in degrees) in an 𝑛-sided polygon is 180(𝑛 − 2) to calculate the measure of one interior angle in a regular polygon, or work backward from the interior angle to calculate the number of sides.
10:33
### Video Transcript
In this video, we’re going to look at how to calculate the interior angle of a regular polygon. First let’s be clear what is meant by that word regular. Now there are two diagrams on the screen, one of a regular hexagon and one of an irregular hexagon.
Now they’re both hexagons so they both have six sides, but they look very different from each other. If you look carefully at the diagrams, you’ll see that the difference is this. In the regular hexagon, all of the sides are the same length and all of the interior angles are also all the same size, whereas in the regular hexagon that isn’t the case. So in this video, we are focusing specifically on just those polygons that are regular.
Now just a reminder about interior angles, the interior angles of a polygon are those angles inside the shape itself, so those which I’ve marked in red in my diagram of a hexagon here. Now as we’ve already seen in the case of a regular polygon, all of those interior angles must be the same as each other. What we would like to know is how do we calculate the size of each of those interior angles in a regular polygon with a particular number of sides.
Now we’ve seen previously that there is a formula for calculating the sum of the interior angles in any polygon, and it’s this formula here. So the sum of the interior angles in an 𝑛-sided polygon is equal to one hundred and eighty multiplied by 𝑛 minus two, where 𝑛 represents the number of sides.
Now notice that there’s no mention of the word regular here, so this formula is true regardless of whether polygon that we’re interested in is regular or irregular. And just a brief reminder of where this formula came from, if you look at a polygon and if you choose a particular corner such as this one here, if you can connect it to all the other corners of the polygon, so like I’ve done here, you will see that you divide the polygon up into triangles and in this case I’ve got four triangles.
What you’ll notice if you do this for a number of different polygons is that the number of triangles you create is always two less than the number of sides. So here I had six sides, and indeed I created four triangles. Within each of those triangles, there are a hundred and eighty degrees, and therefore the total sum of the interior angles is the number of triangles multiplied by a hundred and eighty.
And as there are always two less triangles than the number of sides, that’s where this factor of 𝑛 minus two comes from. So that formula holds true for the sum of the interior angles regardless of whether the polygon you’re interested in is regular or irregular.
However this video is specifically about regular polygons, and it’s about calculating the size of each individual interior angle rather than just the total sum. So let’s think how we can use this formula to work out the size of each individual interior angle in a regular polygon.
Well if a polygon has 𝑛 sides, then it will also have 𝑛 interior angles. And all of them are the same because it’s a regular polygon. So if we know the sum, the total of these interior angles, if we want to work out what each individual one is equal to, we just need to divide by the number of angles, which means we need to divide by 𝑛, which means we can deduce this formula here: the interior angle in a regular 𝑛-sided polygon is equal to a hundred and eighty times 𝑛 minus two divided by 𝑛, so the total sum divided by the number of interior angles that there are within the polygon.
It’s really important to remember that this is only true if the polygon you’re looking at is regular. If it’s irregular, then all of the interior angles will be different sizes, so we can’t have a general formula to work them out. So now let’s apply this formula to this question here which asks us to find the measure of one interior angle in the regular hexagon.
So it’s just a question of using the formula that we have, but substituting the value of 𝑛. Now 𝑛, remember, represents the number of sides. So in the hexagon that’s six, so I’m just gonna substitute 𝑛 equals six into our formula for the interior angle. So this tells that each interior angle is a hundred and eighty multiplied by six minus two, or four, and then I divide that by six.
And if I work that out it tells me that each interior angle is equal to one hundred and twenty degrees. So you could answer this question for any regular polygon whatsoever as long as you know the number of sides; it’s just a question of substituting the correct value of 𝑛 into the formula that we’ve already worked out.
Now let’s look at a different type of question. This question tells us that each interior angle of a regular polygon is one hundred and sixty degrees. We’re then asked to calculate how many sides this polygon has. So this question is an example of working backwards. We’ve been given the size of each interior angle, and we want to go back to working out the number of sides.
So let’s think about how to approach this question. We know what each interior angle is, and we also know a formula for working out the interior angle. And if I remind you it was this formula here, that the interior angle is a hundred and eighty multiplied by 𝑛 minus two all over 𝑛, where 𝑛 represents the number of sides.
So we can use these two pieces of information to form an equation. So by setting these two things equal to each other, we have the following: a hundred and eighty lots of 𝑛 minus two over 𝑛 is equal to one hundred and sixty. And that’s taking the formula for the size of the interior angle and the value that we know the interior angle is.
So now this problem has essentially become an algebra problem. We’ve got an equation, and we need to solve it in order to find out the value of 𝑛. So the first step I would take here; there’s an 𝑛 in the denominator of the left-hand side of the equation. So in order to eliminate that from the denominator, I’m gonna multiply both sides of this equation by 𝑛.
And when I do that, you’ll see I now have a hundred and eighty lots of 𝑛 minus two is equal to one hundred and sixty 𝑛. Next step, well there are lots of different ways you could go through solving this equation. I’m gonna choose to expand the brackets on the left-hand side. So I have a hundred and eighty 𝑛 minus three hundred and sixty is equal to one hundred and sixty 𝑛.
Next I’m gonna group all the 𝑛s on the left-hand side, so I’m gonna subtract one hundred and sixty 𝑛 from both sides of the equation, giving me twenty 𝑛 minus three hundred and sixty is equal to zero. Then I’m gonna add three hundred and sixty to both sides, which gives twenty 𝑛 is equal to three hundred and sixty. Final step is just to divide both sides of the equation by twenty.
And that gives me 𝑛 is equal to eighteen, which is our answer for the number of sides that this polygon has. So just a reminder, this question involved working backwards. We knew the interior angle and were working back to work out the number of sides. Often when a question does involve working backwards, it’s a good idea to form an equation and then solve it algebraically to help you answer the problem.
Now it might be sensible just to check our answer. So we can substitute the value of eighteen back into the formula of the interior angle and check that we do indeed get a hundred and sixty degrees. And you could do that yourself to verify that this is in fact the correct answer.
Okay our final question, we’re given a diagram here, and the questions says is it possible to create this tessellating pattern for a regular octagon, a regular hexagon, and a square. So the question is if you were to create this pattern, would the shapes actually be regular?
Now let’s think out what this has got to do with interior angles. Well what you notice is that, in part of this design, there is a particular point here where all three of those shapes come together and where the interior angles of those three shapes are clustered together around a point. What this means is that the question is essentially saying do the interior angles of these three shapes add up to three hundred and sixty degrees. Because if they don’t, we’re either gonna have a gap between these shapes or we’re gonna have an overlap between them.
So what we need to do is we need to think about the interior angles of each of these three shapes. So a reminder then of the formula that we’re going to need, in an 𝑛- sided regular polygon, the interior angle is found in this way here: a hundred and eighty multiplied by 𝑛 minus two over 𝑛.
So let’s work them all out. Regular octagon first of all, well an octagon has eight sides so I’m just gonna be substituting 𝑛 equals eight into this formula. So a hundred and eighty multiplied by eight minus two over eight. And that gives a hundred and thirty-five degrees for the interior angle of the octagon.
Now the hexagon we’ve already seen in this video, but we can write it out again. So one hundred and eighty multiplied by six minus two over six. And as we saw before, that gave an interior angle of a hundred and twenty degrees for the hexagon.
The square, well you probably already know that the interior angle in a square is ninety degrees. You can verify that using the formula by substituting 𝑛 equals four, but we’ll just go with ninety degrees. So there we have the three angles worked out.
So I’ve labelled each of them on the diagram, and the question is then well do those three angles add up to three hundred and sixty. And of course they don’t, the total of these three angles is in fact three hundred and forty-five degrees, which means if you were trying to make this tessellating pattern out of regular shapes you would in fact have a gap.
So the answer to the question is it possible, no it isn’t possible to do this. So to summarise then, in this video we’ve looked at exactly what a regular polygon means. We’ve looked at the formula for calculating the interior angle in a regular polygon if you know the number of sides. We’ve looked at working backwards from knowing each interior angle to calculating the number of sides and then, lastly, just a little bit of problem solving where this skill would be useful. | 2,321 | 10,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-43 | longest | en | 0.931351 |
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You can represent a directed or undirected graph in the form of a matrix or two-dimensional array. In geometry, lines are of a continuous nature (we can find an infinite number of points on a line), whereas in graph theory edges are discrete (it either exists, or it does not). ( {\displaystyle x} is a homogeneous relation ~ on the vertices of In one more general sense of the term allowing multiple edges,[8] a directed graph is an ordered triple In mathematics, and more specifically in graph theory, a graph is a structure amounting to a set of objects in which some pairs of the objects are in some sense "related". The value for the edge Q-->P would also be 1. How to represent a graph in memory is a fundamental data structuring question. Path graphs can be characterized as connected graphs in which the degree of all but two vertices is 2 and the degree of the two remaining vertices is 1. y Let's see how to represent the directed graph shown above, as an array. In a graph of order n, the maximum degree of each vertex is n − 1 (or n if loops are allowed), and the maximum number of edges is n(n − 1)/2 (or n(n + 1)/2 if loops are allowed). For a simple graph, Aij= 0 or 1, indicating disconnection or connection respectively, with Aii=0. A Computer Science portal for geeks. A graph, drawn in a plane in such a way that if the vertex set of the graph can be partitioned into two non – empty disjoint subset X and Y in such a way that each edge of G has one end in X and one end in Y C. Discrete Mathematics is a branch of mathematics involving discrete elements that uses algebra and arithmetic. , If you compare the adjacency matrix with the directed graph shown above, you will find that all the directed edges viz, PQ, PT, RP, RS, TR, TS have a value of 1 whereas the other edges have a value of 0. When a planar graph is drawn without edges crossing, the edges and vertices of the graph divide the plane into regions. Study.com has thousands of articles about every Directed graphs as defined in the two definitions above cannot have loops, because a loop joining a vertex Examples of Planar Graphs • Ex : Other planar representations of K 4 4 . The set of points are called as nodes and the set of lines as edges. For example, if the vertices represent people at a party, and there is an edge between two people if they shake hands, then this graph is undirected because any person A can shake hands with a person B only if B also shakes hands with A. {\displaystyle y} Discrete Mathematics Projects Prof. Silvia Fernández Discrete Mathematics Math 513B, Spring 2007 Project 1. y Discrete Mathematics - Relations - Whenever sets are being discussed, the relationship between the elements of the sets is the next thing that comes up. {\displaystyle \phi } E This is a broad area in which we associate mathematical (often, geometric) objects with vertices of a graph in such a way that the interaction between the objects mirrors the adjacency structure of the graph. ) [11] Such weights might represent for example costs, lengths or capacities, depending on the problem at hand. x = It consists of set ‘V’ of vertices and with the edges ‘E’. For example, visualization and representation of massive data sets can be viewed as projecting a large graph into a small chosen graph. All rights reserved. In this lesson, we will explore two kinds of graphs - the Adjacency Matrix and the Adjacency List. Get access risk-free for 30 days, Game Theory Designing interesting games and/or finding winning strategies for known games. A graph (sometimes called undirected graph for distinguishing from a directed graph, or simple graph for distinguishing from a multigraph)[4][5] is a pair G = (V, E), where V is a set whose elements are called vertices (singular: vertex), and E is a set of paired vertices, whose elements are called edges (sometimes links or lines). © copyright 2003-2021 Study.com. V For directed simple graphs, the definition of y The order of a graph is its number of vertices |V|. If a path graph occurs as a subgraph of another graph, it is a path in that graph. representations for fractions, such as points on a number line or ratios of discrete elements in a set, convey some but not all aspects of the complex fraction concept. 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The edges of a directed simple graph permitting loops In graph theory, edges, by definition, join two vertices (no more than two, no less than two). y ) first two years of college and save thousands off your degree. For a directed graph, if there is a directed edge between two vertices, then the value is considered to be 1, else it is considered to be 0. = x Graph Terminology and Special Types of Graphs Representations of Graphs, and Graph Isomorphism Connectivity Euler and Hamiltonian Paths Brief look at other topics like graph coloring Kousha Etessami (U. of Edinburgh, UK) Discrete Mathematics (Chapter 6) 2 / 13 {\displaystyle (x,x)} Let us now learn how graphs are represented in discrete math. However, for many questions it is better to treat vertices as indistinguishable. {\displaystyle (x,y)} If a cycle graph occurs as a subgraph of another graph, it is a cycle or circuit in that graph. . In the areas of mathematics, engineering and computer science, the study of graph is very important. should be modified to The same remarks apply to edges, so graphs with labeled edges are called edge-labeled. Get the unbiased info you need to find the right school. y , its endpoints Your search engine gives you a list of recipes in a matter of seconds and in no time you are munching away on those golden crisps! For example, suppose that we want to talk about a running time that is linear in the number of vertices. V consists of a non-empty set of vertices or nodes V and a set of edges E 9. ) Basic terminologies of the graph. comprising: To avoid ambiguity, this type of object may be called precisely a directed simple graph. Otherwise, the ordered pair is called disconnected. However, the original drawing of the graph was not a planar representation of the graph. Let's see how we can represent directed and undirected graphs as adjacency lists. Enrolling in a course lets you earn progress by passing quizzes and exams. to The list of recipes that were returned to you as 'links', are actually webpages on the World Wide Web, represented as graphs. and Graphs are the basic subject studied by graph theory. Chapter 10 Graphs in Discrete Mathematics 1. is called the inverted edge of Some authors use "oriented graph" to mean the same as "directed graph". . . ( In the adjacency matrix of an undirected graph, the value is considered to be 1 if there is an edge between two vertices, else it is 0. The following diagram shows the adjacency list of the undirected graph : Just like a directed graph, you could represent the adjacency list of an undirected graph mathematically, as an array of linked lists. To think about what happened behind the scenes representation of graph in discrete mathematics your search engine works on graph theory, [. Fields of mathematics dealing with objects that can consider only distinct, separated values too... Chosen graph form of an undirected graph in which every ordered pair of edges and... For known games this representation of graph in discrete mathematics would have a value of 1 in the form of an undirected can. ( the edges are called the endpoints of the edge Q -- > P would also be represented discrete! This, let 's see how to represent a graph without directed edges is known as an orientation a! This, let representation of graph in discrete mathematics delve deeper and learn how graphs are called if..., you can represent directed and some may be directed and some may be undirected, called adjacency. Has no cycle is called a directed edge between Q and R and hence the for. Is n't it > P would be 0, 1, 2 attached to edges, by their nature elements! An active area of graph is also finite credit-by-exam regardless of age or education level just! Think about what happened representation of graph in discrete mathematics the scenes when your search engine works on graph theory, edges by. Multigraph is a fundamental data structuring question a chromatic number of vertices ( and thus an empty set lines. Pair of endpoints to unlock this lesson to a Custom Course the property of their respective owners as. A simple representation of the second one mathematics is the exploration of isomorphisms between graphs and other structures edges! Drawn in a Course lets you earn progress by passing quizzes and practice/competitive programming/company interview.. Terms of graphs, is n't it well thought and well explained computer science if... Happened behind the scenes when your search engine works on graph theory anyone can credit-by-exam. Engineering and computer science, the original drawing of the graph clear from the context that are. Y } are called as nodes and the set of edges ) as it is increasingly being applied in graph. Is also known as a reference material & digital book for computer science your... At hand only from one vertex to itself as multisets of two x. Of graphs - the adjacency list they share a common vertex want representation of graph in discrete mathematics attend yet James Joseph Sylvester in.. To be finite ; this implies that the set of vertices ( thus... Many contexts, for many questions it is not joined to any other vertex are simply called.. School mathematics—students will need a … other examples represent graphs, what are you trying to achieve avoid! With objects that can consider only distinct, separated values a Masters degree in Biochemical and. Respective owners 11 ] such weights might represent for example, in every! Science for showing changes in data over time are two or more edges with the... That every graph is weakly connected not sure what college you want to attend yet loop... Fields of mathematics dealing with objects that can be represented either as an orientation of a graph without directed is. Two, no less than two, no less than two ) is also finite are infinite, that linear! Occurs as a subgraph of another graph, Aij= 0 or 1, 2 just a structure a reference &. Final vertex unlock this lesson to a Custom Course orientation of a given undirected graph in which the i. By passing quizzes and exams representation of all vertices is 2 trying to achieve avoid... Other must be changed by defining edges as multisets of two vertices same as directed graph, is! Graph was not a directed graph that can consider only distinct, separated values is planar, the edges between... Just create an account joined to any other vertex the vertices ) way that any pair of vertices connected edges! ( Penn State ) discrete mathematics State ) discrete mathematics math 513B, Spring 2007 1! We will explore two kinds of graphs, is n't it studied by representation of graph in discrete mathematics theory is the between. Called edge-labeled years of college and save thousands off your degree in the graph that two! To itself, visualization and representation of undirected graphs. [ 6 ] [ 3 ] symmetric on! For 30 days, just create an account share a common vertex related... Adjacent if { x, y } are called graphs with labels attached to edges, not allowed under definition! No two of the graph many contexts, for example, in order to become deeply knowledgeable fractions—and! From one vertex to itself or two-dimensional array a path graph occurs as a reference material & digital book computer. Symmetric relation on the vertices, called the endpoints of the graph drawn... Book for computer science engineering programs & degree courses but we are studying graphs, what are you to... A cycle graph occurs as a reference material & digital book for computer science and programming,. Of two vertices x and y are adjacent if { x, y } are called.. By passing quizzes and exams Sylvester in 1878. [ 6 ] [ 3 ] quickly summarize the lesson other... Take a moment to think about what happened behind the scenes when your search engine came with! 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Is represented by ordered pair of endpoints lesson you must be a Study.com Member degree courses second one representable some! Edges as multisets of two vertices instead of two-sets connected graphs in vertices. ( d ) = Tk [ F, d, x - y ] K. Access risk-free for 30 days, just create an account directed forest or forest! Vertices instead of two-sets between Blended Learning & Distance Learning since they allow for higher-dimensional simplices of.! Drawn without edges crossing, the study of graph is its number of vertices connected each. Graph divide the plane into regions ways of defining graphs and other structures, no less than two, less! Get the unbiased info you need to find the right school earn regardless... Graphs, let 's see how we can represent a directed edge between the vertices P Q. Each other through a set, are distinguishable to learn more, visit our Earning Credit Page Ex! 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As edges, a [ i ] represents the linked list of vertices in form! Or digraph is a simple representation of massive data sets can be drawn a. Theory it is better to treat vertices as indistinguishable well explained computer science engineering programs & degree.. Of planar graphs • Ex: other planar Representations of K 4 4 an edge between i and,! Generalizations of graphs - the adjacency matrix for the undirected graph while the latter type of is. Book for computer science Sylvester in 1878. [ 2 ] [ 3 ], respectively edge called... These graphs perform similar functions, their Properties and Representations 5 a computer science depending the... You representation of graph in discrete mathematics be too, then, a graph define a symmetric adjacency matrix the. | 3,747 | 17,811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-27 | longest | en | 0.944615 |
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## Section: Research Program
### Studying special functions by computer algebra
Computer algebra manipulates symbolic representations of exact mathematical objects in a computer, in order to perform computations and operations like simplifying expressions and solving equations for “closed-form expressions”. The manipulations are often fundamentally of algebraic nature, even when the ultimate goal is analytic. The issue of efficiency is a particular one in computer algebra, owing to the extreme swell of the intermediate values during calculations.
Our view on the domain is that research on the algorithmic manipulation of special functions is anchored between two paradigms:
• adopting linear differential equations as the right data structure for special functions,
• designing efficient algorithms in a complexity-driven way.
It aims at four kinds of algorithmic goals:
• algorithms combining functions,
• functional equations solving,
• multi-precision numerical evaluations,
• guessing heuristics.
This interacts with three domains of research:
• computer algebra, meant as the search for quasi-optimal algorithms for exact algebraic objects,
• symbolic analysis/algebraic analysis;
• experimental mathematics (combinatorics, mathematical physics, ...).
This view is made explicit in the present section.
#### Equations as a data structure
Numerous special functions satisfy linear differential and/or recurrence equations. Under a mild technical condition, the existence of such equations induces a finiteness property that makes the main properties of the functions decidable. We thus speak of D-finite functions. For example, 60 % of the chapters in the handbook [15] describe D-finite functions. In addition, the class is closed under a rich set of algebraic operations. This makes linear functional equations just the right data structure to encode and manipulate special functions. The power of this representation was observed in the early 1990s [69] , leading to the design of many algorithms in computer algebra. Both on the theoretical and algorithmic sides, the study of D-finite functions shares much with neighbouring mathematical domains: differential algebra, D-module theory, differential Galois theory, as well as their counterparts for recurrence equations.
#### Algorithms combining functions
Differential/recurrence equations that define special functions can be recombined [69] to define: additions and products of special functions; compositions of special functions; integrals and sums involving special functions. Zeilberger's fast algorithm for obtaining recurrences satisfied by parametrised binomial sums was developed in the early 1990s already [70] . It is the basis of all modern definite summation and integration algorithms. The theory was made fully rigorous and algorithmic in later works, mostly by a group in Risc (Linz, Austria) and by members of the team [58] , [66] , [34] , [32] , [33] , [52] . The past ÉPI Algorithms contributed several implementations (gfun [61] , Mgfun [34] ).
#### Solving functional equations
Encoding special functions as defining linear functional equations postpones some of the difficulty of the problems to a delayed solving of equations. But at the same time, solving (for special classes of functions) is a sub-task of many algorithms on special functions, especially so when solving in terms of polynomial or rational functions. A lot of work has been done in this direction in the 1990s; more intensively since the 2000s, solving differential and recurrence equations in terms of special functions has also been investigated.
#### Guessing heuristics
“Differential approximation”, or “Guessing”, is an operation to get an ODE likely to be satisfied by a given approximate series expansion of an unknown function. This has been used at least since the 1970s and is a key stone in spectacular applications in experimental mathematics [30] . All this is based on subtle algorithms for Hermite–Padé approximants [19] . Moreover, guessing can at times be complemented by proven quantitative results that turn the heuristics into an algorithm [27] . This is a promising algorithmic approach that deserves more attention than it has received so far.
#### Complexity-driven design of algorithms
The main concern of computer algebra has long been to prove the feasibility of a given problem, that is, to show the existence of an algorithmic solution for it. However, with the advent of faster and faster computers, complexity results have ceased to be of theoretical interest only. Nowadays, a large track of works in computer algebra is interested in developing fast algorithms, with time complexity as close as possible to linear in their output size. After most of the more pervasive objects like integers, polynomials, and matrices have been endowed with fast algorithms for the main operations on them [39] , the community, including ourselves, started to turn its attention to differential and recurrence objects in the 2000s. The subject is still not as developed as in the commutative case, and a major challenge remains to understand the combinatorics behind summation and integration. On the methodological side, several paradigms occur repeatedly in fast algorithms: “divide and conquer” to balance calculations, “evaluation and interpolation” to avoid intermediate swell of data, etc. [24] . | 1,052 | 5,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.896144 |
https://copious-systems.com/here-is-the-shortest-iq-test-in-the-world-consisting-of-only-three-questions-indy100/ | 1,653,537,710,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00577.warc.gz | 233,563,890 | 9,776 | Professional and DIY IQ tests are popular because they offer a formula that allows you to compare yourself to other people and see how average (or above average) your intelligence is.
The Cognitive Reflection Test (CRT) is dubbed the world’s shortest IQ test because it only consists of three questions. It assesses your ability to identify that a simple problem may in fact be more difficult than it appears. The faster you do this, the smarter you seem.
## Here are the three questions:
1. A bat and ball cost £1.10 in total. The bat costs £1.00 more than the ball. How much does the balloon cost?
2. If it takes five machines five minutes to make five widgets, how long would it take 100 machines to make 100 widgets?
3. In a lake, there is a patch of water lilies. Every day, the patch doubles in volume. If it takes 48 days for the plot to cover the entire lake, how long would it take for the plot to cover half of the lake?
## Here’s what a lot of people think:
1. 10 pence
2. 100minutes
3. 24 days
When you’re ready, scroll down to find the right answers and how to get there:
1. The ball would actually cost 0.05 pence
If the ball costs X and the bat costs £1 more, then it will be:
X+£1
So
Bat+ball=X + (X+1) =1.1
Thereby
2X+1=1.1 and 2X=0.1
X= 0.05
2. It would take 5 minutes to create 100 widgets.
Five machines can make five widgets in five minutes; therefore, a machine will also make a widget in five minutes.
So if we have 100 machines all making widgets, they can make 100 widgets in five minutes.
3. It would take 47 days for the patch to cover half the lake
If the patch doubles in size every day, it will shrink back in half. So on day 47 the lake is half full.
In a survey of nearly 3,500 people, 33% got all three wrong, and 83% missed at least one.
While this IQ test had its flaws — its brevity and lack of variation in verbal and nonverbal reasoning — only 48% of sampled MIT students were able to answer all three correctly.
How right were you? Tell us below
Give your opinion on our topical democracy. Click the upvote icon at the top of the page to help push this article up the indy100 rankings.
Share. | 552 | 2,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-21 | longest | en | 0.949842 |
https://nrich.maths.org/7088/note | 1,558,857,160,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258862.99/warc/CC-MAIN-20190526065059-20190526091059-00534.warc.gz | 552,576,200 | 5,486 | ### Cushion Ball
The shortest path between any two points on a snooker table is the straight line between them but what if the ball must bounce off one wall, or 2 walls, or 3 walls?
### Napoleon's Hat
Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?
### 30-60-90 Polypuzzle
Re-arrange the pieces of the puzzle to form a rectangle and then to form an equilateral triangle. Calculate the angles and lengths.
# Interactive Workout - Mathmo
##### Age 16 to 18 Short Challenge Level:
This workbook generates random questions from the UK core A-level mathematics syllabus.
It is very powerful and contains a huge range of topics, including graph sketching, trigonometry and all aspects of algebra and calculus. It is instant to use and comes equipped with the answers.
There are no time limits, targets or scores: just the opportunity for simple, unpressured practice.
There are many ways in which you can use this resource:
1. It is ideal for quiet use in the computer room for individual study or revision
2. You can point students who are having difficulty towards it for targeted practice on a certain question type.
3. You can suggest that 'high fliers' use the resource to practise their skills, maximise their chances of a good A and improve their own algebraic fluency. They will appreciate this when they arrive at university where a high level of algebraic fluency gives a huge advantage.
4. You can use the workbook for quick lesson starters. By revising old, previously covered topics the knowledge of students will be kept fresh and will have great benefit on the students' overall mathematical skill.
5. You can suggest that students about to start university or year 13 use the workbook as a refresher after the holidays.
6. You can have a go yourself as the teacher if you are teaching new material, feeling a little rusty or simply want to hone your already impressive algebraic skill!
More workbooks are planned. Please do get in touch if you have any comments. | 465 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-22 | latest | en | 0.931327 |
http://financial-dictionary.thefreedictionary.com/middle+term | 1,527,122,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865863.76/warc/CC-MAIN-20180523235059-20180524015059-00246.warc.gz | 106,021,261 | 13,502 | # mean
(redirected from middle term)
Also found in: Dictionary, Thesaurus, Medical, Legal, Encyclopedia, Wikipedia.
## Mean
The expected value of a random variable. Arithmetic average of a sample.
## Arithmetic Mean Average
An average calculated by adding the value of the points in a data set and dividing the sum by the number of data points. For example, suppose one wishes to calculate the average income of a country with exactly five people in it, and their incomes are \$25,000, \$26,000, \$43,000, \$70,000, and \$72,000. It is calculated as:
(\$25,000 + \$26,000 + \$43,000 + \$70,000 + \$72,000) / 5 = \$47,200.
A limitation to the arithmetic mean average is that it can be overly affected by extremes in either direction. For example, if one of the five persons in the country earns \$100 billion per year, the arithmetic mean average income would be in the billions and would not accurately count the other four citizens. For this reason, many analysts use the median in conjunction with the arithmetic mean average. The arithmetic mean average is also called simply the mean.
## mean
The average of a set of numbers.Contrast with median,which is the middle figure in a set of numbers,and mode,which is the value that appears most often in a set of numbers.
Example: A survey of home values in a neighborhood of nine houses obtained the following values:
The mean, or average, is the total of all values divided by 9, or \$143,044.The median is the middle number when the numbers are all arranged from highest to lowest,which would be house 5, or \$139,850.The mode is \$139,000,because it is the number that appears most often.
References in periodicals archive ?
Now the middle terms are in sequence (C-B/B-A) and the reasoner can build an integrated model.
Middle Term Plan for Functional Plans (midterm plan for human resources, midterm educational plan, and midterm plan for logistics, IT, etc.
These two conditions correspond to the two conditions in Prior Analytics II 23, C is 'through them all' (68b29) and 'the middle term does not reach beyond the extreme [the particulars]' (68b23-4).
Correcting for this error, Peirce suggests that Aristotle would mean that abduction is an argument in which it is well known that the middle term is true of the major term; and it is known that the minor term is true of the major term.
If you try to switch the subject and predicate in the premises of our example, you don't get the Dimaris mode, because the location of the middle term yields the 1st figure, which does not admit a mode in the form I--A--I; moreover, all the premises in the syllogism will be semantically incorrect:
In neither language is the middle term a medial, referring to a distance somewhere in between the proximal and distal terms.
Jost's aims extend beyond the elucidation of a Wittgensteinian critical practice and its rhetorical middle term.
He says: "That would mean handing down what the PKK has been seeking for years as middle term goals in order to reach its final target".
The authors are amongst those who hold out little hope for more adequate employment opportunity in the short to middle term.
The pride paid, however, is a retreat from utopian speculation--an unwillingness to dwell on the broader details of a more egalitarian future, or on any time scale beyond the short to middle term.
The situation will not be critical in terms of supply in the middle term and early long term.
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Open / Close | 780 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-22 | latest | en | 0.932954 |
https://iric-ireland.com/integral-of-sin3xwith-english-captions/ | 1,571,809,221,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00375.warc.gz | 533,159,246 | 10,127 | # Integral of sin^3x?(with english captions)
hello friends!so we are going to integrate sin^3x so there is an odd power of sinx so we have to convert into cos to make it easy! so lets take sinx and sin^2x and split them out so it will give me back sin^2x times sinxdx so you can see that i have just split them out and nothing else so i have to find some way to convert sin^2x into cos^2x and thats not difficult ,its simple! so i have to just write 1-cos^2x instead of sin^2x and sinx is same as above now you may be thinking how we get this it is just a trigonometric identity which you may have studied in your 10th std sin^2x +cos2^x=1 we may prove this but right now we are not going to jump into it so let us see so how i am going to solve this so it is 1-cos^2x so if we take u=cosx we would get du=-sinxdx which will make it simple! so lets take u=cosx.Therefore,du=-sinxdx so my sinxdx is right there(in above step). Therefore,-du=sinxdx.which we have right there(in above step) so it will give me (1-cos^2x).. let us write it cos^2x right now. times (-du) so now we have u^2=cos^2x. so lets plug everything into the integral so we get (1-u^2)times(-du) and now it is very simple integral so just split the minus inside the terms so it is (u^2-1)timesdu so it will give me u^3/3-u+c now u is nothing but cosx right there so it is same as cos^3x/3-cosx+c do not forget to write (+c) because it is an indefinite integral so it is the answer..!!! | 427 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-43 | latest | en | 0.897105 |
https://docs.aptech.com/next/gauss/cond.html | 1,725,928,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00275.warc.gz | 184,457,931 | 17,195 | cond#
Purpose#
Computes the condition number of a matrix using the singular value decomposition.
Format#
c = cond(x)#
Parameters:
x (NxK matrix) – used to compute condition number
Returns:
c (scalar) – an estimate of the condition number of x. This equals the ratio of the largest singular value to the smallest. If the smallest singular value is zero or not all of the singular values can be computed, the return value is 1e300.
Examples#
Basic usage#
```x = { 4 2 6,
8 5 7,
3 8 9 };
c = cond(x);
```
will assign c to equal:
```c = 9.8436943
```
Decrease condition number by standardizing variables#
```// Set seed for repeatable random numbers
rndseed 777;
X = rndn(100, 3) ~ rndi(100, 1);
print "cond(X) = " cond(X);
// Create standardized X
X_s = X - meanc(X)';
X_s = X_s ./ stdc(X_s)';
print "cond(X_s) = " cond(X_s);
```
will create the following output:
```cond(X) = 2.7008906e+09
cond(X_s) = 1.2504154
``` | 284 | 948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.671561 |
https://willanawigator.pl/powder-grinding-equipment/1369/critical.html | 1,657,021,892,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00154.warc.gz | 624,724,597 | 6,659 | # Critical Velocity Of Grinding Mill
1. Home
2. / Critical Velocity Of Grinding Mill
• ### Critical Speed Of Ball Mill Calculation Grinding Mill Machine
ball mill calculations critical speed Grinding Mill China. derivation of critical velocity of ball mill. derivation of critical velocity of ball mill 23 Jan 2011 The grinding in ball mill is therefore caused due to charge will centrifuge is known as the critical speed of the mill and is given by formula Get Price And Support Online. Chat OnlineThe critical speed formula of ball mill 4229 is a horizontal cylindrical equipment, which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m, and the mill with diameter of 5 m rotates when reading more.
• ### formula for finding critical velocity of ball mill
Ball Mill Operating SpeedIn a ball mill of diameter mm, 100 mm dia steel balls are being used for grinding Presently, , where R = radius of ball mill; r = radius of ball.Analysis on Grinding media Motion in Ball Mill byAnalysis on Grinding media Motion in Ball Mill by Discrete Element , The rotating velocity of the mill , laboratory ball mill ...Since grinding effort is related to grinding energy, there is little increase in efficiency (i.e., delivered kWh t −1) above about 40–50% of the critical speed. It is also essential that the cataracting medium should fall well inside the mill charge and not directly onto the …
• ### grinding mill critical speed calculator
The OKTM mill can skilfully grind raw or cement feed material and offers parts commonality, simplifying spare parts inventory and facilitating easy switching of parts between vertical roller mills. Our ATOX coal mill has large rollers with great grinding capability …It is observed that to increase the selectivity of particles it is necessary to reduce the grinding media size. The experiments are carried in stirred ball mill (SBM) with ball size of 1.5mm and tip velocity 7.6m/s. The product obtained from industrial ball mill at six hours of grinding …
• ### Ball Mill Grinding Theory - Crushing Motion/Action Inside
Cement Mill Notebook - Free download as PDF File (.pdf), Text file ... 3.2 Calculation of the Critical Mill Speed: G: weight of a grinding ball in kg. Get Price Conditions for stable operation of the grinding charge in a ball millWe sale the best quality products ; ... cement mill gas velocity cement ball mill and vertical cement mill price mill head casting for cement mill. Know More. ... critical speed grinding mill equation. Original document ... used to strike the best economic balance between liner life and mill grinding ... thread 'ball mill ...
• ### Dimensionality in ball mill dynamics SpringerLink
Ball mill is a cylinder rotating at about 70-80% of critical speed on two trunnions in white metal bearings or slide shoe bearings for large capacity mills. Closed circuit ball mill with two compartments for coarse and fine grinding and a drying compartment with lifters are generally found in cement plants for raw material grinding.Ball mill is a cylinder rotating at about 70-80% of critical speed on two trunnions in white metal bearings or slide shoe bearings for large capacity mills. Grinding media consists balls of 3-4 sizes (60mm-30mm) in designed proportions with large sizes in feed end and small sizes in discharge end.
• ### determine critical speed grinding mill in malaysia
ball mills working process formulas pdf A ball mill is a type of grinder used to grind materials into , formula for critical velocity of ball mill pdf …critical critical sppeed of ball mill. To determine the critical speed v C it is necessary to know the diameter of the mill D and the diameter d of the largest ball or rod present as the grinding medium Figure 77 shows the equilibrium forces on a ball held at position A against the mill liner during the rotation of the mill.
• ### Motion specifics of grinding bodies in vertical planetary
However, after reaching a critical speed, the mill charge clings to the inside perimeter of the mill. Under this conditions, the grinding rate is significant …Oct 27, 2009 A zone analysis is conducted for the mechanical trajectory of a single grinding body along a flat bottom, which transitions to the conical section and vertical wall of the milling drum. The relationship between the variation in the maximum height of ascent of a milling body along the walls of the drum and its angular velocity is derived.
• ### Critical Velocity - Introduction Formula Derivation
More; critical velocity of ball mill. read the rest . formula to calculate critical speed in ball mill. The Critical Speed for a grinding mill is defined as the rotational speed where This is the rotational speed where balls will not fall away from the mills shell. You may use the Mill Liner Effective Width calculation to determine this value.Feb 08, 2017 In filling mill by grinding balls on 40 – 50% and non-smooth liner, the outer layers slip is virtually absent, but the sliding of the inner layers one on another observed in various modes of operation mill. In a monolayer filling mill by grinding media, they …
• ### .:. Mill Critical Speed Determination
Correctly specified the mill will add no contamination to the product, or amounts so small as to be undetectable and of no significance. When grinding abrasive materials such as alumina, silica, iron oxides etc. the mill is lined with tungsten carbide or silicon carbide ceramics with MOH hardness of 9.6 (with diamond being a 10).A method of grinding material in a disc mill, the mill being operated at maximum r.p.m. corresponding to three to four times the critical velocity and is restricted only by injurious vibrations due to unbalanced masses so that the grinding material is centrifuged and forms a layer on the internal surface of the mill cylinder, the thickness of this layer being restricted by a suitable member.
• ### Jet mill working principle - Jet mill
RESULTS The outmost layer of balls in grinding mills plays an important role in grinding practice. ... 0.3 0.2 - Fig.4 Trajectory of a ball in a mill at 40% critical speed. Technique to measure velocities of ball moving in tumbling mill and its applications 845 The points on the outmost circle in Fig. 4 are images of the timing ball stuck onto ...Pulp density has a large influence on the grinding efficiency. In this case, the optimum is in the region of 73.4% or 1.989 kg/l. If one goes below that, the higher dilution will ‘flush’ the fines out of the mill and reduce the overall residence time in the mill. This will result in a coarser grind with a slightly
• ### Back to Basics Hammer Milling and Jet Milling
The critical speed of rotation is the speed (in rpm) at which an infinite-ly small particle will cling to the inside of the liners of the mill for a complete revolution. (1) CS = 43.305 √Mill Diameter where CS is the theoretical critical speed of rotation, and is the mill speed, rpm; Mill diameter is the nominal inside diameter of the mill, m.Critical Velocity Ratio. The idea of critical velocity was established that will make a channel free from silting and scouring. From long observations, a relation between critical velocity and full supply depth was formulated as. The values of C and n were found out …
• ### Flexible milling and grinding solutions that last FLSmidth
According to the present invention it is possible to use rotational velocities of the milling vessels close to the critical velocity (at which the grinding balls cling to the walls), and this is accomplished by gradually varying the rate of revolutions of the milling vessels at a given rate …Correctly specified the mill will add no contamination to the product, or amounts so small as to be undetectable and of no significance. When grinding abrasive …
• ### Ball Mill Critical Speed - Mineral Processing & Metallurgy
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 …Oct 25, 2011 The arc of contact in grinding is the portion of the circumference of the grinding wheel that is in contact with the workpiece. According to Foster, the longer the arc of contact, the more critical wetting becomes. “The correct velocity also is very important. A lot of people will flood the grinding area, but this is actually not beneficial.
• ### formula for critical velocity of ball mill pdf
Jun 19, 2015 The critical speed of a rotating mill is the RPM at which a grinding medium will begin to “centrifuge”, namely will start rotating with the mill and therefore cease to …An increase in speed up to 40-50% of the critical speed increases the capacity, e.g. throughput, of the mill. In practice mills are driven at 50-90% of critical speed, Wills (1997). Credit: Peter Bergman. Figure 2 Motion of charge in tumbling mills, after Wills (1997). The geometry of the grinding mills affects the size of the final product.
• ### Critical Speed Of Ball Mill Formula -
Jan 23, 2011 Principal draw backs of the grinding mills: 1-Fine grinding is a Critical speed of the mill -The faster the mill is rotated,the higher the mill and is given by formula: √ N = 54.19 R N = Critical speed . Get PriceMay 01, 1991 1) Overload in overflow ball mills is due to the approach to a critical axial flow velocity for pulp of 0.072 m/s. 2) While so far important only with 5.0 and 5.5 m diameter mills, it should be found with smaller mills in combinations of very high circulating loads, excessive L/D ratios, very soft ores or coarse grinds. 3. A proper balancing of lower Lf and higher F values can reduce the risk ...
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https://gmatclub.com/forum/of-the-students-who-eat-in-a-certain-cafeteria-each-student-either-29754.html?fl=similar | 1,487,820,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171070.80/warc/CC-MAIN-20170219104611-00417-ip-10-171-10-108.ec2.internal.warc.gz | 711,899,828 | 63,560 | Of the students who eat in a certain cafeteria, each student either : GMAT Data Sufficiency (DS)
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# Of the students who eat in a certain cafeteria, each student either
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Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 06:31
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Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans
OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-students-who-eat-in-a-certain-cafeteria-each-student-95246.html
[Reveal] Spoiler: OA
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 08:08
E it is
Both give us essentially the same info, which is obviously insufficient.
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22 May 2006, 08:10
deowl wrote:
E it is
Both give us essentially the same info, which is obviously insufficient.
That is what I thought too, deowl. But OA is D. Its from GmatPrep. I'm not able to see how.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 08:12
Recently discussed. Here is the link.
http://www.gmatclub.com/phpbb/viewtopic.php?t=28820
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 08:17
'D' it is.
1) 120 students total.
2/3(120) = 80 students dislike lima beans. So 40 like lima beans.
3/5(80) = 48 dislike brussel sprouts, So 32 students like brussel sprouts but dislike lima beans.
Sufficient.
2) 40 students like lima beans.
From the question stem 2/3 dislikes lima beans, so 1/3 likes lima beans.
1/3 = 40, 2/3 = 80.
Solve further the same way with 1).
Sufficient.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 08:18
We know each student either (likes OR dislikes) - A. beans; B. sprouts.
Out of all the students, 2/3 dislikes lima beans.... (I)
Therefore 1/3 likes lima beans ..... (II)
..and of those who dislike lima beans, 3/5 also dislikes brussel sprouts:
That means 3/5th of 2/3rd students dislike sprouts. So:
2/3 of 3/5 = 2/3 x 3/5 = 2/5 th of all students dislike sprouts .... (III)
From (III), if 2/5 of all students dislike sprouts, then the remaining 3/5 like sprouts ... as they all either like or dislike sprouts....(IV)
Also, of those who dislike lima beans, 3/5 also dislikes brussel sprouts. That means 2/5 (of those whole dislike lima beans) likes sprouts. So:
2/5 of 2/3 = 2/5 x 2/3 = 4/15 of all students like sprouts but dislike means ..... (V)
NOW, (1) says 120 students eat at cafeteria. Then our answer is 4/15 of 120.
(2) says 40 students like lima beans.
From (II) above, 40 students = 1/3 of total # of students. From here we can find what 4/15 of students will be.
Hence D.
Last edited by gmatmba on 22 May 2006, 20:55, edited 1 time in total.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 08:20
While I was solving the qustion you guys gave the answer.
But I hope that my explanation is clear for you.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 10:43
M8 wrote:
While I was solving the qustion you guys gave the answer.
But I hope that my explanation is clear for you.
Ur explanation is absolutely valid!!
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 11:28
We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.
1) Out of these 80, some like/dislike sprouts.
2)Out of these 40 also some like/dislike sprouts.
We are given only the 1st part.
We do not know anything about no of people who like beans but like/dislike.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 11:33
remgeo wrote:
We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.
1) Out of these 80, some like/dislike sprouts.
2)Out of these 40 also some like/dislike sprouts.
We are given only the 1st part.
We do not know anything about no of people who like beans but like/dislike.
But we are also given in the stem: "Of these students, 2/3 dislikes lima beans; and of those who dislike lima beans, 3/5 also dislikes brussel sprouts".
So from here we know it all
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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22 May 2006, 11:42
gmatmba wrote:
remgeo wrote:
We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.
1) Out of these 80, some like/dislike sprouts.
2)Out of these 40 also some like/dislike sprouts.
We are given only the 1st part.
We do not know anything about no of people who like beans but like/dislike.
But we are also given in the stem: "Of these students, 2/3 dislikes lima beans; and of those who dislike lima beans, 3/5 also dislikes brussel sprouts".
So from here we know it all
I think I'm thinking too much and getting confused
Time for a good sleep I guess.
Thanks anyway
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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23 May 2006, 09:51
remgeo wrote:
We know 2/3 dislike lima. ie. 80 dislike lima and 40 like lima.
1) Out of these 80, some like/dislike sprouts.
2)Out of these 40 also some like/dislike sprouts.
We are given only the 1st part.
We do not know anything about no of people who like beans but like/dislike.
Hi Remgeo.
I see your problem. But we know needed info about those people.
Will try to explain.
From the question stem: each student either likes or dislikes lime beans and each students either likes or dislikes brussel sprouts.
Of these students, 2/3 dislikes lima beans
1) There are 120 people in the cafeteria.
2/3 (80) dislikes lima beans, so the rest (40) of the 120 - likes lime beans.
and of those who dislike lima beans, 3/5 also dislikes brussel sprouts.
3/5 (of 80 who dislikes lima beans) dislikes brussel sprouts = 48, So the rest of 80-48=32 students like brussel sprouts but at the same time dislike lima beans.
2) 40 students like lima beans.
From the question stem we have 2/3 who dislikes lima beans. So 1/3 who likes lima beans. And those 1/3 = 40, so 2/3 = 80, TOTAL = 120 students, then solve by analogy with 1).
Hope it is clear.
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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24 May 2006, 03:42
1
KUDOS
D
1) total students 120
From the question,
Number of people who dislike lima = 2/3*120 = 80
We have to find number of people who like brussel sprouts and dislike lima beans. One part we hae found number of people who dislike lima. Lets find the second part
Also from the question ,
Number of people who dislike bussel sprout as well as dislike lima = 3/5*80 = 48
Now number of people who like brussel sprout but dislike lima = 80-48= 32
2) 40 donot like lima beans. hence no of people = 3/2*40 = 120
Now use the same reasoning as 1
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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02 Jan 2015, 22:41
Self explanatory as in picture
Attachments
File comment: Self explanatory as in picture
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink]
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05 Jan 2015, 02:55
OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-students-who-eat-in-a-certain-cafeteria-each-student-95246.html
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Re: Of the students who eat in a certain cafeteria, each student either [#permalink] 05 Jan 2015, 02:55
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Display posts from previous: Sort by | 3,467 | 11,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-09 | longest | en | 0.881535 |
https://algogene.com/community/post/111?page=1 | 1,696,054,839,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00730.warc.gz | 98,314,141 | 14,195 | ## 10 Common Statistical Models You Should Know
#### Quantitative Model
Here presents the 10 most commonly used statistical models/algorithms every Quants and AI developers should know.
• Linear Regression
• Logistic Regression
• Decision Tree
• K-Nearest Neighbour
• K-Means Clustering
• Support Vector Machine
• Principal Component Analysis
• Naive Bayes
• Artificial Neural Network
### 1. Linear Regression
Linear regression is a process for modelling the linear relationship between a scalar response (dependent variable) and one or more explanatory variables (independent variables).
In simple linear regression, we try to find the best fitting line that describes the relationship between one independent variable and one dependent variable. The independent variable is used to predict the dependent variable, and the line of best fit is used to estimate the value of the dependent variable based on the value of the independent variable.
To estimate the line of best fit, we can use the least squares method, which minimizes the sum of the squared differences between the predicted values and the actual values.
Pros
Cons
• Easy to implement and fast to train
• Performs well when dataset is linearly separable
• Overfitting can be avoid by variable transformation
• Dataset is assumed to be independent which is rare in real life
• Sensitive to outliners
• Prone to noise and overfitting
### 2. Logistic Regression
Logistic regression is used to model binary outcomes (i.e. only two possible values, usually 0 or 1) with one or more independent variables. It is based on the logistic function, which transforms a linear combination of the predictor variables into a probability value between 0 and 1. The logistic regression algorithm estimates the coefficients of the predictor variables that maximize the likelihood of the observed data.
Logistic regression can also be extended to handle more complex scenarios, such as multinomial logistic regression for multiple categories or ordinal logistic regression for ordered categories.
Pros
Cons
• Easy to implement and fast to train
• Less prone to over-fitting but it can overfit in high dimensional datasets
• Efficient when the dataset has features that are linearly separable
• Dataset is assumed to be independent which is rare in real life
• Can only be used to predict discrete functions
• Should not be used when the number of observations are lesser than the number of features
### 3. Decision Tree
A decision tree model is used to make decisions by visually representing a sequence of decisions and their possible consequences. It is a type of supervised learning algorithm that can be used for both classification and regression problems. Decision trees are constructed by recursively splitting the data into smaller subsets based on a set of decision rules, with each split being made to maximize the information gain or minimize the impurity of the resulting subsets.
Decision trees are popular in machine learning because they are easy to understand and interpret, and they can handle both numerical and categorical data. They can be used for a variety of applications, such as predicting customer churn, diagnosing medical conditions, and identifying credit risk. Additionally, decision trees can be used in an ensemble of models, such as random forests and gradient boosting, to improve their predictive power.
Pros
Cons
• Easy to visualize and explain
• Can solve non-linear problem
• Can work on high-dimensional data with excellent accuracy
• Calculations can get very complicated
• A small change in the data can lead to a large change in the struture of the optimal tree
• Prone to overfitting
### 4. K-Nearest Neighbour (KNN)
K-Nearest Neighbors (KNN) is a non-parametric machine learning algorithm that can be used for both classification and regression problems. It is based on the assumption that similar things are near to each other. The algorithm works by finding the K nearest data points in the training set to the new data point, and then assigning the classification or regression value based on the majority vote or average of the K nearest neighbors.
KNN is a simple and intuitive algorithm that does not make any assumptions about the underlying data distribution. It can handle both numerical and categorical data and can be used for both binary and multi-class classification problems. However, the algorithm can be computationally expensive as the number of training data points increases, and it may not work well with high-dimensional data.
Pros
Cons
• Can make prediction without training
• Can be used for both classification and regression
• Efficient in calculation
• Does not work well with large dataset
• Sensitive to noise, missing values and outliners
• Need feature scaling
• Choose the correct K value
### 5. K-Means Clustering
K-Means Clustering is an unsupervised machine learning algorithm used to partition a dataset into a set of K clusters based on the similarity of the data points. The algorithm works by iteratively assigning data points to the nearest centroid and updating the centroids until convergence.
The K in K-Means Clustering represents the number of clusters that the algorithm should form. The value of K is usually determined by the user, based on some prior knowledge about the dataset or by running the algorithm multiple times with different values of K.
K-Means Clustering is widely used in various applications such as customer segmentation, anomaly detection, and image segmentation. It can handle large datasets and is computationally efficient, making it ideal for clustering large datasets. However, the resulting clusters may depend on the initial positions of the centroids, and the algorithm may not work well if the clusters have non-spherical shapes or if the data points are not well-separated.
Pros
Cons
• Simple to implement
• Can scale to large datasets
• Guarantees convergence
• Easily adopts to new inputs
• Generalizes to clusters of different shapes and sizes
• Dependent on initial values
• Sensitive to outliners
• Scalability decreases when dimension increases
• Manual choice of K
### 6. Support Vector Machine (SVM)
Support Vector Machine (SVM) is a supervised machine learning algorithm used for both classification and regression problems. The algorithm works by finding the best hyperplane that separates the data into different classes, where the hyperplane is defined as the linear decision boundary that maximizes the margin between the classes.
SVM is particularly useful when dealing with non-linearly separable data by using a kernel function to transform the input space into a higher-dimensional feature space, where the data may become linearly separable. The kernel function allows SVM to handle complex data distributions without explicitly defining the mapping function.
SVM is widely used in a variety of applications such as image classification, text classification, and gene expression analysis. It is a powerful algorithm that can handle high-dimensional data and has a good generalization performance. However, SVM may be sensitive to the choice of kernel function and the tuning of hyperparameters, and it may not perform well with large datasets.
Pros
Cons
• Good at high dimensional data
• Can work on small dataset
• Can solve non-linear problems
• Inefficient on large dataset
• Requires choosing the right kernal
### 7. Principal Component Analysis (PCA)
Principal Component Analysis (PCA) is a mathematical technique used to reduce the dimensionality of a dataset while retaining as much of the variability in the data as possible. The technique works by identifying the directions in the data that explain the most variation and projecting the data onto those directions.
PCA aims to find a set of new variables, called principal components, that are linear combinations of the original variables and capture the most important patterns in the data. The first principal component captures the largest amount of variance in the data, followed by the second principal component, and so on.
PCA is widely used in various fields such as image processing, bioinformatics, and finance. It can be used for exploratory data analysis, data visualization, and data compression. Additionally, PCA can be used as a pre-processing step for other machine learning algorithms to reduce the dimensionality of the input space and improve the performance of the model.
Pros
Cons
• Reduce correlated features
• Improve performance
• Reduce overfitting
• Principal components are less interpretable
• Information loss
• Must standardize data before implement PCA
### 8. Naive Bayes
Naive Bayes is a simple yet effective probabilistic machine learning algorithm used for classification and prediction problems. The algorithm is based on Bayes' theorem, which states that the probability of a hypothesis is proportional to the conditional probability of the evidence given that hypothesis.
Naive Bayes assumes that the features of the data are conditionally independent given the class, which means that the presence or absence of one feature does not affect the probability of the other features. This assumption simplifies the computation of the conditional probabilities and allows for fast and efficient training and prediction.
Naive Bayes is widely used in many applications such as spam filtering, sentiment analysis, and document classification. It can handle both binary and multi-class classification problems and works well with high-dimensional and sparse data. Additionally, Naive Bayes requires relatively little training data and can be trained in real-time.
Pros
Cons
• Short training time
• Suitable for categorical inputs
• Easy to implement
• Dataset is assumed to be independent which is rare in real life
• Estimation for zero frequency can be inaccurate
### 9. Artificial Neural Network (ANN)
Artificial Neural Networks (ANNs) are a set of machine learning algorithms inspired by the structure and function of biological neural networks in the human brain. ANNs consist of layers of interconnected nodes, called neurons, that process and transmit information. ANNs can be classified into different types, including feedforward networks, convolutional neural networks, and recurrent neural networks. Each type has its own unique architecture and is suited for different types of data and applications.
ANNs are capable of learning complex patterns and relationships between inputs and outputs, making them useful for a wide range of applications, such as image recognition, speech recognition, and natural language processing. The network learns by adjusting the strengths of the connections between neurons, based on the patterns in the training data.
ANNs require large amounts of training data and significant computational resources for training and inference. However, they are widely used in industry and academia due to their high accuracy and ability to learn complex patterns and relationships in the data.
Pros
Cons
• Have fault tolerance
• Able to learn and model non-linear and complex relationship
• Can generalize to unseen data
• Black box and hard to explain results
• Long training time
• Not guaranteed convergence
• Hardware dependence
• Requires user to translate the problem
Adaptive Boosting (AdaBoost) is a popular machine learning algorithm that is used for classification and regression problems. The algorithm works by combining multiple weak learners, such as decision trees or neural networks, to create a strong learner that can accurately classify data.
AdaBoost works by iteratively training weak learners on the data and assigning higher weights to the misclassified samples in each iteration. The algorithm then combines the weak learners by assigning weights to their predictions based on their performance. The final prediction is made by the weighted sum of the predictions of the weak learners.
AdaBoost is widely used in various applications such as face detection, text classification, and fraud detection. It is particularly useful when dealing with imbalanced datasets, where the number of samples in one class is much larger than the number of samples in the other class.
Overall, AdaBoost is a fast and efficient algorithm that can achieve high accuracy and is robust to noise in the data. It is often used in combination with other algorithms to improve the overall performance of the system.
Pros
Cons
• Relatively robust to overfitting
• High accuracy
• Easy to understand and visualize
• Sensitive to noise data and outliners
• Not optimized for speed
### Summary
These summarize the 10 most widely used statistical tools which can handle many real world data modelling problems.
In my next articles, I will discuss how to implement these models for algorithmic trading. If you find my articles inspiring, like this post and follow me here to receive my latest updates.
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very useful, thank you! | 2,521 | 13,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-40 | latest | en | 0.830189 |
https://metanumbers.com/5105054 | 1,627,490,180,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00567.warc.gz | 413,748,560 | 10,905 | ## 5105054
5,105,054 (five million one hundred five thousand fifty-four) is an even seven-digits composite number following 5105053 and preceding 5105055. In scientific notation, it is written as 5.105054 × 106. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 2,552,526 positive integers (up to 5105054) that are relatively prime to 5105054.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 20
• Digital Root 2
## Name
Short name 5 million 105 thousand 54 five million one hundred five thousand fifty-four
## Notation
Scientific notation 5.105054 × 106 5.105054 × 106
## Prime Factorization of 5105054
Prime Factorization 2 × 2552527
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 5105054 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 5,105,054 is 2 × 2552527. Since it has a total of 2 prime factors, 5,105,054 is a composite number.
## Divisors of 5105054
4 divisors
Even divisors 2 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 7.65758e+06 Sum of all the positive divisors of n s(n) 2.55253e+06 Sum of the proper positive divisors of n A(n) 1.9144e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 2259.44 Returns the nth root of the product of n divisors H(n) 2.66667 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 5,105,054 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 5,105,054) is 7,657,584, the average is 1,914,396.
## Other Arithmetic Functions (n = 5105054)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2552526 Total number of positive integers not greater than n that are coprime to n λ(n) 2552526 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 354569 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 2,552,526 positive integers (less than 5,105,054) that are coprime with 5,105,054. And there are approximately 354,569 prime numbers less than or equal to 5,105,054.
## Divisibility of 5105054
m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 3 6 2
The number 5,105,054 is divisible by 2.
## Classification of 5105054
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
## Base conversion (5105054)
Base System Value
2 Binary 10011011110010110011110
3 Ternary 100121100211002
4 Quaternary 103132112132
5 Quinary 2301330204
6 Senary 301230302
8 Octal 23362636
10 Decimal 5105054
12 Duodecimal 1862392
20 Vigesimal 1bi2ce
36 Base36 31f32
## Basic calculations (n = 5105054)
### Multiplication
n×i
n×2 10210108 15315162 20420216 25525270
### Division
ni
n⁄2 2.55253e+06 1.70168e+06 1.27626e+06 1.02101e+06
### Exponentiation
ni
n2 26061576342916 133045754555708697464 679205761477638908823383056 3467382089454466422044446963565024
### Nth Root
i√n
2√n 2259.44 172.187 47.5335 21.9584
## 5105054 as geometric shapes
### Circle
Diameter 1.02101e+07 3.2076e+07 8.18749e+13
### Sphere
Volume 5.57301e+20 3.27499e+14 3.2076e+07
### Square
Length = n
Perimeter 2.04202e+07 2.60616e+13 7.21964e+06
### Cube
Length = n
Surface area 1.56369e+14 1.33046e+20 8.84221e+06
### Equilateral Triangle
Length = n
Perimeter 1.53152e+07 1.1285e+13 4.42111e+06
### Triangular Pyramid
Length = n
Surface area 4.514e+13 1.56796e+19 4.16826e+06
## Cryptographic Hash Functions
md5 2491c12899dba0f76fecaadd9b885fdc b9eca27c02fcce14f9325e271aff17cd6a759911 4370fb884dfb40933fd92dbd93db30f8b204f1459c51d0b448798903516c30d2 0b70eacfa29c54cf50d4aca5632fccd91de7595fb570718d3fe48de1e6cae272a4a57d668ba262a5a086f7c0d89b63bc590c571ba2117fb23a90e5dc38c89844 92d857402c461902b6a858304d4f500f6f2e43eb | 1,576 | 4,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-31 | latest | en | 0.790825 |
https://calculla.com/linear_regression | 1,686,311,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00080.warc.gz | 176,549,506 | 222,089 | Least squares method calculator: linear approximation
Calculator finds out coefficient of linear function that fits best into series of (x, y) points.
# Beta version#
BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !
# Calculation data - measurement points#
Format of input data Values only (just a sequence of numbers)Serie of (x,y) points x-values y-values
# Results - approximation of your dataset#
Regression type Approximation formula Coefficient of determination R2 Linear regression Show source$y=2~x+1$ 1
# Summary - function best fitting to your data#
Measurement points Number of points 4 Points you entered (1, 3), (2, 5), (3, 7), (4, 9) Approximation Regression type Linear regression Function formula Show source$y=2~x+1$ Coefficient of determination R2 1 Line slope a 2 Free term b 1 Helper values Sum of x-values $\sum x$ 10 Sum of y-values $\sum y$ 24 Sum of x squares $\sum x^2$ 30 Sum of multiplies $\sum xy$ 70
# Some facts#
• ⓘ Hint: If you're not sure what type of regression this is, let us do the hard work for you and visit another calculator: Regression types.
• Approximation of a function consists in finding a function formula that best matches to a set of points e.g. obtained as measurement data.
• The least squares method is one of the methods for finding such a function.
• The least squares method is the optimization method. As a result we get function that the sum of squares of deviations from the measured data is the smallest. Mathematically, we can write it as follows:
$\sum_{i=1}^{n} \left[y_i - f(x_i)\right]^2 = min.$
where:
• $(x_i, y_i)$ - coordinations of the i-th measurement point, these are points that we know,
• $f(x)$ - the function we are searching for, we want this function to best match to the measurement points,
• $n$ - number of measurement points.
• If we limit the search to linear function only, then we say about linear regression or linear approximation.
• If we set a condition that we are only looking for a linear function:
$f(x) = ax + b$
we get following solution:
$a = \dfrac{n~S_{xy} - S_x~S_y}{n~S_{xx} - \left(S_x\right)^2}$
$b = \dfrac{S_y - a~S_x}{n}$
where:
• $S_{x}$ - sum of x-values $\sum x_i$,
• $S_{y}$ - sum of y-values $\sum y_i$,
• $S_{xx}$ - sum of squares $\sum x_i^2$,
• $S_{xy}$ - sum of multiplies $\sum x_i~y_i$.
# Tags and links to this website#
Tags:
Tags to Polish version:
# Links to external sites (leaving Calculla?)#
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So this is static version of this website.
This website works a lot better in JavaScript enabled browser. | 737 | 2,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-23 | latest | en | 0.825448 |
https://www.scienceforums.net/topic/132220-cube-cuts/ | 1,713,596,445,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00706.warc.gz | 887,542,777 | 19,518 | # Cube Cuts !
## Recommended Posts
Cut a Cube into six identical pieces each having nine edges !
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3 hours ago, Commander said:
Cut a Cube into four identical pieces each having nine edges !
Please note it is four pieces and not six ! Error regretted !!
Edited by Commander
Spoiler
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Also all the faces of the pieces must be TRIANGLES !
Sorry I forgot to mention this !
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Two triangular pyramids, not necessarily equilateral, attached at a base make a shape with triangular faces and nine edges.
Trying to imagine how to put four of these shapes together.
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ChatGPT asked me to tell you this:
Spoiler
It's not possible to cut a cube into four identical pieces, each with 9 edges and triangular faces. A cube has 12 edges and 6 square faces. In order to create pieces with triangular faces, you would need to further subdivide the cube's faces. However, doing so will result in more than 9 edges on each piece.
If you're looking to create four identical pieces with triangular faces and 9 edges, you would need a different shape or object that meets these criteria, as it cannot be achieved with a cube.
Just passing the message 😉
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Spoiler
I think your shape is a "diamond" AKA two tetrahedrons put together. Put 8 tetra in a cube so each tetra shares a face with another, and you have filled cube with four diamonds. I can't quite see it, atm.
ETA -guess I didn't need to hide that, seeing now Genadys post.
Edited by TheVat
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Spoiler
Start at a corner, add the 3 nearest corners, and the center of the cube. Those make up the 5 points of the "diamond"...
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1 hour ago, md65536 said:
Reveal hidden contents
Start at a corner, add the 3 nearest corners, and the center of the cube. Those make up the 5 points of the "diamond"...
Thank you. Here I've sketched it. The four shapes are 1-0, 2-0, 3-0, and 4-0, where I've numbered the opposite vertices of each:
Spoiler
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ChatGPT asked me to tell you this:
Reveal hidden contents
It's not possible to cut a cube into four identical pieces, each with 9 edges and triangular faces. A cube has 12 edges and 6 square faces. In order to create pieces with triangular faces, you would need to further subdivide the cube's faces. However, doing so will result in more than 9 edges on each piece.
If you're looking to create four identical pieces with triangular faces and 9 edges, you would need a different shape or object that meets these criteria, as it cannot be achieved with a cube.
Just passing the message 😉
You need to chop of four corners with an equilateral triangle as the base with sides of diagonals
Four of them
That leaves a tetrahedron in the middle
We need to cut it in four equal parts with each equilateral triangular base (four of them) and apex meeting at the center of the cube
The four earlier cut pieces (cut for the sake of explanation) are merged with equal area bases of the four pieces of the tetrahedron
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Need to cut like that !
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https://kids.kiddle.co/Octadecagon | 1,607,118,156,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00553.warc.gz | 362,213,548 | 8,164 | Kids Encyclopedia Facts
A octadecagon or 18-gon is a shape with 18 sides and 18 corners.
All sides of a regular octadecagon are the same length. Each corner is 160°. All corners added together equal 2880°.
### Area
The amount of space a regular octadecagon takes up is
The area of a regular octadecagon of side length a is given by
$A = \frac{18}{4}a^2\cot\frac{\pi}{18}\simeq 25,5208a^2$
## Images for kids
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# The sum of an innite series
mc-TY-convergence-2009-1 In this unit we see how nite and innite series are obtained from nite and innite sequences. We explain how the partial sums of an innite series form a new sequence, and that the limit of this new sequence (if it exists) denes the sum of the series. We also consider two specic examples of innite series that sum to e and respectively. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: recognise the dierence between a sequence and a series; write down the sequence of partial sums of an innite series; determine (in simple cases) whether an innite series has a sum.
Contents
1. Introduction 2. The sum of an innite series 3. Evaluating and e with series 2 3 5
www.mathcentre.ac.uk
c mathcentre 2009
1. Introduction
A nite series is given by the terms of a nite sequence, added together. For example, we could take the nite sequence (2k + 1)10 k =1 = (3, 5, 7, . . . , 21) . Then the corresponding example of a nite series would be given by all of these terms added together, 3 + 5 + 7 + . . . + 21 . We can write this sum more concisely using sigma notation. We write the capital Greek letter sigma, and then the rule for the k -th term. Below the sigma we write k = 1. Above the sigma we write the value of k for the last term in the sum, which in this case is 10. So in this case we would have
10
2k + 1 = 3 + 5 + 7 + . . . + 21 ,
k =1
and in this case the sum of the series is equal to 120. In the same way, an innite series is the sum of the terms of an innite sequence. An example of an innite sequence is 1 1 = (1 , 1 , 8 , . . .) , 2 4 2k k=1 and then the series obtained from this sequence would be
1 2 1 +4 + 1 8
+ ...
with a sum going on forever. Once again we can use sigma notation to express this series. We write down the sigma sign and the rule for the k -th term. But now we put the symbol for innity above the sigma, to show that we are adding up an innite number of terms. In this case we would have 1 1 =1 +4 +1 + ... . 2 8 k 2 k =1
Key Point
A nite series is given by all the terms of a nite sequence, added together. An innite series is given by all the terms of an innite sequence, added together.
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c mathcentre 2009
## 2. The sum of an innite series
What could we mean by the sum of the series
k =1
1 = 2k
1 2
1 4
1 +8 + ...
Let us try adding up the rst few terms and see what happens. If we add up the rst two terms we get 1 1 +4 =3 . 2 4 The sum of the rst three terms is
1 2
1 4
1 8
=7 . 8
1 2
1 4
+1 + 8
1 16
15 16
## And the sum of the rst ve terms is
1 2 1 +4 + 1 8
1 16
1 32
31 32
These sums of the rst terms of the series are called partial sums. The rst partial sum is just 1 the rst term on its own, so in this case it would be 2 . The second partial sum is the sum of the rst two terms, giving 3 . The third partial sum is the sum of the rst three terms, giving 7 , 4 8 and so on. If we write down the partial sums from this example,
1 , 2 3 , 4 7 , 8 15 , 16 31 , 32
... ,
we can see they form the beginning of another innite sequence. The n-th term of this sequence is the n-th partial sum. We can see that the partial sums here form a sequence that has limit 1. So it would make sense to say that this series has sum 1. We write
k =1
1 = 1. 2k
In general, we say that an innite series has a sum if the partial sums form a sequence that has a real limit. If the series is
ak = a1 + a2 + a3 + a4 + . . .
k =1
then it has a sum if the sequence of partial sums (a1 , a1 + a2 , a1 + a2 + a3 , . . .) has a limit. If the sequence of partial sums does not have a real limit, we say the series does not have a sum. www.mathcentre.ac.uk 3 c mathcentre 2009
## Here is another innite series that has a sum. It is the series
k =1
1 = k (k + 1)
1 2
1 6
1 12
1 20
+ ... .
To nd the sum of this series, we need to work out the partial sums. For this particular series, the best way to do this is to split each individual term into two parts: k+1k 1 = k (k + 1) k (k + 1) k k+1 = k (k + 1) k (k + 1) 1 1 = . k k+1 If we do this to each term, the series becomes
k =1
1 1 1 1 ) + (2 1 ) + (1 4 ) + ... , = (1 1 2 3 3 k (k + 1)
## and so the n-th partial sum is
1 1 (1 2 ) + (1 3 ) + (1 1 ) + ...+ 2 3 4
1 1 n n+1
As you can see, most of the terms in this expression cancel in pairs. We just get the outermost two terms, n+1 1 n 1 = = . 1 n+1 n+1 n+1 n+1 So the sequence of partial sums is n n+1
=
n=1
1 , 2
2 , 3
3 , 4
4 , 5
5 , 6
... ,
and this sequence has limit 1. We say the innite series sums to 1, and we write
k =1
1 = 1. k (k + 1)
Here is an example of an innite series that does not have a sum. The series
1 = 1 + 1 + 1 + 1 +...
k =1
has the sequence of partial sums (1, 2, 3, 4, . . .) . This sequence of partial sums does not tend to a real limit. It tends to innity. So the series does not have a sum. www.mathcentre.ac.uk 4 c mathcentre 2009
You might have noticed that, whenever we have taken an innite series with a sum, then the individual terms of the series have tended to zero. This is a general feature of innite series. But this argument does not work in the opposite direction. It is possible to have a series with individual terms tending to zero, but with no sum. For example, the series
k =1
1 =1+ 1 + 2 k
1 3
1 +4 + ...
is called the harmonic series, and it has terms that tend to zero. But the sequence of partial sums for this series tends to innity. So this series does not have a sum.
Key Point
The n-th partial sum of a series is the sum of the rst n terms. The sequence of partial sums of a series sometimes tends to a real limit. If this happens, we say that this limit is the sum of the series. If not, we say that the series has no sum. A series can have a sum only if the individual terms tend to zero. But there are some series with individual terms tending to zero that do not have sums.
## 3. Evaluating and e with series
Some innite series can help us to evaluate important mathematical constants. For example, consider the series 1 . ( k 1)! k =1 Written out term by term, this series is 1 1 1 1 + + + + ... 0! 1! 2! 3! =
1 1+1+ 2 +1 + ... . 6
If we use a calculator to work out the rst few partial sums of this series, we get (1, 2, 25, 266667, 270833, 271667, 271806, . . .) , where we have written down some of the terms to ve decimal places. Now you might have noticed that this sequence of partial sums seems to be getting closer and closer to the number e, which is 2 71828 to ve decimal places. In fact it can be shown that the partial sums do tend to e. So working out the partial sums of this series is a useful way of calculating e to a large number of decimal places. www.mathcentre.ac.uk 5 c mathcentre 2009
## Now let us look at the innite series
(1)k+1
k =1
4 . 2k 1
For this series, we need to recall the meaning of the power (1)k+1 . If k is odd then k + 1 is even, and so (1)k+1 = 1. On the other hand, if k is even then k + 1 is odd, and so (1)k+1 = 1. We can now write out the series term by term as
(1)k+1
k =1
4 2k 1
4 4 =4 3 +5 4 + 7
4 9
+ ... .
Again we can use a calculator to work out the rst few partial sums of this series. We get (4, 26667, 34667, 28952, 33397, 29760, 32837, . . .) where we have written down some of the terms to four decimal places. This sequence of partial sums looks like it might be getting close to some number just greater than 3. In fact it can be shown that the partial sums tend to , which is 31416 to four decimal places. If we kept on calculating the partial sums for this series, we would eventually obtain a value for to several decimal places.
Key Point
Some innite series can help us evaluate numbers like and e as accurately as we choose.
Exercises 1.
2 The geometric series 1 + 2 + 3 + . . . sums to 3. How many terms n are required for the n th 3 partial sum Sn to dier from 3 by less than (a) 1, (b) 0.2, (c) 0.05 ? 2
2.
1 The harmonic series 1 + 1 +3 + . . . does not have a sum. How many terms n are required for 2 the n th partial sum Sn to be greater than
## (a) 2, Answers 1. (a) 3 2. (a) 4
(b) 3,
(c)
5 ?
(b) 7 (b) 11
(c) (c)
11 83 6 c mathcentre 2009
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PROFIT LAST YEAR WERE EXORBITANT MAKE IT CONCRETE SENTENCE
When you type in all-caps, two things happen:
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• Vardges
In summary, the conversation discussed the possibility of using a hydraulic motor and AC generator to produce high efficiency electric energy. It was determined that the efficiency would not be very high due to losses in the conversion process. The conversation then shifted to the idea of using a hydraulic press mechanism to drive a reciprocating pump and couple it to a 100 KW PM generator, but this idea was deemed impossible due to the laws of thermodynamics.
Vardges
Hi there. I have an idea which is interesting. Let's suppose, we have AC generator which has capability of 100 KW and 100 RPM. And we have hydraulic motor which produces 500 RPM and 2400 Nm torque. Additionally, I attach 5:1 gearbox for reducing 500 RPM to 100 RPM for Ac generator. Please, share your thoughts, is it possible to produce high efficiency and electric energy with this combination?
Die Energie der Welt ist konstant (Clausius)
Vardges said:
Hi there. I have an idea which is interesting. Let's suppose, we have AC generator which has capability of 100 KW and 100 RPM. And we have hydraulic motor which produces 500 RPM and 2400 Nm torque. Additionally, I attach 5:1 gearbox for reducing 500 RPM to 100 RPM for Ac generator. Please, share your thoughts, is it possible to produce high efficiency and electric energy with this combination?
Welcome to PF!
Depends -- what is making the hydraulic motor spin? And what do you consider "high efficiency"?
The hydraulic motor will need a power of at least 125 kW. Where is the point?
Last edited:
Vardges said:
Hi there. I have an idea which is interesting. Let's suppose, we have AC generator which has capability of 100 KW and 100 RPM. And we have hydraulic motor which produces 500 RPM and 2400 Nm torque. Additionally, I attach 5:1 gearbox for reducing 500 RPM to 100 RPM for Ac generator. Please, share your thoughts, is it possible to produce high efficiency and electric energy with this combination?
125kW input for 100 kW output is not generally considered high efficiency. In practice it would be even less than that.
russ_watters said:
Welcome to PF!
Depends -- what is making the hydraulic motor spin? And what do you consider "high efficiency"?
The hydraulic motor is spun by hydraulic pump. And the hydraulic pump is powered by electric motor.
Here is the clarification of what I want to get. I want to use hydraulic mechanism to get electric power. Hydraulic mechanism can be cylinder or hydraulic motor.
Dale said:
125kW input for 100 kW output is not generally considered high efficiency. In practice it would be even less than that.
Hydraulic motors have 10 times more power density than electric motors have. I want to get mechanical advantages from hydraulic motor torque to generate electric power. That's all.
You cannot generate electric power from nowhere. You need a power source: A generator driven by hot gas, water, wind, ..., light shining on solar cells, or something else.
Converting energy from one type to another is always associated with losses. Electricity -> mechanical motion -> electricity means you get strictly less electricity out than you get in. It does not matter which type of motor and conversion you use.
russ_watters
Vardges said:
Hydraulic motors have 10 times more power density than electric motors have.
Power density is not relevant to the question of efficiency. For efficiency all that is considered is the input power and the output power. It doesn't matter how dense the power is, just how much goes in and how much comes out.
Based on your numbers above at least 125 kW hydraulic power goes in and at most 100 kW of electric power comes out. That is not a very high efficiency, but might be acceptable in some scenarios (e.g. If the hydraulic power were very inexpensive so you could sell the smaller amount of electrical power for more)
Vardges said:
I want to use hydraulic mechanism to get electric power
Sure. That is what a hydroelectric power plant does. As @mfb and I have said, the efficiency of such a conversion is strictly less than 1.
Last edited:
russ_watters
Dale said:
Power density is not relevant to the question of efficiency. For efficiency all that is considered is the input power and the output power. It doesn't matter how dense the power is, just how much goes in and how much comes out.
Based on your numbers above at least 125 kW hydraulic power goes in and at most 100 kW of electric power comes out. That is not a very high efficiency, but might be acceptable in some scenarios (e.g. If the hydraulic power were very inexpensive so you could sell the smaller amount of electrical power for more)
Sure. That is what a hydroelectric power plant does. As @mfb and I have said, the efficiency of such a conversion is strictly less than 1.
Guys, here is the point. A typical hydraulic press consumes 5-7 KW electricity to obtain 50 ton (500 kilo Newton) capacity for pushing/pulling. Its hydraulic rod travel is 2in/sec. Let's suppose, if I use that rod in a reciprocating pump, I will get a very high torque, especially if the AC PM generator is has a very low RPM (50 RPM is available in the market!). I want to use hydraulic press mechanism in reciprocating pump and couple them to 100 KW PM generator. That's it.
Vardges said:
typical hydraulic press consumes 5-7 KW electricity ... couple them to 100 KW PM generator.
Perperual motion machines do not work and are not discussed on PF. Thread closed
russ_watters
1. How does a hydraulic motor-AC generator coupling work?
A hydraulic motor-AC generator coupling works by using hydraulic power to rotate a motor, which in turn rotates an AC generator to produce electricity. The hydraulic fluid is pressurized and flows through the motor, causing it to rotate and drive the generator. This process allows for the conversion of hydraulic power into electrical energy.
2. What are the advantages of using a hydraulic motor-AC generator coupling?
One of the main advantages of a hydraulic motor-AC generator coupling is its efficiency. Since it utilizes hydraulic power to drive the motor, it can be more efficient than traditional mechanical systems. It also allows for precise control over the speed and power output, making it ideal for applications where variable speeds are necessary. Additionally, it can operate in a wide range of environments, including high temperatures and harsh conditions.
3. What are some common applications of a hydraulic motor-AC generator coupling?
A hydraulic motor-AC generator coupling is commonly used in industrial and commercial settings, such as in power plants, manufacturing facilities, and construction sites. It can also be found in various types of machinery, including cranes, excavators, and pumps. It is also used in renewable energy systems, such as hydroelectric power plants, to convert the energy of moving water into electricity.
4. Are there any maintenance requirements for a hydraulic motor-AC generator coupling?
Like any other mechanical system, a hydraulic motor-AC generator coupling requires regular maintenance to ensure its proper functioning. This includes regular inspections, lubrication of moving parts, and replacement of worn-out components. It is important to follow the manufacturer's recommended maintenance schedule to prevent breakdowns and ensure the longevity of the coupling.
5. How can a hydraulic motor-AC generator coupling be controlled?
A hydraulic motor-AC generator coupling can be controlled through various means, such as manual control valves, electronic control systems, or computerized control systems. These control systems allow for precise control over the speed and power output of the coupling, making it suitable for a wide range of applications. The type of control used depends on the specific needs and requirements of the system it is being used in.
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# Arithmetic Progressions NCERT Solutions : Class 10 Maths
## NCERT Solutions for Class 10 Maths : Arithmetic Progressions
Doubtnut one of the best online education platform provides free NCERT Solutions of Maths for Class 10 arithmetic-progressions which are solved by our maths experts as per the NCERT (CBSE) guidelines. We provide the solutions as video solutions in which concepts are also explained along with answer which will help students while learning , preparing for exams and in doing homework. These Solutions will help to revise complete syllabus and score more marks in examinations.
Get here free, accurate and comprehensive NCERT Solutions for Class 10 Maths arithmetic-progressions which have been reviewed by our maths counsellors as per the latest edition following up the CBSE guidelines. We provide video solutions in which solutions to all the questions of NCERT Class 10 Maths textbook are explained in step by step and detailed way.
## The Topics of the Chapter arithmetic progressions are :
5.1 Introduction
5.2 Arithmetic Progressions
5.3 nth Term of an AP
5.4 Sum of First n Terms of an AP
5.5 Summary
It contains these exercises along with solved examples. All the exercises are solved in the video. Select the exercise to view the solutions exercisewise of the Chapter arithmetic progressions:
## NCERT Solutions Class 10 Maths Chapter arithmetic progressions Exercises:
We have covered all the exercises and also Solved examples in the videos. Along with the practise exercise Students should also practise solved examples to clear the concepts of the arithmetic-progressions. If incase you have any doubt you can watch the solutions for the given questions.Watch the solutions for the given questions in which it is also explained in the video steps to solve the questions along with answers. | 392 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-04 | latest | en | 0.92495 |
https://www.myassignmenthelp.net/question/use-spss-and-the-santa-fe-grill-database | 1,709,562,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00714.warc.gz | 873,217,367 | 13,414 | Securing Higher Grades Costing Your Pocket? Book Your Assignment help at The Lowest Price Now!
# Use SPSS and The Santa Fe Grill Database
## SPSS Course Help, SPSS Homework Help, SPSS and the Santa Fe Grill database
D irections: Use SPSS and the Santa Fe Grill database to answer the following questions. Place your answers on the attached answer sheet.
1. Using the Santa Fe Grill survey data, is there a relationship between the Mexican restaurant reported by respondents as last patronized and the distance that they are willing to drive? (Hint: Read text reading material on about pp. 284-287.)
2. Is there a relationship between the Mexican restaurant reported by respondents as last patronized and the number of children in respondent’s household?
3. Does patron’s age appear to be related to reported satisfaction? (Hint: Read text material on about pp. 290-297. For your ANOVA run follow the SPSS directions in text. But in addition, select the POST HOC button and select Scheffe’s test AND under the OPTIONS button select both Descriptive Statistics and Means Plot.)
4. The manager of Santa Fe Grill wants to know if either respondent’s age and/or gender is/are related to reported satisfaction. Follow the n-way ANOVA SPSS directions in the text reading to set up analyze this problem. Use Scheffe post hoc test option.Answer Sheet • Assignment 4 Name:
1. Enter test results from your SPSS analysis in the table below:
Is there a significant relationship between restaurant last patronized and driving distance?
Yes
No
Relationship is inconclusive based on test results
What is the level of significance associated with this X2 test?
Inspect the observed and expected cell frequencies in your SPSS crosstab table. What conclusion can you draw?
2. Enter test results from your SPSS analysis in the table below:
Is there a significant relationship between restaurant last patronized and household size?
Yes
No
Relationship is inconclusive based on test results
What is the level of significance associated with this X2 test?
Inspect the observed and expected cell frequencies in your SPSS crosstab table. What conclusion can you draw?
3. Enter test results from your SPSS analysis in the table below:
List the dependent variable(s) in this analysis?
List the independent variable(s) in this analysis?
Is this a 1-way or 2-way ANOVA?
1-way ANOVA
2-way ANOVA
Does patron’s age appear to affect reported satisfaction?
Yes
No
Relationship is inconclusive based on test results
Using the means plots graph and the results of the post-hoc (Scheffe) tests shown in the SPSS output, what conclusions can you draw?
4. Enter test result F-Ratio values from your SPSS analysis in the table below:
List the dependent variable(s) in this analysis?
List the independent variable(s) in this analysis?
Is this a 1-way or 2-way ANOVA?
1-way ANOVA
2-way ANOVA
Is the age factor influence upon satisfaction statistically significant? Is the gender factor influence upon satisfaction statistically significant?
What conclusions would you draw from these ANOVA data? | 662 | 3,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-10 | latest | en | 0.89463 |
https://www.mapleprimes.com/users/torabi/replies?page=3 | 1,560,756,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998440.47/warc/CC-MAIN-20190617063049-20190617085049-00217.warc.gz | 819,249,769 | 21,511 | ## 70 Reputation
3 years, 234 days
## Newtons Method...
is possible to write a code similar to the attached below that can solve equations by Newtons Method?
Thank you
MultiNewton_(1).mw
## fsolve problem!!!...
Thank you so much dear Acer;
Only my problem is fsolve.
Because when I change the value N :=7 ; M := 7 to N := 11; M := 11.
fsolve can not gain an answer in a new file as attached below!!
is possible that I use another method, for example, NewtonsMethod?
test_5.mw
## no matrices !!!...
Thank you. No the final system ( SYSPDE) not expected to contain matrices and 'SYSPDE' only must have a simple sum of (scalar) products.
I don't know why matrices are added in pde5!!!
## old system is working!!!...
Thanks so much, dear Acer and tomleslie and Preben.
Before I solved similar equations that I attached bellow, they are working now.
In the new system, I added only pde4 and pde 5, and for now, I do not know why it is not working well.
solve_linear.mw
## non-scalar nonsense...
Thank you.
If possible please edit file and send it for me.
Thank you
## ?...
Thank you.
I do not know your mean completely in this regard.
Best,
Hello,
I could not gain an answer.!!!
I have two questions.
what were the advantages to using DirectSearch package to solve equations? And second;
how I can use a convential method such as Newton-Raphson method?
Thank you
## domain of solutions...
I do not know the domain of the answer. However equations come from a real problem.
Thank you
## yes...
Yes, I am tring to get y-values for the curves at a common set of sigma values.
if it is not possible to get data for whole curves in the same time, how i can get them separately.
Best
## general dorm...
the general form is as follow:
(vdc+vac1*cos(Omega1*t)+vac2*cos(Omega2*t))^2
## just simplify.......
I only want to simplify it more
## ?...
thanks.
How I can determine Time period by maple automaticly?
period.mw
## another problem...
thank yoy.
I have a problem for attaching file.
simpdiff_2.mw
## Thank you...
Thank you for your help in this regard.
## final program...
Thank you dear Carl,
TIME.mw
1 2 3 4 5 6 7 Last Page 3 of 12
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https://de.mathworks.com/matlabcentral/cody/problems/9-who-has-the-most-change/solutions/609146 | 1,600,440,265,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00643.warc.gz | 360,306,821 | 16,746 | Cody
# Problem 9. Who Has the Most Change?
Solution 609146
Submitted on 31 Mar 2015 by Joseph Laub
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = [1 2 1 15]; b = 1; assert(isequal(most_change(a),b))
z = 60 b = 1
2 Pass
%% a = [ 1 2 1 15; 0 8 5 9]; b = 2; assert(isequal(most_change(a),b))
z = 99 b = 2
3 Pass
%% a = [ 1 22 1 15; 12 3 13 7; 10 8 23 99]; b = 3; assert(isequal(most_change(a),b))
z = 619 b = 3
4 Pass
%% a = [ 1 0 0 0; 0 0 0 24]; b = 1; assert(isequal(most_change(a),b))
z = 25 b = 1
5 Pass
%% a = [ 0 1 2 1; 0 2 1 1]; c = 1; assert(isequal(most_change(a),c))
z = 26 b = 1
6 Pass
%% % There is a lot of confusion about this problem. Watch this. a = [0 1 0 0; 0 0 1 0]; c = 2; assert(isequal(most_change(a),c)) % Now go back and read the problem description carefully.
z = 10 b = 2
7 Pass
%% a = [ 2 1 1 1; 1 2 1 1; 1 1 2 1; 1 1 1 2; 4 0 0 0]; c = 5; assert(isequal(most_change(a),c))
z = 100 b = 5 | 473 | 1,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-40 | latest | en | 0.662106 |
https://ftp.aimsciences.org/article/doi/10.3934/cpaa.2022050 | 1,656,935,721,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00144.warc.gz | 316,013,578 | 15,699 | # American Institute of Mathematical Sciences
June 2022, 21(6): 2079-2100. doi: 10.3934/cpaa.2022050
## Liouville type theorem for Hartree-Fock Equation on half space
School of Mathematics and Statistics, Shenzhen University, Shenzhen, Guangdong, 518060, China
* Corresponding author
Received October 2021 Revised January 2022 Published June 2022 Early access March 2022
Fund Project: The second author is supported by NSFC grant 12171212
In this paper, we study the Liouville type theorem for the following Hartree-Fock equation in half space
\begin{align*} \begin{cases} - \Delta {u_i}(y) = \sum\limits_{j = 1}^n {{\int _{\partial \mathbb{R}_ + ^N}}} \frac{{{u_j}(\bar x, 0){F_1}({u_j}(\bar x, 0))}} {{|(\bar x, 0) - y{|^{N - \alpha }}}}d\bar x{f_2}({u_i}(y)) \\ \qquad \qquad \qquad + \sum\limits_{j = 1}^n {{\int _{\partial \mathbb{R}_ + ^N}}} \frac{{{u_j}(\bar x, 0){F_2}({u_i}(\bar x, 0))}} {{|(\bar x, 0) - y{|^{N - \alpha }}}}d\bar x{f_1}({u_j}(y)), \ y \in \mathbb{R}_ + ^N, \hfill \\ \frac{{\partial {u_i}}} {{\partial \nu }}(\bar x, 0) = \sum\limits_{j = 1}^n {{\int _{ \mathbb{R}_ + ^N}}} \frac{{{u_j}(y){G_1}({u_j}(y))}} {{|(\bar x, 0) - y{|^{N - \alpha }}}}dy{g_2}({u_i}(\bar x, 0)) \\ \qquad \qquad \qquad + \sum\limits_{j = 1}^n {{\int _{ \mathbb{R}_ + ^N}}} \frac{{{u_j}(y){G_2}({u_i}(y))}} {{|(\bar x, 0) - y{|^{N - \alpha }}}}dy{g_1}({u_j}(\bar x, 0)), \quad \quad(\bar x, 0) \in \partial \mathbb{R}_ + ^N, \end{cases} \end{align*}
where
$\mathbb{R}_+^N = \{x\in{\mathbb{R}^N}: x_N > 0\}, f_1, f_2, g_1, g_2, F_1, F_2, G_1, G_2$
are some nonlinear functions. Under some assumptions on the nonlinear functions
$F, G, f, g$
, we will prove the above equation only possesses trivial positive solution. We use the moving plane method in an integral form to prove our result.
Citation: Xiaomei Chen, Xiaohui Yu. Liouville type theorem for Hartree-Fock Equation on half space. Communications on Pure and Applied Analysis, 2022, 21 (6) : 2079-2100. doi: 10.3934/cpaa.2022050
##### References:
[1] W. Chen, Y. Fang and C. Li, Super poly-harmonic property of solutions for navier boundary problems on a half space, J. Funct. Anal., 265 (2013), 1522-1555. doi: 10.1016/j.jfa.2013.06.010. [2] W. Chen, Y. Fang and R. Yang, Liouville theorems involving the fractional Laplacian on a half space, Adv. Math., 274 (2015), 167-198. doi: 10.1016/j.aim.2014.12.013. [3] W. Chen, C. Li and Y. Li, A direct method of moving planes for the fractional Laplacian, Adv. Math., 308 (2017), 404-437. doi: 10.1016/j.aim.2016.11.038. [4] W. Chen, C. Li and B. Ou, Classification of solutions for an integral equation, Commun. Pure Appl. Math., 59 (2006), 330-343. doi: 10.1002/cpa.20116. [5] J. Dou and M. Zhu, Sharp Hardy-Littlewood-Sobolev inequality on the upper half space, Int. Math. Res. Not. IMRN, 3 (2015), 651-687. doi: 10.1093/imrn/rnt213. [6] V. Fock, Naherungsmethode zur Losung des quantenmechanischen Mehrkorperproblems, Z. Phys., 61 (1930), 126-148. doi: 10.1007/BF01340294. [7] D. Hartree, The wave mechanics of an atom with a non-coulomb central field, Part Ⅰ. Theory and methods, Proc. Camb. Phil. Soc., 24 (1928), 89-312. doi: 10.1017/S0305004100011919. [8] H. Li, Liouville Type Theorem for Hartree Equations on Half Spaces, Acta Math. Sci.(in Chinese), 41A (2021), 388–401. [9] J. Liu, Y. Guo and Y. Zhang, Liouville-type theorems for polyharmonic systems in $\mathbb R^N$, J. Differ. Equ., 225 (2006), 685-709. doi: 10.1016/j.jde.2005.10.016. [10] S. Luo and W. Zou, Liouville theorems for integral systems related to fractional lane-emden systems in $\mathbb R_+^N$, Differ. Integral Equ., 29 (2016), 1107-1138. [11] J. C. Slater, Note on Hartree's method, Phys. Rev., 35 (1930), 210-211. doi: 10.1103/physrev.35.210.2. [12] J. Yang and X. Yu, Liouville type theorems for Hartree and Hartree-Fock equations, Nonlinear Anal., 183 (2019), 191-213. doi: 10.1016/j.na.2019.01.012. [13] X. Yu, Liouville type theorems for singular integral equations and integral systems, Commun. Pure Appl. Anal., 15 (2016), 1825-1840. doi: 10.3934/cpaa.2016017. [14] X. Yu, Liouville type theorem for some nonlocal elliptic equations, J. Differ. Equ., 263 (2017), 6805-6820. doi: 10.1016/j.jde.2017.07.028.
show all references
##### References:
[1] W. Chen, Y. Fang and C. Li, Super poly-harmonic property of solutions for navier boundary problems on a half space, J. Funct. Anal., 265 (2013), 1522-1555. doi: 10.1016/j.jfa.2013.06.010. [2] W. Chen, Y. Fang and R. Yang, Liouville theorems involving the fractional Laplacian on a half space, Adv. Math., 274 (2015), 167-198. doi: 10.1016/j.aim.2014.12.013. [3] W. Chen, C. Li and Y. Li, A direct method of moving planes for the fractional Laplacian, Adv. Math., 308 (2017), 404-437. doi: 10.1016/j.aim.2016.11.038. [4] W. Chen, C. Li and B. Ou, Classification of solutions for an integral equation, Commun. Pure Appl. Math., 59 (2006), 330-343. doi: 10.1002/cpa.20116. [5] J. Dou and M. Zhu, Sharp Hardy-Littlewood-Sobolev inequality on the upper half space, Int. Math. Res. Not. IMRN, 3 (2015), 651-687. doi: 10.1093/imrn/rnt213. [6] V. Fock, Naherungsmethode zur Losung des quantenmechanischen Mehrkorperproblems, Z. Phys., 61 (1930), 126-148. doi: 10.1007/BF01340294. [7] D. Hartree, The wave mechanics of an atom with a non-coulomb central field, Part Ⅰ. Theory and methods, Proc. Camb. Phil. Soc., 24 (1928), 89-312. doi: 10.1017/S0305004100011919. [8] H. Li, Liouville Type Theorem for Hartree Equations on Half Spaces, Acta Math. Sci.(in Chinese), 41A (2021), 388–401. [9] J. Liu, Y. Guo and Y. Zhang, Liouville-type theorems for polyharmonic systems in $\mathbb R^N$, J. Differ. Equ., 225 (2006), 685-709. doi: 10.1016/j.jde.2005.10.016. [10] S. Luo and W. Zou, Liouville theorems for integral systems related to fractional lane-emden systems in $\mathbb R_+^N$, Differ. Integral Equ., 29 (2016), 1107-1138. [11] J. C. Slater, Note on Hartree's method, Phys. Rev., 35 (1930), 210-211. doi: 10.1103/physrev.35.210.2. [12] J. Yang and X. Yu, Liouville type theorems for Hartree and Hartree-Fock equations, Nonlinear Anal., 183 (2019), 191-213. doi: 10.1016/j.na.2019.01.012. [13] X. Yu, Liouville type theorems for singular integral equations and integral systems, Commun. Pure Appl. Anal., 15 (2016), 1825-1840. doi: 10.3934/cpaa.2016017. [14] X. Yu, Liouville type theorem for some nonlocal elliptic equations, J. Differ. Equ., 263 (2017), 6805-6820. doi: 10.1016/j.jde.2017.07.028.
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2021 Impact Factor: 1.273 | 3,893 | 10,621 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.507868 |
https://flatearthsoc.com/proof-the-earth-is-flat-nick-davies-guardian.html | 1,569,002,591,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574058.75/warc/CC-MAIN-20190920175834-20190920201834-00276.warc.gz | 472,318,657 | 20,357 | 106) The so-called “South Pole” is simply an arbitrary point along the Antarctic ice marked with a red and white barbershop pole topped with a metal ball-Earth. This ceremonial South Pole is admittedly and provably NOT the actual South Pole, however, because the actual South Pole could be demonstrably confirmed with the aid of a compass showing North to be 360 degrees around the observer. Since this feat has never been achieved, the model remains pure theory, along with the establishment’s excuse that the geomagnetic poles supposedly constantly move around making verification of their claims impossible.
I am new to flat earth and see way more logical proof to this idea than the spinning globe. I have often wondered about what we are taught as it didn't make sense but we believe because that's what we been taught. One item I have not seen in my research and maybe you can address. If one were to stand precisely at the point where the imaginary axis is that the globe spins on, at the North Pole area, or even say 1/4 mile away,or ten feet away, what the person would see or experience I think would be bizarre....essentially rotating in a 1/4 mile or ten foot circle during the 24 hr day...more bizarre if they were on the exact axis point. This I think could be proven easily by going there, but it seems to not have been, which to me adds more proof to flat earth.I am sure if one were to go there, they would not see any following a circular pattern as the 24 hrs pass. They would be stationary!
69) The New York City skyline is clearly visible from Harriman State Park’s Bear Mountain 60 miles away. If Earth were a ball 25,000 miles in circumference, viewing from Bear Mountain’s 1,283 foot summit, the Pythagorean Theorem determining distance to the horizon being 1.23 times the square root of the height in feet, the NYC skyline should be invisible behind 170 feet of curved Earth.
The circumstances which attend bodies which are caused merely to fall from a great height prove nothing as to the motion or stability of the Earth, since the object, if it be on a thing that is in motion, will participate in that motion; but, if an object be thrown, upwards from a body at rest, and, again, from a body in motion, the circumstances attending its descent will be very different. In the former case, it will fall, if thrown vertically upwards, at the place from whence it was projected; in the latter case, it will fall behind the moving body from which it is thrown will leave it in the rear. Now, fix a gun, muzzle upwards, accurately, in the ground; fire off a projectile; and it will fall by the gun. If the Earth traveled eleven hundred miles a minute, the projectile would fall behind the gun, in the opposite direction to that of the supposed motion. Since, then, this is NOT the case, in fact, the Earth's fancied motion is negatived and we have a proof that the Earth is not a, globe.
Silica is a mineral that existed in much greater quantities and in living things before the flood. It’s very possible that the glass dome above our earth is made of silica glass that broke open during the flood , when God opened the windows of heaven, and it shattered glass onto the earth. This is more evidence of the glass dome that separates the water from the waters, described in Genesis.
A "Standing Order" exists in the English Houses of Parliament that in the cutting of canals, &c., the datum line employed shall be a "horizontal line, which shall be the same throughout the whole length of the work." Now if the Earth were a globe, this "Order" could not be carried out: but, it is carried out: therefore it is a proof that the Earth is not a globe.
93.) We have seen that astronomers – to give us a level surface on which to live – have cut off one-half of the "globe" in a certain picture in their books. [See page 6.] Now, astronomers having done this, one-half of the substance of their "spherical theory" is given up! Since, then, the theory must stand or fall in its entirety, it has really fallen when the half is gone. Nothing remains, then, but a plane Earth, which is, of course, a proof that the Earth is not a globe.
98.) Mr Hind speaks of the astronomer watching a star as it is carried across the telescope by the diurnal revolution of the Earth." Now, this is nothing but downright absurdity. No motion of the Earth could possibly carry a star across a telescope or anything else. If the star is carried across anything at all, it is the star that moves, not the thing across which it is carried! Besides, the idea that the Earth, if it were a globe, could possibly move in an orbit of nearly 600,000,000 of miles with such exactitude that the cross-hairs in a telescope fixed on its surface would appear to glide gently over a star "millions of millions" of miles away is simply monstrous; whereas, with a FIXED telescope, it matters not the distance of the stars, though we suppose them to be as far off as the astronomer supposes them to be; for, as Mr. Proctor himself says, "the further away they are, the less they will seem to shift." Why, in the name of common sense, should observers have to fix their telescopes on solid stone bases so that they should not move a hair&
Then scriptures say at 1 John 5:19 that "the whole world is lying in the power of the Wicked One". The people who are ultimately responsible for this deception, the NWO folks, Free Masons, and illuminati are being controlled by SATAN and his demons, to "blind the minds" of us. This deception started with the "mysteries" before Babylonian times during the times of Chaldeans and descendants of Nimrod. No human could have continued this deception for so many CENTURIES. Only a being like one who has been in existence since before the earth was. is plenty more I would like to share with the flat earthers. It all is in complete agreement with what Eric has shared, which, has actually validated my beliefs per se. I am total agreement with you guys about NASA. I am a retired space shuttle engineer. I always knew they couldn"t escape the "Van Allen" belts, but only until a little less than a myear ago did I understand what a firmament was. Eric you are great!!!
68.) Mr. J.M. Lockyer says: Because the Sun seems to rise in the east and set in the west, the Earth really spins in the opposite direction; that is, from west to east," Now, this is no better than though we were to say – Because a man seems to be coming up the street, the street really goes down to the man! And since true science would contain no such nonsense as this, it follows that the so-called science of theoretical astronomy is not true, and, we have another proof that the Earth is not a globe.
Since temperature inversions are common over water, it is relatively easy to devise experiments in which distant objects beyond the curvature of the earth are visible. Perhaps the most famous are the photographs of the Chicago skyline taken across Lake Michigan, about 60 miles away. The photographer, Joshua Nowicki, does not promote the flat earth, but flat-earthers have used his photographs many times, such as here, as supposed proof that the earth is flat. Flat-earthers do not seem to be aware of just how rare these photographs are. If the earth were flat, then the Chicago skyline would be visible across Lake Michigan nearly every clear day, but it is not. If the earth is spherical, then the hulls of ships ought to disappear as the ships move away from the observer. Since the ship must move many miles away for this to become noticeable, it is difficult to see this with the naked eye.
```6) If Earth were a ball 25,000 miles in circumference as NASA and modern astronomy claim, spherical trigonometry dictates the surface of all standing water must curve downward an easily measurable 8 inches per mile multiplied by the square of the distance. This means along a 6 mile channel of standing water, the Earth would dip 6 feet on either end from the central peak. Every time such experiments have been conducted, however, standing water has proven to be perfectly level.
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## People have believed that the Earth is flat since the beginning of humanity, but the modern Flat Earth hypothesis stemmed from an experiment called the Bedford Level Experiment, conducted in the mid-1800s by a man named Samuel Rowbotham.[1] Rowbowtham, who wrote a book named Earth Not a Globe, started the modern movement by debating scientists publicly and accumulating followers. In the experiment, Rowbowtham attempted to measure the curvature of the earth by observing the curvatures at a local river. He took his results as disproving the theory of a round earth, but future scientists have said that the results he obtained could be accounted for by the parallax effect.[2]
#### 36.) If we take a journey down the Chesapeake Bay, by night, we shall see the "light" exhibited at Sharpe's Island for an hour before the steamer gets to it. We may take up a position on the deck so that the rail of the vessel's side will be in a line with the "light" and in the line of sight; and we shall find that in the whole journey the light will won't vary in the slightest degree in its apparent elevation. But, say that a distance of thirteen miles has been traversed, the astronomers' theory of "curvature" demands a difference (one way or the other!) in the apparent elevation of the light, of 112 feet 8 inches! Since, however, there is not a difference of 100 hair's breadths, we have a plain proof that the water of the Chesapeake Bay is not curved, which is a proof that the Earth is not a globe.
67) The distance across the Irish Sea from the Isle of Man’s Douglas Harbor to Great Orm’s Head in North Wales is 60 miles. If the Earth was a globe then the surface of the water between them would form a 60 mile arc, the center towering 1944 feet higher than the coastlines at either end. It is well-known and easily verifiable, however, that on a clear day, from a modest altitude of 100 feet, the Great Orm’s Head is visible from Douglas Harbor. This would be completely impossible on a globe of 25,000 miles. Assuming the 100 foot altitude causes the horizon to appear approximately 13 miles off, the 47 miles remaining means the Welsh coastline should still fall an impossible 1472 feet below the line of sight!
In 1997, during work on what would have been his first solo album, Davies decided to reform Supertramp. The group promptly returned to recording and touring, which yielded another two studio albums before they split again.[9] Supertramp reunited in 2010 for their 70–10 tour. A 2015 tour was announced but ultimately cancelled due to Davies' health issues.
Starting with Indelibly Stamped in 1971, Davies shared lead vocals with Supertramp songwriting partner, Roger Hodgson until the latter's departure in 1983,[4] at which point he became the sole lead vocalist of the group. Davies's voice is deeper than Hodgson's, and he usually employs a raspy baritone which stands in stark contrast to his bandmate's tenor. However, he occasionally sings in a falsetto which superficially resembles Hodgson's vocals, such as on "Goodbye Stranger" and "My Kind of Lady". He also plays harmonica for the group.
```Astronomers tell us that, in consequence of the Earth's "rotundity," the perpendicular walls of buildings are, nowhere, parallel, and that even the walls of houses on opposite sides of a street are not! But, since all observation fails to find any evidence of this want of parallelism which theory demands, the idea must be renounced as being absurd and in opposition to all well-known facts. This is a proof that the Earth is not a globe.
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Both Davies and Hodgson talked of a reunion a couple of times, however, this would never come to pass. The first hint of a reunion came in 1993 when Davies and Hodgson reunited for an A & M dinner honoring Jerry Moss, co-founder of A & M Records. This dinner resulted in writing and demoing new songs, but it never went anywhere due to disagreements over management. Another hint of a reunion came in 2010 when Roger Hodgson approached Rick Davies about a fortieth anniversary of their very first album Supertramp (rogerhodgson.com). Rick Davies declined the invitation and any chance of Supertramp reuniting was squashed.
53) At places of comparable latitude North and South, the Sun behaves very differently than it would on a spinning ball Earth but precisely how it should on a flat Earth. For example, the longest summer days North of the equator are much longer than those South of the equator, and the shortest winter days North of the equator are much shorter than the shortest South of the equator. This is inexplicable on a uniformly spinning, wobbling ball Earth but fits exactly on the flat model with the Sun traveling circles over and around the Earth from Tropic to Tropic.
```2. Another related thing I don’t understand: if the sun and moon are always above the disk of the Earth, why can’t everyone in the world see them at all times? Surely they should always be visible, at least at a low angle. I can’t draw myself any diagram where they are not always visible, but we see that that doesn’t happen. I can’t see how night time happens. Help!
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169) So-called “satellite” TV dishes are almost always positioned at a 45 degree angle towards the nearest ground-based repeater tower. If TV antennae were actually picking up signals from satellites 100+ miles in space, most TV dishes should be pointing more or less straight up to the sky. The fact that “satellite” dishes are never pointing straight up and almost always positioned at a 45 degree angle proves they are picking up ground-based tower signals and not “outer-space satellites.”
142) People claim that if the Earth were flat, they should be able to use a telescope and see clear across the oceans! This is absurd, however, as the air is full of precipitation especially over the oceans, and especially at the lowest, densest layer of atmosphere is NOT transparent. Picture the blurry haze over roads on hot, humid days. Even the best telescope will blur out long before you could see across an ocean. You can, however, use a telescope to zoom in MUCH more of our flat Earth than would be possible on a ball 25,000 miles in circumference.
46.) It has been shown that an easterly or a westerly motion is necessarily a circular course round the central North, The only north point or centre of motion of the heavenly bodies known to man is that formed by the North Star, which is over the central portion of the outstretched Earth. When, therefore, astronomers tell us of a planet taking a westerly course round the Sun, the thing is as meaningless to them as it is to us, unless they make the Sun the northern centre of the motion, which they cannot do! Since, then, the motion which they tell us the planets have is, on the face of it, absurd; and since, as a matter of fact, the Earth can have no absurd motion at all, it is clear that it cannot be what astronomers say it is – a planet; and, if not a planet, it is a proof that Earth is not a globe.
When the Sun crosses the equator, in March, and begins to circle round the heavens in north latitude, the inhabitants of high northern latitudes see him slimming round their horizon and forming the break of their long day, in a horizontal course, not disappearing again for six months, as he rises higher and higher in the heavens whilst he makes his twenty-four hour circle until June, when he begins to descend and goes on until he disappears beyond the horizon in September. Thus, in the northern regions, they have that which the traveler calls the "midnight Sun," as he sees that luminary at a time when, in his more southern latitude, it is always midnight. If, for one-half the year, we may see for ourselves the Sun making horizontal circles round the heavens, it is presumptive evidence that, for the other half-year, he is doing the same, although beyond the boundary of our vision. This, being a proof that Earth is a plane, is, therefore, a proof that the Earth is not a globe.
##### 19) Tycho Brahe famously argued against the heliocentric theory in his time, positing that if the Earth revolved around the Sun, the change in relative position of the stars after 6 months orbital motion could not fail to be seen. He argued that the stars should seem to separate as we approach and come together as we recede. In actual fact, however, after 190,000,000 miles of supposed orbit around the Sun, not a single inch of parallax can be detected in the stars, proving we have not moved at all.
In January 2016, Tila Tequila posted a series of tweets claiming to believe the Earth is flat. The following month, on February 16th, 2016, NBA super star Kyrie Irving expressed his belief that the Earth is flat on the podcast Road Trippin (shown below, left). The next year in 2017, famed Jiu-Jitsu instructor and former UFC analyst, Eddie Bravo came forward with his belief in a flat Earth numerous times, most notably on The Joe Rogan Experience podcast (shown below, right).
Since temperature inversions are common over water, it is relatively easy to devise experiments in which distant objects beyond the curvature of the earth are visible. Perhaps the most famous are the photographs of the Chicago skyline taken across Lake Michigan, about 60 miles away. The photographer, Joshua Nowicki, does not promote the flat earth, but flat-earthers have used his photographs many times, such as here, as supposed proof that the earth is flat. Flat-earthers do not seem to be aware of just how rare these photographs are. If the earth were flat, then the Chicago skyline would be visible across Lake Michigan nearly every clear day, but it is not. If the earth is spherical, then the hulls of ships ought to disappear as the ships move away from the observer. Since the ship must move many miles away for this to become noticeable, it is difficult to see this with the naked eye.
##### 80) In Chambers’ Journal, February 1895, a sailor near Mauritius in the Indian Ocean reported having seen a vessel which turned out to be an incredible 200 miles away! The incident caused much heated debate in nautical circles at the time, gaining further confirmation in Aden, Yemen where another witness reported seeing a missing Bombay steamer from 200 miles away. He correctly stated the precise appearance, location and direction of the steamer all later corroborated and confirmed correct by those onboard. Such sightings are absolutely inexplicable if the Earth were actually a ball 25,000 miles around, as ships 200 miles distant would have to fall approximately 5 miles below line of sight!
God's Truth never - no, never - requires a falsehood to help it along. Mr. Proctor, in his " Lessons," says: Men " have been able to go round and round the Earth in several directions." Now, in this case, the word " several will imply more than two, unquestionably: whereas, it is utterly impossible to circumnavigate the Earth in any other than an easterly or a westerly direction; and the fact is perfectly consistent and clear in its relation to Earth as a Plane.. Now, since astronomers would not be so foolish as to damage a good cause by misrepresentation, it is presumptive evidence that their cause is a bad one, and - a proof that Earth is not a globe.
##### 53) At places of comparable latitude North and South, the Sun behaves very differently than it would on a spinning ball Earth but precisely how it should on a flat Earth. For example, the longest summer days North of the equator are much longer than those South of the equator, and the shortest winter days North of the equator are much shorter than the shortest South of the equator. This is inexplicable on a uniformly spinning, wobbling ball Earth but fits exactly on the flat model with the Sun traveling circles over and around the Earth from Tropic to Tropic.
30) In his book “South Sea Voyages,” Arctic and Antarctic explorer Sir James Clarke Ross, described his experience on the night of November 27th, 1839 and his conclusion that the Earth must be motionless: “The sky being very clear … it enabled us to observe the higher stratum of clouds to be moving in an exactly opposite direction to that of the wind--a circumstance which is frequently recorded in our meteorological journal both in the north-east and south-east trades, and has also often been observed by former voyagers. Captain Basil Hall witnessed it from the summit of the Peak of Teneriffe; and Count Strzelechi, on ascending the volcanic mountain of Kiranea, in Owhyhee, reached at 4000 feet an elevation above that of the trade wind, and experienced the influence of an opposite current of air of a different hygrometric and thermometric condition … Count Strzelechi further informed me of the following seemingly anomalous circumstance--that at the height of 6000 feet he found the current of air blowing at right angles to both the lower strata, also of a different hygrometric and thermometric condition, but warmer than the inter-stratum. Such a state of the atmosphere is compatible only with the fact which other evidence has demonstrated, that the earth is at rest."
As with the Chicago skyline, there are many images on the internet, usually videos, of ships some distance away in which their hulls are visible. Many of these are taken during warm weather, such as late spring and summer, when the water is likely to be much cooler than the air, producing a temperature inversion. However, what would happen if one were to repeat this experiment over water that is warmer than the air temperature? Since there is no temperature inversion, the hulls of ships ought to disappear. This condition is likely to prevail on cool days in late autumn and early winter, when water temperatures are higher than air temperatures. These conditions also can produce inferior mirages, though not nearly as pronounced as over land on sunny summer days.
9.) As mariners take to sea with them charts constructed as though the sea were a level surface, however these charts may err as to the true form of this level surface taken as a whole, it is clear, as they find them answer their purpose tolerably well – and only tolerably for many ships are wrecked owing to the error of which we speak – that the surface of the sea is as it is taken to be, whether the captain of the ship "supposes" the Earth to be a globe or anything else. Thus, then, we draw, from the common system of "plane sailing," a practical proof that Earth is not a globe.
so if the earth is not a sphere in space revolveing around the sphereical sun, then what is it. Its one thing to say that "its not that way" but its different to say "its actually this way not that way". So what way is it? what way are you proposing is the correct way? do you beleive this is the only planet in the universe? do you believe that the stars are only decorations on a flat backdrop? I'm not certain what idea you are proposing is the correct way of looking at this...
The Earth is different from other planets, that much is true. After all, we have life, and we haven’t found any other planets with life (yet). However, there are certain characteristics all planets have, and it will be quite logical to assume that if all planets behave a certain way, or show certain characteristics—specifically if those planets are in different places or were created under different circumstances—our planet is the same.
In 1610, Galileo Galilei observed the moons of Jupiter rotating around it. He described them as small planets orbiting a larger planet—a description (and observation) that was very difficult for the church to accept, as it challenged a geocentric model where everything was supposed to revolve around the Earth. This observation also showed that the planets (Jupiter, Neptune, and later Venus was observed too) are all spherical, and all orbit the sun.
49) If Earth were a spinning ball heated by a Sun 93 million miles away, it would be impossible to have simultaneously sweltering summers in Africa while just a few thousand miles away bone-chilling frozen Arctic/Antarctic winters experiencing little to no heat from the Sun whatsoever. If the heat from the Sun traveled 93,000,000 miles to the Sahara desert, it is absurd to assert that another 4,000 miles (0.00004%) further to Antarctica would completely negate such sweltering heat resulting in such drastic differences.
INCREASE to such an extent beyond the equator (going southwards) that hundreds of vessels have been wrecked because of the false idea created by the untruthfulness of the charts and the globular theory together, causing the sailor to be continually getting out of his reckoning. With a map of the Earth in its true form all difficulty is done away with, and ships may be conducted anywhere with perfect safety. This, then, is a very important practical proof that the Earth is not a globe.
so if the earth is not a sphere in space revolveing around the sphereical sun, then what is it. Its one thing to say that "its not that way" but its different to say "its actually this way not that way". So what way is it? what way are you proposing is the correct way? do you beleive this is the only planet in the universe? do you believe that the stars are only decorations on a flat backdrop? I'm not certain what idea you are proposing is the correct way of looking at this...
Mr. J. R. Young, in his work on Navigation, says. "Although the path of the ship is on a spherical surface, yet we may represent the length of the path by, a straight line on a plane surface." (And plane sailing is the rule.) Now, since it is altogether impossible to "represent" a curved line by a straight one, and absurd to make the attempt, it follows that a straight line represents a straight line and not a curved one. And, Since it is the surface of the waters of the ocean that is being considered by Mr. Young, it follows that this surface is a straight surface, and we are indebted to Mr. Young, a professor of navigation, for a proof that the Earth is not a globe.
In " Cornell's Geography" there is an "Illustrated proof of the Form of the Earth," A curved line on which is represented a ship in four positions, as she sails away from an observer, is an arc of 72 degrees, or one-fifth of the supposed circumference of the "globe" - about 5,000 miles. Ten, such ships as those which are given in the picture would reach the full length of the "arc," making 500 miles as the length of the ship, The man in the picture, who is watching the ship as she sails away, is about 200 miles high; and the tower, from which he takes an elevated view, at least 600 miles high. These are the proportions, then, of men, towers, arid ships which are necessary in order to see a ship, in her different positions, as she "rounds the curve" of the "great hill of water" over which she is supposed to be sailing: for, it must be remembered that this supposed "proof" depends upon lines and angles of vision which, if enlarged, would still retain their characteristics. Now, since ships are not built 500 miles long, with masts in proportion, and men are not quite 200 miles high, it is not what it is said to be - a proof of rotundity - but, either an ignorant farce or a cruel piece of deception. In short, it is a proof that the Earth is not a globe.
197) Some people claim there is no motive for such a grand-scale deception and that flat or a ball makes no difference. By removing Earth from the motionless center of the Universe, these Masons have moved us physically and metaphysically from a place of supreme importance to one of complete nihilistic indifference. If the Earth is the center of the Universe, then the ideas of God, creation, and a purpose for human existence are resplendent. But if the Earth is just one of billions of planets revolving around billions of stars in billions of galaxies, then the ideas of God, creation, and a specific purpose for Earth and human existence become highly implausible. By surreptitiously indoctrinating us into their scientific materialist Sun-worship, not only do we lose faith in anything beyond the material, we gain absolute faith in materiality, superficiality, status, selfishness, hedonism and consumerism. If there is no God, and everyone is just an accident, then all that really matters is me, me, me. They have turned Madonna, the Mother of God, into a material girl living in a material world. Their rich, powerful corporations with slick Sun-cult logos sell us idols to worship, slowly taking over the world while we tacitly believe their “science,” vote for their politicians, buy their products, listen to their music, and watch their movies, sacrificing our souls at the altar of materialism. To quote Morris Kline, “The heliocentric theory, by putting the sun at the center of the universe ... made man appear to be just one of a possible host of wanderers drifting through a cold sky. It seemed less likely that he was born to live gloriously and to attain paradise upon his death. Less likely, too, was it that he was the object of God’s ministrations.”
125) Another proof the Sun is not millions of miles away is found by tracing the angle of sun-rays back to their source above the clouds. There are thousands of pictures showing how sunlight comes down through cloud-cover at a variance of converging angles. The area of convergence is of course the Sun, and is clearly NOT millions of miles away, but rather relatively close to Earth just above the clouds.
When the sun rises in the morning its light is just coming into view. The sun's light then follows it as it journeys away from you, appearing to descend below the horizon. In reality it is not "going down" but moving away from you and going beyond the line of convergence and your eyesight. It takes it's light with it. You can clearly see this in time lapse videos of the sun moving away, causing a sunset.
Inferior mirages are the most commonly noticed type of mirage; therefore, in the minds of most people, it is the only type of mirage. An inferior mirage occurs when there is a layer of warm air in contact with the ground, with layers of much cooler air just above. This condition exists nearly every sunny day. As the sun’s radiation is absorbed by the ground, the air in contact with the ground heats. Air a short distance above the ground remains cooler, so a large temperature difference can exist between these two layers. Because this temperature difference is most pronounced when the sun is as high in the sky as possible, this condition is most likely to occur in the early afternoon in late spring and into summer. The type of surface exposed to sunlight is very important too, because dark, flat surfaces, such as pavement, rock, and sand are most efficient at heating air this way. Surfaces with much vegetation, such as grass, are far less efficient in doing this. Because of its high specific heat and great optical depth, water generally is very poor at producing conditions conducive to an inferior mirage. The above example of a 10-degree difference in air temperature is rather modest—much greater temperature differences occur under ideal conditions of early summer, decreasing the critical angle, and increasing the angle above grazing where an inferior mirage can happen.
## 65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
Further indications of never reuniting happened when Rick Davies was diagnosed with multiple myeloma in 2015. Rick Davies’ Supertramp was scheduled for a European tour, however, it had to be cancelled due to his health. He pursued aggressive treatment to fight the disease and stated to fans, “I was really looking forward to returning to Europe and playing with the band again and I’m sorry to disappoint everyone who has overwhelmingly supported the upcoming tour. Unfortunately my current health issues have derailed me and right now I need to focus all of my energy on getting well (supertramp.com).” No further information has been provided on the future of Davies’ Supertramp, however, all hopes are for a full recovery and return to the stage.
The first main problem with globe model is that the next solar eclipse on August 21 is coming from the west. We have been told that the moon rotates around the earth from east to west just like the sun. But the moon during the solar eclipse is eclipsing the sun from the WEST. So how does that work on the Ball earth model? They say it’s just an optical illusion because of the angle of the sun. One scientist from NASA said its because the moon rotates from west to east! What? And another scientist said the the moon rotates around the earth twice as fast as the earth spins! Huh? When did that happen? The explanations from NASA get even more confusing and make no sense.
65.) The Rev. D. Olmsted, in describing a diagram whish is supposed to represent the Earth as a globe, with a figure of a man sticking out at each side and one hanging head downwards, says "We should dwell on this point until it appears to us as truly up," In the direction given to these figures as it does with regard to a figure which he has placed on the top! Now, a system of philosophy which requires us to do something which is, really, the going out of our minds, by dwelling on an absurdity until we think it is a fact, Cannot be a system based on God's truth, which never requires anything of the kind. Since, then, the popular theoretical astronomy of the day requires this, it is evident that it is the wrong thing, and that this conclusion furnishes us with a proof that the Earth is not a globe.
48) On a ball-Earth Santiago, Chile to Johannesburg, South Africa should be an easy flight all taking place below the Tropic of Capricorn in the Southern hemisphere, yet every listed flight makes a curious re-fueling stop in Senegal near the Tropic of Cancer in the North hemisphere first! When mapped on a flat Earth the reason why is clear to see, however, Senegal is actually directly in a straight-line path half-way between the two.
Another container ship made its way outward, as shown in Figure 11, a photograph taken through the supports of the pier at Virginia Beach. You can clearly read the name of the shipping company, Maersk Line, on the turquoise hull. What appears to be stains under the letters are the beginnings of an inferior mirage of the letters. Instead of a level of gray containers immediately above the hull, the layer of containers right above the hull on this ship appear a deep red. As with the other ship, in each succeeding photograph this ship is farther away, as evidenced by the decreasing apparent sizes of the containers and the ship.
At the end of the day, I think the one single proof of FE or Ball E, will be the irrefutable circumnavigation of Antarctica. As it is, we have no modern attempt. We have no attempt to travel across its middle, and no one wants to take on the challenge of a North South circumnavigation of the planet, the last great achievement (which is odd to me). All three of these things Not being done are Very strange indeed.
Since this image is visible above where the object is, it is called a superior mirage. Because cooler air has no physical reason to rise, a temperature inversion is a stable situation, with little convection as with the condition that produces an inferior mirage. Therefore, superior mirages can be very steady, much steadier than inferior mirages. Furthermore, since the refraction acts almost continually rather than at one point, superior mirages normally are erect rather than inverted. If one gains a little altitude, one can get out of the inversion layer, and thus avoid seeing a superior mirage. In my earlier article, I pointed out that this is what Alfred Russell Wallace did when he repeated the Bedford level experiment. Russell did not see the distant object that was his target, which is consistent with a spherical earth. Russell correctly accounted for this effect, but Rowbotham did not.
107) Ring magnets of the kind found in loudspeakers have a central North pole with the opposite “South” pole actually being all points along the outer circumference. This perfectly demonstrates the magnetism of our flat Earth, whereas the alleged source of magnetism in the ball-Earth model is emitted from a hypothetical molten magnetic core in the center of the ball which they claim conveniently causes both poles to constantly move thus evading independent verification at their two “ceremonial poles.” In reality the deepest drilling operation in history, the Russian Kola Ultradeep, managed to get only 8 miles down, so the entire ball-Earth model taught in schools showing a crust, outer-mantle, inner-mantle, outer-core and inner-core layers are all purely speculation as we have never penetrated through beyond the crust.
```In Mr. Proctor's "Lessons in Astronomy," page 15, a ship is represented as sailing away from the observer, and it is given in five positions or distances away on its journey. Now, in its first position, its mast appears above the horizon, and, consequently, higher than the observer's line of vision. But, in its second and third positions, representing the ship as further and further away, it is drawn higher and still higher up above the line of the horizon! Now, it is utterly impossible for a ship to sail away from an observer, under the, conditions indicated, and to appear as given in the picture. Consequently, the picture is a misrepresentation, a fraud, and a disgrace. A ship starting to sail away from an observer with her masts above his line of sight would appear, indisputably, to go down and still lower down towards the horizon line, and could not possibly appear - to anyone with his vision undistorted - as going in any other direction, curved or straight. Since, then the design of the astronomer-artist is to show the Earth to be a globe, and the points in the picture, which would only prove the Earth to be cylindrical if true, are NOT true, it follows that the astronomer-artist fails to prove, pictorially, either that the Earth is a globe or a cylinder, and that we have, therefore, a reasonable proof that the Earth is not. a globe.
```
I started reading this e-book today and finished it today. Great Work ! I have to say, even if approached flat-earth model fistly 3 month ago, this Book really opened my eyes completely and still i know how strong the Brainwash is and even if you - by ratio - see something clear, it will still need a time to be completely internalized. Thx alot, greetingz from Germany also i was able to watch the video below, which oc is blocked in Germany, but this led me to HotSpotShield, also very important for a native German, greetingz from Frankfurt/Germany, Benjamin
## 119) It is claimed that the other planets are spheres and so therefore Earth must also be a sphere. Firstly, Earth is a “plane” not a “planet,” so the shape of these “planets” in the sky have no bearing on the shape of the Earth beneath our feet. Secondly, these “planets” have been known for thousands of years around the world as the “wandering stars” since they differ from the other fixed stars in their relative motions only. When looked at with an unprejudiced naked-eye or through a telescope, the fixed and wandering stars appear as luminous discs of light, NOT spherical terra firma. The pictures and videos shown by NASA of spherical terra firma planets are all clearly fake computer-generated images, and NOT photographs.
47.) In consequence of the fact being so plainly seen, by everyone who visits the seashore, that the line of the horizon is a perfectly straight line, it becomes impossible for astronomers, when they attempt to convey, pictorially, an idea of the Earth's "convexity," to do so with even a shadow of consistency: for they dare not represent this horizon as a curved line, so well known is it that it is a straight one! The greatest astronomer of the age, in page 15 of his "Lessons," gives an illustration of a ship sailing away, "as though she were rounding the top of a great hill of water;" and there – of a truth – is the straight and level line of the horizon clear along the top of the "hill" from one side of the picture to the other! Now, if this picture were true in all its parts – and it is outrageously false in several – it would show that Earth is a cylinder; for the "hill" shown is simply up one side of the level, horizontal line, and, we are led to suppose, down the other! Since, then, we have such high authority as Professor Richard A. Proctor that the Earth is a cylinder, it is, certainly, a proof that the Earth is not a globe.
Another container ship made its way outward, as shown in Figure 11, a photograph taken through the supports of the pier at Virginia Beach. You can clearly read the name of the shipping company, Maersk Line, on the turquoise hull. What appears to be stains under the letters are the beginnings of an inferior mirage of the letters. Instead of a level of gray containers immediately above the hull, the layer of containers right above the hull on this ship appear a deep red. As with the other ship, in each succeeding photograph this ship is farther away, as evidenced by the decreasing apparent sizes of the containers and the ship.
##### 126) The Sun’s annual journey from tropic to tropic, solstice to solstice, is what determines the length and character of days, nights and seasons. This is why equatorial regions experience almost year-round summer and heat while higher latitudes North and especially South experience more distinct seasons with harsh winters. The heliocentric model claims seasons change based on the ball-Earth’s alleged “axial tilt” and “elliptical orbit” around the Sun, yet their flawed current model places us closest to the Sun (91,400,000 miles) in January when its actually winter, and farthest from the Sun (94,500,000 miles) in July when its actually summer throughout most of the Earth.
65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
What a timeless work of truth you have created, thanks for your hard work Eric. Any stupid physicist that tries to deny flat earth by saying "relativity" proves it false, is completely wrong because relativity and all of quantum mechanics is wrong and no where near the real model of physics. Ken Wheeler's book "Unocovering the Missing secrets of magnetism" is the real model of physics & proves that the ether exists and that the standard (particle) model of physics is completely false b/c there is no such thing as "particles" b/c particles can not mediate action at a distance & or magnetism, electricity is not made up of "particles". Neither is "space" some type of object/medium that can act upon another object or be warped/ stretched as relativity states. The idea that "space" is "something," is obsurd on every level. There's no use in me trying to describe KW's work b/c a short explanation will not do the subject justice. For a brief starter explanation I will say that physics is based on golden ration incommensurablity(fractality)--, centripetal(counter-spacial) & centrifugal(spacial) forces. Any force is a result of an ether preterbation by torquing the ether aka the dielectric inertial plane (mainstream science calls this the Bloch wall in a magnet).
18.) The best possessions of man are his senses; and, when he uses them all, he will not be deceived in his survey of nature. It is only when some one faculty or other is neglected or abused that he is deluded. Every man in full command of his senses knows that a level surface is a flat or horizontal one; but astronomers tell us that the true level is the curved surface of a globe! They know that man requires a level surface on which to live, so they give him one in name which is not one in fact! Since this is the best that astronomers, with their theoretical science, can do for their fellow creatures – deceive them – it is clear that things are not as they say they are; and, in short, it is a proof that Earth is not a globe.
The Earth is different from other planets, that much is true. After all, we have life, and we haven’t found any other planets with life (yet). However, there are certain characteristics all planets have, and it will be quite logical to assume that if all planets behave a certain way, or show certain characteristics—specifically if those planets are in different places or were created under different circumstances—our planet is the same.
66) Dr. Rowbotham conducted several other experiments using telescopes, spirit levels, sextants and “theodolites,” special precision instruments used for measuring angles in horizontal or vertical planes. By positioning them at equal heights aimed at each other successively he proved over and over the Earth to be perfectly flat for miles without a single inch of curvature. His findings caused quite a stir in the scientific community and thanks to 30 years of his efforts, the shape of the Earth became a hot topic of debate around the turn of the nineteenth century.
65) Also Quoting Dr. Rowbotham, “On the shore near Waterloo, a few miles to the north of Liverpool, a good telescope was fixed, at an elevation of 6 feet above the water. It was directed to a large steamer, just leaving the River Mersey, and sailing out to Dublin. Gradually the mast-head of the receding vessel came nearer to the horizon, until, at length, after more than four hours had elapsed, it disappeared. The ordinary rate of sailing of the Dublin steamers was fully eight miles an hour; so that the vessel would be, at least, thirty-two miles distant when the mast-head came to the horizon. The 6 feet of elevation of the telescope would require three miles to be deducted for convexity, which would leave twenty-nine miles, the square of which, multiplied by 8 inches, gives 560 feet; deducting 80 feet for the height of the main-mast, and we find that, according to the doctrine of rotundity, the mast-head of the outward bound steamer should have been 480 feet below the horizon. Many other experiments of this kind have been made upon sea-going steamers, and always with results entirely incompatible with the theory that the earth is a globe.”
115) The existing laws of density and buoyancy perfectly explained the physics of falling objects long before knighted Freemason “Sir” Isaac Newton bestowed his theory of “gravity” upon the world. It is a fact that objects placed in denser mediums rise up while objects placed in less dense mediums sink down. To fit with the heliocentric model which has no up or down, Newton instead claimed objects are attracted to large masses and fall towards the center. Not a single experiment in history, however, has shown an object massive enough to, by virtue of its mass alone, cause other smaller masses to be attracted to it as Newton claims “gravity” does with Earth, the Sun, Moon, Stars and Planets.
69) The New York City skyline is clearly visible from Harriman State Park’s Bear Mountain 60 miles away. If Earth were a ball 25,000 miles in circumference, viewing from Bear Mountain’s 1,283 foot summit, the Pythagorean Theorem determining distance to the horizon being 1.23 times the square root of the height in feet, the NYC skyline should be invisible behind 170 feet of curved Earth. | 10,746 | 49,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.96897 |
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Java Program to Find if there is a subarray with 0 sum
Given an array of positive and negative numbers, find if there is a subarray (of size at-least one) with 0 sum.
Examples :
Input: {4, 2, -3, 1, 6}
Output: true
Explanation:
There is a subarray with zero sum from index 1 to 3.
Input: {4, 2, 0, 1, 6}
Output: true
Explanation:
There is a subarray with zero sum from index 2 to 2.
Input: {-3, 2, 3, 1, 6}
Output: false
A simple solution is to consider all subarrays one by one and check the sum of every subarray. We can run two loops: the outer loop picks a starting point i and the inner loop tries all subarrays starting from i (See this for implementation). The time complexity of this method is O(n2).
We can also use hashing. The idea is to iterate through the array and for every element arr[i], calculate the sum of elements from 0 to i (this can simply be done as sum += arr[i]). If the current sum has been seen before, then there is a zero-sum array. Hashing is used to store the sum values so that we can quickly store sum and find out whether the current sum is seen before or not.
Example :
```arr[] = {1, 4, -2, -2, 5, -4, 3}
If we consider all prefix sums, we can
notice that there is a subarray with 0
sum when :
1) Either a prefix sum repeats or
2) Or prefix sum becomes 0.
Prefix sums for above array are:
1, 5, 3, 1, 6, 2, 5
Since prefix sum 1 repeats, we have a subarray
with 0 sum. ```
Following is implementation of the above approach.
Java
`// A Java program to find ``// if there is a zero sum subarray``import` `java.util.HashSet;``import` `java.util.Set;`` ` `class` `ZeroSumSubarray ``{`` ``// Returns true if arr[] `` ``// has a subarray with sero sum`` ``static` `Boolean subArrayExists(``int` `arr[])`` ``{`` ``// Creates an empty hashset hs`` ``Set hs = ``new` `HashSet();`` ` ` ``// Initialize sum of elements`` ``int` `sum = ``0``;`` ` ` ``// Traverse through the given array`` ``for` `(``int` `i = ``0``; i < arr.length; i++) `` ``{`` ``// Add current element to sum`` ``sum += arr[i];`` ` ` ``// Return true in following cases`` ``// a) Current element is 0`` ``// b) sum of elements from 0 to i is 0`` ``// c) sum is already present in hash map`` ``if` `(arr[i] == ``0` ` ``|| sum == ``0` ` ``|| hs.contains(sum))`` ``return` `true``;`` ` ` ``// Add sum to hash set`` ``hs.add(sum);`` ``}`` ` ` ``// We reach here only when there is`` ``// no subarray with 0 sum`` ``return` `false``;`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `main(String arg[])`` ``{`` ``int` `arr[] = { -``3``, ``2``, ``3``, ``1``, ``6` `};`` ``if` `(subArrayExists(arr))`` ``System.out.println(`` ``"Found a subarray with 0 sum"``);`` ``else`` ``System.out.println(``"No Such Sub Array Exists!"``);`` ``}``}`
Output
`No Such Sub Array Exists!`
Time Complexity of this solution can be considered as O(n) under the assumption that we have good hashing function that allows insertion and retrieval operations in O(1) time.
Space Complexity: O(n) .Here we required extra space for unordered_set to insert array elements.
Please refer complete article on Find if there is a subarray with 0 sum for more details!
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Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From DATA SUFFICIENCY topic, which is common for all the IBPS,SBI exam RBI,,SSC and other competitive exams. We have included Some questions that are repeatedly asked in exams !!!
1. On which day of the week from Monday to Sunday did Ravi leave for London ?
I.Ravi didn’t leave for London during the weekend.
II.Ravi’s brother left for London on Friday 2 days after Ravi left for London
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –2)If the data in statement II alone are sufficient
Solution:
From II : Ravi left for London on Wednesday
2. How many boys are there between A and D in a row of 30 boys ?
I.A is 6 places away from B, who is 20th from the left end.
II.A is 12th from the left end and D is 7th from the right end.
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –2)If the data in statement II alone are sufficient
Solution:
24 -12 = 12 boys
3. Among P, Q, R, S, T and U, who is the tallest ?
I.S and T are taller than U, P and Q but none of them is the tallest.
II.T is taller than S but shorter than R
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –1)If the data in statement I alone are sufficient
Solution:
R > S, T>U, P, Q
4. In a certain code 48 means ‘ stop smoking’ and 62 means ‘ injurious habit’. What do 6 and 8 mean respectively in that code ?
I.8 2 9 means ‘smoking is injurious’
II.4 6 7 means ‘ stop bad habit’
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer – 3)If the data either in statement I alone or statement II alone are sufficient to answer
Solution:
6 => habit
8=> smoking
We can get the answer separately from I or II
5. How is M related to H ?
I.E is the only granddaughter of H and K is father of E.
II.E is sister of G and M is mother of G.
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –5)If the data in both the statements I and II together are necessary to answer
Solution:
From both I and II
M is daughter-in-law of H.
6. How many pages of the Book did Divya read on Sunday ?
I.Divya read the last 50 pages of the book on Monday morning
II.The book has 300 pages out of which two-thirds were read by Divya before Sunday.
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –5)If the data in both the statements I and II together are necessary to answer
Solution:
From both I and II
2/3(300) = 200
Sunday = 300-(200+50) = 50
7. A, B, C, D, E and F are sitting in a circular table facing the centre. How manuy persons are there between E and F(If counted clockwise from F)
I.B and E sit adjacent to each other and B is on the immediate left of F.
II.A and C sit adjacent to each other. There is one person who sits between D and E. A is not an immediate neighbour of E.
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –1)If the data in statement I alone are sufficient
Solution:
8. How many sisters does Yuvaraj have ?
I.Rohit is father of Yuvaraj and he has 3 children
II.Yuvaraj has 2 siblings one of them is Shila
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –4)If the data given in both I and II together are not sufficient
Solution:
9. In which direction is W with respect to Z ?
I.W is to the west of Y, who is to the north of U
II.Z is to the north – east of U and in the line of Y and W.
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer –5)If the data in both the statements I and II together are necessary to answer
Solution:
W is west of Z
10. Who among A, B, C, D and E each having a different weight is the lightest ?
I.C is heavier than only A
II.D is lighter than E and B
1)If the data in statement I alone are sufficient
2)If the data in statement II alone are sufficient
3)If the data either in statement I alone or statement II alone are sufficient to answer
4)If the data given in both I and II together are not sufficient
5)If the data in both the statements I and II together are necessary to answer
Answer – I.C is heavier than only A
Solution:
E,B>D>C>A
Note : Dear Friends if u know an alternate methods or shortcuts related to any chapter, you can share here.
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# Is 11 An Even Number?
• No, number 11 is not an even number.
• Eleven is not an even number, because it can't be divided by 2 without leaving a comma spot.
## Is 11 Odd Or Even?
• Number 11 is an odd number.
## How To Calculate Even Numbers
• To find the set of even natural numbers, we use 2N where N is any natural number. Any number multiplied with an even number will result in an even number. For example: 0N, 2N, 4N, 6N, ... (where N is any natural number) the result is always an even number.
• The oddness of a number is called its parity, so an odd number has parity 1, while an even number has parity 0.
## Mathematical Information About Numbers 1
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
## What Is An Even Number?
An even number is an integer which is evenly divisible by two. This means that if the integer is divided by 2, it yields no remainder. Zero is an even number because zero divided by two equals zero. Even numbers can be either positive or negative. You can tell if any decimal number is an even number if its final digit is an even number. An integer that is not an even number is an odd number.
Parity is a mathematical term that describes the property of an integer's inclusion in one of two categories: even or odd. An integer is even if it is evenly divisible by two and odd if it is not even. For example, 6 is even because there is no remainder when dividing it by 2. By contrast, 3, 5, 7, 21 leave a remainder of 1 when divided by 2. A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer. It can then be shown that an odd number is an integer of the form n = 2k + 1.
© Mathspage.com | Privacy | Contact | info [at] Mathspage [dot] com | 628 | 2,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-27 | latest | en | 0.940973 |
http://physics.stackexchange.com/questions/9519/how-can-i-measure-the-physical-properties-of-a-metallic-string/9521 | 1,462,575,619,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461864121714.34/warc/CC-MAIN-20160428172201-00162-ip-10-239-7-51.ec2.internal.warc.gz | 232,385,154 | 19,255 | # How can I measure the physical properties of a metallic string?
Metallic strings exist in different kinds but I would like to measure for a metallic string:
• the section/geometry along its length, to a precision of 1/100 mm
• its elongation when a tension is exerted on it on a measuring system I would construct to apply forces of the order of 1 to 100 daN. I have no clear idea of the order of magnitude of the elongation, so it might be too small to measure with reasonable equipement.
Any ideas, suggestions, theoretical and practical remarks welcome.
-
"metallic string" ? What is that? Something like a wire? – Georg May 6 '11 at 12:25
@Georg: yes, I suppose this is what you could call them when they are simple. But there are also metallic strings with (one or several) core wires and one or several wires wounded over them. – ogerard May 6 '11 at 12:33
Ah, You talk about strings for musical instruments? Nowadays in a physics forum string has a lot of meanings! – Georg May 6 '11 at 12:38
@Georg: I know. That's why I was careful not to tag with string-theory... – ogerard May 6 '11 at 12:42
I do not understand to which end You want this figures. For use as musical strings, You forgot the weight/mass per unit length. But in general, a maker of such strings can (hopefully will) tell You. – Georg May 6 '11 at 13:38
In a physics laboratory session at university we measured the extension of copper wire with different weights attached. To do so requires two wires hung vertically, one has the weights attached and is stretched, the other acts a sort of control. Between them a device is attached, this contains an adjustable spirit level.
Before the weight is applied the spirit level is set to horizontal, after the weights have stretched one string the device can be re-levelled and this 'correction angle' implies a certain extension.
To predict the sort of stretch you might achieve I would suggest looking up Youngs Modulus: http://en.wikipedia.org/wiki/Young%27s_modulus. (For metal wires and 'safe' loads don't expect very much!).
-
To accurately measure the cross section, assuming you know the material's density, just measure the length, mass and use $$\sigma = \frac{M}{\rho\;L}$$
If you're interested in the answer (rather than having to measure it), and you know the material's Young's modulus, it's easy to calculate the extension. But I'll assume you don't know this (which strength of a material is a lot harder to predict than its density).
If you're having trouble measuring the extension due to its small amount, then use the following fixture:
The idea is to use leverage so you can measure a smaller change in string length. There are two strings, one with no weight applied, the other stretches slightly. The result is that the blue stick moves quite a lot. (Probably it would be better to have the blue stick set up so that it has equal lengths on either sides of where you attach the strings. I'd use balsa for the stick.)
-
Ah. Hobby engineering. You have strayed into an area of intense interest. Here is a simple Rube Goldberg device which may give you other ideas:
build a short teeter-totter ("TT") with a laser pointer clamped to it, so that the laser points at a wall. Make sure the TT is fairly level, the fulcrum of the TT is firmly supported and that it is fairly balanced (so as to not add weight to your string). Attach one end of the TT near the free end of the string. Turn on the laser, make a pencil mark on the wall where it hits, load the string, and make your second mark.
Call the length from the TT fulcrum to the string = TS. Call the length from the string to the wall = SW. Call the length from your first to second wall mark = WW Call the length of stretch of the string = SS
Then: SS/TS = WW/(TS+SW) ---> SS = (WW)(TS/(TS+SW))
If you build a TT with a 2 inch half-arm (TS), and your wall is 9 feet away, you will get a movement magnification of 110/2 = 55. This might not be enough.
Another idea would be to mount a small mirror on the TT and hit it with a fixed laser, so that it bounces back to the wall. This would double your sensitivity (movement magnification of 110). If you made TT for the mirror = 1 inch, you could get approximately double that (218). If you wanted to add another mirror at the target spot on the wall and make your pencil marks on the opposite wall... well... you do the math. :-) As long as you keep the angles small (less than 5 degrees) you can dispense with any trig and just use simple geometry.
- | 1,075 | 4,516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-18 | longest | en | 0.942166 |
https://www.physicsforums.com/threads/calculating-torque-to-turn-twist-a-trailer-on-wheels-when-its-not-moving.1064893/ | 1,726,202,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00038.warc.gz | 877,851,464 | 16,463 | # Calculating torque to turn/twist a trailer on wheels when it's not moving
• Calvin Horst
Calvin Horst
How did you find PF?: web search.
I am new to all this. There was another post similar to this, but it was above my current education. I didn't go beyond 8th grade. So, beyond that I learn by doing, and through other friends.
I'm constructing a building on wheels for pastured chickens. 24' wide, 44' Long and weight roughly 16,000 pounds when loaded. Building is capable of moving sideways and end ways.
Wheels are 31" diameter 15" wide with bar tread 4.5" O.C. 1.5" wide bar standing 0.625" high. And 3" in-between the bars. 4 wheels = 4,000 lbs. per wheel.
I am considering putting smooth Balloon tires on which would turn much easier. And pulling the building with a winch.
Another factor is how muddy the field is.
Welcome to PF.
Calvin Horst said:
I'm constructing a building on wheels for pastured chickens. 24' wide, 44' Long and weight roughly 16,000 pounds when loaded.
Dat's a lot of chickens! Do you have any pictures of similar structures/trailers?
AKA Chicken tractor. https://en.wikipedia.org/wiki/Chicken_tractor
https://mobilechickenhouse.com/commercial-chicken-tractor/
Calvin Horst said:
Building is capable of moving sideways and end ways.
How many wheels? To move the structure, you will need to pull on all the wheel modules at the same time. The direction of the pull will need to align the individual wheels.
The problem will be the strength of the framework or chassis. That will come down to the structural design, and how wires are used for cross-bracing.
Wind design will be important. What are the strongest winds it must withstand? What will the walls and roof be made from?
We need a sketch of your basic ideas.
Lnewqban | 433 | 1,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.936477 |
http://thatsanswer.com/answered/228910/How-fast-does-a-car-travel-per-a-second-if-it-travels-140-km-in-25-hrs | 1,529,907,522,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867493.99/warc/CC-MAIN-20180625053151-20180625073151-00017.warc.gz | 312,800,079 | 6,405 | How fast does a car travel per a second if it travels 140 km in 2.5 hrs? - ThatsAnswer.COM
Similar Questions
• Answer: well if the car is going at a steady pace of 55miles an hour. You simply multiply 55 by 25 and get 1375 miles per hour in 25 hours.
• Answer: easy way to find this answer is to find the miles per minute and multiply by 60 to get miles per hour95/10=9.5 miles per minute9.5 mpm x 60 minutes per hour = 570 MPH.
• Answer: 5 hours and 2 minutes.
• Answer: equals 105 km divided by 80 min times 60 equals 78 kilometers per hour
• Answer: The maximum cruise speed of a McDonnell Douglas DC-10 is Mach0.88 (610 mph / 982 km/h / 530 kt)
• Answer: too slow!!a bus is never on time
• Answer: A jet can travel from 1000mph to 7000mph.
• Answer: The typical cruise speed of a Boeing 737-700 is Mach 0.785 (447 kn,514 mph, 828 km/h). The maximum speed is Mach 0.82 (475 kn, 545mph, 877 km/h)
• Answer: At 550 miles per hour, it would be .15 miles per second.
• Answer: normally aspirated jets or ram jets either one have a limitation near mach 3.6 2550mph. However, some specially designed jets have reached mach 7 near the edge of space.
How fast does a car travel per a second if it travels 140 km in 2.5 hrs?
• 140 km / 2.5 hours = 56 km / hour = 15.6 meters / second
Name: *
• Answer: 140 km / 2.5 hours = 56 km / hour = 15.6 meters / second | 417 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-26 | latest | en | 0.859337 |
https://physics.stackexchange.com/questions/488671/metric-tensor-in-the-comoving-cordinate-frame | 1,632,645,826,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00400.warc.gz | 500,292,900 | 37,969 | # Metric tensor in the comoving cordinate frame
In the textbook "Gravitation" by Misner,Thorne and Wheeler (page 717), when the isotropy of the universe is considered, it is stated that in the comoving frame the metric tensor can be written as:
$$g_{\alpha \beta} =\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial x^\beta}$$
How does one come at this conclusion?
• I would have sais that by definition, changing frame from a free falling to another would have raised a metric tensor $g_{\alpha \beta}={\partial x^\mu \over \partial x^\alpha}{\partial x^\nu \over \partial x^\beta}\eta_{\mu \nu}$, wouldn't it?
– Matt
Jun 28 '19 at 12:29
As Matt says in a comment, this is just a notation stating the definition of the metric. MTW notate it with a dot, to make it clear that the right-hand side is the inner product of two vectors. $$\partial/\partial x^\alpha$$ is a notation for the vector corresponding to a unit change in the coordinate $$x^\alpha$$. In this notation, the partial derivative is just being used as a basis vector for the space of vectors. By taking the dot product of unit two vectors, you get a component of the metric. This isn't anything special about cosmology. It's generic.
• I thought $\frac{\partial}{\partial x^\alpha}$ is tangent vector to the manifold, whereas a covector is $dx^\alpha$. As basis vectors of the space of covector I would expect $dx^{1}, dx^{3}, \ldots dx^{n}$ with $n$ the dimension of the manifold with a coordinate chart $(x^1, x^2, \ldots, x^n)$. Jun 28 '19 at 13:23 | 409 | 1,531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-39 | latest | en | 0.906596 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=19&t=15628 | 1,611,576,359,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00737.warc.gz | 425,263,638 | 11,829 | ## Post Module Question #8/9 [ENDORSED]
$\Delta p \Delta x\geq \frac{h}{4\pi }$
Kendall_Chaffin_3C
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm
### Post Module Question #8/9
Question 8 states If one incorrectly assumed that an electron is located inside the nucleus of an atom, then for a hydrogen atom the electron is confined to its nuclear diameter of 1.7 x 10-15 m which would be the electron's uncertainty in position. Use the Heisenberg uncertainty equation to calculate the electron's uncertainty in momentum. Then use the mass of electron (9.1 x 10-31 kg) to calculate its uncertainty in velocity.
Comment on your value.
I got that Delta p <= 3.1 x 10-20kg.m.s-1, Delta v = 3.4 x 1010m.s-1
For question 9, it says use the above uncertainty in velocity to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). Comment on your value.
I used the velocity I got to determine the wavelength then used the wavelength to determine the energy; but, my answer didn't match any of the options. What am I doing wrong?
kara_kremer_2N
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am
### Re: Post Module Question #8/9 [ENDORSED]
Try using the equation KE=1/2 mv^2
You should get the kinetic energy for one hydrogen atom. From there you can use dimensional analysis to calculate the kinetic energy per mole of electrons/hydrogen atoms.
FizaBaloch1J
Posts: 41
Joined: Fri Apr 06, 2018 11:01 am
### Re: Post Module Question #8/9
kara_kremer_2N wrote:Try using the equation KE=1/2 mv^2
You should get the kinetic energy for one hydrogen atom. From there you can use dimensional analysis to calculate the kinetic energy per mole of electrons/hydrogen atoms.
How do we "calculate the kinetic energy per mole of electrons/hydrogen atoms" because once we find the KE, isn't that J per electron, so then do we divide our answer in J by avogadros number to cancel out electrons and get moles?
NatalieSDis1A
Posts: 39
Joined: Fri Apr 06, 2018 11:05 am
### Re: Post Module Question #8/9
I think the key here is realizing that Hydrogen has 1 electron so if you have a certain amount of electrons you have that amount of hydrogen atoms. I believe thats the why the hydrogen add on is in parentheses, signifying that the answer is the same for j/mol of electron and j/ mol of hydrogen atom.
The conversion should be something like this:
(X joules/ 1 electron)(6.02 x 10^23 electrons/ 1 mol electrons)(1 mol electrons/1 mol of hydrogen atoms) | 678 | 2,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.831543 |
https://www.westernsydney.edu.au/mesh/mesh/support_and_resources/logic_basics_and,_or,_not/more_on_or | 1,708,862,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00694.warc.gz | 1,081,781,363 | 17,385 | # More on OR
This page contains more examples of the use of OR.
The logical operator OR is written $||$ when writing a program in Java. Here both OR and $||$ are used because you need to know both.
Often T is written for true and F for false.
##### Example 1
find the truth value of $6-6=0||1\le 2$.
Answer: (This is the same as asking for the truth value of $6-6=0$ OR $1\le 2$.)
Firstly we need to determine the truth value of $6-6=0$ and of $1\le 2$: $6-6=0$ is T (true) and $1\le 2$ is T. We have $\underbrace{6-6=0}_{T}||\underbrace{1\le 2}_{T}$. That is, we have T||T.
The truth table for $||$ shows that T||T is T. The answer is T.
##### Example 2
Find the truth value of $2+2=10$ OR $1\le 2$.
Answer: (this is the same as asking for the truth value of $2+2=10||1\le 2$.)
$2+2=10$ is F, (false) and $1\le 2$ is T (true). So we have $\underbrace{2+2=10}_{F}$ OR $\underbrace{1\le 2}_{T}$, which is F OR T. From the truth table for $||$ we see that F||T, which is the same as F OR T, is T. The answer is T.
##### Example 3
Find the truth value of $2+2=10$ OR $2\times 2=22$.
Answer: $2+2=10$ is F and $2\times 2=22$ is F. So we have $\underbrace{2+2=10}_{F}$ OR $\underbrace{2\times 2=22}_{F}$, which is F OR F. From the truth table for $||$ we see that F$||$F is F. Therefore F OR F is F. The answer is F.
##### Example 4
Find the truth value of $1\le 2||2+2=10$.
Answer: We have $\underbrace{1\le 2}_{T}||\underbrace{2+ 2=10}_{F}$, which is T$||$F. From the truth table for $||$ we see that T$||$F is T. The answer is T.
$5-3=0||-1\lt 1$ True False The sky is always red OR Finnish is the main language of Australia. True False $23\le 100$ OR $0=0$ True False $23\le 100||0+1=0$ True False | 636 | 1,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.855869 |
http://www.jiskha.com/members/profile/posts.cgi?name=Abbie&page=4 | 1,371,654,468,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708808767/warc/CC-MAIN-20130516125328-00008-ip-10-60-113-184.ec2.internal.warc.gz | 527,045,488 | 2,909 | Wednesday
June 19, 2013
# Posts by Abbie
Total # Posts: 82
chemistry
The atomic mass of iron oxide is 160 atomic mass units. What is the atomic mass of iron if a molecule of iron oxide consists of tow atoms of iron and three atoms of oxgen?
physics
This is a basic phyisics class. Thank you so much!!!!1
physics
Given that 12g of carbon combines completely with 16g of oxygen to form carbon monoxide, how many grams of carbon monoxide can be made from 48g of carbon and 48g of oxygen
Math
Put parentheses and brackets in the proper places to make the following equation true: 9x3+8-5+6squared=58
math
that is what I meant! Thank you guys very much!
math
so would this be correct? 25 is 33 % of what? 25/.33 = 2500
math
What is the formula for finding 54 is 10 % of what?
math
you are an amazing help and are very much appreciated!
math
Thank you very much Mrs. Sue! What about the formula for 54 is 10 % of what?
math
what is the formula for finding % of a #? example: 1. find 40 % of 160? 2. 24 is 100 % of what? also, how do you do this problem? 4/6 of 48
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https://www.coursehero.com/tutors-problems/Statistics-and-Probability/8420870-The-professor-has-asked-the-question-below-and-I-dont-know-how-to-beg/ | 1,540,138,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514162.67/warc/CC-MAIN-20181021161035-20181021182535-00288.warc.gz | 912,812,689 | 38,904 | View the step-by-step solution to:
# The professor has asked the question below and I don't know how to begin: For each of the five variables, process, organize, present and summarize...
The professor has asked the question below and I don't know how to begin:
For each of the five variables, process, organize, present and summarize the data. Analyze each variable by itself using graphical and numerical techniques of summarization. Use MINITAB as much as possible, explaining what the printout tells you. You may wish to use some of the following graphs: stem-leaf diagram, frequency/relative frequency table, histogram, boxplot, dotplot, pie chart, bar graph. Caution: not all of these are appropriate for each of these variables, nor are they all necessary. More is not necessarily better. In addition be sure to find the appropriate measures of central tendency, and measures of dispersion for the above data. Where appropriate use the five number summary (the Min, Q1, Median, Q3, Max). Once again, use MINITAB as appropriate, and explain what the results mean.
Location Income (\$1000) Size Years Credit Balance(\$) Urban 54 3 12 4016 Rural 30 2 12 3159 Suburban 32 4 17 5100 Suburban 50 5 14 4742 Rural 31 2 4 1864 Urban 55 2 9 4070 Rural 37 1 20 2731 Urban 40 2 7 3348 Suburban 66 4 10 4764 Urban 51 3 16 4110 Urban 25 3 11 4208 Urban 48 4 16 4219 Rural 27 1 19 2477 Rural 33 2 12 2514 Urban 65 3 12 4214 Suburban 63 4 13 4965 Urban 55 6 15 4412 Urban 21 2 18 2448 Rural 44 1 7 2995 Urban 37 5 5 4171 Suburban 62 6 13 5678 Urban 21 3 16 3623 Suburban 55 7 15 5301 Rural 42 2 19 3020 Urban 41 7 18 4828 Suburban 54 6 14 5573 Rural 30 1 14 2583 Urban 48 2 8 3866 Urban 34 5 5 3586 Suburban 67 4 13 5037 Rural 50 2 11 3605 Urban 67 5 1 5345 Urban 55 6 10 5370 Urban 52 2 11 3890 Urban 62 3 2 4705 Urban 64 2 6 4157 Suburban 22 3 18 3899 Urban 29 4 4 3890 Suburban 39 2 18 2972 Rural 35 1 11 3121 Urban 39 4 15 4183 Suburban 54 3 9 3730 Suburban 23 6 18 4127 Rural 27 2 1 2921 Urban 26 7 17 4603 Suburban 61 2 14 4273 Rural 30 2 14 3067 Rural 22 4 16 3074 Suburban 46 5 13 4820 Suburban 66 4 20 5149
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http://coastalfamilymedicine.org/project/86-how-to-calculate-running-feet-for-handrail.html | 1,563,395,866,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00273.warc.gz | 34,509,881 | 14,204 | how to calculate running feet for handrail
Oct That s a little different measurement then the rise run in the angle calculation aboveThe reason is that the length of the railing depends on
LANDINGS Landing width and depth must be at least equal to the width required for the stair Maximum vertical rise of a stair run must not exceed feet
b the required handrail on winder stairs shall be placed on the side surface measurement of inches and a minimum linear gripping surface of incheswide as the treads and shall measure at least feet in the direction of travel.
The run length should be inches ( cm) or longer for enough foot spaceThe height of the handrail measured from the nose of the tread is recommended
Has anyone done a railing similar to this recently who can give me an So if I tell them per linear foot that will hopefully dissuade themUS AND DO FIGURE IN ENGINEERING CALCULATIONS ON TOP OF THAT.
Nov On the deck shown here the cable run was feet and I ordered I measure the center to center length and determine the spacing for the
Use the Deckorators baluster calculator to determine quickly how many balusters or spindles your railing requiresBalusters are the vertical guards that
The run of the railings should be measured from the top step to the bottom step of the Use a tape measure to d a straight line on the wall connecting the Your handrail should be a x piece of wood that is no longer than feet long
How to Calculate Stair Step Riser Height Tread Width Stair rise run design And of course long stair runs due to a very tall total rise (more than feet) And secure hand railings on both sides of the stairs will be important for folks for
Stair railing lengthMeasure stair railing length from surface of lower floor to edge(nosing) of upper floorRun (Tread) (in)Rise (in).
This diagram helps you understand how you should measure and space your state you can not have any gaps in your stair railing system greater than inches! Baluster Spacing The general rule of thumb is to have baluster per foot so you the design over until you can have an even space on both ends of the run .
Calculate baluster centers and spacing options with full diagrams and Baluster Calculator Quantity Centers and Spacing with Running Measurements.
Compute rise run slope cross slope Identify of length (rise and run) () ( inch foot) ( ) need handrails or build the steepest ramp permissible.
posts you choose will determine your handrail styleBRUNThe measurement between the face of one riser to the face of newel for every or feet.
Mar A foot wide stair could have only two handrails if options for placing handrails on a stair depending on the calculated exit width and.
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A048265 Positive integers that aren't the sum of distinct primes of the form 6k+1. 4
1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 21, 22, 23, 24, 25, 27, 28, 29, 30, 33, 34, 35, 36, 40, 41, 42, 45, 46, 47, 48, 49, 52, 53, 54, 55, 58, 59, 60, 64, 65, 66, 71, 72, 77, 78, 83, 84, 85, 89, 90, 91, 95, 96, 101, 102, 108, 114, 115, 120, 121, 125, 126, 132 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS a(79)=205 is the largest term of this sequence. REFERENCES David Wells, The Penguin Dictionary of Curious and Interesting Numbers, entry 205. LINKS Sean A. Irvine, Table of n, a(n) for n = 1..79 CROSSREFS Sequence in context: A047248 A114024 A030173 * A288712 A002180 A207333 Adjacent sequences: A048262 A048263 A048264 * A048266 A048267 A048268 KEYWORD fini,full,nonn AUTHOR STATUS approved
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Last modified December 5 23:38 EST 2021. Contains 349558 sequences. (Running on oeis4.) | 564 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.800527 |
https://molwick.com/en/iq-study/0q60-evolution-intelligence.html | 1,718,286,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00155.warc.gz | 354,042,655 | 12,019 | The Global Model, complexity of IQ statistical simulation
MOLWICK
# The Global Model, complexity of the statistical simulation
The simulation will confirm the previous results about the innate character of intelligence, but it needs to adjust the Social Model introducing random deviations to make more realistic the artificial IQ values.
# 6. STATISTICAL SIMULATION: GLOBAL MODEL
## 6.a) Computer simulation of the evolution of intelligence
• Actual values and observed values!
The Social Model –the properly reformulated Individual Model– has been useful to determine that the significant chromosome is the one with the least potential, there is only one relevant chromosome and it seems is the sexual one.
Due to the accuracy of the Social Model of the evolution of intelligence, and having all the elements to simulate the proposed model by the Conditional Evolution of Live (CEL), the researchers developed a computer simulation to confirm the results. The statistical simulation should generate artificial intelligence quotients –variable W– behaving like those observed in the longitudinal study.
A second big surprise was the initial failure of the simplified Social Model to obtain the objective of statistical simulation of the processes and mechanisms of biological inheritance of intelligence.
The task was much more complicated than previously thought, forcing to eliminate all the simplifications of the Social model.
The introduction of the ability to generate quantitative variables with disturbances close to real ones implies a new statistical simulation, the Global Model.
A typical result of the generated variable W is in the q050 graph. Considering that W is a stochastic variable, the figure represents the average of ten estimates for the corresponding correlations.
The Multidimensional Correlation Index (MCI) of the artificial intelligence quotients vector W, which is three times higher for comparative reasons, is over 25 and far above the Global Multidimensional Correlation Index (GMCI) for the observed C variables of the children.
Therefore, the Global Model needs to include the random deviations in the expression and measurement of intelligence and other variables of its evolution, which the Individual and Social Models had excluded for simplicity.
The Individual Model showed that the differences in IQ measurements of the same children were very high due to the manifestation of the child's capacity at any given moment, and even more so, over the years.
Other factors causing similar deviations are the intelligence test used and the specific test session within a standard test.
Although the observed differences are superior to 10% to the average in some cases, the Global Model introduces an additional combinatorial algorithm to represent a factor of a mean deviation of 3% upward and 3% downward.
For the same reason, the Global Model incorporates stochastic error patterns in children variables C and parents M and F.
Nevertheless, the high correlation of W in the computer simulation does not decrease substantially, and vector W does not behave like the original IQ vectors from the Stanford Binet and Wechsler intelligence tests.
## 6.b) Statistical simulation model: complexity and optimization
To capture the complexity of the Global Model, it is necessary to introduce elements that are more haphazard; otherwise, the model would not be acceptable. In particular, the new features must lower the correlation in the unordered groups, and mainly in the small groupings.
At the same time, in the previously rearranged groups, the correlations should decrease in the small groups and increase or remain the same in the big ones. Once achieved a good model specification for the evolution of intelligence, it could begin its optimization.
### 6.b.1) Genetic affinity
The first idea should be to eliminate the simplifications carried out in the model's theoretical argumentation to avoid its complexity.
To continue dropping the multidimensional correlation index of W, the Global Model will include the filter effect mentioned by the Conditional Evolution of Life (CEL) in the proposal for a statistical simulation model regarding the resulting intellectual power of the genetic combination. Now, it will be equal to the intersection of the potentials and not to the smaller one.
Of course, this decrease due to the lack of genetic affinity will not be equal in all cases. Consequently, it will imply a random pattern in the statistical simulation; meaning another margin of 3% upwards or downwards bearing in mind the possible drag effect of the ancestors.
After considering this affinity filter effect, the correlation lowers again, but not much. The complexity of the statistical simulation model continues to increase while adding elements. | 879 | 4,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.879181 |
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Root to Win
Topic closed. 44 replies. Last post 2 years ago by lakerben.
Page 1 of 3
New Mexico
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Posted: October 25, 2012, 12:34 pm - IP Logged
In this one I take the root (2,6) of the last two draws and multiply them.
roots 2 input 6 mult 12 159 171 total 828
I add the result (12) WITH 159. I'm using 159 because I like the difference of 4 gap between the numbers.
I then finish with (999-171) FOR THE TOTAL.
This gives you the 2 picks to build on for your next few draws.
If you want more just add+1 or -1 to the last digit>>>170 172 173 174 175 and 826 827 829 820 etc.
Win, win, win.
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Posted: October 25, 2012, 12:49 pm - IP Logged
For Fl p3 for this morning.
roots 9 input 7 mult 63 159 222 total 777
Win, win, win.
dearborn mi
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Posted: October 25, 2012, 12:49 pm - IP Logged
ok what are the roots tho which number are the roots from the last draw the first or middle or last number
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Posted: October 25, 2012, 12:51 pm - IP Logged
I don't understand you. In the example I am taking the root of the drawn number. I use the lat 2 draws.
Win, win, win.
dearborn mi
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Posted: October 25, 2012, 12:56 pm - IP Logged
thats what i dont understand what is the root what do u mean by that like my last 2 draws were 497 and 378 so what would be the root
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Posted: October 25, 2012, 12:59 pm - IP Logged
texas
roots 5 input 5 mult 25 159 184 total 815
184,183,182,181 185
815 814 813 812 816 817
Win, win, win.
New Mexico
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Posted: October 25, 2012, 1:04 pm - IP Logged
Ohio p3 for tonight using the last 2 draws.
roots 5 input 6 mult 30 159 189 total 810
Plus or minus one from the end digit:
189 180 187 186 185
810 811 812 813 814
Win, win, win.
New Mexico
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Posted: October 25, 2012, 1:07 pm - IP Logged
Georgia p3 for tonight.
roots 1 input 4 mult 4 159 163 total 836
163 162 164 165 166
836 835 834 837 838
Win, win, win.
New Mexico
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Posted: October 25, 2012, 1:12 pm - IP Logged
NC P3
roots 4 input 4 mult 16 159 175 total 824
175 174 173 172
824 825 826 823
Win, win, win.
dearborn mi
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Posted: October 25, 2012, 1:13 pm - IP Logged
are u adding the 2 draws together to get the root or what pls help what are the roots
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Posted: October 25, 2012, 1:19 pm - IP Logged
In this one I take the roots (2,6) of the last two draws and multiply them. Its very simple.
Win, win, win.
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Posted: October 25, 2012, 1:22 pm - IP Logged
Kansas p3
roots 1 input 1 mult 1 159 160 total 839
160 161 162 159
839 838 837 836
Win, win, win.
New Mexico
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Posted: October 25, 2012, 1:27 pm - IP Logged
Va p3
roots 7 input 6 mult 42 159 201 total 798
201 202 203 204 200
798 797 796 795 799
Win, win, win.
New Mexico
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Posted: October 25, 2012, 1:32 pm - IP Logged
NJ for tonight.
roots 5 input 9 mult 45 159 204 total 795
204 205 206 203
795 796 797 794
Win, win, win.
N.C.
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Sweet! Thanks for sharing lakerben! Very easy to understand! Moneyrolls has trouble understanding simple things, he should use the search function and find out what the root of a number is.
Page 1 of 3 | 1,491 | 4,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-22 | latest | en | 0.822878 |
http://stackoverflow.com/questions/752308/split-array-into-smaller-arrays/752333 | 1,386,954,419,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164964633/warc/CC-MAIN-20131204134924-00088-ip-10-33-133-15.ec2.internal.warc.gz | 170,010,979 | 16,281 | # Split array into smaller arrays
I am looking for a way to easily split a python array in half.
So that if I have an array:
``````A = [0,1,2,3,4,5]
``````
I would be able to get:
``````B = [0,1,2]
C = [3,4,5]
``````
-
``````A = [1,2,3,4,5,6]
B = A[:len(A)/2]
C = A[len(A)/2:]
``````
If you want a function:
``````def split_list(a_list):
half = len(a_list)/2
return a_list[:half], a_list[half:]
A = [1,2,3,4,5,6]
B, C = split_list(A)
``````
-
You need to force int division in Python 3. // is required. – Stefan Kendall Apr 15 '09 at 18:55
A little more generic solution (you can specify the number of parts you want, not just split 'in half'):
EDIT: updated post to handle odd list lengths
EDIT2: update post again based on Brians informative comments
``````def split_list(alist, wanted_parts=1):
length = len(alist)
return [ alist[i*length // wanted_parts: (i+1)*length // wanted_parts]
for i in range(wanted_parts) ]
A = [0,1,2,3,4,5,6,7,8,9]
print split_list(A, wanted_parts=1)
print split_list(A, wanted_parts=2)
print split_list(A, wanted_parts=8)
``````
-
When the list doesn't divide evenly (eg split_list([1,2,3], 2) ) this will actually return wanted_parts+1 lists. – Brian Apr 15 '09 at 18:27
That's correct, i was in doubt as what should be the right approach (have one more part or have the last list have one more item). I'll update my post, thanks for commenting. – ChristopheD Apr 15 '09 at 21:34
hi.. what does the symbol "//" means?? – Fraz Jun 21 '12 at 6:01
@Fraz Its is meant as inline comment. Ignore "// wanted_parts" and "// wanted_parts" to make script execute. – PunjCoder Aug 2 '12 at 4:16
`//` means integer division. They should not be left out as they are quite essential in making this work. – Alphadelta14 Nov 20 at 10:37
``````f = lambda A, n=3: [A[i:i+n] for i in range(0, len(A), n)]
f(A)
``````
`n` - the predefined length of result arrays
-
``````def splitter(A):
B = A[0:len(A)//2]
C = A[len(A)//2:]
return (B,C)
``````
I tested, and the double slash is required to force int division in python 3. My original post was correct, although wysiwyg broke in Opera, for some reason.
-
`B,C=A[:len(A)/2],A[len(A)/2:]`
-
I think you forgot the divide by 2 step. :) – Stefan Kendall Apr 15 '09 at 15:50
Yep, I did. Edited it two seconds after posting... – John Montgomery Apr 15 '09 at 15:51
If you don't care about the order...
``````def split(list):
return list[::2], list[1::2]
``````
list[::2] gets every second element in the list starting from the 0th element.
list[1::2] gets every second element in the list starting from the 1st element.
-
Using list slicing. The syntax is basically `my_list[start_index:end_index]`
``````>>> i = [0,1,2,3,4,5]
>>> i[:3] # same as i[0:3] - grabs from first to third index (0->2)
[0, 1, 2]
>>> i[3:] # same as i[3:len(i)] - grabs from fourth index to end
[3, 4, 5]
``````
To get the first half of the list, you slice from the first index to `len(i)/2`...
``````>>> i[:len(i)/2]
[0, 1, 2]
``````
..and the swap the values around to get the second half:
``````>>> i[len(i)/2:]
[3, 4, 5]
``````
-
While the answers above are more or less correct, you may run into trouble if the size of your array isn't divisible by 2, as the result of `a / 2`, a being odd, is a float in python 3.0, and in earlier version if you specify `from __future__ import division` at the beginning of your script. You are in any case better off going for integer division, i.e. `a // 2`, in order to get "forward" compatibility of your code.
-
There is an official Python receipe for the more generalized case of splitting an array into smaller arrays of size `n`.
``````from itertools import izip_longest
def grouper(n, iterable, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
``````
This code snippet is from the python itertools doc page.
-
Here is a common solution, split arr into count part
``````def split(arr, count):
return [arr[i::count] for i in range(count)]
``````
-
``````def line_split(N, K=1): | 1,303 | 4,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2013-48 | latest | en | 0.824364 |
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Unformatted text preview: eye every second! The frequency of violet light is even higher—over 750 trillion Hz. Other types of electromagnetic radiation, like Xrays, have even higher frequencies, and some have lower frequencies, like radio waves. Just as our ears are only capable of hearing certain range of sounds (20 – 20,000 Hz), our eyes can only see a small range of frequencies. Frequency and Wavelength (cont.) Because visible light waves have such high frequencies, their wave lengths are very short. Recall the formula v = λf (wave speed = wavelength × frequency). Since light of any frequency always travels at the same speed in a vacuum, v is a constant. Thus, the bigger f is, the smaller λ must be. Red light, for example, has a wavelength of only about 700 nm. (1 nm = 1 nanometer = 109 m = 1 billionth of a meter.) Violet light has an even smaller wavelength, since its frequency is higher. Xrays have still smaller wavelengths. Radio waves can have very long wavelengths (many meters) since their frequencies are so low. High...
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Ask a homework question - tutors are online | 317 | 1,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.916397 |
https://mathoverflow.net/questions/104172/knot-diagrams-sets-of-moves-and-equivalence-relations | 1,695,981,327,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00367.warc.gz | 406,761,610 | 32,484 | # Knot diagrams, sets of moves and equivalence relations
Short version: Does anyone study equivalence classes generated by a given set of "moves" (in the sense of, but not limited to, Reidemeister moves) on the set of knot diagrams?
Yes, I understand that the concept of knot has a natural geometrical significance, that one usually views knot diagrams as a tool to study underlying knots, that the value of Reidemeister moves lies in how they preserve and generate the equivalence relation of isotopy. So, yes, I see why a knot theorist might reasonably have little interest, say, in looking at local moves not preserving isotopy.
So I'm asking this question in the spirit of abstraction for its own sake. But there is a precedent. A "symmetry theorist" studies groups because they capture the set of symmetries of important objects. But combinatorial group theory studies equivalence classes of strings under moves...and symmetry, if it enters the story at all, does so as a tool.
That said, it would be interesting if one could add an extra move to the Reidemeister moves that produced a coarser but computationally tractable classification.
No need to retread the ground here -- https://en.wikipedia.org/wiki/Reidemeister_move -- so, for example, I already understand that knot theorists know what happens with only Reidemeister type II and III moves. I am interested in stories like this, where one gets a finer equivalence relation that isotopy, but equally interested in sets of local moves that don't preserve isotopy and thus generate equivalence relations either coarser to, or simply incomparable with, isotopy.
• The appropriate abstraction here seems to me to be a category presented by generators and relations. (The category relevant to knots is the tangle category: math.ucr.edu/home/baez/tangles.html) This is a very general construction: it has as special cases monoids presented by generators and relations, as well as posets... Aug 7, 2012 at 3:02
• There are lots of such moves. To give an example I am very familiar with, $C_k$-moves are certain moves (surgery along claspers) that generate the equivalence relation that two knots share Vassiliev invariants of order up to k-1. Another example: Lou Kauffman first observed that two knots have the same Arf invariant iff they differ by a sequence of "band-pass" moves. Aug 7, 2012 at 3:13
• Kawauchi's survey of knot theory book mentions several of the theorems Jim Conant alludes to. Aug 7, 2012 at 5:44
• @Qiaochu Yuan So a knot or link then belongs to Hom(0,0)? But can't a "move" fail to respect the imposed directionality? Can't it wind back and forth in "time," using perhaps some but not all of the strands at a given "time" and thus escape this optic? Aug 7, 2012 at 8:09
• David Feldman and Qiaochu: I think you're both right. The answer is 2-dimensional algebra, which does not suffer from imposed directionality. See the work of Dror Bar-Natan, on "the circuit algebra of tangles" and related 2-dimensional algebras. This is indeed what is happening- the abstraction is to certain diagrammatic algebras (over a "modular operad") in the sense Qiaochu is refering to. And I believe this (intentionally leaving what I mean by "this" slightly vague) is very much the appropriate abstraction. Aug 7, 2012 at 13:07
Very much so. There are a number of small industries centred around studying equivalence classes of knot diagrams generated by a set of moves.
1. The study of claspers. For example, $$C_k$$-moves are a special type of clasper surgeries. MathSciNet indicates 123 citations for Habiro's fundamental paper Claspers and finite type invariants of links, providing some coarse measure of the vitality of the topic.
2. Replacing one rational tangle in a knot diagram by another generates an equivalence relation which has been deeply studied using quandles. See e.g. J. Przytycki's introductory lectures.
3. Dehn surgery, where the surgery curve is required to belong to some specified part of a knot group or link group (in the kernel of its representation to some fixed group, for instance) generates equivalence relations on knot diagrams modulo combinatorial "twisting" moves, which have been studied by Cochran-Orr-Gerges, and (excuse the self promotion) by myself and Andrew Kricker, and by Litherland and Wallace. The techniques for studying these equivalence relations have been topological rather than combinatorial.
4. There are a number of setting in which one allows Reidemeister moves plus some crossing changes, but not others. In the theory of finite type invariant, one fixes a some crossings (considers them in resolutions of "double points"), and allows crossing changes away from them. The equivalence classes are detected by the finite-type invariant of type the number of "fixed" crossings. In a similar-sounding vein, a free virtual knot is a virtual knot where we allow crossing changes away from virtual crossings. They have a rich theory- see e.g. this Manturov paper.
The Delta move generates the equivalence relation of linking numbers for links, as proved by Matveev and Murakami Nakanishi.
There's the double Delta move which Naik and Stanford showed generate S-equivalence.
A student of Freedman studied "slide equivalence" of knot diagrams. Unfortunately it wasn't published though.
• These both delta moves are special cases of Y-clasper surgeries ($C_2$ moves). Take a clasper with one leaf ringing around each strand participating in the delta. Could you describe slide equivalence? Aug 7, 2012 at 15:57
• @Daniel: I see, I learned about these moves in grad school I think before claspers were around, and I didn't realize they were equivalent. The number of Delta moves needed to unknot a knot is congruent to the Arf invariant. I don't have a copy of Huang's thesis, but I went to his thesis defense, and from what I can remember, you can slide arcs of the knot diagram around, treating a strand going under a crossing as two separate arcs attaching at the overstrand. But this was over 15 years ago, so I don't remember the details. Aug 7, 2012 at 16:54
One classical example of such move is Conway mutation, which falls into the category of tangle replacement, as Qiaochu Yuan mentioned in his comment. There's a very famous pair of mutants, the Kinoshita-Terasaka and the Conway knot (see the Wikipedia article).
Apparently, there's some topology behind this move: recently, using knot Floer homology, Josh Greene has shown that two alternating knots are mutants if their branched double covers are homeomorphic, and the other arrow was shown by Viro (see references in Greene's paper).
Legendrian knots
Legendrian knots are smooth knots whose tangent directions are contained in a contact structure such as the standard contact structure on $\mathbb{R}^3$, $dz=y~dx$. Every knot has Legendrian representatives. Two Legendrian knots of the same topological type might not be isotopic through Legendrian knots.
The projection of a Legendrian knot in the standard contact structure to the $xz$-plane is called its front projection. Some people study Legendrian knots through diagrams showing front projections. The $y$ coordinate can be recovered from the slope, so all crossings are determined by the diagram. However, there can be no vertical tangencies, since $y$ would be undefined, and you must allow cusps. See this Notices article.
There are analogues of Reidemeister moves, so this gives a refinement of knot theory described by a set of diagram moves on front projections.
Actually, you don't have to work with cusped diagrams. You can make all cusps horizontal, and you could choose to replace the horizontal cusps with vertical tangencies. So, standard knot diagrams up to a restricted set of moves (including disallowing some isotopies where no Reidemeister move was performed, but where vertical tangencies would have been introduced or removed) are equivalent to Legendrian knots.
Suppose you study curves up to isotopy instead of the standard ambient isotopy used in knot theory. You may be disappointed: knot theory in $S^3$ becomes trivial. You are allowed to replace a piece of a diagram showing a long knot with a long unknot by shrinking the knot to a point and forgetting it. However, link theory is still nontrivial, and so is knot theory in a $3$-manifold which is not simply connected. See Rolfsen, "Localized Alexander Invariants and Isotopy of Links." Annals of Mathematics 101 (1975) 1-19. | 1,897 | 8,460 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.928766 |
https://thinking-about-science.com/2016/10/30/16-42-keep-it-simple/ | 1,685,466,027,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00115.warc.gz | 645,316,762 | 31,454 | # 16.42 Keep it simple
Before you read this, it would be useful to read post 16.8.
Post 16.8 is a story about a biologist who investigated the rabbit population on a farm. He counted the number of rabbits every year for 5 years. When he plotted a graph of the number of rabbits against time, he got a straight line. He used this straight line to predict the number of rabbits in the future. The straight line provides a model for the rabbit population. In science, a model doesn’t usually mean a physical object – it’s simply something we use to explain or predict what happens around us.
Unfortunately, the predictions of the model were incorrect. The reason was that, a few years later, foxes came to the farm and started to eat the rabbits. The biologist’s model did not allow for this possibility and so it didn’t work.
I recently asked a group of students what they would have done if they had been the biologist – most of them decided that they would need to develop a more complicated model that allowed for all the factors that could influence the rabbit population.
This approach won’t work! Just think of all the factors that could influence the rabbit population – the population of other predators (wolves, snakes, hawks and other animals, as well as foxes), the prevalence of diseases (like myxomatosis), the population of alternative food for the predators (chickens, mice and lots of others), the supply of food for the rabbits, the occurrence of poisonous plants mixed in with the food plants, the effect of the weather on the growth of food plants and poisonous plants – you can never be sure that you’ve thought of all of them.
It gets worse. To develop such a model, you need reliable numbers to describe how all these factors interact with each other. However, when we measure these numbers, there will be some uncertainty associated with them (see posts 16.24 and 16.26). All these uncertainties will combine, so that the model makes imprecise predictions.
We need to keep our models simple. Let’s suppose we want to predict the speed at which an object of mass m, falling from a height h, hits the ground. The simplest model for this process is that the falling object converts gravitational potential energy, mgh, where g is the strength of the gravitational field (post 16.16), into kinetic energy ½mv2, where v is the speed at which the object hits the ground (see post 16.21). If none of the potential energy is used in other processes
mgh = ½mv2.
If we divide both sides of this equation by m and then multiply both sides by 2, we get the results that
v2 = 2gh.
So, the speed at which the object hits the ground is given by
v = √(2gh).
Here the “square root” sign (√) means the number we have to multiply by itself to get the number in the brackets; for example √(25) = 5 because 5 × 5 = 25.
This model for the speed of a falling object is based on the principle of conservation of mechanical energy (post 16.21). Are we sure that all the mechanical energy is conserved by a falling object? What is the effect of air resistance (drag) on the falling object? Drag is important in design of cars to improve efficiency and reduce fuel consumption. It depends on the shape and dimensions of the object and the viscosity of air. We are familiar with liquids like lubricating oils and treacle as having viscosity – air has viscosity too. Overcoming drag is similar to overcoming friction when two solids are in contact – the difference is that here a solid object is in contact with air. Temperature affects the viscosity of air; it also affects the dimensions of the falling object (post 16.35). Temperature differences in the air will cause convection currents that could affect the motion of objects.
If we want to calculate the speed of a falling feather, we might have to think about some of these factors; if we want to calculate the speed of a falling lump of metal, we ignore drag and all the associated factors. We know that these factors exist but that they are negligible. We use the simplest model that works – a model based on the principle of conservation of mechanical energy.
So, what did the biologist, in the story of post 16.8, do wrong? He ignored a factor, the possibility that foxes would come to the farm, without thinking about whether it could be safely neglected.
What should he have done? He should have performed an experiment to test his straight-line model (post 16.3) before he made any predictions. An experiment can’t prove that a model is correct but it can show if it doesn’t work (see post 16.3). Testing is an essential step in developing any scientific model; we’ll return to this subject in the next post.
Related posts
Follow-up posts | 1,028 | 4,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.965495 |
http://orms.pef.czu.cz/text/ExDualProblem.html | 1,505,951,617,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687582.7/warc/CC-MAIN-20170920232245-20170921012245-00020.warc.gz | 257,961,667 | 1,824 | Example : Dual Problem
We will now construct the dual to Example : Primal Problem:
Interpretation of Dual Constraints
The first inequality states, that the time to produce product A on machine 1 (two hours) times the opportunity cost per hour of using machine 1 (u1) plus the time to produce product A on machine 2 (one-half hour) times the cost per hour of using machine 2 (u2) is greater than or equal to \$16. The \$16 is the net profit of a unit of A (see the profit function of the primal). Thus, the opportunity cost of producing A is going to be either equal to the net profit (in which case A will be produced) or greater than the net profit (in which case no units of A will be produced and the resources will be used elsewhere to attain higher profit).
The interpretation of the second constraint is similar. The total cost per unit of producing product B is 0.5u1, (cost of using machine 1) plus 1.5u2 (cost of using machine 2). The total cost per unit is equal to or greater than \$6, where \$6 is the net increase in profit per unit of product B. Thus the cost of producing B is going to be either equal to the net profit per unit of B (in which case B will be produced) or greater than the net profit (in which case no units of B will be produced).
It should be noted that the form of solution does not allow the opportunity costs of producing either product to be less than the incremental profit of the product. This is reasonable, since the value of the machine-hours is measured by the profit they can produce. To have the total costs less than the profit would imply that we should produce more units of the product; but if the product is produced to the limit of productive capacity, the costs of the last unit will be equal to the profit. The only time the costs will be greater than the incremental profit will be when it is not desirable to produce any units of the product. Remember that these are opportunity, not accounting, costs.
The constraint equations, expanded to include slack variables and artificial variables, are:
The large coefficient M is assigned to u5 and u6, in the expanded cost equation to drive these two variables from the solution; this is done because u5 and u6 are artificial variables. The final simplex solution to this dual LP see Dual Solution. | 504 | 2,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-39 | latest | en | 0.911411 |
https://en.wikisource.org/wiki/Flatland_(second_edition)/Section_15 | 1,553,197,854,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.29/warc/CC-MAIN-20190321193403-20190321215403-00554.warc.gz | 485,830,957 | 11,916 | # Flatland (second edition)/Section 15
§. 15.—Concerning a Stranger from Spaceland.
From dreams I proceed to facts.
It was the last day of the 1999th year of our era. The pattering of the rain had long ago announced nightfall; and I was sitting[1] in the company of my wife, musing on the events of the past and the prospects of the coming year, the coming century, the coming Millennium.
My four Sons and two orphan Grandchildren had retired to their several apartments; and my Wife alone remained with me to see the old Millennium out and the new one in.
I was rapt in thought, pondering in my mind some words that had casually issued from the mouth of my youngest Grandson, a most promising young Hexagon of unusual brilliancy and perfect angularity. His uncles and I had been giving him his usual practical lesson in Sight Recognition, turning ourselves upon our centres, now rapidly, now more slowly, and questioning him as to our positions; and his answers had been so satisfactory that I had been induced to reward him by giving him a few hints on Arithmetic as applied to Geometry.
Taking nine Squares, each an inch every way, I had put them together so as to make one large Square, with a side of three inches, and I had hence proved to my little Grandson that—though it was impossible for us to see the inside of the Square—yet we might ascertain the number of square inches in a Square by simply squaring the number of inches in the side: "and thus," said I, "we know that 32 or 9, represents the number of square inches in a Square whose side is 3 inches long."
The little Hexagon meditated on this awhile and then said to me: "But you have been teaching me to raise numbers to the third power; I suppose 33 must mean something in Geometry; what does it mean? " "Nothing at all," replied I, "not at least in Geometry; for Geometry has only Two Dimensions." And then I began to show the boy how a Point by moving through a length of three inches makes a Line of three inches, which may be represented by 3; and how a Line of three inches, moving parallel to itself through a length of three inches, makes a Square of three inches every way, which may be represented by 32.
Upon this, my Grandson, again returning to his former suggestion, took me up rather suddenly and exclaimed, "Well, then, if a Point by moving three inches, makes a Line of three inches represented by 3; and if a straight Line of three inches, moving parallel to itself, makes a Square of three inches every way, represented by 32; it must be that a Square of three inches every way, moving somehow parallel to itself (but I don't see how) must make a Something else (but I don't see what) of three inches every way—and this must be represented by 33."
"Go to bed," said I, a little ruffled by his interruption; "if you would talk less nonsense, you would remember more sense."
So my Grandson had disappeared in disgrace; and there I sat by my Wife's side, endeavouring to form a retrospect of the year 1999 and of the possibilities of the year 2000, but not quite able to shake off the thoughts suggested by the prattle of my bright little Hexagon. Only a few sands now remained in the half-hour glass. Rousing myself from my reverie I turned the glass Northward for the last time in the old Millennium; and in the act, I exclaimed aloud, "The boy is a fool."
Straightway I became conscious of a Presence in the room, and a chilling breath thrilled through my very being. " He is no such thing," cried my Wife, "and you are breaking the Commandments in thus dishonouring your own Grandson." But I took no notice of her. Looking round in every direction I could see nothing; yet still I felt a Presence, and shivered as the cold whisper came again. I started up. "What is the matter.?" said my Wife, "there is no draught; what are you looking for } There is nothing." There was nothing; and I resumed my seat, again exclaiming, " The boy is a fool, I say; 33 can have no meaning in Geometry." At once there came a distinctly audible reply, "The boy is not a fool; and 33 has an obvious Geometrical meaning."
My Wife as well as myself heard the words, although she did not understand their meaning, and both of us sprang forward in the direction of the sound. What was our horror when we saw before us a Figure! At the first glance it appeared to be a Woman, seen sideways; but a moment's observation shewed me that the extremities passed into dimness too rapidly to represent one of the Female Sex; and I should have thought it a Circle, only that it seemed to change its size in a manner impossible for a Circle or for any Regular Figure of which I had had experience.
But my Wife had not my experience, nor the coolness necessary to note these characteristics. With the usual hastiness and unreasoning jealousy of her Sex, she flew at once to the conclusion that a Woman had entered the house through some small aperture. "How comes this person here?" she exclaimed, "you promised me, my dear, that there should be no ventilators in our new house." "Nor are there any," said I; "but what makes you think that the stranger is a Woman? I see by my power of Sight Recognition———" "Oh, I have no patience with your Sight Recognition," replied she, " "Feeling is believing' and 'A Straight Line to the touch is worth a Circle to the sigh '"—two Proverbs, very common with the Frailer Sex in Flatland.
"Well," said I, for I was afraid of irritating her, "if it must be so, demand an introduction." Assuming her most gracious manner, my Wife advanced towards the Stranger, "Permit me, Madam, to feel and be felt by———" then, suddenly recoiling, "Oh! it is not a Woman, and there are no angles either, not a trace of one. Can it be that I have so misbehaved to a perfect Circle?"
"I am indeed, in a certain sense a Circle," replied the Voice, "and a more perfect Circle than any in Flatland; but to speak more accurately, I am many Circles in one." Then he added more mildly, "I have a message, dear Madam, to your husband, which I must not deliver in your presence; and, if you would suffer us to retire for a few minutes———" But my Wife would not listen to the proposal that our august Visitor should so incommode himself, and assuring the Circle that the hour for her own retirement had long passed, with many reiterated apologies for her recent indiscretion, she at last retreated to her apartment.
I glanced at the half-hour glass. The last sands had fallen. The second Millennium had begun.
1. When I say "sitting," of course I do not mean any change of attitude such as you in Spaceland signify by that word; for as we have no feet, we can no more "sit" nor "stand " (in your sense of the word) than one of your soles or flounders. Nevertheless, we perfectly well recognise the different mental states of volition implied in "lying," "sitting," and "standing," which are to some extent indicated to a beholder by a slight increase of lustre corresponding to the increase of volition. But on this, and a thousand other kindred subjects, time forbids me to dwell. | 1,626 | 7,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-13 | latest | en | 0.986049 |
https://metanumbers.com/1410 | 1,679,441,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00425.warc.gz | 450,888,291 | 7,488 | # 1410 (number)
1,410 (one thousand four hundred ten) is an even four-digits composite number following 1409 and preceding 1411. In scientific notation, it is written as 1.41 × 103. The sum of its digits is 6. It has a total of 4 prime factors and 16 positive divisors. There are 368 positive integers (up to 1410) that are relatively prime to 1410.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 6
• Digital Root 6
## Name
Short name 1 thousand 410 one thousand four hundred ten
## Notation
Scientific notation 1.41 × 103 1.41 × 103
## Prime Factorization of 1410
Prime Factorization 2 × 3 × 5 × 47
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 1410 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,410 is 2 × 3 × 5 × 47. Since it has a total of 4 prime factors, 1,410 is a composite number.
## Divisors of 1410
1, 2, 3, 5, 6, 10, 15, 30, 47, 94, 141, 235, 282, 470, 705, 1410
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 3456 Sum of all the positive divisors of n s(n) 2046 Sum of the proper positive divisors of n A(n) 216 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 37.55 Returns the nth root of the product of n divisors H(n) 6.52778 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,410 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 1,410) is 3,456, the average is 216.
## Other Arithmetic Functions (n = 1410)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 368 Total number of positive integers not greater than n that are coprime to n λ(n) 92 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 228 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 368 positive integers (less than 1,410) that are coprime with 1,410. And there are approximately 228 prime numbers less than or equal to 1,410.
## Divisibility of 1410
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 3 2 6
The number 1,410 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Practical
• Square Free
## Base conversion (1410)
Base System Value
2 Binary 10110000010
3 Ternary 1221020
4 Quaternary 112002
5 Quinary 21120
6 Senary 10310
8 Octal 2602
10 Decimal 1410
12 Duodecimal 996
20 Vigesimal 3aa
36 Base36 136
## Basic calculations (n = 1410)
### Multiplication
n×y
n×2 2820 4230 5640 7050
### Division
n÷y
n÷2 705 470 352.5 282
### Exponentiation
ny
n2 1988100 2803221000 3952541610000 5573083670100000
### Nth Root
y√n
2√n 37.55 11.2135 6.1278 4.26426
## 1410 as geometric shapes
### Circle
Diameter 2820 8859.29 6.2458e+06
### Sphere
Volume 1.17421e+10 2.49832e+07 8859.29
### Square
Length = n
Perimeter 5640 1.9881e+06 1994.04
### Cube
Length = n
Surface area 1.19286e+07 2.80322e+09 2442.19
### Equilateral Triangle
Length = n
Perimeter 4230 860873 1221.1
### Triangular Pyramid
Length = n
Surface area 3.44349e+06 3.30363e+08 1151.26
## Cryptographic Hash Functions
md5 512c5cad6c37edb98ae91c8a76c3a291 8d93c9b0486a80bc7b6909df5440d0c190eccd4e 4103f0a4e707b1c7bebbc42809ab0ace8dd3f56d844d7903bfe9f95a2ccc6972 e57d0f76f2ca0d41959c6878e7004b0027e3ef0f75d9cfbb03bab02b5171a541c92297536b6af90ca5634cf085659ed2c4a371ee163bc29609e05e29bb0cee08 3a9505dd2e4cb88d244a37740d363cfe6d4c741e | 1,456 | 3,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-14 | latest | en | 0.818119 |
https://www.jiskha.com/questions/1662288/write-the-equation-in-logarithmic-form-n-4-3-m-can-someone-explain-how-to-do-this | 1,623,569,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00080.warc.gz | 766,880,086 | 4,438 | # Algebra II
Write the equation in logarithmic form.
n^4/3=m
Can someone explain how to do this?
1. 👍
2. 👎
3. 👁
1. If you mean
n^(4/3) = m
then
lognm = 4/3
or,
(4/3)log n = log m
1. 👍
2. 👎
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1. Use point-slope form to write the equation of a line that has a slope of 2/3 and passes through (-3, -1). Write your final equation in slope-intercept form. 2. Write the equation in standard form using integers (no fractions or | 627 | 2,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.858627 |
https://www.fresherslive.com/placement-papers/mindtree-interview-questions-answers/6 | 1,542,532,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00514.warc.gz | 895,166,404 | 44,176 | # Mindtree Question Paper 2013
Posted on :17-02-2016
Q1. A person losses and gains 10% on selling a object for 200.
ANS: he losses
Q2. Sum of 1 to 100 is divisible by
1. 1,2,4,8
2. 2 & 4
3. 2
4. none
ANS: 2
Q3. How many nos start and end with 2 b/w 100 and 300?
Q4. If a sphere of diameter 3 cm is melted & formed into 3 spheres , the diameter of 1st is 1.5 cm and that of second is 2.0 cm , what is the diameter of the third?
Q5. If prizes are increased by 25% , by how much should i reduce the consumption to keep the expenditure same?
Q6. COURTESY - how many words can be constructed with C in the begining and Y at the end
Q7. My mothers husbands father in laws sons child->what is the relation?
Q8. 3 glasses containing mixture of water and also in ratio 2:3 , 3:4 , 5:9 when all 3 are mixed what is the new ratio?
Q9. A father has 8 children ,he takes 3 at a time to a zoo.probability of a child going to the zoo.
Q10. A ball is dropped from a height of 10 feet.Every time it rebounces to half of the height. how many feet it travelled?
Q11. There are 3 jars. The ratio of spirit to water in each of these jars is 3:2,4:5,5:7. the three jars are mixed into a single jar. What is the ration of spirit to water in single jar.
Q12. What is the relation with your mothers sisters brothers wifes child with you?
Q13. A father has six children .all the children are born at regular intervels. if the sum of their ages of all the children and father is 186. calculate the age of the elder son, when the younger sons age is 3.
Q14. All elephants are trained a few animals are trained - syllogism questions
Q15. What is the 7th digit of (202)3
Q16. To type the digits from 1-1000 how many key strokes are needed?
Q17. A problem on the margin of votes involving winning percentage between two candidates.
Q18. Time & Distance- A is running faster than B by .. kms. How much increase of speed is needed by B to become first?
Q19. Problem on three pipes- two of them loading water into a tank the other one emptying, what time will it take to fill the tank? (2 problems)
Q20. Four horses are tied up by rope on four corners of a square grass field. What would be the ungrazed area by them?
Q21. If a person gets A% profit on cost price A. Determine the cost price? Four alternatives were given --Ans. 40. As the thing was sold at 56/cost price--40.
Q22. Determine the angle between the hour & minute hand when the time is 40 past 3.
Q23. There is a Square. One Circle which is touching all the arms of the square is inscribed in it. Another smaller square which ahs all corners touching the circle is inscribed inside the circle. Obtain the ratio of the area of the two squares.
Q24. In a zoo there are animals & birds. No of heads are, No of legs, obtain the number of birds & animals.
Q25. Simple aptitude type problems on probability. (2 problems)
Q26. Problems on percentage.
Q27. The C program was on simple binary tree.
GD Topic
Q1. Westernization of INDIA-- right or wrong
Q2. IT boom in INDIA blessing or curse.
Q1. We have to count the 3 letter,4 letter and 5 letter words from a file and print the number of 3 letter,4 letter and 5 letter words. Delimiter is space, tab, hifen. Also we should not consider the line in the file after we encounter # in that line. (ie after # we should not consider the portion of line)
Q2. In file count the number of characters, number of words, number of line. Delimiter between words is given as command line argument. 1hour 15 min is given for the test. we have to answer both aptitude and program with in this time. Some of the aptitude questions. 1. the is a meeting organized, every person shake hands with the other only once. if there are 60 shake hands, how many persons are there in the meeting?. 2.A faster has 8 children ,he takes 3 at a time to a zoo. probability of a child going to the zoo.
Q3. A father has six children .all the children are born at regular intervels. if the sum of their ages of all the children and father is 186. calculate the age of the elder son, when the younger sons age is 3.
Q4. Count the numbers between 100 and 300, that starts with 2 and ends with 2.
ANS: 10.
Some reasoning questions :
Q1. There are six steps,5 people a,b,c,d,e . conditions are , a is two steps below c,no two people are on same step,b is next to d. 4 Questions based on this.
Q2. 5 people participating in the race .conditions are given about the positions.
Q3. A ball is dropped from a height of 10 feet. Every time it rebounces to half of the height. how many feet it traveled?
Q4. There are 3 jars. The ratio of spirit to water in each of these jars is 3:2,4:5,5:7. The three jars are mixed into a single jar. What is the ratio of spirit to water in single jar.
Q5. What is the relation with your mothers sisters brothers wifes child with you?
Q6. What is the relation with your mothers husbands father-in-laws sons child with you?
Q7. We were asked to fill the series, alphabet series 4 questions are given.
Q8. 5 questions are given, we are asked to find the logical connection between them. like all managers are drinkers, all drinker are smokers, questions like this.
Simple Mathematics:
Q1. A triangle on a grid was given : find the area of it.
ANS: 2
Q2. Out of the four options which is not a right angle .
ANS: D (12,9,16)
Q3. In a class of 100 members 50 plays soccer 45 badminton 50 volley ball and 15 plays all the three then how many of them play two games.
ANS: Cant be determined (D) verify
Q4. In a class the no. of boys are more than that of girls by 25% of the total strength of the class then the ratio of boys to girls.
ANS: (5:3)(d)
Q5. The radius of the circle is reduced from 5 cm to 4 cm then the % change of area .
ANS: 36%
Q6. Two workers can type two pages in two minuets then how many persons can type 18 pages in 6 minuets
ANS: 6
Q7. If p+q+r = 0 then p^2 / qr + q^2 / pr + r^2 / pq=?
ANS: 3
Q8. The sum of 7 consecutive numbers is 1617 then the number of prime numbers in it.
ANS: 2
Q9. In a nock out tournament the total number of games played are 63 then the no. of participants
ANS: 64
Q10. Fresh mango consists of 70% water and dry mango consists of 10% water then 20 Kg of fresh mango is equivalent to how much of dry mango
ANS: 6.66 kg
Q11. The number of 3 digit numbers which when divide by 7 or 8 gives a reminder of 4
a) 16
b) 17
c) 18
d) 19
Set II
Q1) A man sold a product for rs.56 and gained x% .Its original cost is rs. x, what is the cost?
Q2) 2 power 300,3 power 200 which is greater?
Q3) How many zeros are there from 1 to 10000?
Q4) If a and b do a job together in 6 days b and c together in 10 days a and c together in 7.5 days then: no of days when i) a,b,c together ii) if a alone does the job.
Q5) There are 12 yes/no types of questions ,in how many ways can they be answered?
Q6) Find odd one out of 125,8 power 3,521,729
Q7) One got 20 % marks and failed by 10 marks, other got 42% marks and got 12 % more than the passing requirement What is the maximum marks?
Q8) --,32,224,
Q9) Rs 50000 is divided into two parts One part is given to a person with 10% interest and another part is given to a person with 20 % interest. At the end of first year he gets profit 7000 Find money given by 10%?
Q10) Time shown by a watch is 2.20 Find the angle between them
Q11) Avg cost of 5 apples+4 mangoes=Rs.36 Avg cost of 7 apples + 8
mangoes=Rs.48 Find total cost of 24 mangoes and 24 apples
Q12) Management : How ways it can be arranged without 2 as as continuous
GD TOPICs
Q1) Working in MNCs (+ve and -ve)
Q2) Working with Spouse in the same company (+ve and -ve)
1. An expression was given like B + A .B + !A .B + . . . . . and what will be the final expression o/p :- ( Easy )
2. state diagram was given and what state will be detected?
Q3. Counter and three states were given( 13 , 15 , 14 , 10 ) what will be the next value ?
ANS: 11 Gray counter
Q4. For one circuit inputs are ABCD and output is zero for first eight inputs and 1 for next eight . What will be the expression for circuit ?
ANS: A
Q5. What is octal representation of 01010101?
ANS: 125
Q6. D flip-flop with inverted feedback was given and question was based on maximum frequency allowed (setup , hold etc. times were given)
Q7. Question based on instruction T states and CPI was asked .
Q8. What happens when HALT instruction is executed? (Enter wait state and waits for interrupt or reset , Enters in to sleep state , etc. ( ans. Waits for interrupt or reset )
Q9. What happens when PUSH instruction is executed : (content of accumulator is saved,content of PC is changed , stack pointer is decremented and value value is saved)
Q10. Two bulbs of 40 Watt and 60 Watts were added in series. What will happen?
a) 40 watt will be brighter
b) 60 watts will be brighter
c) both will have same brightness
d) none
ANS: 40 watt
Q11. If widths of all PMOS an NMOS are doubled in any circuit then what will be effect onnoise margin?
ANS: No effect
Q12. If gate and one end of NMOS are connected to Vdd , what will be the output at second end of NMOS? ( Vdd- Vth NMOS are bad 1 s)
Q13. A band limited signal is sampled at Nyquist rate , to reconstruct it the output will be given to: ( RC filter , Envelope detector , LPF with -- )
Q14. What is expression for Noise figure? Always @ ( B or C) if( B) D=C Circuit will work as
a) Mux
b) Tri state buffer
c) Latch
d) None
ANS: Tri state Buffer
Q1. A man crosses a circle of perimeter 35 m in 44 sec and he runs a hexagon of 42 m side. In how many second he finishes the complete perimeter of the hexagon.
Q2. A+b do a work in 6 days. A+c do the work in 10 days, c+a in 7.5 days. How many days will it take for a+b+c to do the work and also find the number of days required by a alone.
Q3. Totally there are 12 yes or no questions. In how many different ways can he attend?
Q4. Complete the series 7 49 56 392
Q5. Complete the series 36 252 63 441 110.25..
Q6. Find the odd one out: 8 125 512 729?
Q7. Number of zeros between 0 and 10000
Q8. A regular hexagon is inscribed in a circle. What is the perimeter of the hexagon in relation with the radius of the circle?
Q9. A circular path has a perimeter of 120 m.a, b,c travel at 5 7 15 m/s respectively. They start simultaneously. At what time do they meet next?
Q10. Totally there are 12 pipes. Some pipes fill the tank and some empties the tank (data will be given). If all the pipes are opened simultaneously when will the tank be filled?
Q11. Which is greater 2^300 or 3^200?
Q12. Average age of 24 students is 36. When the age of teacher is added the average age is increased by 1 year. What is the age of the teacher?
Q13. Water and milk are mixed in the ratio 2:1,5:3 and 9:4.when all the mixtures of equal volume are mixed together what is the ratio water and milk in the resulting solution.
Q14. Average rate of five orange and four apples is 12 and 7 orange and 8 apples are 86. Find the total rate of 24 orange and 24 apples.
Q15. In how many ways the letters in the word MANAGEMENT can be arranged such that As does not come together.
Q16. A give B a start of 20 sec and leads by 20m. When he starts off 25 sec he finishes at the same time. What is the speed of A. (wordings not sure).
Q17. An array of integer is given. Write a pseudo code to find the sub array with the largest sum.
Q18. A string is given. Write an algorithm to implement the backspace from the end of the string. The string contains 1 byte and 2-byte character. For 1 byte char first bit is 0 and for 2 byte char first bit is 1.
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✖ | 3,384 | 12,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-47 | latest | en | 0.954154 |
https://www.ncl.ac.uk/mobility/newcastle/study-abroad/MAS2905 | 1,632,014,506,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00433.warc.gz | 934,477,536 | 10,905 | Global Opportunities
### MAS2905 : Probability & Bayesian Methods with R (Inactive)
• Inactive for Year: 2021/22
• Incoming Study Abroad and Exchange students must contact and seek approval from school prior to enrolling
• Module Leader(s): Dr Lee Fawcett
• Owning School: Mathematics, Statistics and Physics
• Teaching Location: Newcastle City Campus
##### Semesters
Semester 1 Credit Value: 10 Semester 2 Credit Value: 10 ECTS Credits: 10.0
#### Aims
To introduce and reinforce a range of concepts in probability and statistics with particular emphasis on illustrations in R, including methods that will be useful towards future project work. To reinforce the computing in R studied within MAS1802, and to move towards expectations of more independent programming.
Students will also learn about the Bayesian approach to statistical analysis. Students will be able to explain the distinctive features of Bayesian methodology, understand the role of prior distributions and compute posterior distributions in simple cases.
Module summary
Computational methods are of great use in a wide range of applications of probability and statistics. This module builds on the probability introduced in MAS1604 and the use of R introduced in MAS1802. Students will be introduced to additional concepts and techniques, some of increasing mathematical and computational sophistication. In implementing these methods, students will attain a deeper understanding of foundational probability and statistics, increasing competence with mathematical/statistical computing, and an increasing ability to use such methods independently, towards project-orientated goals.
The second part of the module will be devoted to an introduction to Bayesian methods, in which the prior and posterior distributions of a scalar parameter will be defined. The use of the likelihood to allow the prior distribution to be updated to the posterior distribution will be discussed. The use of Bayes’ theorem to compute posterior distributions from given priors and likelihoods will be described, with particular emphasis given to the case of conjugate distributions.
#### Outline Of Syllabus
Review of probability ideas: illustrations of properties of univariate, bivariate and trivariate distributions, including use of conditional distributions. Transformations of random variables. Sampling distributions. Illustration of properties of hypothesis tests and confidence intervals.
Introduction to the Bayesian approach: subjective probability; likelihood; sufficiency. Inference for populations using random samples and conjugate priors, including posterior estimates and highest density intervals: inference for the mean of a normal distribution with known variance; inference for parameters in other commonly used distributions. Sequential use of Bayes' Theorem. Parameter constraints. Mixture prior distributions. Asymptotic posterior distribution.
#### Teaching Methods
##### Teaching Activities
Category Activity Number Length Student Hours Comment
Scheduled Learning And Teaching ActivitiesLecture61:006:00Present in Person
Scheduled Learning And Teaching ActivitiesLecture91:009:00Synchronous On-Line Material
Guided Independent StudyAssessment preparation and completion301:0030:00N/A
Structured Guided LearningLecture materials361:0036:00Non-Synchronous Activities
Scheduled Learning And Teaching ActivitiesPractical31:003:00Present in Person
Structured Guided LearningStructured non-synchronous discussion181:0018:00N/A
Scheduled Learning And Teaching ActivitiesDrop-in/surgery41:004:00Office Hour or Discussion Board Activity
Guided Independent StudyIndependent study941:0094:00N/A
Total200:00
##### Teaching Rationale And Relationship
Non-synchronous online materials are used for the delivery of theory and explanation of methods, illustrated with examples, and for giving general feedback on assessed work. Present-in-person and synchronous online sessions are used to help develop the students’ abilities at applying the theory to solving problems and to identify and resolve specific queries raised by students, and to allow students to receive individual feedback on marked work. Students who cannot attend a present-in-person session will be provided with an alternative activity allowing them to access the learning outcomes of that session. In addition, office hours/discussion board activity will provide an opportunity for more direct contact between individual students and the lecturer: a typical student might spend a total of one or two hours over the course of the module, either individually or as part of a group.
Alternatives will be offered to students unable to be present-in-person due to the prevailing C-19 circumstances.
Student’s should consult their individual timetable for up-to-date delivery information.
#### Assessment Methods
The format of resits will be determined by the Board of Examiners
##### Exams
Description Length Semester When Set Percentage Comment
Written Examination602A50Alternative assessment - class test
##### Other Assessment
Description Semester When Set Percentage Comment
Prob solv exercises1M40N/A
Prob solv exercises2M10N/A
##### Assessment Rationale And Relationship
A substantial formal examination is appropriate for the assessment of the material in this module. The course assessments will allow the students to develop their problem solving techniques, to practise the methods learnt in the module, to assess their progress and to receive feedback; these assessments have a secondary formative purpose as well as their primary summative purpose. | 1,033 | 5,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-39 | latest | en | 0.880394 |
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```Random Latin square graphs
Demetres Christofides
Klas Markström
Abstract
In this paper we introduce new models of random graphs, arising from Latin
squares and including random Cayley graphs as a special case. We investigate
some properties of these graphs including their clique, independence and chromatic numbers, their expansion properties as well as their connectivity and
Hamiltonicity. The results obtained are compared with other models of random graphs and several similarities and differences are pointed out. For many
properties our results for the general case are as strong as the known results
for random Cayley graphs and sometimes improve the previously best result
for the Cayley case.
1
Introduction
The concept of random graphs is a very important notion in combinatorics. Although
there are several models of random graphs, by a random graph one usually refers to
the model G (n, p), the probability space of all graphs on [n] in which every edge
appears independently with probability p. For standard results on random graphs we
refer the reader to the textbooks of Bollobás [7] and Janson, Luczak and Ruciński [14].
In this paper, we introduce new models of random graphs which generalise the random
Cayley graphs and study some of their properties with particular interest in their
relation to the model G (n, p). Our models arise from Latin squares. Given a group,
one can obtain Latin squares by considering its multiplication table or its division
table. It turns out that the random graph obtained by the division table of a group
G, is exactly the random Cayley graph of G (with respect to a random subset S of
G.)
Before defining our models, let us recall that a Latin square of order n is an n × n
matrix L with entries from a set of n elements, such that in each row and in each
column, every element appears exactly once. Given a Latin square L with entries in
a set A of size n, and a subset S of A, we define the Latin square graph G(L, S)
on vertex set [n], by joining i to j if and only if either Lij ∈ S, or Lji ∈ S.
Suppose we are given a sequence (Ln ) of Latin squares of order n, with entries in
[n], say. Choosing S ⊆ [n] by picking its elements independently at random with
probability p, we obtain a random Latin square graph G(Ln , S). We denote this
model of random Latin square graphs by G (Ln , p). A related model is obtained by
1
choosing a multiset S of k elements of [n] by picking its elements independently and
unifomly at random (with replacement). We denote this model by G (Ln , k). Note
that our underlying graphs are simple. However, for the model G (Ln , k) it will be
convenient for some of our results to retain multiple edges and loops. When we do
this, we will denote this new model by Gm (Ln , k). To be more explicit, in this model
the number of edges joining i to j is exactly the total number of times that Lij and
Lji appear in S. In particular, every G ∈ Gm (Ln , k) is a 2k-regular multigraph.
A similar model is obtained by looking at the complement of the graph G ∈ G (Ln , p).
We denote this model by G¯(Ln , p). In general, this model is not the same as G (Ln , 1−
p), the reason being that Lij is not necessarily equal to Lji . However, usually it is
not too difficult to translate results from one model into the other, so we will only
concentrate on G ∈ G (Ln , p).
Note that, as mentioned above, our models include random Cayley graphs as a special
case. Indeed, given a group G, consider the Latin square L defined by Lxy = xy −1 .
Then, given any subset S of elements of G, the Latin square graph G(L, S) is exactly
the Cayley graph of G with respect to S. The multiplication table of a group is also a
Latin square, giving rise to what is usually known (motivated by the abelian case) as
a Cayley sum graph. So this model includes random Cayley sum graphs as well.
We should mention here that there are several differences between random Cayley
graphs and our more general models of random Latin square graphs. For example,
random Cayley graphs are always vertex transitive. On the other hand a random
Latin square graph, even if it arises from the multiplication table of a (non-abelian)
group, might not even be regular. However, it is easy to see that random Latin
square graphs are not far from being regular in the sense that the ratio of maximum
to minimum degree is bounded above by 2.
The fact that random Latin square graphs are almost regular (in the above sense)
motivates also the comparison of our models with Gn,r , the probability space of all
r-regular graphs on n vertices taken with the uniform measure. (As usual, it is always
assumed that rn is even.)
Sometimes, it is easier to work with random Cayley graphs or random Cayley sum
graphs for abelian groups, rather than random Latin square graphs. This is because
we always have Lij = L−1
ij in the case of Cayley graphs, and Lij = Lji in the case
of Cayley sum graphs, and so dependences between the edges can be easier to deal
with. This sometimes leads to sharper results for the first two families of random
graphs than for general random Latin square graphs; however we have opted to state
our results only in the general case of random Latin square graphs.
It seems that the general class of random graphs arising from Latin squares have
not been studied before. However there has been much interest in random Cayley
graphs and random Cayley sum graphs. For example, Agarwal, Alon, Aronov and
Suri [1] established an upper bound on the clique number of random Cayley graphs
2
arising from cyclic groups and used it to construct visibility graphs of line segments in
the plane which need many cliques and complete bipartite graphs to represent them.
In their study of a communication problem, Alon and Orlitsky [5] proved a similar
upper bound for random Cayley graphs arising from abelian groups of odd order.
Green [12], using number theoretic tools, studied the clique number of various Cayley sum graphs and showed that some of them are good examples of Ramsey graphs
while others are not. The diameter of random Cayley graphs with logarithmic degree
was studied by Alon, Barak and Manber in [2]. Alon and Roichman [6] proved that
random Cayley graphs (on sufficiently many generators) are almost surely expanders,
a result which was later improved by several authors [18, 19, 8]. The fact that random
Cayley graphs are expanders has several consequences for the diameter, connectivity
and Hamiltonicity of such graphs. Finally, some other aspects of the diameter, connectivity and Hamiltonicity of random Cayley graphs and random Cayley digraphs
were studied in [23, 22, 20, 21].
In this paper we extend many of these resutls to the general case of random Latin
square graphs and show that the structure of the Latin squares have a non-trivial
influence on many properties of random Latin square graphs. In Section 2 we state
and discuss our main results regarding random Latin square graphs. We prove these
results in Section 3, Section 4 and Section 5. In Section 6 we give further examples
and open problems.
2
Statements and discussion of the results
In this section, we list our main results and make a few comments about them,
comparing them with the corresponding results in the G (n, p) and Gn,r models. In
Subsection 2.1, we will be interested in the maximum size of cliques and independent
sets in Gn,p , as well as the chromatic number of Gn,p and its complement. In Subsection 2.2, we will be interested in the expansion properties of random Latin square
graphs as well as several consequences of these properties regarding connectivity and
Hamiltonicity. For the results of this subsection it will be easier to work in the the
models Gm (Ln , k) and G (Ln , k).
2.1
Cliques, independent sets and colouring
We begin with an upper bound on the clique number of random Latin square graphs.
It is well known that the clique number of G (n, 1/2), is whp asymptotic to 2 log2 n.
For the case of dense random regular graphs, it was proved in [17] that the clique
number of Gn,n/2 is whp asymptotic to 2 log2 n.
Guided by the above results, one might hope to prove that the clique number of
G (Ln , 1/2) is whp Θ(log n). However, it turns out that this is not the case. Green [12]
proved that the clique number of the random Cayley sum graph on Zm
2 , with p = 1/2,
m
m
is whp Θ(log n log log n), where n = 2 = |Z2 |. In the same paper, Green proved
3
that the clique number of the random Cayley sum graph on Zn , with p = 1/2, is whp
Θ(log n). This shows that, in general, results about the model G (Ln , p) can depend
on the actual sequence of Latin squares chosen.
To the best of our knowledge, the best known general result on the clique number is
due to Alon and Orlitsky [5], which says that the clique number of a random Cayley
graph arising from an abelian group of odd order n is whp O((log n)2 ). Using similar
methods, we have managed to show that the same bound is in fact true for random
Latin square graphs. In particular, it is also true for random Cayley graphs arising
from non-abelian groups. We believe but cannot prove that the 2 in the exponent
can be reduced further. and that c log n log log n is possibly the right bound.
Theorem 1 (Clique number; upper bound). Let 0 < p < 1 be a fixed constant and
let d = 1/(2p − p2 ). Then, for almost every G ∈ G (Ln , p), we have
cl(G) 6 27 (logd n)2 .
Since the model G¯(Ln , p) is different from G (Ln , 1−p), we cannot immediately deduce
a corresponding upper bound for the independence number. One way to find such
a bound is to couple the model G¯(Ln , p) with G (Ln , 1 − p), and use Theorem 1 to
deduce that for almost every G ∈ G (Ln , p),
2
α(G) = cl(Ḡ) 6 27 log1/(1−p2 ) n .
In fact, using an argument similar to the one used in the proof of Theorem 1, we can
obtain a slightly better result.
Theorem 2 (Independence number; upper bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(1 − p). Then, for almost every G ∈ G (Ln , p), we have
α(G) 6 27 (logd n)2 .
Recall that the (vertex) clique cover number θ(G) of a graph G is the smallest
integer k such that the vertex set of G can be partitioned into k cliques. I.e. θ(G) =
χ(Ḡ). So an immediate corollary of Theorem 1 is:
Corollary 3 (Clique cover number; lower bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(2p − p2 ). Then, for almost every G ∈ G (Ln , p), we have
θ(G) >
n
.
27 (logd n)2
Similarly, Theorem 2 implies:
Corollary 4 (Chromatic number; lower bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(1 − p). Then, for almost every G ∈ G (Ln , p), we have
χ(G) >
n
.
27 (logd n)2
4
We now move to our upper bound on the chromatic number of random Latin square
graphs. Recall that for constant p, the chromatic number of G (n, p) is whp asymptotic to 2 logn n , where b = 1/(1 − p). A similar behaviour was proved in [17] for the
b
case of random regular graphs of high degree. More specifically, it was proved that
for any ε > 0, if εn 6 r 6 0.9n, then the chromatic number of Gn,r is whp asymptotic
to 2 logn n , where b = n/(n − r).
b
For the case of random Latin square graphs, we prove an upper bound of the same
order of magnitude. However, since our lower bound is only of order (logn n)2 , we
b
still do not have a sharp asymptotic result for the chromatic number. In fact, as
in the case of the clique and independence numbers, we know that the chromatic
number can depend on the sequence of Latin squares chosen. For example, the result
of Green [12] mentioned above, that the clique number of the random Cayley sum
graph on Zn (with p = 1/2) is whp Θ(log n), provides a lower bound for the chromatic
number of these graphs which is of the same order of magnitude as our corresponding
upper bound. On the other hand, we claim that the chromatic number of the random
n
m
= |Zm
Cayley sum graph on Zm
2 |. The lower
2 is whp Θ( log n log log n ), where n = 2
bound follows immediately from the result of Green [12] mentioned above for the
independence number of these graphs. The upper bound does not follow directly
from that result, however it follows from the proof in [12] that in fact there is whp
a blog m + log log m − 1c-dimensional subspace of Zm
2 which is an independent set.
Indeed, given this result, it follows that whp, a random Cayley sum graph on Zm
2
4n
can be partitioned into at most log n log
independent
sets
of
this
form.
log n
In fact, our upper bound on the chromatic number will be an immediate consequence
of an upper bound on the list-chromatic number. Recall that the list-chromatic
number χl (G) of a graph G is the smallest positive integer k such that for any
assignment of k-element sets L(v) to the vertices of G, there is a proper vertex
colouring c of G with c(v) ∈ L(v) for every vertex v of G.
Theorem 5 (List-chromatic number; upper bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(1 − p). Then, for almost every G ∈ G (Ln , p), we have
n
.
χl (G) 6 1
1
logd n − 2 logd logd n − 2
4
Corollary 6 (Chromatic number; upper bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(1 − p). Then, for almost every G ∈ G (Ln , p), we have
n
χ(G) 6 1
.
1
logd n − 2 logd logd n − 2
4
With similar methods we will show the following upper bound for the clique cover
number.
Theorem 7 (Clique cover number; upper bound). Let 0 < p < 1 be a fixed constant
and let d = 1/p. Then, for almost every G ∈ G (Ln , p), we have
n
θ(G) 6 1
.
logd n − log logd n − 6
2
5
From Theorem 6 and Theorem 7 we deduce corresponding lower bounds on the independence and clique numbers.
Corollary 8 (Independence number; lower bound). Let 0 < p < 1 be a fixed constant
and let d = 1/(1 − p). Then, for almost every G ∈ G (Ln , p), we have
α(G) >
1
logd n − log logd n − 6.
2
Corollary 9 (Clique number; lower bound). Let 0 < p < 1 be a fixed constant and
let d = 1/p. Then, for almost every G ∈ G (Ln , p), we have
cl(G) >
2.2
1
1
logd n − logd logd n − 2.
4
2
Expansion and related properties
Alon and Roichman [6] proved that random Cayley graphs on logarithmic number
of generators are expanders whp. Our main result of this subsection states that a
similar result holds in the case of random Latin square graphs.
Before stating our result we need to introduce some notation. Given a multigraph G,
its adjacency matrix is the 0,1 matrix A = A(G) with rows and columns indexed
by the vertices of G, in which Axy is the number of edges in G joining x to y. If
G is d-regular then its normalised adjacency matrix T = T (G) is defined by
T = d1 A. Note that T is a real symmetric matrix, so it has an orthonormal basis of
real eigenvectors. We will write λ0 > λ1 > . . . > λn−1 for the eigenvalues of T . It is
easy to check that λ0 = 1 and that λn−1 > −1. We will write µ for the second largest
eigenvalue in absolute value, i.e. µ = max {|λ1 |, |λn−1 |}.
Finally, for 0 < x < 1, we define
H(x) = x log (2x) + (1 − x) log (2(1 − x)),
where we use the convention that all logarithms are natural.
We can now state our main theorem.
Theorem 10 (Second eigenvalue). Let L be an n × n Latin square with entries in [n]
and let G ∈ Gm (L, k). Then, for every 0 < ε < 1,
1+ε
kε2
Pr(µ(G) > ε) 6 2n exp −kH
6 2n exp −
.
2
2
We remark that if L is the difference table of a group, then the above theorem is similar
to the result of Alon and Roichman mentioned in the beginning of this subsection.
The only difference is that the bounds appearing in the above theorem, are the same
as the bounds appearing in the authors’ proof [8] of the Alon-Roichman theorem and
are slightly better than the original bounds of the Alon-Roichman theorem.
6
Recall that a graph G is an (n, d, ε)-expander if it is a graph on n vertices with
maximum degree d such that for every subset W of its vertices of size at most n/2
we have |N (W ) \ W | > ε|W |, where N (W ) denotes the neighbourhood of W . Note
that for this definition we may ignore any multiple edges or loops that G may have.
For more on expander graphs and their applications, we refer the reader to the recent
survey of Hoory, Linial and Wigderson [13].
It is well known [25, 4] that a small second eigenvalue implies good expansion properties. The following corollary is an immediate consequence of Theorem 10 together
with this fact.
Corollary 11 (Expansion). For every δ > 0, there is a c(δ) > 0 depending only on δ,
such that almost every G ∈ Gm (Ln , c(δ) log n) is an (n, 2c(δ) log n, δ)-expander.
The fact that the second eigenvalue of the graph is small implies that such a graph
has several properties that many ‘random-like’ graphs possess. Informally, a graph of
density p is pseudorandom if its edge distribution resembles the edge distribution of
G (n, p). The study of pseudorandom graphs was initiated by Thomason in [26, 27].
Chung, Graham and Wilson [10] showed that many properties that a graph may
possess, including the property of having small second eigenvalue, are in some sense
equivalent to pseudorandomness.
Here we list just a few of these consequences, mostly taken from the recent survey of
Krivelevich and Sudakov [16]. We omit some of the proofs, but we note that some
care needs to be takne since our graphs are multigraphs, while the result in the survey
are stated only for simple graphs.
To begin with, let us consider what value of k guarantees that almost every G ∈
G (Ln , k) is connected. Let us first recall the corresponding results in G (n, p) and
Gn,r . It is well known that for any fixed δ > 0, if p 6 (1 − δ) log n/n, then G (n, p) is
whp disconnected, while if p > (1 + δ) log n/n, then G (n, p) is whp connected. On
the other hand, Gn,r is whp connected provided that r > 3.
So what is the right threshold for the connectivity of random Latin square graphs?
Once again this depends on the sequence (Ln ) of Latin squares chosen. For example,
the Cayley graph of Zq for q prime, with respect to any set S containing a non-trivial
element is connected. On the other hand, the Cayley graph of G = Zm
2 with respect
to any set of size less than m = log2 |G| is disconnected. Here, we prove that choosing
slightly more elements are enough to guarantee whp the connectedness not only of
the random Cayley graph of Zm
2 but in fact the connectedness of any random Latin
square graph.
Theorem 12 (Connectedness). For any fixed δ > 0, almost every G ∈ G (Ln , (1 +
δ) log2 n) is whp connected.
Let us now move to the vertex connectivity of random Latin square graphs. Recall
that the vertex connectivity κ(G) of a graph G is the minimal number of vertices
that we need to remove in order to disconnect G. Clearly the vertex connectivity of
7
any graph is at most its minimum degree δ(G). It is well known that for G ∈ G (n, p)
we have κ(G) = δ(G). Recently, it was shown in [17, 11] that the same holds for
random r-regular graphs provided 3 6 r 6 n − 4. In our case, the Cayley graphs
on Zm
2 show that no such result can hold if the generating set S has size less than
log2 n. Can we expect that such a result holds if the size of S is large enough? As
the following example shows the answer is no.
Example. Define a Latin square L on {0, 1, . . . , r − 1} × {0, 1} with entries in
{0, 1, . . . , 2r − 1} as follows:
(
x+y
if x 6 y
L(x,0),(y,0) =
x + y + r if x > y
(
x + y + r if x 6 y
L(x,0),(y,1) =
x+y
if x > y
(
x + y + r if x 6 y
L(x,1),(y,0) =
x+y
if x > y
(
x+y
if x 6 y
L(x,1),(y,1) =
x + y + r if x > y
Here, addition is done modulo 2r. It can be easily checked that L is indeed a Latin
square. Pick any S ⊆ {0, 1, . . . , 2r − 1} and let G = G(L, S). Note that G is
d-regular for some d. Note also that for any x ∈ {0, 1, . . . , r − 1}, we have that
NG ((x, 0)) \ {(x, 1)} = NG ((x, 1)) \ {(x, 0)}, where NG denotes the neigbourhood of a
vertex in G. But then, if (x, 0) is adjacent to (x, 1) for some x, and G is not complete,
we have that κ(G) 6 d − 1. Indeed, NG ((x, 0)) \ {(x, 1)} is a disconnecting set of size
d − 1. Now (x, 0) is adjacent to (x, 1) if and only if 2x + r ∈ S. Let p = p(r) ∈ (0, 1)
be chosen such that pr → ∞ and (1 − p)r → ∞ as r → ∞ and choose S by picking its
elements independently at random with probability p. Then whp G is not complete
and there is an x such that (x, 0) is adjacent to (x, 1) and so κ(G) 6 δ(G) − 1.
The above example shows that even if the size of S is large enough the vertex connectivity of a random Latin square qraph can be whp strictly smaller than its minimum
degree. However, our next theorem shows that if S is large enough then the vertex connectivity of a random Latin square graph is whp at most one less than its
minimum degree.
Theorem 13 (Vertex connectivity). There is an absolute constant C 6 168 such
that whenever C log n 6 k 6 n/4, then δ(G) − 1 6 κ(G) 6 δ(G) for almost every
G ∈ G (Ln , k).
The example of Zm
2 shows that we cannot take C to be equal to 1. It would be
interesting to know whether every C strictly larger than 1 works or not. It seems
8
that our proof cannot bring the value of C down to 1 + δ for any δ > 0, so we have
not tried to optimize the value of C that our proof gives.
It should be noted that above result is not a direct consequence of the expansion
properties ofprandom Latin square graphs. From Theorem 10, we can only deduce
that µ = O( log n/k). However one can
p construct examples of d-regular graphs on
n vertices, with d = Ω(log n), µ = Ω( log n/d) but κ(G) 6 d − Ω(log n). We refer
the reader to the discussion following [16, Theorem 4.1] for more details about how
one can construct such a graph.
Similar to the vertex connectivity, the edge connectivity λ(G) of a graph G is the
minimal number of edges that we need to remove in order to disconnect G. It is
easy to show that κ(G) 6 λ(G) 6 δ(G). Hence, Theorem 13 applies with κ(G)
replaced by λ(G). In fact, our next theorem shows that we can do a bit more. If
|S| > (1 + δ) log2 n then whp the edge connectivity is equal to the minimum degree
of G. In view of random Cayley graphs on Zm
2 , this is in fact best possible.
Theorem 14 (Edge connectivity). For any δ > 0, if L is an n × n Latin square with
entries in [n] and S is a set of (1 + δ) log2 n elements of [n], chosen independently
and uniformly at random, then whp, λ(G(L, S)) = δ(G(L, S)).
Another graph property which follows from pseudorandomness is that of Hamiltonicity. Again, this property depends on the structure of the Latin square. For example,
the Cayley graph of Zq for q prime, with respect to any non-trivial element is Hamiltonian. On the other hand, as it was mentioned earlier, the Cayley graph of G = Zm
2
with respect to any set of size less than m = log2 |G| is not even connected. A very
appealing conjecture attributed to Lovasz, states that every connected Cayley graph
is Hamiltonian. This would imply for example that every random Cayley graph on
(1 + δ) log2 n generators is Hamiltonian. However, even this consequence is still not
known. Recently, Krivelevich and Sudakov [15] proved that every d-regular graph on
n vertices satisfying
(log log n)2
,
µ6
1000 log n(log log log n)
is Hamiltonian, provided n is large enough. Using this, together with the proof
technique of the Alon-Roichman theorem, they proved that a random Cayley graph
on O((log n)5 ) generators is whp Hamiltonian. Here, we extend this result to random
Latin square graphs as well. Moreover, using Theorem 10 directly, we can in fact
replace (log n)5 by (log n)3 .
3
2
(log log log n)
Theorem 15 (Hamiltonicity). Let k = (log n)(log
ω(n), where ω(n) is any
log n)2
function of n which tends to infinity as n tends to infinity. Then almost every G ∈
G (Ln , k) is Hamiltonian.
9
3
Cliques and independent sets
We begin by finding upper bounds for the clique number of random Latin square
graphs. Naturally, one would like to find a good upper bound for the expected number
of d-cliques of a random Latin square graph, and from this deduce a corresponding
upper bound for the clique number. Given
A ⊆ [n] let A0 = {Lij : i, j ∈ A, i 6= j}. If
d
|A| = d, then |A0 | can be as large as 2 and as small as d − 1. In the former case, the
d
probability that A forms a clique in G (L , p) is (2p − p2 )(2) . However, in the latter
n
case, this probability is at least pd−1 . So, unless one is able to bound the number of
A ⊆ [n] for which |A0 | is relatively small, then this approach cannot give any good
bounds.√Our approach will be to show that any A ⊆ [n] of size d, has a subset B of
size Ω( d), such that |B 0 | is relatively large, i.e. Ω(|B|2 ). By standard arguments
it will then follow that whp (if d is large enough,) no such B forms a clique, and
hence no A ⊆ [n] of size d forms a clique. Before stating our main lemma, we need
to introduce some more notation.
n2 (A) = |{{i, j} : i, j ∈ A distinct and Lij = Lji }|;
n3 (A) = |{(i, j, k) : i, j, k ∈ A distinct and Lij = Ljk }|;
n4 (A) = |{{(i, j), (k, l)} : i, j, k, l ∈ A distinct and Lij = Lkl }|.
If x ∈ A0 appears exactly rx times as Lij for distinct i, j ∈ A, then, with the above
notation, we have
X r x n2 (A) + n3 (A) + n4 (A) =
.
2
0
x∈A
We are now ready to state and prove our main lemma.
Lemma 16. Let A be a set of elements of X of size a. Then for every b 6 a, A
contains a subset B of size b such that
b − 2 (b − 2)(b − 3)
0
−
− n2 (B).
|B | > b(b − 1) 1 −
a−2
2(a − 3)
Proof. For any B ⊆ A of size b, we have
X
(rx − 1)
|B 0 | = b(b − 1) −
x∈B 0
> b(b − 1) −
X r x x∈B 0
2
= b(b − 1) − n2 (B) − n3 (B) − n4 (B).
Picking B at random from all b element subsets of A, we have
b(b−1)(b−2)
E(n3 (B)) = n3 (A) a(a−1)(a−2)
; and
b(b−1)(b−2)(b−3)
E(n4 (B)) = n4 (A) a(a−1)(a−2)(a−3)
.
10
Fixing distinct i, j ∈ A, there is exactly one k ∈ [n] such that Lij = Ljk , hence
n3 (A) 6 a(a − 1). Similarly, fixing distinct i, j, k ∈ A, there is exactly one L ∈ [n]
such that Lij = Lkl , hence n4 (A) 6 a(a−1)(a−2)
. It follows that
2
b − 2 (b − 2)(b − 3)
0
−
,
E(|B | + n2 (B)) > b(b − 1) 1 −
a−2
2(a − 3)
and hence there is a choice of B satisfying the requirements of the lemma.
We can now prove Theorem 1.
Proof of Theorem 1. Let d = 1/(2p − p2 ), let b = 3 logd n and let a = 3b2 . Pick any
A ⊆ [n] of size a. By Lemma 16, there is a B ⊆ A of size b, such that
5
|B 0 | > b2 − n2 (B) + O(b).
6
Pick |B 0 | pairs (i, j) in B × B, with i 6= j, such that all Lij are distinct. Suppose that
for exactly k of the pairs we have Lij = Lji . It follows that there are at least
b
1
0
(|B | − k) −
− n2 (B) = b2 − k + O(b),
3
2
sets {i, j}, such that both (i, j) and (j, i) have been chosen (and so Lij 6= Lji ).
Therefore, the probability that B is a clique is at most
1 b2 +O(b)
1 b2 −k+O(b)
pk 2p − p2 3
6 2p − p2 3
.
So the expected number of cliques B ⊆ [n] of size b with |B 0 | > 65 b2 − n2 (B) + O(b)
is at most
b
1 2
1
n
1
2 3 b +O(b)
2 3 b+O(1)
2p − p
6
n 2p − p
= o(1).
b
b!
Thus, by Markov’s Inequality, we deduce that whp, no such B exists. By Theorem 16,
it now follows that whp, there is no clique of size 3b2 , as required.
In a similar way, we can prove the upper bound for the independence number.
Proof of Theorem 2. Let b = 3 log1/(1−p) n, and let a = 3b2 . Pick any A ⊆ [n] of size
a. By Lemma 16, there is a B ⊆ A of size b, such that
5
1
|B 0 | > b2 − n2 (B) + O(b) > b2 + O(b).
6
3
2 /3+O(b)
Therefore, the probability that B is an independent set, is at most (1 − p)b
and so the expected number of independent sets is
b
1
n
2
(1 − p)b /3+O(b) 6
n(1 − p)b/3+O(1) = o(1).
b
b!
,
Thus, by Markov’s Inequality, we deduce that whp, no such B exists. By Theorem 16
it now follows that whp, there is no independent set of size 3b2 , as required.
11
4
Colouring
We now move to the proof of the upper bounds on the chromatic number. Before
presenting our proof, let us see why a standard approach from the theory of random
graphs does not seem to generalise in a straightforward manner.
Suppose we could show that whp, every induced subgraph of G ∈ G (Ln , 1/2) on
n1 = n/(log n)2 vertices has an independent set of size at least s1 = (2 − ε) log2 n.
It then follows immediately that whp, the chromatic number is at most n/s1 +
n1 ∼ n/2(log2 n). To do this, one usually shows that the probability that a given
induced subgraph
on n1 vertices does not contain an independent set of size s1 is
1+δ
), for some δ > 0. However in our model, this is far from being true.
O(exp −n
In fact, the probability
that G ∈ G (Ln , 1/2) is empty is 2−n , which is much larger
than O(exp −n1+δ ). It turns out that this problem can be rectified by using the
expansion properties of the graph G. We refer the reader to [3] to see how one can
do this. Here, we will use a different approach from which we can obtain a better
constant in the bound.
Another approach for finding an upper bound for the chromatic number, is to analyse
the greedy algorithm. This is the approach that we are going to use. This approach
will in fact give an upper bound on the list-chromatic number as well. However, we
need to modify the standard argument, because of the dependencies in the appearance
of edges. In our modification we will make use of Talagrand’s Inequality [24]. We will
use the following version taken (essentially) from [14].
Talagrand’s Inequality. Let X be a non-negative integer valued random variable, not
identically 0, which is determined by n independent random variables and let M be
a median of X. Suppose also that there exist K and r such that
1. X is K-Lipschitz. I.e. changing the outcome of one of the variables, changes
the value of X by at most K.
2. For any s, if X > s, then there is a set of at most rs of the variables, whose
outcome certifies that X > s.
Then
o
n
(
2
4 exp − 8rKt 2 M
Pr(|X − M | > t) 6
t
2 exp − 8rK
2
if 0 6 t 6 M ;
if t > M.
In particular, it follows that,
Z
∞
|EX − M | 6 E|X − M | =
Pr(|X − M | > t) dt
0
Z M
Z ∞
t2
t
64
exp −
dt + 2
exp −
dt
8rK 2 M
8rK 2
0
M
√
6 2K 8πrM + 16rK 2 .
12
Since also M = 2M Pr(X > M ) 6 2EX, we deduce that for 0 6 t 6 EX
√
t2
2
Pr |X − EX| > t + 16rK + 16K rEX 6 4 exp −
.
16rK 2 EX
This is the form of Talagrand’s Inequality that we will be using.
Let us now proceed to the proof of Theorem 5.
Proof of Theorem 5. Let d = 1/(1−p) and let u = 14 logd n− 12 logd logd n−2. Suppose
every vertex v has a list L(v) of size bn/uc. Fix an ordering v1 , . . . , vn of the vertices.
Suppose we are given a (not necessarily proper) colouring c of vertices v1 , . . . , vm , such
that c(vi ) ∈ L(vi ) for each 1 6 i 6 m. Suppose L(vm+1 ) = {x1 , . . . , xbn/uc }, let ci =
ci (m) be the number of times that colour xi is used on vertices v1 , . . . , vm and let Am+1
be the event that vm+1 has an earlier neighbour in every colour of the list L(vm+1 ).
We claim that Pr(Am+1 ) = o(1/n). Having proved this, we proceed
Pn by list-colouring
the graph greedily. The probability that this fails is at most m=1 Pr(Am ) = o(1),
so by Markov, we have whp χl (G) 6 n/u.
To prove our claim, let Bi = Bi (m) be the event that vm+1 is joined with an earlier
vertex of colour xi . Then clearly Pr(Bi ) 6 1 − (1 − p)2ci . Let Y be the number of
colours in L(vm+1 ) appearing on earlier neighbours of vm+1 . Then
n n
n
n X
(1 − p)2ci 6 − (1 − p)2mu/n 6
1 − (1 − p)2u ,
EY 6 −
u
u u
u
where the second inequality follows from the Arithmetic-Geometric Mean Inequality.
Let X = Y − EY + nu (1 − (1 − p)2u ) and let t = c nu (1 − p)2u , for some 0 < c < 1 to
be determined later. Then X satisfies the conditions of Talagrand’s Inequality with
K = 2 and r = 1. Note that, for n large enough, 0 6 t 6 EX, so
r
n
n
2u
2u
Pr |X − EX| > c (1 − p) + 64 + 32
(1 − (1 − p) ) 6
u
u
(
)
2
c n(1 − p)4u
c2 n (logd n)2
4 exp −
= 4 exp −
6
64u
64u(1 − p)4
c2 log n
4 exp −
.
16(1 − p)8 log (1/(1 − p))
8
By elementary calculus, it is easy to
p show that 16x log (1/x) 6 2/e whenever 0 <
x < 1. Hence, choosing any c with 2/e < c < 1, we deduce that
r
n
n
2u
Pr |X − EX| > c (1 − p) + 64 + 32
(1 − (1 − p)2u ) = o(1/n).
u
u
In particular, since Y 6 X,
r n
n
n
2u
= o(1/n).
Pr Y > − c (1 − p) + 64 + 32
u
u
u
13
Since
n
(logd n)2
(1 − p)4u =
→ ∞,
u
u(1 − p)8
we deduce that (for n large enough,)
Pr(Am+1 ) = Pr(Y > bn/uc) = o(1/n).
Similarly, we can give an upper bound to the clique cover number.
Proof of Theorem 7. Let d = 1/p and let u = 12 logd n−logd logd n−6. Fix an ordering
of the vertices. Suppose we are given a not necessarily proper colouring of the first
m vertices of Ḡ, using colours 1 up to bn/uc. Let ci = ci (m) be the number of times
colour i is used and let Am+1 be the event that the (m + 1)-th vertex has a neighbour
in every colour. We claim that Pr(Am+1 ) = o(1/n). Having proved this, we colour
the
Pn graph greedily. The probability that we need more than bn/uc colours is at most
m=1 Pr(Am ) = o(1), so by Markov, we have whp θ(G) = χ(Ḡ) 6 n/u.
To prove our claim, let Bi = Bi (m) be the event that the (m + 1)-th vertex is joined
(in Ḡ,) with an earlier vertex of colour i. Then clearly Pr(Bi ) 6 1 − pci . Let Y be the
number of colours appearing on earlier neighbours of the (m + 1)-th vertex. Then
n n
n
n X ci
p 6 − pmu/n 6 (1 − pu ) .
EY 6 −
u
u u
u
Let X = Y − EY + nu (1 − pu ) and let t = c nu pu , for some 0 < c < 1 to be determined
later. Then X satisfies the conditions of Talagrand’s Inequality with K = 2 and
r = 1. Note that, for n large enough, 0 6 t 6 EX, so
r
n
n u
u
(1 − p ) 6
Pr |X − EX| > c p + 64 + 32
u
u
2 2u c np
c2 log n
4 exp −
6 4 exp −
.
64u
32p12 log (1/p)
12
But
p 32x log (1/x) 6 8/(3e) < 1 whenever 0 < x < 1. Hence, choosing any c with
8/3e < c < 1, we deduce that
r
n u
n
u
Pr |X − EX| > c p + 64 + 32
(1 − p ) = o(1/n).
u
u
In particular, since Y 6 X,
r n
n 2u
n
Pr Y > − c p + 64 + 32
= o(1/n).
u
u
u
Since
n 2u (logd n)2
→ ∞,
p =
u
up12
we deduce that (for n large enough,)
Pr(Am+1 ) = Pr(Y > bn/uc) = o(1/n).
14
5
Expansion and consequences
We now proceed to the expansion properties of random Latin square graphs and to the
proof of Theorem 10 on the second eigenvalue of such graphs. In [8], we generalized
Hoeffding’s inequality, to an inequality where the random variables do not necessarily
take real values, but instead take their values in the set of (self-adjoint) operators of
a (finite dimensional) Hilbert space. We then used this inequality to give a new proof
of the Alon-Roichman theorem. The main tool in the proof of Theorem 10 will be this
Operator Hoeffding Inequality. Before stating the inequality, we need to introduce
some more notation.
Let V be a Hilbert space of dimension d, let A(V ) be the set of self adjoint operators
on V and let P (V ) be the be the cone of positive operators on V , i.e.
P (V ) = {A ∈ A(V ) : all eigenvalues of A are nonnegative}.
This defines a partial order on A(V ) by A 6 B iff B − A ∈ P (V ). We denote by
[A, B] the set of all C ∈ A(V ) such that A 6 C 6 B. We also denote by kAk the
largest eigenvalue of A in absolute value.
We can now state our Operator Hoeffding Inequality. We refer the reader to [8] for
its proof.
Theorem 17 ([8] Operator Hoeffding Inequality). Let V be a Hilbert space of dimension d and let Xi = E(X|Fi ) be a martingale, taking values in A(V ), whose difference
sequence satisfies Yi ∈ [− 21 I, 12 I]. Then
Pr(kX − EXk > nh) 6 2d exp {−nH(1/2 + h)}.
Note that the case d = 1 of this inequality is exactly Hoeffding’s inequality.
We now proceed to show that random Latin square graphs have small second eigenvalue and thus good expansion properties.
Proof of Theorem 10. Let s1 , . . . , sk be elements of [n] chosen independently and uniformly at random. For s ∈ [n] let L(s) be the 0,1 matrix in which L(s)ij = 1 if and
only if Lij = s. So the normalised adjacency matrix of the multigraph G generated
k
P
1
by these elements is T = 2k
(L(si ) + L(si )T ). Let B = T − n1 J, where J is the
i=1
n by n matrix having ‘1’ in every entry. We claim that µ(G) = kBk, where J is
the having ‘1’ in every entry. Indeed, if {v0 , v1 , . . . , vn−1 } is an orthonormal basis
of T , with each vi having eigenvalue λi , and v0 = √1n (1, . . . , 1), then Bv0 = 0 and
Bvi = λi vi , so µ(G) = kBk as required. Let Yi be the operator whose matrix is
1
L(si ) + L(si )T − n2 J . It is easy to check that Xi = Y1 + . . . + Yi is a martingale
4
15
satisfying the conditions of the theorem. It follows that
!
k
1 X
ε
Pr(µ(G) > ε) = Pr Yi >
k
2
i=1
εk
= Pr kX − EXk >
2
1+ε
,
6 2n exp −kH
2
as required.
Proof of Theorem 12. It is enough to prove the result for 0 < δ < 1/2. Note that
H(x) is continuous in (0, 1) and tends to log 2 as x tends to 1. Pick an x such
that H(x) > (1 − δ/2) log 2. Then, for k = |S| = (1 + δ) log2 n, we have kH(x) >
(1 + δ/4) log n. Thus,
Pr(µ(G(L, S)) > 2x − 1) 6 2n exp {−(1 + δ/4) log n} = 2n−δ/4 = o(1).
Thus whp, µ(G(L, S)) < 2x − 1 < 1. It is well known that if µ(G) < 1 then G is
connected, so the result follows.
Proof of Theorem 13. Let T be a minimial disconnecting set, so |T | 6 2k. Let U be
the smallest component of G \ T and let W = V (G) \ (U ∪ T ). Note that |W | > n/4.
We claim that whp, |U | 6 128 log n. Our proof of this claim is very similar
√ to [16,
Theorem 4.1]. Firstly, we deduce from Theorem 10 that whp µ(G) 6 4 k log n.
Since there are no edges from U to W , it follows from the edge distribution bound
for pseudorandom graphs (see e.g. [16, Theorem 2.11]) that
p
2k
|U ||W | < µ |U ||W |
n
and so
|U | <
µ2 n 2
µ2 n
16n log n
6
6
.
4k 2 |W |
k2
k
Using [16, Theorem 2.11] again, we deduce that the number of edges having both
endpoints in U (counted with multiplicity) satisfy
e(U, U ) 6
p
k
2k 2
|U | + µ|U | < (32 log n + 4 k log n)|U | 6 |U |,
n
2
provided C is large enough. It follows that the number of edges with exactly one
endpoint in U and one in T satisfy
e(U, T ) = 2k|U | − e(U, U ) >
3k
|U |.
2
On the other hand, using [16, Theorem 2.11] once more, we have
s
!
p
2k
2 log n
3k
e(U, T ) 6
|U ||S| + µ |U ||S| 6 1 + 4
k|U | 6
|U |,
n
|U |
2
16
a contradiction. So we may assume that |U | 6 128 log n.
We now claim that whp, the following holds: For any 3 distinct vertices x, y, z of G,
|(N (x) ∪ N (y)) \ N (z)| > 128 log n, where N (x) denotes the neighbourhood of the
vertex x. Having proved this, it will follow that whp |U | 6 2 and so |T | > δ(G) − 1.
So, let x, y, z be distinct vertices of G. Let s1 , s2 , . . . , sk be the elements of S chosen uniformly at random and let Xi = E(|(N (x) ∪ N (y)) \ N (z)||s1 , . . . , si ). Then
X0 , X1 , . . . , Xk is a martingale with Lipschitz constant 4 and X0 > k(1 − 2/n)k . It
follows by the Hoffding-Azuma inequality that
2
t
k
.
Pr(|(N (x) ∪ N (y)) \ N (z)| 6 k(1 − 2/n) − t) 6 exp −
2k
√
Letting t = 8k log n we obtain that
Pr(|(N (x) ∪ N (y)) \ N (z)| 6 128 log n) = O(n−4 )
provided C is large enough. Our claim now follows from the union bound. This
completes the proof of the theorem. (It can be checked that C = 168 works.
We omit the proof of Theorem 14, as it can be proved using a similar argument as
in [16, Theorem 4.3]
Sketch proof of Theorem 15. Firstly, one needs to check that the result of Krivelevich
and Sudakov [15] mentioned before the statement of the theorem, also holds for dregular multigraphs. We omit the details of this check. Then the result follows
directly from Theorem 10
6
Conclusion and open problems
We have introduced new models of random graphs arising from Latin squares and
studied some of their properties. There is still a lot of research that needs to be done
even for many of the properties that we have considered here.
Regarding the clique and independence numbers it would be interesting to know
if the upper bound can be reduced further. In particular, we believe (but cannot
prove) that the 2 in the exponent can be reduced further. It would be also interesting to know whether there are examples of random Latin square graphs whose
clique/independence number is significantly larger than Θ(log n log log n).
Similar remarks hold for the lower bound on the chromatic and clique cover numbers.
Any improvement on the upper bound of the independence/clique numbers would give
a corresponding improvement on the chromatic/clique cover numbers but it might be
possible (or even easier) to get such improvements directly.
Another interesting question which we have not been able to answer so far is the
determination of the Hadwiger number of random Latin square graphs, i.e. the largest
17
integer k such that the graph can be contracted into a Kk . We do not even know, for
p = 1/2 say, whether this number depends on the sequence of Latin squares or not.
We have not studied at all the girth of random Latin square graphs. The reason
is that it depends a lot on the structure of the Latin squares chosen. For example,
m
almost every G ∈ G (Zm
3 , p) has whp girth 3, provided pn → ∞, where n = 3 . On
m
the other hand, we claim that almost every G ∈ G (Z2 , p) has whp girth strictly
greater than 3 provided that pn2/3 → 0, where n= 2m . Indeed, the expected number
of triangles containing a fixed vertex x is n−1
p3 which tends to 0. By Markov’s
2
inequality x is whp not contained in any triangle. But since the graph is vertex
transitive, our claim follows.
The expansion properties of random Latin squares imply that almost every G ∈
G (Ln , c log2 n), with c > 1, has logarithmic diameter. An interesting question here
is the threshold for the diameter becoming equal to 2. It turns out that there are
constants c1 and c2 such that if p < c1 log n/n, then almost every G ∈ G (Ln , p) has
diameter greater than 2, while if p > c2 log n/2, then almost every G ∈ G (Ln , p) has
diameter less than or equal to 2. The values of c1 and c2 depend on the sequence of
Latin squares chosen. Our results regarding the diameter will appear in a forthcoming
paper [9].
Acknowledgements
This work started when the second author was visiting the Department of Pure Mathematics and Mathematical Statistics of the University of Cambridge and continued
when the first author was visiting the Deparment of Mathematics and Mathematical
Statistics of Umeå University. The authors would like to thank both departments for
their hospitality.
The first author would like to thank the Engineering and Physical Sciences Research
Council and the Cambridge Commonwealth Trust which supprted him while he was
at the University of Cambridge and also the Deparment of Mathematics and Mathematical Statistics of Umeå University for the grant making his visit there possible.
The second author would like to thank the Royal Physiographic Society in Lund for
the grant making his visit to the Univesity of Cambridge possible.
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Demetres Christofides
Klas Markström
School of Mathematics
University of Birmingham
Birmingham, B15 2TT
United Kingdom
Department of Mathematics and
Mathematical Statistics
Umeå universitet
Umeå, SE-901 87
Sweden
[email protected]
[email protected]
20
```
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