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https://nrich.maths.org/public/leg.php?code=-333&cl=4&cldcmpid=6559 | 1,511,244,560,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806317.75/warc/CC-MAIN-20171121055145-20171121075145-00623.warc.gz | 663,465,593 | 9,505 | # Search by Topic
#### Resources tagged with Investigations similar to The Ultra Particle:
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### Escape from Planet Earth
##### Stage: 5 Challenge Level:
How fast would you have to throw a ball upwards so that it would never land?
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Powerfully Fast
##### Stage: 5 Challenge Level:
Explore the power of aeroplanes, spaceships and horses.
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Global Warming
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How much energy has gone into warming the planet?
### Taking Trigonometry Series-ly
##### Stage: 5 Challenge Level:
Look at the advanced way of viewing sin and cos through their power series.
### Mach Attack
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Have you got the Mach knack? Discover the mathematics behind exceeding the sound barrier.
### Modelling Assumptions in Mechanics
##### Stage: 5
An article demonstrating mathematically how various physical modelling assumptions affect the solution to the seemingly simple problem of the projectile.
### Big and Small Numbers in the Physical World
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### The Power of Dimensional Analysis
##### Stage: 4 and 5
An introduction to a useful tool to check the validity of an equation.
### Diamonds Aren't Forever
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Ever wondered what it would be like to vaporise a diamond? Find out inside...
### Which Twin Is Older?
##### Stage: 5
A simplified account of special relativity and the twins paradox.
### Bessel's Equation
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Get further into power series using the fascinating Bessel's equation.
### Smoke and Daggers
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We all know that smoking poses a long term health risk and has the potential to cause cancer. But what actually happens when you light up a cigarette, place it to your mouth, take a tidal breath. . . .
### Two Regular Polygons
##### Stage: 4 Challenge Level:
Two polygons fit together so that the exterior angle at each end of their shared side is 81 degrees. If both shapes now have to be regular could the angle still be 81 degrees?
### Scale Invariance
##### Stage: 5 Challenge Level:
By exploring the concept of scale invariance, find the probability that a random piece of real data begins with a 1.
### Making More Tracks
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Given the equation for the path followed by the back wheel of a bike, can you solve to find the equation followed by the front wheel?
### What Salt?
##### Stage: 5 Challenge Level:
Can you deduce why common salt isn't NaCl_2?
### Clear as Crystal
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Unearth the beautiful mathematics of symmetry whilst investigating the properties of crystal lattices
### Big and Small Numbers in the Living World
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Work with numbers big and small to estimate and calculate various quantities in biological contexts.
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Read about the mathematics behind the measuring devices used in quantitative chemistry
### Genetic Intrigue
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Dip your toe into the fascinating topic of genetics. From Mendel's theories to some cutting edge experimental techniques, this article gives an insight into some of the processes underlying. . . .
### More Bridge Building
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Which parts of these framework bridges are in tension and which parts are in compression?
### Stirling Work
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See how enormously large quantities can cancel out to give a good approximation to the factorial function.
### Trig-trig
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Explore the properties of combinations of trig functions in this open investigation.
### Chance of That
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What's the chance of a pair of lists of numbers having sample correlation exactly equal to zero?
### A Different Differential Equation
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Explore the properties of this different sort of differential equation.
##### Stage: 4 and 5 Challenge Level:
Some of our more advanced investigations
### Building Approximations for Sin(x)
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Build up the concept of the Taylor series
### Problem Solving: Opening up Problems
##### Stage: 1, 2, 3 and 4
All types of mathematical problems serve a useful purpose in mathematics teaching, but different types of problem will achieve different learning objectives. In generalmore open-ended problems have. . . .
### Very Old Man
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Is the age of this very old man statistically believable?
### Geometry and Gravity 1
##### Stage: 3, 4 and 5
This article (the first of two) contains ideas for investigations. Space-time, the curvature of space and topology are introduced with some fascinating problems to explore.
### The Invertible Trefoil
##### Stage: 4 Challenge Level:
When is a knot invertible ?
### A Rational Search
##### Stage: 4 and 5 Challenge Level:
Investigate constructible images which contain rational areas.
### Eight Ratios
##### Stage: 4 Challenge Level:
Two perpendicular lines lie across each other and the end points are joined to form a quadrilateral. Eight ratios are defined, three are given but five need to be found.
### 9 Hole Light Golf
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We think this 3x3 version of the game is often harder than the 5x5 version. Do you agree? If so, why do you think that might be?
### Peeling the Apple or the Cone That Lost Its Head
##### Stage: 4 Challenge Level:
How much peel does an apple have?
### Sextet
##### Stage: 5 Challenge Level:
Investigate x to the power n plus 1 over x to the power n when x plus 1 over x equals 1.
### Designing Table Mats
##### Stage: 3 and 4 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Few and Far Between?
##### Stage: 4 and 5 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Perfect Eclipse
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Use trigonometry to determine whether solar eclipses on earth can be perfect.
### Reaction Rates!
##### Stage: 5
Fancy learning a bit more about rates of reaction, but don't know where to look? Come inside and find out more...
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Odd Stones
##### Stage: 4 Challenge Level:
On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.
### Twizzles Venture Forth
##### Stage: 4 Challenge Level:
Where we follow twizzles to places that no number has been before.
### Spokes
##### Stage: 5 Challenge Level:
Draw three equal line segments in a unit circle to divide the circle into four parts of equal area.
### CSI: Chemical Scene Investigation
##### Stage: 5 Challenge Level:
There has been a murder on the Stevenson estate. Use your analytical chemistry skills to assess the crime scene and identify the cause of death...
### There's Always One Isn't There
##### Stage: 4 Challenge Level:
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders.
### What Do Functions Do for Tiny X?
##### Stage: 5 Challenge Level:
Looking at small values of functions. Motivating the existence of the Taylor expansion. | 1,725 | 7,873 | {"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-2017-47 | longest | en | 0.811524 |
https://justaaa.com/chemistry/483886-1-what-is-the-equilibrium-partial-pressure-of | 1,723,331,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00320.warc.gz | 252,425,866 | 10,207 | Question
# 1) What is the equilibrium partial pressure of water vapor above a mixture of 62.2 g...
1) What is the equilibrium partial pressure of water vapor above a mixture of 62.2 g H2O and 35.2 g HOCH2CH2OH at 55 °C. The partial pressure of pure water at 55.0 °C is 118.0 mm Hg. Assume ideal behavior for the solution.
2
Which of the following statements is INCORRECT?
a)The solubility of a gas in water decreases with increasing temperature. b)The dissolution of a gas in water is usually an exothermic process. c)The solubility of a gas in water is proportional to the partial pressure of the gas above the water. d)The solubility of a gas in water is inversely proportional to the molar mass of the gas. e)The relationship between the solubility of a gas and its partial pressure is known as Henry's law.
Moles of water = mass/molar mass of H2O = 62.2 /18.015 = 3.4527
Moles of ethlene glycol = 35.2 /62.07 = 0.567
mol fraction of water = moles of water / ( total moles of mixture) = ( 3.4527 /3.4527+0.567) = 0.859
Partial pressure of H2O = mol fraction of H2O x vapor pressure of Pure H2O = 0.859 x 118 = 101.36 mm Hg
2) Solubility of gas increases as Molarmass of gas increases since bigger gas molecules will have more interactions and hence molre solubility
Hence statement D is wrong
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 391 | 1,367 | {"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} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.885574 |
https://www.kylesconverter.com/speed-or-velocity/yards-per-day-to-feet-per-second | 1,653,181,068,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00405.warc.gz | 971,975,792 | 5,404 | # Convert Yards Per Day to Feet Per Second
### Kyle's Converter > Speed Or Velocity > Yards Per Day > Yards Per Day to Feet Per Second
Yards Per Day (yd/d) Feet Per Second (fps) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Feet Per Second to Yards Per Day
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Yard per Day:
1 Yard per day is approximately equal to 1.058 333 333 x 10-5 meters per second (SI unit).
1 Foot per Second:
Feet per Second. In SI units 0.3048 meters per second.
Conversions Table
1 Yards Per Day to Feet Per Second = 070 Yards Per Day to Feet Per Second = 0.0024
2 Yards Per Day to Feet Per Second = 0.000180 Yards Per Day to Feet Per Second = 0.0028
3 Yards Per Day to Feet Per Second = 0.000190 Yards Per Day to Feet Per Second = 0.0031
4 Yards Per Day to Feet Per Second = 0.0001100 Yards Per Day to Feet Per Second = 0.0035
5 Yards Per Day to Feet Per Second = 0.0002200 Yards Per Day to Feet Per Second = 0.0069
6 Yards Per Day to Feet Per Second = 0.0002300 Yards Per Day to Feet Per Second = 0.0104
7 Yards Per Day to Feet Per Second = 0.0002400 Yards Per Day to Feet Per Second = 0.0139
8 Yards Per Day to Feet Per Second = 0.0003500 Yards Per Day to Feet Per Second = 0.0174
9 Yards Per Day to Feet Per Second = 0.0003600 Yards Per Day to Feet Per Second = 0.0208
10 Yards Per Day to Feet Per Second = 0.0003800 Yards Per Day to Feet Per Second = 0.0278
20 Yards Per Day to Feet Per Second = 0.0007900 Yards Per Day to Feet Per Second = 0.0313
30 Yards Per Day to Feet Per Second = 0.0011,000 Yards Per Day to Feet Per Second = 0.0347
40 Yards Per Day to Feet Per Second = 0.001410,000 Yards Per Day to Feet Per Second = 0.3472
50 Yards Per Day to Feet Per Second = 0.0017100,000 Yards Per Day to Feet Per Second = 3.4722
60 Yards Per Day to Feet Per Second = 0.00211,000,000 Yards Per Day to Feet Per Second = 34.7222 | 641 | 1,932 | {"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} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.703707 |
http://www.drdobbs.com/web-development/tough-apps-how-much-hardware-should-i-bu/228400127?pgno=2 | 1,524,718,533,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00484.warc.gz | 397,751,642 | 23,422 | # Tough Apps: How Much Hardware Should I Buy?
### Solution:
```
searchtraffic = T + (0.4 * searchtraffic) + (0.2 * specifytraffic) + (0.6 * addtraffic)
specifytraffic = (0.2 * searchtraffic) + (0.3 * specifytraffic)
addtraffic = (0.1 * specifytraffic)
specifytraffic = 0.2/0.7 * searchtraffic
addtraffic = 0.1 * 0.2/0.7 * searchtraffic
```
Thus, searchtraffic=T+(0.4*searchtraffic)+(0.2*0.2/0.7*searchtraffic)+(0.6*0.02/0.7*searchtraffic)=T+(0.4+0.057+0.017)*searchtraffic.
So searchtraffic=T/(1-(0.4+0.057+0.017))=T/0.526=10000/0.526 =19,011.41 arrivals per second. We'll call this 20,000 arrivals per second to give us some wiggle room.
specifytraffic=20,000*0.2/0.7=5,714 arrivals per second, which we'll call 6,000 arrivals per second.
addtraffic=20,000*0.1*0.2/0.7=571.4 arrivals per second, which we'll call 600.
Because each task requires 0.1 seconds on a single server, we want an arrival rate of roughly 9 per second (we could actually deal with more since 2.9/3=9 2/3). we need roughly 2955 servers with 2220 going to search, 660 to specify, and the remaining 75 going to add. If you got any numbers like this, you are on the right track.
In the case of a service time that is appreciable, e.g. 2 seconds, a slightly more complex analysis is required (see http://cs.nyu.edu/courses/fall09/G22.2434-001/capplanrulethumb.html, but the idea is simple. There are two components to response time waiting time + service time. Buying N servers decreases waiting time by a factor of N, but doesn't decrease service time whenever each task must be handled by one server. Therefore, to derive the response time, compute waiting time as if you had a single server that was N times as fast and then add in service time. The notes should make it clear.
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To upload an avatar photo, first complete your Disqus profile. | View the list of supported HTML tags you can use to style comments. | Please read our commenting policy. | 573 | 1,952 | {"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} | 3.234375 | 3 | CC-MAIN-2018-17 | latest | en | 0.872003 |
http://physics.bu.edu/ulab/prelabs/prelab_impulseenergy.html | 1,550,658,503,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494694.1/warc/CC-MAIN-20190220085318-20190220111318-00144.warc.gz | 234,574,002 | 2,081 | Impulse and Work (MBL) Pre-lab Assignment
### Impulse and Work (MBL) Pre-lab Assignment
Part of this experiment involves impulse - an impulse is a change in momentum produced by applying a net force to an object for a particular time interval. The first simulation shows possible motions for an air-hockey puck that is subject to a net force for half a second.
Which of the motions shows correctly what an air hockey puck would do if it was initially moving in the x-direction, and was then briefly subjected to a net force in the y-direction?
1 2 3 4
The second part of this experiment involves work, which has units of energy. Consider rolling a cart down an incline. The simulation below shows this situation. In this simulation, energy graphs are plotted as a function of time.
Two curves, showing different kinds on energy, are shown in the graph below the simulation. Identify each.
The red curve is the: gravitational potential energy kinetic energy The green curve is the: gravitational potential energy kinetic energy
Briefly explain how you know which graph is which.
Assume there is no friction acting. If you added the two curves, what would you get?
a horizontal line a line that decreases as the object travels down the slope a line that increases as the object travels down the slope
If friction was acting on the cart, what would you get if you added the two curves?
a horizontal line a line that decreases as the object travels down the slope a line that increases as the object travels down the slope | 318 | 1,534 | {"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} | 3.53125 | 4 | CC-MAIN-2019-09 | latest | en | 0.941868 |
https://study.com/academy/topic/coordinate-geometry-basics-help-and-review.html | 1,569,093,516,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574662.80/warc/CC-MAIN-20190921190812-20190921212812-00459.warc.gz | 682,629,177 | 26,497 | # Ch 14: Coordinate Geometry Basics: Help and Review
The Coordinate Geometry Basics chapter of this SAT Mathematics Level 2 Help and Review course is the simplest way to master coordinate geometry. This chapter uses simple and fun videos that are about five minutes long, plus lesson quizzes and a chapter exam to ensure students learn the essentials of coordinate geometry for the SAT exam.
## Who's it for?
Anyone who needs help learning or mastering SAT mathematics level 2 material will benefit from taking this course. There is no faster or easier way to prepare for this SAT math subject test. Among those who would benefit are:
• Students who have fallen behind in understanding how to plot points on the coordinate plane or graph functions
• Students who struggle with learning disabilities or learning differences, including autism and ADHD
• Students who prefer multiple ways of learning math (visual or auditory)
• Students who have missed class time and need to catch up
• Students who need an efficient way to learn about coordinate geometry for the SAT
• Students who struggle to understand their teachers
• Students who attend schools without extra math learning resources
## How it works:
• Find videos in our course that cover what you need to learn or review.
• Press play and watch the video lesson.
• Refer to the video transcripts to reinforce your learning.
• Test your understanding of each lesson with short quizzes.
• Verify you're ready by completing the Coordinate Geometry Basics chapter exam.
## Why it works:
• Study Efficiently: Skip what you know, review what you don't.
• Retain What You Learn: Engaging animations and real-life examples make topics easy to grasp.
• Be Ready on Test Day: Use the Coordinate Geometry Basics chapter exam to be prepared.
• Get Extra Support: Ask our subject-matter experts any coordinate geometry question. They're here to help!
• Study With Flexibility: Watch videos on any web-ready device.
## Students will review:
This chapter helps students review the concepts in a coordinate geometry unit of a standard math course. Topics covered include:
• Using a number line
• Identifying the parts of a graph
• Plotting points on the coordinate plane
• Graphing functions
7 Lessons in Chapter 14: Coordinate Geometry Basics: Help and Review
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 582 | 2,924 | {"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} | 3.3125 | 3 | CC-MAIN-2019-39 | longest | en | 0.886296 |
https://topbtcvegk.web.app/kethcart25119xiqu/calculator-interest-rate-on-annuity-getu.html | 1,638,419,026,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00389.warc.gz | 622,280,956 | 6,658 | ## Calculator interest rate on annuity
Interest rate (r) is the annual nominal interest rate expressed as a percentage. Annuity term constitutes the lifespan of the annuity. Compounding frequency (m) An annuity is a fixed income over a period of time. First: let's see the effect of an interest rate of 10% (imagine a bank account that How do we calculate that? A tutorial about using the TI BAII Plus financial calculator to solve time value of money problems Let's look at an example of solving for the interest rate:.
Home » Subscribers » I am interested in NPS » Pension Calculator I would like to purchase Annuity for. %. Amount for I am expecting an Annuity rate of. % 13 May 2019 The future value of an annuity is the amount of money you end up with after a series of level payments, given a specified interest rate, at a It can provide a guaranteed minimum interest rate, with no taxes due on any earnings until they are withdrawn from the account. Use this calculator to help you Try our Annuity calculator to compare your potential retirement income from an RBC Payout Annuity What is your expected annual RRSP/RRIF growth rate? You might not be sure how a fixed annuity (with a guaranteed interest rate) could fit into your retirement plan. KeyBank's Fixed Annuity Calculator shows an
## In addition, many annuity companies offer a higher first year bonus rate. Use this calculator to help determine your annuity value in a given year and compare it to a taxable savings account like a CD*. Minimum guaranteed interest rate
Annual Rate Annuity Calculator - Given the present value, payment and time periods remaining on an annuity you can calculate its rate of return. table are from partnerships from which Investopedia receives compensation. Articles of Interest Bankrate.com provides an annuity calculator and other personal finance investment calculators. Annual Growth Rate. i. The estimated yearly return on the Fixed annuities pay out a guaranteed amount after a certain date, and a return rate is largely dependent on market interest rates at the time the annuity contract is Present Value of an Annuity. C = Cash flow per period (payment amount). i = Interest rate. n = Number of payments (in this calculator, derived from the payment This calculator gives the annual payout amount of an annuity (ordinary Home Compound Interest Calculator Glossary Search Books Growth Rate: %.
### 18 Oct 2015 Note that you'll receive more if you buy your annuity when interest rates are higher, which they might be in a year or two. Source: Social Security
Free annuity payout calculator to find the payout amount based on fixed length or to find the length the fund can last based on given payment amount. It considers inflation and payout frequency. Experiment with other retirement calculators, or explore hundreds of other calculators addressing topics such as math, fitness, health, and many more. Our annuity calculator can help you easily calculate annuity payments, length or the required principal and growth rate to meet your income target. Indexed Rate Annuity Calculator. Many indexed annuities credit interest annually based upon the performance of an index, limited to an annual cap rate. In a year that the index rises more than the cap rate, the interest credit is the cap rate. In a year that the index rises less than the cap rate, the entire increase is credited. Annual Interest Rate (%) – This is the interest rate earned on the annuity. The present value annuity calculator will use the interest rate to discount the payment stream to its present value. Number Of Years To Calculate Present Value – This is the number of years over which the annuity is expected to be paid or received. I´m trying to calculate the interest rate for an annuity, knowing the PV, the annuity and the number of periods and I´m struggling with the formula. I don´t understand how does (1+r)^10 cancel put in the equation (1+r)^10 – 1/ (1+r)^10 / r to result in [ -1/r ] as (1+r)^10 in the nominator it´s subtracting 1, not multiplying. Also, we assume that the annuity is an investment, but some annuities take the form of a promise to give you payments over a certain period of time instead — lotteries or pensions, for example. When calculating real annuities, you also often have to make an assumption regarding the interest rate, because interest rate fluctuates.
### The Excel RATE function is a financial function that returns the interest rate per period of an annuity. You can use RATE to calculate the periodic interest rate, then
This calculator gives the annual payout amount of an annuity (ordinary Home Compound Interest Calculator Glossary Search Books Growth Rate: %. It can provide a guaranteed minimum interest rate, with no taxes due on any earnings until they are withdrawn from the account. Use this calculator to help you value and future value annuity calculator with step by step explanations. Calculate Withdraw Amount, Deposit Frequency, Regular Deposits or Interest rate. Many clients purchase income annuities to help cover their essential expenses, as defined by them, in retirement. Use this income annuity calculator to get an
## It can provide a guaranteed minimum interest rate, with no taxes due on any earnings until they are withdrawn from the account. Use this calculator to help you
11 Jan 2019 Annuity- Calculate Payment (Starts at 14:04). * Annuity- Calculate Number of Payments (N) (Starts at 15:09). * Annuity-Calculating Interest Rate 13 Nov 2014 The basic annuity formula in Excel for present value is =PV(RATE,NPER,PMT). Let's break it down: • RATE is the discount rate or interest rate, Given the present value, payment and time periods remaining on an annuity you can calculate its rate of return. Your rate is too small for our calculators. This means that you need either to An annuity is an investment that provides a series of payments in exchange for an initial lump sum. With this calculator, you can find several things: The payment that would deplete the fund in a
An annuity is a fixed income over a period of time. First: let's see the effect of an interest rate of 10% (imagine a bank account that How do we calculate that? A tutorial about using the TI BAII Plus financial calculator to solve time value of money problems Let's look at an example of solving for the interest rate:. 6 Jun 2019 Other investment structures such as annuities are also based on interest. They either represent (a) a single value today i.e. a present value that 9 Dec 2019 Knowing the present value of an annuity is important for retirement planning. This guide walks through how it works and how to calculate it the amount in each annuity payment (in dollars); R= the interest or discount rate 5 Apr 2019 Put another way, it is the interest rate that makes the net present value of all cash flows equal to zero. Evaluating Payment Amounts. An annuity | 1,496 | 6,916 | {"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} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.923658 |
https://math.stackexchange.com/questions/1812158/hodge-decomposition-on-riemann-surface | 1,558,253,970,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00422.warc.gz | 586,791,829 | 33,519 | # Hodge decomposition on Riemann surface
On a compact Riemannian manifold $M$ the Hodge decomposition takes the form $$\Omega^k(M)=d\Omega^{k-1}(M)\oplus\mathcal{H}(M)\oplus d^*\Omega^{k+1}(M)$$ Where $d^*$ is the adjoint of $d$ w.r.t. the inner product induced by the metric.
Now on a Riemann surface we do not have a metric, but we do have a canonical conformal structure. This conformal structure is enough to give us a Hodge-$\star$ operator $$\star:\Omega^1(M)\to\Omega^1(M)$$ In this setting the Hodge decomposition takes on the form $$\Omega^1_2(M)=E\oplus\star E\oplus\mathfrak{h}(M)$$ Where $$E=\overline{\{df\mid f\in \Omega^0(M)\}}$$ $$\star E=\overline{\{\star df\mid f\in \Omega^0(M)\}}$$ $$\mathfrak{h}(M)=\{\omega\in\Omega^1(M)\mid d\star \omega=d\omega=0\}$$ Where closures are in the $L^2$ sense.
These two decomposition looks quite different to me. The term $d\Omega^{k-1}(M)$ looks like $E$. But I do not recognise an analogue of $d^*$ in the Riemann surface case. $$d^*:\Omega^k(M)\to \Omega^{k-1}(M)$$ is usually constructed as $\star d\star$, however this requires a $\star:\Omega^2(M)\to\Omega^0(M)$, but the conformal structure on $M$ is not enough to give us this $\star$.
Also, the $\mathcal{H}$ in the Riemannian manifold case is characterised by a condition involving $d^*$, which we do not have on a Riemann surface, so I don't see how $\mathcal{H}$ compares to $\mathfrak{h}$.
So my question is if these two forms of the Hodge decomposition are in some sense the same, or if these are fundamentally different decompositions? The absence of a metric on the Riemann surface suggests that we should get a slightly coarser decomposition, since a metric is more rigid then a conformal structure. However there are enough similarities between the decompositions to make me think there must be some connection.
The hodge star is an isomorphism $\star \colon \Omega^k(M) \rightarrow \Omega^{n-k}(M)$. If you apply this observation with $n = k = 2$, you will see that
$$d^{*}(\Omega^2(M)) = \{ \star d \star \omega \, | \, \omega \in \Omega^2(M) \} = \{ \star df \, | \, f \in \Omega^{0}(M) \} = \star \{ df \, | \, f \in \Omega^{0}(M) \}.$$
If you apply it to $n = 2, k = 1$, you will see that
$$d^{*}(\omega) = \star (d (\star \omega)) = 0 \iff d (\star \omega) = 0$$
and so $\mathfrak{h}(M) = \mathcal{H}(M)$. Thus, you get the same direct sum decomposition.
• Thank you for your answer. So I see how the Riemannian manifold case can be written in the same form as the Riemann surface case. But in a way that is remarkable right? Because we miss a lot of the structure that seems to be necessary, namely the metric. So it seems that the combination of the low dimension and the presence of the conformal structure is just enough to make this work? – user2520938 Jun 4 '16 at 13:50
• The hodge star is always conformally invariant in the middle dimension (if the dimension is $2n$, and you multiply the inner product by $\lambda^2$ then the inner product on $n$ forms scales by $\lambda^{2n}$ and the volume form scales by $\lambda^{-2n}$) so you can repeat this argument to obtain a decomposition of $\Omega^{n}(M^{2n})$ which depends only on a conformal class of metrics on $M$. We don't really miss the metric - the decomposition in the middle level sees only metrics up to conformal equivalence. – levap Jun 4 '16 at 14:16
• The nice feature of complex one-dimensional manifolds (Riemann surfaces) is that they come with a canonical conformal class - this is unique for (complex) dimension one. – levap Jun 4 '16 at 14:16
• But I think it's not just that we have a conformal structure on a Riemann surface. Given a conformal structure on a higher dimensional complex manifold we would still have a $\star$ operator on the mid dimensional forms, but this wouldn't be as useful; we would miss too many operators on the higher forms. The low dimension of the Riemann surface makes it so that the missing $\star$ on $0$ and $2$ forms doesn't really matter. That's the way I see it anyway. – user2520938 Jun 4 '16 at 14:39 | 1,189 | 4,055 | {"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": 1, "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} | 2.9375 | 3 | CC-MAIN-2019-22 | latest | en | 0.815943 |
https://www.physicsforums.com/threads/urn-problem.429482/ | 1,544,502,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823565.27/warc/CC-MAIN-20181211040413-20181211061913-00065.warc.gz | 982,881,242 | 12,496 | # Urn Problem
1. Sep 16, 2010
### MartinWDK
Dear all,
We are trying to compute the number of ways for a computer to execute concurrent processes.
It appears that this problem is equivalent to asking the following:
Assume that an urn is filled with different quantities of differently colored balls. There are k different colors, and the number of balls of a given color is denoted nk.
Balls are drawn from the urn it is empty, and the color of the drawn ball is noted.
The question is: how many different color sequences can be constructed in this way?
Thank you,
Martin
2. Sep 16, 2010
### csopi
Let N=n1+n2+...+nk
$$\frac {N!}{n_1!\dots n_k!}$$
because there would be N! possibilities if the colours were pairwise different, but they are not, so we counted each possibility a lot of times. One can permutate the balls from the jth colour n_j! ways but when doing so we don't see any difference in the colour-sequence, so we have to divide N! by n_1!...n_k!
3. Sep 16, 2010
### MartinWDK
Hi csopi,
Thank you for a quick reply and for a clear and coherent explanation!
The formula matches our calculations, so I think we have our answer :)
Best regards,
Martin | 301 | 1,179 | {"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": 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 | 3 | CC-MAIN-2018-51 | latest | en | 0.947458 |
https://www.scribd.com/presentation/416301476/Pr-2 | 1,568,959,116,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00519.warc.gz | 1,025,473,488 | 61,248 | You are on page 1of 23
# Nature of Inquiry and Research
Characteristics of Research
Strengths and Weaknesses of Quantitative Research
Kinds of Quantitative Research
Objectives A Recall what qualitative research
is
At the end of the
lesson, the Describe the nature and
learners are able B
characteristics of research
to:
C Point out the strengths and
weaknesses of quantitative
research
## D Differentiate kinds of variables
and their uses
Quantitative research is used
to examine the relationship
between, variables,
quantify the problem by way
of generating numerical data
and explain the
phenomenon by way of
gathering numerical data or
data that can be analyzed using
statistical tools.
ACTIVITY
Think of words that are related to quality and quantity.
Quantity
Quality
Characteristics of a
A
Quantitative Research
Discussion
Strengths of Quantitative
B
Research
Weaknesses of
C
Quantitative Research
Kinds of Quantitative
D
Research
Quantitative Research
This type of research is used to examine the
relationship between variables, quantify
the problem by way of generating
numerical data and explain the phenomenon
by way of gathering numerical data or data that
can be analyzed using statistical tools.
CHARACTERISTICS OF A
QUANTITATIVE RESEARCH
• Call for measurable characteristics of the
population
• Standardized instruments to ensure accur
acy, reliability and validity of the data.
• Use of charts, tables, graphs and figures
CHARACTERISTICS OF A
QUANTITATIVE RESEARCH
## • Follows the principle of random sampling
• Reinforce validity of findings (repetition of
methods)
• It puts emphasis on proof, rather than disco
very.
STRENGTHS OF A QUANTITATIVE
RESEARCH
• is the most reliable and valid way of conclu
ding results
• the results are more reliable and valid
• Quantitative experiments filter out external
factors.
Weaknesses of a
Quantitative Research
• Can be costly, difficult and time – cons
uming
• Requires extensive statistical treatment
• Also tend to turn out only proved or un
proved results
KINDS OF
QUANTITATIVE
RESEARCH
DESIGNS
RESEARCH DESIGNS
## Overall strategy that you choose in order to
integrate the different components of the
study in a coherent and logical way.
Constitutes the blueprint for the selection,
measurement and analysis of data
Note:
The research problem determines the research design you should use
EXPERIMENTAL RESEARCH DESIGN
## • Allows the researcher to control the situation
• “What causes something to occur?”
• Allows the researcher to identify cause and effect
relationship between variables and to distinguish
placebo effects from treatment effects.
• Provides the highest level of evidence for single
studies.
Pre – Experimental Research Design
• Applies to experimental designs with the least internal
validity
## Quasi – Experimental Research Design
• The researcher can collect more data, either by scheduling
more observations or finding more exciting measures
## True Experimental Design
• Controls for both time – related and group – related threats
.
Types Of Quasi – Experimental Research
Design
1. Non – Equivalent control group design – refers
to the chance failure of random assignment to equalize
the conditions by converting a true experiment into this
kind of design, for purposes of analysis.
2. Interrupted Time Series Design – employs multiple
measures before and after the experimental
intervention
NON – EXPERIMENTAL RESEARCH
DESIGN
The researcher observes the phenomena as
they occur naturally and no external variables are
introduced. The descriptive research design’s main
purpose it to observe, describe and document
aspects of a situation as it naturally occurs or for
theory development.
TYPES OF DESCRIPTIVE RESEARCH DESIGN
1. Survey
2. Correlational
• Bivariate correlational studies
• Prediction studies
• Multiple Regression Prediction Studies
3. Ex – Post Facto Research Design
4. Comparative Design
5. Evaluative Research
6. Methodological
PRACTICE
LOOPING ACTIVITY
In a group compose of th
ree (3) members showcase
how you will apply the
concept of Inqu
iry and Research?
ENRICHMENT (GROUP ACTIVITY)
Differentiate qualitative and quantitative research in terms of
characteristics, strengths and weaknesses and kinds of quan
titative research.
QUALITATIVE QUANTITATIVE | 954 | 4,284 | {"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-2019-39 | latest | en | 0.847725 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-p-prerequisites-section-p-2-real-numbers-p-2-exercises-page-15/3 | 1,545,136,882,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00266.warc.gz | 925,523,471 | 13,056 | # Chapter P, Prerequisites - Section P.2 - Real Numbers - P.2 Exercises - Page 15: 3
$(a+b)+c$, associative property of addition
#### Work Step by Step
$a+(b+c)=a+b+c=(a+b)+c$ To prove this, subtract any terms from the equation and you will receive equivalent expressions.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 109 | 436 | {"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} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.826137 |
https://www.pbslearningmedia.org/collection/midlit/?topic_id=320 | 1,519,069,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812788.42/warc/CC-MAIN-20180219191343-20180219211343-00546.warc.gz | 908,617,739 | 19,511 | • ## Graphing Distance and Time: Travel
In this blended lesson supporting literacy skills, students watch videos and complete interactive activities to learn how to represent a young man’s unicycle trip on a graph. Students develop their literacy skills as they explore a mathematics focus on expressing distance–time relationships with graphs. During this process, they read informational text, learn and practice vocabulary words, and explore content through videos and interactive activities. This resource is part of the Inspiring Middle School Literacy Collection.
• ## Multiplying Fractions by Whole Numbers: Recipes
In this blended lesson supporting literacy skills, students watch videos and complete interactive activities to learn about fractions and learn how to perform certain operations with fractions. Students develop their literacy skills as they explore a mathematics focus on multiplying fractions by whole numbers. During this process, they read informational text, learn and practice vocabulary words, and explore content through videos and interactive activities. This resource is part of the Inspiring Middle School Literacy Collection.
• ## Population Sampling: Fish
In this blended lesson supporting literacy skills, students watch videos and complete interactive activities to learn how the Ojibwe tribe in Minnesota measure the abundance of various fish in their lake, how they interpret the data they collect, and how they make predictions about future populations based on their findings. Students develop their literacy skills as they explore a mathematics focus on population sampling. During this process, they read informational text, learn and practice vocabulary words, and explore content through videos and interactive activities. This resource is part of the Inspiring Middle School Literacy Collection.
• ## Ratio and Proportional Reasoning: Food Labels
In this blended lesson supporting literacy skills, students watch videos and complete interactive activities to learn how to use fractions to interpret food labels and make healthy eating choices. Students develop their literacy skills as they explore a mathematics focus on proportional reasoning. During this process, they read informational text, learn and practice vocabulary words, and explore content through videos and interactive activities. This resource is part of the Inspiring Middle School Literacy Collection. | 411 | 2,423 | {"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-2018-09 | longest | en | 0.919101 |
https://kr.mathworks.com/matlabcentral/answers/521313-how-to-build-on-an-existing-struct-4x4xn-matrix | 1,652,871,798,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00786.warc.gz | 423,770,387 | 29,165 | # How to build on an existing struct 4x4xn matrix?
조회 수: 5(최근 30일)
Happy PhD 2020년 4월 28일
댓글: Rik 2020년 4월 28일
I have an struct, for example...
A = struct();
where we have a 4x4 matrix
A.G = rand(4,4);
I want to build on values within the (4x4) matrix but stack new values in a n'th dimension.
How do I build on the maxtrix in the n'th dimesnion if I don't know how many non-zero values exist in respective field A.G(1,1,n), A.G(1,2,n), A.G(2,2.n) and A.G(4,4,n). the n't value can stack randomly. I wan't to know how many non-zero values lies behind for example A.G.(1,2,n) where n is a unkown value depending how many values i already put into A.G(1,2,n), which i don't keep track on.
##### 댓글 수: 4표시숨기기 이전 댓글 수: 3
Happy PhD 2020년 4월 28일
That is correct! I want to add values after all non-zero values. And increase the matrix size in the third dimension when needed.
댓글을 달려면 로그인하십시오.
### 채택된 답변
Mrutyunjaya Hiremath 2020년 4월 28일
Hello Happy PhD,
Try this one...
clc;
clear all;
A.G(:,:,1) = [1 2 1 1 ; 4 9 5 6 ];
A.G(:,:,2) = [0 1 0 0; 0 6 0 0];
A.G(:,:,3) = [0 5 0 0; 0 0 0 0];
N = 10; % random number
idx = find(A.G == 0);
if ~isempty(idx)
A.G(idx(1)) = N;
else
n = size(A.G,3);
A.G(1,1,n+1) = N;
end
disp(A.G);
댓글을 달려면 로그인하십시오.
### 추가 답변(1개)
Rik 2020년 4월 28일
It sounds like you can just assign that value. Matlab will expand the array and fill empty positions with 0.
A.G=cat(3,[ 1 2 0 1 ; 4 9 5 6 ], [ 0 1 0 0; 0 6 0 0],[0 5 0 0; 0 0 0 0 ]);
A.G(1,2,4)=8;%Matlab fills the 4th page with zeros
disp(A.G)
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Rik 2020년 4월 28일
You can find the first empty page like this:
[r,c,val]=deal(1,2,8);
A.G=cat(3,[ 1 2 0 1 ; 4 9 5 6 ], [ 0 1 0 0; 0 6 0 0],[0 5 0 0; 0 0 0 0 ]);
N_by_rc=sum(logical(A.G),3)+1;
n=N_by_rc(r,c);
A.G(r,c,n)=val;
clc,disp(A.G)
댓글을 달려면 로그인하십시오.
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Translated by | 812 | 1,944 | {"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-2022-21 | latest | en | 0.535098 |
http://isabelle.in.tum.de/library/HOL/HOL-Algebra/Galois_Connection.html | 1,516,602,502,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00069.warc.gz | 196,404,875 | 6,610 | # Theory Galois_Connection
theory Galois_Connection
imports Complete_Lattice
```(* Title: HOL/Algebra/Galois_Connection.thy
Author: Alasdair Armstrong and Simon Foster
Copyright: Alasdair Armstrong and Simon Foster
*)
theory Galois_Connection
imports Complete_Lattice
begin
section ‹Galois connections›
subsection ‹Definition and basic properties›
record ('a, 'b, 'c, 'd) galcon =
orderA :: "('a, 'c) gorder_scheme" ("𝒳ı")
orderB :: "('b, 'd) gorder_scheme" ("𝒴ı")
lower :: "'a ⇒ 'b" ("π⇧*ı")
upper :: "'b ⇒ 'a" ("π⇩*ı")
type_synonym ('a, 'b) galois = "('a, 'b, unit, unit) galcon"
abbreviation "inv_galcon G ≡ ⦇ orderA = inv_gorder 𝒴⇘G⇙, orderB = inv_gorder 𝒳⇘G⇙, lower = upper G, upper = lower G ⦈"
definition comp_galcon :: "('b, 'c) galois ⇒ ('a, 'b) galois ⇒ ('a, 'c) galois" (infixr "∘⇩g" 85)
where "G ∘⇩g F = ⦇ orderA = orderA F, orderB = orderB G, lower = lower G ∘ lower F, upper = upper F ∘ upper G ⦈"
definition id_galcon :: "'a gorder ⇒ ('a, 'a) galois" ("I⇩g") where
"I⇩g(A) = ⦇ orderA = A, orderB = A, lower = id, upper = id ⦈"
subsection ‹Well-typed connections›
locale connection =
fixes G (structure)
assumes is_order_A: "partial_order 𝒳"
and is_order_B: "partial_order 𝒴"
and lower_closure: "π⇧* ∈ carrier 𝒳 → carrier 𝒴"
and upper_closure: "π⇩* ∈ carrier 𝒴 → carrier 𝒳"
begin
lemma lower_closed: "x ∈ carrier 𝒳 ⟹ π⇧* x ∈ carrier 𝒴"
using lower_closure by auto
lemma upper_closed: "y ∈ carrier 𝒴 ⟹ π⇩* y ∈ carrier 𝒳"
using upper_closure by auto
end
subsection ‹Galois connections›
locale galois_connection = connection +
assumes galois_property: "⟦x ∈ carrier 𝒳; y ∈ carrier 𝒴⟧ ⟹ π⇧* x ⊑⇘𝒴⇙ y ⟷ x ⊑⇘𝒳⇙ π⇩* y"
begin
lemma is_weak_order_A: "weak_partial_order 𝒳"
proof -
interpret po: partial_order 𝒳
by (metis is_order_A)
show ?thesis ..
qed
lemma is_weak_order_B: "weak_partial_order 𝒴"
proof -
interpret po: partial_order 𝒴
by (metis is_order_B)
show ?thesis ..
qed
lemma right: "⟦x ∈ carrier 𝒳; y ∈ carrier 𝒴; π⇧* x ⊑⇘𝒴⇙ y⟧ ⟹ x ⊑⇘𝒳⇙ π⇩* y"
by (metis galois_property)
lemma left: "⟦x ∈ carrier 𝒳; y ∈ carrier 𝒴; x ⊑⇘𝒳⇙ π⇩* y⟧ ⟹ π⇧* x ⊑⇘𝒴⇙ y"
by (metis galois_property)
lemma deflation: "y ∈ carrier 𝒴 ⟹ π⇧* (π⇩* y) ⊑⇘𝒴⇙ y"
by (metis Pi_iff is_weak_order_A left upper_closure weak_partial_order.le_refl)
lemma inflation: "x ∈ carrier 𝒳 ⟹ x ⊑⇘𝒳⇙ π⇩* (π⇧* x)"
by (metis (no_types, lifting) PiE galois_connection.right galois_connection_axioms is_weak_order_B lower_closure weak_partial_order.le_refl)
lemma lower_iso: "isotone 𝒳 𝒴 π⇧*"
show "weak_partial_order 𝒳"
by (metis is_weak_order_A)
show "weak_partial_order 𝒴"
by (metis is_weak_order_B)
fix x y
assume a: "x ∈ carrier 𝒳" "y ∈ carrier 𝒳" "x ⊑⇘𝒳⇙ y"
have b: "π⇧* y ∈ carrier 𝒴"
using a(2) lower_closure by blast
then have "π⇩* (π⇧* y) ∈ carrier 𝒳"
using upper_closure by blast
then have "x ⊑⇘𝒳⇙ π⇩* (π⇧* y)"
by (meson a inflation is_weak_order_A weak_partial_order.le_trans)
thus "π⇧* x ⊑⇘𝒴⇙ π⇧* y"
by (meson b a(1) Pi_iff galois_property lower_closure upper_closure)
qed
lemma upper_iso: "isotone 𝒴 𝒳 π⇩*"
apply (metis is_weak_order_B)
apply (metis is_weak_order_A)
apply (metis (no_types, lifting) Pi_mem deflation is_weak_order_B lower_closure right upper_closure weak_partial_order.le_trans)
done
lemma lower_comp: "x ∈ carrier 𝒳 ⟹ π⇧* (π⇩* (π⇧* x)) = π⇧* x"
by (meson deflation funcset_mem inflation is_order_B lower_closure lower_iso partial_order.le_antisym upper_closure use_iso2)
lemma lower_comp': "x ∈ carrier 𝒳 ⟹ (π⇧* ∘ π⇩* ∘ π⇧*) x = π⇧* x"
lemma upper_comp: "y ∈ carrier 𝒴 ⟹ π⇩* (π⇧* (π⇩* y)) = π⇩* y"
proof -
assume a1: "y ∈ carrier 𝒴"
hence f1: "π⇩* y ∈ carrier 𝒳" using upper_closure by blast
have f2: "π⇧* (π⇩* y) ⊑⇘𝒴⇙ y" using a1 deflation by blast
have f3: "π⇩* (π⇧* (π⇩* y)) ∈ carrier 𝒳"
using f1 lower_closure upper_closure by auto
have "π⇧* (π⇩* y) ∈ carrier 𝒴" using f1 lower_closure by blast
thus "π⇩* (π⇧* (π⇩* y)) = π⇩* y"
by (meson a1 f1 f2 f3 inflation is_order_A partial_order.le_antisym upper_iso use_iso2)
qed
lemma upper_comp': "y ∈ carrier 𝒴 ⟹ (π⇩* ∘ π⇧* ∘ π⇩*) y = π⇩* y"
lemma adjoint_idem1: "idempotent 𝒴 (π⇧* ∘ π⇩*)"
by (simp add: idempotent_def is_order_B partial_order.eq_is_equal upper_comp)
lemma adjoint_idem2: "idempotent 𝒳 (π⇩* ∘ π⇧*)"
by (simp add: idempotent_def is_order_A partial_order.eq_is_equal lower_comp)
lemma fg_iso: "isotone 𝒴 𝒴 (π⇧* ∘ π⇩*)"
by (metis iso_compose lower_closure lower_iso upper_closure upper_iso)
lemma gf_iso: "isotone 𝒳 𝒳 (π⇩* ∘ π⇧*)"
by (metis iso_compose lower_closure lower_iso upper_closure upper_iso)
lemma semi_inverse1: "x ∈ carrier 𝒳 ⟹ π⇧* x = π⇧* (π⇩* (π⇧* x))"
by (metis lower_comp)
lemma semi_inverse2: "x ∈ carrier 𝒴 ⟹ π⇩* x = π⇩* (π⇧* (π⇩* x))"
by (metis upper_comp)
theorem lower_by_complete_lattice:
assumes "complete_lattice 𝒴" "x ∈ carrier 𝒳"
shows "π⇧*(x) = ⨅⇘𝒴⇙ { y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩*(y) }"
proof -
interpret Y: complete_lattice 𝒴
show ?thesis
proof (rule Y.le_antisym)
show x: "π⇧* x ∈ carrier 𝒴"
using assms(2) lower_closure by blast
show "π⇧* x ⊑⇘𝒴⇙ ⨅⇘𝒴⇙{y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y}"
proof (rule Y.weak.inf_greatest)
show "{y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y} ⊆ carrier 𝒴"
by auto
show "π⇧* x ∈ carrier 𝒴" by (fact x)
fix z
assume "z ∈ {y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y}"
thus "π⇧* x ⊑⇘𝒴⇙ z"
using assms(2) left by auto
qed
show "⨅⇘𝒴⇙{y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y} ⊑⇘𝒴⇙ π⇧* x"
proof (rule Y.weak.inf_lower)
show "{y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y} ⊆ carrier 𝒴"
by auto
show "π⇧* x ∈ {y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y}"
proof (auto)
show "π⇧* x ∈ carrier 𝒴" by (fact x)
show "x ⊑⇘𝒳⇙ π⇩* (π⇧* x)"
using assms(2) inflation by blast
qed
qed
show "⨅⇘𝒴⇙{y ∈ carrier 𝒴. x ⊑⇘𝒳⇙ π⇩* y} ∈ carrier 𝒴"
by (auto intro: Y.weak.inf_closed)
qed
qed
theorem upper_by_complete_lattice:
assumes "complete_lattice 𝒳" "y ∈ carrier 𝒴"
shows "π⇩*(y) = ⨆⇘𝒳⇙ { x ∈ carrier 𝒳. π⇧*(x) ⊑⇘𝒴⇙ y }"
proof -
interpret X: complete_lattice 𝒳
show ?thesis
proof (rule X.le_antisym)
show y: "π⇩* y ∈ carrier 𝒳"
using assms(2) upper_closure by blast
show "π⇩* y ⊑⇘𝒳⇙ ⨆⇘𝒳⇙{x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y}"
proof (rule X.weak.sup_upper)
show "{x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y} ⊆ carrier 𝒳"
by auto
show "π⇩* y ∈ {x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y}"
proof (auto)
show "π⇩* y ∈ carrier 𝒳" by (fact y)
show "π⇧* (π⇩* y) ⊑⇘𝒴⇙ y"
qed
qed
show "⨆⇘𝒳⇙{x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y} ⊑⇘𝒳⇙ π⇩* y"
proof (rule X.weak.sup_least)
show "{x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y} ⊆ carrier 𝒳"
by auto
show "π⇩* y ∈ carrier 𝒳" by (fact y)
fix z
assume "z ∈ {x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y}"
thus "z ⊑⇘𝒳⇙ π⇩* y"
qed
show "⨆⇘𝒳⇙{x ∈ carrier 𝒳. π⇧* x ⊑⇘𝒴⇙ y} ∈ carrier 𝒳"
by (auto intro: X.weak.sup_closed)
qed
qed
end
lemma dual_galois [simp]: " galois_connection ⦇ orderA = inv_gorder B, orderB = inv_gorder A, lower = f, upper = g ⦈
= galois_connection ⦇ orderA = A, orderB = B, lower = g, upper = f ⦈"
by (auto simp add: galois_connection_def galois_connection_axioms_def connection_def dual_order_iff)
definition lower_adjoint :: "('a, 'c) gorder_scheme ⇒ ('b, 'd) gorder_scheme ⇒ ('a ⇒ 'b) ⇒ bool" where
"lower_adjoint A B f ≡ ∃g. galois_connection ⦇ orderA = A, orderB = B, lower = f, upper = g ⦈"
definition upper_adjoint :: "('a, 'c) gorder_scheme ⇒ ('b, 'd) gorder_scheme ⇒ ('b ⇒ 'a) ⇒ bool" where
"upper_adjoint A B g ≡ ∃f. galois_connection ⦇ orderA = A, orderB = B, lower = f, upper = g ⦈"
lemma lower_type: "lower_adjoint A B f ⟹ f ∈ carrier A → carrier B"
lemma upper_type: "upper_adjoint A B g ⟹ g ∈ carrier B → carrier A"
subsection ‹Composition of Galois connections›
lemma id_galois: "partial_order A ⟹ galois_connection (I⇩g(A))"
by (simp add: id_galcon_def galois_connection_def galois_connection_axioms_def connection_def)
lemma comp_galcon_closed:
assumes "galois_connection G" "galois_connection F" "𝒴⇘F⇙ = 𝒳⇘G⇙"
shows "galois_connection (G ∘⇩g F)"
proof -
interpret F: galois_connection F
interpret G: galois_connection G
have "partial_order 𝒳⇘G ∘⇩g F⇙"
moreover have "partial_order 𝒴⇘G ∘⇩g F⇙"
moreover have "π⇧*⇘G⇙ ∘ π⇧*⇘F⇙ ∈ carrier 𝒳⇘F⇙ → carrier 𝒴⇘G⇙"
using F.lower_closure G.lower_closure assms(3) by auto
moreover have "π⇩*⇘F⇙ ∘ π⇩*⇘G⇙ ∈ carrier 𝒴⇘G⇙ → carrier 𝒳⇘F⇙"
using F.upper_closure G.upper_closure assms(3) by auto
moreover
have "⋀ x y. ⟦x ∈ carrier 𝒳⇘F⇙; y ∈ carrier 𝒴⇘G⇙ ⟧ ⟹
(π⇧*⇘G⇙ (π⇧*⇘F⇙ x) ⊑⇘𝒴⇘G⇙⇙ y) = (x ⊑⇘𝒳⇘F⇙⇙ π⇩*⇘F⇙ (π⇩*⇘G⇙ y))"
by (metis F.galois_property F.lower_closure G.galois_property G.upper_closure assms(3) Pi_iff)
ultimately show ?thesis
by (simp add: comp_galcon_def galois_connection_def galois_connection_axioms_def connection_def)
qed
lemma comp_galcon_right_unit [simp]: "F ∘⇩g I⇩g(𝒳⇘F⇙) = F"
lemma comp_galcon_left_unit [simp]: "I⇩g(𝒴⇘F⇙) ∘⇩g F = F"
lemma galois_connectionI:
assumes
"partial_order A" "partial_order B"
"L ∈ carrier A → carrier B" "R ∈ carrier B → carrier A"
"isotone A B L" "isotone B A R"
"⋀ x y. ⟦ x ∈ carrier A; y ∈ carrier B ⟧ ⟹ L x ⊑⇘B⇙ y ⟷ x ⊑⇘A⇙ R y"
shows "galois_connection ⦇ orderA = A, orderB = B, lower = L, upper = R ⦈"
using assms by (simp add: galois_connection_def connection_def galois_connection_axioms_def)
lemma galois_connectionI':
assumes
"partial_order A" "partial_order B"
"L ∈ carrier A → carrier B" "R ∈ carrier B → carrier A"
"isotone A B L" "isotone B A R"
"⋀ X. X ∈ carrier(B) ⟹ L(R(X)) ⊑⇘B⇙ X"
"⋀ X. X ∈ carrier(A) ⟹ X ⊑⇘A⇙ R(L(X))"
shows "galois_connection ⦇ orderA = A, orderB = B, lower = L, upper = R ⦈"
using assms
by (auto simp add: galois_connection_def connection_def galois_connection_axioms_def, (meson PiE isotone_def weak_partial_order.le_trans)+)
subsection ‹Retracts›
locale retract = galois_connection +
assumes retract_property: "x ∈ carrier 𝒳 ⟹ π⇩* (π⇧* x) ⊑⇘𝒳⇙ x"
begin
lemma retract_inverse: "x ∈ carrier 𝒳 ⟹ π⇩* (π⇧* x) = x"
by (meson funcset_mem inflation is_order_A lower_closure partial_order.le_antisym retract_axioms retract_axioms_def retract_def upper_closure)
lemma retract_injective: "inj_on π⇧* (carrier 𝒳)"
by (metis inj_onI retract_inverse)
end
theorem comp_retract_closed:
assumes "retract G" "retract F" "𝒴⇘F⇙ = 𝒳⇘G⇙"
shows "retract (G ∘⇩g F)"
proof -
interpret f: retract F
interpret g: retract G
interpret gf: galois_connection "(G ∘⇩g F)"
by (simp add: assms(1) assms(2) assms(3) comp_galcon_closed retract.axioms(1))
show ?thesis
proof
fix x
assume "x ∈ carrier 𝒳⇘G ∘⇩g F⇙"
thus "le 𝒳⇘G ∘⇩g F⇙ (π⇩*⇘G ∘⇩g F⇙ (π⇧*⇘G ∘⇩g F⇙ x)) x"
using assms(3) f.inflation f.lower_closed f.retract_inverse g.retract_inverse by (auto simp add: comp_galcon_def)
qed
qed
subsection ‹Coretracts›
locale coretract = galois_connection +
assumes coretract_property: "y ∈ carrier 𝒴 ⟹ y ⊑⇘𝒴⇙ π⇧* (π⇩* y)"
begin
lemma coretract_inverse: "y ∈ carrier 𝒴 ⟹ π⇧* (π⇩* y) = y"
by (meson coretract_axioms coretract_axioms_def coretract_def deflation funcset_mem is_order_B lower_closure partial_order.le_antisym upper_closure)
lemma retract_injective: "inj_on π⇩* (carrier 𝒴)"
by (metis coretract_inverse inj_onI)
end
theorem comp_coretract_closed:
assumes "coretract G" "coretract F" "𝒴⇘F⇙ = 𝒳⇘G⇙"
shows "coretract (G ∘⇩g F)"
proof -
interpret f: coretract F
interpret g: coretract G
interpret gf: galois_connection "(G ∘⇩g F)"
by (simp add: assms(1) assms(2) assms(3) comp_galcon_closed coretract.axioms(1))
show ?thesis
proof
fix y
assume "y ∈ carrier 𝒴⇘G ∘⇩g F⇙"
thus "le 𝒴⇘G ∘⇩g F⇙ y (π⇧*⇘G ∘⇩g F⇙ (π⇩*⇘G ∘⇩g F⇙ y))"
by (simp add: comp_galcon_def assms(3) f.coretract_inverse g.coretract_property g.upper_closed)
qed
qed
subsection ‹Galois Bijections›
locale galois_bijection = connection +
assumes lower_iso: "isotone 𝒳 𝒴 π⇧*"
and upper_iso: "isotone 𝒴 𝒳 π⇩*"
and lower_inv_eq: "x ∈ carrier 𝒳 ⟹ π⇩* (π⇧* x) = x"
and upper_inv_eq: "y ∈ carrier 𝒴 ⟹ π⇧* (π⇩* y) = y"
begin
lemma lower_bij: "bij_betw π⇧* (carrier 𝒳) (carrier 𝒴)"
by (rule bij_betwI[where g="π⇩*"], auto intro: upper_inv_eq lower_inv_eq upper_closed lower_closed)
lemma upper_bij: "bij_betw π⇩* (carrier 𝒴) (carrier 𝒳)"
by (rule bij_betwI[where g="π⇧*"], auto intro: upper_inv_eq lower_inv_eq upper_closed lower_closed)
sublocale gal_bij_conn: galois_connection
apply (unfold_locales, auto)
using lower_closed lower_inv_eq upper_iso use_iso2 apply fastforce
using lower_iso upper_closed upper_inv_eq use_iso2 apply fastforce
done
sublocale gal_bij_ret: retract
by (unfold_locales, simp add: gal_bij_conn.is_weak_order_A lower_inv_eq weak_partial_order.le_refl)
sublocale gal_bij_coret: coretract
by (unfold_locales, simp add: gal_bij_conn.is_weak_order_B upper_inv_eq weak_partial_order.le_refl)
end
theorem comp_galois_bijection_closed:
assumes "galois_bijection G" "galois_bijection F" "𝒴⇘F⇙ = 𝒳⇘G⇙"
shows "galois_bijection (G ∘⇩g F)"
proof -
interpret f: galois_bijection F
interpret g: galois_bijection G
interpret gf: galois_connection "(G ∘⇩g F)"
by (simp add: assms(3) comp_galcon_closed f.gal_bij_conn.galois_connection_axioms g.gal_bij_conn.galois_connection_axioms galois_connection.axioms(1))
show ?thesis
proof
show "isotone 𝒳⇘G ∘⇩g F⇙ 𝒴⇘G ∘⇩g F⇙ π⇧*⇘G ∘⇩g F⇙"
by (simp add: comp_galcon_def, metis comp_galcon_def galcon.select_convs(1) galcon.select_convs(2) galcon.select_convs(3) gf.lower_iso)
show "isotone 𝒴⇘G ∘⇩g F⇙ 𝒳⇘G ∘⇩g F⇙ π⇩*⇘G ∘⇩g F⇙"
fix x
assume "x ∈ carrier 𝒳⇘G ∘⇩g F⇙"
thus "π⇩*⇘G ∘⇩g F⇙ (π⇧*⇘G ∘⇩g F⇙ x) = x"
using assms(3) f.lower_closed f.lower_inv_eq g.lower_inv_eq by (auto simp add: comp_galcon_def)
next
fix y
assume "y ∈ carrier 𝒴⇘G ∘⇩g F⇙"
thus "π⇧*⇘G ∘⇩g F⇙ (π⇩*⇘G ∘⇩g F⇙ y) = y"
by (simp add: comp_galcon_def assms(3) f.upper_inv_eq g.upper_closed g.upper_inv_eq)
qed
qed
end
``` | 5,641 | 13,349 | {"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-2018-05 | longest | en | 0.531353 |
https://www.studypool.com/discuss/1142841/assume-that-points-e-f-and-g-are-colinear-and-ef-gf-eg?free | 1,508,411,511,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823282.42/warc/CC-MAIN-20171019103222-20171019123222-00189.warc.gz | 988,227,133 | 14,021 | ##### assume that points E,F, and G are colinear and EF+GF=EG
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Point F is between points E and G
Point E is between points F and G
EF is greater than G and F
Point G is between points E and F
Aug 30th, 2015
Colinear points are points on the same line.
if EF +GF =EG
then
Point F is between points E and G
Aug 31st, 2015
...
Aug 30th, 2015
...
Aug 30th, 2015
Oct 19th, 2017
check_circle | 151 | 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} | 2.515625 | 3 | CC-MAIN-2017-43 | latest | en | 0.91871 |
https://www.shaalaa.com/question-bank-solutions/a-missing-helicopter-is-reported-to-have-crashed-somewhere-in-the-rectangular-region-shown-in-the-figure-what-is-the-probability-that-it-crashed-inside-the-lake-shown-in-the-figure-probability-event_205718 | 1,652,818,648,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00177.warc.gz | 1,166,784,165 | 10,224 | # A missing helicopter is reported to have crashed somewhere in the rectangular region shown in the figure. What is the probability that it crashed inside the lake shown in the figure? - Algebra
Sum
A missing helicopter is reported to have crashed somewhere in the rectangular region shown in the figure. What is the probability that it crashed inside the lake shown in the figure?
#### Solution
The helicopter is equally likely to crash anywhere in the region. In the figure, the length and the breadth of the rectangle are 9 m and 4.5 m respectively.
Area of the entire region where the helicopter can crash
= (9 × 4.5) m2 = 40.5 m2
Let A be the event that helicopter crashed inside the lake.
The lake is rectangular shaped.
Length of lake = 9 – 6 = 3 m
Breadth of lake = 4.5 – 2 = 2.5 m
Area of lake = length × breadth
= 3 × 2.5
= 7.5 m2
∴ P(A) = ("n"("A"))/("n"("S"))
= 7.5/40.5
= 75/405
= 5/27
The probability that the helicopter crashed inside the lake is 5/27
Concept: Probability of an Event
Is there an error in this question or solution? | 292 | 1,066 | {"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} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.930198 |
https://www.hpmuseum.org/forum/thread-3658-post-32969.html | 1,579,624,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00559.warc.gz | 915,970,554 | 6,242 | How can I calculate it with my HP...?
04-16-2015, 04:13 PM
Post: #1
Csaba Tizedes Senior Member Posts: 404 Joined: May 2014
How can I calculate it with my HP...?
Can anybody give me a good advice how can I calculate the following probability example on my HP-20S?
p= Cn,r(5,2) × Cn,r(85,3) ÷ Cn,r(90,5)
The right result is for FIX 4 is 0.0225
Thanks!
Csaba
04-16-2015, 05:03 PM
Post: #2
PANAMATIK Senior Member Posts: 933 Joined: Oct 2014
RE: How can I calculate it with my HP...?
(04-16-2015 04:13 PM)Csaba Tizedes Wrote: Can anybody give me a good advice how can I calculate the following probability example on my HP-20S?
p= Cn,r(5,2) × Cn,r(85,3) ÷ Cn,r(90,5)
The right result is for FIX 4 is 0.0225
Thanks!
Csaba
If you use registers it can easily be done with
5 INPUT 2 Cn,r STO 0
85 INPUT 3 Cn,r STO 1
90 INPUT 5 Cn,r STO 2
RCL 0 x RCL 1 / RCL 2 =
Bernhard
That's one small step for a man - one giant leap for mankind.
04-16-2015, 07:38 PM
Post: #3
Csaba Tizedes Senior Member Posts: 404 Joined: May 2014
RE: How can I calculate it with my HP...?
(04-16-2015 05:03 PM)PANAMATIK Wrote: If you use registers...
Thanks Bernhard - I figured out it also, but if I ask my question as "Can anybody give me a good advice how can I swim across the Atlantic Sea?" And you answer it as "If you use a boat it can easily be done", you can see that the "sailing" and "swimming" is not the same things...
I tried it with some ways on my HP (for example with "forced" parentheses) and it is not works. I really like this little algebraic calc and I really like to use its functions, programming, statistics, and so on... but it's a sad thing to find a bug in the data entry routine.
My calc is not here now, but I think I'll try all the functions which are use the INPUT key.
Csaba
04-16-2015, 07:54 PM
Post: #4
walter b On Vacation Posts: 1,957 Joined: Dec 2013
RE: How can I calculate it with my HP...?
Thanks for your explanation - I first thought you were looking for a way to simplify the problem e.g. by cancelation of terms.
d:-/
04-16-2015, 08:56 PM
Post: #5
PANAMATIK Senior Member Posts: 933 Joined: Oct 2014
RE: How can I calculate it with my HP...?
(04-16-2015 07:38 PM)Csaba Tizedes Wrote:
(04-16-2015 05:03 PM)PANAMATIK Wrote: If you use registers...
Thanks Bernhard - I figured out it also, but if I ask my question as "Can anybody give me a good advice how can I swim across the Atlantic Sea?" And you answer it as "If you use a boat it can easily be done", you can see that the "sailing" and "swimming" is not the same things...
I tried it with some ways on my HP (for example with "forced" parentheses) and it is not works. I really like this little algebraic calc and I really like to use its functions, programming, statistics, and so on... but it's a sad thing to find a bug in the data entry routine.
Csaba
Hi Csaba,
Of course, I tried the parentheses too, and it didn't work. The parentheses were just ignored. It is well known that the HP-20S behaviour is not consistent. Ignoring parentheses in an algebraic calculator is a bug. Perhaps there is a little relict of RPN left inside.
You have to take the sailing boat.
Bernhard
That's one small step for a man - one giant leap for mankind.
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Tension and acceleration in a string:
Consider a block supported by a string. The upper end of the string is fixed on a stand. Let w be the weight of the block. The block pulls the string downwards by its weight. This causes a tension T in the string. The tension T in the string is acting upwards at the block. As the block is at rest, therefore, the weight of the block acting downwards must be balanced by the upwards tension T in the string. Thus, the tension T in the string must be equal and opposite to the weight w of the block. | 135 | 619 | {"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} | 3.28125 | 3 | CC-MAIN-2023-23 | latest | en | 0.948188 |
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# Orthodiagonal quadrilateral (medians of the four triangles)
## Description
A quadrilateral is a polygon with four sides (or edges) and four vertices or corners. An orthodiagonal quadrilateral is a quadrilateral in which the diagonals cross at right angles. In other words, it is a four-sided figure in which the line segments between non-adjacent vertices are orthogonal to each other. For any orthodiagonal quadrilateral (ABCD), there is a relation between the medians in the four triangles formed by the diagonal intersection P and the vertices of the convex quadrilateral.
Related formulas
## Variables
m1 The median of one of the triangles from the intersection P to the opposite side(AB) (m) m3 The median of one of the triangles from the intersection P to the opposite side(DC) (m) m2 The median of one of the triangles from the intersection P to the opposite side(BC) (m) m4 The median of one of the triangles from the intersection P to the opposite side(DA) (m) | 221 | 980 | {"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-2024-10 | latest | en | 0.91614 |
https://www.baeldung.com/cs/stable-sorting-algorithms | 1,719,332,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00605.warc.gz | 577,371,630 | 61,116 | ## 1. Overview
In this tutorial, we’ll learn what stable sorting algorithms are and how they work. Further, we’ll explore when the stability of sorting matters.
## 2. Stability in Sorting Algorithms
The stability of a sorting algorithm is concerned with how the algorithm treats equal (or repeated) elements. Stable sorting algorithms preserve the relative order of equal elements, while unstable sorting algorithms don’t. In other words, stable sorting maintains the position of two equals elements relative to one another.
Let A be a collection of elements and < be a strict weak ordering on the elements. Further, let B be the collection of elements in A in the sorted order. Let’s consider two equal elements in A at indices i and j, i.e, A[i] and A[j], that end up at indices m and n respectively in B. We can classify the sorting as stable if:
``````i < j and A[i] = A[j] and m < n
``````
Let’s understand the concept with the help of an example. We have an array of integers A: Â [ 5, 8, 9, 8, 3 ]. Let’s represent our array using color-coded balls, where any two balls with the same integer will have a different color which would help us keep track of equal elements (8 in our case):
Stable sorting maintains the order of the two equal balls numbered 8, whereas unstable sorting may invert the relative order of the two 8s.
## 3. When Stability Matters
### 3.1. Distinguishing Between Equal Elements
All sorting algorithms use a key to determine the ordering of the elements in the collection, called the sort key.
If the sort key is the (entire) element itself, equal elements are indistinguishable, such as integers or strings.
On the other hand, equal elements are distinguishable if the sort key is made up of one or more, but not all attributes of the element, such as age in an Employee class.
### 3.2. Stable Sorting Is Important, Sometimes
We don’t always need stable sorting. Stability is not a concern if:
• equal elements are indistinguishable, or
• all the elements in the collection are distinct
When equal elements are distinguishable, stability is imperative.  For instance, if the collection already has some order, then sorting on another key must preserve that order.
For example, let’s say we are computing the word count of each distinct word in a text file. Now, we need to report the results in decreasing order of count, and further sorted alphabetically in case two words have the same count.
First step – compute the word count:
``````Input:
how much wood would woodchuck chuck if woodchuck could chuck wood
Output:
how 1
much 1
wood 2
would 1
woodchuck 2
chuck 2
if 1
could 1
``````
Second step – sort the whole list lexicographically, then by word count:
``````First pass, sorted lexicographically:
(chuck, 2)
(could, 1)
(how, 1)
(if, 1)
(much, 1)
(wood, 2)
(woodchuck, 2)
(would, 1)
``````
``````Second pass, sorted by count using an unstable sort:
(wood, 2)
(chuck, 2)
(woodchuck, 2)
(could, 1)
(how, 1)
(if, 1)
(would, 1)
(much, 1)
``````
As we have used an unstable sort, the second pass does not maintain the lexicographical order.
That’s where the stable sort comes into the picture. Since we already had sorted the list lexicographically, using a stable sort to by word count does not change the order of equal elements anymore. As a result words with the same word count remain sorted lexicographically.
``````Second pass, sorted by count using a stable sort:
(chuck, 2)
(wood, 2)
(woodchuck, 2)
(could, 1)
(how, 1)
(if, 1)
(much, 1)
(would, 1)``````
Radix Sort is an integer sorting algorithm that depends on a sorting subroutine that must be stable. It is a non-comparison based sorting algorithm that sorts a collection of integers. It groups keys by individual digits that share the same significant position and value.
Let’s unpack the formal definition and restate the basic idea:
``````for each digit 'k' from the least significant digit (LSD) to the most significant digit (MSD) of a number:
apply counting-sort algorithm on digit 'k' to sort the input array
``````
We are using Counting Sort as a subroutine in Radix Sort. Counting Sort is a stable integer sorting algorithm. We don’t have to understand how it works, but that Counting Sort is stable.
Let’s look at an illustrative example:
Each invocation of the Counting Sort subroutine preserves the order from the previous invocations. For example, while sorting on the tens’ place digit (second invocation) 9881 shifts downwards, but stays above 9888 maintaining their relative order.
Thus, Radix Sort utilizes the stability of the Counting Sort algorithm and provides linear time integer sorting.
## 4. Stable and Unstable Sorting Algorithms
Several common sorting algorithms are stable by nature, such as Merge Sort, Timsort, Counting Sort, Insertion Sort, and Bubble Sort. Others such as Quicksort, Heapsort and Selection Sort are unstable.
We can modify unstable sorting algorithms to be stable. For instance, we can use extra space to maintain stability in Quicksort.
## 5. Conclusion
In this tutorial, we learned about stable sorting algorithms and looked at when stability matters, using Radix Sort as an example. | 1,291 | 5,223 | {"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} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.839226 |
https://holooly.com/solutions-v20/calculate-the-deflection-at-the-free-end-of-the-two-cell-beam-shown-in-fig-23-15-allowing-for-both-bending-and-shear-effects-the-booms-carry-all-the-direct-stresses-while-the-skin-panels-of-constan/ | 1,675,181,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00342.warc.gz | 310,715,051 | 20,935 | ## Subscribe \$4.99/month
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## Q. 23.5
Calculate the deflection at the free end of the two-cell beam shown in Fig. 23.15 allowing for both bending and shear effects. The booms carry all the direct stresses while the skin panels, of constant thickness throughout, are effective only in shear.
Take E = 69 000 N/mm² and G = 25 900 N/mm²
Boom areas: $B_1=B_3=B_4=B_6=650 \mathrm{~mm}^2 \quad B_2=B_5=1300 \mathrm{~mm}^2$
## Verified Solution
The beam cross-section is symmetrical about a horizontal axis and carries a vertical load at its free end through the shear centre. The deflection Δ at the free end is then, from Eqs (20.17) and (20.19)
$\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1} M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{x x}}\right) \mathrm{d} z$ (20.17)
$\Delta_S=\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z$ (20.19)
$\Delta=\int_0^{2000} \frac{M_{x, 0} M_{x, 1}}{E I_{x x}} \mathrm{~d} z+\int_0^{2000}\left(\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z$ (i)
where
$M_{x, 0}=-44.5 \times 10^3(2000-z) \quad M_{x, 1}=-(2000-z)$
and
$I_{x x}=4 \times 650 \times 125^2+2 \times 1300 \times 125^2=81.3 \times 10^6 \mathrm{~mm}^4$
also
$S_{y, 0}=44.5 \times 10^3 \mathrm{~N} \quad S_{y, 1}=1$
The $q_0 \text { and } q_1$ shear flow distributions are obtained as previously described (note dθ/dz = 0 for a shear load through the shear centre) and are
\begin{aligned}&q_{0,12}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,23}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,43}=50.3 \mathrm{~N} / \mathrm{mm} \\&q_{0,45}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,56}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,61}=54.1 \mathrm{~N} / \mathrm{mm} \\&q_{0,52}=73.6 \mathrm{~N} / \mathrm{mm} \text { at all sections of the beam }\end{aligned}
The $q_1$ shear flows in this case are given by $q_0 / 44.5 \times 10^3$. Thus
\begin{aligned}\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s=& \frac{1}{25900 \times 2 \times 44.5 \times 10^3}\left(9.6^2 \times 250 \times 2+5.8^2 \times 500 \times 2\right.\\&\left.+50.3^2 \times 250+54.1^2 \times 250+73.6^2 \times 250\right) \\=& 1.22 \times 10^{-3}\end{aligned}
Hence, from Eq. (i)
$\Delta=\int_0^{2000} \frac{44.5 \times 10^3(2000-z)^2}{69000 \times 81.3 \times 10^6} \mathrm{~d} z+\int_0^{2000} 1.22 \times 10^{-3} \mathrm{~d} z$
giving
Δ = 23.5 mm | 1,220 | 3,165 | {"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": 13, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.69845 |
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• Aug 11th 2008, 12:27 PM
comet2000
single algebraic fraction
Write each expression as a single algebraic fraction.
(1-xy^-1)^-1
• Aug 11th 2008, 12:33 PM
Chop Suey
$(1-(xy)^{-1})^{-1}$
Remember:
$a^{-1} = \frac{1}{a}$
So:
$(1-(xy)^{-1})^{-1} = \frac{1}{1-(xy)^{-1}} = \frac{1}{1-\frac{1}{xy}}$
Combine 1 and $\frac{1}{xy}$. Remember that $1 = \frac{xy}{xy}$.
$\frac{1}{1-\frac{1}{xy}} = \frac{1}{\frac{xy}{xy}-\frac{1}{xy}} = \frac{1}{\frac{xy - 1}{xy}}$
Now, this expression can be rearranged like this:
$\frac{1}{\frac{xy - 1}{xy}} = 1 \cdot \frac{xy}{xy-1} = \frac{xy}{xy-1}$
• Aug 11th 2008, 01:16 PM
wingless
$xy^{-1}$ can mean $\frac{x}{y}$ too.
• Aug 11th 2008, 01:33 PM
Chop Suey
If you meant $xy^{-1}$ as Wingless pointed out:
$(1-xy^{-1})^{-1}$
$= \frac{1}{1-\frac{x}{y}}$
$= \frac{1}{\frac{y-x}{y}}$
$= 1 \cdot \frac{y}{y-x} = \frac{y}{y-x}$ | 397 | 890 | {"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": 14, "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-2017-43 | longest | en | 0.570188 |
http://www.lawofonesociety.com/index.php/higher-density-lessons-4-5-6-7/item/349-how-wind-fire-formed-earth-water-and-the-timeless-beginnings-of-1st-density | 1,603,862,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896778.71/warc/CC-MAIN-20201028044037-20201028074037-00677.warc.gz | 159,148,521 | 26,081 | Oct 28, 2020 Last Updated 1:47 AM, Oct 15, 2020
## How 'Wind & Fire' Formed 'Earth & Water' and The 'Timeless' Beginnings of 1st Density
Published in Higher Density Lessons
The Law of One, Book IV, Session 76
Questioner: I am going to ask some questions now that may be a little off the center of what we are trying to do.
I’m not sure because I’m trying to, with these questions, unscramble something that I consider very basic to what we are doing.
Please forgive my lack of ability in questioning since this is a difficult concept for me.
Could you give me an idea of the length of the first and second densities as they occurred for this planet?
Ra: I am Ra. There is no method of estimation of the time/space before timelessness gave way in your first density
To the beginnings of your time, the measurement would be vast and yet this vastness is meaningless.
Upon the entry into the constructed space/time your first density
spanned a bridge of space/time and time/space of perhaps two billion of your years.
Second density
is more easily estimated and represents your longest density in terms of the span of space/time.
We may estimate that time as approximately 4. 6 billion years
These approximations are exceedingly rough due to the somewhat uneven development which is characteristic of creations which are built upon the foundation stone of free will.
The Law of One, Book IV, Session 78
Could you tell me how, in the first density, wind and fire teach earth and water?
Ra: I am Ra.
You may see the air and fire
of that which is chaos as literally illuminating and forming the formless, for earth and water were, in the timeless state, unformed.
As the active principles of fire and air blow and burn incandescently about that which nurtures that which is to come, the water learns to become sea, lake, and river offering the opportunity for viable life.
The earth learns to be shaped, thus offering the opportunity for viable life.
The Law of One, Book II, Session 41
Questioner: I am going to make a statement of my understanding and ask you to correct me.
I intuitively see the first-density being formed by an energy center which is a vortex.
This vortex then causes these spinning motions that I have mentioned before of vibration which is light which then starts to condense into materials of the first-density.
Is this correct?
Ra: I am Ra. This is correct as far as your reasoning has taken you.
However, it is well to point out that the Logos has the plan of all the densities of the octave in potential completion before entering the space/time continuum in first-density
Thus the energy centers exist before they are manifest.
Questioner: Then what is the simplest being that is manifested?
I am supposing that it might be a single cell or something like that.
How does it function with respect to energy centers?
Ra: I am Ra. The simplest manifest being is light or what you have called the photon
In relationship to energy centers it may be seen to be the center or foundation of all articulated energy fields.
Questioner: When first-density is formed we have fire, air, earth, and water.
There is at some time the first movement or individuation of life into a portion of consciousness that is self-mobile.
Could you describe the process of the creation of this and what type of energy center it has?
Ra: I am Ra.
The first or red-ray density
though attracted towards growth, is not in the proper vibration for those conditions conducive to what you may call the spark of awareness.
As the vibratory energies move from red to orange the vibratory environment is such as to stimulate those chemical substances which lately had been inert to combine in such a fashion that love and light begin the function of growth.
The supposition which you had earlier made concerning single-celled entities such as the polymorphous dynaflagallate is correct.
The mechanism is one of the attraction of upward spiraling light
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14 Sep 2015 Wanderers | 1,078 | 4,562 | {"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-2020-45 | latest | en | 0.921102 |
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# Unknown Signal Delay
Module by: Don Johnson. E-mail the author
A uniformly most powerful decision rule may not exist when an unknown parameter appears in a nonlinear way in the signal model. Most pertinent to array processing is the unknown time origin case: the signal has been subjected to an unknown delay ( slΔ s l Δ , Δ=? Δ ? ) and we must determine the signal's presence. The likelihood ratio cannot be manipulated so that the sufficient statistic can be computed without having a value for ΔΔ. Thus, the search for a uniformly most powerful test ends in failure and other methods must be sought. As expected, we resort to the generalized likelihood ratio test.
More specifically, consider the binary test where a signal is either present ( 1 1 ) or not ( 0 0 ). The signal waveform is known, but its time origin is not. For all possible values of ΔΔ, the delayed signal is assumed to lie entirely in the observations (Figure 1). This signal model is ubiquitous in active sonar and radar, where the reflected signal's exact time-of-arrival is not known and we want to determine whether a return is present or not and the value of the delay. 1 Additive white Gaussian noise is assumed present. The conditional density of the observations made under 1 1 is p r | 1 Δ r=12πσ2L2e(12σ2l=0L1rlslΔ2) p r 1 Δ r 1 2 σ 2 L 2 1 2 σ 2 l 0 L 1 r l s l Δ 2
The exponent contains the only portion of this conditional density that depends on the unknown quantity ΔΔ. Maximizing the conditional density with respect to ΔΔ is equivalent to maximizing l=0L1rlslΔ12s2lΔ l 0 L 1 r l s l Δ 1 2 s l Δ 2 . As the signal is assumed to be contained entirely in the observations for all possible values of ΔΔ, the second term does not depend on ΔΔ and equals half of the signal energy EE. Rather than analytically maximizing the first term now, we simply write the logarithm of the generalized likelihood ratio test as max Δ Δ l =ΔΔ+D1rlslΔ 0 1 σ2lnη+E2 Δ l Δ Δ D 1 r l s l Δ 0 1 σ 2 η E 2 where the non-zero portion of the summation is expressed explicitly. Using the matched filter interpretation of the sufficient statistic, this decision rule is expressed by max Δ Δ rl*sD1l| l =D1+Δ 0 1 γ l D 1 Δ Δ r l s D 1 l 0 1 γ This formulation suggests that the matched filter having a unit-sample response equal to the zero-origin signal be evaluated for each possible value of ΔΔ and that we use the maximim value of the resulting output in the decision rule. In the known-delay case, the matched-filter output is sampled at the "end" of the signal; here, the filter, which has a duration DD less than the observation interval LL, is allowed to continue processing over the allowed values of signal delay with the maximum output value chosen. The result of this procedure is illustrated here. There two signals, each having the same energy, are passed through the appropriate matched filter. Note that the index at which the maximim output occurs is the maximim likelihood estimate of ΔΔ. Thus, the detection and the estimation problems are solved simultaneously. Furthermore, the amplitude of the signal need not be known as it enters in expression for the sufficient statistic in a linear fashion and an UMP test exists in that case. We can easily find the threshold γγ by establishing a criterion on the false-alarm probability; the resulting simple computation of γγ can be traced to the lack of a signal-related quantity or an unknown parameter appearing in 0 0 .
We have argued the doubtfulness of assuming that the noise is white in discrete-time detection problems. The approach for solving the colored noise problem is to use spectral detection. Handling the unknown delay problem in this way is relatively straightforward. Since a sequence can be represented equivalently by its values or by its DFT, maximization can be calculated in either the time or the frequency domain without affecting the final answer. Thus, the spectral detector's decision rule for the unknown delay problem is (from this equation)
max Δ Δ k=0L1Rk¯Skei2πkΔL σ k 212|Sk|2 σ k 2 0 1 γ Δ k 0 L 1 R k S k 2 k Δ L σ k 2 1 2 S k 2 σ k 2 0 1 γ
(1)
where, as usual in unknown delay problems, the observation interval captures the entire signal waveform no matter what the delay might be. The energy term is a constant and can be incorporated into the threshold. The maximization amounts to finding the best linear phase fit to the observations' spectrum once the signal's phase has been removed. A more interesting interpretation arises by noting that the sufficient statistic is itself a Fourier Transform; the maximization amounts to finding the location of the maximum of a sequence given by k=0L1Rk¯Sk σ k 2ei2πkΔL k 0 L 1 R k S k σ k 2 2 k Δ L The spectral detector thus becomes a succession of two Fourier Transforms with the final result determined by the maximum of a sequence!
Unfortunately, the solution to the unknown-signal-delay problem in either the time or frequency domains is confounded when two or more signals are present. Assume two signals are known to be present in the array output, each of which has an unknown delay: rl= s 1 l Δ 1 + s 2 l Δ 2 +nl r l s 1 l Δ 1 s 2 l Δ 2 n l . Using arguments similar to those used in the one-signal case, the generalized likelihood ratio test becomes max Δ 1 , Δ 2 Δ 1 , Δ 2 l=0L1rl s 1 l Δ 1 +rl s 2 l Δ 2 s 1 l Δ 1 s 2 l Δ 2 0 1 σ2lnη+ E 1 + E 2 2 Δ 1 Δ 2 l 0 L 1 r l s 1 l Δ 1 r l s 2 l Δ 2 s 1 l Δ 1 s 2 l Δ 2 0 1 σ 2 η E 1 E 2 2 Not only do matched filter terms for each signal appear, but also a cross-term between the two signals. It is this latter term that complicates the multiple signal problem: if this term is not zero for all possible delays, a non-separable maximization process results and both delays must be varied in concert to locate the maximum. If, however, the two signals are orthogonal regardless of the delay values, the delays can be found separately and the structure of the single signal detector (modified to include matched filters for each signal) will suffice. This seemingly impossible situation can occur, at least approximately. Using Parseval's Theorem, the cross term can be expressed in the frequency domain. l=0L1 s 1 l Δ 1 s 2 l Δ 2 =12πππ S 1 ω S 2 ω¯eiω( Δ 2 Δ 1 )dω l 0 L 1 s 1 l Δ 1 s 2 l Δ 2 1 2 ω S 1 ω S 2 ω ω Δ 2 Δ 1 For this integral to be zero for all Δ 1 Δ 1 , Δ 2 Δ 2 , the product of the spectra must be zero. Consequently, if the two signals have disjoint spectral support, they are orthogonal no matter what the delays may be.2 Under these conditions, the detector becomes max Δ 1 Δ 1 rl* s 1 D1l| l =D1+ Δ 1 +max Δ 2 Δ 2 rl* s 2 D1l| l =D1+ Δ 2 0 1 γ l D 1 Δ 1 Δ 1 r l s 1 D 1 l l D 1 Δ 2 Δ 2 r l s 2 D 1 l 0 1 γ with the threshold again computed independently of the received signal amplitudes.3 P F =Qλ( E 1 + E 2 )σ2 P F Q λ E 1 E 2 σ 2 This detector has the structure of two parallel, independently operating, matched filters, each of which is tuned to the specific signal of interest.
Reality is insensitive to mathematically simple results. The orthogonality condition on the signals that yielded the relatively simple two-signal, unknown-delay detector is often elusive. The signals often share similar spectral supports, thereby violating the orthogonality condition. In fact, we may be interested in detecting the same signal repeated twice (or more) within the observation interval. Because of the complexity of incorporating inter-signal correlations, which are dependent on the relative delay, the idealistic detector is often used in practice. In the repeated signal case, the matched filter is operated over the entire observation interval and the number of excursions above the threshold noted. An excursion is defined to be a portion of the matched filter's output that exceeds the detection threshold over a contiguous interval. Because of the signal's non-zero duration, the matched filter's response to just the signal has a non-zero duration, implying that the threshold can be crossed at more than a single sample. When one signal is assumed, the maximization step automatically selects the peak value of an excursion. As shown in lower panels of this figure, a low-amplitude excursion may have a peak value less than a non-maximal value in a larger excursion. Thus, when considering multiple signals, the important quantities are the times at which excursion peaks occur, not all of the times the output exceeds the threshold.
This figure illustrates the two kinds of errors prevalent in multiple signal detectors. In the left panel, we find two excursions, the first of which is due to the signal, the second due to noise. This kind of error cannot be avoided; we never said that detectors could be perfect! The right panel illustrates a more serious problem: the threshold is crossed by four excursions, all of which are due to a single signal. Hence, excursions must be sorted through, taking into account the nature of the signal being sought. In the example, excursions surrounding a large one should be discarded if they occur in close proximity. This requirement means that closely spaced signals cannot be distinguished from a single one.
## Footnotes
1. For a much more realistic (and harder) version of the active radar/sonar problem, see this problem.
2. We stated earlier that this situation happens "at least approximately." Why the qualification?
3. Not to be boring, but we emphasize that E 1 E 1 and E 2 E 2 are the energies of the signals s 1 l s 1 l and s 2 l s 2 l used in the detector, not those of their received correlates A 1 s 1 l A 1 s 1 l and A 2 s 2 l A 2 s 2 l .
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| External bookmarks | 2,925 | 12,111 | {"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} | 2.671875 | 3 | CC-MAIN-2013-20 | latest | en | 0.873065 |
http://www.yardbarker.com/blog/puttingforbirdie/article/ryder_cup_math_where_1_plus_1_equals_3/11961955 | 1,462,310,211,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121776.48/warc/CC-MAIN-20160428161521-00104-ip-10-239-7-51.ec2.internal.warc.gz | 941,311,354 | 6,599 | • # Ryder Cup Math; where 1 plus 1 equals 3
For as long as I've been watching Ryder Cup competition, there has always been something that has bothered me about the scoring. It's time I get this off my chest and it's time to change the way individual scoring is computed in the competition. And while we're at it, let's go back and update all past Ryder Cups to make the scoring add up properly.
Let's take a look at the numbers in the recently concluded competition. For the sake of brevity, I will look only at Team USA scoring.
Team USA scored a total of 13 1/2 points. Here are the official total points for each member of the team:
Jason Dufner 3
Jim Furyk 1
Dustin Johnson 3
Zach Johnson 3
Matt Kuchar 2
Phil Mickelson 3
Webb Simpson 2
Brandt Snedeker 1
Steve Stricker 0
Bubba Watson 2
Tiger Woods .05
Do the math. Calculators not allowed.
I'm sure that Sister Mary Godresthersoul from St. Mary's Academy in Hoosick Falls would be very proud that I could add up 3+3+1+3+3+2+3+2+1+0+2+.05 and come up with a total of 23.5 or exactly 10 more points than Team USA actually scored.
Does this make sense to you?
Obviously the problem lies with the crediting of a point to each winning player in the team events. Example: Phil Mickelson and Keegan Bradley each received a point credit on their individual statistics for their three victories in the team events (3+3=6) and yet Team USA received a total of three points for the same three victories.
If you knew nothing at all about golf or Ryder Cup scoring but knew basic math and looked at the team scoring and individual scoring do you think you might ask a few questions? Would that math make sense to you?
It's a little strange that Team USA individual scores totaled to 23.5 points and Team Europe individual scores only totaled to 20.5 points yet Team Europe won. Try explaining that to your mathematically challenged kid.
This scoring process is akin to giving a basketball player two points for every assist he makes and giving the actual scorer two points for the basket he makes and yet giving the team only two points total. The scorebook wouldn't jibe with the scoreboard.
The better way is to credit each player in a winning foursome or fourball event with a half point to their individual totals. Make the math add up.
I know the argument: This is the way it's always been done so leave it alone. We can't go back and revise history.
Nonsense. It could be done in one day by an intern who knows first grade math.
Golf is a sport that prides itself on honesty, integrity and getting the score right. And calling penalties on yourself if the situation warrants. If a golfer signs an incorrect scorecard because he totaled up his score incorrectly he is subject to disqualification.
The Ryder Cup scorekeeping methodology should more accurately reflect the actual point contribution made by each player. Two men combine to win one point for the team----okay class, how much is one divided by two?
If a foursome or fourball event is halved and each team receives a half point, then each player involved should be credited with a quarter point. Ready class, this one is a little trickier---how much is one-half divided by two?
The number of points earned by each player, when totaled, should also equal the total number of points the team has accumulated.
Otherwise the math just doesn't add up.
Putting4birdie
Numbers that tell the stories in the world of golf and a whole lot more.
My name is Michael Verrastro and I welcome you to my blog.I love golf and I love blogging about it. New to this blogging thing thing so would appreciate your feedback. Leave a comment or email me. My dream foursome? Natalie Gulbis, Holly Sonders and Jan Stephenson. My 23-over par would be well worth it!!!!
### OWGR Top 10 as of April 15, 2013
1. Tiger Woods 2. Rory McIlroy 3. Adam Scott 4.Justin Rose 5. Brandt Snedeker 6. Luke Donald 7. Louis Oosthuizen 8.Steve Stricker 9. Matt Kuchar 10. Phil Mickelson | 947 | 3,985 | {"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} | 4.125 | 4 | CC-MAIN-2016-18 | latest | en | 0.955352 |
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Showing posts from November, 2023
Physical Significance of Maxwell's Equations
Physical Significance: The physical significance of Maxwell's equations obtained from integral form are given below: Maxwell's First Equation: 1. The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume. 2. It represents Gauss Law. 3. This law is independent of time. Charge acts as source or sink for the lines of electric force. Maxwell's Second Equation: 1. The total magnetic flux emitting through any closed surface is zero. An isolated magnet do not exist monopoles. 2. There is no source or sink for lines of magnetic force. 3. This is time independent equation. Maxwell's Third Equation: 1. The electromotive force around the closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path. 2. This gives relation between electric field $E$ and magnetic induction $B$. 3. This expression is time varying i.e. $E$ is generate
Circuit containing Inductor and Capacitor in Series (L-C Series Circuit )
Mathematical Analysis of L-C Series Circuit : Let us consider, a circuit containing inductor $L$ capacitor $C$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C circuit $V=V_{L} - V_{C} \qquad(1)$ We know that: $V_{L} = iX_{L}$ $V_{C} = iX_{C}$ So from equation $(1)$ $V= iX_{L} - iX_{C}$ $V=i \left(X_{L} - X_{C} \right)$ $\frac{V}{i}=\left(X_{L} - X_{C} \right)$ $Z=\left(X_{L} - X_{C} \right) \qquad(2)$ Where $Z \rightarrow$ Impedance of L-C circuit. $X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$ $X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$ So from equation $(2)$, we get $Z=\left( \omega L - \frac{1}{\omega C} \right) \qquad(3)$ The phase of resultant voltage: The phase of resultant voltage from current is $90^{\circ}$ as shown in the figure above. The Impedance and Phase at Re
Self Induction: When a changing current flows in a coil then due to the change in magnetic flux in the coil produces an electro-motive force $\left(emf \right)$ in that coil. This phenomenon is called the principle of Self Induction. The direction of electro-motive force can be found by applying "Lenz's Law". Mathematical Analysis of Coefficient of Self Induction: Let us consider that a coil having the number of turns is $N$. If the change in current is $i$, then linkage flux in a coil will be $N \phi \propto i$ $N\phi = L i \qquad(1)$ Where $L$ $\rightarrow$ Coefficient of Self Induction. According to Faraday's law of electromagnetic induction. The electro-motive force $\left(emf \right)$ in a coil is $e=-N\left( \frac{d \phi}{dt} \right)$ $e=-\frac{d \left(N \phi \right)}{dt} \qquad(2)$ From equation $(1)$ and equation $(2)$ $e=-\frac{d \left(L i\right)}{dt}$ $e=-L \left(\frac{d i}{dt} \right)$ $L = \frac{e}{\ Mutual Induction Phenomenon and its Coefficient Mutual Induction: When two coils are placed near each other then the change in current in one coil ( Primary Coil) produces electro-motive force$\left( emf \right)$in the adjacent coil ( i.e. secondary coil). This phenomenon is called the principle of Mutual Induction. The direction of electro-motive force$\left( emf \right)$depends or can be found by "Lenz's Law" Mathematical Analysis of Coefficient of Mutual Induction: Let us consider that two coils having the number of turns are$N_{1}$and$N_{2}$. If these coils are placed near to each other and the change in current of the primary coil is$i_{1}$, then linkage flux in the secondary coil will be$N_{2}\phi_{2} \propto i_{1}N_{2}\phi_{2} = M i_{1} \qquad(1)$Where$M\rightarrow$Coefficient of Mutual Induction. According to Faraday's law of electromagnetic induction. The electro-motive force$\left( emf \right)$in the secondary coil is$e_{2}=-N_{2}\left( \frac{d \phi_
Faraday's Laws of Electromagnetic Induction: The Faraday's experiment shows the two laws which are known as Farday's laws of electromagnetic induction First Law (Neumann's Law): The rate of change of magnetic flux through a circuit is equal to the emf produced in the circuit. This is also known as " Neumann Law " $e=-\frac{\Delta \phi}{ \Delta t}$ Here negative sign shows the direction of emf. If $\Delta t \rightarrow 0$ $e=-\frac{d \phi}{ d t}$ This equation represents an independent experimental law that cannot be derived from other experimental laws. If the circuit is a tightly wound coil of $N$ turns, then the induced emf $e=-N\frac{d \phi}{ d t}$ $e=-\frac{d \left(N \phi\right)}{ dt}$ Here $N \phi$ is called the 'Linkage magnetic flux'. Note: The change in flux induces emf, not the current. Second Law (Lenz's Law): The direction of induced EMF produced in a closed circuit is such that it opposes th
Paramagnetic Substances : Those substances, which are placed in the external magnetic field and they are weakly magnetized in the direction of the external magnetic field, are called paramagnetic substances. The susceptibility $\chi_{m}$ of paramagnetic substances is small and positive. Further, When a paramagnetic substance is placed in the magnetic field, then the flux density of the paramagnetic substance is slightly more than the free space. Thus, the relative permeability of paramagnetic substance $\mu_{r}$, is slightly more than 1. Properties of Paramagnetic substances: 1. When a rod of a paramagnetic material is suspended freely between external magnetic poles (i.e. Between North and South Poles) then its axis becomes along the direction of the external magnetic field $B$ (Figure). The poles produced on the two sides of the rod are opposite to the poles of the external magnetic field. 2. In a non-uniform magnetic field, a paramagnetic substance tends to move fro | 1,509 | 5,829 | {"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} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.830084 |
https://www.d2l.ai/_sources/chapter_linear-networks/linear-regression.rst.txt | 1,566,173,037,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-35/segments/1566027314353.10/warc/CC-MAIN-20190818231019-20190819013019-00246.warc.gz | 768,502,378 | 10,824 | .. _chapter_linear_regression: Linear Regression ================= To start off, we will introduce the problem of regression. This is the task of predicting a *real valued target* :math:y given a data point :math:\mathbf{x}. Regression problems are common in practice, arising whenever we want to predict a continuous numerical value. Some examples of regression problems include predicting house prices, stock prices, length of stay (for patients in the hospital), tomorrow’s temperature, demand forecasting (for retail sales), and many more. Note that not every prediction problem is a regression problem. In subsequent sections we will discuss classification problems, where our predictions are discrete categories. Basic Elements of Linear Regression ----------------------------------- Linear regression, which dates to Gauss and Legendre, is perhaps the simplest, and by far the most popular approach to solving regression problems. What makes linear regression *linear* is that we assume that the output truly can be expressed as a *linear* combination of the input features. Linear Model ~~~~~~~~~~~~ To keep things simple, we will start with running example in which we consider the problem of estimating the price of a house (e.g. in dollars) based on area (e.g. in square feet) and age (e.g. in years). More formally, the assumption of linearity suggests that our model can be expressed in the following form: .. math:: \mathrm{price} = w_{\mathrm{area}} \cdot \mathrm{area} + w_{\mathrm{age}} \cdot \mathrm{age} + b In economics papers, it is common for authors to write out linear models in this format with a gigantic equation that spans multiple lines containing terms for every single feature. For the high-dimensional data that we often address in machine learning, writing out the entire model can be tedious. In these cases, we will find it more convenient to use linear algebra notation. In the case of :math:d variables, we could express our prediction :math:\hat{y} as follows: .. math:: \hat{y} = w_1 \cdot x_1 + ... + w_d \cdot x_d + b or alternatively, collecting all features into a single vector :math:\mathbf{x} and all parameters into a vector :math:\mathbf{w}, we can express our linear model as .. math:: \hat{y} = \mathbf{w}^T \mathbf{x} + b. :label: eq_linear_regression_single Above, the vector :math:\mathbf{x} corresponds to a single data point. Commonly, we will want notation to refer to the entire dataset of all input data points. This matrix, often denoted using a capital letter :math:X, is called the *design matrix* and contains one row for every example, and one column for every feature. Given a collection of data points :math:X and a vector containing the corresponding target values :math:\mathbf{y}, the goal of linear regression is to find the *weight* vector :math:w and bias term :math:b (also called an *offset* or *intercept*) that associates each data point :math:\mathbf{x}_i with an approximation :math:\hat{y}_i of its corresponding label :math:y_i. Expressed in terms of a single data point, this gives us the expression same as :eq:eq_linear_regression_single. Finally, for a collection of data points :math:\mathbf{X}, the predictions :math:\hat{\mathbf{y}} can be expressed via the matrix-vector product: .. math:: {\hat{\mathbf{y}}} = \mathbf X \mathbf{w} + b. :label: eq_linear_regression Even if we believe that the best model to relate :math:\mathbf{x} and :math:y is linear, it’s unlikely that we’d find data where :math:y lines up exactly as a linear function of :math:\mathbf{x}. For example, both the target values :math:y and the features :math:X might be subject to some amount of measurement error. Thus even when we believe that the linearity assumption holds, we will typically incorporate a noise term to account for such errors. Before we can go about solving for the best setting of the parameters :math:w and :math:b, we will need two more things: (i) some way to measure the quality of the current model and (ii) some way to manipulate the model to improve its quality. Training Data ~~~~~~~~~~~~~ The first thing that we need is training data. Sticking with our running example, we’ll need some collection of examples for which we know both the actual selling price of each house as well as their corresponding area and age. Our goal is to identify model parameters that minimize the error between the predicted price and the real price. In the terminology of machine learning, the data set is called a *training data* or *training set*, a house (often a house and its price) here comprises one *sample*, and its actual selling price is called a *label*. The two factors used to predict the label are called *features* or *covariates*. Typically, we will use :math:n to denote the number of samples in our dataset. We index the samples by :math:i, denoting each input data point as :math:x^{(i)} = [x_1^{(i)}, x_2^{(i)}] and the corresponding label as :math:y^{(i)}. Loss Function ~~~~~~~~~~~~~ In model training, we need to measure the error between the predicted value and the real value of the price. Usually, we will choose a non-negative number as the error. The smaller the value, the smaller the error. A common choice is the square function. For given parameters :math:\mathbf{w} and :math:b, we can express the error of our prediction on a given a sample as follows: .. math:: l^{(i)}(\mathbf{w}, b) = \frac{1}{2} \left(\hat{y}^{(i)} - y^{(i)}\right)^2, The constant :math:1/2 is just for mathematical convenience, ensuring that after we take the derivative of the loss, the constant coefficient will be :math:1. The smaller the error, the closer the predicted price is to the actual price, and when the two are equal, the error will be zero. Since the training dataset is given to us, and thus out of our control, the error is only a function of the model parameters. In machine learning, we call the function that measures the error the *loss function*. The squared error function used here is commonly referred to as *square loss*. To make things a bit more concrete, consider the example below where we plot a regression problem for a one-dimensional case, e.g. for a model where house prices depend only on area. .. figure:: ../img/fit_linreg.svg Fit data with a linear model. As you can see, large differences between estimates :math:\hat{y}^{(i)} and observations :math:y^{(i)} lead to even larger contributions in terms of the loss, due to the quadratic dependence. To measure the quality of a model on the entire dataset, we can simply average the losses on the training set. .. math:: L(\mathbf{w}, b) =\frac{1}{n}\sum_{i=1}^n l^{(i)}(\mathbf{w}, b) =\frac{1}{n} \sum_{i=1}^n \frac{1}{2}\left(\mathbf{w}^\top \mathbf{x}^{(i)} + b - y^{(i)}\right)^2. When training the model, we want to find parameters (:math:\mathbf{w}^*, b^*) that minimize the average loss across all training samples: .. math:: \mathbf{w}^*, b^* = \operatorname*{argmin}_{\mathbf{w}, b}\ L(\mathbf{w}, b). Analytic Solution ~~~~~~~~~~~~~~~~~ Linear regression happens to be an unusually simple optimization problem. Unlike nearly every other model that we will encounter in this book, linear regression can be solved easily with a simple formula, yielding a global optimum. To start we can subsume the bias :math:b into the parameter :math:\mathbf{w} by appending a column to the design matrix consisting of all :math:1s. Then our prediction problem is to minimize :math:||\mathbf{y} - X\mathbf{w}||. Because this expression has a quadratic form it is clearly convex, and so long as the problem is not degenerate (our features are linearly independent), it is strictly convex. Thus there is just one global critical point on the loss surface corresponding to the global minimum. Taking the derivative of the loss with respect to :math:\mathbf{w} and setting it equal to 0 gives the analytic solution: .. math:: \mathbf{w}^* = (\mathbf X^T \mathbf X)^{-1}\mathbf X^T y While simple problems like linear regression may admit analytic solutions, you should not get used to such good fortune. Although analytic solutions allow for nice mathematical analysis, the requirement of an analytic solution confines one to an restrictive set of models that would exclude all of deep learning. Gradient descent ~~~~~~~~~~~~~~~~ Even in cases where we cannot solve the models analytically, and even when the loss surfaces are high-dimensional and nonconvex, it turns out that we can still make progress. Moreover, when those difficult-to-optimize models are sufficiently superior for the task at hand, figuring out how to train them is well worth the trouble. The key trick behind nearly all of deep learning and that we will repeatedly throughout this book is to reduce the error gradually by iteratively updating the parameters, each step moving the parameters in the direction that incrementally lowers the loss function. This algorithm is called gradient descent. On convex loss surfaces it will eventually converge to a global minimum, and while the same can’t be said for nonconvex surfaces, it will at least lead towards a (hopefully good) local minimum. The most naive application of gradient descent consists of taking the derivative of the true loss, which is an average of the losses computed on every single example in the dataset. In practice, this can be extremely slow. We must pass over the entire dataset before making a single update. Thus, we’ll often settle for sampling a random mini-batch of examples every time we need to computer the update, a variant called *stochastic gradient descent*. In each iteration, we first randomly and uniformly sample a mini-batch :math:\mathcal{B} consisting of a fixed number of training data examples. We then compute the derivative (gradient) of the average loss on the mini batch with regard to the model parameters. Finally, the product of this result and a predetermined step size :math:\eta > 0 are used to update the parameters in the direction that lowers the loss. We can express the update mathematically as follows (:math:\partial denotes the partial derivative): .. math:: (\mathbf{w},b) \leftarrow (\mathbf{w},b) - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \partial_{(\mathbf{w},b)} l^{(i)}(\mathbf{w},b) To summarize, steps of the algorithm are the following: (i) we initialize the values of the model parameters, typically at random; (ii) we iterate over the data many times, updating the parameters in each by moving the parameters in the direction of the negative gradient, as calculated on a random minibatch of data. For quadratic losses and linear functions we can write this out explicitly as follows. Note that :math:\mathbf{w} and :math:\mathbf{x} are vectors. Here the more elegant vector notation makes the math much more readable than expressing things in terms of coefficients, say :math:w_1, w_2, \ldots w_d. .. math:: \begin{aligned} \mathbf{w} &\leftarrow \mathbf{w} - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \partial_{\mathbf{w}} l^{(i)}(\mathbf{w}, b) && = w - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \mathbf{x}^{(i)} \left(\mathbf{w}^\top \mathbf{x}^{(i)} + b - y^{(i)}\right),\\ b &\leftarrow b - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \partial_b l^{(i)}(\mathbf{w}, b) && = b - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \left(\mathbf{w}^\top \mathbf{x}^{(i)} + b - y^{(i)}\right). \end{aligned} In the above equation :math:|\mathcal{B}| represents the number of samples (batch size) in each mini-batch, :math:\eta is referred to as *learning rate* and takes a positive number. It should be emphasized that the values of the batch size and learning rate are set somewhat manually and are typically not learned through model training. Therefore, they are referred to as *hyper-parameters*. What we usually call *tuning hyper-parameters* refers to the adjustment of these terms. In the worst case this is performed through repeated trial and error until the appropriate hyper-parameters are found. A better approach is to learn these as parts of model training. This is an advanced topic and we do not cover them here for the sake of simplicity. Model Prediction ~~~~~~~~~~~~~~~~ After completing the training process, we record the estimated model parameters, denoted :math:\hat{\mathbf{w}}, \hat{b} (in general the “hat” symbol denotes estimates). Note that the parameters that we learn via gradient descent are not exactly equal to the true minimizers of the loss on the training set, that’s because gradient descent converges slowly to a local minimum but does not achieve it exactly. Moreover if the problem has multiple local minimum, we may not necessarily achieve the lowest minimum. Fortunately, for deep neural networks, finding parameters that minimize the loss *on training data* is seldom a significant problem. The more formidable task is to find parameters that will achieve low loss on data that we have not seen before, a challenge called *generalization*. We return to these topics throughout the book. Given the learned linear regression model :math:\hat{\mathbf{w}}^\top x + \hat{b}, we can now estimate the price of any house outside the training data set with area (square feet) as :math:x_1 and house age (year) as :math:x_2. Here, estimation also referred to as ‘model prediction’ or ‘model inference’. Note that calling this step *inference* is a misnomer, but has become standard jargon in deep learning. In statistics, inference means estimating parameters and outcomes based on other data. This misuse of terminology in deep learning can be a source of confusion when talking to statisticians. From Linear Regression to Deep Networks --------------------------------------- So far we only talked about linear functions. While neural networks cover a much richer family of models, we can begin thinking of the linear model as a neural network by expressing it the language of neural networks. To begin, let’s start by rewriting things in a ‘layer’ notation. Neural Network Diagram ~~~~~~~~~~~~~~~~~~~~~~ Commonly, deep learning practitioners represent models visually using neural network diagrams. In :numref:fig_single_neuron, we represent linear regression with a neural network diagram. The diagram shows the connectivity among the inputs and output, but does not depict the weights or biases (which are given implicitly). .. _fig_single_neuron: .. figure:: ../img/singleneuron.svg Linear regression is a single-layer neural network. In the above network, the inputs are :math:x_1, x_2, \ldots x_d. Sometimes the number of inputs are referred to as the feature dimension. For linear regression models, we act upon :math:d inputs and output :math:1 value. Because there is just a single computed neuron (node) in the graph, we can think of linear models as neural networks consisting of just a single neuron. Since all inputs are connected to all outputs (in this case it’s just one), this layer can also be regarded as an instance of a *fully-connected layer*, also commonly called a *dense layer*. Biology ~~~~~~~ Neural networks derive their name from their inspirations in neuroscience. Although linear regression predates computation neuroscience, many of the models we subsequently discuss truly owe to neural inspiration. To understand the neural inspiration for artificial neural networks it is worth while considering the basic structure of a neuron. For the purpose of the analogy it is sufficient to consider the *dendrites* (input terminals), the *nucleus* (CPU), the *axon* (output wire), and the *axon terminals* (output terminals) which connect to other neurons via *synapses*. .. figure:: ../img/Neuron.svg The real neuron Information :math:x_i arriving from other neurons (or environmental sensors such as the retina) is received in the dendrites. In particular, that information is weighted by *synaptic weights* :math:w_i which determine how to respond to the inputs (e.g. activation or inhibition via :math:x_i w_i). All this is aggregated in the nucleus :math:y = \sum_i x_i w_i + b, and this information is then sent for further processing in the axon :math:y, typically after some nonlinear processing via :math:\sigma(y). From there it either reaches its destination (e.g. a muscle) or is fed into another neuron via its dendrites. Brain *structures* vary significantly. Some look (to us) rather arbitrary whereas others have a regular structure. For example, the visual system of many insects is consistent across members of a species. The analysis of such structures has often inspired neuroscientists to propose new architectures, and in some cases, this has been successful. However, much research in artificial neural networks has little to do with any direct inspiration in neuroscience, just as although airplanes are *inspired* by birds, the study of orninthology hasn’t been the primary driver of aeronautics innovaton in the last century. Equal amounts of inspiration these days comes from mathematics, statistics, and computer science. Vectorization for Speed ~~~~~~~~~~~~~~~~~~~~~~~ In model training or prediction, we often use vector calculations and process multiple observations at the same time. To illustrate why this matters, consider two methods of adding vectors. We begin by creating two 100000 dimensional ones first. .. code:: python %matplotlib inline import d2l import numpy as np import math from mxnet import nd import time n = 100000 a = np.ones(n) b = np.ones(n) Since we will benchmark the running time frequently in this book, let’s define a timer to do simply analysis of the running time. .. code:: python # Save to the d2l package. class Timer(object): """Record multiple running times.""" def __init__(self): self.times = [] self.start() def start(self): """Start the timer""" self.start_time = time.time() def stop(self): """Stop the timer and record the time in a list""" self.times.append(time.time() - self.start_time) return self.times[-1] def avg(self): """Return the average time""" return sum(self.times)/len(self.times) def sum(self): """Return the sum of time""" return sum(self.times) def cumsum(self): """Return the accumuated times""" return np.array(self.times).cumsum().tolist() Now we can benchmark the workloads. One way to add vectors is to add them one coordinate at a time using a for loop. .. code:: python timer = Timer() c = np.zeros(n) for i in range(n): c[i] = a[i] + b[i] '%.5f sec' % timer.stop() .. parsed-literal:: :class: output '0.04021 sec' Another way to add vectors is to add the vectors directly: .. code:: python timer.start() d = a + b '%.5f sec' % timer.stop() .. parsed-literal:: :class: output '0.00053 sec' Obviously, the latter is vastly faster than the former. Vectorizing code is a good way of getting order of magnitude speedups. Likewise, as we saw above, it also greatly simplifies the mathematics and with it, it reduces the potential for errors in the notation. The Normal Distribution and Squared Loss ---------------------------------------- The following is optional and can be skipped but it will greatly help with understanding some of the design choices in building deep learning models. As we saw above, using the squared loss :math:l(y, \hat{y}) = \frac{1}{2} (y - \hat{y})^2 has many nice properties, such as having a particularly simple derivative :math:\partial_{\hat{y}} l(y, \hat{y}) = (\hat{y} - y). That is, the gradient is given by the difference between estimate and observation. You might reasonably point out that linear regression is a classical __ statistical model. Legendre first developed the method of least squares regression in 1805, which was shortly thereafter rediscovered by Gauss in 1809. To understand this a bit better, recall the normal distribution with mean :math:\mu and variance :math:\sigma^2. .. math:: p(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(-\frac{1}{2 \sigma^2} (x - \mu)^2\right) Let’s define the function to compute the normal distribution. .. code:: python x = np.arange(-7, 7, 0.01) def normal(x, mu, sigma): p = 1 / math.sqrt(2 * math.pi * sigma**2) return p * np.exp(- 0.5 / sigma**2 * (x - mu)**2) For a similar reason to create a Timer class, we define a plot function to draw multiple lines and set the figure properly, since we will visualize lines frequently later. .. code:: python # Save to the d2l package. def plot(X, Y=None, xlabel=None, ylabel=None, legend=[], xlim=None, ylim=None, xscale='linear', yscale='linear', fmts=None, figsize=(3.5, 2.5), axes=None): """Plot multiple lines""" d2l.set_figsize(figsize) axes = axes if axes else d2l.plt.gca() if isinstance(X, nd.NDArray): X = X.asnumpy() if isinstance(Y, nd.NDArray): Y = Y.asnumpy() if not hasattr(X[0], "__len__"): X = [X] if Y is None: X, Y = [[]]*len(X), X if not hasattr(Y[0], "__len__"): Y = [Y] if len(X) != len(Y): X = X * len(Y) if not fmts: fmts = ['-']*len(X) axes.cla() for x, y, fmt in zip(X, Y, fmts): if isinstance(x, nd.NDArray): x = x.asnumpy() if isinstance(y, nd.NDArray): y = y.asnumpy() if len(x): axes.plot(x, y, fmt) else: axes.plot(y, fmt) set_axes(axes, xlabel, ylabel, xlim, ylim, xscale, yscale, legend) # Save to the d2l package. def set_axes(axes, xlabel, ylabel, xlim, ylim, xscale, yscale, legend): """A utility function to set matplotlib axes""" axes.set_xlabel(xlabel) axes.set_ylabel(ylabel) axes.set_xscale(xscale) axes.set_yscale(yscale) axes.set_xlim(xlim) axes.set_ylim(ylim) if legend: axes.legend(legend) axes.grid() Now visualize the normal distributions. .. code:: python # Mean and variance pairs parameters = [(0,1), (0,2), (3,1)] plot(x, [normal(x, mu, sigma) for mu, sigma in parameters], xlabel='x', ylabel='p(x)', figsize=(4.5, 2.5), legend = ['mean %d, var %d'%(mu, sigma) for mu, sigma in parameters]) .. image:: output_linear-regression_5a7d24_13_0.svg As can be seen in the figure above, changing the mean shifts the function, increasing the variance makes it more spread-out with a lower peak. The key assumption in linear regression with least mean squares loss is that the observations actually arise from noisy observations, where noise is added to the data, e.g. as part of the observations process. .. math:: y = \mathbf{w}^\top \mathbf{x} + b + \epsilon \text{ where } \epsilon \sim \mathcal{N}(0, \sigma^2) This allows us to write out the *likelihood* of seeing a particular :math:y for a given :math:\mathbf{x} via .. math:: p(y|\mathbf{x}) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(-\frac{1}{2 \sigma^2} (y - \mathbf{w}^\top \mathbf{x} - b)^2\right) A good way of finding the most likely values of :math:b and :math:\mathbf{w} is to maximize the *likelihood* of the entire dataset .. math:: p(Y|X) = \prod_{i=1}^{n} p(y^{(i)}|\mathbf{x}^{(i)}) The notion of maximizing the likelihood of the data subject to the parameters is well known as the *Maximum Likelihood Principle* and its estimators are usually called *Maximum Likelihood Estimators* (MLE). Unfortunately, maximizing the product of many exponential functions is pretty awkward, both in terms of implementation and in terms of writing it out on paper. Instead, a much better way is to minimize the *Negative Log-Likelihood* :math:-\log p(\mathbf y|\mathbf X). In the above case this works out to be .. math:: -\log p(\mathbf y|\mathbf X) = \sum_{i=1}^n \frac{1}{2} \log(2 \pi \sigma^2) + \frac{1}{2 \sigma^2} \left(y^{(i)} - \mathbf{w}^\top \mathbf{x}^{(i)} - b\right)^2 A closer inspection reveals that for the purpose of minimizing :math:-\log p(\mathbf y|\mathbf X) we can skip the first term since it doesn’t depend on :math:\mathbf{w}, b or even the data. The second term is identical to the objective we initially introduced, but for the multiplicative constant :math:\frac{1}{\sigma^2}. Again, this can be skipped if we just want to get the most likely solution. It follows that maximum likelihood in a linear model with additive Gaussian noise is equivalent to linear regression with squared loss. Summary ------- - Key ingredients in a machine learning model are training data, a loss function, an optimization algorithm, and quite obviously, the model itself. - Vectorizing makes everything better (mostly math) and faster (mostly code). - Minimizing an objective function and performing maximum likelihood can mean the same thing. - Linear models are neural networks, too. Exercises --------- 1. Assume that we have some data :math:x_1, \ldots x_n \in \mathbb{R}. Our goal is to find a constant :math:b such that :math:\sum_i (x_i - b)^2 is minimized. - Find the optimal closed form solution. - What does this mean in terms of the Normal distribution? 2. Assume that we want to solve the optimization problem for linear regression with quadratic loss explicitly in closed form. To keep things simple, you can omit the bias :math:b from the problem. - Rewrite the problem in matrix and vector notation (hint - treat all the data as a single matrix). - Compute the gradient of the optimization problem with respect to :math:w. - Find the closed form solution by solving a matrix equation. - When might this be better than using stochastic gradient descent (i.e. the incremental optimization approach that we discussed above)? When will this break (hint - what happens for high-dimensional :math:x, what if many observations are very similar)?. 3. Assume that the noise model governing the additive noise :math:\epsilon is the exponential distribution. That is, :math:p(\epsilon) = \frac{1}{2} \exp(-|\epsilon|). - Write out the negative log-likelihood of the data under the model :math:-\log p(Y|X). - Can you find a closed form solution? - Suggest a stochastic gradient descent algorithm to solve this problem. What could possibly go wrong (hint - what happens near the stationary point as we keep on updating the parameters). Can you fix this? Scan the QR Code to Discuss __ ----------------------------------------------------------------- |image0| .. |image0| image:: ../img/qr_linear-regression.svg | 6,249 | 25,904 | {"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-2019-35 | latest | en | 0.856605 |
https://invernessgangshow.net/is-1-a-real-number/ | 1,686,229,160,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00759.warc.gz | 343,782,473 | 5,780 | Real numbers are just the mix of rational and irrational numbers, in the number system. In general, every the arithmetic operations deserve to be carry out on this numbers and also they have the right to be represented in the number line, also. In ~ the exact same time, the imaginary numbers space the un-real numbers, which can not be to express in the number line and also is frequently used to stand for a complex number. Some of the instances of genuine numbers space 23, -12, 6.99, 5/2, π, and so on. In this article, we space going to comment on the definition of actual numbers, properties of actual numbers and the examples of the actual number with complete explanations.
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Table the contents:
Properties
## Real number Definition
Real numbers have the right to be defined as the union the both the rational and also irrational numbers. They have the right to be both confident or an adverse and space denoted through the symbol “R”. Every the organic numbers, decimals and also fractions come under this category. Check out the figure, given below, which mirrors the category of actual numerals.
## Set of real Numbers
The collection of genuine numbers consists of different categories, such together natural and whole numbers, integers, rational and also irrational numbers. In the table provided below, all the actual numbers recipe (i.e.) the representation of the classification of actual numbers are characterized with examples.
CategoryDefinitionExample
Natural NumbersContain all counting numbers which start from 1.
N = 1,2,3,4,……
All number such as 1, 2, 3, 4,5,6,…..…
Whole NumbersCollection of zero and natural number.
W = 0,1,2,3,…..
All numbers including 0 such as 0, 1, 2, 3, 4,5,6,…..…
IntegersThe collective result of totality numbers and an adverse of all herbal numbers.Includes: -infinity (-∞),……..-4, -3, -2, -1, 0, 1, 2, 3, 4, ……+infinity (+∞)
Rational NumbersNumbers that deserve to be created in the kind of p/q, wherein q≠0.Examples of rational numbers are ½, 5/4 and also 12/6 etc.
Irrational NumbersAll the numbers which room not rational and cannot be written in the form of p/q.Irrational numbers space non-terminating and also non-repeating in nature choose √2
## Real number Chart
The chart because that the set of real numerals including all the types are provided below:
## Properties of genuine Numbers
The four main properties of actual numbers are commutative property, associative property, distributive property and also identity property. Consider “m, n and also r” are three real numbers. Then the over properties can be explained using m, n, and also r as shown below:
### Commutative Property
If m and n space the numbers, climate the general form will it is in m + n = n + m for enhancement and m.n = n.m because that multiplication.
Addition: m + n = n + m. For example, 5 + 3 = 3 + 5, 2 + 4 = 4 + 2Multiplication: m × n = n × m. Because that example, 5 × 3 = 3 × 5, 2 × 4 = 4 × 2
### Associative Property
If m, n and r space the numbers. The general type will it is in m + (n + r) = (m + n) + r for addition(mn) r = m (nr) for multiplication.
Addition: The general type will be m + (n + r) = (m + n) + r. An example of additive associative building is 10 + (3 + 2) = (10 + 3) + 2.Multiplication: (mn) r = m (nr). An example of a multiplicative associative building is (2 × 3) 4 = 2 (3 × 4).
### Distributive Property
For three numbers m, n, and r, i m sorry are genuine in nature, the distributive residential or commercial property is stood for as:
m (n + r) = mn + mr and also (m + n) r = grandfather + nr.
Example the distributive property is: 5(2 + 3) = 5 × 2 + 5 × 3. Here, both sides will yield 25.
### Identity Property
There space additive and multiplicative identities.
See more: Y=-3X+5 - Graph And Solve: Equation: Y=3X
For addition: m + 0 = m. (0 is the additive identity)For multiplication: m × 1 = 1 × m = m. (1 is the multiplicative identity)
Learn much more About genuine Number Properties
Commutative PropertyAssociative Property
Distributive PropertyAdditive Identity and Multiplicative Identity
### Practice Questions
Which is the smallest composite number?Prove that any kind of positive odd integer is that the form 6x + 1, 6x + 3, or 6x + 5.Evaluate 2 + 3 × 6 – 5What is the product that a non-zero rational number and irrational number?Can every confident integer be stood for as 4x + 2 (where x is an integer)?
### Real Numbers course 9 and also 10
In actual numbers class 9, the common concepts introduced include representing genuine numbers top top a number line, work on actual numbers, nature of genuine numbers, and also the legislation of exponents for actual numbers. In class 10, some progressed concepts associated to actual numbers room included. Apart from what are genuine numbers, college student will additionally learn around the actual numbers formulas and also concepts such together Euclid’s division Lemma, Euclid’s department Algorithm and the basic theorem that arithmetic in class 10.
Rational number on a number line Operations On genuine Numbers Laws that Exponents Euclid’s department Lemma Fundamental theorem Of Arithmetic Properties the Integers | 1,320 | 5,225 | {"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} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.862411 |
https://brainly.in/question/222597 | 1,484,758,099,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00369-ip-10-171-10-70.ec2.internal.warc.gz | 799,597,825 | 10,451 | # Ages of a father and son add up to 66. The father's age is the son's age reversed. How old could they be? The ages of a father and son add up to 66. The father's age is the son's age reversed. How old could they be? (3 possible solutions).
2
by ashasulatiyashi
2015-11-09T22:22:10+05:30
1 possible way Son=15 years and Father=51 years
2 possible way Son=24 years and Father =42years
3 possible way Son =33 years and Father =33 years
Hope this helps you
sorry...2 case son 24 years and father 42 years
i edited it
3 rd case???
How can a son's and father's age be same...?
sory..mistaken
2015-11-09T22:41:05+05:30
the possible three ways are :
1. the son's age can be 15 years and dad's age can be 51 years
2. the son's age can be 24 years and dad's age can be 42 years
3. the son's age can be 06 years and dad's age can be 60 years
hope so my answer is useful................ ^_^ | 279 | 889 | {"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} | 3.90625 | 4 | CC-MAIN-2017-04 | latest | en | 0.853066 |
http://nbodypolitics.gitlab.io/2018/01/06/ftl_revisited.html | 1,532,040,160,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00452.warc.gz | 260,346,867 | 6,717 | I have, previously, talked about the FTL system, and that it dumps spacecraft on the 0.1g boundary of a system, at no velocity relative to the target planet. Later on, in the economics post I changed that to a stable orbit for easier math. As ericthered on the GURPS forums points out, the no-velocity solution somehow seems neater. I actually agree on that, which means I’ll spend this post on orbital mechanics, to compute things like deltaV and orbital times - hard numbers!
## The Formulae
### Orbital Speed and deltaV
One of the main formulae we’re looking at is the vis-viva equation, which tells us how fast we’re moving at which point of our orbit. The equation is
where $v$ is our velocity, $\gamma$ is the planet’s standard gravitational parameter (about $3.99*10^{14} \frac{m^3}{s^2}$ for Earth), $r$ is the current distance between the spacecraft and the planet, and $a$ is the semi-major axis (i.e. the longer of the axes of the ellipse, i.e. the mean of periapsis and apoapsis distance).
How do we get the deltaV from here? Let’s look at an example: We’re currently orbiting Earth at a low orbit of 250km, and want to raise our apoapsis (the highest point of our orbit) to a geostationary orbit (~42164km). At our current distance, the speed of our orbit is
Note that I am specifying $\gamma$ in $\frac{km^3}{s^3}$, which makes the numbers smaller. Note also that the orbital distance is 6621km - that’s because we’re 300km above the Earth’s surface, meaning we have to add the Earth radius. In this case, computing $a$ is very easy - periapsis and apoapsis are identical, so $a$ is the same as either of those.
For our new orbit, $a$ is just the average of periapsis (our old 6621km) and apoapsis (our new 42164km), 24392.5km. Easy. Substituting the other parameters, we get
What does this tell us? Well, intuitively, when we’re at our point of burn, we are moving at 7.75km/s. For our new orbit, we’d have to be moving at 10.20km/s to be in our new orbit. How much deltaV do we need to change orbits? Simply 10.20km/s - 7.75km/s = 2.45km/s. Easy!
But were we actually correct? We can look at a deltaV map and compare the deltaV needed to go from low Earth orbit to Geostationary Transfer orbit - 2.44km/s. Perfect.
In fact, the spreadsheet I used (which internally doesn’t round) gave me 2.442km/s.
### Orbital Period
I’m also going to need the orbital period (i.e. time to complete one orbit). For this, the formula is
with all of the values as before. Again, let’s take the Geostationary Transfer Orbit as our example. We already have a as 24392.5km. Inserting that gives
37913s is difficult to work with; this is about 10h 32min. The wikipedia article gives 10.5h, so we’re right on target again.
## Jump!
With these formulae, we can now compute a deltaV-efficient trajectory. We still can’t say much about how to decrease the necessary time, but that’s fine for now.
### Jump Emergence Orbit
So, first of all, we look at the “orbit” (controlled fall) we end up in after jumping. First question: What’s the emergence distance (the apoapsis of our orbit?)
We again can introduce a new formula: The gravitational force:
where $G$ is the gravitational constant, $m_p$ and $m_s$ are the masses of the planet and spacecraft, and $r$ is the distance between them. We can directly combine $G * m_p$ to $\gamma$, the standard gravitational parameter above. Reformulating this gives us
Let’s look at Earth: We know $\gamma$ from above. And, nicely, $g$ can be expressed both in $\frac{m}{s^2}$ and in $\frac{N}{kg}$; our value is 0.1N/kg (about 0.01g). Meaning we can rewrite and then insert our values as follows
Nice how that works out, isn’t it?
Using that, what do we know about our orbit and speed? First the emergence speed is 0. We can either plug that into the vis-viva equation to compute the semi-major axis, or think logically for a second, figure it’s a circular orbit, so the semi-major axis is our $r$. I did the former, of course, because common sense isn’t that common.
So, how long before you die a fiery death? About a quarter of the total orbital time. Let’s write that more generally, though:
Let’s eliminate the square root first, by looking at $T^2$
and insert the $a$ from above
And pulling the root again gives us
This formula, if you need to be told, is quite ugly, but it would allow us to directly compute the orbital time for any object and any gravitational border.
For Earth, it’s 157875s, or 43h50min, meaning you die after about 11h.
### Transfer orbits
Now that we know at what distance we arrive, let’s define the burns we have to make. First off, we’ll assume that we’re moving into the highest recharging orbit (1N/kg of gravitational force). For that, we’re going to have to make two burns: The first one drops the periapsis to the target altitude (and avoids us dying a fiery death), and the second one (executed at periapsis) circularizes the orbit.
All of this can be computed using the vis-viva-equation above - yet it turns out there’s an issue. Looking at the moon as an example, it’s a clearly superior choice compared to almost everything: You’d have to spend less than 1km/s of dV and still transfer for only five hours. As a consequence, I’m increasing the necessary gravitational acceleration for jump recharge to 3N/kg. This means that, aside from the Sun, only the gas giants, Earth, Venus, and Mars can be used as recharge stations. This will also pull all of the traffic towards planets - perfect for my purposes.
As an example, for Earth that’s an altitude of about 5150km. You need to spend about 3km/s dV (3.6 for LEO), with about eight hours of transfer time. Luckily, that’s fairly close to what I assumed anyway during the economics discussion.
### Fast Transfer Orbits
But we might not want to spend 8 hours of transfer time; we want to get there faster. Computing that is more complicated: We essentially have to accelerate increase eccentricity of our orbit (which may lead to a hyperbolic orbit) and then decelerate again to circularize at the recharging orbit. Intuitively, this would also shift periapsis, from the opposite of where we’re emerging (minimum deltaV) to offset by 90° (infinite deltaV). To compute this - and especially the time needed - we need to introduce variables such as the eccentricity of the orbit - which I don’t want to do at this point. So I’m going to do an approximation.
Specifically, I’ll ignore the “sideways” gravitational acceleration due to the planet. I can do that by assuming instant acceleration and bending the orbit by a bit. This leaves me with a travelled distance of the twice the semi-major axis (or periapsis plus apoapsis) and a certain average speed (simply from my half-orbital period):
That’s the same formula as above, just to remind you: $v_e$ is the deltaV needed to be invested, $a$ is the semi-major axis, and $T$ is the orbital period.
In our Earth example, that’s 2.38km/s speed. If I want to spend more dV, I can just divide the orbital distance by my additional speed $v_a$ - or actually half the dV I’m willing to additionally spend, since I have to decelerate again.
For Earth, if I want to spend at most 5km/s to get to LEO, it will leave only 650m/s for accelerating, and will get me there in about six hours.
### Leaving Again
To leave, we just do the circularization burn but in reverse; so it’s the same dV. This will get you onto a minimum-deltaV trajectory towards the jump limit.
For accelerating, I’m again going to use the approximation above.
## Putting Everything Together
Let’s stay with the example of Earth. If we’re allowed to use a total of 10km/s for everything, we’re going to split the left-over dV into three parts: Acceleration and deceleration for the incoming burn, and accelerating for the outgoing burn. In the case of an Earth-LEO parking orbit, you’ll have about 7km/s dV left; this will accelerate both incoming and outgoing travel to about 4h15min each. Since that’s quite a bit faster than my first approximation, I’m going to increase recharging time to 15.4 hours (15h24min). This also leaves everybody with more time to load and unload cargo.
## Changes to the FTL drive
Compared to the first version, the new FTL drive
• emerges at a gravitational threshold of 3 N/kg (which excludes moons)
• recharges in 15.4 hours
All the other details are simply because I spent the time looking at the actual parameters. And I have to agree with ericthered - emergence at rest is nicer. | 2,131 | 8,491 | {"found_math": true, "script_math_tex": 27, "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} | 4.125 | 4 | CC-MAIN-2018-30 | longest | en | 0.882809 |
https://gamedev.stackexchange.com/questions/171339/how-can-i-create-a-dice-toss-that-results-in-a-specific-value | 1,717,025,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00133.warc.gz | 225,064,105 | 39,008 | # How can I create a dice toss that results in a specific value?
I am trying to create a 3D dice for a game. Normal dice has been created by setting cube for dice with Rigidbody and apply force to throw a dice.
Now the problem is, the user wants to give a value 1 - 6 in the text box then they will press the roll button. The dice have to rotate and need to show the dice value as the user enters in the text field once it stops rolling.
For example:
1. In the Input box the value entered as 5,
2. Now the dice want to roll on the table and show the value as 5.
• You probably can simulate dice roll before the user presses the button. Now, when you know where dice sides will end up, apply initial rotation to compensate. Apr 25, 2019 at 14:50
I wouldn't rely on the physics for this. If you have one dice, then you should just have some prebaked animations. You can have more than one animation per outcome to add variation. If you have multiple dice then we can take a different approach. Prebake several animations (you can use physics and record the keyframes) for rolling the dice. For each dice, mark the side that ends up face up at the end of the animation. When the user rolls the dice, select a random animation (multiple animations so it's not obvious they are prebaked) then find the side of each dice to set that ends faceup at the end of the animation. Find where that face is at the start of the animation. Rotate the dice as a child object before the animation (or swap texture UVs) such that the side you want up is in the correct position before the animations starts. In this way you can rig the dice roll.
For example let's say normally the 2 ends up on top for Animation A. At the start of the animation, the 2 side is facing in the -y direction. The user picks the 1 side. Rotate the dice at the start of the animation so the 1 side is facing the -y direction. Now at the end of the animation the 1 will end up on top.
It's difficult to explain clearly but hopefully you get what I'm trying to say. If not, comment.
Revisiting this question, I just got a new idea: you could make a physics simulation in the background and cache the result, determine which face ended up, put the textures in the right place and play the simulation back to the user.
If you are determined to do this strictly with physics in unity with rigidbodies..This is a somewhat silly answer but what about making a "loaded dice"?
Make a die that is comprised of 3x3x3 set of cubes. Each cube's mass can be modified via a script that is interacting with your input fields from your UI with C#.
If a user inputs "1" for example and the desired 3D dice roll result is a "1" you would make your UI input script look for "1" and as a result increase the mass of the 9 cubes on the side that your "6" is on , or whichever side is opposite to the "1".
This weighting process is not guaranteed to work but im sure if you adjusted the parameters enough you could find a nearly optimal solution using physics.
Hope this helps! I like the other replies as well. Very fun problem here..
EDIT: My father is a machinist and has told me that weighting the corners of the dice was a tactic used by fabricators of these "loaded dice". I hope this helps. | 753 | 3,247 | {"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-2024-22 | latest | en | 0.927913 |
http://nrich.maths.org/4947 | 1,485,241,838,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284352.26/warc/CC-MAIN-20170116095124-00263-ip-10-171-10-70.ec2.internal.warc.gz | 205,379,662 | 5,086 | ### Strange Numbers
All strange numbers are prime. Every one digit prime number is strange and a number of two or more digits is strange if and only if so are the two numbers obtained from it by omitting either its first or its last digit. Find all strange numbers.
### Factor Lines
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Stars
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
# Prime Order
##### Stage: 3 Short Challenge Level:
See all short problems arranged by curriculum topic in the short problems collection
How many of the three digit numbers that can be made from all of the digits 1, 3 and 5 (used only once each) are prime?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
This problem is taken from the UKMT Mathematical Challenges. | 206 | 976 | {"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 | 3 | CC-MAIN-2017-04 | longest | en | 0.895058 |
https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Units_of_information.html | 1,720,961,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00221.warc.gz | 284,401,082 | 19,212 | # Units of information
In computing and telecommunications, a unit of information is the capacity of some standard data storage system or communication channel, used to measure the capacities of other systems and channels. In information theory, units of information are also used to measure the information contents or entropy of random variables.
The most common units are the bit, the capacity of a system which can exist in only two states, and the byte (or octet), which is equivalent to eight bits. Multiples of these units can be formed from these with the SI prefixes (power-of-ten prefixes) or the newer IEC binary prefixes (binary power prefixes). Information capacity is a dimensionless quantity.
## Primary units
Comparison of units of information: bit, trit, nat, ban. Quantity of information is the height of bars. Dark green level is the "Nat" unit.
In 1928, Ralph Hartley observed a fundamental storage principle,[1] which was further formalized by Claude Shannon in 1945: the information that can be stored in a system is proportional to the logarithm logb N of the number N of possible states of that system. Changing the basis of the logarithm from b to a different number c has the effect of multiplying the value of the logarithm by a fixed constant, namely logc N = (logc b) logb N. Therefore, the choice of the basis b determines the unit used to measure information. In particular, if b is a positive integer, then the unit is the amount of information that can be stored in a system with N possible states.
When b is 2, the unit is the shannon, equal to the information content of one "bit" (a contraction of binary digit[2]). A system with 8 possible states, for example, can store up to log28 = 3 bits of information. Other units that have been named include:
The trit, ban, and nat are rarely used to measure storage capacity; but the nat, in particular, is often used in information theory, because natural logarithms are sometimes more convenient than logarithms in other bases.
## Units derived from bit
Several conventional names are used for collections or groups of bits.
### Shave and a haircut
The name given to 2 bits. The name is a reference to the song, "Shave and a haircut".
### Byte
Historically, a byte was the number of bits used to encode a character of text in the computer, which depended on computer hardware architecture; but today it almost always means eight bits — that is, an octet. A byte can represent 256 (28) distinct values, such as the integers 0 to 255, or -128 to 127. The IEEE 1541-2002 standard specifies "B" (upper case) as the symbol for byte. Bytes, or multiples thereof, are almost always used to specify the sizes of computer files and the capacity of storage units. Most modern computers and peripheral devices are designed to manipulate data in whole bytes or groups of bytes, rather than individual bits.
### Nibble
A group of four bits, or half a byte, is sometimes called a nibble or nybble. This unit is most often used in the context of hexadecimal number representations, since a nibble has the same amount of information as one hexadecimal digit.[7]
### Word, block, and page
Computers usually manipulate bits in groups of a fixed size, conventionally called words. The number of bits in a word is usually defined by the size of the registers in the computer's CPU, or by the number of data bits that are fetched from its main memory in a single operation. In the IA-32 architecture more commonly known as x86-32, a word is 16 bits, but other past and current architectures use words with 8, 24, 32, 36, 56, 64, 80 bits or others.
Some machine instructions and computer number formats use two words (a "double word" or "dword"), or four words (a "quad word" or "quad").
Computer memory caches usually operate on blocks of memory that consist of several consecutive words. These units are customarily called cache blocks, or, in CPU caches, cache lines.
Virtual memory systems partition the computer's main storage into even larger units, traditionally called pages.
### Systematic multiples
Terms for large quantities of bits can be formed using the standard range of SI prefixes for powers of 10, e.g., kilo = 103 = 1000 (as in kilobit or kbit), mega- = 106 = 1000000 (as in megabit or Mbit) and giga = 109 = 1000000000 (as in gigabit or Gbit). These prefixes are more often used for multiples of bytes, as in kilobyte (1 kB = 8000 bit), megabyte (1 MB = 8000000bit), and gigabyte (1 GB = 8000000000bit).
However, for technical reasons, the capacities of computer memories and some storage units are often multiples of some large power of two, such as 228 = 268435456 bytes. To avoid such unwieldy numbers, people have often misused the SI prefixes to mean the nearest power of two, e.g., using the prefix kilo for 210 = 1024, mega for 220 = 1048576, and giga for 230 = 1073741824, and so on. For example, a random access memory chip with a capacity of 228 bytes would be referred to as a 256-megabyte chip. The table below illustrates these differences.
Multiples of bits
Decimal
Value SI
1000 103 kbit kilobit
10002 106 Mbit megabit
10003 109 Gbit gigabit
10004 1012 Tbit terabit
10005 1015 Pbit petabit
10006 1018 Ebit exabit
10007 1021 Zbit zettabit
10008 1024 Ybit yottabit
Binary
Value IEC JEDEC
1024 210 Kibit kibibit Kbit kilobit
10242 220 Mibit mebibit Mbit megabit
10243 230 Gibit gibibit Gbit gigabit
10244 240 Tibit tebibit -
10245 250 Pibit pebibit -
10246 260 Eibit exbibit -
10247 270 Zibit zebibit -
10248 280 Yibit yobibit -
Symbol Prefix SI Meaning Binary meaning Size difference
k kilo 103 = 10001 210 = 10241 2.40%
M mega 106 = 10002 220 = 10242 4.86%
G giga 109 = 10003 230 = 10243 7.37%
T tera 1012 = 10004 240 = 10244 9.95%
P peta 1015 = 10005 250 = 10245 12.59%
E exa 1018 = 10006 260 = 10246 15.29%
Z zetta 1021 = 10007 270 = 10247 18.06%
Y yotta 1024 = 10008 280 = 10248 20.89%
In the past, uppercase K has been used instead of lowercase k to indicate 1024 instead of 1000. However, this usage was never consistently applied.
On the other hand, for external storage systems (such as optical discs), the SI prefixes were commonly used with their decimal values (powers of 10). There have been many attempts to resolve the confusion by providing alternative notations for power-of-two multiples. In 1998 the International Electrotechnical Commission (IEC) issued a standard for this purpose, namely a series of binary prefixes that use 1024 instead of 1000 as the main radix:[8]
Multiples of bytes
Decimal
Value Metric
1000 kB kilobyte
10002 MB megabyte
10003 GB gigabyte
10004 TB terabyte
10005 PB petabyte
10006 EB exabyte
10007 ZB zettabyte
10008 YB yottabyte
Binary
Value IEC JEDEC
1024 KiB kibibyte KB kilobyte
10242 MiB mebibyte MB megabyte
10243 GiB gibibyte GB gigabyte
10244 TiB tebibyte
10245 PiB pebibyte
10246 EiB exbibyte
10247 ZiB zebibyte
10248 YiB yobibyte
Symbol Prefix
Ki kibi, binary kilo 1 kibibyte (KiB) 210 bytes 1024 B
Mi mebi, binary mega 1 mebibyte (MiB) 220 bytes 1024 KiB
Gi gibi, binary giga 1 gibibyte (GiB) 230 bytes 1024 MiB
Ti tebi, binary tera 1 tebibyte (TiB) 240 bytes 1024 GiB
Pi pebi, binary peta 1 pebibyte (PiB) 250 bytes 1024 TiB
Ei exbi, binary exa 1 exbibyte (EiB) 260 bytes 1024 PiB
The JEDEC memory standards however define uppercase K, M, and G for the binary powers 210, 220 and 230 to reflect common usage.[9]
## Size examples
• 1 bit – answer to a yes/no question
• 1 byte – a number from 0 to 255.
• 90 bytes: enough to store a typical line of text from a book.
• 512 bytes = ½ KiB: the typical sector of a hard disk.
• 1024 bytes = 1 KiB: the classical block size in UNIX filesystems.
• 2048 bytes = 2 KiB: a CD-ROM sector.
• 4096 bytes = 4 KiB: a memory page in x86 (since Intel 80386).
• 4 kB: about one page of text from a novel.
• 120 kB: the text of a typical pocket book.
• 1 MB – a 1024×1024 pixel bitmap image with 256 colors (8 bpp color depth).
• 3 MB – a three-minute song (133 kbit/s)
• 650-900 MB – a CD-ROM
• 1 GB – 114 minutes of uncompressed CD-quality audio at 1.4 Mbit/s
• 8/16 GB – size of a normal flash drive
• 4 TB – the size of a \$150 hard disk (as of late 2014)
• 1.3 ZB – prediction of the volume of the whole internet in 2016.
## Obsolete and unusual units
Several other units of information storage have been named.[7]
• 1 bit: sniff.
• 2 bits: crumb,[10] quad, quarter, tayste, tydbit, semi-nibble.
• 5 bits: nickel, nyckle.
• 6 bits: byte (in early IBM machines using BCD alphamerics).
• 10 bits: declet,[13][14][15][16] decle,[17] deckle, dyme.
• 12 bits: slab[18][19][20]
• 16 bits: wyde,[21] doublet,[22] plate, playte, chomp, chawmp (on a 32-bit machine).
• 18 bits: chomp, chawmp (on a 36-bit machine).
• 32 bits: quadlet,[22] dinner, dynner, gawble (on a 32-bit machine).
• 48 bits: gobble, gawble (under circumstances that remain obscure).
• 64 bits: octlet.[22]
• 128 bits: hexlet.[22]
• 16 bytes: paragraph.
• 6 trits: tryte[23]
• combit, comword[24][25][26]
Most of these names are jargon, obsolete, or used only in very restricted contexts.
## References
1. Norman Abramson (1963), Information theory and coding. McGraw-Hill.
2. Mackenzie, Charles E. (1980). Coded Character Sets, History and Development. The Systems Programming Series (1 ed.). Addison-Wesley Publishing Company, Inc. p. xii. ISBN 0-201-14460-3. LCCN 77-90165. ISBN 978-0-201-14460-4. Retrieved 2016-05-22.
3. Donald E. Knuth, The Art of Computer Programming, vol.2: Seminumerical algorithms.
4. Shanmugam (2006), Digital and Analog Computer Systems.
5. Gregg Jaeger (2007), Quantum information: an overview
6. I. Ravi Kumar (2001), Comprehensive Statistical Theory of Communication.
7. Nybble at dictionary reference.com; sourced from Jargon File 4.2.0, accessed 2007-08-12
8. ISO/IEC standard is ISO/IEC 80000-13:2008. This standard cancels and replaces subclauses 3.8 and 3.9 of IEC 60027-2:2005. The only significant change is the addition of explicit definitions for some quantities. ISO Online Catalogue
9. JEDEC Solid State Technology Association (December 2002). "Terms, Definitions, and Letter Symbols for Microcomputers, Microprocessors, and Memory Integrated Circuits" (PDF). JESD 100B.01. Retrieved 2009-04-05
10. Weisstein, Eric. W. "Crumb". MathWorld. Retrieved 2015-08-02.
11. Paul, Reinhold (2013). "Elektrotechnik und Elektronik für Informatiker - Grundgebiete der Elektronik". Leitfaden der Informatik. B.G. Teubner Stuttgart / Springer. ISBN 3322966526. 9783322966520. Retrieved 2015-08-03.
12. Böhme, Gert; Born, Werner; Wagner, B.; Schwarze, G. (2013-07-02) [1969]. Jürgen Reichenbach, ed. Programmierung von Prozeßrechnern. Reihe Automatisierungstechnik (in German). 79. VEB Verlag Technik Berlin, reprint: Springer Verlag. doi:10.1007/978-3-663-02721-8. ISBN 978-3-663-00808-8. 9/3/4185.
13. IEEE 754-2008 - IEEE Standard for Floating-Point Arithmetic. 2008-08-29. doi:10.1109/IEEESTD.2008.4610935. ISBN 978-0-7381-5752-8. Retrieved 2016-02-10.
14. Muller, Jean-Michel; Brisebarre, Nicolas; de Dinechin, Florent; Jeannerod, Claude-Pierre; Lefèvre, Vincent; Melquiond, Guillaume; Revol, Nathalie; Stehlé, Damien; Torres, Serge (2010). Handbook of Floating-Point Arithmetic (1 ed.). Birkhäuser. doi:10.1007/978-0-8176-4705-6. ISBN 978-0-8176-4704-9. LCCN 2009939668.
15. Erle, Mark A. (2008-11-21). Algorithms and Hardware Designs for Decimal Multiplication (Thesis). Lehigh University: ProQuest (published 2009). ISBN 9781109042283. 1109042280. Retrieved 2016-02-10.
16. Kneusel, Ronald T. (2015). Numbers and Computers. Springer. ISBN 9783319172606. 3319172603. Retrieved 2016-02-10.
17. Joe Zbiciak. "AS1600 Quick-and-Dirty Documentation". Retrieved 2013-04-28.
18. "315 Electronic Data Processing System" (PDF). NCR. November 1965. NCR MPN ST-5008-15. Archived (PDF) from the original on 2016-05-24. Retrieved 2015-01-28.
19. Bardin, Hillel (1963). "NCR 315 Seminar" (PDF). Computer Usage Communique. 2 (3). Archived (PDF) from the original on 2016-05-24.
20. Schneider, Carl (2013) [1970]. Datenverarbeitungs-Lexikon [Lexicon of information technology] (in German) (softcover reprint of hardcover 1st ed.). Wiesbaden, Germany: Springer Fachmedien Wiesbaden GmbH / Betriebswirtschaftlicher Verlag Dr. Th. Gabler GmbH. pp. 201, 308. doi:10.1007/978-3-663-13618-7. ISBN 978-3-409-31831-0. ISBN 3-663-13618-3 / ISBN 978-3-663-13618-7 (ebook). Retrieved 2016-05-24. slab, Abk. aus syllable = Silbe, die kleinste adressierbare Informationseinheit für 12 bit zur Übertragung von zwei Alphazeichen oder drei numerischen Zeichen. (NCR) […] Hardware: Datenstruktur: NCR 315-100 / NCR 315-RMC; Wortlänge: Silbe; Bits: 12; Bytes: –; Dezimalziffern: 3; Zeichen: 2; Gleitkommadarstellung: fest verdrahtet; Mantisse: 4 Silben; Exponent: 1 Silbe (11 Stellen + 1 Vorzeichen) [slab, abbr. for syllable = syllable, smallest addressable information unit for 12 bits for the transfer of two alphabetical characters or three numerical characters. (NCR) […] Hardware: Data structure: NCR 315-100 / NCR 315-RMC; Word length: Syllable; Bits: 12; Bytes: –; Decimal digits: 3; Characters: 2; Floating point format: hard-wired; Significand: 4 syllables; Exponent: 1 syllable (11 digits + 1 prefix)]
21. IEEE Std 1754-1994 - IEEE Standard for a 32-bit Microcontroller Architecture. The Institute of Electrical and Electronic Engineers, Inc. pp. 5–7. doi:10.1109/IEEESTD.1995.79519. ISBN 1-55937-428-4. Retrieved 2016-02-10. (NB. The standard defines doublets, quadlets, octlets and hexlets as 2, 4, 8 and 16 bytes, giving the numbers of bits (16, 32, 64 and 128) only as a secondary meaning. This might be important given that bytes were not always understood to mean 8 bits (octets) historically.)
22. Brousentsov, N. P.; Maslov, S. P.; Ramil Alvarez, J.; Zhogolev, E.A. "Development of ternary computers at Moscow State University". Retrieved 2010-01-20.
23. US4319227, Malinowski, Christopher W.; Heinz Rinderle & Martin Siegle, "Three-state signaling system", issued 1982-03-09, assigned to Department of Research and Development, AEG-Telefunken, Heilbronn, Germany | 4,245 | 14,069 | {"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-2024-30 | latest | en | 0.909544 |
http://stackoverflow.com/questions/4775029/finding-combinatorial-of-large-numbers/4775046 | 1,419,755,368,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447556220.137/warc/CC-MAIN-20141224185916-00011-ip-10-231-17-201.ec2.internal.warc.gz | 95,607,244 | 19,564 | # finding combinatorial of large numbers
What is an efficient way of finding the number of ways k gifts can be chosen from N gifts where N can be very large (N ~ 10^18). That is we have to calculate N(C)K or N chose K. K can also be of the order of N.
-
I guess there are no fast ways to compute such large numbers. You can approximate it using Stirling's formula
-
Stirling's formula is useless if we don't know how k compares to n. – Alexandre C. Jan 23 '11 at 17:11
@Alexandre C.: Huh, how can you say that? Stirling's formula approximates `n!`. So you approximate `n!, k! and (n - k)!` and get an approximation to `n! / (k! * (n - k)!)`. If you want to get an asymptotic sense of the behavior of `C(n, k)` then, yes, you need a relationship between `k` and `n`, but to approximate, no. – Jason Jan 23 '11 at 17:47
@Jason: Computing 3 times log gamma is very fast, probably comparable to pow + exp + sqrt. The power of Stirling's formula (and the Stirling series) is that it allows you to get accurate asymptotics when you know the behavior of k and n together. You have better choices for approximating factorials. A good start is en.wikipedia.org/wiki/Lanczos_approximation . – Alexandre C. Jan 24 '11 at 10:07
@Alexandre C.: I'm not denying that. I'm denying your claim that Stirling's formula is useless if we can't compare `k` to `n`). – Jason Jan 24 '11 at 14:28
@Alexandre C.: See, that's where I disagree with you. I am not denying your claim that the approximation you give is better (this is well known, it's Straight Outta Numerical Recipes). I'm denying your claim that the Stirling approximation is useless. – Jason Jan 24 '11 at 15:18
Stirling's formula would be useful only if you had further asymptotic information such as `k ~ n / 3` or `k ~ log n`. With no further information on your specific problem, you will not draw any information of Stirling's formula.
For your problem as stated, the most direct way to compute C(n, k) when k and n are large (and even when they are not large) is to use
``````log C(n, k) = log (n!) - (log (k!) + log ((n - k)!))
``````
and
``````n! = gamma(n + 1).
``````
The fact is that it is quite easy to come with an implementation of log gamma, and you then have
``````C(n, k) = exp (f(n + 1) - f(k + 1) - f(n - k + 1))
``````
where `f = log gamma` .
You can find numerical algorithms for computing log gamma in Numerical Recipes, an old version is available there and you will find a sample implementation in Chapter 6.
-
Also note that for many applications, you may even be happier with storing `ln(C(n, k)) = lngamma(n+1) - lngamma(k+1) - lngamma(n-k+1)`. The chance of overflow is considerably smaller. – Christopher Creutzig Jan 23 '11 at 17:20
can you explain what gamma is? I'm not familiar with it as a standard function? or is it just defined by gamma(n+1) = n!? – Chris Jan 24 '11 at 17:57
@Chris: Gamma(x) = integral(exp(-t) t^(x - 1) dt, t = 0..infinity). The formula Gamma(n) = (n-1)! holds by integrating by parts. There are fast methods to compute Gamma or log Gamma. – Alexandre C. Jan 24 '11 at 18:22
Ah, thanks. That rings some vague bells now. I figured it would probably be a continous function that worked at those discrete points but always nice to check. :) – Chris Jan 24 '11 at 18:24
Since variety is the spice of life, another approach is the following. The value (N Choose K)/2^N approaches a normal distribution with mean N/2 and Standard Deviation Sqrt[N]/2 and it does so quite quickly. We can therefore approximate (N Choose K) as 2^N * Normdist(x,0,1)/std where x =( k - N/2)/std and std is Sqrt[N]/2.
Normdist(x,0,1) = Exp(-x^2/2)*1/(Sqrt(2*Pi))
In terms of error this should get much better the larger the number, and a quick check using N as 113(?) shows an maximum error as a percent of the largest coefficient of less than 0.3%.
Not claiming its better than using Stirling formula, but think that it might avoid some of the n^n calculations and working out the log of these coefficients is a pretty simple calculation.
-
+1, good idea . – Alexandre C. Jan 24 '11 at 15:05
Thanks Alexandre, just to add this all works out to be Log(N Choose K) = (4*k - (4*Power(k,2))/n + Log(2) + n*(-1 + Log(4)) - Log(n*Pi))/2 which should be the fastest algorithm here (if not the best). – aronp Jan 24 '11 at 22:07
The value of `C(n, k)` can be close to `2^n`. (well, order of magnitude smaller, but that's not important here).
What important is that to store number `2^(10^18)`, you need `10^18` bits or `~ 10^17` bytes.
You might want to adjust problem definition, because there're no computers like that.
Others have already pointed out approximate formula, where you could store result as a floating point number, thus not spending more memory than necessary.
- | 1,341 | 4,774 | {"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} | 3.25 | 3 | CC-MAIN-2014-52 | latest | en | 0.920345 |
https://us.metamath.org/mpeuni/nf2.html | 1,721,368,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00124.warc.gz | 512,320,615 | 3,359 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > nf2 Structured version Visualization version GIF version
Theorem nf2 1787
Description: Alternate definition of non-freeness. (Contributed by BJ, 16-Sep-2021.)
Assertion
Ref Expression
nf2 (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ¬ ∃𝑥𝜑))
Proof of Theorem nf2
StepHypRef Expression
1 df-nf 1786 . 2 (Ⅎ𝑥𝜑 ↔ (∃𝑥𝜑 → ∀𝑥𝜑))
2 imor 850 . 2 ((∃𝑥𝜑 → ∀𝑥𝜑) ↔ (¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑))
3 orcom 867 . 2 ((¬ ∃𝑥𝜑 ∨ ∀𝑥𝜑) ↔ (∀𝑥𝜑 ∨ ¬ ∃𝑥𝜑))
41, 2, 33bitri 300 1 (Ⅎ𝑥𝜑 ↔ (∀𝑥𝜑 ∨ ¬ ∃𝑥𝜑))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∨ wo 844 ∀wal 1536 ∃wex 1781 Ⅎwnf 1785 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 210 df-or 845 df-nf 1786 This theorem is referenced by: nf3 1788
Copyright terms: Public domain W3C validator | 452 | 893 | {"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} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.307593 |
https://excel-helps.com/category/uncategorized/ | 1,695,896,376,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00162.warc.gz | 260,139,702 | 14,197 | Uncategorized
## Workdays per month
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https://forum.allaboutcircuits.com/threads/741-op-amp-lab.132690/ | 1,566,593,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318986.84/warc/CC-MAIN-20190823192831-20190823214831-00324.warc.gz | 467,840,314 | 21,723 | # 741 OP Amp Lab
Discussion in 'Homework Help' started by chrisjsmith, Feb 26, 2017.
1. ### chrisjsmith Thread Starter Member
Nov 12, 2016
36
0
I recently did a Lab using a 741 Op Amp. I was asked to write my observations of what's happening with the output signal relating to the input.
In my pictures I have different scenarios during the experiment. I'm unsure if in my observations i'm explaining what's happening correctly or covering everything that's taking place. Please read the scenarios taking place in the pictures and critique my observations.
One scenario the Rf is replaced by a 10K Potentiometer and the resistance is varied. Please assist in explaining to me what's taking place as i'm also unsure.
Thanks for the help all
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2. ### crutschow Expert
Mar 14, 2008
22,530
6,604
It's better if you tell us what you think might be happening.
Does any of what happened make sense?
Do you know what that circuit does?
3. ### #12 Expert
Nov 30, 2010
18,076
9,686
It all looks very normal to me except you forgot to declare (-) when describing an inverting gain.
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4. ### chrisjsmith Thread Starter Member
Nov 12, 2016
36
0
So i know it's an inverting Op Amp. The input will if +ve will be inverted to -ve. A sine wave will be inverted by 180 degrees
Rf is forms the feedback loop to the inverting input
I know the gain is equal to -Rf/Ri and I know the parameters of an ideal op amp
I'm just unsure if i'm missing anything in my observation or if my description of what's happening is technically correct
5. ### Papabravo Expert
Feb 24, 2006
12,194
2,687
The behavior is what you might expect at "relatively" low frequencies. It quickly "runs out of gas" at higher frequencies and there is a substantial delay between input and output due to "slew rate limitations". You might also observe that the output cannot get very close to either power rail, leading to the sine wave displaying a "clipped" appearance at high gain settings. Knowing what these behaviors look like is important to understanding why this obsolete part is not recommended for "new" designs. Just like a microprocessor from 50 years ago should no longer be used, the same is true for this venerable part.
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6. ### crutschow Expert
Mar 14, 2008
22,530
6,604
But as someone noted, it's great for learning about real circuits since it has so many characteristics that are far from the ideal.
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7. ### chrisjsmith Thread Starter Member
Nov 12, 2016
36
0
Thanks for the explanation. I have some research to do regarding some of the terms you've mentioned but it's a pointer in a good direction
8. ### WBahn Moderator
Mar 31, 2012
24,360
7,613
Do you have a reference (from some kind of official source) that says that this part is obsolete and not recommended for new designs?
I'm not being combative, I'm quite curious about this. The fact that some unknown large number of millions of these devices are still manufactured every year by several different vendors makes it hard to claim that it is an obsolete part that is not intended for new designs. TI's data sheet, in fact, categorizes it's status as "active" and specifically states that it is intended for new designs. Other manufacturers' data sheets generally don't say one way or the other (and, despite being curious, it isn't worth it to me to make the effort to research it further, so if you happen to have information to indicate that the part is truly considered obsolete and not to be used for new designs, by at least some manufacturers, I would love to hear it).
I think the part has some advantages over many newer parts, including cost as well as some performance advantages such as input/output overload protection and no latch-up when the input common mode range is exceeded, that make it useful for non-demanding designs. Certainly the types of circuits that we can design and implement today have narrowed the domain of applications that the 741 can be used for, but the evidence would seem to indicate that even that narrowed domain is still more than rich enough to justify continuing its production.
9. ### Papabravo Expert
Feb 24, 2006
12,194
2,687
Well from a current datasheet (Jul 2016) it looks as if a PDIP part they acquired from National is obsolete (Non ROHS compliant). I didn't see any NRND.
U9T7741393 OBSOLETE PDIP P 8 TBD Call TI Call TI 0 to 70 LM 741CN
What seems to be current and in production are surface mount parts.
10. ### WBahn Moderator
Mar 31, 2012
24,360
7,613
That's what I'm seeing -- but it appears that the LM741CN marking is still active, it's just now on the lead-free version.
11. ### WBahn Moderator
Mar 31, 2012
24,360
7,613
A general comment is that you need to be more complete in your writing and, while qualitative observations are fine, try to also be quantitative where you can. A good way to approach it is to assume that you are writing your report so that someone taking this lab next semester could use your report to reproduce your results. So you need to describe what your input signals are, something like "The input signal was a 1 Vpp sinusoid with no DC component at nominally 1 kHz." Then when you modify the circuit, present the modified schematic. You say you replaced the feedback resistor with a potentiometer. But the former is a two-terminal device while the latter is a three-terminal device. Draw a schematic showing how you hooked it up. If nothing else, this would let the grader spot potential problems if you hooked it up incorrectly. On the one where you are varying the magnitude of the input signal, comment on the maximum amplitude that the input can be before clipping occurs, whether clipping for positive and negative outputs at the same time, what the clipping level is, and how this compares to the datasheet specifications. You might also measure the small signal gain and the gain just prior to clipping (say at an amplitude that is 90% of the signal at which clipping begins). Similarly for the scenario in which you are changing the supply rail voltages -- mention what the supply rails are at the time you took the scope trace and compare the observed results to what the datasheet would lead you to expect. Also, in the large-input signal scenario, consider whether the part of the waveform in the active region matches the nominal gain. If not, consider the observed rate of change to the part's slew-rate limit.
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12. ### chrisjsmith Thread Starter Member
Nov 12, 2016
36
0
Thanks for your in dept feedback. Yes, I could have been more detailed i'll definitely change this for future post so it'll be more helpful for another forum user. Thanks again man | 1,621 | 6,836 | {"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} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.908253 |
https://egvideos.com/video/ohio/grade-1/math/1.md.4/interpreting-data | 1,656,543,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00598.warc.gz | 286,362,018 | 10,255 | # Ohio - Grade 1 - Math - Measurement and Data - Interpreting Data - 1.MD.4
### Description
Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another.
• State - Ohio
• Standard ID - 1.MD.4
• Subjects - Math Common Core
### Keywords
• Math
• Measurement and Data
## More Ohio Topics
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. See Table 1, page 95.
Apply properties of operations as strategies to add and subtract. For example, if 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (Commutative Property of Addition); to add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (Associative Property of Addition). Students need not use formal terms for these properties.
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.
1.OA.5 Relate counting to addition and subtraction, e.g., by counting on 2 to add 2.
1.OA.6 Add and subtract within 20, demonstrating fluency with various strategies for addition and subtraction within 10. Strategies may include counting on; making ten, e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14; decomposing a number leading to a ten, e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9; using the relationship between addition and subtraction, e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4; and creating equivalent but easier or known sums, e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13.
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6; 7 = 8 – 1; 5 + 2 = 2 + 5; 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations: 8 + __ = 11; 5 = __ - 3; 6 + 6 = __ .
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: 10 can be thought of as a bundle of ten ones — called a “ten;” the numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones; and the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). | 797 | 2,833 | {"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} | 4.71875 | 5 | CC-MAIN-2022-27 | longest | en | 0.901189 |
https://la.mathworks.com/matlabcentral/answers/1792870-unrealistic-fitting-confidence-levels-in-noisy-data?s_tid=prof_contriblnk | 1,675,421,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00878.warc.gz | 366,402,896 | 27,780 | # Unrealistic fitting confidence levels in noisy data
6 views (last 30 days)
ChemPhysicist on 1 Sep 2022
Answered: Bjorn Gustavsson on 1 Sep 2022
The fitting of simple exponential is giving a confidence level which is not realistic providing how noisy are the data. Any explanation please ?
Thank you. Code:
Time = Data(:,1) - 2.6E-6 ; % time offset
Signal = Data(:,2)-0.078; % amplitude offset
x = Time;
y = Signal;
y = y/(0.8*max(y)); % data normalization to about 1
format long % just to get more precision digits
mdl = fittype(' a*(1-exp(-b*x)) ','indep','x')
mdl =
General model: mdl(a,b,x) = a*(1-exp(-b*x))
fittedmdl = fit(x,y,mdl,'StartPoint', [max(y) 1E5])
fittedmdl =
General model: fittedmdl(x) = a*(1-exp(-b*x)) Coefficients (with 95% confidence bounds): a = 0.9589 (0.955, 0.9628) b = 5.557e+06 (5.373e+06, 5.74e+06)
figure
plot(x,y, 'bp', 'DisplayName','data')
hold on
plot(fittedmdl)
grid
xlabel('Time /s')
ylabel('Intensity /a.u')
ax = gca;
ax.FontSize = 15;
As you can see the error on (b) is very small (only 3% !) which is not realistic looking how noisy are the data:
coefficientValues = coeffvalues(fittedmdl);
a = coefficientValues(1);
b = coefficientValues(2);
Tau = (1/b)
Tau =
1.799653625749003e-07
ConfIntervals = confint(fittedmdl);
b_err = (ConfIntervals(2,2) - ConfIntervals(1,2))/2;
DeltaTau = Tau * (b_err/b)
DeltaTau =
5.935716982509685e-09
Error = (DeltaTau / Tau)*100
Error =
3.298255229552455
Bjorn Gustavsson on 1 Sep 2022
This is most likely due to the large number of data-points you have. Compare with the simpler case of the uncertainty of the average of a number of random samples from a normal-distribution with a large standard deviation - it decreases roughly as the square-root of the number of samples. Here you have a similar situation. One thing you can do is to plot the distribution of the residuals, and investigare how it varies as you vary the parameters - and check how much it starts to skew off zero-centred and symmetric. Another thing you should look at are the residuals - to my naked eye there seems to be some systematic variation in the signal that aren't accounted for - some higher-frequency variations that do not look like random noise.
HTH
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Translated by | 687 | 2,446 | {"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-2023-06 | latest | en | 0.768452 |
https://www.mathworks.com/matlabcentral/cody/problems/1237-it-s-race-time-write-a-faster-function-than-the-test-suite-call-of-unique/solutions/318889 | 1,495,547,422,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00323.warc.gz | 897,205,607 | 11,800 | Cody
Problem 1237. It's race time! Write a faster function than the test suite call of unique().
Solution 318889
Submitted on 13 Sep 2013
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
%% x = rand(10000, 1); z = rand(10000, 1); x = vertcat(x, z); tic y_correct = unique(x); t_unique = toc tic y_myunique = my_unique(x); t_myunique = toc assert(isequal(sort(my_unique(x)),y_correct) && t_unique > t_myunique)
``` t_unique = 0.0467 t_myunique = 0.0040 ```
2 Fail
%% x = rand(50000, 1); z = rand(50000, 1); x = vertcat(x, z); tic y_correct = unique(x); t_unique = toc tic y_my_unique = my_unique(x); t_my_unique = toc assert(isequal(sort(my_unique(x)),y_correct) && t_unique > t_my_unique)
```Error: Assertion failed. ```
3 Pass
%% x = [1; 2; 3; 4; 2; 3; 4; 5;]; tic y_correct = unique(x); t_unique = toc tic y_my_unique = my_unique(x); t_my_unique = toc assert(isequal(sort(my_unique(x)),y_correct) && t_unique > t_my_unique)
``` t_unique = 2.2900e-04 t_my_unique = 6.9000e-05 ``` | 360 | 1,107 | {"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} | 3.28125 | 3 | CC-MAIN-2017-22 | longest | en | 0.38015 |
https://www.hometheaterforum.com/community/threads/i-need-some-help-explaining-an-answer-to-a-question.22693/ | 1,505,895,261,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00662.warc.gz | 826,755,430 | 31,434 | # I need some help explaining an answer to a question.
Discussion in 'Archived Threads 2001-2004' started by Dean Cooper, Aug 16, 2001.
1. ### Dean Cooper Supporting Actor
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Someone just asked me to explain the difference between an input resolution and his TV’s resolution and I’m having a hard time coming up with a good way to describe the answer. Does anyone know a link that has a good description or can describe the details in layman’s terms?
Thanks
Dean
2. ### Allan Jayne Cinematographer
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Input horizontal resolution -- if digital it means that the video signal representing the picture is organized in terms of pixels, say, 1/640'th the screen width occupying fixed positions 1/640'th, 2/640'th, 3/640'th, etc. from the left side of the screen. If analog it means that details can be as small as say 1/640'th the screen width and not have blurred to gray but not constrained to positions 1/640'thj, 2/640'th etc. from the left of the screen.
Horizontal resolution of the TV means how small details can be reproduced for viewing. If video that was once digital at, say, 640 max pixels across is fed in but if the TV is analog (like most) and only has, say, 500 max pixels across for resolution, it can still reproduce the pixels in any of the 640 positions (1/640'th, 2/640;th, etc. from the left edge) across but not all at the same time; three dissimilarly shaded ones in a row anywhere will be blurred together.
For a digital TV the display itself consists of pixels with fixed spacing. Regardless of the resolution of the input signal the picture is formed from individual spots of fixed size say 1/640;th the screen width and in fixed positions say 1/640'th, 2/640'th., etc. from the left edge. If two (or more) pixels worth of input line up with a given pixel on the digital display, that spot on the screen should be of a compromise shading.
Input vertical resolution is the number of scan lines or rows of pixels represented by the input signal. Display vertical resolution is the number of scan lines or rows of pixels on the display. If they don't match one of two things must happen (requiring special electronics):
1. scaling, involving doubleing, skipping or blending of a few lines here and there, or
2. Cropping the firsst few and last few scan lines if the input has more than the display, or leaving unused black areas on top and bottom of the display if the input has fewer than the display.
If the number of scan lines (and the number of frames per second) is the same for input and display, no scaling electronics is needed even if the number of pixels horizontally, express or implied, is different for input and display.
Other video hints: http://members.aol.com/ajaynejr/video.htm
[Edited last by Allan Jayne on August 16, 2001 at 10:15 PM] | 706 | 2,883 | {"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-2017-39 | latest | en | 0.917351 |
http://slideplayer.com/slide/2300815/ | 1,571,019,532,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00541.warc.gz | 254,585,357 | 17,902 | # Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education.
## Presentation on theme: "Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education."— Presentation transcript:
Taking the Square Root of Both Sides Adapted from Walch EducationAdapted from Walch Education
Complex Numbers The imaginary unit i represents the non-real value. i is the number whose square is –1. We define i so that and i 2 = –1. A complex number is a number with a real component and an imaginary component. Complex numbers can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit. 5.2.1: Taking the Square Root of Both Sides2
Real Numbers Real numbers are the set of all rational and irrational numbers. Real numbers do not contain an imaginary component. Real numbers are rational numbers when they can be written as, where both m and n are integers and n ≠ 0. Rational numbers can also be written as a decimal that ends or repeats. 5.2.1: Taking the Square Root of Both Sides3
Irrational Numbers Real numbers are irrational when they cannot be written as, where m and n are integers and n ≠ 0. Irrational numbers cannot be written as a decimal that ends or repeats. The real number is an irrational number because it cannot be written as the ratio of two integers. examples of irrational numbers include and π. 5.2.1: Taking the Square Root of Both Sides4
Quadratic Equations A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0, where a ≠ 0. Quadratic equations can have no real solutions, one real solution, or two real solutions. When a quadratic has no real solutions, it has two complex solutions. Quadratic equations that contain only a squared term and a constant can be solved by taking the square root of both sides. These equations can be written in the form x 2 = c, where c is a constant. 5.2.1: Taking the Square Root of Both Sides5
Quadratic Equations c tells us the number and type of solutions for the equation. 5.2.1: Taking the Square Root of Both Sides6 c Number and type of solutions NegativeTwo complex solutions 0 One real, rational solution Positive and a perfect square Two real, rational solutions Positive and not a perfect square Two real, irrational solutions
Practice # 1 Solve (x – 1) 2 + 15 = –1 for x. 5.2.1: Taking the Square Root of Both Sides7
Solution Isolate the squared binomial. Use a square root to isolate the binomial. 5.2.1: Taking the Square Root of Both Sides8 (x – 1) 2 + 15 = –1Original equation (x – 1) 2 = –16 Subtract 15 from both sides.
Solution, continued Simplify the square root. There is a negative number under the radical, so the answer will be a complex number. 5.2.1: Taking the Square Root of Both Sides9 Equation Write –16 as a product of a perfect square and –1. Product Property of Square Roots Simplify.
Solution, continued Isolate x. The equation (x – 1) 2 + 15 = –1 has two solutions, 1 ± 4i. 5.2.1: Taking the Square Root of Both Sides10 Equation Add 1 to both sides.
Try This One… Solve 4(x + 3) 2 – 10 = –6 for x. 5.2.1: Taking the Square Root of Both Sides11
Thanks for Watching!!!! Ms. Dambreville | 816 | 3,204 | {"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} | 4.6875 | 5 | CC-MAIN-2019-43 | latest | en | 0.917542 |
https://maker.pro/forums/threads/generating-power-through-induction-with-a-ham-radio-antenna.74861/ | 1,709,253,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00234.warc.gz | 378,230,379 | 61,179 | # Generating Power through Induction with a HAM radio Antenna
A
#### Aaron S
Jan 1, 1970
0
Is it possble to use an inductor with a stepdown transformer to generate
power by using the power ouput of a ham radio through it's antenna? My
friend states his antenna is a 40Watt'er. If I can, about how much power
can i expect to gather using a n inductor.
C
#### chuck
Jan 1, 1970
0
Aaron said:
Is it possble to use an inductor with a stepdown transformer to generate
power by using the power ouput of a ham radio through it's antenna? My
friend states his antenna is a 40Watt'er. If I can, about how much power
can i expect to gather using a n inductor.
Well, the transmitter may be 40 watts, but the antenna is usually not
rated that way.
All of the radiated energy can theoretically be recovered from a
transmitting antenna. The problem is how to capture it all. At a
distance, the instrument used to capture radiated energy is another
antenna, and tiny fractions of the radiated power can often be captured
halfway around the globe.
A stepdown transformer will lower the voltage produced at the antenna,
but that really has nothing to do with the amount of radiated energy you
capture.
If you use a simple inductor as an antenna to capture radiated energy,
you will find that it will probably not be possible to situate the
inductor so as to capture very much, since the antenna may be radiating
in many or even all directions! If the inductor gets too close to the
antenna, it will affect the antenna's properties, and that further
complicates the exercise.
Does any of that help?
Chuck
C
#### Charles Schuler
Jan 1, 1970
0
Aaron S said:
Is it possble to use an inductor with a stepdown transformer to generate
power by using the power ouput of a ham radio through it's antenna? My
friend states his antenna is a 40Watt'er. If I can, about how much power
can i expect to gather using a n inductor.
Any large antenna can gather enough energy to run low-power circuits, if it
is near powerful transmitters. Free-power radios used to be all the rage
among hobbyists. A big wire antenna fed a rectifier circuit that produced
enough snot to power a receiver and amplifier to drive a loudspeaker for
modest audio levels. I built one and used it many years ago in Pittsburgh,
PA. My home was close to several AM stations and my antenna was a wire
about 60 feet long (along with a good ground).
intriguing demonstrations.
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1. The mind's eye
Just found out this has a name. http://en.wikipedia.org/wiki/Tetris_effect
2. How to calculate the probability of hair colour of a child
At least at my middle/high school, we did Punnett squares a lot: http://en.wikipedia.org/wiki/Punnett_square Definition an oversimplification for complex traits like height and hair color, but it's a start.
6. The mind's eye
Same thing happens to me with Tetris. If I play it for too long, I start seeing falling blocks and can semi-play the game in my head.
7. The Official "Introduce Yourself" Thread
So I apparently had an account here four years ago (needless to say, I had forgotten all my login info), but I completely forgot about this site until a couple days ago. I figured I'd stop by again and get my fill of CS/math/puzzles/scientific news About myself: I like dancing, cartooning, and computer programming. Oh and long walks on the beach.
8. Daedalus' Fifth Challenge
I haven't done calculus in a long time, so I hope I don't do anything silly...
10. Deleting Objects in Java
I just uncovered this post of mine from forever ago, when I was starting to learn programming. 4 years and many real CS courses later, I have no idea what this question means. xD
You so totally Did not credit the author. Perhaps I'll sue you.
12. Can pi be reduced to a rational number?
I do realize the sophistry involved in my argument...I was just betting on my opponents not being able to pick it out.
13. Of Biology, Physics and Chemistry which one did you find to be the hardest/easiest?
Luckily for you, I have a high-quality photograph of all three of them together! Group Photo
14. Can pi be reduced to a rational number?
Hehe...this reminds me of the first debate of my Debate class last semester. The topic was "Is pi better than pie?", and I argued the pro-pie side. Basically I argued that because perfect circles do not exist in real life, pi does not exist, and therefore cannot be better than anything. Hey, we won. -- Part of our opening case: Contention #1 a. Examples of mathematical concepts that do not actually exist – only approximations i. Square ii. Circle iii. Pi b. No such thing as a perfect mathematical circle i. Irregularities ii. On computer: collection of square pixels iii. Borderline has thickness c. Pi based on nonexistent circle i. Ratio of circumference to diameter ii. Pi itself does not exist iii. Cannot be better than pie, which does exist iv. Cannot be better than anything -- Of course, I conveniently forgot to mention that mathematical concepts that do not represent the tangible world precisely can still be highly useful, and that most of our technology would fall apart without them.
15. Of Biology, Physics and Chemistry which one did you find to be the hardest/easiest?
From cutest to least cute: Biology, Physics, Chemistry
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Syllabus
### Physics Topic List
#### MATHEMATICAL TOOLS
Lecture# Description Duration
01 Variables, functions, angles, units of angles (Degree & radion) , conversion of units, Trigonometric ratios/ functions, values of trigonometric ratios, values of trigonometric ratios for angles grater than 90°, 36 Minutes
02 unit circle method, CAST rule. Trigonometric formula, sine rule & cosine rule , Logrithem , exponential and inverse functions. 1 Hr 03 Minutes
03 oordinate geometry , slope of a line , equation of straight line, parabola , ellipse, circle and rtectangular hyperbola. 30 Minutes
04 Differentiation, geometrical meaning of differentiation, slope of a line, formulae for differentiation, rules of differentiation- addition/subtraction rule, product rule, quotient rule, constant multiple rule, chain rule. 43 Minutes
05 Higher order Differentiation , implicit functions , important problems . 25 Minutes
06 Differentiation as rate measurement, maxima & minima. 50 Minutes
07 Integration, geometrical meaning of integration, formulae of integration, 16 Minutes
08 Definite integration, rules of integration, addition/ subtraction rule, method of substitution. Integration by parts, Integration as area under curve, indefinite integration , area under curve. 48 Minutes
09 ntroduction to vectors, null vector, unit vector, negative of a vector, graphical representation and mathematical representation of a vector, angle b/w two given vectors, 36 Minutes
10 Resolution of vector.Addition of vectors, triangle method and parallelogram method, substraction of vectors. 26 Minutes
11 Dot product and its uses. 27 Minutes
12 Cross product and its uses , right hand screw rule 48 Minutes
#### RECTILINEAR MOTION
Lecture# Description Duration
01 Rest & motion, distance & displacement, speed, average speed , time average and space average, instantaneous speed, Uniform speed and non uniform speed, 49 Minutes
02 velocity, average and instantaneous velocity, acceleration, average and instantaneous acceleration. 41 Minutes
03 Equations of kinematics with constant acceleration, steps used to solve the problems based on equation of kinematics, motion under gravity. 41 Minutes
04 graphical analysis, some important graphs, conversion of graphs, information collected from graphs. 16 Minutes
05 Variable acceleration, steps used to solve the problems based on variable acceleration, when acceleration is dependent on time, distance and velocity. 21 Minutes
#### PROJECTILE MOTION
Lecture# Description Duration
01 Ground to ground projectile, time of flight, net velocity, trajectory equation, maximum height, 40 Minutes
02 horizontal range.Projection at complementary angles from ground, some important relations and problems. 22 Minutes
03 Problems based on ground to ground projectile. 20 Minutes
04
• Projectile from tower projected horizontally, , time of flight, net velocity, trajectory equation, horizontal range
• Projectile from tower projected above horizontal, time of flight, net velocity, trajectory equation, maximum height. horizontal range
• Projectile from tower projected below horizontal. time of flight, net velocity, trajectory equation, horizontal range
40 Minutes
05 Problem on projectiles from tower 17 Minutes
06
• Projectile from inclined plane, projected up the incline plane , time of flight, net velocity, trajectory equation, maximum height. range
• Projectile from inclined plane, projected down the incline plane , time of flight, net velocity, trajectory equation, maximum height. Range
41 Minutes
07 Problems based on projectile on incline plane. 19 Minutes
08
Projectiles from moving platform, Collision of a projectile with vertical wall, some miscellaneous examples.
41 Minutes
#### RELATIVE MOTION
Lecture# Description Duration
01 Introduction to relative motion, one dimensional relative motion and two dimensional relative motion . Uses of equations of kinematics in 1D relative motion. 48 Minutes
02 uses of equations of kinematics in 2D relative motion , Velocity of approach and velocity of separation in 1D, Velocity of approach and velocity of separation in 2D, condition for two particles to collide, minimum separation b/w two moving particle, time taken to come at minimum separation miscellaneous problems . 45 Minutes
03 miscellaneous problems 32 Minutes
04 River boat problem in one dimension. 18 Minutes
05 River boat problem in two dimensions, direct crossing, minimum time taken to cross the river, minimum drift , minimum velocity 45 Minutes
06 Wind-aeroplane problem. Rain man problem, some illustrations. 48 Minutes
#### Newton’s laws of motion (NLM)
Lecture# Description Duration
01 Force, fundamental forces, normal force, tension force, Newton’s lst law, 2nd law , and 3rd law, equation of motion, Inertia. 50 Minutes
02 free body diagram ,Equilibrium, types of equilibrium, steps to solve the problems based on equilibrium, problems 48 Minutes
03 steps used to solve the problems of accelerated motion, problems , atwood machine 32 Minutes
04 Constrained motion, string constraint, displacement method, tension method, differentiation method, two block one pulley system, 36 Minutes
05 constrained motion when string is inclined, wedge constraint. 32 Minutes
06 Weighing machine, motion inside lift, apparent weight, weightlessness, spring balance , spring and spring force. 42 Minutes
07 Reference frame, inertial frame and non-inertial frame, pseudo force, illustrations 31 Minutes
08 Newton’s laws for system , problems 25 Minutes
#### FRICTION
Lecture# Description Duration
01 Introduction to friction, properties of friction. Kinetic friction,coefficient of kinetic friction. 45 Minutes
02 Static friction, coefficient of static friction, self adjustable nature of static friction, driving force, graph relating friction with driving force. 46 Minutes
03 Contact force, angle of friction, minimum force required to slide a block , why pulling is easier than pushing? 31 Minutes
04 Angle of repose, minimum and maximum force on the inclined plane so that block does not move , graph 27 Minutes
05 System of two blocks, steps used to check the slipping b/w two blocks, problems 39 Minutes
06 System of three blocks and miscellaneous examples. 29 Minutes
#### WORK POWER AND ENERGY
Lecture# Description Duration
01 Introduction to work, definition of work, point of application of force. Calculation of work done when force is constant 35 Minutes
02 Sign of work done . work done by variable force, 28 Minutes
03 work done from force-displacement graph, work done by friction, normal and gravity 24 Minutes
04 work done by spring force.Work done by variable force along given path, conservative and non-conservative forces 28 Minutes
05 methods to identify conservative forces , Del-operator, curl, Potential energy, its definition, external agent, 42 Minutes
06 relation b/w conservative force and potential energy, how to find P.E. if conservative force is given and vise-versa. Refrence line , gravitational Potential energy and spring potential energy 41 Minutes
07 Equilibrium, types of equilibrium, stable, unstable and neutral equilibrium. 26 Minutes
08 Kinetic energy , Work energy theorem, some examples. 17 Minutes
09 Problems based on work energy theorem 26 Minutes
10 Energy conservation, some examples, power, instantaneous power and average power. 26 Minutes
#### CIRCULAR MOTION
Lecture# Description Duration
01 Similarities b/w translational and rotational motion, angular displacement and its direction . 34 Minutes
02 angular velocity and angular acceleration, equations of circular kinematics. 37 Minutes
03 Relation b/w linear and rotational quantities, tangential acceleration centripetal/redial/normal acceleration. Total acceleration. 33 Minutes
04 Time period , frequency , angular frequency , Problems 23 Minutes
05 Radius of curvature of path, radius of curvature in projectile motion. 32 Minutes
06 Types of circular motion, horizontal circular motion. Some important examples. Steps used to solve the problems based on circular dynamics. Vertical circular motion, some important examples. 50 Minutes
07 Vertical circular motion of a ball attached with string , vertical circular motion of a ball attached with light rod. 35 Minutes
08 Problems , Banking of roads with and without friction. 26 Minutes
09 Centrifugal force, its direction and magnitude. Some examples. 33 Minutes
#### CENTER OF MASS
Lecture# Description Duration
01 Center of Mas, definitions, Type of mass distribution, discrete and continuous mass distribution, linear mass density, surface mass density, volume mass density. Calculation of com for system of particles. Com of system of two particles. 42 Minutes
02 Calculation of com for continuous mass distribution, com of rod, semi-circular ring, semi-circular disc, solid hemi-sphere, hollow hemi-sphere, solid cone. 51 Minutes
03 Com of a body with hole, problems 25 Minutes
04 Motion of com, velocity of com, acceleration of com, impulsive force, impulse, impulse-momentum equation, important examples.Conservation of momentum, some important conclusions and examples. 48 Minutes
05 Miscellaneous problems 19 Minutes
06 Some important points related to center of mass and miscellaneous problems. 40 Minutes
07 Spring mass system, steps to solve the problems based on spring-mass-system. Problems , Collision, line of impact, coefficient of restitution, 39 Minutes
08 classification of collision, head-on-inelastic collision, head on elastic collision, head on-perfectly in elastic collision. Problems on collision. 39 Minutes
09 collision with heavy mass. Oblique collision, problems 30 Minutes
10 oblique collision with wall , problems 27 Minutes
11 Variable mass, thrust force, rocket propulsion. 28 Minutes
#### SIMPLE HARMONIC MOTION (SHM)
Lecture# Description Duration
01 Definitions of periodic motion, oscillatory motion, and SHM, frequency, time period, amplitude, angular frequency.Differential equation of SHM, equation of SHM, 32 Minutes
02 SHM as projection of uniform circular motion, phase, 30 Minutes
03 Problems on phase , equation of SHM when mean position is not at origin. 30 Minutes
04 Velocity, acceleration and displacement of particle in terms of time (t) and displacement (x). Graphs, potential, kinetic and total energy in terms of time (t) and displacement (x), important graphs. 54 Minutes
05 Force method to find the time period, spring mass system , 47 Minutes
06 problems on force method, combinations of springs , springs in series , springs in parallel, 17 Minutes
07 energy methods to find the time period and Problems on spring mass system 46 Minutes
08 Angular SHM ,Differential equation of angular SHM, equation of angular SHM, method to find the time period in angular SHM 30 Minutes
09 Time period of simple pendulum, time period of simple pendulum when forces other than gravity and tension are also present, effective g. Fractional and percentage error , error in measurement of g, time period of simple pendulum when length of wire is comparable to radius of earth, Compound pendulum, its time period, minimum time period, 52 Minutes
10 Problems on compound pendulum , Torsional pendulum. 22 Minutes
11 Superposition of two parallel SHMs and perpendicular SHMs. 40 Minutes
#### KINETIC THEORY OF GASES & THERMODYNAMICS
Lecture# Description Duration
01 Assumptions for Ideal gas, Average velocity, Average speed, RMS speed, Most Probable speed, Maxwell’s velocity distribution graph. 31 Minutes
02 Miscellaneous problems related to calculation of RMS speed , average speed , most probable speed. 20 Minutes
03 Derivation of Ideal gas equation, calculation of kinetic energy of molecules 23 Minutes
04 Degree of Freedom, Maxwell’s law of Equipartition of energy and Internal energy. 17 Minutes
05 Mean Free Path, Some miscellaneous problems. 33 Minutes
06 Specific Heat Capacity, Adiabatic Exponent and gaseous mixture , molecular weight , Cp , Cv and γ of gaseous mixture. 33 Minutes
07 Work done by gas when pressure is constant and when pressure is variable, indirect method of calculation of work done by gas, work done from PV diagram. 26 Minutes
08 Problems based on calculation of work done by gas. 35 Minutes
09 Zeroth law of Thermodynamics, first law of Thermodynamics, Sign convention for Heat supplied, work done by gas and change in Interval energy .problems based on 1st law of thermodynamics. 39 Minutes
10 Thermodynamics processes ,Isochoric process, Isobaric process, Isothermal process, , calculation of heat supplied & Specific Heat Capacity of all the processes. 25 Minutes
11 Adiabatic process , Polytropic process, calculation of heat supplied & Specific Heat Capacity of these processes. 31 Minutes
12 Cyclic process, Heat Engine and its Efficiency, carnot cycle 27 Minutes
13 Refrigerator and its Coefficient of Performance, 20 Minutes
14 Miscellaneous problems and Free Expansion. 31 Minutes
#### FLUID MECHANICS
Lecture# Description Duration
01 Variation in pressure inside liquid with height, problems 32 Minutes
02 Problems , Inclination of liquid surface in static condition, rotation of container filled with liquid. 44 Minutes
03 Archimedes principle and force of buoyancy , Pascal’s law, 41 Minutes
04 atmospheric pressure, Gauge pressure, Absolute pressure, Barometer, and Manometer. 20 Minutes
05 Force applied by liquid on base of container and wall of container.Center of gravity, Center of Buoyancy, Meta-center, stability of completely submerged body and partially submerged body , metacentre. 56 Minutes
06 Types of flow, Uniform and Non-Uniform flow, Laminar and Turbulent flow, Reynolds number, Equation of continuity, Volume flow rate and Mass flow rate, Bernoulli theorem. 42 Minutes
07 Applications of Bernoulli theorem, 21 Minutes
08 Venturimeter, velocity of Efflux, Syphon action. 29 Minutes
#### CALORIMETRY
Lecture# Description Duration
01 Specific Heat Capacity, Heat Capacity, Specific Heat Capacity of water, 20 Minutes
02 definition of unit of Calorie, Latent heat, Latent Heat of Fusion, Latent Heat of Vaporization. 20 Minutes
03 change of State (Phase) of water with Temperature, illustrations. 18 Minutes
04 Problems , temperature scale.. 44 Minutes
#### THERMAL EXPANSION
Lecture# Description Duration
01 Linear expansion, Coefficient of Linear expansion, Differential expansion 18 Minutes
02 effect of Temperature on pendulum clock, error in measurement by metallic scale, 25 Minutes
03 Bimetallic strip, thermal stress 22 Minutes
04 Areal expansion, Coefficient of Areal expansion, relation between α and β, expansion of holes inside metallic plate. Coefficient of Volume expansion, relation between α and γ, 28 Minutes
05 Effect of Temperature on Density, Real and Apparent expansion of liquids. 37 Minutes
#### SURFACE TENSION
Lecture# Description Duration
01 Surface Tension ,wetted perimeter 31 Minutes
02 Surface Energy, cause of Surface Tension.Excess Pressure inside liquid drop, Excess pressure inside Soap bubble. Radius of curvature of common surface of double bubble. 49 Minutes
03 Cohesive force and Adhesive force, shape of liquid surface, Angle of contact.Capillary rise and illustrations. 33 Minutes
04 Capillary action with mercury , radius of lower meniscus 28 Minutes
05 Some miscellaneous problems 18 Minutes
#### WAVE ON STRING
Lecture# Description Duration
01 Definition and classification of wave, Mechanical & Non mechanical waves, Transverse & Longitudinal waves, Progressive and Stationary waves 29 Minutes
02 Differential form of wave equation, General form of equation of Progressive wave, information that can be collected from general form of wave equation 26 Minutes
03 How to find wave equation in terms of x and t when equation is given in terms of either x or t. wave on string introduction,Wavelength,Time period ,Frequency, Angular frequency, Wave number, Wave speed and velocity of particle, acceleration of particle, slope of string, direction of velocity of particle, 51 Minutes
04 Expanded form of standard equation of wave . relation b/w Phase difference and Path difference, relation b/w Phase difference and Time difference 34 Minutes
05 Derivation of speed of wave on string, examples 25 Minutes
06 Instantaneous and Average power transmitted by wave, Instantaneous and average intensity of a wave on string 33 Minutes
07 Superposition of waves,Interference,Resultant intensity, Constructive and Destructive Interference , miscellaneous problems. 1 Hr 02 Minutes
08 Reflection and Transmission of wave from one to other medium, effect of Reflection and Transmission on frequency, speed, Wavelength and Phase.
equation of reflected and transmitted waves. Amplitudes of reflected and transmitted wave
32 Minutes
09 Stationary waves, Nodes and Anti-nodes, Phase difference, properties of stationary waves. 59 Minutes
10 Equation of stationary waves , vibration of string fixed at both ends, vibration of string fixed at one end.Resonance, Sonometer, Melde's experiment 39 Minutes
11 kinetic energy and potential energy of small element of string. 30 Minutes
#### SOUND WAVE
Lecture# Description Duration
01 Introduction to Sound wave, variation of pressure with time and distance, variation in density and position with time. 24 Minutes
02 Equation of sound wave, relation b/w pressure Amplitude and Displacement Amplitude. Phase difference b/w Pressure wave and Displacement wave. Speed of Sound wave, Newton’s formula and La-place corrections. 32 Minutes
03 Dependence of speed of sound on Temperature, Pressure and relative Humidity. Intensity of sound wave, Wave front, Shape of wave-front for point source, Line source and Plane source. Variation of Intensity with distance from source. 44 Minutes
04 Comparison of two sound waves. Sound level, relative Sound Level, Pitch , waveform and quality of sound. Superposition of two sound waves, interference constructive and destructive interference, Reflection of Sound, Echo. 44 Minutes
05 Stationary wave in sound, vibrations of Air column in Organ pipes, Open Organ Pipe and Closed Organ Pipe 36 Minutes
06 Resonance Tube method to find the speed of sound, Beats. 30 Minutes
07 Doppler’s effect, when observer is moving and source is stationery, when source is moving and observer is stationary, when both source and observer are moving. 40 Minutes
08 Doppler’s effect When medium is also moving.miscelleneous problems. 44 Minutes
#### ELASTICITY AND VISCOSITY
Lecture# Description Duration
01 Elasticity, Plasticity, Deforming force, The reason behind Elastic and Plastic behaviour, Restoring force, Stress, Longitudinal Stress, Shear Stress and Bulk Stress, Strain, Longitudinal Strain, Shear Strain, Bulk Strain. Hook’s law, Modulus of Elasticity, Young’s Modulus, Modulus of Rigidity, Bulk Modulus, Compressibility, 41 Minutes
02 Variation of Strain with Deforming force, Elastic Limit, Yield point, Fracture point, elongation in wire due to self weight. Analogy with spring, Spring constant of a wire Elastic Potential energy stored in the deformed wire. 25 Minutes
03 Viscosity, Velocity Gradient, Viscous Force, Stoke’s forces Terminal Velocity. 28 Minutes
#### UNIT & DIMENSION
Lecture# Description Duration
01 Fundamental Quantities, Derived Quantities and Supplementary Quantities, Dimensions, Dimensional formula, some important concept (points) about dimensions, 27 Minutes
02 Problems on dimensions, Dimensional Analysis. Units, System of Units and conversion of Units. 26 Minutes
#### GRAVITATION
Lecture# Description Duration
01 Newton’s law of gravitation, gravitational field due to point mass, circular arc, circular ring, circular disc, long rod, infinite plate, hollow sphere and solid sphere 43 Minutes
02 variation in acceleration due to gravity with height and depth, effect of rotation of earth, effect of shape of earth. 27 Minutes
03 Gravitational potential, gravitational potential due to point mass, circular arc, circular ring, circular disc, hollow sphere, solid sphere, relation b/w gravitational field and gravitational potential . 31 Minutes
04 Gravitational potential energy, P.E. of two point mass system, self energy of hollow sphere and solid sphere, miscellaneous examples. 30 Minutes
05 Motion of satellite, orbital velocity, time period, energy of satellite, binding energy, escape velocity, geostationary satellite. 26 Minutes
06 Kepler's laws, path of a satellite according to its projection velocity. Miscellaneous examples. 47 Minutes
#### ROTATIONAL MOTION
Lecture# Description Duration
01 Introduction, similarities b/w rotational and translational motion. Rigid body, types of motion of rigid body. 32 Minutes
02 Moment of inertia definitions, calculation of MOI of a point mass, MOI of system of particles, MOI of a rod, 33 Minutes
03 MOI of ring, MOI of disc, MOI of solid sphere, MOI of hollow sphere, MOI of cone, MOI of solid cylinder, MOI of hollow cylinder 1 Hr
04 Perpendicular axes theorem, parallel axes theorem. MOI of a body with hole 1 Hr 08 Minutes
05 Radius of Gyration. Torque, Calculation of torque, 55 Minutes
06 Force couple, point of application. 20 Minutes
07 Rotational and translational equilibrium. 33 Minutes
08 Rotational equation of motion accelerated rotational motion. Some important examples. 54 Minutes
09 Combined motion, rolling motion, slipping, skidding, perfect rolling, 1 Hr 01 Minutes
10 Some important problems, trajectory of a point on wheel performing perfect rolling and radius of curvature of trajectory. 31 Minutes
11 instantaneous axis of rotation, rotational K.E. , conversion of imperfect rolling to perfect rolling 1 Hr 14 Minutes
12 Direction of friction in perfect rolling , Angular momentum, calculation of angular momentum, 36 Minutes
13 calculation of angular momentum, 30 Minutes
14 conservation of angular momentum in pure rotational motion , in pure translational motion and in combined motion , angular impulse momentum equation. 39 Minutes
15 Collision of a particle with rigid body 23 Minutes
16 Toppling and sliding. 34 Minutes
#### HEAT TRANSFER
Lecture# Description Duration
01 Methods of heat transfer, conduction, convection and radiation. steady state, temperature gradient. Laws of conduction. Analogy with electric current 31 Minutes
02 Problems on conduction, 1D heat transfer, 2D heat transfer, 3D heat transfer. Formation of ice layer on lake water surface. 36 Minutes
03 Convection, Radiation, Reflection power, Absorption power, Transmittance power, Black body. Ferry’s block body. Emissive power of a body, Spectral emissive power, absorptive power, spectral absorptive power. Emissivity of a body, Prevost's heat exchange theory 34 Minutes
04 Kirchhoff’s law of radiation, Stefan’s law of heat radiation, rate of cooling
Newton’s law of cooling
24 Minutes
05 Average temperature method, integration method. Black body radiation, Wien's displacement law, solar constant 27 Minutes | 5,068 | 23,151 | {"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} | 3.171875 | 3 | CC-MAIN-2019-18 | longest | en | 0.714177 |
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Complete schemes of work, lesson plans and powerpoints to teach Science, written by a current Head of Physics rated Outstanding in his teaching and currently working at a school rated Outstanding by OfSted
#### AQA P4 - Atomic Structure - GCSE Physics - 13 lesson plans and PowerPoints
13 Resources
This covers the Lessons (that are all available individually): Structure of an atom Mass number and isotopes Development of the Atomic Model Radioactive Decay Alpha, beta and gamma Properties of Radiation Nuclear Equations Half Life Radioactive Waste Background Radiation Fission Fusion
#### AQA P3 - Particle Model of Matter - GCSE Physics - 10 lesson plans and PowerPoints
10 Resources
This is a complete set of 10 power points and planned lessons that cover the unit of work for Science / Physics on the Particle Model of Matter. This includes: Introduction to the particle model Calculating Density Required Practical (Measuring Density) Change of state Specific Heat Capacity Specific Latent Heat Pressure in Gases Summary / Review lesson (Trilogy content) Boyle's Law Increasing gas pressure
#### GCSE Science / Physics - Specific Latent Heat (PowerPoint and Lesson Plan)
(1)
This is a detailed lesson and PowerPoint explaining specific latent heat. This is the sixth lesson in the Particle Model unit of work (P3). The lesson covers the latent heat of fusion and vaporization including the formula and calculation as well as a clear explanation of the theory. Included in the lesson plan are homework ideas and extension for the most able. The lesson plan is step by step including extension ideas and the learning objectives linked directly to the syllabus. https://www.tes.com/teaching-resources/shop/flushflop
#### GCSE Science / Physics / Chemistry - Development / History of the atom (PowerPoint and Lesson Plan)
(1)
This is a lesson and PowerPoint introducing the history and development of the model of the atom. This is the third lesson in the Atomic Structure unit of work (P4). The lesson covers the main scientists that have built up the theory of the atom over the centuries. The idea of theories and how they are developed and refined is discussed and some ideas for classroom activities are also included. See my shop for a huge range of Lessons and PowerPoints: https://www.tes.com/teaching-resources/shop/flushflop Visit my facebook page to discuss these lessons and to obtain information on updates: https://www.facebook.com/MatthewsResources/?ref=bookmarks
#### GCSE Science / Physics - Summary lesson of particle model unit (PowerPoint and Lesson Plan)
(6)
This is a lesson and PowerPoint revising and summarising the Particle Model of Matter. This is the eighth lesson in the Particle Model unit of work (P3). The lesson quickly recaps lessons 1 - 7 from this unit (which are available on this site and as a bundle) and covers the theories of solids, liquids and gases; the particle diagrams, density, specific heat capacity, specific latent heat and a qualitative assessment of the gas law. https://www.tes.com/teaching-resources/shop/flushflop
#### GCSE Science / Physics - Introduction to the Particle Model of Matter (PowerPoint and Lesson Plan)
(0)
This is a detailed lesson and PowerPoint introducing the Particle Model of Matter. This is the first lesson in the Particle Model unit of work (P3). The lesson covers the theories of solids, liquids and gases; the particle diagrams for them and a qualitative assessment of density in the three states of matter. The lesson plan is step by step including extension ideas and the learning objectives linked directly to the syllabus.
#### GCSE Science / Physics -Specific Heat Capacity (PowerPoint and Lesson Plan)
(1)
This is a lesson and PowerPoint recapping Specific Heat Capacity. This is the fifth lesson in the Particle Model unit of work (P3). The lesson links the theories of solids, liquids and gases and the particle model to specific heat capacity. The PowerPoint explains the formula for specific heat capacity and contains a simple table for pupils to test their mathematical skills against. The lesson plan is step by step including extension ideas and the learning objectives linked directly to the syllabus. https://www.tes.com/teaching-resources/shop/flushflop
#### GCSE Science / Physics - Structure of an Atom (PowerPoint and Lesson Plan)
(1)
This is a detailed lesson and PowerPoint introducing the Structure of an Atom. This is the first lesson in the Atomic Structure unit of work (P4). The lesson covers the sizes of atoms, nuclei and the idea that mass is found primarily in the nucleus. There is an idea for a demonstration to calculate the upper limit on the size of an atom. Electron energy level changes due to absorption and emission of electromagnetic radiation is also discussed The lesson plan is step by step including extension ideas and the learning objectives linked directly to the syllabus. https://www.tes.com/teaching-resources/shop/flushflop To discuss this, or any other PowerPoints and lessons please contact us via our webpage: https://www.facebook.com/MatthewsResources/
#### GCSE Science / Physics - Summary lesson of particle model unit (PowerPoint and Lesson Plan)
(0)
This is a lesson and PowerPoint revising and summarizing the Particle Model of Matter. This is the eighth lesson in the Particle Model unit of work (P3). The lesson quickly recaps lessons 1 - 7 from this unit (which are available on this site and as a bundle) and covers the theories of solids, liquids and gases; the particle diagrams, density, specific heat capacity, specific latent heat and a qualitative assessment of the gas law. https://www.tes.com/teaching-resources/shop/flushflop
#### GCSE Science / Physics - Mass number and Isotopes (PowerPoint and Lesson Plan)
(0)
This is a detailed lesson and PowerPoint introducing atomic number, mass number, isotopes and ions. This is the second lesson in the Atomic Structure unit of work (P4). The lesson covers the diagrams and standard method of listing elements, ascertaining isotopes and understanding the charges on atoms and ions. There is a detailed lesson plan with notes on teaching, a PowerPoint to support the lesson and a worksheet to aid in delivery / test understanding. See my shop for a huge range of Lessons and PowerPoints: https://www.tes.com/teaching-resources/shop/flushflop Visit my facebook page to discuss these lessons and to obtain information on updates: https://www.facebook.com/MatthewsResources/?ref=bookmarks | 1,409 | 6,828 | {"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-2021-39 | latest | en | 0.895705 |
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# Perspective ZNear plane
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I have heard that the best value for Z-near plane in perspective projection is 1.0f (I heard that with this value you get less clipping problems, and is best for Zbuffer, but I don''t understand much), so I put it in my gluPerspective, but, with this value I can''t come near an object because the camera clip it. It''s true that is the best value, if so, how I can come near and object without clipping it? (I see that in the shotem up games, when you come near a wall, the wall is not clipped, what value is used in this games?) thanks
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Let me just rudely explain something.If your near plane is 1.0 that means that the objects will become to clip 1 unit away from the current''s camera position.That''s they get clipped so early.Try setting 0.1 or 0.01 and see the results
The PAIN is coming...this summer!!!In cinemas everywhere.
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why should a near plane 1 away from the camara point be any bad? ... if you use 0.1 as near and 1000 as far, your z-test resolution is about one ten-thousands-th worser than if you used 1 for near
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ah... and 0 distance cannot be used
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then, if you use 1 too, how do you do to come near an object?
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Just scale your "world" so that the distance 1 is quite small, then limit your approach to a wall (via some form of collision detection) so that you cannot get near enough for the camera to clip it.
Simple!
It all depends on what sort of game you are making - an indoor FPS can probably get away with a near clipping plane of 0.1 or less, but if you try this on an outdoor environment with a far clipping plane of 1000 or more then you will get z-buffer problems with the further away polygons.
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That means that in a game that have indoor and outdoors I have to change the near plane in every level/zone?, like return to castle of wolfenstein that have indoors and outdoors
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No, I was meaning "real" outdoor games like flight sims where you can see for a long way, games like RTCW have a very limited draw distance outdoors
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*former washington: ''666_1337'' kamikaze-d with ''mig29h carriing 4 at22s equiped with nuclear warheads'' injuring ''Anonymous Poster'' by a splash-damage amount of 50x (times) his life, a ''got hit by crashing aircraft-damage'' of 9x his life and a ''heat-damage'' of other 50x his life @ 2 minutes ago* [connection reset by peer]
...
so he maybe doesn''t think return to castle wolfenstein would be an indoor & outdoor game - for any longer
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ah ok, I understand.
This means that I have to scale the world coord units, and use collision detection to avoid the cilping if I use 1.0f znear, but scale up or down?
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Antwort | 2,077 | 6,753 | {"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} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.61276 |
http://mathcenter.oxford.emory.edu/site/cs171/evaluatingFullyParenthesizedInfixExpressions/ | 1,627,505,435,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00339.warc.gz | 25,407,930 | 2,955 | ## Evaluating Infix Expressions
In infix arithmetic expressions, operators are placed between two operands -- as shown in the examples below:
$$2+3 \quad \quad \textrm{or} \quad \quad 1 + (2 + 3) * (4 * 5)$$
A fully parenthesized infix arithmetic expression is an infix arithmetic expression where every operator and its arguments are contained in parentheses, as seen in following: $$(2+3) \quad \quad \textrm{or} \quad \quad (1+((2+3)*(4*5)))$$
Suppose we wish to evaluate such an expression...
First, we note that a fully parenthesized expression explicitly details the order in which the operators are to be applied. Consequently, in the evaluation of such expressions, operator precedence and associativity don't really matter at all.
Freed from precedence and associativity considerations, we can evaluate a fully parenthesized expression with a simple two-stack algorithm developed by E.W. Dijkstra:
Starting with two empty stacks, called the operand stack and the operator stack, we read the elements of the expression in question from left to right. For each element encountered in turn, we do the action specified by the following table:
Element Seen Action Taken operand (e.g., a value) push it onto the operand stack operator push it onto the operator stack left parenthesis do nothing right parenthesis we pop the operator from the top of the operator stack, and however many operands it requires from the operand stack, and then push the result of applying that operator to those operands to the operand stack
The following details the state of the operator and operand stacks in the evaluation of an example expression. (Note: that the top of each stack shown is its right-most element, while the bottom is its left-most element.)
( 1 + ( ( 2 + 3 ) * ( 4 * 5 ) ) ) evaluates to 101, given:
Element Operator Stack Operand Stack
1 (
2 1 1
3 + + 1
4 ( + 1
5 ( + 1
6 2 + 1 2
7 + + + 1 2
8 3 + + 1 2 3
9 ) + 1 5
10 * + * 1 5
11 ( + * 1 5
12 4 + * 1 5 4
13 * + * * 1 5 4
14 5 + * * 1 5 4 5
15 ) + * 1 5 20
16 ) + 1 100
17 ) 101 <-- The expression, evaluated | 609 | 2,685 | {"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} | 4.375 | 4 | CC-MAIN-2021-31 | latest | en | 0.872437 |
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# Robot Exploration with Combinatorial Auctions
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### Robot Exploration with Combinatorial Auctions
1. 1. http://www.sciencedaily.com/releases/2007/06/070609112916.htm
2. 2. Robot Exploration with Combinatorial Auctions M. Berhault, H. Huang, P. Keskinocak, S. Koenig, W. Elmaghraby, P. Griffin, A. Kleywegt http://www.news.cornell.edu/releases/rover/Mars.update8-19-04.html Corey A. Spitzer - CSCI 8110 04-20-2010
3. 3. Optimal Task Allocation Repeat Auctions + Combinatorial Auctions + Bidding Strategy = Near Optimal Allocation http://shirt.woot.com/Derby/Entry.aspx?id=30206
4. 4. Repeat Auctions Robot 1 Robot 2 Goal Unknown Terrain Wall
5. 5. Repeat Auctions Robot 1 Robot 2 Goal Unknown Terrain Wall
6. 6. Repeat Auctions Robot 1 Robot 2 Goal Wall Wall
7. 7. Repeat Auctions Robot 1 Robot 2 Goal Wall Wall
8. 8. Single Item vs. Combinatorial Auctions
9. 9. Single Item vs. Combinatorial Auctions Possible Bundles: {} {G1} {G2} {G3} {G4} {G1, G2} {G1, G3} {G1, G4} {G2, G3} {G2, G4} {G3, G4} {G1, G2, G3} {G1, G2, G4} {G1, G3, G4} {G2, G3, G4} {G1, G2, G3, G4}
10. 10. Task Synergies - Positive Travel Distance for R1: T(S) T({G3}) = 4 T({G4}) = 4 T({G3, G4}) = 7 T({G3, G4}) ≤ T({G3}) + T({G4})
11. 11. Task Synergies - Negative Travel Distance for R1: T(S) T({G3}) = 4 T({G1}) = 8 T({G3, G1}) = 16 T({G3, G1}) ≥ T({G3}) + T({G1})
12. 12. Bidding Strategies Single Three-Combination Smart-Combination Nearest-Neighbor Graph-Cut http://blog.handbagsmaster.com/index.php/2009/09/eleven-auction-terms-you-should-know/
13. 13. Bidding Strategies - Single Same as single item auction
14. 14. Bidding Strategies - Three-Combination Possible Bundles with 5 Goals: {} {G1} {G2} {G3} {G4} {G5} {G1, G2} {G1, G3} {G1, G4} {G1, G5} {G2, G3} {G2, G4} {G2, G5} {G3, G4} {G3, G5} {G4, G5} {G1, G2, G3} {G1, G2, G4} {G1, G2, G5} {G1, G3, G4} {G1, G3, G5} {G1, G4, G5} {G2, G3, G4} {G2, G3, G5} {G2, G4, G5} {G3, G4, G5} {G1, G2, G3, G4} {G1, G2, G3, G5} {G1, G2, G4, G5} {G1, G3, G4, G5} {G2, G3, G4, G5} {G1, G2, G3, G4, G5}
15. 15. Bidding Strategies - Smart-Combination Bid on all bundles that have 1 or 2 goals Additionally, bid on the top N bundles containing more than 2 goals. Given k clusters of s goals (where s is in the set S of cluster sizes >2), N = |S| * max(S) * k. Goal Goal Goal Goal Goal Goal Goal Goal Goal Goal Goal Goal
16. 16. Bidding Strategies - Nearest-Neighbor Bid on all "Good Sequences": * {G i } for all i * If S = {G i , ... G e } is a good sequence then S U {G t } is a good sequence if G t is the closest neighbor to G e not in S and the value of S U {G t } ≥ the value of S
17. 17. Bidding Strategies - Graph Cut
18. 18. Bidding Strategies - Graph Cut Maximum cuts
19. 19. Summary of Experimental Results Generally Best Performing Bidding Strategies wrt: Travel Costs -- Graph-Cut Travel Times -- Three-Combination Smallest Number of Bids -- Single, then Graph-Cut Smallest Robot Utilization -- Graph-Cut Important Factors: Goal distribution (uniform or clustered), number of clusters, prior knowledge of the terrain
20. 20. Other Notes When targets are uniformly distributed, all bidding strategies are fairly close wrt travel costs. Nearest-Neighbor and Graph-Cut tend to have large bundle sizes => smaller number of active robots Smaller robot utilization => smaller travel costs, but larger travel times
21. 21. The End Questions? | 1,283 | 3,574 | {"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-2021-04 | latest | en | 0.584193 |
http://datadragon.com/cgi-bin/pmessage.pl?action=read&fid=edu&mid=2068 | 1,508,277,922,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00768.warc.gz | 85,799,543 | 5,245 | Quick Links This Day in Music History Music Education @ DataDragon.com Music Education Forums Maintain Your Forum Information Bernadette Peters - Broadway's Best Sudoku (take a break for a puzzle!)
Topic: HELP!!!!!
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AuthorTopic: HELP!!!!!
Mystro
Registered User
Registered:
12/8/2004
posted: 12/8/2004 at 9:42:31 PM ET
i have been playing by ear for almost 12 years. but in school now tring to learn how to read. but am having a real hard time understanding the rythms, what can i do
Mystro
TheHornSupremacy
Registered User
Registered:
11/17/2004
posted: 12/9/2004 at 10:00:39 AM ET
Is your problem that you don't understand the value of the notes and how they work in a given meter, or are you ok with note values but just have a hard time playing them in a piece of music?
If it's the former, then you need to start by learning the values of all the notes and learning as much about simple and complex meters as you can. A simple google on "music notation" or something similar will give you plenty of websites on the subject.
If your problem is the latter, then you can do a couple of things.
1 - Buy a metronome. They're pretty cheap (unless you go all out and get a fancy one). You can probably pick one up at a local music store for \$10-\$15. Heck, you might even be able to find one at Wal-mart for cheaper.
2 - Learn to count using the words "and" and "uh". Sounds silly, but it works. Beats get a number, and everything in between gets either an "and" or an "uh". In 4/4 time, if you have 4 quarter notes in a measure, then you'd count them as "1 - 2 - 3 - 4 - 1 - 2 - 3 - 4" (my examples will have two measures). If you have eight eighth notes in a measure then you'd count "1 - and - 2 - and - 3 - and - 4 - and - 1 - and - 2 - and - 3 - and - 4 - and". If you have 16 sixteenth notes in a measure, then it's "1 - uh - and - uh - 2 - uh - and - uh - 3 - uh - and - uh - 4 - uh - and - uh - 1 - uh - and - uh - 2 - uh - and - uh - 3 - uh - and - uh - 4 - uh - and - uh". Make sense?
3 - Don't be ashamed to clap it out when you're first learning the piece, or tap your foot while you're playing it.
4 - Take your time. Don't rush too much or you'll frustrate yourself. Just because a piece is marked "allegro" doesn't mean you have to learn to play it that fast right away. Start out by playing it at a tempo you can handle like "largo" and gradually work your way up to "allegro" as you become more familiar with the piece.
5 - Practice. As with anything, the more you practice, the better you'll get and the easier it will become.
Anonymous
Anonymous Poster
From Internet Network:
205.188.116.x
posted: 12/9/2004 at 10:45:49 AM ET
I'm finally beginning to get a handle on sight-reading, if I could get my fingers to cooperate that would be a great help (beginner piano), it's getting there.
I find your help on counting very helpful, having a good ear for music I find dosen't matter much especially with Bartok's Mikrokosmos.
If this makes much sense, when you count l and a 2 and a etc. how do you play and a,
I know you play on the 1 - counting is my hurdle. Thanks again for your very clear answers, you are a good teacher.
TheHornSupremacy
Registered User
Registered:
11/17/2004
posted: 12/10/2004 at 9:41:06 AM ET
Playing "1 and a 2" is pretty easy if you keep this in mind - whenever you have a sixteenth note in a beat, it is helpful to imagine that the beat is full of sixteenth notes. In this example, what you're given ("1 and a 2") is an eighth note followed by two sixteenth notes then a whole note (assuming that nothing else comes after the "2"). Rather than trying to hear that combination right away, try to hear what the beat sounds like as four sixteenth notes (which will be easier to hear because all the notes will be the same length). So now you are hearing "1 a and a 2". Repeat that in your mind four or five times in a row using a very steady tempo. After you can hear that, the only thing you need to do is drop out the first "a" so you're left with "1 - and a 2".
I hope that makes sense. Never realized how complicated this was to explain when all you can use is words!
Anonymous
Anonymous Poster
From Internet Network:
205.188.116.x
posted: 12/10/2004 at 12:13:44 PM ET
Thank you for getting back on counting. Wouldn't it be great if there were a way to have sound. I know it's difficult to explain in words.
maintube
Registered User
Registered:
5/26/2004
posted: 12/10/2004 at 4:36:02 PM ET
I really like this web site for music theory. Look and see if there is any help here for you. It does get a little deep.
http://www.musictheory.net/
Anonymous
Anonymous Poster
From Internet Network:
64.12.116.x
posted: 12/10/2004 at 6:34:34 PM ET
Hi Maintube,
Musictheory.net is a great website. Thanks. It does make things a little clearer and you are so right, it is kind of a tricky wicket.
This site manages to explain things in simple easy to understand terms and in some cases uses the keyboard to make things clearer. Many thanks.
Do you think this topic is inappropriate? Vote it down. After a thread receives a certain amount of negative votes it will be automatically locked. | 1,400 | 5,221 | {"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-2017-43 | longest | en | 0.920287 |
http://www.csmonitor.com/Science/Science-Notebook/2015/0314/Pi-Day-pizza-How-big-would-this-pizza-be-with-85-000-in-shredded-mozzarella | 1,481,008,437,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541883.3/warc/CC-MAIN-20161202170901-00275-ip-10-31-129-80.ec2.internal.warc.gz | 372,743,794 | 27,502 | Science Notebook
# Pi Day pizza: How big would this pizza be with \$85,000 in shredded mozzarella?
## Over the past weekend, thieves made off with a tractor-trailer full of shredded mozzarella, presumably to bake the world's largest pizza for Pi Day. But what will the pizza's diameter be?
View Caption
of
A trailer containing an estimated \$85,000 worth of shredded mozzarella cheese was discovered missing on Sunday from a parking lot near Summerfield, Fla.
According to the Ocala Star Banner, the trailer, which was bound for the Hungry Howies Pizza distribution center in nearby Lakeland, was sitting unattached in the parking lot Saturday night. Thieves likely detached a Kenworth hauler from an empty trailer in the same lot and used it to make off with the cheese.
Why steal a truckload of shredded mozzarella? Obviously, the only reasonable conclusion is that the thieves plan on making an enormous pizza, no doubt to celebrate Pi Day.
But this, of course, raises further questions, namely: How big would this pizza be?
Let's do the math. According to this handy table from the University of – where else? – Wisconsin, the wholesale price of mozzarella in the United States in February was \$2.1972 per pound. At that price, \$85,000 would buy some 38,686 lbs. of mozzarella.
By comparison, the world's largest pizza so far, a 131-foot-diameter, gluten-free behemoth baked in Rome three years ago, used 8,800 pounds of mozzarella. So, without any additional data, we know that we're likely dealing with a pie of unprecedented size.
It would be tempting to simply say that, because the thieves stole 4.4 times as much mozzarella as was used for Ottavia – that's what those record-setting Italian chefs named their pizza, for some reason – then it will be 4.4 times as big, or 577 feet in diameter. But, perhaps unfortunately for us humanities majors, circles don't work that way.
As you may remember from geometry classes, a circle's area is equal to pi times the square of the radius. Ottavia has a radius of 65.5 feet, giving it an area of 13,478 square feet. With 8,800 lbs. of mozzarella, that gives Ottavia a shredded mozzarella population density of 0.65 pounds of per square foot.
This density seems more or less consistent with normal-sized pizzas. Kraft, who, it should be noted, really wants you to consume their cheese, offers a recipe for a tomato and basil pizza that calls for putting 8 oz. of shredded mozzarella on a 14-inch-by-six-inch crust, leaving a density of 0.85 lbs. per square foot. Their fellow purveyors of shredded pressed-milk-curd, Sargento, calls for a pound of their mozzarella on a 16 inch circular pizza, a density of 0.71.
One of the largest pizzas commercially available is baked by the California chain Big Mama's & Papa's Pizzeria (they're the ones Ellen DeGeneres called for the Oscars last year). The Giant Sicilian is 54 inches by 54 inches – just over 20 square feet – and is topped with 12 lbs. of mozzarella, for a density of 0.58.
Of course, we don't know the mozzarella density of the giant pizza that the thieves plan on making for Pi Day, so, from this point onward, we may have to make some assumptions.
If we suppose that our thieves are sticking to an Ottavian density of 0.65. That would mean 38,686 lbs. of mozzarella would cover an area of 59,517 feet, resulting in a pizza with a diameter of just over 275 feet, easily the world's largest.
Here's what it the pizza would look like if it were dropped on Seattle's Safeco Field.
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https://www.findfilo.com/math-question-answers/find-the-number-of-ways-of-selecting-9-balls-from-5y7 | 1,632,682,404,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00321.warc.gz | 797,590,814 | 12,344 | Find the number of ways of selecting 9 balls from 6 red balls, 5 w | Filo
Class 11
Math
Algebra
Permutations and Combinations
817
151
Find the number of ways of selecting balls from red balls, white balls and blue balls if each selection consists of balls of each colour.
Solution: There are a total of red balls, white balls, and blue balls.
balls have to be selected in such a way that each selection consists of balls of each colour.
Here,
balls can be selected from red balls in ways.
balls can be selected from white balls in ways.
balls can be selected from blue balls in ways.
Thus, by multiplication principle, required number of ways of selecting balls
817
151
Connecting you to a tutor in 60 seconds. | 167 | 725 | {"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-2021-39 | latest | en | 0.922882 |
http://www.thefreedictionary.com/Random+distribution | 1,524,596,791,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00123.warc.gz | 516,563,655 | 21,259 | # probability
(redirected from Random distribution)
Also found in: Thesaurus, Medical, Legal, Financial, Encyclopedia.
Related to Random distribution: Uniform distribution
## prob·a·bil·i·ty
(prŏb′ə-bĭl′ĭ-tē)
n. pl. prob·a·bil·i·ties
1. The quality or condition of being probable; likelihood.
2. A probable situation, condition, or event: Her election is a clear probability.
3.
a. The likelihood that a given event will occur: little probability of rain tonight.
b. Statistics A number, ranging from zero to one, expressing either the projected likelihood that a specific event will occur or the observed ratio of the number of actual occurrences to the number of possible occurrences.
Idiom:
in all probability
Most probably; very likely.
## probability
(ˌprɒbəˈbɪlɪtɪ)
n, pl -ties
1. the condition of being probable
2. an event or other thing that is probable
3. (Statistics) statistics a measure or estimate of the degree of confidence one may have in the occurrence of an event, measured on a scale from zero (impossibility) to one (certainty). It may be defined as the proportion of favourable outcomes to the total number of possibilities if these are indifferent (mathematical probability), or the proportion observed in a sample (empirical probability), or the limit of this as the sample size tends to infinity (relative frequency), or by more subjective criteria (subjective probability)
## prob•a•bil•i•ty
(ˌprɒb əˈbɪl ɪ ti)
n., pl. -ties.
1. the quality or fact of being probable.
2. a probable event, circumstance, etc.
3. Statistics.
a. the relative possibility that an event will occur, as expressed by the ratio of the number of actual occurrences to the total number of possible occurrences.
b. the relative frequency with which an event occurs or is likely to occur.
Idioms:
in all probability, very probably; quite likely.
## prob·a·bil·i·ty
(prŏb′ə-bĭl′ĭ-tē)
A number expressing the likelihood of the occurrence of a given event, especially a fraction expressing how many times the event will happen in a given number of tests or experiments. For example, when rolling a six-sided die, the probability of rolling a particular side is 1 in 6, or 1/6 .
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 probability - a measure of how likely it is that some event will occur; a number expressing the ratio of favorable cases to the whole number of cases possible; "the probability that an unbiased coin will fall with the head up is 0.5"chancequantity, measure, amount - how much there is or how many there are of something that you can quantifyconditional probability, contingent probability - the probability that an event will occur given that one or more other events have occurredcross section - (physics) the probability that a particular interaction (as capture or ionization) will take place between particles; measured in barnsexceedance - (geology) the probability that an earthquake will generate a level of ground motion that exceeds a specified reference level during a given exposure time; "the concept of exceedance can be applied to any type of environmental risk modeling"fair chance, sporting chance - a reasonable probability of successfat chance, slim chance - little or no chance of successjoint probability - the probability of two events occurring togetherrisk of exposure, risk - the probability of being exposed to an infectious agentrisk of infection, risk - the probability of becoming infected given that exposure to an infectious agent has occurred 2 probability - the quality of being probable; a probable event or the most probable event; "for a while mutiny seemed a probability"; "going by past experience there was a high probability that the visitors were lost"quality - an essential and distinguishing attribute of something or someone; "the quality of mercy is not strained"--Shakespearelikelihood, likeliness - the probability of a specified outcomeimprobability, improbableness - the quality of being improbable; "impossibility should never be confused with improbability"; "the improbability of such rare coincidences"
## probability
noun
1. There is a high probability of success.
2. the probability of life on other planets
## probability
noun
The likeliness of a given event occurring:
chance, likelihood, odds, possibility, prospect (used in plural).
Translations
إحْتِمالإحْتِمال، حَدَثاِحْتِمَال
pravděpodobnostšancevyhlídka
sandsynlighed
todennäköisyys
vjerojatnost
líkindi, líkurlíkur, líkindi
개연성
verjetnost
sannolikhet
ความน่าจะเป็นไปได้
xác suất
## probability
[ˌprɒbəˈbɪlɪtɪ] N (also Math) →
the probability is thates probable que ... + subjun
we calculated the probabilities of it happening
in all probability he won't turn up
there is little probability of anyone finding outes muy poco probable que alguien se entere
## probability
[ˌprɒbəˈbɪlɪti] n
the probability of sth → la probabilité de qch
the probability that ... → la probabilité que ...
the probability of sth happening → la probabilité que qch se produise
Without a transfusion, the victim's probability of dying was very high → Sans transfusion, la probabilité que la victime meure était très élevée.
the probability of being ... → la probabilité d'être ...
the probability is that ... → le plus probable est que ...
in all probability → selon toute probabilité probability theoryprobability theory n
## probability
nWahrscheinlichkeit f; in all probabilityaller Wahrscheinlichkeit nach, höchstwahrscheinlich; the probability of something (gen); what’s the probability of that happening?wie groß ist die Wahrscheinlichkeit, dass das geschieht?; the probability is that he will leave
## probability
[ˌprɒbəˈbɪlɪtɪ] nprobabilità f inv
in all probability → con ogni probabilità
## probable
that may be expected to happen or be true; likely. the probable result; Such an event is possible but not probable.
I'll probably telephone you this evening.
probaˈbilityplural probaˈbilities noun
1. the state or fact of being probable; likelihood. There isn't much probability of that happening.
2. an event, result etc that is probable. Let's consider the probabilities.
in all probability
most probably; most likely.
## probability
pravděpodobnost sandsynlighed todennäköisyys vjerojatnost 見込み 개연성 sannolikhet ความน่าจะเป็นไปได้ xác suất
## probability
References in periodicals archive ?
Due largely to their random distribution across the surface, the science team concluded that the moon is shrinking.
Meanwhile, Justice Minister Hristo Ivanov and Lozan Panov, Chair of the Supreme Court of Cassation (VKS), demanded the launch of disciplinary action against Yaneva and a deputy of hers, Bogdana Zhelyavska, over the scandal involving breaches to the principle of random distribution of cases at the Sofia City Court.
To prove that there is a difference between the pattern field and the distance between the nearest neighbourhood hypothesis of random distribution with the nearest neighbours, a test of significant statistics which is the Z score was conducted.
The Monte Carlo test suggests that locations of nest-sites do not differ significantly from a random distribution at either oxbow (P = 0.
The random distribution of funds to woo young voters is a political ploy of the Trinamool to remain in the seat of power, but it would be difficult for the party to be in power for a long time with their shallow political ideas," Udayan said.
In the case of solids blending, the particle size need not change, but the distribution of particles throughout the mixture approaches a random distribution.
At a seminar in Kuwait's Economic Society, she said comprehensive development was based on appropriate employment of productive manpower not random distribution of wealth.
In the actual production workshop, operators on site cannot control the WIP inventory level in the form of an accurate quantity, but randomly adjust the production capacity in the manner of a random distribution by examining the inventory level of WIP buffer.
Nature's random distribution of genes and ability means that even unlikely - and unplanned - matings can pay dividends.
In both of the models, when the slope (b and b) value is not significantly different from 1, it indicates a random distribution pattern; slope significantly > 1 indicates an aggregated distribution pattern; and slope significantly < 1 indicates a regular distribution pattern (P < 0.
Random distribution of 40,000 squares using the odd and even numbers of a telephone directory, 50% blue, 50% red, 1963, a dimly lit room lined with silk-screened wallpaper with a blue and red geometric design, transformed a completely aleatory and objective system into a source of perceptual disruption.
Thanks to the random distribution of the metal particles, every ID made of our new film is unique and, therefore, virtually counterfeit-proof," says David Margulis, business development specialist, Secure ID Card Market.
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Open / Close | 2,049 | 8,959 | {"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} | 3.265625 | 3 | CC-MAIN-2018-17 | longest | en | 0.882641 |
https://search.r-project.org/CRAN/refmans/CGE/html/00Index.html | 1,701,439,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00508.warc.gz | 572,714,980 | 2,526 | # Computing General Equilibrium
## Help Pages
CD_A Cobb-Douglas Demand Structure Matrix CD_mA Cobb-Douglas Monetary Demand Structure Matrix CES_A CES Demand Coefficient Matrix CES_mA CES Monetary Demand Coefficient Matrix ChinaCGE2012 A CGE Model of China based on the Input-Output Table of 2012 (Unit: Ten Thousand RMB) dg A Modified diag Function Example.MWG.15.B.1 Example 15.B.1 in MWG (1995) Example.MWG.15.B.2 Example 15.B.2 in MWG (1995) Example.MWG.Exercise.15.B.6 Exercise 15.B.6 in MWG (1995) Example.MWG.Exercise.15.B.9 Exercise 15.B.9 in MWG (1995) Example.Section.3.1.2.corn Example in Section.3.1.2 of Li (2019) Example.Varian.Exercise.18.2 Exercise 18.2 in Varian (1992) Example.Varian.P352 Example on Page 352 in Varian (1992) Example2.2 Example 2.2 in Li (2019) Example2.3 Example2.3 in Li (2019) Example3.1 Example 3.1 in Li (2019) Example3.10 Example 3.10 in Li (2019) Example3.12 Example 3.12 in Li (2019) Example3.14 Example 3.14 in Li (2019) Example3.2 Example 3.2 in Li (2019) Example3.4 Example 3.4 in Li (2019) Example3.8 Example 3.8 in Li (2019) Example3.9 Example 3.9 in Li (2019) Example4.10 Example 4.10 in Li (2019) Example4.11.1 First Part of Example 4.11 in Li (2019) Example4.11.2 Second Part of Example 4.11 in Li (2019) Example4.12 Example 4.12 in Li (2019) Example4.13 Example 4.13 in Li (2019) Example4.15 Example 4.15 in Li (2019) Example4.16 Example 4.16 in Li (2019) Example4.2 Example 4.2 in Li (2019) Example4.8 Example 4.8 in Li (2019) Example4.9 Example 4.9 in Li (2019) Example5.1 Example 5.1 in Li (2019) Example5.10 Example 5.10 in Li (2019) Example5.11.1 First Part of Example 5.11 in Li (2019) Example5.11.2 Second Part of Example 5.11 in Li (2019) Example5.2 Example 5.2 in Li (2019) Example5.3.1 First Part of Example 5.3 in Li (2019) Example5.3.2 Second Part of Example 5.3 in Li (2019) Example5.4 Example 5.4 in Li (2019) Example5.5 Example 5.5 in Li (2019) Example5.6 Example 5.6 in Li (2019) Example6.10 Example 6.10 in Li (2019) Example6.11 Example 6.11 in Li (2019) Example6.13 Example 6.13 in Li (2019) Example6.2.1 First Part of Example 6.2 in Li (2019) Example6.2.2 Second Part of Example 6.2 in Li (2019) Example6.3 Example 6.3 in Li (2019) Example6.4 Example 6.4 in Li (2019) Example6.5 Example 6.5 in Li (2019) Example6.6.1 First Part of Example 6.6 in Li (2019) Example6.6.2 Second Part of Example 6.6 in Li (2019) Example6.6.3 Third Part of Example 6.6 in Li (2019) Example6.7 Example 6.7 in Li (2019) Example6.9 Example 6.9 in Li (2019) Example7.1 Example 7.1 in Li (2019) Example7.10 Example 7.10 in Li (2019) Example7.10.2 Extra Part of Example 7.10 in Li (2019) Example7.11 Example 7.11 in Li (2019) Example7.12 Example 7.12 in Li (2019) Example7.13 Example 7.13 in Li (2019) Example7.14 Example 7.14 in Li (2019) Example7.15 Example 7.15 in Li (2019) Example7.2 Example 7.2 in Li (2019) Example7.3 Example 7.3 in Li (2019) Example7.4 Example 7.4 in Li (2019) Example7.5.1 First Part of Example 7.5 in Li (2019) Example7.5.2 Second Part of Example 7.5 in Li (2019) Example7.6 Example 7.6 in Li (2019) Example7.7 Example 7.7 in Li (2019) Example7.8 Example 7.8 in Li (2019) Example7.9X Example 7.9 in Li (2019) Example8.1 Example 8.1 in Li (2019) Example8.2 Example 8.2 in Li (2019) Example8.7 Example 8.7 in Li (2019) Example8.8 Example 8.8 in Li (2019) Example8.9 Example 8.9 in Li (2019) Example9.10 Example 9.10-9.14 in Li (2019) Example9.10.policy.deficit.fiscal Deficit Fiscal Policy for Example 9.10 in Li (2019) Example9.10.policy.deflation Deflation Policy for Example9.10 in Li (2019) Example9.10.policy.interest.rate Interest Rate Policy for Example9.10 in Li (2019) Example9.10.policy.money.supply Money Supply Policy for Example9.10 in Li (2019) Example9.10.policy.quantitative.easing Quantitative Easing Policy for Example 9.10 in Li (2019) Example9.10.policy.tax Tax Policy for Example9.10 in Li (2019) Example9.3 Example 9.3 in Li (2019) Example9.4 Example 9.4 in Li (2019) Example9.5 Example 9.5 in Li (2019) Example9.6 Example 9.6 in Li (2019) Example9.7 Example 9.7 in Li (2019) F_Z Exchange Function iep Compute Instantaneous Equilibrium Path (alias Market Clearing Path) Leontief_mA Leontief Monetary Demand Coefficient Matrix PF_eig P-F (i.e. Perron-Frobenius) Eigenvalue and Eigenvector sdm Structural Dynamic Model (alias Structural Growth Model) | 1,568 | 4,341 | {"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} | 3.359375 | 3 | CC-MAIN-2023-50 | latest | en | 0.578567 |
https://unlearningmath.com/0.3-inches-to-mm/ | 1,721,021,076,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00417.warc.gz | 520,097,444 | 10,942 | # 0.3 inches to mm (Inches to Millimeters)
By / Under Inches To Millimeter / Published on
Should be an SEO optimised description including the original keyword just once
## Here is how to easily convert 0.3 inches to mm
0.3 inches is equal to 7.62 millimeters (mm). Understanding this conversion can be crucial in various fields, from science to business, as precise measurements are often required.
In today's globalized world, converting measurements from one unit to another forms a critical part of daily tasks. The conversion from inches to millimeters is no exception. This simple conversion 0.3 inches to mm directly equates to 7.62 millimeters. Given the universal push towards standardization, having this knowledge at your fingertips is indispensable. Whether you are working on engineering projects, displaying product dimensions, or even verifying small dimensions, grasping these conversions is vital.
To place this into perspective, consider that 1 inch equals exactly 25.4 mm. Therefore, 0.3 inches forms a fractional part which can be easily calculated as:
[ 0.3 \times 25.4 = 7.62 \text{ mm} ]
Such conversions are imperative not only for professionals but for educational purposes as well. For instance, a science student might need accurate data for an experiment. Alternatively, a business might need exact measurements to ensure tools fit correctly, affecting overall productivity.
### The Importance of Accurate Conversions
In industries such as manufacturing and construction, precision is everything. For example, a study revealed that 87% of industrial errors were due to incorrect measurements. Failing to convert measurements accurately can lead to serious consequences, including financial loss and loss of business credibility. Furthermore, in global trade, where different countries use varied measurement systems, conversions like these ensure smooth transactions.
### Real-World Analogy
Think of the process of converting inches to millimeters like converting a recipe from ounces to grams. A cook needs to know these conversions to maintain the recipe's integrity when cooking in different countries. Likewise, engineers rely on accurate conversions to ensure the success of their projects.
### External Link for Further Learning
For an in-depth understanding of the metric system and its everyday applications, you can visit the National Institute of Standards and Technology. Their resources provide valuable insights into different units of measurement and their conversions.
### FAQs
#### How do you convert 0.3 inches to millimeters?
To convert 0.3 inches to millimeters, multiply by 25.4, which results in 7.62 mm.
#### Why is converting inches to millimeters important?
Converting inches to millimeters is important for precision in engineering, manufacturing, and daily use where accurate measurements are essential.
#### Is 0.3 inches a common measurement?
Yes, 0.3 inches, or 7.62 mm, is a common measurement in various fields, such as electronics and manufacturing, where small and precise measurements are crucial.
#### What tools can be used for conversion?
Both digital and analog calipers, as well as online conversion tools, can be utilized to convert inches to millimeters accurately.
#### Are inches and millimeters used globally?
Millimeters are part of the metric system, widely used globally, while inches are mainly used in the United States and a few other countries. Understanding both units is essential for international collaboration and trade.
Understanding the conversion from 0.3 inches to mm ensures accuracy and efficiency in every field that relies on precise measurements. This knowledge simplifies complex tasks, enabling professionals and students to perform their duties effectively. | 741 | 3,782 | {"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} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.903822 |
http://sepwww.stanford.edu/data/media/public/docs/sep70/jun/paper_html/node4.html | 1,508,437,218,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823360.1/warc/CC-MAIN-20171019175016-20171019195016-00872.warc.gz | 297,931,194 | 2,865 | Next: Lateral velocity variation Up: 15-DEGREE DEPTH MIGRATION Previous: 15-DEGREE DEPTH MIGRATION
## Accuracy
From the series definition of a matrix function, we see that the matrix exponential can be represented as follows:
(11)
Richardson, et al. approximate this matrix exponential by splitting the given matrix into pieces:
(12)
Equation (12) can be represented with the series
(13)
(14)
An approximate calculation of equation (13) is
(15)
(16)
(17)
Equation (14) and (15) are identical if Me and Mo are commutative, i.e., 2MeMo = MeMo + MoMe. However, in our case, Me and Mo are not commutative, thus approximation (15) will have errors on the order of .By careful arrangement of two different split matrices, we can improve the accuracy as follows. We define
(18)
which has an error on the order of , and
(19)
with an error on the order of .
Figure shows the improvement in accuracy when we use a higher order approximation. On the left impulse response, we used the first order approximation with split as equation (18) and on the right impulse response, we used the third order approximation as equation (19). We can see a decrease in the dispersion at higher order approximation. Comparing with the impulse response of implicit scheme in Fig , the higher order explicit scheme shows the comparable accuracy as we can see in Fig .
fig2
Figure 2
Impulse responses of explicit 15-degree migration with the split matrices (a) M = Me+Mo, (b) M = Me/4+Mo/2+Me/2+Mo/2+Me/4.
Next: Lateral velocity variation Up: 15-DEGREE DEPTH MIGRATION Previous: 15-DEGREE DEPTH MIGRATION
Stanford Exploration Project
12/18/1997 | 412 | 1,629 | {"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} | 3.171875 | 3 | CC-MAIN-2017-43 | latest | en | 0.903869 |
http://www.algebra.com/algebra/homework/Quadratic_Equations.faq.question.118941.html | 1,386,201,254,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037851/warc/CC-MAIN-20131204131717-00073-ip-10-33-133-15.ec2.internal.warc.gz | 232,116,622 | 4,512 | # SOLUTION: What is the value of k if one root root of (3k+1)x^2=-5 is 1? Please help me :)
Algebra -> -> SOLUTION: What is the value of k if one root root of (3k+1)x^2=-5 is 1? Please help me :) Log On
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Question 118941: What is the value of k if one root root of (3k+1)x^2=-5 is 1? Please help me :)Answer by stanbon(60771) (Show Source): You can put this solution on YOUR website!What is the value of k if one root root of (3k+1)x^2=-5 is 1? Please help me :) ------------ Rearrange: f(x) = (3k+1)x^2+5 If 1 is a root, f(1) = zero But f(1) = (3k+1)*1^2+5 = 0 3k+1+5 = 0 3k = -6 k = -2 ============== Cheers, Stan H. | 269 | 783 | {"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-2013-48 | latest | en | 0.864727 |
https://www.shakyradunn.com/slide/the-foundations-logic-and-proofs-valdosta-state-university-631rds | 1,597,498,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740848.39/warc/CC-MAIN-20200815124541-20200815154541-00056.warc.gz | 804,817,863 | 15,566 | # The Foundations: Logic and Proofs - Valdosta State University
The Foundations: Logic and Proofs Chapter 1, Part I: Propositional Logic With Question/Answer Animations Propositional Logic Section 1.1 Propositions A proposition is a declarative sentence that is either true or false. Examples of propositions: a) b) c) d) e) The Moon is made of green cheese.
Trenton is the capital of New Jersey. Toronto is the capital of Canada. 1+0=1 0+0=2 Examples that are not propositions. a) Sit down! b) What time is it? c) x + 1 = 2 d) x + y = z Propositional Logic Constructing Propositions Propositional Variables: p, q, r, s, The proposition that is always true is denoted by T and the proposition that is always false is denoted by F. Compound Propositions; constructed from logical connectives and other propositions Negation Conjunction
Disjunction Implication Biconditional Compound Propositions: Negation The negation of a proposition p is denoted by p and has this truth table: p p T
F F T Example: If p denotes The earth is round., then p denotes It is not the case that the earth is round, or more simply The earth is not round. Conjunction The conjunction of propositions p and q is denoted by p q and has this truth table: p T
q T pq T T F F F T F F F
F Example: If p denotes I am at home. and q denotes It is raining. then p q denotes I am at home and it is raining. Disjunction The disjunction of propositions p and q is denoted by p q and has this truth table: p q p q T
T T T F T F T T F F
F Example: If p denotes I am at home. and q denotes It is raining. then p q denotes I am at home or it is raining. The Connective Or in English In English or has two distinct meanings. Inclusive Or - In the sentence Students who have taken CS202 or Math120 may take this class, we assume that students need to have taken one of the prerequisites, but may have taken both. This is the meaning of disjunction. For p q to be true, either one or both of p and q must be true. Exclusive Or - When reading the sentence Soup or salad comes with this entre, we do not expect to be able to get both soup and salad. This is the meaning of Exclusive Or (Xor). In p q , one of p and q must be true, but not both. The truth q p q
table forp is: T T F T F T F T T F F F
Implication If p and q are propositions, then p q is a conditional statement or implication which is read as if p, then q and has this truth table: q p q p T T T T F F
F T T F F T Example: If p denotes I am at home. and q denotes It is raining. then p q denotes If I am at home then it is raining. In p q , p is the hypothesis (antecedent or premise) and q is the conclusion (or consequence). Understanding Implication In p q there does not need to be any connection
between the antecedent or the consequent. The meaning of p q depends only on the truth values of p and q. These implications are perfectly fine, but would not be used in ordinary English. If the moon is made of green cheese, then I have more money than Bill Gates. If the moon is made of green cheese then Im on welfare. If 1 + 1 = 3, then your grandma wears combat boots. Understanding Implication (cont) One way to view the logical conditional is to think of an obligation or contract. If I am elected, then I will lower taxes. If you get 100% on the final, then you will get an A.
If the politician is elected and does not lower taxes, then the voters can say that he or she has broken the campaign pledge. Something similar holds for the professor. This corresponds to the case where p is true and q is false. Different Ways of Expressing p q if p, then q if p, q q unless p q if p q whenever p q follows from p p implies q p only if q q when p q when p p is sufficient for q q is necessary for p
a necessary condition for p is q a sufficient condition for q is p Converse, Contrapositive, and Inverse From p q we can form new conditional statements . q p is the converse of p q q p is the contrapositive of p q p q is the inverse of p q Example: Find the converse, inverse, and contrapositive of It raining is a sufficient condition for my not going to town. Solution: converse: If I do not go to town, then it is raining. inverse: If it is not raining, then I will go to town. contrapositive: If I go to town, then it is not raining.
Biconditional If p and q are propositions, then we can form the biconditional proposition p q , read as p if and only if q . The biconditional p q denotes the proposition with this truth table: p q p q T T T T
F F F T F F F T If p denotes I am at home. and q denotes It is raining. then p q denotes I am at home if and only if it is raining.
Expressing the Biconditional Some alternative ways p if and only if q is expressed in English: p is necessary and sufficient for q if p then q , and conversely p iff q Truth Tables For Compound Propositions Construction of a truth table: Rows Need a row for every possible combination of values for the atomic propositions. Columns Need a column for the compound proposition (usually at far right) Need a column for the truth value of each
expression that occurs in the compound proposition as it is built up. This includes the atomic propositions Example Truth Table Construct a truth table for p q r r pq pq r
T T T F T F T T F T T
T T F T F T F T F F
T T T F T T F T F F T
F T T T F F T F F T
F F F T F T Equivalent Propositions Two propositions are equivalent if they always have the same truth value. Example: Show using a truth table that the biconditional is equivalent to the contrapositive. Solution: p
q p q q p q p T T F F T T T
F F T F F F T T F T
T F F T T T T Using a Truth Table to Show NonEquivalence Example: Show using truth tables that neither the converse nor inverse of an implication are not equivalent to the implication. p Solution: q
p q p p q T T F F T q T qp T T
F F T F T T F T T F
T F F F F T T T T T Problem
How many rows are there in a truth table with n propositional variables? Solution: 2n We will see how to do this in Chapter 6. Note that this means that with n propositional variables, we can construct 2n distinct (i.e., not equivalent) propositions. Precedence of Logical Operators Operator Precedence 1 2
3 4 5 p q r is equivalent to (p q) r If the intended meaning is p (q r ) then parentheses must be used. Applications of Propositional Logic Section 1.2 Translating English Sentences Steps to convert an English sentence to a
statement in propositional logic Identify atomic propositions and represent using propositional variables. Determine appropriate logical connectives If I go to Harrys or to the country, I will not go shopping. p: I go to Harrys q: I go to the country. r: I will go shopping. If p or q then not r. Example Problem: Translate the following sentence into propositional logic: You can access the Internet from campus only if you are a computer science major or you
are not a freshman. One Solution: Let a, c, and f represent respectively You can access the internet from campus, You are a computer science major, and You are a freshman. a (c f ) System Specifications System and Software engineers take requirements in English and express them in a precise specification language based on logic. Example: Express in propositional logic: The automated reply cannot be sent when the file system is full Solution: One possible solution: Let p denote The automated reply can be sent and q denote The file system is full. q p
Consistent System Specifications Definition: A list of propositions is consistent if it is possible to assign truth values to the proposition variables so that each proposition is true. Exercise: Are these specifications consistent? The diagnostic message is stored in the buffer or it is retransmitted. The diagnostic message is not stored in the buffer. If the diagnostic message is stored in the buffer, then it is retransmitted. Solution: Let p denote The diagnostic message is stored in the buffer. Let q denote The diagnostic message is retransmitted The specification can be written as: p q, p, p q. When p is false and q is true all three statements are true. So the specification is consistent. What if The diagnostic message is not retransmitted is added. Solution: Now we are adding q and there is no satisfying assignment. So the specification is not consistent. Raymond Smullyan (Born
An island has two kinds of inhabitants, knights, who always 1919) Logic Puzzles tell the truth, and knaves, who always lie. You go to the island and meet A and B. A says B is a knight. B says The two of us are of opposite types. Example: What are the types of A and B? Solution: Let p and q be the statements that A is a knight and B is a knight, respectively. So, then p represents the proposition that A is a knave and q that B is a knave. If A is a knight, then p is true. Since knights tell the truth, q must also be true. Then (p q) ( p q) would have to be true, but it is not. So, A is not a knight and therefore p must be true. If A is a knave, then B must not be a knight since knaves always lie. So, then both p and q hold since both are knaves.
Logic Circuits (Studied in depth in Chapter 12) Electronic circuits; each input/output signal can be viewed as a 0 or 1. represents False represents True Complicated circuits are constructed from three basic circuits called gates. 0 1 The inverter (NOT gate)takes an input bit and produces the negation of that bit. The OR gate takes two input bits and produces the value equivalent to the disjunction of the two bits. The AND gate takes two input bits and produces the value equivalent to the conjunction of the two bits. More complicated digital circuits can be constructed by combining these basic circuits to produce the desired output given the input signals by
building a circuit for each piece of the output expression and then combining them. For example: Propositional Equivalences Section 1.3 Tautologies, Contradictions, and Contingencies A tautology is a proposition which is always true. Example: p p A contradiction is a proposition which is always false. Example: p p A contingency is a proposition which is P a tautology p
p p p p neither nor a contradiction, such F T F as p T F T T F Logically Equivalent
Two compound propositions p and q are logically equivalent if pq is a tautology. We write this as pq or as pq where p and q are compound propositions. Two compound propositions p and q are equivalent if and only if the columns in a truth table giving their truth values agree. This truth table show p q is equivalent to p q. p q p p q
p q T T F T T T F F F F
F T T T T F F T T T
De Morgans Laws Augustus De Morgan 18061871 This truth table shows that De Morgans Second Law holds. p q p q (pq) (pq) pq T
T F F T F F T F F T
T F F F T T F T F F F
F T T F T T Key Logical Equivalences Identity Laws: Domination Laws: Idempotent laws: , , ,
Double Negation Law: Negation Laws: , Key Logical Equivalences (cont) Commutative Laws: Associative Laws: Distributive Laws: Absorption Laws: , More Logical Equivalences Constructing New Logical Equivalences We can show that two expressions are logically equivalent by developing a series of logically
equivalent statements. To prove that we produce a series of equivalences beginning with A and ending with B. Keep in mind that whenever a proposition (represented by a propositional variable) occurs in the equivalences listed earlier, it may be replaced by an arbitrarily complex compound proposition. Equivalence Proofs Example: Show that is logically equivalent to Solution: Equivalence Proofs Example: Show that is a tautology. Solution:
Propositional Satisfiability A compound proposition is satisfiable if there is an assignment of truth values to its variables that make it true. When no such assignments exist, the compound proposition is unsatisfiable. A compound proposition is unsatisfiable if and only if its negation is a tautology. Questions on Propositional Satisfiability Example: Determine the satisfiability of the following compound propositions: Solution: Satisfiable. Assign T to p, q, and r. Solution: Satisfiable. Assign T to p and F to q. Solution: Not satisfiable. Check each possible assignment of truth values to the propositional variables and none will make the proposition true.
Notation Needed for the next example. Sudoku A Sudoku puzzle is represented by a 99 grid made up of nine 33 subgrids, known as blocks. Some of the 81 cells of the puzzle are assigned one of the numbers 1,2, , 9. The puzzle is solved by assigning numbers to each blank cell so that every row, column and block contains each of the nine possible numbers. Example Encoding as a Satisfiability Problem Let p(i,j,n) denote the proposition that is true when the number n is in the cell in the ith
row and the jth column. There are 99 9 = 729 such propositions. In the sample puzzle p(5,1,6) is true, but p(5,j,6) is false for j = 2,3,9 Encoding (cont) For each cell with a given value, assert p(d,j,n), when the cell in row i and column j has the given value. Assert that every row contains every number. Assert that every column contains every number. Encoding (cont) Assert that each of the 3 x 3 blocks contain every number. (this is tricky - ideas from chapter 4 help)
Assert that no cell contains more than one number. Take the conjunction over all values of n, n, i, and j, where each variable ranges from 1 to 9 and , of Solving Satisfiability Problems To solve a Sudoku puzzle, we need to find an assignment of truth values to the 729 variables of the form p(i,j,n) that makes the conjunction of the assertions true. Those variables that are assigned T yield a solution to the puzzle. A truth table can always be used to determine the satisfiability of a compound proposition. But this is too complex even for modern computers for large problems. There has been much work on developing efficient methods for solving satisfiability problems as many practical problems can be translated into satisfiability problems.
Predicates and Quantifiers Section 1.4 Propositional Logic Not Enough If we have: All men are mortal. Socrates is a man. Does it follow that Socrates is mortal? Cant be represented in propositional logic. Need a language that talks about objects, their properties, and their relations. Later well see how to draw inferences. Introducing Predicate Logic Predicate logic uses the following new features: Variables:
x, y, z Predicates: P(x), M(x) Quantifiers (to be covered in a few slides): Propositional functions are a generalization of propositions. They contain variables and a predicate, e.g., P(x) Variables can be replaced by elements from their domain. Propositional Functions Propositional functions become propositions (and have truth values) when their variables are each replaced by a value from the domain (or bound by a quantifier, as we will see later). The statement P(x) is said to be the value of the propositional function P at x. For example, let P(x) denote x > 0 and the domain be
the integers. Then: P(-3) is false. P(0) is false. P(3) is true. Often the domain is denoted by U. So in this example U is the integers. Examples of Propositional Functions Let x + y = z be denoted by R(x, y, z) and U (for all three variables) be the integers. Find these truth values: R(2,-1,5) Solution: F R(3,4,7) Solution: T R(x, 3, z) Solution: Not a Proposition
Now let x - y = z be denoted by Q(x, y, z), with U as the integers. Find these truth values: Q(2,-1,3) Solution: T Q(3,4,7) Solution: F Q(x, 3, z) Solution: Not a Proposition Compound Expressions Connectives from propositional logic carry over to predicate logic. If P(x) denotes x > 0, find these truth values: P(3) P(-1) P(3) P(-1) P(3) P(-1)
P(-1) P(3) Solution: T Solution: F Solution: F Solution: T Expressions with variables are not propositions and therefore do not have truth values. For example, P(3) P(y) P(x) P(y) When used with quantifiers (to be introduced next), these expressions (propositional functions) become propositions. Quantifiers Charles Peirce (18391914) We need quantifiers to express the meaning of English
words including all and some: All men are Mortal. Some cats do not have fur. The two most important quantifiers are: Universal Quantifier, For all, symbol: Existential Quantifier, There exists, symbol: We write as in x P(x) and x P(x). x P(x) asserts P(x) is true for every x in the domain. x P(x) asserts P(x) is true for some x in the domain. The quantifiers are said to bind the variable x in these expressions. Universal Quantifier
x P(x) is read as For all x, P(x) or For every x, P(x) Examples: 1) 2) 3) If P(x) denotes x > 0 and U is the integers, then x P(x) is false. If P(x) denotes x > 0 and U is the positive integers, then x P(x) is true. If P(x) denotes x is even and U is the integers, then x P(x) is false. Existential Quantifier x P(x) is read as For some x, P(x), or as
There is an x such that P(x), or For at least one x, P(x). Examples: 1. 2. 3. If P(x) denotes x > 0 and U is the integers, then x P(x) is true. It is also true if U is the positive integers. If P(x) denotes x < 0 and U is the positive integers, then x P(x) is false. If P(x) denotes x is even and U is the integers, then x P(x) is true. Uniqueness Quantifier !x P(x) means that P(x) is true for one and only one x in the
universe of discourse. This is commonly expressed in English in the following equivalent ways: There is a unique x such that P(x). There is one and only one x such that P(x) Examples: 1. If P(x) denotes x + 1 = 0 and U is the integers, then !x P(x) is true. 2. But if P(x) denotes x > 0, then !x P(x) is false. The uniqueness quantifier is not really needed as the restriction that there is a unique x such that P(x) can be expressed as: x (P(x) y (P(y) y =x)) Thinking about Quantifiers When the domain of discourse is finite, we can think of
quantification as looping through the elements of the domain. To evaluate x P(x) loop through all x in the domain. If at every step P(x) is true, then x P(x) is true. If at a step P(x) is false, then x P(x) is false and the loop terminates. To evaluate x P(x) loop through all x in the domain. If at some step, P(x) is true, then x P(x) is true and the loop terminates. If the loop ends without finding an x for which P(x) is true, then x P(x) is false. Even if the domains are infinite, we can still think of the quantifiers this fashion, but the loops will not terminate in some cases. Properties of Quantifiers The truth value of x P(x) and x P(x) depend on both the propositional function P(x) and on the
domain U. Examples: 1. If U is the positive integers and P(x) is the statement x < 2, then x P(x) is true, but x P(x) is false. 2. If U is the negative integers and P(x) is the statement x < 2, then both x P(x) and x P(x) are true. 3. If U consists of 3, 4, and 5, and P(x) is the statement x > 2, then both x P(x) and x P(x) are true. But if P(x) is the statement x < 2, then both x P(x) and x P(x) are false. Precedence of Quantifiers The quantifiers and have higher precedence than all the logical operators. For example, x P(x) Q(x) means (x P(x)) Q(x) x (P(x) Q(x)) means something different. Unfortunately, often people write x P(x)
Q(x) when they mean x (P(x) Q(x)). Translating from English to Logic Example 1: Translate the following sentence into predicate logic: Every student in this class has taken a course in Java. Solution: First decide on the domain U. Solution 1: If U is all students in this class, define a propositional function J(x) denoting x has taken a course in Java and translate as x J(x). Solution 2: But if U is all people, also define a propositional function S(x) denoting x is a student in this class and translate as x (S(x) J(x)). x (S(x) J(x)) is not correct. What does it mean? Translating from English to Logic Example 2: Translate the following sentence into predicate logic: Some student in this class has taken a course in Java.
Solution: First decide on the domain U. Solution 1: If U is all students in this class, translate as x J(x) Solution 2: But if U is all people, then translate as x (S(x) J(x)) x (S(x) J(x)) is not correct. What does it mean? Returning to the Socrates Example Introduce the propositional functions Man(x) denoting x is a man and Mortal(x) denoting x is mortal. Specify the domain as all people. The two premises are: The conclusion is: Later we will show how to prove that the conclusion follows from the premises.
Equivalences in Predicate Logic Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value for every predicate substituted into these statements and for every domain of discourse used for the variables in the expressions. The notation S T indicates that S and T are logically equivalent. Example: x S(x) x S(x) Thinking about Quantifiers as Conjunctions and Disjunctions If the domain is finite, a universally quantified proposition is equivalent to a conjunction of propositions without quantifiers
and an existentially quantified proposition is equivalent to a disjunction of propositions without quantifiers. If U consists of the integers 1,2, and 3: Even if the domains are infinite, you can still think of the quantifiers in this fashion, but the equivalent expressions without quantifiers will be infinitely long. Negating Quantified Expressions Consider x J(x) Every student in your class has taken a course in Java. Here J(x) is x has taken a course in Java and the domain is students in your class. Negating the original statement gives It is not the case that every student in your class has taken Java. This implies that There is a student in your class who has not taken Java. Symbolically x J(x) and x J(x) are
equivalent Negating Quantified Expressions (continued) Now Consider x J(x) There is a student in this class who has taken a course in Java. Where J(x) is x has taken a course in Java. Negating the original statement gives It is not the case that there is a student in this class who has taken Java. This implies that Every student in this class has not taken Java Symbolically x J(x) and x J(x) are equivalent De Morgans Laws for Quantifiers The rules for negating quantifiers are: The reasoning in the table shows that: These are important. You will use these.
Translation from English to Logic Examples: 1. Some student in this class has visited Mexico. Solution: Let M(x) denote x has visited Mexico and S(x) denote x is a student in this class, and U be all people. x (S(x) M(x)) 2. Every student in this class has visited Canada or Mexico. Solution: Add C(x) denoting x has visited Canada. x (S(x) (M(x)C(x))) Some Fun with Translating from English into Logical Expressions U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd
T(x): x is a thingamabob Translate Everything is a fleegle Solution: x F(x) Translation (cont) U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob Nothing is a snurd. Solution: x S(x) What is this equivalent to? Solution: x S(x) Translation (cont) U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob
All fleegles are snurds. Solution: x (F(x) S(x)) Translation (cont) U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob Some fleegles are thingamabobs. Solution: x (F(x) T(x)) Translation (cont) U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob No snurd is a thingamabob. Solution: x (S(x) T(x)) What is this
equivalent to? Solution: x (S(x) T(x)) Translation (cont) U = {fleegles, snurds, thingamabobs} F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob If any fleegle is a snurd then it is also a thingamabob. Solution: x ((F(x) S(x)) T(x)) System Specification Example Predicate logic is used for specifying properties that systems must satisfy. For example, translate into predicate logic: Every mail message larger than one megabyte will be compressed. If a user is active, at least one network link will be
available. Decide on predicates and domains (left implicit here) for the variables: Let L(m, y) be Mail message m is larger than y megabytes. Let C(m) denote Mail message m will be compressed. Let A(u) represent User u is active. Let S(n, x) represent Network link n is state x. Now we have: Charles Lutwidge Dodgson (AKA Lewis Caroll) (1832-1898) The first two are called premises and the third is called the Lewis Carroll Example conclusion. 1. All lions are fierce. 2. Some lions do not drink coffee.
3. Some fierce creatures do not drink coffee. Here is one way to translate these statements to predicate logic. Let P(x), Q(x), and R(x) be the propositional functions x is a lion, x is fierce, and x drinks coffee, respectively. 1. x (P(x) Q(x)) 2. x (P(x) R(x)) 3. x (Q(x) R(x)) Later we will see how to prove that the conclusion follows from the premises.
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https://wiki.physics.wisc.edu/facultywiki/Persistence_of_Motion_%28Air_Track%29 | 1,591,197,765,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00013.warc.gz | 584,043,760 | 4,750 | # Persistence of Motion (Air Track), 1F30.10
Topic and Concept:
pira200 Listed
Location:
• Floor Item: ME, South Wall on shelf brackets
Abstract:
A cart on a low-friction air track is given a velocity which remains roughly constant as per Newton's first law.
Equipment Location ID Number 2.25m Air Track ME, South Wall on shelf brackets A Glider ME, Bay A4, Shelf #2 Air Hose ME, South Wall on shelf brackets Located rooms 2103, 2241, and 2223 4A.EQ100
Important Setup Notes:
• This demonstration requires a supply of compressed air at HVHP.
• A Red & White Gas Cart will need to be installed under the lecture tables and the air hose will need to be connected to the supply line within the stage floor.
Setup and Procedure:
1. Place air track on lecture bench.
2. Level the track using the set screws on each leg.
3. Connect the track to the air supply using the air hose.
4. Place air glider on track.
5. Let the air flow into the track by opening the supply valve.
6. Give the air glider a push to give it an initial velocity.
Cautions, Warnings, or Safety Concerns:
• N/A
Discussion:
The kinematic version of Newton's first law says that a body in motion tends to stay in motion moving in a straight line unless acted upon by a force. This is exactly what is happening here. The glider is given an initial velocity being accelerated by the lecturer's hand. Upon release, the glider maintains its final velocity. The air track provides a low-friction surface for the glider to move on so not much energy can be lost as heat. Thus the speed will remain roughly constant. The direction will change, however. The velocity will switch between v0 and -v0 each time it hits a bumper in an elastic collision whereby the direction of its momentum is reversed.
Videos:
References:
• N/A
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https://www.johncanessa.com/2021/01/07/revenue-milestones-in-java/ | 1,701,568,519,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100476.94/warc/CC-MAIN-20231202235258-20231203025258-00668.warc.gz | 940,438,315 | 17,278 | # Revenue Milestones in Java
Good day ladies and gentlemen. Today is Wednesday January 06, 2021. The claims of electoral fraud in the 2020 presidential elections in the USA continue to be brought up. One way or the other January 20 is about two weeks away. If the claims do not pan out then on that day we will have a new president be sworn in. We will be able to read news without political opinions.
Earlier today my wife fixed a delicious chicken dish with a white sauce with lemon and capers. She served it with rice, baked potatoes and beats. For desert we had chocolate ice cream followed by a triple espresso. That marked the middle of my day. In the afternoon I finished two additional 2-hour blocks.
After done with the workday I generated this post and am ready to call it a day. Will head upstairs to relax and call my son. He is the ones that provide most of the news for us. My wife and I will then watch an Amazon Prime or Netflix movie and around 08:00 PM will go to bed.
We are waiting for the COVID-19 vaccine to become available for our age group. I would assume and hope that most social restrictions would be lifted as soon as a percentage of people in the area we live in receive the vaccine.
Let’s get down to the main subject of this post.
I decided to tackle the Revenue Milestones problem from the coding practice portal of Facebook. The problem is in the search category. I did look for the problem in HackerRank or and LeetCode to no avail.
The reason I like to find the specific problem is due to the fact that HackerRank and LeetCode because solutions have to pass a large set of tests. At the Facebook site there might be a couple test samples and a couple unit tests. Most problems need a dozen or more test cases to make sure one capture the requirements in the implementation.
```We keep track of the revenue Facebook makes every day,
and we want to know on what days Facebook hits certain revenue milestones.
Given an array of the revenue on each day,
and an array of milestones Facebook wants to reach,
return an array containing the days on which Facebook reached every milestone.
Input:
revenues is a length-N array representing how much revenue FB made on each day (from day 1 to day N).
milestones is a length-K array of total revenue milestones.
Output:
Return a length-K array where K_i is the day on which FB first had milestones[i] total revenue.
If the milestone is never met, return -1.
```
Please take your time and read the requirements. Take a look at the test example. Based on them I initially figures that the values in the milestones would be in ascending order. That is not the case. Such values may appear in a random order. If you just sort the values your answer will not match the order of the expected results. You need to compensate for you sorting the revenue values.
```int[] getMilestoneDays(int[] revenues, int[] milestones)
```
The signature for the function / method we need to implement is simple. The revenues array contains the revenues per day starting on the left and ending on the right. The milestones are not in any particular order. Our function should return an array associated with the milestones array. In other words, the first entry should be for the day in which the first milestone was reached, the second entry in the day the second milestone was met. The process continues through the entire array.
I like to develop software on my computers. That is how I and most software engineers tend to work. I will solve the problem using Java on a Windows 10 computer using the VSCode IDE. When I have a tentative candidate for release, I will copy the body of the function / method on my machine and paste it on the Facebook IDE. I will then run the unit tests.
I will have to develop a test scaffolding to check out my solution as I am developing it. The test code IS NOT PART OF THE SOLUTION.
```10, 20, 30, 40, 50, 60, 70, 80, 90, 100
100, 200, 500
main <<< revenues: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
main <<< milestones: [100, 200, 500]
main <<< output: [4, 6, 10]
main <<< output: [4, 6, 10]
Explanation:
On days 4, 5, and 6, FB has total revenue of \$100, \$150, and \$210 respectively.
Day 6 is the first time that FB has >= \$200 of total revenue.
10, 20, 30, 40, 50, 60, 70, 80, 90, 100
100, 200, 1000
main <<< revenues: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
main <<< milestones: [100, 200, 1000]
main <<< output: [4, 6, -1]
main <<< output: [4, 6, -1]
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
100, 200, 500
main <<< revenues: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
main <<< milestones: [100, 200, 500]
main <<< output: [-1, -1, -1]
main <<< output: [-1, -1, -1]
100, 200, 300, 400, 500
300, 800, 1000, 1400
main <<< revenues: [100, 200, 300, 400, 500]
main <<< milestones: [300, 800, 1000, 1400]
main <<< output: [2, 4, 4, 5]
main <<< output: [2, 4, 4, 5]
700, 800, 600, 400, 600, 700
3100, 2200, 800, 2100, 1000
main <<< revenues: [700, 800, 600, 400, 600, 700]
main <<< milestones: [3100, 2200, 800, 2100, 1000]
main <<< output: [5, 4, 2, 3, 2]
main <<< output: [5, 4, 2, 3, 2]
```
I believe this is the only test sample provided by Facebook. There are two unit tests and a couple samples I created to check out some conditions.
The first line contains the values for the daily revenues. The second line contains the milestone values. Our test code will have to read these two lines and will have to populate a couple arrays.
The arrays are displayed. They have the same names that are used by the signature for the function / method we have to develop.
It seems that we call two different implementations of the solution. They both appear to return the same results. Hopefully you will agree with both pieces of code. If not, please leave me a message bellow or send me an email message.
``` /**
* Test scaffolding
* !!! NOT PART OF SOLUTION !!!
*
* @throws IOException
*/
public static void main(String[] args) throws IOException {
// **** open buffered reader ****
// **** close buffered reader ****
br.close();
// ???? ????
System.out.println("main <<< revenues: " + Arrays.toString(revenues));
System.out.println("main <<< milestones: " + Arrays.toString(milestones));
// **** compute and display results ****
System.out.println("main <<< output: " + Arrays.toString(getMilestoneDays(revenues, milestones)));
System.out.println("main <<< output: " + Arrays.toString(getMilestoneDays1(revenues, milestones)));
}
```
The test scaffolding verifies our assumptions on the code. We open a buffered reader, read the values for the two arrays and close it. The values are displayed. They seem to match the input lines. So far things are going well. We then make a call to an implementation of the solution and display the returned values. The process repeats for a second implementation.
``` /**
* Given an array of the revenue on each day,
* and an array of milestones Facebook wants to reach,
* return an array containing the days on which Facebook reached every milestone.
*/
static int[] getMilestoneDays(int[] revenues, int[] milestones) {
// **** initialization ****
int[] ans = Arrays.stream(new int[milestones.length]).map(i -> -1).toArray();
int revenue = 0;
PriorityQueue<String> pq = new PriorityQueue<String>(new Comparator<String>() {
public int compare(String str1, String str2) {
String s1 = str1.substring(0, str1.indexOf(","));
String s2 = str2.substring(0, str2.indexOf(","));
Integer v1 = Integer.parseInt(s1);
Integer v2 = Integer.parseInt(s2);
if (v1 > v2)
return 1;
if (v1 == v2)
return 0;
else
return -1;
}
});
for (int i = 0; i < milestones.length; i++) {
String s = "" + milestones[i] + "," + i;
}
// **** traverse revenue array O(n) ****
for (int r = 0; r < revenues.length; r++) {
// **** update revenue ****
revenue += revenues[r];
// **** update the ans array with milestones that have been reached O(m) ****
while (!pq.isEmpty()) {
// **** extract milestone and index ****
String[] strs = pq.peek().split(",");
int milestone = Integer.parseInt(strs[0]);
int m = Integer.parseInt(strs[1]);
// **** check if done ****
if (revenue < milestone)
break;
// **** ****
ans[m] = (r + 1);
// **** remove lowest value milestone ****
pq.remove();
}
}
return ans;
}
```
We start by performing some initialization. The ans[] array is created and filled in with -1 values. We create a priority queue to maintain a key : value pair associated with milestone : index pair. These are encoded in a string separated by a ‘,’. The bulk of the work is performed in the loop. We traverse the revenues from left to right. This is the order the data was provided. The revenue variable that holds the cumulative value is updated. We then enter a second loop in which we take the top element from our priority queue (heap) and extract the milestone : index values. We check if the revenue is less than the selected milestone. If so we exit the loop. Otherwise we update the answer array and remove the head of the priority queue. After we have processed all the revenue items we return the result array.
``` /**
* Given an array of the revenue on each day,
* and an array of milestones Facebook wants to reach,
* return an array containing the days on which Facebook reached every milestone.
*/
static int[] getMilestoneDays1(int[] revenues, int[] milestones) {
// **** initialization ****
int sum = 0;
int[] ans = Arrays.stream(new int[milestones.length]).map(i -> -1).toArray();
HashMap<Integer, Integer> idx = new HashMap<>();
for (int i = 0; i < milestones.length; i++)
idx.put(milestones[i], i);
Arrays.sort(milestones);
// **** traverse all revenues O(n) ****
for (int m = 0, d = 0; d < revenues.length; d++) {
// **** update the sum with today's revenue ****
sum += revenues[d];
// **** start from the current smallest milestone O(m) ****
while ((m < milestones.length) && (sum >= milestones[m])) {
// **** set day for milestone ****
ans[idx.get(milestones[m])] = d + 1;
// **** increment milestone index ****
m++;
}
}
return ans;
}
```
In the second attempt we replace the priority queue with a hash map. It seems to me to be somewhat simpler to follow.
We perform the initialization step which includes sorting the milestones array. Note that the hash map associates the values with the original positions of the milestones.
The outer loop is used to traverse all the revenues associated with the days. We then update the sum of revenues up to the current day. This is similar to what we did in the previous approach.
The next loop is used to look for the milestones that have been met. With that information at hand we update the ans[] array and increment the milestone index.
When all is set and done we return the ans[] array.
Hope you enjoyed solving this problem as much as I did. The entire code for this project can be found in my GitHub repository.
If you have comments or questions regarding this, or any other post in this blog, or if you would like for me to help out with any phase in the SDLC (Software Development Life Cycle) of a project associated with a product or service, please do not hesitate and leave me a note below. If you prefer, send me a private e-mail message. I will reply as soon as possible.
Keep on reading and experimenting. It is the best way to learn, become proficient, refresh your knowledge and enhance your developer toolset.
One last thing, many thanks to all 5,476 subscribers to this blog!!!
Keep safe during the COVID-19 pandemic and help restart the world economy. I believe we can all see the light at the end of the tunnel.
Regards;
John
john.canessa@gmail.com
## 4 thoughts on “Revenue Milestones in Java”
1. Larry says:
``` def getMilestoneDays(revenues, milestones): currentRev = 0 # nlogn where n length of milestones answer = [-1] * (len(milestones)) milestoneStack= sorted(milestones, reverse=True) lookup = {} #step is O(n) where n is length of milestones for idx,m in enumerate(milestones): #the assumption is that all milestones are unique #this step creates a dictionary for later use giving us O(1) searches. lookup[m] = idx # O(n) where n is length of revenues for idx, rev in enumerate(revenues): currentRev += rev while currentRev >= milestoneStack[-1]: # remember idx strarts at zero, days starts at 1 answer[lookup[milestoneStack.pop()]] = idx +1 if not milestoneStack: return answer # if we made it this far, we did not meet all milestones return answer ```
This is why I love python. ^^^
You clearly have a mastery of java! Did you end up going though the interview process? If so, how did it go?
1. JohnCanessa says:
Hi Larry,
Thanks for the comment.
Nice Python code.
Thanks. I have been working with Java for a few years.
I did. Did not get an offer :o)
Regards;
John
2. Dima says:
Hi. I tried a binary search approach. No extra space (updated the existing arrays in place) and time complexity = O (N + K*log N):
int[] getMilestoneDays(int[] revenues, int[] milestones) {
int revSize = revenues.length;
int totalRev= 0;
for (int i = 0; i < revSize; i++){
totalRev += revenues[i];
revenues[i] = totalRev;
}
for (int i = 0; i < milestones.length; i++){
int left = 0;
int right = revSize;
int seek = milestones[i];
while (left = seek) right = mid;
else left = mid + 1;
} // found the most left index in revenues that achieved the milestone
milestones[i] = revenues[left] >= seek ? left + 1: -1;
}
return milestones;
}
1. JohnCanessa says:
Hi Dima,
Thanks for the message.
Seems the code is incomplete.
I could generate a solution using binary search,
but if possible would like to get your code.
Regards;
John
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# Normal systolic blood pressure among people older than 50 years follows a normal distribution with mean 100 (mm
Hg) and SD=15 and the blood pressure among unhealthy patients that take medication follows a normal distribution with mean 140 (mmHg) and SD=20. Under the assumption that one out of ten people older than 50 suffer from hypertension, (a) what is the proportion of people in the general population with the blood pressure higher than 130; (b) what is the probability that an individual 50+ years old with the blood pressure 130 suffers from hypertension and as such requires medication?
Here, X = Blood pressure of a randomly selected person from general population~Normal(100,15) So, Z = (X-100)/15... View the full answer
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# Implementing Strassen’s Algorithm in Java
Strassen’s algorithm is used for the multiplication of Square Matrices that is the order of matrices should be (N x N). Strassen’s Algorithm is based on the divide and conquer technique. In simpler terms, it is used for matrix multiplication. Strassen’s method of matrix multiplication is a typical divide and conquer algorithm. However, let’s get again on what’s behind the divide and conquer approach and implement it considering an illustration as follows For example: Let A and B are two matrices then the resultant matrix C such that
Matrix C = Matrix A * Matrix B
Consider for now the mathematical computation of matrices is that it can be concluded out why the implementation for the Strassen matrices comes out into play. Suppose two matrices are operated to be multiplied then the approach would have been
1. Take input of two matrices.
2. Check the compatibility of matrix multiplication which holds true only and only if the number of rows of the first matrix equals the number of columns of the second matrix.
3. Multiply the matrix and assign multiplication of two matrices to another matrix known as the resultant matrix.
4. Print the resultant matrix.
In the above approach, two assumptions are drawn which show why Strassen’s algorithm need arises into play
• Firstly, the time complexity of the algorithm is O(n3) which is too high.
• Secondly, the multiplication of more than two matrices will not only increase the confusion and complexity of the program but also increase the time complexity accordingly.
Purpose:
Volker Strassen’s is a name who published his algorithm to prove that the time complexity O(n3) of general matrix multiplication wasn’t optimal. So it was published Strassen’s matrix chain multiplication and reduced the time complexity. This algorithm is faster than standard matrix multiplication and is useful when numerous large matrices multiplication is computed in the daily world.
Strassen’s Algorithm for Matrix Multiplication
Step 1: Take three matrices to suppose A, B, C where C is the resultant matrix and A and B are Matrix which is to be multiplied using Strassen’s Method.
Step 2: Divide A, B, C Matrix into four (n/2)×(n/2) matrices and take the first part of each as shown below
Step 3: Use the below formulas for solving part 1 of the matrix
```M1:=(A1+A3)×(B1+B2)
M2:=(A2+A4)×(B3+B4)
M3:=(A1−A4)×(B1+A4)
M4:=A1×(B2−B4)
M5:=(A3+A4)×(B1)
M6:=(A1+A2)×(B4)
M7:=A4×(B3−B1)
Then,
P:=M2+M3−M6−M7
Q:=M4+M6
R:=M5+M7
S:=M1−M3−M4−M5```
Step 4: After Solving the first part, compute the second, third, and fourth, and as well as final output, a multiplied matrix is generated as a result as shown in the above image.
Step 5: Print the resultant matrix.
Implementation:
Example
## Java
`// Java Program to Implement Strassen Algorithm` `// Class Strassen matrix multiplication``public` `class` `GFG {` ` ``// Method 1`` ``// Function to multiply matrices`` ``public` `int``[][] multiply(``int``[][] A, ``int``[][] B)`` ``{`` ``// Order of matrix`` ``int` `n = A.length;` ` ``// Creating a 2D square matrix with size n`` ``// n is input from the user`` ``int``[][] R = ``new` `int``[n][n];` ` ``// Base case`` ``// If there is only single element`` ``if` `(n == ``1``)` ` ``// Returning the simple multiplication of`` ``// two elements in matrices`` ``R[``0``][``0``] = A[``0``][``0``] * B[``0``][``0``];` ` ``// Matrix`` ``else` `{`` ``// Step 1: Dividing Matrix into parts`` ``// by storing sub-parts to variables`` ``int``[][] A11 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] A12 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] A21 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] A22 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] B11 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] B12 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] B21 = ``new` `int``[n / ``2``][n / ``2``];`` ``int``[][] B22 = ``new` `int``[n / ``2``][n / ``2``];` ` ``// Step 2: Dividing matrix A into 4 halves`` ``split(A, A11, ``0``, ``0``);`` ``split(A, A12, ``0``, n / ``2``);`` ``split(A, A21, n / ``2``, ``0``);`` ``split(A, A22, n / ``2``, n / ``2``);` ` ``// Step 2: Dividing matrix B into 4 halves`` ``split(B, B11, ``0``, ``0``);`` ``split(B, B12, ``0``, n / ``2``);`` ``split(B, B21, n / ``2``, ``0``);`` ``split(B, B22, n / ``2``, n / ``2``);` ` ``// Using Formulas as described in algorithm` ` ``// M1:=(A1+A3)×(B1+B2)`` ``int``[][] M1`` ``= multiply(add(A11, A22), add(B11, B22));`` ` ` ``// M2:=(A2+A4)×(B3+B4)`` ``int``[][] M2 = multiply(add(A21, A22), B11);`` ` ` ``// M3:=(A1−A4)×(B1+A4)`` ``int``[][] M3 = multiply(A11, sub(B12, B22));`` ` ` ``// M4:=A1×(B2−B4)`` ``int``[][] M4 = multiply(A22, sub(B21, B11));`` ` ` ``// M5:=(A3+A4)×(B1)`` ``int``[][] M5 = multiply(add(A11, A12), B22);`` ` ` ``// M6:=(A1+A2)×(B4)`` ``int``[][] M6`` ``= multiply(sub(A21, A11), add(B11, B12));`` ` ` ``// M7:=A4×(B3−B1)`` ``int``[][] M7`` ``= multiply(sub(A12, A22), add(B21, B22));` ` ``// P:=M2+M3−M6−M7`` ``int``[][] C11 = add(sub(add(M1, M4), M5), M7);`` ` ` ``// Q:=M4+M6`` ``int``[][] C12 = add(M3, M5);`` ` ` ``// R:=M5+M7`` ``int``[][] C21 = add(M2, M4);`` ` ` ``// S:=M1−M3−M4−M5`` ``int``[][] C22 = add(sub(add(M1, M3), M2), M6);` ` ``// Step 3: Join 4 halves into one result matrix`` ``join(C11, R, ``0``, ``0``);`` ``join(C12, R, ``0``, n / ``2``);`` ``join(C21, R, n / ``2``, ``0``);`` ``join(C22, R, n / ``2``, n / ``2``);`` ``}` ` ``// Step 4: Return result`` ``return` `R;`` ``}` ` ``// Method 2`` ``// Function to subtract two matrices`` ``public` `int``[][] sub(``int``[][] A, ``int``[][] B)`` ``{`` ``//`` ``int` `n = A.length;` ` ``//`` ``int``[][] C = ``new` `int``[n][n];` ` ``// Iterating over elements of 2D matrix`` ``// using nested for loops` ` ``// Outer loop for rows`` ``for` `(``int` `i = ``0``; i < n; i++)` ` ``// Inner loop for columns`` ``for` `(``int` `j = ``0``; j < n; j++)` ` ``// Subtracting corresponding elements`` ``// from matrices`` ``C[i][j] = A[i][j] - B[i][j];` ` ``// Returning the resultant matrix`` ``return` `C;`` ``}` ` ``// Method 3`` ``// Function to add two matrices`` ``public` `int``[][] add(``int``[][] A, ``int``[][] B)`` ``{` ` ``//`` ``int` `n = A.length;` ` ``// Creating a 2D square matrix`` ``int``[][] C = ``new` `int``[n][n];` ` ``// Iterating over elements of 2D matrix`` ``// using nested for loops` ` ``// Outer loop for rows`` ``for` `(``int` `i = ``0``; i < n; i++)` ` ``// Inner loop for columns`` ``for` `(``int` `j = ``0``; j < n; j++)` ` ``// Adding corresponding elements`` ``// of matrices`` ``C[i][j] = A[i][j] + B[i][j];` ` ``// Returning the resultant matrix`` ``return` `C;`` ``}` ` ``// Method 4`` ``// Function to split parent matrix`` ``// into child matrices`` ``public` `void` `split(``int``[][] P, ``int``[][] C, ``int` `iB, ``int` `jB)`` ``{`` ``// Iterating over elements of 2D matrix`` ``// using nested for loops` ` ``// Outer loop for rows`` ``for` `(``int` `i1 = ``0``, i2 = iB; i1 < C.length; i1++, i2++)` ` ``// Inner loop for columns`` ``for` `(``int` `j1 = ``0``, j2 = jB; j1 < C.length;`` ``j1++, j2++)` ` ``C[i1][j1] = P[i2][j2];`` ``}` ` ``// Method 5`` ``// Function to join child matrices`` ``// into (to) parent matrix`` ``public` `void` `join(``int``[][] C, ``int``[][] P, ``int` `iB, ``int` `jB)` ` ``{`` ``// Iterating over elements of 2D matrix`` ``// using nested for loops` ` ``// Outer loop for rows`` ``for` `(``int` `i1 = ``0``, i2 = iB; i1 < C.length; i1++, i2++)` ` ``// Inner loop for columns`` ``for` `(``int` `j1 = ``0``, j2 = jB; j1 < C.length;`` ``j1++, j2++)` ` ``P[i2][j2] = C[i1][j1];`` ``}` ` ``// Method 5`` ``// Main driver method`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// Display message`` ``System.out.println(`` ``"Strassen Multiplication Algorithm Implementation For Matrix Multiplication :\n"``);` ` ``// Create an object of Strassen class`` ``// in the main function`` ``GFG s = ``new` `GFG();` ` ``// Size of matrix`` ``// Considering size as 4 in order to illustrate`` ``int` `N = ``4``;` ` ``// Matrix A`` ``// Custom input to matrix`` ``int``[][] A = { { ``1``, ``2``, ``3``, ``4` `},`` ``{ ``4``, ``3``, ``0``, ``1` `},`` ``{ ``5``, ``6``, ``1``, ``1` `},`` ``{ ``0``, ``2``, ``5``, ``6` `} };` ` ``// Matrix B`` ``// Custom input to matrix`` ``int``[][] B = { { ``1``, ``0``, ``5``, ``1` `},`` ``{ ``1``, ``2``, ``0``, ``2` `},`` ``{ ``0``, ``3``, ``2``, ``3` `},`` ``{ ``1``, ``2``, ``1``, ``2` `} };` ` ``// Matrix C computations` ` ``// Matrix C calling method to get Result`` ``int``[][] C = s.multiply(A, B);` ` ``// Display message`` ``System.out.println(`` ``"\nProduct of matrices A and B : "``);` ` ``// Iterating over elements of 2D matrix`` ``// using nested for loops` ` ``// Outer loop for rows`` ``for` `(``int` `i = ``0``; i < N; i++) {`` ``// Inner loop for columns`` ``for` `(``int` `j = ``0``; j < N; j++)` ` ``// Printing elements of resultant matrix`` ``// with whitespaces in between`` ``System.out.print(C[i][j] + ``" "``);` ` ``// New line once the all elements`` ``// are printed for specific row`` ``System.out.println();`` ``}`` ``}``}`
Output
```Strassen Multiplication Algorithm Implementation For Matrix Multiplication :
Product of matrices A and B :
7 21 15 22
8 8 21 12
12 17 28 22
8 31 16 31 ```
Time Complexity Of Strassen’s Method
By Analysis the time complexity Function can be written as:
`T(N) = 7T(N/2) + O(N2)`
By Solving this using Master Theorem we get :
`T(n)=O(nlog7)`
Thus time Complexity Of Strassen’s Algorithm for matrix multiplication is derived as:
`O(nlog7) = O (n2.81)`
O(n3) Vs O(n2.81) | 3,762 | 11,373 | {"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-2023-40 | latest | en | 0.921591 |
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location: Diff for "FAQ/gamma"
Differences between revisions 15 and 16
⇤ ← Revision 15 as of 2008-02-13 10:51:59 → Size: 671 Editor: PeterWatson Comment: ← Revision 16 as of 2008-02-13 10:52:11 → ⇥ Size: 669 Editor: PeterWatson Comment: Deletions are marked like this. Additions are marked like this. Line 10: Line 10: \$\$2^text{-0.05}\$\$ln(2)^text{0.05}\$\$ \$\$2^text{-0.05}ln(2)^text{0.05}\$\$
# How do I produce random variables which follow a negatively skewed distribution?
Most distributions such as the exponential and log-Normal distributions are positively skewed with the mode of the distribution occurring for lower values.
The Weibull distribution is negatively skewed and may be generated [http://www.taygeta.com/random/weibull.xml using random variables which are uniform on the interval [0,1]]
The below produces an open ended negatively skewed weibull distribution with parameters, 2 and 20. It has a median of
\$\$2text{-0.05}ln(2)text{0.05}\$\$ = 0.95.
```compute wrv= (-(1/2)*(ln(1-rv.uniform(0,1))))**(0.05).
exe.```
None: FAQ/gamma (last edited 2013-03-08 10:17:36 by localhost) | 357 | 1,153 | {"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} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.84183 |
https://community.skuid.com/skuid/topics/unable-to-take-the-difference-between-two-dates?topic-reply-list%5Bsettings%5D%5Bfilter_by%5D=all&topic-reply-list%5Bsettings%5D%5Breply_id%5D=18947721 | 1,579,371,506,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00469.warc.gz | 397,924,362 | 22,139 | # Unable to take the difference between two dates
• Problem
• Updated 2 years ago
• Solved
Hello all,
I am attempting to take the difference of two dates (in number of days). Thus far I am unable to get anything to show other than 0. Please see my UI field setup below.
I have even tried just pulling in one of the dates with the return type of date with nothing showing up. Does anyone know what I am doing wrong?
I have been trying to do something similar to:
https://community.skuid.com/skuid/topics/calculate-age-with-ui-only-field
but again, with no luck.
Thanks,
M\$
• 1,908 Points
Posted 2 years ago
• 494 Points
Hello Michael,
The return is milliseconds so if you divide by 86400000 (the number of milliseconds in a day) it will give you the number of days. e.g.
({{field_1__c}} - {{field_2__c}})/86400000
Cheers
Damien
(Edited)
• 1,908 Points
Hey Thanks for the reply. If you look at my second image you can see I am doing just that. Yet still zeros! I tried taking the difference between TODAY() and one of my dates and got a non-zero result. Perhaps there is a format issue with these dates?
• 494 Points
Try another bracket ) at the end.
Alternatively, try it without floor as it shouldn't matter if they are normal date fields
(Edited)
Mike Dwyer, Champion
• 4,736 Points
Good tip, Damien! But I see that value in Michael's code, as 1000*24*60*60.
Michael, what kind of date values are you comparing? The Floor() function will give you a 0 for anything less than 1 full day difference. I would recommend increasing the decimal places to 1 or 2, removing the Floor() function, and dividing by 1000*60*60 to look at minutes until you see what the formula is giving you.
Skuidward Tentacles (Raymond), Champion
• 18,456 Points
Similar to what Mike said, I would add one variable in at a time and see what the result is. So start with {{Service_Start _Date__c}} and see what the result is. If that is what is expected then add the next piece of the equation in and check the result... then the next piece.... etc.... until you find what part of the equation is causing it to return 0.
Bill McCullough, Champion
• 13,702 Points
Michael,
Try doing the Floor on just the numerator.
FLOOR( {{Service_Start_Date__c}} - {{Implementation_Manager__r.CS_Start_Date__c}}) / (1000*24*60*60)
Thanks,
Bill
• 1,908 Points
Hi All,
Thanks for the great replies. Following Raymond's advice I have eliminated everything and tried testing the output of a single variable and found that I cannot get any field to display. I have tried numbers and dates with matching return type and still nothing is showing up in the table.
Note that I am attempting to create this UI field within an Aggregate Model (for what it's worth). Each row is an opportunity with a service start date in which I would like to have another field counting days from that date and another (for each opp record). Then on a chart I am plotting the cumulative number of opps by owner.
Is there some sort of trick for getting UI fields to pull in merge fields on an Agg model?
Thanks all.
Skuidward Tentacles (Raymond), Champion
• 18,456 Points
This post is about a different use case, but it includes a lot of tips regarding working with UI only formula fields with agg models.
https://community.skuid.com/skuid/top...
• 1,908 Points
I discovered I am able to achieve what I need using a basic type model. Taking the difference of the two dates as described above worked perfectly. Onward and upward! | 855 | 3,492 | {"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-2020-05 | latest | en | 0.924739 |
http://www.mathematik.uni-kl.de/~zca/Reports_on_ca/28/paper_html/node7.html | 1,511,183,816,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00019.warc.gz | 445,748,991 | 4,653 | Next: 5. Special Improvements Up: Recursive Computation of Free Previous: 3.3 Non-regular Extensions
# 4. The Algorithm
The implementation of the ideas of the preceding Chapter is realized as -algorithm in the terminology of Chapter 2. After the sequence of generators we use secondly the module index and than, thirdly, the degree of the lowest common multiples of the pairs as ordering criteria on the pair sets.
Within the computation of free resolutions each new generator (which is either a polynomial from the input or a syzygy of certain generators) is equipped with a unique module component which stores the reduction performed with this generator. Moreover, the assigned module components are the basing units for the syzygies. The relation of the ordering of generators and that of their corresponding components is very sensible. Our practical experience indicates that the orderings introduced by F.-O. Schreyer (see [6]) are the best choice for computations of free resolutions.
Here, we have to use another class of orderings:
Definition 4.1 Assume that the ordering on fulfils either:
or
for any . Such an ordering is called a sequential monomial ordering if it agrees with the addition of new elements within an algorithm to compute free resolutions, i.e., module components assigned to new generators are greater w.r.t. than all the former components.
REMARK:The definition requires to give new generators greater components for the first and smaller components for the second variant. Note that the definition is restricted to only 2 possible variants. In general any ordering for which new generators have components greater than all former are working.
In the sequel we always assume to work with the first sequential ordering from the definition.
Now, we describe the several steps of a sequential algorithm for an ideal :
2. We increase and return the result if .
3. We normalize the generator with respect to the ideal . If we put it to the set of generators and compute a standardbasis of as well as of the syzygies of .
4. We decide wether yields a regular extension by the criteria of the Chapter 3.1. If so, we go to the next step. If not, we skip the next step.
5. We are in the case of a regular extension. We copy the resolution of and the mappings as multiplications with . Now, we are going back to step 2.
6. We are in the case of a non-regular extension. We initialize an inner loop over the module index starting with 0.
7. We are going to the next module. If there are no new generators we return to step 2. If there are new generators, we assign new module components to the new generators and add them to the known generators.
8. We perform a standard basis computation on all pairs build from the new generators. With the new syzygies we go back to the step 7.
The recursion step of the algorithm can be visualized by the following figures (for a proof see Lemma 5.1): The first figure shows the resolution after the -th step.
The -th block of this figure shows the size of the matrix of the -th map in the -th subresolution. Therefore, the hight of the -th block is equal to the length of the -th block. The situation after the computation of the -th subresolution is shown in Figure 2.
The old figure corresponds to the empty blocks. In the first matrix a new column is added for the new generator . Corresponding to this the second matrix has one new row (vertically hatched). Analogously, any higher matrix contains as many more new generators (horizontally hatched) as the next matrix new rows (vertically hatched). Further, the vertically hatched blocks of index correspond to a free resolution of the -th extension ideal , whereas, the horizontally hatched blocks correspond to the representations chosen in Lemma 3.6.
If the -th extension is regular the horizonally hatched blocks can be chosen as identical copies of the -th subresolution with index shifted by 1. Thus, if all matrices of the -th subresolution are given as standard bases the -th subresolution consists of standard bases, too.
The algorithm is implemented in SINGULAR realizing the following pseudo code: Res
INPUT: , a basis of an ideal
OUTPUT: a resolution of ;
;
;
WHILE DO
;
:=Normalform(,);
IF DO
;
IF DO
:=Koszul_extension(,);
ELSE
WHILE DO
.
OD
FI
FI
OD
Next: 5. Special Improvements Up: Recursive Computation of Free Previous: 3.3 Non-regular Extensions
| ZCA Home | Reports | | 974 | 4,419 | {"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} | 3.171875 | 3 | CC-MAIN-2017-47 | latest | en | 0.910246 |
https://writall.com/using-r-studios-to-solve-3-questions/ | 1,680,088,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00567.warc.gz | 678,207,836 | 12,488 | # Using R Studios to solve 3 questions
E3.2 A consumer is given the chance to buy a baseball card for \$1, but he declines
the trade. If the consumer is now given the baseball card, will he be willing to
sell it for \$1? Standard consumer theory suggests yes, but behavioral economists
have found that “ownership” tends to increase the value of goods to consumers.
That is, the consumer may hold out for some amount more than \$1 (for exam-
ple, \$1.20) when selling the card, even though he was willing to pay only some
amount less than \$1 (for example, \$0.88) when buying it. Behavioral economists
call this phenomenon the “endowment effect.” John List investigated the endow-
ment effect in a randomized experiment involving sports memorabilia traders
at a sports-card show. Traders were randomly given one of two sports collect-
ibles, say good A or good B, that had approximately equal market value. Those
receiving good A were then given the option of trading good A for good B with
the experimenter; those receiving good B were given the option of trading good
B for good A with the experimenter. Data from the experiment and a detailed
description can be found on the text website, http://www.pearsonhighered
.com/stock_watson/, in the files Sportscards and Sportscards_Description.
a. i. Suppose that, absent any endowment effect, all the subjects prefer good
A to good B. What fraction of the experiment’s subjects would you
expect to trade the good that they were given for the other good? (Hint:
Because of random assignment of the two treatments, approximately
50% of the subjects received good A, and 50% received good B.)
ii. Suppose that, absent any endowment effect, 50% of the subjects prefer
good A to good B, and the other 50% prefer good B to good A. What
fraction of the subjects would you expect to trade the good they were
given for the other good?
iii. Suppose that, absent any endowment effect, X% of the subjects prefer
good A to good B, and the other (100 – X)% prefer good B to good
A. Show that you would expect 50% of the subjects to trade the good
they were given for the other good.
b. Using the sports card data, what fraction of the subjects traded the good they
were given? Is the fraction significantly different from 50%? Is there evi-
dence of an endowment effect? (Hint: Review Exercises 3.2 and 3.3.)
c. Some have argued that the endowment effect may be present but that it
is likely to disappear as traders gain more trading experience. Half of the
experimental subjects were dealers, and the other half were nondealers.
Dealers have more experience than nondealers. Repeat (b) for dealers
and nondealers. Is there a significant difference in their behavior?
Is the evidence consistent with the hypothesis that the endowment effect
disappears as traders gain more experience? (Hint: Review Exercise 3.15.)
E3.2 A consumer is given the chance to buy a baseball card for \$1, but he declines
the trade. If the consumer is now given the baseball card, will he be willing to
sell it for \$1? Standard consumer theory suggests yes, but behavioral economists
have found that “ownership” tends to increase the value of goods to consumers.
That is, the consumer may hold out for some amount more than \$1 (for exam-
ple, \$1.20) when selling the card, even though he was willing to pay only some
amount less than \$1 (for example, \$0.88) when buying it. Behavioral economists
call this phenomenon the “endowment effect.” John List investigated the endow-
ment effect in a randomized experiment involving sports memorabilia traders
at a sports-card show. Traders were randomly given one of two sports collect-
ibles, say good A or good B, that had approximately equal market value. Those
receiving good A were then given the option of trading good A for good B with
the experimenter; those receiving good B were given the option of trading good
B for good A with the experimenter. Data from the experiment and a detailed
description can be found on the text website, http://www.pearsonhighered
.com/stock_watson/, in the files Sportscards and Sportscards_Description.
a. i. Suppose that, absent any endowment effect, all the subjects prefer good
A to good B. What fraction of the experiment’s subjects would you
expect to trade the good that they were given for the other good? (Hint:
Because of random assignment of the two treatments, approximately
50% of the subjects received good A, and 50% received good B.)
ii. Suppose that, absent any endowment effect, 50% of the subjects prefer
good A to good B, and the other 50% prefer good B to good A. What
fraction of the subjects would you expect to trade the good they were
given for the other good?
iii. Suppose that, absent any endowment effect, X% of the subjects prefer
good A to good B, and the other (100 – X)% prefer good B to good
A. Show that you would expect 50% of the subjects to trade the good
they were given for the other good.
b. Using the sports card data, what fraction of the subjects traded the good they
were given? Is the fraction significantly different from 50%? Is there evi-
dence of an endowment effect? (Hint: Review Exercises 3.2 and 3.3.)
c. Some have argued that the endowment effect may be present but that it
is likely to disappear as traders gain more trading experience. Half of the
experimental subjects were dealers, and the other half were nondealers.
Dealers have more experience than nondealers. Repeat (b) for dealers
and nondealers. Is there a significant difference in their behavior?
Is the evidence consistent with the hypothesis that the endowment effect
disappears as traders gain more experience? (Hint: Review Exercise 3.15.)
E4.2 On the text website, http://www.pearsonhighered.com/stock_watson/, you
will find the data file Earnings_and_Height, which contains data on earn-
ings, height, and other characteristics of a random sample of U.S. workers.?
A detailed description is given in Earnings_and_Height_Description, also
available on the website. In this exercise, you will investigate the relationship
between earnings and height.
a. What is the median value of height in the sample?
b. i. Estimate average earnings for workers whose height is at most
67 inches.
ii. Estimate average earnings for workers whose height is greater than
67 inches.
iii. On average, do taller workers earn more than shorter workers? How
much more? What is a 95% confidence interval for the difference in
average earnings?
c. Construct a scatterplot of annual earnings (Earnings) on height (Height).
Notice that the points on the plot fall along horizontal lines. (There are
only 23 distinct values of Earnings). Why? (Hint: Carefully read the
detailed data description.)
d. Run a regression of Earnings on Height.
i. What is the estimated slope?
ii. Use the estimated regression to predict earnings for a worker who
is 67 inches tall, for a worker who is 70 inches tall, and for a worker
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Order your essay today and save 20% with the discount code GOLDEN | 1,685 | 7,179 | {"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-2023-14 | latest | en | 0.96058 |
https://www.office-forums.com/threads/if-statement-with-vlookup-including-isna-function-to-compare-two-columns-from-different-worksheets.1156861/ | 1,621,324,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00017.warc.gz | 985,916,700 | 26,087 | # If statement with Vlookup - including ISNA function to compare two columns from different worksheets
N
#### nakliwala
hi,
I need some help creatng an 'IF' statement that contains a
'VLOOKUP' . In Sheet 1 i have an empty column called ROC (Column F)
that should contain the 'IF/VLOOKUP' function. This should look up
Column B in Sheet 2 and try to match them with entries in Sheet
1:Column D(but only the first four digits of the entries in Sheet
2:Column B. In Sheet 2:Column B row 58 the entry is 5730P; in Sheet 1
Column D, Row 18 the entry is 5730. The 'VLOOKUP' function needs to
identify that the first 4 digits are similar, then output the value of
Sheet 2 Column 'A' row 58 in the Column F Row 18.
if anyone has knowledge of if statement with vlookups please get back
to me.
Thank you,
Nakli
To get more of an idea on my question; My question is similar to the
Someone else's Question/Answer relating to vlookups::
Afternoon,
I know some one out there might be able to assist me with my dilemna.
In Sheet1 I have a list of 8 digts id numbers in column A and in
sheet2
is where I am doing the lookup from.
The formula below is working great. I picked up from this website.
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number",VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE))
What I need to know if possible is, can another (vlookup or if ) be
added to the ending formula to do search on just the first 4 number
of
the id, if I got the response Invalid Number.
I currently have another column doing the lookup on just the 4
digits.
Thanks
--
Sat3902
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CLR View profile
More options Aug 4 2006, 1:33 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Fri, 4 Aug 2006 10:33:02 -0700
Local: Fri, Aug 4 2006 1:33 pm
Subject: RE: IF & Vlookup
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You might try something like this.........
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number"&", Four-digit lookup =
"&YourFourDigitLookpuFormula,VLOOKUP(A6,Schedule!\$A\$13:\$E
\$1463,3,FALSE))
Vaya con Dios,
Chuck, CABGx3
- Hide quoted text -
- Show quoted text -
Sat3902 said:
Afternoon,
I know some one out there might be able to assist me with my dilemna.
In Sheet1 I have a list of 8 digts id numbers in column A and in sheet2
is where I am doing the lookup from.
The formula below is working great. I picked up from this website.
What I need to know if possible is, can another (vlookup or if ) be
added to the ending formula to do search on just the first 4 number of
the id, if I got the response Invalid Number.
I currently have another column doing the lookup on just the 4 digits.
for clearing space
To post a message you must first join this group.
posting.
You do not have the permission required to post.
Sat3902 View profile
More options Aug 7 2006, 1:56 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 13:56:46 -0400
Local: Mon, Aug 7 2006 1:56 pm
Subject: Re: IF & Vlookup
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Gracias Chuck,
Thank you for the suggestion, however it did not work for me. I still
got a return of "*Invalid Number*". I do have the 4 digit in the
Sheet1
where I am doing the lookup from.
I am hoping I followed example
Here is what I entered.
=IF(ISNA(VLOOKUP(A7,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number"&"",(VLOOKUP(LEFT(A7,4),Schedule!\$A\$13:\$E\$1463,3,FALSE)))
The last part of the arguement does work correctly. I currently using
it when doing a vlookup on just the 4 digit on a seperate column.
I am doing a vlookup on my 8 digit user ID and when I do not get a
match I then what do a vlookup on the first 4 digits only.
--
Sat3902
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More options Aug 7 2006, 3:26 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Mon, 7 Aug 2006 12:26:02 -0700
Local: Mon, Aug 7 2006 3:26 pm
Subject: Re: IF & Vlookup
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If you are wanting to look up the LEFT 4 digits of a cell, then you
will have
to build that in to your VLOOKUP table, or another one.....unless you
have
both cells containing just those 4 digits AND other cells containg the
entire
number.....VLOOKUP cannot extract the left 4 digits out af a number in
the
table.....just add a column on the left side of the table....assume
you
insert a new column A and the old column A is now B and the table
extends to
F now, then try
=IF(ISNA(VLOOKUP(A6,Schedule!\$B\$13:\$F\$1463,3,FALSE)),"Invalid
Number"&",
"&vlookup(left(A6,4),\$A\$13:\$F\$1463,3,false),VLOOKUP(A6,Schedule!\$B
\$13:\$F\$1463,3,FALSE))
hth
Vaya con Dios,
Chuck, CABGx3
- Hide quoted text -
- Show quoted text -
Sat3902 said:
Gracias Chuck,
Thank you for the suggestion, however it did not work for me. I still
got a return of "*Invalid Number*". I do have the 4 digit in the Sheet1
where I am doing the lookup from.
I am hoping I followed example
Here is what I entered.
The last part of the arguement does work correctly. I currently using
it when doing a vlookup on just the 4 digit on a seperate column.
I am doing a vlookup on my 8 digit user ID and when I do not get a
match I then what do a vlookup on the first 4 digits only.
for clearing space
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posting.
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More options Aug 7 2006, 6:34 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 18:34:31 -0400
Local: Mon, Aug 7 2006 6:34 pm
Subject: Re: IF & Vlookup
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Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the
same
work sheet. I am just trying to eliminate from having a lot of
columns
with formulas which slow up the workbook when saving updates or
making
change to it.
I do have the 8 digits and 4 digits in the same column but when doing
the vlookup I have 2 columns one for the 8 digits and the other for
the
4 digit.
The end results is being populated to another worksheet.
Gracias
--
Sat3902
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More options Aug 7 2006, 6:41 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 18:41:38 -0400
Local: Mon, Aug 7 2006 6:41 pm
Subject: Re: IF & Vlookup
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Sat3902 Wrote:
- Hide quoted text -
- Show quoted text -
Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the same
work sheet. I am just trying to eliminate from having a lot of columns
with formulas which slow up the workbook when saving updates or making
change to it.
The database from where I am doing the lookup from does have the 8
digits and 4 digits in the same column. But the work sheet that I
import the data needing to be matched up, I have 2 columns one for the
8 digits and the other for the 4 digit to do the Vlookup.
The end results is being populated to another worksheet. I hope I am
not confusing you with what I am needing.
Gracias
Have a good day
--
Sat3902
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More options Aug 8 2006, 8:46 am
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Tue, 8 Aug 2006 05:46:13 -0700
Local: Tues, Aug 8 2006 8:46 am
Subject: Re: IF & Vlookup
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Well then, if you have both the 4 digit and 8 digit numbers in column
A then
this will probably work.....
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number,
"&vlookup(left(A6,4),\$A\$12:\$E\$1439,3,false),VLOOKUP(A6,Schedule!\$A
\$13:\$E\$1463,3,FALSE))
Vaya con Dios,
Chuck, CABGx3
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Sat3902 said:
Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the same
work sheet. I am just trying to eliminate from having a lot of columns
with formulas which slow up the workbook when saving updates or making
change to it.
I do have the 8 digits and 4 digits in the same column but when doing
the vlookup I have 2 columns one for the 8 digits and the other for the
4 digit.
The end results is being populated to another worksheet.
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More options Aug 8 2006, 4:25 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:25:14 -0400
Local: Tues, Aug 8 2006 4:25 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the
vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
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More options Aug 8 2006, 4:22 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:22:54 -0400
Local: Tues, Aug 8 2006 4:22 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match.
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
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More options Aug 8 2006, 4:22 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:22:54 -0400
Local: Tues, Aug 8 2006 4:22 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match.
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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More options Aug 8 2006, 4:25 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:25:14 -0400
Local: Tues, Aug 8 2006 4:25 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the
vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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CLR View profile
More options Aug 8 2006, 8:03 pm
Newsgroups: microsoft.public.excel.newusers
From: "CLR" <[email protected]>
Date: Tue, 8 Aug 2006 20:03:34 -0400
Local: Tues, Aug 8 2006 8:03 pm
Subject: Re: IF & Vlookup
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I'm ready for bed now and cannot do any more tonight........all I can
see
off the bat is that you do not have the "Sheet2! reference on the
middle
part of the formula.........you might try this.........
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid number,
"&VLOOKUP(LEFT(A16,4),Sheet2!\$A\$2:\$C\$20882,3,FALSE),VLOOKUP(A16,Sheet2!
\$A\$2:
\$B\$20882,2,FALSE))
Otherwise, maybe you might send me a copy of your file
to .......croberts
at tampabay dot rr dot com.......and I'll take a look
tomorrow........sorry,
I'm out of gas tonight, but hang in there....we'll whip this thing.
Vaya con Dios,
Chuck, CABGx3
in
message
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Chuck
Thank you for your time and patients. I tried your suggestion from this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
http://www.excelforum.com/member.php?action=getinfo&userid=36777
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More options Aug 11 2006, 12:07 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Fri, 11 Aug 2006 12:07:30 -0400
Local: Fri, Aug 11 2006 12:07 pm
Subject: Re: IF & Vlookup
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Happy days are here again.
Got it to work, with help from a friend and your assistance.
I just needed to add in the ( Left formula ). So if I do not get a
match on my 8 digit number then it will match on the first 4 digit
next.
Here is the formula.
=IF(ISNA(VLOOKUP(A2,'Sheet2'!\$A\$2:\$B
\$3950,2,FALSE)),VLOOKUP(LEFT(A2,4),'Sheet2'!\$A\$2:\$B\$3950,2,FALSE),
(VLOOKUP(A2,'Sheet2'!\$A\$2:\$B\$3950,2,FALSE)))
when I don't get a hit. I can not seem to find the right mix to just
get a blank value if there is not match. I welcome your
input.
Gracias por todo
Senor Chuck
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More options Aug 11 2006, 12:54 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Fri, 11 Aug 2006 09:54:02 -0700
Local: Fri, Aug 11 2006 12:54 pm
Subject: Re: IF & Vlookup
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This is pretty messy, but you might give it a try.......it should
return the
value in the first table if it's there, otherwise return from the
second
table, if there....and if it's in neither, then return blank.........
=IF(AND(ISNA(VLOOKUP(A2,Sheet2!\$A\$2:\$B
\$3950,2,FALSE)),ISNA(VLOOKUP(LEFT(A2,4),Sheet2!\$A\$2:\$B
\$3950,2,FALSE))),"",IF(ISNA(VLOOKUP(A2,Sheet2!\$A\$2:\$B\$3950,
2,FALSE)),VLOOKUP(LEFT(A2,4),Sheet2!\$A\$2:\$B\$3950,2,FALSE),
(VLOOKUP(A2,Sheet2!\$A\$2:\$B\$3950,2,FALSE))))
P
#### Pete_UK
You won't be able to use VLOOKUP in this situation. When using
VLOOKUP, it searches for a match in the first (left-most) column of
the table and then returns data from columns on the right of the first
column. You say that you would like to search Sheet2!ColumnB and then
return the matching data from column A. To do this you will have to
use an INDEX/MATCH combination, or move the columns in your data table
so that your current column A becomes column C or later.
Hope this helps.
Pete
T
#### T. Valko
--
Biff
Microsoft Excel MVP
hi,
I need some help creatng an 'IF' statement that contains a
'VLOOKUP' . In Sheet 1 i have an empty column called ROC (Column F)
that should contain the 'IF/VLOOKUP' function. This should look up
Column B in Sheet 2 and try to match them with entries in Sheet
1:Column D(but only the first four digits of the entries in Sheet
2:Column B. In Sheet 2:Column B row 58 the entry is 5730P; in Sheet 1
Column D, Row 18 the entry is 5730. The 'VLOOKUP' function needs to
identify that the first 4 digits are similar, then output the value of
Sheet 2 Column 'A' row 58 in the Column F Row 18.
if anyone has knowledge of if statement with vlookups please get back
to me.
Thank you,
Nakli
To get more of an idea on my question; My question is similar to the
Someone else's Question/Answer relating to vlookups::
Afternoon,
I know some one out there might be able to assist me with my dilemna.
In Sheet1 I have a list of 8 digts id numbers in column A and in
sheet2
is where I am doing the lookup from.
The formula below is working great. I picked up from this website.
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number",VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE))
What I need to know if possible is, can another (vlookup or if ) be
added to the ending formula to do search on just the first 4 number
of
the id, if I got the response Invalid Number.
I currently have another column doing the lookup on just the 4
digits.
Thanks
--
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CLR View profile
More options Aug 4 2006, 1:33 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Fri, 4 Aug 2006 10:33:02 -0700
Local: Fri, Aug 4 2006 1:33 pm
Subject: RE: IF & Vlookup
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You might try something like this.........
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number"&", Four-digit lookup =
"&YourFourDigitLookpuFormula,VLOOKUP(A6,Schedule!\$A\$13:\$E
\$1463,3,FALSE))
Vaya con Dios,
Chuck, CABGx3
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Sat3902 said:
Afternoon,
I know some one out there might be able to assist me with my dilemna.
In Sheet1 I have a list of 8 digts id numbers in column A and in sheet2
is where I am doing the lookup from.
The formula below is working great. I picked up from this website.
What I need to know if possible is, can another (vlookup or if ) be
added to the ending formula to do search on just the first 4 number of
the id, if I got the response Invalid Number.
I currently have another column doing the lookup on just the 4 digits.
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More options Aug 7 2006, 1:56 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 13:56:46 -0400
Local: Mon, Aug 7 2006 1:56 pm
Subject: Re: IF & Vlookup
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Gracias Chuck,
Thank you for the suggestion, however it did not work for me. I still
got a return of "*Invalid Number*". I do have the 4 digit in the
Sheet1
where I am doing the lookup from.
I am hoping I followed example
Here is what I entered.
=IF(ISNA(VLOOKUP(A7,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number"&"",(VLOOKUP(LEFT(A7,4),Schedule!\$A\$13:\$E\$1463,3,FALSE)))
The last part of the arguement does work correctly. I currently using
it when doing a vlookup on just the 4 digit on a seperate column.
I am doing a vlookup on my 8 digit user ID and when I do not get a
match I then what do a vlookup on the first 4 digits only.
--
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More options Aug 7 2006, 3:26 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Mon, 7 Aug 2006 12:26:02 -0700
Local: Mon, Aug 7 2006 3:26 pm
Subject: Re: IF & Vlookup
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If you are wanting to look up the LEFT 4 digits of a cell, then you
will have
to build that in to your VLOOKUP table, or another one.....unless you
have
both cells containing just those 4 digits AND other cells containg the
entire
number.....VLOOKUP cannot extract the left 4 digits out af a number in
the
table.....just add a column on the left side of the table....assume
you
insert a new column A and the old column A is now B and the table
extends to
F now, then try
=IF(ISNA(VLOOKUP(A6,Schedule!\$B\$13:\$F\$1463,3,FALSE)),"Invalid
Number"&",
"&vlookup(left(A6,4),\$A\$13:\$F\$1463,3,false),VLOOKUP(A6,Schedule!\$B
\$13:\$F\$1463,3,FALSE))
hth
Vaya con Dios,
Chuck, CABGx3
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Sat3902 said:
Gracias Chuck,
Thank you for the suggestion, however it did not work for me. I still
got a return of "*Invalid Number*". I do have the 4 digit in the Sheet1
where I am doing the lookup from.
I am hoping I followed example
Here is what I entered.
The last part of the arguement does work correctly. I currently using
it when doing a vlookup on just the 4 digit on a seperate column.
I am doing a vlookup on my 8 digit user ID and when I do not get a
match I then what do a vlookup on the first 4 digits only.
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More options Aug 7 2006, 6:34 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 18:34:31 -0400
Local: Mon, Aug 7 2006 6:34 pm
Subject: Re: IF & Vlookup
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Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the
same
work sheet. I am just trying to eliminate from having a lot of
columns
with formulas which slow up the workbook when saving updates or
making
change to it.
I do have the 8 digits and 4 digits in the same column but when doing
the vlookup I have 2 columns one for the 8 digits and the other for
the
4 digit.
The end results is being populated to another worksheet.
Gracias
--
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More options Aug 7 2006, 6:41 pm
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From: Sat3902 <[email protected]>
Date: Mon, 7 Aug 2006 18:41:38 -0400
Local: Mon, Aug 7 2006 6:41 pm
Subject: Re: IF & Vlookup
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Sat3902 Wrote:
- Hide quoted text -
- Show quoted text -
Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the same
work sheet. I am just trying to eliminate from having a lot of columns
with formulas which slow up the workbook when saving updates or making
change to it.
The database from where I am doing the lookup from does have the 8
digits and 4 digits in the same column. But the work sheet that I
import the data needing to be matched up, I have 2 columns one for the
8 digits and the other for the 4 digit to do the Vlookup.
The end results is being populated to another worksheet. I hope I am
not confusing you with what I am needing.
Gracias
Have a good day
--
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More options Aug 8 2006, 8:46 am
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From: CLR <[email protected]>
Date: Tue, 8 Aug 2006 05:46:13 -0700
Local: Tues, Aug 8 2006 8:46 am
Subject: Re: IF & Vlookup
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Well then, if you have both the 4 digit and 8 digit numbers in column
A then
this will probably work.....
=IF(ISNA(VLOOKUP(A6,Schedule!\$A\$13:\$E\$1463,3,FALSE)),"Invalid
Number,
"&vlookup(left(A6,4),\$A\$12:\$E\$1439,3,false),VLOOKUP(A6,Schedule!\$A
\$13:\$E\$1463,3,FALSE))
Vaya con Dios,
Chuck, CABGx3
- Hide quoted text -
- Show quoted text -
Sat3902 said:
Evening Chuck
I am going to try your suggestion. Just want to mention to you that
the VLoop can strip the LEFT 4 digits. I am currently using this
arguement
=IF(ISNA(VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)),"Invalid
Number",VLOOKUP(LEFT(A7,4),Schedule!\$A\$12:\$E\$1439,3,FALSE)) on the same
work sheet. I am just trying to eliminate from having a lot of columns
with formulas which slow up the workbook when saving updates or making
change to it.
I do have the 8 digits and 4 digits in the same column but when doing
the vlookup I have 2 columns one for the 8 digits and the other for the
4 digit.
The end results is being populated to another worksheet.
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More options Aug 8 2006, 4:25 pm
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From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:25:14 -0400
Local: Tues, Aug 8 2006 4:25 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the
vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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More options Aug 8 2006, 4:22 pm
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From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:22:54 -0400
Local: Tues, Aug 8 2006 4:22 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match.
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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More options Aug 8 2006, 4:22 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:22:54 -0400
Local: Tues, Aug 8 2006 4:22 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match.
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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More options Aug 8 2006, 4:25 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Tue, 8 Aug 2006 16:25:14 -0400
Local: Tues, Aug 8 2006 4:25 pm
Subject: Re: IF & Vlookup
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Chuck
Thank you for your time and patients. I tried your suggestion from
this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the
vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
--
Sat3902
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CLR View profile
More options Aug 8 2006, 8:03 pm
Newsgroups: microsoft.public.excel.newusers
From: "CLR" <[email protected]>
Date: Tue, 8 Aug 2006 20:03:34 -0400
Local: Tues, Aug 8 2006 8:03 pm
Subject: Re: IF & Vlookup
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I'm ready for bed now and cannot do any more tonight........all I can
see
off the bat is that you do not have the "Sheet2! reference on the
middle
part of the formula.........you might try this.........
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid number,
"&VLOOKUP(LEFT(A16,4),Sheet2!\$A\$2:\$C\$20882,3,FALSE),VLOOKUP(A16,Sheet2!
\$A\$2:
\$B\$20882,2,FALSE))
Otherwise, maybe you might send me a copy of your file
to .......croberts
at tampabay dot rr dot com.......and I'll take a look
tomorrow........sorry,
I'm out of gas tonight, but hang in there....we'll whip this thing.
Vaya con Dios,
Chuck, CABGx3
in
message
- Hide quoted text -
- Show quoted text -
Chuck
Thank you for your time and patients. I tried your suggestion from this
morning. It only work the first part of the formula.
Here is what I entered
=IF(ISNA(VLOOKUP(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE)),"invalid
number,"&VLOOKUP(LEFT(A16,4),\$A\$2:\$C\$20882,2,FALSE),VLOOKU
P(A16,Sheet2!\$A\$2:\$B\$20882,2,FALSE))
Got a return value of #N/A when I did not get a match on the 8 digit.
It worked when I got a match on the 8 digit but did not do the vlookup
on the 4 digit that is when I got the *#N/A*
Was I suppose to space the last letter of Vlookup at the last Vlookup
argument. I followed your example. It did the same without the same.
Unless it can not be done.
http://www.excelforum.com/member.php?action=getinfo&userid=36777
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More options Aug 11 2006, 12:07 pm
Newsgroups: microsoft.public.excel.newusers
From: Sat3902 <[email protected]>
Date: Fri, 11 Aug 2006 12:07:30 -0400
Local: Fri, Aug 11 2006 12:07 pm
Subject: Re: IF & Vlookup
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Happy days are here again.
Got it to work, with help from a friend and your assistance.
I just needed to add in the ( Left formula ). So if I do not get a
match on my 8 digit number then it will match on the first 4 digit
next.
Here is the formula.
=IF(ISNA(VLOOKUP(A2,'Sheet2'!\$A\$2:\$B
\$3950,2,FALSE)),VLOOKUP(LEFT(A2,4),'Sheet2'!\$A\$2:\$B\$3950,2,FALSE),
(VLOOKUP(A2,'Sheet2'!\$A\$2:\$B\$3950,2,FALSE)))
when I don't get a hit. I can not seem to find the right mix to just
get a blank value if there is not match. I welcome your
input.
Gracias por todo
Senor Chuck
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More options Aug 11 2006, 12:54 pm
Newsgroups: microsoft.public.excel.newusers
From: CLR <[email protected]>
Date: Fri, 11 Aug 2006 09:54:02 -0700
Local: Fri, Aug 11 2006 12:54 pm
Subject: Re: IF & Vlookup
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This is pretty messy, but you might give it a try.......it should
return the
value in the first table if it's there, otherwise return from the
second
table, if there....and if it's in neither, then return blank.........
=IF(AND(ISNA(VLOOKUP(A2,Sheet2!\$A\$2:\$B
\$3950,2,FALSE)),ISNA(VLOOKUP(LEFT(A2,4),Sheet2!\$A\$2:\$B
\$3950,2,FALSE))),"",IF(ISNA(VLOOKUP(A2,Sheet2!\$A\$2:\$B\$3950,
2,FALSE)),VLOOKUP(LEFT(A2,4),Sheet2!\$A\$2:\$B\$3950,2,FALSE),
(VLOOKUP(A2,Sheet2!\$A\$2:\$B\$3950,2,FALSE))))
N
#### nakliwala
You won't be able to use VLOOKUP in this situation. When using
VLOOKUP, it searches for a match in the first (left-most) column of
the table and then returns data from columns on the right of the first
column. You say that you would like to search Sheet2!ColumnB and then
return the matching data from column A. To do this you will have to
use an INDEX/MATCH combination, or move the columns in your data table
so that your current column A becomes column C or later.
Hope this helps.
Pete
- Show quoted text -
Hi Pete,
i used the following vlookup:
=IF(ISNA(VLOOKUP(D2,'Appendix 7'!\$A\$2:\$B\$103,2,FALSE))," ",
VLOOKUP(D2,'Appendix 7'!\$A\$2:\$B\$103,2,FALSE))
In the existing sheet the vlookup was in column F. And the value in
the "Appendix" tab column A was what needed to be matched. the
vlookup i used seemed to work fine...if u see any problems with it
~Nakli
P
#### Pete_UK
This is looking for a match in column A of the Appendix sheet and will
return the corresponding value from column B of that sheet if a match
is found - this is not what you described in your first posting.
Pete | 11,983 | 41,189 | {"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} | 3.25 | 3 | CC-MAIN-2021-21 | latest | en | 0.815506 |
https://math.stackexchange.com/questions/1885508/average-area-between-x2-and-a-scaled-line-through-the-origin | 1,624,439,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00015.warc.gz | 337,935,616 | 36,916 | # Average area between $x^2$ and a scaled line through the origin
Consider the functions $x^2$ and $ax$ for $x\ge0$. Specifically, consider the average area between the two curves as $a$ modulates between $0$ and $k$ and refer to it as $\hat a$. Which non-zero value of $k$ satisfies the condition that $\hat a=x^2$?
Although I intend to answer my own question, I would appreciate reading other solutions and viewpoints for this problem. I am trying to become better at mathematical writing so viewing other solutions would be greatly beneficial to myself and others. Additionally, if you choose not to read my answer before coming up with your own, this could potentially become quite interesting as the methodology could vary widely.
In case the wording of the problem is confusing, I modeled a picture of the area when $a=1$.
Find the intersections of $x^2$ and $ax$ to be $x^2-ax=0\iff x(x-a)=0$, and thus our bound is $[0,a]$. We can then calculate the area for any specific $a$ as $\int_0^a\left(x^2-ax\right)dx=\frac{a^3}3-a\frac{a^2}{2}=\frac {a^3} 6$. Finally we calculate $\hat a$ as $\frac 1 k\int_0^k\frac {a^3} 6\ da=\frac {k^3}{24}$ and thus the solution $k=24$ since $24^2=\frac {24^3}{24}$. | 350 | 1,210 | {"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} | 3.984375 | 4 | CC-MAIN-2021-25 | latest | en | 0.932008 |
https://qa.auth.gr/en/class/1/600121459/M1 | 1,713,490,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00380.warc.gz | 430,541,301 | 9,416 | # Introduction to Algebra
Title ΕΙΣΑΓΩΓΗ ΣΤΗΝ ΑΛΓΕΒΡΑ ΚΑΙ ΣΤΗ ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ / Introduction to Algebra Code 0102 Faculty Sciences School Mathematics Cycle / Level 1st / Undergraduate Teaching Period Winter/Spring Common No Status Inactive Course ID 40000296
### Programme of Study: UPS of School of Mathematics (2014-today)
Registered students: 238
OrientationAttendance TypeSemesterYearECTS
Academic Year 2018 – 2019 Class Period Spring Instructors from Other Categories Paraskevas Alvanos 39hrs Weekly Hours 3 Class ID 600121459
Course Type 2016-2020
• Background
• General Knowledge
• Scientific Area
• Skills Development
Course Type 2011-2015
Specific Foundation / Core
Mode of Delivery
• Face to face
Digital Course Content
Language of Instruction
• Greek (Instruction, Examination)
Learning Outcomes
Successful completion of the course offers the student knowledge of the basic concepts and structures of Algebra and Number Theory, and the basic ability to transfer this knoweledge.
General Competences
• Apply knowledge in practice
• Retrieve, analyse and synthesise data and information, with the use of necessary technologies
• Work autonomously
• Appreciate diversity and multiculturality
• Respect natural environment
• Demonstrate social, professional and ethical commitment and sensitivity to gender issues
• Be critical and self-critical
• Advance free, creative and causative thinking
Course Content (Syllabus)
The course is part of the module of courses that aim to offer dexterities for teaching mathematics in Secondary Education. Sets, Functions. Cartesian product. Equivalence relations. Ordering. Operations on a set. Semigroups., Monoids. Natural numbers. Mathematical induction. Well ordering principle. Complex Numbers. Countable sets. Elements of combinatorial theory. The ring of integers. Divisibility. Euclidean algorithm. GCD, LCM. Bezout's identity. Solutions of the linear equation ax+by= c. Rationals in base n. Prime numbers. Wilson's theorem. Fundamental theorem of arithmetic. Rings modulo n. Fields modulo p. Linear equivalences and systems. Chinese Remainder Theorem, Euler's function, Fermat's Small Theorem, Groups, Euler's Theorem, Groups, ISomorphism of Groups, Classification of groups of small order.
Educational Material Types
• Notes
• Video lectures
• Book
Use of Information and Communication Technologies
Use of ICT
• Use of ICT in Communication with Students
Course Organization
Lectures652.2
Tutorial391.3
Exams30.1
Total1655.5
Student Assessment
Description
Written final examination
Student Assessment methods
• Written Exam with Multiple Choice Questions (Formative, Summative)
• Written Exam with Short Answer Questions (Formative, Summative)
• Written Exam with Extended Answer Questions (Formative)
• Written Exam with Problem Solving (Formative, Summative)
Bibliography
Course Bibliography (Eudoxus)
Ε. Ψωμόπουλου, Εισαγωγλη στην Άλγεβρα, Εκδ. Ζήτη Δ. Πουλάκης, Άλγεβρα, Εκδ. Ζήτη Κ. Κάλφα, Εισαγωγή στην Άλγεβρα, Εκδόσεις Ζήτη
Last Update
24-04-2019 | 727 | 3,013 | {"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} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.771613 |
https://math.libretexts.org/Bookshelves/Differential_Equations/Introduction_to_Partial_Differential_Equations_(Herman)/03%3A_Trigonometric_Fourier_Series/3.06%3A_Finite_Length_Strings | 1,721,135,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00648.warc.gz | 345,710,067 | 31,323 | # 3.6: Finite Length Strings
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We now return to the physical example of wave propagation in a string. We found that the general solution can be represented as a sum over product solutions. We will restrict our discussion to the special case that the initial velocity is zero and the original profile is given by $$u(x, 0)=f(x)$$. The solution is then $u(x, t)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi x}{L} \cos \frac{n \pi c t}{L}\label{eq:1}$ satisfying $f(x)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi x}{L} .\label{eq:2}$ We have seen that the Fourier sine series coefficients are given by $A_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} d x .\label{eq:3}$
We can rewrite this solution in a more compact form. First, we define the wave numbers, $k_{n}=\frac{n \pi}{L}, \quad n=1,2, \ldots,\nonumber$ and the angular frequencies, $\omega_{n}=c k_{n}=\frac{n \pi c}{L} .\nonumber$ Then, the product solutions take the form $\sin k_{n} x \cos \omega_{n} t\nonumber$ Using trigonometric identities, these products can be written as $\sin k_{n} x \cos \omega_{n} t=\frac{1}{2}\left[\sin \left(k_{n} x+\omega_{n} t\right)+\sin \left(k_{n} x-\omega_{n} t\right)\right] .\nonumber$ Inserting this expression in the solution, we have $u(x, t)=\frac{1}{2} \sum_{n=1}^{\infty} A_{n}\left[\sin \left(k_{n} x+\omega_{n} t\right)+\sin \left(k_{n} x-\omega_{n} t\right)\right] .\label{eq:4}$
Since $$\omega_{n}=c k_{n}$$, we can put this into a more suggestive form: $u(x, t)=\frac{1}{2}\left[\sum_{n=1}^{\infty} A_{n} \sin k_{n}(x+c t)+\sum_{n=1}^{\infty} A_{n} \sin k_{n}(x-c t)\right] .\label{eq:5}$
We see that each sum is simply the sine series for $$f(x)$$ but evaluated at either $$x+c t$$ or $$x-c t$$. Thus, the solution takes the form $u(x, t)=\frac{1}{2}[f(x+c t)+f(x-c t)] .\label{eq:6}$ If $$t=0$$, then we have $$u(x, 0)=\frac{1}{2}[f(x)+f(x)]=f(x)$$. So, the solution satisfies the initial condition. At $$t=1$$, the sum has a term $$f(x-c)$$.
Recall from your mathematics classes that this is simply a shifted version of $$f(x)$$. Namely, it is shifted to the right. For general times, the function is shifted by $$c t$$ to the right. For larger values of $$t$$, this shift is further to the right. The function (wave) shifts to the right with velocity $$c$$. Similarly, $$f(x+c t)$$ is a wave traveling to the left with velocity $$-c$$.
Thus, the waves on the string consist of waves traveling to the right and to the left. However, the story does not stop here. We have a problem when needing to shift $$f(x)$$ across the boundaries. The original problem only defines $$f(x)$$ on $$[0, L]$$. If we are not careful, we would think that the function leaves the interval leaving nothing left inside. However, we have to recall that our sine series representation for $$f(x)$$ has a period of $$2 L$$. So, before we apply this shifting, we need to account for its periodicity. In fact, being a sine series, we really have the odd periodic extension of $$f(x)$$ being shifted. The details of such analysis would take us too far from our current goal. However, we can illustrate this with a few figures.
We begin by plucking a string of length $$L$$. This can be represented by the function $f(x)=\left\{\begin{array}{cc} \frac{x}{a} & 0 \leq x \leq a \\ \frac{L-x}{L-a} & a \leq x \leq L \end{array}\right.\label{eq:7}$ where the string is pulled up one unit at $$x=a$$. This is shown in Figure $$\PageIndex{1}$$.
Next, we create an odd function by extending the function to a period of $$2 L$$. This is shown in Figure $$\PageIndex{2}$$.
Finally, we construct the periodic extension of this to the entire line. In Figure $$\PageIndex{3}$$ we show in the lower part of the figure copies of the periodic extension, one moving to the right and the other moving to the left. (Actually, the copies are $$\frac{1}{2} f(x \pm c t)$$.) The top plot is the sum of these solutions. The physical string lies in the interval $$[0,1]$$. Of course, this is better seen when the solution is animated.
The time evolution for this plucked string is shown for several times in Figure $$\PageIndex{4}$$. This results in a wave that appears to reflect from the ends as time increases.
The relation between the angular frequency and the wave number, $$\omega=$$ $$c k$$, is called a dispersion relation. In this case $$\omega$$ depends on $$k$$ linearly. If one knows the dispersion relation, then one can find the wave speed as $$c=\frac{\omega}{k}$$. In this case, all of the harmonics travel at the same speed. In cases where they do not, we have nonlinear dispersion, which we will discuss later.
This page titled 3.6: Finite Length Strings is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform. | 3,206 | 9,195 | {"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": 1, "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} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.198121 |
http://www.codeforge.com/read/125870/Example_2.m__html | 1,455,187,308,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701161942.67/warc/CC-MAIN-20160205193921-00059-ip-10-236-182-209.ec2.internal.warc.gz | 350,991,727 | 13,847 | Home » Source Code » » Example_2.m
## Example_2.m ( File view )
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clear;
alpha=0.5; % Stepsize of NLMS
mu_lms=0.3; % Stepsize of LMS
M=5;
k=0:99;
x=cos(pi*k/100).*sin(pi*k/5);
b=[1,0,-1];
a=[1,1,0.9];
d=filter(b,a,x);
delta=0.05;
[e,w,mu_k]=nlms(alpha,M,x,d,delta);
[e_lms,w_lms]=lms(mu_lms,M,x,d);
[c,i]=max(mu_k);
%------------------------------------------------
subplot(3,1,1);
stem(k,x,'.');
hold on;
plot(k,cos(pi*k/100),'--','color','m');
title('Normalized LMS Method','fontsize',18)
ylabel('x(k)','fontsize',14);
grid on;
subplot(3,1,2);
stem(k,mu_k,'.');
text(i,c,'\leftarrow Peak Value');
ylabel('\mu(k)','fontsize',14);
grid on;
subplot(3,1,3);
stem(k,abs(e),'.');
ylabel('e(k)','fontsize',14);
xlabel('k','fontsize',14);
grid on;
%-------------------------------------------
mu_lms_vec=mu_lms*ones(1,100);
figure(2);
subplot(2,1,1);
stem(k,mu_k,'.');
ylim([0,10]);
title('Difference between NLMS and LMS','fontsize',18);
text(i,c,'\leftarrow Peak Value');
ylabel('\mu(k)','fontsize',14);
xlabel('NLMS','fontsize',14);
grid on;
subplot(2,1,2);
stem(k,mu_lms_vec,'.');
ylim([0,10]);
ylabel('\mu','fontsize',14);
xlabel('LMS','fontsize',14);
grid on;
%-------------------------------------------
y_nlms=filter(w,1,x);
y_lms=filter(w_lms,1,x);
% Compute the Frequency Response
D=fft(d);
Y_nlms=fft(y_nlms);
Y_lms=fft(y_lms);
f=0:0.01:0.99;
figure(3)
subplot(2,1,1);
plot(k,d,k,y_nlms,'r',k,y_lms,'g');
title('The waveform in time domain','fontsize',16);
ylabel('Amplitude','fontsize',14);
subplot(2,1,2);
plot(f,abs(D),f,abs(Y_nlms),'r',f,abs(Y_lms),'g');
title('Magnitude Response','fontsize',16);
xlabel('Normalized frequence','fontsize',12);
grid on;
figure(4)
subplot(2,1,1);
stem(k,abs(e),'.');
title('Error Comparison between NLMS and LMS','fontsize',18);
ylabel('e(k)','fontsize',14);
xlabel('NLMS','fontsize',14);
grid on;
subplot(2,1,2)
stem(k,abs(e_lms),'.')
ylabel('e(k)','fontsize',14);
xlabel('LMS','fontsize',14);
grid on;
...
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Favorite by Ctrl+D | 930 | 2,767 | {"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} | 2.8125 | 3 | CC-MAIN-2016-07 | longest | en | 0.115256 |
https://www.theaverageteacher.com/tag/math-activities/ | 1,656,793,143,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00479.warc.gz | 1,093,101,211 | 37,812 | • Fun Statistics Activities for Kids I know you probably read the title of this post and asked yourself why elementary students would ever need to know statistics. Well, they don’t exactly, but that doesn’t mean that they wouldn’t still benefit from some subtle exposure to the concept! April is National Statistics & Mathematics Awareness Month, and is the perfect opportunity to add in some fun statistics activities to your instruction. If this is something you would like to do, here are a few (age appropriate) ideas! 1. Play SNAKE. SNAKE is a fun math dice game that loosely involves probability, as well as a little bit of risk! This game…
• Easter Activities for Kids Easter is one of my favorite times in the classroom! I am 100% here for all the Easter eggs and candy and Peeps. I know Peeps are controversial, but I LOVE them. Speaking of Peeps…did you know you can use them in several Easter activities for kids? Yep! So if you’re gifted a bunch in the next couples weeks but don’t want to eat them, you can turn them into a fun activity instead! If you do like Peeps, don’t worry – there are activities you can do that don’t involve Peeps, as well! So you can have your Peeps and eat them, too. 🙂 Anyway,…
• Celebrating Pi Day at School One of my favorite math holidays is Pi Day (March 14th, in honor of 3.14). Although, in my house we treat it more like PIE Day! Typically, we have chicken pot pie (our favorite!) for dinner, and then some other form of pie for dessert. However, for teachers, celebrating Pi Day can be fun in the classroom as well. Since this fun day is just around the corner, today I wanted to share some Pi Day ideas for the classroom. If your students are too young to learn about the actual number pi (which, if you’re a regular reader of this blog, they probably are!),…
• Building Problem Solving Skills for Kids Problem solving is one of the key foundational aspects of math. At its core, math isn’t about numbers – it’s about solving problems! Naturally, kids with strong problem solving skills have an easier time in math class. Problem solving comes naturally to some students, but not for others. However, there are things we as teachers can do to help build strong problem solving skills in our students. If this is something you think would benefit your students, here are a few ideas for carrying this out. Problem Solving Skills for Kids Activities Each of the following activities are helpful for building students’ problem solving…
• Valentine’s Day Activities for Kids Valentine’s Day is just around the corner, and if you’re like me, you can’t wait to do all the heart-themed activities! If you know me, you know I’m a big fan of easy, low-prep, and hands-on activities. So, to help you out this February, I compiled a list of a few of my favorite Valentine’s Day activities for kids that are up to my standards! 3 Valentine’s Day Activities for Kids 1. Candy Heart Towers This game/STEM activity is SO simple, and all you need are some of those jumbo sized candy hearts! (You can do them with small ones, too, but I personally prefer…
• Are you looking for fun, versatile games for the classroom? I know I always am! My favorite thing ever is taking a super simple, low-prep game and turning it into something educational. The BEST games are ones that can be easily adapted to use for any subject. Today I wanted to share with you one of my favorite, versatile games for the classroom… Musical chairs! Yes! Musical chairs! Everyone knows how to play it, it’s SO low-prep (just need some chairs, music, and a worksheet). And the best part is you can play this game in ANY subject! I obviously used it in math, but since it’s a game that’s…
• Do you use math stories in your instruction? We all know the importance of reading for our students. Obviously, teachers spend a lot of time reading in subjects like language arts, and even social studies – but math is probably the most challenging subject when it comes to integrating reading. However, it isn’t impossible! One of my favorite ways to integrate reading with math is through the use of math stories. There are so many amazing math story books out there already, or you can always write your own! (Or have your students write some). If this is something that interests you, keep reading for some ideas for using math…
• Brilliant Elementary Math Ideas Are you on teacher Instagram? If not, you should be! Instagram is one of my favorite ways to find amazing ideas for teaching elementary math. When you connect with hundreds (or thousands) of teachers across the globe, you learn so much! Today, I wanted to share some of the most GENIUS elementary teaching tips and ideas for math that I’ve found on Instagram. If you like the ideas, too, then click the links for each one and give them a follow! 1. Teaching Times Tables I love this tip from @perfectforprimary! Instead of teaching times tables in order from 1-12, she teaches them in a more…
• Using Math Bingo in the Classroom I don’t know about you, but my students always went crazy over bingo! It’s such a simple game, but what can I say? They loved it. Bingo was a regular go-to to help review educational concepts, or just for fun during holidays are class parties! I especially loved using math bingo to support my students in math. Obviously, bingo is a fun game – but can it really be all that educational? Or is it all just fluff? Personally, I believe that bingo can be an incredibly rigorous activity – as long as it’s used correctly! Let’s explore how we can use bingo in…
• Fun Math Activities with Cups There are so many great math activities out there for students. Between Amazon, the Target Dollar Spot, and your local teacher store, it isn’t hard to spend a few hundred bucks on math games and activities for your students. Unfortunately, our teacher salary doesn’t exactly give us the flexibility to drop that kind of cash…so, finding budget-friendly solutions for fun math activities is important! Fortunately, there are many activities you can make yourself for a fraction of the cost of a store-bought equivalent! I call these “math DIYs.” Previously, I shared some activities that you can make out of paper plates. Today, I want to… | 1,351 | 6,228 | {"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-2022-27 | longest | en | 0.951526 |
http://reference.wolfram.com/language/ref/JacobiP.html | 1,490,849,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00553-ip-10-233-31-227.ec2.internal.warc.gz | 298,426,979 | 10,164 | # JacobiP
JacobiP[n,a,b,x]
gives the Jacobi polynomial .
# Details
• Mathematical function, suitable for both symbolic and numerical manipulation.
• Explicit polynomials are given when possible.
• satisfies the differential equation .
• The Jacobi polynomials are orthogonal with weight function .
• For certain special arguments, JacobiP automatically evaluates to exact values.
• JacobiP can be evaluated to arbitrary numerical precision.
• JacobiP automatically threads over lists.
• JacobiP[n,a,b,z] has a branch cut discontinuity in the complex z plane running from to .
# Examples
open allclose all
## Basic Examples(2)
Compute the 2 Jacobi polynomial:
In[1]:=
Out[1]=
Plot :
In[1]:=
Out[1]= | 166 | 712 | {"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} | 2.5625 | 3 | CC-MAIN-2017-13 | longest | en | 0.636424 |
https://livetvguide.net/free-body-diagram-static.html | 1,560,681,342,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00181.warc.gz | 507,371,151 | 8,254 | # Free Body Diagram Static
Free body diagram static Autodesk forceeffect is a new mobile simulation app for ipad that forceeffect provides an intuitive environment for drawing constraining and simulating concepts using free body diagrams by simply The ladder has a mass of 16 kg. The floor has friction with static friction coefficient between wall and ladder of 0.4. A man of mass 70 kg climbs the ladder. A draw a free body diagram of the The best way to look at the problem is to create a force diagram also called a free body diagram. In physics a free body diagram shows but i will first assume that the static friction pushing up.
Free body diagram static Fundamental concepts and conditions of static free body diagram is essential in the solution of problems concerning the equilibrium. Let us consider the equilibrium of a body in space. In this Dublin business wire research and marketshttpwww.researchandmarkets.comresearchedbea6fundamentalsofma has announced the addition of john wiley and sons ltds new book quotfundamentals of The problem can be readily solved using simple free body diagrams and summation of forces in two directions. But developing the solution using fea can be tricky. How do you create an fea idealization.
Free body diagram static The free body diagram method is emphasized. Topics include vector algebra force moment of force couples static equilibrium of rigid bodies trusses friction properties of areas shear and moment Statics is one main branch of mechanics and deals with forces on bodies which are at rest static equilibrium drawn closing the vector diagram and connecting the finishing point to the starting Const http requirehttp express requireexpress app express server requirehttp.serverapp bodyparser requirebody parser webpageport app.usebodyparser.json.
A group of chinese researchers reckon theyre close to a practical method of harvesting the static electricity you surface triboelectric generator the body is used as an electrode as per this Draw a free body diagram for the block. Make the eq mus eq is the coefficient of static friction between the block and the incline. It can be found by solving for static equilibrium made easier with free body diagrams for key points of force interaction at the box slider and belt and head pulley. Loads in our example are due.
It's possible to get or download caterpillar-wiring diagram from several websites. If you take a close look at the diagram you will observe the circuit includes the battery, relay, temperature sensor, wire, and a control, normally the engine control module. With an extensive collection of electronic symbols and components, it's been used among the most completed, easy and useful wiring diagram drawing program. Free Body Diagram Static. The wiring diagram on the opposite hand is particularly beneficial to an outside electrician. Sometimes wiring diagram may also refer to the architectural wiring program. The simplest approach to read a home wiring diagram is to begin at the source, or the major power supply. Basically, the home wiring diagram is simply utilized to reveal the DIYer where the wires are.
If you can't locate the information, get in touch with the manufacturer. The info in the diagram doesn't indicate a power or ground supply. The intention of the fuse is to safeguard the wiring and electrical components on its circuit. A typical watch's basic objective is to tell you the good time of day. When selecting the best type of computer cable to fulfill your requirements, it is very important to consider your upcoming technology plans.
Installing a tachometer on your Vehicles can assist in preventing critical repair problems, however. You might have a weak ground issue. The way the brain learns is a subject that still requires a good deal of study. How it learns can be associated by how it is able to create memories.
In a parallel circuit, each unit is directly linked to the power supply, so each system gets the exact voltage. There are 3 basic sorts of standard light switches. The circuit needs to be checked with a volt tester whatsoever points. Free Body Diagram Static. Each circuit displays a distinctive voltage condition. You are able to easily step up the voltage to the necessary level utilizing an inexpensive buck-boost transformer and steer clear of such issues. The voltage is the sum of electrical power produced by the battery. Be sure that the new fuse isn't blown, and carries the very same amperage.
Each fuse is going to have a suitable amp rating for those devices it's protecting. The wiring is merely a bit complicated. Our automotive wiring diagrams permit you to relish your new mobile electronics in place of spend countless hours attempting to work out which wires goes to which Ford part or component. Overall the wiring is really straight forward. There's a lot wiring that you've got to tie into your truck's wiring harness, but it's much easier to do than it seems. A ground wire offers short circuit protection and there's no neutral wire used. There's one particular wire leading from the distributor which may be used for the tachometer.
When you have just a single cable going into the box, you're at the close of the run, and you've got the simplest scenario possible. All trailer plugs and sockets are extremely easy to wire. The adapter has the essential crosslinks between the signals. Wiring a 7-pin plug on your truck can be a bit intimidating when you're looking at it from beyond the box.
The control box may have over three terminals. After you have the correct size box and have fed the cable to it, you're almost prepared to permit the wiring begin. Then there's also a fuse box that's for the body controls that is situated under the dash. Free Body Diagram Static. You will find that every circuit has to have a load and every load has to have a power side and a ground side. Make certain that the transformer nameplate power is enough to supply the load that you're connecting.
The bulb has to be in its socket. Your light can be wired to the receiver and don't require supply additional capacity to light as it can get power from receiver. In the event the brake lights aren't working, a police officer may block the vehicle and issue a warning to create the repair within a particular time limit. Even though you would still must power the relay with a power source or battery. Verify the power is off before trying to attach wires. In case it needs full capacity to begin, it won't operate in any way.
Replacing thermostat on your own without a Denver HVAC technician can be quite harrowing if you don't hook up the wiring correctly. After the plumbing was cut out, now you can get rid of the old pool pump. It's highly recommended to use a volt meter to make sure there is no voltage visiting the motor, sometimes breakers do not get the job done properly, also you might have turned off the incorrect breaker. Remote distance is left up to 500m. You may use a superior engine ground. The second, that's the most frequently encountered problem, is a weak ground in the computer system. Diagnosing an electrical short can be extremely tough and costly.
Solving free-body diagrams statics free body diagram cartesian kinematics free body diagram wrench statics free-body diagrams static friction free body diagram free body diagram symbols free body diagram statics bridge statics free-body diagrams frame. | 1,457 | 7,428 | {"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-2019-26 | latest | en | 0.849325 |
https://grahamhancock.com/phorum/read.php?1,1109890,1149675 | 1,627,811,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00618.warc.gz | 298,562,343 | 13,461 | Mysteries : The Official GrahamHancock.com forums
For serious discussion of the controversies, approaches and enigmas surrounding the origins and development of the human species and of human civilization. (NB: for more ‘out there’ posts we point you in the direction of the ‘Paranormal & Supernatural’ Message Board).
From the previous post;
13.965, -87.9 (NGC 2169)
13965 - 879 = "13086"
(13)_ (965 - 879) = "86"
(13_86) = "1386"
The #s in front of the decimal points are; 13_87
13 + 87
= 100 _ (1)
965 + 879 = 1844
1844 + 1 = "1845"..............(of course 1 added to the # 8 would make "1944")
1386 + 1845 = "3231"
32 * 31 = 992
Reverse the #s and multiply as above, the answer is reversed.
(13 * 23) = 992
992 + 299 = 1291
1291 * 1921 = 2480011
(248 + 11) = 259
25 + 59 = " 84"
(95 + 52) = "147"
84 + 147 = "231"
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 986 DUNE 24-Jul-17 15:58
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 929 Horatech 25-Jul-17 06:32
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 906 Horatech 10-Aug-17 11:36
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 758 Horatech 11-Sep-17 07:23
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 697 Horatech 27-Oct-17 15:27
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 838 Horatech 27-Oct-17 18:31
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 884 Horatech 10-Oct-17 05:12
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 766 Horatech 30-Oct-17 06:11
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 844 Horatech 21-Nov-17 09:46
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 749 Horatech 07-Dec-17 09:49
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 645 Horatech 12-Apr-18 02:00
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 802 Horatech 29-Nov-17 05:30
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 772 Horatech 07-Jan-18 13:48
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 621 Horatech 31-Jan-18 11:57
wake up 826 DavidK 04-Feb-18 22:53
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Re: wake up 625 molder 05-Feb-18 12:53
Re: wake up 632 DavidK 05-Feb-18 14:55
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Re: wake up 654 Horatech 05-Feb-18 21:18
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Re: wake up 343 Horatech 16-May-19 23:44
Re: wake up (Attention Eddie Larry) 311 Horatech 19-May-19 07:11
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 717 DavidK 31-Jan-18 16:58
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 692 Horatech 03-Feb-18 07:26
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 603 Horatech 13-Feb-18 04:29
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 628 Horatech 07-Feb-18 05:33
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 636 DavidK 07-Feb-18 09:23
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 661 Horatech 07-Feb-18 09:26
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 637 Horatech 21-Feb-18 03:04
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Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 592 Horatech 21-Feb-18 23:31
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 571 Horatech 25-Feb-18 08:23
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 613 Horatech 26-Feb-18 00:22
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 606 Horatech 27-Feb-18 07:27
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 625 Horatech 28-Feb-18 04:14
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2/SOLight 606 Horatech 09-Mar-18 08:16
Using Scottish units - the fall 18.5 units 696 DavidK 09-Mar-18 12:16
Re: Using Scottish units - the fall 18.5 units 581 Horatech 16-Mar-18 09:29
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 632 Horatech 16-Mar-18 14:00
Re: GEOMETRICAL INSPIRATIONS THE CODE _ 2 601 Horatech 16-Mar-18 17:56
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Sorry, you can't reply to this topic. It has been closed. | 3,855 | 9,420 | {"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-2021-31 | longest | en | 0.518818 |
https://number.academy/111284 | 1,720,822,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00861.warc.gz | 338,384,770 | 11,493 | # Number 111284 facts
The even number 111,284 is spelled 🔊, and written in words: one hundred and eleven thousand, two hundred and eighty-four. The ordinal number 111284th is said 🔊 and written as: one hundred and eleven thousand, two hundred and eighty-fourth. Color #111284. The meaning of the number 111284 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 111284. What is 111284 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 111284.
## Interesting facts about the number 111284
### Asteroids
• (111284) 2001 XH42 is asteroid number 111284. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 12/9/2001.
## What is 111,284 in other units
The decimal (Arabic) number 111284 converted to a Roman number is (C)(X)MCCLXXXIV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
111284 seconds equals to 1 day, 6 hours, 54 minutes, 44 seconds
111284 minutes equals to 2 months, 3 weeks, 6 hours, 44 minutes
### Codes and images of the number 111284
Number 111284 morse code: .---- .---- .---- ..--- ---.. ....-
Sign language for number 111284:
Number 111284 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 111284 is not a prime number. The closest prime numbers are 111271, 111301.
#### Factorization and factors (dividers)
The prime factors of 111284 are 2 * 2 * 43 * 647
The factors of 111284 are 1, 2, 4, 43, 86, 172, 647, 1294, 2588, 27821, 55642, 111284.
Total factors 12.
Sum of factors 199584 (88300).
#### Powers
The second power of 1112842 is 12.384.128.656.
The third power of 1112843 is 1.378.155.373.354.304.
#### Roots
The square root √111284 is 333,592566.
The cube root of 3111284 is 48,099908.
#### Logarithms
The natural logarithm of No. ln 111284 = loge 111284 = 11,619841.
The logarithm to base 10 of No. log10 111284 = 5,046433.
The Napierian logarithm of No. log1/e 111284 = -11,619841.
### Trigonometric functions
The cosine of 111284 is -0,804141.
The sine of 111284 is 0,594439.
The tangent of 111284 is -0,739223.
## Number 111284 in Computer Science
Code typeCode value
PIN 111284 It's recommended that you use 111284 as your password or PIN.
111284 Number of bytes108.7KB
CSS Color
#111284 hexadecimal to red, green and blue (RGB) (17, 18, 132)
Unix timeUnix time 111284 is equal to Friday Jan. 2, 1970, 6:54:44 a.m. GMT
IPv4, IPv6Number 111284 internet address in dotted format v4 0.1.178.180, v6 ::1:b2b4
111284 Decimal = 11011001010110100 Binary
111284 Decimal = 12122122122 Ternary
111284 Decimal = 331264 Octal
111284 Decimal = 1B2B4 Hexadecimal (0x1b2b4 hex)
111284 BASE64MTExMjg0
111284 MD547bb42d5daaab5bfea17a96e858b5a7f
111284 SHA224d28a9e2ac273391370d287bbd209a8df4cbcabd3d1cfd242e3c8892c
111284 SHA2568565f92ba791364fd2c438c5197eb9e19373444de27894482e1eb4214fa98010
More SHA codes related to the number 111284 ...
If you know something interesting about the 111284 number that you did not find on this page, do not hesitate to write us here.
## Numerology 111284
### Character frequency in the number 111284
Character (importance) frequency for numerology.
Character: Frequency: 1 3 2 1 8 1 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 111284, the numbers 1+1+1+2+8+4 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## № 111,284 in other languages
How to say or write the number one hundred and eleven thousand, two hundred and eighty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 111.284) ciento once mil doscientos ochenta y cuatro German: 🔊 (Nummer 111.284) einhundertelftausendzweihundertvierundachtzig French: 🔊 (nombre 111 284) cent onze mille deux cent quatre-vingt-quatre Portuguese: 🔊 (número 111 284) cento e onze mil, duzentos e oitenta e quatro Hindi: 🔊 (संख्या 111 284) एक लाख, ग्यारह हज़ार, दो सौ, चौरासी Chinese: 🔊 (数 111 284) 十一万一千二百八十四 Arabian: 🔊 (عدد 111,284) مائة و أحد عشر ألفاً و مئتان و أربعة و ثمانون Czech: 🔊 (číslo 111 284) sto jedenáct tisíc dvěstě osmdesát čtyři Korean: 🔊 (번호 111,284) 십일만 천이백팔십사 Danish: 🔊 (nummer 111 284) ethundrede og ellevetusindtohundrede og fireogfirs Hebrew: (מספר 111,284) מאה ואחד עשר אלף מאתיים שמונים וארבע Dutch: 🔊 (nummer 111 284) honderdelfduizendtweehonderdvierentachtig Japanese: 🔊 (数 111,284) 十一万千二百八十四 Indonesian: 🔊 (jumlah 111.284) seratus sebelas ribu dua ratus delapan puluh empat Italian: 🔊 (numero 111 284) centoundicimiladuecentottantaquattro Norwegian: 🔊 (nummer 111 284) en hundre og elleve tusen to hundre og åttifire Polish: 🔊 (liczba 111 284) sto jedenaście tysięcy dwieście osiemdziesiąt cztery Russian: 🔊 (номер 111 284) сто одиннадцать тысяч двести восемьдесят четыре Turkish: 🔊 (numara 111,284) yüzonbinikiyüzseksendört Thai: 🔊 (จำนวน 111 284) หนึ่งแสนหนึ่งหมื่นหนึ่งพันสองร้อยแปดสิบสี่ Ukrainian: 🔊 (номер 111 284) сто одинадцять тисяч двісті вісімдесят чотири Vietnamese: 🔊 (con số 111.284) một trăm mười một nghìn hai trăm tám mươi bốn Other languages ...
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https://gitlab.mpi-sws.org/swasey/coq-stdpp/-/commit/82853b401ca2640fcdaddd41fb4c4e50b5b564e0 | 1,618,347,889,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038074941.13/warc/CC-MAIN-20210413183055-20210413213055-00593.warc.gz | 389,254,087 | 33,384 | Commit 82853b40 by Hai Dang Committed by Robbert Krebbers
### generalize filter from gmap to fin_map
parent eecf7526
... ... @@ -19,6 +19,13 @@ Class FinMapDom K M D `{∀ A, Dom (M A) D, FMap M, Section fin_map_dom. Context `{FinMapDom K M D}. Lemma dom_map_filter {A} (P : K * A → Prop) `{!∀ x, Decision (P x)} (m : M A): dom D (filter P m) ⊆ dom D m. Proof. intros ?. rewrite 2!elem_of_dom. destruct 1 as [?[Eq _]%map_filter_lookup_Some]. by eexists. Qed. Lemma elem_of_dom_2 {A} (m : M A) i x : m !! i = Some x → i ∈ dom D m. Proof. rewrite elem_of_dom; eauto. Qed. Lemma not_elem_of_dom {A} (m : M A) i : i ∉ dom D m ↔ m !! i = None. ... ...
... ... @@ -130,6 +130,9 @@ is unspecified. *) Definition map_fold `{FinMapToList K A M} {B} (f : K → A → B → B) (b : B) : M → B := foldr (curry f) b ∘ map_to_list. Instance map_filter `{FinMap K M} {A} : Filter (K * A) (M A) := λ P _, map_fold (λ k v m, if decide (P (k,v)) then <[k := v]>m else m) ∅. (** * Theorems *) Section theorems. Context `{FinMap K M}. ... ... @@ -1002,6 +1005,67 @@ Proof. assert (m !! j = Some y) by (apply Hm; by right). naive_solver. Qed. (** ** The filter operation *) Section map_Filter. Context {A} (P : K * A → Prop) `{!∀ x, Decision (P x)}. Lemma map_filter_lookup_Some: ∀ m k v, filter P m !! k = Some v ↔ m !! k = Some v ∧ P (k,v). Proof. apply (map_fold_ind (λ m1 m2, ∀ k v, m1 !! k = Some v ↔ m2 !! k = Some v ∧ P _)). - setoid_rewrite lookup_empty. naive_solver. - intros k v m m' Hm Eq k' v'. case_match; case (decide (k' = k))as [->|?]. + rewrite 2!lookup_insert. naive_solver. + do 2 (rewrite lookup_insert_ne; [|auto]). by apply Eq. + rewrite Eq, Hm, lookup_insert. split; [naive_solver|]. destruct 1 as [Eq' ]. inversion Eq'. by subst. + by rewrite lookup_insert_ne. Qed. Lemma map_filter_lookup_None: ∀ m k, filter P m !! k = None ↔ m !! k = None ∨ ∀ v, m !! k = Some v → ¬ P (k,v). Proof. intros m k. rewrite eq_None_not_Some. unfold is_Some. setoid_rewrite map_filter_lookup_Some. naive_solver. Qed. Lemma map_filter_lookup_equiv m1 m2: (∀ k v, P (k,v) → m1 !! k = Some v ↔ m2 !! k = Some v) → filter P m1 = filter P m2. Proof. intros HP. apply map_eq. intros k. destruct (filter P m2 !! k) as [v2|] eqn:Hv2; [apply map_filter_lookup_Some in Hv2 as [Hv2 HP2]; specialize (HP k v2 HP2) |apply map_filter_lookup_None; right; intros v EqS ISP; apply map_filter_lookup_None in Hv2 as [Hv2|Hv2]]. - apply map_filter_lookup_Some. by rewrite HP. - specialize (HP _ _ ISP). rewrite HP, Hv2 in EqS. naive_solver. - apply (Hv2 v); [by apply HP|done]. Qed. Lemma map_filter_lookup_insert m k v: P (k,v) → <[k := v]> (filter P m) = filter P (<[k := v]> m). Proof. intros HP. apply map_eq. intros k'. case (decide (k' = k)) as [->|?]; [rewrite lookup_insert|rewrite lookup_insert_ne; [|auto]]. - symmetry. apply map_filter_lookup_Some. by rewrite lookup_insert. - destruct (filter P (<[k:=v]> m) !! k') eqn: Hk; revert Hk; [rewrite map_filter_lookup_Some, lookup_insert_ne; [|by auto]; by rewrite <-map_filter_lookup_Some |rewrite map_filter_lookup_None, lookup_insert_ne; [|auto]; by rewrite <-map_filter_lookup_None]. Qed. Lemma map_filter_empty : filter P ∅ = ∅. Proof. apply map_fold_empty. Qed. End map_Filter. (** ** Properties of the [map_Forall] predicate *) Section map_Forall. Context {A} (P : K → A → Prop). ... ...
... ... @@ -227,81 +227,6 @@ Proof. - by rewrite option_guard_False by (rewrite not_elem_of_dom; eauto). Qed. (** Filter *) (* This filter creates a submap whose (key,value) pairs satisfy P *) Instance gmap_filter `{Countable K} {A} : Filter (K * A) (gmap K A) := λ P _, map_fold (λ k v m, if decide (P (k,v)) then <[k := v]>m else m) ∅. Section filter. Context `{Countable K} {A} (P : K * A → Prop) `{!∀ x, Decision (P x)}. Implicit Type (m: gmap K A) (k: K) (v: A). Lemma gmap_filter_lookup_Some: ∀ m k v, filter P m !! k = Some v ↔ m !! k = Some v ∧ P (k,v). Proof. apply (map_fold_ind (λ m1 m2, ∀ k v, m1 !! k = Some v ↔ m2 !! k = Some v ∧ P _)). - naive_solver. - intros k v m m' Hm Eq k' v'. case_match; case (decide (k' = k))as [->|?]. + rewrite 2!lookup_insert. naive_solver. + do 2 (rewrite lookup_insert_ne; [|auto]). by apply Eq. + rewrite Eq, Hm, lookup_insert. split; [naive_solver|]. destruct 1 as [Eq' ]. inversion Eq'. by subst. + by rewrite lookup_insert_ne. Qed. Lemma gmap_filter_lookup_None: ∀ m k, filter P m !! k = None ↔ m !! k = None ∨ ∀ v, m !! k = Some v → ¬ P (k,v). Proof. intros m k. rewrite eq_None_not_Some. unfold is_Some. setoid_rewrite gmap_filter_lookup_Some. naive_solver. Qed. Lemma gmap_filter_dom m: dom (gset K) (filter P m) ⊆ dom (gset K) m. Proof. intros ?. rewrite 2!elem_of_dom. destruct 1 as [?[Eq _]%gmap_filter_lookup_Some]. by eexists. Qed. Lemma gmap_filter_lookup_equiv m1 m2: (∀ k v, P (k,v) → m1 !! k = Some v ↔ m2 !! k = Some v) → filter P m1 = filter P m2. Proof. intros HP. apply map_eq. intros k. destruct (filter P m2 !! k) as [v2|] eqn:Hv2; [apply gmap_filter_lookup_Some in Hv2 as [Hv2 HP2]; specialize (HP k v2 HP2) |apply gmap_filter_lookup_None; right; intros v EqS ISP; apply gmap_filter_lookup_None in Hv2 as [Hv2|Hv2]]. - apply gmap_filter_lookup_Some. by rewrite HP. - specialize (HP _ _ ISP). rewrite HP, Hv2 in EqS. naive_solver. - apply (Hv2 v); [by apply HP|done]. Qed. Lemma gmap_filter_lookup_insert m k v: P (k,v) → <[k := v]> (filter P m) = filter P (<[k := v]> m). Proof. intros HP. apply map_eq. intros k'. case (decide (k' = k)) as [->|?]; [rewrite lookup_insert|rewrite lookup_insert_ne; [|auto]]. - symmetry. apply gmap_filter_lookup_Some. by rewrite lookup_insert. - destruct (filter P (<[k:=v]> m) !! k') eqn: Hk; revert Hk; [rewrite gmap_filter_lookup_Some, lookup_insert_ne; [|by auto]; by rewrite <-gmap_filter_lookup_Some |rewrite gmap_filter_lookup_None, lookup_insert_ne; [|auto]; by rewrite <-gmap_filter_lookup_None]. Qed. Lemma gmap_filter_empty `{Equiv A} : filter P (∅ : gmap K A) = ∅. Proof. apply map_fold_empty. Qed. End filter. (** * Fresh elements *) (* This is pretty ad-hoc and just for the case of [gset positive]. We need a notion of countable non-finite types to generalize this. *) ... ...
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http://mattegi.com/x-works-motion | 1,597,126,351,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738735.44/warc/CC-MAIN-20200811055449-20200811085449-00440.warc.gz | 73,474,128 | 5,811 | X Works Motion » mattegi.com
Digital Motion X-ray.
Motion X-Support These are MotionX's headquarters in Santa Cruz, California, Surf City. That's where you'll find us all building our technology and supporting our customers. Digital Motion X-ray - DMX is a medical imaging technology that visualizes the human body real-time while in motion. This tutorial will go over how to set up and run a basic motion simulation in Solidworks, how to use position plots, and export and use the data in other programs such as Excel and MATLAB. First we will start with opening the assembly and setting up the simulation workbench. 1. Open funnel and ball.asm 2. An unconstrained rigid body in space has six degrees of freedom: three translational and three rotational. It can move along its X, Y, and Z axes and rotate about its X, Y, and Z axes. Integration Methods. A set of coupled differential and algebraic equations define the equations of motion of a. X10 Protocol Basics: Wireless & Wired.Wireless: The heartbeat of a wireless X10 Home Automation set-up is the Wireless Transceiver TM751. The Transceiver receives Wireless RF signals from wireless remotes like a Slimline Switch SS13A then sends that signal through your homes existing powerlines.
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This effect can be seen by following the direction of motion of the two cables which comprise the mechanism. Note that all of the vertical arrows point in the same direction. You could imagine attaching a stepper motor to one of the pulleys. Now, the horizontal bar. Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. To solve projectile motion problems, perform the following steps: 1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. | 443 | 2,147 | {"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} | 3.09375 | 3 | CC-MAIN-2020-34 | latest | en | 0.902768 |
https://www.azcalculator.com/calc/distance-between-two-points(3-dimension)-.php | 1,708,532,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00743.warc.gz | 690,575,175 | 4,005 | Distance Between Two Points(3 Dimension) Calculator
Posted by Dinesh on
This calculator calculates the distance using x1, x2, y1, y2, z1, z2 values.
Distance Between Two Points(3 Dimension) Calculation
Formula:
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http://www.360doc.com/content/06/0323/09/872_84229.shtml | 1,579,387,311,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00348.warc.gz | 184,126,750 | 11,237 | 0 0
## 神奇的数字
2006-03-23
“数学是科学之母” 3 x 37 = 1116 x 37 = 2229 x 37 = 33312 x 37 = 44415 x 37 = 55518 x 37 = 66621 x 37 = 77724 x 37 = 88827 x 37 = 999 Nice one: 111,111,111 x 111,111,111 =12,345,678,987,654,321 Trapeze:1 x 9 + 2 = 1112 x 9 + 3 = 111123 x 9 + 4 = 11111234 x 9 + 5 = 1111112345 x 9 + 6 = 111111123456 x 9 + 7 = 11111111234567 x 9 + 8 = 1111111112345678 x 9 + 9 = 111111111 another Trapeze: 1 x 8 + 1 = 912 x 8 + 2 = 98123 x 8 + 3 = 9871234 x 8 + 4 = 987612345 x 8 + 5 = 98765123456 x 8 + 6 = 9876541234567 x 8 + 7 = 987654312345678 x 8 + 8 = 98765432123456789 x 8 + 9 = 987654321 and another one: 0 x 9 + 8 = 89 x 9 + 7 = 8898 x 9 + 6 = 888987 x 9 + 5 = 88889876 x 9 + 4 = 8888898765 x 9 + 3 = 888888987654 x 9 + 2 = 88888889876543 x 9 + 1 = 8888888898765432 x 9 + 0 = 888888888987654321 x 9 - 1 = 88888888889876543210 x 9 - 2 = 88888888888 祝 大 家 都 发！发！发！发！发！
0条评论 | 483 | 881 | {"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-2020-05 | latest | en | 0.392942 |
https://stats.stackexchange.com/questions/144335/the-effect-of-covariate-measurement-error-on-coefficients-in-regression-with-dum | 1,721,135,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00548.warc.gz | 481,874,761 | 43,137 | # The effect of covariate measurement error on coefficients in regression with dummy variables
I am trying to understand if I should be more concerned with covariate measurement error in linear models including dummy variables, than with all continuous predictors. Say I have a simple linear model with one independent variable (IV) and one covariate:
\begin{align} Y_i = \beta_0 +\beta_1X_i+\beta_2Z_i+\epsilon_i \end{align}
where $Y_i$ is the dependent variable, my error term $\epsilon_i \sim N(0,\sigma_{\epsilon}^2)$, $X_i$ is my independent variable, and $Z_i$ my covariate measured with a stochastic measurement error $\eta_i \sim N(0,\sigma_{\eta}^2)$. Does the measurement error affect the estimates of the effect of my IV: $\beta_1$ any worse for a dichotomous vs. continuous IV?
Bonus. How are the IV estimates affected by measurement errors when there is 1) a systematic measurement error of the covariate 2) more than one IV 3) more than one covariate with independent errors 4) more than one covariate but with correlated errors.
It seems to me there should not be a qualitative difference, but a colleague told me that ANCOVA results are very unstable with respect to measurement errors in the covariates, as opposed to regression. This worries me because I often use dummy variables in a regression where the covariates have a sizable measurement error.
My attempt at a solution.
say $y$, $x_1...x_{n-1}$ and $z = z^*+u$ are particular realizations of the variables in the model, and u is the measurement error. Using the standard formula for the estimates $$\hat\beta = (X^TX)^{-1}X^Ty$$ so $$\hat\beta_i= \frac{1}{2|\text{cov}(X,X)|}\sum_{j=1}^n \epsilon_{jpq}\epsilon_{ikl}\text{cov}_{pk}(X,X)\text{cov}_{ql}(X,X)\text{cov}_{j1}(X,y)$$
where the permutation symbol $$\epsilon_{ijk} = \left\{ \begin{array} \\ 1 & i,j,k = 1,2,3; 2,3,1 \text{or} 3,1,2 \\ -1 & i,j,k = 3,2,1; 2,1,3 \text{or} 1,3,2 \\ 0 & \text{else}\end{array} \right.$$
now measurement error will only affect the $\hat \beta_i$ where the covariance matrix element cov$_{jk}(X,\cdot)$ includes the covariate with the error terms. That doesn't seem to be affected by whether we use a dichotomous or continuous variable. So the only thing that could be affected is the determinant in the denominator, so we probably should be weary of collinearity that can arise with many dummy variables, other than that I don't see how using dichotomous variables can be any worse.
## How does measurement error in a binary predictor affect bias?
Starting from the single variable case, we observe a binary variable $x$ which is the true predictor measured with error, $X = x + u$ (I drop the $i$ subscripts for convenience). For the sake of the illustration and because much of this work was done in that field, let's suppose $X$ is a disease and $x$ is a doctor's diagnosis. Let's define the following quantities:
• $P$ is the proportion of people in the population who truly have the disease
• $\tilde{P}$ is the proportion of people diagnosed with the disease according to our doctor, then $\tilde{Q} = (1-\tilde{P})$ is the proportion of those diagnosed as healthy
• $\eta$ is the proportion of people who truly have the disease but are classified as not having the disease
• $\nu$ is the proportion of people who are truly healthy but who are classified as having the disease
The errors-in-variables framework is then $$P = (1-\nu)\tilde{P} + \eta \tilde{Q}$$ in order to allow misclassification into both directions. From the set-up the marginal distributions of $X$ and $x$ are Bernoulli with parameter $P$ and $\tilde{P}$, respectively.
Savoca (2000) derives the quantities needed for evaluating the bias of OLS, \begin{align} E(u) &= \nu - (\eta + \nu)P \\ Var(u) &= \nu + (\eta - \nu)P - \left[\nu - (\eta + \nu)P\right]^2 \\ Cov(x,u) &= -(\eta + \nu)P(1-P) \end{align}
So as compared to the classical measurement error in a continuous explanatory variable the error here does not have zero mean unless $E(X) = E(x) = P$ - but this would mean that there is no misclassification in our diagnosis.
The corresponding coefficient from the above regression with one binary regressor with error would be $$\widehat{\beta} = \beta \left[ \frac{P(1-P)(1-\nu-\eta)}{\tilde{P}(1-\tilde{P})} \right]$$
The resulting bias, as in any other measurement error case, is towards zero. This has been shown as early as Aigner (1973).
## Is measurement error worse for binary or continuous predictors?
In terms of whether measurement error in a binary variable is worse than that in a continuous variable it is not necessarily obvious which case has a larger bias. Consider the following simulation exercise (using Stata). First we try the errors-in-variables framework with a binary predictor:
set seed 777
set obs 1000
* suppose the true P = 0.44
* generate our true X
gen X = rbinomial(1, 0.44)
* generate some error to be used in constructing the observed x
gen e = rnormal(0,1)
gen error = (e>2) | (e<-1.5)
* generate the observed x (with error)
gen x = X
replace x = 0 if X==1 & err==1
replace x = 1 if X==0 & err==1
* generate the dependent variable with true beta = 1.2
gen eps = rnormal(0,1)
gen y = 1 + 1.2*X + eps
* regression with measurement error
reg y x
The result is
Source | SS df MS Number of obs = 1000
-------------+------------------------------ F( 1, 998) = 180.45
Model | 208.290565 1 208.290565 Prob > F = 0.0000
Residual | 1152.00081 998 1.15430943 R-squared = 0.1531
Total | 1360.29137 999 1.36165303 Root MSE = 1.0744
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .9171922 .0682789 13.43 0.000 .7832055 1.051179
_cons | 1.109043 .0458538 24.19 0.000 1.019062 1.199024
------------------------------------------------------------------------------
So that's pretty far off the true value. The correlation between $X$ and $x$ here is 0.8 and if we consider a similar correlation for their continuous versions,
*specify a correlation matrix
matrix C = (1, 0.8, 0 \ 0.8, 1, 0 \ 0, 0, 1)
* simulate the data (fortunately here we can use a Stata function rather than doing all by hand)
corr2data X x e, n(10000) means(0.5 0.5 0) sds(0.5 0.5 1) corr(C)
gen y = 1 + 1.2*X + e
the result is
. reg y x
Source | SS df MS Number of obs = 10000
-------------+------------------------------ F( 1, 9998) = 2039.25
Model | 2303.76962 1 2303.76962 Prob > F = 0.0000
Residual | 11294.8704 9998 1.12971298 R-squared = 0.1694
Total | 13598.64 9999 1.36 Root MSE = 1.0629
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .96 .0212587 45.16 0.000 .9183288 1.001671
_cons | 1.12 .0150318 74.51 0.000 1.090535 1.149465
------------------------------------------------------------------------------
which is only marginally closer to the true value than what we had in the binary case. However, what will happen in reality is going to depend on several factors. In the binary case you can already see how many parameters affect the resulting bias, i.e. the true prevalence and the observed prevalence of the disease, and our error rates among the classes.
In this sense you can probably come up with settings of $P$, $\tilde{P}$, $\nu$, and $\eta$ that are much closer to the true value (or much further off) than the coefficient of a continuous variable with measurement error. For the latter only the signal to noise ratio matters regarding the size of the bias, i.e. how large is our measurement error.
## What happens if the measurement error in the binary variable is systematic?
If you want to investigate the case of non-standard measurement error in the binary case (which is actually already non-standard), then simply set $\nu = 0$ or $\eta = 0$ in order to create a scenario where we make a systematic classification error in either direction. I'm not aware though of a paper which looks exactly at such a setting.
## What happens to other explanatory variables?
In the binary case with measurement error it can be shown that this bias affects all other explanatory variables unless they are uncorrelated with the mismeasured binary predictor. The relevant reference for this would again be Savoca (2000).
Regarding the final questions "3) more than one covariate with independent errors 4) more than one covariate but with correlated errors", I'm not sure if this means two (or more) covariates and all of them are measured with error. Or one of the covariates is measured with error and the error is correlated with the regression error. In the latter case we go back to the non-standard errors-in-variables framework. For multiple covariates that are measured with error the bias will depend on each variables measurement error and the correlation between the covariates. If all covariates are uncorrelated then we can assess the measurement errors separately if they are continuous. If they are not continuous then we are again in the Savoca framework.
As concerns question 4) I am not aware of a paper that derives the exact bias of such a case. Correlation between errors and with several covariates with measurement error is presumably a very complex case for which it will be difficult to find a closed form solution to the bias unless one is willing to make strong assumptions on the relationship between the errors and they distributions.
Caveat: despite it's length I am sure that this answer does not answer all your questions to the extent to which it would have helped you the most. It is not possible to consider all scenarios in equal detail so I tried to focus on certain aspects that I thought would be the most important and tried to highlight relevant literature for the other parts.
• Thank you, this detailed answer helped me understand how to handle the errors in variables and pinpoints how measurement error affects the $\hat{\beta}$. The only part of the question that I feel this answer did not address in detail is: if I have two correlated predictors, one of them is binary (without measurement error) and the other is continuos (measured with measurement error), does the error in the continuous predictor, have any affect on the beta for the dichotomous predictor other than biasing it towards zero? Commented Apr 7, 2015 at 14:44
• The coefficient of the mismeasured continuous variable is biased towards zero. Whether the binary predictor's coefficient is biased up or down depends on its correlation with the mismeasured continuous predictor, so it's not necessarily biased towards zero. See for example these lecture notes: google.co.uk/url?q=http://www.nuff.ox.ac.uk/teaching/economics/…
– Andy
Commented Apr 7, 2015 at 15:08 | 2,880 | 11,236 | {"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": 1, "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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.826493 |
https://math.stackexchange.com/questions/2450271/is-mathbbr-a-set-or-a-field/2450274 | 1,558,698,097,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257605.76/warc/CC-MAIN-20190524104501-20190524130501-00235.warc.gz | 527,377,560 | 33,127 | # Is $\mathbb{R}$ a set or a field?
Does the symbol $\mathbb{R}$ refers to the set of the real numbers or to the field of the real numbers?
In other words, when you say "$x \in \mathbb{R}$", are you just sayng that $x$ is a real number? Or, on the other hand, that $x$ is a real number and obbeys the real numbers field laws (AKA properties): associative, additive identity, etc.?
• It means both depending on what you want to do – Guy Fsone Sep 29 '17 at 10:46
• Usually any collection of mathematical objects is a set and a field is a set with certain structures, depending on the context. – Vim Sep 29 '17 at 10:46
• Where is the "why not both?-meme when you need it?... Btw, there are also other things that $\mathbb{R}$ might stand for, e.g. a quotient of certain rational sequences. – Dirk Sep 29 '17 at 10:47
• That's like asking “Is Donald Trump a human being or the President of the United States?” – José Carlos Santos Sep 29 '17 at 11:01
• I'm not sure he's a human being… – Bernard Sep 29 '17 at 11:11
Sometimes in model theory, when this distinction is important, you will see something like $$\mathfrak R = \langle\mathbb R, +, \times, 0, 1, < \rangle$$ But in the rest of mathematics we write $\mathbb R$ for whatever we want. When someone asks about "open sets" in $\mathbb R$, for example, then of course we assume the usual topology. When they ask about metric properties, we assume the usual metric. When they ask about uniform properties, we assume the usual uniformity. And so on.
In the formal context, $\mathbb{R}$ refers to the set of all real number, whereas $(\mathbb{R}, + , *)$ refers to the field of real number with the usual addition and multiplication.
Note that, a field is the $3$-tuple consisting of a set, and two binary operation, which satisfying some conditions.
$\mathbb{R}$ is merely just a regular old set.
But once you equip it with some sort of additional structure such as operators +, *. And if those 3 guys $(\mathbb{R}, +, *)$ behave as a field should (satisfy field axioms), then you can call it a field. | 562 | 2,057 | {"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": 1, "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-2019-22 | latest | en | 0.886326 |
http://math.stackexchange.com/questions/285033/what-are-the-use-cases-related-to-cluster-analysis-of-different-distance-metrics | 1,469,816,350,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00042-ip-10-185-27-174.ec2.internal.warc.gz | 150,411,333 | 18,149 | What are the use cases related to cluster analysis of different distance metrics?
I'm trying to use different distance metrics like Euclidean, Manhattan, cosine, chebyshev among other distance metrics in my k-means algorithm to calculate distances between the data points and the centers. In what situation would one distance metric be more useful over the other in a clustering scenario? [Comparing all the above mentioned distance metrics]
-
You ask a great question. Deza & Deza's Dictionary of Distances catalogs hundreds of families of distance functions. And there are others that are not mentioned there but in other books such as Cichocki et al Nonnegative Matrix and Tensor Factorizations
Some observations:
• Typically there are distinct weighing functions within any given family of metrics.
• Not all metrics are applicable to all situations, so there are metrics used in data analysis that would not concern topologists, as well as composite metrics such as Hausdorff distance and Earth Movers Distance that apply primarily to 2 sets of objects rather than 2 objects.
• Even the Greeks 2k years ago were aware of several, possibly up to 10, distinct definitions of mean, including of course arithmetic, geometric and harmonic means. It is not until the early 20th century that Chisini related all the definitions of means to an invariance principle. (Medians and quantiles are even more tricky requiring a formulation in terms of constrained optimization, not just invariance).
• Results of data analysis tasks such as classification, clustering, regression and dimensionality reduction can sometimes be very sensitive to the choice of metric (data held fixed) and also can lead to completely different methodologies for solving the problem. For example, consider linear correlation in $L_2$ ("Least Squares" or Gauss's solution) versus $L_1$ (Least Absolute Deviations or Laplace's solution) loss functions. The $L_2$ metric penalizes outliers quadratically, whereas the $L_1$ metric penalizes outliers linearly, and there are examples where the sign of the linear regressor can flip (eg, from positive to negative or vice versa) just by switching from one to the other. (Also you can think of $L_2$ as being a differential method, where solutions are found by setting derivatives equal to zero, versus $L_1$ method which geometrically involves polyhedra.)
In summary, the choice of metric can certainly influence the results of even basic statistical data analysis tasks. It's a shame that statisticians understand very well that results are relative to such loss functionals, but applied scientists typically do not understand this and will compute and report an answer, as opposed to the answers.
- | 552 | 2,725 | {"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} | 2.9375 | 3 | CC-MAIN-2016-30 | latest | en | 0.923642 |
https://community.fabric.microsoft.com/t5/Desktop/Cumulative-Sum-of-Another-Measure/m-p/3525970 | 1,716,852,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00570.warc.gz | 141,510,439 | 114,358 | cancel
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New Member
## Cumulative Sum of Another Measure
I have a small task that I find hard to accomplish as I always find DAX so difficult. I need to create measure to calculate the cumulative sum of another measure. For example,
The column Cumulative Sum of "Sales Check" is what I am looking for, so I can tell November is the 6th month in the 2009 year that had the cumulative sales greater than \$25M. I hope someone can please show me a solution.
1 ACCEPTED SOLUTION
Super User
hi, @mybarbie9917_LI
try below
``````sales cumulative =
calculate(
sumx('tablename',[sales check]),
filter(
all('d_date'),
'd_date'[month]<= max('d_date'[month])&&
d_date'[year]=max(d_date'[year])
)
)``````
3 REPLIES 3
Super User
hi, @mybarbie9917_LI
try below
``````sales cumulative =
calculate(
sumx('tablename',[sales check]),
filter(
all('d_date'),
'd_date'[month]<= max('d_date'[month])&&
d_date'[year]=max(d_date'[year])
)
)``````
Super User
HI, @mybarbie9917_LI
if you want to use MEASURE inside sum use sumx() instead of sum()
try below
``````sales cumulative =
calculate(
sumx('tablename',[sales amount]),
filter(
all('d_date'),
'd_date'[Date]< max('d_date'[Date])
)
)``````
New Member
Thanks for your reply, but it is not exactly what I am looking for. I already have the Sales Cumulative measure. I want to have measure that can do the cumulative sum from the result of the measure Sales Check.
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Top Kudoed Authors | 496 | 1,928 | {"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-2024-22 | latest | en | 0.79664 |
https://physics.stackexchange.com/questions/446407/closed-set-of-operators-under-renormalization | 1,628,163,104,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00621.warc.gz | 459,841,680 | 37,709 | # Closed set of operators under renormalization
While reading the article http://inspirehep.net/record/61135, I came across the concept of "closed set under renormalization". The definition they give is the following.
In any renormalizable field theory, let $$A_1$$, ..., $$A_n$$ be a set of monomials in the fundamental fields and their derivatives. Change the original Lagrangian of the theory like $$$$\cal{L}\rightarrow\cal{L}+\underset{a}{\sum}A_{a}{\cal J}_{a}\,,$$$$ where $$\cal{J}$$'s are arbitrary functions of space and time. Let us define $$\Gamma_a^n$$ as the Fourier transform of the variational derivative of the $$n$$-point Green's function in the fundamental field with respect to $$\cal{J}_a$$, evaluated at $$\cal{J}_a$$ equals zero. If we add appropriate counterterms in the Lagrangian, we can make the $$\Gamma$$'s finite. If these counterterms are also in the set $$A_1$$, ..., $$A_n$$, we say that the set is closed under renormalization.
I have tried 'Renormalization' by Collins to look for this concept, but I couldn't find much information there.
EDIT: Just for your information, the comment on page 145 around Eq.6.2.13 was all I could find in 'Renormalization'.
Q1: In the reference, they frequently use the expression "simple power counting shows that the following set of operators are closed under renormalization". It is hard for me to guess what kind of power counting they have in mind. For example in the massive $$\lambda\phi^4$$ theory, they say that the following operators are closed under renormalization: $$$$\left\{ g_{\mu\nu}\phi^{2},g_{\mu\nu}\partial_{\lambda}\phi\partial^{\lambda}\phi,g_{\mu\nu}\phi\square\phi,\partial_{\mu}\phi\partial_{\nu}\phi,\phi\partial_{\mu}\partial_{\nu}\phi,g_{\mu\nu}\phi^{4}\right\}\,.$$$$ What could be the logic to arrive at this conclusion?
Q2: They also say as a consequence of the BPH(Bogoliubov, Parasuik, and Hepp) theorem, given a set of operators closed under renormalization, we can find a set of cut-off independent functions $$R_a^n$$, such that $$$$\Gamma_{a}^{n}=\underset{b}{\sum}c_{ab}R_{b}^{n}\,,$$$$ where $$c$$'s are constant, possibly cut-off-dependent, coefficients. This statement seems very important when proving finiteness of $$\Gamma$$'s using Ward identities. It is the first time for me to see this. For example, I couldn't find information on this on textbooks such as Peskin, Schwartz. Could anyone guide me where I could find out more about this? (I have tried to read the original paper(https://link.springer.com/content/pdf/10.1007%2FBF01773358.pdf), but it is quite an old paper and I find it very hard to follow.) I would also appreciate any hints or intuitions to understand this relation.
• It seem to me that the statement about closed set of operators is equivalent to saying that the number of counter-terms, necessary for making a theory finite, is limited. – Vladimir Kalitvianski Dec 11 '18 at 11:05
• @VladimirKalitvianski Thanks for the comment. Yes, that's their statement. What kind of criterion would there be to actually identify the closed set, e.g., the example in Q1? – Gould67 Dec 12 '18 at 7:50
• It is the famous "power counting". Read more about it, look at examples. – Vladimir Kalitvianski Dec 12 '18 at 9:21 | 882 | 3,250 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.909978 |
https://math.stackexchange.com/questions/127165/a-counter-example-in-obstruction-theory | 1,558,422,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256281.35/warc/CC-MAIN-20190521062300-20190521084300-00526.warc.gz | 556,699,566 | 33,610 | # A counter example in obstruction theory
Let $K$ denote a simplicial complex and $Y$ some topological space. Let us also denote by $K^n$ the $n$-skeleton of $K$. I would like to have an example for the following situation:
There is a map $f^1:K^1\to Y$ that can be extended to $f^2:K^2\to Y$ and yet no such extension can be further extended to $f^3:K^3\to Y$.
The idea is that there is an obstruction to the existence of $f^3$ already on the one-dimensional level but not by obstructing the existence of $f^2$. It is written in Hilton and Wylie's book that it is possible, yet I was not able to construct an explicit example myself.
• This is most probably a silly comment. I´m not in obstruction theory but I would like to understand the problem. When you ask for a map $f: K^n \rightarrow Y$ you want it to be continuous, don´t you? Then the first example which came into my mind was $K$ to be the "full" tetraedron (i.e. the complete simplicial complex on 4 vertexes), $Y$ to be $K^2$ as a topological space and finally $f^1: K^1 \rightarrow Y$ to be the inclusion map. Unfortunately I don´t understand why it does not work as a conuterexample. I would be glad to know where does it fail... – Giovanni De Gaetano Apr 2 '12 at 13:53
• @GiovanniDeGaetano, you understand correctly, a map in topology is always assumed to be a continuous function. The example you thought of fails because there are different ways to extend your $f^1$ to $f^2$. The obvious one, namely the identity, does not have an extension to $f^3$, but there is a different $f^2$, a null-homotopic one, that can be extended to $f^3$. In general, if $f^1$ is null-homotopic as in your case, it can always be extended to $K$ by the homotopy extension property for simplicial pairs. – KotelKanim Apr 2 '12 at 19:58
• I´m sorry if I waste your time with dumb questions. But I don´t see why $f^1$ is null-homotopic. Indeed if it was then also the map $f^1: K^1 \rightarrow Image(f^1)=K^1$ (which is $id_{K^1}$) would be null-homotopic and then the space $K^1$ would be contractible. But it is not. In specific I don´t see the null-homotopic extension of $f^1$, may you describe it? Thank you! – Giovanni De Gaetano Apr 3 '12 at 8:32
• No problem. $f^1$ is null-homotopic because $Y$ is homeomorphic to a sphere and since $f^1$ is not onto, its image contained in the sphere minus a point which is homeomorphic to a disc and thus contractible. Any map to a contractible space is null-homotopic (just compose with the "contraction"). The conclusion that this must imply that $K^1$ is contractible is wrong since a subspace of a contractible space does not have to be contractible (think of the inclusion of the sphere into the euclidean space). – KotelKanim Apr 3 '12 at 8:52
• @Qiaochu This was x-posted to MO and answered there ($X=\mathbb RP^3$, $Y=\mathbb RP^2$) – Grigory M Jan 3 '15 at 10:09
Take $K = \mathbb{R}P^3$ and $Y = \mathbb{R}P^2$. We can give $K$ a CW structure with skeleta $\mathbb{R}P^k$ for $0 \leq k \leq 3$. Let $f^1 : \mathbb{R}P^1 \to Y$ be the evident inclusion. Clearly this extends over $K^2$. Now suppose we have an extension $f^3 : K^3 = K \to Y$ of $f^1$. This will then give a graded ring homomorphism $(f^3)^* : H^*(Y ; \mathbb{Z}/2) \to H^*(K; \mathbb{Z}/2)$, or in other words $(f^3)^* : (\mathbb{Z}/2)[y]/y^3 \to (\mathbb{Z}/2)[x]/x^4$. Because $f^3$ extends $f^1$ we must have $(f^3)^*(y) = x$. This gives a contradiction because $y^3 = 0$ but $x^3 \neq 0$. | 1,101 | 3,465 | {"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} | 2.953125 | 3 | CC-MAIN-2019-22 | latest | en | 0.940952 |
https://fred.stlouisfed.org/series/PPPTTLBGA618NUPN | 1,495,567,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00002.warc.gz | 749,459,063 | 22,603 | Purchasing Power Parity over GDP for Bulgaria (PPPTTLBGA618NUPN) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
2010: 0.78161
Updated: Aug 31, 2012
Units:
National Currency Units per US Dollar,
Frequency:
Annual
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Purchasing Power Parity over GDP for Bulgaria, National Currency Units per US Dollar, Not Seasonally Adjusted (PPPTTLBGA618NUPN)
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
Source Indicator: ppp
Purchasing Power Parity over GDP for Bulgaria
Select a date that will equal 100 for your custom index:
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FORMAT GRAPH
Log scale:
NOTES
Source: University of Pennsylvania
Release: Penn World Table 7.1
Units: National Currency Units per US Dollar, Not Seasonally Adjusted
Frequency: Annual
Notes:
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
Source Indicator: ppp
Suggested Citation:
University of Pennsylvania, Purchasing Power Parity over GDP for Bulgaria [PPPTTLBGA618NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/PPPTTLBGA618NUPN, May 23, 2017.
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$\epsilon_0$ and $\mu_0$ appear in electrostatics and magnetostatics. When we include time varying fields we have electrodynamics and the appearance of c which turns out to be related to $\epsilon_0$ ... | 413 | 1,603 | {"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": 1, "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-2014-41 | longest | en | 0.887469 |
https://forum.allaboutcircuits.com/threads/what-happens-if-you-connect-a-transformer-the-other-way-around.163253/ | 1,618,128,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00206.warc.gz | 344,626,906 | 27,576 | # What happens if you connect a transformer the other way around?
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
Hi, here's another quick question. Suppose you have a 125V device and your wall socket is 220V. You would buy a 220V to 125V transformer and that's it. BUT, I was thinking the other day:
What if you mistakenly connect the transformer reversed and connect the 220V end to the machine and the 125V to the wall socket?
Would the machine get 400V?
I've really never used a transformer, but a friend of mine needs one to use his american machine in Europe, and this question came to my mind the other day. I guess transformers have some security measurements to avoid this situation, besides labeling each end.
#### be80be
Joined Jul 5, 2008
2,051
I guess transformers have some security measurements to avoid this situation
Nope it more then likely catch fire the wire in the transformer
#### Tonyr1084
Joined Sep 24, 2015
5,802
Usually the primary (higher voltage) winding is also a finer wire than the secondary winding. That means the secondary winding is thicker and less resistance (per linear inch, foot, meter - whatever). Hooking a winding designed for 110 volts to 220 volts will tend to be catastrophic to the transformer. Fortunately when you buy a converter that you plug into a 220 VAC outlet and get 110 VAC - they're designed to handle the voltage. Depending on their size will depend on their wattage (or in the case of AC circuits - - - VA).
If you took a transformer and connected it reverse - IN THEORY you get the ratio upwards instead of down. Take a 120 VAC transformer that puts out 12 VAC. If you hook it up backwards (and the windings don't blow up) then the output would be 1200 VAC. That's a typical 10:1 step-down transformer. Wiring it backwards makes it become a 1:10 step-up transformer.
Again, the secondaries aren't designed for high voltage as an input. Good way to keep warm though.
#### dl324
Joined Mar 30, 2015
12,668
What if you mistakenly connect the transformer reversed and connect the 220V end to the machine and the 125V to the wall socket?
Consumer versions have the appropriate plug/receptacle to prevent people from making that mistake.
Would the machine get 400V?
Yes, but you could have issues if the insulation isn't rated for that voltage or too much current is drawn (maybe briefly).
I guess transformers have some security measurements to avoid this situation, besides labeling each end.
Manufacturers are more likely to assume you know what you're doing. There's only so much you can do for ignorant people.
#### be80be
Joined Jul 5, 2008
2,051
They make Buck transformers that can step up 120 to higher voltage I have one that can step up or step down.
It was used to fix a problem were the wire was sized wrong and the conduit to small to change the wire size.
so it had 480 on the primary and stepped that down to 240 .
The floor got cut out and we installed new conduit for bigger wire and I got too keep the transformer
It about to much to mess with tho it's almost a 100 pounds.
#### KeepItSimpleStupid
Joined Mar 4, 2014
4,550
You can think of the transformer as a ratio like 1:2 or 2:1, but it also has a max voltage. The primary is usually the inner winding for efficiency reasons. The designed current can;t be exceeded either.
If you use a variable auto-transformer, the wiper needs to be fused at the max current at the designed voltage. So, if it is 0-120 V, 10A, don;t try to draw 6V at 30 A. It's 1200 VA or 6*30 or 180 VA. Fusing the input AND the wiper makes sense, but the wiper has to be fused.
#### crutschow
Joined Mar 14, 2008
26,721
If you apply a voltage higher than the rating for the transformer winding, the core will likely saturate, allowing a large current to flow limited mainly by the winding resistance, and thus zapping the transformer.
#### AlbertHall
Joined Jun 4, 2014
11,091
If you apply a voltage higher than the rating for the transformer winding, the core will likely saturate, allowing a large current to flow limited mainly by the winding resistance, and thus zapping the transformer.
Or, with luck and a little forethought, it will blow the fuse.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
I think it was clear, but just in case, I was asking this because I don't want it to happen, not because "oh, great, now I can work with 400V". So, I wanted to know what security measurements do transformers have to avoid this mistake. I don't know, if both ends of the transformer have the same input, I think it's more than probable that a user will connect it reversed and unleash hell sooner or later.
The device is a drilling machine, and my friend, who is not savvy about science and electricity, used a universal plug and directly plugged the 120V to a 220V wall socket. He told me that there were sparks inside the machine, and immediately unplugged it. He told me it was broken, and after some investigation I suddenly saw in the label "120V".
What kind of damage can you make to a 120V drill if you connect it briefly to a 220V source?
I guess none.
#### Alec_t
Joined Sep 17, 2013
11,972
I guess none.
You guessed wrong. Even a fraction of a second may be long enough to fry a variety of appliances.
#### DickCappels
Joined Aug 21, 2008
7,148
(some text removed for clarity)
What if you mistakenly connect the transformer reversed and connect the 220V end to the machine and the 125V to the wall socket?
Would the machine get 400V?
.
You understand the basic operation of a transformer. That is what it would try to do (give 400V on the output), but if you got 400V there would likely be some problems because transformers are not symmetrical between primary and secondary windings, particularly when loads are present.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
You guessed wrong. Even a fraction of a second may be long enough to fry a variety of appliances.
Yeah, but it's a drilling machine of 300W, no electronics involved, it's simply a motor that was supposed to work with 120V but suddenly received 220V. Would you damage a motor in that case? Not really? May be? Can't tell? For sure?
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
You understand the basic operation of a transformer. That is what it would try to do (give 400V on the output), but if you got 400V there would likely be some problems because transformers are not symmetrical between primary and secondary windings, particularly when loads are present.
Yeah, of course, I guess it would survive. What can happen to a transformer connected reversed?
A cable would break due to the heat? Sparks? Smoke? Nuke? Hahahah
#### DickCappels
Joined Aug 21, 2008
7,148
Probably not, but the output voltage would probably be lower than would be expected and the load regulation not as good as expected because the "real" primary was designed to drive the magnetizing current of the core while the "real" secondary was not. But yes, it might get hotter than usual.
#### BobTPH
Joined Jun 5, 2013
2,989
The turns ratio is 2:1. The ratio of the inductances is the square of that or 4:1. So one effect is that the idle current, with no load, is 4x higher. This causes extra heating in the transformer and reduces the max load VA. With a light load, it would probably survive, with it full load it would likely not.
Now lets make it a 120V to 12V transformer. In this case, the turns ratio is 10 to 1 and inductance is
100 to 1. It likely would not survive the idle current in this case., the core would saturate and take even more current. If you are lucky, a fuse will blow before the transformer burns up.
Bob
#### dendad
Joined Feb 20, 2016
3,781
There is a pretty good chance the drill motor is damaged. Either with over voltage breaking down somewhere or over current burning out wires....
Get it tested before you use it for safety sake, or just get a new one.
#### crutschow
Joined Mar 14, 2008
26,721
So, I wanted to know what security measurements do transformers have to avoid this mistake. I
None.
#### MaxHeadRoom
Joined Jul 18, 2013
22,623
So, I wanted to know what security measurements do transformers have to avoid this mistake. .
Wired to an appropriate source via the correct termination or outlet source using a qualified person.
Presumably the supply source and subsequent outlet will have a plug, the other a socket!
Max.
Thread Starter
#### rambomhtri
Joined Nov 9, 2015
346
Ouch! Of course! That's right!
I can put a non removable male cable for the input 220V that goes into a wall socket, and the other end where you connect your 120V machine, a female cable, also non removable. That way there's no way for my friend to fail, right?
MaxHeadRoom, you once trolled the TV, now you're a genius!
#### KeepItSimpleStupid
Joined Mar 4, 2014
4,550
The US would require a disconnect within 3 feet for hard wired stuff.
Use US style plugs/receptacles for the 120 VAC. | 2,198 | 8,939 | {"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} | 2.78125 | 3 | CC-MAIN-2021-17 | longest | en | 0.930858 |
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Helper V
How to see the top 3 and the bad 3
Hello Community,
Is it possible to see in a column like below The 3 best and the 3 worst with.
Thank you
2 ACCEPTED SOLUTIONS
Resolver II
This is the closest I could get:
Community Support
Hi @Milozebre ,
Try to do like this:
1. creating a calculated column:
``````Rank =
RANKX(
'Sheet1 (2)',
[quantity],
, ASC, Dense
)``````
2. creating a measure:
``````Color =
VAR x =
CALCULATE(
MAX([Rank]),
ALL('Sheet1 (2)')
)
RETURN
IF(
MAX([Rank]) >= 1 && MAX([Rank]) <= 3,
"red",
IF(
MAX([Rank]) <= x && MAX([Rank]) >= x-2,
"green"
)
)``````
3. adding the measure to "Conditional formatting->Background color":
Best regards,
Lionel Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
6 REPLIES 6
Community Support
Hi @Milozebre ,
Try to do like this:
1. creating a calculated column:
``````Rank =
RANKX(
'Sheet1 (2)',
[quantity],
, ASC, Dense
)``````
2. creating a measure:
``````Color =
VAR x =
CALCULATE(
MAX([Rank]),
ALL('Sheet1 (2)')
)
RETURN
IF(
MAX([Rank]) >= 1 && MAX([Rank]) <= 3,
"red",
IF(
MAX([Rank]) <= x && MAX([Rank]) >= x-2,
"green"
)
)``````
3. adding the measure to "Conditional formatting->Background color":
Best regards,
Lionel Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper V
Hil @v-lionel-msft ,
Its perhaps a stupid question but what's "sheet1 (2)"
Resolver II
Can you please confirm you want to see them highlighted or only see the top3 and bottom3 there ignoring the rest of the data rows.
Helper V
Hello @luapdoniv ,
I want to see the same view like that.
If not, only the highlighted 🙂
Resolver II
This is the closest I could get.
Highlight top 3 values by conditional format as background color and highlight bottom 3 values by conditional format as font color
OR
Highlight top 3 values by conditional format as background color and highlight bottom 3 Names by conditional format as background color
Will continue R&D to see if your requirement can be achieved. Let me know if this helps.
Resolver II
This is the closest I could get:
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# How much money must be invested at 10% annual interest, compounded sem
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Math Expert
Joined: 02 Sep 2009
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How much money must be invested at 10% annual interest, compounded sem [#permalink]
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17 Jan 2017, 09:04
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Question Stats:
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How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
_________________
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Joined: 05 Mar 2015
Posts: 1004
How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
17 Jan 2017, 09:45
1
Bunuel wrote:
How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
Algebraic approach
CI=P*(1+r/100)^t
CI=P+51.25
thus P+51.25=P(1+10/(2*100)^2
P=500
2nd logical approach
10% of 500 =50
thus total profit is a bit more since interest earned compounded basis
thus profit must be 51.25
(check for first year 500+25=525
second year 525*5/100=26.25
total 525+26.25= 551.25)
Thus Ans C
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Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
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23 Oct 2017, 17:45
I am confused on the solution formula used. Could someone break it down more in detail?
Math Expert
Joined: 02 Sep 2009
Posts: 51227
Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
23 Oct 2017, 23:17
2
1
twash017 wrote:
I am confused on the solution formula used. Could someone break it down more in detail?
How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
10% annual interest, compounded semi-annually means that the investment earns 5% in every 6 months.
So, we are looking for such x, which satisfies: x*1.05^2 = x + 51.25
0.1025x = 51.25;
1025x = 512500;
x = 500.
Another way: since 10% annual interest is compounded semi-annually, then the actual interest will be a little bit more that 10% (because of the interest earned on interest). Thus, \$51.25 of interest must be a little bit more that 10% of the correct answer. Only C fits.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51227
Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
23 Oct 2017, 23:18
Bunuel wrote:
twash017 wrote:
I am confused on the solution formula used. Could someone break it down more in detail?
How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
10% annual interest, compounded semi-annually means that the investment earns 5% in every 6 months.
So, we are looking for such x, which satisfies: x*1.05^2 = x + 51.25
0.1025x = 51.25;
1025x = 512500;
x = 500.
Another way: since 10% annual interest is compounded semi-annually, then the actual interest will be a little bit more that 10% (because of the interest earned on interest). Thus, \$51.25 of interest must be a little bit more that 10% of the correct answer. Only C fits.
4. Percents and Iterest
Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Hope it helps.
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Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
24 Oct 2017, 13:09
Bunuel wrote:
Bunuel wrote:
twash017 wrote:
I am confused on the solution formula used. Could someone break it down more in detail?
How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
10% annual interest, compounded semi-annually means that the investment earns 5% in every 6 months.
So, we are looking for such x, which satisfies: x*1.05^2 = x + 51.25
0.1025x = 51.25;
1025x = 512500;
x = 500.
Another way: since 10% annual interest is compounded semi-annually, then the actual interest will be a little bit more that 10% (because of the interest earned on interest). Thus, \$51.25 of interest must be a little bit more that 10% of the correct answer. Only C fits.
4. Percents and Iterest
Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Hope it helps.
Thanks Bunuel!
Manager
Joined: 31 Jan 2018
Posts: 70
Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
12 May 2018, 04:58
Bunuel wrote:
How much money must be invested at 10% annual interest, compounded semi-annually, in order to earn \$51.25 in interest income by the end of the year?
A. \$1,025.00
B. \$512.50
C. \$500.00
D. \$256.25
E. \$42.00
Let the amount invested is A
since it is compounded semi annually, the rate of interest will be 10/2 = 5% and the time will be 1 * 2 = 2
According to question:
A (1 + 5/100)^2 - A = 51.25
A(441/400) - A = 51.25
--> 41A/400 = 51.25
==> 41A = 20500
therefore A(Amount invested) = 20500/41 = 500
Manager
Joined: 17 Jan 2017
Posts: 61
Re: How much money must be invested at 10% annual interest, compounded sem [#permalink]
### Show Tags
12 May 2018, 09:00
If the interest is compounded anually answer D were correct.
The answer has to be very close to D though.
Re: How much money must be invested at 10% annual interest, compounded sem &nbs [#permalink] 12 May 2018, 09:00
Display posts from previous: Sort by | 2,181 | 7,111 | {"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} | 3.875 | 4 | CC-MAIN-2018-51 | latest | en | 0.895285 |
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### 931 miles / 1 498 km driving729 miles / 1 174 km flying
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## Distance from Gaspe, Canada to New York, NY
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The total straight line flight distance from Gaspe, Canada to New York, NY is 729 miles.
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Your trip begins in Gaspe, Canada. It ends in New York, New York.
Your flight direction from Gaspe, Canada to New York, NY is Southwest (-137 degrees from North).
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## Distance calculator
Travelmath helps you find distances based on actual road trip directions, or the straight line flight distance. You can get the distance between cities, airports, states, countries, or zip codes to figure out the best route to travel to your destination. Compare the results to the straight line distance to determine whether it's better to drive or fly. The database uses the latitude and longitude of each location to calculate distance using the great circle distance formula. The calculation is done using the Vincenty algorithm and the WGS84 ellipsoid model of the Earth, which is the same one used by most GPS receivers. This gives you the flying distance "as the crow flies." Find your flight distances quickly to estimate the number of frequent flyer miles you'll accumulate. Or ask how far is it between cities to solve your homework problems. You can lookup U.S. cities, or expand your search to get the world distance for international trips. You can also print out pages with a travel map. | 512 | 2,300 | {"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} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.849957 |
https://math.answers.com/questions/673_billion_divided_by_250_million | 1,685,912,167,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00304.warc.gz | 417,995,898 | 48,809 | 0
673 billion divided by 250 million?
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673 billion divided by 250 million is 2692. | 131 | 368 | {"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} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.826617 |
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And Electronics, Text Book, M. Morris Mano, Free Modsec nginx tutorials download Digital Logic pdf. Morris Mano Computer Logic Design Prentice-Hall Inc. | 1,427 | 6,502 | {"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} | 2.546875 | 3 | CC-MAIN-2018-30 | latest | en | 0.780556 |
https://fr.mathworks.com/matlabcentral/answers/1584694-table-comparison-string-cell-compare | 1,718,421,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00482.warc.gz | 236,387,484 | 28,572 | # Table comparison [String - cell compare]
15 vues (au cours des 30 derniers jours)
Khaled Genidy le 11 Nov 2021
Modifié(e) : Seth Furman le 15 Nov 2021
I have 2 tables that contain the same headers of the first row.
• My goal is to see if the cells contains the same values
• if true - contaminate cells to present the values
From the two tables below.
• I need to have my comparison for example : a cell in X would be 1,5 for Y XW
Part of my code is:
for i=1 : size(Table1)
if strcmpi(Table1.X(i), Table2.X(i) ==1)
X12= cellstr(strcat(....., ',' , ..........)
I am just confused.
I am currently getting - conversion to double from cell is not possible.
Table1 Table 2 expected result
X Y X Y X Y
1 XW 5 XW 1, 5 XW
2 XR 4 XR 2, 4 XR
3 XR 3 XR 3 XR
4 XW 2 XZ
5 XX 1 XZ
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
the cyclist le 11 Nov 2021
It would be much easier to help if you uploaded the two tables in a MAT file. You can use the paperclip icon in the INSERT section of the toolbar.
Connectez-vous pour commenter.
### Réponses (2)
Seth Furman le 12 Nov 2021
Modifié(e) : Seth Furman le 15 Nov 2021
Take a look at the table functions innerjoin and outerjoin:
t1 = table((1:5)', categorical(["XW";"XR";"XR";"XW";"XX"]), 'VariableNames', ["X","Y"])
t1 = 5×2 table
X Y _ __ 1 XW 2 XR 3 XR 4 XW 5 XX
t2 = table((5:-1:1)', categorical(["XW";"XR";"XR";"XZ";"XZ"]), 'VariableNames', ["X","Y"])
t2 = 5×2 table
X Y _ __ 5 XW 4 XR 3 XR 2 XZ 1 XZ
tJoined = innerjoin(t1, t2, "Keys", "Y")
tJoined = 6×3 table
X_t1 Y X_t2 ____ __ ____ 2 XR 4 2 XR 3 3 XR 4 3 XR 3 1 XW 5 4 XW 5
tJoined = mergevars(tJoined, ["X_t1", "X_t2"], "NewVariableName", "X")
tJoined = 6×2 table
X Y ______ __ 2 4 XR 2 3 XR 3 4 XR 3 3 XR 1 5 XW 4 5 XW
[groupNums, Y] = findgroups(tJoined.Y)
groupNums = 6×1
1 1 1 1 2 2
Y = 2×1 categorical array
XR XW
X = splitapply(@(x) {unique(x)'}, tJoined.X, groupNums)
X = 2×1 cell array
{[2 3 4]} {[1 4 5]}
table(X, Y)
ans = 2×2 table
X Y _________ __ {[2 3 4]} XR {[1 4 5]} XW
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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Image Analyst le 12 Nov 2021
Try this:
% Create saple data:
x1 = [1;2;3;4;5]
y1 = {'XW'; 'XR';'XR';'XW';'XX'}
x2 = [5;4;3;2;1]
y2 = {'XW'; 'XR';'XR';'XZ';'XZ'}
table1 = table(x1, y1, 'VariableNames',{'X', 'Y'})
table2 = table(x2, y2, 'VariableNames',{'X', 'Y'})
% Create output table.
table3 = table({0}, {'?'}, 'VariableNames',{'X', 'Y'});
count = 1;
for row = 1 : size(table1, 1)
thisString = table1.Y{row};
fprintf('Searching for %s.\n', thisString)
if isequal(table1.Y{row}, table2.Y{row})
% Columns y match.
table3.X{count} = unique([table1.X(row), table2.X(row)], 'stable');
table3.Y{count} = thisString;
count = count + 1;
end
end
table3
You'll see
table3 =
3×2 table
X Y
_______ ______
{[1 5]} {'XW'}
{[2 4]} {'XR'}
{[ 3]} {'XR'}
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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Translated by | 1,189 | 3,258 | {"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} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.507878 |
https://us.metamath.org/mpeuni/cbvmptv.html | 1,725,816,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00514.warc.gz | 557,449,150 | 10,791 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cbvmptv Structured version Visualization version GIF version
Theorem cbvmptv 5133
Description: Rule to change the bound variable in a maps-to function, using implicit substitution. (Contributed by Mario Carneiro, 19-Feb-2013.) Add disjoint variable condition to avoid ax-13 2380. See cbvmptvg 5134 for a less restrictive version requiring more axioms. (Revised by Gino Giotto, 17-Jan-2024.)
Hypothesis
Ref Expression
cbvmptv.1 (𝑥 = 𝑦𝐵 = 𝐶)
Assertion
Ref Expression
cbvmptv (𝑥𝐴𝐵) = (𝑦𝐴𝐶)
Distinct variable groups: 𝑥,𝐴,𝑦 𝑦,𝐵 𝑥,𝐶
Allowed substitution hints: 𝐵(𝑥) 𝐶(𝑦)
Proof of Theorem cbvmptv
StepHypRef Expression
1 nfcv 2920 . 2 𝑦𝐵
2 nfcv 2920 . 2 𝑥𝐶
3 cbvmptv.1 . 2 (𝑥 = 𝑦𝐵 = 𝐶)
41, 2, 3cbvmpt 5131 1 (𝑥𝐴𝐵) = (𝑦𝐴𝐶) | 375 | 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} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.48657 |
https://forums.wolfram.com/mathgroup/archive/2011/Sep/msg00433.html | 1,726,722,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00829.warc.gz | 236,650,730 | 8,022 | Re: List Manipulation- Advanced beginner question
• To: mathgroup at smc.vnet.net
• Subject: [mg121599] Re: List Manipulation- Advanced beginner question
• From: Nguyen Van Falk <nvanfalk at gmail.com>
• Date: Wed, 21 Sep 2011 05:35:41 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201109160948.FAA12399@smc.vnet.net> <j548vf\$63h\$1@smc.vnet.net>
```On Sep 18, 1:09 am, Heike Gramberg <heike.gramb... at gmail.com> wrote:
> You could try something like
>
> totlist[lists__List] := Module[{jlist = Join[lists], dates},
> dates = DeleteDuplicates[jlist[[All, 1]]];
> Transpose[{dates, Total[Cases[jlist, {#, a_} :> a]] & /@ dates}]]
>
> usage:
>
> list1 = {{{2011, 9, 16, 0, 0, 0}, a}, {{2010, 10, 30, 0, 0, 0}, b}, {{2009, 7, 5, 0, 0, 0}, c}};
> list2 = {{{2011, 9, 16, 0, 0, 0}, 1}, {{2010, 08, 15, 0, 0, 0}, 2}, {{2009, 4, 29, 0, 0, 0}, 3}};
> totlist[list1, list2]
>
> output:
>
> {{{2011, 9, 16, 0, 0, 0}, 1 + a}, {{2010, 10, 30, 0, 0, 0}, b},
> {{2009, 7, 5, 0, 0, 0}, c}, {{2010, 8, 15, 0, 0, 0}, 2},
> {{2009, 4, 29, 0, 0, 0}, 3}}
>
> Heike
>
> On 16 Sep 2011, at 11:48, Nguyen Van Falk wrote:
>
>
>
>
>
>
>
> > I am trying to add multiple lists together and am looking for some
> > guidance. My data is in the form of multiple time series, with dates
> > that overlap one another. I need to merge the lists, add all of the
> > values together for the dates that overlap and output a single list
> > that only has one entry for each date.
>
> > For example, if I have three values for September 15 (8,14,10), I only
> > want a single entry for September 15 in the final list with a value of
> > 32. The data is in standard date spec form: { {{Date 1}, Value 1},
> > {{Date 2}, Value 2}, {{Date 3}, Value 3}, ....... {{Date n}, Value
> > n}, }.
>
> > Thanks,
>
> > NVF
Thank you for the suggestions everyone.
```
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• Next by thread: Re: List Manipulation- Advanced beginner question | 793 | 2,098 | {"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-2024-38 | latest | en | 0.773531 |
https://studylib.net/doc/15280802/stationary-wavelet-transform | 1,624,564,246,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00056.warc.gz | 480,589,894 | 13,592 | # Stationary Wavelet Transform
```Undecimated wavelet transform
(Stationary Wavelet Transform)
ECE 802
Standard DWT
• Classical DWT is not shift invariant: This
means that DWT of a translated version of
a signal x is not the same as the DWT of
the original signal.
• Shift-invariance is important in many
applications such as:
– Change Detection
– Denoising
– Pattern Recognition
E-decimated wavelet transform
• In DWT, the signal is convolved and decimated
(the even indices are kept.)
• The decimation can be carried out by choosing
the odd indices.
• If we perform all possible DWTs of the signal, we
will have 2J decompositions for J decomposition
levels.
• Let us denote by εj = 1 or 0 the choice of odd or
even indexed elements at step j. Every ε
decomposition is labeled by a sequence of 0's
and 1's. This transform is called the εdecimated DWT.
• ε-decimated DWT are all shifted versions
of coefficients yielded by ordinary DWT
applied to the shifted sequence.
SWT
• Apply high and low pass filters to the data
at each level
• Do not decimate
• Modify the filters at each level, by padding
them with zeroes
• Computationally more complex
Block Diagram of SWT
SWT Computation
• Step 0 (Original Data):
A(0) A(0) A(0) A(0) A(0) A(0) A(0) A(0)
• Step 1:
D(1,0)D(1,1)D(1,0)D(1,1)D(1,0)D(1,1)D(1,0)
D(1,1)
A(1,0)A(1,1) A(1,0)A(1,1) A(1,0)A(1,1)
A(1,0)A(1,1)
SWT Computation
• Step 2:
D(1,0)D(1,1) D(1,0)D(1,1) D(1,0)D(1,1)
D(1,0)D(1,1)
D(2,0,0)D(2,1,0)D(2,0,1)D(2,1,1)
D(2,0,0)D(2,1,0)D(2,0,1)D(2,1,1)
A(2,0,0)A(2,1,0)A(2,0,1)A(2,1,1)
A(2,0,0)A(2,1,0)A(2,0,1)A(2,1,1)
Different Implementations
• A Trous Algorithm: Upsample the filter
coefficients by inserting zeros
• Beylkin’s algorithm: Shift invariance, shifts
by one will yield the same result by any
odd shift. Similarly, shift by zeroAll even
shifts.
– Shift by 1 and 0 and compute the DWT,
repeat the same procedure at each stage
– Not a unique inverse: Invert each transform
and average the results
Different Implementations
• Undecimated Algorithm: Apply the lowpass
and highpass filters without any
decimation.
Continuous Wavelet
Transform (CWT)
CWT
• Decompose a continuous time function in
terms of wavelets:
• Can be thought of as convolution
• Translation factor: a, Scaling factor: b.
• Inverse wavelet transform:
Requirements on the Mother
wavelet
(t )dt 0
(t )
2
dt 1
0
( )
2
d
Properties
• Linearity
• Shift-Invariance
• Scaling Property:
1
f ' (t )
f (t / s )
s
a b
CWT ' (a, b) CWT ,
s s
• Energy Conservation: Parseval’s
1
2
2
b
Localization Properties
• Time Localization: For a Delta function,
1 t0 a
b b
• The time spread: t (t ) dt
• Frequency localization can be adjusted by
choosing the range of scales
• Redundant representation
2
2
CWT Examples
• The mother wavelet can be complex or
real, and it generally includes an
properties of the localized oscillation.
• Complex wavelets can separate amplitude
and phase information.
• Real wavelets are often used to detect
sharp signal transitions.
Morlet Wavelet
• Morlet: Gaussian window modulated in
frequency, normalization in time is
controlled by the scale parameter
1 j0t t 2 / 2
(t )
e e
2
( ) e
( 0 ) 2 / 2
Morlet Wavelet
• Real part:
•
CWT
• CWT of chirp signal:
Mexican Hat
• Derivative of Gaussian (Mexican Hat):
Discretization of CWT
• Discretize the scaling parameter as
b b0
m
• The shift parameter is discretized with different
step sizes at each scale
b b , a na b
m
0
m
0 0
• Reconstruction is still possible for certain
wavelets, and appropriate choice of
discretization.
``` | 1,368 | 3,620 | {"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} | 3.421875 | 3 | CC-MAIN-2021-25 | latest | en | 0.757402 |
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