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https://forum.freecodecamp.org/t/for-loops-to-test-an-prime-number/653124
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# For loops to test an prime number
Please, I need some help to solve this problem!
Challenge is:
What should be the expressions in the instruction below to test whether a given number is prime (divisible only by 1 and itself) or not?
In the code image, you can see the space where I can make changes to better understand the challenge. I can only make changes in the [initialization]; [condition] as you see in the example on the attached image.
beneath you can see my tried code
``````let n = 17; // the commands below can work for any values of n
let prime = true;
for(let i = 1; i > n; i--)
if(n%i == 0) // Tests whether n is divisible by i
prime = false;
console.log(`\${n} \${!prime?"no ":""} is a prime number.`);
``````
Note - its best to use {} with your loops and if statements
hi ! so I have tried many ways but none of them print to me the correct answer.
So the first try was for ( let i = 0 ; i > n ; i–) and it did not work.
That was my last try.
however the answer has to print prime and non-prime numbers when I change the variable N
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https://it.mathworks.com/matlabcentral/cody/problems/58-tic-tac-toe-ftw/solutions/1872678
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Cody
# Problem 58. Tic Tac Toe FTW
Solution 1872678
Submitted on 12 Jul 2019 by Aidan Kwok
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [ 1 0 1 0 -1 0 -1 -1 1]; b = [4 8]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
2 Pass
a = [ 1 0 0 0 -1 0 -1 0 1]; b = [0]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
3 Pass
a = [ 1 0 0 0 1 -1 1 -1 -1]; b = [2 7]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
4 Pass
a = [ 1 0 0 -1 1 -1 1 -1 0]; b = [7 9]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
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http://booksiread.org/pdf/electricians-calculations-manual-second-edition/
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## Electrician S Calculations Manual Second Edition
Author: Nick Fowler
Publisher: McGraw Hill Professional
ISBN: 0071770178
Size: 74.30 MB
Format: PDF, ePub
View: 5660
THE ULTIMATE ON-THE-JOB COMPANION--FULLY UPDATED Thoroughly revised to reflect the 2011 National Electrical Code (NEC) and the latest industry advances, Electrician's Calculations Manual, Second Edition gives you quick access to the basic calculations needed for any given job. The book also serves as an ideal review for license preparation. End-of-chapter questions plus an end-of-book final test help reinforce the material covered. Written by a Master Electrician with more than 40 years of experience, this practical guide helps you: Find answers for both AC and DC circuits Solve problems related to motor circuits and transformers Calculate single-dwelling and multifamily loads Accurately figure requirements for commercial jobs Perform conduit-bending math Handle service entrance problems Understand the math behind electrical solutions And much more
## Mathematics Manual For Water And Wastewater Treatment Plant Operators Second Edition
Author: Frank R. Spellman
Publisher: CRC Press
ISBN: 1482224259
Size: 67.51 MB
Format: PDF, ePub, Docs
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To properly operate a waterworks or wastewater treatment plant and to pass the examination for a waterworks/wastewater operator’s license, it is necessary to know how to perform certain calculations. All operators, at all levels of licensure, need a basic understanding of arithmetic and problem-solving techniques to solve the problems they typically encounter in the workplace. Hailed on its first publication as a masterly account written in an engaging, highly readable, user-friendly style, the Mathematics Manual for Water and Wastewater Treatment Plant Operators, Second Edition has been expanded and divided into three specialized texts that contain hundreds of worked examples presented in a step-by-step format. They are ideal for all levels of water treatment operators in training and practitioners studying for advanced licensure. In addition, they provide a handy desk reference and handheld guide for daily use in making operational math computations. This first volume, Basic Mathematics for Water and Wastewater Operators, introduces and reviews fundamental concepts critical to qualified operators. Presented at a basic level, this volume reviews fractions and decimals, rounding numbers, significant digits, raising numbers to powers, averages, proportions, conversion factors, flow and detention time, and the areas and volumes of different shapes. It also explains how to keep track of units of measurement (such as inches, feet, and gallons) during the calculations. After building a strong foundation based on theoretical math concepts, the text moves on to applied math—basic math concepts applied in solving practical problems for both water and wastewater operations. The material is presented using clear explanations in manageable portions to make learning quick and easy, and illustrative real-world problems are provided that correlate to modern practice and design.
## Hvac Procedures Forms Manual Second Edition
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Developed over the course of many years of on-the-job projects involving HVAC energy auditing, testing/balancing and cost estimating, and refined through feedback from thousands of engineers and technicians who have used them, the forms contained in this manual are concise, comprehensive, and optimally organized for easy reference. Complete sets of forms are provided for all aspects of testing and balancing, energy auditing, indoor quality diagnosis, and load calculations. The first edition, entitled HVAC Energy Audit & Balancing Forms Manual compiled these time-saving forms for the first time in a single reference. This enhanced second edition adds a new chapter on technical management, providing procedures for achieving thorough, systematic and accurate problem solving, troubleshooting and decision making in building systems management and contracting.
## Steam Plant Calculations Manual Second Edition Revised And Expanded
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Maintaining a question-and-answer format, this second edition provides simplified means of solving nearly 200 practical problems that confront engineers involved in the planning, design, operation and maintenance of steam plant systems. Calculations pertaining to emissions, boiler efficiency, circulation and heat transfer equipment design and performance are provided. Solutions to 70 new problems are featured in this edition.
## Electronic And Electrical Engineering Solutions Manual S M Second Edition
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## The Electrical Engineering Handbook Second Edition
Author: Richard C. Dorf
Publisher: CRC Press
ISBN: 9781420049763
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In 1993, the first edition of The Electrical Engineering Handbook set a new standard for breadth and depth of coverage in an engineering reference work. Now, this classic has been substantially revised and updated to include the latest information on all the important topics in electrical engineering today. Every electrical engineer should have an opportunity to expand his expertise with this definitive guide. In a single volume, this handbook provides a complete reference to answer the questions encountered by practicing engineers in industry, government, or academia. This well-organized book is divided into 12 major sections that encompass the entire field of electrical engineering, including circuits, signal processing, electronics, electromagnetics, electrical effects and devices, and energy, and the emerging trends in the fields of communications, digital devices, computer engineering, systems, and biomedical engineering. A compendium of physical, chemical, material, and mathematical data completes this comprehensive resource. Every major topic is thoroughly covered and every important concept is defined, described, and illustrated. Conceptually challenging but carefully explained articles are equally valuable to the practicing engineer, researchers, and students. A distinguished advisory board and contributors including many of the leading authors, professors, and researchers in the field today assist noted author and professor Richard Dorf in offering complete coverage of this rapidly expanding field. No other single volume available today offers this combination of broad coverage and depth of exploration of the topics. The Electrical Engineering Handbook will be an invaluable resource for electrical engineers for years to come.
## Lineman And Cablemans Field Manual Second Edition
Author: Thomas Shoemaker
Publisher: McGraw Hill Professional
ISBN: 0071621229
Size: 42.53 MB
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A Compact, On-the-Job Reference for Linemen and Cablemen Fully updated with the latest NEC and OSHA standards, this one-stop portable guide contains the crucial electrical data, formulas, calculations, and safety information essential at any jobsite. The Lineman's and Cableman's Field Manual, Second Edition, provides easy-to-follow details on constructing, operating, and maintaining both overhead and underground electric distribution and transmission lines. Helpful charts, tables, diagrams, equations, and definitions are included throughout this handy resource. The new edition of the manual covers: Line conductors * Cable, splices, and terminations * Distribution voltage transformers * Wood-pole structures * Guying * Lightning and surge protection * Fuses * Inspection and maintenance plans * Tree trimming * Rope, knots, splices, and gear * Grounding * Protective grounds * Safety equipment and rescue
## Struktur Und Interpretation Von Computerprogrammen
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Publisher: Springer-Verlag
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## Offshore Electrical Engineering Manual
Author: Geoff MacAngus-Gerrard
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ISBN: 0123854997
Size: 43.78 MB
Format: PDF, ePub
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Offshore Electrical Engineering Manual, Second Edition, is for electrical engineers working on offshore projects who require detailed knowledge of an array of equipment and power distribution systems. The book begins with coverage of different types of insulation, hot-spot temperatures, temperature rise, ambient air temperatures, basis of machine ratings, method of measurement of temperature rise by resistance, measurement of ambient air temperature. This is followed by coverage of AC generators, automatic voltage regulators, AC switchgear transformers, and programmable electronic systems. The emphasis throughout is on practical, ready-to-apply techniques that yield immediate and cost-effective benefits. The majority of the systems covered in the book operate at a nominal voltage of 24 y dc and, although it is not necessary for each of the systems to have separate battery and battery charger systems, the grouping criteria require more detailed discussion. The book also provides information on equipment such as dual chargers and batteries for certain vital systems, switchgear tripping/closing, and engine start batteries which are dedicated to the equipment they supply. In the case of engines which drive fire pumps, duplicate charges and batteries are also required. Packed with charts, tables, and diagrams, this work is intended to be of interest to both technical readers and to general readers. It covers electrical engineering in offshore situations, with much of the information gained in the North Sea. Some topics covered are offshore power requirements, generator selection, process drivers and starting requirements, control and monitoring systems, and cabling and equipment installation Discusses how to perform inspections of electrical and instrument systems on equipment using appropriate regulations and specifications Explains how to ensure electrical systems/components are maintained and production is uninterrupted Demonstrates how to repair, modify, and install electrical instruments ensuring compliance with current regulations and specifications Covers specification, management, and technical evaluation of offshore electrical system design Features evaluation and optimization of electrical system options including DC/AC selection and offshore cabling designs
## Electrical Engineering Reference Manual For The Electrical And Computer Pe Exam
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Publisher: Ppi
ISBN: 9781591260967
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Electrical Engineering Reference Manual is the most comprehensive reference available for the electrical and computer engineering PE exam.
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https://community.airtable.com/t/nps-excluding-blank-cells/34191
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NPS Excluding Blank Cells
Hi! I’m going crazy trying to come up with an NPS formula. I am planning to have a formula return -100, 0, or 100 depending on the value so I can then average that column on dashboards.
I can get my NPS just fine, but I can’t figure out how to exclude blank cells in an IF statement.
In words, I want the following:
If {Program NPS Value} is between 0 (inclusive) and 6 (inclusive), -100
If {Program NPS Value} is 7 or 8, 0
If {Program NPS Value} is 9 or 10, 100
If {Program NPS Value} is BLANK, “”
There’s gotta be a way to factor out these blank cells.
I’m getting -100s for all my blank cells with this formula:
IF({NPS Value} < 7, -100, IF({NPS Value} > 8,100,0))
I appreciate your help!!
Welcome to the community, @Amy_Herman!
Ha, please don’t get me started on the madness of Airtable’s numerical evaluations. Airtable is the only programming language I’ve ever worked with that evaluates a completely empty cell as a valid number to be evaluated! It’s very odd.
In any case, here’s what you would need to use for your formula:
``````IF(
{NPS Value}=BLANK(),BLANK(),
IF({NPS Value} < 7,-100,
IF({NPS Value} > 8,100,
0)))
``````
Hope this helps! Let me know if it works for you!
And, if this answers your question, could you please mark this comment as the solution to your question? This will help other people who have a similar question.
1 Like
Lifesaver!! Thank you!!
1 Like
This topic was solved and automatically closed 3 days after the last reply. New replies are no longer allowed.
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https://se.tradingview.com/script/ieVGZXw5-APA-Adaptive-Ehlers-Early-Onset-Trend-Loxx/
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APA-Adaptive, Ehlers Early Onset Trend [Loxx]
Uppdaterad
APA-Adaptive, Ehlers Early Onset Trend is Ehlers Early Onset Trend but with Autocorrelation Periodogram Algorithm dominant cycle period input.
What is Ehlers Early Onset Trend?
The Onset Trend Detector study is a trend analyzing technical indicator developed by John F. Ehlers , based on a non-linear quotient transform. Two of Mr. Ehlers' previous studies, the Super Smoother Filter and the Roofing Filter, were used and expanded to create this new complex technical indicator. Being a trend-following analysis technique, its main purpose is to address the problem of lag that is common among moving average type indicators.
The Onset Trend Detector first applies the EhlersRoofingFilter to the input data in order to eliminate cyclic components with periods longer than, for example, 100 bars (default value, customizable via input parameters) as those are considered spectral dilation. Filtered data is then subjected to re-filtering by the Super Smoother Filter so that the noise (cyclic components with low length) is reduced to minimum. The period of 10 bars is a default maximum value for a wave cycle to be considered noise; it can be customized via input parameters as well. Once the data is cleared of both noise and spectral dilation, the filter processes it with the automatic gain control algorithm which is widely used in digital signal processing. This algorithm registers the most recent peak value and normalizes it; the normalized value slowly decays until the next peak swing. The ratio of previously filtered value to the corresponding peak value is then quotiently transformed to provide the resulting oscillator. The quotient transform is controlled by the K coefficient: its allowed values are in the range from -1 to +1. K values close to 1 leave the ratio almost untouched, those close to -1 will translate it to around the additive inverse, and those close to zero will collapse small values of the ratio while keeping the higher values high.
Indicator values around 1 signify uptrend and those around -1, downtrend.
What is an adaptive cycle, and what is Ehlers Autocorrelation Periodogram Algorithm?
From his Ehlers' book Cycle Analytics for Traders Advanced Technical Trading Concepts by John F. Ehlers , 2013, page 135:
"Adaptive filters can have several different meanings. For example, Perry Kaufman’s adaptive moving average ( KAMA ) and Tushar Chande’s variable index dynamic average ( VIDYA ) adapt to changes in volatility . By definition, these filters are reactive to price changes, and therefore they close the barn door after the horse is gone.The adaptive filters discussed in this chapter are the familiar Stochastic , relative strength index ( RSI ), commodity channel index ( CCI ), and band-pass filter.The key parameter in each case is the look-back period used to calculate the indicator. This look-back period is commonly a fixed value. However, since the measured cycle period is changing, it makes sense to adapt these indicators to the measured cycle period. When tradable market cycles are observed, they tend to persist for a short while.Therefore, by tuning the indicators to the measure cycle period they are optimized for current conditions and can even have predictive characteristics.
The dominant cycle period is measured using the Autocorrelation Periodogram Algorithm. That dominant cycle dynamically sets the look-back period for the indicators. I employ my own streamlined computation for the indicators that provide smoother and easier to interpret outputs than traditional methods. Further, the indicator codes have been modified to remove the effects of spectral dilation.This basically creates a whole new set of indicators for your trading arsenal."
Release Notes:
Small fix.
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Average Stock Price: What is Average Share Price & How to Calculate
Last updated:
Average Price Introduction
We all have heard the concept of average in our school or in daily life. Suppose, if we buy 3 items for ₹30, ₹50 and ₹40 each, then the average price will be ₹40. In simple words, the average price is nothing but the total amount divided by the number of items. However, the concept of the average price formula in the stock market is not that simple. The stock price average formula rather uses the weighted average method.
What is the average share price?
As an investor, you always wish to pay less and earn more. But in the share market, you are not able to predict the accurate share price and the returns. Like everyone, you also end up buying the stock at its high and then keep waiting for the right time for an exit. Here, judicious averaging of price can help you. If you are considering a simple average for calculation of average price of a stock then you end up making a loss.
Volatility and uncertainty in share prices are the regular features of the stock market. As a result, price fluctuation of securities is a normal thing and you also want to desire to buy a stock at a reasonable price, with a long-term perspective.
Appropriate average in stock market might be the answer to arrive at a rational price of a stock for better returns. Therefore, average pricing might be the best way to take the effective decision for long term buying and selling.
How to Calculate average share price?
The concept of the average pricing in the stock market is valid in buying as well as selling transactions. The concept of an average price of share is critical because you normally buy one company shares at different prices at different times. In reality, you must weigh these prices by the quantities purchased to get the right average price stocks. Here is an illustration where you buy ABC Ltd. 50 shares at the price of ₹120 each. The total spend is ₹6,000. The price of the stock falls to ₹100 next week after your initial purchase. You incur a loss of ₹20 per share. Now, you have two options. You can either wait for the stock's price to increase or invest ₹7,500 more to buy 75 shares.
If you does the latter, the calculation of average price stock is as follows:
• Total share price = ₹220 (₹120 + ₹100)
• Total number of transactions = 2
• Average stock price in the simple average method = ₹110
• Total invested amount = ₹13,500 (₹6,000 + ₹7,500)
• Total number of shares = 125 (50 + 75)
• Average stock price in the weighted average method = ₹108
There isn't a significant difference between the simple average and the weighted average in this example. But, when you trade in larger volumes with higher volatility in prices, use weighted average price to determine your total cost of purchase. The weighted average price per unit would then come down to ₹108. This way, your price of buying will come down to a much safer level from a long-term perspective.
Next, focus on the sell-side calculation of stock average price. Here is an illustration where you sell ABC Ltd. 50 shares at the price of ₹120 each. The total amount earned is ₹6,000. The price of the stock rises to ₹140 next week after your initial selling. Additionally, you incur a loss of ₹20 per share in the selling process. Now, you have two options. You can either wait for the stock's price to increase further or either sell more 75 shares of the same company to earn ₹10,500.
If you does the latter, the calculation of average price stock is as follows:
• Total share price = ₹260 (₹120 + ₹140)
• Total number of transactions = 2
• Average stock price in the simple average method = ₹130
• Total earned amount = ₹16,500 (₹6,000 + ₹10,500)
• Total number of shares = 125 (50 + 75)
• Average stock price in the weighted average method = ₹132
In the sell-side case also, there isn't a significant difference between the simple average and the weighted average share price. But, when you trade in larger volumes with higher volatility in prices, use weighted average price to determine your total earning. The weighted average price per unit would then rise to ₹132. This way, your price of selling will rise to a higher level.
How to use the average share price to your benefit?
Averaging is beneficial in both rising and falling markets. Averaging helps you accumulate more profits, if you buy stocks in rising markets. Similarly, in declining markets, it aids in lowering the average purchase price. In selling, it helps you to earn more average profits in case of rising markets.
1. Averaging works best when a company's fundamentals are not that bad but its stock performance is influenced by industry-specific conditions or overall hurting market sentiments.
2. Sometimes the stock of a promising company is under pressure and prices begin to fall. To make averaging profitable, an investor should consider the quality of management, the balance sheet, valuation parameters and financial ratios of such companies.
3. When stock prices are rising, many investors prefer to do the averaging considering that now it is a good time to buy more stock. However, the average cost will rise further as a result of using this strategy. If the stock price continues to rise, this strategy will yield good profits when you sell these shares.
Factors that can impact the stock average price
There are four factors that will affect the averaging of a stock price.
1. Fundamentals of a stock
Averaging works only for those stocks which have good fundamentals, belong to a market favorable sector and are much below their highest price level. Otherwise, for a fundamentally poor stock, it could add stress in becoming a liability with deepening losses as you end up buying or selling of these poor stocks.
2. Volatility
Shares average may appear desirable when a fundamentally sound stock and its price is showing great fluctuations and becoming volatile.
3. Futuristic trend
Any stock that exhibits a futuristic trend towards growth for example, digitisation and Information Technology companies stocks are well-suited for price averaging.
4. To correct the wrong in stock picking
This is the basic reason and need for averaging, i.e., to favorably minimize the price of your total holding in buy and sell-side. This implies that for a stock purchased at its highest price, which dipped fast in future averaging could be proved profitable.
Ways to lower your averaging price
These are three ways in which you could lower share avg price in total holdings.
1. Systematic investment plan (SIP)
This may be regarded as an automated mode of averaging price provided by the AMC. This implies buying your favorite stock in smaller lots regularly at every fall, paying in smaller amounts every month and thereby lowering your total average share price of holdings.
2. Random averaging
You have two options if you have purchased a relatively larger lot of a stock and the market suddenly drops which leads to the fall in the price of your total holdings. One option is to sell the shares in order to stop the further loss and the second option might be to believe in the value of a stock to rise further, then use the averaging technique and buy more of that.
3. Imprudent averaging
These could be the stocks that you purchased based on your instinct or emotions that means you end up buying a fundamentally weaker stock or a stock that was once a hit in the market but has now become a penny stock. Averaging in this scenario can be harmful because you end up buying more of a loss making stock and rather you should sell these shares immediately to minimize the loss.
Conclusion on Average Share Price
Knowing how to calculate average price of shares using the weighted average price method helps in determining how your investment is performing as a whole relative to the current share price. The share average pricing is a common strategy used in stock trading, wherein the investor scales up or scales down its total holdings depending on the share price and thereby minimizing the effects of market volatility.
Share:
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# Makey Makey- Storyboard
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This puzzle is used to build a storyboard with different 3D printed puzzle pieces. Each piece is uniquely identified by a different resistor. This is placed on the board, which completes a voltage divider circuit and is recognized by the MakeyMakey through an analogRead function. It used this to create a story telling platform that fills in words of a story/poem based on where a piece is placed on the board.
Required Materials:
1. 6 models to 3D print
2. MakeyMakey board
3. 1 kilo Ohm resistor - 6 nos.
4. 24 magnets
5. 470 Ohm resistor -1
6. 680 Ohm resistor -1
7. 1000 Ohm resistor -1
8. 2200 Ohm resistor -1
9. 3300 Ohm resistor -1
10. 4700 Ohm resistor -1
11. USB cable
12. Copper tape
13. Wires
14. Acrylic Sheet
15. Solder
## Step 1: Build Your Own Storyboard
To build your own storyboard you will have to make six different puzzle pieces. Each of them has a picture on it. Here we use:
i) cloud
ii) moon
iii) star
iv) house
v) tree
vi) flower
Make 3D models of these things and add them on top of the square puzzle pieces(each of them is 40 x 40 mm). Model the puzzle pieces in such a way so that it has place to hold two magnets and a resistor(as in image 3). Now 3D print these puzzle pieces.
## Step 2: Make the Voltage Divider Circuit.
With the puzzle pieces printed now build the voltage divider circuit.
1. Draw your own board for to build the voltage divider circuit. You can use Rhino, Adobe Illustrator, Inkscape or other softwares to do so else use puzzleframe.3dm and Story Puzzle.3dm files attached herewith.
2. Your board should be able to hold the magnets and the voltage divider circuit in place.
## Step 3: Build the Voltage Divider Circuit
Once you are done with the 3D print and laser cutting now build the voltage divider circuit:
1. The circuit should have six resistors(1 kilo Ohm each) and 12 magnets.
2. Place a resistor in between two magnets.
3. Now take wires and connect one end of each of the resistors to ground. In picture the black wire is connected to ground of the MakeyMakey board.
4. Take the other end of the resistor and connect it to a wire which int turn should be connected to an AnalogPin of the MakeyMakey board.
5. Follow the pattern as shown in Image 1 to build the voltage divider circuit.
Your voltage divider should look something like Image 2.
## Step 4: Connecting to MakeyMakey Board
Connect the wires to the MakeyMakey board.
1. The wire connecting all the grounds should be plugged into the GND pin of MakeyMakey board.
2. Connect other ends of the resistors to Analogs pins A0,A1,A2,A3,A4,A5.
Make a connection from the MakeyMakey board to the Voltage divider frame so that when the puzzle pieces are placed on the frame a connection is established.
## Step 5: Completing the Puzzle Pieces
1. Add a different resistor to each of the 6 puzzle pieces. Use 470, 680, 1000, 2200, 3300 and 4700 Ohm resistor and place them inside the gap of the puzzle pieces.
2. Add two magnets on either side of the resistor.
It should look like the image above.
## Step 6: Reprogram MakeyMakey
Once done with the connections and building the circuit you should reprogram your MakeyMakey board. Download the code from Link to the code for reprogramming MakeyMakey.
## Step 7: Making Your Storyboard With the Website
Now you are good to play around with the puzzle pieces.
2. Use the website MakeyMakey Storyboard.
3. Go to the StoryBoard tab.
4. Select any story you would like to build.
5. Place a puzzle piece on the Storyboard frame.
6. Hear the website read it off for you and also see the text change in the website from blank to the puzzle piece you have placed in.
7. Repeat the same for all the other puzzle pieces.
8. Now select Read Story and the website should read out the entire story for you.
## Recommendations
• ### Internet of Things Class
18,914 Enrolled
## 3 Discussions
What kind of magnets did you use for this project? How did you create the "voltage divider circuit"? Can you share the programming you used to create the website/stories?
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## Narrow Search
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Filters: Your search found 5 results.
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# MRC: How Do I Measure This? (Grades 6-8)
This is a lesson about measurement and cratering. Learners will read about the origin of the foot as a standardized unit of measure, work collaboratively to conduct an experiment about cratering, and collect and record data to draw logical and... (View More)
# MRC: How Do I Measure This? (Grades 3-5)
This is a lesson about measurement and cratering. Learners will read about the origin of the foot as a standardized unit of measure, work collaboratively to conduct an experiment about cratering, and collect and record data to draw logical and... (View More)
# Does The Sun Move?
This is an activity about modeling the apparent motion of the Sun as seen from Earth. Learners will use a flashlight, toothpick, and styrofoam model Sun to mimic the relative shadow motion produced by a sundial. The activity will help learners... (View More)
Audience: Informal education
Materials Cost: \$1 - \$5
# How Does Light Travel?
This is an activity about how light travels. Learners will perform two experiments. The first explores blocking light to create shadows. The second asks learners to use mirrors to figure out that light travels in a straight line. This is Activity 4... (View More)
Audience: Informal education
Materials Cost: \$5 - \$10
# Freshwater Macroinvertebrates Protocol
This activity guides students through sampling, identification and counting of macroinvertebrates sampled in a GLOBE hydrology study site, and understand how the taxa composition found in the sample can be an indicator of water quality and ecosystem... (View More)
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https://mathematica.stackexchange.com/questions/267766/coupled-pde-using-ndsolve
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# Coupled PDE using NDSolve
The problem is to study the solution of a biharmonic which can be decomposed into a coupled PDE system $$\Delta f - \nabla f\cdot\Gamma_{1} = g \\ \Delta g = \nabla f\cdot\Gamma_{2}$$where $$f, g$$ are smooth in $$\Omega = [-1, 1]\times[-1, 1]$$, and $$\Gamma_ 1$$ and $$\Gamma_ 2$$ are continuous in $$\Omega$$ and
$$\Gamma_ 1 = \frac{1}{(x^2+y^2)^{3/2}}(x, y)$$ and $$\Gamma_ 2 = \frac{1}{(x^2+y^2)^{3/2}}(-y, x)$$ .The boundary conditions are : $$\psi (-1, y) = 0, \quad\psi (1, y) = 1, \quad\hat {n}\cdot \nabla\psi | _ {y = \pm 1} = 0$$ I am not sure if I get the right solution because it gives me $$f = x - y$$ and $$g = x + y$$ which does not satisfy the PDE system. I don' t trust the solutions I get, because they remain unchanged when I change the vectors $$\Gamma$$ .
eqn1 = {Laplacian[psi[x, y], {x, y}] -
1/(x^2 +
y^2) (D[psi[x, y], {x, 1}] x/Sqrt[x^2 + y^2] +
D[psi[x, y], {y, 1}] y/Sqrt[x^2 + y^2]) == u1[x, y],
Laplacian[u1[x, y], {x, y}] - 1/(
x^2 + y^2) (D[psi[x, y], {x, 1}] y/Sqrt[x^2 + y^2] -
D[psi[x, y], {y, 1}] x/Sqrt[x^2 + y^2]) == 0 };
bcs = {psi[-1, y] == 1, psi[1, y] == 0}
sol = NDSolveValue[{eqn1, bcs}, {u1, psi}, {x, -1, 1}, {y, -1, 1}]
Exeprts, I am not in need of the full solution or anything. I am just in need of a hint as to whether this is the right approach to solve this problem, or is finite difference the way to go about solving a fourth order PDE. Thanks for the help!
EDIT :: (Apologies for editing this question, instead of starting a new one if that's the policy.)
I am confused on how to implement boundary conditions on a coupled system of PDEs. Let me state the physical problem that I am solving which will help explain my boundary conditions for the problem. The PDE system stated in the first equation describes the hydrodynamic flow where $$f$$ is the stream function and the fluid velocity is related to the stream function $$f$$ by $$v = \hat{z}\times \nabla f$$.This is done to impose the incompressibility condition that $$\nabla\cdot v=0$$ My boundary conditions are that the fluid enters from the right and exits from the left, and that on the top and bottom of the box the fluid does not slip, so that is the tangential component of $$v$$ is 0.
My idea was that I can set a Dirichlet condition on the top and bottom of the box so that the equipotential lines are horizontal in the direction of the fluid flow, and on the left and right hand side I can set a uniform flux in and out parallel to the horizontal axis, as shown in the figure in 1. How do I go about setting a Neumann condition on $$u_1$$. If I set the second equation equal to a Neumann value Mathematica ignores it and tells me that no boundary conditions have been set. Perhaps then I would need to set a Dirichlet condition on the left and right also, that is $$f(x=\pm L_x/2,y) = c_0 y + c_1$$? A hint in the right direction would be enough. Thanks for the help.
EDIT2 : Poiseuille flow :
T = 500; \[Nu] = 0.1 ; t0 = 0.01;
\[Psi]1[0][x_, y_] := 0 ;
Do[{\[Psi]1[i], \[Xi]1[i]} =
NDSolveValue[{{-(\[Psi][x, y] - \[Psi]1[i - 1][x, y])/
t0 + \[Nu] Laplacian[\[Psi][x, y], {x, y}] - \[Xi][x, y],
Laplacian[\[Xi][x, y], {x, y}]} == {0,
0}, {DirichletCondition[\[Psi][x, y] == 1,
y == Ly/2 || -Lx/2 <= x <= Lx/2],
DirichletCondition[\[Psi][x, y] == -1,
y == -Ly/2 || -Lx/2 <= x <= Lx/2],
DirichletCondition[\[Xi][x, y] == 0, x == Lx/2],
DirichletCondition[\[Xi][x, y] == 0, x == -Lx/2],
DirichletCondition[\[Psi][x, y] == 2 y/Ly,
x == Lx/2]}}, {\[Psi], \[Xi]}, {x,
y} \[Element] \[CapitalOmega],
Method -> {"FiniteElement",
"InterpolationOrder" -> {\[Psi] -> 2, \[Xi] -> 2},
"MeshOptions" -> {"MaxCellMeasure" -> 0.001}}], {i, 1, T}];
How to take care of $$\xi$$ ?
• In Latex you define solution on $\Omega$=Rectangle[{-1,-1},{1,1}], while in NDSolve you use one part $\Omega$ only {x, 0, 1}, {y, 0, 1}. Is it typo? May 4, 2022 at 2:14
• @AlexTrounev yes sorry for the typo. May 4, 2022 at 6:13
• I don't understand how you model has been derived from NSE. Could you give a link to the paper where this model described? May 7, 2022 at 5:24
• here's a reference - escholarship.org/uc/item/8kw631r4 May 7, 2022 at 8:03
• Ah, this is MHD, and therefore part of this system describes magnetic field. Could you show what part of your model follows from equations (2), (3) of the paper linked? May 7, 2022 at 9:36
To solve this problem we can use FEM as follows
Clear["Global*"]
Needs["NDSolveFEM"]
reg = Rectangle[{-1, -1}, {1, 1}]; mesh =
ToElementMesh[reg, MaxCellMeasure -> 0.001]
eqn1 = {Laplacian[psi[x, y], {x, y}] -
1/(x^2 + y^2) (D[psi[x, y], {x, 1}] x/Sqrt[x^2 + y^2] +
D[psi[x, y], {y, 1}] y/Sqrt[x^2 + y^2]) == u1[x, y],
Laplacian[u1[x, y], {x, y}] -
1/(x^2 + y^2) (D[psi[x, y], {x, 1}] y/Sqrt[x^2 + y^2] -
D[psi[x, y], {y, 1}] x/Sqrt[x^2 + y^2]) == 0};
bcs = {DirichletCondition[psi[x, y] == 0, x == -1],
DirichletCondition[psi[x, y] == 1, x == 1]};
sol = NDSolve[{eqn1, bcs}, {u1, psi}, Element[{x, y}, mesh]]
Since there is no any boundary condition for u1 we have a message
NDSolve::femibcnd: No DirichletCondition or Robin-type NeumannValue was specified for {u1}; the result may not be unique.
Visualization
{Plot3D[Evaluate[psi[x, y] /. sol[[1]]], Element[{x, y}, mesh],
ColorFunction -> "Rainbow", MeshStyle -> White,
PlotTheme -> "Marketing", Boxed -> False, AxesLabel -> Automatic,
PlotLabel -> "psi"],
Plot3D[Evaluate[u1[x, y] /. sol[[1]]], Element[{x, y}, mesh],
ColorFunction -> "Rainbow", MeshStyle -> White,
PlotTheme -> "Marketing", Boxed -> False, AxesLabel -> Automatic,
PlotLabel -> "u1"]}
Update 1. In a case of homogenous flow on the right and left borders we have
Clear["Global*"]
Needs["NDSolveFEM"]
reg = Rectangle[{-1, -1}, {1, 1}]; mesh =
ToElementMesh[reg, MaxCellMeasure -> 0.001];
eqn1 = {Laplacian[psi[x, y], {x, y}] -
1/(x^2 + y^2) (D[psi[x, y], {x, 1}] x/Sqrt[x^2 + y^2] +
D[psi[x, y], {y, 1}] y/Sqrt[x^2 + y^2]) - u1[x, y] == 0,
Laplacian[u1[x, y], {x, y}] -
1/(x^2 + y^2) (D[psi[x, y], {x, 1}] y/Sqrt[x^2 + y^2] -
D[psi[x, y], {y, 1}] x/Sqrt[x^2 + y^2]) == 0};
bcs = {DirichletCondition[{psi[x, y] == y,
u1[x, y] == -y/(x^2 + y^2)^(3/2)}, y == -1 || y == 1],
DirichletCondition[psi[x, y] == y, x == 1 || x == -1]};
sol = NDSolve[{eqn1, bcs}, {u1, psi}, Element[{x, y}, mesh]]
Visualization
StreamDensityPlot[
Evaluate[{-Derivative[0, 1][psi][x, y],
Derivative[1, 0][psi][x, y]} /. sol[[1]]], Element[{x, y}, mesh],
ColorFunction -> Hue, PlotLegends -> Automatic]
• Thanks for the answer Alex. I had a question related to this problem, when imposing boundary conditions for a coupled system, is it possible to define a Dirichlet and Neumann boundary condition simultaneously for the two dependent variables in the system. What I mean is that in the problem above the psi is the stream function that imposes an incoming and outgoing stream flux on the top and bottom of the box and u1 is the laplacian of this quantity which is zero, so is it possible to impose a Dirichlet boundary condition of u1=0 and a Neumann boundary condition of constant on psi ? May 6, 2022 at 19:53
• @Charlie Do you mean condition for velocity $\vec{v}=0$ at $y=\pm 1$? May 9, 2022 at 10:53
• @ Alex, the boundary conditions on $y=\pm 1$ are the no-slip conditions May 9, 2022 at 15:54
• @Charlie It is clear from Figure 1 you shown that you mean slip conditions $\vec{v}.\vec{n}=0$. For the no-slip conditions in standard approach we need to add $f=0, \partial^2 f/\partial n^2=0$ on the border. But your equations are not in standard form. Can you show your no-slip boundary conditions in mathematical form? Above you suppose $f=0, u1=0$, but it is not working as no-slip conditions. May 10, 2022 at 3:41
• apologies I didn't specify the boundary conditions properly, I think the boundary conditions should be - at the top and bottom $\psi=1$ and $\psi=0$ respectively, which implies $v=\hat{z}\times\nabla\psi = 0$ and thus $u_1 = 0$which imposes the no-slip condition. At the left and right, flow is homogenous and $\psi = y$ and $u_1$ is found from second PDE in the coupled PDE system. May 10, 2022 at 16:46
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# MCCB Selection
I have 3 phase 220V domestic power supply to my home by 16 mm aluminium 4 core armored cable. I have a sanction of 15KW load. With a concern for safety I intend to retrofit a MCCB of suitable rating. Request you to please guide me with respect to Amperage, Icu 240V, Icu 410V & Ics 415 V ratings required for the MCCB. Let me know if you require any additional information. asked 4/19/2016 Sandeep Dogra2
kVA = kW/0.8P.F kVA = 15/0.8 kVA = 18.75kVA I = kVA/(1.732x415V) I = 18.75/718.78 I = 0.0260kA I = 26A I = 26A x 1.2 (overload factor) I (MCCB,TP) = 32A Icu = 10kA and Ics = 6kA will be more then enough. 32A MCCB, TP as Incoming breaker Further outgoing breakers for 220V system could be MCCB's, DP or SP as per requirement. answered 1/29/2017 حسان نوابی 2 Hi friend, As per your given details MCCB is not required MPCB is enough. But you have MCCB for this load you chose the 50 A MCCB for 220 V. For 415 V 32 A MCCB is enough. I=P/(SQRT(3)*220*0.85)) You take your loads I=15000/(1.732*220*0.85) it comes 46 A you take 15% extra 50 A MCCB is enough. answered 6/16/2016 Dinesh137 2
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https://www.csunplugged.org/en/topics/binary-numbers/programming/binary-cards-representing-any-number/
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# Display the binary cards needed to represent any decimal number
Learning outcomes
Students will be able to:
## Requirement:
Write a program that asks the user to enter any decimal number as the input and displays the binary cards representing that number (B for displaying black cards with no dots, and W for displaying white cards with dots) as the output.
## Testing examples:
Your program should display the outputs shown in this table for the given inputs provided:
Input Output
11 The binary representation for the number 11 is WBWW
255 The binary representation for the number 255 is WWWWWWWW
256 The binary representation for the number 256 is WBBBBBBBB
## Languages
Scratch
#### What it should look like
Click on the green flag, enter the inputs provided in the “testing examples” to see the expected output of your program.
Recommended blocks
when green flag clicked
say (join (join (join [The binary representation for the number ] (answer)) [ is ]) (cards))
repeat until <(number) < (bit value)>
end
if <<(number) > (bit value)> or <(number) = (bit value)>> then
else
end
repeat until <(bit value) = [1]>
end
set [number v] to (answer)
set [bit value v] to [1]
set [cards v] to []
set [bit value v] to ((bit value) * (2))
set [bit value v] to ((bit value) / (2))
set [cards v] to (join (cards) [W])
set [number v] to ((number) - (bit value))
set [cards v] to (join (cards) [B])
Hints
• Make variables called:
• “number” and set its value to the input number given by the end user.
• “bit value” and set its value to ‘1’. Find the smallest bit value which is larger than the “number” by doubling value of “bit value” until it is bigger than the “number”.
• “cards” is a string variable and stores the binary cards needed (‘B’ for black cards and ‘W’ for white cards).
• Set the variable “bit value” to 1 and find the smallest “bit value” which is larger than “number” by multiplying “bit value” by 2 (Use the () * () block under “Operators” to multiply the “bit value” by 2) until it is larger than “number”. You can do this by using a repeat until <> loop.
• Now divide the “bit value” by 2 and check if “number” is greater than or equal to “bit value”. If it is, add ‘W’ to string variable “cards” and subtract “bit value” from the “number”. If not, add ‘B’ to string variable “cards”. Repeat until “bit value” is equal to 1. Display the value of “cards” as the output.
• Test your program with some values on the boundaries (for example test it with numbers 255 and 256).
Python
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## Ball mill_Henan Mine Machinery Co., Ltd- Mill Speed - Critical Speed - Paul O. Abbe
Aug 02, 2013· In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum/ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do. Often, the ball used in ball mills is oversize "just in case". Well, this safety factor can cost you much in recovery and/or mill liner wear and tear.2.4 Effect of ball size 29 2.4.1 Empirical approaches 29 2.4.2 Probabilistic approaches 33 2.5 Abnormal breakage 36 2.6 Effect of ball mixture 37 2.6.1 Ball size distribution in tumbling mills 37 2.6.2 Milling performance of a ball size distribution 40 2.7 Summary 41 Chapter 3 …
## Ball mill - Wikipedia - Ball Mill Explained - saVRee
Ball mill is suitable for grinding various ores and other materials. It is widely used in mineral processing, building materials and chemical industry. It can be divided into dry and wet grinding methods. According to the different ways of discharging, it can be divided into two types: grid type and overflow type.We investigated tlte effects of ball diameter and feed size on the rate constant when the ball mass, feed mass, and the mill's rotational speed were constant. The results indicated that the grinding rate constant can be expressed by modifying the equation that was proposed by Snow as the function of ball diameter and feed size. 1.
## Recommended Ball Mill Speed & Liner Configuration - Ball Nose Finishing Mills Speed & Feed Calculator - DAPRA
B Recommended make-up ball size, inches. F80 Circuit feed 80 percent passing size, microns. K A constant, 350 for wet overflow mills, etc. WI Bond test Work Index, kWh/short ton. SG Ore specific gravity. Cs Mill speed in percent of critical. ID Mill inside (liners, working) diameter, feet. It pre-dated (semi-)autogenousMill Speed - Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product.
## EFFECTS OF GRINDING MEDIA SHAPES ON BALL MILL … - Calculate and Select Ball Mill Ball Size for Optimum Grinding
High-Speed Steel Ball End Mills with Two Milling Ends When one end wears out, switch to the opposite end for two times the life of a standard high-speed steel end mill. Made of high-speed steel, these end mills are for general purpose milling in most material, such as aluminum, brass, bronze, iron, and steel.大量生產
## Mill Speed - an overview | ScienceDirect Topics - Factors Affecting Ball Mill Grinding Efficiency
Oct 25, 2017· The following are factors that have been investigated and applied in conventional ball milling in order to maximize grinding efficiency: a) Mill Geometry and Speed – Bond (1954) observed grinding efficiency to be a function of ball mill diameter, and established empirical relationships for recommended media size and mill speed that take this factor into account.Dipak K. Sarkar, in Thermal Power Plant, 2015 4.6.1 Low-speed mill. Mills operating below 75 rpm are known as low-speed mills.Low-speed units include ball or tube or drum mills, which normally rotate at about 15–25 rpm.Other types of mills, e.g., ball-and-race and roll-and-race mills, that generally fall into the medium-speed category may also be included in this category provided their ...
## Silicon Powder Properties Produced in a Planetary Ball - The critical speed of the ball mill is less than 45% ...
Jul 18, 2013· (2000). Ball-mill rotation speed and rate of particle breakage: results for a full-scale unit. Mineral Processing and Extractive Metallurgy: Vol. 109, No. 3, pp. 161-164.Major parameters for ball milling Temperature Size and Number of the balls Nature of the balls Rotation speed 8 . ... Ball milling should be considered as a potentially attractive solution for solvent-free synthesis. 21 Solvent free One-pot
## Ball Nose Milling Strategy Guide - In The Loupe - How to Size a Ball Mill -Design Calculator & Formula
EFFECTS OF GRINDING MEDIA SHAPES ON BALL MILL PERFORMANCE Niyoshaka Nistlaba Stanley Lameck A dissertation submitted to the Faculty of Engineering and The Built Environment,Steel Ball Mills. Benefits of Tumble Milling: 1. HIGH EFFICIENCY – Due to the relatively slow rotational speed but large mass of media, more of the energy goes into milling and less wasted as heat. 2. NARROW PARTICLE DISTRIBUTION – Solids milled in tumble mills are normally so fine and consistent in size that it's rare to require ...
## Ball-mill rotation speed and rate of particle breakage - Ball Mill Critical Speed & Working Principle - YouTube
2), the filling ratio in the volume of milling ball and container (j), milling ball size (d B), milling revolution speed (nR), and initial mass of bulk MoS 2 (m MoS 2). The yield of exfoliation is found to be 95% at the optimum ball milling conditions (SC/MoS 2 = 0.75, j = 50%, m MoS 2 = 0.20g). In addition, yield and size of the exfoliated MoSIf a ball mill uses water during grinding, it is a 'wet' mill. A typical ball mill will have a drum length that is 1 or 1.5 times the drum diameter. Ball mills with a drum length to diameter ratio greater than 1.5 are referred to as tube mills. Ball mills may be primary or secondary grinders.
## Ball Milling - University of Massachusetts Boston - EFFECT OF PARTICLE FILLING AND SIZE ON THE BEHAVIOUR …
Jun 16, 2021· The effects of ball milling parameters on the composites and particle size of TiB2 powder were investigated. It was shown that the average particle size of TiB2 powder decreased from 5.8 to 1.59 µm and the wear rate of WC balls was 1.58 wt%, when the ball-to-powder weight ratio (BPR), the rotary speed and milling time and were 10:1, 600 rpm ...Ball end mills, also known as ball nose end mills, are used for milling contoured surfaces, slotting and pocketing. A ball end mill is constructed of a round cutting …
## Milling Services: Jet, Ball & Attrition Milling | Steward - Ball End Mills - MSC Industrial Supply
Nov 20, 2019· The conditions of the ball milling such as SC/MoS 2 weight ratio, ball filling ratio (), initial amount of MoS 2 (), revolution speed (n R), ball milling duration (t), ball size (d), and exfoliant were optimized in terms of yield, size and thickness of the MoS 2 nanosheet. As a result, the size and thickness were controlled by tuning the ...Oct 19, 2006· Posted 21 October 2006 - 04:24 PM. Congratulations on your ball mill! You'll want to further reduce your motor speed by 3:1 to mill more efficiently. Right now, your RPM is just slightly under the speed where the media won't do any work. The optimum speed for your jar to turn is around 90 RPM like Frozentech said.
## Size-controlled MoS2 nanosheet through ball milling - Ball Mill Parameter Selection & Calculation - Power ...
Correlation between process parameters and milling efficiency - DiVA. well as the kinetic energy of the milling ball affected the size reduction; more collisions or higher energy resulted in a higher milling efficiency. ..... to high speed steel in high quantity production because it allows faster .... glycol, PEG, and grinding liquid, a mixture of ethanol and water, in a cylindrical ball mill ...The particles size distribution slope is not affected by the mill speed (see Graph 2) as the curves are mostly overlapping. Density These tests were done with a ∅0.82 x 1 m grate discharge mill with a 30 mm graded ball charge and a 30% filling degree. The mill discharge pulp density was increased from 68.8% to …
# LCDY
خلال 30 عامًا من العمل الشاق ، بنى موظفو LCDY تفوقًا في المصداقية والجودة الممتازة وخدمة العلامة التجارية "LCDY"
#### ابقى على تواصل
رقم 1688 ، طريق Gaoke شرق ، حي بودونغ الجديد ، شنغهاي ، الصين.
#### النشرة الإخبارية
تقوم الشركة بشكل أساسي بتصنيع الكسارات المتنقلة والكسارات الثابتة وآلات صنع الرمل
حقوق النشر © 2023.LCDY كل الحقوق محفوظة.خريطة الموقع
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# Distance between Luxi (LUM) and Skardu (KDU)
Flight distance from Luxi to Skardu (Dehong Mangshi Airport – Skardu Airport) is 1567 miles / 2522 kilometers / 1362 nautical miles. Estimated flight time is 3 hours 27 minutes.
Driving distance from Luxi (LUM) to Skardu (KDU) is 2723 miles / 4383 kilometers and travel time by car is about 54 hours 6 minutes.
## Map of flight path and driving directions from Luxi to Skardu.
Shortest flight path between Dehong Mangshi Airport (LUM) and Skardu Airport (KDU).
## How far is Skardu from Luxi?
There are several ways to calculate distances between Luxi and Skardu. Here are two common methods:
Vincenty's formula (applied above)
• 1566.861 miles
• 2521.618 kilometers
• 1361.565 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1565.639 miles
• 2519.652 kilometers
• 1360.503 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Dehong Mangshi Airport
City: Luxi
Country: China
IATA Code: LUM
ICAO Code: ZPLX
Coordinates: 24°24′3″N, 98°31′54″E
B Skardu Airport
City: Skardu
Country: Pakistan
IATA Code: KDU
ICAO Code: OPSD
Coordinates: 35°20′7″N, 75°32′9″E
## Time difference and current local times
The time difference between Luxi and Skardu is 3 hours. Skardu is 3 hours behind Luxi.
CST
PKT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 184 kg (405 pounds).
## Frequent Flyer Miles Calculator
Luxi (LUM) → Skardu (KDU).
Distance:
1567
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1567
Round trip?
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https://onemathematicalcat.org/Math/Precalculus_obj/extValues.htm
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Extreme Values of Functions (Max/Min)
# EXTREME VALUES OF FUNCTIONS (MAX/MIN)
LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Thanks for your support!
• PRACTICE (online exercises and printable worksheets)
Roughly, ‘extreme values’ are the $\,y\,$-values of high/low points on a graph.
You also usually want to know the $\,x\,$-value(s) where these high/low points occur.
The $\,y\,$-value of a highest point is called a MAXIMUM value.
The $\,y\,$-value of a lowest point is called a MINIMUM value.
As illustrated below,
the ONLY places that extreme values can occur
are at horizontal tangent lines, kinks/breaks, and endpoints.
maximum value is $\,7\,$;
it occurs at $\,x = 3\,$
minimum value is $\,2\,$;
it occurs at $\,x = 3\,$
The plural of maximum is maxima.
The plural of minimum is minima.
maxima/minima can occur at horizontal tangent lines
maximum value is $\,7\,$;
it occurs at $\,x = 3\,$
minimum value is $\,2\,$;
it occurs at $\,x = 3\,$
A ‘kink’ is a sudden change in direction.
maxima/minima can occur at kinks
maximum value is $\,7\,$;
it occurs at $\,x = 3\,$
minimum value is $\,2\,$;
it occurs at $\,x = 3\,$
Imagine tracing a graph with a pencil.
A ‘break’ is a place where you have to pick up the pencil.
maxima/minima can occur at breaks
maximum value is $\,7\,$;
it occurs at $\,x = 5\,$
minimum value is $\,2\,$;
it occurs at $\,x = 1\,$
maxima/minima can occur at endpoints
The ONLY places that extreme values can occur
are at horizontal tangent lines, kinks/breaks, and endpoints.
But, you don't HAVE to have an extreme value at such places!
In all the examples below, there are greater and lesser values arbitrarily close by.
horizontal tangent line; no extreme value here kink; no extreme value here break; no extreme value here
endpoint;
no extreme value here
To the left of the endpoint, the function oscillates:
it's $\,2\,$ for rational inputs, and $\,-2\,$ for irrational inputs.
You have to work hard to get an example of an endpoint that's not a max or min!
## A Discussion of Extreme Values from a Non-Graphical Viewpoint
Suppose you have a function, $\,f\,$. A function is a rule—it takes an input, does something to it, and gives a corresponding output. The function $\,f\,$ can be viewed as a box: drop an input $\,x\,$ in the top $\,f\,$ does something to the input the corresponding output $\,f(x)\,$ drops out the bottom
For extreme value questions, you have the following scenario: You're interested in a particular set of inputs—it might be the entire domain; it might be some interval. You drop these inputs of interest into the function box, and get a corresponding pile of outputs. Take a look at this output pile: Is there a greatest number in the output pile? That is, can you pick up a number that is greater than or equal to all the other numbers in the pile? If so, we call this number an extreme value—in this case, a maximum. Is there a least number in the output pile? That is, can you pick up a number that is less than or equal to all the other numbers in the pile? If so, we call this number an extreme value—in this case, a minimum. In both cases, we likely want to know what input(s) gave rise to these ‘extreme’ outputs.
extreme value questions
have to do with finding the greatest/least values of the output of a function,
and figuring out what input(s) they came from
## Greatest/Least overall?Or just close by?(Global versus Local Max/Min)
Sometimes you're interested in highest/lowest values overall (like at B, E, and F). These are called global (or absolute) max/min. Sometimes you're interested in ‘local’ highest/lowest values (like at A, C, and D). These are called local (or relative) max/min. These are highest/lowest, provided you don't look too far away! [Note: You'll learn in Calculus that a global max is also a local max, and a global min is also a local min.] Some people like the terminology global/local. Others like absolute/relative. You choose! Calculus provides language and tools to enable precise discussions of extreme values and related concepts. The informal language used here (e.g., ‘high/low points’, ‘breaks’, ‘kinks’) will all be made precise in Calculus. This section is only an informal introduction.
The field of Operations Research is devoted to maximizing or minimizing real-world objectives.
Maximize profits. Minimize costs. Minimize the amount of material needed to construct a box with a specified volume.
Extreme values have extreme applicability!
## And Finally—You May Not Have Extreme Values!
You may not have any extreme values. You might have a max, but not a min. You might have a min, but not a max. In the graph shown, the number $\,5\,$ is ‘trying to be’ the maximum value. But, there's no $\,5\,$ in the output pile, because the point isn't there! For example, you could pick up a $\,4.9\,$ from the output pile—but this wouldn't be greatest, because there's a $\,4.99\,$ there! And $\,4.99\,$ isn't greatest, because there's a $\,4.999\,$ there (and so on). There's an important theorem in Calculus, called the Max/Min Theorem (or the Extreme Value Theorem) that guarantees conditions under which you'll have both a max and a min. Calculus is so wonderful!
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
review of quadratic functions:
vertex form, max/min, intercepts, more
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
AVAILABLE MASTERED IN PROGRESS
(MAX is 17; there are 17 different problem types.)
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https://thelongtrek.com/subnet/subnetting-intro.htm
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20 February 2013
Ken Long - kenglong@gmail.com
# A subnetting problem using all I’ve learned so far.
## 29 . 99 . 37 . 88 /17
We know this is a class A and we know the default mask for a class A is 255 . 0 . 0 . 0
First, write out your binary charts for easy reference while you work on the problem.
### 1. Find the subnet mask in binary and decimal.
First, write out the mask using 255 for the default portion, 1’s for all the borrowed bits and zeros for all the host bits.
255 . 1 1 1 1 1 1 1 1 . 1 0 0 0 0 0 0 0 . 0 n . s s s s s s s s . s h h h h h h h . h
Converting from binary to decimal, the mask is 255.255.128.0. The magic number is 128.
### 2. Find the subnetwork number.
The magic number is in the third octet so we need to convert that octet in the IP, 37, to decimal. Write out your conversion table then do the conversion.
128 64 32 16 8 4 2 1
37 in decimal = 0 0 1 0 0 1 0 1 in binary. Now perform a bit-wise AND with the binary number you just found and the binary version of the third octet in the mask.
1 0 0 0 0 0 0 0
0 0 1 0 0 1 0 1
---------------
0 0 0 0 0 0 0 0
That gives us the network number for the given IP and mask.
29 . 99 . 0 . 0
Note that we simply brought down the 99 from the second octet because that octet in the mask was all ones. This example can be extra confusing because the borrowed bits cross an octet boundary. That’s why I used it here.
### 3. Find the first useable host address.
The first host address is always the subnetwork number plus 1.
29 . 99 . 0 . 1
### 4. Find the next subnetwork number.
This is always the current network number plus the magic number. In this case it is:
29 . 99 . 128 . 0
Note that the 128 is in the 3rd octet because that’s where the magic number is.
The BC is always the next subnetwork number minus one and is always odd.
29 . 99 . 127 . 255
### 6. Find the last useable host address.
The last host address is always the BC minus one and is always even.
29 . 99 . 127 . 254
The total number of bits is 32. The mask is using 17. That leaves 15 for the host portion. Now raise 2 to the 15th power to get the total number of IPs.
215 = 32768
### 8. Find the total number of valid host addresses.
This is always the total number of IPs minus 2.
32768 - 2 = 32766
### 9. Find the total number of subnetworks.
Take the difference between the default mask, /8, and this mask, /17, to get 9. Now raise 2 to the power of 9.
29 = 512
### 10. Find the first two subnetwork addresses.
The first subnetwork number will have all the borrowed bits set to zero *and* all the host bits set to zero. Looking back at the binary mask in #1 above, we can see that if we set all the borrowed bits to zero we get the following:
First subnetwork number:
29 . 0 . 0 . 0 (Magic# = 128; 128 x 1 – 128)
Now we add the magic number to that, in the correct octet, to get the second subnetwork number:
29 . 0 . 128 . 0 (aka: Magic number x 1)
And so on…
29 . 1 . 0 . 0 29 . 1 . 128 . 0 29 . 2 . 0 . 0
### 11. Find the last two subnetwork addresses.
Here’s a trick. The very last subnetwork number will be the mask! Here’s how that works. Keep the first octet of the IP because it’s a class A. If it were a class B, you would keep the first two and for a class C, the first three. Then fill in the rest of the octets from the mask.
29 . 255 . 128 . 0
Next to last subnetwork:
29 . 255 . 0 . 0
And so on…
29 . 254 . 128 . 0 29 . 254 . 0 . 0 29 . 253 . 128 . 0
### In summary.. .
Given IP address and CIDR: 29 . 99 . 37 . 88 /17 Subnet mask: 255 . 255 . 128 . 0 Subnetwork address: 29 . 99 . 0 . 0 Valid host address range: 29.0.0.1 to 29.99.127.254 Broadcast address: 29 . 99 .127 . 255 Total number of IPs: 32768 Total number of valid host addresses: 32766 Subnets: 29 . 0 . 0 . 0 to 29 . 255 . 128 . 0, incrementing by 128 Total number of subnetworks: 512
## Next:
Now, let's apply all of this to a practical subnetting problem.
Thanks to Mr. Hart and Mr. Clauss at Central New Mexico Community College for taking me this far. I'm sure I'll refine this as more of the steps become second nature and I more fully understand the underlying mechanics of it all.
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# Solve this
Question:
If $\cos \theta=\frac{4}{5}$ then $\tan \theta=?$
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(C) $\frac{3}{5}$
(d) $\frac{5}{3}$
Solution:
Given: $\cos \theta=\frac{4}{5}$
Since, $\cos \theta=\frac{B}{H}$
$\Rightarrow B=4$ and $H=5$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow P^{2}+4^{2}=5^{2}$
$\Rightarrow P^{2}=25-16$
$\Rightarrow P^{2}=9$
$\Rightarrow P=3$
Therefore,
$\tan \theta=\frac{P}{B}=\frac{3}{4}$
Hence, the correct option is (a).
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# 20 Top Electrical Wiring Residential 18Th Edition Blueprints Images
## Other recommended diagram ideas:
Top Electrical Wiring Residential 18Th Edition Blueprints Images - A.) How excessive above the finished ground within the dwelling room are receptacles positioned? 12in, (pg. Forty nine) outdoor receptacle retailers on this living are positioned 18 in. Above grade. (Pg. 49) inside the areas provided, draw the perfect image for every of the descriptions listed in (a) via (r).??not covered within the studying undertaking the the front fringe of a box installed in a combustible wall need to be flush with the finished surface. (Pg. 35) listing the most range of 12 awg conductors authorised in a: (pg. Forty four—desk 314.Sixteen(a) a.) 4 in. X 1½ in. Octagon box. ----------------6 b.) Four-eleven/16 in. X 1½ in. Rectangular field.-----------13 c.) 3 in. X 2 in. X 3½ in. Tool container. ----------Eight.
Two 12/2 and 14/2 nonmetallic-sheathed awg cables enter a container. Each cable has an gadget grounding conductor. The 12 awg conductors are related to a receptacle. Two of the 14 awg conductors are linked to a toggle transfer. The other two 14 awg conductors are spliced together seeing that they function a switch loop. The container consists of two cable clamps. Calculate the minimal cubic-inch quantity required for the container. Solution: ground cord----#12 (2.25cu. In.)??1 x 2.25 ----------2.25 cu. In. Conductors: 2—#12/2 ---- 2 x 2 x 2.25 ------------------------------9.00 cu. In. 2—#14/2 ---- 2 x 2 x 2.00 ------------------------------8.00 cu. In. Devices: 1—toggle transfer –1 x 2 x 2.25 ------------------------4.50 cu. In. 1—receptacle –1 x 2 x 2.25 ----------------------------4.50 cu. In. Fittings (clamps) 2—cable clamps—1 x 2.25 ----------------------------2.25 cu. In. ------------------------------------------------------------- 31.50 cu. In. Using the identical quantity and size of conductors as in questions 28, however using electrical steel tubing, calculate the minimum cubic-inch volume required for the box. There can be no separate gadget grounding conductors, nor will there be any clamps within the box. Answer: floor twine—should figure on one ground wire because of the nearby code requirement of all conduit structures requiring a floor cord. The answers might be the equal because the electric steel tubing would require using metallic connectors that should enter the box, you'll be counted them as fittings, which means that that irrespective of how many fittings are entering the box the you may parent it at ( 1-times the cubic inch of the largest conductor that enters the container. To permit for adequate conductor duration at electric outlet and tool boxed to make up connections, 300.14 requires that no longer much less than [ 3 in.(75 mm). 6 in. (150 mm), 9 in. (225 mm)] of conductor period be supplied. This duration is measured from wherein the conductor emerges from the cable or raceway to the end of the conductor. For box openings having any measurement less than 8in. (200mm), the minimal period of conductor measured shape the box establishing in the wall to the end of the conductor is [ 3 in. (Seventy five mm), 6 in. (150 mm), 9 in. (225 mm). Circle the appropriate solution. Solution web page 21 figure 2-2 whilst wiring a residence, what should be considered while putting in wall bins on both aspects of a commonplace partition that separates the storage and a habitable room? The hearth score of the walls should be maintained which means that that packing containers won't be set up “again to back” or in the same air void. A separation of 24 inches horizontal have to be maintained except the bins are listed as being fireplace resistant. The boxes that are at the fire rated wall need to be nicely blanketed or encased with a fire puddy. Does the nec® permit metal raceways for use with nonmetallic boxes? Sure _____x______ no__________ nec®______nec® 314.3 (exception) –previded that every one metal raceways or cables coming into the box are bonded together to maintanin the integrity of the grounding route to other device within the installation. P. 33.
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Maths of Planet Earth | Limitless Applications
Puzzles
Published on June 12th, 2013 | by Emily Corbett
0
# Puzzle Challenge 5
This months puzzles have been provided by Michael Evans from AMSI
The winner of this months puzzle of the month will receive a signed copy of a Keith Devlin book and a \$50 voucher, to enter email full working to all puzzles to mpe@amsi.org.au by 31 August. Entries will be judged on number of correct answers, originality and presentation of ideas.
Where is the water?
1. There is a fixed volume of water available. There are n jugs, each big enough to hold all of the water. Initially all of the jugs contain the same amount of water. Water can be poured from any jug A to any other jug B so that the amount poured from jug A to jug B is the same as the amount of water in jug B. For which values of n is it possible to collect all the water in one jug?
Intriguing integer
2. Show that for every sufficiently large integer n, it is possible to split the integers 1, 2,…, n into two disjoint subsets such that the sum of the elements in one set equals the product of the elements in the other.
Hitting the social scene
3. There are 2N people at a party. Each knows at least N others. Prove that one can always choose four people and place them at a round table so that each person knows both neighbours.
Colourful learning
4. In a regular octagon each side is coloured blue or yellow. From such a colouring a new colouring is obtained in one step as follows: If two neighbours of a side have a different color the new colour of that side will be blue, otherwise it will be yellow (the colours are modified simultaneously).
a. Show that after finitely many moves all sides will be coloured yellow.
b. What is the maximum number of moves that may be needed to achieve this state?
[subscribe2]
Tags: , , ,
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quadratic equation completing the square
Related topics:
excel simultaneous calculation | 2nd order ode matlab | fractions/mixed numbers/decimals/integers/percentages | holt math algebra 1 workbook answers | common usages for radical expressions | review of math 300 | visual algebra ks2 | how to plug in a cubic root in ti-83 | glencoe/mcgraw-hill il algebra 1 chapter 9, section2 pratice sheets | rewrite a division as a multiplication | algebra e pra tice work sheet for 9th grade | quick images math cubes sheets
Author Message
Coffie-n-Toost
Registered: 25.10.2002
From: Rainy NW ::::
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oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Sunday 07th of Aug 08:52 Algebrator is one of the best resources that can offer help to a person like you. When I was a newbie , I took support from Algebrator. Algebrator covers all the principles of Basic Math. Rather than using the Algebrator as a step-by-step guide to solve all your homework assignments, you can use it as a coach that can give the basics of leading coefficient, 3x3 system of equations and trinomials. Once you understand the basics , you can go ahead and work out any tough assignments on Intermediate algebra within minutes.
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pcaDFX
Registered: 03.07.2001
From:
Posted: Sunday 07th of Aug 18:39 Hey there! I tried Algebrator last year when I was having problems with my college math. This program made solving problems so easy. Since then, I always keep a copy of it on my laptop.
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Dolknankey
Registered: 24.10.2003
From: Where the trout streams flow and the air is nice
Posted: Monday 08th of Aug 09:04 A great piece of math software is Algebrator. Even I faced similar problems while solving graphing, ratios and sum of cubes. Just by typing in the problem workbookand clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several algebra classes - Algebra 2, Pre Algebra and Algebra 2. I highly recommend the program.
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# How to calculate difference from initial value for each group in R?
I have data arranged like this in R:
``````indv time val
A 6 5
A 10 10
A 12 7
B 8 4
B 10 3
B 15 9
``````
For each individual (`indv`) at each time, I want to calculate the change in value (`val`) from the initial time. So I would end up with something like this:
``````indv time val val_1 val_change
A 6 5 5 0
A 10 10 5 5
A 12 7 5 2
B 8 4 4 0
B 10 3 4 -1
B 15 9 4 5
``````
Can anyone tell me how I might do this? I can use
``````ddply(df, .(indv), function(x)x[which.min(x\$time), ])
``````
to get a table like
``````indv time val
A 6 5
B 8 4
``````
However, I cannot figure out how to make a column `val_1` where the minimum values are matched up for each individual. However, if I can do that, I should be able to add column `val_change` using something like:
``````df['val_change'] = df['val_1'] - df['val']
``````
EDIT: two excellent methods were posted below, however both rely on my time column being sorted so that small time values are on top of high time values. I'm not sure this will always be the case with my data. (I know I can sort first in Excel, but I'm trying to avoid that.) How could I deal with a case when the table appears like this:
``````indv time value
A 10 10
A 6 5
A 12 7
B 8 4
B 10 3
B 15 9
``````
-
Here's a plyr solution using `ddply`
``````ddply(df, .(indv), transform,
val_1 = val[1],
change = (val - val[1]))
indv time val val_1 change
1 A 6 5 5 0
2 A 10 10 5 5
3 A 12 7 5 2
4 B 8 4 4 0
5 B 10 3 4 -1
6 B 15 9 4 5
``````
To get your second table try this:
``````ddply(df, .(indv), function(x) x[which.min(x\$time), ])
indv time val
1 A 6 5
2 B 8 4
``````
# Edit 1
To deal with unsorted data, like the one you posted in your edit try the following
``````unsort <- read.table(text="indv time value
A 10 10
A 6 5
A 12 7
B 8 4
B 10 3
do.call(rbind, lapply(split(unsort, unsort\$indv),
function(x) x[order(x\$time), ]))
indv time value
A.2 A 6 5
A.1 A 10 10
A.3 A 12 7
B.4 B 8 4
B.5 B 10 3
B.6 B 15 9
``````
Now you can apply the procedure described above to this sorted dataframe
# Edit 2
A shorter way to sort your dataframe is using `sortBy` function from doBy package
``````library(doBy)
orderBy(~ indv + time, unsort)
indv time value
2 A 6 5
1 A 10 10
3 A 12 7
4 B 8 4
5 B 10 3
6 B 15 9
``````
# Edit 3
You can even sort your df using `ddply`
``````ddply(unsort, .(indv, time), sort)
value time indv
1 5 6 A
2 10 10 A
3 7 12 A
4 4 8 B
5 3 10 B
6 9 15 B
``````
-
Dear Jilber, thanks so much for your help! However, this would only work if the "time" column has values sorted from low to high. I'm not sure if this is always the case in my data, although I could sort it first in Excel (which i'm trying to avoid using however). Would there be a method when the values are not sorted, like in the following table: indv time value A 10 10 A 6 5 A 12 7 B 8 4 B 10 3 B 15 9 – Thomas Nov 14 '12 at 21:21
Yes, See my edit. Hope you find it useful – Jilber Nov 14 '12 at 21:29
Dear Jilber, your edit seems to work great, thank you so much! I had been trying all day to figure this out! Now I will try it with my real data, thanks again for your help!!! – Thomas Nov 14 '12 at 21:38
Glad to be useful. See my second edit for a shorter way to sort your data.frame. – Jilber Nov 14 '12 at 21:41
Here is a `data.table` solution that will be memory efficient as it is setting by reference within the data.table. Setting the key will sort by the key variables
``````library(data.table)
DT <- data.table(df)
# set key to sort by indv then time
setkey(DT, indv, time)
DT[, c('val1','change') := list(val[1], val - val[1]),by = indv]
# And to show it works....
DT
## indv time val val1 change
## 1: A 6 5 5 0
## 2: A 10 10 5 5
## 3: A 12 7 5 2
## 4: B 8 4 4 0
## 5: B 10 3 4 -1
## 6: B 15 9 4 5
``````
-
You can do this with the base functions. using your data
``````df <- read.table(text = "indv time val
A 6 5
A 10 10
A 12 7
B 8 4
B 10 3
B 15 9", header = TRUE)
``````
We first `split()` `df` on the `indv` variable
``````sdf <- split(df, df\$indv)
``````
Next we transform each component of `sdf` adding in the `val_1` and `val_change` variables in a manner similar to how you suggest
``````sdf <- lapply(sdf, function(x) transform(x, val_1 = val[1],
val_change = val - val[1]))
``````
Finally we arrange for the individual components to be bound row wise into a single data frame:
``````df <- do.call(rbind, sdf)
df
``````
Which gives:
``````R> df
indv time val val_1 val_change
A.1 A 6 5 5 0
A.2 A 10 10 5 5
A.3 A 12 7 5 2
B.4 B 8 4 4 0
B.5 B 10 3 4 -1
B.6 B 15 9 4 5
``````
## Edit
To address the sorting issue the OP raises in the comments, modify the `lapply()` call to include a sorting step prior to the `transform()`. For example:
``````sdf <- lapply(sdf, function(x) {
x <- x[order(x\$time), ]
transform(x, val_1 = val[1],
val_change = val - val[1])
})
``````
In use we have
``````## scramble `df`
df <- df[sample(nrow(df)), ]
## split
sdf <- split(df, df\$indv)
## apply sort and transform
sdf <- lapply(sdf, function(x) {
x <- x[order(x\$time), ]
transform(x, val_1 = val[1],
val_change = val - val[1])
})
## combine
df <- do.call(rbind, sdf)
``````
which again gives:
``````R> df
indv time val val_1 val_change
A.1 A 6 5 5 0
A.2 A 10 10 5 5
A.3 A 12 7 5 2
B.4 B 8 4 4 0
B.5 B 10 3 4 -1
B.6 B 15 9 4 5
``````
-
Dear Gavin, thanks for your help! That method indeed works for my data, however only when the the time column is sorted so that the first values appear before the later values. I'm not sure my data will always be sorted in that way. Is there a similar method for when the times are out of order, like in the following table: indv time value A 10 10 A 6 5 A 12 7 B 8 4 B 10 3 B 15 9 – Thomas Nov 14 '12 at 21:26
Then sort `val` first or sort by `time` whichever it is that you want. It is much easier to sort the data frame first that to worry about handling it in the differencing operation. – Gavin Simpson Nov 14 '12 at 21:31
I have suggested a solution to that. In future, it helps to formulate the question fully to avoid extended updating of answers etc. Hopefully the edit I made to my question helps? – Gavin Simpson Nov 14 '12 at 21:37
Dear Gavin, thanks. Yes, the edit helps. I'm sorry I didn't formulate the question fully, I didn't realize I had mis-formulated it until I read the answers. I will try to be more careful next time! – Thomas Nov 14 '12 at 21:42
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Posted by IGoR on July 14, 2001 7:56 PM
This is my formula : =RANK(\$E\$1:\$E\$12,\$E\$1:\$E\$12,1)
but when there is no data entered on the required
cell to be rank, it will show #N/A or other number.
How can the rank formula will show nothing when there
is no data entered. Thanks in advance
Posted by Aladin Akyurek on July 14, 2001 11:20 PM
Make that
=IF(ISBLANK(E1),"",RANK(E1,\$E\$1:\$E\$12,1))
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# How to identify the age of a "living" person?
An Indian old man claims he is [180 years old] now.
We are not able to tell he is 180 or 181 probably. But can we tell he is 90, 120, 100 or 150?
Different people may age differently.
A 60-year old person may appear to be 80, while somebody else could appear younger than their mathematical age. Some other people may have progeria.
What is the most accurate way to identify a living person's age nowadays? Are there reliable parameters to measure?
C-14 needs a long time base line. Hence, it is suitable for the bone of dead people.
What kind of method can guarantee it can distinguish a 30y and a 80y old person? What's the error bar of that method?
• Cut off a limb and count the rings Aug 5, 2015 at 13:24
• possible duplicate of Can we determine a person's age by dating methods or other means?
– AliceD
Aug 5, 2015 at 14:22
• Birth certificate, I presume :P Aug 5, 2015 at 14:31
• @WYSIWYG By the way, a person celebrates his/her birthday on the correct day only for (max) 4 years after his/her birth. Aug 7, 2015 at 5:24
• You have the wrong impression about carbon-14 dating, the long time horizon doesn't have anything to do with not being able to use it to date the age of a living organism. Living things are constantly replenishing the very limited supply of carbon-14 while they are alive. It is only after the organism dies that the carbon-14 clock starts to tick. A dead organism no longer replenishes its supply of carbon-14 and the carbon-14 it did have begins to decay into nitrogen-14. We can then measure the percentage of carbon-14 left in the remains and date them to when they died.
– AMR
Dec 24, 2015 at 4:53
The most accurate method known so far may be Horvath's methylation dating algorithm, which uses 353 CpG sites (Genome Biology research article, Nature news article).
From the Nature article, this method was validated with at least one population.
[After the publication in Genome Biology] Marco Boks at the University Medical Centre Utrecht in the Netherlands applied [the algorithm] to blood samples collected from 96 Dutch veterans of the war in Afghanistan aged between 18 and 53. The correlation between predicted and actual ages was 99.7%, with a median error measured in months.
However, I did not read the original research article and it seems that the dating method works poorly on cancer patients.
• DNA methylation is susceptible to environmental factors. Aug 5, 2015 at 14:31
• Good article. This may replace teeth as the gold standard. Aug 5, 2015 at 18:45
• @anongoodnurse, I was actually hoping someone would argue that dating using teeth would be the best way to estimate the age, since I'm not familiar with this literature. Aug 6, 2015 at 9:29
• @Jonathan - See the possible duplicate. I think it's close. I almost posted an answer, but the Nature article surprised me. Plus, you don't need to pull a tooth. Aug 6, 2015 at 14:36
• My question is about the error bar in fact. Look at the boldface sentence please. Dec 23, 2015 at 13:06
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# Physics MCQ's with answers and Solution for AMC/NUMS Test preparation
To get admission in Army Medical College (AMC) and other affiliated colleges through NUMS is very challenging due to less number of seats available. Every year thousands of students appears in NUMS test only few secure admission in AMC and other affiliated colleges of NUMS. Entrance test for NUMS/ AMC is very technical and competitive and only those students can get good marks in NUMS (AMC) entrance test who have good conceptual knowledge of entry test subjects. National University of Medical Sciences (NUMS) entry test is based on MCQ’s type questions conducted by NTS. NUMS test include Physics, Chemistry, English and Biology MCQ’s. Due to less number of seats competation is high, we recommend students to prepare from www.pakprep.com for NUMS entry test. www.pakprep.com provides more than 15000+ fully explained MCQ’s with answers and solution. For your preparation click here. Here are some MCQ’s of Physics with answers and solution.
Topic: Fluid & Dynamics
Question:
If weight of the body immersed in water exceeds the buoyant force, then the body will?
• Answer A : Rise until the weight equals the buoyant force
• Answer B : Moves downwards and may eventually
• Answer D : None of the above
Correct: B
Explanation
Buoyant force is like the drag force of water on the body. If weight is more than the drag force the body will sink.
Topic: Current Electricity
Question:
In order to measure the resistance of an unknown emf source the reading of galvanometer in the given arrangement must be equal to? Assume that AB is a resistor of uniform cross sectional area
Correct: D
Explanation
The reading of galvanometer in the given potentiometer arrangement must be equal to zero. The EMF= E* r/R or E* AC/AB.
Topic: Modern Physics
Question:
Which of the following is most easily detected as quantum radiation?
• Answer C : Ultra violet
• Answer D : Gamma Rays
Correct: D
Explanation
P= h/ λ, so momentum is higher for shortest wavelengths thus higher probability of detection as quantum particle. Gamma rays have shortest wavelength thus are most easily detected as quantum radiation.
## 6 thoughts on “Physics MCQ's with answers and Solution for AMC/NUMS Test preparation”
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2014-04-12T10:26:28-04:00
In a right triangle, the square of the hypotenuse is equal to the sum of squares of the other 2 sides. The hypotenuse is the side that is opposite to the right angle.
2014-04-12T10:48:06-04:00
The Pythagorean theorem is:
a^2 + b^2 = c^2
Therefore
c= sqroot of (a^2 + b^2)
And
C^2 - a^2= b^2
Where:
c is the length of the hypotenuse of a triangle
a & b is the length of the two other sides
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# Lim solver
This Lim solver helps to quickly and easily solve any math problems. Our website can solve math problems for you.
## The Best Lim solver
Lim solver can be found online or in math books. I'm not quite sure what you're asking for here. Do you need help with your algebra homework? If so, there are plenty of resources available to help you out. Try looking online for algebraicy tutorials or search for algebra homework help on a search engine. You should also be able to find helpful resources in your textbook or from your teacher. If you're still struggling, you can always ask a friend or family member for help. Good luck!
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If you're working with an arithmetic sequence, there's a simple formula you can use to find the value of any element in the sequence. Just plug in the element's position in the sequence (its index), and you'll get the value. This formula is especially useful if you're working with a very long sequence and need to find specific values quickly.
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There are a few things you can do to make solving math word problems easier. First, read the problem carefully and make sure you understand what it is asking. Once you know what the problem is asking, try to identify any key information or relevant equations that you will need to solve it. Once you have all of the necessary information, you can begin solving the problem. If you get stuck, don't be afraid to ask for help from a friend or a teacher.
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# Posts Tagged ‘ Le Monde ’
## Le Monde puzzle [#5]
February 10, 2011
By
Another Sudoku-like puzzle from the weekend edition of Le Monde. The object it starts with is a 9×9 table where each entry is an integer and where neighbours take adjacent values. (Neighbours are defined as north, west, south and east of an entry.) The question is about whether or not it is possible to find
## Le Monde puzzle [#4]
February 4, 2011
By
A fairly simple puzzle in this weekend Le Monde magazine: given five points on a line such that their pairwise distances are 1,2,4,…,14,18,20, find the respective positions of the five points over the line and deduce the missing distances. Without loss of generality, we can set the first point at 0 and the fifth point
## R exam
January 30, 2011
By
$R exam$
I spent most of my Saturday perusing R codes to check the answers written by my students to the R exam I gave two weeks ago… The outcome is mostly poor, even though some managed to solve a fair part of the long problem. Except for the few hopeless cases who visibly never wrote a
## Le Monde puzzle [1]
January 10, 2011
By
Following the presentation of the first Le Monde puzzle of the year, I tried a simulated annealing solution on an early morning in my hotel room. Here is the R code, which is unfortunately too rudimentary and too slow to be able to tackle n=1000. #minimise \sum_{i=1}^I x_i #for 1\le x_i\le 2n+1, 1\e i\le I
## Le Monde puzzle [52|solution]
January 1, 2011
By
$Le Monde puzzle [52|solution]$
I have now received the first issue of Le Monde magazine, including the solution to puzzle #52 I solved just in time by simulated annealing! The trick is in using the following theorem: Iter(1,x,y) is divisible by 10x-1 if and only if y is divisible by 10x-1. Then the value to be found is divisible
## Le Monde puzzle [49]
December 7, 2010
By
Here is a quick-and-dirty solution to Le Monde puzzle posted a few days ago: the R code counts the number of winning tickets between 1 and N, and stops when there is a proportion of 10% of winning tickets. #winning ticket win=function(n){ #decimal digits decomposition x=rep(0,4) x=n%%10 m=(n-x)/10 x=m%%10 m=(m-x)/10 x=m%%10 m=(m-x)/10 x=m%%10 tic=0 for
## Le Monde puzzle [48: resolution]
December 4, 2010
By
$Le Monde puzzle [48: resolution]$
The solution to puzzle 48 given in Le Monde this weekend is rather direct (which makes me wonder why the solution for 6 colours is still unavailable..) Here is a quick version of the solution: Consider one column, 1 say. Since 326=5×65+1, there exists one value c with at least 66 equal to c. Among
## Le Monde puzzle [48]
December 1, 2010
By
$Le Monde puzzle [48]$
This week(end), the Le Monde puzzle can be (re)written as follows (even though it is presented as a graph problem): Given a square 327×327 symmetric matrix A, where each non-diagonal entry is in {1,2,3,4,5} and , does there exist a triplet (i,j,k) such that Solving this problem in R is very easy. We can create
## Random graphs with fixed numbers of neighbours
November 24, 2010
By
In connection with Le Monde puzzle #46, I eventually managed to write an R program that generates graphs with a given number n of nodes and a given number k of edges leaving each of those nodes. (My early attempt was simply too myopic to achieve any level of success when n was larger than
## Le Monde puzzle [43]
November 7, 2010
By
Here is the puzzle in Le Monde I missed last week: Given a country with 6 airports and a local company with three destinations from each of the six airports, is it possible to find a circular trip with three intermediate stops from one of the airports? From all of the airports? One more airport
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# Algorithm for computing the Arf invariant of a knot
According to "The knot book", by Colin Adams, two knots are pass equivalent if they are related by a finite sequence of pass-moves. Moreover every knot is pass-equivalent to either the unknot or the trefoil knot and these two knots are not pass-equivalent.
We define Arf invariant of a knot to be 0 if the know is pass equivalent to unknot and to be 1 if it is pass equivalent to the trefoil knot.
Now here is my question: I am not interested to know how a knot is pass-equivalent to unknot or the trefoil knot. I just want to know if there is an easy algorithm for computing the Arf invariant of a knot based on its projection (for example, like the algorithm that we use to compute the linking number of links)?
Can we move along a knot and compute its Arf invariant by taking its crossings somehow into consideration (maybe by counting its positive and negative crossings)?
• Why do you call it $\mathrm{Arf}$-invariant? Is it really the $\mathrm{Arf}$-invariant of a certain form? Then, maybe, you can compute the form? – Alex Degtyarev Mar 19 '15 at 7:39
• Alex, this is one of the standard definitions of Arf invariant for knots (see "The knot book", by Colin Adams or Arf invariant of a knot. – Hooman Mar 19 '15 at 8:14
• All I want to say is that this name should be for a reason. Try to search for a better definition elsewhere :) – Alex Degtyarev Mar 19 '15 at 8:34
• Alex: The Arf invariant of a knot is the usual Arf invariant of the quadratic form on the Z/2 homology of a Seifert surface, defined as the self-linking number of a curve representing the homology class. This is the same as the Seifert form. The associated bilinear form is just the intersection pairing, also determined by the Seifert form. Since there's an algorithm for finding a Seifert surface from a projection, there is an algorithm for finding the Arf invariant. I think the OP was looking for something more diagrammatic in nature. – Danny Ruberman Mar 19 '15 at 12:03
The well-known (see here) relations between the Arf invariant and the Alexander/Jones polynomials give you algorithms for computing the Arf invariant, if that's all you want.
But it sounds like what you're after is a 'crossing-linear' algorithm, in the sense that we go along the knot diagram, examine each crossing once, update some internal state, and at the end spit out the Arf invariant.
I don't know of such an algorithm. But it's not terrifically hard to come up with a close approximation to this.
We'll use the property mentioned in the above link (there's a proof Kauffman's Formal Knot Theory, where this is cited as a folklore theorem) that for a standard smoothing $K_+, K_-, K_0$ of a crossing, where $K_0 = L_1 \cup L_2$ is a two-component link, we have $$\mathrm{Arf}(K_+) + \mathrm{Arf}(K_-) \equiv \mathrm{lk}(L_1, L_2) \mod{2}.$$
We can definitely compute the right-hand side by passing along the crossings of $K_0$, so all that remains is to choose crossings cleverly so that whichever one of $K_+, K_-$ is not our original knot is a bit simpler. (Alternately, if you're happy with an algorithm that's exponential-time in the number of crossings you could do some state-sum-style approach here.) This territory is a bit more well-known; I'd look at the more efficient algorithms for the Jones polynomial implemented by Bar-Natan in his Knot Theory package for Mathematica. Ultimately this should get you something that's 'crossing-polynomial', maybe even 'crossing-quadratic'.
Arf invariant is modulo two reduction of the Vassiliev invariant $$v_2$$ (the coefficient of $$z^2$$ in the Conway polynomial). Thus a well-known Gauss diagram formula for $$v_2$$ can be used for its computation. Namely, fix a knot diagram and choose a base point distinct from the crossings. Traverse the diagram starting from the base point. Each crossing is visited twice; write down the sequence in which we visit these crossings, denoting a passage on an overpass by $$O_i$$ and on an underpass by $$U_i$$. The result is the Gauss code of the knot diagram. For example, the standard diagram of a trefoil will be encoded by $$U_1 O_2 U_3 O_1 U_2 O_3$$. Then the Arf invariant is the number of "linked" pairs of crossings $$...U_i...O_k...O_i...U_k...$$ in the Gauss code.
This is too long for a comment, it's just an idea for this computation. Start with a Seifert surface $F \subset \Bbb R^3$ for the knot $K$. Up to isotopy, we can assume that $F$ projects regularly to the plane $\Bbb R^2 \subset \Bbb R^3$. Now, $F$ can be presented by bands attached to a disk. Up to pass-moves, we can unlink and unknot the bands of $F$, as these moves correspond to Reidemeister moves applied to the core graph of $F$ with its band decomposition. Now, the bands are trivial but can be twisted. By further pass-moves we can cancel any two full twists in any band, giving a new surface $F'$ such that any band has 0 or 1 full twists. So, the Arf invariant should be the total number of the full twists of $F'$ (mod 2). I don't have a proof now, it's just intuition, the motivation being that the pass-moves preserve the Arf invariant. By the way, this should imply that the Arf invariant is the total twisting of the bands of the original $F$ (mod 2).
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > repswlsw Structured version Visualization version GIF version
Theorem repswlsw 13729
Description: The last symbol of a nonempty "repeated symbol word". (Contributed by AV, 4-Nov-2018.)
Assertion
Ref Expression
repswlsw ((𝑆𝑉𝑁 ∈ ℕ) → (lastS‘(𝑆 repeatS 𝑁)) = 𝑆)
Proof of Theorem repswlsw
StepHypRef Expression
1 nnnn0 11491 . . . 4 (𝑁 ∈ ℕ → 𝑁 ∈ ℕ0)
2 repsw 13722 . . . 4 ((𝑆𝑉𝑁 ∈ ℕ0) → (𝑆 repeatS 𝑁) ∈ Word 𝑉)
31, 2sylan2 492 . . 3 ((𝑆𝑉𝑁 ∈ ℕ) → (𝑆 repeatS 𝑁) ∈ Word 𝑉)
4 lsw 13538 . . 3 ((𝑆 repeatS 𝑁) ∈ Word 𝑉 → (lastS‘(𝑆 repeatS 𝑁)) = ((𝑆 repeatS 𝑁)‘((♯‘(𝑆 repeatS 𝑁)) − 1)))
53, 4syl 17 . 2 ((𝑆𝑉𝑁 ∈ ℕ) → (lastS‘(𝑆 repeatS 𝑁)) = ((𝑆 repeatS 𝑁)‘((♯‘(𝑆 repeatS 𝑁)) − 1)))
6 simpl 474 . . 3 ((𝑆𝑉𝑁 ∈ ℕ) → 𝑆𝑉)
71adantl 473 . . 3 ((𝑆𝑉𝑁 ∈ ℕ) → 𝑁 ∈ ℕ0)
8 repswlen 13723 . . . . . 6 ((𝑆𝑉𝑁 ∈ ℕ0) → (♯‘(𝑆 repeatS 𝑁)) = 𝑁)
91, 8sylan2 492 . . . . 5 ((𝑆𝑉𝑁 ∈ ℕ) → (♯‘(𝑆 repeatS 𝑁)) = 𝑁)
109oveq1d 6828 . . . 4 ((𝑆𝑉𝑁 ∈ ℕ) → ((♯‘(𝑆 repeatS 𝑁)) − 1) = (𝑁 − 1))
11 fzo0end 12754 . . . . 5 (𝑁 ∈ ℕ → (𝑁 − 1) ∈ (0..^𝑁))
1211adantl 473 . . . 4 ((𝑆𝑉𝑁 ∈ ℕ) → (𝑁 − 1) ∈ (0..^𝑁))
1310, 12eqeltrd 2839 . . 3 ((𝑆𝑉𝑁 ∈ ℕ) → ((♯‘(𝑆 repeatS 𝑁)) − 1) ∈ (0..^𝑁))
14 repswsymb 13721 . . 3 ((𝑆𝑉𝑁 ∈ ℕ0 ∧ ((♯‘(𝑆 repeatS 𝑁)) − 1) ∈ (0..^𝑁)) → ((𝑆 repeatS 𝑁)‘((♯‘(𝑆 repeatS 𝑁)) − 1)) = 𝑆)
156, 7, 13, 14syl3anc 1477 . 2 ((𝑆𝑉𝑁 ∈ ℕ) → ((𝑆 repeatS 𝑁)‘((♯‘(𝑆 repeatS 𝑁)) − 1)) = 𝑆)
165, 15eqtrd 2794 1 ((𝑆𝑉𝑁 ∈ ℕ) → (lastS‘(𝑆 repeatS 𝑁)) = 𝑆)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1632 ∈ wcel 2139 ‘cfv 6049 (class class class)co 6813 0cc0 10128 1c1 10129 − cmin 10458 ℕcn 11212 ℕ0cn0 11484 ..^cfzo 12659 ♯chash 13311 Word cword 13477 lastSclsw 13478 repeatS creps 13484 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-8 2141 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 ax-rep 4923 ax-sep 4933 ax-nul 4941 ax-pow 4992 ax-pr 5055 ax-un 7114 ax-cnex 10184 ax-resscn 10185 ax-1cn 10186 ax-icn 10187 ax-addcl 10188 ax-addrcl 10189 ax-mulcl 10190 ax-mulrcl 10191 ax-mulcom 10192 ax-addass 10193 ax-mulass 10194 ax-distr 10195 ax-i2m1 10196 ax-1ne0 10197 ax-1rid 10198 ax-rnegex 10199 ax-rrecex 10200 ax-cnre 10201 ax-pre-lttri 10202 ax-pre-lttrn 10203 ax-pre-ltadd 10204 ax-pre-mulgt0 10205 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1073 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-eu 2611 df-mo 2612 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-ne 2933 df-nel 3036 df-ral 3055 df-rex 3056 df-reu 3057 df-rab 3059 df-v 3342 df-sbc 3577 df-csb 3675 df-dif 3718 df-un 3720 df-in 3722 df-ss 3729 df-pss 3731 df-nul 4059 df-if 4231 df-pw 4304 df-sn 4322 df-pr 4324 df-tp 4326 df-op 4328 df-uni 4589 df-int 4628 df-iun 4674 df-br 4805 df-opab 4865 df-mpt 4882 df-tr 4905 df-id 5174 df-eprel 5179 df-po 5187 df-so 5188 df-fr 5225 df-we 5227 df-xp 5272 df-rel 5273 df-cnv 5274 df-co 5275 df-dm 5276 df-rn 5277 df-res 5278 df-ima 5279 df-pred 5841 df-ord 5887 df-on 5888 df-lim 5889 df-suc 5890 df-iota 6012 df-fun 6051 df-fn 6052 df-f 6053 df-f1 6054 df-fo 6055 df-f1o 6056 df-fv 6057 df-riota 6774 df-ov 6816 df-oprab 6817 df-mpt2 6818 df-om 7231 df-1st 7333 df-2nd 7334 df-wrecs 7576 df-recs 7637 df-rdg 7675 df-1o 7729 df-er 7911 df-en 8122 df-dom 8123 df-sdom 8124 df-fin 8125 df-card 8955 df-pnf 10268 df-mnf 10269 df-xr 10270 df-ltxr 10271 df-le 10272 df-sub 10460 df-neg 10461 df-nn 11213 df-n0 11485 df-z 11570 df-uz 11880 df-fz 12520 df-fzo 12660 df-hash 13312 df-word 13485 df-lsw 13486 df-reps 13492 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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# Diff Amps
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ALL ABOUT HOW IT WORKS AND ITS ANALYSIS
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### Diff Amps
1. 1. EE-4232 Differential Amplifiers 0
2. 2. Different modes of operation of the differential pair: The differential pair with a common- mode input signal vCM. 1
3. 3. Different modes of operation of the differential pair: The differential pair with a “large” differential input signal. 2
4. 4. Different modes of operation of the differential pair: The differential pair with a large input signal of polarity opposite to that in (b). 3
5. 5. Different modes of operation of the differential pair: The differential pair with a small differential input signal vi. 4
6. 6. Transfer characteristics of the BJT differential pair of previous figure assuming α ≅ 1. 5
7. 7. The current and voltages in the differential amplifier when a small difference signal vd is applied. 6
8. 8. A simple technique for determining the signal currents in a differential amplifier excited by a differential voltage signal vd; dc quantities are not shown. 7
9. 9. A differential amplifier with emitter existence. Only signal quantities are shown (on color). 8
10. 10. Equivalence of the differential amplifier (a) to the two common-emitter amplifiers in (b). This equivalence applies only for differential input signals. Either of the two common-emitter amplifiers in (b) can be used to evaluate the differential gain, input differential resistance, frequency response, and so on, of the differential amplifier. 9
11. 11. (a) The differential amplifier fed by a common-mode voltage signal. (b) Equivalent “half-circuits” for the common-mode calculations. 10
12. 12. Analysis of the current mirror taking into account the finite β of the BJTs. 11
13. 13. Generation of a number of cross currents. 12
14. 14. A current mirror with base-current compensation. 13
15. 15. The Wilson current mirror. 14
16. 16. The Widlar current source. 15
17. 17. 16
18. 18. A differential amplifier with an active load. 17
19. 19. Small-signal model of the differential amplifier. 18
20. 20. The MOSFET differential pair. 19
21. 21. Normalized plots of the currents in a MOSFET differential pair. Note that VGS is the gate-to-source voltage when the drain current is equal to the dc bias current (I/2). 20
22. 22. MOS current mirrors: basic 21
23. 23. MOS current mirrors: cascode 22
24. 24. MOS current mirrors: Wilson 23
25. 25. MOS current mirrors: modified Wilson 24
26. 26. A multistage amplifier circuit. 25
27. 27. References • Electronics by A. Hambley • Microelectronics Circuits by Sedra & Smith • Other books on Electronics 26
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# Need help with basic calculus
• Aug 1st 2009, 02:59 AM
honestliar
Need help with basic calculus
How can I integrate the following given?
int[x^2(1-x^3)/(x^2-1)]dx ans. -1/4x^4-1/2x^2+x-ln(x+1)+c
int[cos[1/2x]dx/3/4-2/3 sin[1/2x] ans. -3ln(3/4-2/3sin[1/2x]+c
int=integral symbol
Thanks
• Aug 1st 2009, 03:12 AM
songoku
Hi honestliar
1.
$\displaystyle \int \frac{x^2(1-x^3)}{x^2-1}dx$ ; the numerator divided by denominator will be :
$\displaystyle =\int(-x^3 - x + 1 + \frac{1-x}{x^2-1})dx$
Solve $\displaystyle \int \frac{1-x}{x^2-1}dx$ using partial fraction
2.
hint : $\displaystyle \frac{d}{dx}sec (x) = tan(x) sex(x)$
• Aug 1st 2009, 03:27 AM
DeMath
Quote:
Originally Posted by honestliar
How can I integrate the following given?
int[x^2(1-x^3)/(x^2-1)]dx
int[cos[1/2x]dx/3/4-2/3 sin[1/2x]
int=integral symbol
Thanks
$\displaystyle \int {\frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}}dx} .$
$\displaystyle \frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}} = - \frac{{{x^2}\left( {1 - x} \right)\left( {1 + x + {x^2}} \right)}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} = - \frac{{{x^2}\left( {1 + x + {x^2}} \right)}}{{1 + x}} = - \frac{{{x^2}\left( {1 + x} \right) + {x^4}}}{{1 + x}} =$
$\displaystyle = - {x^2} - \frac{{{x^4}}}{{1 + x}} = - {x^2} + \frac{{1 - {x^4} - 1}}{{1 + x}} = - {x^2} + \frac{{1 - {x^4}}}{{1 + x}} - \frac{1} {{1 + x}} =$
$\displaystyle - {x^2} + \frac{{\left( {1 - {x^2}} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} = - {x^2} + \frac{{\left( {1 - x} \right)\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}{{1 + x}} - \frac{1}{{1 + x}} =$
$\displaystyle = - {x^2} + \left( {1 - x} \right)\left( {1 + {x^2}} \right) - \frac{1} {{1 + x}} = 1 - x - {x^3} - \frac{1}{{1 + x}}.$
$\displaystyle \int {\frac{{{x^2}\left( {1 - {x^3}} \right)}}{{{x^2} - 1}}dx} = \int {\left( {1 - x - {x^3} - \frac{1}{{1 + x}}} \right)dx} .$
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```Research in business studies
Department of Business Administration
SPRING 2009-10
Quantitative and Qualitative Data Analysis
by
Assoc. Prof. Sami Fethi
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Quantitative data analysis
Examining differences
Relationship between variables
Explaining and predicting relationship between variables
Data reduction, structure and dimension
Additional methods
Characteristic of qualitative research
Qualitative data
Analytical procedure
Interpretation
Strategies for qualitative analysis
Quantify qualitative data
Validity in qualitative research
2
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Examining differences
Research in business studies
Hypotheses about one mean
In research we often have to make statements about the mean. When the
population variance is unknown, the stadard error of the mean is also
unknown. The standard error of the mean must be estimated from sample
data.
e.g. SDX= SD‘/ N
where
SDX= standard error of mean
SD‘= estimated standard deviation
N= sample size
N
( xi X )
i 1
2
SD‘=
N 1
N-1 is degrees of freedom
Example 1: For a supermarket chain to add a new product, at least 100
units must be sold per week. The new product is tested in ten randomly
selected stores for a limited time.
Apply a test such as one-tailed t test and answer the question that will the
new product sell more than 100 unit per week?
a) construct hypothesis
b) calculate mean and standard deviation if they are not given.
c) calculate standart error of mean
3
d) find t- value
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Examining differences
Research in business studies
a) H0: X<=100
H1: X>100
b) X and SD are given 109.4 and 14.90 respectively.
c) SDX = 14.90/ 10 1 =4.55
d) t= (X-µ)/SDX=(109.4-100)/4.55=2.07
Where t-table is 1.83 at 5% significant level.
We reject the null
Hypotheses about two means
This is usually associated with such a question: Are
the tastes in region A different from the tastes in
region B?
( X 1 X 2 ) (1 2 )
Z
e.g.
SD
X1 X 2
Where
X1= sample mean for the first sample
X2= sample mean for the second sample
Research Methods in Business Studies
4
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Examining differences
SDX 1 X 2 = the standard eror of differences in means
µ1 and µ2 are the unknown population means and
the general estimate of:
SD 21 SD 2 2
SDX1 X 2 SDX2 1 SDX2 2
N1
N2
In assuming the two population variances to be equal, the
common population variance can be generated by pooling the
samples. When the variances are unknonw and the standard
errors of means must be estimated, then the t represents an
adequate test statistics, distributed with v= N1+ N2-2- degrees
of freedom.
Example2: A manufacturer has developed a new product and
wonders whether the label of the package should be red or
blue. The new products with two different labels are tested in
ten randomly selected stores. The means sales obtained for
the red package are 403.0 and for the blue package 390.3. The
standard error of estimate for the difference means is 8.15.
Research Methods in Business Studies
5
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Examining differences
a) construct hypothesis
b) find t- value
a) H0: (µ1- µ2 )=0
H1: (µ1- µ2 )≠0
or
H0: (µ1- µ2 )<=0
H1: (µ1- µ2 )>0
b)
t
( X 1 X 2 ) (1 2 )
=((403.0-390.3)-0)/8.15=1.56
SDX1 X 2
V=10+10-2=18 degrees of freedom...5% and df 18 so
critical value from the table is 2.101. This means that null
hypothesis is accepted.. H0: (µ1- µ2 )=0. This means that the two
unknown population means are assumed to be same.
Research Methods in Business Studies
6
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Useful alternative tests
o
In problems involving one or two population means, t-methods are
usually appropriate, but often non-parametric methods are good
alternatives.
e.g. Non-parametric methods have advantage of requiring less in
terms of assumptions and less powerful than t-methods (see siegel
and Castella; 1998).
e.g. The main difference between them is that t-method associates
with means while non-parametric methods are concerned with
medians.
ANOVA- analysis of variance measures comparisons of more than two
groups simultaneously. This method rests on comparing the ratio of
systematic variance to unsystematic variance.
In ANOVA, the following is computed:
Total variation by comparing each observation with the grand mean.
The between-group variation by comparing the treatment means with
the grand mean.
The within-group variation by comparing each score in the group with
the group mean.
Recall-MANOVA-multivariate analysis of variance. This has more than
one dependent variable compared to ANOVA:
Research Methods in Business Studies
7
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Comparison of more than two group
Example 3: In the
following table,
three advertising
campaigns tested
in 24 randomly
selected cities
comparable in size
and demographics.
The following
output is an anova
analysis results:
Source Sum Degree Mean F-ratio
of
of
sq.
freedom
sq.
Between 49.0
2
24.1 5.88
group
Within
group
87.5
21
total
136.5
23
4.17
8
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Example 3
Research in business studies
a) construct hypothesis
b) find F- value whether significant or not
c) Comment on the F-values
a) H0: G1= G2= G3
H1: G1≠ G2 ≠ G3
d.f= 24-1=23, between group 3-1=2 within group 232=21.
b) Fcalculated=24.1/4.17=5.88
Fcritical=n-k,k-1=24-3,3-1=(21,2). From F-distribution,
Fcritical is 3.47.
c) Since 5.88 is greater than 3.47, we reject the null
hypothesis, that is, the group means are equal and
accept the alternative hypothesis that the advertising
campaigns vary in effectiveness.
Research Methods in Business Studies
9
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Relationship between variables
In research, we are often preoccupied with whether
there is a relationship or two or more variables covary.
o Correlation coefficient
Based on the Pearson criterion, it examines the strength
of linear relationship between two variables, for example
x and y.
o Theoretically, the Correlation coefficient can take the
values from -1 to 1. A correlation coefficient of 1 tells us
that two variables perfectly covary positively whereas -1
shows that two variables perfectly inversely related.
Close to 0 indicates that the variables are unrelated.
The formula of the Correlation coefficient as fololw:
Where X and Y represent the sample means of X and Y.
rXY
( x X )( y Y )
(x X ) ( y Y )
i
i
2
i
10
2
i
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Relationship between variables
o Correlation coefficient
A Correlation coefficient shows covariation between two variables,
and not that the variables are causally related.
The square of the Correlation coefficient is the coefficient of
determination.
R2=Explained variation/Total variation
o Example 4- partial correlation
Using the following table (Table 1) and calculate the relationship
between advertisement recognition, appeal and sex. In other words,
Is the relationship between advertisement recognition and appeal
inluenced by controling for sex?
11
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Example 4
o This is partial correlation and can be formulated as follow based on
partial Correlation coefficient r123 as such ad.roc, appeal, sex
r123
r12 (r13 ) (r23 )
1 r13
2
1 r23
2
rAd .roc, appeal, sex
0.24 (0.33) (0.09)
1 (0.33)
2
1 (0.09)
2
0.29
o This shows that controlling for sex the observed relationship
between ad.roc, and appeal positive and strengthened.
12
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Explaining and predicting relationship between variables
o Explaining and predicting relationship between variables are
important tasks in business research. One of the most applied and
useful approaches to examining relationships between variables is
regression analysis. In regression analysis, we want to fit a model
that best describes the data which is done in regression analysis by
applying the method of least squares. More precisely, this is done by
fitting a straight line that minimizes the squared vertical deviations
from that line as shown in following figure.
o Single Linear Regression
Y= a0+a1xi+ei
Where Y=the outcome variable, X=predictor variable, a1=slope of the
straight line fitted to the data and a0=intercept of the line and
ei=difference between the score predicted and the score actually
obtained. This is called residual.
Research Methods in Business Studies
13
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Single Linear Regression
Explaining and Predicting Relationship between Variables
Figure 1 The linear model
Research Methods in Business Studies
14
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Single Linear Regression
Example 5
o Assume that a car dealer collects data for six months on four
variables; Tv advertising, printing advertising, competitors’
advertising and sales. Y is sales. The car dealer expects carsales to
be positively correlated with TV-ads and Print-ads and negatively
correlated with competitors’ ads.
Table 2 Data matrix
Research Methods in Business Studies
15
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Simple Mean Regression-output
Example 5
o Assume that a car dealer collects data for six months on four
variables; Tv advertising, printing advertising, competitors’
advertising and sales. Y is sales. The car dealer expects carsales to
be positively correlated with TV-ads and Print-ads and negatively
correlated with competitors’ ads. Based on the information below,
comment on the estimated coefficinent and T-ratio as well as R2
on Tv-Ads.
Table 3 Simple mean regression-output
Research Methods in Business Studies
16
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Simple Mean Regression-output
Example 5-Answer
o The estimated constant term 0.7 shows that If the dealer does not
use Tv-ads at all (Tv-ads=0), the estimated expected value of
carsale is 0.7 unit that is 7 car. The estimated regression coefficient
of sales on Tv-Ads is 0.9. This coefficient shows that if the variable
Tv-ads is increased by 1 unit, the estimated expected value of
carsales increases by 0.9 units, that is nine car. The result, Rsquare, R2 that is 85.3 percent shows that the sample determination of
coefficient is equal to 0.853. Practically speaking, this means that the
variation in the variable Tv-ads has explained 85.3 percent of the
variations in the dependent variable carsales. Estimated t-value on Tvads is 4.81 which is greater than 2 (tabular value from t-distribution) or
rule of thumb so it is signficant 5% and 1% levels. This means that we
can reject the null hypothesis that is the corresponding population
regression coefficient is equal to zore. The conclusion then is that Tvads and sales are significantly related to each other or Tv-ads has
positive impact on sales.
17
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Assumptions in Regression analysis
o The expected value of the error term is zero
o The variance for the error term for each X is constant.
This term homoscedasticity. If the variance to e varies
with X, this is termed heteroscedasticity.
o The error for the observations are uncorrelated.
o e should be normally distributed for each X.
o The error term should not be correlated with x-corr(e,
x)=0
o It is also a common assumption that the regression
model should be linear in its parameters.
18
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Correlation Coefficients-output
Example 6
o Assume that a car dealer collects data for six months
on four variables; Tv advertising, printing advertising,
competitors’ advertising and sales. Y is sales. The car
dealer expects carsales to be positively correlated with
TV-ads and Print-ads and negatively correlated with
competitors’ ads. Use the concept of correlation
coefficient and explain the relationships between the
variable under inspection based on the information given
in table 4.
Table 4 Correlation coefficients-output
Research Methods in Business Studies
19
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Correlation Coefficients-output
Example 6 -Answer
Research in business studies
o The relationship between carsales (dependent) and Tv
advertising, printing advertising, competitors’ advertising
(explanatory) are expected to be high. The relationship
between the explanatory variables as such Tv
advertising, printing advertising, competitors’ advertising
are expected to be low. So high correlation coefficient
between for example Tv advertising and printing
advertising shows a high degree of multicollinearity.
This influences the estimates results badly. To remedy
this situation, the relevant variable can be dropped from
the regression equation. For example between sales
and Tv-ads is 0.92 which is highly reasonable score or
between sales and Comp-ads is 0.155 which is very low
score .
Research Methods in Business Studies
20
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Multiple Regression
Research in business studies
o In multiple regression, at least two or more independent or
explanatory variables are applied to explain/predict the dependent
variable. The purpose is to make the model more realistic, control for
other variables, and explain more of the variance in the dependent
variable as well as reduce the residuals. The following is a typical
example output for a multiple regression.
21
Table 5 Multiple regression – output
Research Methods in Business Studies
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Research in business studies
Dummy Variables
o In a multiple regression, dummy variable can be used in two ways.
As a dependent variables where its values take 1 or 0 that is also
called dichotomous. The other type can be used as independent
variable which takes the value 0 or 1. The dummy variable used in
an analysis when there does not exist as numerical values. For
example, in the following table that is a nominal scaled variable that
can not be ranked so to be applied in a regression analysis, the
seasons need to be assigned numbers
Table 6 Coding of dummy variable
Research Methods in Business Studies
22
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Dummy variables
Example 7
Research in business studies
o In the following table, there three new variables A, B and C and indicates
that the four seasons are different combinations of zeros and ones. Assume
that the following regression model for sales of women’s clothing where the
price (P) is also included, has been estimated:
Sale=1000 - 0.5P+100A - 20B - 50C
a) Calculate the sales in the summer by considering dummy variables as
well (i.e. p=\$200 ).
b) Calculate the sales in the autumn by considering dummy variables as well
(i.e. p=\$200 ).
c) Compare the sales in winter and spring by keeping the same price.
Table 6 Coding of dummy variable
Research Methods in Business Studies
23
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
Dummy variables
Example 7-Answer
o
Research in business studies
In the following table, there three new variables A, B and C and
indicates that the four seasons are different combinations of zeros and
ones. Assume that the following regression model for sales of women’s
clothing where the price (P) is also included, has been estimated:
Sale=1000 - 0.5P+100A - 20B - 50C
a) Calculate the sales in the summer by considering dummy variables as
well (i.e. p=\$200 ).
Sale=1000 - 0.5 (200)+100(1) – 20(0) – 50(0)=\$1000
b) Calculate the sales in the autumn by considering dummy variables as
well (i.e. p=\$200 ).
Sale=1000 - 0.5 (200)+100(0) – 20(1) – 50(0)= \$880
c) Compare the sales in winter and spring by keeping the same price.
Winter- Sale=1000 - 0.5 (200)+100(0) – 20(0) – 50(1)= \$950
spring- Sale=1000 - 0.5 (200)+100(0) – 20(0) – 50(0)= \$900
Research Methods in Business Studies
24
© 2009/10, Sami Fethi, EMU, All Right Reserved, Pearson Education, 2005, 3. Ed.
```
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# Thread: Binary Sequence combinatorics
1. ## Binary Sequence combinatorics
The problem simply states that:
show that there cannot be 171 binary sequences such that each differ in at least four digits.
So far I've deducted that there should be some pigeonhole principle in here (I'm assuming, at least). So far I've written down that we should take the 2^12 different sequences and subtract the ones which only differ by one and those that differ only by two and then by three. I've tried this but with little success. Could I please have some guidance on how to go about this question?
thank you
2. ## Re: Binary Sequence combinatorics
Originally Posted by pikachu26134
The problem simply states that:
show that there cannot be 171 binary sequences such that each differ in at least four digits.
So far I've deducted that there should be some pigeonhole principle in here (I'm assuming, at least). So far I've written down that we should take the 2^12 different sequences and subtract the ones which only differ by one and those that differ only by two and then by three. I've tried this but with little success. Could I please have some guidance on how to go about this question?
As far as I know the is no universal agreement of the definition of Binary Sequence.
Look at this reference. As you can see there they assume an 8-bit string.
In that case your $\displaystyle 2^{12}$ makes no sense.
So you should define the terms in this posting.
For me a Binary Sequenc is any sequence of zeros and ones.
Thus you had best tell us what you are on about, i.e. define your terms,
3. ## Re: Binary Sequence combinatorics
By Binary Sequence, I mean any sequence including only the numbers 0 and 1 (base 2). When I say that each differs in at least four digits, I mean that if we have 0000 and 1111, they differ in four digits, as do 00000 and 11110. On the other hand, 0000 and 1110 do not differ by four digits.
I apologise about the previous ambiguity.
4. ## Re: Binary Sequence combinatorics
The OP was:
Originally Posted by pikachu26134
The problem simply states that:
show that there cannot be 171 binary sequences such that each differ in at least four digits.
Originally Posted by pikachu26134
By Binary Sequence, I mean any sequence including only the numbers 0 and 1 (base 2). When I say that each differs in at least four digits, I mean that if we have 0000 and 1111, they differ in four digits, as do 00000 and 11110. On the other hand, 0000 and 1110 do not differ by four digits.
You still have not clarified what kind of bit-strings the question is about: What is the length?
For example if the strings are of length seven then there are only $\displaystyle 2^7=128$ total strings.
So there is nothing to prove.
On the other hand, if the strings are of length $\displaystyle 100$ the statement is false.
5. ## Re: Binary Sequence combinatorics
The strings are of length 12 and hence there are 2^12 total strings
6. ## Re: Binary Sequence combinatorics
In terms of this paper (PDF), the problem asks to show that $\displaystyle A_2(12,4)<171$. By Corollary 1.3.1 on p. 12, $\displaystyle A_2(12,4)=A_2(11,3)$. In fact, Lemma 1.1 is sufficient to conclude that $\displaystyle A_2(12,4)\le A_2(11,3)$. Then Theorem 1.4 on p. 14 established the Hamming bound saying that $\displaystyle A_2(11,3)\le\frac{2^{11}}{\binom{11}{0}+\binom{11} {1}}$, which happens to be just smaller than 171.
This paper is a nice introduction to error-correcting codes and is a pleasant reading, so I don't think the time spent on understanding the results above would be lost.
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# Quantum Error Correction Michele Mosca. Quantum Error Correction: Bit Flip Errors l Suppose the environment will effect error (i.e. operation ) on our.
## Presentation on theme: "Quantum Error Correction Michele Mosca. Quantum Error Correction: Bit Flip Errors l Suppose the environment will effect error (i.e. operation ) on our."— Presentation transcript:
Quantum Error Correction Michele Mosca
Quantum Error Correction: Bit Flip Errors l Suppose the environment will effect error (i.e. operation ) on our quantum computer when the environment is in state
Analysis of 6-qubit system which uses syndrome bits to correct errors l Then we get (assuming ) Correct information Bits with errors
Quantum Error Correction l Then we get (assuming )
Quantum Error Correction l Then we get (assuming )
Quantum Error Correction l Then we get (assuming )
Quantum Error Correction l More generally, if the error effected on the system in state is of the form
Quantum Error Correction l and if the state only consists of mixtures of superpositions of codewords and (that is,) then the correction procedure (call it ) will map
Main Error Correction Theorem l Theorem 10.2: Suppose C is a quantum code and is the error-correction operation constructed in the proof of Theorem 10.1 to recover from a noise process with operation elements. Suppose is a quantum operation with elements which are linear combinations of the. Then the error correction operation also corrects the effects of the noise process on the code C.
Quantum Error Correction l E.g. the error correction procedure that we have described for the 3-qubit code will correct any combination of error operators
Quantum Error Correction l Notice that if lives in a Hilbert space of dimension then the operators Form a basis for the set of possible operation elements
Download ppt "Quantum Error Correction Michele Mosca. Quantum Error Correction: Bit Flip Errors l Suppose the environment will effect error (i.e. operation ) on our."
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Spike's Calculators
# Ellipse - Metric
Calculate the area and perimeter of an ellipse.
This calculator uses metres for measurements.
Ellipse = squashed circle.
### Ellipse Area and Perimeter - Metric
Length a m
Length b m
Decimal Precision
#### Results:
###### Area of the Ellipse
Square Metres
Square Feet ft²
Square Inches in²
Metres m
Feet ft
Inches in
#### Calculation
1. length a in metres
2. length b in metres
3. decimal precision, the number of digits after the decimal point
#### Results
1. the area of the ellipse in square metres
2. area in square feet
3. the area in square inches
4. the perimeter of the ellipse in metres
5. the perimeter in feet
6. the perimeter in inches
##### Formula
```A = ab/4
P = 𝛑R(1+(3H)/(10+√(4-3*H)))
where A is the area of the ellipse
a length of the short diameter
b length of the l diameter
P the perimeter of the ellipse
𝛑 = 3.14159265
R = (a+b)/2
S = (a-b)/2
H = R²/S²
```
##### Conversions
```one metre (m) = 3.28083989501312 feet (ft)
one metre (m) = 39.3700787401575 inches (in)
one square metre (m²) = 10.7639104167097 square feet (ft²)
one square metre (m²) = 1550.0031000062 square inches (in²)
```
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### Symmetry, Integrability and Geometry: Methods and Applications (SIGMA)
SIGMA 3 (2007), 001, 12 pages cond-mat/0701075 https://doi.org/10.3842/SIGMA.2007.001
Contribution to the Proceedings of the O'Raifeartaigh Symposium
### Non-Local Finite-Size Effects in the Dimer Model
Nickolay Sh. Izmailian a, b, c, Vyatcheslav B. Priezzhev d and Philippe Ruelle e
a) Institute of Physics, Academia Sinica, Nankang, Taipei 11529, Taiwan
b) Yerevan Physics Institute, Alikhanian Brothers 2, 375036 Yerevan, Armenia
c) National Center of Theoretical Sciences at Taipei, Physics Division, National Taiwan University, Taipei 10617, Taiwan
d) Bogolyubov Laboratory of Theoretical Physics, Joint Institute for Nuclear Research, 141980 Dubna, Russia
e) Institut de Physique Théorique, Université catholique de Louvain, 1348 Louvain-La-Neuve, Belgium
Received September 29, 2006, in final form December 12, 2006; Published online January 04, 2007
Abstract
We study the finite-size corrections of the dimer model on ∞ × N square lattice with two different boundary conditions: free and periodic. We find that the finite-size corrections depend in a crucial way on the parity of N, and show that, because of certain non-local features present in the model, a change of parity of N induces a change of boundary condition. Taking a careful account of this, these unusual finite-size behaviours can be fully explained in the framework of the c = -2 logarithmic conformal field theory.
Key words: dimer model; finite-size corrections; conformal field theory.
pdf (351 kb) ps (235 kb) tex (190 kb)
References
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2. Kasteleyn P.W., The statistics of dimers on a lattice. I. The number of dimer arrangements on a quadratic lattice, Physica 27 (1961), 1209-1225.
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5. Temperley H.N.V., Fisher M.E., Dimer problem in statistical mechanics - an exact result, Philos. Mag. (8) 6 (1961), 1061-1063.
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7. Hartwig R.E., Monomer pair correlations, J. Math. Phys. 7 (1966), 286-299.
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11. Kong Y., Logarithmic corrections in the free energy of monomer-dimer model on plane lattices with free boundaries, Phys. Rev. E 74 (2006), 011102, 8 pages.
12. Wu F.Y., Pfaffian solution of a dimer-monomer problem: single monomer on the boundary, Phys. Rev. E 74 (2006), 020140(R), 4 pages, Erratum, Phys. Rev. E 74 (2006), 020104(E), cond-mat/0607647.
13. Ferdinand A.E., Statistical mechanics of dimers on a quadratic lattice, J. Math. Phys. 8 (1967), 2332-2339.
14. McCoy B.W., Wu T.T., The two-dimensional Ising model, Harvard University Press, Cambridge, MA, 1973.
15. Bhattacharjee S.M., Nagle F.F., Finite-size effect for the critical point of an anisotropic dimer model of domain walls, Phys. Rev. A 31 (1985), 3199-3213.
16. Brankov J.G., Priezzhev V.B., Critical free energy of a Möbius strip, Nuclear Phys. B 400 (1993), 633-652.
17. Lu W.T., Wu F.Y., Dimer statistics on the Möbius strip and the Klein bottle, Phys. Lett. A 259 (1999), 108-114, cond-mat/9906154.
18. Lu W.T., Wu F.Y., Close-packed dimers on nonorientable surfaces, Phys. Lett. A 293 (2002), 235-246, Erratum, Phys. Lett. A 298 (2002), 293, cond-mat/0110035.
19. Ivashkevich E., Izmailian N.Sh., Hu C.-K., Kronecker's double series and exact asymptotic expansion for free models of statistical mechanics on torus, J. Phys. A: Math. Gen. 35 (2002), 5543-5561.
20. Izmailian N.Sh., Oganesyan K.B., Hu C.-K., Exact finite-size corrections of the free energy for the square lattice dimer model under different boundary conditions, Phys. Rev. E 67 (2003), 066114, 14 pages.
21. Izmailian N.Sh., Priezzhev V.B., Ruelle P., Hu C.-K., Logarithmic conformal field theory and boundary effects in the dimer model, Phys. Rev. Lett. 95 (2005), 260602, 4 pages, cond-mat/0512703.
22. Itzykson C., Saleur H., Zuber J.-B., Conformal invariance of nonunitary 2d-models, Europhys. Lett. 2 (1986), 91-96.
23. Blote H.W.J., Hilhorst H.J., Roughening transitions and the zero-temperature triangular Ising antiferromagent, J. Phys. A: Math. Gen. 15 (1982), L631-L637.
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25. Baxter R.J., Exactly solved models in statistical mechanics, Academic Press, New York, 1982.
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27. Priezzhev V.B., The dimer problem and the Kirchhoff theorem, Sov. Phys. Usp. 28 (1985), 1125-1135.
28. Blöte H.W., Cardy J.L., Nightingale M.P., Conformal invariance, the central charge, and universal finite-size amplitudes at criticality, Phys. Rev. Lett. 56 (1986), 742-745.
29. Affleck I., Universal term in the free energy at a critical point and the conformal anomaly, Phys. Rev. Lett. 56 (1986), 746-748.
30. Cardy J.L., Effect of boundary conditions on the operator content of two-dimensional conformally invariant theories, Nuclear Phys. B 275 (1986), 200-218.
31. Brankov J.G., Isomorphism of dimer configurations and spanning trees on finite square lattices, J. Math. Phys. 36 (1995), 5071-5083.
32. Majumdar S.N., Dhar D., Equivalence between the Abelian sandpile model and the q ® 0 limit of the Potts model, Phys. A 185 (1992), 129-145.
33. Ruelle P., A c = -2 boundary changing operator for the Abelian sandpile model, Phys. Lett. B 539 (2002), 172-177, hep-th/0203105.
34. Piroux G., Ruelle P., Pre-logarithmic and logarithmic fields in a sandpile model, J. Stat. Mech. Theory Exp. (2004), P10005, 24 pages, hep-th/0407143.
35. Piroux G., Ruelle P., Logarithmic scaling for height variables in the Abelian sandpile model, Phys. Lett. B 607 (2005), 188-196, cond-mat/0410253.
36. Jeng M., Piroux G., Ruelle P., Height variables in the Abelian sandpile model: scaling fields and correlations, J. Stat. Mech. Theory Exp. (2006), P10015, 63 pages, cond-mat/0609284.
37. Ghosh A., Dhar D., Jacobsen J.L., Random trimer tilings, cond-mat/0609322.
38. Priezzhev V.B., Ruelle P., Boundary monomers in the dimer model, in preparation.
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# Re: st: bootstrapping
From Maarten buis To statalist@hsphsun2.harvard.edu Subject Re: st: bootstrapping Date Thu, 10 Jul 2008 21:04:32 +0100 (BST)
```I would do the following before starting to mess with stuff like the
bootstrap:
expand nnoti
proportion esenti
-- Maarten
--- Vincenzo Carrieri <vincenzo.carrieri@gmail.com> wrote:
> I have the following stata output:
>
> > esenti = 0
>
> Variable | Obs Mean Std. Dev. Min
> Max
> ______________________________________________________
> nnotti | 1032 4.545543 4.936243 1
> 80
>
> -> esenti = 1
>
> Variable | Obs Mean Std. Dev. Min
> Max
> __________________________________________________________
> nnotti | 3125 9.78528 12.73163 1
> 90
>
>
> "Esenti" is a dummy variable that indicates if a person copayment
> free
> or not. "nnotti" are the numbers of days spent in a hospital.
> So, the total number of nights, are 1032 (Obs) x 4.545543 + 3125 X
> 9.78528 = 35270
> I wish to calculate the proportion of nights spent by esenti=1 out
> of
> total nights. I compute simply this proportion, but It's just a
> sample
> proportion. Is it meaningful to randomly replicate my sample 1000
> times and compute my proportion for each of these samples? It allows
> me to have a range of values of my proportion and not only a single
> sample value. I think this should be bootstrapping, but I don't have
> a
>
> I appreciate any advice or comment.
> Many thanks,
>
> Vincenzo
> *
> * For searches and help try:
> * http://www.stata.com/support/faqs/res/findit.html
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
-----------------------------------------
Maarten L. Buis
Department of Social Research Methodology
Vrije Universiteit Amsterdam
Boelelaan 1081
1081 HV Amsterdam
The Netherlands
Buitenveldertselaan 3 (Metropolitan), room Z434
+31 20 5986715
http://home.fsw.vu.nl/m.buis/
-----------------------------------------
__________________________________________________________
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1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# The set of real values of t∈[−π2,π2] satisfying 2sint=1−2x+5x23x2−2x−1, ∀x∈R−{−13,1} lies in the interval
A
[π2,π10][3π10,π2]
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B
[3π10,π2]
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C
[π2,π10]
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D
[π2,π2]
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Open in App
Solution
## The correct option is A [−π2,−π10]∪[3π10,π2]2sint=1−2x+5x23x2−2x−1⇒2sint=1−2x+5x2(3x+1)(x−1)⇒x2(6sint−5)+x(2−4sint)−(1+2sint)=0∵x∈R⇒Δ≥0⇒(2−4sint)2+4(6sint−5)(1+2sint)≥0⇒4(4sin2t−2sint+1)+4(12sin2t−4sint−5)≥0⇒16sin2t−8sint−4≥0⇒4sin2t−2sint−1≥0⇒[sint−(1−√54)][sint−(1+√54)]≥0⇒sint≤1−√54 or sint≥1+√54 Now using the graph, we get ∴t∈[−π2,−π10]∪[3π10,π2]
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
I’ve had several requests for an R function to simulate future values from a TBATS model. We will eventually include TBATS in the fable package, and the facilities will be added there. But in the meantime, if you are using the forecast package and want to simulate from a fitted TBATS model, here is how do it.
## Simulating via one-step forecasts
Doing it efficiently would require a more complicated approach, but this is super easy if you are willing to sacrifice some speed. The trick is to realise that a simulation can be handled easily for almost any time series model using residuals and one-step forecasts. Note that a residual is given by $e_t = y_t – \hat{y}_{t|t-1}$ so we can write
$$y_t = \hat{y}_{t|t-1} + e_t.$$
Therefore, given data to time $T$, we can simulate iteratively using $$y_{T+i}^* = \hat{y}_{T+i|T+i-1} + \varepsilon_{T+i}, \qquad i=1,\dots,h,$$ where $\varepsilon_{T+i}$ is randomly generated from the error distribution, or bootstrapped by randomly sampling from past residuals. The value of $\hat{y}_{T+i|T+i-1}$ can be obtained by applying the model to the series $\{y_1,\dots,y_T,y^*_{T+1},\dots,y^*_{T+i-1}\}$ (without re-estimating the parameters) and forecasting one-step ahead. This is the same trick we use to get prediction intervals for neural network models.
## Implementation
Because simulate() is an S3 method in R, we have to make sure the corresponding simulate.tbats() function has all the necessary arguments to match other simulate functions. It’s also good to make it as close as possible to other simulate functions in the forecast package, to make it easier for users when switching between them. The real work is done in the last few lines.
simulate.tbats <- function(object, nsim=length(object$y), seed = NULL, future=TRUE, bootstrap=FALSE, innov = NULL, ...) { if (is.null(innov)) { if (!exists(".Random.seed", envir = .GlobalEnv)) { runif(1) } if (is.null(seed)) { RNGstate <- .Random.seed } else { R.seed <- .Random.seed set.seed(seed) RNGstate <- structure(seed, kind = as.list(RNGkind())) on.exit(assign(".Random.seed", R.seed, envir = .GlobalEnv)) } } else { nsim <- length(innov) } if (bootstrap) { res <- residuals(object) res <- na.omit(res - mean(res, na.rm = TRUE)) e <- sample(res, nsim, replace = TRUE) } else if (is.null(innov)) { e <- rnorm(nsim, 0, sqrt(object$variance))
} else {
e <- innov
}
x <- getResponse(object)
y <- numeric(nsim)
if(future) {
dataplusy <- x
} else {
# Start somewhere in the original series
dataplusy <- ts(sample(x, 1), start=-1/frequency(x),
frequency = frequency(x))
}
fitplus <- object
for(i in seq_along(y)) {
y[i] <- forecast(fitplus, h=1)\$mean + e[i]
dataplusy <- ts(c(dataplusy, y[i]),
start=start(dataplusy), frequency=frequency(dataplusy))
fitplus <- tbats(dataplusy, model=fitplus)
}
return(tail(dataplusy, nsim))
}
I’ve added this to the forecast package for the next version.
Something similar could be written for any other forecasting function that doesn’t already have a simulate method. Just swap the tbats call to the relevant modelling function.
## Example
library(forecast)
library(ggplot2)
fit <- tbats(USAccDeaths)
p <- USAccDeaths %>% autoplot() +
labs(x = "Year", y = "US Accidental Deaths",
title = "TBATS simulations")
for (i in seq(9)) {
p <- p + autolayer(simulate(fit, nsim = 36), series = paste("Sim", i))
}
p
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Outlook: Maiden Holdings Ltd. 6.625% Notes due 2046 is assigned short-term B1 & long-term B2 estimated rating.
AUC Score : What is AUC Score?
Short-Term Revised1 :
Dominant Strategy : Sell
Time series to forecast n: for Weeks2
Methodology : Multi-Instance Learning (ML)
Hypothesis Testing : Paired T-Test
Surveillance : Major exchange and OTC
1The accuracy of the model is being monitored on a regular basis.(15-minute period)
2Time series is updated based on short-term trends.
## Summary
Maiden Holdings Ltd. 6.625% Notes due 2046 prediction model is evaluated with Multi-Instance Learning (ML) and Paired T-Test1,2,3,4 and it is concluded that the MHLA stock is predictable in the short/long term. Multi-instance learning (MIL) is a machine learning (ML) problem where a dataset consists of multiple instances, and each instance is associated with a single label. The goal of MIL is to learn a model that can predict the label of a new instance based on the labels of the instances that it is similar to. MIL is a challenging problem because the instances in a dataset are not labeled individually. This means that the model cannot simply learn a mapping from the features of an instance to its label. Instead, the model must learn a way to combine the features of multiple instances to predict the label of a new instance.5 According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Sell
## Key Points
1. Multi-Instance Learning (ML) for MHLA stock price prediction process.
2. Paired T-Test
3. Is it better to buy and sell or hold?
4. Can statistics predict the future?
5. Reaction Function
## MHLA Stock Price Forecast
We consider Maiden Holdings Ltd. 6.625% Notes due 2046 Decision Process with Multi-Instance Learning (ML) where A is the set of discrete actions of MHLA stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
Sample Set: Neural Network
Stock/Index: MHLA Maiden Holdings Ltd. 6.625% Notes due 2046
Time series to forecast: 8 Weeks
According to price forecasts, the dominant strategy among neural network is: Sell
F(Paired T-Test)6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Multi-Instance Learning (ML)) X S(n):→ 8 Weeks $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of MHLA stock
j:Nash equilibria (Neural Network)
k:Dominated move of MHLA stock holders
a:Best response for MHLA target price
Multi-instance learning (MIL) is a machine learning (ML) problem where a dataset consists of multiple instances, and each instance is associated with a single label. The goal of MIL is to learn a model that can predict the label of a new instance based on the labels of the instances that it is similar to. MIL is a challenging problem because the instances in a dataset are not labeled individually. This means that the model cannot simply learn a mapping from the features of an instance to its label. Instead, the model must learn a way to combine the features of multiple instances to predict the label of a new instance.5 A paired t-test is a statistical test that compares the means of two paired samples. In a paired t-test, each data point in one sample is paired with a data point in the other sample. The pairs are typically related in some way, such as before and after measurements, or measurements from the same subject under different conditions. The paired t-test is a parametric test, which means that it assumes that the data is normally distributed. The paired t-test is also a dependent samples test, which means that the data points in each pair are correlated.6,7
For further technical information as per how our model work we invite you to visit the article below:
How do Predictive A.I. algorithms actually work?
### MHLA Stock Forecast (Buy or Sell) Strategic Interaction Table
Strategic Interaction Table Legend:
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
### Financial Data Adjustments for Multi-Instance Learning (ML) based MHLA Stock Prediction Model
1. To calculate the change in the value of the hedged item for the purpose of measuring hedge ineffectiveness, an entity may use a derivative that would have terms that match the critical terms of the hedged item (this is commonly referred to as a 'hypothetical derivative'), and, for example for a hedge of a forecast transaction, would be calibrated using the hedged price (or rate) level. For example, if the hedge was for a two-sided risk at the current market level, the hypothetical derivative would represent a hypothetical forward contract that is calibrated to a value of nil at the time of designation of the hedging relationship. If the hedge was for example for a one-sided risk, the hypothetical derivative would represent the intrinsic value of a hypothetical option that at the time of designation of the hedging relationship is at the money if the hedged price level is the current market level, or out of the money if the hedged price level is above (or, for a hedge of a long position, below) the current market level. Using a hypothetical derivative is one possible way of calculating the change in the value of the hedged item. The hypothetical derivative replicates the hedged item and hence results in the same outcome as if that change in value was determined by a different approach. Hence, using a 'hypothetical derivative' is not a method in its own right but a mathematical expedient that can only be used to calculate the value of the hedged item. Consequently, a 'hypothetical derivative' cannot be used to include features in the value of the hedged item that only exist in the hedging instrument (but not in the hedged item). An example is debt denominated in a foreign currency (irrespective of whether it is fixed-rate or variable-rate debt). When using a hypothetical derivative to calculate the change in the value of such debt or the present value of the cumulative change in its cash flows, the hypothetical derivative cannot simply impute a charge for exchanging different currencies even though actual derivatives under which different currencies are exchanged might include such a charge (for example, cross-currency interest rate swaps).
2. The assessment of whether an economic relationship exists includes an analysis of the possible behaviour of the hedging relationship during its term to ascertain whether it can be expected to meet the risk management objective. The mere existence of a statistical correlation between two variables does not, by itself, support a valid conclusion that an economic relationship exists.
3. In some jurisdictions, the government or a regulatory authority sets interest rates. For example, such government regulation of interest rates may be part of a broad macroeconomic policy or it may be introduced to encourage entities to invest in a particular sector of the economy. In some of these cases, the objective of the time value of money element is not to provide consideration for only the passage of time. However, despite paragraphs B4.1.9A–B4.1.9D, a regulated interest rate shall be considered a proxy for the time value of money element for the purpose of applying the condition in paragraphs 4.1.2(b) and 4.1.2A(b) if that regulated interest rate provides consideration that is broadly consistent with the passage of time and does not provide exposure to risks or volatility in the contractual cash flows that are inconsistent with a basic lending arrangement.
4. A similar example of a non-financial item is a specific type of crude oil from a particular oil field that is priced off the relevant benchmark crude oil. If an entity sells that crude oil under a contract using a contractual pricing formula that sets the price per barrel at the benchmark crude oil price minus CU10 with a floor of CU15, the entity can designate as the hedged item the entire cash flow variability under the sales contract that is attributable to the change in the benchmark crude oil price. However, the entity cannot designate a component that is equal to the full change in the benchmark crude oil price. Hence, as long as the forward price (for each delivery) does not fall below CU25, the hedged item has the same cash flow variability as a crude oil sale at the benchmark crude oil price (or with a positive spread). However, if the forward price for any delivery falls below CU25, the hedged item has a lower cash flow variability than a crude oil sale at the benchmark crude oil price (or with a positive spread).
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
### MHLA Maiden Holdings Ltd. 6.625% Notes due 2046 Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*B1B2
Income StatementBa2C
Balance SheetCaa2B3
Leverage RatiosCaa2Baa2
Cash FlowBa1B3
Rates of Return and ProfitabilityBaa2C
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
## References
1. L. Busoniu, R. Babuska, and B. D. Schutter. A comprehensive survey of multiagent reinforcement learning. IEEE Transactions of Systems, Man, and Cybernetics Part C: Applications and Reviews, 38(2), 2008.
2. Bell RM, Koren Y. 2007. Lessons from the Netflix prize challenge. ACM SIGKDD Explor. Newsl. 9:75–79
3. Bessler, D. A. T. Covey (1991), "Cointegration: Some results on U.S. cattle prices," Journal of Futures Markets, 11, 461–474.
4. Athey S. 2017. Beyond prediction: using big data for policy problems. Science 355:483–85
5. Canova, F. B. E. Hansen (1995), "Are seasonal patterns constant over time? A test for seasonal stability," Journal of Business and Economic Statistics, 13, 237–252.
6. Dudik M, Langford J, Li L. 2011. Doubly robust policy evaluation and learning. In Proceedings of the 28th International Conference on Machine Learning, pp. 1097–104. La Jolla, CA: Int. Mach. Learn. Soc.
7. M. Benaim, J. Hofbauer, and S. Sorin. Stochastic approximations and differential inclusions, Part II: Appli- cations. Mathematics of Operations Research, 31(4):673–695, 2006
Frequently Asked QuestionsQ: Is MHLA stock expected to rise?
A: MHLA stock prediction model is evaluated with Multi-Instance Learning (ML) and Paired T-Test and it is concluded that dominant strategy for MHLA stock is Sell
Q: Is MHLA stock a buy or sell?
A: The dominant strategy among neural network is to Sell MHLA Stock.
Q: Is Maiden Holdings Ltd. 6.625% Notes due 2046 stock a good investment?
A: The consensus rating for Maiden Holdings Ltd. 6.625% Notes due 2046 is Sell and is assigned short-term B1 & long-term B2 estimated rating.
Q: What is the consensus rating of MHLA stock?
A: The consensus rating for MHLA is Sell.
Q: What is the forecast for MHLA stock?
A: MHLA target price forecast: Sell
What did you think about the prediction? (Insufficient-Outstanding)
Tell us how we can improve PredictiveAI
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https://www.mytutor.co.uk/answers/60185/GCSE/Physics/Draw-a-graph-depicting-a-skydivers-speed-against-time-when-jumping-from-a-plane-until-he-deploys-his-parachute-explaining-the-logic-of-your-answer-through-the-forces-applicable-to-the-body/
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# Draw a graph depicting a skydivers speed against time when jumping from a plane, until he deploys his parachute, explaining the logic of your answer through the forces applicable to the body.
Upon jumping from the plane, the diver experiences three forces; a downward force from his weight; opposed by a lift force (equal to the weight of the air he displaces); and an upward drag force = ½ . (CD.(RHO).V2.A).
Initially the largest force is the wight of the jumper, leading an increase in downward velocity due to newtons second law, however as can be seen from the drag equation, this increase in velocity will lead to a larger increase in drag force, slowing the divers acceleration, until the downward forces = the up. At this point the diver is at his terminal velocity. When he releases his parachute, there is again a massive increase in the drag force due to the larger frontal area (A), thereby decelerating the diver until he again reaches a new, reduced, terminal velocity.
Answered by Bronagh R. Physics tutor
1241 Views
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Untitled document (8).docx - A group of medical professionals is considering the construction of a private clinic If the medical demand is high(i.e
# Untitled document (8).docx - A group of medical...
• 2
This preview shows page 1 - 2 out of 2 pages.
A group of medical professionals is considering the construction of a private clinic. If the medical demand is high (i.e., there is a favorable market for the clinic), the physicians could realize a net profit of \$100,000. If the market is not favorable, they could lose \$40,000. Of course, they don’t have to proceed at all, in which case there is no cost. In the absence of any market data, the best the physicians can guess is that there is a 50–50 chance the clinic will be successful. Question 9 (20 points)Question 9 options:Construct a decision tree by fill-in the blanks below in reference to the following chart.The decision choice at Decision 1 is ____ and that at Decision 2 is ____ Event 1 is ____ and Event 2 is ____ .The probability for Prob1 is ____ and that for Prob2 is ____ .Payoff 1 is ____ and Payoff 2 is ____ . EMV 1 is ____ and EMV 2 is ____ .and that at Decision 2 isEvent 1 isand Event 2 is.
The probability for Prob1 isand that for Prob2 is. Payoff 1 isand Payoff 2 is. EMV 1 isand EMV 2 is.
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## Ozelenerji I trust Worksheet.
Published at Monday, April 15th 2019, 07:06:05 AM. Worksheet. By Margaux Poulain.
The present generation seems to be blessed immensely with intellect and the benefits of mastering math are something worth considering. It is a well-known fact that math is not a subject that one learns by simply reading the problems and its solutions. In order to master the subject, earnest practice on multiple problems is the best way to go. However, not every person is bestowed with required materials like math worksheets to receive adequate amount of practice.
If you cannot purchase a math work sheet because you think you may not have time to, then you can create on using your home computer and customize it for your kid. Doing this is easy. All you need is Microsoft word application in your computer to achieve this. Just open the word application in your computer and start a new document. Ensure that the new document you are about to create is based on a template. Then, ensure that your internet connection is on before you can search the term ”math worksheet” from the internet. You will get templates of all kinds for your worksheet. Choose the one you want and then download.
### 3 Digit Addition And Subtraction With Regrouping Word Problems
##### Free Printable Reading Comprehension Worksheets For High School
###### Writing Equations From Word Problems Worksheet 6th Grade
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom’s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn’t. The bottom line is that kids learn math much better when it makes sense.
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom’s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn’t. The bottom line is that kids learn math much better when it makes sense.
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### Three Squares
What is the greatest number of squares you can make by overlapping three squares?
### Two Dice
Find all the numbers that can be made by adding the dots on two dice.
### Biscuit Decorations
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
# Count the Digits
##### Age 5 to 11Challenge Level
Count the Digits printable sheet
We can do all sorts of things with numbers - add, subtract, multiply, divide, ...
Most of us start with counting when we are very little. We usually count things, objects, people, etc. In this activity we are going to count the number of digits that are the same.
Rule 1 - The starting number has to have just three different digits chosen from $1, 2, 3, 4$.
Rule 2 - The starting number must have four digits - thousands, hundreds, tens and ones.
For example, we could choose $2124$ or $1124$.
When we've got our starting number we'll do some counting. Here is an example.
Starting Number:
4 1 3 4
We will count in order the number of $1$s, then the number of $2$s, then $3$s and lastly $4$s, and write it down as shown here.
4 1 3 4 1 1 1 3 2 4 3 1 1 2 1 3 1 4 4 1 1 2 2 3 1 4 3 1 2 2 1 3 2 4
The first count gave one $1$, one $3$ and two $4$s.
We have continued this underneath, so the third line shows that the line above had three $1$s, one $2$, one $3$ and one $4$.
The fourth line counts the digits in the line above it, giving four $1$s, one $2$, two $3$s and one $4$.
And so it goes on until... until when?
Your challenge is to start with other four-digit numbers which satisfy the two rules and count the digits the way we did in the picture above.
What do you notice?
What happens if you have five digits in the starting number instead?
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# MAP
Formulas / MAP
Apply a custom function to an array.
`MAP(array1, [array2], lambda)`
• array1 - required, an array to be mapped
• lambda_or_array<#> - required, an LAMBDA which must be the last argument in the function call, or another array to be mapped
## Examples
• `=MAP({1,2,3},LAMBDA(a,a+1))`
The MAP function can be used to apply a formula to every value in an array. For example, this formula adds one to each value of the array {1,2,3}, and returns {2,3,4}. This is a useful way to apply a formula to a range of values quickly and easily.
• `=MAP(A1:A5,LAMBDA(a,a*2))`
The MAP function can also be used to apply a formula to a range of cells. For example, this formula multiplies each value in the range A1:A5 by two, and returns the resulting array. This can be a useful way to transform a range of values quickly and easily.
• `=MAP({1,2,3},{4,5,6},LAMBDA(a,b,a+b))`
The MAP function can also be used to combine two arrays. For example, this formula adds each value in the first array to the corresponding value in the second array, and returns {5,7,9}. This can be a useful way to combine two arrays quickly and easily.
• `=MAP({1,2,3},{4,5,6},{7,8,9},LAMBDA(a,b,c,a+b+c))`
The MAP function can also be used to perform calculations on multiple arrays, like the example above.
## Summary
The MAP function provides an easy way to visualize data. It can be used to show values or categories through variations of colors or by different colors. This makes it a useful tool to quickly analyze data.
• The MAP function maps a custom LAMBDA function to each value in an array and returns an array of values. The LAMBDA function is applied to each value in the array and the result is an array of the same dimensions as the original array.
• The MAP function is useful when processing each item in an array one by one and for complex formula logic that should be managed in a single location. It is also possible to reuse the same code elsewhere by using a named LAMBDA function with MAP.
• The MAP function requires two required arguments: array1 and lambda. The array1 argument is the array to be mapped and the lambda argument is the custom lambda function.
What is the MAP function?
The MAP function is a function that applies a lambda to each value in an array to create a new value.
What are the arguments for the MAP function?
The MAP function takes two arguments:
• array1 - required argument
• lambda_or_array<#> - required argument, this is the lambda to apply to each array passed
What does the MAP function do?
The MAP function applies a lambda to each value in an array to create a new value.
## Make Better Decisions With Data
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The path of a firework is described by the equation: y= – 4.9(x-5) 2+124, where y represents the height in meters and x represents the time
Question
The path of a firework is described by the equation: y= – 4.9(x-5) 2+124, where y represents the height in meters and x represents the time in seconds… What is the maximum height of the firework. how many seconds does it take for the firework to reach maximum height. calculate the height 3.5 after the launch.
in progress 0
6 months 2021-08-12T15:00:27+00:00 1 Answers 9 views 0
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#### Balance Chemical Equation - Online Balancer
Balanced equation:
HCOOH + NaOH = HCOONa + H2O
Reaction type: double replacement
Reaction stoichiometry Limiting reagent
CompoundCoefficientMolar MassMolesWeight
HCOOH146.02538
NaOH139.99710928
HCOONa168.00720928
H2O118.01528
Units: molar mass - g/mol, weight - g.
Direct link to this balanced equation:
Instructions on balancing chemical equations:
• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
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Back to Online Chemical Tools Menu
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A287870 The extended Wythoff array (the Wythoff array with two extra columns) read by antidiagonals downwards. 2
0, 1, 1, 1, 3, 2, 2, 4, 4, 3, 3, 7, 6, 6, 4, 5, 11, 10, 9, 8, 5, 8, 18, 16, 15, 12, 9, 6, 13, 29, 26, 24, 20, 14, 11, 7, 21, 47, 42, 39, 32, 23, 17, 12, 8, 34, 76, 68, 63, 52, 37, 28, 19, 14, 9, 55, 123, 110, 102, 84, 60, 45, 31, 22, 16, 10, 89, 199, 178, 165, 136, 97, 73, 50, 36, 25, 17, 11 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 REFERENCES Conway, John; Ryba, Alex; The extra Fibonacci series and the Empire State Building. Math. Intelligencer 38 (2016), no. 1, 41-48. LINKS EXAMPLE The extended Wythoff array is the Wythoff array with two extra columns, giving the row number n and A000201(n), separated from the main array by a vertical bar: 0 1 | 1 2 3 5 8 13 21 34 55 89 144 ... 1 3 | 4 7 11 18 29 47 76 123 199 322 521 ... 2 4 | 6 10 16 26 42 68 110 178 288 466 754 ... 3 6 | 9 15 24 39 63 102 165 267 432 699 1131 ... 4 8 | 12 20 32 52 84 136 220 356 576 932 1508 ... 5 9 | 14 23 37 60 97 157 254 411 665 1076 1741 ... 6 11 | 17 28 45 73 118 191 309 500 809 1309 2118 ... 7 12 | 19 31 50 81 131 212 343 555 898 1453 2351 ... 8 14 | 22 36 58 94 152 246 398 644 1042 1686 2728 ... 9 16 | 25 41 66 107 173 280 453 733 1186 1919 3105 ... 10 17 | 27 44 71 115 186 301 487 788 1275 2063 3338 ... 11 19 | 30 49 79 ... 12 21 | 33 54 87 ... 13 22 | 35 57 92 ... 14 24 | 38 62 ... 15 25 | 40 65 ... 16 27 | 43 70 ... 17 29 | 46 75 ... 18 30 | 48 78 ... 19 32 | 51 83 ... 20 33 | 53 86 ... 21 35 | 56 91 ... 22 37 | 59 96 ... 23 38 | 61 99 ... 24 40 | 64 ... 25 42 | 67 ... 26 43 | 69 ... 27 45 | 72 ... 28 46 | 74 ... 29 48 | 77 ... 30 50 | 80 ... 31 51 | 82 ... 32 53 | 85 ... 33 55 | 88 ... 34 56 | 90 ... 35 58 | 93 ... 36 59 | 95 ... 37 61 | 98 ... 38 63 | ... ... CROSSREFS Cf. A035513 (the Wythoff array), A287869, A000201. Sequence in context: A111739 A182214 A216161 * A235354 A058743 A117970 Adjacent sequences: A287867 A287868 A287869 * A287871 A287872 A287873 KEYWORD nonn,tabl AUTHOR N. J. A. Sloane, Jun 14 2017 STATUS approved
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Last modified January 21 13:44 EST 2019. Contains 319350 sequences. (Running on oeis4.)
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# A bullet goes through a ballistic pendulum, what is it's initial velocity?
• Nibbz
In summary, a 7.0-g bullet with a speed of 200 m/s is fired into a 1.5-kg ballistic pendulum. The block rises to a maximum height of 12 cm. To find the initial speed of the bullet, the Conservation of Momentum for Glancing Collisions with both X and Y components can be used. However, the angle at which the pendulum rises may need to be taken into account. Alternatively, the Conservation of Energy equation may also be used.
Nibbz
## Homework Statement
A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.
## Homework Equations
Conservation of Momentum
## The Attempt at a Solution
I first converted all of my units into SI units. I then went to use the Conservation of Momentum for Glancing Collisions, with X and Y components, but I am confused as to what the components of the bullet are. All I know is that the pendulum rises at some angle to .12 m above it's original position and I'm not sure how to interpret this into an equation. Any assistance will be greatly appreciated.
Is it necessary to do this with Cons of momentum as Cons of energy seems easier?
## 1. What is a ballistic pendulum?
A ballistic pendulum is a device used to measure the velocity of a projectile by observing the motion of a pendulum after it has been struck by the projectile.
## 2. How does a ballistic pendulum work?
A ballistic pendulum works by taking advantage of the principle of conservation of momentum. When a projectile is fired into a pendulum, it transfers some of its momentum to the pendulum. By measuring the motion of the pendulum after the impact, the initial velocity of the projectile can be calculated.
## 3. Why is a ballistic pendulum used to measure velocity?
A ballistic pendulum is used to measure velocity because it is a simple and accurate method. It also allows for the measurement of high velocities that may be difficult to measure directly.
## 4. What factors can affect the accuracy of a ballistic pendulum?
The accuracy of a ballistic pendulum can be affected by factors such as air resistance, friction in the pivot point of the pendulum, and the elasticity of the pendulum arm. These factors can cause the pendulum to lose some of its kinetic energy, leading to a lower calculated initial velocity for the projectile.
## 5. How can the initial velocity of a projectile be calculated using a ballistic pendulum?
The initial velocity of a projectile can be calculated by using the formula v = √(2gh), where v is the initial velocity of the projectile, g is the acceleration due to gravity, and h is the height of the pendulum's swing. This formula is derived from the conservation of momentum principle and can be used to calculate the initial velocity of the projectile after it has been fired into the pendulum.
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# Yahoo Web Search
1. ### Physics Question?
...0.77 m/s² ◄ The direction of the acceleration (and force) relative to West is Θ = arctan(12/17) = 35º S of West ◄ ...
1 Answers · Science & Mathematics · 13/11/2020
... W of North (or 64º N of West ) B. |a| = |F| / m = 3.95 m/s² same direction...please select Favorite Answer. You get points too!
1 Answers · Science & Mathematics · 20/10/2020
3. ### Basic physics question?
blue cart initial position xb = 1.54 m East acceleration ab = 0.14 m/sec^2 West red cart initial position xr = 0.12 m East initial velocity Vr = 0.45...
1 Answers · Science & Mathematics · 05/10/2020
4. ### What is Green Goblin’s speed relative to Spiderman so Spiderman can accurately shot his web?
... to the ground, GG is moving at 5.0 m/s West and 23.0 m/s North. From Spidey's point of view, GG is moving towards him at (25.0+5.0)m/s...
1 Answers · Science & Mathematics · 28/09/2020
5. ### Two balls, X and Y, move along a horizontal frictionless surface, as illustrated below ( A LEVEL PHYSICS QUESTION )?
...b) what is the final velocity (V) (a) conserve momentum east- west px = m₁*3.0 - 2.5*9.6*cos60º = 0 = 3.0*m₁...helpful, please select Favorite Answer. You get points too!
1 Answers · Science & Mathematics · 06/11/2020
...3.18 m (E) 3rd Kick: 2.10 m (30* West of South) = 2.10 x cos30* (S) & 2.10 x... Final displacement from starting point = [6m (N) + 3.18m (N) + 1.82m (S)] &...
2 Answers · Science & Mathematics · 17/07/2020
7. ### Marry and Chorge are talking about their friendship with a little tension between them. The little tension becomes huge fight after a while,?
1 Answers · Science & Mathematics · 31/05/2020
8. ### 5: Balance the following face plate if the masses are as follows:A2.8kgB3.5kgC4.2kgD4.8kgE = ?
...right, then you want to balance the moments. balancing east west : 3.5kg*0.35*sin45º + 4.2kg*0.315*cos30º - 4.8kg*0...helpful, please select Favorite Answer. You get points too!
1 Answers · Science & Mathematics · 22/04/2020
9. ### A bicyclist rides 2.94 km due east, while the resistive force from the air has a magnitude of 7.47 N and points due west .?
work is a scalar quantity, therefore W = 7.47*2.94*2 = 43.9 kjoule
3 Answers · Science & Mathematics · 06/11/2019
10. ### A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are?
... of each stroke and sum. Let's call north and east positive; south and west negative. The first stroke has a north component of 4.04 m; the...
1 Answers · Science & Mathematics · 19/09/2019
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Specificity Sensitivity Question! - USMLE Forums
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USMLE Forums Specificity Sensitivity Question!
USMLE Step 1 Forum USMLE Step 1 Discussion Forum: Let's talk about anything related to USMLE Step 1 exam
#1
03-25-2012
USMLE Forums Veteran Steps History: 1+CK+CS Posts: 257 Threads: 49 Thanked 75 Times in 50 Posts
Specificity Sensitivity Question!
Assuming that mammography has a sensitivity of 90% and a specificity of 98% and that consecutive tests are independent, what is the probability that a woman with breast cancer will have a negative yearly screening mammogram for
two consecutive years?
a. 1/10
b. 2/10
c. 4/10
d. 1/100
e. 4/100
#2
03-25-2012
USMLE Forums Guru Steps History: Step 1 Only Posts: 452 Threads: 11 Thanked 358 Times in 200 Posts Reputation: 368
D.........
#3
03-25-2012
USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 529 Threads: 57 Thanked 291 Times in 157 Posts Reputation: 301
D
They are asking about the probability of FALSE NEGATIVE results of mammogram for 2 consecutive events. FALSE NEGATIVE RATE = 1-seNsitivity = 1-0.9 = 0.1
Now the probability of having 2 false negative results in 2 consecutive yrs are independent events so u multiply the probabilities 0.1*0.1 or 1/10*1/10 =1/100 (option d)
The above post was thanked by: Dr. Mexito (03-25-2012), excellence (05-18-2014), Hope2Pass (03-26-2012), Kruno (03-15-2016), taher (11-16-2013)
#4
03-26-2012
USMLE Forums Veteran Steps History: 1+CK+CS Posts: 257 Threads: 49 Thanked 75 Times in 50 Posts
#5
11-16-2013
USMLE Forums Scout Steps History: Not yet Posts: 65 Threads: 9 Thanked 15 Times in 9 Posts Reputation: 25
Guys, I donot understand why do we have to multiply 1/10 of the first year with the 1/10 of next year as those two tests are independent. Why the answer is not just 1/10. Please explain?
If sensitivity is 90% means, out of 100 diseased 90 were correctly identified as diseased (True Positives) and 10% were missed (False negatives) &
If the specificity is 98% means, out of 100 healthy people 98 were correctly identified as normal (True negatives) and 2 were missed (False positives)
False Negative rate = False negatives/ No. of diseased = 10/100 = 1/10. I can get this far. But the multiplying thing. Elaborate please..........
#6
08-10-2016
USMLE Forums Newbie Steps History: Not yet Posts: 3 Threads: 0 Thanked 0 Times in 0 Posts Reputation: 10
Quote:
Originally Posted by devareddy Guys, I donot understand why do we have to multiply 1/10 of the first year with the 1/10 of next year as those two tests are independent. Why the answer is not just 1/10. Please explain? If sensitivity is 90% means, out of 100 diseased 90 were correctly identified as diseased (True Positives) and 10% were missed (False negatives) & If the specificity is 98% means, out of 100 healthy people 98 were correctly identified as normal (True negatives) and 2 were missed (False positives) False Negative rate = False negatives/ No. of diseased = 10/100 = 1/10. I can get this far. But the multiplying thing. Elaborate please..........
You have to multiply because we have to find the <probability that a woman with breast cancer will have a negative yearly screening mammogram for two consecutive year>. So she has to have 2 negative yearly screening mammograms for 2 consecutive years.
This type of probability is called COMPOUND PROBABILITY and it can be calculated by multiplying the porbabilities of each independent event.
So we have two independets events with a probability of 1/10 each one; we multiply 1/10*1/10 and we find 1/100 (answer D).
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# Allowable Blood Loss (ABL) Calculator
Uses patient weight, ABV and haematocrit to estimate the lost blood allowance during surgery.
In the text below the calculator there is more information about the blood loss compensation.
The allowable blood loss calculator estimates the ABL based on patient data in order to facilitate blood provision during surgery and other procedures.
It accounts for the type of patient (with a choice from premature neonate to adult man or woman), patient weight, initial and final haematocrit and average blood volume.
Please note that the ABV values used for each type of patient can be found in the text below the tool.
The first step in finding ABL is retrieving the estimated blood volume:
■ EBV in mL = Weight in kg x Average blood volume in mL/kg
EBV is then multiplied by hematocrit values, to determine the allowable blood loss:
■ ABL = (EBV x (Hi-Hf))/Hi
Where:
- Hi – initial hematocrit;
- Hf – final hematocrit.
Patient
Weight
Initial hematocrit
Final hematocrit
Average blood volume (ABV)
Embed Print Share
## Send Us Your Feedback
Steps on how to print your input & results:
1. Fill in the calculator/tool with your values and/or your answer choices and press Calculate.
2. Then you can click on the Print button to open a PDF in a separate window with the inputs and results. You can further save the PDF or print it.
Please note that once you have closed the PDF you need to click on the Calculate button before you try opening it again, otherwise the input and/or results may not appear in the pdf.
## ABL variables and formula
This ABL calculator determines the allowance for blood loss during surgery and other invasive procedures. The variables used are explained in the table below:
Variable Description Type of patient Premature neonates Full term neonate Infant Adult woman Adult man Weight Patient weight in either kg or lbs Initial hematocrit Percentage blood composition before surgery Final hematocrit Percentage blood composition expected after surgery Average blood volume Quantity of blood in mL per kg (per guidelines) Premature neonates (95) Full term neonate (85) Infant (80) Adult woman (65) Adult man (75)
Allowable blood loss is calculated after the Gross formula and can provide assistance in regard to haemorrhage management in both internal and external blood loss.
There are two steps to perform when calculating ABL:
■ First step is to establish the estimated blood volume:
EBV in mL = Weight in kg x Average blood volume in mL/kg
■ Second step is to use the EBV and haematocrit values to determine the allowable blood loss:
ABL = (EBV x (Hi-Hf))/Hi
Where:
- Hi – initial hematocrit;
- Hf – final hematocrit.
When the maximum ABL value is reached, haemorrhage complication become increasingly more likely.
## Blood loss compensation
Using an accutate recording of blood loss means that replacement therapy (crystalloid or colloid solutions) can be started in a timely manner to reobtain normovolemia (normal intravascular volumes).
A haemoglobin between 7 and 10 mg/dL (which corresponds to a haematocrit of 21-30%) marks the threshold from which IV solutions are replaced with blood transfusions.
However, it is also important to note that in acute haemorrhage, haematocrit changes do not reflect the true extent of blood loss, for example 1000 mL blood loss only marks a 3% drop in Ht during the first hour.
ABL comes to usage especially because in many clinical settings (e.g. obstetrics), blood loss is underestimated.
The following introduces the main stages of blood loss and the associated blood pressure changes:
Stage Blood loss Blood pressure changes Compensation BL 500 – 1,000 mL No change in BP Mild BL 1,000 – 1,5000 mL Slight fall in BP (to 80-100 mmHg) Moderate BL 1,500 – 2,000 mL Marked fall in BP (to 70-80 mmHg) Severe BL More than 2,000 mL Profound fall in BP (to 50-70 mmHg)
## References
1. Hahn RG. Estimating allowable blood loss with correction for variations in blood volume. Acta Anaesthesiologica Scandinavica. 1989; 33(6), 508–512.
2. Iijima T, Brandstrup B, Rodhe P, Andrijauskas A, Svensencor CH. The maintenance and monitoring of perioperative blood volume. Perioper Med (Lond). 2013; 2(9).
3. Seigne R, Puri GD, Niraj G, Arun D, Chakravarty V, Aveek J, Chari P. Intraoperative blood transfusions. Oxford Journals, Medicine & Health BJA. 2004; 92(4).
Specialty: Hematology
System: Cardiovascular
Objective: Determination
Type: Calculator
No. Of Variables: 5
Abbreviation: ABL
Article By: Denise Nedea
Published On: May 16, 2017
Last Checked: May 16, 2017
Next Review: May 16, 2023
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Python: Create 12 fixed dates from a specified date over a given period - w3resource
# Python: Create 12 fixed dates from a specified date over a given period
## Python Datetime: Exercise-27 with Solution
Write a Python program to create 12 fixed dates from a specified date over a given period. The difference between two dates is 20.
Sample Solution:
Python Code:
# Import the datetime module
import datetime
# Define a function named every_20_days that takes a date object (date) as input
def every_20_days(date):
# Print the starting date
print('Starting Date: {d}'.format(d=date))
# Print a message indicating the next 12 days
print("Next 12 days :")
# Execute a loop that repeats 12 times
for _ in range(12):
# Add 20 days to the current date
date = date + datetime.timedelta(days=20)
# Print the resulting date
print('{d}'.format(d=date))
# Define a date object representing August 1, 2016
dt = datetime.date(2016, 8, 1)
# Call the every_20_days function with the specified date object
every_20_days(dt)
Output:
Starting Date: 2016-08-01
Next 12 days :
2016-08-21
2016-09-10
2016-09-30
2016-10-20
2016-11-09
2016-11-29
2016-12-19
2017-01-08
2017-01-28
2017-02-17
2017-03-09
2017-03-29
Explanation:
In the exercise above,
• The code imports the "datetime" module.
• Define the function:
• It defines a function named "every_20_days()" that takes a date object (date) as input.
• Within this function,
• It prints the starting date passed to the function.
• It prints a message indicating the next 12 days.
• It executes a loop that repeats 12 times.
• Inside the loop, it adds 20 days to the current date (date) using the "timedelta()" function from the "datetime" module.
• It prints the resulting date.
• Initializing the date object:
• It initializes a date object 'dt' representing August 1, 2016.
• Call the function:
• It calls the "every_20_days()" function with the specified date object 'dt' as an argument.
Flowchart:
Python Code Editor:
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Total:
\$0.00
Subjects
Resource Types
Product Rating
4.0
File Type
PDF (Acrobat) Document File
7.44 MB | 9 pages
### PRODUCT DESCRIPTION
FRACTIONS Add and Subtract Unlike Denominators (2 Players)
PRINT and GO with black and white
-or-
MULTIPLE USE with color
Board 1: Skill Level: Addition Fraction-ator
Students add the fractions and justify the answer before marking their X or O. NO SIMPLIFYING.
Board 2: Skill Level: Addition and Subtraction Fraction-ator
Addition and subtraction are mixed throughout the boards. ALL ANSWERS NEED SIMPLIFYING.
Includes 2 full color boards, 2 BW boards, answer page for student self-monitoring, teacher prep instructions, optional Wacky Bird tokens, optional color and bw white boards with rectangles and number lines for visual models and room for numerical calculations.
Teacher Prep: Provide game board and 9 counters. Students will use scrap paper (or dry erase boards) to justify the answers.
While these games are fully functional with the minimal prep instructions, additional OPTIONAL materials are also provided: printable game tokens and printable white board for justifying work (place it in a sheet protector and use with dry erase markers).
Other Fraction Activities
Greatest Common Factors
Equivalent Fractions: Missing Numerator
Equivalent Fractions
Least Common Multiples
Least Common Denominator
FRACTION BUNDLE
HERE!
Find all other math activities 3-5
HERE!
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-or- Contact me at PortigoPublications@gmail.com
CCSS.MATH.CONTENT.3.NF.A.3
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
CCSS.MATH.CONTENT.4.NF.A.2
Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2.
CCSS.MATH.CONTENT.5.NF.A.1
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)
Total Pages
9
N/A
Teaching Duration
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(691 Followers)
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# Centripetal and Centrifugal Acceleration.
• ZxcvbnM2000
In summary, the conversation discusses the concept of accelerations in circular motion, including linear acceleration and centripetal acceleration. The concept of centrifugal force is also brought up, which is a fictitious force that only exists in rotating reference frames. This force is necessary to make Newton's laws applicable in non-inertial frames. It is not a real force and is only felt by someone at the center of rotation. The centrifugal force is actually the resistance of the body to changes in velocity.
ZxcvbnM2000
I am a bit confused about these accelerations.
Anything that rotates has a linear acceleration which is a vector in the same direction as linear velocity . There is also a centripetal acceleration which "pulls" wants to pull the objects towards the axis of rotation. So the sum of these two vectors will be another total acceleration which will be equal to the square root of the sum of a linear and a angular . Therefore, if there is a body of mass m rotating about an axis,it would have a total force of m*(a total) which would always be perpendicular to the distance from the centre of rotation and thus explaining the circular motion.So if we are on a frictionless turntable rotating at a speed ω the centripetal acceleration will have no means (friction or tension )to keep us towards the centre and as a result we would follow the path of the tangential velocity ( V linear ).
What i cannot understand is what is a centrifugal force and how it is created and by whom .
Does it always exist in a circular motion just like the centripetal acceleration ?
My intuition whether it's true or not says that the centrifugal acceleration must always be smaller than the centripetal acceleration otherwise the resultant force would be directed outwards and instead of following the direction of the linear velocity we would "leave" at angle φ to it .
What about Newton's laws ? i thought they were always valid !
ZxcvbnM2000 said:
I am a bit confused about these accelerations.
Anything that rotates has a linear acceleration which is a vector in the same direction as linear velocity . There is also a centripetal acceleration which "pulls" wants to pull the objects towards the axis of rotation. So the sum of these two vectors will be another total acceleration which will be equal to the square root of the sum of a linear and a angular . Therefore, if there is a body of mass m rotating about an axis,it would have a total force of m*(a total) which would always be perpendicular to the distance from the centre of rotation and thus explaining the circular motion.So if we are on a frictionless turntable rotating at a speed ω the centripetal acceleration will have no means (friction or tension )to keep us towards the centre and as a result we would follow the path of the tangential velocity ( V linear ).
If some mass is moving in a circle, it must have a radial (centripetal) component of acceleration. If it's also changing speed as it moves, it will have a tangential component of acceleration as well.
What i cannot understand is what is a centrifugal force and how it is created and by whom .
Does it always exist in a circular motion just like the centripetal acceleration ?
Centrifugal force is a 'fictitious' force (as opposed to a 'real' interaction force) that only appears when viewing things from a rotating reference frame. (Such 'forces' are also called pseudoforces or inertial forces.) As long as you stay with an inertial frame of reference, you won't need any such 'force'.
What about Newton's laws ? i thought they were always valid !
Newton's laws are valid in a non-accelerating, inertial frame. In order to apply them to accelerating frames, you must add various 'fictitious' forces to make things work out.
ZxcvbnM2000 said:
What i cannot understand is what is a centrifugal force and how it is created and by whom .
Does it always exist in a circular motion just like the centripetal acceleration ?
Centrifugal force is something that exists only in the eye of the beholder.
Think of a hammer throw.
From the perspective of someone watching the event, the hammer thrower and the hammer are rotating about one another. There's a tension on the line that is responsible for the circular motion of the hammer. This force is directed inward; it's a centripetal force. There is no centrifugal force from this point of view.
Now think of what the hammer thrower sees. From his perspective, the hammer is not undergoing circular motion. It is instead just hanging there, motionless, right there in front of him. That tension on the line still exists. So how to explain that the hammer isn't moving? Simple: Invent a fictitious force that counters the centripetal force.
Note that there are no centrifugal forces in inertial frames of reference. They are a device that makes Newton's first two laws applicable in non-inertial frames.
Thank you very much !
"Note that there are no centrifugal forces in inertial frames of reference. They are a device that makes Newton's first two laws applicable in non-inertial frames." What do you mean by that ? ( English is not my first language).
ZxcvbnM2000 said:
Thank you very much !
"Note that there are no centrifugal forces in inertial frames of reference. They are a device that makes Newton's first two laws applicable in non-inertial frames." What do you mean by that ? ( English is not my first language).
An inertial frame is a frame of reference which is not itself undergoing acceleration.
Two inertial frames may be moving relative to each other, but only with constant relative velocity. Physics experiments conducted in two such frames should arrive at the same conclusions.
Accelerating reference frames (including rotating ones) can lead to curious results. E.g. Foucault's pendulum - if you thought the Earth was not rotating you'd have trouble explaining the pendulum's precession.
Hi ZxcvbnM2000!
ZxcvbnM2000 said:
So the centrifugal force is a fictitious force that is felt only when you are in the centre of rotation and it only exists so that the Newton's Laws of motion can be valid.So an outside observer you not experience this force.
What we think is a centrifugal force if we are at the centre of rotation is actually the resistance of the body to changes in velocity , so what i am saying is that in reality what we call centrifugal force opposite and equal to the centripetal force is really a tangential force equal to m*alinear , right ?
you usually aren't at the centre
the classic case is you're in a car moving in a circle, clockwise say
in the frame of the car, there is a centrifugal force to the left …
if your seat was frictionless, you would slide left …
if there's friction, the friction force (to the right) balances the centrifugal force
in a stationary frame, there's only friction …
the friction force (to the right) equals the centripetal acceleration times the mass
That's how a washing mashine works , the wheel spins and the water is resisting to the continuous change in motion , so if it's spinning fast enough it will travel in a straight line and get out from the holes .
So how artificial gravity works ? I watched this video
22:40 - 32:00 . And i am really confused .
"the water is resisting to the continuous change in motion" … what does that mean?
in the frame of a "car" spinning with the wheel, the water will slide sideways off the seat because of the centrifugal force
(and there's no video attached )
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-5/ :p
"the water is resisting to the continuous change in motion" … what does that mean?
If you are in a car and you are moving fast , if you hit the brakes then your body will move forward because it wants to keep going forward , it doesn't want to change .
In circular motion the mass wants to keep moving in a fixed direction but this doesn't happen because the velocity is changing the whole time so that's the " true centrifugal" force .
ZxcvbnM2000 said:
"the water is resisting to the continuous change in motion" … what does that mean?
If you are in a car and you are moving fast , if you hit the brakes then your body will move forward because it wants to keep going forward , it doesn't want to change .
oh i see …
yes, the water doesn't want to change
but that's not "resisting the continuous change in motion" (of the car or of the wheel), it's ignoring it!
In circular motion the mass wants to keep moving in a fixed direction but this doesn't happen because the velocity is changing the whole time so that's the " true centrifugal" force .
which frame are you in now?
i don't think that explanation works in either frame
ZxcvbnM2000 said:
What i cannot understand is what is a centrifugal force and how it is created and by whom .
[...]
My intuition whether it's true or not says that the centrifugal acceleration must always be smaller than the centripetal acceleration [...]
You are mentioning two different things:
- centrifugal force
- centrifugal acceleration
That's two different things.Now, let's look at some forces.
Let's say you are running along, and you use a sturdy pole to help you make a very fast U-turn. As you are about to pass the pole at arms length you extend your arm, you grab the pole, and you swing around 180 degrees.
In that situation, you provide the required centripetal force yourself. The pole is rigid in the ground, the pole will not budge. So the exerted force acts as a centripetal force, rotating you 180 degrees around the pole. Or, for that matter, you may just hold on, and keep circling the pole.
Is there a centrifugal force involved? Well, clearly you are exerting a force in centrifugal direction upon the pole. But since that pole is anchored to the entire Earth there is no way you're going to move it.
Here's the thing: the centripetal force, and the force-in-centrifugal-direction, are equal in strength. That's a physics principle. (Codified as Newton's third law of motion.)
The difference is that you are unattached, and the pole is anchored to the entire Earth. So there is centripetal acceleration in proportion to the centripetal force, but there is in the example I give no centrifugal acceleration.
One remark, the pole may bend a little, and you can attribute that bending to a force exerted upon the pole, a force in centrifugal direction
But now another example:
You team up with somebody, you hold hands and then you start circling each other, moving as fast as the two of you can, so that the two of you are both circling the common center of mass. That fast circling requires a lot of centripetal force; you have to maintain a strong grip. You are exerting a centripetal force upon your teammate, and your teammate is exerting a centripetal force upon you.
In that case there is no point in attributing any of the motion to some "centrifugal force". A third example, now just linear accceleration.
You are standing upright, in a bus that is accelerating. You hold on to a pole in that bus. Since you grip the pole firmly the pole is exerting a force upon you, accelerating you. You are exerting the same force upon that pole, but since that pole is secured to that bus-as-a-whole, and the big tires of that bus grip the road, you are not going to move that pole; that pole is moving you.
So it's always a case of a force pair, (as codified in the form of Newton's third law). The two forces of the force pair are equal in strength, as a matter of principle. It's just that the object that is not attached to anything else is the one that is accelerated.
Last edited:
Cleonis said:
Is there a centrifugal force involved? Well, clearly you are exerting a force in centrifugal direction upon the pole. But since that pole is anchored to the entire Earth there is no way you're going to move it.
Here's the thing: the centripetal force, and the force-in-centrifugal-direction, are equal in strength. That's a physics principle. (Codified as Newton's third law of motion.)
Please do not confuse your 'force-in-centrifugal-direction' with the standard centrifugal force that we've been talking about in this thread. What you are illustrating can be called the reactive centrifugal force (if you insist on giving it a name--I wouldn't), but that is a real interaction force that obeys Newton's 3rd law. On the other hand, the standard centrifugal force under discussion is a fictitious force that does not obey Newton's 3rd law.
thank you very much :) Much appreciated !
ZxcvbnM2000 said:
[...] What i cannot understand is what is a centrifugal force and how it is created and by whom. [...]
Doc Al said:
[...] the standard centrifugal force that we've been talking about in this thread. [...]
Indeed as the thread developed what the replies were about was the fictitious centrifugal force.
But the original question was quite un-specific, it referred to 'a centrifugal force'. The original question implied a notion of a force that is actually exerted. So I chose to concentrate on a force that is actually exerted, referring to it generally as 'a force-in-centrifugal direction'.
## What is centripetal acceleration?
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is caused by the centripetal force, which is necessary to keep the object moving in a circular path.
## What is centrifugal acceleration?
Centrifugal acceleration is a perceived acceleration that appears to push an object away from the center of a circular path. However, it is not a real force and is actually the result of the inertia of the object trying to move in a straight line.
## What is the difference between centripetal and centrifugal acceleration?
The main difference between centripetal and centrifugal acceleration is that centripetal acceleration is a real force acting on an object, while centrifugal acceleration is a perceived force due to the inertia of the object.
## How are centripetal and centrifugal acceleration related?
Centripetal and centrifugal acceleration are related by Newton's third law of motion. According to this law, for every action, there is an equal and opposite reaction. This means that the centripetal force causing the centripetal acceleration also produces an equal and opposite centrifugal force.
## What are some examples of centripetal and centrifugal acceleration in everyday life?
Some examples of centripetal acceleration are the motion of a satellite orbiting the Earth, a car rounding a curve, and a rotating amusement park ride. Examples of centrifugal acceleration include the feeling of being pushed outwards when making a turn in a car or riding a rollercoaster, and the formation of a whirlpool in a body of water.
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# What is the difference between linear regression and logistic regression?
What is the difference between linear regression and logistic regression?
When would you use each?
• In the linear regression model the dependent variable $y$ is considered continuous, whereas in logistic regression it is categorical, i.e., discrete. In application, the former is used in regression settings while the latter is used for binary classification or multi-class classification (where it is called multinomial logistic regression). May 28, 2012 at 18:33
• Although written in a different context, it may help you to read my answer here: Difference between logit and probit models, which contains a lot of information about what's happening in logistic regression that may help you understand these better. Jul 25, 2013 at 21:53
• All the previous answers are right, but there are reasons you might favor a linear regression model even when your outcome is a dichotomy. I've written about these reasons here: statisticalhorizons.com/linear-vs-logistic Jul 7, 2015 at 19:08
Linear regression uses the general linear equation $Y=b_0+∑(b_i X_i)+\epsilon$ where $Y$ is a continuous dependent variable and independent variables $X_i$ are usually continuous (but can also be binary, e.g. when the linear model is used in a t-test) or other discrete domains. $\epsilon$ is a term for the variance that is not explained by the model and is usually just called "error". Individual dependent values denoted by $Y_j$ can be solved by modifying the equation a little: $Y_j=b_0 + \sum{(b_i X_{ij})+\epsilon_j}$
Logistic regression is another generalized linear model (GLM) procedure using the same basic formula, but instead of the continuous $Y$, it is regressing for the probability of a categorical outcome. In simplest form, this means that we're considering just one outcome variable and two states of that variable- either 0 or 1.
The equation for the probability of $Y=1$ looks like this: $$P(Y=1) = {1 \over 1+e^{-(b_0+\sum{(b_iX_i)})}}$$
Your independent variables $X_i$ can be continuous or binary. The regression coefficients $b_i$ can be exponentiated to give you the change in odds of $Y$ per change in $X_i$, i.e., $Odds={P(Y=1) \over P(Y=0)}={P(Y=1) \over 1-P(Y=1)}$ and ${\Delta Odds}= e^{b_i}$. $\Delta Odds$ is called the odds ratio, $Odds(X_i+1)\over Odds(X_i)$. In English, you can say that the odds of $Y=1$ increase by a factor of $e^{b_i}$ per unit change in $X_i$.
Example: If you wanted to see how body mass index predicts blood cholesterol (a continuous measure), you'd use linear regression as described at the top of my answer. If you wanted to see how BMI predicts the odds of being a diabetic (a binary diagnosis), you'd use logistic regression.
• This looks like a fine answer, but could you explain what the $\epsilon_i$ stand for and--in particular--why you include them within the summations? (What is being summed over, anyway?)
– whuber
May 28, 2012 at 18:55
• It looks to me Bill that he meant to write i.e. (Latin abbreviation for that is) rather than e. i. May 28, 2012 at 20:02
• But the εi in the summation of the exponent shouldn't be there. It looks like the noise term in the model was accidentally carried there. The only summing should be over the bis that represent the p coefficients for the p covariates. May 28, 2012 at 20:09
• There's an error in your expression for $P(Y=1)$. You should have $$P(Y=1) = \frac{1}{1 + \exp \{-X \boldsymbol{\beta} \} },$$ not $$P(Y=1) = \frac{1}{1 + \exp \{ -(X \boldsymbol{\beta}+\varepsilon) \} }$$ The randomness in a logistic regression model comes from the fact that these are bernoulli trials, not from there being errors in the success probabilities (which is how you're written it). May 28, 2012 at 20:51
• @samthebrand logistic regression is not binary per se. It can be used to model data with a binary response via probabilities that range between 0 and 1. Going to shamelessly plug my blog post on this which should clear your confusion.
– Ben
Aug 24, 2015 at 3:21
Linear Regression is used to establish a relationship between Dependent and Independent variables, which is useful in estimating the resultant dependent variable in case independent variable change. For example:
Using a Linear Regression, the relationship between Rain (R) and Umbrella Sales (U) is found to be - U = 2R + 5000
This equation says that for every 1mm of Rain, there is a demand for 5002 umbrellas. So, using Simple Regression, you can estimate the value of your variable.
Logistic Regression on the other hand is used to ascertain the probability of an event. And this event is captured in binary format, i.e. 0 or 1.
Example - I want to ascertain if a customer will buy my product or not. For this, I would run a Logistic Regression on the (relevant) data and my dependent variable would be a binary variable (1=Yes; 0=No).
In terms of graphical representation, Linear Regression gives a linear line as an output, once the values are plotted on the graph. Whereas, the logistic regression gives an S-shaped line
Reference from Mohit Khurana.
• Re: "Linear Regression is used to establish a relationship between Dependent and Indipendent variables" - this is also true about logistic regression - it's just that the dependent variable is binary. May 29, 2012 at 0:46
• Logistic Regression isn't only for predicting a binary event ($2$ classes). It can be generalized to $k$ classes (multinomial logistic regression) Sep 16, 2016 at 7:05
The differences have been settled by DocBuckets and Pardis, but I want to add one way to compare their performance not mentioned.
Linear regression is usually solved by minimizing the least squares error of the model to the data, therefore large errors are penalized quadratically. Logistic regression is just the opposite. Using the logistic loss function causes large errors to be penalized to an asymptotically constant.
Consider linear regression on a categorical {0,1} outcomes to see why this is a problem. If your model predicts the outcome is 38 when truth is 1, you've lost nothing. Linear regression would try to reduce that 38, logistic wouldn't (as much).
• Wre then, the situations/cases that are penalized in a logistic, i.e., in what cases would we have a poor fit?
– MSIS
Aug 25, 2016 at 14:30
• Just the opposite: whenever larger deviations from the fit actually do incur worse results. For instance, logistic regression is good at keeping you at hitting a dart board, but can't make a bullseye look nice. Or, similarly, thinks that a near miss of the board is the same as sticking your neighbor. Aug 25, 2016 at 21:54
• Great answer. Was there any research done how much does it hurt model's performance? I mean if a linear regression was used to predict response={0,1} instead of a logistic regression. Jan 25, 2018 at 22:56
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# Heat capacity of a solid using the Einstein model
by Antti
Tags: capacity, einstein, heat, model, solid
P: 25 1. The problem statement, all variables and given/known data Plot the heat capacity as a function of temperature for aluminium, using the Einstein model. 2. Relevant equations The Einstein model for heat capacity: 3. The attempt at a solution The model assumes that the frequencies of all the harmonic oscillators in the material are the same. But does it mean "same for all temperatures" or "same for a given temperature"? And if the latter is true, will I need to find a function $$\epsilon (T)$$? Because otherwise the right hand side of the Cv equation wil have two free variables. Thankful for help, my textbook is very unclear about this.
HW Helper PF Gold P: 3,440 In this model, the frequency of each oscillator is the same at all temperatures, but the energy varies with temperature. Define $$\frac{1}{x}=\frac{\epsilon}{kT}$$ Plot C(1/x) / 3Nk vs. x which is a dimensionless plot, good for any choice of specific values. Here, it's the shape that counts.
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Terminal Velocity In Quran
Highest speed in free-fall
Terminal Velocity
During a free-fall terminal velocity is reached a few seconds after the drop.
Based on wind resistance, for example, the terminal speed of a skydiver in a belly-to-earth (i.e., face down) free fall position is about 195 km/h (120 mph; 54 m/s). This speed is the asymptotic limiting value of the speed, and the forces acting on the body balance each other more and more closely as the terminal speed is approached. In this example, a speed of 50% of terminal speed is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.
Wikipedia, Terminal Velocity, 2019.
Skydivers have high terminal speed however there are birds that can easily overtake any skydiver.
World Atlas, The Fastest Birds In The World, 2019.
The Peregrine Falcon reaches terminal speeds of 389 km/hr and the Golden Eagle reaches terminal speeds of 240 - 320 km/hr. So these two birds can easily overtake any skydiver in a free-fall. They also turned-out to be raptors (flesh eating birds). This was only recently, however 1400 years ago this was portrayed in the Quran:
[Quran 22:31] Being true to Allah, without associating anything with Him. Whoever associates anything with Allah it is as though he has fallen from the sky, and was snatched by the birds, or was taken down by the wind to a deep place.
"Was snatched by the birds" these birds are attacking humans means they are flesh eating raptors. But to catch humans in mid-air they have to dive faster than the skydiver. Today we know that there are two raptors that can easily overtake any skydiver.
How could an illiterate man who lived 1400 years ago have known that some raptors are faster than skydivers?
Miracles Of Quran
The Quran (Koran, the book of Islam) contains scientific knowledge that could not have been known 1400 years ago. It ranges from basic arithmetics to the most advanced topics in astrophysics. You are invited to go through those miracles and judge for yourself.
STUDY OF THE EARTH
Geology
Coal, Landslide, Weathering - Erosion, Porous Rocks, Mass Extinction, Steam Explosions, Mountains, Internal Waves, Earth, Pumice, Photic Zone, Tectonics, Dead Sea, Sinkhole, Volcano, Shorelines, Minerals, Hydrothermal Vents, Freshwater, Fire Whirl, Rocks Crack, Subduction, Fault Lines, Earthquake, Soil Expansion, Flash Flood, Desertification.
SCIENCE OF LIVING ORGANISMS
Biology
Fetal Development, Honey, Cholesterol, Meiosis, Antioxidants, Fats, Hypoxia, Miscarriage, Gardens, Photosynthesis, White Hair, Pollination, Lightning Strike, Inner Ear, Cataracts, Vision, Bones, Milk, Brain Functions, Evolution, Honey Bees, Human Senses, Frost, Gender, Eye Pupil, Womb, Fingerprints, Skin Nerves, Fasting, Breastfeeding, Eardrum, Fetus, Fluorescence, Human Embryo, Bedsores, Perspiration, Bacteria, Plant Stress.
SCIENCE OF Quantity, SPACE, AND CHANGE
Mathematics
Distance, Prime Numbers, Pi, Arithmetics, Relational Algebra, Base-19.
study of the past
History
Paper Money, Hieroglyphs, Digital Books, Calendar, Moses, Pharaoh's Mummy, Haman, Karnak Temple, Pharaoh, Flight, Noah, Dead Sea, Pompeii, North, Petra, Ubar.
ANIMAL SCIENCE
Zoology
Raptors, Spider Web, Ants, Honey Bees, Colonies, Animal Languages, Mosquito, Crow, Nocturnal Animals, Housefly.
STUDY OF THE ATMOSPHERE
Meteorology
Sea Breeze, Orographic Effect, Wind, Volcanic Gases, Mass Extinction, Microburst, Cloud Seeding Dew, Shorelines, Fire Whirl, Freshwater, Atmospheric Pressure, Flash Flood, Weight Of Clouds.
SCIENCE OF CELESTIAL OBJECTS
Astronomy
Sunlight, Full Moon, Meteorites, Exoplanets, Planetary Orbits, Magnetosphere, Moonlight, Day, Multi-Star System, Iron, Starlight.
SCIENCE OF ELEMENTS AND COMPOUNDS
Chemistry
Viscosity, Superionic Water, Photosynthesis, Volcanic Gases, Mass Extinction, Bones, Steam Explosions, Internal Waves, Pyramids, Hydrogen, Water Stratification, Cloud Seeding, Fluorescence, Pompeii, Rust.
SCIENCE OF MOTION AND BEHAVIOR THROUGH SPACE AND TIME
Physics
Work, Light, Atoms, Pairs, Speed Of Light, String Theory, Rayleigh Scattering, Terminal Velocity, Time Relative, Wormholes, Pulsars, Gravity, Spacetime, Armor Peircing, Gravitational Waves, Sonic Weapons, Mass, String Theory, Equivalence Principle, Red Shifting.
SCIENCE OF THE ORIGIN AND EVOLUTION OF THE UNIVERSE
Cosmology
Shape Of Universe, Age Of Universe, Seven Heavens, Big Bang, Expanding Universe, Sound Waves, Isotropy, Primordial Smoke, Dark Energy, Galaxy Filaments. Dark Energy
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# Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$ - not able to evaluate residue
Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$,
Here is my attempt:
Replace cos (ax) with $e^{iaz}$, then take the real part such that:
$\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx = R.P[\int_{c}^{} \frac{e^{iaz}}{z^2+z+1} dz]$ where c is the upper half circle.
$\int_{c}^{} \frac{e^{iaz}}{z^2+z+1} dz = 2\pi i Res[\frac{e^{iaz}}{z^2+z+1}, -1+\frac{i\sqrt{3}}{2}]$ (considering only the poles inside my contour )..
then I got stuck , how can I follow from here?
• What is your definition of residues? – yngabl Dec 4 '16 at 7:33
In this case, you can take the integral to be
$$\operatorname{Re}{\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^2+x+1}}$$
Assume $a \gt 0$ for now. The roots of the denominator are at $x_{\pm} = -\frac12 \pm i \frac{\sqrt{3}}{2} = e^{\pm i 2 \pi/3}$.
Because we assumed $a \gt 0$, we consider the contour integral
$$\oint_C dz \frac{e^{i a z}}{z^2+z+1}$$
where $C$ is a semicircle in the upper half plane. By the residue theorem, we need only consider the pole at $x_+$. As the radius of the semicircle goes to infinity, we find that
$$\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^2+x+1} = i 2 \pi \frac{e^{-\sqrt{3} a/2} e^{-i a/2}}{i \sqrt{3}}$$
Thus, when $a \gt 0$,
$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{x^2+x+1} = \frac{2 \pi}{\sqrt{3}} e^{-\sqrt{3} a/2} \cos{\left ( \frac{a}{2} \right )}$$
When $a \lt 0$, we must use a contour in the lower-half plane. In this case, we use the residue at the pole at $x_-$. Thus, for all values of $a$:
$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{x^2+x+1} = \frac{2 \pi}{\sqrt{3}} e^{-\sqrt{3} |a|/2} \cos{\left ( \frac{a}{2} \right )}$$
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Related Articles
# Length of longest straight path from a given Binary Tree
• Difficulty Level : Medium
• Last Updated : 21 Jun, 2021
Given a Binary Tree, the task is to find the length of the longest straight path of the given binary tree.
Straight Path is defined as the path that starts from any node and ends at another node in the tree such that the direction of traversal from the source node to the destination node always remains the same i.e., either left or right, without any change in direction that is left->left ->left or right->right->right direction.
Examples:
Input:
Output: 2
Explanation:
The path shown in green is the longest straight path from 4 to 6 which is of length 2.
Input:
Output: 3
Explanation:
The path shown in green is the longest straight path from 5 to 0 which is of length 3.
Approach: The idea is to use the Postorder traversal. Follow the steps below to solve the problem:
1. For each node, check the direction of the current Node (either left or right) and check which direction of its child is providing the longest length below it to that node.
2. If the direction of the current node and the child giving the longest length is not the same then save the result of that child and pass the length of the other child to its parent.
3. Using the above steps find the longest straight path at each node and save the result to print the maximum value among all the straight paths.
4. Print the maximum path after the above steps.
Below is the implementation of the above approach:
Code block
Output:
`2`
Time Complexity: O(N)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
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# Physics Simulation Forum
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Post subject: Giant static forestPosted: Sun Apr 01, 2012 4:00 pm
Joined: Sun Jan 29, 2012 10:01 pm
Posts: 49
I have a giant forest with trees to collide with but just adding them as static rigid bodies gives poor performance because they are in the same broad phase as dynamic bodies.
Can I modify the height field class to return cones and other shapes from each tile in a low resolution grid?
If not, I could let the modified height field point to triangle shapes to transform per vertex but taking any shape would be more reusable.
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Post subject: Re: Giant static forestPosted: Wed Apr 04, 2012 12:38 pm
Joined: Fri Dec 10, 2010 3:39 am
Posts: 15
wait, so are they all trimeshes right now?
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Post subject: Re: Giant static forestPosted: Wed Apr 04, 2012 1:17 pm
Joined: Sun Jan 29, 2012 10:01 pm
Posts: 49
They are cone shapes with hand written dimensions right now. I have a list of triangles in the graphical model that I can read directly inside a collision shape but a cone shape would avoid multiplying each vertex with a transformation matrix each time they are used for collision.
My idea is to look up all shapes in the cells that overlap the query AABB and return all shapes who's AABB intersect with the dynamic body's AABB. The Query AABB is the dynamic body's world space AABB + tolerance for how much the static bodies are outside of their cells in the grid.
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Post subject: Re: Giant static forestPosted: Sun Apr 22, 2012 11:48 am
Joined: Sun Jan 29, 2012 10:01 pm
Posts: 49
Now when I have my debug drawing working, I see that the reason for the bad performance is invalid calculation of world space bounding boxes. Every tree's bounding box is stretched to the center of the world so that the tree for static bodies do nothing for the speed. I don't use compound shapes for the cones and I place them at their initial locations before adding them to the world.
Edit:
Calling setInterpolationWorldTransform after calling setWorldTransform with the same transformation when placing static bodies solved the bug.
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Activity02 Soln
# Activity02 Soln - Activity#02(SOLUTIONS Last Name First...
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ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2001 Page 1 of 6 Activity #02 (SOLUTIONS) Last Name First Name Student ID Number 1.1 1.2 Total Grader Initials 50 points 50 points 100 points Activity 1.1 (First hour) A. (35 points) Consider a simple CPU with just the instructions listed below Instructio n Mnemoni c Operation Number of Cycles Number of Bytes LDAA Load Accumulator from Memory 4 3 STAA Store Accumulator to Memory 4 3 ADDA Add Memory to Accumulator 4 3 BEQ Branch on Equal to 0 3 2 CMPA Compare accumulator to Memory 4 3 NEGA Negate Accumulator 2 1 BRA Branch always 3 2 MUL Multiply Accumulator 10 1 SUBA Subtract Memory from Accumulator 4 3 ANDA AND Accumulator with Memory 4 3 ORAA OR accumulator A with Memory 4 3 HALT Halt 1 1 NOP No Operation 1 1 a. (5 points) What is the average number of cycles per instruction (CPI) for this CPU, assuming that all instructions are executed equally often? Answer: n instructio cycles ns instructio cycles / 69 . 3 13 48 = Grading Policy: 5 pts correct answer, 2 pts partially correct answer, 0 pts for not trying.
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ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2001 Page 2 of 6 b. (5 points) If the CPU cycle time is 20ns, what is the average number of instructions executed per second, assuming that all instructions are executed equally often? Answer: Remember that 1ns = 10 -9 sec. It takes 3.69 × 20 × 10 -9 secs per instruction By taking the reciprocal, this translates to 13.55 × 10 6 instructions/sec, which is 13.55 MIPS . Grading Policy: 5 pts correct answer, 2 pts partially correct answer, 0 pts for not trying. c. (5 points) What would the highest "peak MIPS" that could be (misleadingly) claimed, and what operation would it use? Answer: The most favorable assumption is that the machine executes only 1-cycle instructions such as NOPs all the time. This implies that it takes 20ns per instruction, which translates to 50 MIPS . Grading Policy: 5 pts correct answer, 2 pts partially correct answer, 0 pts for not trying.
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## The Definite Integral
A definite integral quite simply returns the area under the graph of a function - the area between the graph of a function and the x - axis - between two x values.
The diagram shows the graph of
$y=4-x^2$
.
The curve intersects the x - axis at
$x=-2$
and
$x=2$
.
The area between the curve and the x -axis between the x values -2 and 2 is
$\int^2_{-2} x^2 dx = [\frac{x^3}{3}]^2_{-2} = \frac{2^3}{3}- \frac{(-2)^3}{3}=\frac{8}{3}- (- \frac{8}{3})=\frac{16}{3}$
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##### 7.2.1.5. Generating all set partitions
Now let’s shift gears and concentrate on a rather different kind of partition. The partitions of a set are the ways to regard that set as a union of nonempty, disjoint subsets called blocks. For example, we listed the five essentially different partitions of {1, 2, 3} at the beginning of the previous section, in 7.2.1.4–(2) and 7.2.1.4–(4). Those five partitions can also be written more compactly in the form
using a vertical line to separate one block from another. In this list the elements of each block could have been written in any order, and so could the blocks themselves, because ‘13|2’ and ‘31|2’ and ...
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Peterson-Codazzi equations
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Equations constituting together with Gauss' equation (see Gauss theorem) necessary and sufficient conditions for integrability of the system of partial differential equations to which the problem of recovering a surface from its first and second fundamental forms reduces (see Bonnet theorem). The Peterson–Codazzi equations take the form
$$\frac{\partial b_{i1}}{\partial u^2}+\Gamma_{i1}^1b_{12}+\Gamma_{i1}^2b_{22}=\frac{\partial b_{i2}}{\partial u^1}+\Gamma_{i2}^1b_{11}+\Gamma_{i2}^2b_{21},$$
where the $b_{ij}$ are the coefficients in the second fundamental form and the $\Gamma_{ij}^k$ are the Christoffel symbols of the second kind.
The equations were first discovered by K.M. Peterson in 1853 and were rediscovered by G. Mainardi in 1856 and D. Codazzi in 1867.
References
[1] P.K. Rashevskii, "A course of differential geometry" , Moscow (1956) (In Russian)
In Western literature these equations are usually called the Mainardi–Codazzi equations, or Gauss–Mainardi–Codazzi equations.
Generally, for hypersurfaces in Euclidean $n$-space the Gauss equations and the Mainardi–Codazzi equations are obtained by decomposing the (vanishing) curvature tensor of the ambient space into tangential and normal parts and expressing these parts in surface terms. The Mainardi–Codazzi equations have the following form in this terminology:
$$D_XL(Y)-D_YL(X)-L([X,Y])=0,$$
where $L$ is the Weingarten mapping (shape operator) of the hypersurface, $D$ the induced covariant derivation and $X,Y$ are smooth tangent vector fields.
References
[a1] W. Blaschke, K. Leichtweiss, "Elementare Differentialgeometrie" , Springer (1973) [a2] N.J. Hicks, "Notes on differential geometry" , v. Nostrand (1965) [a3] H.W. Guggenheimer, "Differential geometry" , Dover, reprint (1977)
How to Cite This Entry:
Peterson-Codazzi equations. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Peterson-Codazzi_equations&oldid=32919
This article was adapted from an original article by A.B. Ivanov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# earth
On planet Tehar, the free-fall acceleration is the same as that on Earth but there is also a strong downward electric field that is uniform close to the planet's surface. A 2.00 kg ball having a charge of 5.00 ยตC is thrown upward at a speed of 29.2 m/s and it hits the ground after an interval of 4.40 s. What is the potential difference between the starting point and the top point of the trajectory?
1. ๐ 0
2. ๐ 0
3. ๐ 31
## Similar Questions
1. ### earth
On planet Tehar, the free fall acceleration is the same as that on the Earth, but there is also a strong downward electric field that is uniform close to the planets surface. A 2.00k ball having a charge of 5.00C is thrown upward
asked by Anonymous on December 14, 2010
2. ### physics
On planet x, an object weighs 10.4N. On planet B where the magnitude of the free fall acceleration is 1.48g (where g=9.8 m/s^2 is the gravitational acceleration on Earth),the object weighs 24.6N. The acceleration of gravity is 9.8
asked by alexa on July 3, 2011
3. ### Pyhsics
A) On planet X, an object weighs 19.2 N. On planet B where the magnitude of the free-fall acceleration is 1.17 g (where g = 9.8 m/s2 is the gravitational acceleration on Earth), the object weighs 34.9 N. The acceleration of
asked by Kim on November 29, 2014
4. ### physics
Planet Z is 1.00ร10^4 km in diameter. The free-fall acceleration on Planet Z is 10.0 m/s^2. a)What is the mass of Planet Z? b)What is the free-fall acceleration 9000 km above Planet Z's north pole?
asked by kelly on December 1, 2009
5. ### college physics
A new planet has been discovered that has a mass one-tenth that of Earth and a radius that is one-sixth that of Earth. Determine the free fall acceleration on the surface of this planet. Express your answer in the appropriate mks
asked by Joe on September 24, 2012
6. ### physics
On planet X, an object weighs 18.1 N. On planet B where the magnitude of the free-fall acceleration is 1.28 g (where g = 9.8 m/s 2 is the gravitational acceleration on Earth), the object weighs 37.9 N. The acceleration of gravity
asked by zach on December 8, 2015
7. ### Physics
The free-fall acceleration at the surface of planet 1 is 18 m/s^2. The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?
asked by Alex on April 21, 2012
8. ### physics
On planet Tehar, the acceleration of gravity is the same as that on Earth but there is also a strong downward electric field with the field being uniform close to the planet's surface. A 2.16 kg ball having a charge of 4.80 uC is
asked by VJC on September 9, 2010
9. ### physics
The acceleration due to gravity on planet X is one fifth that on the surface of the earth. If it takes 4.6 s for an object to fall a certain distance from rest on earth, how long would it take to fall the same distance on planet
asked by Anonymous on September 26, 2011
10. ### physics
The acceleration due to gravity on planet X is one fifth that on the surface of the earth. If it takes 3.5 s for an object to fall a certain distance from rest on earth, how long would it take to fall the same distance on planet
asked by Anonymous on September 20, 2010
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Nulls, Three-Valued Logic, and Missing Information » GFXhome for Graphic Designers
» » Nulls, Three-Valued Logic, and Missing Information
Information of news
• Author: avelin
• Date: 22-12-2014, 03:16
22-12-2014, 03:16
### Nulls, Three-Valued Logic, and Missing Information
Category: Tutorials
Nulls, Three-Valued Logic, and Missing Information | 1.65 GB
Genre: E-Learning
The logic on which the relational model is based is a two valued logic (2VL). In its attempt to deal with the so called “missing information” problem, however, SQL—more specifically, SQL’s support for “nulls”—is based on a three valued logic (3VL) instead. This session explains in detail why any such approach is doomed to failure. More specifically, it shows why 3VL (a) doesn’t solve the problem, (b) isn’t useful, and (c) can actually be dangerous.
The presentation is divided into three modules. Module I reviews conventional two-valued logic (2VL), examines some of the most immediate differences between it and three-valued logic (3VL), and shows how 3VL inevitably gives rise to errors. Module II goes into more depth on why such errors are inevitable. It also considers the question: Which 3VL are we talking about, anyway? Module III then considers many-valued logics in general, and four-valued logic (4VL) in particular, and discusses a variety of further related points. It also briefly describes an approach to missing information based on 2VL and classical relational theory.
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# ＜LeetCode天梯＞Day029环形链表（哈希表+双指针） | 初级算法 | Python
?作者简介：大家好，我是车神哥，府学路18号的车神?
?个人主页：应无所住而生其心的博客_府学路18号车神_CSDN博客
?点赞评论收藏 == 养成习惯（一键三连）?
?本系列主要以刷LeetCode力扣）网站的各类题为标准，实现自我能力的提升为目标⚡
⚡希望大家多多支持?~一起加油 ?
# 题干
## 哈希表
``````class Solution:
def hasCycle(self, head: ListNode) -> bool:
hashmap = []
return True
return False
``````
``````class Solution:
def hasCycle(self, head: ListNode) -> bool:
hashmap = set()
return True
return False
``````
## 双指针
``````class Solution:
def hasCycle(self, head: ListNode) -> bool:
# 双指针
return False
# 设置两个指针快和慢
while fast != None and fast.next != None:
slow = slow.next # 慢走一步
fast = fast.next.next # 快走两步
# 设置判断，如果相遇则说明有环
if fast == slow:
return True
# 否则False
return False
``````
# Reference
## 加油！！！
❤坚持读Paper，坚持做笔记，坚持学习，坚持刷力扣LeetCode❤！！！
To Be No.1
⚡⚡
ღ( ´･ᴗ･` )
THE END
https://blog.csdn.net/baidu_39105563/article/details/121429734
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## Interview Question for Software Engineer / Developers
Comment hidden because of low score. Click to expand.
1
of 1 vote
How about sorting the array and making every element equal to the mid point of the array? That way, it is a trade-off for elements on either side (greater and smaller) and no element has to be incremented a large number of times ?
Also, with the average, we face the risk of the average not being an integer, and thus the deferment may not be an integer either. In which case, we would never be able to achieve that value using the ++/-- operations allowed.
Comment hidden because of low score. Click to expand.
0
If the average is not an interger, doesn't that mean we would never be able to achieve the goal?
Comment hidden because of low score. Click to expand.
0
of 0 vote
A[0] will be either -1 or 1
Check A[0] and put that value in rest of the array (the requirement is to make all elements of A equal).
Comment hidden because of low score. Click to expand.
0
I don't quite follow what you posted. There is no need to make any of A[i] to be 1 or -1. All elements can as well be made equal.
Comment hidden because of low score. Click to expand.
0
of 0 vote
If u take the average and then add / deferment all other elements by the difference from average , would that work?
Comment hidden because of low score. Click to expand.
0
of 0 vote
O(N) solution:
1. In a single scan find the min and max elements of the array.
2. If their difference is more than 2 then its impossible .
3. If less than 2 then either all the elements are to be made equal to min/max/ min+1
4. In a single scan just increase/decrease each element accordingly
Comment hidden because of low score. Click to expand.
0
of 0 vote
O(N) solution:
1. In a single scan find the min and max elements of the array.
2. If their difference is more than 2 then its impossible .
3. If less than 2 then either all the elements are to be made equal to min/max/ min+1
4. In a single scan just increase/decrease each element accordingly
Comment hidden because of low score. Click to expand.
0
of 0 vote
consider the array of two elements A[0]=7 , A[1]=8
---------------
Consider any operation of unit one inc/decrement ... they can not be equal ???
Comment hidden because of low score. Click to expand.
0
of 0 vote
I am sorry , i thought we can also keep an element unaltered. In which case my soln will work
Comment hidden because of low score. Click to expand.
0
of 0 vote
1.take the avg of all no. if it is an integer then it possible else -1.
Comment hidden because of low score. Click to expand.
0
of 0 vote
And the ans will be Max(min-avg,max-avg)
Comment hidden because of low score. Click to expand.
0
of 0 vote
Soln:
1. In a scan, find the max, min element and also see if all the elements are odd or even or there is a mixture
2. if the array contains a mixture of odd, even then return -1
3. Minimum number of operations is (Max - Min)/2.
here I am trying to bring each number close to the avg
Comment hidden because of low score. Click to expand.
1
of 1 vote
How about bringing all the numbers to median
Comment hidden because of low score. Click to expand.
0
of 0 vote
while solving this prob, I noticed that a solution is possible if and only if:
1. either all elements are odd
2. or all elements are even
otherwise no solution..
can some one validate this understanding?
Comment hidden because of low score. Click to expand.
0
thanks harleen... you helped me solve the problem
// if all are even elements or if all are odd elements, proceed, else return impossible
// avg = max + min / 2
// if element is less than avg, increment
// else, decrement
// do until max = min
``````oddCount = 0; evenCount = 0 , i=0;
while(i<A.length)
{
if(A[i]%2==0)
evenCount++;
else
oddCount++;
}
if(oddCount>0 && evenCount>0)
{
return -1;
}
max = a[i];
min = a[i];
for(int i = 1; i<A.length; i++)
{
if(a[i] > max)
max = a[i];
if(a[i] < min)
min = a[i];
}
noOperations = 0;
while(max!=min)
{
noOperations++;
avg = (max + min) / 2;
for(i=0;i<A.length;i++)
{
if(a[i]<avg)
a[i]++;
else
a[i]--;
}
max--;
if(avg > min)
min++;
else
min--;
}``````
Name:
Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
### Books
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Finally learned how to make graphs in R & Adobe Illustrator and decided to try out a few data visualization techniques using some of the most common statistical distributions. DISCLAIMER - While I hope it's informative, this is by no means a deep dive, as only so much can be captured in 20 seconds. But I do hope this helps give an idea what these are.
Probability distributions or densities represent the relative likelihood that a certain experiment/event would be equal to a certain occurrence. What is the probability of getting a 6 in a dice roll? What is the probability of winning in a game of baccarat? What is the probability of getting a positive return from a certain stock? Depending on the underlying behavior of these events, different distributions are used to analyze and understand these events. The animations below simply aim to give a quick understanding on four of the more common distributions and their characteristics.
- - - - - - - - - - - - - -
THE NORMAL / GAUSSIAN DISTRIBUTION
Parameters: Mean, Standard Deviation
A continuous probability density distribution where observations tend towards a central value. Mean determines the position of the center of the graph, while Standard Deviation determines the spread, or how far from the mean values fall. Other features are that it is bell-shaped and symmetric.
- - - - - - - - - - - - - -
THE UNIFORM DISTRIBUTION
Parameters: Max, Min
A probability distribution where the individual values (if discrete) or intervals (if continuous) have the same probability (ie. rolling any number from a 6 sided dice, drawing a number or suit from a deck of cards). It is symmetric and the probabilities are determined by the maximum and minimum possible values.
- - - - - - - - - - - - - -
THE STUDENT'S T DISTRIBUTION
Parameters: v (degrees of freedom)
Similar to the Normal Distribution, the T Distribution is bell-shaped and symmetric as well. However, the tails of the distribution are heavier, which means that the values tend to fall farther away from the mean than a Normal Distribution. Tail heaviness is determined by the degrees of freedom of the distribution, with the T tending towards the normal curve as degrees of freedom become larger. Used in risk analysis as extreme values would give more information than the center.
- - - - - - - - - - - - - -
THE BINOMIAL DISTRIBUTION
Parameters: n (trials), p (probability of success)
Given a certain experiment with two possible outcomes (say, flipping a coin), the probability of one outcome successfully happening can be defined by p, and the other outcome is 1-p. This is called a Bernoulli Trial. A binomial distribution is a discrete probability distribution that gives the probability of experiencing k successes in n outcomes (i.e. in 10 coin flips, what is the probability that you can get 5 heads? -- 24.6%, assuming a fair coin.)
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Home > Relative Error > Relative Error Algebra 1
# Relative Error Algebra 1
## Contents
Generated Wed, 26 Oct 2016 23:17:10 GMT by s_wx1126 (squid/3.5.20) Measuring to the nearest meter means the true value could be up to half a meter smaller or larger. Back to Top To calculate the relative error use the following way:Observe the true value (x) and approximate measured value (xo). you didn't measure it wrong ... Check This Out
how many serving will I get? Howcast 4.953 προβολές 1:47 Physics 002 Ch01 Pt02 Relative error and Absolute error - Διάρκεια: 9:49. Quick Tips Related ArticlesHow to Compare and Order FractionsHow to Find the Area of a Square Using the Length of its DiagonalHow to Calculate PercentagesHow to Calculate Slope and Intercepts of Many scientific tools, like precision droppers and measurement equipment, often has absolute error labeled on the sides as "+/- ____ " 3 Always add the appropriate units. Get More Info
## How To Calculate Relative Error
So you know that your measurement is accurate to within + or - 1 mm; your absolute error is 1 mm. Repeat the same measure several times to get a good average value. 4. Accuracy is a measure of how close the result of the measurement comes to the "true", "actual", or "accepted" value. (How close is your answer to the accepted value?) Tolerance is Machines used in manufacturing often set tolerance intervals, or ranges in which product measurements will be tolerated or accepted before they are considered flawed.
In this case to measure the errors we use these formulas. Quick and Fast Learn 1.037 προβολές 5:24 Error and Percent Error - Διάρκεια: 7:15. This is the experimental value. Relative Error Definition Video Tips Make sure that your experimental value and real value are all expressed in the same unit of measurement.
The three measurements are: 24 ±1 cm 24 ±1 cm 20 ±1 cm Volume is width × length × height: V = w × l × h The smallest possible Volume FAST!? CNAQMath 3.090 προβολές 3:05 Be Human Calculator - Διάρκεια: 3:48. Bonuses Learn more about Algebra Sources: mathworld.wolfram.com azformula.com Related Questions Q: What is the tangent formula?
An expected value is usually found on tests and school labs. Absolute Error Example Full Answer > Filed Under: Algebra Q: What is the formula for calculating the slope of a line? That is the "real" value. b.) the relative error in the measured length of the field.
## Relative Error Formula
TEDx Talks 4.092.009 προβολές 9:14 Differentials - Error analysis - Διάρκεια: 3:05. Find the absolute error, relative error and percent of error of the approximation 3.14 to the value , using the TI-83+/84+ entry of pi as the actual value. How To Calculate Relative Error I'm taking it too in a couple of hours... Absolute Error Formula Two-Point-Four 33.187 προβολές 2:12 The surprising beauty of mathematics | Jonathan Matte | TEDxGreensFarmsAcademy - Διάρκεια: 9:14.
Due to his negligence he takes the value as 50.32 m whereas the actual precise value is 50.324 m. his comment is here How to Calculate the Relative Error? This is your absolute error![2] Example: You want to know how accurately you estimate distances by pacing them off. You can, however, say you had "10% relative error."[10] Community Q&A Unanswered Questions When a measured value is negative how do I determine the exact value and the relative value? Relative Error Chemistry
By continuing to use our site, you agree to our cookie policy. numericalmethodsguy 19.533 προβολές 3:47 10 Relative Error - August 2008 #28 - Διάρκεια: 2:24. If a number is a ± b, then the relative error is b/a Example If 200 is correct to 2 significant figures, what is the relative error? this contact form Just like before, make sure you let the audience know what you were measuring in otherwise a simple "2" doesn't mean anything.
http://mathworld.wolfram.com/RelativeError.html Wolfram Web Resources Mathematica» The #1 tool for creating Demonstrations and anything technical. Absolute And Relative Error In Numerical Methods this is about accuracy. In plain English: The absolute error is the difference between the measured value and the actual value. (The absolute error will have the same unit label as the measured quantity.) Relative
## It takes 10ml drops of water to cause a reaction, but his dropper claims it is "+/- .5ml." The Absolute Error in his measurements must be: +/- .5ml 5 Understand what
Percent of Error: Error in measurement may also be expressed as a percent of error. When weighed on a defective scale, he weighed 38 pounds. (a) What is the percent of error in measurement of the defective scale to the nearest tenth? (b) If Millie, the Please try the request again. Type Of Error In Measurement The smaller the unit, or fraction of a unit, on the measuring device, the more precisely the device can measure.
Relative Error =|Measured−Actual|Actual{\displaystyle ={\frac {|{\mathrm {Measured} }-{\mathrm {Actual} }|}{\mathrm {Actual} }}} Multiply the whole thing by 100 to get Relative Error Percentage all at once.[9] 4 Always provide units as context. Q: What is the set of all points called? Ways to Improve Accuracy in Measurement 1. http://supercgis.com/relative-error/relative-error-formula-algebra.html Absolute errors do not always give an indication of how important the error may be.
Once you understand the difference between Absolute and Relative Error, there is really no reason to do everything all by itself. In plain English: 4. For example, you measure a length to be 3.4 cm. In general, if a number is a ± b, then the absolute error is b .
Imperfect equipment used either for measurement or studies, such as very small, precise measurements or burners that provide uneven heat.[6] Method 2 Calculating Relative Error 1 Divide the Absolute Error by If the object you are measuring could change size depending upon climatic conditions (swell or shrink), be sure to measure it under the same conditions each time. Absolute, Relative and Percentage Error The Absolute Error is the difference between the actual and measured value But ... What is the Formula for Relative Error?
Relative Errors The absolute error doesn"t really tell us much about how big the error really is. About this wikiHow 106reviews Click a star to vote Click a star to vote Thanks for voting! Take a stab at the following problems, then highlight the space after the colon (:) to see your answer. Please I need the right one.
Expand» Details Details Existing questions More Tell us some more Upload in Progress Upload failed. If you are measuring a 200 foot boat, and miss the measurement by 2 feet, your percentage error will be much lower than missing the 20 foot tree measurement by 2 These approximation values with errors when used in calculations may lead to larger errors in the values. Calculate the absolute error and relative error.
Use the same unites as the ones in your measurements.[4] 4 Practice with several examples. Back to Top Suppose the measurement has some errors compared to true values.Relative error decides how incorrect a quantity is from a number considered to be true. The precision of a measuring instrument is determined by the smallest unit to which it can measure. Apply correct techniques when using the measuring instrument and reading the value measured.
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Is there a plainer english explanation of bravery and poltroonery?
Despite having reread the rules section multiple times, I don't understand how death numbers, death rolls and bravery / poltroonery settings other than flat numerical modifiers work. Is there an idiots guide? How do they work?
Limited Bravery and how you get to modify your death rolls.
You place in your orders Bravery: 8 (we're going to be silly brave)
Scenario 1: The death number set by the GM is equal to or lower than 8.
What happens: Nothing, he rolls 2d6 and if that number is equal to or higher than the set death number, you die. If it is less, your character lives and then rolls are made for loot, MiD, etc.
Scenario 2: The death number is Higher than 8.
What happens: The GM takes the death number minus your bravery number, then adds that to his death roll. So if the death number was 10. He would take 10-8= 2 and add 2 to his death roll. He would then roll 2d6+2 and if that number was equal to or higher than 10, your character would die. If it was lower, then your character would live. AND get a +2 to all his loot, MiD, etc.
For Completion
Poltroonery goes to 11 (yes, I'll make that pun)
Scenario 1: The Death number set by GM is 11 or higher
What happens: Nothing, rolls 2d6 if number is equal to or greater than, you die. Same as scenario 1 on the bravery side.
Scenario 2: Death number is 10 or lower.
What happens: GM takes your Poltroonery number minus the death number, then subtracts that number from the death roll. So if the death number was 2. you take 11-2=9. Then roll 2d6-9 and if that number was equal to/greater than 2 you then die. If it was lower your character would still be alive, but you would probably be called a coward for running away.Disgraced, kicked out of your unit, etc etc.
• This helps incredibly! I was unable to parse the rules as written, whereas this makes sense! – Samuel Russell Feb 23 '13 at 1:21
• Just to be sure: this means that the lower the death number is, the more likely you are to die, right? – Cristol.GdM Feb 23 '13 at 2:52
• @Scrollmaster The more reckless bravery there is than necessary, the more likely the fool is going to go and make the situation dangerous enough to get killed, is how I understand it. Of course, fortune favours the bold, so if you win the gamble and don't get yourself killed you get the rewards of going so far above and beyond. – SevenSidedDie Feb 23 '13 at 3:16
• Correct, To be more D&D non-thac0 the Death number is your AC and the death roll is his attack against your AC. – Ben Hardy Feb 23 '13 at 3:39
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square root method
Related topics:
Easy To Understand Algebra 1 | Software Algerbra Tutorial | Purdue University Tutor | Inequalities Algebra 1 | Used Mcdougal Littell Algebra 1 An Integrated Approach | Gustafson Frisk And Tussy | dividing fractions with variables calculator | Math Quiz Yr 10 | algebra pizzazz workbooks
Author Message
Angole
Registered: 24.10.2006
From:
Posted: Wednesday 22nd of Dec 10:41 I'm having great difficulty knowing the logic behind the question about square root method . Can somebody please assist me to know how to come up with a detailed answer and explanation about square root method especially in topic of radicals? I was taught how to solve this before but now I forgot and confused how to answer it. I find it difficult to understand it alone so I believe I need help since I think I can’t do this on my own. If someone knows about square root method can you please help me? Thanks!
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
Posted: Friday 24th of Dec 09:27 I know how hard it can be if you are a bit lost with square root method . It’s a bit hard to give you advice without more information of your problems. But if you rather not get a tutor, then why not just use some computer program and see how it goes. There are so many programs out there, but one you should get a hold of would be Algebra Helper . It is pretty useful plus it is worth the money.
Vild
Registered: 03.07.2001
From: Sacramento, CA
Posted: Sunday 26th of Dec 12:03 I myself have been using this software since a year now, and it has never let me down. It won’t just solve a problem for you, but it’ll also give details of every step that was taken to arrive at a particular solution. And that’s the best feature in my opinion.I used to face a lot of problems tackling questions based on square root method but ever since I bought this software, math has beenreally easy for me.
SjberAliem
Registered: 06.03.2002
From: Macintosh HD
Posted: Monday 27th of Dec 10:46 I remember having difficulties with dividing fractions, interval notation and absolute values. Algebra Helper is a really great piece of math software. I have used it through several algebra classes - Basic Math, Algebra 2 and Pre Algebra. I would simply type in the problem from a workbook and by clicking on Solve, step by step solution would appear. The program is highly recommended.
Fearliss
Registered: 31.03.2003
From:
Posted: Tuesday 28th of Dec 07:55 You have given an amazing solution to the problem. Please recommend a site from where I can purchase the software.
TC
Registered: 25.09.2001
From: Kµlt °ƒ Ø, working on my time machine
Posted: Thursday 30th of Dec 09:52 Sure, why not! You can grab a copy of the software from http://www.algebra-answer.com/faq.shtml. You are bound to get addicted to it. Best of Luck.
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How do I make cement
Mixing concrete yourself is not a challenge - as long as you know the correct mixing ratios. These in turn depend on what is to be concreted. Will the concrete later only be exposed to mechanical loads or will it also have to withstand frost and chemical influences? In order to achieve a robust result, the ratio of the individual components must be optimally coordinated.
mixture
Concrete consists of three components: cement, water and the so-called aggregate. The surcharge is sand, gravel or grit.
The ratio between cement and aggregate is 1: 4. Concrete therefore consists of one part cement and four parts aggregate. Then there is the water. How much water is added, however, depends on the exposure class. This indicates how heavily and how the concrete will be loaded later. However, the amount of water can also be easily calculated.
Density and quantity
Concrete has an average density of 2.4 to 2.5 kg / dmĀ³ (kilograms per cubic decimeter). For a volume of one cubic meter, 2,400 to 2,500 kg of concrete are required. From this it is very easy to calculate the amount of cement required.
Concrete consists of one part cement and four parts aggregate and one fifth of cement. The total amount of concrete required is therefore only divided by five to get the amount of cement required. For one cubic meter this means:
• 2400 kg / 5 = 480 kg
• 2500 kg / 5 = 500 kg
For one Cubic meters of concretewill therefore 480 to 500 kg of cementneeded.
Of course, the calculation can also be carried out in the other direction in order to calculate how much concrete can be made from the existing cement. To do this, the amount of cement is simply multiplied by five. This results in the standard packaging size of 25 kg of cement:
• 25 kg cement x 5 = 125 kg concrete
W / C value
The W / C value indicates how much water has to be added to the cement and the aggregate in order to obtain the most resistant result possible. The value depends on the stresses and environmental influences the concrete is to withstand later. The W / C value is in each of the following cases:
Mechanical wear
• Very strong 0.40
• Strong 0.45
• Mediocre 0.55
frost
• With high water saturation 0.50
• With moderate water saturation 0.60
Chemical influences
• Weak 0.60
• Mediocre 0.50
• Strong 0.45
Calculate ratios
In order to calculate the need for cement, aggregate and water, the necessary total amount in kilograms must first be calculated. To do this, the volume is first calculated, whereby length, width and height are multiplied with each other. For example, if a path one meter wide and ten meters long is to be provided with a ten centimeter thick layer of concrete, the required concrete volume is:
• 1 meter (width) x 10 meters (length) = 10 square meters
• 10 square meters x 0.1 meters (height or thickness of the concrete layer) = 1 cubic meter
Based on the average mass of 2,450 kg of concrete per cubic meter, 2,400 to 2,500 kg of concrete are required. This in turn results in a need for cement of:
• 2,500 kg concrete / 5 = 500 kg cement (20 bags of 25 kg each)
For the surcharge, the total amount of concrete is multiplied by 4/5 or 0.8 or the difference between the total amount and concrete is used.
• 2,500 kg concrete x 0.8 = 2,000 kg surcharge
• 2,500 kg concrete - 500 kg cement = 2,000 kg surcharge
Finally, the required amount of water is calculated using the W / C value. For example, if the concrete is exposed to strong mechanical wear, the W / C value is 0.40. The ratio of cement and water is calculated by simple multiplication:
• 500 kg cement x 0.40 W / C value = 200 liters of water
For the example given with an amount of 2,500 kilograms of concrete, 500 kg of cement, 2,000 kg of aggregate and 200 liters of water are required.
costs
For a standard packaging size of 25 kg of cement, costs of 3 to 6 euros should be expected. The above example requires 500 kg of cement. The costs for this are calculated as follows:
• 500 kg total amount / 25 kg sack = 20 sacks of cement
• 3 euros per sack x 20 sacks = 60 euros
• 6 euros per sack x 20 sacks = 120 euros
The average costs can therefore be between 60 and 120 euroslie.
In addition, there are the costs for the respective surcharge. For concrete gravel, chippings and sand, depending on the size, you can expect 30 to 50 euros per ton, which results in a price of 60 to 100 euros for the calculation example.
So for the raw materials it should be with 120 to 220 eurosbe expected.
It should be ensured that the delivery is often higher than the price of cement and aggregate due to the high weights. This information is therefore not the total price.
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Sunday
January 22, 2017
# Posts by Chris
Total # Posts: 2,670
chemistry
2.56 grams of zinc react with excess phosphoric acid to produce how many molecules of hydrogen gas?
March 8, 2015
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All of them are correct...
March 4, 2015
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a parent pushes a kid on a swing, A) If the child on the swing rises 2m above the lowest point during each cycle, what is her speed at the lowest point of the swing? B)If the beginning of a cycle (2m above the lowest point) the child is given a push so that her starting speed ...
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A 39-turn circular coil of radius 3.20 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 0t + 0.040 0t2, where B is in teslas and...
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A 24.8-g bullet is fired from a rifle. It takes 2.15 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 778 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
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The brakes of a truck cause it to slow down by applying a retarding force of 7760 N to the truck over a distance of 870 m. What is the work done by this force on the truck?
March 1, 2015
tablun elementary school
There is no way to solve this with this information.
March 1, 2015
Physics
Power is Work/Time, or Force*Distance/Time. You have a velocity, which is Distance over time, so you need to find the force that the winch needs to pull the block upwards. That force is the frictional force (coefficient of friction* normal force) + the force of gravity down ...
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in number 13,407.036 there are two 3's explain how the values of the 3's are related?
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in the number 13,407.036 the are two 3's explain how the values of 3's are related?
February 26, 2015
trig with enough info?
Two oil Wells are 4 km and 6 km from the river. The points along the river are 8 km apart. Find the common loading point on the river, using the shortest amount of pipe.
February 25, 2015
trig
Two oil Wells are 4 km and 6 km from the river. The points along the river are 8 km apart. Find the common loading point on the river.
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MATH
determine the average savings balance on a particular day. error of tolerance is 50. what is the minimum number of accounts he must sample to construct a 95% confidence interval. the standard deviation is 350
February 19, 2015
Prob/Stats
determine the average savings balance on a particular day. error of tolerance is 50. what is the minimum number of accounts he must sample to construct a 95% confidence interval. the standard deviation is 350
February 19, 2015
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There are a few different horsepower definitions - imperial, metric, DIN etc but an average conversion factor is 1hp = 745W Generator input = 1500 * 745 W = 1117500W Generator output = 1117500W * 80.0% = 894000W (3 sig figs) Current = 894000W / 1910V = 468.06A
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A weight suspended from a spring is seen to bob up and down over a distance of 11 cm twice each second. What is its frequency? Answer in units of Hz. 002 (part 2 of 3) 10.0 points What is its period? Answer in units of s. 003 (part 3 of 3) 10.0 points What is its amplitude? ...
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A weight suspended from a spring is seen to bob up and down over a distance of 11 cm twice each second.
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art
number 3 is not C...It is D...just took the test and got #3 wrong...
February 5, 2015
English
I don't know but i have this same quiz tommorrow please help
February 2, 2015
Math
In “Adventures of Isabel,” how is the doctor different from Isabel’s other enemies—the bear, the witch, and the giant? He is Nice to Her He Frighten Her He gets killed He may be Real Witch one pls? I think It's The first one..
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college mathematics 1
describe the graph of g(x)=f(x)+1 relitave to f(x)
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Calculus
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In an action movie, the heroine is supposed to jump with a motorcycle from the roof of one skyscraper to another one. The roof of the second skyscraper is h=9 meters lower, and the gap between the buildings is d=15 meters wide. Both roofs are horizontal and flat. Neglecting ...
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A juggler tosses a ball h = 2 m into the air with one hand and catches it with the other hand at the same level. Neglecting air resistance, how long is the ball in the air?
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During a long airport layover, a physicist father and his 8-year-old daughter try a game that involves a moving walkway. They have measured the walkway to be 41.5 m long. The father has a stopwatch and times his daughter. First, the daughter walks with a constant speed in the ...
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A man in search of his dog drives first 13.3 mi northeast, then 17.3 mi straight south, and finally 6.85 mi in a direction 30.3° north of west. 1. a) What is the magnitude of his resultant displacement? 2. b) What is the direction of his resultant displacement? Express ...
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A 26.3 g marble sliding to the right at 21.0 cm/s overtakes and collides elastically with a 13.3 g marble moving in the same direction at 12.2 cm/s. After the collision, the 13.3 g marble moves to the right at 23.7 cm/s. Find the velocity of the 26.3 g marble after the collision.
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January 26, 2015
Physics
A 26.3 g marble sliding to the right at 21.0 cm/s overtakes and collides elastically with a 13.3 g marble moving in the same direction at 12.2 cm/s. After the collision, the 13.3 g marble moves to the right at 23.7 cm/s. Find the velocity of the 26.3 g marble after the ...
January 26, 2015
Physics
Two carts with masses of 4.3 kg and 3.9 kg move toward each other on a frictionless track with speeds of 5.6 m/s and 5.0 m/s, respec- tively. The carts stick together after colliding head-on. Find their final speed.
January 26, 2015
Physics
A 2170 kg car moving east at 10.5 m/s collides with a 3230 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 6.00 m/s. a) What is the velocity of the 3230 kg car before the collision?
January 26, 2015
Physics
A football punter accelerates a 0.53 kg foot- ball from rest to a speed of 8.6 m/s in 0.24 s. What constant force does the punter e
January 25, 2015
Physics
A 83.8 kg astronaut is working on the en- gines of a spaceship that is drifting through space with a constant velocity. The astronaut turns away to look at Earth and several sec- onds later is 39.6 m behind the ship, at rest relative to the spaceship. The only way to re- turn ...
January 25, 2015
math
The area is the trinomial and the side lengths are the binomials
January 25, 2015
Static
1.0x10^-6
January 24, 2015
Science
Every individual including young people can make decisions to use resources wisely use the terms reduce reuse and recycle to explain how the students in the image above can help minimize solid waste. HELP LAST DAY TO DO THIS I NEED AN ANSWER.
January 22, 2015
physics
A 10.7 kg bowling ball moves in a straight line at 4.26 m/s. How fast must a 2.55 g Ping-Pong ball move in a straight line so that the two balls have the same momentum?
January 21, 2015
12 Math
slope: 22 instantaneous: 22 1/2
January 21, 2015
Physics
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10−27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
January 19, 2015
Physics
Calculate the magnitude of the linear momen- tum for each of the following casesEarth (m = 5.98 × 1024 kg) moving with an orbital speed equal to 26200 m/s. Answer in units of kg · m/s.
January 19, 2015
Physics
A railroad car with a mass of 2.33 × 104 kg moving at 3.71 m/s collides and joins with two railroad cars already joined together, each with the same mass as the single car and initially moving in the same direction at 1.44 m/s.What is the final speed of the three joined ...
January 19, 2015
Superhero by.Kiana S. This student most likely creates art based on all of the following except A.personal interests B.personal style C.skill set D.knowledge of art history
January 14, 2015
english
Hello Ski! did you by the way completed the test? if you did can you post answers that was correct?
January 13, 2015
Math
Eryn and seven of her friends go out to lunch. The total bill comes to \$143.35. They decide to leave a 15% tip. Each person will contribute an equal amount to the total tip. Estimate what each person should contribute.
January 7, 2015
math
a piece of ribbon is cut into two shorter pieces in the ratio 2.8 : 1.25 The difference in the length of the shorter pieces is 80.6 cm. What is the length of the original piece?
January 6, 2015
Math
First do 28-7=21 and do it so on and on until you get to 0.
January 5, 2015
math
Martha wrote the following inequality statement about the temperatures: -4.3<-2.91<5.7
January 3, 2015
Math
The mean of Susan's math and science scores is 74 points. The mean of her math and English scores is 83 points. How many more points did Susan score in English than in science?
December 17, 2014
Math
25(5) to 38(10) (5)(5)(5) to (2)(19)(2)(5) (5)(5) to (4)(19) 25 to 76
December 11, 2014
Psychology Questions
ur right!
December 7, 2014
trig
She means it intersects at (-20/29,-21/29) x=-20/29 y=-21/29
December 6, 2014
Statistics
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X = 2), n = 8, p = 0.2
December 2, 2014
physics-finding velocity
A small block of mass .450 kg is sliding down a ramp inclined at an angle of 53 degrees to the horizontal. it starts to slide from a vertical height of .65 meters. Find the Velocity when it slides off the ramp. The way I approached this problem was by sing Vf^2-VI^2=2ad. ...
December 1, 2014
physics
You take a pendulum clock to the moon, Assuming you do not change anything about the clock, will it still keep the right time? why?
December 1, 2014
physics
You have a pendulum clock which keeps using a bob on a long shaft. The clock is running slow, You can add or subtract mass from the bobs, or move the bob up and down on the shaft? What should you do to correct the time on the clock?
December 1, 2014
physics
You test a moon buggy on earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds. Will the period be the same when buggy is used on the moon?
December 1, 2014
chemistry
If you place 10.0 L of propanol (C3H8O) in a sealed room that is 3 m long, 2.75 m wide, and 2.5 m high, will all the propanol evaporate? If some liquid remains, how much will there be? The vapor pressure of propanol is 10.7 torr at 25 °C, and the density of the liquid at ...
November 30, 2014
Math
What is the riddle for moving words
November 29, 2014
Chemistry
I must be going wrong somewhere. 1.3x.250 = .325 Mols KOH .325 Mols KOH = .325 Mols HBr Then I'm taking 350ml (.350) and multiplying it by .325 Mols HBr and I get .11375... That doesn't seem quite right. I really do appreciate your help. I never took chemistry in high ...
November 29, 2014
Chemistry
Let’s assume you have 250 ml of a 1.3M KOH solution and you want to neutralize it with the acid HBr. It takes 350 ml of the HBr solution to do so. What must have been the molarity of the HBr solution? HBr(aq) + KOH(aq) => KBr(aq) + H2O(l) The stoichiometry is already ...
November 29, 2014
Chemistry
Ok, thank you!
November 29, 2014
Chemistry
For an H+ concentration, [H+], of 4.3 x 10-4 M, would this be acidic, neutral or basic? Why? I'm assuming that because it is an H+ it would be acidic and H+ means acidic and OH- means base. Is this correct? And is there a better formula for this equation?
November 29, 2014
Chemistry
To make 250 ml of a 5%(w/v) calcium carbonate solution, how many grams of calcium carbonate will I need?
November 29, 2014
physics
we did a lab and my observation chart is like this when my mass is 1.4 kgs (height is 0.12m,distance is 1.0m,time is 0.87s,velocity is 1.1 m/s ,Eg is 1.6 and the Ek is 0.85) when my mass is 2.14 kgs(height is 0.12m,distance is 1.0m,time is 0.88s,velocity is 1.1 m/s ,Eg is 2.8 ...
November 24, 2014
Maths
I have 40 questions, 4 marks given for a correct answer, 2 marks deducted for each wrong answer and 1 mark deducted for for each answer not attempted I attempt 19 and score 25 marks.how many questions did I get right
November 18, 2014
algebra
student takes out two loans 1 at 10% and the other at 12 percent interest the 10 percent loan is \$2000 more then the 12% loan and the total interest for one year is \$430 how much is each loan?
November 17, 2014
Science
True/false Earth's atmosphere oceans and continents began to form during the first several hundred million years of the Precambrian time.
November 17, 2014
Chemistry
Dr. Bob, I am unable to see the answer you have provided? I'm not sure why, but I see "Bob Pursley answered your first post. It's below." The problem is, the is no post below.
November 13, 2014
Chemistry
I missed a few days of school due to a sick child. I'm reviewing for an exam and the following questions are giving me trouble as I wasn't there when they were reviewed in class. If possible, I need the answer and step-by-step process so I can see what what to do and ...
November 13, 2014
Chemistry
Thank you so much. This is sure to be very helpful!
November 13, 2014
Chemistry
Ah, this is what it means "below". In the second question I posted (above I guess) I rephrased my question for answer and all work. I'm really confused as to what your provided answer means as I missed these classes. I honestly have no idea what to make of the ...
November 13, 2014
Chemistry
I missed a few days of school due to a sick child. I'm reviewing for an exam and the following questions are giving me trouble as I wasn't there when they were reviewed in class. If possible, I need the step-by-step process. Thanks. 1.) If you put hydrogen peroxide on ...
November 13, 2014
How to express 12/15 as a fraction, percent, and decimal. Someone please show me how to solve.
November 12, 2014
The poster shows 12/15 games won so far. I need help on how to express as a fraction, percent, and as a decimal.
November 12, 2014
The poster shows how many of its games the football team has won so far. Express this information as a fraction, a percent, and as a decimal. Can someone show me how to solve this problem?
November 12, 2014
physics
A 4.5 kg block is dropped onto a spring of spring constant 1861 N/m from a height of 1500 cm find the speed of the block when the compression of the spring is 12 cm. The acceleration of gravity is 9.81 m/s
November 12, 2014
History
What is the national monument of America?
November 12, 2014
Physics-coefficient of friction
What's the minimum coefficient of friction between the ladder (of length d) and the ground will keep it from slipping? (consider the wall frictionless) (θ=56.5 degrees) I think the formula is supposed to be mg L/2 sin y, but it does not provide a length, nor a mass. ...
November 11, 2014
Physics- tension
An 63 kg construction worker is sitting 2.0 m from the end of a 1435 kg steel beam to think about the state of the world. Find the tension in the cable that is supporting the beam. Am I required to use the formula T1=sin(theta)*m*g? I am not sure if the problem is similar to a...
November 11, 2014
Physics- spring constant
Astronauts are in space so there is no gravity, but they wish to weight themselves. The spring constant is 265 N/m. The question shows a graph; the y is the position in meters while the x is the seconds. I do not wish for an answer, but the way in which i'm supposed to ...
November 11, 2014
Math
A sea turtle laid 95eggs,each weighing 1 ounce.what is the total amount of weight in pound?
November 10, 2014
Science
heated 100 grams of powdered mercuric oxide to produce 93 grams of liquid mercury, how much oxygen would be released?
November 10, 2014
English
What is a metaphor for improvement?
November 10, 2014
Physics
A. You drag a suitcase of mass 19 kg with a force of F at an angle 41.3 degrees with respect to the horizontal along a surface with kinetic coefficient of friction 0.54. The acceleration due to gravity is 9.8 m/s^2. What is the normal force on the suitcase? Answer in units of ...
November 6, 2014
physics-distance
Here's a stationary bike. If you pedal such that the front wheel rotates at 6.9 rad/s, and that wheel's radius is half a meter, how far would you have gone in 35 minutes, if the bike where a non-stationary bike? (Assume a constant speed of rotation) I am not sure where...
November 4, 2014
science
Using the systematic method and information in the Appendices F and J (note that log value is given in Appendix J), determine [SCN-], [Hg2+], and [HgSCN+] at equilibrium when Hg(SCN)2(s) is dissolved in water.
November 4, 2014
chemistry
Propose methods for protecting the sodium hydroxide solution in the bottle and the buret from exposure to carbon dioxide in the air.
November 3, 2014
Math
A woman has money in two accounts. One account pays 7% annual interest, whereas the other pays 15% annual interest. If she has \$1,500 more invested at 15% than she does at 7% and her total interest for a year is \$1,545, how much does she have in each account? Please show the ...
November 1, 2014
physics
A skier leaves the starting gate at the top of a ski ramp with an initial speed of 4.00 m/s (Figure 6-41). The starting position is 120 m higher than the end of the ramp, which is 3.00 m above the snow. Find the final speed of the skier if he lands 145 m down the 20.0° ...
October 31, 2014
english
the air was thick and grey what else???
October 29, 2014
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A trapezoid has sides of 13, 5, 11, and 8. The diagonals also have integer length. What are they? Answer and solvers. I'm fairly certain that this is the smallest quadrilateral with 6 different integers for the sides and diagonals. Robert Reid sent me the 25, 39, 60, 52 quadrilateral, which has perpendicular integer diagonals. He hypothesized, in 1972, that this was the smallest such. Claudio Baiocchi disagreed, and sent me a smaller nonconvex example.
Robert Reid: "On the island of Plunk the only straight edges they have are on quadrilateral pieces of slate. Strangely enough, the side lengths are always integral multiples of their unit of measurement. The slate is extremely hard, so they make rulers with only 6 marks, (one on one edge, two on the next, three on the third and none on the fourth). Amazingly, they can measure all lengths from one to twenty. What are the possible quadrilaterals, and how are they marked?" Answer.
At the La Jolla Covering center, I noticed cyclic coverings. A cyclic cover starts with a polygon where the sides and diagonals all have different length. I hand-solved the quadrilateral in a 13-gon, it is a nice puzzle. Try it! Here are the solutions for the pentagon in a 21-gon, and a hexagon in a 31-gon. Repeat 21 or 31 times, and you get the full graph. This, and Reid's problem, are related to Golomb Rulers.
A quickie by Michael Riemann: 69 squared is 4761. 69 cubed is 328509. Each number from 0..9 appears exactly once. Now, find a number so that the third and fourth powers have this property. Answer. Igor Krivokon: I solved it using this Perl one-liner:
++\$i while length(\$_=\$i*\$i*\$i.\$i*\$i*\$i*\$i) != 10 || /(.).*\1/; print \$i
An antique with lots of secret compartments is demonstrated at Clip 4, here. You can't buy that, but there are a lot of things you can buy at Mr. Puzzle, Kadon Enterprises, Cleverwood, and Funagain Games. The latter has the complete Games 100 on sale.
One of the champions of Mathematical Recreations is G P Jelliss, so you might like his website.
I mentioned weeks ago that Patrick Hamlyn's 30-60-90 partridge puzzle was offered by Walter Hoppe. A paper version is here -- arrange the 10 triangles to make a 30-60-90 triangle. Patrick sent me a nice picture of his copy. If you'd like one, send a check to: Walter H. Hoppe, 3648 Tolland Road, Shaker Heights, OH 44122-5141. You can see samples of his work at Puzzle World . His aol.com address is LaserLynne.
PQRST 03 has just started. You have six days to send answers and ratings of 10 puzzles. At Cihan Altay's WPC-2002 page, you can find links to 5 trip reports.
Fifteen golfers want to get to know each other, so they play in groups of three for a week. (15-3-7) By the end of the week, all the golfers have played once with each other. This particular setup is identical to the Kirkman Schoolgirl Problem. It is also an example of a Golf Tournament Problem. I realized that the answer to this problem might make a nice picture. For example, 16 golfers playing in groups of four for five days (16-4-5) can be done as follows:
Andy Liu and Michael Reid let me know that the above can be solved as a finite affine plane. If you look for similar designs in the La Jolla Covering Repository, you'll see these particular systems are solved via affine geometry. Twenty five golfers playing in groups of 5 for 6 days (25-5-6) can be done as follows:
I thought that 36 golfers might be able to play in groups of 6 for 7 days, but that winds up being impossible. If it were, then the first 4 days could be used as a solution for the 36 Officer Problem. Euler's Graeco-Roman square conjecture is related. Twenty seven golfers playing in groups of three for 13 days (27-3-13) can be done as follows (the thirteenth picture has groups 1-2-3, 4-5-6, etc):
As a puzzle, you might try to find a symmetric picture for 9 golfers in threesomes for 4 days. Here is the answer. Now, you might be wondering about the 15-3-7 problem I started with. Despite my efforts, I could not find a symmetrical picture for it. Let me know if you can find one. Below is my best effort. Other groups that seem to be unsolved are 21-3-10, 28-4-9, 33-3-16, 40-4-13, 45-3-22, 45-5-11, 49-7-8, 52-4-17, 64-4-21. If you can solve any of these, or improve the bounds, let Warwick know.
The Kirkman Schoolgirl Graph. Can it be made symmetrical?
David Wilson recently asked me whether a square or general triangle could be divided into acute isosceles triangles. I managed to divide the 2-3-4 triangle into 17 such triangles, and I divided the square into 16 acute isosceles triangles. Can these be beat? Let me know. Maybe these can be improved. If you are new to this type of problem, you might like trying to divide a square into 8 acute triangles, or 9 (harder). (Gardner presented solutions for both of these.) Footnote: Dan Hennessy let me know that V.E. Hoggatt Jr., and R. Denman, "American Mathematical Monthly," Nov. 1961, p. 912-913 shows that any obtuse triangle can be divided into 8 acute isosceles triangles. R. William Gosper found better solutions for both. His solution for the square in 10 acute isosceles triangles appears to be unique, and involves an order 12 polynomial. He verified the 2-3-4 triangle can be divided into 7 acute isosceles triangles.
The Periodic Table Table that I helped to build and promote has won the Ignobel Prize in Chemistry. Theo Gray wrote up our trip to Harvard. At the awards ceremony, by an odd coincidence I was seated right in front of a delegation from the National Puzzler's League. A few days later, I had a chance to see MIT's infinite corridor, and the Harold Edgerton exhibit. If you've ever seen the splash of a milk drop, or a bullet cutting a playing card, you've seen Edgerton's work. On Saturday, Theo's father pointed out the Deutsche Bank Conceptual Word Search that filled up many pages of the New York Times (I found it fairly pointless).
I rather like the Chromatron series of mirror-based puzzles. I just started using the Mozilla editor -- let me know if you see anything that needs fixing (I love Mozilla as a browser). Many people sent answers to my F26 question. My own method for solving was to three-color the edges. Since most of the figure is a hexagonal grid, three-coloring is easy. That leads to the first solution below. The circular embeddings are what you'll see if your straighten out the hexagonal embeddings.
I like symmetry, such as those seen in the graphs at the Foster Census. But they don't have pictures. You can see some of these graphs in A New Kind of Science, and I've started putting more of them on Mathworld. Graph F26 led to a neat little puzzle ... is it Hamiltonian? Yes, it is. Can you find a cycle that goes through all the vertices, without reusing an edge, and ending where you started? With the right approach (just four words!), this is very easy, and the solution is gorgeous. Answer. If you figure out my approach, just send me that.
The World Puzzle Championships were won by Japan this year. Will Shortz is getting a column in next issue of the Reader's Digest, at least until December. I'll be going to Harvard this weekend, to pick up an award at the Ignoble Prize ceremony.
Theo Gray's Wooden Periodic Table project that I Igored (which won an ignoble prize), has details of a Sodium party. Not long ago, while standing far, far away, we learned that a large piece of sodium turns into a hopping/explosive/caustic projectile, that bounces 40 feet at a time. On the whole, I preferred magnesium. I'll get a new acrylic version of the periodic table if it's available in November.
Patrick Hamlyn recently ignored a (poor) impossibility proof, and found a way to cover a cylinder with the 5 tetrominoes. Can you figure out how? Here is the answer, by Clinton Weaver. Many people solved this (it's easy).
Jim Propp (discoverer of the "highway" in Langton's Ant) tells me that I was likely the first to notice the binary counting turmite. Normally, the "turmite mound" starts out with all zeroes, but I decided to try starting with {{0, 1}, {1, 0}}. Here is what happened. I let it run for awhile -- simple counting after a detour.
The American Mathematical Society Notices has my review of Martin Gardner's Colossal Book of Mathematics.
In a three-coloring, such as the di-L-tetrominoes solution by Patrick Hamlyn (below), only three colors are used, and no two regions of the same color share an edge. The Penrose Tiling can be 3-colored -- This was proven by Robert Babilon [Discrete Mathematics 235(1-3):137-143, May 2001] Each kite and dart touches exactly 4 others, so I tried to expand the problem with a little conjecture: A map is 3-colorable if each interior region is bounded by an even number of sides. That's false. Dean Hickerson found a very nice counterexample. It's an 8 region map, where each region shares an edge with exactly 4 other regions. Hickerson's map requires 4 colors, so my conjecture was wrong. Can you find it, or a different counterexample? Answer. A hint: In the figure below, no more than 3 regions meet at any point, so it's called a cubic map. A cubic map is 3-colorable if and only if each interior region is bounded by an even number of sides. This is easily proven. Dean Hickerson, Clinton Weaver, Stephen Kloder and Jaap Scherphuis solved it. Another picture.
Dr. Sebastian Wedeniwski of zetagrid.net has collected the first 110 billion zeroes of the Zeta Function. He is offering various prizes that you can win by joining the grid and making mathematical discoveries. (I made a similar announcement for Eternity, and readers of this page won that prize.)
One of my favorite puzzle games back when I had a Mac+ was The Fools Errand by Cliff Johnson. I just learned that fools-errand.com is an active site by the original programmer, he's providing a number of free downloads, with PC versions of the original games.
Erich Friedman: Arrange 14 cubes so that each cube touches 6 others along some portion of a face (not just an edge or a corner). Answer. It was solved by Erwin Eichner.
A curiousity I found with Mathematica: 1183893^2 = 1!+2!+3!+7!+8!+9!+10!+11!+12!+13!+14!+15! . I greatly enjoyed reading the antics at Scam-o-rama.
From Federico Kereki: This problem comes "from real life" -- it was proposed to my nephew in High School, and he couldn't solve it. FIRST PROBLEM: Set 102 straight lines on a plane, so they intersect in exactly 2002 points. As it is, it's quite difficult to solve -- I guess the professor made a typo when he wrote the problem, so here is a... SECOND PROBLEM: What number would have made that a reasonable High School level problem? Pictures. Answers.
From Erich Friedman: The twelve different pentominoes have been paired up. Unpair them. (Solved by Erwin Eichner, Darrel C Jones, Clinton Weaver, and Koshi Arai)
The game of awarihas been solved by John W. Romein and Henri E. Bal , through an analysis of all 889,063,398,406 positions. It's a draw. The "perfect" playing applet is quite interesting.
From Jeff Barnes: SONGBIRD divided by BIRDSONG = OWL (a recurring decimal). What is the 8 digit number SONGBIRD? Bob Kraus sent the solution. (He used a program, just as I did).
64 squared is 4096. If you put a zero in front of that, it can be the middle entry in a 5 by 5 square of square numbers. Basically, a crossword with five square numbers reading across, five square numbers reading down, and 04096 as the third down entry. What is this grid? Answer. Matthew Bjorge solved it by hand.
George Maydwell has found a cellular automata that forms odd "iceballs", when allowed to run for a long time.
Eric Weisstein has graciously added my write-up of Turmites and Guilloche patterns (found on money, tickets, passports, and licenses worldwide) to his encyclopedia. Feel free to write me for details of the code.
Jean-Charles Meyrignac has put together a page of Solitaire problems, with a javascript implementation.
The Wooden Periodic Table project has turned up more surprises. I learned thatsodium attracts droves of sulfur butterflies. We have good samples of almost all the safe elements, now, including scandium. We still need lanthanum, terbium, homium, ytterbium, and lutetium. If you have a kilogram of 99.99% lutetium sitting around in your basement, we'd love to have it.
The German equivalent of Scientific American (www.wissenschaft-online.de/spektrum/) has a good math column. (translated by google).
John Gowland(at ampnet.co.za, jgowland, would love to see your comments) presents All 9's Squared: "You may like to consider my latest puzzle. I sent it to Oyler out of interest to get an opinion of the standard of puzzle." (Oyler liked it a lot.)
Don Knuthhas made available a work on Combinations, to join his preleases on Permutations and n-tuples. All of them can be viewed by using Ghostview and Ghostscript. All of these were updated 24 August.
Jorge L. Mireles has updated his gorgeous applet based on rotating decagons and stars. Claudio Baiocchi comments: "Thank you for signalling this. From my point of view your evaluation could in fact appear as an understatement, the puzzle being quite impressive. I noticed that a new release of the applet just appeared, offering a lot of new possibilities."
On another topic, dividing a triangle into similar but non-congruent pieces has come up. Here's a picture of the 4-5-6 triangle in 6 pieces. Michael Reid tells me the reference is h. kaiser, perfekte dreieckzerlegungen. elemente der mathematik 46 (1991), pp. 106-111. mr 92 d:51019.
Results of the International Puzzle Design competition are now available at www.johnrausch.com/DesignCompetition/default.htm. All of the main winners used clever mathematical tricks.
Puzzle of the week: The Lonely Unit Cube. Arrange 1 side-1 cube, 54 side-2 cubes, 24 side-3 cubes, and 2 side-5 cubes to make a side-11 cube. Answer.Second answer. (Placements of key pieces is enough) Patrick Hamlyn discovered this after I posed the harder question of whether a cube could be dissected into smaller cubes with only one unit cube. If anyone can make this puzzle out of appropriately sized dice or other objects, I'd very much appreciate a picture. Patrick found with fewer than 8 unit cubes, and a side length below 12, the set of cubes is unique.
I was looking through The Ins and Outs of Peg Solitaire, and found this gem: . John Beasley claims that any cell can be vacated, and for each such position, the final peg can rest on that spot. Further, he claims no smaller square-grid board with this property is possible.
The reason I was looking was the Lamp problem (Martin Gardner, Further Mathematical Diversions, p 125). The best that was done at the time was a 10-move solution. Jean-Charles Meyrignac has found a 9-move solution, where the pegs start on the x's, and the final peg is at the center of the board. Can you find it?
A book recommendation -- the next time you're travelling, ask a money-changer if they have an old issue of the MRI Banker's Guide to Foreign Currency that you may have. Another obscure favorite: Triangle Centers by Clark Kimberling. A taste for this excellent book can of course be seen at the accompanying website.
A network security book by Kaufman, Perlman, and Speciner has the dedication "Si spy net work, big fedjaw iog link kyxogy". Perlman says that it's a straight substitution cypher. Answers.
Older Material - 16 Jul 02 to 26 Aug 02
Older Material - 13 May 02 to 9 Jul 02
Older Material - 4 Mar 02 to 6 May 02
Older Material - 17 Feb 01 to 5 Aug 01
Previous Puzzles of the week are here.
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Solutions
Algebra 1 - Concepts and Skills
Algebra 1 - Concepts and Skills (0th Edition) Edit edition Solutions for Chapter 9.2
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 1
The objective is to decide whether the equation is quadratic equation or not.
The general quadratic equation is, where a, b and c are known values and a can’t be zero.
Clearly the equation is not a quadratic equation. Because it has no term of That is.
Therefore the equation is not a quadratic equation.
Corresponding Textbook
Algebra 1 - Concepts and Skills | 0th Edition
9780547008332ISBN-13: 0547008333ISBN: Authors:
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# sG: Synthetic Goals
##### August 3, 2023, Micah Blake McCurdy, @IneffectiveMath
Every NHL player affects their team's results in a number of different ways. Comparing the impact of two different players at the same skill is fairly straightforward, but it is much trickier to compare the value of two different skills. For example, Alexander Ovechkin is possessed of a very remarkable finishing talent, turning the shots that he and his teammates generate for him into many more goals than an ordinary shooter would. On the other hand, he's also well-known to be weak defensively, causing his opponents to take more shots than a league-average skater would. Which of these two effects is stronger? Once this question is asked, of course, we would like to have a comprehensive way of valuing all of the different ways that a player can help or harm their team, all on the same scale.
The guiding principle behind all of my player evaluation models is that they should estimate a player's intrinsic ability, that is, isolated from any effects outside of their control. In practice, this means accounting for teammates, opponents, game state, where relevant, coaching systems, and, vitally, minutes played, since playing time is determined by (somewhat constrained) coaching choices, and is not in any player's ability to affect, no matter how strong or weak they may be.
The elements of player ability that I include here are:
• Impact on 5v5 shot rates, for and against, as measured by my 5v5 shot rate model.
• Impact on power-play shot rates (for) and on penalty-kill shot rates (against), as measured by my special teams shot rate model.
• Impact on goal scoring, most obviously by finishing talent but also through setting talent; both of these are measured via my xG model.
• Impact on team penalties drawn and taken, as measured by my penalty model.
While I am reasonably happy with each of these models, they were all developed separately and each one has its own units: the impact is calculated per 5v5 hour, or per power-play hour, or per penalty-kill hour, or per shot, or per all-situation hour, as appropriate. In other to synthesise a total impact we must bring all of these skill estimates together in a way that is independent of context, so that all players are treated the same.
Naturally, the units of our result will be goals, but these goals are not any particular goals that have been scored, for which credit is being divvied up; they are instead notional goals that a player would cause to be scored, in contrived circumstances, rather than their past circumstances or any particular future circumstances. For both of these reasons, I call the resulting unit "synthetic goals", or sG for short. Furthermore, a given skill may impact goals scored by a player's team, or by their opponents, or in some cases, both. Sometimes it will be convenient to distinguish offensive sG from defensive sG, sometimes it will be convenient to combine them.
Since we cannot use a player's past icetime, we must choose a "reference basket" of icetime, which I have chosen to be:
• One thousand minutes of 5v5 time.
• One hundred minutes of power-play time.
• One hundred minutes of penalty-kill time.
These proportions are chosen to be close to what we see overall in the league in recent years.
While sG is an "all-in-one" stat, that reductively simplifies a player's manifold abilities into a single number, it is not designed to be similar to a WAR/GAR-style stat. It is different in two important ways: first, sG is not descriptive of the past; like all of the constituent model measurements that go into sG, it is an estimate of ability at a given moment in time. In this sense sG is somewhat closer to the "predictive" side of the (slightly over-accentuated) descriptive-predictive "dichotomy". Second, sG is not a counting stat; it is instead a rate stat. You should think of it like you would think of a speedometer reading on your car, where instead of distance per hour the units are goals per reference basket of icetime.
## 5v5 Shot Rates
The conversion for 5v5 shot rate impact is the simplest. For example, at time of writing, Sidney Crosby's isolated impact on 5v5 offence is 0.31 xG per hour (that is, per 5v5 hour). Thus, in a thousand 5v5 minutes, this amounts to 5.17 xG; leaving aside for the moment that some of these shots will be taken by Crosby himself, assuming the shots are taken by league average shooters, with league average setters, and against league average goaltenders, thee 5.17 xG should become 5.17 goals. Thus, this skill is worth 5.17 sG. His defensive impact is currently more modest, just -0.05 xG per hour, so his defensive ability over a thousand minutes is -0.83 xG, that is, Crosby's imagined opponents can be expected to score 0.83 fewer goals than they would have scored had Crosby been replaced with a league-average player at this skill. These two impacts are felt in two different areas (goals scored and goals allowed) but they can be gathered together to give a total of almost exactly 6.00 sG.
## Power-play and Penalty-Kill Shot Rates
Continuing to use Crosby-as-of-the-northern-summer-of-2023 as our running example, my estimate of his impact on his team's power-play shot rates as 0.42 xG per hour, that is, per power-play hour. Again leaving aside the fact that many of these shots that Crosby generates for his team will be taken by him, we expect that over one hundred minutes of power-play time in average conditions this will amount to 0.70 sG. Crosby's estimated impact on penalty-kill shot rates against is -0.05 xG per hour, that is, per penalty-kill hour, so we expect this will amount to -0.08 goals over the course of 100 short-handed minutes. Combining the two impacts together, we have an impact of 0.62 sG.
Crosby's estimate is close to zero in part because he is not observed killing penalties very much. In his most recent season (22-23), he played 322 minutes on the power-play and just 4 short-handed. This eminently wise coaching decision is deliberately excluded from consideration here, because we are trying to measure Crosby himself, not the synergy between the player and their coach's decisions, be they wise or unwise.
## Finishing and Setting
Impact on shot quality is somewhat more complicated to value than impact on shot quantity. First, we compute a distribution of all shots taken by all NHL teams over a broad swath of time, specifically all shots taken in the 2018-2023 regular seasons. For all of these shots, I can compute the probability of the shot being blocked, being missed if not blocked, and being scored if not missed; that is, each shot can be assigned to a point in a three-dimensional configuration space. By discretizing this space, we can make a tractable distribution of what kinds of shots are taken in NHL games. The setting and shooting estimates for each player that I produce from my xG model have units of logits, that is, impact on goal probability when expressed on the logit scale. For every element in our distribution of shots, we can compute the change in goal odds that would result from Crosby taking on a specific role in that shot. For instance, suppose a shot is described by conditional sucess probabilities $$(p_b,p_m,p_g)$$, that is, a probability $$p_b$$ of being unblocked, a probability $$p_m$$ of being on net rather than missed, given that it is unblocked, and a probability $$p_g$$ of being a goal given that the shot is on net; and suppose that Crosby's shooter talent estimate is to have logit-scale impacts of $$c_b$$, $$c_m$$, and $$c_g,$$ at each stage. Then Crosby acting as shooter shifts those probabilities to $$(p_b',p_m',p_g') = (l(l^{-1}(p_b)+c_b), l(l^{-1}(p_m) + c_m), l(l^{-1}(p_g) + c_g) )$$ where $$l$$ is the logistic function that maps probabilities from the logit scale to the unit interval. Then the impact on this particular shot can be computed as $$p_b'p_m'p_g' - p_bp_mp_g$$. Then, by forming the weighted sum of this using the weights in the distribution of shots we computed earlier, we can compute the impact of Crosby's shooting on an "average shot". Finally, we can multiply this per-shot impact by 0.94, the all-situations average number of shots taken per minute in 2018-2023, and then multiply the result by 1200, the total number of minutes in our reference basket of minutes. In the future, we could consider how likely specific players are to be the shooter of a given shot, but for now let us assume that forwards take two thirds of the shots and defenders one third, in line with recent tendencies. Then multiplying Crosby's shooter impact by 2/9 (since there are usually three forwards on the ice), we obtain a final value of +2.6 sG through shooting talent. A similar calculation for setting, where we assume that 80% of shots have setters, and 7/9 of those setters are forwards, gives a setting impact of +3.5 sG.
## Penalties Drawn and Taken
The chance of a 5v5 goal being scored in a given two-minute stretch in the last several years is 8.5%, at 5v4 the chance rises to 24.0%, and at 4v5 it falls to 3.3%. Thus drawing a penalty has an offensive value of 24.0% - 8.5% = +0.155 goals for and a defensive value of 3.3% - 8.5% = -0.052 goals against, for a net benefit of 0.207 goals; similarly, taking a penalty has a net detriment of 0.207 goals. The outputs of my penalty model are impacts on team penalties drawn and taken, per thousand all-situations minutes; multiplying these estimates by 0.207 and then by 1200/1000, to stretch the impact to our 1200 minutes, gives the estimates of sG for drawing and taking penalties. For Crosby, whose presence on the ice is associated with the Penguins both taking and drawing slightly more penalties than average, the sG values are +0.7 (drawing) and -0.3 (taking).
## Synergy
For the moment I've kept all of the individual contributions to sG strictly separated—so, for instance, a player who creates offence and who shoots well gets credit for both of those things, but in fact such a player has an even larger positive impact on their team's scoring because some of the extra shots that they create will be taken by them. I've decided not to include these "synergy" effects at the moment, for reasons of complexity, but I might add them in the future.
## Future Skills
As long as skills are reasonably independent of one another, they can be added to a player's sG. I have one model in preparation that I mean to add, measuring a player's tendency to gain or lose territory in their shifts. In principle any number of things could be added as we learn more.
Similarly, sG values could be computed for goaltenders and for coaches, which I mean to do over the course of the coming months.
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# Bootstrapping standard normal data does not produce a mean of zero and standard deviation of 1
Please let me know my mistake. I expected bootstrapping a standard normal data produces a mean of zero and standard deviation of 1, but my example does not. rm(list=ls())
Create random data with a mean of zero and standard deviation of one
V1 <- rnorm(9, mean=0, sd=1); V2 <- rnorm(9, mean=0, sd=1); V3 <- rnorm(9, mean=0, sd=1)
V4 <- rnorm(9, mean=0, sd=1); V5 <- rnorm(9, mean=0, sd=1); V6 <- rnorm(9, mean=0, sd=1)
V7 <- rnorm(9, mean=0, sd=1); V8 <- rnorm(9, mean=0, sd=1)
rename the columns
colnames(charDataDiff) <- c("s380","s390","s400","s410","s420","s430","s440","s450"); charDataDiff
patchId <- c("C", "B", "A", "A", "B", "C", "A", "B", "C"); patchId
provide an ordering variable
idColor <- c("C", "B", "A")
Put all the pieces together to create the data
charDataDiff <- cbind.data.frame(patchId,charDataDiff); charDataDiff
Split the data by patchId
patchSpectrumBiasSplit <- split(charDataDiff, charDataDiff\$patchId)
Order the data by idColor
patchSpectrumBiasOrdered <- patchSpectrumBiasSplit[idColor]
Remove the first column
patchSpectrumBias <- lapply(patchSpectrumBiasOrdered, "[", 2:9)
Bootstrap
sampleOne <- function(x) x[sample(seq_len(nrow(x)), replace = TRUE), ]
sampleBoot <- function(x, n) replicate(n, sampleOne(x), simplify = FALSE)
applyMedian <- function(l) do.call(rbind, lapply(l, apply, 2, median))
k <- lapply(lapply(patchSpectrumBias, sampleBoot, n = 1000), applyMedian)
Calculate the mean, I expected the mean to be zero following the law of large numbers.
bootMeansBias <- do.call(rbind, lapply(k, apply, 2, mean));bootMeansBias
Calculate the standard deviation, I expected the standard deviation to be one
standardError <- do.call(rbind, lapply(k, apply, 2, sd));standardError
-
Crossposted at stackoverflow.com/questions/13532997/… – MattBagg Nov 23 '12 at 18:06
@Matt The cross-post now has been closed and at least one statistically interesting answer has been offered here, so this looks like the right place for this thread to remain. Thanks for flagging it! – whuber Nov 23 '12 at 20:54
You could never expect the mean to be exactly 0 if you were using the law of large numbers. You're not using large numbers in your sample, only your bootstrap. Your bootstrap will build up a representation out of the sample. You're just highlighting a problem in bootstrapping, that the results you get depend on the sample. So, a sample of 9 is very unlikely to have a mean of 0 and can deviate quite a bit from it. A sample of 3 is even worse. Bootstrapping won't fix this.
-
Thank all, I modified my code and realized my mistake. the new code is as follows:
rm(list=ls())
n <-2000
Create random data with a mean of zero and standard deviation of one
a <- lapply(1:8, function(x)cbind(rnorm(n)))
b <- as.data.frame(do.call(cbind,a ))
rename the columns
names(b) <- c("s380","s390","s400","s410","s420","s430","s440","s450")
Create a second random data
c <- lapply(1:8, function(x)cbind(rnorm(n)))
rename the columns
d <- as.data.frame(do.call(cbind, c))
names(d) <- c("s380","s390","s400","s410","s420","s430","s440","s450"); str(d)
Create a list of the above data
e <- list(b,d)
Bootstrap
sampleOne <- function(x) x[sample(seq_len(nrow(x)), replace = TRUE), ]
sampleBoot <- function(x, nn) replicate(n, sampleOne(x), simplify = FALSE)
applyMean <- function(l) do.call(rbind, lapply(l, apply, 2, mean))
k <- lapply(lapply(e, sampleBoot, nn = 100), applyMean)
calculate the mean, it is almost zero
bootMeansBias <- round(do.call(rbind, lapply(k, apply, 2, mean)),3); bootMeansBias
calculate the standard deviation, it is almost 1
standardError <- round(do.call(rbind, lapply(k, apply, 2, sd)),3);standardError
standardDeviation <- standardError*sqrt(n); standardDeviation
Thank you all
Ragy Isaac
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## Drake – Gods Plan
After dropping the music video to Gods Plan, Drake is sitting in 2nd place on Youtube’s trending videos. You have got to love Drake! His albums have brought us nothing but pure joy and in this video, you see exactly why we love Drake. #ForTheOVOLovers
## BAFTA 2018- Red Carpet To The Winners
The EE British Academy Film Awards took place last night and we spotted some beautiful and handsome looks on the red carpet and those that took home some awards. We’re in love with Katie’s look on the red carpet.
Below are some looks, let us know what you think.
## The Biggest Loser Challenge
Anyone remembers The Biggest Loser Challenge? of course you do! For those who never saw the show; “The show features obese or overweight contestants competing to win a cash prize by losing the highest percentage of weight relative to their initial weight” (Wikipedia)
A lot of people like the show, in fact some so much that they couldn’t help but start their own Biggest Loser challenge. We found this to be so much better than having a gym partner, because we are all competitive in nature and competitions usually push us to the limit.
We going to tell you all about it so you can grab friends, family and colleagues to join you.
### A few steps to get you started:
Step 1: Gather a group of friends; this can be your close circles or colleagues depending on who you are comfortable with.
Step 2: Set up rules – nothing is as annoying as rules made up along the way. You want to make sure that everyone is fair, and no one is cheating or disadvantaged.
Step 3: Set a price – everyone has to contribute towards the price, whether it’s winner takes all or there’s 1st, 2nd and 3rd place. This must happen before the challenge begins, people will put in as much effort as they have invested towards what they are doing.
Step 4: Record everyone’s weight; Record on the day you start.
Step 5: Let the games begin. Note down everyone’s weight daily and have them calculated at the end of each round. You want to review and give prices weekly, that way everyone is always pushing and competitive.
Here’s a formula you can use to calculate the percentage: (Remember the winner is determined by the weight percentage they’ve lost)
% = Weight lost / Original weight X 100
E.g. if Matilda weighed 80kg originally and now weighs 75kg (lost 5kg)
5/75 x 100 = 6.67%
The one with the highest percentage for round wins.
Who said keeping fit and losing weight doesn’t have to be fun?
## What are you wearing to Black Panther?
As we all know and have been anticipating the Black Panther movie will be on show today.
Everyone spoke about representing the Kingdom of Wakanda by dressing in their traditional attires. They all looked spectacular!! Talk about unity, what are you wearing to watch Black Panther? Here are some ideas for your looks below…
Courtesy of: www.fashionista.com
## Office Looks
Hey Ladies, we’re back again but this time we are hooking you up with Office looks.
We know that dressing for office is not as easy as it looks. Whether your office is casual and creative, or buttoned-up and sophisticated, get inspired and shop high style work-appropriate looks. It still takes time to get the best look. So below we have a few outfits that we think you should try out.
## The Body transformation we all need
It’s been a while since you started your weight loss program, the gym has been taking it’s tall and the scale is taking forever to move to the lower digits. If that’s you we have some inspiration for you, to help you go through this part of your journey. We decided to share some inspirational pictures, showing celebrities who you can relate to.
## Sophie Ndaba
Sophie is famously known for her role as queen in Generations on SABC1.
## Tumi Morake
She is currently the #WTFTumi host.
## Julius Malema
Julius “Juju” Malema the leader of the EFF and former ANC youth leader.
## Zolani Mahola
Who could forget the melodic Freshly Ground star Zolani Mahola.
## Unathi Msengana
Singer and Idols judge, her weight loss has to be the greatest transformation ever. She looks beautiful and she knows it!
## New York Fall Fashion Week 2018 – Run Way to Street Style
New York fashion week (which started on the 8th of Feb and ends on the 16th) has us all excited about the looks we’re seeing on the runway and out in the streets. From our favourite designers such as Micheal Kors to Prabal Gurung, each have amazing looks that stand out from their Fall collections 2018.
We’ve put together looks that we think you would love…comment below and let us know what you think?
Source: www.harpersbazaar.com
## Best Weekend Looks
Hello Raven Divas
We hope that the first week of February has been nothing but fabulous. To top it all off below we have some looks that you could possibly try out If are planning on going out with your friends, family or on a date!.
## The wisdom of ‘the other’
Success is seldom created in a vacuum, we are inspired and challenged by what we see around us. We are challenged to move when we have seen others move, and when we have seen that its possible to move. This makes us intentional about the people we surround ourselves with, who are they, what is it they carry, why should they be in our lives. How can we be changed and challenged when with them, do they motivate us to be better humans, or they are there is echo our fears & insecurities.
Here is some encouragement from one awesome Oprah Winfrey. She is one to keep as a source for encouragement, making you believe anything is possible. As she says, “You get in life what you have the courage to ask for”.
It’s not that I’ve always known who I would be. It was just very clear to me from an early age who I wouldn’t be.
The opportunities for a girl born black in Mississippi in 1954 were limited. You could teach in a segregated school. Or be a maid. A cook. A dishwasher. A servant. I never thought that would be the life for me.
I vividly remember standing on my grandmother’s small screened-in back porch, churning butter while she boiled clothes in a big black cast-iron pot in the yard. As she pulled the steaming clothes from the pot to hang on the line to dry, she called to me, “Oprah Gail, you better watch me now, ’cause one day you gon’ have to know how to do this for yourself.”
I did what she told me. I watched carefully as she pulled the clothespins from her apron, held them two at a time between her lips, and placed one and then the other on opposite ends of the sheets and towels and shirts and dresses she hung on the line.
A still, small voice inside me, really more a feeling than a voice, said, “This will not be your life. Your life will be more than hanging clothes on a line.”
The certainty of that divine assurance got me through many a difficult moment during my growing years.
I wanted to be a teacher. And to be known for inspiring my students to be more than they thought they could be. I never imagined it would be on TV.
I believe there’s a calling for all of us. I know that every human being has value and purpose. The real work of our lives is to become aware. And awakened. To answer the call.
## Black Panther World Premier
In case you missed the Black Panther World Premier last week, that took place in LA, we spotted some looks that were show-stopping and jaw-dropping. Who do you think was best dressed?
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# Chapter 4: Reaction and Exercises
## Alexis Nogelo
Reaction:
This chapter dealt primarily with drawing cubes, hypercubes and their shadows, as well as counting various elements of higher-dimensional cubes and simplexes such as edges, vertices, faces and n-cubes. There was also quite a thorough treatment of one application of our class's newly acquired methods of slicing and projection that is used by the Geological Sciences Department at Brown to study the climate changes over thousands of years in an area of the Midwestern United States. I enjoyed reading a more detailed analysis of how one can count all of the elements of higher-dimensional objects, since the mathematical beauty in the patterns is significant. I also appreciated the small reference to group theory, because I need those refreshers to remind me to consider the geometric connections to algebra that I have studied. (Or vice versa--algebraic elaborations on geometry I am studying.)
I do have one request to ask of Professor Banchoff--could we view some more movies in class? I would like to see The Hypercube: Projections and Slicing as well as some more of your world-famous geometric films.
Exercises:
Familiarize yourself with counting using combinations. (a)In how many ways can a teacher arrange 28 students into groups with 4 students per group? Suppose there are 12 girls and 16 boys in the class. (b)In how many ways can the teacher form 4 groups with 3 girls and 4 boys in each group? (c)In how many ways can just one group of 4 students be formed in a class of 28? (d)How about the teacher's options for the first of the coed groups of 7 students?
I know that we are to do our own exercises, but I was losing confidence in my solutions to these problems and would like to check them outside of the CIT where the air is more clear and I will be less likely to make a foolish mistake. I will add my answers A.S.A.P.
Alexis Nogelo
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Decision statement problems erfanakbari1 Silly Frenchman Posts: 20 Threads: 14 Joined: Sep 2018 Reputation: 0 Likes received: 0 #1 Mar-04-2019, 07:10 PM (This post was last modified: Mar-04-2019, 08:21 PM by erfanakbari1. Edited 1 time in total.) Hey guys , I'm trying to solve some exercises about digits and string and whole decision statement and I don't know actually how to check whether a input that is been taken by user consists digits or strings or both! inptUser = input('Please enter a variable : ') print(inptUser) for x in inptUser: if x.isdigit(): print(' The input is full of digits ') This is my code but I don't know what's wrong with it . I appreciate if you can help me with this . Thank you ichabod801 Number Four Posts: 4,103 Threads: 90 Joined: Sep 2016 Reputation: 254 Likes received: 1224 #2 Mar-04-2019, 08:44 PM The value of inptUser is a string. Looping over a string, as you do on line 4, loops over each character in the string. So you are checking each digit of the string, and for each one that is a digit, you are printing that the input is all digits. Note that isdigit will check to see if a string is all digits: it will check each character by itself. However, it would return False if you only have some digits, like 'ichabod801', so it's not good for checking if there are any digits in the string. It's not clear to me exactly what you supposed to do for this exercise, so I'm not sure how I would correct your code. Craig "Ichabod" O'Brien - xenomind.com I wish you happiness. Recommended Tutorials: BBCode, functions, classes, text adventures woooee Minister of Silly Walks Posts: 453 Threads: 0 Joined: Feb 2018 Reputation: 33 Likes received: 36 #3 Mar-06-2019, 12:27 AM (This post was last modified: Mar-06-2019, 12:28 AM by woooee. Edited 1 time in total.) You want to do something when it is not a digit def is_it_digit(inptUser): for x in inptUser: if not x.isdigit(): print('a not digit found') return False print(' The input is full of digits ') return True inptUser = input('Please enter a variable : ') print(inptUser) print(is_it_digit(inptUser)) And when you have some time, take a look at the Python Style Guide; variables are lower_case_with_underscores https://www.python.org/dev/peps/pep-0008/ « Next Oldest | Next Newest »
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Marginal Revenue and Marginal Cost
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Module 1: Understanding Fixed, Variable and Marginal Costs in Economics
# Marginal Revenue and Marginal Cost
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Karam A. 0 0 here i would like to make a relation ship between Marginal Revenue and Marginal Cost,,, The marginal cost of production measures the change in total cost of a good that arises from producing one additional unit of that good. The marginal cost is calculated by dividing the change in the total cost by the change in quantity. Using calculus, the marginal cost is calculated by taking the first derivative of the total cost function with respect to the quantity: MC = dTC/dQ. For example, the total cost of producing 100 units of a good is \$200. The total cost of producing 101 units is \$204. The average cost of producing 100 units is \$2, or \$200/100; however, the marginal cost for producing the 101st unit is \$4, or (\$204 - \$200)/(101-100). The marginal revenue measures the change in the revenue that arises when one additional unit of a product is sold. The marginal revenue is calculated by dividing the change in the total revenue by the change in the quantity. In calculus terms, the marginal revenue is the first derivative of the total revenue function with respect to the quantity: MR = dTR/dQ. For example, suppose the price of a product is \$10 and a company produces 20 units per day. The total revenue is calculated by multiplying the price by the quantity produced. In this case, the total revenue is \$200, or \$10*20. The total revenue from producing 21 units is \$205. The marginal revenue is calculated as \$5, or (\$205 - \$200)/(21-20). When marginal revenue and the marginal cost of production is equal, profit is maximized at that level of output and price. In terms of calculus, the relationship is stated as: dTR/dQ = dTC/dQ When marginal revenue is less than the marginal cost of production, a company is producing too much and should decrease its quantity supplied until marginal revenue equals the marginal cost of production. When the marginal revenue is greater than the marginal cost, the firm is not producing enough goods and should increase its output until profit is maximized. Read more: How is marginal revenue related to the marginal cost of production? | Investopedia http://www.investopedia.com/ask/answers/041315/how-marginal-revenue-related-marginal-cost-production.asp#ixzz3vHDSNM4E Follow us: Investopedia on Facebook
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# Logical Operators and Functions Matlab Help
Logical Operators and Functions
MATLAB has five logical operators, which are sometimes called Boolean operators (see Table 4.3-1). These operators perform element-by-element operations. With the exception of the NOT operator (-), they have a lower precedence than the arithmetic and relational operators (see Table 4.3-2). The NOT symbol is called the tilde. The NOT operation -A returns an array of the same dimension as A; the new array has ones where A is zero and zeros where A is nonzero. If A is logical, then -A replaces ones with zeros and zeros with ones. For example
Logical operators
if X = [0,3,9] and y = [14,-2,9], then z = -x returns the array z = [1,0,0] and the statement u == -x > y returns the result u = [0,1,0]. This expression is equivalent to u = ( – x) > y, whereas v = – (x > y) gives the result v = [1, °,1] .This expression is equivalent to v = (x <= y). The & and I operators compare two arrays of the same dimension. The only exception, as with the relational operators, is that an array can be compared to a scalar. The AND operation A&B returns ones where both A and B have nonzero elements and zeros where any element of A or B is zero. The expression z = 0&3 returns z = 0; z = 2&3 returns z = 1; z = 0&0 returns z = O,and z = [5,-3,0,0]&[2,4,0·,5] returns z = [l,l,O,O].Because of operator precedence, z =’ 1& 2 +3 is equivalent to z = 1& ( 2 +3 ) • which returns z = 1. Similarly, z = 5<6&1 is equivalent to z (5<6) &1,\
which returns z = 1.
Let x = [ 6 , 3 , 9] and y = [ 14 , 2 , 9] and let a = [ 4 , 3 , 12 ]. The expression
z = (x>y) & a
gives z = [0′, 1, 0], and
z = (x>y}&(x>a)
returns the result z [‘0 , °,°J. This is equivalent to
z = x>y&x>a
Be careful when using the logical operators with inequalities. For example, note that – (x .> y) is equivalent to x <= y. It is not equivalent to x < y. As another example, the relation 5 < x < 10 must be written as
(5 < x) & (x < 10)
in MATLAB.
The OR operation A&B returns ones where at least one of A and B has nonzero elements and zeros where both A and Bare zero. The expression z = 013 returns z = 1; the expression z = °10 returns z = 0; and
z = [5,-3,0,0] I [2,4,0,5]
returns z = [ 1 , 1 , °,1] . Because of operator precedence,
z = 3<514==7
is equivalent to
z =(3<5)1(4==7)
which returns z = 1. Similarly, Z = 110&11s equivalent to Z =~ (110&1,
which returns z = 1, while z = 110&0 returns z = 0, and z = 0&011 returns z = 1.
Because of the precedence of the NOT operator, the statement
z = -3==714==6
.returns the result z = 0, which is equivalent to
– ((-3)==7)1(4==6)
The exclusive OR function xor (A, B) returns zeros where A and B are either both nonzero or both zero, and ones where either A or B is nonzero, but not both.The function is defined in terms of the AND, OR, and NOT operators as follows.
function z = xor(A,B)
z = (A&B) & -(A&B);
The expression
z = xor ( [3,0,6], [5, 0, 0] )
returns z = [0, 0, 1], whereas
z = [3,0,6]1 [5,0,0]
returns z = [ 1 , a r 1 ]
Table 4.3-3 is a so-called truth table that defines the operations of the logical operators and the function xor. Until you acquire more experience with the logical operators, you should use this table to check your statements. remember that true is equivalent to logical I, and false is equivalent to logical O. We can test the truth table .by building its numerical equivalent as follows. Let x and y represent the first two columns of the truth table in terms of ones and zeros.
Truth table
The following MATLAB session generates the truth table in terms of ones and zeros.
»x = [1,1,0,0]’;
»y = [1,0,1, 0] ,;
»Truth_Table = [x,y,-x,xly,x&y,xor(x,y)]
Truth_Table .=
110 110
100 1 ° 1
011 1 ° 1
001 ° ° °
Starting with MATLAB 6, the AND operator (s) was given a higher precedence than the OR operator. ( I). This was not true in earlier versions of MATLAB, so if you are using code created in an earlier version, you should make the necessary changes before using it in MATLAB 6 or higher. For example, now the statement y = 1 I 5& 0 is evaluated as y = 1 I (5 & 0 ) , yielding the result y = 1, whereas in MA1LAB 5.3 and earlier, the statement would have been evaluated as y = (115) &0, yielding the result y = O. To avoid potential problems due to precedence, it is important to use parentheses in statements containing arithmetic, relational, or logical operators, even where parentheses are optional. MATLAB now provides a feature to enable the system to produce either an error message or a warning for any expression containing & and I that would be evaluated differently than in earlier versions. If you do not use this feature, MATLAB will issue a warning as the default. To activate the error feature, type feature (‘OrAndError’ ,1) .To reinstate the default, type feature(‘OrAndError’ ,O).
Short-Circuit Operators
The following operators perform AND and OR operations on logical expressions -containing scalar values only. They are called short-circuit operators because they evaluate their second operand only 1 then the result is not fully determined by the first operand. They are defined as follows in terms of the two logical variables A
and B.
A&&B Returns true logical if both ,”\and B evaluate to true, and false (logical 0) if they do not.
A I IB Returns true (logical I) l’ either A or B, or both, evaluate to true, and false (logical 0) if they do not.
Thus in the statement A&&B, if A equals logical zero, then the entire expression will evaluate to false. regardless of the value of B, and therefore there IS no need to evaluate B.
For AI I B. if A is true, regardless of the value of B, the statement will evaluate to true.
Logical functions
Table 4.3-4lists several useful logical functions. You learned about the find function in Chapter 1
Logical Operators and the find Function
The find function is very useful for creating decision-making programs, especially when combined with the relational or logical operators. The function find x)”computes an array containing the indices of the nonzero elements of the array x. We saw examples of its use with relational operators in Chapter 1. It is also useful when combined with the logical operators. For example, consider the session
»x = [5, -3, 0, 0, 8); y = [2,4,0,5,7);
»z = find (x&y)
z =
1 2 5
The resulting array z ,= [1, 2, 51 indicates that ‘the first. second. and fifth elements of x and y are both nonzero. Note that the find function returns the indices. and not the’values. In the following session. note the difference between the result obtained by y (X&y) and the result o,b~ned by find (x&y) above
»x = [5, -3, 0, 0, 8]; y = [2, 4, 0, 5, 7];
»values = y (x&y)
values =
2 4 7
» how_many = length. (vaIues)
how_many =
3
Thus there are three nonzero values in the, array y that correspond to nonzero values in the array x. They are the first, second, and fifth values, which are 2, 4,
and 7.
In the above example, there were only a few numbers in the arrays x and y, and thus we could have obtained the answers by visual inspection, However. these MATLAB methods are very useful either where there is so much data that visual inspection ‘would be very time-consuming, or where the values are generated internally in a program.
14.3-1 If x = [ 5, – 3, 18, 41 and y – [- 9, 13 , 7, 4]. what will be the
a. z =’ -y > x “/ ‘
b. z = x&y / .
C. Z = xly
d. z = xor(x,y)
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## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body
We start from the definition of the kinetic energy $T$,
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2,$$ and insert the expression (4.42) for the velocity:
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2+\sum_i m_i\left(\omega \times \mathbf{r}_i\right) \cdot \dot{\mathbf{r}}_0 .$$
The third term is a scalar triple product and can therefore be rewritten as follows:
$$\sum_i m_i \mathbf{r}_i \cdot\left(\dot{\mathbf{r}}_0 \times \omega\right)$$
There are two typical cases for the discussion of the rigid body:
One point of the body remains space-fixed, while the body rotates with the angular velocity $\omega$. Then it appears absolutely reasonable to choose this point as the origin $S$ of $\Sigma$ and in general also as the origin of $\widehat{\Sigma}$. One then speaks of a spinning top for which holds:
$$\mathbf{r}_0=\mathbf{0}, \quad \dot{\mathbf{r}}_0=\mathbf{0}$$
If no point is space-fixed one usually chooses the origin $S$ at the center of mass and that means:
$$\sum_i m_i \mathbf{r}i=\mathbf{0}$$ We see that these two cases, the only relevant ones, both let the third term in (4.43) disappear. We therefore apply from the beginning the kinetic energy in the form: $$T=\frac{1}{2} M \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2=T_T+T{\mathrm{R}}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Properties of the Inertial Tensor
Strictly speaking it is nothing other than a proper extension of the term ‘vector’. By a
tensor of $k$-th rank in an $n$-dimensional space
one understands an $n^k$ number of elements
$$\left(F_{i 1, i_2, \ldots, i_k}\right) ; \quad i_j=1, \ldots, n,$$
which for coordinate rotations transform linearly satisfying certain rules. The elements are called the components of the tensor. They carry $k$ indexes each of which runs from 1 to $n$. The rules are chosen just so that the ‘normal’ vectors are first-rank tensors. One requires that in connection with coordinate rotations a tensor of $k$-th rank transforms itself with respect to all $k$ indexes like a ‘normal’ vector. According to our underlying physical problems of course only the cases $n=1,2,3$ are interesting. Furthermore, in physics we can restrict ourselves to $k=0,1,2$.
$\mathbf{k}=\mathbf{0}:$ scalar: $\quad \bar{x}=x$
$\mathbf{k}=\mathbf{1}$ : vector, $n=3$ components (in the three-dimensional space), for which, according to (1.309), it holds after a coordinate rotation:
$$\bar{x}i=\sum_j d{i j} x_j$$
$\left(d_{i j}\right.$ : components of the rotation matrix (1.307)),
$\mathbf{k}=\mathbf{2}:\left(F_{i j}\right){i, j=1,2,3}: n^2=9$ components with $$\bar{F}{i j}=\sum_{l, m} d_{i l} d_{j m} F_{l m}$$
and so on.
Second-rank tensors can always be written as square matrices. However, in contrast to normal matrices which are represented by collections of elements (numbers), which may behave arbitrarily with coordinate transformations, the above-mentioned transformation behavior is absolutely mandatory for the elements of a tensor.
Why is it necessary that the system of coefficients (4.47) does exhibit tensor properties? The components of the inertial tensor in a given system of coordinates are uniquely determined by the mass distribution of the rigid body. But with a rotation of the system of coordinates the components will change. Furthermore, of course also the components of the angular velocity $\omega$ will undergo a change. However, it is clear that a rotation of the coordinate system should not influence the (measurable) rotational kinetic energy $T_{\mathrm{R}}$.
# 理论力学代写
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinetic Energy of the Rigid Body
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2$$
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2+\sum_i m_i\left(\omega \times \mathbf{r}_i\right) \cdot \dot{\mathbf{r}}_0 .$$
$$\sum_i m_i \mathbf{r}_i \cdot\left(\dot{\mathbf{r}}_0 \times \omega\right)$$
$$\mathbf{r}_0=\mathbf{0}, \quad \dot{\mathbf{r}}_0=\mathbf{0}$$
$$\sum_i m_i \mathbf{r} i=\mathbf{0}$$
$$T=\frac{1}{2} M \dot{\mathbf{r}}_0^2+\frac{1}{2} \sum_i m_i\left(\omega \times \mathbf{r}_i\right)^2=T_T+T \mathrm{R}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Properties of the Inertial Tensor
$$\left(F_{i 1, i_2, \ldots, i_k}\right) ; \quad i_j=1, \ldots, n$$
$\mathbf{k}=\mathbf{0}$ :标量: $\quad \bar{x}=x$
$\mathbf{k}=\mathbf{1}$ : 矢量， $n=3$ 分量（在三维空间中），根据（1.309)，它在坐标旋转后成立:
$$\bar{x} i=\sum_j d i j x_j$$
$\left(d_{i j}:\right.$ 旋转矩阵的分量 (1.307)),
$\mathbf{k}=\mathbf{2}:\left(F_{i j}\right) i, j=1,2,3: n^2=9$ 组件与
$$\bar{F} i j=\sum_{l, m} d_{i l} d_{j m} F_{l m}$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion
As a further important example of a rigid body with only one rotational degree of freedom we consider the
homogeneous cylinder rolling off an inclined plane
Though the rotation axis is again body-fixed it is not space-fixed. It is shifting in parallel to itself (Fig. 4.11). The velocity of each of the cylinder points is composed by two contributions, a rotational contribution due to the rotation around the cylinder axis during the rolling motion and a translational contribution which is the same for all points of the cylinder and happens in $s$ direction:
$$\dot{\mathbf{r}}i=\dot{\mathbf{r}}{i R}+\dot{\mathbf{r}}{i T}$$ The rotational contribution we have already calculated in (4.8): $$\dot{\mathbf{r}}{i R}=\left(\omega \times \overline{\mathbf{r}}i\right)$$ The translational contribution is obtained from the rolling off condition $$\Delta s=R \Delta \varphi \Longrightarrow\left|\dot{\mathbf{r}}{i T}\right|=|\dot{\mathbf{s}}|=R|\dot{\varphi}|$$
The cylinder shall roll, not slide.
(a) Kinetic Energy
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2=\frac{1}{2} \sum_i m_i\left[\left(\omega \times \overline{\mathbf{r}}_i\right)^2+2 \dot{\mathbf{s}} \cdot\left(\omega \times \overline{\mathbf{r}}_i\right)+\dot{s}^2\right]$$
The mixed term disappears because in a homogeneous cylinder two volume elements located diametrally opposite to the rotation axis have the same mass but rotation velocities are in opposite directions (Fig. 4.12). The sum over all elements is therefore zero. It can of course be shown also by a direct calculation that
$$\sum_i m_i\left(\omega \times \overline{\mathbf{r}}_i\right)=0$$
must hold.
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinematics of the Rigid Body
In our introductory Sect. 4.1 we had already decomposed the general motion of a rigid body into
1. the translation of an arbitrarily chosen point $S$ of the body and
2. the rotation around an axis through this point $S$.
We now introduce two reference systems which are initially both Cartesian:
$\widehat{\Sigma}$ : space-fixed reference system with a space-fixed origin of coordinates $\mathcal{O}$. It is assumed to be an inertial system. Axis : $\hat{\mathbf{e}}\alpha, \alpha=1,2,3$. $\Sigma$ : body-fixed reference system with the body-fixed origin $S$. Axes: $\mathbf{e}\alpha(t), \alpha=$ $1,2,3$
The point $S$ has the position vector $\mathbf{r}0(t)$ as seen from $\widehat{\Sigma}$. Then it holds for the points of the rigid body: $$\begin{array}{ll} \hat{\mathbf{r}}_i(t)=\sum{\alpha=1}^3 \hat{x}{i \alpha}(t) \hat{\mathbf{e}}\alpha & (\text { in } \widehat{\Sigma}), \ \mathbf{r}i(t)=\sum{\alpha=1}^3 x_{i \alpha} \mathbf{e}\alpha(t) & \text { (in } \Sigma) \end{array}$$ with the obvious relation: $$\hat{\mathbf{r}}_i(t)=\mathbf{r}_0(t)+\mathbf{r}_i(t)$$ The coordinates $x{i \alpha}$ in the body-fixed system $\Sigma$ are by the definition of the rigid body time-independent quantities. The position of the rigid body is therewith completely given by the position of $\Sigma$ relative to $\widehat{\Sigma}$.
We are now interested in the velocities of the mass points of the rigid body (Fig. 4.13). These we find rather easily with the general theory of arbitrarily relative to each other moving reference systems that we derived in Sect. 2.2.5. The full time derivative of a vector represented in $\Sigma$ seen from $\widehat{\Sigma}$ can be written as the operator
The first term on the right-hand side plays by definition no role for the rigid body. Thus it remains:
$$\dot{\mathbf{r}}_i=\left(\omega \times \mathbf{r}_i\right)$$
or with (4.40):
$$\dot{\hat{\mathbf{r}}}_i(t)=\dot{\mathbf{r}}_0(t)+\left(\omega \times \mathbf{r}_i\right) .$$
This is an important result. It signifies that at any moment of time the motion of a rigid body can be resolved into the translational motion $\mathbf{r}_0(t)$ of the origin of the body-fixed system and the rotation around the momentary rotation axis $\omega(t)$ where the latter always passes through the origin $S$ of the body-fixed system.
# 理论力学代写
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rolling Motion
，尽管旋转轴同样是物体固定的，但它不是空间固定的。它与自身平行移动（图 4.11)。每个圆柱点的速 度由两个贡献组成，一个是由于在滚动运动期间绕圆柱轴的旋转引起的旋转贡献，另一个是对圆柱的所有 点都相同的平移贡献，并且发生在 $s$ 方向:
$$\dot{\mathbf{r}} i=\dot{\mathbf{r}} i R+\dot{\mathbf{r}} i T$$
$$\dot{\mathbf{r}} i R=(\omega \times \overline{\mathbf{r}} i)$$
$$\Delta s=R \Delta \varphi \Longrightarrow|\dot{\mathbf{r}} i T|=|\dot{\mathbf{s}}|=R|\dot{\varphi}|$$
(a) 动能
$$T=\frac{1}{2} \sum_i m_i \dot{\mathbf{r}}_i^2=\frac{1}{2} \sum_i m_i\left[\left(\omega \times \overline{\mathbf{r}}_i\right)^2+2 \dot{\mathbf{s}} \cdot\left(\omega \times \overline{\mathbf{r}}_i\right)+\dot{s}^2\right]$$
$$\sum_i m_i\left(\omega \times \overline{\mathbf{r}}_i\right)=0$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Kinematics of the Rigid Body
1. 任意选择点的平移 $S$ 身体的和
2. 通过该点绕轴旋转 $S$.
我们现在介绍两个最初都是笛卡尔的参考系统:
$\widehat{\Sigma}$ : 具有空间固定坐标原点的空间固定参考系 $\mathcal{O}$. 假设它是一个惯性系统。轴: $\hat{e} \alpha, \alpha=1,2,3 . \Sigma$ : 具有身体固定原点的身体固定参考系 $S$. 轴: $\mathrm{e} \alpha(t), \alpha=1,2,3$
重点 $S$ 有位置向量 $\mathbf{r} 0(t)$ 从 $\widehat{\Sigma}$. 然后它适用于刚体的点:
$$\left.\hat{\mathbf{r}}i(t)=\sum \alpha=1^3 \hat{x} i \alpha(t) \hat{\mathbf{e}} \alpha \quad(\text { in } \widehat{\Sigma}), \mathbf{r} i(t)=\sum \alpha=1^3 x{i \alpha} \mathbf{e} \alpha(t) \quad \text { (in } \Sigma\right)$$
具有明显的关系：
$$\hat{\mathbf{r}}_i(t)=\mathbf{r}_0(t)+\mathbf{r}_i(t)$$
我们现在对刚体质点的速度感兴趣（图 4.13) 。这些我们很容易通过我们在第 1 节中导出的任意相对于彼
根据定义，右侧的第一项对刚体没有任何作用。因此它仍然是:
$$\dot{\mathbf{r}}_i=\left(\omega \times \mathbf{r}_i\right)$$
或 $(4.40)$ :
$$\dot{\hat{\mathbf{r}}}_i(t)=\dot{\mathbf{r}}_0(t)+\left(\omega \times \mathbf{r}_i\right)$$
这是一个重要的结果。它表示在任何时刻，刚体的运动都可以分解为平移运动 $\mathbf{r}_0(t)$ 身体固定系统的原点 和绕瞬时旋转轴的旋转 $\omega(t)$ 后者总是通过原点 $S$ 身体固定系统。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Central Forces
A force-type of the profile
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{e}_r$$
is called a ‘central force’. Thus the force is directed along the radial rays which start from the center (origin) of the force (Fig. 2.45). For such forces the angular momentum $\mathbf{L}$, if referred to the force center, is constant, according to (2.245). Central forces in the general form (2.248) are not necessarily conservative. It rather holds:
Central force $\mathbf{F}$ conservative $\Longleftrightarrow \mathbf{F}=f(r) \mathbf{e}_r$.
It is clear that $\mathbf{F}$ must not depend on $\dot{\mathbf{r}}$ and $t$ to be conservative. For a proof of (2.249) we therefore can restrict ourselves to forces $\mathbf{F}$ of the form:
$$\mathbf{F}=f(\mathbf{r}) \mathbf{e}_r .$$
According to (2.234) the force $\mathbf{F}$ is conservative if and only if the curl of $\mathbf{F}$ vanishes. That we inspect with (1.289):
$$\nabla \times \mathbf{F}=\frac{f(\mathbf{r})}{r} \nabla \times \mathbf{r}+\left[\left(\nabla \frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] .$$
After (1.292) we can exploit $\nabla \times \mathbf{r}=0$, so it remains to require:
$$0 \stackrel{!}{=}\left[\nabla\left(\frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] \text {. }$$
Hence the two vectors in the square bracket have to be parallel. In view of (1.271) and the subsequent discussion one realizes that the gradient vector is orthogonal to the planes $f(\mathbf{r}) / r=$ const. Hence these planes must simultaneously be orthogonal to $\mathbf{r}$. That means, however, that $f(\mathbf{r}) / r$ has to be constant on the surface of a sphere. This is possible only if $f(\mathbf{r})=f(r)$. That proves (2.249)!
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Integration of the Equations of Motion
If the law of conservation of angular-momentum
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=\text { const }$$
or the energy conservation law
$$E=\frac{m}{2} \dot{\mathbf{r}}^2+V(\mathbf{r})=\mathrm{const}$$
are valid then one speaks of
first integrals of motion
The original equations of motion are always differential equations of second order, the conservation laws, on the other hand, are only of first order. Furthermore, on the basis of the conservation laws a general procedure for the complete solution of the equations of motion can be developed.
We have shown that the angular-momentum conservation law holds if and only if the acting force is a central force:
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{r}$$
(The trivial case $\mathbf{F} \equiv 0$ shall be excluded!)
If simultaneously the energy conservation law holds then definitely a potential must exist. Hence, the central force is conservative and must be of the form:
$$\mathbf{F}=f(r) \mathbf{r} .$$
Moreover we know that in such a case the potential can depend only on the magnitude of $\mathbf{r}$ :
$$V=V(r) .$$
Therewith we will further evaluate the conservation laws. Because of the constancy of the angular momentum the motion will happen in a fixed plane. Let this be the $x y$ plane.
# 理论力学代写
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Central Forces
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{e}_r$$
$$\mathbf{F}=f(\mathbf{r}) \mathbf{e}_r .$$
$$\nabla \times \mathbf{F}=\frac{f(\mathbf{r})}{r} \nabla \times \mathbf{r}+\left[\left(\nabla \frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] .$$
$$0 \stackrel{!}{=}\left[\nabla\left(\frac{f(\mathbf{r})}{r}\right) \times \mathbf{r}\right] \text {. }$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Integration of the Equations of Motion
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=\text { const }$$
$$E=\frac{m}{2} \dot{\mathbf{r}}^2+V(\mathbf{r})=\text { const }$$
$$\mathbf{F}=f(\mathbf{r}, \dot{\mathbf{r}}, t) \mathbf{r}$$
（琐碎的情况 $\mathbf{F} \equiv 0$ 应被排除!）
$$\mathbf{F}=f(r) \mathbf{r} .$$
$$V=V(r) .$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYC90007
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Angular Momentum and Torque
If we multiply the basic dynamical equation (2.43) vectorially by $\mathbf{r}$,
$$m(\mathbf{r} \times \ddot{\mathbf{r}})=(\mathbf{r} \times \mathbf{F}),$$ then there appears on the left-hand side the time-derivative of an important physical quantity:
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=(\mathbf{r} \times \mathbf{p}) \quad \text { angular momentum } .$$
Since both position $\mathbf{r}$ and momentum $\mathbf{p}$ are polar vectors the resulting $\mathbf{L}$ must be an axial vector oriented perpendicularly to the plane spanned by $\mathbf{r}$ and $\mathbf{p}$. With the further definition,
$$\mathbf{M}=(\mathbf{r} \times \mathbf{F}) \quad \text { torque (moment) },$$
it follows from (2.241):
$$\frac{d}{d t} \mathbf{L}=\mathbf{M} .$$
This equation represents the angular-momentum law:
The time rate of the change of angular momentum is equal to the applied torque. If the torque is identical to zero then this theorem becomes the
Law of Conservation of Angular-Momentum
$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Law of Conservation of Angular-Momentum
$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$
There are two possibilities for getting $\mathbf{M}=\mathbf{0}$ :
1) $\mathbf{F} \equiv \mathbf{0} \quad$ (trivial case) ,
2) $\mathbf{F} \uparrow \uparrow \mathbf{r} \quad$ (central field) .
$(2.246)$
Case (1) is identical to the uniform straight-line motion of the mass point:
$$\dot{\mathbf{r}}=\mathbf{v}=\text { const . }$$
At first glance it appears astonishing that a uniform straight-line movement possesses any, even if constant, angular momentum. In Fig. $2.44 \mathbf{L}$ is perpendicular to the plane of the paper with the magnitude $m v d$. Only if the reference point (origin of coordinates) lies on the straight line then $\mathbf{L}$ indeed disappears. That gives evidence that the angular momentum is not at all a genuine particle property, but rather depends on the choice of the reference point.
A shift of the origin of coordinates by the constant vector a,
$$\mathbf{r}^{\prime}=\mathbf{r}+\mathbf{a} ; \quad \dot{\mathbf{r}}^{\prime}=\dot{\mathbf{r}} \Longrightarrow \mathbf{p}^{\prime}=\mathbf{p}$$
means for the angular momentum:
$$\mathbf{L}^{\prime}=\left(\mathbf{r}^{\prime} \times \mathbf{p}^{\prime}\right)=(\mathbf{r} \times \mathbf{p})+(\mathbf{a} \times \mathbf{p})=\mathbf{L}+(\mathbf{a} \times \mathbf{p}) .$$
If $\mathbf{L}$ is constant then $\mathbf{L}^{\prime}$ is also constant only if simultaneously the conservation of momentum also holds $\mathbf{p}=$ const. Furthermore, it does not necessarily follow from $\mathbf{L}=\mathbf{0}$ that also $\mathbf{L}^{\prime}=\mathbf{0}$. In general that is indeed not the case.
The second possibility for $\mathbf{M}=\mathbf{0}$ in (2.246) shall be discussed in a separate section.
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Angular Momentum and Torque
$$m(\mathbf{r} \times \ddot{\mathbf{r}})=(\mathbf{r} \times \mathbf{F}),$$
$$\mathbf{L}=m(\mathbf{r} \times \dot{\mathbf{r}})=(\mathbf{r} \times \mathbf{p}) \quad \text { angular momentum } .$$
$$\mathbf{M}=(\mathbf{r} \times \mathbf{F}) \quad \text { torque }(\text { moment }),$$
$$\frac{d}{d t} \mathbf{L}=\mathbf{M}$$
$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Law of Conservation of Angular-Momentum
$$\mathbf{M}=0 \Longleftrightarrow \frac{d}{d t} \mathbf{L}=0 ; \quad \mathbf{L}=\text { const }$$
$$\mathbf{r}^{\prime}=\mathbf{r}+\mathbf{a} ; \quad \dot{\mathbf{r}}^{\prime}=\dot{\mathbf{r}} \Longrightarrow \mathbf{p}^{\prime}=\mathbf{p}$$
$$\mathbf{L}^{\prime}=\left(\mathbf{r}^{\prime} \times \mathbf{p}^{\prime}\right)=(\mathbf{r} \times \mathbf{p})+(\mathbf{a} \times \mathbf{p})=\mathbf{L}+(\mathbf{a} \times \mathbf{p})$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS3020
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Classical Particle Paths
Our very general considerations already permit us to draw far reaching conclusions about possible particle paths. Since $T$ is non-negative it follows from (2.213):
classically allowed region of motion : $E \geq V(x)$,
(2.215)
classically forbidden region of motion: $E<V(x)$,
(2.216)
classical turning points : $E=V(x)$.
The supplement classical is important since the above statement has to be commented on when dealing with the all-embracing quantum theory.
Examples
(a) Harmonic oscillator:
Because of (2.215) it is to be expected that an oscillatory motion takes place between the two turning points $\pm x_0$. The distance between $E=E_0$ and $V(x)$ is a measure for the velocity of the mass point (Fig. 2.35). At the turning points the velocity of the particle is zero. The direction of motion reverses.
(b) General state dependence of potential:
For $x \leq x_1$ no movement is possible, and so is the case between $x_2$ and $x_3$, also. Between $x_1$ and $x_2$ an oscillatory behavior takes place, whilst a particle coming from $+\infty$ is reflected at $x_3$ (Fig. 2.36). Possible equilibrium positions of the particle are those points where no forces act. Obviously these are the extremal values of the potential $V$ :
$$F=0=-\frac{d V}{d x} \Longleftrightarrow V \text { extremal } .$$
In case of a maximum the particle is in an unstable equilibrium. The smallest position change lets it fall down the potential wall. In case of a minimum the particle finds itself in a stable equilibrium.
Finally we add a remark about the dimension, which is the same for $T, W, V$ and $E$ :
$$[E]=k g m^2 s^{-2}=\text { Joule . }$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Work, Power, and Energy
We start with the term ‘work’ which has to be generalized for arbitrary force fields,
$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t)$$
in analogy and compared to (2.204). To produce an infinitesimal displacement $d \mathbf{r}$ the work
$$\delta W=-\mathbf{F} \cdot d \mathbf{r}$$
has to be invested. The sign convention is the same as explained after (2.205). The symbol $\delta$ is chosen consciously since this expression does not necessarily represent a total differential as we will see in the following. Here it merely denotes an infinitesimally small quantity.
For finite pathways (Fig. 2.40) it holds:
$$W_{21}=-\int_{P_1}^{P_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot d \mathbf{r} .$$
This quantity normally depends on:
1) force field $\mathbf{F}$,
2) endpoints $P_1, P_2$,
3) path $C$,
4) temporal course of movement.
If $\mathbf{F}=\mathbf{F}(\mathbf{r})$ then of course point 4) becomes meaningless, i.e. $W_{21}$ depends only on the shape of the path and no longer on the temporal course of motion of the mass point along the trajectory. The integration in (2.220) represents a socalled curvilinear (line) integral. One evaluates such line integrals by tracing them back, in some way, to normal Riemann-integrals. That can be done with the parametrization of the space curve $C$ introduced in Sect. 1.4.1 (Fig. 2.41). The parameter $\alpha$ can but need not necessarily be the time $t$ :
$$\begin{gathered} C: \mathbf{r}=\mathbf{r}(\alpha) ; \quad \alpha_1 \leq \alpha \leq \alpha_2 ; \ d \mathbf{r}=\frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha . \end{gathered}$$
Therewith Eq. (2.220) can also be written as follows:
$$W_{21}=-\int_{\alpha_1}^{\alpha_2} \mathbf{F}\left(\mathbf{r}_{,} \dot{\mathbf{r}}, t\right) \cdot \frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Classical Particle Paths
(2.215)
(2.216)
(a) 谐振子:
(b) 势能的一般状态依赖性:
$$F=0=-\frac{d V}{d x} \Longleftrightarrow V \text { extremal } .$$
$$[E]=k \mathrm{~km}^2 \mathrm{~s}^{-2}=\text { Joule } .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Work, Power, and Energy
$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t)$$
$$\delta W=-\mathbf{F} \cdot d \mathbf{r}$$
$$W_{21}=-\int_{P_1}^{P_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot d \mathbf{r} .$$
1) 力场 $\mathbf{F}$,
2) 端点 $P_1, P_2$ ，
3) 路径 $C$ ，
4) 时间运动过程。
$$C: \mathbf{r}=\mathbf{r}(\alpha) ; \quad \alpha_1 \leq \alpha \leq \alpha_2 ; d \mathbf{r}=\frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha$$
$$W_{21}=-\int_{\alpha_1}^{\alpha_2} \mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) \cdot \frac{d \mathbf{r}(\alpha)}{d \alpha} d \alpha$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYSICS2532
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Inverse Matrix
Definition 1.6.7 $A=\left(a_{i j}\right)$ is a given $(n \times n)$-matrix. Then one denotes as its inverse matrix
$$A^{-1}=\left(\left(a^{-1}\right){i j}\right)$$ just the $(n \times n)$-matrix, for which holds: $$A^{-1} A=A A^{-1}=\mathbb{1} .$$ Theorem 1.6.2 $A^{-1}$ exists only when $\operatorname{det} A \neq 0$. The elements are then found by: $$\left(a^{-1}\right){i j}=\frac{U_{j i}}{\operatorname{det} A} .$$
(Note the order of the indexes!)
Proof Let $\widehat{A}=\left(\alpha_{i j}=U_{j i}\right)$ be an $(n \times n)$-matrix. With the expansion theorems (1.327) and (1.332) we find:
\begin{aligned} &\operatorname{det} A=\sum_j a_{i j} U_{i j}=\sum_j a_{i j} \alpha_{j i}=(A \cdot \widehat{A}){i i}, \ &\operatorname{det} A=\sum_i a{i j} U_{i j}=\sum_i \alpha_{j i} a_{i j}=(\widehat{A} \cdot A){i j} . \end{aligned} The diagonal elements of the product matrices $A \cdot \widehat{A}$ and $\widehat{A} \cdot A$ are thus all identical to $\operatorname{det} A$. What about the non-diagonal elements? With (1.336) one finds: $$(A \cdot \widehat{A}){i j}=\sum_k a_{i k} \alpha_{k j}=\sum_k a_{i k} U_{j k}=0 \quad \text { for } i \neq j$$
It follows that $A \cdot \widehat{A}$ and $\widehat{A} \cdot A$ are diagonal matrices with
$$A \cdot \widehat{A}=\widehat{A} \cdot A=\operatorname{det} A \cdot \mathbb{1} \text {. }$$
With $\operatorname{det} A \neq 0$ and by comparison with (1.337) the theorem is proved:
$$\frac{\widehat{A}}{\operatorname{det} A}=A^{-1} \Longleftrightarrow \frac{U_{j i}}{\operatorname{det} A}=\left(a^{-1}\right)_{i j}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rotation Matrix
We remember the question which came up in connection with (1.321). Under what conditions an arbitrary matrix $D$ based in a given $\operatorname{CONS}\left{\mathbf{e}i\right}$ is a rotation matrix? At first it must satisfy the orthonormality relations (1.308) and (1.316): \begin{aligned} &\sum_m d{i m} d_{j m}=\delta_{i j}, \ &\sum_m d_{m i} d_{m j}=\delta_{i j} . \end{aligned}
What is more, the new basis system $\left{\overline{\mathbf{e}}{\mathbf{j}}\right}$ originating from the original system $\left{\mathbf{e}_i\right}$ by rotation shall again be a right-handed trihedron, i.e. (1.342) must also be valid for the $\overline{\mathbf{e}}{\mathbf{j}}$. That is not yet guaranteed by the conditions (1.308) and (1.316). For instance, if we replace in the $i$-th row of $D$ the $d_{i j}$ by $\left(-d_{i j}\right)$, the orthonormality relations will still be valid. On the other hand, however, according to (1.305) $\left{\overline{\mathbf{e}}_i\right}$ transfers into $\left(-\overline{\mathbf{e}}_i\right)$. Thus the right-handed trihedron becomes a left-handed one. However, we notice with (1.305):
\begin{aligned} \overline{\mathbf{e}}1 \cdot\left(\overline{\mathbf{e}}_2 \times \overline{\mathbf{e}}_3\right) &=\sum{m, n, p} d_{1 m} d_{2 n} d_{3 p} \mathbf{e}m \cdot\left(\mathbf{e}_n \times \mathbf{e}_p\right)=\ &=\sum{m, n, p} \varepsilon_{m n p} d_{1 m} d_{2 n} d_{3 p}=\operatorname{det} D \end{aligned}
That means that besides the orthonormality of rows and columns a rotation matrix $D$ still must fulfill:
$$\operatorname{det} D=1$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Inverse Matrix
$$A^{-1}=\left(\left(a^{-1}\right) i j\right)$$
$$A^{-1} A=A A^{-1}=1 .$$
$$\left(a^{-1}\right) i j=\frac{U_{j i}}{\operatorname{det} A} .$$
(注意索引的顺序!)
$\operatorname{det} A=\sum_j a_{i j} U_{i j}=\sum_j a_{i j} \alpha_{j i}=(A \cdot \widehat{A}) i i, \quad \operatorname{det} A=\sum_i a i j U_{i j}=\sum_i \alpha_{j i} a_{i j}=(\widehat{A} \cdot A) i j .$
$$(A \cdot \widehat{A}) i j=\sum_k a_{i k} \alpha_{k j}=\sum_k a_{i k} U_{j k}=0 \quad \text { for } i \neq j$$
$$A \cdot \widehat{A}=\widehat{A} \cdot A=\operatorname{det} A \cdot 1 .$$
$$\frac{\widehat{A}}{\operatorname{det} A}=A^{-1} \Longleftrightarrow \frac{U_{j i}}{\operatorname{det} A}=\left(a^{-1}\right)_{i j}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Rotation Matrix
loperatorname{CONS}\left{\mathbf{e}i\right } } \text { 是旋转矩阵? 首先它必须满足正交关系 (1.308) 和 (1.316) : }
$$\sum_m d i m d_{j m}=\delta_{i j}, \quad \sum_m d_{m i} d_{m j}=\delta_{i j} .$$
Veft{\overline{\mathbf{e}}_i\right } } \text { 䉽入 } ( – \overline { \mathbf { e } } _ { i } ) \text { . 因此右手三面体变成左手三面体。但是，我们注意到 (1.305): }
$$\overline{\mathbf{e}} 1 \cdot\left(\overline{\mathbf{e}}2 \times \overline{\mathbf{e}}_3\right)=\sum m, n, p d{1 m} d_{2 n} d_{3 p} \mathbf{e} m \cdot\left(\mathbf{e}n \times \mathbf{e}_p\right)=\quad \sum m, n, p \varepsilon{m n p} d_{1 m} d_{2 n} d_{3 p}=\operatorname{det} D$$
$\operatorname{det} D=1$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Calculation Rules for Matrices
Let us first agree upon what we want to understand as the sum of two matrices:
Definition 1.6.3 If $A=\left(a_{i j}\right), B=\left(b_{i j}\right)$ are two $(m \times n)$-matrices then the sum is given as the matrix $C=A+B=\left(c_{i j}\right)$ with the elements:
$$c_{i j}=a_{i j}+b_{i j}, \quad \forall i, j .$$
$C$ is again a $(m \times n)$-matrix.
Example
\begin{aligned} &A=\left(\begin{array}{lll} 6 & 3 & 0 \ 1 & 4 & 5 \end{array}\right) \ &B=\left(\begin{array}{lll} 1 & 3 & 5 \ 2 & 4 & 6 \end{array}\right) \Longrightarrow C=A+B=\left(\begin{array}{ccc} 7 & 6 & 5 \ 3 & 8 & 11 \end{array}\right) . \end{aligned}
The so defined addition is obviously commutative as well as associative.
The next step concerns the multiplication of a matrix by a real number:
Definition 1.6.4 If $A=\left(a_{i j}\right)$ is a $(m \times n)$-matrix then the matrix $\lambda A(\lambda \in \mathbb{R})$ is to understand as the $(m \times n)$-matrix:
$$\lambda A=\left(\lambda a_{i j}\right) .$$
Hence each matrix element is multiplied by $\lambda$
Example
$$3\left(\begin{array}{ccc} 5 & -3 & 1 \ 0 & 2 & -1 \end{array}\right)=\left(\begin{array}{ccc} 15 & -9 & 3 \ 0 & 6 & -3 \end{array}\right) \text {. }$$
We know from normal vectors, which represent nothing else than special matrices, namely $(n \times 1)$ – and $(1 \times n)$-matrices, respectively, that they can be multiplicatively connected in form of scalar products. That is generalized correspondingly for matrices.
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Transformation of Coordinates
Let $\Sigma, \bar{\Sigma}$ be two systems of coordinates specified by the orthonormal basis vectors (Fig. 1.72):
$\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ and $\overline{\mathbf{e}}_1, \overline{\mathbf{e}}_2, \overline{\mathbf{e}}_3$, respectively
Translations are relatively uninteresting. We therefore assume that the origins of $\Sigma$ and $\bar{\Sigma}$ coincide. Let us now consider an arbitrarily chosen position vector $\mathbf{r}$ :
$$\begin{array}{ll} \mathbf{r}=\left(x_1, x_2, x_3\right) \text { in } \Sigma & {[\mathbf{r}(\Sigma)]} \ \mathbf{r}=\left(\bar{x}_1, \bar{x}_2, \bar{x}_3\right) \text { in } \bar{\Sigma} & {[\mathbf{r}(\bar{\Sigma})]} \end{array}$$
Let us presume that the elements $x_i$ in $\Sigma$ are known while the elements $\bar{x}_j$ in $\bar{\Sigma}$ are to be determined. $\mathbf{r}$ itself is of course independent of the special choice of the system of coordinates, both with respect to direction as well as magnitude.
Therefore:
$$\sum_{j=1}^3 x_j \mathbf{e}j=\sum{j=1}^3 \bar{x}j \overline{\mathbf{e}}_j$$ The basis vectors $\overline{\mathbf{e}}_j$ can be represented in $\Sigma$ : $$\overline{\mathbf{e}}_j=\sum_k d{j k} \mathbf{e}k$$ We determine the expansion coefficients $d{j k}$ by scalar multiplication of this equation by $\mathbf{e}m$ : $$d{j m}=\overline{\mathbf{e}}j \cdot \mathbf{e}_m=\cos \varphi{j m}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Calculation Rules for Matrices
$$c_{i j}=a_{i j}+b_{i j}, \quad \forall i, j .$$
$C$ 又是一个 $(m \times n)$-矩阵。
$$\lambda A=\left(\lambda a_{i j}\right) .$$
$$3\left(\begin{array}{lllll} 5 & -3 & 10 & 2 & -1 \end{array}\right)=\left(\begin{array}{llllll} 15 & -9 & 30 & 6 & -3 \end{array}\right) .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Transformation of Coordinates
$\mathbf{e}1, \mathbf{e}_2, \mathbf{e}_3$ 和 $\overline{\mathbf{e}}_1, \overline{\mathbf{e}}_2, \overline{\mathbf{e}}_3$ ，分别 翻译比较无趣。因此，我们假设 $\Sigma$ 和 $\bar{\Sigma}$ 重合。现在让我们考虑一个任意选择的位置向量 $\mathbf{r}$ : $$\mathbf{r}=\left(x_1, x_2, x_3\right) \text { in } \Sigma \quad[\mathbf{r}(\Sigma)] \mathbf{r}=\left(\bar{x}_1, \bar{x}_2, \bar{x}_3\right) \text { in } \bar{\Sigma} \quad[\mathbf{r}(\bar{\Sigma})]$$ 我们假设元素 $x_i$ 在 $\Sigma$ 已知元素 $\bar{x}_j$ 在 $\bar{\Sigma}$ 有待确定。 $\mathbf{r}$ 它本身当然独立于坐标系统的特殊选择，无论是方向还是幅度。 所以： $$\sum{j=1}^3 x_j \mathbf{e} j=\sum j=1^3 \bar{x} j \overline{\mathbf{e}}_j$$
$$\overline{\mathbf{e}}_j=\sum_k d j k \mathbf{e} k$$
$$d j m=\overline{\mathbf{e}} j \cdot \mathbf{e}_m=\cos \varphi j m$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS3020
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
With the aid of the partial derivative we have the possibility to find out how a field alters as we proceed along one of the axis of coordinates. We want to investigate now how a scalar field changes along an arbitrary (!) space direction e, i.e. we are interested in the term
\begin{aligned} &\Delta \varphi=\varphi(\mathbf{r}+\Delta \mathbf{r})-\varphi(\mathbf{r}), \ &\Delta \mathbf{r}=\left(\Delta x_1, \Delta x_2, \Delta x_3\right) \uparrow \uparrow \mathbf{e} . \end{aligned}
If $\Delta \mathbf{r}$ were, e.g., parallel to the 1-axis then for sufficiently small changes $\Delta \mathbf{r}=$ $\Delta x_1 \mathbf{e}_1$ we would have:
$$\Delta \varphi=\frac{\partial \varphi}{\partial x_1} \Delta x_1 \quad\left[\varphi(\mathbf{r}+\Delta \mathbf{r})=\varphi\left(x_1+\Delta x_1, x_2, x_3\right)\right] .$$
This presumption is not in general fulfilled (Fig. 1.71). It is, however, possible to realize it by a proper rotation of the coordinate axes. The physical field $\varphi$ is of course not affected by such a redefinition of the axes directions. We execute the rotation in such a way that the new 1 axis coincides with the e direction. Then we must have:
$$\Delta \varphi=\frac{\partial \varphi}{\partial \bar{x}_1} \Delta \bar{x}_1 .$$
We now can express $\Delta \mathbf{r}$ in the new and old system of coordinates, respectively, as follows:
$$\Delta \mathbf{r}=\Delta \bar{x}_1 \overline{\mathbf{e}}_1=\Delta x_1 \mathbf{e}_1+\Delta x_2 \mathbf{e}_2+\Delta x_3 \mathbf{e}_3 .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Divergence and Curl
The gradient, nabla operator introduced in the last section acts exclusively on scalar fields $\varphi$, while the resulting gradient field $\operatorname{grad} \varphi=\nabla \varphi$ is itself a vector. An obvious question then is whether it is possible to apply the nabla operator $\nabla$, formally defined in (1.269) as vector-differential operator, also to vectors. The answer is yes! There are again two kinds of application, similar to the previously discussed multiplicative connection of two ordinary vectors, one in the sense of a scalar product, the other in the sense of a vector product.
Definition 1.5.6 Let a $(\mathbf{r}) \equiv\left(a_1(\mathbf{r}), a_2(\mathbf{r}), a_3(\mathbf{r})\right)$ be a continuously differentiable vector field.
Then one calls
$$\sum_{j=1}^3 \frac{\partial a_j}{\partial x_j} \equiv \operatorname{div} \mathbf{a}(\mathbf{r}) \equiv \nabla \cdot \mathbf{a}(\mathbf{r})$$
the divergence (the source field) of $\mathbf{a}(\mathbf{r})$.
By this definition, to a given vector field $\mathbf{a}(\mathbf{r})$ a new scalar field diva(r) is assigned. The illustrative interpretation of div $\mathbf{a}(\mathbf{r})$ as a source field of $\mathbf{a}(\mathbf{r})$ will become understandable later by some examples from physics.
The reader should prove as an exercise the following calculation rules:
\begin{aligned} \operatorname{div}(\mathbf{a}+\mathbf{b}) &=\operatorname{diva}+\operatorname{divb}, \ \operatorname{div}(\gamma \mathbf{a}) &=\gamma \operatorname{diva} ; \quad \gamma \in \mathbb{R}, \ \operatorname{div}(\varphi \mathbf{a}) &=\varphi \operatorname{diva}+\mathbf{a} \cdot \operatorname{grad} \varphi \end{aligned}
( $\varphi$ : scalar field; a: vectorial field).
## 理论力学代写
$$\Delta \varphi=\varphi(\mathbf{r}+\Delta \mathbf{r})-\varphi(\mathbf{r}), \quad \Delta \mathbf{r}=\left(\Delta x_1, \Delta x_2, \Delta x_3\right) \uparrow \uparrow \mathbf{e} .$$
$$\Delta \varphi=\frac{\partial \varphi}{\partial x_1} \Delta x_1 \quad\left[\varphi(\mathbf{r}+\Delta \mathbf{r})=\varphi\left(x_1+\Delta x_1, x_2, x_3\right)\right] .$$
$$\Delta \varphi=\frac{\partial \varphi}{\partial \bar{x}_1} \Delta \bar{x}_1 .$$
$$\Delta \mathbf{r}=\Delta \bar{x}_1 \overline{\mathbf{e}}_1=\Delta x_1 \mathbf{e}_1+\Delta x_2 \mathbf{e}_2+\Delta x_3 \mathbf{e}_3$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Divergence and Curl
$$\sum_{j=1}^3 \frac{\partial a_j}{\partial x_j} \equiv \operatorname{div} \mathbf{a}(\mathbf{r}) \equiv \nabla \cdot \mathbf{a}(\mathbf{r})$$
$$\operatorname{div}(\mathbf{a}+\mathbf{b})=\operatorname{diva}+\operatorname{divb}, \operatorname{div}(\gamma \mathbf{a})=\gamma \operatorname{diva} ; \quad \gamma \in \mathbb{R}, \operatorname{div}(\varphi \mathbf{a})=\varphi \operatorname{diva}+\mathbf{a} \cdot \operatorname{grad} \varphi$$
( $\varphi$ : 标量场；a：矢量场)。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYS2041
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Elements of Integral Calculus
The technique of ‘differentiation’, which we discussed in the previous section, follows the scope of work:
given: $\quad y=f(x)$
finding: $\quad f^{\prime}(x)=\frac{d f}{d x}:$ ‘derivation’,
The reverse program, namely
given: $\quad f^{\prime}(x)=\frac{d f}{d x}$
finding: $\quad y=f(x)$
leads to the technique of ‘integration’. Consider for example
$$f^{\prime}(x)=c=\text { const }$$
then we remember according to (1.77) that
$$y=f(x)=c \cdot x$$
fulfills the condition $f^{\prime}(x)=c$.
Definition $F(x)$ is the ‘antiderivative (primitive function)’ of $f(x)$, if it holds:
$$F^{\prime}(x)=f(x) \quad \forall x .$$
In this connection the above example means:
$$f(x) \equiv c \quad \curvearrowright \quad F(x)=c \cdot x+d .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus
We consider the definite integral over a continuous function $f(t)$, but now with variable upper limit:
$$F(x)=\int_{a}^{a} f(t) d t \quad \text { ‘area function’ }$$
The area under the curve $f(t)$ in this case is not constant but a function of $x$ (Fig. 1.22). If the upper bound of integration is shifted by $\Delta x$ the area will change by:
$$\Delta F=F(x+\Delta x)-F(x)=\int_{a}^{x+\Delta x} f(t) d t-\int_{a}^{x} f(t) d t=\int_{x}^{x+\Delta x} f(t) d t$$
In the last step we have used the rule (1.114). Without explicit proof we accept the important
‘mean value theorem of integral calculus’
This theorem implies:
$$\exists \hat{x} \in[x, x+\Delta x] \text { with } \Delta F=\Delta x \cdot f(\hat{x}) .$$
Although not exactly proven the theorem appears rather plausible according to Fig. 1.23. So we can further conclude:
$$F^{\prime}(x)=\lim {\Delta x \rightarrow 0} \frac{\Delta F}{\Delta x}=\lim {\Delta x \rightarrow 0} f(\hat{x})=f(x) .$$
Thus after (1.105), the area function is the antiderivative of $f(x)$ ! Furthermore, the equivalence of the definitions (1.105) and (1.110) for the antiderivative, which remained unsettled in Sect. $1.2 .1$, is now settled.
‘fundamental theorem of calculus’
$$\frac{d}{d x} F(x) \equiv \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Elements of Integral Calculus
$$f^{\prime}(x)=c=\text { const }$$
$$y=f(x)=c \cdot x$$
$$F^{\prime}(x)=f(x) \quad \forall x .$$
$$f(x) \equiv c \quad \curvearrowright \quad F(x)=c \cdot x+d .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Theorem of Calculus
$$F(x)=\int_{a}^{a} f(t) d t \quad \text { ‘area function’ }$$
$$\Delta F=F(x+\Delta x)-F(x)=\int_{a}^{x+\Delta x} f(t) d t-\int_{a}^{x} f(t) d t=\int_{x}^{x+\Delta x} f(t) d t$$
$$\exists \hat{x} \in[x, x+\Delta x] \text { with } \Delta F=\Delta x \cdot f(\hat{x}) .$$
$$F^{\prime}(x)=\lim \Delta x \rightarrow 0 \frac{\Delta F}{\Delta x}=\lim \Delta x \rightarrow 0 f(\hat{x})=f(x) .$$
1.2.1，现已解决。
‘微积分基本定理’
$$\frac{d}{d x} F(x) \equiv \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 物理代写|理论力学作业代写Theoretical Mechanics代考|PHYC90007
statistics-lab™ 为您的留学生涯保驾护航 在代写理论力学Theoretical Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写理论力学Theoretical Mechanics代写方面经验极为丰富，各种代写理论力学Theoretical Mechanics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Differential Quotient
The ‘slope (gradient)’ of a straight line is the quotient of ‘height difference’ $\Delta y$ and ‘base line’ $\Delta x$ (see Fig. 1.10). For the gradient angle $\alpha$ we obviously have:
$$\tan \alpha=\frac{\Delta y}{\Delta x} .$$
Analogously one defines the slope (gradient) of an arbitrary function $f(x)$ at a point $P$ (see Fig. 1.11). The secant $\overline{P Q}$ has the increase
$$\frac{\Delta y}{\Delta x}=\tan \alpha^{\prime}$$
One dennotês
$$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
as ‘difference quotient’. If we now shift the point $Q$ along the curve towards the point $P$ then the increase of the secant becomes the increase of the tangent on the curve $f(x)$ at $P$ (broken line in Fig. 1.11),
$$\tan \alpha=\lim {\alpha^{\prime} \rightarrow \alpha} \tan \alpha^{\prime}=\lim {\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$$
and one arrives at the ‘differential quotient’
$$\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \equiv \frac{d y}{d x} .$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Taylor Expansion
Occasionally it is unavoidable for a physicist to digress from rigorous mathematical exactness in order to come by adopting some ‘reasonable’ mathematical simplifications to concrete physical results. In this respect, the so-called ‘Taylor expansion (series)’ of a mathematical function $y=f(x)$ represents a very important and frequently used auxiliary means. We assume that this function possesses arbitrarily many continuous derivatives at $x=x_{0}$. Then the following power series expansion is valid what is explicitly proved as Exercise $1.1 .9$ :
\begin{aligned} f(x) &=f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1 !}\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\ldots \ &=\sum_{n=0}^{\infty} \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n} \ f^{(n)}\left(x_{0}\right) &=\left.f^{(n)}(x)\right|{x=x{0}} . \end{aligned}
The assumption $\left|x-x_{0}\right|<1$ guarantees the convergence of the series. Then one can assume that the terms of the series become smaller and smaller with increasing index $n$, so that it should be allowed, in the sense of a controlled approximation, to cut the series after a finite number of summands. The error can strictly be estimated as will be demonstrated in Sect. $1.2$ of volume 3 .
However, the Taylor expansion can also be used for the derivation of exact series as is shown by the following examples:
$1 .$
$$f(x)=\frac{1}{1+x} ; x_{0}=0 ;|x|<1$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Differential Quotient
$$\tan \alpha=\frac{\Delta y}{\Delta x} .$$
$$\frac{\Delta y}{\Delta x}=\tan \alpha^{\prime}$$
$$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
$$\tan \alpha=\lim \alpha^{\prime} \rightarrow \alpha \tan \alpha^{\prime}=\lim \Delta x \rightarrow 0 \frac{\Delta y}{\Delta x}$$
$$\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} \equiv \frac{d y}{d x}$$
## 物理代写|理论力学作业代写Theoretical Mechanics代考|Taylor Expansion
$$f(x)=f\left(x_{0}\right)+\frac{f^{\prime}\left(x_{0}\right)}{1 !}\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}+\ldots \quad=\sum_{n=0}^{\infty} \frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n} f^{(n)}\left(x_{0}\right)$$
$$f(x)=\frac{1}{1+x} ; x_{0}=0 ;|x|<1$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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# Events & Promotions
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Open Detailed Calendar
# INSEAD -January 2014 Intake - Calling All Applicants
Author Message
Intern
Joined: 18 Dec 2012
Posts: 9
Concentration: Marketing, Strategy
GMAT 1: 720 Q47 V41
GPA: 3.5
Followers: 0
Kudos [?]: 6 [0], given: 2
### Show Tags
22 Aug 2013, 07:30
docks007 wrote:
Are they going to check Partner's assets by any chance ? Do we need to give proofs for that as well ?
Quick question - How are you guys uploading additional documentation for the scholarships? I don't see a link where you can upload either a text or a PDF document...
Intern
Joined: 06 Jun 2011
Posts: 42
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
22 Aug 2013, 07:54
bensonzhsl wrote:
bensonzhsl wrote:
Any R3 candidate has received the confirmation letter that mentioned on the website?
Thanks.
Seems R3 candidates are so quiet...
Just got the email from INSEAD stating that the application is complete and update on next steps, if any, will be given latest by 13th September.
Intern
Joined: 12 Dec 2012
Posts: 22
Concentration: Social Entrepreneurship, Technology
WE: Management Consulting (Consulting)
Followers: 0
Kudos [?]: 8 [0], given: 17
### Show Tags
22 Aug 2013, 08:20
The supporting documentation needs to be mailed to the following address: Finance-MBA.FB@insead.edu - they will also send a confirmation email once received.
sniggykols wrote:
Quick question - How are you guys uploading additional documentation for the scholarships? I don't see a link where you can upload either a text or a PDF document...
Intern
Joined: 18 Dec 2012
Posts: 9
Concentration: Marketing, Strategy
GMAT 1: 720 Q47 V41
GPA: 3.5
Followers: 0
Kudos [?]: 6 [0], given: 2
### Show Tags
22 Aug 2013, 08:51
Nazz wrote:
The supporting documentation needs to be mailed to the following address: Finance-MBA.FB@insead.edu - they will also send a confirmation email once received.
sniggykols wrote:
Quick question - How are you guys uploading additional documentation for the scholarships? I don't see a link where you can upload either a text or a PDF document...
Sweet! Thanks for the info
Manager
Joined: 03 Mar 2013
Posts: 62
Followers: 0
Kudos [?]: 2 [0], given: 0
### Show Tags
22 Aug 2013, 18:23
paragcs wrote:
bensonzhsl wrote:
bensonzhsl wrote:
Any R3 candidate has received the confirmation letter that mentioned on the website?
Thanks.
Seems R3 candidates are so quiet...
Just got the email from INSEAD stating that the application is complete and update on next steps, if any, will be given latest by 13th September.
Hi, from whom? Who is the coordinator?
As one of me recommendation letter is sent by the pdf version, they are still retrieving now so that at this stage, they cannot confirm my application.
Hope it's OK.
Intern
Joined: 06 Jun 2011
Posts: 42
Followers: 0
Kudos [?]: 0 [0], given: 1
### Show Tags
22 Aug 2013, 18:37
bensonzhsl wrote:
Hi, from whom? Who is the coordinator?
As one of me recommendation letter is sent by the pdf version, they are still retrieving now so that at this stage, they cannot confirm my application.
Hope it's OK.
Stefanie Hanny.
Intern
Joined: 23 Aug 2013
Posts: 25
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
23 Aug 2013, 14:55
Can someone post the nationalities number of middle eastern students (lebanon, syria, iraq, iran, jordan)? Thank you!
Senior Manager
Joined: 14 Feb 2012
Posts: 488
Concentration: Finance, Entrepreneurship
GPA: 3.76
WE: Engineering (Other)
Followers: 4
Kudos [?]: 46 [0], given: 20
### Show Tags
24 Aug 2013, 20:28
Quote:
lebanon, syria, iraq, iran, jordan
lebanon: 10 records
syria: 0
iraq: 0
iran: 1
jordan: 0
_________________
Day 0: application submitted
Day 3: application confirm
Day 26: Pre-selected interview (Note: A bit earlier than other applicant in this forum.)
Day 40: Interview information
Day 50: First interview
Day 54: Second interview
Intern
Joined: 26 Apr 2013
Posts: 3
Location: United States
Concentration: Entrepreneurship, General Management
GPA: 3.23
WE: Corporate Finance (Internet and New Media)
Followers: 0
Kudos [?]: 10 [0], given: 0
### Show Tags
25 Aug 2013, 21:31
Intern
Joined: 06 Feb 2013
Posts: 42
Location: France
WE: Engineering (Other)
Followers: 0
Kudos [?]: 8 [0], given: 8
### Show Tags
25 Aug 2013, 22:48
vivitao wrote:
According to the R1 applicants, we have to wait untill the very last week, somewhere between Tuesday and Thursday.
Wouldn't mind it to receive the "yes" or "no" this week already though.
Intern
Joined: 12 Dec 2012
Posts: 22
Concentration: Social Entrepreneurship, Technology
WE: Management Consulting (Consulting)
Followers: 0
Kudos [?]: 8 [0], given: 17
### Show Tags
26 Aug 2013, 01:21
robisrojis wrote:
Wouldn't mind it to receive the "yes" or "no" this week already though.
You and me both! I'm sure I'm not the only one that can't concentrate on anything but the decision. I foresee the next week and a bit dragging quite a bit...
Intern
Joined: 12 Aug 2013
Posts: 9
Followers: 0
Kudos [?]: 2 [0], given: 0
### Show Tags
27 Aug 2013, 12:02
Nazz wrote:
robisrojis wrote:
Wouldn't mind it to receive the "yes" or "no" this week already though.
You and me both! I'm sure I'm not the only one that can't concentrate on anything but the decision. I foresee the next week and a bit dragging quite a bit...
Just taking way too long - is there anyone from R2 with admissions decision?
Intern
Joined: 12 Aug 2013
Posts: 9
Followers: 0
Kudos [?]: 2 [0], given: 0
### Show Tags
27 Aug 2013, 12:05
ducceus wrote:
Quote:
lebanon, syria, iraq, iran, jordan
lebanon: 10 records
syria: 0
iraq: 0
iran: 1
jordan: 0
Can someone please post the nationalities numbers for CEE - Bulgaria, Romania, Czech and Poland - tnx lots
Senior Manager
Joined: 14 Feb 2012
Posts: 488
Concentration: Finance, Entrepreneurship
GPA: 3.76
WE: Engineering (Other)
Followers: 4
Kudos [?]: 46 [0], given: 20
### Show Tags
27 Aug 2013, 12:38
dtveli04 wrote:
ducceus wrote:
Quote:
lebanon, syria, iraq, iran, jordan
lebanon: 10 records
syria: 0
iraq: 0
iran: 1
jordan: 0
Can someone please post the nationalities numbers for CEE - Bulgaria, Romania, Czech and Poland - tnx lots
Bulgaria: 1
Romania: 1
Czech: 0
Poland 0
_________________
Day 0: application submitted
Day 3: application confirm
Day 26: Pre-selected interview (Note: A bit earlier than other applicant in this forum.)
Day 40: Interview information
Day 50: First interview
Day 54: Second interview
Manager
Joined: 06 Sep 2011
Posts: 55
GPA: 3.8
Followers: 0
Kudos [?]: 6 [0], given: 0
### Show Tags
27 Aug 2013, 14:21
Can someone please post the nationalities numbers for CEE - Bulgaria, Romania, Czech and Poland - tnx lots [/quote]
Bulgaria: 1
Romania: 1
Czech: 0
Poland 0[/quote]
hey ducceus ! i am in as well !!! i was waitlisted and just got my final offer today !!! have added myself to the group !
Senior Manager
Joined: 14 Feb 2012
Posts: 488
Concentration: Finance, Entrepreneurship
GPA: 3.76
WE: Engineering (Other)
Followers: 4
Kudos [?]: 46 [0], given: 20
### Show Tags
27 Aug 2013, 14:25
joey88 wrote:
Can someone please post the nationalities numbers for CEE - Bulgaria, Romania, Czech and Poland - tnx lots
Bulgaria: 1
Romania: 1
Czech: 0
Poland 0[/quote]
hey ducceus ! i am in as well !!! i was waitlisted and just got my final offer today !!! have added myself to the group ![/quote]
welcome to the family, join facebook group
_________________
Day 0: application submitted
Day 3: application confirm
Day 26: Pre-selected interview (Note: A bit earlier than other applicant in this forum.)
Day 40: Interview information
Day 50: First interview
Day 54: Second interview
Intern
Joined: 06 Feb 2013
Posts: 42
Location: France
WE: Engineering (Other)
Followers: 0
Kudos [?]: 8 [0], given: 8
### Show Tags
27 Aug 2013, 14:37
joey88 wrote:
hey ducceus ! i am in as well !!! i was waitlisted and just got my final offer today !!! have added myself to the group !
Hello Joey88, did you do anything between receiving your waitlist confirmation and your final admission? Did you contact the admission office, or sent in an extra recommendation letter, or something like that?
Intern
Joined: 05 Aug 2013
Posts: 5
Concentration: Strategy, Entrepreneurship
Followers: 0
Kudos [?]: 5 [0], given: 0
### Show Tags
27 Aug 2013, 15:41
Ducceus: could you post current admits from Nordics and Baltics: Denmark, Estonia, Finland, Latvia, Lithuania, Sweden, Norway....
Intern
Joined: 27 Aug 2013
Posts: 6
Location: United States
Followers: 0
Kudos [?]: 6 [0], given: 0
### Show Tags
27 Aug 2013, 16:01
Waiting is excruciating! Will hear back by the 6th...
Senior Manager
Joined: 14 Feb 2012
Posts: 488
Concentration: Finance, Entrepreneurship
GPA: 3.76
WE: Engineering (Other)
Followers: 4
Kudos [?]: 46 [0], given: 20
### Show Tags
27 Aug 2013, 17:26
garsh84 wrote:
Ducceus: could you post current admits from Nordics and Baltics: Denmark, Estonia, Finland, Latvia, Lithuania, Sweden, Norway....
Denmark: 1
Estonia: 0
Finland: 0
Latvia: 1
Lithuania: 1
Sweden: 0
Norway: 1
_________________
Day 0: application submitted
Day 3: application confirm
Day 26: Pre-selected interview (Note: A bit earlier than other applicant in this forum.)
Day 40: Interview information
Day 50: First interview
Day 54: Second interview
Re: INSEAD -January 2014 Intake - Calling All Applicants [#permalink] 27 Aug 2013, 17:26
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# Padé Delay is Okay Today
This article is going to be somewhat different in that I’m not really writing it for the typical embedded systems engineer. Rather it’s kind of a specialized topic, so don’t be surprised if you get bored and move on to something else. That’s fine by me.
Anyway, let’s just jump ahead to the punchline. Here’s a numerical simulation of a step response to a $p=126, q=130$ Padé approximation of a time delay:
Impressed? Maybe you should be. This is a goal that’s been bugging the heck out of me for a few months, and I’ve finally reached something useful.
## 1. Denial
If you’ve done any studying of linear time-invariant (LTI) systems, you know that the transfer function of a time delay T is $H(s) = e^{-sT}$. An exponential function. Very simple, but it’s kind of an oddball. That’s because most transfer functions are written as a ratio of polynomials $H(s) = \frac{P(s)}{Q(s)}$ — in other words a rational function — and analyzed by standard techniques to find their poles and zeros and do some sort of numerical calculation with them. So we can’t find the “poles” of $H(s) = e^{-sT}$, because there aren’t any. Huh.
Okay, so what do we do instead?
If you’re using Simulink to model something, it’s fine to just add a time delay. Simulink has some secret recipe for handling time delays and it works just fine.
But what if you want to perform some kind of continuous-time analysis where you need a rational transfer function? Well, you could try to convert everything to a discrete-time approximation, where a time delay is very easy and becomes a shift in a state variable array. Or you can approximate a continuous-time delay with a rational function. And whenever you see the words “approximate… rational function”, the first thing that should come to mind is Padé approximation. This is a technique to compute the “closest” rational function to a given function using power series. It’s kind of the rational-function equivalent to a Taylor series — it’s based upon the function and its derivatives at a single point — and if you’ve read my article on Chebyshev approximation, you’ll know I don’t particularly like Taylor series, but it has its place, and so does Padé approximation. Padé approximation has a reputation for being a “magic” solution that somehow captures more of the essence of a smooth function than a power series, with a wider range of convergence. Let’s look at a mildly interesting example: $f(x) = 10 \tan^{-1} (x-2) + \frac{2}{x}$ near $x=1$.
import numpy as np
import scipy
import matplotlib.pyplot as plt
%matplotlib inline
def f(x):
return 10*np.arctan(x-2) + 2.0/x
x = np.arange(-3,5,0.002)
# fixup plot so it doesn't draw an extra line across a discontinuity
def asymptote_nan(y,x,x0list):
x0list = np.atleast_1d(x0list)
for x0 in x0list:
y[np.argmin(abs(x-x0))] = float('nan')
return y
y = asymptote_nan(f(x),x,x0list=[0])
fig = plt.figure(figsize=(8,6))
ax.plot(x,y)
ax.set_ylim(-40,40)
ax.grid('on')
We can use scipy to figure out numerical coefficients of a Taylor series and the Padé approximations, using scipy.interpolate.approximate_taylor_polynomial and scipy.misc.pade: (warning: scipy.misc.pade expects its input coefficients to be in order of ascending degree, whereas the return values of both functions are numpy.poly1d objects that yield coefficients in order of descending degree. Hence the T10.coeffs[::-1] expression to reverse the order.)
import scipy.interpolate
import scipy.misc
x0 = 1.0
T10poly = scipy.interpolate.approximate_taylor_polynomial(f,x=x0,degree=10, scale=0.05)
T10 = lambda x: T10poly(x-x0)
R5_5 = lambda x: P5poly(x-x0)/Q5poly(x-x0)
fig = plt.figure(figsize=(8,6))
ax.plot(x,y,label='f(x)')
ax.plot(x0,f(x0),'.k',markersize=8)
yR5 = R5_5(x)
# use real roots of Q5(x-x0) for finding asymptotes
yR5 = asymptote_nan(yR5, x, [r+x0 for r in roots(Q5poly) if np.abs(np.imag(r)) < 1e-8])
ax.plot(x,T10(x),label='T10(x)')
ax.plot(x,yR5, label='R5,5(x)')
ax.set_ylim(-40,40)
ax.legend(loc='lower right',labelspacing=0)
ax.grid('on')
So there’s an example of Padé magic. The rational function does a better job than the polynomial. Why is that a big deal?
Well, the two functions $T_{10}(x)$ and $R_{5,5}(x) = P_5(x)/Q_5(x)$ each have 11 degrees of freedom: $T_{10}$ has 11 coefficients, $P_5$ and $Q_5$ each have 6 coefficients but we lose one degree of freedom since we can scale $P_5$ and $Q_5$ by the same arbitrary constant $K$ and leave the resulting rational function unchanged. Moreover, we derived $T_{10}$ directly from $f(x)$, but derived $R_{5,5}$ from $T_{10}$. So there’s no additional “information” in $R_{5,5}$ beyond what’s in $T_{10}$, just some freedom of structure, since $T_{10}$ is required to be a polynomial whereas $R_{5,5}$ is a rational function.
But $R_{5,5}$ has a much wider range of applicability than $T_{10}$: while $T_{10}$ is a decent approximation to $f(x)$ over a range of about $x \in [0.2, 1.9]$, the function $R_{5,5}$ looks very good over the range $x \in [-3, 2.3]$. And here’s the kicker: somehow the function $R_{5,5}(x)$ “knows” about the discontinuity at $x=0$ just from looking at the polynomial $T_{10}(x)$. We lost the pole at $x=0$ going from $f(x)$ to the Taylor series $T_{10}(x)$, but somehow got it back when we used Padé approximation to go from $T_{10}(x)$ to the rational function $R_{5,5}(x)$. Magic!
In the words of the authors of Numerical Recipes (see page 202):
Padé has the uncanny knack of picking the function you had in mind from among all the possibilities. Except when it doesn’t!
### Time Delays and Padé Approximants
The Padé approximation $R_{p,q}(x) = \frac{N_{p,q}(x)}{D_{p,q}(x)} \approx e^{-x}$ is well-known, and its coefficients can be determined exactly using these formulas (see Moler and Van Loan for a reputable reference for $e^A$):
$$\begin{eqnarray} N_{p,q}(x) &=& \sum\limits_{j=0}^p\frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}(-x)^j \cr D_{p,q}(x) &=& \sum\limits_{j=0}^q\frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}x^j \cr \end{eqnarray}$$
Here the numerator $N_{p,q}(x)$ has degree $p$ and denominator $D_{p,q}(x)$ has degree $q$. The expression $n!$ denotes n factorial, namely the product of all positive integers between 1 and $n$, so $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. The various combinations of $p$ and $q$ can be used to show the various approximations in tabular form, known as a Padé table. Just for some examples,
$$\begin{array}{ll} R_{1,1}(x) = \frac{1 - \frac{1}{2}x} {1 + \frac{1}{2}x} & R_{1,2}(x) = \frac{1 - \frac{1}{3}x} {1 + \frac{2}{3}x + \frac{1}{6}x^2} & R_{1,3}(x) = \frac{1 - \frac{1}{4}x} {1 + \frac{3}{4}x + \frac{1}{4}x^2 + \frac{1}{24}x^3} \cr R_{2,1}(x) = \frac{1 - \frac{2}{3}x + \frac{1}{6}x^2} {1 + \frac{1}{3}x} & R_{2,2}(x) = \frac{1 - \frac{1}{2}x + \frac{1}{12}x^2} {1 + \frac{1}{2}x + \frac{1}{12}x^2} & R_{2,3}(x) = \frac{1 - \frac{2}{5}x + \frac{1}{20}x^2} {1 + \frac{3}{5}x + \frac{3}{20}x^2 + \frac{1}{60}x^3} \cr R_{3,1}(x) = \frac{1 - \frac{3}{4}x + \frac{1}{4}x^2 - \frac{1}{24}x^3} {1 + \frac{1}{4}x} & R_{3,2}(x) = \frac{1 - \frac{3}{5}x + \frac{3}{20}x^2 - \frac{1}{60}x^3} {1 + \frac{2}{5}x + \frac{1}{20}x^2} & R_{3,3}(x) = \frac{1 - \frac{1}{2}x + \frac{1}{10}x^2 - \frac{1}{120}x^3} {1 + \frac{1}{2}x + \frac{1}{10}x^2 + \frac{1}{120}x^3} \cr \end{array}$$
which can be seen in the following semilogarithmic graph:
fig = plt.figure(figsize=(10,6))
x = np.arange(-5,5,0.0005)
ax.semilogy(x,np.exp(-x),'--',label='e^-x')
ax.grid('on')
'''
Calculate the numerator and denominator
approximation to e^-x
'''
n = max(p,q)
c = 1
d = 1
clist = [c]
dlist = [d]
for k in xrange(1,n+1):
c *= -1.0*(p-k+1)/(p+q-k+1)/k
if k <= p:
clist.append(c)
d *= 1.0*(q-k+1)/(p+q-k+1)/k
if k <= q:
dlist.append(d)
return np.array(clist[::-1]),np.array(dlist[::-1])
def argbox(y,ymin,ymax,imin,imax):
'''
find limits (we hope)
where y[i] is between ymin and ymax
'''
ii = np.argwhere(np.logical_and(y>ymin,y<ymax))
ii = ii[ii >= imin]
ii = ii[ii <= imax]
return np.min(ii),np.max(ii)
nx = len(x)
xmin = min(x)
xmax = max(x)
xlim = (xmin,xmax)
ylim = (1e-4,1e4)
for p in [1,2,3]:
for q in [1,2,3]:
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
num = P(x)
den = Q(x)
ax.semilogy(x,num/den,label='p=%d, q=%d' % (p,q))
# now label the end lines
it = argbox(num/den, ylim[0], ylim[1], 0, nx/2)[0]
atend = it == 0
xt = x[it] * (0.99 if atend else 1)
yt = P(xt)/Q(xt) if atend else ylim[1]*0.95
ax.text(xt,yt,' (%d,%d)' % (p,q), va='top',
rotation='horizontal' if atend else -90.0)
it = argbox(num/den, ylim[0], ylim[1], nx/2, nx)[1]
atend = it == (nx-1)
xt = x[it] * (0.99 if atend else 1)
yt = P(xt)/Q(xt) if atend else ylim[0]*1.2
ax.text(xt,yt,'(%d,%d)' % (p,q), va='bottom',
ha='right' if atend else 'left',
rotation='horizontal' if atend else -90.0)
ax.set_xlim(xlim)
ax.set_ylim(ylim)
ax.set_xlabel('x')
ax.set_ylabel('y = f(x)')
ax.legend(labelspacing=0)
<matplotlib.legend.Legend at 0x107935810>
Increasing either $p$ or $q$ expands the region of close convergence by a little bit.
But we were talking about simulating time delays. So we don’t really care too much about $e^{-x}$ for real values of $x$; instead we care about Bode plots of $e^{-s}$ for $s=j\omega$.
w = np.arange(0,10,0.001)
s = 1j*w
fig = plt.figure(figsize=(10,6))
for k in [1,2]:
if k == 1:
f = lambda H: 20*np.log10(np.abs(H))
ylabel = '$20\\,\\log_{10} |H(j\\omega)|$'
else:
f = lambda H: np.unwrap(np.angle(H))*180/np.pi
ylabel = '$\\angle H(j\\omega)$'
ax.set_yticks(np.arange(-720,0.01,45))
ax.set_xlabel('$\omega$', fontsize=16)
ax.plot(w,-w*180/np.pi,'--',label='e^jw')
for q in [1,2,3]:
for p in np.arange(1,q+1):
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = P(s)/Q(s)
ax.plot(w,f(H),label='p=%d,q=%d' % (p,q))
ax.set_ylabel(ylabel, fontsize=16)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
These are Bode plots, but linear rather than logarithmic in frequency, so that we can compare to $e^{-s}$ which has a unity gain and linear phase responses. Here we are only using $p \leq q$, otherwise the gain increases as $\omega \rightarrow \infty$, which is not realizable. When $p = q$, the gain is 1, independent of frequency. The higher $p$ and $q$ are, the more the phase response approaches the ideal $e^{-s}$ for larger frequencies. Looks great, right?
Let’s try using the Padé coefficients in the time domain, with scipy.signal.lti to simulate the step responses:
import scipy.signal
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [1,2,3]:
for p in np.arange(1,q+1):
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
_,y = H.step(T=t)
ax.plot(t,y,label='p=%d,q=%d' % (p,q))
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
<matplotlib.legend.Legend at 0x105479e10>
Well, that doesn’t look very satisfying. Not even close to a step. This is the problem with time ↔ frequency transforms; if something looks great in the time domain it usually looks bad in the frequency domain, and vice-versa.
Let’s just look at the $p=q$ case since the degree of the transfer function denominator is the main determining factor in realizability; once we’ve decided, for example, that we’re going to use a 7th degree polynomial denominator, it doesn’t really matter much for realization purposes whether the numerator is just 1 or it’s a 6th or 7th degree polynomial. And for a fixed-degree denominator, the higher degree numerator does a better job of approximating $e^{-s}$. Besides, when you see implementations of Padé approximations, like in MATLAB, they’re usually just using $p=q$. This gives a unity gain at all frequencies (“all-pass”) without any low-pass-filtering.
Here’s the step responses for the first 10 $p=q$ Padé approximations to a unit delay:
import scipy.signal
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in np.arange(1,11):
p = q
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
_,y = H.step(T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
<matplotlib.legend.Legend at 0x1087c7b10>
They get better as $q$ increases, but there’s still this nasty jump near $t=0$, and ringing between $t=0$ and $t=1$ that’s kind of like the phenomenon of Gibbs ears in partial Fourier sums to approximate a square wave.
But still, that’s kind of cool; we can approximate a time delay with a rational transfer function.
Great! So let’s use just keep on going in degree:
import scipy.signal
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [5,10,15,20,21,22]:
p = q
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
_,y = H.step(T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.set_ylim(-0.5,2.5)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
<matplotlib.legend.Legend at 0x1055b8d50>
Uh oh. Things look good until about $q=21$, and then BAM! we get instability. The footnote in MATLAB’s Padé approximation says this:
High-order Padé approximations produce transfer functions with clustered poles. Because such pole configurations tend to be very sensitive to perturbations, Padé approximations with order N>10 should be avoided.
And maybe four or five years ago, when I was last messing around with Padé approximations for a time delay, I left it at that. Because why would you need a high-order Padé approximation anyway? It’s rare to have systems where there’s a time delay but no additional high-frequency rolloff, and when you add that, the ringing doesn’t look as bad. Here’s the step response of a unit time delay cascaded in series with a 1-pole low-pass filter. We’ll show time constants 0.1, 0.2, 0.5, and 1.0, and we’ll look at the exact answer and some Padé approximations of order 5, 10, and 15.
t = np.arange(0,4,0.001)
Tdelay = 1.0
for tau in [0.1, 0.2, 0.5, 1.0]:
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
y_exact = (t>Tdelay)*(1-np.exp(-(t-Tdelay)/tau))
ax.plot(t,y_exact,'--k',label='exact')
for q in [5,10,15]:
p = q
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
_,y = H.step(T=t)
# 1-pole LPF
_,y2,_ = scipy.signal.lsim([[1/tau],[1, 1/tau]],y,t)
ax.plot(t,y2,label='p=q=%d' % q)
ax.set_ylim(-0.1,1.1)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.text(1.1*Tdelay,0,'tau=%.2f' % tau, fontsize=16)
The longer the time constant, the less relevant are those ripples from the time-delay approximation.
See? There’s no problem here. No problem at all.
## 2. Anger
But then a few months ago I needed to get the step response (in a Simulink-free manner) for some systems that had time delays that weren’t very small, and I needed to know if the ripples were due to numerical simulation issues or they were real, so a clean response was important. Grrr.
Something’s fishy here. MathWorks says we should avoid orders more than 10, but the step responses of $q=15$ and $q=20$ look reasonable. Let’s take a look at this “clustered poles” explanation by plotting the poles and zeros:
import scipy.signal
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,10))
for q in [5,10,15,20,25]:
p = q
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
#poles = np.roots(qcoeffs)
#zeros = np.roots(pcoeffs)
poles = H.poles
zeros = H.zeros
for z,m in [(poles,'x'),(zeros,'o')]:
ax.plot(np.real(z),np.imag(z),'.k',marker=m,mfc='none')
plt.axis('equal')
ax.set_xlim(-50,50)
ax.set_ylim(-50,50);
Huh — that doesn’t look bad. The poles and zeros are symmetric for $p=q$ and line up along arcs that get further away from the origin as the value of $q$ increases. I don’t see any “clustering” here, and we’re showing values up to $p=q=25$.
So why do we get instability in simulation?
It really depends on how simulations are implemented. For scipy.signal.lti, the key is that time-domain simulations (lsim, step, impulse, etc.) depend on state-space representation of an LTI system. This means we rely on converting from transfer function form (arrays of numerator and denominator polynomial coefficients) to state-space via tf2ss. How does tf2ss work?
### Pigs in State-Space
Here’s a sample system: let’s look at $H(s) = \frac{P(s)}{Q(s)} = \frac{1}{(s+1)(0.5s+1)(0.2s+1)(0.1s+1)} = \frac{100}{s^4 + 18s^3 + 97s^2 + 180s + 100}$:
ABCD = scipy.signal.tf2ss([100],[1,18,97,180,100])
for k,suffix in enumerate('ABCD'):
print '%s=\n%s' % (suffix, ABCD[k])
A=
[[ -18. -97. -180. -100.]
[ 1. 0. 0. 0.]
[ 0. 1. 0. 0.]
[ 0. 0. 1. 0.]]
B=
[[ 1.]
[ 0.]
[ 0.]
[ 0.]]
C=
[[ 0. 0. 0. 100.]]
D=
[ 0.]
What’s all this A B C D stuff anyway?
State-space is a representation for LTI systems using matrices, that has a couple of advantages over the rational transfer function approach we tend to learn in introductory control systems courses. One is that it can be used to model multi-input multi-output (MIMO) systems. Another is that we can transform a multi-order scalar differential equation (e.g. $\frac{d^4y}{dt^4} + 18\frac{d^3y}{dt^3} + 97\frac{d^2y}{dt^2} + 180\frac{dy}{dt} + 100y = 100u$) into a first-order matrix differential equation. It can simulate aspects of systems (non-controllable / non-observable systems) that aren’t easily modeled with transfer functions. It’s also more amenable to computerized analysis, and it has some numerical advantages over the transfer function approach that we’ll look at in a bit.
An LTI state-space system looks like this:
$$\begin{eqnarray} \frac{d}{dt}\mathbf x &=& A\mathbf x + B\mathbf u \cr \mathbf y &=& C\mathbf x + D\mathbf u \end{eqnarray}$$
where $\mathbf u$ is a vector of input signals, $\mathbf x$ is a vector of internal state signals (which is why it’s called state-space) and $\mathbf y$ is a vector of output signals. You can still use state-space in the SISO case, in which case $u$ and $y$ are scalar values but $\mathbf x$ remains a vector of length equal to the order of the system. In our example case, we have
$$\begin{eqnarray} A &=& \begin{bmatrix} -18 & -97 & -180 & -100 \cr 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr \end{bmatrix} \cr B &=& \begin{bmatrix} 1 & 0 & 0 & 0 \cr \end{bmatrix}^T \cr C &=& \begin{bmatrix} 0 & 0 & 0 & 100 \end{bmatrix} \cr D &=& 0 \end{eqnarray}$$
The $A$ matrix here has a particular structure. It’s a variant of what’s called a companion matrix of the monic denominator polynomial $Q(x)$ of the transfer function (monic just means that the term of highest degree has a coefficient of 1); what’s special about it is
• the first row is -1 times the coefficients of $Q(x)$ aside from the leading coefficient
• the subdiagonal elements are 1, and the remaining elements are 0, forming a shift matrix in the lower rows
• the characteristic polynomial of the companion matrix is equal to $Q(x)$, meaning its eigenvalues are equal to the roots of $Q(x)$
Furthermore — and this applies to any square matrix $A$ — if all the eigenvalues, which are equal to the roots of $Q(x)$, are distinct, then $A$ is similar to any other matrix $M$ with the same eigenvalues, meaning that there is some transformation matrix $T$ such that $M = T^{-1}AT$, and $A$ is diagonalizable, meaning it can be factored as $A = V\Lambda V^{-1}$ where $V$ is a matrix of eigenvectors and $\Lambda$ is the diagonal matrix of corresponding eigenvalues.
Matrix matrix eigenblahblahblah — if you’re not familiar or comfortable with linear algebra, this stuff will just make your eyes glaze over. (Especially when you ask how to obtain the transfer function, which ends up being $H(s) = C(sI-A)^{-1}B + D$; computing the inverse term gets ugly very quickly, and this equation always makes me think of some obscure Hungarian mathematician for some reason.) Don’t worry; I’ll walk you through the implications.
In our particular example, the state equation can be written out as
$$\frac{d\mathbf x}{dt} = \frac{d}{dt} \begin{bmatrix} \frac{d^3x}{dt^3} \cr \frac{d^2x}{dt^2} \cr \frac{dx}{dt} \cr x \end{bmatrix} = \begin{bmatrix} -18 & -97 & -180 & -100 \cr 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr \end{bmatrix} \begin{bmatrix} \frac{d^3x}{dt^3} \cr \frac{d^2x}{dt^2} \cr \frac{dx}{dt} \cr x \end{bmatrix} + \begin{bmatrix}1 \cr 0 \cr 0 \cr 0\end{bmatrix} u = A\mathbf x + B u$$
so in this representation of the transfer function, the elements of the state vector are just a scalar state variable $x$ with transfer function $\frac{1}{Q(s)} = \frac{x(s)}{u(s)}$, and the first $n-1$ derivatives of $x$. The $n-1$ bottom rows fall out from the relation between successive elements of the state vector: each element is the derivative of the next element. And the top row is self-evident from the differential equation relating $x$ and $u$.
This is called the controllable canonical form, which tf2ss uses, and its main virtue is that it is easy to construct the $A$, $B$, $C$, and $D$ matrices directly from the transfer function. It has some lousy properties, though.
### Betrayal: The Perils of Polynomial Coefficients and the Companion Matrix
For symbolic analysis of LTI systems, use of the companion matrix is fine. But things can get ugly if we try to use them in numerical analysis.
The problems start because of finite computational precision and something called conditioning, which measures the relative error gain from “input” to “output”, where “input” and “output” here are interpreted loosely, depending on the context. Usually when someone talks about ill-conditioning, it’s in the context of solving the linear matrix equation $Ax=b$ for $x$, in which case the condition number is $\kappa(A) = \|A\| \cdot \|A^{-1}\|$ — this involves matrix norms, which you can read about at your leisure. When solving $Ax=b$, it means that if I have some relative error $\epsilon_b$ in $b$, the relative error $\epsilon_x$ in the solution $x$ can be bounded by $\epsilon_x \leq \kappa(A) \epsilon_b$, so the condition number $\kappa(A)$ represents a error magnification ratio.
For example, if you have the ill-conditioned equation
$$Ax = \begin{bmatrix} 2&1 \cr 4&1.998 \end{bmatrix}\begin{bmatrix}x_1 \cr x_2\end{bmatrix} = b$$
where $\kappa(A) \approx 6248$, the solution for $b = [1,2]^T$ is $x_1 = 0.5, x_2 = 0$. If we have $b = [0.99,2]^T$, a small change, the solution shifts drastically to $x_1 = -4.495, x_2 = 10$. The problem input $b$ only changed by a relative error of $0.01/\sqrt{5} \approx 0.004472$, whereas the problem output $x$ changed by a relative error of $\sqrt{4.995^2 + 10^2}/0.5 \approx 22.3562$ — a magnification factor of 4999.
Ill-conditioning in numerical analysis is kind of like the biomagnification of mercury in fish, only worse. If you eat a lot of tuna, the mercury content in your body can accumulate to a higher concentration than it was in the tuna, which in turn had higher concentrations than in the fish eaten by the tuna.
There’s not a practical way of removing the mercury once it’s in the fish. But you can control how much of it you eat, and the human body does get rid of it (albeit very slowly). Whereas once you have introduced numerical errors into mathematical calculations, there’s rarely any way to get rid of them. Let me say this again: Numerical errors are even worse than toxic poison! The best way to avoid their problems is to keep them out of your system in the first place; once they’ve made their way into calculations it’s too late. And unfortunately, ill-conditioned numerical problems are error magnets that turn the very small but unavoidable errors of floating-point calculations into large errors.
Anyway, the sensitivity of solving linear systems is only one way to define condition numbers. Another is to look at sensitivity of the roots of a polynomial given its coefficients.
For example, let’s take $Q(s) = s^4 + 18s^3 + 97s^2 + 180s + 100$ and find the sensitivity of its roots to the non-leading coefficients. There are four roots and four non-leading coefficients, so we can find a 4 × 4 matrix of sensitivities. If we increase the last coefficient by $1+\delta$ with $\delta = 0.0001$ (from 100 to 100.01), the roots change from $[-10, -5, -2, -1]$ to approximately $[-9.99997222, -5.00016666, -1.99958324, -1.00027788]$, which is a relative change of approximately $\delta \times [-0.02777811, 0.33331204, -2.08380389, 2.77882873]$. This is the first column of a sensitivity matrix, and we can repeat the same thing for the other three columns:
Q = np.poly1d([1,18,97,180,100])
def find_root_sensitivities(Q, delta=1e-10, extras=False):
"""
Returns a relative sensitivity matrix S from coefficients to roots.
If extras is True, returns S, kappa, v, with:
kappa = the norm of the sensitivity matrix
v = a unit vector of maximum sensitivity which produces sensitivity gain kappa
"""
Q = np.poly1d(Q)
n = Q.order
roots = np.roots(Q)
S = np.zeros((n,n), dtype=roots.dtype) # allow for complex numbers
for k in xrange(n):
# find approx numerical derivative
# from symmetric differences
Qplus = Q*1.0
Qminus = Q*1.0
Qplus[k] *= (1+delta/2.0)
Qminus[k] *= (1-delta/2.0)
rplus = np.roots(Qplus)
rminus = np.roots(Qminus)
if np.any(np.iscomplex(rplus)) or np.any(np.iscomplex(rminus)):
S = S.astype(np.complex128)
S[:,k] = (np.roots(Qplus) - np.roots(Qminus)) / roots / delta
if not extras:
return S
# extras: find a direction of maximum sensitivity
u,s,v = np.linalg.svd(S, compute_uv=True)
# largest singular direction in reverse order
# to match polynomial coefficients n-1:0
return S,s[0],v[0,::-1]
S,kappa,v = find_root_sensitivities(Q, extras=True)
S,kappa,v
(array([[-0.02777334, 0.50001248, -2.69444733, 4.99999508],
[ 0.33334402, -3.00002156, 8.08333667, -7.49996509],
[-2.08332684, 7.50000284, -8.08332734, 2.99997693],
[ 2.7777769 , -5.00000152, 2.69445133, -0.49999782]]),
17.015155693892876,
array([-0.48543574, 0.70522631, -0.49695306, 0.14158266]))
# double-check that this direction vector v works:
# if we perturb the coefficients towards v and away from v,
# we should be able to see that the magnitude of change
# in the coefficient vector is amplified by kappa to yield
# a larger magnitude of change in the root vector
Qplus = Q*1.0
Qminus = Q*1.0
r = np.roots(Q)
delta = 1e-10
Qplus.coeffs[1:] *= 1+v*delta/2.0
Qminus.coeffs[1:] *= 1-v*delta/2.0
rplus = np.roots(Qplus)
rminus = np.roots(Qminus)
dr = (rplus - rminus)/r/delta
print dr
np.linalg.norm(dr,2)
[ -4.57977833 10.8794147 -11.17900439 5.02097586]
17.015180620416512
We can adjust each non-leading coefficient $c_k$ by some relative factor $1 + v_k\delta$, where $\sum |v_k|^2 = 1$ and $\delta$ is a magnitude of adjustment. The resulting measure of relative change of the roots $R = \sqrt{\sum |\delta r_k|^2}$ where $\delta r_k$ is the relative change in each root. If we choose the $v_k$ appropriately (and it turns out that we can obtain $v_k$ by calculating the singular value decomposition of the sensitivity matrix $S = U\Sigma V$, and then take $v_k$ as the first row of $V$), then we can get the maximum value of $R = \kappa \delta$: the root-sum-square of relative changes in the roots, is equal to the root-sum-square of relative changes in the coefficients, magnified by this factor $\kappa$ which is the Euclidean norm of the sensitivity matrix, and can be considiered as a condition number because it tells the relative gain of input error to output error.
For our example, $Q(s) = s^4 + 18s^3 + 97s^2 + 180s + 100$, $\kappa \approx 17.015$. Numerical errors in the coefficients produces numerical errors in the roots that, depending on the direction, can be as much as just over 17.015 times as large.
All monomials (1st-degree polynomials) have $\kappa = 1$, but as the degree of a polynomial increases, the condition number of its roots generally increases to a very large number. For example, the 10th-degree polynomial $P_{10}(x) = \prod\limits_{k=1}^{10} (x-\frac{1}{k})$, the condition number $\kappa$ is over a million. In fact, it’s difficult to estimate this condition number accurately.
P10 = 1.0
for k in xrange(1,11):
P10 = P10*np.poly1d([1, -1.0/k])
P10
poly1d([ 1.00000000e+00, -2.92896825e+00, 3.51454365e+00,
-2.31743276e+00, 9.41614308e-01, -2.48582176e-01,
4.34780093e-02, -5.00165344e-03, 3.63756614e-04,
-1.51565256e-05, 2.75573192e-07])
S,kappa,v = find_root_sensitivities(P10, extras=True)
kappa
1586991.6991524738
This matters because small changes in polynomial coefficients can produce large changes in the roots. In fact, polynomial coefficients expressed as floating-point numbers are a really lousy way to analyze and evaluate polynomials unless the polynomial degree is small. Let that sink in for a moment: Large degree polynomials specified by floating-point coefficients are poorly conditioned for numerical analysis. (And by “large”, we’re talking above the 5-10 degree range, depending on the application.)
In terms of simulation, this is important: the roots of the transfer function’s denominator polynomial determine the dynamics of the system, and high sensitivity of these roots to errors in coefficients (which is how we specify the $A$ matrix in tf2ss) make simulating high-degree transfer functions very error-prone.
We can look at the root sensitivity condition number for the $(p,q)$ Padé approximation of a unit time delay, as a function of the polynomial degree:
for n in [5,10,15,20,25]:
Q = np.poly1d(qcoeffs)
S,kappa,v = find_root_sensitivities(Q, extras=True)
print 'n=%d, kappa=%g' % (n,kappa)
n=5, kappa=71.3885
n=10, kappa=28069.6
n=15, kappa=1.43265e+07
n=20, kappa=8.28673e+09
n=25, kappa=1.52504e+11
OUCH! No wonder the transfer function approach fails for large degree polynomials. This really stinks. In fact, if we continue to try to compute the poles and zeros of the Padé approximations from their coefficients, the results start to become inaccurate above about degree 25. Plotting those results looks like a marching band about to go berserk:
import scipy.signal
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,10))
for q in [22,24,26,28,30,32]:
p = q
P = np.poly1d(pcoeffs)
Q = np.poly1d(qcoeffs)
H = scipy.signal.lti(P,Q)
#poles = np.roots(qcoeffs)
#zeros = np.roots(pcoeffs)
poles = H.poles
zeros = H.zeros
for z,m in [(poles,'x'),(zeros,'o')]:
ax.plot(np.real(z),np.imag(z),'.k',marker=m,mfc='none')
plt.axis('equal')
ax.set_xlim(-60,60)
ax.set_ylim(-60,60);
The scipy simulation functions (lsim, step, impulse, etc.) for scipy 0.15.x and earlier are even worse in this regard than finding roots of polynomials, because they try to diagonalize the $A$ matrix as $A = V\Lambda V^{-1}$, which has its own problems at large degree. For example, if we diagonalize the companion matrix of the 10th degree polynomial $P_{10}(x) = \prod\limits_{k=1}^{10} (x-\frac{1}{k})$, we get an eigenvector matrix $V$ that has a condition number of more than $10^{11}$, whereas the eigenvalue sensitivity we computed earlier was only about 1.59 million.
A10=np.polynomial.polynomial.polycompanion(P10.coeffs[::-1])
Lambda,V = np.linalg.eig(A10)
print '%g' % np.linalg.cond(V)
1.36681e+11
Drat. Drat drat drat.
## 3. Bargaining
### Throwing Polynomial Coefficients out the Window
Well, if one major cause of problems is the expression of transfer functions as a pair of polynomials by listing their coefficients, then maybe there’s another way to make them work.
There are other ways to describe polynomials. Two of them which don’t suffer from the ill-conditioned problems of polynomial coefficients are
• factoring polynomials — write a polynomial as a product $P(x) = A\prod\limits_{k=0}^{n}(x-r_k)$ rather than a sum $P(x) = \sum\limits_{k=0}^{n} a_kx^k$
• sum of Chebyshev polynomials — write $P(x) = \sum\limits_{k=0}^{n} c_kT_k(u)$ with $u = \frac{2x-a-b}{b-a}$ to scale $x \in [a,b]$ to $u \in [-1,1]$
Chebyshev polynomials $T_k(u)$ work well because they are bounded between $\pm 1$ and don’t run into the numerical problems we get when we evaluate large powers of $x$ that partially cancel each other out. Evaluating them from the recurrence relation $T_n(u) = 2uT_{n-1}(u) - T_{n-2}(u)$ is numerically stable. But they only work well in a bounded range $x \in [a,b]$. If we have an unbounded range then they won’t help.
So factoring is another approach. But then we have to find the roots of the polynomials. And the formulas we have for the Padé approximation to $e^{-s}$ give us the coefficients; we already know finding the roots from coefficients is a badly-conditioned problem. If we have to start from coefficients, the poison is already there, and we’re dead in the water.
### Zeros of the Hypergeometric Function ${}_1F_1$
Both the numerator and the denominator of the Padé approximation to $e^{-s}$ have the same form, which can be written in terms of the hypergeometric function ${}_1F_1(a;b;z)$:
$$\begin{eqnarray} N_{p,q}(x) &=& \sum\limits_{j=0}^p\frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}(-x)^j &=& {} _ 1F_1 (-p;-p-q;-x)\cr D_{p,q}(x) &=& \sum\limits_{j=0}^q\frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}x^j &=& {} _ 1F_1 (-q;-p-q;x)\cr \end{eqnarray}$$
Now we just need a way to find the zeros of these functions so we can dispense with having to deal with their coefficients.
There are tons of obscure mathematical papers on the zeros of special functions; Saff and Varga’s 1975 paper On the Zeros and Poles of Padé Approximants to $e^x$ prove theorems about the locations of the poles and zeros in the complex plane, but say nothing about how to numerically compute them.
After a bunch of floundering around, I found two very interesting papers that are relevant to finding these poles and zeros:
Campos and Calderón’s paper on the Bessel polynomials explore a different special function, but they have the same general problem and they make a rather off-hand remark about finding zeros using a property of the Bessel polynomials:
$$\sum\limits_{k\neq j}\frac{1}{z_j - z_k} + \frac{az_j + 2}{2z_j^2} = 0$$
This set of nonlinear equations can be solved by standard methods. We have used a Newton method to solve them up to $n=500$ and $a=100$.
This intrigued me — they found roots of a 500th degree polynomial using some equations and Newton’s method — but it was unclear how to repeat the approach or adapt it to the Padé approximants.
Grünbaum’s article states that for any polynomials with simple (nonrepeated) roots,
If $$P(x)=(x-x_1)(x-x_2)\ldots (x-x_n)$$ then at any root $x_k$ of $P$ we have $$P’‘(x_k) = 2P’(x_k)\sum\limits_{j \neq k}\frac{1}{x_k-x_j}$$
This means that at any root $x_k$, then
$$\frac{P’‘(x_k)}{2P’(x_k)} = \sum\limits_{j \neq k}\frac{1}{x_k-x_j}$$
For our Padé polynomials, the Wikipedia page on ${}_1F_1$ has $w={}_1F_1(a;b;z)$ satisfying the differential equation
$$z \frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0$$
and at the roots where $w=0$,
$$z \frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} = 0$$
so that
$$\frac{\frac{d^2w}{dz^2}}{2\frac{dw}{dz}} = \frac{z-b}{2z}$$
Therefore, at all zeros $x_k$ of the denominator polynomial $D_{p,q}(x) = w={}_1F_1(a=-q;b=-p-q;x)$, we have
$$0 = -\frac{1}{2} + \frac{-p-q}{2x_k} + \sum\limits_{j\neq k}\frac{1}{x_k-x_j}$$.
There are $q$ of these equations and they can each be treated as one component of a vector equation $\mathbf F(\mathbf v) = \mathbf 0$ where $\mathbf v$ is the set of the $x_k$ roots. The Newton-Raphson method applied to vector equations is that if we have some estimate $\mathbf v[m]$, we can calculate the next estimate $\mathbf v[m+1]$ as follows:
• calculate $\mathbf y = \mathbf F(\mathbf v[m])$
• calculate the Jacobian matrix $\mathbf J = \frac{\partial \mathbf F}{\partial \mathbf v}$ evaluated at $\mathbf v[m]$
• solve for $\Delta \mathbf v$ where $\mathbf J \Delta \mathbf v = \mathbf y$
• calculate $\mathbf v[m+1] = \mathbf v[m] - \Delta \mathbf v$
We also need an initial guess; some empirical messing around trying to approximate the zeros for $D_{p,q}(x)$ in the q=10-20 range led me to the initial guess
$$\mathbf v[0] = -p - \frac{q}{3} + \frac{2q}{3}\mathbf u^2 + \frac{2j(p+q)}{3}\mathbf u$$
where $\mathbf u$ is the vector with the $k$th component $u_k = \frac{2k - (q-1)}{q}$ for $k = 0 \ldots q-1$ — this leads to the $u_k$ nearly spanning the range $(-1,1)$ — and $\mathbf u^2$ is the element-by-element square of the vector $\mathbf u$.
This approach actually surprisingly well. What I don’t understand is why it converges to the right answer, rather than getting stuck at some other solution or running into some numerical problems due to ill-conditioning; Newton’s methods can be very iffy and don’t always converge well. But in this case it can easily be used to get the poles and zeros for $p$ and $q$ in the range of several hundred. (The numerator $N_{p,q}(x)$ has the same exact method and its vector of zeros $v$ are equal to -1 times the zeros of the denominator polynomial with $p$ and $q$ swapped.)
And here’s an example of it in action with $p=200, q=200$:
import sspade
def plotpolezero(z,p,ax=None):
if ax is None:
fig = plt.figure(figsize=(8,8))
ax.plot(np.real(z),np.imag(z),'ok',markersize=3,mfc='none')
ax.plot(np.real(p),np.imag(p),'xk',markersize=3)
return ax
ax=None
for q in [25,50,75,100,125,150,175,200]:
zeros, poles, k = pe.zpk
ax=plotpolezero(zeros,poles,ax)
Things get more interesting if we look at $p<q$; if $p$ is small enough, some of the poles can drift into the right half-plane, indicating an unstable system:
ax=None
for p in [15,20,25,30,35,40]:
zeros, poles, k = pe.zpk
ax=plotpolezero(zeros,poles,ax)
ax.grid('on')
The minimum value $p_{min}(q)$ that produces all poles in the left half-plane is less than but close to $q$; if we graph $q - p_{min}(q)$ we get values that gradually increase:
pminlist = []
qlist = np.arange(250)
p = 0
for q in qlist:
pprev = p
while p < q:
zeros,poles,k = pe.zpk
if np.all(np.real(poles) < 0):
break
else:
p += 1
if (q-1 - pprev) < (q-p):
print p,q
pminlist.append(p)
plt.plot(qlist, qlist-pminlist,'.')
plt.xlabel('q')
plt.ylabel('q - p_min')
0 0
0 1
0 2
0 3
0 4
1 6
3 9
7 14
11 19
17 26
25 35
34 45
45 57
59 72
75 89
93 108
114 130
139 156
166 184
196 215
<matplotlib.text.Text at 0x104c57950>
Let’s shift our attention back to actually trying to use the Padé approximation. Now that we can find the poles and zeros using Newton’s method, we can avoid the ill-conditioned poison of the polynomial coefficient expression, and start using it in the frequency domain even for large values of $p$ and $q$, since it’s just a matter of evaluating the transfer function at the frequencies of our choice. The phase angle of the transfer function $\frac{N_{p,q}(s)}{D_{p,q}(s)}$ roughly matches a pure unit time delay $e^{-s}$ up to a frequency of approximately $\omega = p+q$; here’s a graph showing the $p=q$ case up to $p=q=120$:
w = np.arange(0,300,0.01)
s = 1j*w
plt.plot(w,-w, ':', label='e^-s')
for q in [20,40,60,80,100,120]:
plt.plot(w,np.unwrap(np.angle(pe(s))), label='(p=q=%d)' % q)
plt.xlabel('$\\omega$', fontsize=16)
plt.ylabel('$\\angle H(j\\omega)$, radians', fontsize=16)
plt.legend(loc='lower left', labelspacing=0);
The tough part is going to be dealing with our system in the time domain. We have the technology, though, now that we can get the poles and zeros without having to go through the polynomial coefficients.
## 4. Depression
Okay, so now the question is how we setup a linear system for time-domain simulation. If you look at the documentation for scipy.signal.lti, you’ll see that
The lti class can be instantiated with either 2, 3 or 4 arguments. The following gives the number of elements in the tuple and the interpretation: - 2: (numerator, denominator) - 3: (zeros, poles, gain) - 4: (A, B, C, D)
Each argument can be an array or sequence.
So we’ll just use the 3-argument constructor so we can utilize the zero, poles, gain triple directly:
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [18,20,22]:
zeros, poles, k = pe.zpk
H = scipy.signal.lti(zeros, poles, k)
_,y = H.step(T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.5,1.2);
Huh? We’re in trouble again at $p=q=22$. What’s going on?
The key here is how scipy.signal.lti handles simulations. This behavior changed between version 0.15 and version 0.16, but in both cases simulation always utilizes the (A,B,C,D) state-space form, and if the system is specified in one of the other two methods (numerator/denominator coefficient vectors, or ZPK = zero-pole-gain), it is converted to state-space first.
In version 0.15 the calculation of the state-space matrices in the scipy.signal.lti class goes through the _update method which, in the case of ZPK, calls zpk2ss:
def _update(self, N):
if N == 2:
self._zeros, self._poles, self._gain = tf2zpk(self.num, self.den)
self._A, self._B, self._C, self._D = tf2ss(self.num, self.den)
if N == 3:
self._num, self._den = zpk2tf(self.zeros, self.poles, self.gain)
self._A, self._B, self._C, self._D = zpk2ss(self.zeros,
self.poles, self.gain)
if N == 4:
self._num, self._den = ss2tf(self.A, self.B, self.C, self.D)
self._zeros, self._poles, self._gain = ss2zpk(self.A, self.B,
self.C, self.D)
This is promising, but then if we look at zpk2ss, we see this:
def zpk2ss(z, p, k):
"""Zero-pole-gain representation to state-space representation
Parameters
----------
z, p : sequence
Zeros and poles.
k : float
System gain.
Returns
-------
A, B, C, D : ndarray
State space representation of the system, in controller canonical
form.
"""
return tf2ss(*zpk2tf(z, p, k))
We call zpk2tf which goes from zero-pole-gain form to transfer function coefficients, and then call tf2ss. Horrors! That’s right — we went to great lengths to avoid polynomial coefficients because of their ill-conditioning, and instead work directly with the poles and zeros, but zpk2ss throws all that care away and goes right back to the error-prone polynomial coefficient approach.
Version 0.16 has been refactored to separate classes for each form of construction, and the ZeroPolesGain class has a to_ss method that’s used in simulation:
def to_ss(self):
"""
Convert system representation to StateSpace.
Returns
-------
sys : instance of StateSpace
State space model of the current system
"""
return StateSpace(*zpk2ss(self.zeros, self.poles, self.gain))
And in version 0.16 it’s the same old sloppy zpk2ss function that goes through transfer function coefficients:
def zpk2ss(z, p, k):
"""Zero-pole-gain representation to state-space representation
Parameters
----------
z, p : sequence
Zeros and poles.
k : float
System gain.
Returns
-------
A, B, C, D : ndarray
State space representation of the system, in controller canonical
form.
"""
return tf2ss(*zpk2tf(z, p, k))
It’s the same in scipy 0.17. So we lose! Dammit. Controller canonical form really sucks for numerical purposes.
### The Right Way to Implement zpk2ss
We need a better way to implement zpk2ss, and since scipy won’t do it for us, we’ll have to do it ourselves. A little research on matrices easily shows that the way to keep the condition number of the $A$ matrix to a minimum is to make it a pure diagonal matrix. Let’s go back to the 4th-order system we talked about earlier:
$$H(s) = \frac{P(s)}{Q(s)} = \frac{1}{(s+1)(0.5s+1)(0.2s+1)(0.1s+1)} = \frac{100}{s^4 + 18s^3 + 97s^2 + 180s + 100}$$
which has a controller canonical form of
$$\begin{eqnarray} A &=& \begin{bmatrix} -18 & -97 & -180 & -100 \cr 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr \end{bmatrix} \cr B &=& \begin{bmatrix} 1 & 0 & 0 & 0 \cr \end{bmatrix}^T \cr C &=& \begin{bmatrix} 0 & 0 & 0 & 100 \end{bmatrix} \cr D &=& 0 \end{eqnarray}$$
where the $A$ matrix has a condition number of just over 521:
A,B,C,D = scipy.signal.tf2ss([100],[1,18,97,180,100])
np.linalg.cond(A)
521.33808185890325
We could diagonalize $A = V\Lambda V^{-1}$ numerically:
Lambda,V = np.linalg.eig(A)
print "Lambda = ", Lambda
print "V ="
print V
print "V*diag(Lambda)*V.inv ="
print np.dot(np.dot(V,np.diag(Lambda)),np.linalg.inv(V))
Lambda = [-10. -5. -2. -1.]
V =
[[ 9.94987442e-01 -9.79797151e-01 8.67721831e-01 -5.00000000e-01]
[ -9.94987442e-02 1.95959430e-01 -4.33860916e-01 5.00000000e-01]
[ 9.94987442e-03 -3.91918861e-02 2.16930458e-01 -5.00000000e-01]
[ -9.94987442e-04 7.83837721e-03 -1.08465229e-01 5.00000000e-01]]
V*diag(Lambda)*V.inv =
[[ -1.80000000e+01 -9.70000000e+01 -1.80000000e+02 -1.00000000e+02]
[ 1.00000000e+00 4.27435864e-15 1.19904087e-14 -3.55271368e-15]
[ 1.04083409e-17 1.00000000e+00 1.02140518e-14 3.55271368e-15]
[ 2.42861287e-17 5.55111512e-17 1.00000000e+00 -8.88178420e-16]]
But then we’ve already picked up the poison of ill-conditioning, by working with controller canonical form. It works okay for small matrices, but we shouldn’t have much hope above $n=10$ or so.
Instead we have to diagonalize things analytically, and that means working through the modal form, which uses a diagonal $A$ matrix with elements equal to the transfer function poles:
$$A = \begin{bmatrix} -10 & 0 & 0 & 0 \cr 0 & -5 & 0 & 0 \cr 0 & 0 & -2 & 0 \cr 0 & 0 & 0 & -1 \cr \end{bmatrix}$$
Now the only hard part is coming up with $B$ and $C$ matrices to make things work properly. There isn’t a unique choice here; we just need to make sure that the net transfer function $H(s) = C(sI-A)^{-1}B + D$ is what we need, and for a diagonal matrix, the $(sI-A)^{-1}$ term is also a diagonal matrix with terms $\frac{1}{s+p}$ along the diagonal:
$$(sI-A)^{-1} = \begin{bmatrix} \frac{1}{s+10} & 0 & 0 & 0 \cr 0 & \frac{1}{s+5} & 0 & 0 \cr 0 & 0 & \frac{1}{s+2} & 0 \cr 0 & 0 & 0 & \frac{1}{s+1} \cr \end{bmatrix}$$
The reason to use modal form is that with modes as state variables, there is no cross-coupling between modes, and in the time domain each state variable represents an independent first-order system. Very easy to analyze! As far as figuring out the $B$ and $C$ matrices goes, since $(sI-A)^{-1}$ is diagonal with terms $\frac{1}{s+p_k}$, any choice of $B$ and $C$ is fine that satisfies $H(s) = \sum\limits_k \frac{b_kc_k}{s+p_k}$. We need to compute the residue coefficients $r_k$ such that $H(s) = \sum\limits_k \frac{r_k}{s+p_k}$ and divide up factors $r_k$ between $B$ and $C$ as desired: for example, taking $B$ as all ones and $c_k = r_k$, or $C$ as all ones and $b_k = r_k$, or make them identical as $b_k = c_k = \sqrt{r_k}$.
In our specific fourth-order case, we can use the Heaviside method to determine the residues as
$$\begin{eqnarray} a _ 0 &=& \left.\frac{100}{(s+1)(s+2)(s+5)}\right| _ {s=-10} = -\frac{5}{18} \cr a _ 1 &=& \left.\frac{100}{(s+1)(s+2)(s+10)}\right| _ {s=-5} = \frac{5}{3} \cr a _ 2 &=& \left.\frac{100}{(s+1)(s+5)(s+10)}\right| _ {s=-2} = -\frac{25}{6} \cr a _ 3 &=& \left.\frac{100}{(s+2)(s+5)(s+10)}\right| _ {s=-1} = \frac{25}{9} \end{eqnarray}$$
and therefore if we take $B = \begin{bmatrix}1 & 1 & 1 & 1\end{bmatrix}^T$, then $C = \begin{bmatrix}-\frac{5}{18} & \frac{5}{3} & -\frac{25}{6} & \frac{25}{9}\end{bmatrix}$, which we can verify:
H2 = scipy.signal.lti(diag([-10,-5,-2,-1]),
[[1]]*4,
[-5.0/18, 5.0/3, -25.0/6, 25.0/9],
0
)
print 'num=',H2.num
print 'den=',H2.den
print 'cond(A)=',np.linalg.cond(H2.A)
num= [[ 0.00000000e+00 -1.06581410e-14 -8.52651283e-14 -1.13686838e-13
1.00000000e+02]]
den= [ 1. 18. 97. 180. 100.]
cond(A)= 10.0
The modal form expresses any state-space system as the parallel sum of separate first-order systems; the condition number of the $A$ matrix is just the ratio of the largest and smallest eigenvalue magnitudes. Actual physical systems rarely have only real poles, but rather at least one pair of complex conjugate eigenvalues, so we have to be cautious with this approach, because it means the $A$, $B$, and $C$ matrices will have complex coefficients; an alternative is to group any complex poles in conjugate pairs and use a block-diagonal $A$ matrix with block elements that are either 1x1 for real poles, or 2x2 of the form $\begin{bmatrix}-\sigma&\omega\cr-\omega&-\sigma\end{bmatrix}$, which has eigenvalues $-\sigma\pm j\omega$. This yields real-valued matrices and can be used with all of the scipy functions; some of them don’t work well with complex-valued state-space matrixes.
Okay, so let’s do it; in the sspade.PadeExponential class I’ve included properties lti_ssmodal (pure diagonal state-space matrix) and lti_ssrealmodal (block diagonal state-space matrix):
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [18,20,22]:
zeros, poles, k = pe.zpk
H = pe.lti_ssrealmodal
_,y = H.step(T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.5,1.2);
Wheeeeeee!!!!!!!!! It works! Let’s march onward to higher degree:
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [20,22,24,26,28]:
zeros, poles, k = pe.zpk
H = pe.lti_ssrealmodal
_,y = H.step(T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.5,1.2);
Oh, it doesn’t work. Argh.... after a bunch of looking around, I noticed the lsim2 and step2 functions, which use ODE solvers to simulate; the lsim and step functions try to be smart and use closed-form solutions to linear systems (relying on matrix exponentials), and this works better for some systems but causes numerical problems in others, and high-order systems is one of those times.
So let’s use scipy.signal.step2 instead:
t = np.arange(0,3,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [20,22,24,26,28]:
zeros, poles, k = pe.zpk
H = pe.lti_ssrealmodal
H.D = np.array(1) # TODO: FIX FIX FIX
_,y = scipy.signal.step2(H,T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.5,1.2);
We managed to push things out further by another 4 degrees, to $p=q=24$; it’s borderline stable at $p=q=26$, and then fails at higher values.
Damn, damn, damn.
## 5. Acceptance…?
After I managed to figure out how to compute poles and zeros of high-order Padé time delay approximations using Newton’s method, I spent a very long day trying to get the ZPK implementation working, first stymied by the implementation of scipy‘s zpk2ss, and then in a failed attempt at a modal form implementation.
It took me a long time to figure out why the modal form doesn’t work, even though the condition number of the $A$ matrix is much lower:
pe = sspade.PadeExponential(12,12)
zeros,poles,k = pe.zpk
H1 = scipy.signal.lti(zeros, poles, k)
H2 = pe.lti_ssrealmodal
print "controller canonical form K(A)=%g" % np.linalg.cond(H1.A)
print "modal form K(A)=%g" % np.linalg.cond(H2.A)
controller canonical form K(A)=1.63807e+15
modal form K(A)=1.28474
Wow. The modal form has a condition number of 1.28 for $p=q=12$, whereas the controller canonical form has a condition number of over $10^{15}$. The modal form shouldn’t be causing us numerical problems at all.... but the problem is that the ill-conditioning has been “swept aside” into the $B$ and $C$ matrices:
print "B="
print H2.B
print "C="
print H2.C
B=
[[ 18.05591692]
[ 132.91771897]
[ 851.69993802]
[ 304.37549706]
[ 2894.68914148]
[ 2500.50514786]
[ 3177.55610818]
[-1303.54031239]
[ 1898.61079121]
[ 126.23776885]
[ -223.46066064]
[ 37.6807615 ]]
C=
[[ 18.05591692 132.91771897 851.69993802 304.37549706
2894.68914148 2500.50514786 -3177.55610818 1303.54031239
-1898.61079121 -126.23776885 223.46066064 -37.6807615 ]]
My implementation uses $b_k = c_k = \sqrt{r_k}$ to reduce magnitudes, but there’s still a delicate cancellation between modes. Each of the modes is oscillating at its own characteristic frequency, most of them with a relatively large amplitude, and they are supposed to sum up to small values. This doesn’t work too well once the system order gets fairly large:
for q in [16,18,20,22,24]:
zeros,poles,k = pe.zpk
H2 = pe.lti_ssrealmodal
print "C_%d=" % q
print H2.C
C_16=
[[ 1.55868479e+01 5.19857547e+02 2.04887023e+03 2.57337115e+03
1.82882935e+04 7.10695442e+03 4.31275008e+04 3.60662991e+04
-4.50120393e+04 2.09511572e+04 -3.21075818e+04 -6.80395204e+02
7.59499100e+03 -1.63614662e+03 2.48124131e+02 6.35855671e+01]]
C_18=
[[ 6.59640301e+01 1.13979997e+02 3.42432615e+03 1.23189110e+04
1.02201903e+04 7.85142459e+04 3.08958948e+04 1.65125772e+05
1.36577005e+05 -1.69418842e+05 8.21961937e+04 -1.28145915e+05
-3.50145127e+02 3.68157292e+04 -8.01311715e+03 2.46639032e+03
-8.02882151e+02 -4.43317147e+01]]
C_20=
[[ 7.71477361e+01 1.09480088e+03 2.20976889e+03 1.89068039e+04
6.52104021e+04 3.93204917e+04 3.27775187e+05 1.29770199e+05
6.30151994e+05 5.16584425e+05 -6.37675797e+05 3.19757062e+05
-5.05485067e+05 6.73972798e+03 1.68214908e+05 -3.63903841e+04
1.72063240e+04 -6.14389658e+03 -1.86962826e+02 5.54803332e+01]]
C_22=
[[ 2.92848728e+01 6.92739882e+02 9.76114462e+03 1.95615111e+04
9.45543017e+04 3.19193112e+05 1.48199173e+05 1.34170005e+06
5.32633296e+05 2.39897177e+06 1.95222307e+06 -2.40016773e+06
1.23634882e+06 -1.97709719e+06 5.41285003e+04 7.38679949e+05
-1.57696485e+05 1.01020692e+05 -3.78633338e+04 5.91146021e+02
1.31571948e+03 -1.08300118e+02]]
C_24=
[[ 1.30410652e+02 1.35369198e+03 3.24570356e+03 6.70668054e+04
1.32333628e+05 4.44087115e+05 1.48231772e+06 5.50774714e+05
5.41332494e+06 2.15076021e+06 9.11611817e+06 7.37282635e+06
-9.03411488e+06 4.75871679e+06 -7.68407951e+06 3.05267070e+05
3.15357164e+06 -6.62292235e+05 5.35078004e+05 -2.06529518e+05
1.42833365e+04 1.38783403e+04 -1.40768237e+03 -1.29015506e+01]]
The modal form has good numerical stability in the $A$ matrix, but sacrifices numerical stability in the system as a whole, so unfortunately it’s not a good choice either.
The right choice is a cascade (series) implementation. If we can break down the system into 1st- and 2nd-order systems with appropriate eigenvalues that can be cascaded into a high-order system, then we don’t require additive cancellation between modes.
The algebra of cascaded state-space implementations is as follows:
If we have two state-space systems $S_1 = (A_1, B_1, C_1, D_1)$ and $S_2 = (A_2, B_2, C_2, D_2)$ such that the inputs of $S_2$ are the outputs of $S_1$, then the combined system $S_{12} = (A_{12}, B_{12}, C_{12}, D_{12})$ can be computed as
$$\begin{eqnarray} A_{12} &=& \begin{bmatrix}A_1 & 0 \cr B_2C_1 & A_2\end{bmatrix} \cr B_{12} &=& \begin{bmatrix}B_1 \cr B_2D_1 \end{bmatrix} \cr C_{12} &=& \begin{bmatrix}D_2C_1 & C_2\end{bmatrix} \cr D_{12} &=& D_2D_1 \cr \end{eqnarray}$$
This cascade combination can be continued to more than two systems; for example, the three-system cascade looks like this:
$$\begin{eqnarray} A_{123} &=& \begin{bmatrix}A_1 & 0 & 0 \cr B_2C_1 & A_2 & 0 \cr B_3D_2C_1 & B_3C_2 & A_3\end{bmatrix} \cr B_{123} &=& \begin{bmatrix}B_1 \cr B_2D_1 \cr B_3D_2D_1\end{bmatrix} \cr C_{123} &=& \begin{bmatrix}D_3D_2C_1 & D_3C_2 & C_3\end{bmatrix} \cr D_{123} &=& D_3D_2D_1 \cr \end{eqnarray}$$
And the four-system cascade looks like this:
$$\begin{eqnarray} A_{1234} &=& \begin{bmatrix}A_1 & 0 & 0 & 0\cr B_2C_1 & A_2 & 0 & 0 \cr B_3D_2C_1 & B_3C_2 & A_3 & 0 \cr B_4D_3D_2C_1 & B_4D_3C_2 & B_4C_3 & A_4 \end{bmatrix} \cr B_{1234} &=& \begin{bmatrix}B_1 \cr B_2D_1 \cr B_3D_2D_1 \cr B_4D_3D_2D_1\end{bmatrix} \cr C_{1234} &=& \begin{bmatrix}D_4D_3D_2C_1 & D_4D_3C_2 & D_4C_3 & C_4\end{bmatrix} \cr D_{1234} &=& D_4D_3D_2D_1 \cr \end{eqnarray}$$
In general, the resulting cascaded state-space system has an $A$ matrix which is block lower triangular; if the $D$ terms are zero then it’s block lower bidiagonal. The condition number of this matrix isn’t as low as with the modal form (though still much lower than controller canonical form!), but overall it’s much more stable:
pe = sspade.PadeExponential(12,12)
zeros,poles,k = pe.zpk
print "cascade form K(A) = ", np.linalg.cond(H3.A)
print H3.C
cascade form K(A) = 7056.31427616
[[ -66.02737609 -628.04637117 -63.9781648 -196.10488621 -59.72456992
-102.02762662 -52.888034 -56.65414163 -42.63766873 -28.21917019
-26.74418646 -8.83087886]]
So finally we can try our time-delay step response with high-order systems using a cascaded implementation:
t = np.arange(0,2,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [20,40,60,80,100]:
zeros, poles, k = pe.zpk
_,y = scipy.signal.step2(H,T=t)
ax.plot(t,y,label='p=q=%d' % q)
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.1,1.1);
SUCCESS! We did it!
We can reduce the ringing during the delay time (with a minor increase in ringing after the delay time) if we use a slightly lower order polynomial in the numerator of the transfer function:
t = np.arange(0,2,0.001)
fig = plt.figure(figsize=(10,6))
ax.set_xlabel('t')
ax.plot(t,t>=1,'--k')
for q in [20,40,60,80,100]:
p = q-4
zeros, poles, k = pe.zpk
_,y = scipy.signal.step2(H,T=t)
ax.plot(t,y,label='p=%d, q=%d' % (p,q))
ax.grid('on')
ax.legend(loc='best', labelspacing=0)
ax.set_ylim(-0.1,1.1);
And that’s pretty much all there is to say here....
### Taking it further
Hmmm. The overshoot is kind of annoying, though, isn’t it? What if, instead of taking all 100 poles and putting them into a time delay, we put some of them into a very slight low-pass filter to attenuate the ripples?
We have to be careful here, because most sharp-cutoff low-pass filters introduce more overshoot rather than less overshoot. The best one to use is probably a Bessel filter, for two reasons:
• it has very little overshoot
• it has a very flat group delay; in the passband the frequency-dependence of time delay is very low
There’s another reason too; the transfer function of a Bessel filter happens to be the same as the $p=q$ Padé approximation to a time delay, but with a numerator of 1, and the cutoff frequency scaled by a factor of $\frac{1}{2}$. (There’s those pesky hypergeometric functions again!) So we already have all the machinery to construct a state-space system for Bessel filters.
B5 = sspade.Bessel(5, 1.0)
print "5th order unit Bessel filter denominator polynomial"
Bp5 = np.poly(B5.poles)
assert np.all(np.imag(Bp5)<1e-12)
print np.real(Bp5)
_,y = scipy.signal.step2(H,T=t)
plt.plot(t,y);
5th order unit Bessel filter denominator polynomial
[ 1. 15. 105. 420. 945. 945.]
A unit Bessel filter has a low-frequency group delay of 1.0, so what we can do is allocate some of the time delay to a Bessel filter and some of the time delay to a Padé delay:
a = 0.05 # 5% of the delay goes to Bessel, the rest to Pade
n = 100 # order of the whole system
m = 10 # order of the Bessel filter
t = np.arange(0,2,0.001)
# System #3: Bessel only
fig=plt.figure(figsize=(10,10))
for label, H in [
('Bessel %d $\\rightarrow$ Pade %d,%d' % (m,pe.p,pe.q), H1),
('Bessel %d' % n, H3)
]:
_,y = scipy.signal.step2(H,T=t)
for axk in ax:
axk.plot(t,y,label=label)
for k in xrange(2):
if k > 0:
axk.set_xlim(0.9,1.1)
ax[k].legend(loc='best')
ax[k].set_ylim(-0.1,1.1)
The pure Bessel filter of order 100 is very smooth but not very steep. The pure Padé filter of order 100 (with numerator degree 90) is very steep but has quite a bit of ripple. The hybrid cascade isn’t quite as steep as the Padé filter but it greatly attenuates the ripple.
The thing about the Padé filter is that it optimizes the frequency response, and comes closest to any other filter of the same numerator/denominator at meeting the frequency response of a pure time delay. But time domain and frequency domain characteristics conflict to some degree: a low-pass filter with a perfectly sharp cutoff in the frequency domain implies ripple in the time domain, and vice versa. If we were optimizing, we might want to find the order-100 IIR filter that is the best least-squared approximation to the step response of a time-shifted version of a Gaussian filter:
$$h(t) = \frac{1}{\sqrt{2\pi}\tau}e^{-\frac{t^2}{2\tau^2}} \leftrightarrow H(s) = \frac{1}{\sqrt{2\pi}}e^{s^2\tau^2}$$
This is a noncausal filter, which has a step response $\newcommand{\erf}{\mathop{\rm erf\,}\nolimits} y(t) = \frac{1}{2}\left(1 + \erf{\frac{t}{\sqrt{2}\tau}}\right)$ where $\erf{x}$ is the error function; it ramps up smoothly from 0 to 1 with maximum slope $\frac{1}{\sqrt{2\pi}\tau}$:
import scipy.special
t = np.arange(-6,6,0.001)
tau = 2.0
plt.plot(t,0.5 + 0.5*scipy.special.erf(t/np.sqrt(2)/tau))
slope = 1.0/np.sqrt(2*np.pi)/tau
plt.plot([-0.5/slope, 0.5/slope],[0,1],'--r')
plt.ylim(-0.02,1.02);
For certain particular combinations of delay/order/steepness, we can come pretty close to a Gaussian filter. The Bessel filter of order $N$ and cutoff frequency $f_0$ is supposedly the best approximation to the Gaussian with time constant $\tau = T_0/\sqrt{2N}$ and an added time delay $T_0 = 1/f_0$:
n=15
f0=2.5
T0=1.0/f0
tau=T0/np.sqrt(2*n)
t = np.arange(0,2,0.001)*T0
_,y = scipy.signal.step2(Hb,T=t)
plt.figure(figsize=(8,6))
plt.plot(t,0.5 + 0.5*scipy.special.erf((t-T0)/np.sqrt(2)/tau))
plt.plot(t,y)
plt.ylim(-0.05,1.05)
plt.legend(['Gaussian filter tau=%.3f' % tau, 'Bessel (n=%d, f0=%.3f)' % (n,f0)],
loc='upper left');
The thing is, I’m not sure if there’s a straightforward procedure for finding the best nth-order IIR filter approximation to a Gaussian filter with arbitrary time delay — where “best” here means least-squared error in the time domain. It’s one of those things that Gauss or Fourier or Legendre probably wrote a paper about some general topic, at the age of 23, of which my problem is just one special case, and they stuck it in a drawer because at the time there weren’t any applications.
So I’m just going to leave it at that.
## Wrapup
• The Padé approximation to $e^{-sT}$ (PAEST) is a rational function with specified degree $p$ in the numerator and $q$ in the denominator, that comes very close to matching a time delay in the frequency domain, and can be used as a rational transfer function for simulating a delay of time T.
• Padé approximation in general is similar to a Taylor series approach, but it uses rational functions instead of polynomials, and can produce a better approximation for certain functions that aren’t well-suited to polynomial approximation.
• Polynomial coefficients for this approximation are integers (for T=1 at least) and can be computed in terms of factorials or combinations using recurrence formulas.
• scipy.signal.lti can be used with numerator and denominator polynomials to simulate a Padé time delay.
• The low-order approximations don’t look very good (especially with $p=q$; if $p$ is slightly less than $q$ they’re not too bad)
• Padé time delays don’t necessarily have to be very accurate, if they are combined with other systems that have similar or slower dynamics.
• Using scipy.signal.lti with polynomial coefficients of PAEST breaks down just above order 20.
• The reason this approach breaks down is because numerical conditioning of polynomial coefficients becomes TOXIC as the degree increases. (And, no, TOXIC isn’t some clever acronym, I’m just emphasizing it loudly.)
• State-space formulations of a linear system are not unique, and have several forms. The default state-space form used by scipy.signal.lti is the controller canonical form, which expresses a state-space system directly from the TOXIC polynomial coefficients of the transfer function numerator and denominator.
• Analysis of the poles and zeros of PAEST is an end-run around ill-conditioned polynomial coefficients, and can be accomplished using Newton-Raphson iteration to obtain poles and zeros directly, with high-accuracy, based on residue sums using the second-order differential equation form for the hypergeometric function ${}_1F_1$.
• PAEST with $p<q$ is unstable if $p$ is too small, because of the presence of right-half-plane poles.
• scipy.signal.lti has a constructor that can use ZPK form (zeros, poles, DC gain) but it fails miserably because scipy.signal.zpk2ss converts ZPK form to TOXIC polynomial coefficients as an intermediate form before converting to state-space
• We can convert poles and zeros of PAEST to state-space, but we have to take care in doing so.
• The best-conditioned state-space $A$ matrix uses a diagonalized (“modal”) form, but this fails on most high-order systems as well, because the ill-conditioning is pushed out to the $B$ and $C$ matrices, and convergence relies on delicate cancellation of sums of first-order systems.
• The cascaded series form appears to be the best overall method of formulating a high-order state-space system from poles and zeros, by first grouping them into order-1 (for real poles and zeros) or order-2 (for complex conjugate poles and zeros) systems and then forming overall state-space matrices using combination formulas.
• PAEST of order 100+ works well using the cascade form with scipy.signal.step2 and scipy.signal.lsim2 to simulate time-domain response.
• The ripple of PAEST can be reduced greatly for a given order system, by sacrificing some bandwidth and combining with a Bessel filter.
## References and other stuff
I’ve included the sspade Python module on my bitbucket account; it is free to use under the Apache license, and has a number of features I’ve used in this article, including these:
• cascade(sys1,sys2) — computes the state-space system that is the series cascade of systems sys1 and sys2
• zpk2ss_cascade(z,p,k) — converts ZPK = a list of zeros, poles, and DC gain, to a state-space LTI system using the cascade approach outlined in this article
• zpk2ss_modal(z,p,k) — converts ZPK to a state-space LTI system using a diagonal A matrix
• zpk2ss_realmodal(z,p,pk) — converts ZPK to a state-space LTI system using a real block diagonal A matrix composed of at most 2x2 blocks
• PadeExponential(p,q,f0) — utility class for creating Padé approximations of numerator order $p$ and denominator order $q$ for a time delay
• Bessel(n,f0) — utility class for creating Bessel filters
### Technical references
#### Numerical Analysis
The classic numerical analysis books I have on my bookshelf are:
• Numerical Recipes in C, Press, Teukolsky, Vetterling, and Flannery — the math explanations are invaluable and fairly easy to understand. The C code is awful, a straight port of Fortran from the days when alphanumeric characters were a scarce resource and so variable names had to be cryptic in order to fit in memory.
• The Art of Computer Programming, Vol. 2, Donald A. Knuth. This doesn’t get into linear algebra, but the foundations of scalar floating-point arithmetic are covered here. If you can understand a section of TAOCP fully without your brain hurting, you must be a mathematician.
• Matrix Computations, Gene H. Golub and Charles Van Loan. (Also known as “Golub and Van Loan”) I have the 3rd edition. This book is dense, and if you don’t understand the basics of linear algebra you will get lost quickly, but it includes error analysis bounds and order of operation counts on many matrix algorithms.
It’s also worth reading the paper What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg, which is a little bit more accessible (and more in-depth) than Knuth’s treatment, in my opinion.
These books may be useful as well to the advanced reader, and are on my to-get list — I can’t vouch for them yet, but their authors are well-reknowned.
#### Linear Systems and State-Space
Sorry, I don’t have anything I can recommend here; what I’ve read is either mediocre or falls deeply into theoretical matrix hell without really coming up for air. (The Dahlehs at MIT fall into this latter category.)
Linear Systems Theory by João P. Hespanha may be of use. Chapter 1 is freely available online, and I found it useful for the various state-space interconnections. But it looks like a typical textbook overreliance on MATLAB. Oh, and I can’t stand the heading typography; textbooks shouldn’t look like ships’ logs from Star Trek.
### Other stuff
Dedicated to A.R., who was frequently upset.
[ - ]
Comment by May 22, 2016
Really impressive. One of the best papers regarding Padè approximation. Thanks.
[ - ]
Comment by February 29, 2016
Small typo: in the extract from the Grunbaum article, where I assume you want the second derivative, you've got
where I think you want
This happens twice.
(This is what it looks like when I inspect the HTML source in my browser, after mathjax has done its work.)
[ - ]
Comment by February 29, 2016
Oops, the comment box ate my escaping. I'll try again.
you've got
’‘
where I think you want
’’
[ - ]
Comment by February 29, 2016
Yeah, I noticed it too; "escaping" was the problem in this case also (I use markdown, and the markdown converter sometimes eats my MathJax). thanks, I'll fix it.
[ - ]
Comment by July 30, 2020
Which Hungarian mathematician are you referring to?
[ - ]
Comment by July 30, 2020
Not sure. :-) The CsI sequence of letters makes me think I guess of Mihaly Csikszentmihaly.
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
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# What Are The Laws Of Motion?
The laws of motion are a set of scientific principles that explain the movement of objects. These laws were first proposed by English physicist Isaac Newton in 1687.
Checkout this video:
## What are the laws of motion?
The laws of motion are a set of scientific principles that explain the movement of objects. These laws were first proposed by English physicist Isaac Newton in 1687 and are still used today to describe the motion of objects. The three laws of motion are:
-An object at rest will stay at rest unless acted on by an external force.
-An object in motion will stay in motion in a straight line unless acted on by an external force.
-For every action, there is an equal and opposite reaction.
These laws explain why objects move the way they do, and they can be used to predict the future behavior of moving objects.
## What are the three laws of motion?
The three laws of motion were first formulated by Isaac Newton in 1687. They describe the relationship between a moving object and the forces that act upon it. The laws are:
-Law of Inertia: An object at rest will remain at rest unless acted upon by an outside force. An object in motion will continue in motion in a straight line unless acted upon by an outside force.
-Law of Acceleration: The rate of change of momentum is proportional to the force applied.
-Law of Action and Reaction: For every action, there is an equal and opposite reaction.
## What is the first law of motion?
In physics, the first law of motion is the law that states that an object at rest will stay at rest, and an object in motion will stay in motion, unless acted on by an unbalanced force. This law is also known as the law of inertia.
## What is the second law of motion?
The second law of motion is the force that is needed to move an object. This force is affected by the mass of the object and the acceleration of the object.
## What is the third law of motion?
The third law of motion is that for every action, there is an equal and opposite reaction.
## How do the laws of motion apply to everyday life?
The laws of motion are a set of scientific principles that describe the movement of objects. These laws were first formulated by English physicist Isaac Newton in the 17th century and they have been refined over the centuries by other scientists. The three most famous laws are the law of inertia, the law of acceleration, and the law of universal gravitation.
Inertia is the tendency of an object to resist changes in its motion. This is why it is difficult to get a heavy object moving, but once it is moving, it is difficult to stop. The law of acceleration says that the greater the force applied to an object, the greater its acceleration will be. The law of universal gravitation says that all objects in the universe are attracted to each other by a force called gravity.
These laws may seem theoretical and abstract, but they actually have many practical applications in everyday life. For example, when you drive a car, you rely on the law of acceleration to speed up and slow down. And when you drop something, you can thank gravity for making it fall to the ground!
## What are some real-world examples of the laws of motion?
The laws of motion are a set of scientific principles that describe the physical behavior of objects in motion. These laws are often used to predict how objects will move in real-world situations.
There are three laws of motion:
-The law of inertia: An object at rest will stay at rest, and an object in motion will stay in motion, unless acted upon by an external force.
-The law of acceleration: The rate of change of velocity is proportional to the force applied to an object.
-The law of action and reaction: Every action has an equal and opposite reaction.
All three laws are valid in both classical mechanics (which deals with objects moving at speeds much slower than the speed of light) and special relativity (which deals with objects moving at speeds approaching the speed of light). However, the law of inertia is modified in special relativity to take into account the effects of time dilation.
## What are some fun facts about the laws of motion?
The laws of motion are a set of three physical laws that govern the way objects move. The laws are:
-Law of Inertia: An object will stay at rest or continue moving in a straight line unless acted on by an outside force.
-Law of Acceleration: The rate of change of velocity is proportional to the force applied to an object.
-Law of Action and Reaction: For every action, there is an equal and opposite reaction.
These laws were first formulated by Isaac Newton in his 1687 book Philosophiæ Naturalis Principia Mathematica. Newton’s laws of motion are still used today to explain the movement of objects.
## What are some interesting applications of the laws of motion?
There are many interesting applications of Newton’s laws of motion. Here are just a few examples:
-An object in motion will stay in motion unless an external force acts on it. This is why it is difficult to stop a moving train or car. The momentum of the object keeps it moving until a force (such as friction) slows it down.
-A ball thrown into the air will come back down to the ground because the force of gravity is pulling it down.
-A rocket can be launched into space because the engines create a force that is greater than the force of gravity.
## Did you know?
Newton’s three laws of motion are the foundation for classical mechanics. These laws describe the relationship between a body and the forces acting upon it, and its motion in response to those forces.
The first law, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This means that an object will not start moving on its own, and will continue moving in a straight line until another force acts upon it.
The second law explains what happens when a force is applied to an object. It states that the object will accelerate in the direction of the force, and that the amount of acceleration is proportional to the strength of the force. This means that if you double the force applied to an object, it will accelerate twice as much.
The third law states that for every action there is an equal and opposite reaction. This means that if one object exerts a force on another object, the second object will exert an equal and opposite force back on the first.
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# Most counted object in the world
Fisher Flour has had scone stands at the Western Washington Fair for decades. They announced last week that they had sold their 100 millionth scone. I sure that at some point over the years someone accidentally put 14 scones into a baker’s dozen bag. Mistakes happen; you couldn’t definitively point to number 100,000,000. But I’ll give them the benefit of the doubt and say they were pretty close.
Which got me wondering, what is the most counted object in the world? I’m talking individual objects, so the value of currency printed by the U.S. Mint doesn’t count, but the number of bills of any particular denomination would.
I don’t know that we’d ever find a certain answer for this, but can anyone suggest any possible answers?
McDonald’s hamburgers?
Light years.
No, wait, centi light years!
No, wait!
That’s possible. Are they still counting, and are the “billions and billions served” all just hamburgers?
Doesn’t count; not a physical object.
Also, distance isn’t countable. I think we’ll have to rule out continuous quantities, but that breaks down at some point. Mass isn’t countable in any everyday sense, but ultimately you can only divide it into a countable number of atoms…
According to this, they stopped in 1994. Didn’t know that. Changed the signs to “Billions and Billions Served”.
To some degree of certainty, we know how many atoms are in a given mass of material. But that number is derived mathematically. Even if we have the technology to count them as distinct objects, that hasn’t been done as far as I know.
Votes? In a typical U.S. Election , well over a hundred million are tallied. An Indian election can top 700 million.
Steps? I may be a little OCD but i count steps almost every time I go up a flight of stairs, even if I climbed them a thousand times before. I don’t normally count them on the way down.
Anybody’s right thumb.
Hmm, that’s an interesting idea. But a “vote” isn’t necessarily the same object from one jurisdiction to another, or may not be a physical object at all. My area used to vote by machine; I assume it just kept some mechanical count and those numbers were added up at the end of the day.
I don’t know about you, but I only have one.
But how often does it get counted?
According to this site, there have been 455,627,740,918 Lincoln Cents minted since 1909.
How about coins from other countries?
Do the number of decimals in pi count?
Sheep?
Okay, that’s definitely a contender.
I assume even the Mint can’t count to 455 billion without a mistake or two along the way; I wonder how close to the real number that is.
How many people have been reading this thread with a little voice in their head going “*one, ah ah ahhh… two, ah ah ahhh… *”?
You forgot the flashes of lightning and thunder. Every. Damn. Time.
Bullets. Armies like to keep track of those - I bet you could find a count of every round of ammo purchased by the U.S. military in the 20th century, and that’s just one country.
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View Full Version : Four Color Theorem
Pages : [1] 2
Matthew
2005-Sep-06, 07:26 AM
So you only need four color's to color a map? What if all the localaties met at a single point? See the attached diagram. Lets a ssume the lines are perfect and borders intersect at a single point. Would that mean that more than 4 colors are needed?
01101001
2005-Sep-06, 08:05 AM
Wikipedia: Four color theorem (http://en.wikipedia.org/wiki/Four_color_theorem)
The four color theorem states that any plane separated into regions, such as a political map of the counties of a state, can be colored using no more than four colors in such a way that no two adjacent regions receive the same color. Two regions are called adjacent if they share a border segment, not just a point.
jfribrg
2005-Sep-06, 12:50 PM
There is a book, called Four Colours Suffice (http://www.amazon.com/exec/obidos/tg/detail/-/014100908X/qid=1126010749/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/102-9099553-7088128?v=glance&s=books&n=507846) that gives the history of the problem. It is a layman's book, but the math level varies. Sometimes it is too simplistic. Other times it goes too fast, but in all, I have a better understanding of the proof than I did before I read it.
worzel
2005-Sep-06, 01:55 PM
But have they proved it for all colours?
jfribrg
2005-Sep-06, 02:13 PM
But have they proved it for all colours?
I'm not sure I understand what you mean. It is trivial to prove that 3 or fewer colors is not sufficient. All you need is to show that there is a map where 3 colors aren't enough. Here is a simple counterexample: Start with a thick circle with a hole in the center. Section the circle into 3 parts, each of which must be a different color. The hole in the center must also be different from the first 3 colors used on the circle, which proves that 3 colors are too few. For 5 colors, take a 4 color map and arbitrarily change one of the areas to a color that isn't anywhere else on the map. For 6 colors, take the above mentioned 5 color map and change one area to a color that isn't yet used. Do this for as many colors as you have in your crayon box.
hhEb09'1
2005-Sep-06, 02:39 PM
I know what he means. **bap** now, say you're sorry
It has long been known that a map on a torus requires seven colors, and seven always suffices. To see that seven is necessary, take a torus, and divide it into bands of seven different regions encircling the torus through its hole. Now, cut a line around the top of the "donut" and slide the outside parts along the line 2 1/2 regions. All seven regions will share a boundary with all six other regions.
worzel
2005-Sep-06, 03:23 PM
What I acutally meant was although it is true for the traditional colours used on a political map, has it also been proven for, say, 4 slightly different shades of pink.
I know what he means. **bap** now, say you're sorrySorry. :o
zrice03
2005-Sep-06, 09:27 PM
It doesn't matter which colors you use, as long as they are all different. So, yes, it would work for 4 different shades of pink because they are all different. It would also work for four shades of blue, green, maybe two of red and two of grey, etc.
Basically, when you have four countries that all touch each other, one of those contries has to be completely surrounded. Notice in jfribrg's diagram, the brown country is completely surrounded. Go ahead and try it yourself on a piece of paper.
There is, however, an unstated premise: All the countries have to be in one piece. For example, the United States is not in one piece because of Alaska. There are two large "pieces" of the U.S. (and various islands)... If you tried this on the actual map of the Earth, you might not find the 4 color theorem is true (then again, it might be. I've never tried it).
worzel
2005-Sep-06, 11:33 PM
It doesn't matter which colors you use, as long as they are all different. So, yes, it would work for 4 different shades of pink because they are all different. It would also work for four shades of blue, green, maybe two of red and two of grey, etc.
Ok, thanks for the clarification ;)
There is, however, an unstated premise: All the countries have to be in one piece. For example, the United States is not in one piece because of Alaska. There are two large "pieces" of the U.S. (and various islands)... If you tried this on the actual map of the Earth, you might not find the 4 color theorem is true (then again, it might be. I've never tried it).
I think you would find that the 4 color theorem is true irrespective of the potential problem you raise, it just isn't the same thing you're describing.
hhEb09'1
2005-Sep-07, 09:08 AM
LOL. Yep. Just take any map of any set of N countries and put an "embassy" for each one, in each one. Then, you need N colors.
zrice03
2005-Sep-08, 05:20 AM
LOL. Yep. Just take any map of any set of N countries and put an "embassy" for each one, in each one. Then, you need N colors.
Exactly.
jfribrg
2005-Sep-08, 06:56 PM
Basically, when you have four countries that all touch each other, one of those contries has to be completely surrounded.
Basically that is the key, but proving it is the problem.
There is, however, an unstated premise: All the countries have to be in one piece.
Actually, this is stated explicitly in the formal definition of the problem.
For example, the United States is not in one piece because of Alaska.
Hawaii also, but the problem is limited to contiguous areas on a flat (two dimensional) surface that share a border. IIRC, for these purposes, Michigan is a problem too.
worzel
2005-Sep-08, 10:06 PM
There is, however, an unstated premise: All the countries have to be in one piece.Actually, this is stated explicitly in the formal definition of the problem.
Exactly. ;)
peter eldergill
2005-Sep-08, 11:01 PM
Are there any practical applications for the four colour theorem or is it just one of those "neat things" to know like Fermat's Last Theorem and Relativity (Ha! Kidding!)
Pete
worzel
2005-Sep-08, 11:05 PM
Are there any practical applications for the four colour theorem or is it just one of those "neat things" to know like Fermat's Last Theorem and Relativity (Ha! Kidding!)
Pete
Maybe it's why there are CMYK printers - but that would suggest that there is a three colour theorem waiting to be proved.
Halcyon Dayz
2005-Sep-08, 11:15 PM
It is obviously useful for (game)mapmakers.
BenM
2005-Sep-08, 11:16 PM
I spent an entire summer as a kid trying to draw maps that would disprove the theorem. It wasn't a complete waste of time, because it did get me started down the path of scientific inquiry.
worzel
2005-Sep-09, 12:25 AM
I was so naive when I first saw the problem that I figured that as it was obvious, I would be able to come up with a simple proof and dazzle the world :o
peter eldergill
2005-Sep-09, 03:44 AM
My Uncle thought he had a proof of Fermat's Last theorem, less than a page. Even I found his mistake quickly.
He also had a "proof" that an angle can be trisected. I couldn't follow what he was doing, as it was purely geometrical, which I'm not familiar with
Well, gotta go!
Pete
montebianco
2005-Sep-09, 04:03 AM
IIRC, for these purposes, Michigan is a problem too.
I guess the problem would be not that Michigan comes in two pieces (they're easy enough to connect, at least on the map), but that neither Michigan and Illinois nor Indiana and Wisconsin share a border...
montebianco
2005-Sep-09, 04:05 AM
My Uncle thought he had a proof of Fermat's Last theorem, less than a page.
Maybe it was the same proof Fermat had :)
Enzp
2005-Sep-09, 04:16 AM
And we try to keep as much water as possible between us and Canada. (Here being Michigan) If the water is blue, does that count as a color?
I'm with Worzel here.
And what if the colors run?
Who among us before they were old enough to know how smart they weren't didn't figure, "Oh, I can beat that, lemme at it..."?
montebianco
2005-Sep-09, 05:12 AM
And we try to keep as much water as possible between us and Canada. (Here being Michigan)
Was just there a couple of weeks ago, didn't look like a particularly long bridge :)
hhEb09'1
2005-Sep-09, 10:03 AM
He also had a "proof" that an angle can be trisected. I couldn't follow what he was doing, as it was purely geometrical, which I'm not familiar withHere's a way (http://mentock.home.mindspring.com/trisect.htm) to trisect an angle. :)
jfribrg
2005-Sep-09, 10:54 AM
Here's a way (http://mentock.home.mindspring.com/trisect.htm) to trisect an angle. :)
From page 101 of the book "Elements of Abstract Algebra", which provides an interesting use of field theory to provide straightedge-and-compass proofs, this trisection is addressed.
First let us show how angles may be trisected easily if we allow and incorrect usage of the straightedge. (Apparently this practical construction was known to ancient geometers.)
As with other math problems, the "rules" are clearly defined. From the same book, here are the rules:
(1)the points (0,0) and (1,0) are constructible. Any two points of the plane may be chosen for (0,0) and (1,0) and the distance between them taken as the unit length.
(2) A circle with a constructible point as center and a constructible length as radius is constructible. A constructible length is the distance between two constructible points.
(3) The intersection of two constructible lines is a constructible point.
(4) The points (or point) of intersection of a constructible line and a constructible circle are constructible
(5) The points (or point) of intersection of two constructible circles are constructible
Step 3 in the link from the previous post is the illegal operation. The result is also that you get an angle that is 1/3 of the original, but the angles are in different places. When you bisect an angle using "legal" operations, you split the original angle in half, but here you generate an angle elsewhere, which is a different issue.
The book then provides a single counterexample. First, the constructible points and operations are restated in terms of complex numbers, and a proof is given that these operations and points form a field. It then shows using field theory that it is impossible to trisect a 60 degree angle.
worzel
2005-Sep-09, 11:06 AM
Step 3 in the link from the previous post is the illegal operation. The result is also that you get an angle that is 1/3 of the original, but the angles are in different places. When you bisect an angle using "legal" operations, you split the original angle in half, but here you generate an angle elsewhere, which is a different issue.
You could use the compass to measure the chord and then mark that off from E or F.
hhEb09'1
2005-Sep-09, 01:56 PM
Step 3 in the link from the previous post is the illegal operation.Step 2 is also "illegal". :)
However, I use the straightedge legally, throughout.
The result is also that you get an angle that is 1/3 of the original, but the angles are in different places. When you bisect an angle using "legal" operations, you split the original angle in half, but here you generate an angle elsewhere, which is a different issue.No, it's not. As worzel points out, you can use that angle to easily construct a copy anywhere else, including between the original angle.
Enzp
2005-Sep-10, 03:53 AM
Montebianco - which bridge? Try walking that Mackinac bridge, she is longer than she looks. Every so often a car or truck blows over on it. We have our ways...
worzel
2005-Sep-10, 11:03 AM
Two questions.
The first to jfribrg or anyone else who knows the answer. What purpose are those 5 rules of geometric construction supposed to serve given that they appear not to define what can actually be done, geometrically, with a straight edge and compass?
The second to hhEb09'1. Did you come up with that trisection yourself?
hhEb09'1
2005-Sep-10, 02:34 PM
The second to hhEb09'1. Did you come up with that trisection yourself?The neusis or verging "cheat" trisections have been known for over two thousand years. That particular one is my modification of Archimedes's--which, according to The Book of Numbers by Conway and Guy, uses a marked ruler--which is an incorrect use of a ruler, as jfribrg's link says. Apparently, a lot of people know about the incorrect use of rulers, so I changed it to an incorrect use of a compass instead. :)
montebianco
2005-Sep-11, 01:34 AM
Montebianco - which bridge? Try walking that Mackinac bridge, she is longer than she looks. Every so often a car or truck blows over on it. We have our ways...
I was talking about the one between Detroit and Windsor, where one can drive in a southward direction from the US into Canada :)
peter eldergill
2005-Sep-11, 04:30 AM
Wow, guys & gals, I didn't mean to start a conspiracy (ahem) about trisecting an angle. There seems to be some gaps in the general idea. Specifically, I don't actually know what the original theorem says (trisecting angles) and why there is a "fake" proof
Could someone please state for me exactly what the theorem is, and thy there is a "fake" proof for it? (As far as I know, you can't trisect an angle with a stright edge and compass) Is there an exception maybe for a specific case?
L8R
Pete
hhEb09'1
2005-Sep-11, 04:43 PM
Could someone please state for me exactly what the theorem is, and thy there is a "fake" proof for it? (As far as I know, you can't trisect an angle with a stright edge and compass) The construction I linked to is an actual trisection, but the classical construction uses a rule with no marks, and a compass that collapses as soon as you lift it up from drawing a circle--so that construction is illegal according to those criteria. And so is worzel's idea of transferring a chord length and marking it off from another point--but there are alternate ways of achieving the same thing.
Enzp
2005-Sep-13, 06:04 AM
Ah, the Ambassador Bridge. Longest suspension bridge in the world...in 1929. 1850 foot span and 7490 feet long. And a lovely bridge.
The Big Mac has an 8614 foot span and is five miles long.
Maddad
2005-Sep-14, 04:42 PM
Basically that [when you have four countries that all touch each other, one of those contries has to be completely surrounded.] is the key, but proving it is the problem.I proved it 34 years ago in high school with pencil and paper before anyone had computers. I've since given talks on why the current attempts to solve the problem are not trying to solve the problem, but rather trying to rename it. I've also taken the obligatory cheap shots at computers used in a supposed proof. An image of the proof is the only tatoo that I've ever considered getter, but I chickened out.
I spent an entire summer as a kid trying to draw maps that would disprove the theorem. It wasn't a complete waste of time, because it did get me started down the path of scientific inquiry.Yep. That how I started too.
David Madison
Maddad@Hotmail.com?Subject=FourColorMapTheorem
worzel
2005-Sep-14, 05:17 PM
I proved it 34 years ago in high school with pencil and paper before anyone had computers.Wow! I don't suppose you've got your proof on the web somewhere, have you?
Maddad
2005-Sep-14, 06:56 PM
No sir, not on the web. I've been thinking of trying to find someone to officially recognize a solution. I remember talking to a guy in the science department (head?) at Ohio State University several years ago. He said, "Yeah, sure, send in your proof so that I can disprove it" or something like that. He also was worried that I'd submit reams of computer data, and he didn't want to have to read through it all. He considered himself a lightning rod for people working on the puzzle.
It's not really hard at all. I should have spoken with him again, but I was a bit frosted by his demeanor, and just never got around to it. Think I'll write it up tonight and put it on my website. My temporary website ( http://www.geocities.com/davidmadison01 ); my regular one ( http://www.maddad.org ) is down.
worzel
2005-Sep-14, 09:10 PM
Has anyone checked it over for you? (not that I'm offering, not much of theorem cruncher me)
hhEb09'1
2005-Sep-15, 04:37 PM
I proved it 34 years ago in high school with pencil and paper before anyone had computers. I've since given talks on why the current attempts to solve the problem are not trying to solve the problem, but rather trying to rename it. I've also taken the obligatory cheap shots at computers used in a supposed proof. An image of the proof is the only tatoo that I've ever considered getter, but I chickened out.Post it! if we can't find the error within a day, I'll eat my hat. :)
Maddad
2005-Sep-15, 11:51 PM
Wow! I don't suppose you've got your proof on the web somewhere, have you?I just asked Phil in an email if he would support my authorship if I posted the solution here. If he agrees, I'll step you through it.
peter eldergill
2005-Sep-16, 01:42 AM
Maddad said:
I proved it 34 years ago in high school with pencil and paper before anyone had computers.
Are you referring to the four colour theorem in total, or just a part of it. It would be a remarkable achievement to do something like that in high school, although I am a bit skeptical that you have a valid proof....
As far as I know, the computer proof of this takes into consideration all 2000 (or so) possible cases that arise from the theorem. I'm curious to see how you could reduce that into something simple
Looking forward to it! I'm assuming that I'll be able to follow your proof as you had only highschool training at the time
I'll also look forward to sharing it with my grade 12 Data Management class, where I actually discuss the theorem with the class (briefly)
L8R
Pete
hhEb09'1
2005-Sep-16, 02:18 PM
I've since given talks on why the current attempts to solve the problem are not trying to solve the problem, but rather trying to rename it. I've also taken the obligatory cheap shots at computers used in a supposed proof. I notice in your resume (http://www.geocities.com/davidmadison01/david500.htm) that you say yours is the only proof so far. Why do you discount the one by Appel and Haken (cite (http://www.math.gatech.edu/~thomas/FC/fourcolor.html)--and I notice they say they have another proof)?
worzel
2005-Sep-16, 03:57 PM
I notice in your resume (http://www.geocities.com/davidmadison01/david500.htm) that you say yours is the only proof so far. Oh that's up now. I came across it in google the other day but it was down.
Wish I had a resume like that, you know, pretty normal really, up until the last line of other interests, "Oh yeah, and I'm the only person to have proved fermat's last theorem in the space of a margin."
Hope Phil agrees to supporting your authorship, Madad.
Maddad
2005-Sep-16, 11:59 PM
I notice in your resume (http://www.geocities.com/davidmadison01/david500.htm) that you say yours is the only proof so far. Why do you discount the one by Appel and Haken (cite (http://www.math.gatech.edu/~thomas/FC/fourcolor.html)--and I notice they say they have another proof)?Valid question. As mentioned in that page, that proof is not satisfactory.
There are a number of reasons that a computer cannot support any proof of the theorem. The problem starts because the logic is faulty. If I told you that my good friend Fred who lives in the apartment down the street is really smart, has worked on the thorem a long time, and has been unable to solve it, would you accept that as proof that a map needs no more than four colors? I don't think so. You would say that you have no way of knowing what methods Fred tried, why he even thinks it can't be done, and since you have no way of knowing the method he used, you cannot ensure that he didn't skip some important step, smart as he is or not (or so I claim). There is no way to get the same answer that Fred did, so the reproducibility requirement of science is left unsatisfied. Even if he's tried and failed, we have no way to tell that he he finished looking, so we have no way to tell if he would solve it by putting in another five minutes of study.
A computer proof suffers from exactly the same problem, plus some others. Who wrote the program that performed the solution? Did he write it accurately, or did he fall victim to GIGO? Remember that nobody has checked the answer because it is too big, so in effect we have no reproducibility. Was the method used by the programmer capable of testing the problem? All you know is that some progammer somewhere thinks so. Has there ever been an error in programming anywhere in the history of computing? How do you know one of them didn't show up here? Was his computer search exhaustive enough? The programmer claims so, but how do you actually know since his work cannot be checked? And what about hardware failures? Some of you may remember the difficulty with the original pentium that inaccurately performed some obscure mathematical calculations. Intel had to recall those processors and replace them with ones that had been fixed. Could such a problem ever happen agin? You say no? How do you know that, for certain, when you must know that for certain to say that the problem has been proved? Suppose that this is not the first time that microprocessors have been manufactured and sold that incorrectly performed math operations? If the original Pentium was the second offence, not the first, would you worry a bit more that such a problme MIGHT arise again? While you probable are familiar with the inadequacy of the original Pendium, few people remember that the 386 chip had a similar problem that went undetected until the release of Autocad 10. Intel went to work, fixing the problem, and released a replacement processor, which makes the Pentium problem the second time around. But it's not even the second time, because Intel botched the repair. Their fix was still broken, so they had to release yet a third version of the 386 to get it to quit making math errors.
This track record with computer hardware is stark explanation of why a computer cannot be used to prove the Four Color Map Theorem. The problem though is even bigger. Had the computer drawn the desired map in which five territories each touched another, it still would not have prooved the problem. Can anyone guess why?
Maddad
2005-Sep-17, 12:01 AM
Hope Phil agrees to supporting your authorship, Madad.He strongly advised against it. Suggested that I submit it to an official publication somewhere for peer review. I don't know who might take it though.
cfgauss
2005-Sep-17, 05:44 AM
There are a number of reasons that a computer cannot support any proof of the theorem. The problem starts because the logic is faulty. If I told you that my good friend Fred [..............]
Ya, too bad we don't, you know, write down what we tell our computers to do. If only there was some kind of "code" we could write, that was algorithmic, and other people could check...
A computer proof suffers from exactly the same problem, plus some others. Who wrote the program that performed the solution? Did he write it accurately, or did he fall victim to GIGO?
If only other mathematicians would develop and learn this "code" they could check it so easily...
Has there ever been an error in programming anywhere in the history of computing? How do you know one of them didn't show up here? Was his computer search exhaustive enough? The programmer claims so, but how do you actually know since his work cannot be checked?
And if only the program was only made to show us its work instead of just printing "proved" or "not proved".
And what about hardware failures? Some of you may remember the difficulty with the original pentium that inaccurately performed some obscure mathematical calculations. Intel had to recall those processors and replace them with ones that had been fixed.
Hardly obscure... they knew about it when they began making them. They just assumed (rightly) that the error would be small enough that it wouldn't effect most people.
Could such a problem ever happen agin? You say no? How do you know that, for certain, when you must know that for certain to say that the problem has been proved?
[.................................................]
And if only they could some how use this hypothetical "code" to check if the program gave the same "proved" or "not proved" answer on systems with different components...
Maddad
2005-Sep-17, 04:38 PM
Ok, the last part of that long post asked the question, "If someone drew a satisfactory map with five colors, why would it not prove the theorem?"
jfribrg
2005-Sep-18, 02:07 AM
Ok, the last part of that long post asked the question, "If someone drew a satisfactory map with five colors, why would it not prove the theorem?"
Because the theorem states that it can be done with 4. It had been known for a century or more IIRC that 5 is sufficient. Just because it can be done with 5 doesnt prove that it can be done with 4.
worzel
2005-Sep-18, 09:00 AM
Ok, the last part of that long post asked the question, "If someone drew a satisfactory map with five colors, why would it not prove the theorem?"Actually, what you asked was:
Had the computer drawn the desired map in which five territories each touched another, it still would not have prooved the problem.If the computer, or anyone else for that matter, drew a map where five territories each touched one another then wouldn't that disprove the four color theorem.
hhEb09'1
2005-Sep-18, 10:57 AM
If the computer, or anyone else for that matter, drew a map where five territories each touched one another then wouldn't that disprove the four color theorem.Yes. A map with all five countries touching each other would mean that five colors would be needed, and the theorem would be false. "Touching" of course means sharing a length of border, not just a point (as all pieces of a pie do).
Martin Gardner once wrote a short story where the main character was told about an island where all five divisions of the island touched each other. He was OK with that since the Four Color Theorem had not been proved (even though it had been shown without computers that just five countries could not touch each other), but after exploring the island and verifying the claim, he was shocked when he suddenly realized that the ocean was a sixth region that touched all the other five.
Maddad
2005-Sep-18, 09:11 PM
If the computer, or anyone else for that matter, drew a map where five territories each touched one another then wouldn't that disprove the four color theore.The problem is that the person doing so would immediately be asked, "Hey, if you can draw a map with five colors, could you do it with six?" The new celebrity would scatch his head, mutter something about five being hard enough, and maybe give it a try. People would be talking about the new Five Color Map Theorem
The problem is that success at drawing the map doesn't solve the real problem; it just changes the name. The real problem is to establish the upper limit in colors.
The irony is that both success and failure in drawing the map result in failure of a meaningful proof.
ASEI
2005-Sep-18, 09:19 PM
If you want to prove the four color theorem, you could try to construct a proof as follows:
Assuming that a map requires more than four colors, it would have to have five seperate regions each capable of touching each of the other four regions by an independent route on a 2-d surface.
An analogy would be using five points to symbolize the regions, and non-intersecting lines to symbolize the connections touching each region. Can you get all five to tough all other four points? (4 connections/point, 20 total?)
I doodled a lot of these, but couldn't find any way to get all five points to connect to all their necessary neighbors.
A proof of the four color theorem could work the same way, assuming all possible combinations are tried.
worzel
2005-Sep-18, 11:44 PM
Maddad, I am totally confused about what you are trying to say [ subtext being I think you are totally confused :) ].
The problem is that the person doing so would immediately be asked, "Hey, if you can draw a map with five colors, could you do it with six?" The new celebrity would scatch his head, mutter something about five being hard enough, and maybe give it a try. People would be talking about the new Five Color Map TheoremThat's just silly. If you have proved it for 5 colours then you have proved it for 5 or more colours. The theorem is that at most 4 colours are needed, not that you can't do it with more colours if you like. If it is proven that 5 suffice, then we know that we can take any map and randomly select one region and instead colour it with a 6th colour. Ditto for 7, all the way up to n colours for n regions, which is trivial of course.
The problem is that success at drawing the map doesn't solve the real problem; it just changes the name. The real problem is to establish the upper limit in colors.What do you mean by "success at drawing the map"? The theorem is not that there exists a map such that... and the proof, therefore, is not simply an example of such a map. The theorem is that for all maps, 4 colours suffice. The real problem was to establish the lower limit of colours needed, which has now been proven to be 4 (it's easy to come up with a counter example for 3 colours).
The irony is that both success and failure in drawing the map result in failure of a meaningful proof.Let's get this straight:
A proof that 5 colours suffices for any map proves that 6,7,...,n colours also suffice but says nothing about 4 colours.
A proof that 4 colours suffice for any map proves that 5,6,7,...n colours also suffice and subsumes any proof for any number > 4.
An example of a map coloured with 5 colours (which is what you seem to be getting at) says nothing except that 5 colours suffice for that one example.
A map which has five regions which all share common boundaries with each other would disprove the 4 colour theorem because it is a counter example that requires at least 5 colours. As the 4 colour theorem has been proved, no such map exists.
Do you disagree with any of those?
I'm sorry if I seem to be stating the blindingly obvious, but for someone who claims to have proven the four colour theorem you don't seem to have much of a grasp of what the theorem says. Hopefully I just misunderstood you.
worzel
2005-Sep-18, 11:51 PM
Martin Gardner once wrote a short story where the main character was told about an island where all five divisions of the island touched each other. He was OK with that since the Four Color Theorem had not been proved (even though it had been shown without computers that just five countries could not touch each other), but after exploring the island and verifying the claim, he was shocked when he suddenly realized that the ocean was a sixth region that touched all the other five.Was the island inhabited by Tauruslanders? :)
Maddad
2005-Sep-20, 10:48 PM
Maddad, I am totally confused about what you are trying to say. . . If you have proved it for 5 colours then you have proved it for 5 or more colours.Yes, proving the FCMT for four colors also proves it for five or more. However, current attempts to deal with the FCMT involve trying to draw the map. Should that attempt succeed, then it would disprove the FCMT. I can see why you were confused by my post; I wasn't clear on that. If you have disproved the FCMT in four colors by drawing such a map, it does not necessarily disprove it for five or more.
What do you mean by "success at drawing the map"?People are trying to draw a map that does display five territories, each of which touches the other four in an arc or line segment. If they succeed, then they have disproved the FCMT. If they fail, then the results are indeterminant.
Now, are we both on the same page, and do you agree with the following:
1. A computer program that says it proves you cannot draw a five color map is insufficient proof of the problem.
2. Success at drawing a map with five territories simply renames the problem to the Five Color Map Theorem rather than establishing the maximum number of colors that you might have to use.
worzel
2005-Sep-20, 11:57 PM
Yes, proving the FCMT for four colors also proves it for five or more. However, current attempts to deal with the FCMT involve trying to draw the map. Should that attempt succeed, then it would disprove the FCMT. I can see why you were confused by my post; I wasn't clear on that. If you have disproved the FCMT in four colors by drawing such a map, it does not necessarily disprove it for five or more.
People are trying to draw a map that does display five territories, each of which touches the other four in an arc or line segment. If they succeed, then they have disproved the FCMT. If they fail, then the results are indeterminant.
Now, are we both on the same pageYes, I agree with all of that so far, if we assume for the moment that the FCMT is not yet proved.
and do you agree with the following:
1. A computer program that says it proves you cannot draw a five color map is insufficient proof of the problem.No, I strongly disagree. cfguass made some good points about this.
I would go as far as to say that a formal system of logic is, by definition, a system that can be computerised. The axioms and rules of inference give us all we need to write a theorem prover. If the code is correct (it can be checked by mathematicians just like any formal system can) and the computer executes it correctly (the result can be checked on many different architechtures) then it is safe. You may have some philosophical problems with it, but for me it is no different to using a calculator to compute some of the arithmetic in the proof. Indeed, Godel showed how one can turn each step of a proof into an arithmetical statement. That is not to say that finding the proof is algorithmic, Godel had something to say about that as well :)
IIRC computerised theorem proving wasn't even what was used for the FCMT. The non-computerised part (the bit I presume you'd be happy with) was to show that any map can be reduced to an equivalent map (for the purposes of the theorem) from a finte set. The computer then simply checked that each one of these maps could be coloured with only four colours. Proving something for a finite number of cases can be done by brute fource and is very different to proving something for an infinte number of cases. Assuming the code has been checked and the results verified, the program did undeniably colour each case with only four colours.
2. Success at drawing a map with five territories simply renames the problem to the Five Color Map Theorem rather than establishing the maximum number of colors that you might have to use.Yeah, I guess it would. But that's rather moot as the Four Colour Map Theorem has already been proven - so any search for a map with five pairwise neighbouring regions is doomed to failure. As you claim to have proven the FCMT yourself, presumably you agree with this even if you are not convinced by the generally accepted proof.
hhEb09'1
2005-Sep-21, 05:42 PM
People are trying to draw a map that does display five territories, each of which touches the other four in an arc or line segment. If they succeed, then they have disproved the FCMT. If they fail, then the results are indeterminant.They will always fail. It is fairly easy to prove that five territories cannot each touch the other four.
Now, are we both on the same page, and do you agree with the following:
1. A computer program that says it proves you cannot draw a five color map is insufficient proof of the problem.
2. Success at drawing a map with five territories simply renames the problem to the Five Color Map Theorem rather than establishing the maximum number of colors that you might have to use.Not "renames" the problem. It would no longer be a problem, since it was proven that five is sufficient long before computers even. :)
But, as I said, it would take much more than just a map of five territories. Link (http://mathworld.wolfram.com/Four-ColorTheorem.html)
Maddad
2005-Sep-22, 03:02 AM
the Four Colour Map Theorem has already been proven - so any search for a map with five pairwise neighbouring regions is doomed to failure. . . . presumably you agree with this even if you are not convinced by the generally accepted proof.It's finding a starting point in your post is difficult because I have a great many choices. I'm going to pick a couple.
The problem with a computer proof is not philosophy; it's that we are unable to know, to the level necessary to claim proof of the Theorem, that the computer did in fact perform it's calculation correctly. Maybe you quickly passed over my original reasoning, but we are unable to know that the code is correct because we are unable to be sure that the microcode in the processor is not flawed. Remember, we have not one, but three prior instances in which this microcode later turned out to be defective.
I am glad that you brought up reduction of a map from a finite set as part of a proof. I used a reverse process in developing my own proof. I do not need a computer; I never exceed two possibilities for where and how a new teritority might be positioned and connected. One side of an 8½ x 11 sheet of paper will do. If you're using a computer for selecting one of two possibilities, you're in league with my mother-in-law.
worzel
2005-Sep-22, 08:43 AM
The problem with a computer proof is not philosophy; it's that we are unable to know, to the level necessary to claim proof of the Theorem, that the computer did in fact perform it's calculation correctly. Maybe you quickly passed over my original reasoning, but we are unable to know that the code is correct because we are unable to be sure that the microcode in the processor is not flawed. Remember, we have not one, but three prior instances in which this microcode later turned out to be defective.You've restated what you said originally but haven't answered cfgauss's points. If a microcode bug affected the program in anyway then it wouldn't run consistently on different platforms.
I am glad that you brought up reduction of a map from a finite set as part of a proof. I used a reverse process in developing my own proof. I do not need a computer; I never exceed two possibilities for where and how a new teritority might be positioned and connected. One side of an 8½ x 11 sheet of paper will do.Well it shouldn't be too hard to knock up a webpage explaining it then :)
Maddad
2005-Sep-22, 06:57 PM
I've addressed them several times now. The code can run consistently on several platforms; it can also run consistently incorrectly. We are unable to know that such code contains no error. Yes, mathematicians could check the code easily, if you say to them, "We think this is the specific microcode error that exists." However, if you instead say to them, "Verify that no error whatsoever exists in this microcode", then it's no longer so easy to verify. If it was so easy to verify, why did Intel have such an error three times? And if they had it three times, how do you know that there is not a fourth example of it?
If you cannot accept that computer participation in the Four Color Map Theorem invalidates the solution, then there is not point in my continuing to demonstrate a proof here that avoids the problem.
worzel
2005-Sep-22, 10:28 PM
I've addressed them several times now. The code can run consistently on several platforms; it can also run consistently incorrectly. We are unable to know that such code contains no error. Yes, mathematicians could check the code easily, if you say to them, "We think this is the specific microcode error that exists." However, if you instead say to them, "Verify that no error whatsoever exists in this microcode", then it's no longer so easy to verify. If it was so easy to verify, why did Intel have such an error three times? And if they had it three times, how do you know that there is not a fourth example of it?Remember that the microcode is not part of the proof, it does not port with the code from one platform to another. A mathematician checking the code is checking the algorithm that does port, not the microcode of each chip it is ported to. The algorithm has not only been checked, but improved and reimplemented in different languages, and then run with equivalent output.
If I were charged with programming this task, I would have the code output how it dealt with each configuration, not just a "yes" or "no". If the output matched on different architectures then the chances of there being an error in the microcode of each chip used in just such a way that it affected each execution of the code on those different platforms in just such a way as to produce identical output is astronomically small. I'm sure the smart people who have done all the checking did something at least as thorough, why wouldn't they? Finding a genuine error in the proof would gain one far more notoriety than confirming it.
If you cannot accept that computer participation in the Four Color Map Theorem invalidates the solution, then there is not point in my continuing to demonstrate a proof here that avoids the problem.If you won't present your proof then there is little point carrying on the discussion. An elegant proof would obviously be highly desirable whatever one thinks of the current proof - sounds to me like you're trying to weasel out because you know your proof is flawed. If I'm wrong, prove me wrong, present your proof!
peter eldergill
2005-Sep-23, 01:19 AM
Could it be argued then along the same lines that Euclidean Geometry is not valid either because he based it on "postulates" (or assumptions)? Like parallel lines never meet and others (Sorry, I can't remember...I thought there a five postulates he had, some of which have been shown to be equivalent?
I agree with Worzel about a traditional proof. This would be fantastic, but since none exists (until Maddad supplies us with one, that is :surprised ), I will accept the computer proof.
Pete
montebianco
2005-Sep-23, 01:29 AM
Could it be argued then along the same lines that Euclidean Geometry is not valid either because he based it on "postulates" (or assumptions)?
Won't get far without postulates :)
Like parallel lines never meet and others (Sorry, I can't remember...I thought there a five postulates he had, some of which have been shown to be equivalent?
Some info here.
http://www.health.uottawa.ca/biomech/csb/laws/euclid.htm
There have been attempts to prove that the fifth postulate is redundant, but it is now known that this cannot be done, and systems of geometry in which the fifth postulate is false have been developed.
I agree with Worzel about a traditional proof. This would be fantastic, but since none exists (until Maddad supplies us with one, that is :surprised ), I will accept the computer proof.
Pete
It fits on a couple of sheets of paper? If such a thing exists, then who cares about a computer proof.
peter eldergill
2005-Sep-23, 01:42 AM
Quote:
Originally Posted by peter eldergill
Could it be argued then along the same lines that Euclidean Geometry is not valid either because he based it on "postulates" (or assumptions)?
Won't get far without postulates
That's what I was trying to get at I think, don't we have to assume that the computer works well enough to prove this? I would trust the word of the chip designer if that's their claim (i.e. I will take the word of the expert)
Pete
peter eldergill
2005-Sep-23, 01:45 AM
Some info here.
http://www.health.uottawa.ca/biomec...laws/euclid.htm
There have been attempts to prove that the fifth postulate is redundant, but it is now known that this cannot be done, and systems of geometry in which the fifth postulate is false have been developed.
Are Hilbert's spaces then more complete than Euclids (in terms of missing assumption in the axioms?)
Pete
montebianco
2005-Sep-23, 02:45 AM
That's what I was trying to get at I think, don't we have to assume that the computer works well enough to prove this? I would trust the word of the chip designer if that's their claim (i.e. I will take the word of the expert)
Pete
Well, we would have to assume that, but I think that's the issue here. At least one person here objects to this as an axiom (I don't particularly like it myself :))
montebianco
2005-Sep-23, 02:51 AM
Are Hilbert's spaces then more complete than Euclids (in terms of missing assumption in the axioms?)
Pete
It would appear so, but I'm way out of my element here. A Hilbert space to me means something rather different (an infinite-dimensional vector space with an inner-product, maybe some other requirement that I've forgotten) :)
peter eldergill
2005-Sep-23, 03:33 AM
I thought Hilbert Spaces were finite dimensional (usually). Or perhaps hyperdimensional? Ha!
I recall a geometry prof starting many theorems with "Let V be a finite dimensional Hilbert space with an inner product", but I never even knew that Hilbert had more accurate axioms for geometry
Pete
peter eldergill
2005-Sep-23, 03:35 AM
Well, we would have to assume that, but I think that's the issue here. At least one person here objects to this as an axiom (I don't particularly like it myself )
Reply With Quote
Isn't a large chunk of chaos theory (dynamical systems) done with the aid of computers?
Should we also abandon all of that?
Pete
montebianco
2005-Sep-23, 04:06 AM
Isn't a large chunk of chaos theory (dynamical systems) done with the aid of computers?
Should we also abandon all of that?
Not of any particular interest to me, but why do you seem to suspect that I feel it should be abandoned?
montebianco
2005-Sep-23, 04:14 AM
I thought Hilbert Spaces were finite dimensional (usually). Or perhaps hyperdimensional? Ha!
I don't usually think explicitly in Hilbert space terms, but I do recall reading a book (it was March 2003, I remember where I was when I read it and I've only been there once) which defined them as necessarily infinite-dimensional. I also recall thinking that that was odd...but certainly, the space of L-2 integrable functions on the real line is a common example of a Hilbert space, and that's infinite-dimensional.
I recall a geometry prof starting many theorems with "Let V be a finite dimensional Hilbert space with an inner product", but I never even knew that Hilbert had more accurate axioms for geometry
I didn't either. He did quite a lot though, including, I think coming up with the first solution to the field equation of Einstein, even before Einstein did - but perhaps I am mistaken, perhaps it was derivation of the equation itself. Or maybe I'm completely misremembering...
HenrikOlsen
2005-Sep-23, 05:11 AM
(Sorry, I can't remember...I thought there a five postulates he had, some of which have been shown to be equivalent?
He had 5 postulates.
You can draw a straight line from any point to any point
you can continue a line to infinity
you can contruct a circle with any center and any radius
all right angles are equal to one another
If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if extended indefinitely, meet on the side on which are the angles less than two right angles.
None of these can be derived from the others.
People had been trying to derive the 5th from the others for millenia until it was shown that it couldn't be done by generating internally consistent geometries using other propositions
peter eldergill
2005-Sep-23, 01:10 PM
Not of any particular interest to me, but why do you seem to suspect that I feel it should be abandoned?
I mentioned this due to the recent discussion about possible flaws in the computer processors
Pete
hhEb09'1
2005-Sep-23, 06:41 PM
I don't usually think explicitly in Hilbert space terms, but I do recall reading a book (it was March 2003, I remember where I was when I read it and I've only been there once) which defined them as necessarily infinite-dimensional. I also recall thinking that that was odd...but certainly, the space of L-2 integrable functions on the real line is a common example of a Hilbert space, and that's infinite-dimensional.Hilbert spaces can be infinite dimensional, but there are finite dimensional Hilbert spaces (cite (http://mathworld.wolfram.com/HilbertSpace.html)).
Maddad
2005-Sep-25, 11:32 PM
Remember that the microcode is not part of the proofIt most certainly is! The person running the progrogam interfaces to what the programer wrote. That sits on top of the program that compiled the program, which in turn rests on the operating system. The entire shooting match talks to the microcode, which means that the microcode is the most fundamental part of the proof offered by a computer program.
Porting to a new chip bring up a new microcode set, to be sure, but now you're needing to examine the validity under multiple systems for errors of unknown quality. Not only have we not positively researched the first microcode set; we sure as shootin' haven't compared the results against a second set. While much of the microcode is identical from one manufacturer to the next, there are differences if for no other reason than to avoid copyright infringment issues. Howsomever, because the ultimate job of the microcod of one processor is essentially identical to the job of another, it must be very close. If errors exist, they may be conceptual errors that both systems share.
Worzel, you're a smart guy, and I'm not telling you anything you don't already know. You have to be operating on an emotional inability to deal with the issue to keep missing such a basic issue.
there is little point carrying on the discussion.I agree. If we cannot get past these basics, then the logic reversals in the proof will indeed be too confusing to understand.
Could it be argued then along the same lines that Euclidean Geometry is not valid either because he based it on "postulates" (or assumptions)?No sir.
That's what I was trying to get at I think, don't we have to assume that the computer works well enough to prove this? I would trust the word of the chip designer if that's their claim (i.e. I will take the word of the expert)Establishing oroofs is not about accepting the word of an expert.
montebianco
2005-Sep-26, 12:50 AM
Hilbert spaces can be infinite dimensional, but there are finite dimensional Hilbert spaces (cite (http://mathworld.wolfram.com/HilbertSpace.html)).
That's what I would have thought, but after the post I made above, I even went and checked my book, and it does indeed say necessarily infinite-dimensional! I remember thinking this was odd at the time.
Arg, now I have to check again because I can't remember who the author is...
montebianco
2005-Sep-26, 12:56 AM
I mentioned this due to the recent discussion about possible flaws in the computer processors
Pete
I'm not suggesting abandoning anything. I just don't like adding "The computer works properly" to a list of axioms. In a rigorous logical system with a well-defined notion of truth and falsity, then the axioms determine the truth or falsity of a statement, independently of any proof. The proof tells us whether it is true or false, but it was true or false before we came up with the proof; we just didn't know it. I just wouldn't want to add "The computer works properly" to a list a fundamental axioms, anymore than, for a proof done without a computer, I would want to add "The person who wrote the proof and the people who have checked it didn't make any mistakes."
This was the intended meaning of my earlier statement.
Grey
2005-Sep-26, 01:47 AM
I agree. If we cannot get past these basics, then the logic reversals in the proof will indeed be too confusing to understand.
Regardless of what Worzel thinks, I'd be very interested to see your proof. Either it has a flaw in it, which will be interesting to track down, or it's valid, in which case I'd fly out to see Pace eat his hat. :)
peter eldergill
2005-Sep-26, 02:07 AM
Bring it on! Then I can pretend to understand it!
Pete
hhEb09'1
2005-Sep-26, 04:59 AM
If you're so willing to spring for the fare, I'd fly there :)
worzel
2005-Sep-26, 08:51 AM
Worzel, you're a smart guy, and I'm not telling you anything you don't already know. You have to be operating on an emotional inability to deal with the issue to keep missing such a basic issue.I'm afraid you're missing a very basic point. The microcode on different architechtures is completely different. The same program written in 'c', say, would even compile to different machine codes on different architectures - how could they possibly share microcode? If the same program written in a high level language is compiled for, and executed on, differing archictures and produces identical output, then if that output is quite verbose the odds of there being a microcode bug on each architecture which affected the outputs identically is astronomically small. Moreover, the algorithm has been reimplemented in different languages, and improved algorithms have also been developed. Guess what, they all confirm the result.
I put it to you, Maddad, that you have an emotional inability to let go of what was a temporary doubt with the current proof because you wish to cling on to your fantasy that you are the first to prove it.
I agree. If we cannot get past these basics, then the logic reversals in the proof will indeed be too confusing to understand.Nonsense. The only thing stopping us from discussing your proof is the lack of your proof. As both I and Grey have stated, what I think of the current proof is irrelevant to your proof. You claim to have come up with it while stil at school and claim that it fits on a sheet of A4 so I am pretty confident that my feeble mind can deal with it. But even if it is beyond me, I can assure you that both Grey and hhEb09'1 are more than up to the task and are both eager to see your proof.
Celestial Mechanic
2005-Sep-26, 05:50 PM
The problem with a computer proof is not philosophy; it's that we are unable to know, to the level necessary to claim proof of the Theorem, that the computer did in fact perform it's calculation correctly. Maybe you quickly passed over my original reasoning, but we are unable to know that the code is correct because we are unable to be sure that the microcode in the processor is not flawed. Remember, we have not one, but three prior instances in which this microcode later turned out to be defective.
I've addressed them several times now. The code can run consistently on several platforms; it can also run consistently incorrectly. We are unable to know that such code contains no error. Yes, mathematicians could check the code easily, if you say to them, "We think this is the specific microcode error that exists." However, if you instead say to them, "Verify that no error whatsoever exists in this microcode", then it's no longer so easy to verify. If it was so easy to verify, why did Intel have such an error three times? And if they had it three times, how do you know that there is not a fourth example of it?
But what assurance do we have that mathematicians have performed their calculations and logic correctly? In the case of the four-color theorem, there was a proof published by Kempe in 1879 that stood for 11 years before Heawood gave a counter-example and re-opened the problem in 1890. The chances are very good that your proof likewise has a small logic error, something easily overlooked. No one will ever know until your proof is submitted to review.
As for your three examples, I know that one of them was in floating-point division (something not likely to be used in a combinatorics problem like the four-color theorem), but I don't think the others were in arithmetic. (But please correct me if they were!) I think the others were technical things that affected operating system functionality but not arithmetic. The odds of a computational bug is very small precisely because so many results are known precisely and we can verify the results we get.
jfribrg
2005-Sep-26, 07:04 PM
One thing I haven't seen mentioned is that there is ongoing research into mathematical proofs of program correctness. I haven't dealt with it since grad school, so I can't give an example until I dig up my old books on the subject, but in general it is possible to prove that certain types of programs are correct. The problem here is that it has been proven that real-time systems cannot be proven correct, so the issue of the microcode becomes the main theoretical stumbling block. I have not been able to find any references to prove the correctness of the 4 color theorem ( I didn't look too hard either). I think this can be done in 3 steps:
1) Prove that the algorithm is correct ;
2) Prove that the RISC implementation of the program matches the algorithm
3) build a special purpose RISC computer that runs the program and nothing else. Specifically, this computer can have no interrupts, and the only output is a binary value. This last step may seem far-fetched but is necessary in order to complete the proof.
I am not sure if the theory of program correctness proofs has matured enough to handle something as complex as the 4 color theorem, but I don't see any reason why it couldn't be done, especially if we use an automated theorem proving program to do step 1 for us. That may sound like a circular argument (using a computer to prove that a computer proof is correct), but it isn't since the ATP output can be hand checked.
I know this is a rambling post, and I will make additional posts in the near future to clarify these ideas.
Maddad
2005-Sep-26, 09:15 PM
Regardless of what Worzel thinks, I'd be very interested to see your proof. Either it has a flaw in it, which will be interesting to track down, or it's valid, in which case I'd fly out to see Pace eat his hat. :)Ok. I told myself that I'd post it if there was one person interested. You're it.
Our initial world will be a limited two dimensional surface--a sheet of paper. We will arrange territories so that they do not border each other directly, but rather we represent their border by drawing a line to connect them. A border between two territories may not cross the border between another two. We'll represent the territories by simplifying their shapes to a circle. When we add the first territory to the new map, there is only one possible place to put it. On the map.
Grey, are you Ok with my reduction of territory shapes to circles and representing borders between them with connecting lines? I don't know if I need more explanation here or not.
hhEb09'1
2005-Sep-26, 10:18 PM
Grey, are you Ok with my reduction of territory shapes to circles and representing borders between them with connecting lines? I don't know if I need more explanation here or not.I don't think there should be any problem with that. I think that's actually the usual way of representing the regions.
Grey
2005-Sep-27, 12:48 AM
Grey, are you Ok with my reduction of territory shapes to circles and representing borders between them with connecting lines? I don't know if I need more explanation here or not.That's just fine. As Pace pointed out, most mathematical treatments of the question generally use the same method.
snarkophilus
2005-Sep-27, 11:27 AM
You can do simple tests to see whether or not the basic electronics function properly. It is not hard. I remember doing it in a second year hardware course. This is always done at the factory, of course.
Of course, you might argue that random data corruption might skew your results. This happens rather frequently, in fact. Little static discharges in the air and across circuit boards change data all the time. However, we have error-correcting mechanisms built into machines that make the chances of an error in a processor not being picked up very small indeed. Also, these errors are not reproducible, and are weeded out when you run the programme again.
Or maybe your OS is bad, and it changes your data in some way. Run under two OS's, because there is almost no way that such an error would exist in two separate products. If you're really paranoid, build your own OS, one whose correctness can be proven (and this can be done).
All that's left is the algorithm and the logic of the proof itself, which have been rigorously checked.
Look at it this way. Treat the computer as a scientifically studied object. If, after determining that the chance of getting bad results is less than, say, 10^-12 (which is much worse than data corruption going undetected at the hardware level, and much worse than the same corruption happening the same way on two machines), you can't trust your computer to do its calculations correctly, then you can't trust anything. You can't trust pictures you get from Hubble. You can't trust the clock on your wall. You can't trust that the food in your stove is being cooked at a high enough temperature, or that you are even reading this right now and not imagining it. All of those things have errors worse than those of a properly built computer. Very smart people have thought of these things, and they've built in safeguards.
All in all, a proof being false as a result of massive, widespread machine failure is far less likely than my going insane and just imagining that the proof was false. I'm willing to go out on a limb here and say that it's probably fine.
(As an aside, the Pentium arithmetic problem was a big deal for a couple of people I know... they did not know about it prior to buying machines, and ended up having to re-run several thousand hours of simulations. So now they build pretty comprehensive test programmes to run on any machine before using it for scientific purposes.)
Maddad
2005-Sep-27, 08:52 PM
Snark
That approach works if we are looking for the kind of error the system exhibits. If it has a different kind of error then we would not be successful.
Grey
Ok. When you introduce the second territory on the map, there is also only one possible place you can put it. Somewhere else. I like to keep the separation between them at least several territory diameters, but that is only an issue of taste. If you make the distance much less, then as your map becomes triangular it becomes unnecessarily crowded when you add another internal territory. If you make the disconnection too great, then the diagram can become ungainly. I also like to think of the line connecting the two territories as slightly thicker than a simple pencil line, but again this does not affect the outcome of the problem whatsoever. Thin works just fine.
What you wind up with is a dumbbell appearance. Two circles separated by a bar. I think of the dumbbell as being horizontal, roughly centered on the page, but it will function identically if you draw it vertically or even diagonally.
Ok, before we introduce our third territory, do you have any questions?
Grey
2005-Sep-28, 01:34 AM
Ok, before we introduce our third territory, do you have any questions?Nope. Looks good so far.
jfribrg
2005-Sep-28, 01:44 PM
I recall a few years ago someone an empirical demonstration of the 4 color theorem by using a "chemical computer", using amino acids. By combining different amino acids in a solution, if four colors were not sufficient, then certain types of proteins would be produced. After combining them, no such proteins were found. Its not exactly a proof, but since many trillions of molecules were used in the solution, the correctness of the theorem is very compelling. Combine that with the other proofs and it becomes very difficult to argue that the theorem is wrong. One may still argue that it is not rigorously proven (others would disagree), but to argue that it is false is a very difficult position to support.
Grey
2005-Sep-28, 02:02 PM
I recall a few years ago someone an empirical demonstration of the 4 color theorem by using a "chemical computer", using amino acids. By combining different amino acids in a solution, if four colors were not sufficient, then certain types of proteins would be produced.Do you have a reference for this anywhere? I'm fascinated by the notion of being able to correlate the creation of certain types of proteins to the falsity of the four color theorem, and I'd really like to see how they did it.
Maddad
2005-Sep-28, 09:41 PM
When you add a third territory, you appear to have two choices. You may put the new territory either to one side of the dumbbell, or you may put it above or below. These though will turn out to be the same.
For the first case, place the third territory centered above the dumbell such that the three territories form an equallateral triangle. Draw in the remaining two sides of the triangle, which now represent the borders between this new territory and the first two territories. Note that had you placed the new territory below the dumbbell instead of above it, then you would produce an upside-down triangle which would function as a map identically to being rightside-up.
In the second case, slide the dumbbell over to the left and place the third territory to the right of the dumbbell such that the left side and right side territories are equally as far from the center territory. Draw a line from the new territory to the center territory, representing the border between them. Now draw a curved line between the two outside territories that passes under the center territory without touching it. All three territories now border each other.
Get out a hammer and torch and heat the two straight lines. Then beat the two outside territories with the hammer, driving them down below the horizontal position of the center territory. The two straight lines now angle down and to the sides from the center, and the curve of the bottom line is more exaggerated. Keep hammering until the angle between the two straight lines at the center territory is 60 degrees. Now straighten out the bottom curved line between the two outside territories; it will not touch the center one. You now have another equallateral triangle with circular territories at the corners, which is identical to the first case.
Ready for territory number four?
Maddad
2005-Sep-28, 10:21 PM
At this point we should be confident that we have a diagram representing every possible arrangement of three territories on a limited two dimensional surface. If you have lost this orientation during the building, then let me know so that we can all be on the same page.
01101001
2005-Sep-28, 11:56 PM
Get out a hammer and torch and heat the two straight lines. Then beat the two outside territories with the hammer, driving them down below the horizontal position of the center territory. The two straight lines now angle down and to the sides from the center, and the curve of the bottom line is more exaggerated.
This is a mathematically rigorous proof?
Can we just stipulate that all planar graphs of 4 nodes are four-colorable and move on from there to all remaining planar graphs?
peter eldergill
2005-Sep-29, 02:00 AM
Maddad, have you studied graph theory? You are describing edges and veritices, specifically the complete graph of 3 vertices, K3 (the 3 is subscript). It is a planar praph, meaning that it can be redrawn in a way so that no edges cross.
A complete graph means that every vertex is connected to every other vertex with exactly one edge.
I'm not sure where you're going with this yet, but K4 (complete graph with 4 vertices) is also planar, but K5 is not planar.
If you haven't studied graph theory, I think you would enjoy it, and I suggest you pick up any standard introduction to the subject (Bondy and Murty is a standard as I recall)
Pete
Edit: The graphs you have describes in case 1 and 2 are called isomorphic, and I don't think it is necessary to distinguish them, especially in terms of graph theory. An edge is only defined by it's associated vertices, not length, curvature, etc...
peter eldergill
2005-Sep-29, 02:08 AM
01101001.....(I had to check that I got your name right about 4 times!)
In terms of planar graphs/colouring and in graph theory terms, what exactly are we trying to prove?
Pete
Joff
2005-Sep-29, 02:17 AM
The four colour theorem. Sshhh, it's going to get interesting soon. I think he's going to add a fourth vertex and join it to the other three.
Grey
2005-Sep-29, 12:54 PM
Ready for territory number four?Yes, please. You're welcome to move a little faster, if you'd like. I can always ask you to slow down when things get complicated.
hhEb09'1
2005-Sep-29, 04:20 PM
At this point we should be confident that we have a diagram representing every possible arrangement of three territories on a limited two dimensional surface. If you have lost this orientation during the building, then let me know so that we can all be on the same page.That is not quite true. For instance, one could have three territories side by side. The graph would have three circles in a row, joined by just two lines.
I have a feeling that that is not going to make a difference in your proof, but who knows?
Musashi
2005-Sep-29, 04:27 PM
That is not quite true. For instance, one could have three territories side by side. The graph would have three circles in a row, joined by just two lines.
I have a feeling that that is not going to make a difference in your proof, but who knows?
Don't all the territories need to be connected to each other? In that case, the three side by side would be connected by two lines PLUS another line connectin the outer two territories together.
http://img307.imageshack.us/img307/8245/threev8qc.gif (http://imageshack.us)
For the next step, it doesn't seem to matter where you put the 4th, it is going to isolate (at least) one of the other territories from a future 5th. This is the big step, I imagine, figuring out how to place a 4th so that it doesn't isolate. I haven't been able to do it, and even though I have only been trying since last night, it is not apparent to me that it is even possible.
http://img263.imageshack.us/img263/3024/fourv7ob.gif (http://imageshack.us)
Grey
2005-Sep-29, 05:26 PM
Don't all the territories need to be connected to each other? In that case, the three side by side would be connected by two lines PLUS another line connectin the outer two territories together.I don't think there's any specific requirement that all the territories touch each other, the arrangement Pace suggests is certainly possible. However, it's also true that, since we're trying to establish the maximum number of regions that could in principle be touching, it's pretty likely that simply choosing not to have two regions be adjacent when they could be isn't going to get you far.
For the next step, it doesn't seem to matter where you put the 4th, it is going to isolate (at least) one of the other territories from a future 5th. This is the big step, I imagine, figuring out how to place a 4th so that it doesn't isolate. I haven't been able to do it, and even though I have only been trying since last night, it is not apparent to me that it is even possible.If it were possible, that would of course be a counterexample to the four color theorem. I think, actually, that it's fairly easy to show that no arrangement of just five territories can have all of them touching each other. I think that most of those who think the four color theorem is false (I think there are pretty few such people, actually) think that there might be an arrangement of more than five territories which somehow forces at least five to all be touching each other.
The use of pictures is nice. Maddad, may I recommend doing the same? It will probably save you some time explaining what you mean, and making sure that we're all drawing the same pictures as we follow along at home. :)
hhEb09'1
2005-Sep-29, 05:26 PM
Don't all the territories need to be connected to each other? In that case, the three side by side would be connected by two lines PLUS another line connectin the outer two territories together.It's certainly not true in real life--Virginia, North Caroliina, and South Carolina, USAn states, are connected, but Virginia does not share a border with South Carolina. As I said, it may not make a difference in the proof, but it is not true that all three have to touch the other two, even if they are all connected.
For the next step, it doesn't seem to matter where you put the 4th, it is going to isolate (at least) one of the other territories from a future 5th. This is the big step, I imagine, figuring out how to place a 4th so that it doesn't isolate. I haven't been able to do it, and even though I have only been trying since last night, it is not apparent to me that it is even possible.It's not possible. Or, as peter eldergill said, a complete graph on five points cannot be planar.]
Musashi
2005-Sep-29, 05:36 PM
Well, I am trying to be openminded about Maddad's claim. I hope he comes back and tells me where to put #4 so that I can still place #5 properly.
As far as real life, from what I know, the four color theorem doesn't really have anything to do with real life maps. I also don't see how adding verticies that don't connect to the others helps disprove the theorem (and keep it simple). Would you mind explaining?
Grey
2005-Sep-29, 05:55 PM
Well, I am trying to be openminded about Maddad's claim. I hope he comes back and tells me where to put #4 so that I can still place #5 properly.No, Maddad still thinks the four color theorem is true. He just thinks that a proof using a computer is not a valid proof, and says that he has a simpler one.
By the way, I'm wrong about this:
I think that most of those who think the four color theorem is false (I think there are pretty few such people, actually) think that there might be an arrangement of more than five territories which somehow forces at least five to all be touching each other.A bit of quick reading shows that it's pretty easy to prove that you can't get five regions to all be bordering each other at the same time. However, that by itself doesn't prove the four color theorem. Why not? Well, the largest number of regions that a territory touches isn't necessarily the same as the number of colors needed for the map. For example, a ring of five regions (like the one on the left here (http://www.mathpages.com/home/kmath266/Image4247.gif)) has no region touching more than two other regions, yet it requires three colors.
Musashi
2005-Sep-29, 06:28 PM
Ah! That is what I get for reading the first half and the second half of this thread at different times. Sorry. Things are a bit clearer now. Thank you Grey.
hhEb09'1
2005-Sep-29, 06:28 PM
I also don't see how adding verticies that don't connect to the others helps disprove the theorem (and keep it simple). Would you mind explaining?As Grey points out, and others have said, there is no set of five regions where all five touch each of the other four. That's easy to show. Over a hundred years ago, it was shown that even a map of a few dozen territories was insufficient to force five colors--but obviously, there are territories that don't touch each other in that map. Eventually, you do have to add territories that don't touch each other, in producing such maps.
But, as I say, I don't think that is going to make much difference in this proof.
worzel
2005-Sep-29, 06:50 PM
As I said, it may not make a difference in the proof, but it is not true that all three have to touch the other two, even if they are all connected.But we already know that we may as well only consider maximal planar graphs, because if they are four-colorable, then trivially so are non-maximal ones. I'm having difficulty seeing how non-maximal ones could be crucial to any proof. But then you could say that we also know from Kempe that we might as well start with a vertex with five edges, and assume that the rest of the map is four-colorable, but this is Maddad's proof, so we should stay open minded as you say, and see where he's going. Let's hope this one doesn't take 10 years to disprove :)
Grey
2005-Sep-29, 10:41 PM
But we already know that we may as well only consider maximal planar graphs, because if they are four-colorable, then trivially so are non-maximal ones. I'm having difficulty seeing how non-maximal ones could be crucial to any proof.I don't want to try to speculate much further until we see the next step from Maddad, but there is at least one thing to consider here. It's true that we need only consider maximally connected graphs, but if we're building up the graphs by adding one vertex at a time and then establishing all legitimate connections, I think that doesn't span the space of all maximally connected graphs. That is, I'm fairly certain that there exist maximal graphs for which the removal of any single vertex will leave you a non-maximal graph, and if you were to add all possible connections to this smaller graph (to make it maximally connected), you could no longer create the original graph by the re-addition of a vertex. So a proof that builds up a graph by the addition of vertices one at a time will at least need to allow non-maximal graphs in the construction process. Does that make sense? Can someone who has done more work in graph theory confirm or deny this?
Maddad
2005-Sep-29, 11:10 PM
Maddad, have you studied graph theory? . . . I'm not sure where you're going with this yetHaven't studied it yet, but I get into stuff like that as it crosses what I need. Where I'm going is to construct a diagram representative of any possible arrangement of four territories that allows you to see at a glance that you are unable to add a fifth territory in such a way that it borders all four other territories.
That is not quite true. For instance, one could have three territories side by side. The graph would have three circles in a row, joined by just two lines.It is indeed quite true. Those three territories must all border the other two, so the side of the two outside ones must flow above or below the center one to meet up. After thinning the flow to lines, this becomes exactly the equallateral triangle described in my longer previous post.
Musashi is now on the right track. Thank you very much for the drawings, by the way. Yes, the territories do need to be all connected to each other. You have reduced the territories to black dots, but that works just fine. I'll run with this representation.
"For the next step, it doesn't seem to matter where you put the 4th, it is going to isolate (at least) one of the other territories from a future 5th." Bingo! You got it. You skipped though one possibility. That fourth territory could be added outside the diagram instead of inside it. How about making us one more drawing of a three territory triangle with the fourth territory added outside the diagram? I would appreciate that very much.
"This is the big step, I imagine, figuring out how to place a 4th so that it doesn't isolate. I haven't been able to do it, and even though I have only been trying since last night, it is not apparent to me that it is even possible." You're right on both counts. It's a big step, and it's not possible. In being not possible, it proves the Four Color Map Theorem. However, before we claim solution, we need to see how adding the fourth territory outside the triangle of three territories relates to the diagram of four territories you have already given us. You will see that they are actually identical.
Thank you again Musashi. You've saved me a ton of explanation that I was dreading.
worzel
2005-Sep-29, 11:16 PM
Does that make sense? Yes, good point Grey, I think I've found an example of what you're talking about (see attachment). We have to start with a triangle (say abc), then when we must add another vertex (say d) such that it has an edge with a,b, and c. But there is no vertex in the example that has an edge to each vertex of a triangle.
Sorry if I'm sidetracking the discussion a bit, Maddad, maybe you can help us get back on track and add the fourth vertex.
worzel
2005-Sep-29, 11:29 PM
You're right on both counts. It's a big step, and it's not possible. In being not possible, it proves the Four Color Map Theorem.This sounds like you're saying that proving the theorem involves only proving that you can't draw five territories such that they all touch each each other. Is that the case, or am I just being impatient?
Musashi
2005-Sep-30, 12:51 AM
You skipped though one possibility. That fourth territory could be added outside the diagram instead of inside it. How about making us one more drawing of a three territory triangle with the fourth territory added outside the diagram? I would appreciate that very much.
I am not sure what you mean by outside the diagram. In both of the 4 territory maps, I added the 4th above the sets of three and then connected. Here I will do it again starting with two triangles and putting one 4th in the center (bottom picture) and one below the triangle (top picture).
http://img121.imageshack.us/img121/7245/fourvii6oa.gif (http://imageshack.us)
I am pretty sure that no matter where I put the 4th, as long as it is outside the triangle, it will isolate one territory.
peter eldergill
2005-Sep-30, 02:49 AM
"For the next step, it doesn't seem to matter where you put the 4th, it is going to isolate (at least) one of the other territories from a future 5th." Bingo! You got it. You skipped though one possibility. That fourth territory could be added outside the diagram instead of inside it. How about making us one more drawing of a three territory triangle with the fourth territory added outside the diagram? I would appreciate that very much.
"This is the big step, I imagine, figuring out how to place a 4th so that it doesn't isolate. I haven't been able to do it, and even though I have only been trying since last night, it is not apparent to me that it is even possible." You're right on both counts. It's a big step, and it's not possible. In being not possible, it proves the Four Color Map Theorem. However, before we claim solution, we need to see how adding the fourth territory outside the triangle of three territories relates to the diagram of four territories you have already given us. You will see that they are actually identical.
.
Sounds to me that you've proved that K5 is not planar, which I'm pretty sure doesn't prove the four colour theorem
Pete
ZaphodBeeblebrox
2005-Sep-30, 04:14 AM
Sounds to me that you've proved that K5 is not planar, which I'm pretty sure doesn't prove the four colour theorem
Pete
Isn't that, The Point, though?
Maps, are Two-Dimensional!!!
QED
worzel
2005-Sep-30, 11:53 AM
Isn't that, The Point, though?
Maps, are Two-Dimensional!!!
QEDNo, it's not enough. Here's a map that has no set of 4 mutually connected vertices (doesn't contain k4).
http://mboyd.demon.co.uk/fct-pent.jpg
Yet it requires at least four colors. We need at least three for b,c,d,e,f, because with only two we must alternate and will end up with two bordering with the same color. a must be different to all b,c,d,e,f so it must use a fourth color.
So we have a map that requires four colors even though it doesn't contain k4. How do we know that there aren't similar maps that require 5 colors even though they don't (and inded can't) contain k5?
peter eldergill
2005-Sep-30, 05:18 PM
This is from Wikipedia, with references and all
http://en.wikipedia.org/wiki/Four_color_theorem
Apparantly, The Four Colour Theorem stated in graph theory terms is very simple:
Prove that any planar graph is four colourable (colourable meaning adjacent vertices have different colours)
There is also (an outline of) a proof of the five colour theorem at
http://en.wikipedia.org/wiki/Five_color_theorem
Pete
hhEb09'1
2005-Sep-30, 07:44 PM
It is indeed quite true. Those three territories must all border the other two, so the side of the two outside ones must flow above or below the center one to meet up. After thinning the flow to lines, this becomes exactly the equallateral triangle described in my longer previous post.It may be true in your proof, sure, but my point was just that in general it is not true.
So a proof that builds up a graph by the addition of vertices one at a time will at least need to allow non-maximal graphs in the construction process. Does that make sense? Can someone who has done more work in graph theory confirm or deny this?Yes, that was my point. But it's clear now that it does not make a difference, in this proof.
It's a big step, and it's not possible. In being not possible, it proves the Four Color Map Theorem.That's the wrong Four Color Map Theorem. :)
Grey
2005-Sep-30, 09:06 PM
Yes, good point Grey, I think I've found an example of what you're talking about (see attachment).Yes, I think that's the smallest possible graph that serves as an example. It's maximally connected, but removing any vertex gives a graph which is not.
Sounds to me that you've proved that K5 is not planar, which I'm pretty sure doesn't prove the four colour theorem.You're correct, Peter. That wasn't actually clear to me when this discussion started, but worzel has stated the simplest objection. DeMorgan was able to show that five regions cannot be in contact with each other in 1852, when he was first introduced to the question, but also realized that this doesn't settle the question of whether four colors is always sufficient to color any map.
Here (http://www.mathpages.com/home/kmath266/kmath266.htm) is a site that I found very useful in discussing the history of the four color theorem, discussing some of the subtleties (like this one) involved in proving it, describing the simpler proofs that six or five colors are sufficient (showing that six is sufficient is almost trivial), and giving the general outline of the proof given by Appel and Haken.
worzel
2005-Oct-01, 10:08 AM
Great link, Grey. Here's (http://www.stonehill.edu/compsci/LC/Four-Color/Four-color.htm) a very accessible one I found that explains Kempe's proof, and the flaw in in it.
01101001
2005-Oct-01, 11:01 PM
Where's the rest of Maddad's proof of the Four Color Theorem? How does it proceed from 4- and fewer-node planar graphs being at most four-colorable to all planar graphs?
Maddad
2005-Oct-02, 01:13 AM
This sounds like you're saying that proving the theorem involves only proving that you can't draw five territories such that they all touch each each other. Is that the case, or am I just being impatient?Bingo. Remember, we are now satisfied that the triangle is representative of every possible arrangement of three territories. (I understand that we have one holdout, hhEb09'1's, who rejected my answer to his objection.) Since the triangle represents every possible arrangement of three territories, identifying every possible place that the fourth territory may go would mean we can be confident of diagramming every possible arrangement of four territories. Our objective to prove the theorem is to show what the maximum number of territories will be. We will find that at four because we will be unable to add the fifth territory.
I am pretty sure that no matter where I put the 4th, as long as it is outside the triangle, it will isolate one territory.Yes, when you draw the fourth territory outside the triangle as you have done in placing it underneath, the loop around the left side isolates the left side territory from a fifth outside territory placement in the resulting diagram of four territories. In fact, wherever you place your new territory, the loop to connect it to the last territory will isolate one territory from a new fith outside territory. I does not though cover internal palcement in a four territory map, which we will deal with next. Before it gets too confusing though, we are going to show that those last two diagrams you drew are actually identical.
Isn't that, The Point, though? Maps, are Two-Dimensional!!!
QEDFor the purposes of this proof, yes they are because I have defined them that way. They are not only two dimensional, they are also limited rather than endless. Once we have a clear grasp of the logic behind working the theorem, then it will be an easy matter to extend our definition to an unlimited two dimensional surface or a curved surface.
Yet it requires at least four colors. We need at least three for b,c,d,e,f, because with only two we must alternate and will end up with two bordering with the same color.Actually, your map requires only three colors. Note that territory abc may be the same color as ade, and territory acd may be the same color as aef.
Maddad
2005-Oct-02, 01:31 AM
Ok, we are going to take another look at the two seemingly different drawings that Musashi provided for us. Rember how we bent the bars of the three territories in a row to match the initial three territories arranged in an equallateral triangle? We're going to do the same thing to morph Musashi's two diagrams in his last image until they match each other. We'll leave the bottom one alone to use as a reference for what we want the upper one to become.
First we shorten the center bar, pulling the three territories on the left into a vertical line. The loop on the left is now a bit longer than we need, so we'll just tug on it to shorten it up a bit for neatness's sake. Now we'll scoot the territory on the right and the on in the center downward. We stop when we readch 60 degree angles with the various territories. After pulling the looping connecting borders to a straight line, we should be able to wind up with a diagram that looks just like Musashi's bottom diagram of four territories. This means that every possible arrangement of four territories may be faithfully represented on this map by his bottom image.
Since we now have a single diagram for four territories, all that is left to do is examine where we can add the fifth territory. There are only two possible locations, similar to the two possible locations in the thee territory diagram. We may add the fifth territory to the exterior of the diagram, or we may add it to one of the three identical interior cells. By now some of you can see what is about to happen.
hhEb09'1
2005-Oct-02, 06:57 AM
Actually, your map requires only three colors. Note that territory abc may be the same color as ade, and territory acd may be the same color as aef.The territories are a and b and c and d and e and f. The borders are the lines between them.
worzel
2005-Oct-02, 08:48 AM
This sounds like you're saying that proving the theorem involves only proving that you can't draw five territories such that they all touch each each other. Is that the case, or am I just being impatient?Bingo. Remember, we are now satisfied that the triangle is representative of every possible arrangement of three territories. (I understand that we have one holdout, hhEb09'1's, who rejected my answer to his objection.)Only if the graph is required to be maximally connected. But as Grey pointed out, you can't reach all maximally connected graphs adding one vertex at a time such that each graph along the way is also maximally connected (if you could, the four color theorem is actually quite easy to prove). But when you said, "Those three territories must all border the other two, so the side of the two outside ones must flow above or below the center one to meet up," it sounds like you're arguing from the point of view that a map must completely cover the plane. As hhEb09'1 has said, that may be a requirement of your proof but it is not a requirement of planar maps in general. Moreover, even if that is a requirement of your proof, you are still wrong. Here's a map (an actual map, not a graph) of three territories covering the plane that do not all touch each other. Under it is the graph representing this map which is also the graph hhEb09'1 pointed out to you.
http://mboyd.demon.co.uk/fct-three.gif
None of this is probably relevant to your proof though. How about assuming for the sake of argument that we all understand and agree that one cannot create a map with five mutually bordering regions. How do you get from there to proving the four colour theorem?
mickal555
2005-Oct-02, 11:47 AM
Hmmm, I tried to disprove it in pait- just for fun...
Remarkable how it works....
hhEb09'1
2005-Oct-02, 02:11 PM
Under it is the graph representing this map which is also the graph hhEb09'1 pointed out to you.Thanks worzel, that's a good example map, since it leaves no "room" to create an arbitrary path from a to c.
Maddad
2005-Oct-03, 01:33 AM
The territories are a and b and c and d and e and f. The borders are the lines between them.Hmmm. You still need only four colors. Territory b borders territories f, a, and c which allows you to color c and d the same way you do b. While it does require four colors, it does not show that you cannot arrange five territories in such a way that each one touches all the other four in a line or arc segment. Therefore it does not prove the Theorem.
Only if the graph is required to be maximally connected. But as Grey pointed out, you can't reach all maximally connected graphs adding one vertex at a time such that each graph along the way is also maximally connectedThe term maximally has not been definied. I can indeed satisfactorily connect the territories because I have.
when you said, "Those three territories must all border the other two, so the side of the two outside ones must flow above or below the center one to meet up," it sounds like you're arguing from the point of view that a map must completely cover the plane.I defined the plane as limited. The territories must indeed meet up because we are examining the maximum number of territories that do meet up.
The reason your map of A, B, and C is not relevant for the discussion is because A is isolated from C. No territories may be isolated from any other if you are to proceed.
worzel
2005-Oct-03, 09:20 AM
Hmmm. You still need only four colors. Territory b borders territories f, a, and c which allows you to color c and d the same way you do b. While it does require four colors, it does not show that you cannot arrange five territories in such a way that each one touches all the other four in a line or arc segment. Therefore it does not prove the Theorem.The only purpose in presenting that graph was as an example of a maximally connected graph that can not be constructed one vertex at a time such that each graph thus obtained is also maximally connected.
The term maximally has not been definied. We're all using the standard definition: a maximal planar graph is one where no more edges can be added (assuming only one edge is allowed per pair of vertices). Equivalently, every face of a maximal planar graph is a triangle.
I can indeed satisfactorily connect the territories because I have.Indeed you can, but you claimed that your triangle was the only graph possible for three territories, that was incorrect.
I defined the plane as limited. The territories must indeed meet up because we are examining the maximum number of territories that do meet up.The problem here is that if you're only going to consider graphs that result from you making the maximum number of connections possible each time you add a vertex then your proof won't apply to the graph you referred to above because it can't be constructed that way.
The reason your map of A, B, and C is not relevant for the discussion is because A is isolated from C. No territories may be isolated from any other if you are to proceed.It's only relevance was to show that your statement that your triangle was the only possible graph representing three territories was incorrect. If you require that your graphs be maximally connected then that example is not relevant, but then you can't construct the previous example :) If you require that every territory added must border every other territory then you can't get past four territories so your proof is limited to maps with only four regions.
Of course, none of us can tell if any of these points are relevant to your proof because you haven't finished presenting it yet. So rather than you requiring us to all agree on each step you take and every word you say before moving onto the next one, and us therefore being bound to go over every proviso due to your lack of rigour, why not just present the rest of your proof. Then we can either congratulate you or focus on just the bit where you went wrong.
Maddad
2005-Oct-03, 08:37 PM
you claimed that your triangle was the only graph possible for three territories, that was incorrect.Not quite. I claimed the triangle was representative of every possible way to satisfactorily arrange three territories, which is different. I said nothing about where there could be another way to represent them.
It's only relevance was to show that your statement that your triangle was the only possible graph representing three territoriesAs in the quote above, you are reading something into my statements that is not there. While it is a graph of every possible arrangement, it is not the only possible graph, which is different.
then you can't construct the previous exampleI can't tell you how impressed I am that you think I cannot do what I have done. If I ever need someone to explain something, I'll be sure to come looking for you.
rather than you requiring us to all agree on each stepIf you cannot understand the current step, then you cannot understand the next since each step is built upon the previous one. Since you want to argue, instead of seeing the solution, I have no interest in progressing with you.
worzel
2005-Oct-03, 10:22 PM
Not quite. I claimed the triangle was representative of every possible way to satisfactorily arrange three territories, which is different. I said nothing about where there could be another way to represent them.Well you never defined "satisfactorily", maybe you should, it is certainly not representative of every three territory map. But you've got it the wrong way round anyway. This map
http://mboyd.demon.co.uk/fct-three.gif
is a map who's graph is not a triangle, its graph below it is not another way to represent your triangle graph.
As in the quote above, you are reading something into my statements that is not there. While it is a graph of every possible arrangement, it is not the only possible graph, which is different.Again, you have it the wrong way round. No one is saying that there are other ways to represent what your graph represents. What we are saying is that your graph does not represent every possible three territory map. If you are placing restrictions on the construction such that your triangle is representative of every possible restricted construction then you need to be explicit about those restrictions so that we can evaluate whether or not your proof covers all maps.
then you can't construct the previous exampleI can't tell you how impressed I am that you think I cannot do what I have done. If I ever need someone to explain something, I'll be sure to come looking for you.Thank you for cutting the qualification out of that sentence. In case you misunderstood I'll try to rephrase it for you: If you require that every map in your construction as you iteratively add vertices be fully connected then you can not construct this map:
http://mboyd.demon.co.uk/fct-diamond.gif
If that is your requirement and you don't agree, show us how you can construct the above map one vertex at a tine while maintaining a fully connected graph at each stage. If that is not your requirement, then tell us what your requirement is (there must be one, since you disallow my map above) and move on from this then irrelevant point.
If you cannot understand the current step, then you cannot understand the next since each step is built upon the previous one. Since you want to argue, instead of seeing the solution, I have no interest in progressing with you.I am sorry you feel that way. I do not want to argue for its own sake, but if you require that I agree to every step along the way then I must point out where I don't agree. Otherwise, if the last step is valid, but a previous step that I had agreed to for the sake of argument is not, then I have no choice but to back up and contradict myself by disagreeing with a step that I previously agreed with. It is quite likely, however, that these points may be moot. We can not know until you actually provide your proof. So rather than waste time arguing the toss over points that may well turn out to have no impact on your proof, why don't you just get on with it.
If you take exception to me (sorry guys, I never was any good with rope) then just ignore me for now and get on with it anyway for the benefit of all those eager, and rather patient, observers who are still with us.
Grey
2005-Oct-04, 01:38 AM
Maddad, these are indeed side issues, and perhaps we shouldn't have brought them up while waiting for your next step. But worzel is correct, in that by presenting each step one at a time, it becomes necessary to raise any issues with that step, even if those issues do not have any actual bearing on the proof.
But perhaps we can cut to the chase, as it were. Even though you haven't quite completed your proof, from your comments you seem to be under the impression that merely showing that five territories cannot mutually touch each other is sufficient to prove the Four Color Theorem. Am I correct in that?
peter eldergill
2005-Oct-04, 02:04 AM
Yes, Maddad, I think that also. Your posts are getting very defensive , especially when you say things like
Since you want to argue, instead of seeing the solution, I have no interest in progressing with you.
You made no mention of the sites I suggested and since you're trying to use Graph Theory to prove your result, I think it would be good to at least learn a couple of the terms, and perhaps even go through the prrof of the five colour theorem to get some good ideas.
This sarcasm is also not warranted, in my opinion:
I can't tell you how impressed I am that you think I cannot do what I have done. If I ever need someone to explain something, I'll be sure to come looking for you.
You've spent most of your posts talking about moving territories around. The location of the territories is not relevant, only the colour of the vertex. In graph theory, the only thing that defines an edge is the end vertices, not the distance, placement of veritces, shape of the edge, etc.
Later
Pete
Maddad
2005-Oct-05, 11:46 PM
worzel is correct, in that by presenting each step one at a time, it becomes necessary to raise any issues with that step, even if those issues do not have any actual bearing on the proof.Not quite. If it does not bear on the proof, then it is just a distraction.
you seem to be under the impression that merely showing that five territories cannot mutually touch each other is sufficient to prove the Four Color Theorem. Am I correct in that?Provided that you can show that the diagram represents every possible way of arranging the territories in the universe you have defined, then yes, it proves the theorem.
worzel
2005-Oct-06, 12:02 AM
you seem to be under the impression that merely showing that five territories cannot mutually touch each other is sufficient to prove the Four Color Theorem. Am I correct in that?
Provided that you can show that the diagram represents every possible way of arranging the territories in the universe you have defined, then yes, it proves the theorem.How have you defined your universe of territories? Is is that no territory may be added unless it connects to every other territory already added? You seemed to say so when you said "No territories may be isolated from any other if you are to proceed."
hhEb09'1
2005-Oct-06, 09:57 AM
worzel is correct, in that by presenting each step one at a time, it becomes necessary to raise any issues with that step, even if those issues do not have any actual bearing on the proof.Not quite. If it does not bear on the proof, then it is just a distraction.If it does not bear on the proof, why make the comments that raise the issues? They're distracting :)
Grey
2005-Oct-06, 12:50 PM
Not quite. If it does not bear on the proof, then it is just a distraction.The point was that you weren't presenting your entire proof, just each individual step, so we therefore didn't know what would be relevant to the proof and what would not.
Provided that you can show that the diagram represents every possible way of arranging the territories in the universe you have defined, then yes, it proves the theorem.Ah, well that answers that issue then. DeMorgan realized back in 1852 when he was originally presented with the question that it was impossible to have five regions that mutually touch each other, so your proof is actually nothing new. Unfortunately, it's also not sufficient to prove the Four Color Theorem (which DeMorgan also realized). It might be possible for a single territory to touch four others, which, because of their connections to other various territories, are forced to have four different colors, and hence the territory in question would have to use a fifth one. Now, this may not be possible (indeed, if one accepts the standard proof, it is not), but you haven't shown that.
Look again at worzel's example. I'll redraw it so that it's clear what the territories are, just in case that was a concern (we'll see if I can get the upload files feature to work here). Remember, the dots are the territories, the lines show which ones border which others, which is the standard convention you were using as well. If it's confusing, I can always draw the actual map that corresponds to this graph. Anyway, this is a map which has no set of mutually touching territories larger than three, yet it requires four colors. So we can see immediately that the number of colors required for a map is not necessarily the same as the largest number of territories that touch each other.
Maddad
2005-Oct-06, 03:56 PM
The point was that you weren't presenting your entire proof, just each individual step, so we therefore didn't know what would be relevant to the proof and what would not.I can see your point.
DeMorgan realized back in 1852 when he was originally presented with the question that it was impossible to have five regions that mutually touch each other, so your proof is actually nothing new.The question the theorem tries to settle is why each of the five regions are unable to touch all the other four. Knowing that they cannot wasn't ever an issue.
It might be possible for a single territory to touch four others, which, because of their connections to other various territories, are forced to have four different colors, and hence the territory in question would have to use a fifth one. Now, this may not be possible (indeed, if one accepts the standard proof, it is not), but you haven't shown that.Being forced to use a different color is a result of touching another territory. If they do not touch, then you may use the color of that territory. If they do touch, then you must use another.
I allowed myself to get derailed by irrelevant comments a few days ago. The conclusion of the proof is that since the diagram of four territories, a triangle with the fourth in the center, has only two possible places for the fifth territory, you only have to show that neither choice allows a connection to all four previous territories. If you choose external placement, then you are isolated from the center territory. If you choose one of the three identical internal cells, then you are isolated from the territory at the far corner.
I do not need to look at worzel's example. I am perfectly willing to admit that he is unable to solve the theorem with his approach.
pghnative
2005-Oct-06, 04:15 PM
The question the theorem tries to settle is why each of the five regions are unable to touch all the other four.
From my reading of this thread, that is not what the theorem is trying to settle.
It might be possible for a single territory to touch four others, which, because of their connections to other various territories, are forced to have four different colors, and hence the territory in question would have to use a fifth one. Now, this may not be possible (indeed, if one accepts the standard proof, it is not), but you haven't shown that.
Being forced to use a different color is a result of touching another territory. If they do not touch, then you may use the color of that territory. If they do touch, then you must use another.Incorrect. See worzels example above where no territory touches more than two other territories. By your logic only three colors should be needed. But in the example, four colors must be used. This seems to be the critical point that you are missing.
editted three times due to my incompetance with nested quotes
peter eldergill
2005-Oct-06, 04:22 PM
It's not Worzel who's trying to solve the problem, it's you...and yes, you are claiming that since K5 is not planar, then the 4 colour theorem is proved.
As mentioned by Grey, this is not sufficent.
The question the theorem tries to settle is why each of the five regions are unable to touch all the other four. Knowing that they cannot wasn't ever an issue.
It is precicesly the issue, as you are using it to try to prove your theorem.
I do not need to look at worzel's example. I am perfectly willing to admit that he is unable to solve the theorem with his approach.
Wow...ignoring counterexamples or other cases which can possibly show your proof is incorrect, or needs revising. Is that really how you think mathematics should be? Calling our comments "irrelevant" is not particularly polite, either, and won't convince me that your proof is valid.
Pete
hhEb09'1
2005-Oct-06, 05:16 PM
(color added)
The conclusion of the proof is that since the diagram of four territories, a triangle with the fourth in the center, has only two possible places for the fifth territory, you only have to show that neither choice allows a connection to all four previous territories. If you choose external placement, then you are isolated from the center territory. If you choose one of the three identical internal cells, then you are isolated from the territory at the far corner.That does not prove the Four Color Map theorem, as has been said, but I'm curious about what the tattoo would have looked like.
01101001
2005-Oct-06, 05:28 PM
Invalid. Nice try.
worzel
2005-Oct-07, 12:13 AM
OK Maddad, I'm going to ignore your scathing comments and resist the urge to quote every instance where you claimed my comments were irrelevant, followed by every instance of someone referring to my posts in attempting to convince you that your proof is mistaken (although I couldn't resist making this irrelevant point :)).
By now it is clear to everyone that your proof is wrong, and that the provisos we were discussing were actually quite relevant. All we have left to do is to hopefully convince you that you are wrong. Maybe this will spur you on to investigate further. I, personally, would love to see a simple proof of the FCT - I've had a go at it a few times myself but have never really got anywhere (it's one of those nagging problems that just ought to have a simple proof).
I was quite impressed, by the way, that you figured out that representing territories and borders with graphs was the way to go. I never thought of that before I first read it somewhere (although I probably read it a few pages after first reading about the FCT). You shouldn't take it personally when we criticise your proof - that is how all scientific and mathematical ideas are tested: can they stand up to criticism - not that this is a peer reviewed journal, but it's better than nothing (and there are some frighteningly smart people here).
Right. Take the map of the United States (or something similar). Suppose that we have colored it using only four colors. Now suppose we add the sea and for some strange reason we decide that we should use one of the colors already used. But when we do we find that each of the four colors is already used on at least one coastal state. What can we do? Well we could try changing the color of all the blue coastal ones, say, to free up blue for the sea. But in doing so, what if we find that some of these blue ones border three differently coloured states? After changing a blue coastal state like this we will need to recolor one of its neighbours also. But what if we encounter the same problem with the neighbour we recolor? Well we could just keep on recoloring and see what happens. But what if we end up back on the coast again but this time we find ourselves left with no choice but to color a coastal state that wasn't blue and the only choice we have left is blue due to the recoloring choices we made earlier?* Maybe we could backtrack and try out some different choices. The four color theorem basically says that if we do backtrack eventually we will find a set of choices that will allow us to color the states and the sea using only four colors - but it may be quite difficult to figure out what those choices are.
I hope that you can see that this is far more complex and subtle than simply showing that five states can't all border each other.
*If we make the decision never to choice another coastal state for recoloring after our initial one, how do we know that this process will ever end?
Grey
2005-Oct-07, 01:29 PM
The question the theorem tries to settle is why each of the five regions are unable to touch all the other four. Knowing that they cannot wasn't ever an issue.I see at least two misconceptions here. The first is simply that whether it's possible to have five mutually touching regions in a graph is not the question addressed by the four color theorem. The question is whether it's possible to draw a graph that would require more than four colors to make certain that all adjacent regions are colored differently. To be sure, they are closely related questions, and if it were possible for five regions to touch each other, that would certainly mean that the four color theorem was false. However, the reverse is not true; showing that five regions cannot touch is not sufficient to prove the four color theorem. More about that below.
The second problem here, though, is that a proof in mathematics is never about showing why something is true, when that something was already known to be true. If we know something is true, it's because we have a proof demonstrating that. If we don't have a proof, then we cannot be sure that it's true. DeMorgan knew that it's not possible for five regions to be in mutual contact not because it was obvious, or because nobody was able to come up with a counterexample, but because he was able to demonstrate it, using a proof essentially equivalent to yours.
Being forced to use a different color is a result of touching another territory. If they do not touch, then you may use the color of that territory. If they do touch, then you must use another.True, but remember that regions need not be in direct contact in order to constrain what colors can be used for them. That is, the color of a region directly constrains the colors of regions it borders. Those regions may then constrain colors of regions that border them, and so forth, which means that the choice of color for the first region may have an indirect effect on the available colors for those other regions, even though it does not directly border them. Thus, it may be necessary to consider regions beyond those that immediately border any one region to determine the number of colors required for a map. For example, I've added an outside region to the one worzel used (note that this also makes it a maximally connected graph, just in case that was an issue for some reason). Now, if you start by choosing a color for the central region, the outer region will always end up the same color, no matter what choices you make for all the other regions.
I allowed myself to get derailed by irrelevant comments a few days ago. The conclusion of the proof is that since the diagram of four territories, a triangle with the fourth in the center, has only two possible places for the fifth territory, you only have to show that neither choice allows a connection to all four previous territories. If you choose external placement, then you are isolated from the center territory. If you choose one of the three identical internal cells, then you are isolated from the territory at the far corner.No trouble here. I think we'll all accept that four mutually touching regions is the most possible in a map.
I do not need to look at worzel's example. I am perfectly willing to admit that he is unable to solve the theorem with his approach.He's not trying to prove the four color theorem, he's showing where you've made an unfounded assumption. Let me try to make this explicit. You're saying, "no more than four regions can mutually touch each other in a two dimensional map, therefore no more than four colors are need to color any map such that no two adjacent territories have the same color". You should be able to see that there is an assumption being made here. Namely, that the largest number of regions in mutual contact is the same as the number of colors required for the map. However, if you look carefully at the graph (you can find an example that works in this message as well, if you don't want to go back to worzel's post), you'll see that there is no group of regions in mutual contact larger than three. Let me repeat that: there is no place in this graph where you have four regions touching each other. However, this map requires four colors to paint, not three. Therefore, it is not true that the number of regions in largest group which all border each other in a map is also the same as the minimum number of colors for that map. Without that additional assumption, you cannot get from what you've shown (which you've done quite nicely) to what the four color theorem states.
Maddad
2005-Oct-07, 07:31 PM
I will address one single and especially idiotic statement out of a choice of many. That by pghnative is particularly enlightening because it speaks to the core of our discussion:
"Originally Posted by Maddad
The question the theorem tries to settle is why each of the five regions are unable to touch all the other four. [Knowing that they cannot wasn't ever an issue.]"
From my reading of this thread, that is not what the theorem is trying to settle.Incorrect.Thirty five years ago I picked up the Life Science Library book on Mathematics, copyrighted in 1963 and again in 1969. It presents the Four Color Map Theorem on pages 184 and 185. Since I still have that book in one of my living room bookcases, let me quote you what David Bergamini and the editors of Time said about your post: "no one has been able to prove what map makers have known for ages—that four colors are enough for any flat map or sphere."
Got that pghnative?
Look, I cannot believe that you guys are really that stupid. You are not; you are probably reasonably intelligent. So what is the problem? Why are you having so much trouble understanding such a simple idea? The answer is that you feel threatened by me. I am asking you to think, and thinking of a new idea is difficult.
One of the members here labeled someone an expert because they had supposedly come up with a proof. Never mind that nobody here can understand that proof. Never mind that we know of nobody who has verified it. Never mind that experts have been known to make mistakes. Never mind that the member making this interesting statement does not really know what qualifications this expert has to earn the title of expert. (Yes, I know you can know go look it up now, but you did not know when you were certain this man was the last word on the subject.) We have called them an expert, and from that concluded that the issue is settled and therefore needs no further investigation. What you are doing is insulating yourself from a need to think. Once you call him an expert, you need think no further.
Pghnative, why in God’s name do you want me to examine worzel’s example? I have not even bothered reading his last post. A month ago he jumped into this thread without a clue; look at his first two posts. He still does not get it; look at his last post. Yes, he is dealing with diagrams that are distantly related to mine because they are colored territories, and some of them do border others. If he though, like you, cannot understand the very most fundamental statement of the Four Color Map Theorem given in red above, then what possible hope does he have of contributing anything useful? Why should I waste my time chasing someone else’s approach to arranging territories when it does not bear on the method I am using to limit possibilities to manageability? I developed this approach specifically to deal with this problem. Saying that my approach is different from what worzel heard about is expected. I solved the problem; worzel and company did not. If it bears a resemblance to something worzel’s heard about, then that does not mean that it must share all characteristics with mine. In fact, it should not, because if it did, everybody here would still not understand why we do not need more than four colors to draw a map.
As I had said I would, I have provided the proof of the Four Color Map Theorem. You may accept it or reject it, as you choose. It is not my job to drag you by the scruff of your collective necks to make you think. Thinking is a privilege and an art, but it is not easy. If you are more secure in finding excuses not to think, then by all means, do so. I am done with this thread and will not return to it to read any further comments.
hhEb09'1
2005-Oct-07, 07:49 PM
As I had said I would, I have provided the proof of the Four Color Map Theorem.D*ng. But at least, my hat (http://www.bautforum.com/showpost.php?p=558098&postcount=39) is safe this time. :)
PS: This is part of the maddingly attractive nature of the four color map theorem--it seems like it should be so easy to apply it to a map of any number of territories, once you have shown it for a map with just five. Some of the world's greatest mathematicians have tried and failed to extend it past the point that Maddad achieved. Some of them convinced themselves that they had done it, too, so he's not alone in that regard. It can drive you mad.
pghnative
2005-Oct-07, 08:36 PM
Maddad
Has it occurred to you that your tone is quite belligerant?
Has it occurred to you that if everyone on this thread thinks that you are wrong, that maybe just maybe you are indeed wrong.
Has it occurred to you that if everyone on this thread thinks that you are wrong, that calling everyone "stupid" reflects rather poorly on yourself?
What you have proven (that five regions cannot touch the other four) is necessary to prove the Four Colors Theorem, but it is not sufficient. Do you understand the difference between those two adjectives?
Edited to add: I've now read Grey's post, and he's explained the flaws in Maddad's logic much better than I. I'll keep the following paragraphs in since I do not wish to delete what I've posted, but I'll ask all to basically read Grey's explanation instead of mine.
Worzel's example shows multiple groups of four regions where none of four touches the other three. But three colors is not sufficient to color his map. Four is necessary. The point is that even if no territory touches more than X other territories, X+1 colors may not be enough.
What you need to do is show that multiple groupings of five reqions still can be colored with only four colors. And you need to do that for all types of multiple groupings. You have not done that.
ToSeek
2005-Oct-07, 09:20 PM
Maddad, your post is inappropriate.
You may not call people stupid or idiotic on this forum (or accuse them of not thinking). You have been banned for 24 hours. I would encourage you to use this period to actually read what people are telling you on this thread because - in my personal opinion - you could learn something.
I will also lock the thread for the duration so that the discussion will not continue without you.
ToSeek
2005-Oct-08, 10:37 PM
I've unlocked the thread.
Be nice!
worzel
2005-Oct-09, 08:42 AM
A month ago he jumped into this thread without a clue; look at his first two posts.What, you didn't get that I was joking?:rolleyes:
Putting your hands over your ears and shouting "I have so proved it" does't change the fact that you have not even attempted to show how proving that five regions can't all border each other proves the four color theorem. You seem to think the connection is too obvious to need explaining - well it's not.
All our supposedly irrelevant side tracking has been our well-intentioned attempts to get you to see the difference - but it all boils down to just that one missing connection.
Musashi
2005-Oct-09, 05:21 PM
I admit, when I first looked at the problem I thought that proving 5 territories couldn't all touch each other was sufficient, but through the conversations here and some thought on the issue, I realized that it wasn't.
peter eldergill
2005-Oct-09, 08:21 PM
I've unlocked the thread.
Be nice!
I'm always nice...I can't seem to click on any of the edit buttons right now, including quote and smilies!
Pete
Dragon
2007-Feb-21, 03:08 AM
...
One of the members here labeled someone an expert because they had supposedly come up with a proof. Never mind that nobody here can understand that proof. Never mind that we know of nobody who has verified it. Never mind that experts have been known to make mistakes. Never mind that the member making this interesting statement does not really know what qualifications this expert has to earn the title of expert. (Yes, I know you can know go look it up now, but you did not know when you were certain this man was the last word on the subject.) We have called them an expert, and from that concluded that the issue is settled and therefore needs no further investigation. What you are doing is insulating yourself from a need to think. Once you call him an expert, you need think no further.
...
Who was labelled an expert? (I'm an expert on this stuff and I'd like to read the "proof"; not to imply it's false).
On the other hand, is it just me, or do the dropdown boxes and smileys not work?
hhEb09'1
2007-Feb-21, 03:43 AM
Who was labelled an expert?Andrew Wiles
(I'm an expert on this stuff and I'd like to read the "proof"; not to imply it's false).Have you read his proof?
Dragon
2007-Feb-23, 10:10 PM
Andrew Wiles? In four-colour theorem!? Since when? I thought he stayed with the Taniyama-Shimura conjecture... Oh, well, guess not; but no, I haven't read his proof. Could you please post where I can read it? I'm really curious now, because he is a someone that might actually present a errorless proof.
hhEb09'1
2007-Feb-24, 02:40 AM
Andrew Wiles? Oops, my mistake. For some reason, I thought I remembered Maddad going on about Fermat's theorem. I think he was referring to Appel and Haken. But you better not take my word for it. :)
Dragon
2007-Feb-26, 12:33 AM
Whew. I was afraid my mathematical knowledge had fallen behind current events again. Well, I assume Appel and Haken's proof is correct; even if it isn't, the proof by Robertson, Sanders, Seymour, and Thomas is, and it is more hand-checkable. I bet Maddad would freak out if he reads his "proof" here (http://www.mathpages.com/home/kmath266/kmath266.htm) and the explanation of why it doesn't work, and will claim that it's been ripped off the boards. ;) j/k
Grey
2007-Feb-28, 05:27 PM
It wasn't so much that Maddad though those proofs were flawed, as that he thought they were needlessly complex. In particular, he thought that showing that one cannot have five mutually adjoining regions is sufficient to prove the four color theorem. The site you link to gives a simple explanation of why that is not enough.
peter eldergill
2007-Feb-28, 05:31 PM
Then he got mad at all of us for trying to show him why his proof wasn't sufficient....I don't even know if he still posts here.
I was hoping this thread went into the annals of obscurity
Pete
Ivan Viehoff
2007-Feb-28, 06:15 PM
There are thousands of Maddads out there. They think they have a wondrous proof, and are very unhappy when it is pointed out that they don't.
I've told this story somewhere else on this forum before, but it bears repeating. One of the top maths profs in Britain, I forget which, tells the story of how when first appointed he started getting all these manuscripts from amateurs. Initially he read them, perhaps hoping to find the next Ramanujan. But they were not, they were all erroneous. Initially he replied politely, pointing out the errors, giving constructive criticism and encouraging their interest in maths. He thought they would be pleased by the attention, but in fact they were very angry, and sent abusive letters in return, often accusing the establishment of a conspiracy to keep them down. So they just get a "I don't have time to look at your paper" acknowledgment now.
peter eldergill
2007-Feb-28, 08:59 PM
I think it may have been in this thread! I remember you, anyway
Pete
Dragon
2007-Mar-06, 03:40 AM
There are thousands of Maddads out there. They think they have a wondrous proof, and are very unhappy when it is pointed out that they don't.
I've told this story somewhere else on this forum before, but it bears repeating. One of the top maths profs in Britain, I forget which, tells the story of how when first appointed he started getting all these manuscripts from amateurs. Initially he read them, perhaps hoping to find the next Ramanujan. But they were not, they were all erroneous. Initially he replied politely, pointing out the errors, giving constructive criticism and encouraging their interest in maths. He thought they would be pleased by the attention, but in fact they were very angry, and sent abusive letters in return, often accusing the establishment of a conspiracy to keep them down. So they just get a "I don't have time to look at your paper" acknowledgment now.
Yeah, I've read this story somewhere, except the one I read ended with a "First mistake in the proof pointed out" response. Maybe it changed since then. If it hadn't, on the other hand, I'd like to mail him a proof of mine... too bad I don't have a copyright on it until July though...
peter eldergill
2007-Mar-06, 03:42 AM
Maybe we can all wait and you can post it here? (Not that I'll be able to understand it)
Pete
Dragon
2007-Mar-06, 03:49 AM
Wow, awesomely quick response, maybe I should rate this thread a 5. ;) But anyways, it's surprisingly simple. In fact, it's so simple I'm having people read it to make sure there's not a fatal flaw, which is why my school prohibited me from posting it until after I graduate. So far it's made it past two levels and is currently on the third. The next level is the state level, and after that gets done, I have to wait until July to post it in a journal. However, state level people generally stay busy a lot so it might not return thence much earlier than July.
Just by the way, how do I rate a thread anyways?
hhEb09'1
2007-Mar-06, 04:11 AM
[FONT="Fixedsys"] But anyways, it's surprisingly simple. This is a proof of the four color map theorem?
How many areas do you treat, in your proof?
grav
2007-Mar-06, 06:07 AM
Well, I have spent the last hour or so reading through this thread. It sounds interesting to me. I'm sure I've read about it somewhere before. After reading through what has been said here, I think I have come up with a simple proof. I will leave up to any of you to point out any potential flaws, which is possible, since I've barely had time to really think about it, but it makes sense to me and I don't see why it shouldn't be the case.
Let's say we start with a single territory and color it with color #1. Then we add territories all the way around it and alternate the colors with colors #2 and #3. Now, if the number of surrounding territories is even, then we would only require three colors. If it is odd, then we require an additional color #4 to fill in the gap. That's the first step.
The initial territory is now taken care of, being completely surrounded. So let's begin again with one of the outer ones. Each of the outer ones will be connected to the inner one and two other outer ones, correct? But whatever territories we connect to the main outer one will also be connected to one of the other two outer ones but not the inner one. So again, we just alternate the colors of the other two outer ones all the way around the main outer one. If the number surrounding it is even, then we stop there. If it is odd, then one of them must be a fourth color to fill in the gap. But since none of them will touch the inner one, we just use that color.
So basically, we can just keep adding new territories this way, by surrounding each territory in turn by alternating between it's two outer most adjoining territories' colors, and then adding the fourth "oddball" color if the number that surrounds it is odd, using whichever color remains that is not that of the main territory being surrounded or either of the original two that are alternated around it. And we should do this beginning with the inner-most territories and work our way outwards. We can continue doing this indefinitely on a flat surface. I'm not sure about 3D surfaces, though.
So how's that measure up for a proof? :) Is it sufficient?
grav
2007-Mar-06, 07:04 AM
Okay. Yup. Too simple. One extra rule is required. If the number of additional territories that is added on in order to surround the main outer territory is odd, so that the oddball color must be used, then that oddball color should be buried under any territories that exist past those that encompasses three or more of the ones that are being added on. In other words, if a territory beyond those that are presently being colored in touches three (or more) of them at the same time, then one of them will then become completely surrounded, and it should be the oddball. So all in all, only three of the colors should ever exist on the outer perimeter at any time.
hhEb09'1
2007-Mar-06, 07:16 AM
So how's that measure up for a proof? :) Is it sufficient?:) :) :)
Here's a country map:
6655
7224
7314
7344
8888
If I follow your directions, I think I paint #1 red, then #2 yellow, #3 blue, and #4 green. Continuing with the outer country #2, it is bordered by #4 green and #3 blue, so I start its perimeter by coloring #5 blue, #6 green, and I'm forced to use red for #7. This forces #8 to be yellow, right?
PS: I see you've modified your algorithm already. :)
So all in all, only three of the colors should ever exist on the outer perimeter at any time.Should? :) Now prove that you can do that.
PSS: I'm not sure, but my map may take care of your objection already. See, any finite map can be four-colored, we know that. (It's been proven :) ) The problem is stating an algorithm that is guaranteed to not get us into trouble, that will result in the map being four-colored. How would your new algorithm solve the problem in my map?
worzel
2007-Mar-06, 01:05 PM
In other words, if a territory beyond those that are presently being colored in touches three (or more) of them at the same time, then one of them will then become completely surrounded, and it should be the oddball.
What if there are no such territories?
peter eldergill
2007-Mar-06, 01:36 PM
But anyways, it's surprisingly simple. In fact, it's so simple I'm having people read it to make sure there's not a fatal flaw
Please don't forget that simple is such a relative term. I can assure you that to me, introductory limits and derivatives are extremely simple, as I have been teaching it for 10 years, but to my students....not so simple
Don't sell yourself short about a proof being too simple, although honestly, it does ring alarm bells...
Pete
grav
2007-Mar-06, 03:24 PM
:) :) :)
Here's a country map:
6655
7224
7314
7344
8888
If I follow your directions, I think I paint #1 red, then #2 yellow, #3 blue, and #4 green. Continuing with the outer country #2, it is bordered by #4 green and #3 blue, so I start its perimeter by coloring #5 blue, #6 green, and I'm forced to use red for #7. This forces #8 to be yellow, right?
PS: I see you've modified your algorithm already. :)
Should? :) Now prove that you can do that.
PSS: I'm not sure, but my map may take care of your objection already. See, any finite map can be four-colored, we know that. (It's been proven :) ) The problem is stating an algorithm that is guaranteed to not get us into trouble, that will result in the map being four-colored. How would your new algorithm solve the problem in my map?It does seem that my exception to the general rule has somehow turned my proof that it can be done into an algorithm for how it can be done. I'm not sure that it would be the same thing as a proof, then, unless I can prove that the exception will also never fail.
But as far as the algorithm goes, though, as an expansion to the same rule where one territory touches three or more of the inner ones, it appears that it also applies to any inner territory that touches three or more outer ones. That would be #2 in your map, since it touches #5, #6, and #7 on the third run. So on the last go around, we would color #5 blue and #7 green, making #6 the oddball red, being nestled within the other three. #8 would also be red, then, since we can never have more than three colors on the outer perimeter.
grav
2007-Mar-06, 03:25 PM
Quote:
Originally Posted by grav
In other words, if a territory beyond those that are presently being colored in touches three (or more) of them at the same time, then one of them will then become completely surrounded, and it should be the oddball.
What if there are no such territories?Then there is no problem. A complication only arises if one of the outer most territories touches at least three of the inner ones (or vice versa). But that is taken care of by what what discussed earlier in this thread, that no five territories can all touch each other at the same time, so one of them must be completely cut off. If we use the oddball color for that one, then it is then kept from obstructing the rest of the colors.
hhEb09'1
2007-Mar-06, 03:44 PM
It does seem that my exception to the general rule has somehow turned my proof that it can be done into an algorithm for how it can be done.It's not a proof--notice, you change your reasoning later on, for #2. How do you know that when you go back to change #2, that it doesn't screw up one of the alignments you made earlier?
grav
2007-Mar-06, 03:59 PM
It's not a proof--notice, you change your reasoning later on, for #2. How do you know that when you go back to change #2, that it doesn't screw up one of the alignments you made earlier?Yes, I know. It's not so much a proof now, just a method. I can only prove it so far for when no territory on the outer perimeter touches more than two inner ones. But as far as #2 is concerned here, one doesn't go back to change anything ever, they just kind of watch where they are going. ;) As long as all of the perimeter territories contain only one of three colors at a time, they are fine. I guess that also means starting with the oddball color each time, which makes a total of five rules now. I'll see if I can't make a computer program that will color in a screen full of random territories this way. I suppose the goal is to see if it can color them all in correctly the first time every time, without working backwards through the colors. Then I can show if the algorithm works, even if I can no longer prove it.
hhEb09'1
2007-Mar-06, 04:09 PM
I can only prove it so far for when no territory on the outer perimeter touches more than two inner ones.That fails for even my little map.
But as far as #2 is concerned here, one doesn't go back to change anything ever, they just kind of watch where they are going. ;)"Kinda watch where they are going..." :) :) :)
In other words, your algorithm is basically "We know we can do it, so we look at the whole map where we are going and figure out how to do it and then we do it." That's just worthless. :)
worzel
2007-Mar-06, 04:15 PM
Then there is no problem. A complication only arises if one of the outer most territories touches at least three of the inner ones (or vice versa)..
I see. So if this isn't a problem then we just end up choosing the oddball one at random. So I'd just need to think up one case where a random choice for the oddball would turn out, eventually, to be wrong and then your method would fail, right?
grav
2007-Mar-06, 04:44 PM
That fails for even my little map."Kinda watch where they are going..." :) :) :)
In other words, your algorithm is basically "We know we can do it, so we look at the whole map where we are going and figure out how to do it and then we do it." That's just worthless. :)We only need to watch for the territories that border the ones we are coloring, and we only need one to place the oddball color. After that, everything's fine. It does make things only slightly complicated, yes, and that can't be helped, so mistakes might be made in the process unless I can find a simpler way to explain it, but the proofcheck for our progress with each time we color in the surrounding territories of the main one is that the result would be that not more than three colors will ever exist on the outer perimeter at a time. Maybe I can explain it better after I make a program, since I can see the way I had the computer do it through the language of the program itself.
grav
2007-Mar-06, 04:46 PM
I see. So if this isn't a problem then we just end up choosing the oddball one at random. So I'd just need to think up one case where a random choice for the oddball would turn out, eventually, to be wrong and then your method would fail, right?That's right. It should always work with just the one first rule in that case.
hhEb09'1
2007-Mar-06, 05:04 PM
It does make things only slightly complicated, yes, and that can't be helped, so mistakes might be made in the process unless I can find a simpler way to explain it, but the proofcheck for our progress with each time we color in the surrounding territories of the main one is that the result would be that not more than three colors will ever exist on the outer perimeter at a time.That's not a proofcheck--if you ever get to that point, you will have failed. In other words you will need five colors to continue coloring.
You cannot just say, "we'll do that". We know we can do that because it has been proven, but you have to be specific in what you are going to do. Waving your hands and saying, we'll make sure that there's never four colors on the perimeter is not sufficient.
You can always make sure that there are only three on the perimeter--but that's just a restatement of the four color map theorem! That's what you're trying to prove!
Tog
2007-Mar-06, 05:28 PM
I'm not sure I want to get involved here, as much of this is well beyond me, but I have a drawing that I just did that I can't see a way to color with only 4 colors. Maybe you guys can see how to do it.
Musashi
2007-Mar-06, 05:33 PM
http://img401.imageshack.us/img401/3215/mapzm5.png (http://imageshack.us)
hhEb09'1
2007-Mar-06, 05:36 PM
Maybe you guys can see how to do it.Color #2, #3, #4, and #5--you have to use all four colors, since each of them touches the other three. #1 has to be the same as #4 since it is the only one that it doesn't touch. #8 has to be the same as #3 since it is the only one it doesn't touch. And #7 has to be the same as #2, since it is the only one it doesn't touch.
PS: Just like Mushashi did! :)
Tog
2007-Mar-06, 05:37 PM
Ahh, I got hung up on symmetry. Back to the shadows then.:)
hhEb09'1
2007-Mar-06, 05:41 PM
Ahh, I got hung up on symmetry. Rotate it through 180 degrees, it's the same diagram, with yellow exchanged for red, and red for yellow.
Tog
2007-Mar-06, 05:43 PM
Rotate it through 180 degrees, it's the same diagram, with yellow exchanged for red, and red for yellow.
In my head, 3 and 7 had to be the same color. So did 4 and 8. For some reason nothing that I tried could change that.
hhEb09'1
2007-Mar-06, 05:49 PM
In my head, 3 and 7 had to be the same color. What kind of symmetry is that? :)
worzel
2007-Mar-06, 08:48 PM
I think your map, Tog_, proves grav's current method inadequate.
Starting with #5 as the middle one and grav's method we always end up with #2 and #3 being colored the same despite them bordering each other (there's an even number of adjacent territories so there's no oddball, strictly alternating).
grav, have you looked at Kempe's famously mistaken proof (http://www.stonehill.edu/compsci/LC/Four-Color/Four-color.htm)? It might give you some ideas for your attempt.
peter eldergill
2007-Mar-06, 09:17 PM
Grav, the fact that you're even attempting this is impressive to me
Pete
molinaro
2007-Mar-07, 01:15 PM
This thread sure had a disapointing ending.
So, was it the teachers who failed to teach or was the student not willing to learn? It seems like it would be the later that was the problem.
But would not a better attempt at teaching have resulted in a different outcome? I say this not to point fingers, but rather I just have to hope that it is possible to lead someone to a better understanding.
We know where Maddog went wrong. He made an assertion, that proving the impossibility of a k-5 planer graph is equivalent to proving the 4 color theorem. And when attempts were made to show that this was an assertion he resorted to the all to familiar cry of an unwillingness to learn. It seems that a reaction like that is all too familiar.
When you get to the point of trying to show exactly where they went wrong, they turn it into a fight against the establishment that is too stupid to see things in a new way.
How do we make them see the error of their ways? Is it always possible? Would a better aproach be more successfull? Should we be less eager to jump to the error that we see coming?
I just can't shake the feeling that it is a failure on the part of the smart people. Although it may just be wishfull thinking on my part that success should always be possible.
ASEI
2007-Mar-07, 01:37 PM
Didn't they have some sort of geometry with borders winding together tighter and tighter, with the four color theorum being violated at the singularity in the center?
PS- what about fractal geometries with infinitely convoluted shapes where, given a point, you can only tell if it is color 1 or 2 to some probability based on the number of iterations? :-P
WaxRubiks
2007-Mar-07, 03:02 PM
I came up with a possible sollution a few years ago, it involved just looking at the boarders only not looking at the thing as a 2d map. A map is basically made of countries each with a boarder and all you have to do is look at all the possible boarders and forget about the rest of the map.
I thought that it was the way to go.
http://img253.imageshack.us/img253/9898/4colourqj0.png (http://imageshack.us)
the numbers represent the colours.
It seems that if the number of intersection with boarder X toX is odd then you are going to need 4 colours if it is even all you need is 3.
Since all lines on a map are just boarders then maybe you can apply this idea to it whatever the shape of the map.
It may need some work to become an actual proof.
number 1 is the colour of the country that the boarder X to X encompases.
grav
2007-Mar-07, 03:59 PM
Well, it appears my proof has deteriorated even further from an algorithm to perhaps a few insights on how to possibly obtain an algorithm. I find I have to keep adding more and more rules for more complex situations. A computer might be able to keep up with them, but it's probably becoming almost humanly impossible otherwise. Oh, well. One of these days I might take Nereid's advice and post a reply after I've thought things completely through. Naah... What would be the fun in that? I like putting myself out there on the frontline sometimes. I'll either stand or I'll duck, drop, roll, and run for cover. It's more interesting, and I can learn something either way.
So what I'm doing now, instead of a compilation of rules that are almost impossible to keep track of and follow, is to try to find some mathematical basis for the geometry, which would automatically keep track of its own setup, by beginning with how many connections each territory has to the others, and then accounting for how each of those connected territories are connected also, by following some kind of numerical sequence where the result for each territory can then be hopefully divided by four so that the remainder gives the value for the color to be used, or something like that. It's kind of hit and miss so far, but this thing has caught my interest, and I know it's possible, so I'm going to keep working on it.
Grey
2007-Mar-07, 07:48 PM
I came up with a possible sollution a few years ago, it involved just looking at the boarders only not looking at the thing as a 2d map. A map is basically made of countries each with a boarder and all you have to do is look at all the possible boarders and forget about the rest of the map.
I thought that it was the way to go.Most analyses of the problem move from a map to an equivalent graph, where each territory is a point, and there are lines connecting any two points that share a border.
http://img253.imageshack.us/img253/9898/4colourqj0.png (http://imageshack.us)
the numbers represent the colours.
It seems that if the number of intersection with boarder X toX is odd then you are going to need 4 colours if it is even all you need is 3.I'm afraid it's not quite that simple, since a region's color may be constrained by what's happening at more than one border. For example, let's think about the two lowest regions labelled "2" in your drawing here. What if they touch? Well, then they can't both be color 2, and we'd need a fourth color. And what if they both touch the third region from the bottom labelled "2"? Well, then we'd either need a fifth color, or we'd need to redo the coloring scheme you've got to do it with only four.
Well, it appears my proof has deteriorated even further from an algorithm to perhaps a few insights on how to possibly obtain an algorithm. I find I have to keep adding more and more rules for more complex situations.That's why this is such a fascinating problem. It seems like it ought to be pretty easy to prove. And since we know that it is true, you'll never be able to come up with a counterexample. But to be able to actually demonstrate rigorously that there's no way to construct such a map is surprisingly difficult. It's all too easy to leave something out, or make an assumption that has not been proven, or fail to take into account some way of arranging things. An algorithm that show definitively how to color any map would be acceptable as a proof that it can be done, but only if you can show that the algorithm takes into account every possible case, and as I think you've realized, that's a lot harder than it sounds.
molinaro
2007-Mar-07, 08:48 PM
On that point, didn't the earlier posts talking about the computer proof say that it required 2000 seperate cases to be checked?
And a proper proof would of course need to include a proof that those are all the possible cases. Tough is an understatement!
worzel
2007-Mar-07, 10:14 PM
Any proof would have to prove it for an infinite number of cases to count. The question is, how easily can all infinite cases be proved. The current proof does require many cases to be considered separately (hence the computer), but who knows, many there's a much simpler proof out there waiting to be discovered.
ASEI
2007-Mar-08, 11:10 PM
Any proof would have to prove it for an infinite number of cases to count.
To violate the four color theroum, you would have to have a set of five countries, each of which touch the other four, within a 2D plane. If you cannot arrange five shapes so that they all touch each other in a plane, with non-overlapping territories, then you can't violate the theorum. Is that right?
01101001
2007-Mar-08, 11:47 PM
To violate the four color theroum, you would have to have a set of five countries, each of which touch the other four, within a 2D plane.
No. That would be sufficient to disprove it, but that is not necessary to disprove it. For instance, a contrived set of 105 countries that required 5 colors might also disprove it. So might such a set of 1001 countries.
If you cannot arrange five shapes so that they all touch each other in a plane, with non-overlapping territories, then you can't violate the theorum. Is that right?
No. If that were necessary and sufficient, you'd be home free on a proof. You could just enumerate all possible graphs of 5 nodes and show that all were 4-colorable.
ASEI
2007-Mar-09, 12:18 AM
But to make a 5th color necessary, wouldn't you have to have that specific situation somewhere on your map? It wouldn't matter between which countries, but you would need 5 of them to do what I outlined.
To make the 1st color necessary, you need one country.
To make the 2nd color necessary, you just need two to touch.
Third, three countries, each of which shares borders with all the others.
Four, four countries, each of which shares borders with all the others.
Ect...
Otherwise, you wouldn't have the necessary constraint, would you?
ASEI
2007-Mar-09, 12:29 AM
On different geometries (toruses, ect) you get the 7th, 8th, whateverth color by coming up with a situation where somewhere there exists a set of 7 countries, each of which touch borders with all the others, right?
hhEb09'1
2007-Mar-09, 01:44 AM
But to make a 5th color necessary, wouldn't you have to have that specific situation somewhere on your map?Prove that, and you have a proof :)
worzel
2007-Mar-09, 02:46 AM
To violate the four color theroum, you would have to have a set of five countries, each of which touch the other four, within a 2D plane. If you cannot arrange five shapes so that they all touch each other in a plane, with non-overlapping territories, then you can't violate the theorum. Is that right?
I'm too lazy to google one now, but there are examples of maps that are quite difficult to four color. While that doesn't prove anything, (other than that they are four colorable once you've found a way), trying made me quickly see why it wasn't as simple proving that five countries can't all touch each other.
If you end up four coloring twenty countries only to find that the twenty first needs a fifth color unless you backtrack, recoloring as you go until you reach an early decision that was, in retrospect, the wrong one (and think about how a constructive* proof would incorporate that), then one realizes (that's when I did, anyway) that this isn't going to be as easy one originally thought :)
* of course, the proof doesn't actually have to be constructive to count, but that seems to be the obvious approach to proving it - "look, you can always do it like this..."
worzel
2007-Mar-09, 02:52 AM
On different geometries (toruses, ect) you get the 7th, 8th, whateverth color by coming up with a situation where somewhere there exists a set of 7 countries, each of which touch borders with all the others, right?
If you can get n countries to all border each other then you will need at least n colours. We know you need at least 4 on the 2d plane becase we can make 4 countries all border each other. The theorem is that you will need at most 4 on the 2d plane.
Grey
2007-Mar-09, 02:18 PM
But to make a 5th color necessary, wouldn't you have to have that specific situation somewhere on your map? It wouldn't matter between which countries, but you would need 5 of them to do what I outlined.
To make the 1st color necessary, you need one country.
To make the 2nd color necessary, you just need two to touch.
Third, three countries, each of which shares borders with all the others.
Four, four countries, each of which shares borders with all the others.
Ect...
Otherwise, you wouldn't have the necessary constraint, would you?It's not quite that simple. We've brought up the issue earlier in the thread, but I'll just link again. Take a look at this page (http://www.mathpages.com/home/kmath266/kmath266.htm), and jump down to the fourth image. There you'll see a graph that only has two regions sharing a border with each other, yet requires three colors, and a graph with no more than three regions all sharing borders, and yet it requires four colors. Remember that although the color you use for a given region is only directly constrained by its immediate neighbors, it may be indirectly constrained by their neighbors, and the neighbors of those regions, and so forth.
So you can't assume that the maximum number of colors needed is the same as the largest group of mutually touching regions. Since there's a counterexample that shows that it's clearly not the case for two or three touching regions, you'd have to first prove that it holds for four, and that's not as easy as it sounds. It's a pretty common error, though; that's the exact mistake Maddad was making.
snarkophilus
2007-Mar-09, 10:32 PM
On that point, didn't the earlier posts talking about the computer proof say that it required 2000 seperate cases to be checked?
And a proper proof would of course need to include a proof that those are all the possible cases. Tough is an understatement!
There is a more recent proof (well, modification of the original) that only requires around 630 cases. However, the proof that those are sufficient is sufficiently difficult that it needs a computer. :p
Dragon
2007-Mar-14, 02:49 AM
First of all, I'd like to appologize I haven't been here for a while: I didn't check my email for too long; I need to switch the email address for this forum to a more frequently checked one. But anyways, here's to business.
First of all, thank you Tog_ (http://www.bautforum.com/showthread.php?p=941247)Tog_[/QUOTE] for the diagram: I had a similar one in my proof that I coloured using my theorem, and was going to post it but then I saw your'n. And as Peter Engergillsaid (http://www.bautforum.com/showthread.php?p=941084), simplicity is relativistic. When I said simple, I meant it did not have 2000 possibilities: it had 8 rules and 10 cases. So therefore, grav, you shouldn't give up: even if you get a lot of rules, if you can get your attempt (http://www.bautforum.com/showthread.php?p=942048) to work, you can probably get accredited with a new proof. Granted, it would be difficult, but you've got a nice idea. In fact, were my proof not completed, I might have wasted a summer to try to perfect your proof.
worzel
2007-Mar-14, 12:27 PM
As your proof is completed, Dragon, are you going to share it with us?
Tog
2007-Mar-14, 01:04 PM
First of all, thank you Tog_ (http://www.bautforum.com/showthread.php?p=941247)Tog_ for the diagram: I had a similar one in my proof that I coloured using my theorem, and was going to post it but then I saw your'n. And as Peter Engergillsaid (http://www.bautforum.com/showthread.php?p=941084), simplicity is relativistic. [/quote]
But mine was shown to be invalid about... what, 7 seconds later. I never expected mine to be an example, I just couldn't get it work because I was hung up on something silly. I'm not sure I did much to help in any way at all. I would like to see your proof as well, however.
Dragon
2007-Mar-15, 01:46 AM
I appologize that my last post was so misshapen, but my computer was being difficult, so I was fortunate enough to get it up there as it is.
On a different note, I am not allowed to share my proof until after I publish it, and I am not allowed to publish it until July due to the copyright information agreement of the organization to whom I submitted it for revision. So once again, I appologize.
On a third note (man, I might as well compose a symphony ;)), Tog, your diagramme actually showed that grav's proof idea did not work; I don't see what you meant when you said it did not work.
worzel
2007-Mar-15, 02:29 AM
On a third note (man, I might as well compose a symphony ;))
One more and you'll be rivalling Beethovens famous intro :)
grav
2007-Mar-17, 04:29 AM
So therefore, grav, you shouldn't give up: even if you get a lot of rules, if you can get your attempt to work, you can probably get accredited with a new proof. Granted, it would be difficult, but you've got a nice idea. In fact, were my proof not completed, I might have wasted a summer to try to perfect your proof.
Thank you, Dragon. I appreciate that. Your comment prompted me to re-examine this puzzle more carefully, and then I realized, it's just that. It's a puzzle. As I was working through different possibilities, it reminded me of the number puzzles I like to work on, cross sums, where one tries to place digits in each row or column so that they add up to a specified number for that line. Each digit may only be used once and they must coincide with the numbers in the other lines at the same time. There are other similar puzzles, too, of course, like suduko. This is basically the same thing, but uses only four digits, and the geometry includes more than just columns and rows. Similar rules can therefore be applied, and that's just what I did. We wouldn't just throw down some digits and start backtracking when the number puzzle doesn't work out, so neither should we with this. It has an ordered logic to it.
Okay, so I figured it out, or so I say. :) I've only applied it to a couple of dozen random scenarios so far with about twenty territories each, give or take, but it worked the first time every time for every one. There is only one rule. Well, okay, three. But the first two are just common sense and constitute part of a proof. The third is the one that must be proven or disproven. So here we go (again). All we have to do is to begin by taking three territories that all touch each other with no space in between, and label them A, B and C, each representing a color. From this point on, we are just eliminating possibilities. All other territories are so far unmarked. Any unmarked territory means that four possibilities still exist for it. So we will only mark those for which we have determined that only one, two, or three possibilities remain.
All three rules must be followed in order with each run. We start with the most obvious. 1) If a territory is bordered by any three territories that have different labels that have already been determined, then the territory that borders all three must have the fourth label. See, told you it was common sense. Next rule, 2) If three territories all border each other, and two of them include the same two possibilities for their labels, then the third territory that borders them both does not include either. To demonstrate, if two territories that border each other include the possibilities A and D, then one is A and the other D, or one is D and the other A. Therefore, any territory that borders both of them will border an A and D regardless, so cannot include either. Finally, and this is the one to be proven or disproven, once a run has proceeded as far as it can with no further reduction in possibilities available, then one begins with the outer-most perimeter of territories that has been marked (containing two or three possibilities each) and works their way inward until one comes across a territory with only two possibilities. One may then choose either of the two possibilities for that territory they wish to label that territory. The run then begins again.
So this is how each run is performed. Beginning with our first three territories, we fill in the possibilities for all territories that surround them. There will be three for any that border only one of them, two for those that border two, and only one for any that touch all three, which can then be labeled as such. If any are labeled, then we begin the run again, by finding all of the possibilities for the territories that surround the one we just labeled as well. After we have done this until we get to the point that no more territories can be labeled, so that all territories on the outer perimeter so far contain either two or three possibilities each, we use rule number two. We must be very careful not to miss anything with these first two rules. It is very important. They can be especially difficult to catch sometimes. One must look at all territories that contain two possibilities one by one and be sure that no two bordering territories contain the same two possibilities. If they do, then one examines any territories that simultaneously border both of these and marks off those same two possibilities. If that leaves two left for the third one, then we continue with rule number two, checking also the third one in the same way since it now also contains two possibilities as well. If only one possibility remains for the third territory, then we label it as such and begin the run again (or complete rule #2 first and begin again, either way really in this case).
Finally, once both of the first rules have been exhausted, we find all territories that exist at the outer most perimeter for those that have been marked with two or three possibilities available. If any of those contain only two, then we can just pick one of them randomly to label that territory with and then start another run accordingly. If all of those on the outer-most perimeter contain three each, then we move to the next perimeter inward that doesn't include them and do the same thing, and so on. If no territories contain two possibilites, which would only occur if the territories share the same borders, then one may label any of them they wish, choosing from the possibilities given for the territory, of course.
A couple of points. First, if there is more than one territory containing two possibilities on the outer perimeter, which there usually are, then one may choose from any of them for rule number three (although now that I think about it, these are more steps than rules, since they must be followed in order). Second, when choosing a label for 'step' three, it is best to choose one that is contained in the most territories that border it, in order to eliminate more possibilities for them. But this is just a suggestion, not a rule. Anyway, that's it. One just continues steps one, two, and three, eliminating possibilities, until all territories are labeled. :D
hhEb09'1
2007-Mar-17, 04:39 AM
Anyway, that's it. :DNot quite. Now you have to prove it.
I'm betting that it doesn't work :)
Dragon
2007-Mar-17, 05:33 AM
Here is a graph which shows that your method does not work. Next time, could you be a little more careful when checking your algorithm? KMath266 (http://www.mathpages.com/home/kmath266/kmath266.htm) is a great site for examples (that's where I got this one). I understand you were just a beginner, and so were limited in the foresight of the failure. As soon as I read it, I could come up with an example, but it took like an hour to put it into a bitmap and another half hour to make another diagramme that is small enough to be uploaded.
Anyways, now that you see you have a contradiction, you might try to figure out what's wrong with your proof and correct it: I had to do that several times before I got mine to work on the ultimately difficult picture to colour. (I attached that too for you to try your methods on before you post them for me to check.) As a side note, your proof is starting to resemble my four-colour proof. :p Good luck.
HenrikOlsen
2007-Mar-17, 05:45 AM
Or alternatively, someone has to design a map that fails on the rules, which might be an interesting challenge as well. :)
Dragon
2007-Mar-17, 05:49 AM
OK, I posted a response a little while ago, but my computer is being difficult again, and the post doesn't come up. Here is an example of a counterproof, but upon revising your method, don't post again until they get my previous post moved hither, so that you could read more suggestions about it. Sorry, I don't remember what all the last post said, but I figured out what causes the problem and I hope this website will do something to fix the problem.
Edit: HenrikOlsen, I guess this answers your question. :)
It's just too bad that problem impedes my response that explained a lot of stuff.
peter eldergill
2007-Mar-17, 05:51 AM
But you only need to find one counterexample... my Calculus students often think they can find one example to prove a theorem or statement...I like to tell them that they got that one right, but there's an infinite number of other examples they have to show to be true...so one out of infinity changed to a mark out of 5... they often find that humourous (especially when we do limits)
Pete
Dragon
2007-Mar-17, 05:53 AM
Funny how everyone is up at this tremendously late hour. Or are y'all in a different time zone?
hhEb09'1
2007-Mar-17, 06:08 AM
Edit: HenrikOlsen, I guess this answers your question. Your diagrams seem to have a detached red region at the top. I don't think that grav's algorithm would produce that. Maybe I misunderstand something.
Dragon
2007-Mar-17, 06:34 AM
Oh, sorry, my bad, I misread his definition of outer, which he, by the way, would need to define better were this to be a formal proof. But at this point it does not matter because a counterproof is easy to come up with. I attach another example.
Dragon
2007-Mar-17, 06:38 AM
Why is everyone up so late anyways?
grav
2007-Mar-17, 06:40 AM
Your diagrams seem to have a detached red region at the top. I don't think that grav's algorithm would produce that. Maybe I misunderstand something.You're right, it wouldn't. At least not on the first run like that. The other three are fine to start with though.
grav
2007-Mar-17, 06:45 AM
Oh, sorry, my bad, I misread his definition of outer, which he, by the way, would need to define better were this to be a formal proof. But at this point it does not matter because a counterproof is easy to come up with. I attach another example.Boy, that was quick. The territory that is colored in the second picture should not be so yet. It has two possibilities, red or blue, and neither can be eliminated yet.
By the way, I guess it's about time I learned, how do I put pictures on here? It would probably ease this tremendously.
grav
2007-Mar-17, 06:50 AM
Oh, sorry, my bad, I misread his definition of outer, which he, by the way, would need to define better were this to be a formal proof. But at this point it does not matter because a counterproof is easy to come up with. I attach another example.Can you number the areas for that by any chance? That way, I can explain better.
HenrikOlsen
2007-Mar-17, 06:56 AM
Funny how everyone is up at this tremendously late hour. Or are y'all in a different time zone?
This is a worldwide board, there's people from all timezones here, plus lots who post at times that you wouldn't expect from their timezones.
Eg. I'm probably most prolific between 20:00 and 4:00, but with a large spread.
Incidentally, your use of font does makes your posts recognizable, but also harder to read.
Take a few minutes to consider which is most important.
Dragon
2007-Mar-17, 07:01 AM
Numbering the territories would make the file too big to fit within the limits of the upload, so we'll have to discuss it in terms of order of territories coloured.
...
All three rules must be followed in order with each run. We start with the most obvious. 1) If a territory is bordered by any three territories that have different labels that have already been determined, then the territory that borders all three must have the fourth label. See, told you it was common sense.
I did that for picture 1.
Next rule, 2) If three territories all border each other, and two of them include the same two possibilities for their labels, then the third territory that borders them both does not include either. To demonstrate, if two territories that border each other include the possibilities A and D, then one is A and the other D, or one is D and the other A. Therefore, any territory that borders both of them will border an A and D regardless, so cannot include either.
There are no territories that fall into that category.
Finally, and this is the one to be proven or disproven, once a run has proceeded as far as it can with no further reduction in possibilities available, then one begins with the outer-most perimeter of territories that has been marked (containing two or three possibilities each) and works their way inward until one comes across a territory with only two possibilities. One may then choose either of the two possibilities for that territory they wish to label that territory. The run then begins again.
I did that, and that is how I coloured the picture 2, to answer your question.
...
One must look at all territories that contain two possibilities one by one and be sure that no two bordering territories contain the same two possibilities. If they do, then one examines any territories that simultaneously border both of these and marks off those same two possibilities. If that leaves two left for the third one, then we continue with rule number two, checking also the third one in the same way since it now also contains two possibilities as well. If only one possibility remains for the third territory, then we label it as such and begin the run again (or complete rule #2 first and begin again, either way really in this case).
I could not quite understand this part.
...
Second, when choosing a label for 'step' three, it is best to choose one that is contained in the most territories that border it, in order to eliminate more possibilities for them. But this is just a suggestion, not a rule.
...And, I followed this suggestion when colouring.
Dragon
2007-Mar-17, 07:05 AM
This is a worldwide board, there's people from all timezones here, plus lots who post at times that you wouldn't expect from their timezones.
Eg. I'm probably most prolific between 20:00 and 4:00, but with a large spread.
Incidentally, your use of font does makes your posts recognizable, but also harder to read.
Take a few minutes to consider which is most important.
Harder to read? How? That's the first time I've been told that one.
As a side note, you can rid yourself of my font if you hold CTRL+scroll (decreasing or increasing font sizes on web pages) because my font only has a 10pt and a 16pt size; all other sizes get substituted with other fonts. If I'm not mistaken, Courier New for sizes larger than 10pt and Times New Roman for those smaller.
Dragon
2007-Mar-17, 07:34 AM
I hate to triple-post, but the posts are all addressing a different issue. I was going to wait until someone else posted to post this, but now I am getting ready to go to bed, and still no one posted, so I have little choice. To attach pictures, click on the "Manage Attachments" button in the "Additional Options" below the response box in advanced reply. It is fairly simple thence.
hhEb09'1
2007-Mar-17, 07:49 AM
Harder to read? How? That's the first time I've been told that one.OK, here's a second. :)
grav
2007-Mar-17, 08:13 AM
Sorry I took so long. The first three areas are already colored in, so we don't have to worry about them. So starting with the one to the left of the yellow one and working clockwise around the original three, we'll number them 1 through 6. Then from the one on the far left, 7 through 9. The color choices are represented by R, G, B, and Y.
Okay. So the possibilities for the ones surrounding the first three are:
1) RGB
2) RB
3) RBY
4) BY
5) GBY
6) GB
None of them cotain only one possibility, since none of the adjoining areas border all three, so step one is done. And no two of the six contain the same two color possibilities and border each other, so step two is done. So now we just pick one of the three with two colors and choose one of the two colors possibilities it contains (oh, you did do that right, sorry). So we'll pick 2R, as you have done in the second picture.
We now return to step one. We subtract R from the possibilities for any adjoining areas and include any additional ones in our set that also border it, which is only area 8 at this point. Now we have (not including the ones already colored in):
1) GB
3) BY
4) BY
5) GBY
6) GB
8) GBY
Rule #1 is taken care of, so we move to #2. There are two GB's and two BY's. 1 and 6 are both GB and border area 7, so area 7 becomes RY. 3 and 4 are both BY and border area 9, so area 9 becomes RG. We now have:
1) GB
3) BY
4) BY
5) GBY
6) GB
7) RY
8) GBY
9) RG
Okay. So rules #1 and #2 are now fulfilled. For #3, we start with the outermost areas with two or three possibilities and work inward. Those would be 7-9. If any of those contain two possibilities, then we choose one of them at random. So that third picture is incorrect for this, but I can see how that could be confusing. The outermost areas with two possibilities are 7 and 9. I'll choose 7. Then I'll choose Y for it. So none of the bordering territories can now contain Y as a possibility. That makes our set for those areas that are left uncolored:
1) GB
3) BY
4) BY
5) GB
6) GB
8) GB
9) RG
Moving through rule #1 and to rule #2 again, then, we find nothing changes. So back to rule #3. This time, let's go with 8B. So, back to rule #1, that makes 1G and 3Y. Then, in turn, 1G makes 6B, and 3Y makes 4B. Next, 4B makes 5G, and finally, 5G makes 9R. And that's all of them.
Hope that helps. :)
grav
2007-Mar-17, 08:27 AM
Not quite. Now you have to prove it.
I'm betting that it doesn't work :)Well, the first two rules are self-explanatory, basically common sense. And as far as I can tell, they cover all of the mandatory operations. When those two are no longer applicable, then, we can just about pick any possibility we wish from any of the areas that are left uncolored, and go from there. The only problem that I see with that, though, is why I started adding more rules in the first place when I started this a week ago. It occurs when an area becomes "isolated", buried underneath other areas. But if we work our way inward from the outer perimeter (of two or three possibility areas) when randomly choosing an area to color in, then we can get around that as well, so problem solved. I'm wondering if we could even pick randomly from an area with three possibilities and it still work out, but this is what I tried first, and it works out fine (so far, anyway), so that's what I'll stick with.
hhEb09'1
2007-Mar-17, 08:45 AM
Well, the first two rules are self-explanatory, basically common sense. I wouldn't call them common sense so much as mandatory. It seems to me if you violate them, you've failed.
And as far as I can tell, they cover all of the mandatory operations. When those two are no longer applicable, then, we can just about pick any possibility we wish from any of the areas that are left uncolored, and go from there.Now that seems like common sense--which is why I'm 99.999999% sure that this is doomed to failure. :)
That and your track record. :)
worzel
2007-Mar-17, 12:18 PM
Harder to read? How? That's the first time I've been told that one.
OK, here's a second. :)
And a third. :)
worzel
2007-Mar-17, 12:43 PM
When those two are no longer applicable, then, we can just about pick any possibility we wish from any of the areas that are left uncolored, and go from there.
If I had more time I'd take that example and systematically try every possible choice. If it failed for just one choice that'd prove that your algorithm doesn't work as advertised. Have you tried that?
If your algorithm works for any choice on that example then I'd try to construct a pathological case to make it fail for at least one choice.
But you'll get more kudos for finding a counter example yourself, you know, so I'll leave the checking to you :)
Dragon
2007-Mar-17, 01:57 PM
Well, so I guess you don't like my font. I'm just going to add the post at the bottom in my font then, just so that I could identify my posts easily. Too bad I can't make it show up one way for you and another way for me. But here are no other fonts that look good. Wish they'd let me use some of my other ones. Why is it hard to read?
But back to the proof. So I read his comment, and went back and did it again, and once again, failure. In this example, I coloured anything having a possibility of two colours with those two colours, and if aught had a possibility of three colours, I coloured it with a darker version of the fourth colour. Good luck fixing the problem.
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Well, so I guess you don't like my font. I'm just going to add the post at the bottom in my font then, just so that I could identify my posts easily. Too bad I can't make it show up one way for you and another way for me. But here are no other fonts that look good. Wish they'd let me use some of my other ones. Why is it hard to read?
But back to the proof. So I read his comment, and went back and did it again, and once again, failure. In this example, I coloured anything having a possibility of two colours with those two colours, and if aught had a possibility of three colours, I coloured it with a darker version of the fourth colour. Good luck fixing the problem.
worzel
2007-Mar-17, 02:18 PM
Hey Dragon, why don't you just make yourself an avatar if you want to see your posts quickly?
Dragon
2007-Mar-17, 02:22 PM
Well, you see, every thing that identifies me I want to have some sort of inspiration. For example, the font I use, I've been using ever since the old DOS computers that used it. Even everything on my computer, from default Internet Explorer font to the title bars of windows and names of files, is configured to have that font as a theme created back in the days, and it works still, even though WINDOWS XP technically no longer supports this font. I mean it's not in the dropdowns for the font you want anymore. So I just haven't had a chance to find a significant avatar. But how is my font hard to read?
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Well, you see, every thing that identifies me I want to have some sort of inspiration. For example, the font I use, I've been using ever since the old DOS computers that used it. Even everything on my computer, from default Internet Explorer font to the title bars of windows and names of files, is configured to have that font as a theme created back in the days, and it works still, even though WINDOWS XP technically no longer supports this font. I mean it's not in the dropdowns for the font you want anymore. So I just haven't had a chance to find a significant avatar. But how is my font hard to read?
Dragon
2007-Mar-17, 02:33 PM
Too bad I can't make it show up one way for you and another way for me.
Wait a minute, I was looking through posts; does this show up as my font to you? It does to me, but I suspect it's because it's considered the default font. If it does not for you, I could post everything inside code boxes, and then I'd recognize it's my post and it would appear normal for you.
Please note my post 211; they finally got it through.
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Please note my post 211; they finally got it through.
01101001
2007-Mar-17, 04:36 PM
Well, so I guess you don't like my font. I'm just going to add the post at the bottom in my font then, just so that I could identify my posts easily. Too bad I can't make it show up one way for you and another way for me. But here are no other fonts that look good. Wish they'd let me use some of my other ones. Why is it hard to read?
Please don't post everything twice just for your convenience. Please don't mess with the fonts. Or colors. Forget all that tempting decoration and just type your text in the style everyone is using. Look around. Be like your peers. Stop trying to make your words look weird. Thanks.
The font you chose is bitmapped, in need of anti-aliasing, looked even worse italicized in quoting, it was monospaced. It looked bold. It was ugly. It was hard to read. You don't want your words to look ugly and hard to read do you?
Dragon
2007-Mar-17, 05:03 PM
Well, at first, posting everything twice seemed to solve the problems:
1. You could read it well because I had a regular font too.
2. I could find my posts because they include my font.
3. It was convenient for the both of us, because I can read the regular font too rather than reading the text at the bottom, so as long as both of us read the top, it worked.
However, the more I thought about it, the more it seemed redundant. So I thought about putting my font into a signature. But then the signature needs to be long so as not to be inconspicuous, and I cannot think of anything which long and useful. So any suggestions about what to do?
I can see that people can see it as bold and ugly, but how is it difficult to read?
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Well, at first, posting everything twice seemed to solve the problems:
1. You could read it well because I had a regular font too.
2. I could find my posts because they include my font.
3. It was convenient for the both of us, because I can read the regular font too rather than reading the text at the bottom, so as long as both of us read the top, it worked.
However, the more I thought about it, the more it seemed redundant. So I thought about putting my font into a signature. But then the signature needs to be long so as not to be inconspicuous, and I cannot think of anything which long and useful. So any suggestions about what to do?
I can see that people can see it as bold and ugly, but how is it difficult to read?
grav
2007-Mar-17, 05:50 PM
If I had more time I'd take that example and systematically try every possible choice. If it failed for just one choice that'd prove that your algorithm doesn't work as advertised. Have you tried that?
If your algorithm works for any choice on that example then I'd try to construct a pathological case to make it fail for at least one choice.
But you'll get more kudos for finding a counter example yourself, you know, so I'll leave the checking to you :)Actually, that would be part of a proof for it, by trying every possible choice. That's actually how I found that this works in the first place, so I'll explain how that works as well. I just didn't want to get into all of that right off the bat since I knew rule #3 would be difficult to explain to begin with.
Okay. So here's what I actually did. I wanted a rigid proof, so I only eliminated possibilities according to rules #1 and #2, which are definite. It really doesn't matter what order those two are performed in. But after those two are exhausted, then what? Well, I took an area with two choices and tried a run for each one, placing the one I originally chose in a box and eliminating possibilities for other areas with a left to right diagonal slash through. I generally picked an area that had the most potential of eliminating possibilities for other areas, usually those with the most areas bordering it. Any other areas where the color was determined this way I also used a box around that possibility to show it. When all of the possibilities for rules #1 and #2 have been used, I then started back by setting a triangle around the other possibility for the same original area and doing the same thing with the other areas, this time eliminating possibilities with a right to left diagonal slash and setting triangles around the possibilities for other areas that are determined this way.
Now, after doing all of that, one can tell which possibilities for which areas can be determined to be a specific color, because they will have both a box and a triangle set around the same possibility for that area. So since the original area we started with led to the same possibility for another, and that is all of the possibilities for the original area, then any possibilities for any other areas that are both boxed and triangled must be definite, so the color for that area is now determined. Likewise, any possibilities for any other areas that contain a diagonal slash in both the left to right and right to left directions are therefore not used for either possibility in the original area, so those possibilities for the other areas are now crossed off. After doing all of this, we then start the run over. Now, I could have and probably should have explained it this way to begin with, but when I found a pattern that showed that using any possibility for any area on the outer perimeter with two possibilities came to the same conclusion, I started using that instead, and so far it has worked out every time.
Dragon
2007-Mar-17, 05:54 PM
Grav, did you look at my illustrations?
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Grav, did you look at my illustrations?
grav
2007-Mar-17, 06:00 PM
But back to the proof. So I read his comment, and went back and did it again, and once again, failure. In this example, I coloured anything having a possibility of two colours with those two colours, and if aught had a possibility of three colours, I coloured it with a darker version of the fourth colour. Good luck fixing the problem.I would have to see what you did to see where you may have gone wrong. I'm probably just not explaining it well enough.
hhEb09'1
2007-Mar-17, 06:07 PM
and so far it has worked out every time.Now, all you have to do is prove that it always works :)
Dragon
2007-Mar-17, 06:11 PM
Grav, did you look at my illustrations?
I would have to see what you did to see where you may have gone wrong.
Guess not. Well, here, I'm just going to re-post both of the diagrammes: don't miss the clickable links at the bottom. In the meanwhile, someone suggest what I can do about the font dilemma.
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Guess not. Well, here, I'm just going to re-post both of the diagrammes: don't miss the clickable links at the bottom. In the meanwhile, someone suggest what I can do about the font dilemma.
grav
2007-Mar-17, 06:23 PM
Now, all you have to do is prove that it always works :)You know, now that I think about it, I will just substitute rule #3 for the method I just stated, which is much more rigid and obvious, even if it is the long way around, for now anyway. I can show why choosing a random possibility from an area on the outer perimeter with two possibilities works out at a later time, since I really don't know, I just see that it does. So now all I have to prove is that there will be at least one area with two possibilities and that it will lead to at least one possibility for another area being determined or cancelled out each time. There always will be at least one area with two possibilities unless the inner areas share all of the same "cross-wise" borders with the outer ones, in which case we would have to use rule #3 for all three possibilities.
Dragon
2007-Mar-17, 06:34 PM
And for all the people who ask you to prove stuff, don't worry about it. Once I see that your method works, I can come up with a proof. I am fairly good at making algorithms that would come up with counterexamples, and once your proof works, the algorithm will just lead to a contradiction, meaning that it is impossible to come up with a counterexample. That would be sufficient to show that your proof works.
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And for all the people who ask you to prove stuff, don't worry about it. Once I see that your method works, I can come up with a proof. I am fairly good at making algorithms that would come up with counterexamples, and once your proof works, the algorithm will just lead to a contradiction, meaning that it is impossible to come up with a counterexample. That would be sufficient to show that your proof works.
hhEb09'1
2007-Mar-17, 06:40 PM
So now all I have to prove is that there will be at least one area with two possibilities and that it will lead to at least one possibility for another area being determined or cancelled out each time. There always will be at least one area with two possibilities unless the inner areas share all of the same "cross-wise" borders with the outer ones, in which case we would have to use rule #3 for all three possibilities.You mean like this:
1 1 2 2
1 R R 2
4 G B 3
4 4 3 3
None of the areas 1, 2, 3, or 4 have only two possibilities.
hhEb09'1
2007-Mar-17, 06:42 PM
[FONT=Georgia]And for all the people who ask you to prove stuff, don't worry about it. Once I see that your method works, I can come up with a proof. I am fairly good at making algorithms that would come up with counterexamples, and once your proof works, the algorithm will just lead to a contradiction, meaning that it is impossible to come up with a counterexample. That would be sufficient to show that your proof worksEh?
That's what this whole discussion is about, a proof. :)
grav
2007-Mar-17, 06:47 PM
Guess not. Well, here, I'm just going to re-post both of the diagrammes: don't miss the clickable links at the bottom. In the meanwhile, someone suggest what I can do about the font dilemma.
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Guess not. Well, here, I'm just going to re-post both of the diagrammes: don't miss the clickable links at the bottom. In the meanwhile, someone suggest what I can do about the font dilemma.Sorry. I didn't notice that. In the first attachment, the third image has areas 2 and 3 with the same two possibilities, blue and yellow. That means rule #2 must be applied before we go any further, so area 8 is green or red. Only then can we go to rule #3 and choose a color for an area on the outer perimeter, which would include areas 5, 6, or 8. We cannot choose any for areas 2 or 3 yet, because they then lie further in.
[EDIT-Whoops. Scratch that. I almost did the same thing there. In the third image, areas 5 and 6 also contain the same possibilities, so we must also set area 9 at red or green. That leaves only two possible areas to choose from on the outer perimeter at this piont, areas 8 and 9.]
Dragon
2007-Mar-17, 06:56 PM
I know, but as long as my counterexample stands (post 242, sorry, I'm not allowed to post hyperlinks to posts until I have a total of at least 30 posts), coming up with a proof is pointless, because you know the proof will fail and show that there is a counterexample. Also, that second diagramme on post 242 is for grav to colour with his revised method before he posts it, because that is the diagramme that caused Alfred Kempe's 1890 proof to fail. Chances are, that diagramme can make almost any false proof fail; I attach another diagramme that shows what Alfred Kempe's proof got the fc diagramme down to when it failed.
Edit: As you should be able to tell, he used a method slightly different from yours: he started with the points that have more than 5 areas touching them.
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I know, but as long as my counterexample stands (post 242, sorry, I'm not allowed to post hyperlinks to posts until I have a total of at least 30 posts), coming up with a proof is pointless, because you know the proof will fail and show that there is a counterexample. Also, that second diagramme on post 242 is for grav to colour with his revised method before he posts it, because that is the diagramme that caused Alfred Kempe's 1890 proof to fail. Chances are, that diagramme can make almost any false proof fail; I attach another diagramme that shows what Alfred Kempe's proof got the fc diagramme down to when it failed.
Edit: As you should be able to tell, he used a method slightly different from yours: he started with the points that have more than 5 areas touching them.
grav
2007-Mar-17, 07:04 PM
You mean like this:
1 1 2 2
1 R R 2
4 G B 3
4 4 3 3
None of the areas 1, 2, 3, or 4 have only two possibilities.Yep. That would be it. We would have to run through all of the possibilities for that, with reference to any other surrounding areas.
Dragon
2007-Mar-17, 07:11 PM
...
In the first attachment, the third image has areas 2 and 3 with the same two possibilities, blue and yellow. That means rule #2 must be applied before we go any further, so area 8 is green or red.
I went from image 3 to image 4 by applying rule 2. Then I recoloured the map to show the possibilities for the remaining territories.
...
In the third image, areas 5 and 6 also contain the same possibilities
...
So I applied rule 2 there, and then more of rule 1 to go from 4 to 5.
In case you didn't understand what I did, I attach an image with a more detailed step-by-step procedure of what I did. This time it's going from up to down. If you can colour the map, please attach a step-by-step bitmap (not just text, that gets confusing and takes too long to draw in order to look at it) like I did.
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In case you didn't understand what I did, I attach an image with a more detailed step-by-step procedure of what I did. This time it's going from up to down. If you can colour the map, please attach a step-by-step bitmap (not just text, that gets confusing and takes too long to draw in order to look at it) like I did.
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# Spectrum of Cayley graphs on the symmetric group generated by transpositions
Roi Krakovski
Department of Mathematics
Simon Fraser University
Burnaby, B.C. V5A 1S6
Supported in part by postdoctoral support at the Simon Fraser University.
Bojan Mohar
Department of Mathematics
Simon Fraser University
Burnaby, B.C. V5A 1S6
Supported in part by an NSERC Discovery Grant (Canada), by the Canada Research Chair program, and by the Research Grant P1–0297 of ARRS (Slovenia).On leave from: IMFM & FMF, Department of Mathematics, University of Ljubljana, Ljubljana, Slovenia.
###### Abstract
For an integer , let be the Cayley graph on the symmetric group generated by the set of transpositions . It is shown that the spectrum of contains all integers from to (except 0 if or ).
## 1 Introduction
For a graph , the spectrum of is the spectrum of its adjacency matrix. For a group and a subset , the Cayley graph of generated by , denoted by , is the graph with vertex set and an edge for each and . If is inverse-closed, then is an undirected graph, which we will assume henceforth.
In this paper we are interested in the spectrum of the Cayley graph , where is an integer, is the symmetric group and . Friedman [3] proved that if is any set of transpositions and (up to conjugacy) then the spectrum of is never integral. Abdollahi and Vatandoost [1] conjectured that spectrum of is integral, and contains all integers in the range from to (with the sole exception that when , zero is not an eigenvalue of ). Using a computer, the conjecture was verified for . In this paper we prove the second part of the conjecture.
###### Theorem 1.
Let be an integer and let . For each integer , are eigenvalues of with multiplicity at least . If , then is an eigenvalue of with multiplicity at least .
Note that is a simple eigenvalue of since the graph is -regular, bipartite, and connected.
## 2 Partial permutation graphs
In this section we introduce a family of graphs, called partial permutation graphs. Let and be positive integers such that . Let be the set of all -tuples with entries from the set having no repeated entries, that is,
The partial permutation graph is defined as follows. Its vertex set is , and two -tuples are adjacent if and only if they differ in exactly one coordinate. The following lemma, whose proof is straightforward, lists some basic properties of .
###### Lemma 2.
Let and be positive integers such that . Then satisfies the following properties:
1. .
2. The graph is -regular.
3. The cardinality of every maximal clique in is .
Next we show that is a partial permutation graph.
###### Lemma 3.
Let be an integer. Then is isomorphic to . In particular, is bipartite.
###### Proof.
Set and . For an -tuple of pairwise distinct elements of , let be the permutation defined by ().
Let be the mapping defined by , where . We claim that is a graph isomorphism. Clearly, is bijective. It remains to show that if and only if .
Suppose . Let and . Let so that and differ in the th coordinate (since , exists and it is unique). By definition of , we have and . Then , where is the transposition , hence .
For the converse, observe that if , then and differ in at least two coordinates that are different from the first coordinate. Hence, for any transpositions of the form (). ∎
For integers , let be the set of all subsets of of cardinality . For , let be the set of all -tuples (with no repetitions) with entries from the set . Note that and . Let .
For , we say that a -tuple is unique with respect to , if there exists , such that but , for every . We say that is independent, if every subset of has (at least one) unique -tuple. With this notation we have the following.
###### Lemma 4.
There exists an independent set of cardinality .
###### Proof.
We proceed by induction on . If , let . Note that and , for . Clearly, is an independent set of cardinality and the claim follows. If , then , and the claim follows with .
We may assume that and . Let . By the induction hypothesis, there exists an independent set of cardinality . Next we claim that
• there exists an independent set of cardinality , such that for every subset there is a unique -tuple with respect to , such that one of its entries contains the value .
To prove (1), let be a mapping defined by
ϕ(AI)=AI∖{n}
where and .
Clearly, is bijective. By the induction hypothesis, there is an independent set of cardinality . It is easily seen that is an independent set in with desired property. This proves (1).
By definition, and are disjoint, hence . To conclude the proof it remains to show that is an independent set.
Let . We will show that has a unique -tuple. This is trivially true if . Hence, we may assume that . Let be the unique -tuple with respect to as exists by (1). Then, one of the entries of contains the value . The uniqueness of with respect to follows since is unique with respect to and for each , no -tuple of has an entry with value . This concludes the proof. ∎
###### Lemma 5.
Let be integers and let . Then there exists a family of functions, each with domain and range , such that the following holds:
1. , for every .
2. , for every and every maximal clique .
3. If we view each as a vector in , then contains a linearly independent set of cardinality .
###### Proof.
Consider , for some . Since each element of is a -tuple, we can view as a subset of . Then, (the subgraph of induced on the vertex set ) is isomorphic to . By Lemma 3, is bipartite. Let () be the corresponding bipartition. Let be defined by
ϕI(v)=⎧⎪ ⎪⎨⎪ ⎪⎩−1,v∈A1I,1,v∈A2I,0,v∈V(X)∖AI.
Let . We will show that satisfies (i)–(iii). Property (i) is satisfied trivially, since for every , is bipartite with at least one edge.
For (ii) we argue as follows. We may assume that , for if then is isomorphic to (since then ) and satisfies (ii) as required. Let be a maximal clique in . Since , every edge of is contained in a clique of size at least 3, and so . Since is bipartite, . We claim that
1. .
To prove (1), suppose to the contrary that . Let . Let be such that does not appear in the -tuple ( exists since ). Since is a clique, the -tuples in agree on exactly coordinates and pairwise differ in exactly one coordinate, say the th coordinate (). Now consider the -tuple , obtained from by changing its th entry to . Clearly has no repetitions, , and since is maximal, . But by the definition of , . Hence, ; a contradiction. This proves (1).
By (1) and since , we have either or . In both cases satisfies the equation in (ii), thus (ii) holds.
Finally, part (iii) is a consequence of Lemma 4. First, we take a subset of cardinality which is independent in the sense of Lemma 4. Next, we consider the functions in . The existence of unique -tuples shows that for each , there is a -tuple such that , but for every . Thus we have a linearly independent set in of cardinality . ∎
## 3 Proof of the main result
Let be a group, a subgroup of and . The Schreier coset graph on generated by is the graph with , the set of left cosets of , and there is an edge for each coset and each . If is inverse-closed, then is an undirected multigraph (possibly with loops). Note that if is the identity element of , then is the Cayley graph on generated by . The following lemma is well-known, see, e.g. [3].
###### Lemma 6.
Let be a group, an inverse-closed subset of , and . If is an eigenvalue of of multiplicity , then is an eigenvalue of of multiplicity at least .
The following lemma is straightforward.
###### Lemma 7.
For , let be the subgroup of containing all permutations fixing the elements . For each -tuple , where are pairwise distinct, let be a permutation so that , for . Let . Then, is a disjoint union of all left cosets , .
Let be the Schreier coset graph , where and is inverse-closed. By Lemma 7, we see that is isomorphic to the graph whose vertex set is the set of all -tuples (with no repetition) from the set , in which a -tuple is adjacent to , for each . (This was also observed by Bacher in [2].) Note that elements of may give rise to multiple edges and loops in .
Now suppose . Let be the set of -tuples containing the value . Let . The following lemma is easily verified.
###### Lemma 8.
Let be the partition of defined above. Then the following holds:
• No two distinct vertices of are adjacent in .
• Every has self-loops and neighbors in .
• Let . Then is adjacent in to exactly vertices in , that are obtained from be replacing the entry 1 with a number different from the entries of .
###### Lemma 9.
Let . Then is an eigenvalue of with multiplicity at least .
###### Proof.
Let and let and be as above. We construct the following auxiliary graph . The vertex set of is , and two vertices are adjacent if and only if and there exists such that . Clearly, if then and differ in exactly one coordinate. Hence, is isomorphic to . Note that comes from the fact that the value does not appear in any of the vertices of . Observe that , so is defined.
Let be the set of functions satisfying the properties stated in Lemma 5, and let be a linearly independent set in of cardinality , whose existence is given by Lemma 5(iii).
For each , let be an extension of to defined by:
ϕ′(v)={ϕ(v),v∈¯¯¯¯¯V1,0,v∈V1.
To conclude the proof, it suffices to show that for every , is an eigenvector of with eigenvalue , since is an independent set. This task is equivalent to verifying that, for every , the following eigenvalue equation holds:
(n−k−1)ϕ′(v)=∑vu∈E[k]ϕ′(u) (1)
where the sum is over all edges of incident with including possible self-loops. Recall that . If , then by Lemma 8 (i) and (ii), the only edges contributing non-zero values to the right hand side of (1) are the self-loops at . Hence (1) holds in this case. If , then the left hand side of (1) is equal to 0. Now, by the definition of , the only edges contributing non-zero values to the right hand side of (1) are the edges , where . By Lemma 8 (iii), has precisely neighbors in , and they form a clique in of cardinality . By Lemma 2, is a clique of maximum cardinality in (recall that is isomorphic to ). By Lemma 5 (ii), the sum of the -values of the vertices of every maximum clique is zero and thus (1) holds in this case as well. ∎
The proof of the main result follows at once.
###### Proof of Theorem 1 (for non-zero eigenvalues)..
Since is -regular, is an eigenvalue of of multiplicity 1. Lemma 9 implies that for , is an eigenvalue of with multiplicity at least , and hence by Lemma 6 also of . The same conclusion holds for the negative values since is bipartite, and hence the spectrum is symmetric with respect to 0. ∎
## 4 Eigenvalue zero
In this section we prove that 0 is an eigenvalue of of multiplicity at least .
Let be the graph on vertex set (all pairs of distinct elements from ). Two such pairs are adjacent if and only if either they have the same second coordinate, or they are transpose of each other (one is obtained from the other by interchanging the coordinates).
###### Proposition 10.
The Cayley graph is a cover over .
###### Proof.
For , let . Consider the mapping defined as if . We claim that is a covering projection . Since both graphs are regular of degree , it suffices to see that every is adjacent in to a permutation in for each and is adjacent to a permutation in . This is confirmed below. Clearly, if , where , then and and . Thus . Similarly, and . Thus . This completes the proof since in each case, is a neighbor of in . ∎
The following corollary provides the missing evidence for the completion of the proof of Theorem 1.
###### Proposition 11.
If , then zero is an eigenvalue of and hence also of of multiplicity at least .
###### Proof.
It is well known that eigenvalues of the base graph are also eigenvalues of the cover. By Proposition 10, it suffices to show that 0 is an eigenvalue of with multiplicity at least .
Let be disjoint subsets of , where and . For , let and be real numbers so that for , for , for , and for , such that and . Finally, for each , define
xij=αiβj+αjβi. (2)
Observe that and for any .
We claim that is an eigenvector for eigenvalue 0 in . To see this, we have to show that the sum of the values on all neighbors of in is zero. But this is easy to see:
s=xji+∑l≠i,jxlj=n∑l=1xlj=n∑l=1(αlβj+αjβl)=βjn∑i=1αl+αjn∑i=1βl=0.
The eigenvectors for eigenvalue 0 as defined above span a subspace of dimension at least . The proof is by induction on . For , consider partitions of , where , , or (respectively), and . They give three linearly independent vectors. To see this, note that each of the corresponding eigenvectors defined by (2) has a non-zero value where the other two have value zero. For , consider independent vectors obtained by taking subsets of . They all have coordinate 0 for every (). Finally, we can add other eigenvectors that have precisely one non-zero coordinate for : for the th one, take and , except for , when and ). All together we have independent eigenvectors. ∎
## References
• [1] A. Abdollahi and E. Vatandoost. Which Cayley graphs are integral? Electronic J. Comb., 16(1):R122, 1–17, 2009.
• [2] R. Bacher. Valeur propre minimale du laplacien de Coxeter pour le group symetrique. J. Algebra, 167:460–472, 1994.
• [3] J. Friedman. On Cayley graphs on the symmetric group generated by tranpositions. Combinatorica, 20(4):505–519, 2000.
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Move GeneralRec one chapter slot later, since Subset should be a prereq
author Adam Chlipala Wed, 26 Oct 2011 17:14:28 -0400 ab60b10890ed dc99dffdf20a
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(* Copyright (c) 2006, 2011, Adam Chlipala
*
* Creative Commons Attribution-Noncommercial-No Derivative Works 3.0
* The license text is available at:
*)
(* begin hide *)
Require Import Arith List.
Require Import CpdtTactics Coinductive.
Set Implicit Arguments.
(* end hide *)
(** %\chapter{General Recursion}% *)
(** Termination of all programs is a crucial property of Gallina. Non-terminating programs introduce logical inconsistency, where any theorem can be proved with an infinite loop. Coq uses a small set of conservative, syntactic criteria to check termination of all recursive definitions. These criteria are insufficient to support the natural encodings of a variety of important programming idioms. Further, since Coq makes it so convenient to encode mathematics computationally, with functional programs, we may find ourselves wanting to employ more complicated recursion in mathematical definitions.
What exactly are the conservative criteria that we run up against? For %\emph{%#<i>#recursive#</i>#%}% definitions, recursive calls are only allowed on %\emph{%#<i>#syntactic subterms#</i>#%}% of the original primary argument, a restriction known as %\index{primitive recursion}\emph{%#<i>#primitive recursion#</i>#%}%. In fact, Coq's handling of reflexive inductive types (those defined in terms of functions returning the same type) gives a bit more flexibility than in traditional primitive recursion, but the term is still applied commonly. In Chapter 5, we saw how %\emph{%#<i>#co-recursive#</i>#%}% definitions are checked against a syntactic guardness condition that guarantees productivity.
Many natural recursion patterns satisfy neither condition. For instance, there is our simple running example in this chapter, merge sort. We will study three different approaches to more flexible recursion, and the latter two of the approaches will even support definitions that may fail to terminate on certain inputs, without any up-front characterization of which inputs those may be.
Before proceeding, it is important to note that the problem here is not as fundamental as it may appear. The final example of Chapter 5 demonstrated what is called a %\index{deep embedding}\emph{%#<i>#deep embedding#</i>#%}% of the syntax and semantics of a programming language. That is, we gave a mathematical definition of a language of programs and their meanings. This language clearly admitted non-termination, and we could think of writing all our sophisticated recursive functions with such explicit syntax types. However, in doing so, we forfeit our chance to take advantage of Coq's very good built-in support for reasoning about Gallina programs. We would rather use a %\index{shallow embedding}\emph{%#<i>#shallow embedding#</i>#%}%, where we model informal constructs by encoding them as normal Gallina programs. Each of the three techniques of this chapter follows that style. *)
(** * Well-Founded Recursion *)
(** The essence of terminating recursion is that there are no infinite chains of nested recursive calls. This intuition is commonly mapped to the mathematical idea of a %\index{well-founded relation}\emph{%#<i>#well-founded relation#</i>#%}%, and the associated standard technique in Coq is %\index{well-founded recursion}\emph{%#<i>#well-founded recursion#</i>#%}%. The syntactic-subterm relation that Coq applies by default is well-founded, but many cases demand alternate well-founded relations. To demonstrate, let us see where we get stuck on attempting a standard merge sort implementation. *)
Section mergeSort.
Variable A : Type.
Variable le : A -> A -> bool.
(** We have a set equipped with some %%#"#less-than-or-equal-to#"#%''% test. *)
(** A standard function inserts an element into a sorted list, preserving sortedness. *)
Fixpoint insert (x : A) (ls : list A) : list A :=
match ls with
| nil => x :: nil
| h :: ls' =>
if le x h
then x :: ls
else h :: insert x ls'
end.
(** We will also need a function to merge two sorted lists. (We use a less efficient implementation than usual, because the more efficient implementation already forces us to think about well-founded recursion, while here we are only interested in setting up the example of merge sort.) *)
Fixpoint merge (ls1 ls2 : list A) : list A :=
match ls1 with
| nil => ls2
| h :: ls' => insert h (merge ls' ls2)
end.
(** The last helper function for classic merge sort is the one that follows, to partition a list arbitrarily into two pieces of approximately equal length. *)
Fixpoint partition (ls : list A) : list A * list A :=
match ls with
| nil => (nil, nil)
| h :: nil => (h :: nil, nil)
| h1 :: h2 :: ls' =>
let (ls1, ls2) := partition ls' in
(h1 :: ls1, h2 :: ls2)
end.
(** Now, let us try to write the final sorting function, using a natural number %%#"#[<=]#"#%''% test [leb] from the standard library.
[[
Fixpoint mergeSort (ls : list A) : list A :=
if leb (length ls) 2
then ls
else let lss := partition ls in
merge (mergeSort (fst lss)) (mergeSort (snd lss)).
]]
<<
Recursive call to mergeSort has principal argument equal to
"fst (partition ls)" instead of a subterm of "ls".
>>
The definition is rejected for not following the simple primitive recursion criterion. In particular, it is not apparent that recursive calls to [mergeSort] are syntactic subterms of the original argument [ls]; indeed, they are not, yet we know this is a well-founded recursive definition.
To produce an acceptable definition, we need to choose a well-founded relation and prove that [mergeSort] respects it. A good starting point is an examination of how well-foundedness is formalized in the Coq standard library. *)
Print well_founded.
(** %\vspace{-.15in}% [[
well_founded =
fun (A : Type) (R : A -> A -> Prop) => forall a : A, Acc R a
]]
The bulk of the definitional work devolves to the %\index{accessibility relation}\index{Gallina terms!Acc}\emph{%#<i>#accessibility#</i>#%}% relation [Acc], whose definition we may also examine. *)
Print Acc.
(** %\vspace{-.15in}% [[
Inductive Acc (A : Type) (R : A -> A -> Prop) (x : A) : Prop :=
Acc_intro : (forall y : A, R y x -> Acc R y) -> Acc R x
]]
In prose, an element [x] is accessible for a relation [R] if every element %%#"#less than#"#%''% [x] according to [R] is also accessible. Since [Acc] is defined inductively, we know that any accessibility proof involves a finite chain of invocations, in a certain sense which we can make formal. Building on Chapter 5's examples, let us define a co-inductive relation that is closer to the usual informal notion of %%#"#absence of infinite decreasing chains.#"#%''% *)
CoInductive isChain A (R : A -> A -> Prop) : stream A -> Prop :=
| ChainCons : forall x y s, isChain R (Cons y s)
-> R y x
-> isChain R (Cons x (Cons y s)).
(** We can now prove that any accessible element cannot be the beginning of any infinite decreasing chain. *)
(* begin thide *)
Lemma noChains' : forall A (R : A -> A -> Prop) x, Acc R x
-> forall s, ~isChain R (Cons x s).
induction 1; crush;
match goal with
| [ H : isChain _ _ |- _ ] => inversion H; eauto
end.
Qed.
(** From here, the absence of infinite decreasing chains in well-founded sets is immediate. *)
Theorem noChains : forall A (R : A -> A -> Prop), well_founded R
-> forall s, ~isChain R s.
destruct s; apply noChains'; auto.
Qed.
(* end thide *)
(** Absence of infinite decreasing chains implies absence of infinitely nested recursive calls, for any recursive definition that respects the well-founded relation. The [Fix] combinator from the standard library formalizes that intuition: *)
Check Fix.
(** %\vspace{-.15in}%[[
Fix
: forall (A : Type) (R : A -> A -> Prop),
well_founded R ->
forall P : A -> Type,
(forall x : A, (forall y : A, R y x -> P y) -> P x) ->
forall x : A, P x
]]
A call to %\index{Gallina terms!Fix}%[Fix] must present a relation [R] and a proof of its well-foundedness. The next argument, [P], is the possibly dependent range type of the function we build; the domain [A] of [R] is the function's domain. The following argument has this type:
[[
forall x : A, (forall y : A, R y x -> P y) -> P x
]]
This is an encoding of the function body. The input [x] stands for the function argument, and the next input stands for the function we are defining. Recursive calls are encoded as calls to the second argument, whose type tells us it expects a value [y] and a proof that [y] is %%#"#less than#"#%''% [x], according to [R]. In this way, we enforce the well-foundedness restriction on recursive calls.
The rest of [Fix]'s type tells us that it returns a function of exactly the type we expect, so we are now ready to use it to implement [mergeSort]. Careful readers may have noticed that [Fix] has a dependent type of the sort we met in the previous chapter.
Before writing [mergeSort], we need to settle on a well-founded relation. The right one for this example is based on lengths of lists. *)
Definition lengthOrder (ls1 ls2 : list A) :=
length ls1 < length ls2.
(** We must prove that the relation is truly well-founded. To save some space in the rest of this chapter, we skip right to nice, automated proof scripts, though we postpone introducing the principles behind such scripts to Part III of the book. Curious readers may still replace semicolons with periods and newlines to step through these scripts interactively. *)
Hint Constructors Acc.
Lemma lengthOrder_wf' : forall len, forall ls, length ls <= len -> Acc lengthOrder ls.
unfold lengthOrder; induction len; crush.
Defined.
Theorem lengthOrder_wf : well_founded lengthOrder.
red; intro; eapply lengthOrder_wf'; eauto.
Defined.
(** Notice that we end these proofs with %\index{Vernacular commands!Defined}%[Defined], not [Qed]. Recall that [Defined] marks the theorems as %\emph{transparent}%, so that the details of their proofs may be used during program execution. Why could such details possibly matter for computation? It turns out that [Fix] satisfies the primitive recursion restriction by declaring itself as %\emph{%#<i>#recursive in the structure of [Acc] proofs#</i>#%}%. This is possible because [Acc] proofs follow a predictable inductive structure. We must do work, as in the last theorem's proof, to establish that all elements of a type belong to [Acc], but the automatic unwinding of those proofs during recursion is straightforward. If we ended the proof with [Qed], the proof details would be hidden from computation, in which case the unwinding process would get stuck.
To justify our two recursive [mergeSort] calls, we will also need to prove that [partition] respects the [lengthOrder] relation. These proofs, too, must be kept transparent, to avoid stuckness of [Fix] evaluation. *)
Lemma partition_wf : forall len ls, 2 <= length ls <= len
-> let (ls1, ls2) := partition ls in
lengthOrder ls1 ls /\ lengthOrder ls2 ls.
unfold lengthOrder; induction len; crush; do 2 (destruct ls; crush);
destruct (le_lt_dec 2 (length ls));
repeat (match goal with
| [ _ : length ?E < 2 |- _ ] => destruct E
| [ _ : S (length ?E) < 2 |- _ ] => destruct E
| [ IH : _ |- context[partition ?L] ] =>
specialize (IH L); destruct (partition L); destruct IH
end; crush).
Defined.
Ltac partition := intros ls ?; intros; generalize (@partition_wf (length ls) ls);
destruct (partition ls); destruct 1; crush.
Lemma partition_wf1 : forall ls, 2 <= length ls
-> lengthOrder (fst (partition ls)) ls.
partition.
Defined.
Lemma partition_wf2 : forall ls, 2 <= length ls
-> lengthOrder (snd (partition ls)) ls.
partition.
Defined.
Hint Resolve partition_wf1 partition_wf2.
(** To write the function definition itself, we use the %\index{tactics!refine}%[refine] tactic as a convenient way to write a program that needs to manipulate proofs, without writing out those proofs manually. We also use a replacement [le_lt_dec] for [leb] that has a more interesting dependent type. *)
Definition mergeSort : list A -> list A.
(* begin thide *)
refine (Fix lengthOrder_wf (fun _ => list A)
(fun (ls : list A)
(mergeSort : forall ls' : list A, lengthOrder ls' ls -> list A) =>
if le_lt_dec 2 (length ls)
then let lss := partition ls in
merge (mergeSort (fst lss) _) (mergeSort (snd lss) _)
else ls)); subst lss; eauto.
Defined.
(* end thide *)
End mergeSort.
(** The important thing is that it is now easy to evaluate calls to [mergeSort]. *)
Eval compute in mergeSort leb (1 :: 2 :: 36 :: 8 :: 19 :: nil).
(** [= 1 :: 2 :: 8 :: 19 :: 36 :: nil] *)
(** Since the subject of this chapter is merely how to define functions with unusual recursion structure, we will not prove any further correctness theorems about [mergeSort]. Instead, we stop at proving that [mergeSort] has the expected computational behavior, for all inputs, not merely the one we just tested. *)
(* begin thide *)
Theorem mergeSort_eq : forall A (le : A -> A -> bool) ls,
mergeSort le ls = if le_lt_dec 2 (length ls)
then let lss := partition ls in
merge le (mergeSort le (fst lss)) (mergeSort le (snd lss))
else ls.
intros; apply (Fix_eq (@lengthOrder_wf A) (fun _ => list A)); intros.
(** The library theorem [Fix_eq] imposes one more strange subgoal upon us. We must prove that the function body is unable to distinguish between %%#"#self#"#%''% arguments that map equal inputs to equal outputs. One might think this should be true of any Gallina code, but in fact this general %\index{extensionality}\emph{%#<i>#function extensionality#</i>#%}% property is neither provable nor disprovable within Coq. The type of [Fix_eq] makes clear what we must show manually: *)
Check Fix_eq.
(** %\vspace{-.15in}%[[
Fix_eq
: forall (A : Type) (R : A -> A -> Prop) (Rwf : well_founded R)
(P : A -> Type)
(F : forall x : A, (forall y : A, R y x -> P y) -> P x),
(forall (x : A) (f g : forall y : A, R y x -> P y),
(forall (y : A) (p : R y x), f y p = g y p) -> F x f = F x g) ->
forall x : A,
Fix Rwf P F x = F x (fun (y : A) (_ : R y x) => Fix Rwf P F y)
]]
Most such obligations are dischargable with straightforward proof automation, and this example is no exception. *)
match goal with
| [ |- context[match ?E with left _ => _ | right _ => _ end] ] => destruct E
end; simpl; f_equal; auto.
Qed.
(* end thide *)
(** As a final test of our definition's suitability, we can extract to OCaml. *)
Extraction mergeSort.
(** <<
let rec mergeSort le x =
match le_lt_dec (S (S O)) (length x) with
| Left ->
let lss = partition x in
merge le (mergeSort le (fst lss)) (mergeSort le (snd lss))
| Right -> x
>>
We see almost precisely the same definition we would have written manually in OCaml! It might be a good exercise for the reader to use the commands we saw in the previous chapter to clean up some remaining differences from idiomatic OCaml.
One more piece of the full picture is missing. To go on and prove correctness of [mergeSort], we would need more than a way of unfolding its definition. We also need an appropriate induction principle matched to the well-founded relation. Such a principle is available in the standard library, though we will say no more about its details here. *)
Check well_founded_induction.
(** %\vspace{-.15in}%[[
well_founded_induction
: forall (A : Type) (R : A -> A -> Prop),
well_founded R ->
forall P : A -> Set,
(forall x : A, (forall y : A, R y x -> P y) -> P x) ->
forall a : A, P a
]]
Some more recent Coq features provide more convenient syntax for defining recursive functions. Interested readers can consult the Coq manual about the commands %\index{Function}%[Function] and %\index{Program Fixpoint}%[Program Fixpoint]. *)
(** * A Non-Termination Monad *)
Section computation.
Variable A : Type.
Definition computation :=
{f : nat -> option A
| forall (n : nat) (v : A),
f n = Some v
-> forall (n' : nat), n' >= n
-> f n' = Some v}.
Definition runTo (m : computation) (n : nat) (v : A) :=
proj1_sig m n = Some v.
Definition run (m : computation) (v : A) :=
exists n, runTo m n v.
End computation.
Hint Unfold runTo.
Ltac run' := unfold run, runTo in *; try red; crush;
repeat (match goal with
| [ _ : proj1_sig ?E _ = _ |- _ ] =>
match goal with
| [ x : _ |- _ ] =>
match x with
| E => destruct E
end
end
| [ |- context[match ?M with exist _ _ => _ end] ] => let Heq := fresh "Heq" in
case_eq M; intros ? ? Heq; try rewrite Heq in *; try subst
| [ _ : context[match ?M with exist _ _ => _ end] |- _ ] => let Heq := fresh "Heq" in
case_eq M; intros ? ? Heq; try rewrite Heq in *; subst
| [ H : forall n v, ?E n = Some v -> _,
_ : context[match ?E ?N with Some _ => _ | None => _ end] |- _ ] =>
specialize (H N); destruct (E N); try rewrite (H _ (refl_equal _)) by auto; try discriminate
| [ H : forall n v, ?E n = Some v -> _, H' : ?E _ = Some _ |- _ ] => rewrite (H _ _ H') by auto
end; simpl in *); eauto 7.
Ltac run := run'; repeat (match goal with
| [ H : forall n v, ?E n = Some v -> _
|- context[match ?E ?N with Some _ => _ | None => _ end] ] =>
specialize (H N); destruct (E N); try rewrite (H _ (refl_equal _)) by auto; try discriminate
end; run').
Lemma ex_irrelevant : forall P : Prop, P -> exists n : nat, P.
exists 0; auto.
Qed.
Hint Resolve ex_irrelevant.
Require Import Max.
Ltac max := intros n m; generalize (max_spec_le n m); crush.
Lemma max_1 : forall n m, max n m >= n.
max.
Qed.
Lemma max_2 : forall n m, max n m >= m.
max.
Qed.
Hint Resolve max_1 max_2.
Lemma ge_refl : forall n, n >= n.
crush.
Qed.
Hint Resolve ge_refl.
Hint Extern 1 => match goal with
| [ H : _ = exist _ _ _ |- _ ] => rewrite H
end.
Section Bottom.
Variable A : Type.
Definition Bottom : computation A.
exists (fun _ : nat => @None A); abstract run.
Defined.
Theorem run_Bottom : forall v, ~run Bottom v.
run.
Qed.
End Bottom.
Section Return.
Variable A : Type.
Variable v : A.
Definition Return : computation A.
intros; exists (fun _ : nat => Some v); abstract run.
Defined.
Theorem run_Return : run Return v.
run.
Qed.
Theorem run_Return_inv : forall x, run Return x -> x = v.
run.
Qed.
End Return.
Hint Resolve run_Return.
Section Bind.
Variables A B : Type.
Variable m1 : computation A.
Variable m2 : A -> computation B.
Definition Bind : computation B.
exists (fun n =>
let (f1, Hf1) := m1 in
match f1 n with
| None => None
| Some v =>
let (f2, Hf2) := m2 v in
f2 n
end); abstract run.
Defined.
Require Import Max.
Theorem run_Bind : forall (v1 : A) (v2 : B),
run m1 v1
-> run (m2 v1) v2
-> run Bind v2.
run; match goal with
| [ x : nat, y : nat |- _ ] => exists (max x y)
end; run.
Qed.
Theorem run_Bind_inv : forall (v2 : B),
run Bind v2
-> exists v1 : A,
run m1 v1
/\ run (m2 v1) v2.
run.
Qed.
End Bind.
Hint Resolve run_Bind.
Notation "x <- m1 ; m2" :=
(Bind m1 (fun x => m2)) (right associativity, at level 70).
Definition meq A (m1 m2 : computation A) := forall n, proj1_sig m1 n = proj1_sig m2 n.
Theorem left_identity : forall A B (a : A) (f : A -> computation B),
meq (Bind (Return a) f) (f a).
run.
Qed.
Theorem right_identity : forall A (m : computation A),
meq (Bind m (@Return _)) m.
run.
Qed.
Theorem associativity : forall A B C (m : computation A) (f : A -> computation B) (g : B -> computation C),
meq (Bind (Bind m f) g) (Bind m (fun x => Bind (f x) g)).
run.
Qed.
Section monotone_runTo.
Variable A : Type.
Variable c : computation A.
Variable v : A.
Theorem monotone_runTo : forall (n1 : nat),
runTo c n1 v
-> forall n2, n2 >= n1
-> runTo c n2 v.
run.
Qed.
End monotone_runTo.
Hint Resolve monotone_runTo.
Section lattice.
Variable A : Type.
Definition leq (x y : option A) :=
forall v, x = Some v -> y = Some v.
End lattice.
Hint Unfold leq.
Section Fix.
Variables A B : Type.
Variable f : (A -> computation B) -> (A -> computation B).
Hypothesis f_continuous : forall n v v1 x,
runTo (f v1 x) n v
-> forall (v2 : A -> computation B),
(forall x, leq (proj1_sig (v1 x) n) (proj1_sig (v2 x) n))
-> runTo (f v2 x) n v.
Fixpoint Fix' (n : nat) (x : A) : computation B :=
match n with
| O => Bottom _
| S n' => f (Fix' n') x
end.
Hint Extern 1 (_ >= _) => omega.
Hint Unfold leq.
Lemma Fix'_ok : forall steps n x v, proj1_sig (Fix' n x) steps = Some v
-> forall n', n' >= n
-> proj1_sig (Fix' n' x) steps = Some v.
unfold runTo in *; induction n; crush;
match goal with
| [ H : _ >= _ |- _ ] => inversion H; crush; eauto
end.
Qed.
Hint Resolve Fix'_ok.
Hint Extern 1 (proj1_sig _ _ = _) => simpl;
match goal with
| [ |- proj1_sig ?E _ = _ ] => eapply (proj2_sig E)
end.
Definition Fix : A -> computation B.
intro x; exists (fun n => proj1_sig (Fix' n x) n); abstract run.
Defined.
Definition extensional (f : (A -> computation B) -> (A -> computation B)) :=
forall g1 g2 n,
(forall x, proj1_sig (g1 x) n = proj1_sig (g2 x) n)
-> forall x, proj1_sig (f g1 x) n = proj1_sig (f g2 x) n.
Hypothesis f_extensional : extensional f.
Theorem run_Fix : forall x v,
run (f Fix x) v
-> run (Fix x) v.
run; match goal with
| [ n : nat |- _ ] => exists (S n); eauto
end.
Qed.
End Fix.
Hint Resolve run_Fix.
Lemma leq_Some : forall A (x y : A), leq (Some x) (Some y)
-> x = y.
intros ? ? ? H; generalize (H _ (refl_equal _)); crush.
Qed.
Lemma leq_None : forall A (x y : A), leq (Some x) None
-> False.
intros ? ? ? H; generalize (H _ (refl_equal _)); crush.
Qed.
Definition mergeSort' : forall A, (A -> A -> bool) -> list A -> computation (list A).
refine (fun A le => Fix
(fun (mergeSort : list A -> computation (list A))
(ls : list A) =>
if le_lt_dec 2 (length ls)
then let lss := partition ls in
ls1 <- mergeSort (fst lss);
ls2 <- mergeSort (snd lss);
Return (merge le ls1 ls2)
else Return ls) _); abstract (run;
repeat (match goal with
| [ |- context[match ?E with O => _ | S _ => _ end] ] => destruct E
end; run);
repeat match goal with
| [ H : forall x, leq (proj1_sig (?f x) _) (proj1_sig (?g x) _) |- _ ] =>
match goal with
| [ H1 : f ?arg = _, H2 : g ?arg = _ |- _ ] =>
generalize (H arg); rewrite H1; rewrite H2; clear H1 H2; simpl; intro
end
end; run; repeat match goal with
| [ H : _ |- _ ] => (apply leq_None in H; tauto) || (apply leq_Some in H; subst)
end; auto).
Defined.
Print mergeSort'.
Lemma test_mergeSort' : run (mergeSort' leb (1 :: 2 :: 36 :: 8 :: 19 :: nil))
(1 :: 2 :: 8 :: 19 :: 36 :: nil).
exists 4; reflexivity.
Qed.
Definition looper : bool -> computation unit.
refine (Fix (fun looper (b : bool) =>
if b then Return tt else looper b) _);
abstract (unfold leq in *; run;
repeat match goal with
| [ x : unit |- _ ] => destruct x
| [ x : bool |- _ ] => destruct x
end; auto).
Defined.
Lemma test_looper : run (looper true) tt.
exists 1; reflexivity.
Qed.
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# 83.1 kg in pounds
• 83.1 kilograms is 183.2 pounds.
• tk10publ tk10canl
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# Solved Question Papers - IIT JEE/MathsQP1996
IIT JEE MATHEMATICS QSN Let A={0,1,2,3,4,5,6,7,8,9} Exhibit the set B={X|X € A and x is even}& set C={X|2X € A }by listing memebers of B & C
Next qsn The angle between the pair of tangents drawn from a point P to a circle x^2+y^2+4x-6y+9sin^2 (A) +13cos^2 (A)=0 is 2A. Then the equation of locus of point P:
Next qsn An ellipse has excentricity 1/2 and one focus S (1/2,1). Its one directrix is common tangent (nearer to S) to circe x^2 +y^2=1 and x^2-y^2=1. Then equation of ellipse in standard form is_____
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# QlikView Layout & Visualizations
Discussion Board for collaboration on QlikView Layout & Visualizations.
Highlighted
Not applicable
## Linear regression in straight table
Hi community. I am trying to predict the number of defects in future months based on linear regression of data from prior months. I am trying to do this in a straight table which shows multiple products that live in different business units (BU). My problem is that unless I filter on a single product the formula is returning the same value for all products. I would guess it has something to do with the aggr or total functions in my expression or calculated dimension, but can’t seem to figure out the problem. Can someone please look at the code below or the attached sample QVW and let me know what I’m doing wrong?
Calculated Dimension for Product (used to exclude products that don’t have data relevant to my table):
=aggr(only({<[Product]= P([Product]),[Data Source]= {'Second'},[Data Metric]= {'KPI Two'}>}[Product]),[Product])
Expression:
=linest_m(total aggr(if( [Month Num]>=14 and [Month Num]<=25 ,count({<[Include Flag]={1}>} distinct [Case Number])),[Month Num]),[Month Num])*27
+
linest_b(total aggr(if( [Month Num]>=14 and [Month Num]<=25 ,count({<[Include Flag]={1}>} distinct [Case Number])),[Month Num]),[Month Num])
Output:
Thanks so much.
1 Solution
Accepted Solutions
Not applicable
## Re: Linear regression in straight table
I believe I found the answer in the following thread ... Multi-dimensional regression analysis
Here is my revised expression:
=linest_m(total <BU,Product> aggr(if( [Month Num]>=14 and [Month Num]<=25,count({<[Include Flag]={1}>} distinct [Case Number])),BU,Product,[Month Num]),[Month Num])*27
+
linest_b(total <BU,Product> aggr(if( [Month Num]>=14 and [Month Num]<=25,count({<[Include Flag]={1}>} distinct [Case Number])),BU,Product,[Month Num]),[Month Num])
Here are the revised results:
Not applicable
## Re: Linear regression in straight table
I believe I found the answer in the following thread ... Multi-dimensional regression analysis
Here is my revised expression:
=linest_m(total <BU,Product> aggr(if( [Month Num]>=14 and [Month Num]<=25,count({<[Include Flag]={1}>} distinct [Case Number])),BU,Product,[Month Num]),[Month Num])*27
+
linest_b(total <BU,Product> aggr(if( [Month Num]>=14 and [Month Num]<=25,count({<[Include Flag]={1}>} distinct [Case Number])),BU,Product,[Month Num]),[Month Num])
Here are the revised results:
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# EMU of Capacitance to Gigafarad Converter
1 EMU of Capacitance = 1 Gigafarad
## One EMU of Capacitance is Equal to How Many Gigafarads?
The answer is one EMU of Capacitance is equal to 1 Gigafarad and that means we can also write it as 1 EMU of Capacitance = 1 Gigafarad. Feel free to use our online unit conversion calculator to convert the unit from EMU of Capacitance to Gigafarad. Just simply enter value 1 in EMU of Capacitance and see the result in Gigafarad.
Manually converting EMU of Capacitance to Gigafarad can be time-consuming,especially when you don’t have enough knowledge about Electrostatic Capacitance units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online EMU of Capacitance to Gigafarad converter tool to get the job done as soon as possible.
We have so many online tools available to convert EMU of Capacitance to Gigafarad, but not every online tool gives an accurate result and that is why we have created this online EMU of Capacitance to Gigafarad converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert EMU of Capacitance to Gigafarad (EMU to GF)
By using our EMU of Capacitance to Gigafarad conversion tool, you know that one EMU of Capacitance is equivalent to 1 Gigafarad. Hence, to convert EMU of Capacitance to Gigafarad, we just need to multiply the number by 1. We are going to use very simple EMU of Capacitance to Gigafarad conversion formula for that. Pleas see the calculation example given below.
$$\text{1 EMU of Capacitance} = 1 \times 1 = \text{1 Gigafarad}$$
## What Unit of Measure is EMU of Capacitance?
EMU is a unit of measurement for electrostatic capacitance. One EMU of electrostatic capacitance is equal to 1 abfarad.
## What is the Symbol of EMU of Capacitance?
The symbol of EMU of Capacitance is EMU. This means you can also write one EMU of Capacitance as 1 EMU.
## What Unit of Measure is Gigafarad?
Gigafarad is a unit of measurement for electrostatic capacitance. A capacitor of one gigafarad produces a potential difference of one gigavolt between its plates when it stores an electric charge of one gigacoulomb. Gigafarad is a multiple of electrostatic capacitance unit farad. One gigafarad is equal to 1e+9 farad.
## What is the Symbol of Gigafarad?
The symbol of Gigafarad is GF. This means you can also write one Gigafarad as 1 GF.
## How to Use EMU of Capacitance to Gigafarad Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select EMU of Capacitance and in the first input field, enter a value.
• From the second dropdown, select Gigafarad.
• Instantly, the tool will convert the value from EMU of Capacitance to Gigafarad and display the result in the second input field.
## Example of EMU of Capacitance to Gigafarad Converter Tool
EMU of Capacitance
1
1
# EMU of Capacitance to Other Units Conversion Table
ConversionDescription
1 EMU of Capacitance = 1000000000 Farad1 EMU of Capacitance in Farad is equal to 1000000000
1 EMU of Capacitance = 1e-9 Exafarad1 EMU of Capacitance in Exafarad is equal to 1e-9
1 EMU of Capacitance = 0.000001 Petafarad1 EMU of Capacitance in Petafarad is equal to 0.000001
1 EMU of Capacitance = 0.001 Terafarad1 EMU of Capacitance in Terafarad is equal to 0.001
1 EMU of Capacitance = 1 Gigafarad1 EMU of Capacitance in Gigafarad is equal to 1
1 EMU of Capacitance = 1000 Megafarad1 EMU of Capacitance in Megafarad is equal to 1000
1 EMU of Capacitance = 1000000 Kilofarad1 EMU of Capacitance in Kilofarad is equal to 1000000
1 EMU of Capacitance = 10000000 Hectofarad1 EMU of Capacitance in Hectofarad is equal to 10000000
1 EMU of Capacitance = 100000000 Dekafarad1 EMU of Capacitance in Dekafarad is equal to 100000000
1 EMU of Capacitance = 10000000000 Decifarad1 EMU of Capacitance in Decifarad is equal to 10000000000
1 EMU of Capacitance = 100000000000 Centifarad1 EMU of Capacitance in Centifarad is equal to 100000000000
1 EMU of Capacitance = 1000000000000 Millifarad1 EMU of Capacitance in Millifarad is equal to 1000000000000
1 EMU of Capacitance = 1000000000000000 Microfarad1 EMU of Capacitance in Microfarad is equal to 1000000000000000
1 EMU of Capacitance = 1000000000000000000 Nanofarad1 EMU of Capacitance in Nanofarad is equal to 1000000000000000000
1 EMU of Capacitance = 1e+21 Picofarad1 EMU of Capacitance in Picofarad is equal to 1e+21
1 EMU of Capacitance = 1e+24 Femtofarad1 EMU of Capacitance in Femtofarad is equal to 1e+24
1 EMU of Capacitance = 1e+27 Attofarad1 EMU of Capacitance in Attofarad is equal to 1e+27
1 EMU of Capacitance = 1000000000 Coulomb/Volt1 EMU of Capacitance in Coulomb/Volt is equal to 1000000000
1 EMU of Capacitance = 1 Abfarad1 EMU of Capacitance in Abfarad is equal to 1
1 EMU of Capacitance = 899000042253000000000 Statfarad1 EMU of Capacitance in Statfarad is equal to 899000042253000000000
1 EMU of Capacitance = 899000042253000000000 ESU of Capacitance1 EMU of Capacitance in ESU of Capacitance is equal to 899000042253000000000
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2018-05-03, 17:51 #67 Cruelty May 2005 2·809 Posts status report k=2 and k=4 @ base=3 tested till n=1.72M 127*128^n-1 tested till n=1.8M
2018-09-02, 00:35 #68 Cruelty May 2005 65216 Posts status report k=2 and k=4 @ base=3 tested till n=1.76M 127*128^n-1 tested till n=1.86M
2018-11-01, 12:37 #69 Cruelty May 2005 2·809 Posts status report k=2 and k=4 @ base=3 tested till n=1.78M 127*128^n-1 tested till n=1.88M
2018-11-30, 23:31 #70 Cruelty May 2005 2×809 Posts status report k=2 and k=4 @ base=3 tested till n=1.8M 127*128^n-1 tested till n=1.88M
2019-01-01, 15:39 #71 Cruelty May 2005 161810 Posts status report k=2 and k=4 @ base=3 tested till n=1.81M 127*128^n-1 tested till n=1.93M
2019-03-01, 19:15 #72 Cruelty May 2005 2×809 Posts status report k=2 and k=4 @ base=3 tested till n=1.84M 127*128^n-1 tested till n=1.95M
2019-04-14, 17:28 #73 kar_bon Mar 2006 Germany 23·5·71 Posts Made pages for 2*3^n-1 and 4*3^n-1 in the PrimeWiki, S.Harvey is aware of these, too. For k=4 I've not included n=0 as in the OEIS-entry.
2019-09-05, 13:17 #74 Cruelty May 2005 2×809 Posts status report k=2 and k=4 @ base=3 tested till n=1.93M 127*128^n-1 tested till n=2.07M
2019-10-05, 17:28 #75 Cruelty May 2005 2×809 Posts status report k=2 and k=4 @ base=3 tested till n=1.94M 127*128^n-1 tested till n=2.08M
2020-01-03, 20:42 #76 Cruelty May 2005 65216 Posts status report k=2 and k=4 @ base=3 tested till n=2M (sieving) 127*128^n-1 tested till n=2.15M
2020-04-04, 00:20 #77 Cruelty May 2005 2·809 Posts status report k=2 and k=4 @ base=3 tested till n=2M (sieving) 127*128^n-1 tested till n=2.22M
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# Cart friction problem
1. Dec 1, 2009
### williamx11373
a horizontal force of 200 Newtons is applied to a 55 kg cart across a 10 meter surface. If the cart accelerates at 2 meters/second squared, then what is the work done by the force of friction as it acts to retard the motion of the cart ??
Work= FX
work= maX
55 kg x 2 x 10 = 1100 - 200 N = 900 Joules ??
is this correct ?
2. Dec 2, 2009
### kuruman
Re: friction
It is not correct. The "F" in F = ma is the net force, i.e. the sum of all the forces. In this case, it is the vector sum of the 200 N and the force of friction. What expression do you know that gives you the force of (kinetic) friction? Use that to calculate the work done by it.
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From ae88be93dd8082e711861466cf71e56173ff3866 Mon Sep 17 00:00:00 2001 From: jim Date: Thu, 19 Mar 2015 18:47:08 -0400 Subject: [PATCH] explain mappable laws better --- topics/week7_introducing_monads.mdwn | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/topics/week7_introducing_monads.mdwn b/topics/week7_introducing_monads.mdwn index adcf0cc2..8f320140 100644 --- a/topics/week7_introducing_monads.mdwn +++ b/topics/week7_introducing_monads.mdwn @@ -130,7 +130,7 @@ has to obey the following Map Laws: `map (id : α -> α) = (id : α -> α)` `map (g â f) = (map g) â (map f)` - Essentially these say that `map` is a homomorphism from `(α -> β, â, id)` to `(α -> β, â', id')`, where `â'` and `id'` are `â` and `id` restricted to arguments of type `_`. + Essentially these say that `map` is a homomorphism from `(α -> β, â, id)` to `(α -> β, â', id')`, where `â'` and `id'` are `â` and `id` restricted to arguments of type `_`. That might be hard to digest because it's so abstract. Think of the following concrete example: if you take a `α list` (that's our `α`), and apply `id` to each of its elements, that's the same as applying `id` to the list itself. That's the first law. And if you apply the composition of functions `g â f` to each of the list's elements, that's the same as first applying `f` to each of the elements, and then going through the elements of the resulting list and applying `g` to each of those elements. That's the second law. These laws obviously hold for our familiar notion of `map` in relation to lists. * ***MapNable*** (in Haskelese, "Applicatives") A Mappable box type is *MapNable* -- 2.11.0
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