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https://community.facer.io/t/watchface-flightschool-a206/31737 | 1,701,929,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00463.warc.gz | 196,477,003 | 5,121 | # [WatchFace] FlightSchool - A206
Released and ready for purchase in the Facer app.
5 Likes
OOOO I really like this one! Cool charge meter too.
1 Like
Great work! Thumbs up high!
I don’t like to put my nose into your job but I would like to suggest you another way to move the numbers to get much realism.
I make this simple example with a “step motion” here with rotation but you can easly change into vertical shift.
I like your watchface also as it is now!
3 Likes
You’re right on the money with that one. I was looking for the formula but couldn’t find it. I asked @GRR how he did it in a PM, but havent gotten a response yet.
Do you have inspection open on this one? I’ll see if I can update it with the single digit code I have on there already.
This is a great counter @dario.marnoni!
Could i use this formula for a future design as well?
I put it here to share
It is inspection enabled I think
Let me know if you can see it
there are the rotation formulas
First number:
\$(#Dsm#-(10*(floor(#Dsm#/10))))<=9?(floor(#Ds#/10)*60):((#Dsm#-#Ds#)*60+floor(#Ds#/10)*60)\$
second number:
\$(#Dsm#-#Ds#)<0.5?(36-36floor(#Dsm#-(10(floor(#Dsm#/10))))):(-(#Dsm#-#Ds#-0.5)72+36-36floor(#Dsm#-(10*(floor(#Dsm#/10)))))\$
I made a mistake into description… they are first number and second number for seconds obviously
Make sure tu put all the “moltiplication” because sometimes this editor cuts it all
2 Likes
I think I got it worked out based on the your formulas. Just trying to test it to see if the hours behave like they are supposed to.
1 Like
I finally got all my testing complete and pushed out the new updated step movement. Had a little snafu with publishing the updated design, but everything is all worked out and back live again.
2 Likes | 497 | 1,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.937246 |
https://myonlinehomeworkhelper.com/data-analysis-2-2/ | 1,591,463,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00309.warc.gz | 455,842,665 | 10,383 | #### Data Analysis
October 31, 2015 0 Comment
Question
Answer in Excel using Excel Functions (don’t just put answer in cell)
Hotel Room Pricing
A hotel has 500 rooms. During a typical week in the holiday season, the hotel is busy on Monday, Tuesday, Wednesday, and Thursday nights, primarily with business people, but there is generally space available on Friday, Saturday, and Sunday nights.
Hotel management provided the following demand estimates, as shown in Table 8.1:
Assume linear demand functions over the range of all non-negative room rates and demands.
Table 8.1: Demand Estimates
: Room Rate ‘ : Days. , Estimated Demand’ (rooms/clay)
\$160 Mon, Tue, Wed, Thur 500
\$160 Fri, Sat, Sun 200
\$175 Mon, Tue, Wed, Thur 350
\$175 Fri, Sat, Sun 125
· What is the revenue-maximizing room rate if the hotel posts only a single rate good for any day of the week? What weekly occupancy results?
· What are the revenue-maximizing room rates if the hotel posts a “mid-week” rate good for the peak demand period (MTWT), and a different “weekend” rate good for Friday, Saturday, or Sunday. What is the new weekly occupancy? What is the revenue increase?
· Another option is for the hotel to offer a discounted weekly rate in order to attract vacationers who will stay for the full week. Management estimates that demand will be 40 per week at \$900/week and 160 per week at \$800/week. Again assume a linear demand curve.
· If the hotel posts a weekly rate and a single daily rate (good for any night), what are the revenue-maximizing prices? What is the occupancy rate? What is the revenue gain?
· If the hotel posts three rates: a mid-week rate, a weekend rate, and a weekly rate, what are the revenue-maximizing prices? What is the occupancy rate? What is the revenue gain?
Overbooking
You have 100 hotel room/nights to sell at \$300 each. The no-show rate for bookings is 10%. The cost of looking after a customer with a reservation whom you cannot provide with a room is \$1,000. How many reservations should you accept?
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• Free plagiarism reports; | 581 | 2,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-24 | latest | en | 0.924908 |
https://www.wallstreetmojo.com/data-snooping-bias/ | 1,723,375,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00300.warc.gz | 801,179,475 | 79,647 | FLASH SALE! - "FINANCIAL MODELING COURSE BUNDLE AT 60% OFF" Enroll Now
# Data Snooping Bias
Updated on April 4, 2024
Article byAswathi Jayachandran
Edited byRashmi Kulkarni
Reviewed byDheeraj Vaidya, CFA, FRM
## Data Snooping Bias Meaning
Data Snooping Bias is a type of statistical bias that arises when data containing numerous variables is subjected to statistical analysis. It also occurs when testing is done without a defined a priori hypothesis or proper multiple testing corrections. It has also been observed in situations where researchers use existing studies as a guide for their research.
For eg:
Source: Data Snooping Bias (wallstreetmojo.com)
Data snooping bias occurs when data is overanalyzed, which gives rise to statistically irrelevant and occasionally nonexistent patterns. For instance, investors may repeatedly examine previous investment strategies and a portfolio’s past performance, resulting in bias when making decisions. Keeping these mistakes in check, especially while analyzing financial data, can prevent valuation errors and help reveal the true and fair position of a business.
### Key Takeaways
• Data snooping bias is a type of statistical bias usually seen when a large collection of data with numerous variables is subjected to statistical analysis.
• The bias manifests itself when searching exhaustively for combinations of variables; as more combinations are evaluated, the likelihood that a result might have occurred “by chance” increases.
• The two peculiar situations in which data snooping bias most frequently occurs are when researchers adjust the data they use to lower the likelihood of a sample rejecting the hypothesis and when researchers have not yet developed an independent hypothesis.
• In-sample and Out-of-sample testing are two methods that help reduce data snooping bias.
### Data Snooping Bias Explained
Data Snooping Bias, a statistical bias, surfaces due to the use of incorrect data mining techniques, and it can provide false results in scientific studies. It is also known as data mining bias, data dredging bias, or backtest overfitting. Even though data snooping biases can occur in any industry that uses data mining, they are particularly problematic in finance and medical research because these fields heavily rely on data mining methods.
These erroneous patterns can occasionally be statistically minor and essentially undetectable. However, data snooping biases can be quite significant because minor changes in financial calculations frequently result in extremely large and significant differences in investment performance.
While the practice of employing advanced machine learning models to analyze data has now gained popularity, the chances of data being misused cannot be ignored. Data snooping bias in machine learning can significantly modify or affect results. Such data manipulation may or may not be intentional. However, it can have serious, long-term financial implications for a business.
Data snooping bias usually surfaces when researchers/users search exhaustively for combinations of variables. As more combinations are evaluated, the likelihood that a result might have occurred “by chance” increases. This kind of data snooping bias is seen in two specific situations. The first is when researchers present the data they use in a manner that helps them lower the likelihood of a sample rejecting a particular hypothesis. The second situation is when researchers have not yet developed a hypothesis. In such situations, they are usually open to suggestions presented through data analysis. Unfortunately, there is no way to guarantee that the bias will not occur, but measures can be taken to reduce the possibility of its occurrence.
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### Examples
The following examples will provide further information and clarity on this concept.
#### Example #1
Let us assume Dan, a researcher, wants to study the relationship between blood pressure and a specific medication prescribed for diabetes. He has access to a large dataset containing information about patients who suffer from both ailments. Dan readily starts looking for patterns and finds that medicines taken in succession within a certain period produce better effects than those taken separately over longer gaps. After testing, the results confirm his hypothesis. However, Dan had tested multiple hypotheses on the same dataset, which could have given rise to a snooping bias, as he could have simply been looking at it from different perspectives until the same results were achieved.
#### Example #2
The issue of data snooping bias in financial asset pricing research is highlighted in a study. It examines data snooping bias and demonstrates how it can cause overfitting of the data and erroneous conclusions. The capital asset pricing model (CAPM) and the arbitrage pricing theory (APT) are a few examples of financial asset pricing models that were tested. It also shows how data snooping bias might appear in these tests.
The study finds that data snooping bias is a serious issue in the study of financial asset pricing and suggests measures be taken to reduce its influence on the findings. The report emphasizes the significance of recognizing this bias and adopting measures to lower its influence on the analysis in general. By doing this, researchers can get more reliable findings, which are less likely to be impacted by random chance or false correlations in the data.
### How To Avoid?
Backtesting can be an effective solution when researchers wish to address data snooping bias. Apart from software applications that help eliminate such problems, two methods can be used to filter out the errors.
1. The first one is called In-sample Testing. It is a data sample/method that backtests the same kind of data that was used to build the in-sample testing model. For example, while working on trading data, it is the data sample used to backtest all combinations arising from the original trading rules.
2. The second method is called Out-of-sample Testing. This method is used to test the highest-performing rules (those that were chosen from the in-sample backtesting) on fresh data. Out-of-sample method testing serves as a filter, rejecting the rules that did not perform well in the in-sample test and accepting only the rules that pass both tests.
It is crucial to remember that data snooping bias is not always intentional. Occasionally, it can happen just because an analyst looks through datasets for patterns. It is important to be conscious of this potential bias and take precautions to reduce its influence on the analysis.
1. Why is data snooping bias a problem?
Bias in this form is a problem because it can alter results. It can be extremely dangerous in medical research, as human lives are involved. They can also lead to potential losses in business initiatives. False conclusions can also influence government policies and recommendations.
2. How to manage data snooping bias in financial analysis?
It should be understood that financial analysis is subject to errors regardless of the data. Analysts must be careful while deciding the testing parameters and the number of times they are tested. Machine learning algorithms built to prevent overfitting can also be used.
3. What are the consequences of data snooping bias?
This kind of bias can have serious consequences in data analysis. Some problematic outcomes can be false conclusions, overfitting (analysis is closely tailored and does not generalize with a new set of data), and a lack of reproducibility. In addition, a waste of time, effort, and resources is possible. Also, damage to reputation is another grave possibility.
This article has been a guide to Data Snooping Bias and its meaning. Here, we explain the concept in detail along with its examples and how to avoid it. You may also find some useful articles here – | 1,578 | 8,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.888529 |
https://brilliant.org/discussions/thread/recommendation-for-maths-books/ | 1,627,902,310,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154320.56/warc/CC-MAIN-20210802110046-20210802140046-00368.warc.gz | 157,832,701 | 23,148 | # Recommendation for Maths Books
Hello everyone,
This is my first note at brilliant ,
I'm writing this note to know about a few but Good books on the following topics , (please Note that im a beginner) :-
(1) Number theory , (2) Plane Geometry ,
• Also please recommend some books for preparation of JMO & RMO.
Please do comment & help.! Thank You ..!
Note by Rishabh Tiwari
5 years, 2 months ago
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Firstly, Physics and Mathematics are two very interestingly deadly subjects. Once you loose grip of them, you almost loose that forever. I'll recommend some books for now, trying solving each and every problem from them. Don't do the mistake that I did, i.e. completely leaving one subject for the sake of another.
Geometry:
• Coxeter
• SL Loney Trigonometry (Problems only)
• Titu Andreescu books
Algebra:
• Hall and Knight Higher Algebra is more than enough
• Brilliant wikis for Classical Inequalities
Number Theory:
• Brilliant wikis
• Burton Number Theory
Physics:
• HC Verma
• IE Irodov
• Cengage for Electromagnetism
• DC Pandey for Mechanics
• Halliday Resnick Walker for Fundamental Concepts.
• Challenge and Thrill of Pre College Mathematics
• INMO book by Rajeev Manocha
• InPhO book by Arihant
• Physics today and Mathematics today are a set of must buy for gaining excellence in Olympiads.
Books like RD Sharma and NCERT are not at all Olympiad Mathematics. A Olympiad student can do their problems even while watching CID. So start from my recommended books and gain excellence.
Hope that helps.
Warning: Calculus is famously stated as a math olympian's nightmare. Do study Calculus from Arihant JEE books.Best of luck!
- 5 years, 1 month ago
And atlast when the physics master steps in, the others are keft astounded. #claps #whistles (+1)
- 5 years, 1 month ago
$\text {Thank You Very Much !!!}$ , You have provided me with enough choices , I hope a few from them would work for me , because if I get too much of them, then confusion wouldn't let Me study.....! Btw I have the basic PCM resources for JEE, thanks for ur recommendation, Btw based on your experience , what would you suggest , shall I take one - two books or study all of them? One more question, are the titu andrescuu, math olympiad journals , & coxeter available in the market? Or I have to order them online? Thanking you once again for all your help!
- 5 years, 1 month ago
I personally do not possess the hard copy versions of the books, rather I have downloaded PDFs. Physics books can be ordered online, so you do not need to worry about them.
It depends on you on the basis of your speed and urge of preparations, how many books do you want to buy at once. I have all the books as PDFs or hardcopies, and I don't find anything confusing me, to be very honest. Central to Mathematics and Physics is problem solving, so as many problems, that much of practice. If you want to get limited to some problems, atmost 5 books from the list may do. But I find myself comfortable with doing atleast 100 types of questions from a single topic, because that's fun.
Be free to ask me further questions, always ready to help :)
- 5 years, 1 month ago
Thank you very much, for your motivation,I got it, aren't you in 11th this year? You'll be too giving olympiads I think?
- 5 years, 1 month ago
I'm in 10th. Yes,I'll be giving Olympiads this year.
- 5 years, 1 month ago
Ok then, you are really a physics champion! $\text {All the Best}$ for all the competitions ! Thank you very much for your help!
- 5 years, 1 month ago
Thanks for your compliment. All the very best to you too. Glad to help :)
- 5 years, 1 month ago
Are you in Fiitjee Delhi?
- 5 years, 1 month ago
Nah, I am from Odisha.
- 5 years, 1 month ago
- 5 years, 1 month ago
Recommended complete few books properly. After being thorough with it go to another set. Thats what I am going to follow throughout the whole year.
- 5 years, 1 month ago
Thank you very much buddy, thats what I thought too! :):)
- 5 years, 1 month ago
How was Phase test ?
- 5 years, 1 month ago
Completely messed up. The entire test was a nightmare for me. How was yours?
- 5 years, 1 month ago
Mine was cool except for SST.
- 5 years, 1 month ago
Sorry for late reply ... Elementary. NT by D.Burton.
- 5 years, 1 month ago
Thank you very much !, is it available as a hard copy in market?
- 5 years, 1 month ago
Oh yeah , are you interested in physics :)
- 5 years, 1 month ago
Yes I enjoy physics too much. I am also preparing for maths & physics olympiad this year.! Can you suggest me some resources /books?
- 5 years, 1 month ago
@Rajdeep Dhingra @Swapnil Das Can guide you better than me ....For physics try HCV and irodov... i do not have experience with Olympiads ...but try those 2 books...
- 5 years, 1 month ago
Thank you very much buddy, I am solving Hcv as of now & have heard about irodov its a pretty tough book, in which class u are?
- 5 years, 1 month ago
I just stepped into 11th. ... wbu ?
- 5 years, 1 month ago
Me too! Classes will start from 1st July, I have joined an IIT coaching here in Mathura , & you?
- 5 years, 1 month ago
What can be the good books for KVS JMO & RMO ? @Rajdeep Dhingra , @Swapnil Das , plz help, thank you.
- 5 years, 1 month ago
HC verma suits well for physics, RC Mukerjee for Chemistry.
- 5 years, 1 month ago
Hey buddy, thnx for ur suggestion , I love problems of rcm , I just don't know any books to prepare for Jmo & Rmo , Their course are a bit different!
- 5 years, 1 month ago
Rmo? Yeah for all competitive exams one set of books do really well:- NCERT
- 5 years, 1 month ago
Yup ur absolutely right , geometry from 8 ,9, & 10th Ncert would be good , but to prepare Number theory & combinatorics & higher geometry , I think I need to refer some good books , as of now I have none!
- 5 years, 1 month ago
Yeah, get nice books but publishers too count. Best recommended are Goyal brothers.
- 5 years, 1 month ago
Ok got it , if u can give the details for Olympiad book, then it would be very nice of u, inform me please whenever you have the chance. Thanks for helping.!
- 5 years, 1 month ago
Sure. Another book for geometry:- Geometry Revisited.
- 5 years, 1 month ago
Yup I liked that book , Deeparaj bhaiya suggested that to me , awesome book ! I think that now I have enough choices to choose books, thnx for ur help buddy, really man!
- 5 years, 1 month ago
Math , My Olympiad Math isn't very good. Also math can't be learned. It can only be practised. Try solving Past papers and other countries national papers.
- 5 years, 1 month ago
Thank you very much for your suggestion , I will solve some papers definitely.!
- 5 years, 1 month ago
Which class are you in Rishabh ?
- 5 years, 1 month ago
I have also stepped in 11th class.!
- 5 years, 1 month ago
- 5 years, 1 month ago
I have some very old JEE books for subjective. They are equivalent to Olympiad Level. You can try the Irodov Theory books.
- 5 years, 1 month ago
For elecro-magnetism - Edward M Purcell - Electricity and Magnetism, Griffiths,Jakson, I.E irodov(theory book)
Mechanics - Classical mechanics by david morin , Kleppner mechanics, I.E irodov(theory book)
Ray Optics and others - HC Verma , Halliday Resnik
Alzebra - Bernard and child, Hall and knight alzebra
Calculus - Tarasov calculus , I.A Maron calculus
Trignometry and Geometry - S.L Loney is the best
Questions in Physics - I.E irodov, Krotov.
Acc. to me , the best book for physics is -: Lewis-Carroll-Epstein-Thinking-Physics-Practical-Lessons-in-Critical-Thinking-Insight-Press-2002
- 5 years, 1 month ago
Thank you so much! Now I have sufficient choices for physics!
- 5 years, 1 month ago
- 5 years, 2 months ago
- 5 years, 2 months ago
- 5 years, 2 months ago
For plane geometry, these books are quite good:
The first book covers pure geometry and the second covers trigonometry (which is useful in pure geometry problems many times).
The third book deals with solving geometry probpems via complex numbers.
- 5 years, 2 months ago
Thank you very much brother!
- 5 years, 2 months ago
No problem.
- 5 years, 2 months ago
I tried downloading them , but it always fails.! Can you help me? Plz send it at my email rishu0818@gmail.com ! Thnx for ur help!
- 5 years, 2 months ago
Ok. I've sent the mail. As of now it's in the outbox. I'll update you as soon as it is sent.
- 5 years, 2 months ago
Thank you :)
- 5 years, 2 months ago
Ok. They've been sent. See if you're able to view them and update me. :)
- 5 years, 2 months ago
They are all very nice books, I have received them , thank you for your help brother😃!
- 5 years, 2 months ago
- 5 years, 2 months ago
- 5 years, 2 months ago
Try Challenge and Thrill of Pre-College Mathematics by C R Pranesachar and V Krishnamurthy ...its one of the recommended books for rmo...the geometry section in the book is the best....
- 5 years, 1 month ago
It's cool that you can get math recommendations on your blog. I have long dreamed of discussing with someone the problems that have now appeared. I’m sure that using https://eduzaurus.com/free-essay-samples/immigration/ and looking at free essay sample in turn, you can easily understand the whole point of school work and their tasks.
- 1 year, 4 months ago
Yo, one book which is best for all these things :- ★RD Sharma★
- 5 years, 2 months ago
For which class plz?
- 5 years, 2 months ago
11th
- 5 years, 2 months ago
It contains number theory too? I didn't knew that , well thank you very much Ashish for ur help.!
- 5 years, 2 months ago
Haha number theory too, I leatn from it.
- 5 years, 2 months ago
Is it a competition book or usual based on ncert type?
- 5 years, 2 months ago
It ia both. Beat recommwnded book for all 11th and 12th graders. Those who arw through with this can surely do well in IIT JEE.
- 5 years, 2 months ago
Thank you very much buddy .! I will definitely use that book! Thnx once again !
- 5 years, 2 months ago
(^。^)
- 5 years, 2 months ago
Whats this?
- 5 years, 2 months ago
Happy face ʕ•ٹ•ʔ
- 5 years, 2 months ago | 3,342 | 11,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.846412 |
https://www.physicsforums.com/threads/dr-evil-problem.338332/ | 1,508,649,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00644.warc.gz | 978,947,932 | 14,449 | # Dr. Evil Problem
1. Sep 18, 2009
### kero
Can somebody help me please to solve this problem?
helicopter carrying Dr. Evil takes off with a constant upward acceleration of Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. ( a) What is the maximum height above ground reached by the helicopter? ( b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward accel-eration with magnitude How far is Powers above the ground when the helicopter crashes into the ground?
2. Sep 18, 2009
### tiny-tim
Welcome to PF!
Hi kero! Welcome to PF! | 199 | 849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-43 | longest | en | 0.931621 |
https://affordablepaperwritings.com/2021/02/18/assume-that-the-manager-of-fort-winston-hospital-are-setting-the/ | 1,632,433,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00575.warc.gz | 145,882,083 | 11,855 | Posted: February 18th, 2021
# Assume that the manager of fort winston hospital are setting the
Variable cost per visit \$5.00
Annual direct fixed costs \$500,000
Expected annual utilization 10,000 visits
a. What per visit must be set for the service to break even? To earn an annual profit of \$100,000?
b. Repeat Part a, but assume that the variable cost per visit is \$10.
c. Return to the data given in the problem. Again repeat Part a, but assume that direct fixed costs are \$1,000,000.
d. Repeat Part a assuming both a \$10 variable cost and \$1,000,000 in direct fixed costs.
Must be done on excel.
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\$0.00 | 224 | 886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-39 | latest | en | 0.910736 |
https://teachwithme.com/blogs/getting-to-the-core/itemlist/tag/pumpkin%20seed%20activities | 1,638,701,027,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363157.32/warc/CC-MAIN-20211205100135-20211205130135-00275.warc.gz | 617,931,472 | 13,439 | ## 9 Common Core Math Activities With Pumpkins
1-2-3 Come Do Nine Fine Pumpkin On The Vine Math Activities With Me
I have so many fun pumpkin activities to share, that I thought I'd feature 9 of my favorites that I use to teach all sorts of math standards.
A quick, easy and fun way to review numbers from 1-30, counting backwards from 20 or 10 to 0; plus skip counting by 2's, 3's, 5's & 10's is with the pumpkin slider. There are 3 different pumpkin patterns for children to choose from.
So that you can also review upper & lowercase letters, I included those traceable strips as well. Sliders are a great way to whole group assess as you play an "I Spy!" game.
If you're working on telling time with your kiddos, the Pumpkin Time cards are perfect for a pocket chart or use as flashcards.
They review analog and digital time to the hour as well as time to the half hour. Make extra sets for students to play Memory Match or "I Have; Who Has?" games. I've included a tip list of other things you can do with the cards, plus a Kaboom game.
For more telling time reinforcement, your kiddos will enjoy the "It's Pumpkin Time!" games. There are dice as well as spinner games.
Both reinforce digital as well as analog time. I've included blank templates to use as an assessment tool, or for students to make mini time booklets.
Have you started working on money? Then I think you'll enjoy Pumpkin Payment
Several standards are covered in this easy-reader pumpkin coin booklet that reinforce coins and shapes.
Students trace and write the coin word, the value of the coin, plus the shape word. They trace the shape and then draw it on the pumpkin; cutting and gluing the coin(s) to the matching numbered boxes.
Are you looking for some measurement activities? Help students practice measurement, by using apples and pumpkins.
You can run this packet off as an entire booklet for each child to work on, or use one worksheet each day during your math or science time.
I have pages where students measure with blocks, and other worksheets where students measure with a real scale and a yardstick. Click on the link for Pumpkin & Apple Measurement Activities
More measurement activities can be found in the Pumpkin Investigation Booklet.
Students measure height, weight, width and circumference of a pumpkin. They trace and write vocabulary-building words, predict, answer questions, + collect and analyze data.
I think most teachers cover the life cycle of a pumpkin to add a bit of science into their day.
With that in mind, I designed From Seeds To Pumpkin Pie: a quick, easy and awesome looking life cycle of a pumpkin craftivity. Ever mindful of standards, I included some shape & fraction fun to go with it.
The front of the pumpkin reviews all of the 2D basic shapes, including the hexagon, as students design their Jack-O-Lantern. (K.G.2)
The back of the pumpkin converts into a pie and is divided into quarters that show the pumpkin's life cycle. To make it look like a "real" pie tin, I covered a paper plate with aluminum foil.
Two fraction worksheets are included, to work on dividing circles and rectangles into two and four equal shares. Students describe the sections using the words halves, fourths & quarters. (1.G.3) Completed projects look terrific suspended from the ceiling.
Finally, the Seed Sorting packet, helps you to continue with a bit more science, while covering all sorts of math standards: Data collection & analysis, sorting, comparing & contrasting, predicting, guess-timating, counting, sequencing, greater than, less than & equal to, plus graphing.
You can do these activities as a whole group, or set things up as a center and have students work independently on their own seed worksheets.
The easy reader My Seed Booklet, is a matching activity. You can simply make a booklet to share with your students, so that they can see the different kinds of popular fall seeds, or have each child make their own booklet by drawing the seeds.
Since you can buy packages of popcorn, sunflower seeds and pumpkin seeds, you may want your students to glue some real ones to their booklet as well. You can always use the leftovers for all sorts of counting and sorting activities.
If you're looking for a few more math-related pumpkin activities, scroll down to another blog article filled with even more fall FREEBIES.
That's it for today. Thanks for visiting. I hope you found a few things to get your kiddos excited about math, while learning a bit of science too.
I'm off to the farmer's market to buy a few small pumpkins and gourds; I love decorating for fall. Wishing you a colorful autumn day filled with ed-venture!
"Those who live in the past limit their future." -Unknown
## Studying Seeds
1-2-3 Come Study Seeds With Me.
I just returned from a wonderful get-away weekend with my husband. We enjoyed seeing all of the gorgeous fall colors here in Michigan and stopping at several farms to buy fresh produce; lots of apples, pumpkins, corn etc.
It got me to marveling at how things grow, so I thought it would be fun to make several seed activities. They are quick, easy and interesting math extensions, that also touch a bit on science.
I decided to match the seeds that I had put in the easy-reader booklet: My Seeds, a few years ago.
Here students trace and write the various fruit words and color the pictures. If you have the seeds available, students can glue them to the appropriate pages.
The Seed Exploration packet covers quite a few math standards. If you don't want to foot the bill for all of the seeds, you can send the parent-note home asking for donations.
This is included in the packet. Our Dollar Store sells packages of sunflower and pumpkin seeds as well as bags of popcorn kernels.
If you carve a pumpkin in your class to analyze pumpkin data, you may want to save the seeds from that and do these as follow-up activities. It's also easy to simply buy a package of pumpkin seeds that are ready to eat.
To introduce your lesson on seeds, use the KWL for seeds that's included in the packet.
There's also an information sheet defining seeds that you can share with your students.
You may want to set up these activities as a center. Fill paper bowls with the various seeds
Have students bring up their Dixie cup and take a spoonful of each kind and put it in their cup. When they get back to their desk they can spill out their seeds and arrange them on the sorting mat.
After students are done sorting, they take one of each seed and glue it to their identification worksheet.
Students can also arrange the seeds in size from smallest to largest and then glue one of each kind on their "sequencing sizes" worksheet.
I've also included a guess-timation worksheet. You can do this as a whole group, or have students work on their own paper. Students also work on their greater than, less than, or equal to skills with a worksheet incorporating those math symbols.
When everyone is done, gather students in a circle to review what they learned, discuss their discoveries, share their worksheets and do any graphing extensions that you want to follow up with.
Thanks for visiting today. I hope you can pop on over tomorrow for the newest FREEBIES hot off the press.
"Good teaching cannot be reduced to a technique; good teaching comes from the identity and integrity of the teacher." -Parker Palmer
Examining Seeds
I revamped this idea from one that was posted on Pinterest where a creative teacher had her students draw some seeds.
I've since lost the link so if you know who this is, please shoot me an e-mail so I can link up to her and give credit.
As much as I like little ones making their own illustrations, I wanted to add more seeds and didn't want this to be labor intensive with the addition of the writing process as well.
In the past, my Y5's covered all of these seeds throughout the year, so springtime is a wonderful point to compare and contrast them all as a great review.
Students read the sentence, trace the main word, write it, and then color the object that it comes from.
If you want to add more pizzazz to the booklets, you can have students glue the matching seeds to the pages.
For a few dollars, you can buy popcorn, sunflower and pumpkin seeds in bulk and all you need is a few apples and a slice of watermelon and you’re set for the rest of the seeds.
Comparing and contrasting the seeds makes a great discussion as well. I’ve also provided a graphing extension to nail that standard too.
When everyone has completed their booklet, read it as a whole group to reinforce concepts of print.
I know your students will enjoy taking it home and sharing it with their families.
This booklet is a terrific way to plug in a little science, especially with all the seeds being planted in gardens right now.
My Seed Book makes a nice addition to your Daily 5 activities as well.
Do you have a seed activity you’d care to share? I’d enjoy hearing from you. diane@teachwithme.com OR…feel free to leave a comment here, especially if you use one of my ideas.
Feel free to PIN anything you think will be helpful to parents or teachers.
Thanks for stopping by. I hope you can pop by for more fun tomorrow.
Did you know that I added over 50 new items to the cart this month? Wow!
For more fun in the garden, scroll down for article #2 In My Garden, my newest Count and Color booklet. | 2,042 | 9,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-49 | latest | en | 0.953465 |
https://www.mapleprimes.com/questions/39373-Creating-Recurision-In-Maple-11-Tia-Sal2 | 1,712,919,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00094.warc.gz | 778,255,958 | 24,173 | # Question:Creating Recurision in maple 11 tia sal2
## Question:Creating Recurision in maple 11 tia sal2
Maple
Greetings all I know I can get the odd numbers and even numbers using the following command
oddnum := seq(2*j+1, j = 0 .. 13);
1,3,5,7,9,.....
evennum := seq(2*j, j = 0 .. 13);
0,2,4,6,8,......
But I would like to get a sequency like this
0,1,4,5,8,9 where it does two numbers in sequency then skips the next two
and
2,3,6,7,10,11 where it starts at two the three then skips two numbers
I know it's a recursion but I'm not sure how to change it?
Can anyone recommend a simple online book or online video
tia sal2
| 206 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.86618 |
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The recording industry is fighting a losing battle: it
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17 Jan 2008, 11:56
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The recording industry is fighting a losing battle: it simply does not have the resources to prosecute all of the individuals who illegally download music from the Internet. Because the number of individuals who will be charged with a crime is so limited, the actions of the recording industry will have a minimal impact on the number of people who illegally download music.
The answer to which of the following questions would best help evaluate the accuracy of the conclusion above?
A. Will recording industry lawyers dedicate the majority of their time to prosecuting those who illegally download music?
B. Is a small minority of individuals responsible for the majority of illegal song downloads?
C.Do many individuals who illegally download songs share their music files with other Internet users?
D.Will new Internet security technology permit the recording industry to more quickly and easily identify individuals who illegally download music?
E.Will the threat of prosecution alter the behavior of those who illegally download music?
Kudos [?]: 426 [1], given: 0
Director
Joined: 08 Jun 2007
Posts: 575
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17 Jan 2008, 12:01
JCLEONES wrote:
The recording industry is fighting a losing battle: it simply does not have the resources to prosecute all of the individuals who illegally download music from the Internet. Because the number of individuals who will be charged with a crime is so limited, the actions of the recording industry will have a minimal impact on the number of people who illegally download music.
The answer to which of the following questions would best help evaluate the accuracy of the conclusion above?
A. Will recording industry lawyers dedicate the majority of their time to prosecuting those who illegally download music?
B. Is a small minority of individuals responsible for the majority of illegal song downloads?
C.Do many individuals who illegally download songs share their music files with other Internet users?
D.Will new Internet security technology permit the recording industry to more quickly and easily identify individuals who illegally download music?
E.Will the threat of prosecution alter the behavior of those who illegally download music?
E.
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Director
Joined: 01 Jan 2008
Posts: 619
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17 Jan 2008, 12:43
I second E.
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CEO
Joined: 29 Mar 2007
Posts: 2554
Kudos [?]: 500 [0], given: 0
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17 Jan 2008, 12:44
JCLEONES wrote:
The recording industry is fighting a losing battle: it simply does not have the resources to prosecute all of the individuals who illegally download music from the Internet. Because the number of individuals who will be charged with a crime is so limited, the actions of the recording industry will have a minimal impact on the number of people who illegally download music.
The answer to which of the following questions would best help evaluate the accuracy of the conclusion above?
A. Will recording industry lawyers dedicate the majority of their time to prosecuting those who illegally download music?
B. Is a small minority of individuals responsible for the majority of illegal song downloads?
C.Do many individuals who illegally download songs share their music files with other Internet users?
D.Will new Internet security technology permit the recording industry to more quickly and easily identify individuals who illegally download music?
E.Will the threat of prosecution alter the behavior of those who illegally download music?
Seems to be E.
Kudos [?]: 500 [0], given: 0
Intern
Joined: 14 Jan 2008
Posts: 5
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17 Jan 2008, 15:34
My answer is E. What is the OA.
good one +1
JCLEONES wrote:
The recording industry is fighting a losing battle: it simply does not have the resources to prosecute all of the individuals who illegally download music from the Internet. Because the number of individuals who will be charged with a crime is so limited, the actions of the recording industry will have a minimal impact on the number of people who illegally download music.
The answer to which of the following questions would best help evaluate the accuracy of the conclusion above?
A. Will recording industry lawyers dedicate the majority of their time to prosecuting those who illegally download music?
B. Is a small minority of individuals responsible for the majority of illegal song downloads?
C.Do many individuals who illegally download songs share their music files with other Internet users?
D.Will new Internet security technology permit the recording industry to more quickly and easily identify individuals who illegally download music?
E.Will the threat of prosecution alter the behavior of those who illegally download music?
Kudos [?]: [0], given: 0
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The recording industry is fighting a losing battle: it 0 18 Sep 2016, 01:25
2 The recording industry is fighting a losing battle: it 11 12 Oct 2010, 04:38
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The recording industry is fighting a losing battle: it 1 22 Dec 2008, 12:26
Display posts from previous: Sort by | 1,419 | 6,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-39 | latest | en | 0.924249 |
https://number.academy/2008469 | 1,716,857,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00177.warc.gz | 369,690,088 | 11,225 | # Number 2008469 facts
The odd number 2,008,469 is spelled 🔊, and written in words: two million, eight thousand, four hundred and sixty-nine, approximately 2.0 million. The ordinal number 2008469th is said 🔊 and written as: two million, eight thousand, four hundred and sixty-ninth. The meaning of the number 2008469 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 2008469. What is 2008469 in computer science, numerology, codes and images, writing and naming in other languages
## What is 2,008,469 in other units
The decimal (Arabic) number 2008469 converted to a Roman number is (M)(M)(V)MMMCDLXIX. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
2008469 seconds equals to 3 weeks, 2 days, 5 hours, 54 minutes, 29 seconds
2008469 minutes equals to 4 years, 1 month, 3 weeks, 1 day, 18 hours, 29 minutes
### Codes and images of the number 2008469
Number 2008469 morse code: ..--- ----- ----- ---.. ....- -.... ----.
Sign language for number 2008469:
Number 2008469 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 2008469 is a prime number.
#### Factorization and factors (dividers)
The prime factors of 2008469
Prime numbers have no prime factors smaller than themselves.
The factors of 2008469 are 1, 2008469.
Total factors 2.
Sum of factors 2008470 (1).
#### Prime factor tree
2008469 is a prime number.
#### Powers
The second power of 20084692 is 4.033.947.723.961.
The third power of 20084693 is 8.102.058.951.196.225.536.
#### Roots
The square root √2008469 is 1417,204643.
The cube root of 32008469 is 126,169692.
#### Logarithms
The natural logarithm of No. ln 2008469 = loge 2008469 = 14,512883.
The logarithm to base 10 of No. log10 2008469 = 6,302865.
The Napierian logarithm of No. log1/e 2008469 = -14,512883.
### Trigonometric functions
The cosine of 2008469 is 0,121572.
The sine of 2008469 is -0,992583.
The tangent of 2008469 is -8,164536.
## Number 2008469 in Computer Science
Code typeCode value
2008469 Number of bytes1.9MB
Unix timeUnix time 2008469 is equal to Saturday Jan. 24, 1970, 5:54:29 a.m. GMT
IPv4, IPv6Number 2008469 internet address in dotted format v4 0.30.165.149, v6 ::1e:a595
2008469 Decimal = 111101010010110010101 Binary
2008469 Decimal = 10210001002202 Ternary
2008469 Decimal = 7522625 Octal
2008469 Decimal = 1EA595 Hexadecimal (0x1ea595 hex)
2008469 BASE64MjAwODQ2OQ==
2008469 MD534a34ef8b132254dae01ff1d4fe2a73c
2008469 SHA199831b48a525a199cccf1cc29c9d5f0d08fd6927
2008469 SHA224e8cf0d0e950a8be79bc60b23a2a389e53069302d28eff5e3041388f8
2008469 SHA256acbb30c3291a1de96b38a678223898ddc566d4d21a8621bedb986832b3b90bb3
2008469 SHA384df6b948c41fa66cdd5c5407ba0c1749eafa45204389aa166ef8e0f88e3b2fac1e437000207ebf5c3cac92b8f807de865
More SHA codes related to the number 2008469 ...
If you know something interesting about the 2008469 number that you did not find on this page, do not hesitate to write us here.
## Numerology 2008469
### Character frequency in the number 2008469
Character (importance) frequency for numerology.
Character: Frequency: 2 1 0 2 8 1 4 1 6 1 9 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 2008469, the numbers 2+0+0+8+4+6+9 = 2+9 = 1+1 = 2 are added and the meaning of the number 2 is sought.
## № 2,008,469 in other languages
How to say or write the number two million, eight thousand, four hundred and sixty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 2.008.469) dos millones ocho mil cuatrocientos sesenta y nueve German: 🔊 (Nummer 2.008.469) zwei Millionen achttausendvierhundertneunundsechzig French: 🔊 (nombre 2 008 469) deux millions huit mille quatre cent soixante-neuf Portuguese: 🔊 (número 2 008 469) dois milhões e oito mil, quatrocentos e sessenta e nove Hindi: 🔊 (संख्या 2 008 469) बीस लाख, आठ हज़ार, चार सौ, उनहत्तर Chinese: 🔊 (数 2 008 469) 二百万八千四百六十九 Arabian: 🔊 (عدد 2,008,469) مليونان و ثمانية آلاف و أربعمائة و تسعة و ستون Czech: 🔊 (číslo 2 008 469) dva miliony osm tisíc čtyřista šedesát devět Korean: 🔊 (번호 2,008,469) 이백만 팔천사백육십구 Danish: 🔊 (nummer 2 008 469) to millioner ottetusinde og firehundrede og niogtreds Hebrew: (מספר 2,008,469) שני מיליון ושמונת אלפים ארבע מאות שישים ותשע Dutch: 🔊 (nummer 2 008 469) twee miljoen achtduizendvierhonderdnegenenzestig Japanese: 🔊 (数 2,008,469) 二百万八千四百六十九 Indonesian: 🔊 (jumlah 2.008.469) dua juta delapan ribu empat ratus enam puluh sembilan Italian: 🔊 (numero 2 008 469) due milioni e ottomilaquattrocentosessantanove Norwegian: 🔊 (nummer 2 008 469) to million åtte tusen fire hundre og sekstini Polish: 🔊 (liczba 2 008 469) dwa miliony osiem tysięcy czterysta sześćdziesiąt dziewięć Russian: 🔊 (номер 2 008 469) два миллиона восемь тысяч четыреста шестьдесят девять Turkish: 🔊 (numara 2,008,469) ikimilyonsekizbindörtyüzaltmışdokuz Thai: 🔊 (จำนวน 2 008 469) สองล้านแปดพันสี่ร้อยหกสิบเก้า Ukrainian: 🔊 (номер 2 008 469) два мільйони вісім тисяч чотириста шістдесят дев'ять Vietnamese: 🔊 (con số 2.008.469) hai triệu tám nghìn bốn trăm sáu mươi chín Other languages ...
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I am interested in pursuing an understanding of K-theory. Primarily, the $K_3(\mathbb{Z})$ algebraic K-group over ring of integers of an algebraic number field and its relationship to the $\mathbb{Z}/48$ ring of integers modulo 48.
This is (of course), again, from Terry Gannon's "Moonshine Beyond the Monster" where he talks about many amazing coincidences with the number 24, the Riemann Zeta Function $\Sigma_{n=1}^\infty (1/n)^{-1} = -1/12$, Apery's constant, where $\Sigma_{n=1}^\infty (1/n)^2 = \pi/6$ (which he states are both synonomous in their relationship to $K_3(\mathbb{Z})\leftrightarrow \mathbb{Z}/48$....)
A little harder to discern is the (possible) relationship of the Bimonster, $M (M \times M.x ) \rtimes \mathbb{Z}/2 \to M \wr 2$, and the Incidence Graph of the M-13 pseudogroup with 13 points and lines, (the 13 point, 13 line projective plane, where here, the coincidence would appear to be the number 26, which is the dimension of Bosonic String Theory (2 + 24 dimensions, the quantum harmonic oscillator on a 2-dimbrane, which relates to -1/12 above per John Baez "My Favorite Number is 24")). It's tempting to see the resemblance of 24 relating to the Monster, and 48 to the BiMonster, but that seems to obvious. Finally, is there any relevance in bringing in the M12-Mathieu group here, being so close to the M13-pseudogroup? I apologize ahead of time if this last paragraph is "shooting the moon" but hopefully my first two paragraphs are well-stated questions.
2 latex-ed the question up; added 1 characters in body
I am interested in pursuing an understanding of K-theory. Primarily, the K3(Z) $K_3(\mathbb{Z})$ algebraic K-group over ring of integers of an algebraic number field and its relationship to the Z/48 $\mathbb{Z}/48$ ring of integers modulo 48.
This is (of course), again, from Terry Gannon's "Moonshine Beyond the Monster" where he talks about many amazing coincidences with the number 24, the Riemann Zeta Function Sigma_1_oo $\Sigma_{n=1}^\infty (1/n)^-1 1/n)^{-1} = -1/12, 1/12$, Apery's constant, where Sigma_1_oo $\Sigma_{n=1}^\infty (1/n)^2 = pi/6 \pi/6$ (which he states are both synonomous in their relationship to K3(Z)<-> Z/48....)$K_3(\mathbb{Z})\leftrightarrow \mathbb{Z}/48$....)
A little harder to discern is the (possible) relationship of the Bimonster, $M X \times M .x Z/2 -> \mathbb{Z}/2 \to M wreath 2\wr 2$, and the Incidence Graph of the M-13 pseudogroup with 13 points and lines, (the 13 point, 13 line projective plane, where here, the coincidence would appear to be the number 26, which is the dimension of Bosonic String Theory (2 + 24 dimensions, the quantum harmonic oscillator on a 2-dimbrane, which relates to -1/12 above per John Baez "My Favorite Number is 24")). It's tempting to see the resemblance of 24 relating to the Monster, and 48 to the BiMonster, but that seems to obvious. Finally, is there any relevance in bringing in the M12-Mathieu group here, being so close to the M13-pseudogroup? I apologize ahead of time if this last paragraph is "shooting the moon" but hopefully my first two paragraphs are well-stated questions.
1
# Z/48 and Moonshine Beyond the Monster
I am interested in pursuing an understanding of K-theory. Primarily, the K3(Z) algebraic K-group over ring of integers of an algebraic number field and its relationship to the Z/48 ring of integers modulo 48.
This is (of course), again, from Terry Gannon's "Moonshine Beyond the Monster" where he talks about many amazing coincidences with the number 24, the Riemann Zeta Function Sigma_1_oo (1/n)^-1 = -1/12, Apery's constant, where Sigma_1_oo (1/n)^2 = pi/6 (which he states are both synonomous in their relationship to K3(Z)<-> Z/48....)
A little harder to discern is the (possible) relationship of the Bimonster, M X M .x Z/2 -> M wreath 2, and the Incidence Graph of the M-13 pseudogroup with 13 points and lines, (the 13 point, 13 line projective plane, where here, the coincidence would appear to be the number 26, which is the dimension of Bosonic String Theory (2 + 24 dimensions, the quantum harmonic oscillator on a 2-dimbrane, which relates to -1/12 above per John Baez "My Favorite Number is 24")). It's tempting to see the resemblance of 24 relating to the Monster, and 48 to the BiMonster, but that seems to obvious. Finally, is there any relevance in bringing in the M12-Mathieu group here, being so close to the M13-pseudogroup? I apologize ahead of time if this last paragraph is "shooting the moon" but hopefully my first two paragraphs are well-stated questions. | 1,308 | 4,729 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2013-20 | latest | en | 0.892818 |
https://www.stat.math.ethz.ch/pipermail/r-help/2013-May/353261.html | 1,670,132,524,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710962.65/warc/CC-MAIN-20221204040114-20221204070114-00613.warc.gz | 1,061,620,784 | 1,923 | # [R] How can I extract part of the data with a selection criterion?
David Winsemius dwinsemius at comcast.net
Fri May 10 04:31:13 CEST 2013
```On May 9, 2013, at 7:06 PM, jpm miao wrote:
> Hi,
>
> As an example, how can I get the data such that field a of ab, ab["a"],
> equals 3? I expect the answer to be the union of 2 and 4, as
>
> Thanks,
>
>> a<-c(1,3,4,3,5,6,5)
>> b<-c(2,4,6,7,3,1,2)
>> ab<-data.frame(a,b)
>> ab
> a b
> 1 1 2
> 2 3 4
> 3 4 6
> 4 3 7
> 5 5 3
> 6 6 1
> 7 5 2
>
>> ab[a==3]
> Error in `[.data.frame`(ab, a == 3) : undefined columns selected
>> ab[ab["a"]==3]
> [1] 3 3 4 7
>
Try;
ab[ ab[["a"]]==3, ]
And then read ?Extract to help drive home the point the ab["a"] is a list with one element and ab[["a"]] is a vector with 7 elements.
--
David Winsemius
Alameda, CA, USA
``` | 348 | 809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.789851 |
https://www.chegg.com/homework-help/definitions/heat-transfer-5 | 1,563,309,906,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00045.warc.gz | 650,168,891 | 51,813 | # Heat Transfer
Heat transfer, also referred to simply as heat, is the movement of thermal energy from one thing to another thing of different temperature. These objects could be two solids, a solid and a liquid or gas, or even within a liquid or gas. There are three different ways the heat can transfer: conduction (through direct contact), convection (through fluid movement), or radiation (through electromagnetic waves). Heat transfer occurs when the temperatures of objects are not equal to each other and refers to how this difference is changed to an equilibrium state. Thermodynamics then deals with things that are in the equilibrium state.
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A: | 1,375 | 4,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-30 | latest | en | 0.879932 |
http://www.bosspintar.com/blog/best-smelling-jjl/business-calculus-problems-fd98d9 | 1,627,133,483,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00223.warc.gz | 60,773,870 | 8,930 | What is the marginal cost when $$x = 200$$ and $$x = 500$$? Note that it is important to note that $$C'\left( n \right)$$ is the approximate cost of producing the $${\left( {n + 1} \right)^{{\mbox{st}}}}$$ item and NOT the nth item as it may seem to imply! Look for words indicating a largest or smallest value. The developers had that in mind when they created the calculus calculator, and thatâs why they preloaded it with a handful of useful examples for every branch of calculus. First off, Calculus is the Mathematics of Motion and Change. They know that if the complex contains x apartments the maintenance costs for the building, landscaping etc. Glad to see you made it to the business calculus differentiation rules section. CostFunctions If we assume that a cost function, C(x), has a smooth graph as in Fig. 7. We will revisit finding the maximum and/or minimum function value and we will define the marginal cost function, the average cost, the revenue function, the marginal revenue function and the marginal profit function. How to solve problems in business applications such as maximizing a profit function and calculating marginal profit Now, as long as $$x > 0$$ the second derivative is positive and so, in the range of possible solutions the function is always concave up and so producing 50,000 widgets will yield the absolute minimum production cost. Optimization Problems for Calculus 1 with detailed solutions. Let’s take a quick look at another problem along these lines. What is the marginal cost, marginal revenue and marginal profit when $$x = 200$$ and $$x = 400$$? We can see from this that the average cost function has an absolute minimum. Now, as we noted above the absolute minimum will occur when $$\overline C'\left( x \right) = 0$$ and this will in turn occur when. This function is typically called either the demand function or the price function. Finally, to product the 401st widget it will cost approximately $78. The course covers one semester of Business Calculus for college students and assumes students have had College Algebra. We can’t just compute $$C\left( {301} \right)$$ as that is the cost of producing 301 widgets while we are looking for the actual cost of producing the 301st widget. Antiderivatives in Calculus. Business Calculus Online Practice Exams: Test 1, Test 1 (with solutions) from Spring, 2004 UNCC (pdf) Test 2, Test 2 (with solutions) from Spring, 2004 UNCC (pdf) Test 3, Test 3 (with solutions) from Spring, 2004 UNCC (pdf) Final, Final (with solutions) from Spring, 2004 UNCC (pdf) Test 1, Test 1 (with solutions) from Spring, 2003 UNCC (pdf) ⦠I have additional lecture notes you can read down below under Additional Resource. FX Calculus Solver is a comprehensive math software, based on an automatic mathematical problem solving engine, and ideal for students preparing term math exams, ACT, SAT, and GRE: - ⦠Note that with these problems you shouldn’t just assume that renting all the apartments will generate the most profit. If we assume that the maximum profit will occur at a critical point such that $$P'\left( x \right) = 0$$ we can then say the following. With this analysis we can see that, for this complex at least, something probably needs to be done to get the maximum profit more towards full capacity. Finally, the marginal revenue function is $$R'\left( x \right)$$ and the marginal profit function is $$P'\left( x \right)$$ and these represent the revenue and profit respectively if one more unit is sold. Business Calculus by Dale Hoffman, Shana Calloway, and David Lippman is a derivative work based on Dale Hoffmanâs Contemporary Calculus. Meaning of the derivative in context: Applications of derivatives Straight ⦠Phone support is available Monday-Friday, 9:00AM-10:00PM ET. If you really want to get better at calculus, following these problems is a great way to make yourself practice!Past calculus problems of the week. Bad notation maybe, but there it is. Learn business calculus 1 with free interactive flashcards. Note that to really learn these applications and all of their intricacies you’ll need to take a business course or two or three. Be careful to not confuse the demand function, $$p\left( x \right)$$ - lower case $$p$$, and the profit function, $$P\left( x \right)$$ - upper case $$P$$. Business Calculus (Under Construction) Business Calculus Lecture Slides. For the most part these are really applications that we’ve already looked at, but they are now going to be approached with an eye towards the business world. If they sell x widgets during the year then their profit, in dollars, is given by, ... We learn a new technique, called substitution, to help us solve problems involving integration. Course Summary This Business Calculus Syllabus Resource & Lesson Plans course is a fully developed resource to help you organize and teach business calculus. We can also see that this absolute minimum will occur at a critical point when $$\overline C'\left( x \right) = 0$$ since it clearly will have a horizontal tangent there. This course is built in Ximera. This course teaches all the essential business calculus topics in a simple and fun video format. Intro. By ⦠Choose from 500 different sets of business calculus 1 flashcards on Quizlet. What is the marginal cost when $$x = 175$$ and $$x = 300$$? Let’s now move onto the revenue and profit functions. A company can produce a maximum of 1500 widgets in a year. We then will know that this will be a maximum we also were to know that the profit was always concave down or. All that we’re really being asked to do here is to maximize the profit subject to the constraint that $$x$$ must be in the range $$0 \le x \le 250$$. On a winning streak? 1. Basic fact: If it moves or if it changes it requires calculus to study it! Business Calculus Demystified clarifies the concepts and processes of calculus and demonstrates their applications to the workplace. Notice this particular equation involves both the derivative and the original function, and so we can't simply find $$B(t)$$ using basic integration.. Algebraic equations contain constants and variables, and the solutions of ⦠In this section, we will explore the concept of a derivative, the different differentiation rules and sample problems. The marginal functions when 7500 are sold are. We’ve already looked at more than a few of these in previous sections so there really isn’t anything all that new here except for the fact that they are coming out of the business world. A management company is going to build a new apartment complex. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. Nailed all the derivative calculus problems here on calculus 1? In your first calculus course, you can expect to cover these main topics: 1. What do your answers tell you about the production costs? Working with substitution. and the demand function for the widgets is given by, 4. Recall from the Optimization section we discussed how we can use the second derivative to identity the absolute extrema even though all we really get from it is relative extrema. Things out with a couple of optimization problems is an overview of the function... Has a smooth graph as in Fig a year finally, to product the 401st widget it cost... These problems you shouldn ’ t just assume that a cost function, C ( x = 300\.. The point of this values of the ideas in the final section of this,! From any calculus topic under additional Resource calling 1-800-876-1799 x apartments the maintenance costs calculus needed... New apartment complex these numbers tell you about the cost function has an absolute minimum you ’! Concept of a differential equation derivatives to the business world down below under additional.. Or the business calculus problems function given by calculus Demystified clarifies the concepts and properties of antiderivatives in calculus are presented calculus! Units sold 301st widget will cost approximately$ 10 profit will always be at the upper of! Derivatives to the business world that involve calculus equation will have an equals sign selected problems. Business problems of our customer support team by calling 1-800-876-1799 it is possible. Is business calculus problems the basic derivative rules like the product rule, the different differentiation section... Look for words indicating a largest or smallest value at the following example world!, in any manufacturing business it is usually possible to express profit as of! Problems - this page from the Lamar University website includes business problems that require optimization is by... Assistance from your school if you are having problems entering the answers into your online assignment semester business! Of using these the average cost function, C ( x \right \. Revenue function is then business calculus problems much money is made by selling \ ( (! Money is made by selling \ ( x\ ) items and is point of this chapter ’. Up often is the cost, marginal revenue and profit Barnett &,! C\Left ( x ), we could get the first couple of derivatives to the business field in..., in order to produce the 301st widget will cost approximately $78 expect to cover these topics. The demand function for some item then the average cost function cursory of... Homework and textbook problems, anytime, anywhere the company sells exactly what they produce in order to minimize costs! Differential equation variables, find the constraint equation off by looking at the following example in words... In this section we took a brief discussion on maximizing the profit apply calculu⦠âWill guide how! By calling 1-800-876-1799 has a smooth graph as in Fig Printer-Friendly Documents guide you how solve. Website uses cookies to ensure you get the best experience first thing we need the derivative and then ’! Off, calculus is used in a simple and fun video format cursory discussion some. Problem along these lines not be posted price function in your first calculus,... Overview of the word constraint, and one you will very likely upon! Are sold are calculus example problems - this page from the Lamar University website includes business problems require... Is a derivative work based on Dale Hoffmanâs Contemporary calculus by ⦠business Fall... The chain rule we assume that the constraint equation or if it changes it requires calculus to it! ( b ), we could get the first couple of optimization problems a cost,! Be a maximum we also were to know that the constraint equation will have an equals sign basic... Derivative, the first thing we need the derivative and then find the critical...., marginal revenue and profit finding limits algebraically - when direct substitution is not.. Finding limits algebraically - when direct substitution is not possible the demand function the... Than the sciences marginal profit when 2500 business calculus problems are sold and when 7500 widgets sold... One semester of business calculus 1$ 10 costfunctions if we assume that renting all various! Basic fact: if it changes it requires calculus to reach a.. Items and is rules you business calculus problems see come up often is the rate of of... Let ’ s work a quick look at some of the week could from! Topics: 1 s work a quick look at an example of using these it moves or it... Numbers tell you about the production costs calculu⦠âWill guide you how solve! Also were to know that the company sells exactly what they produce additional Resource answers into online... Per day should they produce if you are having problems entering the answers your... Of lnx essential business calculus example problems - this page from the Lamar University includes... And series step-by-step this website uses cookies to ensure you get the average cost function is typically called either demand. Cookies to ensure you get the best experience new technique, called substitution, to help us solve involving... ) \ ) is the cost function a largest or smallest value looking at the limit! Some applications of the lecture notes will not be posted will cost around $38 of. Textbook: applied calculus with Linear Programming a Special Edition by Barnett & Ziegler, Pearson Custom.! Function from example business calculus problems above ( C\left ( x ), has a smooth as. Note that with these problems you shouldn ’ t just assume that maximum profit will be... Not just assume that maximum profit will always be at the upper limit of the cost, revenue profit! Average cost function for some item then the average cost function has an absolute minimum 1500 widgets in field! That a cost function for the derivative of lnx, anytime, anywhere in the field! Chapter 1: limits in your class this page from the Lamar University website business. A cursory discussion of some basic applications of the derivative will not be posted, in to., and one you will see come up often is the sketch of the word,... Of producing the 301st widget will cost around$ 38 values of the word constraint, and David is...: 1 widget it will cost approximately $10 how to solve business problems that optimization... To solve business problems that require optimization = 300\ ) problem along these lines what we ’ ll close section. And remember that the company sells exactly what they produce Edition by Barnett & Ziegler Pearson... Is given by Change of the derivative and then we ’ re looking for here is b ), will! Can produce a maximum we also were to know that if the complex have in order to maximize profit... 1: limits in your class Programming a Special Edition by Barnett & Ziegler Pearson... Producing the 301st widget is$ 295.91 at some applications of the derivative $.. Discussion on maximizing the profit was always concave down or main topics: 1 that calculus! Think about the production costs with Linear Programming a Special Edition by Barnett &,... Build a new technique, called substitution, to product the 401st it... Calculus topics in a simple and fun video format by Barnett & Ziegler, Custom! Our customer support team by calling 1-800-876-1799 word constraint, and David Lippman is a derivative the... Minimize production costs derivative optimization problems in Economics usually possible to express profit as function of the derivative problems. Constraint equation of a differential equation differentiation rules section Linear Programming a Special Edition by Barnett &,... We learn in this section out with a brief look at some applications of derivatives of the range you! 500 different sets of business calculus Fall Term 2013 ( 2141 ) Printer-Friendly Documents Student Manual! When 7500 widgets are sold this chapter let ’ s take a quick example this! Is usually possible to express profit as function of the lecture notes will not be posted problems that optimization... Use the tools of calculus to study it very likely touch upon in your first calculus course you! Us solve problems involving integration it will cost approximately$ 78 could get the best experience easy! Given by both students and assumes students have had college Algebra solve your calculus homework and textbook,! What do your answers tell you about the cost function calculus calculator calculate! Algebraically - when direct substitution is not possible widget is \$ 295.91 sold when. Cursory discussion of some basic applications of derivatives in the final section of this important in... They rent in order to maximize their profit 401st widget it will cost around 38! We assume that the average cost function is it is usually possible to express profit function... Edition ) this bookcomes highly recommended by both students and assumes students have had college Algebra the company sells what! Finally, to help us solve problems involving integration on the concepts and properties of antiderivatives calculus. The widgets is given by most profit limits algebraically - when direct substitution is not.. Custom Publishing build a new technique, called substitution, to product the 401st widget it will cost approximately 78! Has an absolute minimum what we ’ ll close this section, we need to do is get the. Of the ideas in the business world that involve calculus the ideas in the final section of this students. Require optimization have in order to maximize their profit it requires calculus to study it algebraically! Having problems entering the answers into your online assignment Barnett & Ziegler, Pearson Publishing! ( x \right ) \ ) is the Mathematics of Motion and Change you may speak a. Problems involving integration we ’ ll need item then the average cost function from 4., called substitution, to help us solve problems involving integration profit.. | 3,470 | 16,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-31 | latest | en | 0.90831 |
http://classic.chem.msu.su/cgi-bin/ceilidh.exe/gran/gamess/forum/?C38af602eabEu-9089-825+00.htm | 1,685,667,960,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00248.warc.gz | 9,797,656 | 2,673 | ## Firefly and PC GAMESS-related discussion club
Learn how to ask questions correctly
Re: Performing a transition state calculation
Hi!
wxMacMolPlt nicely reads the firefly output vibrational modes of a hessian calculation (which you perform by adding the keywords hssend=.t. in \$STATPT group of the optimization run and reading a bit more in the manual for understanding pits and caveats).
For wxMacMolPlt compatibility, you just have to use the "-legacy" option. Just run firefly as:
firefly801 -i inp.inp -p -f -run -legacy > out.out
cheers
On Tue Nov 18 '14 6:57pm, Kevin wrote
-------------------------------------
>Hello, fellow fireflies,
>I'm still somewhat new to the quantum chemical scene, so forgive my naïveté.
>I'm attempting a Firefly emulation of Jans Jensen's excellent tutorial on conducting a GAMESS transition state calculation of flouride anion reacting with chloromethane (found here: http://molecularmodelingbasics.blogspot.com/2009/08/finding-transition-state-sn2-reaction.htmlhttp://molecularmodelingbasics.blogspot.com/2009/08/finding-transition-state-sn2-reaction.html). The procedure is essentially as follows:
>1) While holding constant the fluorine to carbon intermolecular distance, optimize the geometry.
>2) Determine the imaginary frequency "frame".
>3) Use this imaginary frequency "frame" data as the initial geometry guess of a saddle point calculation.
>The input and execution of the initial optimization, for the purpose of finding an imaginary frequency, are no problem. Where I get hung up is in taking the output to serve as the initial geometry for a saddle point calculation. I understand that I need both the \$hess as well as the frequency (specifically, the imaginary frequency) data. Dr. Jensen, in his tutorial, simply opens up the GAMESS .log output in wxMacMolPlt, chooses the imaginary frequency "frame" of interest, and uses this to directly produce a new input file for the SADPOINT calculation. By contrast, wxMacMolPlt does not, as far as I can tell, read the "frames" of an optimization of a Firefly .out. Perhaps Avogadro can substitute, but Avogadro crashes whenever I attempt to open even a simple Firefly calculation (does anyone else have this problem?).
>While I can visualize the frames nicely in ViewMol3D, it lacks MacMolPlt's function to produce an input file. This, then, is the crux of my problem: What data from the raw output of optimization can/should I take to implement into a saddle point input, as Dr. Jensen does to perform his second, saddle point calculation, following optimization?
>For instance, while there are over 20 "frames" in the optimization sequence, there are only two \$hess groupings. So which would I choose? Additionally, how do I implement the imaginary frequency information into the saddle point calculation input file?
>I hope I've framed my question clearly enough. I've attached the output file I produced. Please let me know if I can clarify anything. | 672 | 2,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.84683 |
http://www.anandtech.com/comments/5730/antec-eleven-hundred-the-p280s-gloves-come-off | 1,503,067,239,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104681.22/warc/CC-MAIN-20170818140908-20170818160908-00111.warc.gz | 458,801,079 | 22,389 | ### Back to Article
• #### Stahn Aileron - Monday, April 09, 2012 - link
Would it be too much to ask that you actually place the ambient tempurature during testing somewhere on the graphs themselves so we have a readily accessible reference point right there on the graphs? It would also work as a reminder to readers what the conditions are. Reply
• #### Dustin Sklavos - Monday, April 09, 2012 - link
The ambient temperature is going to vary from test system to test system, that's why I switched to listing the delta. Reply
• #### Rick83 - Monday, April 09, 2012 - link
You did check, that the delta is a constant over ambient temperature though, did you? It may not be for every case. Also, of course, fan speeds will be impacted by ambient/internal temperature.
If you can, you should probably run two series of tests, one in the morning and one in the evening, and then either average that or use one measurement, but at least comment the other. At least for noise, we need ambient temperature, as otherwise that value is completely without base for comparisons.
• #### kandrtech - Monday, April 09, 2012 - link
Those familiar with thermodynamics, and the equations utilized, would agree that the delta approach is the best. Variances of a few (or 10) degrees on ambient will not appreciably change the delta results. By appreciably, I mean you'll see differences out to one or more decimal places . . . . Reply
• #### niva - Monday, April 09, 2012 - link
Are you talking 10 deg C or F?
Ideally your ambient temperature should be somewhere in the +/- 5 deg of 70 deg F. These are the normal temperatures most households are kept at. There may be a significant difference between the noise produced by components at 65 deg F, and someone's house which may be normally kept at 85 deg during the summer daytime because of lack of AC?
And I'm talking about idle situation here...
The point was valid, just include the temperature in your test data.
• #### bobbozzo - Monday, April 09, 2012 - link
The article describing their new methodology for case testing seems to indicate that ambient temps are maintained between 71-74F.
http://www.anandtech.com/show/5709/introducing-our...
• #### ShieTar - Tuesday, April 10, 2012 - link
But those familiar with modern PC design are aware that fan-control systems generally try to achieve a constant CPU/GPU-temperature. Thus when you raise the ambient temperature to somewhat higher levels, CPU/GPU fans tend to speed up, giving you lower Delta-T values at increased noise.
Thus it is important to still run these tests at comparable ambient temperatures, and if this is indeed checked at each test, it should be no problem to change the title of the temperature graph to read "Delta over ambient at 20+/-2°C". Or whichever is the range that is controlled and accepted by the tester.
• #### O8h7w - Saturday, April 14, 2012 - link
I feel perfectly good about showing the temperature as delta above ambient instead of absolute temperature. But it seems many readers would like to see the ambient temperature at the time of testing reported as well, and I have to agree.
The way of doing this that would make perfect sense in the graphs is to modify the labels to look like this:
Antec 1100
@ 23°C ambient
• #### Lucian2244 - Monday, April 09, 2012 - link
Good review, i was wondering how it would look with a mATX in there.
Is it just me or their cases get uglier and uglier ?
• #### Iketh - Monday, April 09, 2012 - link
It's not you... this case is ugly... AND stupid...
Enough with filterless side vents already! Why even have side vents? That's a damn 80's design, speaking figuratively...
• #### Sabresiberian - Monday, April 09, 2012 - link
Uh, if you don't get why the side vents are there, YOU are the one that's stupid, not Antec. Reply
• #### Iketh - Monday, April 09, 2012 - link
no no no... if you are still using side vents, YOU are stupid Reply
• #### JarredWalton - Monday, April 09, 2012 - link
Yay for intelligent arguments!
The reason some people like side vents is that if you have two GPUs, especially on a motherboard where they're only two slots apart (e.g. a "GPU sandwich"), putting a couple fans right above the GPUs can be very helpful for temperatures. From a noise and dust standpoint, though, it's not a good thing and aesthetically some will dislike panels as well.
• #### Sabresiberian - Monday, April 09, 2012 - link
It''s you, the case isn't ugly at all to me.
Some people think their concept of beauty in a case should rule over every case a company makes, but, hey, there's a reason that Antec makes so many different types of cases (as well as other manufacturers).
;)
• #### dtolios - Monday, April 09, 2012 - link
What's the point of getting a large or mid-expensive range case to combo with mATX again? I don't understand why "enthusiast" oriented cases should be tested using an mATX mobo the first place....ofc it can do mATX and one GPU...big deal...can it do 2x large GPUs and full ATX good enough is always a WAY more valid question - both for an organizational and thermal performance standpoint.
Guess it is just me...
• #### ClockHound - Monday, April 09, 2012 - link
Would it be too much to ask for proof reading before publishing?
"If you'll let me beat this dead horse one last time, I'm keen to point out what the Antec Eleven Hundred is that the Antec P280 isn't: a cheaper P280."
I'm keen to understand what you meant. Does this mean that the P280 isn't a cheap P280 or did you mean that the Eleven Hundred isn't a cheap P280? And how did the Three Hundred get into the review text? It's in the text of the overclock page.
I do agree the delta is the better number to display....but it does beg the question with this new test system, why you can't test in a temperature-controlled environment? Why not test with different ambient temps, like room temperature and a 'hot' room temperature?
Thanks for the review.
• #### JarredWalton - Monday, April 09, 2012 - link
Reviews do get proofed most of the time (by me for Dustin's reviews), but I try to take a hands off approach and I thought initially he was trying to say something else. I've fixed that. Anyway, while you're happily flogging us for minor typos, you might want to research what it means to "beg the question". ;-)
As for the temperature controlled testing environment, it would be awesome to have such equipment, but we don't. Environmentally regulated test environments don't come cheap, and they also pose a different problem: 70F ambient without a lot of airflow from the AC isn't the same as 70F with an AC moving quite a bit of air. The difference may not be that large, but I'd bet it would be measurable.
• #### Sabresiberian - Monday, April 09, 2012 - link
I get trying to keep the quality of Anandtech high, and I think that pointing out errors in communication is appropriate, but you would make a better point by making a post that is free of spelling, grammar and usage errors itself.
;)
• #### kevith - Monday, April 09, 2012 - link
Why don't you ever bother to experiment a bit with different numbers and placement of fans? In this case, it would have been VERY interesting to know, what impact that fan behind the motherboard has on temps. AND noise, since it's tugged away far from the user. And there's a lot of other empty fan placements, that, filled up with fans might change the performance and accoustics. But I guess I'l have to buy the case to find out. (And what's the purpose of reading reviews then...?)
You'l probably say, that writing a review takes a lot of time, even without digging deeper into fans, their numbers and placement. But why use all that time, and then in the end the review is only half? Who wants to do or read something, that's ALMOST great? The vast majority of people, that would consider buying this case - and other hi-end cases - will definitely want to experiment.
Aand then you spend a lot of time comparing the P280 and the 1100. Except for readings...!? Why don't you show the figures of the P280 in the graphs?
And I don't think you understood Stahn Aileron's question: We all know, why you changed to showing Delta over ambient, but please let us know what the ambient is, so we know how hot the thing is.
• #### PhoenixEnigma - Monday, April 09, 2012 - link
If you read the review, it's noted that the ambient temperature was about 23C for these tests. The 550D was apparently tested in a room about a degree cooler.
Of course, it would make more sense for the reader to use their own ambient temperature - that's the advantege to having the delta and not the final number, it's easier to adjust for your conditions.
• #### Dustin Sklavos - Monday, April 09, 2012 - link
"Why didn't you place the fans here? Why didn't you do x/y/z?"
It's a can of worms that oftentimes isn't worth opening. There are so many different fan configurations many of these cases are capable of that invariably SOMEONE is going to ask for/gripe that the configuration they would've used wasn't tested. It's a slippery slope. From there you also have to ask "what kind of fans." Are we going to use SilverStone APs? What about a pair of Scythes? Or just some regular off the shelf 120mm fans? There's too much variance; simply put it's much more practical to test an enclosure in its stock configuration and then speculate on its potential.
The P280's readings can't be shown in the graphs because they pre-date the current testbed. I can't keep stacks of enclosures on hand just so I can go and retest them later, it's not like when I did the review of the P280 I thought to myself "better keep this around in case I decide to change how we test cases."
And finally, telling you what the ambient temperature is will just tell you how hot the components and case are in my apartment at the time I tested it, not how hot they will be or how efficiently the case will actually remove heat. I don't make it a point to specify exactly what the ambient temperature was because it ultimately isn't relevant to the comparative results; the ambient actually varies even between test runs as the room heats or cools depending on air conditioning, weather, how much heat the testbed spews out, etc. Right now my apartment's been pretty consistently between 22.5C and 24C, but when summer comes that's going to go up.
This is something Anand was concerned about when I discussed these revised procedures with him, that the data wouldn't read as well. But I'm sorry, I'd rather be producing accurate, useful data than something that just reads better. If you have to ask why I'm not making the ambient temperatures evident in the charts, you don't understand why we made the switch.
• #### Iketh - Monday, April 09, 2012 - link
I'm sure he was referring to the uniqueness of the mobo fan and not that you're required to test every possible combination... it would seem a no-brainer to me to test what effect this fan has on the system. Reply
• #### Robert in Calgary - Monday, April 09, 2012 - link
Hello Dustin,
If I restrict myself to just one case, I'm hoping you can bring back the Solo II for testing on the new set-up.
Thanks.
• #### haukionkannel - Monday, April 09, 2012 - link
I would very much see p180 compared to p280 in the new test bench. Does easier set up means any functional differences? Reply
• #### Belard - Monday, April 09, 2012 - link
"The front bezel of the Eleven Hundred is almost completely ventilated, and that includes the shields for the 5.25" drive bays. It's actually a bit surprising that Antec didn't include any front-mounted intake fans," - I see this in many reviews, about the lack of front cooling fans.
I don't think these are really needed in many cases and the top fans are simply over-kill... (not so much for gaming PCs of course)
If the PSU and a large fan in the back are sucking in AIR from the front, then they will do just fine - as long as there is not over-kill in vents, such as on the sides - as on cheaper non-gaming cases. Adding front fans adds noise, cable mess and cramps the space.
I have the Antec P150 case (5 years old) which is a bit smaller doorless version of the P180. it has a single large fan in the back, running on LOW (3 speed fan). The air filters still get dirty and my CPU and GPU stay cool enough to run. Yes, I can save 2-3C in temp by going to MED setting for the fan... but seriously, most people cannot hear my Q6600, 2 3.5" drives running. Yeah, I'm looking to stick a i5-35xx in there next month or so.
I have an old ATI 4670 card, but I specifically bought the HIS with its dual-slot cooler which does NOT dump heat into the case. Its huge fan runs at a slow RPM, so it too is almost silent.
My previous GeForce 7600GT was a \$190 fan-less version (its huge and looks great) because I wanted silence... but I had to run the rear fan on MED because of over-heating of the case... there goes the silence.
So, for low-noise, get a fanless card (non gamers) or a dual-slot that doesn't dump heat all over the inside of your case, making your CPU, memory and drives warmer.
Again, because the single rear fan and the PSU, the case is able to draw in enough air to keep the HDs cool and everything else.
• #### Iketh - Monday, April 09, 2012 - link
You must live in a dust-free environment. Negative case pressure is a magnet for dust in my home. Everything from my optical drive to the case lights to the door hinges gets clogged with dust. Having front fans (with rpms higher than exhaust fans, if any) ensures air enters only through the filters.
While we're on the subject, I'm so SICK of seeing case designs with filterless side vents. That is a 10 year old design. Why in the world are cases still being made this way? You would think AT authors would gripe about this the same way they do laptop keyboards, etc...
• #### Arbie - Monday, April 09, 2012 - link
@Iketh - Write to Antec. I did so on that very subject and got a good reception. Reply
• #### Belard - Monday, April 09, 2012 - link
I see what you mean, I get a bit of dust around the drives. But that seems to be most cases... From experience, more fans = more noise, more cables, more power, more vibrations. My son's computer gets a bit more dustier than mine on the inside, his case is more generic and has a front fan. its a lose-lose situation.
I don't use the cheap home air-filters either (\$1 each), but go with the 3M \$18~24 filters.
In one of the offices I do work, it has excellent filtered air. Over the period of a year, almost no dust... not even the huge 6 fan Mozart TX.
I agree with you about the side-vents. They do sell aftermarket fan-size filters. I usually just try to avoid such cases... or simple tape black cardboard or plastic on the inside.
• #### TrackSmart - Monday, April 09, 2012 - link
For mid-range systems, I'm in strong agreement with Belard. If you seal off extraneous side vents, you can get away with a single, top-rear 120mm fan + the PSU, which has it's own fan.
Summary: The top-rear fan exhausts hot air. The front intakes pull in cool, fresh air. And you get nice flow from front to back without side-vent interference.
Yes, you need to dust the inside of your case every 6 months. So what? That's a small price to pay for a quiet, reasonably cool case. Extra front fans made a difference of about 1-2C under load for my case. Not worth the extra noise or cost.
• #### entity279 - Monday, April 09, 2012 - link
Dustin,
The 1st page table only specifies the available fan mounts. I couldn't find any explicit mention regarding which of the mounts come with pre-installed fans when you buy the case.
• #### jgutz20 - Monday, April 09, 2012 - link
Everyone has the best additions/tests for you to run, yet they arent making their own articles, just criticizing others!
Good job on the review, Ohh and you missed a period after that one sentence, please fix it so i can understand what i'm reading
• #### bhima - Monday, April 09, 2012 - link
Would it help to take one of these cases that have been tested in the newer config to be tested in the older test set up as well so we have a baseline difference between the two testing methods? Would that help us get a reasonable idea of how the older cases would perform with the new testing methodologies? Reply
• #### Arbie - Monday, April 09, 2012 - link
A big top exhaust is great, but it should have been done as 2x120mm as in the Coolermaster CM690. There is far too little choice of 200mm fans, and then you have to rule out all the sleeve bearing models because it's a horizontal mount. Yes, a single big fan is 'better' than two smaller ones in theory, but the market is far from supporting that approach. Reply
• #### cyberguyz - Monday, April 09, 2012 - link
I absolutely hate reviews like this. Anand, fire this guy!
Why do jokers like this guy review fill size cases with mATX motherboards? If you review an mATX case, then fine, use an mATX motherboard.
Guys, these are full size cases, designed to hold full sized ATX motherboards. Any jackass can assemble a 'clean' and uncluttered system using any case like this size and an mATX motherboard. Try it using real full ATX motherboards and then tell us how much room you have in there to assemble your rig.
Don't slap in an mATX board, then say "Oh lookie how roomy this case is!!". To do anything else is to do a half-assed case review that is not worth the few minutes of wasted life it takes to read it.
At least he didn't attempt to pass off uber this case is by mounting a mini ITX board in it.
• #### JarredWalton - Monday, April 09, 2012 - link
Reading comprehension fail, I suppose? We linked our testing methodology article, but the simple reason for using mATX is that mATX can support just as much performance as most ATX. Now we can compare ATX and mATX cases against each other, rather than having to use two different motherboards. And amazingly enough, you CAN tell how spacious a case is without installing a large motherboard -- though I don't seem to recall "roominess" being mentioned as a selling point here. Reply
• #### Twoboxer - Monday, April 09, 2012 - link
Reviewers need to pay more attention to comparisons between negative- and positive-pressure cases. The major benefit to a positive-pressure case does NOT show up in even the most thorough (short-term) review.
A negative-pressure case draws air in from every crack and crevice. These openings cannot be filtered and so inevitably your optical drives, card readers, usb ports, fan blade edges, and cooling coils become clogged with dust.
In a positive-pressure case, each of the intakes can easily be filtered leaving the interior dust free.
I'm not aware of any compelling thermodynamic advantage to a negative-pressure case, either theoretical or practical. There are at least some anecdotal reasons to believe-positive pressure cases have a theoretical advantage in sound dampening.
If that's correct, Reviewers should be helping us all get more positive-pressure case designs by factoring this consideration into their reviews of price/performance. OTOH, if negative-pressure cases do have a thermodynamic advantage, it would be interesting to quantify it in some way.
Because AFAIC, even if a positive-pressure case ran a couple of degrees warmer and cost a few dollars more, that case is by far preferable to a negative-pressure case for almost all users.
• #### quanta - Tuesday, April 10, 2012 - link
Am I the only one who noticed many of the so-called high-end case have been steadily losing external drive bays? There used to be 6 to 7 bays even on medium cases, but lately you'll be lucky to find 4 bays. With the use of optical drive, memory card reader and fan controller, the spare front panels are pretty much gone. The side-loading drive bays are arguably less versatile than front-loading variety, because there are no aftermarket hot swap bay fitting 3.5-inch drive bays. However, front hot swap bay requires no manual disconnection of cables or removing front panel once it is installed, saving time for hardware testing. Besides, the 5.25-in external bays are perfect for cooling bay for hard drives, and handy for converting them into intake fan slots. Reply
• #### ShieTar - Tuesday, April 10, 2012 - link
There is always the Lian Li PC-V343B if you just can not get enough external bays. 18x5.25" should seriously be enough for everybody. But testing it with an mATX board an no watercooling by be somewhat insulting to the case.
But in general you are correct. If you go on any price-comparison site, you should still find that about 10% of all cases have at least six 5.25" bays, but on closer inspection you will see that most of those are somewhat older designs.
I assume the main reason for this is the fact that more and more people get sepparate storage systems and use their main systems with one (or two) SSDs as Game/Work Systems only. And Anandtech, as much as I personally enjoy their tests, are really just picking a small sample of all cases available with their 20 or so tests per year, so it is understandable if they concentrate on the one or two cases per manufacturer which can be expected to be of interest for the majority of customers/readers. | 5,043 | 21,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-34 | latest | en | 0.936588 |
https://www.answerswave.com/ExpertAnswers/use-a-truth-table-to-determine-if-the-following-statement-is-a-tautology-contradiction-or-neither-pv-aw198 | 1,603,308,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00173.warc.gz | 565,194,746 | 6,649 | Home / Answered Questions / Other / use-a-truth-table-to-determine-if-the-following-statement-is-a-tautology-contradiction-or-neither-pv-aw198
# (Solved): Use A Truth Table To Determine If The Following Statement Is A Tautology, Contradiction, Or Neither....
Use a truth table to determine if the following statement is a tautology, contradiction, or neither. (pvq) v (p^~9) Show that the following two statements are equivalent or show that they are not equivalent. (p →9)^(q r)^(r + p)= (p + 9)^( p r ) Write a conditional statement in if...then form in English. Then restate it in at least 3 other equivalent forms. Creativity counts
We have an Answer from Expert | 178 | 671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-45 | longest | en | 0.819991 |
https://www.netlib.org/blas/ssyrk.f | 1,701,873,931,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00501.warc.gz | 1,006,382,833 | 3,050 | *> \brief \b SSYRK * * =========== DOCUMENTATION =========== * * Online html documentation available at * http://www.netlib.org/lapack/explore-html/ * * Definition: * =========== * * SUBROUTINE SSYRK(UPLO,TRANS,N,K,ALPHA,A,LDA,BETA,C,LDC) * * .. Scalar Arguments .. * REAL ALPHA,BETA * INTEGER K,LDA,LDC,N * CHARACTER TRANS,UPLO * .. * .. Array Arguments .. * REAL A(LDA,*),C(LDC,*) * .. * * *> \par Purpose: * ============= *> *> \verbatim *> *> SSYRK performs one of the symmetric rank k operations *> *> C := alpha*A*A**T + beta*C, *> *> or *> *> C := alpha*A**T*A + beta*C, *> *> where alpha and beta are scalars, C is an n by n symmetric matrix *> and A is an n by k matrix in the first case and a k by n matrix *> in the second case. *> \endverbatim * * Arguments: * ========== * *> \param[in] UPLO *> \verbatim *> UPLO is CHARACTER*1 *> On entry, UPLO specifies whether the upper or lower *> triangular part of the array C is to be referenced as *> follows: *> *> UPLO = 'U' or 'u' Only the upper triangular part of C *> is to be referenced. *> *> UPLO = 'L' or 'l' Only the lower triangular part of C *> is to be referenced. *> \endverbatim *> *> \param[in] TRANS *> \verbatim *> TRANS is CHARACTER*1 *> On entry, TRANS specifies the operation to be performed as *> follows: *> *> TRANS = 'N' or 'n' C := alpha*A*A**T + beta*C. *> *> TRANS = 'T' or 't' C := alpha*A**T*A + beta*C. *> *> TRANS = 'C' or 'c' C := alpha*A**T*A + beta*C. *> \endverbatim *> *> \param[in] N *> \verbatim *> N is INTEGER *> On entry, N specifies the order of the matrix C. N must be *> at least zero. *> \endverbatim *> *> \param[in] K *> \verbatim *> K is INTEGER *> On entry with TRANS = 'N' or 'n', K specifies the number *> of columns of the matrix A, and on entry with *> TRANS = 'T' or 't' or 'C' or 'c', K specifies the number *> of rows of the matrix A. K must be at least zero. *> \endverbatim *> *> \param[in] ALPHA *> \verbatim *> ALPHA is REAL *> On entry, ALPHA specifies the scalar alpha. *> \endverbatim *> *> \param[in] A *> \verbatim *> A is REAL array, dimension ( LDA, ka ), where ka is *> k when TRANS = 'N' or 'n', and is n otherwise. *> Before entry with TRANS = 'N' or 'n', the leading n by k *> part of the array A must contain the matrix A, otherwise *> the leading k by n part of the array A must contain the *> matrix A. *> \endverbatim *> *> \param[in] LDA *> \verbatim *> LDA is INTEGER *> On entry, LDA specifies the first dimension of A as declared *> in the calling (sub) program. When TRANS = 'N' or 'n' *> then LDA must be at least max( 1, n ), otherwise LDA must *> be at least max( 1, k ). *> \endverbatim *> *> \param[in] BETA *> \verbatim *> BETA is REAL *> On entry, BETA specifies the scalar beta. *> \endverbatim *> *> \param[in,out] C *> \verbatim *> C is REAL array, dimension ( LDC, N ) *> Before entry with UPLO = 'U' or 'u', the leading n by n *> upper triangular part of the array C must contain the upper *> triangular part of the symmetric matrix and the strictly *> lower triangular part of C is not referenced. On exit, the *> upper triangular part of the array C is overwritten by the *> upper triangular part of the updated matrix. *> Before entry with UPLO = 'L' or 'l', the leading n by n *> lower triangular part of the array C must contain the lower *> triangular part of the symmetric matrix and the strictly *> upper triangular part of C is not referenced. On exit, the *> lower triangular part of the array C is overwritten by the *> lower triangular part of the updated matrix. *> \endverbatim *> *> \param[in] LDC *> \verbatim *> LDC is INTEGER *> On entry, LDC specifies the first dimension of C as declared *> in the calling (sub) program. LDC must be at least *> max( 1, n ). *> \endverbatim * * Authors: * ======== * *> \author Univ. of Tennessee *> \author Univ. of California Berkeley *> \author Univ. of Colorado Denver *> \author NAG Ltd. * *> \ingroup single_blas_level3 * *> \par Further Details: * ===================== *> *> \verbatim *> *> Level 3 Blas routine. *> *> -- Written on 8-February-1989. *> Jack Dongarra, Argonne National Laboratory. *> Iain Duff, AERE Harwell. *> Jeremy Du Croz, Numerical Algorithms Group Ltd. *> Sven Hammarling, Numerical Algorithms Group Ltd. *> \endverbatim *> * ===================================================================== SUBROUTINE SSYRK(UPLO,TRANS,N,K,ALPHA,A,LDA,BETA,C,LDC) * * -- Reference BLAS level3 routine -- * -- Reference BLAS is a software package provided by Univ. of Tennessee, -- * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..-- * * .. Scalar Arguments .. REAL ALPHA,BETA INTEGER K,LDA,LDC,N CHARACTER TRANS,UPLO * .. * .. Array Arguments .. REAL A(LDA,*),C(LDC,*) * .. * * ===================================================================== * * .. External Functions .. LOGICAL LSAME EXTERNAL LSAME * .. * .. External Subroutines .. EXTERNAL XERBLA * .. * .. Intrinsic Functions .. INTRINSIC MAX * .. * .. Local Scalars .. REAL TEMP INTEGER I,INFO,J,L,NROWA LOGICAL UPPER * .. * .. Parameters .. REAL ONE,ZERO PARAMETER (ONE=1.0E+0,ZERO=0.0E+0) * .. * * Test the input parameters. * IF (LSAME(TRANS,'N')) THEN NROWA = N ELSE NROWA = K END IF UPPER = LSAME(UPLO,'U') * INFO = 0 IF ((.NOT.UPPER) .AND. (.NOT.LSAME(UPLO,'L'))) THEN INFO = 1 ELSE IF ((.NOT.LSAME(TRANS,'N')) .AND. + (.NOT.LSAME(TRANS,'T')) .AND. + (.NOT.LSAME(TRANS,'C'))) THEN INFO = 2 ELSE IF (N.LT.0) THEN INFO = 3 ELSE IF (K.LT.0) THEN INFO = 4 ELSE IF (LDA.LT.MAX(1,NROWA)) THEN INFO = 7 ELSE IF (LDC.LT.MAX(1,N)) THEN INFO = 10 END IF IF (INFO.NE.0) THEN CALL XERBLA('SSYRK ',INFO) RETURN END IF * * Quick return if possible. * IF ((N.EQ.0) .OR. (((ALPHA.EQ.ZERO).OR. + (K.EQ.0)).AND. (BETA.EQ.ONE))) RETURN * * And when alpha.eq.zero. * IF (ALPHA.EQ.ZERO) THEN IF (UPPER) THEN IF (BETA.EQ.ZERO) THEN DO 20 J = 1,N DO 10 I = 1,J C(I,J) = ZERO 10 CONTINUE 20 CONTINUE ELSE DO 40 J = 1,N DO 30 I = 1,J C(I,J) = BETA*C(I,J) 30 CONTINUE 40 CONTINUE END IF ELSE IF (BETA.EQ.ZERO) THEN DO 60 J = 1,N DO 50 I = J,N C(I,J) = ZERO 50 CONTINUE 60 CONTINUE ELSE DO 80 J = 1,N DO 70 I = J,N C(I,J) = BETA*C(I,J) 70 CONTINUE 80 CONTINUE END IF END IF RETURN END IF * * Start the operations. * IF (LSAME(TRANS,'N')) THEN * * Form C := alpha*A*A**T + beta*C. * IF (UPPER) THEN DO 130 J = 1,N IF (BETA.EQ.ZERO) THEN DO 90 I = 1,J C(I,J) = ZERO 90 CONTINUE ELSE IF (BETA.NE.ONE) THEN DO 100 I = 1,J C(I,J) = BETA*C(I,J) 100 CONTINUE END IF DO 120 L = 1,K IF (A(J,L).NE.ZERO) THEN TEMP = ALPHA*A(J,L) DO 110 I = 1,J C(I,J) = C(I,J) + TEMP*A(I,L) 110 CONTINUE END IF 120 CONTINUE 130 CONTINUE ELSE DO 180 J = 1,N IF (BETA.EQ.ZERO) THEN DO 140 I = J,N C(I,J) = ZERO 140 CONTINUE ELSE IF (BETA.NE.ONE) THEN DO 150 I = J,N C(I,J) = BETA*C(I,J) 150 CONTINUE END IF DO 170 L = 1,K IF (A(J,L).NE.ZERO) THEN TEMP = ALPHA*A(J,L) DO 160 I = J,N C(I,J) = C(I,J) + TEMP*A(I,L) 160 CONTINUE END IF 170 CONTINUE 180 CONTINUE END IF ELSE * * Form C := alpha*A**T*A + beta*C. * IF (UPPER) THEN DO 210 J = 1,N DO 200 I = 1,J TEMP = ZERO DO 190 L = 1,K TEMP = TEMP + A(L,I)*A(L,J) 190 CONTINUE IF (BETA.EQ.ZERO) THEN C(I,J) = ALPHA*TEMP ELSE C(I,J) = ALPHA*TEMP + BETA*C(I,J) END IF 200 CONTINUE 210 CONTINUE ELSE DO 240 J = 1,N DO 230 I = J,N TEMP = ZERO DO 220 L = 1,K TEMP = TEMP + A(L,I)*A(L,J) 220 CONTINUE IF (BETA.EQ.ZERO) THEN C(I,J) = ALPHA*TEMP ELSE C(I,J) = ALPHA*TEMP + BETA*C(I,J) END IF 230 CONTINUE 240 CONTINUE END IF END IF * RETURN * * End of SSYRK * END | 2,318 | 7,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.605469 |
http://www.chiefdelphi.com/forums/archive/index.php/t-34493.html | 1,369,370,430,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704179963/warc/CC-MAIN-20130516113619-00089-ip-10-60-113-184.ec2.internal.warc.gz | 382,163,998 | 1,945 | PDA
View Full Version : Coefficient of friction of tetra
suneel112
02-12-2005, 12:54 PM
Can anyone please tell me the coefficient of friction from the tetra to the carpet (both mu_S and mu_k please)? Your help is greatly appreciated. :yikes: :D
Leav
02-12-2005, 02:40 PM
Can anyone please tell me the coefficient of friction from the tetra to the carpet (both mu_S and mu_k please)? Your help is greatly appreciated. :yikes: :D
you could determin this experimentally:
get a force meter (mechanical bathroom scales will work too.....)
measure force at the moment it starts to move....
Fmax=mu_s*N
so:
mu_s=Fmax/N
and once it's moving:
mu_k=Fmax/N
theorecticly you should get two different values, but i don't think you could notice the difference....
-Leav | 209 | 765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.882137 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-6-rational-exponents-and-radical-functions-6-6-solve-radical-equations-6-6-exercises-problem-solving-page-458/61b | 1,726,647,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00704.warc.gz | 728,721,086 | 14,911 | ## Algebra 2 (1st Edition)
$\approx80.39$ mph
Plugging in $B=12$ into the formula we get: $12=1.69\sqrt{s+4.25}-3.55\\15.55=1.69\sqrt{s+4.25}\\9.2\approx \sqrt{s+4.25}\\84.64\approx s+4.25\\s\approx80.39$ | 96 | 205 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-38 | latest | en | 0.606902 |
https://www.justintools.com/unit-conversion/area.php?k1=carreau&k2=townships&q=319000 | 1,632,078,290,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00309.warc.gz | 883,467,398 | 27,263 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Carreau] to [Townships], (carreau to township)
## AREA
319000 Carreau
= 44.134694239435 Townships
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: area
Conversion: Carreau to Townships
The base unit for area is square meters (Non-SI/Derived Unit)
[Carreau] symbol/abbrevation: (carreau)
[Townships] symbol/abbrevation: (township)
How to convert Carreau to Townships (carreau to township)?
1 carreau = 0.00013835327347785 township.
319000 x 0.00013835327347785 township = 44.134694239435 Townships.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Carreau (carreau) is equal to 12900 square-meters, while 1 Townships (township) = 93239571.972 square-meters.
319000 Carreau to common area units
319000 carreau = 4115100000 square meters (m2, sq m)
319000 carreau = 41151000000000 square centimeters (cm2, sq cm)
319000 carreau = 4115.1 square kilometers (km2, sq km)
319000 carreau = 44294586827.121 square feet (ft2, sq ft)
319000 carreau = 6378417756835.5 square inches (in2, sq in)
319000 carreau = 4921618639.5336 square yards (yd2, sq yd)
319000 carreau = 1588.8489928241 square miles (mi2, sq mi)
319000 carreau = 6.3784177568355E+18 square mils (sq mil)
319000 carreau = 411510 hectares (ha)
319000 carreau = 1016862.4563242 acres (ac)
(Carreau) to (Townships) conversions | 530 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-39 | latest | en | 0.747765 |
https://bilakniha.cvut.cz/next/en/predmet6756906.html | 1,695,825,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00478.warc.gz | 157,017,112 | 4,302 | CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2023/2024
# Statistical Pattern Recognition and Decision Making Methods
Code Completion Credits Range Language
18SMRR ZK 2 2P+0C Czech
Garant předmětu:
Jaromír Kukal
Lecturer:
Jaromír Kukal
Tutor:
Jaromír Kukal
Supervisor:
Department of Software Engineering
Synopsis:
Collection of recognition and classification methods with accent to mathematical and statistical principles of their design
and functionality.
Requirements:
Syllabus of lectures:
1.Introduction - what is pattern recognition and decision making
2.Statistical (feature-based) and structural (syntactic) pattern recognition
3.Introduction to statistical pattern recognition - supervised and non-supervised classifiers
4.Simple metric classifiers - NN classifier, k-NN classifier, linear classifier
5.Bayesian classifier - the basic principle, parametric and non-parametric B.c., B.c. for normally distributed classes, parameter estimation, necessary conditions of linearity, special cases in two dimensions
6.Non-metric classifiers, decision trees
7.Non-supervised classifiers - cluster analysis in the feature space, iterative and hierarchical methods, criteria of cluster separability
8.k-means iterative algorithm and its modifications
9.Agglomerative hierarchical clustering, inter-cluster metrics, stop conditions, estimating the number of clusters
10.Dimensionality reduction of the feature space, feature extraction and selection, class separability criteria, Mahalanobis distance
11.Principal component transform
12.Optimal and sub-optimal feature selection methods, sequential and floating search
13.Decision making as a discrete optimization problem
14.Basic methods for unconstrained and constrained discrete optimization
Syllabus of tutorials:
Study Objective:
Study materials:
Key references:
[1] Urbanowicz, R. J. J., Browne, W. N. Introduction to Learning Classifier Systems. Berlin: Springer, 2017.
[2] Matloff, N. Statistical Regression and Classification: From Linear Models to Machine Learning. Boca Raton: CRC
press, 2017.
Recommended references:
[3] Duda, R. O., Hart, P. E., Stork, D. G. Pattern Classification. 2nd edition. New York: Willey, 2007.
[4] Scholkopf, B., Smola, A. J. Learning with Kernels. Cambridge: MIT Press, 2001.
[5] Aggarwal, Ch. C. Data Mining: The Textbook. Cham (Switzerland): Springer, 2015.
[6] Izenman, A. J. Modern Multivariate Statistical Techniques: Regression, Classification, and Manifold Learning.
Corr. 2nd printing 2013 edition. New York: Springer, 2013.
[7] Proceedings of the International Workshop on Multiple Classifier Systems (MCS).
Note:
Time-table for winter semester 2023/2024:
Time-table is not available yet
Time-table for summer semester 2023/2024:
Time-table is not available yet
The course is a part of the following study plans:
Data valid to 2023-08-30
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet6756906.html | 702 | 2,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | longest | en | 0.694084 |
https://origametria.com/lessoncat/inservice-training-of-k-6-teachers/inservice-training-for-teachers-of-mathematics/ | 1,590,702,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00311.warc.gz | 494,126,822 | 11,055 | ## Inservice Training for teachers of Mathematics
#### Quadrilaterals B: Square & Rectangle – same and different
• Lesson Aims
• The pupils will identify the square and rectangle and know the same and different properties of each shape.
• The pupils will fold a rectangle from a square and investigate the lengths of the sides.
• Name of Model
• An envelope
• The lesson structure
• Prior knowledge
• The concepts of vertex, side and polygon
• Stages in the curriculum
Themes: Inservice Training of K-6 teachers > Inservice Training for teachers of Mathematics
To enter the rate you need to purchase the appropriate lessons package
#### 1 Quadrilaterals activities
• Lesson Aims
• The pupils will identify the following quadrilaterals: Square,rectangle,trapezoid.
• Name of Model
• Space ship.
•
• Lesson Structure
• In this lesson the pupils will investigate through folding different quadrilaterals and their properties. The final shape will be a surprize for the pupils.While folding the pupils will learn new concepts and revise previously learnt concepts.
• Prior Knowledge
• The concepts of vertices and sides of polygons.
Themes: Inservice Training of K-6 teachers > Inservice Training for teachers of Mathematics
To enter the rate you need to purchase the appropriate lessons package | 271 | 1,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | longest | en | 0.807622 |
http://conversion.org/length/nautical-league/micron | 1,721,911,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00683.warc.gz | 6,885,941 | 7,253 | # nautical league to micron conversion
Conversion number between nautical league [NL; nl] and micron [µ] is 5556000000. This means, that nautical league is bigger unit than micron.
### Contents [show][hide]
Switch to reverse conversion:
from micron to nautical league conversion
### Enter the number in nautical league:
Decimal Fraction Exponential Expression
[NL; nl]
eg.: 10.12345 or 1.123e5
Result in micron
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 nautical league = (exactly) (5556) / (0.000001) = 5556000000 micron
• 1 micron = (exactly) (0.000001) / (5556) = 1.7998560115191 × 10-10 nautical league
• ? nautical league × (5556 ("m"/"nautical league")) / (0.000001 ("m"/"micron")) = ? micron
### High precision conversion
If conversion between nautical league to metre and metre to micron is exactly definied, high precision conversion from nautical league to micron is enabled.
Decimal places: (0-800)
nautical league
Result in micron:
?
gads
### nautical league to micron conversion chart
Start value: [nautical league] Step size [nautical league] How many lines? (max 100)
visual:
nautical leaguemicron
00
1055560000000
20111120000000
30166680000000
40222240000000
50277800000000
60333360000000
70388920000000
80444480000000
90500040000000
100555600000000
110611160000000
Copy to Excel
## Multiple conversion
Enter numbers in nautical league and click convert button.
One number per line.
Converted numbers in micron:
Click to select all
## Details about nautical league and micron units:
Convert Nautical league to other unit:
### nautical league
Definition of nautical league unit: ≡ 3 nmi.
Convert Micron to other unit:
### micron
Definition of micron unit: ≡ 1×10−6 m. Old name for micrometre
← Back to Length units
© 2024 conversion.org Terms of use | 527 | 1,863 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.656257 |
https://study-assistantph.com/math/question2386601 | 1,670,494,720,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00597.warc.gz | 579,181,549 | 61,140 | • Accueil
• Math
• Find the interest and compound amount on a loan of...
# Find the interest and compound amount on a loan of php 54,600 for 5 years and 6 months with an interest of 8% compounded quarterly.
• Réponse publiée par: tayis
Interest = .02
compounded amount = 81, 132.73
Step-by-step explanation:
FutureValue = 54,600
Interest = 8 x 10% ÷ 4 (quarterly =4) = .02
Number of years = 5 x 4 (quarterly =4) = 20
Compound Formula = FV(1+i)^n
54,600(1+.02)^20 = 81,132.73 (rounded off)
• Réponse publiée par: cyrilc310 | 178 | 533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-49 | latest | en | 0.777268 |
http://forums.wolfram.com/student-support/topics/23592 | 1,444,392,982,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737929054.69/warc/CC-MAIN-20151001221849-00245-ip-10-137-6-227.ec2.internal.warc.gz | 123,594,884 | 5,781 | Student Support Forum: 'MapAt with All operator' topicStudent Support Forum > General > "MapAt with All operator"
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Author Comment/Response yehuda ben-shimol 09/27/10 04:28am In Response To 'Re: Re: MapAt with All operator'---------I have found some more spare minutes so I'll elaborate... using Span (;; notation) is available only as indexes of lists and matrices. It is a more "friendly" syntax for Range which always work now, to use MapAt for a sub matrix, or modify the matrix in several places you need to define each such position. This eventually mimics (internally) the behavior I posted earlier, it is not better nor worse using Map rather than MapAt (and vice versa). you also need not "compute" the size of a matrix, there is a Dimensions function for that matter so, combining all of this, I have implemented a small function that will fit your needs MapSpanAt[fun_, mat_, rows_, cols_] := Module[{dim = Dimensions[mat], r, c}, r = Switch[rows, _Integer, Range[rows], All, Range[First[dim]], _Span, Range @@ (rows /. All -> First[dim])]; c = Switch[cols, _Integer, Range[cols], All, Range[Last[dim]], _Span, Range @@ (cols /. All -> Last[dim])]; MapAt[fun, mat, Tuples[{r, c}]] ] now, if you define aaa = {{1, 2 + I, 3 + I, 4 + I}, {5, 6 + I, 7 + I, 8 + I}, {9, 10 + I, 11 + I, 12 + I}}; and call for example MapSpanAt[Im,aaa,All,2;;All] it returns the desired result yehuda URL: ,
Subject (listing for 'MapAt with All operator') Author Date Posted MapAt with All operator krindik 09/23/10 8:02pm Re: MapAt with All operator yehuda ben-s... 09/26/10 02:37am Re: Re: MapAt with All operator krindik 09/26/10 7:19pm Re: Re: Re: MapAt with All operator yehuda ben-s... 09/27/10 00:29am Re: Re: Re: MapAt with All operator yehuda ben-s... 09/27/10 04:28am
< Previous Comment Help | Reply To Comment | Reply To Topic | 565 | 1,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2015-40 | latest | en | 0.765255 |
https://discourse.pymc.io/t/hierarchical-logistic-regression-with-binomial-likelihood-the-chain-reached-the-maximum-tree-depth-the-rhat-statistic-is-larger-than-1-05-for-some-parameters-the-estimated-number-of-effective-samples-is-smaller-than-200-for-some-parameters/8437 | 1,656,839,068,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00650.warc.gz | 263,060,011 | 6,656 | # Hierarchical Logistic Regression with Binomial Likelihood - The chain reached the maximum tree depth, The rhat statistic is larger than 1.05 for some parameters, The estimated number of effective samples is smaller than 200 for some parameters
I have a dataset with 32 diff. networks, only 1 Independent RV - EXP_FLAG_C and Target Variable - CONV_FLAG_ORIGINAL that looks like below:
Firs I build individual model for each NETWORK with pm.Binomial Likelihood and saving traces into dict. using the code below:
run = False
if run:
indiv_traces = {}
for network_name in network_names:
print (f"Sampling for NETWORK: {network_name}")
# Select subset of data belonging to network
n_data = data_compressed.loc[data_compressed.NETWORK == network_name]
n_data = n_data.reset_index(drop=True)
n_exp_flag = n_data.EXP_FLAG_C
n_conv_flag = n_data.nCONV_FLAG_ORIGINALs
n_trials = n_data.ntrials
#building a model for selected network
with pm.Model() as idividual_model:
# priors
alpha = pm.Normal('alpha', mu=0, sd=1)
beta = pm.Normal('beta', mu=0, sd=1)
m = alpha + pm.math.dot(n_exp_flag, beta)
# inverse link function with alias for the Theano function
p = pm.Deterministic('p', pm.math.sigmoid(m))
# boundary decision, which is the value used to separate class of the target
bd = pm.Deterministic('bd', -alpha/beta)
conv = pm.Binomial('conv',
p = p,
n = n_trials,
observed = n_conv_flag)
# posterior/create the trace
trace = pm.sample(chains = 4,
target_accept = 0.9,
return_inferencedata=True)
indiv_traces[network_name] = trace
Sampling goes smooth with no Divergency and beta coeff for each NETWORK are following :
Now I’m trying to build Hierarchical model and defining Hyperpriors for group nodes with code below:
with pm.Model() as hr_model_nc:
# Hyperpriors for group nodes
mu_a = pm.Normal('mu_alpha', mu=0, sigma=1)
sigma_a = pm.HalfCauchy('sigma_alpha', beta = 5)
mu_b = pm.Normal('mu_beta', mu=0, sigma=1)
sigma_b = pm.HalfCauchy('sigma_beta', beta = 5)
# priors
# Intercept for each network, distributed around group mean mu_a
alpha = pm.Normal('alpha',
mu=mu_a,
sd=sigma_a,
shape =len(data_compressed.NETWORK_IDX.unique()))
# Slope for each network, distributed around group mean mu_b
beta = pm.Normal('beta',
mu=mu_b,
sd=sigma_b,
shape =len(data_compressed.NETWORK_IDX.unique()))
m = alpha[network_idx] + pm.math.dot(data_compressed.EXP_FLAG_C.values, beta[network_idx])
# inverse link function with alias for the Theano function
p = pm.Deterministic('p', pm.math.sigmoid(m))
# boundary decision, which is the value used to separate class of the target
bd = pm.Deterministic('bd', -alpha[network_idx]/beta[network_idx])
conv = pm.Binomial('conv',
p=p,
n = data_compressed.ntrials.values,
observed=data_compressed.nCONV_FLAG_ORIGINALs.values)
# priors_hr = pm.sample_prior_predictive()
# posterior/create the trace
trace_hr_nc = pm.sample(chains= 2,
# tune=2000,
target_accept = 0.96,
max_treedepth = 15,
return_inferencedata=True)
As the result of sampling - 0 divergences but following messages:
Increasing tree depth only make longer time for sampling. I already increase target_accept.
My Hyperpriors is pretty informative. I don’t understand how should I reparametrize it.
Fores plot from Hierarchical Model with beta coeff for each NETWORK below:
If anybody can help with any thoughts or examples/ blogs/ articles would be much appreciate. Thank you
Without actually digging into your model, I might suggest reducing the target_accept and increasing the number of tuning steps (using the tune argument). Hitting the max tree depth suggests sub-optimal step sizes (set during tuning) which is going to be exacerbated by asking for very high acceptance rates. Your posterior could definitely be “difficult” and cause these problems on its own, but I would try the easy tweaks first.
Thank you @cluhmann for taking look into this.
Sorry for taking time to get back to you. I started run it on GPU system and had to Install Linux with Ubuntu, set conda and pymc3 env on it to preform better speed.
So my latest modification on this Hierarchical model based on your recommendation is reducing target_accept and increase tune to 8 000, also draws=10000:
trace_hr_nc = pm.sample(chains=2,
draws=10000,
tune=8000,
# max_treedepth = 15,
return_inferencedata=True)
As result - tons of divergences and following : | 1,098 | 4,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-27 | latest | en | 0.677152 |
https://math.stackexchange.com/questions/1862135/bernoulli-trials-law-of-large-numbers-vs-gamblers-fallacy-the-n-paradox | 1,713,154,727,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00876.warc.gz | 354,856,548 | 37,164 | Bernoulli Trials: Law of Large Numbers vs Gambler's Fallacy, the N paradox
I have asked this question before but I think it wasn't clear what I implied with my succinct question, so I will be a bit more verbose this time.
Lets set the following example: Bernoulli trials, K=17 p=0.525 N=20,000
The probability of a streak of at least 17 consecutive successes in 20,000 trials is 15.3% The same but with N ten times larger, the probability of a streak of at least 17 consecutive successes in 200,000 trials is 81.01%
So my question is the following: is the probability of getting 17 consecutive successes still 81.01% if I run 10 independent trials of N 20,000?
If the law of large numbers are correct, nothing should change since N is simply incrementing, 20,000 today and 20,000 tomorrow is the same as running 40,000 straight, right? So what happens when I run 20,000 on the tenth day? Does that last 20,000 really have 81% of winning 17 streaks just because it is totaling 200,000? That definitely sounds like the Gambler's fallacy. If we consider that each trial is random and independent, 20,000 should always represent 15.3% regardless of how many times we run it...
It should be indistinct to be tossing the coin 200,000 nonstop and tossing 10 times groups of 20,000. How on Earth would pausing and resuming tosses change anything? Right? On the other hand each group of 20,000 tosses are independent and random so there is no way its probability of getting 17 streaks should increase.
So what is the right answer?
• This is not clear. Suppose My first block of $20000$ ends with $10\;H's$ in a row, and the next starts with $7\;H's$ in a row. Does that count? If it does, then you are back in the $200000$ case. If it doesn't, you aren't.
– lulu
Jul 17, 2016 at 12:41
• To make the contrast more plain: suppose you split your $200000$ trials into groups of ten. Then there is $0$ probability of getting seventeen $H's$ in a single block.
– lulu
Jul 17, 2016 at 12:42
• I guess that makes sense. I guess I wasn't considering the splitting of a streak between the groups. So there are 10 opportunities of breaking the streaks. But why are you saying there is 0 probability of getting seventeen H's in a single block? Isn't it still 15.3% according to the Bernoulli trials? Jul 17, 2016 at 12:47
• If I have blocks of length ten, then it is impossible to get $17\;H's$ in a row within a single block, obviously.
– lulu
Jul 17, 2016 at 12:50
• My extreme example just amplifies your observation. Using groups of length ten, it is impossible to get the desired streak within a single block, yet of course there is a very high probability that we get the streak if we ignore the separation into blocks. All that means is that, as the block size decreases, the probability that a favorable streak spans multiple blocks increases.
– lulu
Jul 17, 2016 at 12:53
not quite the same, you could get 7 at the end of one trial and 10 at the beginning of the other, but that is a minor error. What you are essentially doing is running 1 bernoulli trial with 20k paths and $p = 15.3\%$, and then repeating it 10 times, so not getting any success has chance of $$(1-p)^{10}$$ which is indeed a reasonably small number... | 856 | 3,219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-18 | latest | en | 0.9508 |
http://www.codingforums.com/javascript-programming/121896-run-recursive-list-update.html | 1,409,388,398,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500834663.62/warc/CC-MAIN-20140820021354-00092-ip-10-180-136-8.ec2.internal.warc.gz | 311,688,543 | 17,342 | Hello and welcome to our community! Is this your first visit?
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# Thread: Run a recursive list to update?
1. ## Run a recursive list to update?
Hello there,
I am trying to recursively (as in a loop) update input boxes to make maintaining code for a calculator easier.
There is basically a Rate value, that gets multiplied (and at the end divided by 2) by input boxes that have the name metreA through to metreE. Then I want to put those values in field names totalA through to totalE.
So the forumula would be
theForm.totalA.value = rate * theForm.mertreA.value / 2
If that works, I'd like to run it for metreA,metreB etc, and put it in the right total.
I've started using jQuery by the way - and love it so far! so if some code looks odd, that might be the reason.
The code itself is:
Code:
``` function CalculateSum(form, PriceRate, Avalue, Bvalue, Cvalue, Dvalue, Evalue)
{
var Rate = parseFloat(PriceRate);
var metreA = parseFloat(Avalue);
var metreB = parseFloat(Bvalue);
var metreC = parseFloat(Cvalue);
var metreD = parseFloat(Dvalue);
var metreE = parseFloat(Evalue);
//var result = Rate*metreA;
// Round to two decimal places
//result = Math.round(10000 * result) / 10000;
result = calculatePersonA(Rate,metreA, metreB, metreC, metreD, metreE);
form.total.value = Math.round(result*100)/100;
calculateOthers(form, Rate, metreB, metreC, metreD, metreE);
}
function calculateOthers(form, Rate, B,C,D,E)
{
// LETTER INCREMENT
//create a string containing all letters
var alphaChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
//initialize a test character to 'A'
var myChar = "A";
//this function accepts a character, increments it
//by one, and returns it
function incrementLetter(letterToIncrement){
//find where the letter is at in the alphaChars string
var indexOfLetter = alphaChars.search(letterToIncrement);
//if it's not the last letter, then return the next
//letter in the string
if (indexOfLetter+1 < alphaChars.length) {
return(alphaChars.charAt(indexOfLetter+1));
}
//otherwise return the input letter
else{
return(letterToIncrement);
}
}
// END OF LETTER INCREMENT
// END OF LETTER INCREMENT
var i=0
while (i<=4)
{
//\$("input[@name=eval("total" + myChar)]").val("55");
\$("input[@name='totalA']).val("50"));
//document.write("The number is " + i + " of " + myChar);
//document.write("<br >");
//eval("form.total" + myChar + ".value") = myChar * Rate;
myChar = incrementLetter(myChar); // phwoar cool
i=i+1
}
// When the loop is done, reset your variables!
myChar = "A";
//var myChar = "A";
}```
the actual info its getting from is a simple html page, with code like:
Code:
`<label class="quote-total">Total <input type="text" name="totalA" value="" >`
The line in bold,
\$("input[@name='totalA']).val("50"));
Is pretty much what I need help with - how can I make it change to totalA, totalB, totalC as the loop runs?
• Code:
` \$("input[@name='totalA']).val("50"))`
? Do you really need the dollar function? And where is the dollar function? Anyway, I guess it is not a correct syntax anyway (because of the wrong and incomplete quotes). What that line suppose to do?
• Originally Posted by Kor
Code:
` \$("input[@name='totalA']).val("50"))`
? Do you really need the dollar function? And where is the dollar function? Anyway, I guess it is not a correct syntax anyway (because of the wrong and incomplete quotes). What that line suppose to do?
The dollar function calls jQuery. It finds a input box with the name of "totalA", and changes its value to 50.
I have the code working by using:
Code:
```form.totalB.value = B*Rate/2;
form.totalC.value = C*Rate/2;
form.totalD.value = D*Rate/2;
form.totalE.value = E*Rate/2;```
But its very clumsy.
• But what is, for instance B in B*rate/2 ? A variable? What variable? Can you clarify again what kind of math operation you intend to do?
• Originally Posted by Kor
But what is, for instance B in B*rate/2 ? A variable? What variable? Can you clarify again what kind of math operation you intend to do?
My apologies, I see what you mean...
B is a variable from a input box, I grab that by using jQuery (bless it):
\$("input[@name='extraA']").val() - i put each letter into a variable to save having to type it many times
The math operation is fairly simple, but might need to be done many times.
rate is like the others, a value typed into a input box (numerical).
I'd like to have a few series of radio boxes, to tell which person (A - E ) has a extra applied to them. There is two potential extras, one thats worked out with x * rate / 2 (x being a person named a,b,c,d or e), and another one which can be on top of it..
At the moment again, I've got it looking very horrible, looking at the selected radio value, then does work with a bunch of if statements
• Is what I'm asking right?
I ask because there's been no replies, so I wonder if what im asking is really "recursive"..... It comes down to that I'd like to run a loop, and set values to a different object based on what another variable is. Say:
theBox = 'a';
numberToPutin = 500 | 1,323 | 5,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-35 | latest | en | 0.702953 |
https://stats.stackexchange.com/questions/407095/probability-of-a-event-that-happens-n-times-in-a-range-of-time-given-others | 1,660,010,875,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00106.warc.gz | 481,378,523 | 65,890 | # Probability of a event that happens N times in a range of time given others
I'm running an application for making shipments to any place around the globe. I have a set of rules which are like:
A customer makes a shipment...
• Once a week (ie, 50 shipments/year)
• At least once a month (ie, 15 shipments/year)
• At least once a trimester (ie, 6 shipments/year)
• At least once a semester (ie, 4 shipments/year)
• At least once a year (ie, 2 shipments/year)
• Occasionally (ie, 1 shipment/year)
Those are the different profiles of our customers... But I'd like to infer from those rules,
• what is the chance that a customer makes a shipment once a month (ie, 12 shipments/year)?
• What is the chance that a customer makes more than one shipment per month?
Assume I have the data, that one can find in the appendix, permiting that to calculate both the mean and standard desviation.
My approach
I thought about using the Normal Distribution, but turned out that, when doing P(X=12) was null (and that obviously, does not make any sense at all).
I have been exploring on the Internet, and I have found the Poisson Dist, which is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume (Wikipedia)
$$P(k\ events\ in\ interval\ t)=e^{-\lambda}\frac{\lambda^k}{k!}$$
where:
• $$\lambda$$ is the average number of events per interval (How should I calculate the mean of my data? $$\lambda=\frac{\sum_{1}^{i} X_i*N_i}{TOTAL}$$)?.
• $$k$$ number of events (in my case 12).
Is this last approach the correct one? does it make sense? Should I use different approaches based on the question?
Any help will be much appreciated.
APPENDIX
Mock data of our customers:
[Xi] Profile (shipments/year) [Ni] Customers
1 261
2 473
4 139
6 419
15 79
50 24
0 6
TOTAL: 1401
In order to assume that a Poisson distribution is a reasonable approximation, we need the mean and the variance to be close (preferably equal), together with the other assumptions.
Using a notation close to what you used, the total number of customers $$\mathcal{N}$$ is: $$\mathcal{N} = \sum_{i=1}^{n} N_i$$
The mean $$\overline{X}$$ of your data can be computed by: $$\overline{X} = \frac{\sum_{i=1}^{n} N_i X_i}{\mathcal{N}}$$ where $$n$$ is the number of shipment categories.
And the variance $$s^2$$ of your data can be computed by: $$s^2 = \frac{\sum_{i=1}^{n} N_i X_i^2 - \mathcal{N} \overline{X}^2}{\mathcal{N} - 1}$$
In the data you presented, you have $$\mathcal{N} = 1401$$, $$\overline{X} = 4.755$$ and $$s^2 = 46.826$$, so I would say that the Poisson distribution may not have a good fit to your data.
However, from such a large number of customers, you might be able to make some reasonable estimates:
• The probability of finding a customer that makes a shipment in category $$i$$ is, approximately: $$P_i = \frac{N_{i}}{\mathcal{N}}$$ For the "once a month" category, we have a probability near 5%: $$P_i = \frac{79}{1401} = 0.05639$$
• And the probability of a customer making a shipment at least as often as in category $$i$$ is: $$P_{\geq i} = \frac{\sum_{j \geq i} N_{j}}{\mathcal{N}}$$ For the at least "once a month" shipment, we have a probability near 7%: $$P_i = \frac{79+24}{1401} = 0.07352$$ | 992 | 3,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-33 | longest | en | 0.936015 |
https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-36/issue-3/On-Some-Robust-Estimates-of-Location/10.1214/aoms/1177700058.full | 1,708,706,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00106.warc.gz | 475,792,193 | 28,500 | June, 1965 On Some Robust Estimates of Location
Peter J. Bickel
Ann. Math. Statist. 36(3): 847-858 (June, 1965). DOI: 10.1214/aoms/1177700058
## Abstract
During the past 15 years various approaches have been proposed to deal with the lack of robustness of the sample mean as an estimate of the population mean when the distribution sampled is contaminated by gross errors, i.e., has heavier tails than the normal distribution. First, Tukey and the Statistical Research Group at Princeton in [9] suggested and investigated the properties of "trimmed" and "Winsorized" means. More recently, Hodges and Lehmann [6], proposed estimates related to the well-known robust Wilcoxon and normal scores tests, among others. Finally Huber in [7] considered essentially the class of maximum likelihood estimates and found those members of this class which minimize the maximum variance over various classes of contaminated distributions. For a review of work in these directions in testing as well as estimation the reader is referred to Elashoff [3]. In Theorems 3.1 and 3.2 we state the main results of the asymptotic theory of the Winsorized and trimmed means and outline the proof. An alternative method of trimming and Winsorizing (not equivalent to that of Tukey) which encompasses the efficient estimates proposed by Huber and which generalizes to higher dimensions is discussed in Section 2. The fourth section (Theorem 4.1) gives the minimum efficiency with respect to the families of all symmetric and symmetric unimodal distributions, of the Winsorized and trimmed means with respect to the mean. The lower bounds found for the trimmed means (for small trimming proportions) in the unimodal case compare well with that found by Hodges and Lehmann in [5] for the median of averages of pairs, the Hodges-Lehmann estimate. However, the Winsorized mean (for unimodal distributions) has minimum efficiency $\frac{1}{3}$ with respect to the mean whatever be the trimming proportion used. For all distributions, the minimum efficiency is 0. Also in the fourth section (Theorem 4.2) we compare the trimmed mean to the $H-L$ estimate and find that while the latter can be infinitely more efficient than the former, the $H-L$ estimate, for small trimming proportions, $\alpha = .05$, is at least 90 per cent (approximately) as efficient. This would suggest that unless the computations involved are prohibitive, the $H-L$ estimate is to be preferred in any situation where the degree of contamination and type of distribution is not known with great precision. The same remarks apply to the Winsorized mean with only somewhat less force since the lower bounds involved are .74 for all symmetric distributions and .79 for symmetric unimodal distributions. Finally we compare the principal estimate proposed by Huber in [7] (Proposal 2) to the mean and the Hodges-Lehmann estimate, both for all symmetric densities and for the symmetric unimodal family. Results similar to those already mentioned in connection with the trimmed mean are obtained in Theorems 5.1 and 5.2.
## Citation
Peter J. Bickel. "On Some Robust Estimates of Location." Ann. Math. Statist. 36 (3) 847 - 858, June, 1965. https://doi.org/10.1214/aoms/1177700058
## Information
Published: June, 1965
First available in Project Euclid: 27 April 2007
zbMATH: 0192.25802
MathSciNet: MR177484
Digital Object Identifier: 10.1214/aoms/1177700058 | 795 | 3,399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.931837 |
https://kr.mathworks.com/matlabcentral/cody/problems/60256-final-stone-weight | 1,726,572,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00372.warc.gz | 314,480,730 | 19,955 | # Problem 60256. Final Stone Weight
You are given an array with weights of stones. The objective is to determine the weight of the final stone remaining after all collisions have occurred. If there are no stones left, the result is 0.
Here's how the process works:
1. Remove the two heaviest stones from the array.
2. Collide these two stones together, which results in a new stone. The weight of this new stone is the difference between the weights of the two original stones.
3. Add the new stone back into the array. After this step, the initial array has one less element.
4. Repeat the process until only one or no stone remains.
Example:
Consider an initial array of stones with weights [9, 15, 20, 25, 30]
• First, take the heaviest stones, 30 and 25. The collision results in a new stone of weight 30-25=5. The updated array of stones is now: [9, 15, 20, 5]
• Next, take the heaviest stones, 20 and 15. The collision results in a new stone of weight 20-15=5. The updated array of stones is now: [9, 5, 5]
• Then, take the heaviest stones, 9 and 5. The collision results in a new stone of weight 9-5=4. The updated array of stones is now: [4, 5]
• Finally, take the heaviest stones, 5 and 4. The collision results in a new stone of weight 5-4=1. The updated array of stones is now: [1]
So, the weight of the final stone is 1.
### Solution Stats
48.15% Correct | 51.85% Incorrect
Last Solution submitted on Jul 22, 2024
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 416 | 1,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-38 | latest | en | 0.847477 |
https://discuss.leetcode.com/topic/92626/my-c-solution-based-on-trie-45ms | 1,516,660,396,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00134.warc.gz | 691,739,774 | 8,835 | # My C++ solution based on trie (45ms)
• ``````struct node{
node* next[26];
char c;
bool isleaf;
bool found;
node():c('\0'), isleaf(false), found(false){
for(int i=0; i<26; ++i) next[i]=NULL;
}
node(char ch):c(ch), isleaf(false), found(false){
for(int i=0; i<26; ++i) next[i]=NULL;
}
};
class Trie{
public:
Trie(){
root = new node();
}
void insert(string word){
node* p = root;
for(int i=0; i<word.size(); ++i){
int idx = word[i]-'a';
if(!p->next[idx])
p->next[idx] = new node(word[i]);
p = p->next[idx];
}
p->isleaf = true;
}
node* root;
};
class Solution {
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
trie = new Trie();
for(int i=0; i<words.size(); ++i) trie->insert(words[i]);
vector<string> res;
node* cur = trie->root;
int m=board.size(), n=board[0].size();
for(int i=0; i<m; ++i)
for(int j=0; j<n; ++j){
string tmp = "";
helper(board, res, tmp, i, j, cur);
}
return res;
}
private:
Trie* trie;
void helper(vector<vector<char>>& board, vector<string>& res, string& tmp, int i, int j, node* cur){
int m=board.size(), n=board[0].size();
if(i<0||i==m||j<0||j==n||board[i][j]=='\0') return;
int idx = board[i][j]-'a';
node* next = cur->next[idx];
if(!next) return;
if(next&&next->isleaf&&!next->found){
tmp.push_back(board[i][j]);
res.push_back(tmp);
next->found=true;
tmp.pop_back();
}
char ch = board[i][j];
board[i][j] = '\0';
tmp.push_back(ch);
helper(board, res, tmp, i-1, j, next);
helper(board, res, tmp, i, j-1, next);
helper(board, res, tmp, i+1, j, next);
helper(board, res, tmp, i, j+1, next);
tmp.pop_back();
board[i][j] = ch;
}
};
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 562 | 1,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.368541 |
https://keys.direct/blogs/blog/how-to-random-sort-in-excel | 1,713,436,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00722.warc.gz | 301,139,082 | 61,511 | Blog
# How to Random Sort in Excel?
Are you trying to randomize data in your Excel spreadsheet? Have you ever wished there was an easier way to randomly sort data? If so, you’re in luck! Random sorting in Excel is not only possible, but it’s also relatively simple to do. In this article, we’ll explain the steps to random sorting in Excel so you can quickly and easily get the random order you need. Let’s get started!
## How to Randomly Sort Data in Excel
Sorting data in Excel is a great way to organize information and make it easier to interpret. Randomly sorting data is an excellent way to ensure that data is unbiased and not affected by any kind of bias or preference. Excel provides several ways to randomly sort data so that it can be used more effectively.
One way to randomly sort data in Excel is to use the RAND function. This function generates a random number that can be used to sort the data in a spreadsheet. The RAND function can be used in conjunction with the SORT function to sort the data in a specific order. The RAND function can also be used to generate random numbers for use in a formula or other calculations.
Another way to randomly sort data in Excel is to use the SORT function. The SORT function allows the user to specify certain criteria for sorting the data. This can include criteria such as sorting by the cell’s value, by column, or by row. The SORT function can also be used to sort data in any order, including random order.
### Using the RAND Function
The RAND function is a simple and effective way to randomly sort data in Excel. To use the RAND function, the user must specify a range of values that the random number should be generated from. The RAND function will then generate a random number from the specified range and use it to sort the data. The RAND function can also be used in conjunction with other functions, such as the SORT function, to sort the data in a specific order.
#### Pros of Using the RAND Function
The RAND function is easy to use and understand. It is also a quick and efficient way to randomly sort data in Excel. The RAND function is also versatile and can be used in conjunction with other functions to sort data in any order.
#### Cons of Using the RAND Function
The RAND function can be difficult to use if the user is unfamiliar with the syntax. The RAND function also requires the specification of a range of values, which can be time-consuming.
### Using the SORT Function
The SORT function is an effective way to randomly sort data in Excel. To use the SORT function, the user must specify certain criteria for sorting the data. This can include criteria such as sorting by the cell’s value, by column, or by row. The SORT function can also be used to sort data in any order, including random order.
#### Pros of Using the SORT Function
The SORT function is easy to use and understand. It is also a quick and efficient way to randomly sort data in Excel. The SORT function is also versatile and can be used to sort data in any order, including random order.
#### Cons of Using the SORT Function
The SORT function can be difficult to use if the user is unfamiliar with the syntax. The SORT function also requires the specification of certain criteria for sorting the data, which can be time-consuming.
## Related FAQ
### 1. What is Random Sort in Excel?
Random sort in Excel is a feature that allows you to randomly rearrange the rows or columns of a data set. This can be useful when you want to analyze data from different points of view, or when you want to create a random sample from a larger data set. It is also useful for creating randomized versions of worksheets for use in experiments or simulations. To use the random sort feature in Excel, you must first select the data you want to sort. Then, click on the “Data” tab and select the “Sort” command. In the Sort dialog box, select “Randomize” as the Sort by option. Once you click OK, Excel will randomly rearrange the rows or columns of data in your selection.
### 2. How do I select data for Random Sort in Excel?
To select data for random sort in Excel, first click on the data you want to sort. You can select a single cell, a range of cells, or a complete row or column. You can also select multiple rows or columns by holding down the “Ctrl” key while you click on them. Once you have your data selected, click on the “Data” tab and select the “Sort” command.
### 3. How do I use the Random Sort feature in Excel?
To use the Random Sort feature in Excel, select the data you want to sort and then click on the “Data” tab and select the “Sort” command. In the Sort dialog box, select “Randomize” as the Sort by option. Once you click OK, Excel will randomly rearrange the rows or columns of data in your selection.
### 4. How do I change the order of the data after using Random Sort in Excel?
Once you have used Random Sort in Excel, the order of the data cannot be changed. If you want to change the order of the data, you will need to select the data again and re-sort using a different method.
### 5. How do I ensure that the data is randomized correctly?
To ensure that the data is randomized correctly in Excel, you can use the “Randomize” option in the Sort dialog box. This will ensure that the data is randomly rearranged each time the sort is performed. Additionally, you can also use the “Randomize with Replacement” option, which will ensure that no two rows or columns are rearranged in the same order each time the sort is performed.
### 6. Are there any limitations to Random Sort in Excel?
Random Sort in Excel is limited to rearranging the data in a single worksheet. You cannot use the Random Sort feature to rearrange data from multiple worksheets or from an external data source. Additionally, you cannot use the Random Sort feature to rearrange data that is filtered or hidden.
### Sorting Data Randomly In Excel
Sorting your data in Excel is an easy and efficient way to organize information. With the random sort feature, you can quickly and easily mix up your data in a way that is not possible with manual sorting. By following the instructions provided above, you can easily learn how to random sort in Excel and make your data more manageable. | 1,321 | 6,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.846938 |
https://www.physicsforums.com/threads/frame-dependency-of-accelerations.867339/ | 1,695,356,201,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00016.warc.gz | 1,061,527,640 | 21,663 | # Frame dependency of accelerations
• I
• Masterov
In summary, the relativistic effects are visual only if you are looking at an object from a specific angle.f
#### Masterov
No. Length contraction can be seen as a result of viewing a 4d object at a different angle. It's closely analogous to the fact that you can slice a sausage perpendicular to its length and get a circular face, or at an angle and get an elliptical face. The sausage hasn't changed (and certainly spacetime hasn't), but the part of it you are looking at has.
Does this mean that the relativistic effects are visual?
Does this mean that the relativistic effects are visual?
No. Any experiment, optical or mechanical or whatever, will yield the same (contracted) length.
Any?
Experiment (for example):
We will measure acceleration a(t):
The acceleration is the absolute,
and it can be measured by
the mass on the spring.
The actual position and speed:
Acceleration is not absolute. Everyone agrees whether someone is undergoing proper acceleration or not. They do not agree on the magnitude of the 3-acceleration, which is what you are integrating.
Acceleration is not absolute?
Mass on the spring shows the same acceleration, regardless of the speed of the inertial system.
Or no?
No. ##F=\gamma^3 ma##, assuming velocity and acceleration are parallel. If what you said were true, a 1N force applied to a 1kg mass going 1m/s below c would exceed light speed in one second (thanks to @Dale for that example).
The acceleration experienced by an astronaut depends on the velocity of the observer.
So?
It's hard to imagine it.
google translate not cope with the transformation.
======================================
Mass on the spring shows the same acceleration, regardless of the speed of the inertial system.
Or no?
A mass on a spring doesn't measure acceleration; it measures force in the frame in which the mass is (momentarily) at rest. You can convert this to proper acceleration if you know the mass. It doesn't measure force in any other frame.
You are mixing up frames.
In the astronaut's rest frame then his acceleration is his proper acceleration and is an invariant.
However, in any other frame, the measured acceleration depends on the astronaut's velocity relative to that frame.
The second one is the relevant point of view if we are discussing length contraction. In the first case we would measure proper length and proper acceleration.
The acceleration experienced by an astronaut depends on the velocity of the observer.
So?
It's hard to imagine it.
Probably hard to image because you are misinterpreting what was said, and setting up a false straw man argument. No one said the astronaut him/herself did not EXPERIENCE a specific acceleration, what was said that other frames of reference do not agree that it is the same. The astronaut doesn't care what they see, just as neither I nor my clock care that we are seen as MASSIVELY time dilated by a particle in CERN.
EDIT: I see Ibix corrected you before I finished typing.
It does not. However, that is not relevant to the thread(Edit: not relevant to the topic of length contraction, which was the topic of the thread this was orignially posted in), nor to the point you were trying to make with your mass on a spring.
Last edited:
Weight of astronaut (in outer space), whith he presses on a seat, do not depend on the velocity of the observer.
With this you agree?
The acceleration experienced by an astronaut depends on the velocity of the observer. Or no?
If by "acceleration" you mean "proper acceleration" (which is what the astronaut feels pushing on him, but has very little to do with his or anyone else's measurement of how his speed is changing) then the answer is "no"
If by "acceleration" you mean coordinate acceleration (how his speed is changing with time) then the answer is "yes".
Does this mean that the relativistic effects are visual?
Any?
Experiment (for example):
We will measure acceleration a(t):
The acceleration is the absolute,
and it can be measured by
the mass on the spring.
The actual position and speed:
======================================
The acceleration experienced by an astronaut depends on the velocity of the observer.
Or no?
Good by...
Obviously ##\vec{a}(t)## is only a function of time.
Actually, on second thought, this isn't so obvious! But we write a(t), and not a(t,v), so in this sense, a(t) is only a function of time.
While it's clear that ##\vec{a}(t)## is in some only a function of time, It's not clear what you mean by "the acceleration experienced by an astronaut". One presumes that we are invited to imagine that you are asking about ##\vec{a}(t)##, but the presence of the unexplained spring and mass in your diagram and the wording of your question suggests you might be interested in something other than ##\vec{a}##.
Several people have talked about things you might have mean by "the acceleration experienced by the astronaut" at length, but you've refused to clarify exactly what you meant, and instead of honoring the repeated requests to clarify your quesiton, you refuse to clarify and demanded a yes or no response to an unclear question. This is rather unfortunate.
Additionally, if we have a frame S, and a frame S', and frame S' is moving relative to frame S with a velocity v, we can talk about a(t) in frame S, and a'(t') in frame S', and ask the question - does the transform between a(t) and a'(t') depend on the relative velocity v between S and S'? The answer in special relativity is yes, unlike the answer in Newtonian mechanics.
Last edited:
Well, let's calculate! To discuss Lorentz-transformation properties of physical observables it's good to use manifestly covariant descriptions. This leads quite naturally to the use of the proper time as time variable. Then momentum is
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
Since by definition (##c=1##, natural units)
$$\mathrm{d} \tau^2 = \mathrm{d} x^{\mu} \mathrm{d} x_{\nu}$$
you have the mass-shell condition
$$p_{\mu} p^{\mu}=m^2=\text{const},$$
and ##m## is the invariant mass (a Lorentz scalar).
The covariant equations of motion reads
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=\dot{p}^{\mu} = K^{\mu},$$
where ##K^{\mu}## is the Lorentz-force four-vector. Because of the mass-shell condition you have
$$p_{\mu} \dot{p}^{\mu}=0 \; \Rightarrow \; K^{\mu} p_{\mu}=0.$$
This constraint makes only three of the four components of the equations of motion independent, as it should be, i.e., you can solve the three spatial equations, and then the temporal one (which is nothing but the relativistic work-energy theorem) is fulfilled automatically.
A typical example for a Lorentz four-force is the force on a charged particle in an electromagnetic field
$$K_{\mu}=\frac{q}{m} F_{\mu \nu} p^{\nu},$$
where ##F_{\mu \nu}## is the antisymmetric field-strength tensor (or Faraday tensor) with the components of the electric and magnetic field. Then of course the constraint is fulfilled.
The covariant acceleration
$$a^{\mu}=\frac{1}{m} \dot{p}^{\mu}=\ddot{x}^{\mu}$$
is obviously a four-vector and transforms as such under Lorentz transformations. Since ##a_{\mu} p^{\mu}=0## and ##p^{\mu}## is always a time-like vector ##a^{\mu}## is spacelike. Nevertheless, contrary to the situation in non-relativistic mechanics, the acceleration also changes its components under rotation-free Lorentz boosts.
Ibix | 1,753 | 7,442 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.923531 |
https://www.airmilescalculator.com/distance/hib-to-hcr/ | 1,627,699,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154042.23/warc/CC-MAIN-20210731011529-20210731041529-00067.warc.gz | 650,344,110 | 80,810 | # Distance between Hibbing, MN (HIB) and Holy Cross, AK (HCR)
Flight distance from Hibbing to Holy Cross (Hibbing Range Regional Airport – Holy Cross Airport) is 2718 miles / 4375 kilometers / 2362 nautical miles. Estimated flight time is 5 hours 38 minutes.
Driving distance from Hibbing (HIB) to Holy Cross (HCR) is 3584 miles / 5768 kilometers and travel time by car is about 114 hours 43 minutes.
## Map of flight path and driving directions from Hibbing to Holy Cross.
Shortest flight path between Hibbing Range Regional Airport (HIB) and Holy Cross Airport (HCR).
## How far is Holy Cross from Hibbing?
There are several ways to calculate distances between Hibbing and Holy Cross. Here are two common methods:
Vincenty's formula (applied above)
• 2718.280 miles
• 4374.647 kilometers
• 2362.120 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
• 2710.121 miles
• 4361.517 kilometers
• 2355.031 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 Hibbing Range Regional Airport
City: Hibbing, MN
Country: United States
IATA Code: HIB
ICAO Code: KHIB
Coordinates: 47°23′11″N, 92°50′20″W
B Holy Cross Airport
City: Holy Cross, AK
Country: United States
IATA Code: HCR
ICAO Code: PAHC
Coordinates: 62°11′17″N, 159°46′29″W
## Time difference and current local times
The time difference between Hibbing and Holy Cross is 3 hours. Holy Cross is 3 hours behind Hibbing.
CDT
AKDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 301 kg (663 pounds).
## Frequent Flyer Miles Calculator
Hibbing (HIB) → Holy Cross (HCR).
Distance:
2718
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2718
Round trip? | 519 | 1,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.855429 |
https://socratic.org/questions/what-is-the-standard-form-of-y-4x-4-x-2-5x-5 | 1,653,608,857,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00085.warc.gz | 597,994,591 | 6,061 | # What is the standard form of y= (4x-4)(x^2+5x-5)?
Dec 27, 2015
$y = 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$
#### Explanation:
For convenience, separate out the scalar factor $4$ temporarily while multiplying out, group the terms in descending degree and combine. For illustration I have shown more steps than normal:
$\left(4 x - 4\right) \left({x}^{2} + 5 x - 5\right)$
$= 4 \left(x - 1\right) \left({x}^{2} + 5 x - 5\right)$
$= 4 \left(x \left({x}^{2} + 5 x - 5\right) - 1 \left({x}^{2} + 5 x - 5\right)\right)$
$= 4 \left(\left({x}^{3} + 5 {x}^{2} - 5 x\right) - \left({x}^{2} + 5 x - 5\right)\right)$
$= 4 \left({x}^{3} + 5 {x}^{2} - 5 x - {x}^{2} - 5 x + 5\right)$
$= 4 \left({x}^{3} + \left(5 {x}^{2} - {x}^{2}\right) + \left(- 5 x - 5 x\right) + 5\right)$
$= 4 \left({x}^{3} + \left(5 - 1\right) {x}^{2} + \left(- 5 - 5\right) x + 5\right)$
$= 4 \left({x}^{3} + 4 {x}^{2} - 10 x + 5\right)$
$= 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$
Alternatively, just look at the combinations of terms to give each power of $x$ in descending order like this:
$\left(4 x - 4\right) \left({x}^{2} + 5 x - 5\right)$
$= 4 {x}^{3} + \left(20 - 4\right) {x}^{2} - \left(20 + 20\right) x + 20$
$= 4 {x}^{3} + 16 {x}^{2} - 40 x + 20$ | 589 | 1,229 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-21 | latest | en | 0.34853 |
http://eisforexplore.blogspot.com/2013/02/plastic-eggs_15.html | 1,656,516,786,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00513.warc.gz | 15,063,165 | 18,019 | ## Friday, February 15, 2013
### Plastic Eggs!
Fraction Eggs
This activity can also be changed to a fraction activity. Eggs are not labeled, but filled with pom pom ball colors. Each student gets one egg. Students write down the fraction for each color pom pom ball in their egg.
Place Value Eggs
Students count out pom pom balls in eggs labeled thousands, hundreds, tens, and ones. They build a number based on how many pom poms are in each "place."
Money Eggs
Put various coins inside numbered plastic eggs. Then, have kids add up the money in each egg and write their answers on a piece of paper. They write the amount next to the corresponding number.
Inference Eggs
Have each student bring a plastic egg to school with a top secret surprise inside! Along with their egg, have them write three clues about what's inside, without giving it away. Tape the clues to the plastic egg. Then, toss the eggs in a bag and have each child pull an egg out. Make sure they don't end up with their own egg! After each child has an egg with the clues attached, he / she will make an inference about what's inside, using the clues that came with the egg. Will their inferencing skills solve the mystery?!
Equation Eggs
Students open an egg and write down the number inside of the egg, using a crayon that matches the egg color. Then, they solve their equation!
#### 2 comments:
1. Great ideas. I used to use these in Kindergarten for word families. Ex, bottom part had ad. Top part had single and double letters combos like h, p, m, br. They would turn the top to line up the letters and list the words they made. They had so much fun with them. | 369 | 1,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-27 | latest | en | 0.953687 |
https://groups.google.com/g/xAct/c/z9biQq2KPHo | 1,719,007,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00232.warc.gz | 261,024,740 | 117,348 | Index order in xCoba
43 views
Salva Mengual Sendra
Feb 8, 2023, 7:56:06 AM2/8/23
to xAct Tensor Computer Algebra
Hello everyone,
I have what I think should be some very basic questions, but I honestly can't figure them out:
Consider that a metric and a chart are defined in a given manifold, and then you define two 2-tensors A and B. In the attached notebook I use the Schwarzschild metric and two random 2-vectors as example. The questions are:
1.- Why does the contraction A[-a,-i] B[i,b] give something of the form CTensor[...][b,-a] instead of CTensor[...][-a,b]? (like in [15] in the notebook)
2.- Why doesn't HeadOfTensor[ A[-a,-i] B[i,b] ,{-a,b}][b,-a] give the same result as before? (see [16] in the notebook)
3.- Why do I need to write HeadOfTensor[ A[-a,-i] B[i,b] ,{-a,b}][-a,b] to get what I expected to obtain from A[-a,-i] B[i,b]? (see [17])
Something similar happens with more complicated contractions of tensors with more indices and I don't know what is going on.
Cheers,
Salva
Example-Indices.nb
Jose
Feb 8, 2023, 8:14:14 PM2/8/23
to xAct Tensor Computer Algebra
Hi,
On Wednesday, February 8, 2023 at 6:56:06 AM UTC-6 salvam...@hotmail.com wrote:
Hello everyone,
I have what I think should be some very basic questions, but I honestly can't figure them out:
Consider that a metric and a chart are defined in a given manifold, and then you define two 2-tensors A and B. In the attached notebook I use the Schwarzschild metric and two random 2-vectors as example. The questions are:
1.- Why does the contraction A[-a,-i] B[i,b] give something of the form CTensor[...][b,-a] instead of CTensor[...][-a,b]? (like in [15] in the notebook)
The order of indices is irrelevant in the following sense: CTensor[{{1, 2}, {3, 4}}, {B, B}][b, a] is the same as CTensor[{{1, 3}, {2, 4}}, {B, B}][a, b]. There is nothing wrong in choosing one or the other, and xCoba chooses one more or less randomly, not worrying about order. If you then want to choose a canonical order, use the function ToCCanonical, which is a bit like ToCanonical in rearranging things into a canonical form, but for CTensor objects. ToCanonical and ToCCanonical do have a preference of order of indices, but most of the rest of the system does not. Worrying about order of indices takes time.
2.- Why doesn't HeadOfTensor[ A[-a,-i] B[i,b] ,{-a,b}][b,-a] give the same result as before? (see [16] in the notebook)
Because HeadOfTensor[ctensor, {a, b}][b, a] is effectively a transposition of ctensor. You are indicating that the matrix of components should be extracted from ctensor with index configuration {a, b}, but then you are reconstructing a different tensor from that array with indices {b, a}. In other words, for a given matrix M, the tensors CTensor[M, {B, B}][a, b] and CTensor[M, {B, B}][b, a] are transposes of each other. Think of the operation T[a, b] - T[b, a] for the same T. Don't we expect to get twice the antisymmetric part of T ? This interpretation is the same for abstract tensors (defined with DefTensor) and component tensors (i.e. CTensor objects).
3.- Why do I need to write HeadOfTensor[ A[-a,-i] B[i,b] ,{-a,b}][-a,b] to get what I expected to obtain from A[-a,-i] B[i,b]? (see [17])
I think this is the same question as 2., and my answer is the same.
In practical terms, when you perform a component computation, you need to think of the order of indices in which you want to get your final result. It is standard to work with a sorted collection of indices, so that the result CTensor[array, bases][a, b, c, ...] contains the array with no extra transpositions. This is what ToCCanonical will achieve at the end of a xCoba computation, again a bit like ToCanonical does at the end of a xTensor computation. But you could have decided to use any other order, so to get the correct array at the end, you would have to extract it with HeadOfTensor[..., indices], specifying what your non-canonical order of indices is.
Hope this clarifies things.
Cheers,
Jose. | 1,115 | 3,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.920508 |
https://de.mathworks.com/matlabcentral/answers/293634-how-to-extrapolate-a-plot | 1,674,868,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00424.warc.gz | 209,732,233 | 31,879 | How to extrapolate a plot?
89 views (last 30 days)
user1996 on 4 Jul 2016
Edited: David Goodmanson on 4 Nov 2022
How can I extrapolate the line so it begins in 0. A0=16.10 A2=7.50 Numbers: A1=[14.25 13.65 13.10 12.45 11.80 11.00 10.50 10.00 9.45 9.00 8.55 8.00] T=[180 360 420 480 540 780 1020 1200 1320 1380 1500 1740] C=log((A0-A1)./(A-A1))
plot(T,C,'*') After that I used polyfit function z=polyfit(T,C,1) And z1=polyval(z,T), and I plot it- plot(T,C,'*',T,z1). So the linear graph beging at same point as the T,C bu I want it to begin from 0, so how can I extrpolate the linear line?
user1996 on 6 Jul 2016
We did this on class so it was in the task.
John D'Errico on 4 Jul 2016
Edited: John D'Errico on 4 Jul 2016
Extrapolating this curve using anything other than a linear model is a foolish task. Even with a linear model, you are asking for a random result.
Worse, attempting to extrapolate the FIRST two points down to zero is absolute foolishness. Sorry, but it is. Given the apparent noise in your data, any prediction down at zero will be hugely uncertain from that.
plot(x,y)
axis(gca,[0,15,0,2000])
In fact, even that is a highly dangerous thing to do. You result will be virtually anything you want it to be.
Mark Twain says in Life on the Mississippi (1884):
“In the space of one hundred and seventy six years the Lower Mississippi has shortened itself two hundred and forty-two miles. That is an average of a trifle over a mile and a third per year. Therefore, any calm person, who is not blind or idiotic, can see that in the Old Oölitic Silurian Period, just a million years ago next November, the Lower Mississippi was upwards of one million three hundred thousand miles long, and stuck out over the Gulf of Mexico like a fishing-pole. And by the same token any person can see that seven hundred and forty-two years from now the Lower Mississippi will be only a mile and three-quarters long, and Cairo [Illinois] and New Orleans will have joined their streets together and be plodding comfortably along under a single mayor and a mutual board of aldermen. There is something fascinating about science. One gets such wholesale returns of conjecture out of such a trifling investment of fact.”
The above is my favorite quote about mathematics. It points out the dangers of long range extrapolation. Here, your data lives on the interval [8,14], so an interval of length 6, and you want to extrapolate all the way down to zero. This verges on something nearly as silly as what Mark Twain said.
So I'll use my own polyfitn tool to look at a linear model.
P = polyfitn(x,y,1)
P =
ModelTerms: [2x1 double]
Coefficients: [-245.52 3605.6]
ParameterVar: [168.62 20992]
ParameterStd: [12.985 144.89]
DoF: 10
p: [3.7142e-09 2.51e-10]
R2: 0.97279
RMSE: 81.601
VarNames: {'X1'}
Polyfitn is on the file exchange. Since you want to extrapolate down to x==0, the constant term is the estimate of the model at x==0, the y-intercept.
We can use the parameter standard deviation, plus or minus twice the standard deviation to see some confidence limits around that point.
P.Coefficients(2) + P.ParameterStd(2)*2*[-1 1]
ans =
3315.8 3895.4
So plot it all together.
xev = 0:14;
yev = polyvaln(P,xev);
plot(x,y,'go',xev,yev,'-b',[0 0],P.Coefficients(2) + P.ParameterStd(2)*2*[-1 1]','rs')
The red squares are estimates of the uncertainty (roughly a 95% interval) around the point at zero, based on a fit from the entire curve. I would hardly trust that extrapolation down that far even with this simple linear model.
Stephen23 on 6 Jul 2016
+1 the Mark Twain quote is brilliant.
David Goodmanson on 4 Nov 2022
Edited: David Goodmanson on 4 Nov 2022
What is being fitted is T vs. C, with the idea of extrapolating to T=0.
The C variable,
C=log((A0-A1)./(A-A1))
contains A, which has not been defined. Also, A2 has not been used. So if what was intended is (A2-A1), (and reversing it to keep a positive argument to the log), the result is shown on the plot. For an extrapolation, this seems reasonable. The added domain is about 12% of the original domain.
A0 = 16.10;
A2 = 7.50;
A1 = [14.25 13.65 13.10 12.45 11.80 11.00 10.50 10.00 9.45 9.00 8.55 8.00];
T = [180 360 420 480 540 780 1020 1200 1320 1380 1500 1740]
C = log((A0-A1)./(A1-A2))
p = polyfit(T,C,1)
Cfit = polyval(p,T);
Cfit0 = polyval(p,0)
figure(1)
plot(T,C,T,Cfit,0,Cfit0,'ok')
grid on
Cfit0 = -1.6067 % quoting -1.6 seems appropriate under the circumstances
z
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# Scored in the 11th percentile
Author Message
Intern
Joined: 10 Oct 2017
Posts: 1
Scored in the 11th percentile [#permalink]
### Show Tags
20 Nov 2017, 07:32
Help! I scored 11 percentile overall, super low quant and IR, does anyone have a recommendation on study aid for these two areas? Verbal is not something of great concern.
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Re: Scored in the 11th percentile [#permalink]
### Show Tags
20 Nov 2017, 07:47
1
nrbaeza wrote:
Help! I scored 11 percentile overall, super low quant and IR, does anyone have a recommendation on study aid for these two areas? Verbal is not something of great concern.
Posted from my mobile device
Hi nrbaeza,
You should solidify your concepts first. It’s a good thing you have taken a GMAT test. You can now know your weaknesses and work on them. If you are willing to study dedicatedly for around 3 months, you are sure to achieve your goal. I believe you may benefit from taking a GMATPREP course. If you are willing, there are some great GMAT prep companies that can help you with your preparation.
In order to make an informed decision I would highly encourage you to go to their websites and try on their free trial and decide for yourself which one do you like better. You try out free access to EmpowerGMAT, Magoosh and Optimus Prep as they have great reviews on GMATCLUB.
Also for verbal, I would highly encourage you to consider e-gmat verbal online or the e-gmat verbal live course. They are both amazing courses especially designed for non-natives. They offer almost 25% of their courses for free so you can try out their free trial to decide which one you want to go for. Plus the e-gmat Scholaranium which is included in both the courses is one of the best verbal practice tools in the market.
You can also try out the MGMAT guides they are phenomenal and cover the entire syllabus really well. I must add that if you are particularly looking to discover and improve on your weak areas in quant; a subscription to GMATCLUB tests is the best way to do that. They are indeed phenomenal and will not only pinpoint your weak areas but also help you improve on them. Further taking multiple mocks might help. Apart from the GMATPREP, Manhattan GMAT tests and Veritas Prep Tests in my experience have a good verbal and Quant section and will certainly help you point out and improve your weak areas.
Further another advantage of taking many mocks is to build up your stamina. Apart from the GMATPREP tests, taking practise tests of any major GMATPREP company ought to do that.
https://gmatclub.com/forum/best-gmat-ve ... 68383.html
Lastly, you can check out a very interesting article by Mike McGarry from Magoosh detailing a 3 month study plan
https://magoosh.com/gmat/2012/3-month-g ... -students/. You will find it very helpful as it gives out a study plan as per your needs.
Hope this helps. All the best.
_________________
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Re: Scored in the 11th percentile [#permalink]
### Show Tags
20 Nov 2017, 16:39
Hi nrbaeza,
It's not clear whether you're referring to an Official GMAT Score or a practice CAT Score. Before I can offer you the specific advice that you’re looking for, it would help if you could provide a bit more information on how you've been studying and your goals:
Studies:
1) How long have you studied?
2) What study materials have you used so far?
3) How have you scored on each of your CATs (including the Quant and Verbal Scaled Scores for each)?
4) Have you taken the Official GMAT (and if so, then how did you score - including the Quant and Verbal Scaled Scores)?
Goals:
5) What is your overall goal score?
6) When are you planning to take the GMAT?
7) When are you planning to apply to Business School?
8) What Schools are you planning to apply to?
GMAT assassins aren't born, they're made,
Rich
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Re: Scored in the 11th percentile &nbs [#permalink] 20 Nov 2017, 16:39
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### Baba-k
Hi,
I'm slowly reading through the book What is Mathematics which asks the following question at the end of its quadratic residues section. I'm not sure how to begin it really, so any hints/suggestions would be greatly appreciated.
1. The problem statement, all variables and given/known data
We have seen that $$x^{2} \equiv (p - x)^{2} \pmod p$$. Where p is a prime > 2 and x is not divisible by p
Show that these are the only congruences among the numbers $$1^{2}, 2^{2}, 3^{2},...,(p-1)^{2}$$
2. Relevant equations
$$(p-x)^{2} = p^{2} - 2px + x^{2}\equiv x^{2} \pmod p$$
3. The attempt at a solution
No idea..
babak
2. Aug 9, 2010
### Petek
I think we have to define what the author means by "the only congruences among the numbers $1^{2}, 2^{2}, 3^{2},...,(p-1)^{2}$." I'd say that he means something like:
If a,b $\in$ {1, 2, ..., p-1}, a $\neq$ b and $a^2 \equiv b^2 (mod p)$, then a = p-b.
Can you prove that?
Petek
3. Aug 10, 2010
### Baba-k
Hi Petek,
Thanks the response. I'm not sure how atm but will give it ago hehe
cheers
babak
4. Aug 22, 2010
### Baba-k
Hi Petek,
Heres what I've come up with, what do you think ?
$$a^{2} \equiv b^{2} \pmod p$$
$$a^{2} - b^{2}\equiv 0 \pmod p$$
$$(a-b)(a+b)\equiv 0 \pmod p$$
$$(a-b)\equiv 0 \pmod p \texttt{ or } (a+b)\equiv 0 \pmod p$$
$$a+b=p \texttt{ or } a-b=p$$
$$a=p-b \texttt{ as } a\in \left \{ 1, 2,...,p-1 \right \} \texttt{ hence } a<p$$
thanks alot,
babak
5. Aug 22, 2010
### Petek
Looks good!
6. Aug 22, 2010
### Dick
That IS a good start. But it's a little sloppy. i) Where did you use that p is prime? ii) If a and b are in {1...p-1} it's pretty unlikely that a-b=p, don't you think? Can you elaborate a little?
Last edited: Aug 22, 2010
7. Aug 23, 2010
### Baba-k
Hi Dick,
i) The product $$(a-b)(a+b)\equiv 0 \pmod p$$ is divisible by the prime p only if one of the factors is, so either $$a-b\equiv 0$$ or $$a+b\equiv 0 \pmod p$$ must be divisible by p.
ii) I chose $$a+b=p$$ as $$a \in \left \{1,..,p-1 \right \}$$ so will always be < p but with $$a-b=p, a=p+b$$ which is >p, which isn't possible.
What do you think ?
cheers
babak
8. Aug 23, 2010
### epenguin
is it not fairly obvious to you that
p2 - 2px = 0 (mod p) (identity)
since p is a factor of that bit?
9. Aug 23, 2010
### Dick
i) is right on. ii) is still a little funny. If a number is divisible by p then it must have the form k*p where k is a integer, right? If a and b are in {1...p-1} and a+b is divisible by p then a+b=p, since it can't be 0 and it can't be as large as 2*p. And if a-b is divisible by p shouldn't a-b=0?
10. Aug 30, 2010
### Baba-k
Hi,
Okay thanks, think I see what you're saying. Is it enough to say..
since $$(a+b)(a-b)\equiv 0 \pmod p$$ then
$$a+b=np$$ or $$a-b=kp$$ where $$a, b \in \left \{ 1, ..., p-1 \right \}$$
if $$a-b=kp$$ then $$a-b$$ will always be < p and hence k is 0. While $$0 < a+b < 2p$$. Hence n is 1 as we've already said that a-b=0.
Pretty much just re-written what you said before I guess hehe. Am not too sure about the logic for getting to n=1 though..
cheers!
babak
11. Aug 31, 2010
### Dick
a+b is a multiple of p. Since it's also between 0 and 2p that means it's equal to p, right? I'm not sure what you are not sure about.
12. Aug 31, 2010
### Baba-k
Riteo dont worry, was just confusing my self hehe. | 1,207 | 3,367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-13 | longest | en | 0.851139 |
https://math.stackexchange.com/questions/3960924/integral-of-legendre-polynomials-with-tangent | 1,713,598,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00899.warc.gz | 345,434,625 | 34,073 | # Integral of Legendre polynomials with tangent
I have encountered the following relationship$$^{[1][2]}$$, stated without proof both times
$$\int_0^\gamma dt \tan(t/2)\cdot [P_n(\cos(t))+P_{n-1}(\cos(t))]=\frac{1}{n}[P_{n-1}(\cos(\gamma))-P_{n}(\cos(\gamma))]$$
Where $$P_n(x)$$ is the $$n^{th}$$ Legendre polynomial, and $$\gamma<\pi$$.
I've tried integration by parts, substitution of $$u=\cos(t)$$, and looked in Gradshteyn. Mathematica doesn't integrate it (at least, not in the current form). I've verified that the relationship is correct for modest values of $$n$$. I suspect that there is some property of the Legendre polynomials that facilitates this, but I don't see what it is.
[1] Mixed boundary value problems, Dean G. Duffy, eq. 3.1.80, p95
Since $$\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$$, by letting $$\cos(\gamma)=T\in[-1,1]$$ your identity can be written in the simplified form
$$\int_{T}^{1}\frac{dx}{1+x}(P_n(x)+P_{n-1}(x))\,dx = \frac{1}{n}(P_{n-1}(T)-P_n(T)).$$
The identity clearly holds at $$T=1$$. By differentiating both sides wrt $$T$$ we find Bonnet's identity and we are done. | 364 | 1,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.856527 |
https://lists.squeakfoundation.org/pipermail/vm-dev/2017-August/025639.html | 1,716,802,480,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00366.warc.gz | 313,916,089 | 3,417 | # [Vm-dev] vm crash when using rairedTo: with fractions
Nicolas Cellier nicolas.cellier.aka.nice at gmail.com
Thu Aug 10 20:42:43 UTC 2017
```2017-08-10 22:19 GMT+02:00 Andrei Chis <chisvasileandrei at gmail.com>:
>
> Hi Nicolas,
>
> Thanks for the info. Indeed sending #asFloat to the operand leads to a
> correct behaviour.
>
> Just then is there a need for this special use case for handling Fractions
> if it can lead to such problematic behaviour? Would the logarithmic way
> ((aNumber * self ln) exp) not be enough? On this example the computation
> takes a few minutes before crashing the image.
>
>
Having (1/1000 raisedTo: 2/3) = (1/100) is a nice thing and we want to
keep it.
But once we detect that we can't have an exact result, we should adopt a
more efficient strategy.
There are corner cases where we want to avoid overflow/underflow in
intermediate values, such as (1<<2000 + 1) raisedTo: 1/200 or ((1<<2000 +
1) reciprocal raisedTo: 1/200, so we must not convert asFloat too soon.
But since ln already handles those edge cases (at least in Squeak, I have
to check Pharo), ((aNumber * self ln) exp) is a good approximation (a few
ulp off, depending on the quality of underlying libm).
Nicolas
> Cheers,
> Andrei
>
> On Thu, Aug 10, 2017 at 9:27 PM, Nicolas Cellier <
> nicolas.cellier.aka.nice at gmail.com> wrote:
>
>>
>> Hi Andrei,
>> indeed, the method does not scale well...
>>
>> If the result is an exact Fraction, then it will answer the Fraction.
>> Else if not exact, it will be converted to a Float.
>>
>> The problem is that it will try to answer nearest Float with a rather
>> naive algorithm.
>> Moreover, computing the nthRoot: first, then raising the result to the
>> power of numerator will cumulate rounding errors.
>> So trying to get a very accurate nthRoot: first in case of Float result
>> is not a good strategy anyway.
>>
>> Why the VM crashes exactly is another problem and remains to see, we'd
>> prefer an Exception.
>>
>> As a workaround, I suggest sending asFloat to the receiver and/or operand.
>>
>>
>> 2017-08-10 12:02 GMT+02:00 Andrei Chis <chisvasileandrei at gmail.com>:
>>
>>>
>>> Hi,
>>>
>>> I was executing this code '(2009/2000) ** (3958333/100000)' with the
>>> Pharo6.1 distribution and the vm crashed with she stack attached below.
>>> Tried it on both mac and windows 10.
>>> Seems that #raisedTo: has a special case for fractions that ends up
>>> calling #nthRoot: like '2009 nthRoot: 100000' leading to the crash.
>>>
>>> Cheers,
>>> Andrei
>>>
>>>
>>> 0xaddeac M LargePositiveInteger(Integer)>quo: 0x314093e8: a(n)
>>> LargePositiveInteger
>>> 0xaddec8 M LargePositiveInteger(LargeInteger)>quo: 0x314093e8: a(n)
>>> LargePositiveInteger
>>> 0xaddee8 M LargePositiveInteger(Integer)>// 0x314093e8: a(n)
>>> LargePositiveInteger
>>> 0xaddf04 M LargePositiveInteger(LargeInteger)>// 0x314093e8: a(n)
>>> LargePositiveInteger
>>> a(n) LargePositiveInteger
>>> a(n) LargePositiveInteger
>>> 0xaddfb4 I Fraction>nthRoot: 0x4f9a940: a(n) Fraction
>>> 0xaddfd8 I Fraction(Number)>raisedTo: 0x4f9a940: a(n) Fraction
>>> 0xaddffc I Fraction(Number)>** 0x4f9a940: a(n) Fraction
>>> 0xade018 M UndefinedObject>DoIt 0x5fe5d00: a(n) UndefinedObject
>>> 0xade048 I OpalCompiler>evaluate 0x4f9a998: a(n) OpalCompiler
>>> 0xade074 I RubSmalltalkEditor>evaluate:andDo: 0x305e5878: a(n)
>>> RubSmalltalkEditor
>>> 0xade09c I RubSmalltalkEditor>highlightEvaluateAndDo: 0x305e5878: a(n)
>>> RubSmalltalkEditor
>>> 0x3062fdc8: a(n) GLMMorphicPharoScri
>>> enderer
>>> 0xade0d8 I MorphicAlarm(MessageSend)>value 0x4f9ab20: a(n) MorphicAlarm
>>> 0xade0f4 M MorphicAlarm>value: 0x4f9ab20: a(n) MorphicAlarm
>>> 0xade114 M WorldState>triggerAlarmsBefore: 0x71bb5e0: a(n) WorldState
>>> 0xade140 M WorldState>runLocalStepMethodsIn: 0x71bb5e0: a(n) WorldState
>>> 0xade164 M WorldState>runStepMethodsIn: 0x71bb5e0: a(n) WorldState
>>> 0xade180 M WorldMorph>runStepMethods 0x6ab7778: a(n) WorldMorph
>>> 0xade198 M WorldState>doOneCycleNowFor: 0x71bb5e0: a(n) WorldState
>>> 0xade1b4 M WorldState>doOneCycleFor: 0x71bb5e0: a(n) WorldState
>>> 0xade1d0 M WorldMorph>doOneCycle 0x6ab7778: a(n) WorldMorph
>>> 0xade1e8 M WorldMorph class>doOneCycle 0x6a9f960: a(n) WorldMorph class
>>> 0xade200 M [] in MorphicUIManager>spawnNewProcess 0x2cc88718: a(n)
>>> MorphicUIManager
>>> 0xade220 I [] in BlockClosure>newProcess 0x2f178150: a(n) BlockClosure
>>>
>>>
>>
>>
>
>
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``` | 1,502 | 4,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.860562 |
http://m.dlxedu.com/m/askdetail/3/e4c6e3e537f6afdd76b2773ac74ec139.html | 1,537,323,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155814.1/warc/CC-MAIN-20180919004724-20180919024724-00403.warc.gz | 159,916,835 | 3,932 | I have a Table Owners
Owner product
``Jhon product1Jhon product2Jhon product3Chris product4``
Another Table: Products
Product QuantitySold
``Product1 3Product2 5Product3 2Product4 7``
How do I do a SQL to come up with the following result:
1. Name of Owner
2. Number of products
3. QuantitySold
NameOfOwner NumberOfProducts QuantitySold
Jhon 3 10 (3+5+2)
Chris 1 7
I tried:
``select Owners.owner, count(distinct Owners.product) as NumberOfProducts, sum(Product.QuantitySold) as QuantitySold from Owners O, Products Pgroup by O.owner``
But that returns the total of quantitySold for any owner (17 = 3+5+2+7) multiple by the number of products.
``NameOfOwner NumberOfProducts QuantitySoldJhon 3 51 = 3 * 17 (3+5+2+7)Chris 1 17 = 1 * 17 (3+5+2+7)``
Thank you very much
``````select o.Name, count(o.Name) NumberOfProducts, sum(QuantitySold) QuantitySold from #Owner o inner join #Product p on p.Name=o.Product
Group By o.Name
``````
You have to use a inner join (or other kind of join):
``````select owner, count(a.product) nr_of_products, sum(quantity) Qnt_Sold from Owners inner join Products on Owners.product = Products.product group by owner
``````
Different SQL JOINs:
INNER JOIN: Returns all rows when there is at least one match in BOTH tables
LEFT JOIN: Return all rows from the left table, and the matched rows from the right table
RIGHT JOIN: Return all rows from the right table, and the matched rows from the left table
FULL JOIN: Return all rows when there is a match in ONE of the tables
``````select a.owner as NameOfOwner, count(a.Product) as NumberOfProducts,sum(QuantitySold) as QuantitySold from Owners as a
INNER JOIN Products on a.Product = Products.Product group by a.owner
``````
Pretty much a repeat here of the other answers, but including a run on Sql Fiddle.
``````SELECT o.owner as Owner, count(o.owner) as NumberofProducts, sum(p.quantitySold) as Quantity
FROM Owners o
inner Join Products p
on o.product = p.product
Group by o.owner
``````
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From: "Ondrej " <ondrej.muransky@ansto.gov.au>
Newsgroups: comp.soft-sys.matlab
Subject: create matrix/array
Date: Mon, 23 Mar 2009 22:39:01 +0000 (UTC)
Organization: The MathWorks, Inc.
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Hi Guys,
I would need a help with creating a matrix/array which is going to have 18 columns and 76 rows,… first 4 rows are “0”, next 4 rows are “5”, next 4 rows are “10” and so on the last 4 rows are “90”,…. for this I wrote the following script:
row1 = 1:4:76;
row2 = 0:5:90;
for ii =1:1:size(row1,2);
for jj = 1:1:size(row2,2);
b(row1(ii),:)= (row2(jj))*ones(1,18);
b((row1(ii)+1),:)= (row2(jj))*ones(1,18);
b((row1(ii)+2),:)= (row2(jj))*ones(1,18);
b((row1(ii)+3),:)= (row2(jj))*ones(1,18);
end
end
but it gives “90” in all 76x18 cells for reason I don’t see,… there must be some problem with jj loop?
Many thanks for the help
Ondrej
``` | 526 | 1,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-52 | latest | en | 0.606745 |
http://scorescience.humboldt.k12.ca.us/fast/teachers/submit/Health-math/mtech7.html | 1,553,343,789,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202804.80/warc/CC-MAIN-20190323121241-20190323143241-00378.warc.gz | 182,256,321 | 1,504 | Aptos High School Global Health Odyssey
Health Professionals Use Math for Problem Solving
Medical Lab Technician 7
Our bodies are about 0.85% salt, or about 0.85 grams salt in 100 ml. Because this is the normal level in our bodies, we call this concentration "normal saline solution".
When people sweat, they lose salt as well as water. So when we feed people intravenous solution for dehydration, we usually give them normal saline solution at this percent, not just water.
How many grams of salt would you add to a liter of water to make up normal saline solution?
Hint: How many cc or ml in a liter?
Hint: in water, grams, ml, and cc are all equal
back to index | 155 | 671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-13 | latest | en | 0.936275 |
https://www.scribd.com/document/132157702/automatic-control | 1,571,073,009,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00074.warc.gz | 1,079,267,444 | 69,359 | You are on page 1of 4
# then also the feedback system x (t ) = Ax(t ) B [Cx(t ), t ] (4)
ABSOLUTE STABILITY
Analysis of dynamical systems and circuits is mostly done under the assumption that the system is linear and time-invariant. Powerful mathematical techniques are then available for analysis of stability and performance of the system, for example, superposition and frequency domain analysis. In fact, even if the system is nonlinear and time-varying, such assumptions can often be used to get a rst estimate of the system properties. The purpose of absolute stability theory is to carry the analysis one step further and get a bound on the possible inuence of the nonlinear or time-varying components. The approach was suggested by the Russian mathematician Lure in the 1940s and has, since then, developed into a cornerstone of nonlinear systems theory. The basic setup is illustrated in Fig. 1, where the linear time-invariant part is represented by the transfer function G(s) and the nonlinear parts of the system are represented by a feedback loop w (v, t). The analysis of the system is based on conic bounds on the nonlinearity (Fig. 2): (v, t )/v for all v = 0 (1)
is exponentially stable. The name of this result comes from its graphical interpretation. The Nyquist plot, that is, the plot of G(i) in the complex plane as R, must not cross or circumscribe the circle centered on the real axis and passing through 1/ and 1/ (Fig. 3). An important aspect of the circle criterion is that it demonstrates how frequency domain properties can be used in a nonlinear setting. It is instructive to compare with the Nyquist criterion, which states that the closed-loop system with linear feedback w(t) kv(t) is stable for all k [, ], provided that G(i) does not intersect the real axis outside the interval [1/ , 1/ ]. The circle criterion replaces the interval condition with a circular disk. As a consequence, the stability assertion is extended from constant feedback to nonlinear and time-varying feedback. The proof of the circle criterion is based on a quadratic Lyapunov function of the form V (x ) = x Px where the matrix P is positive denite. It can be veried that V(x) is decreasing along all possible trajectories of the system, provided that the frequency condition [Eq. (3)] holds. As a consequence, the state must approach zero, regardless of the initial conditions. THE POPOV CRITERION In the case that has no explicit time dependence, the circle criterion can be improved. For simplicity, let 0 and hence 0 (v )/v for all v = 0 (5)
This problem was studied in Russia during the 1950s. In particular, a conjecture by Aiserman was discussed, hoping that the system would be stable for all continuous functions in the cone [Eq. (1)], if and only if it was stable for all linear functions in the cone. This conjecture was nally proved to be false, and it was not until the early 1960s that a major breakthrough was achieved by Popov (1).
THE CIRCLE CRITERION Popov and his colleagues made their problem statements in terms of differential equations. The linear part then has the form
The Popov criterion can then be stated as follows. Theorem 2 (Popov criterion). Suppose that : R R is Lipschitz continuous and satises Eq. (5). Suppose the system x Ax is exponentially stable and let G(i) C(iI A)1B. If there exists R such that R e[(1 + i)G(i)] > then the system x (t ) = Ax(t ) B [Cx(t )] is exponentially stable. Note that the circle criterion is recovered with 0. Also the Popov criterion can be illustrated graphically. Introduce the Popov plot, where ImG(i) is plotted versus ReG(i). Then
1
x = Ax Bw v = Cx
and the corresponding transfer function is G(s ) = C(sI A ) 1 B
(2)
(6)
Theorem 1 (Circle criterion). Suppose that the system x Ax is exponentially stable and that : R2 R is Lipschitz continuous and satises Eq. (1). If G(i) + 1 0 < Re G(i) + 1 R (3)
(7)
J. Webster (ed.), Wiley Encyclopedia of Electrical and Electronics Engineering. Copyright # 1999 John Wiley & Sons, Inc.
ABSOLUTE STABILITY
(, t)
w v
15 10 5
Im
G(s)
0 5 10
1/
1/
Figure 1. Absolute stability deals with linear time-invariant systems in interconnection with nonlinear functions.
G(i)
stability can be concluded from the Popov criterion if and only if there exists a straight line separating the plot from the point 1/ . The slope of the line corresponds to the parameter . See the following example. Example. To apply the Popov criterion to the system x 1 = 5x1 4x2 + (x1 ) x 2 = x1 2x2 + 21(x1 ) let
15 10 0 Re (a) 10
15 10 5
Im
s + 82 G ( s ) = C ( i A ) 1 B = 2 s + 7s + 6 The plot in Fig. 4 then shows that the Popov criterion is satised for 1. The theory of absolute stability had a strong development in the 1960s, and various improvements to the circle and Popov criteria were generated, for example, by Yakubovich. Many types of nonlinearities were considered and stronger criteria were obtained in several special cases (24). Important aspects of the theory were summarized by Willems (5), using the notions of dissipativity and storage function. GAIN AND PASSIVITY In the results described so far, the results were stated in terms of differential equations. A parallel theory was developed by Zames (6) avoiding the state-space structure and studying stability purely in terms of inputoutput relations.
0 5 10 15
1/
1/
G(i)
10
0 Re (b)
10
Figure 3. The circle criterion proves stability as long as the Nyquist plot does not cross or circumscribe the circle corresponding to the conic bounds on the nonlinearity. (a) 0 . (b) 0 .
2 0 2 1/ +
w w= (v)
Im G(i)
4 6 8
v
v
10 12 5 0 5 Re G(i) 10 15
v
Figure 2. The nonlinearity is bounded by linear functions.
Figure 4. The Popov criterion can be applied when there exists a straight line separating the Popov plot from the point 1/ .
ABSOLUTE STABILITY
For this purpose, a dynamical system is viewed as a map F from the input u to the output Fu. The map F is said to be bounded if there exists C 0 such that
T T
The only property of saturation that would be exploited by the Popov criterion is that 0 sat(v)/ v 1 and, consequently,
T
0 |u|2 dt
0
## w(t )[v(t ) w(t )] dt
|Fu|2 dt C2
0 0
for all T 0. The gain of F is denoted F and dened as the minimal such constant C. The map is said to be causal if two inputs that are identical until time T will generate outputs that are also identical until time T. This makes it possible to state the following well-known result: Theorem 3 (Small gain theorem). Suppose that the input output maps F and G are bounded and casual. If F G <1 then the feedback equations v = Gw + f w = Fv + e dene a bounded causal map from the inputs (e, f ) to the outputs (v, w). It is worthwhile to make a comparison with the circle criterion. Consider the case when . Then the condition [Eq. (3)] becomes |G(i)| < 1 [0, ] (8)
for all w sat(v), T 0. However, the inequality will remain valid even if some perturbation of amplitude smaller than one is added to the factor w in the product w(v w). One way to do this is to introduce a function h(t) with the property h(t)dt 1 and replace the previous expression by (w h w) (v w), where h w is a convolution. The integral inequality then becomes
T
0
0
(w + h w )(v w ) dt
(9)
Using this inequality, the Popov criterion [Eq. (6)] can be replaced by the condition Re[(1 + i + H (i))(G(i) + 1 )]] > 0
wR
(10)
This is the same as Eq. (8), since is the gain of and maxG(i) is the gain of the linear part [Eq. (2)]. Another important notion closely related to gain is passivity. The inputoutput map F is said to be passive if
T
where H(i) ei h(t)dt. The factor 1 i H(i) is called multiplier. The theory and applications of absolute stability have recently had a revival since new computer algorithms make it possible to optimize multipliers numerically and to address applications of much higher complexity than previously. The inequality [Eq. (9)] is a special case of what is called an integral quadratic constraint, IQC. Such constraints have been veried for a large number of different model components such as relays, various forms of hysteresis, time delays, time variations, and rate limiters. In principle, all such constraints can be used computationally to improve the accuracy in stability and performance analysis. A unifying theory for this purpose has been developed by Megretski and Rantzer (8), while several other authors have contributed with new IQCs that are ready to be included in a computer library for system analysis. BIBLIOGRAPHY
1. V. M. Popov, Absolute stability of nonlinear systems of automatic control, Autom. Remote Control, 22: 857875, 1962. (Original in Russian, August, 1961.) 2. V. A. Yakubovich, Absolute stability of nonlinear control systems in critical cases, parts 13, Avtomaika i Telemechanika, 24 (3): 293302; 24 (6): 717731, 1963; 25 (25): 601612, 1964. (English translation in Autom. Remote Control.) 3. V. A. Yakubovich, On an abstract theory of absolute stability of nonlinear systems, Vestnik Leningrad Univ. Math., 10: 341361, 1982. (Original in Russian, 1977.) 4. S. Lefschetz, Stability of Nonlinear Control Systems, New York: Academic Press, 1963. 5. J. C. Willems, Dissipative dynamical systems, part 1, General theory; part 2, Linear systems with quadratic supply rates, Arch. Rational Mech. Anal., 45 (5): 321393, 1972. 6. G. Zames, On the inputoutput stability of nonlinear time-varying feedback systemspart 1, Conditions derived using concepts of loop gain; part 2, Conditions involving circles in the frequency plane and sector nonlinearities, IEEE Trans. Autom. Control, 11: 228238, 1966.
0
0
u(t )y(t ) dt
## for all T and all y = Fu
For example, if the input is a voltage and the output is a current, then passivity property means that the system only can consume electrical power, not produce it. Stability criteria can also be stated in terms of passivity. For example, the circle criterion can be interpreted this way, if 0 and is large. MULTIPLIERS AND INTEGRAL QUADRATIC CONSTRAINTS Less conservative stability criteria can often be obtained by exploiting more information about the nonlinearity. One way to do this is to introduce so-called multipliers. Consider, for example, a system with a saturation nonlinearity:
Ax + B x = Ax Bsat(Cx ) = Ax + BCx Ax B
if Cx 1 if |Cx| < 1 if Cx 1
## ABSTRACT DATA TYPES
7. G. Zames and P. L. Falb, Stability conditions for systems with monotone and slope-restricted nonlinearities, SIAM J. Control, 6 (1): 89108, 1968. 8. A. Megretski and A. Rantzer, System analysis via Integral Quadratic Constraints, IEEE Trans. Autom. Control, 47: 819830, 1997.
ANDERS RANTZER
Lund Institute of Technology
ABSORBER. See ELECTROMAGNETIC FERRITE TILE ABSORBER. ABSORPTION MODULATION. See ELECTROABSORPTION. | 2,815 | 10,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-43 | latest | en | 0.923138 |
https://www.ruby-forum.com/t/long-division-180/151163 | 1,660,326,501,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00120.warc.gz | 869,375,141 | 8,784 | # Long Division (#180)
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## Long Division (#180)
Your program should take two arguments: the dividend and the divisor.
Your output should display the long division needed to determine the
quotient and remainder (if it exists). For example, if I run your
program like so:
`````` \$ ruby long_division.rb 11 4096
``````
`````` 372 R4
+----
11|4096
33
--
79
77
--
26
22
--
4
``````
If there is no remainder, do not display anything after the quotient;
that is, do not display R0. As an alternative to the remainder, you
may instead calculate the decimal fraction out to N digits (e.g. use
command-line option --digits=N or similar to switch to decimal
fraction output).
## Long Division (#180)
Here is my solution
divisor,dividend = ARGV
products,bringdown,f,r,quotient = [],[],[],[],[]
(1…dividend.length).each {|x|f << dividend.unpack(“a#{x}”).join}
str = “a#{f.detect{|x| x.to_i >= divisor.to_i}.length}” +
“a1”*dividend.length
u = dividend.unpack(str).select{|x| x != “”}.map{|x| x.to_i}
products << u[0] - u[0] % divisor.to_i
r << u[0] % divisor.to_i
bringdown << ((u[0] % divisor.to_i).to_s + u[1].to_s).to_i
quotient << u[0] / divisor.to_i
(0…u.length-1).each do |x|
products << bringdown[x] - bringdown[x] % divisor.to_i
r << bringdown[x] % divisor.to_i
bringdown << ((bringdown[x] % divisor.to_i).to_s + u[x+2].to_s).to_i
quotient << bringdown[x] / divisor.to_i
end
fmt = dividend.length + divisor.length + 3
print “#{quotient.join.rjust(fmt)}”
print " R#{r[-1]}\n" if r[-1] != 0
print “\n” if r[-1] == 0
print “#{(”-" * (dividend.length + 1)).rjust(fmt)}\n"
print “#{divisor} | #{dividend}\n”
fmt2 = divisor.length + 3 + f.detect{|x| x.to_i >= divisor.to_i}.length
(0…u.length).each do |x|
print “#{products[x].to_s.rjust(fmt2+x)}\n”
print “#{(”-"*divisor.length).rjust(fmt2+x)}\n"
print “#{r[x].to_s.rjust(fmt2+x)}#{u[x+1]}\n”
end
Harry
On Fri, 17 Oct 2008 08:57:01 -0500, Matthew M. wrote:
## Long Division (#180)
22
4
If there is no remainder, do not display anything after the quotient;
that is, do not display R0. As an alternative to the remainder, you may
instead calculate the decimal fraction out to N digits (e.g. use
command-line option --digits=N or similar to switch to decimal fraction
output).
class Integer
def longdiv divisor
begin
# do math
products=[]
dividends=[self] #intermediate dividends – the next number we’ll
divide
exps=[]
dividend=self
quotient=0
while dividend>=divisor
Math.log10(dividend).ceil.downto(0) do |exp|
magnitude=10**exp
trydiv,rest=dividend.divmod(magnitude)
if trydiv>=divisor
exps << exp
dividends[-1]=trydiv
quotient_digit,remainder=trydiv.divmod(divisor)
products << quotient_digitdivisor
quotient+=quotient_digit
magnitude
dividend=(remainder*magnitude+rest)
dividends << remainder
break
end
end
end
``````ensure
#danger of infinite loops, so if I have
#to hit ^C to debug one, I want to be sure to print
#what I've got so I can debug
fmtwidth=self.to_s.size+divisor.to_s.size+1
exps << 0
printf "%#{fmtwidth}d",quotient
print " R#{dividends.last}" if dividends.last > 0
puts ""
puts " "*divisor.to_s.size+"+"+"-"*self.to_s.size
puts divisor.to_s+"|"+self.to_s
i=0
while i<products.size
printf "%#{fmtwidth-exps[i]}d\n",products[i]
printf "%#{fmtwidth-exps[i]}s\n","-"*products[i].to_s.size
i+=1
printf "%#{fmtwidth-exps[i]}d\n",dividends[i]
end
puts
end
[quotient,dividend]
``````
end
end
require ‘test/unit’
require ‘stringio’
class TestLongDiv < Test::Unit::TestCase
def test_division
assert_equal [372,4],4096.longdiv(11)
assert_equal [302,4],(4096-770).longdiv(11)
assert_equal [0,100],100.longdiv(1000)
``````#the following cases showed up as problematic ones when I ran
#the big loop that follows
assert_equal 2205.divmod(10),2205.longdiv(10)
assert_equal 2442.divmod(2),2442.longdiv(2)
#are there problems I didn't think of?
1000.times do
dividend=rand(10000)
divisor=rand(50)+1
printf "%d / %d\n", dividend, divisor
assert_equal dividend.divmod(divisor),dividend.longdiv(divisor)
end
``````
end
## def test_output_format oldstdout=\$stdout begin newstdout=StringIO.new \$stdout=newstdout expected=<<OUTPUT 372 R4 ±— 11|4096 33
``````79
77
--
26
22
--
4
``````
OUTPUT
4096.longdiv(11)
\$stdout=oldstdout
assert_equal expected,newstdout.string
ensure
\$stdout=oldstdout
end
end
end
On Sun, 19 Oct 2008 10:07:34 -0500, Sebastian H. wrote:
## divisor [–base=BASE] [–remainder] BASE is the base as a decimal number Some sample output from irb: 1
7
40
divide(1,6)
36
4
divide(1,6,10,false)
0 R1
±
6|1
0
1
=> nil
I’m posting Sebastian’s solution to the newsgroup, because I don’t know
long pastie will keep the code around. Please don’t use pastebins with
ruby-talk – the code may disappear while the message sticks around,
then
#!/usr/bin/ruby
module LongDivision
module_function
def divide(dividend, divisor, base=10, decimal=true)
result = “”
division = " “*divisor.to_s(base).length
division << “+”.ljust(dividend.to_s(base).length + 1,”-") << “\n”
division << “#{divisor.to_s(base)}|#{dividend.to_s(base)}”
indent = divisor.to_s(base).length + 1
digits = dividend.to_s(base).chars.map {|d| d.to_i(base) }
quotient = 0
remainder = 0
while remainder < divisor && digits.size > 0
remainder = remainder * base + digits.shift
indent += 1
end
``````digits << nil
digits.each do |digit|
quotient = remainder / divisor
result << quotient.to_s(base)
prod = (quotient * divisor).to_s(base)
division << "\n" << prod.rjust(indent) << "\n"
division << ("-" * remainder.to_s(base).length).rjust(indent) <<
``````
“\n”
remainder %= divisor
remstring = “”
if digit
remstring = “0” if remainder == 0
remainder = remainder * base + digit
indent += 1
end
remstring << remainder.to_s(base)
division << remstring.rjust(indent)
end
``````if remainder == 0
puts result.rjust(division.lines.first.length-1), division
elsif decimal
rem_positions = {}
dec_result = ""
periodicity = 0
while remainder > 0
if rem_positions[remainder]
periodicity = rem_positions.size - rem_positions[remainder]
break
end
rem_positions[remainder] = rem_positions.size
division << "0\n"
indent += 1
remainder *= base
quotient = remainder / divisor
dec_result << quotient.to_s(base)
prod = (quotient * divisor).to_s(base)
division << prod.rjust(indent) << "\n"
division << ("-" * remainder.to_s(base).length).rjust(indent) <<
``````
“\n”
remainder %= divisor
division << remainder.to_s(base).rjust(indent)
end
result =
“#{result.rjust(division.lines.first.length-1)}.#{dec_result}”
if periodicity > 0
puts ("_"*periodicity).rjust(result.size)
end
puts result, division
else
puts “#{result.rjust(division.lines.first.length-1)}
R#{remainder.to_s(base)}”, division
end
end
end
if \$0 == FILE
base = ARGV.find {|arg| arg =~ /^–base/}
if base
base = base.split("=",2).last.to_i(10)
else
base = 10
end
dividend = ARGV.shift.to_i(base)
divisor = ARGV.shift.to_i(base)
decimal = !ARGV.include?("–remainder")
LongDivision.divide(dividend, divisor, base, decimal)
end
Matthew M. wrote:
## Long Division (#180)
This isn’t thoroughly tested but it seems to work. It supports
specifying a
base and it can display the result either as integer+remainder or as a
decimal. There’s no option to limit the digits as I didn’t see the
point.
If the number is periodic, it draws a vinculum above the apropriate
digits.
Usage: long_divide dividend divisor [–base=BASE] [–remainder]
BASE is the base as a decimal number (both dividend and divisor are
specified
as BASE)
If --remainder is specified, it will print the integer part of the
division
and the remainder, instead of printing the result as a decimal. If the
division has no remainder, there is no difference.
The code can be used from irb (or another script, if you should want to
do
that for some reason), by requireing it and then using the divide
method.
Here’s the code:
http://pastie.org/295741
Some sample output from irb:
divide(1,3)
_
0.3
±
3|1
0
1
=> nil
divide(1,3,2)
__
0.01
±
11|1
0
## 100 11
`````` 1
``````
=> nil
divide(10,7)
`````` ______
``````
## 30 28
``````20
14
--
60
56
--
40
35
--
50
49
--
10
7
--
3
``````
=> nil
divide(1,6)
_
0.16
±
6|1
0
## 40 36
``````4
``````
divide(1,6,10,false)
0 R1
±
6|1
0
1
=> nil
Apologies for lack of summary… mid-terms and stuff. Will try to have
it and new quiz done before the weekend.
## Long Division (#180)
This is about the same as my first solution.
I just cleaned up the code a little.
divisor,dvd = ARGV[0].to_i,ARGV[1]
f, quotient, r, products = [],[""],[""],[""]
(1…dvd.length).each {|x|f << dvd.unpack(“a#{x}”).join}
g = f.detect{|x| x.to_i >= divisor}.length
str = “a#{g}” + “a1”*(dvd.length-g)
dividend = dvd.unpack(str).unshift("")
(0…dividend.length-1).each do |x|
quotient << (r[x] + dividend[x+1]).to_i / divisor
r << ((r[x] + dividend[x+1]).to_i % divisor).to_s
products << (r[x] + dividend[x+1]).to_i - r[x+1].to_i
end
fmt = dvd.length + 3 + divisor.to_s.length
print “#{quotient.join.rjust(fmt)}”
print " R#{r[-1]}\n" if r[-1].to_i != 0
print “\n” if r[-1].to_i == 0
print “#{(”-" * (dvd.length + 1)).rjust(fmt)}\n"
print “#{divisor.to_s} | #{dvd}\n”
fmt2 = divisor.to_s.length + 3 + g
(1…dividend.length).each do |x|
print “#{products[x].to_s.rjust(fmt2+x-1)}\n”
print “#{(”-"*(divisor.to_s.length+1)).rjust(fmt2+x-1)}\n"
print “#{r[x].rjust(fmt2+x-1)}#{dividend[x+1]}\n”
end
Harry | 3,114 | 9,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.546886 |
https://www.bartleby.com/questions-and-answers/use-the-following-definitions-to-express-each-union-or-intersection-given.-you-can-use-roster-or-set/14501c07-ad4d-4dbc-a74a-f2d2bf3e4162 | 1,571,086,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00137.warc.gz | 822,302,885 | 21,158 | # Use the following definitions to express each union or intersection given. You can use roster or set builder notation in your responses, but no set operations. Ai={i0,i1,i2}(Recall that for any numberx, x0=1.) Bi={x∈R:−i≤x≤1/i} Ci={x∈R:−1/i≤x≤1/i}(a) ⋂i=25Ai(b) ⋃i=25Ai(c) ⋂i=1100Bi(d) ⋃i=1100Bi(e) ⋂i=1100Ci(f) ⋃i=1100Ci
Question
Use the following definitions to express each union or intersection given. You can use roster or set builder notation in your responses, but no set operations.
Ai={i0,i1,i2}
(Recall that for any number
x
,
x0=1
.)
Bi={x∈R:−i≤x≤1/i}
Ci={x∈R:−1/i≤x≤1/i}
(a)
⋂i=25Ai
(b)
⋃i=25Ai
(c)
⋂i=1100Bi
(d)
⋃i=1100Bi
(e)
⋂i=1100Ci
(f)
⋃i=1100Ci
Step 1
Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.
a. Ai={i^0,i^1,i^2}
Ai= {1,i,i^2}
Step 2
### Want to see the full answer?
See Solution
#### Want to see this answer and more?
Our solutions are written by experts, many with advanced degrees, and available 24/7
See Solution
Tagged in | 432 | 1,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.683057 |
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# ROI, NPV and PP
## by Rahmad Kurniawan, Lecturer Of Informatics Engineering at UIN Sultan Syarif Kasim Riau on Mar 24, 2013
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## ROI, NPV and PPPresentation Transcript
• ASSALAMU’ALAIKUM INFORMATION TECHNOLOGY MANAGEMENT ROI, NVP, AND PAYBACK PERIODReturn on Investment, Net Present Value and Payback Period Disajikan Oleh : RAHMAD KURNIAWAN P68500
• Contents1. ROI2. NPV 1. NPV on “Chapter 17, Exercises and Project, No.2”3. Payback Period 2 RAHMAD KURNIAWAN P68500
• What is ROI?ROI can be defined as: One of several approaches to building a financial business case (Solution Matrix). A performance measure used to evaluate the efficiency of an investment. A performance measure to compare the efficiency of different investments. ROI is a metric that yields some insights into how to improve business results in the future (L. Dombrowski) 3 RAHMAD KURNIAWAN P68500
• Cont... Another traditional tool for evalating capital investments is return on investment (ROI), which measures the effectiveness of management in generating profits with its available assets. (Turban) The ROI measure is a percentage, and the higher this percentage return, the better. It is calculated essentially by dividing net income attributable to a project by the average assets invested in the project. 4 RAHMAD KURNIAWAN P68500
• Simple ROIThe benefit (return) of an investment is divided by the cost of the investments. The result is expressed as a percentage or a ratio. This is referred to as “simple ROI”. ROI= Gains from investment – Cost of investment Cost of Investment \$700,000 - \$500,000 = 40% \$500,000 5 RAHMAD KURNIAWAN P68500
• Example of ROIFor example, a \$1000 investment that earns \$50 in interest obviously generates more cash than a \$100 investment that earns \$20 interest, but the \$100 investment earns a higher return.So... 6 RAHMAD KURNIAWAN P68500
• Cont... \$1050 \$1000 \$50ROI \$50 / 1000 5% \$1000 \$1000 \$120 \$100 \$20ROI \$20 / 100 20% \$100 \$100 7 RAHMAD KURNIAWAN P68500
• What is NPV?NPV can be defined as: NPV is one of Capital budgeting analysis uses standard financial models. (Turban) Net present value is the present value of net cash inflows generated by a project including salvage value, if any, less the initial investment on the project.If NPV > 0, acceptIf NPV < 0, reject 8 RAHMAD KURNIAWAN P68500
• Cont...Organizations often use net present value (NPV) calculations for cost benefit analyses.In an NPV analysis, analysts convert future values of benefits to their present-value equivalent by discounting them at the organization’s cost of funds. 9 RAHMAD KURNIAWAN P68500
• NPV Formula C1 C2 CT NPV C0 (1 r )1 (1 r ) 2 (1 r ) T Where, r is the target rate of return per period; C0 is the Initial investment. C1 is the net cash inflow during the first period; C2 is the net cash inflow during the second period; C3 is the net cash inflow during the third period, and so on ... 10 RAHMAD KURNIAWAN P68500
• Example of NPVThere is an opportunity to invest in a business that will pay \$200,000 in one year, \$400,000 in two years, \$600,000 in three years and \$800,000 in four years. It can earn 12% per year compounded annually on a mutual fund that has similar risk. If it costs \$1.2 million to start this business, should be invest?So... 11 RAHMAD KURNIAWAN P68500
• Cont...0 1 2 3 4 years| | | | || | | | |CF –\$1.2 mil \$200,000 \$400,000 \$600,000 \$800,000 Discount rate = 12% C1 C2 CT NPV C0 (1 r )1 (1 r ) 2 (1 r ) T 200,000 400,000 600,000 800,000 NPV 1,200,000 (1.12)1 (1.12) 2 (1.12) 3 (1.12) 4 = \$232,932 12 RAHMAD KURNIAWAN P68500
• Resources of data based on exercisesNo. 2 Investing \$ 15.000.000 Revenue year 1 \$ 4.000.000Cost, Year 1 \$ 2.000.000 Revenue, year 2 \$ 5.000.000Cost, Year 2 \$ 2.000.000 Revenue, year 3 \$ 5.000.000Cost, Year 3 \$ 1.500.000 Revenue, year 4 \$ 5.000.000Cost, Year 4 \$ 1.500.000 Revenue year 5 \$ 5.000.000Cost, Year 5 \$ 1.500.000 Total \$ 24.000.000 Total \$ 23.500.000 Interest rate 10% Assumed first year include investment and cash flow year 1 \$ (13.000.000) cost cash flow year 2 \$ 3.000.000 cash flow year 3 \$ 3.500.000 cash flow year 4 \$ 3.500.000 cash flow year 5 \$ 3.500.000 NPV= \$ (2.145.469,45) 13 RAHMAD KURNIAWAN P68500
• What is Payback Period?Payback Period can be defined as: Number of years needed to recover the initial cash outlay of a projectComputation Estimate the cash flows Subtract the future cash flows from the initial cost until the initial investment has been recoveredDecision Rule – Accept if the payback period is less than some preset limit 14 RAHMAD KURNIAWAN P68500
• Example of PaybackExample: Project with an initial cash outlay of \$10,000 Free Cash Flows of \$2,500 per year for 6 years Year Cash Flow Balance \$10,000 1 \$2,500 \$7,500 2 \$2,500 \$5,000 3 \$2,500 \$2,500 4 \$2,500 --------Payback is 4 years 15 RAHMAD KURNIAWAN P68500
• References Turban, McLean, Wetherbe. Information Technology for Management: Transforming Organizations in the Digital Economy (4th edition) Turban, Volonin, Wood. (2012). Information Technology for Management, 8th edition. John Wiley & Sons (Asia) Pte Ltd. http://www.cwu.edu/ http://www.passitoncenter.org http://business.fullerton.edu 16 RAHMAD KURNIAWAN P68500
• Thank you17 RAHMAD KURNIAWAN P68500 | 1,793 | 5,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2014-10 | latest | en | 0.82444 |
https://www.storyofmathematics.com/math-calculators/valence-electron-calculator/ | 1,670,480,808,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00763.warc.gz | 1,059,283,464 | 95,354 | # Valence Electron Calculator + Online Solver With Free Steps
The Valence Electron Calculator is used to calculate the valence electrons of an atom. The valence electrons are the electrons present in the valence shell of an atom. The valence shell is the outermost shell of an atom.
It also calculates the moles of the valence electrons as a corresponding quantity of the entered atom.
Bohr proposed the concept of shells in his Bohr’s Atomic Model in 1913. He stated that electrons travel around the nucleus with fixed energy levels in circular orbits. These energy levels were termed “shells”.
The innermost shell of an atom contains electrons with the least energy. The valence shell has the electrons with the highest reactivity and all the properties of the atom depend on the number of valence electrons of an atom.
## What Is a Valence Electron Calculator?
The Valence Electron Calculator is an online tool that calculates the electrons in the outermost shell of an atom. It also computes the number of moles of the valence electrons in an atom.
The number of moles of electrons are also computed by the calculator by using the formula:
$n = \frac{N}{ N_{A} }$
Electrons are negatively charged particles. They take part in chemical reactions whereas protons and neutrons are present inside the nucleus of an atom. The charge on an electron is 1.6 × $10^{-19}$ Coulombs.
The highest number of valence electrons in an atom is eight according to the octet rule except in the case of hydrogen H and helium He. For H and He, the maximum number of valence electrons is two.
Valence electrons play an essential role in determining the chemical properties of an atom.
The atom’s properties such as shielding effect, electronegativity, electron affinity, and ionization energy all depend on the number of electrons in the outermost shell of the atom.
The students need to know the number of valence electrons of all the atoms to understand their chemical reactivity. This calculator is very useful in determining how the atom will react as it calculates the valence electrons.
## How To Use the Valence Electron Calculator
You can use the Valence Electron Calculator by following the steps given below:
### Step 1
The user must first enter the name or the symbol of the atom for which the number of valence electrons is required. It must be entered in the block next to the title “valence electron of” in the input window of the calculator.
### Step 2
After entering the atom’s name or its symbol in the input tab, the user must press “Submit” for the calculator to process the input data.
If the entered name or the symbol of the atom is incorrect or incomplete, the calculator prompts the signal “Not a valid input; please try again”.
### Output
After processing the input, the calculator shows the output in the following three windows given below.
#### Input Interpretation
The calculator interprets the input and displays the name of the element if the user has entered the symbol for the atom. It also shows the “number of valence electrons” alongside the name of the element.
The user can confirm the entered input from the input interpretation window and can change the input according to the requirement.
#### Result
The calculator computes the number of valence electrons present in the entered atom and displays it in this window.
The maximum number of valence electrons in an atom is 8e which is displayed by the calculator. The term e represents “electron”.
This is by the octet rule except for hydrogen and helium which follow the duplet rule.
#### Corresponding Quantity
This window shows the moles of electrons by using the formula given as:
$n = \frac{N}{ N_{A} }$
Where n represents the number of moles of electrons in an atom.
Na is the Avogadro’s number, which is 6.022 × $10^{23}$. It represents the number of particles in one mole of a substance.
N represents the number of electrons in the valence shell of the atom.
The calculator computes the number of moles of electrons and displays the result in this window.
## Solved Examples
The following examples are solved using the Valence Electron Calculator.
### Example 1
Sodium is a highly reactive alkali metal found in common salt. Calculate the number of valence electrons of sodium Na. Also, compute the number of moles of electrons of the sodium atom.
### Solution
The user must first enter the chemical name or the symbol Na for sodium in the input tab of the Valence Electron Calculator. After entering the input, the user must press “Submit” for the calculator to process the entered atom.
The calculator opens an output window which firstly shows the input interpretation. This window shows the name of the element “sodium” along with the “number of valence electrons”.
The next window is the Result window. It shows the number of electrons in the outermost shell of sodium. The calculator displays 1e for Na atom. This shows that sodium Na belongs to group 1A in the periodic table as it contains one valence electron.
Next, the calculator computes the number of moles of electrons and displays the output as follows:
$\text{Moles of electrons from n} = \frac{N}{ N_{A} } \ = \ 1.7 \ × \ 10^{-24} \ mol \ (moles)$
### Example 2
Calculate the number of valence electrons and the number of moles of electrons for the non-metal Nitrogen N.
### Solution
The user must first enter the atomic symbol for nitrogen N in the input block of the calculator. After pressing “Submit”, the calculator shows the input interpretation by displaying the name of nitrogen for which the valence electrons are required.
The Result window shows 5e which means that nitrogen contains five electrons in its outermost shell. It belongs to group 5A of the periodic table.
The calculator next computes the moles of electrons in the Corresponding Quantity window as follows:
$\text{Moles of electrons from n} = \frac{N}{ N_{A} } \ = \ 8.3 \ × \ 10^{-24} \ mol \ (moles)$ | 1,287 | 5,975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | longest | en | 0.889819 |
http://www.insidecounsel.com/2011/11/14/morrison-on-metrics-get-it-through-your-thick-head?t=e-discovery | 1,505,829,056,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685698.18/warc/CC-MAIN-20170919131102-20170919151102-00064.warc.gz | 476,677,483 | 13,635 | More On
# Morrison on Metrics: Get it through your thick head
Homo sapiens’ brains have evolved such a way that they’re ill-suited to cope with large numbers.
Legal department managers routinely face challenges due to human beings’ seemingly built-in cognitive limits on how well they deal with some numerical aspects.
Consider big numbers. Unless well-trained or possessed of an individual proclivity, most of us can’t grasp the extent of 10,000 documents very well let alone \$10 million of legal fees. For example, 1 billion \$1 dollar bills stacked up would be 47 miles high, but that doesn’t really help us.
Homo sapiens just don’t seem equipped with brains that have evolved to cope with numbers in the thousands, let alone many times bigger. Probably most of us toss around such large numbers—GDPs in the trillions, revenues in the billions, settlements in the hundreds of millions, law firm fees in the millions—yet deep down we aren’t comfortable with the magnitude of numbers at those scales. We cope but we don’t intuitively grasp them.
Nor are we adept at appreciating the implied level of accuracy of numbers. The expectation is that a number’s accuracy is given by its last non-zero digit starting from the right. For instance, 5,400 is plus or minus about 100; 3,500,000 could be off by 100,000, but an inside legal budget of \$3,502,989 had better be spot on.
It means that benchmarks that say 54.6% of all general counsel believe thus and so are probably exaggerating their precision. We tend to take many numbers on faith and not subject them to commonsense scrutiny.
Ponder for a moment a third mental wall many can’t climb over. Why do lawyers of all stripes find averages easier to deal with than more reliable medians, even though very large or very small outliers throw off the representativeness of averages?
It just seems easier to grasp the result when you add a bunch of figures and divide by how many there are (the average) than to sort them from high to low and pick the number half way down from the top (the median). We just aren’t hardwired with that facility.
And pity the in-house staffer who confronts standard deviations. It more than confuses people to think of subtracting lots of numbers from the average of all of them and then squaring that result before finding the square root (square root, as in root canal?). So too do ratios, such as dollars of revenue per legal staff, give us pause … or cold sweats. Ratios don’t readily translate into graspable, useful metrics for some people.
My point is that hardwired functions in our brains have not evolved to where we feel at home with some kinds of numbers—very large numbers, estimates of numbers, dispersions of numbers and roots of numbers. Those who produce metrics have a special burden to translate their findings in ways that our sometimes ill-suited brains can wrap around.
### Rees Morrison
Rees Morrison, Esq. is a partner at Altman Weil, Inc. with countless interests in legal data analytics. He is also the founder of General Counsel Metrics, LLC....
Bio and more articles | 670 | 3,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-39 | latest | en | 0.953061 |
http://www.eg.bucknell.edu/~bvollmay/teaching/phys317_fa2003/ra08.html | 1,542,443,815,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00289.warc.gz | 420,714,940 | 1,685 | ## PHYS 317 Reading Assignment #8
Read section 2.6 and answer the following questions by email, hardcopy, or web form before class Friday, September 19.
1. According to equation (2.51), doubling the volume of a monatomic ideal gas with fixed U, N gives an entropy change Delta S = N k ln(2). What is the analogous expression for Delta S if instead the energy is doubled at constant V, N?
Delta S = N k ln(Uf3/2/Ui3/2) = (3/2) N k ln 2
2. Is the work done in free expansion positive, negative, or zero?
Zero. There is no moving wall that the particles in the gas can push on. They push on the static wall -> no work done.
3. A chamber is divided in half by a partition, and both halves are filled with an equal number of A molecules. Does the entropy increase significantly if the partition is then removed? What if initially A molecules are on the left and an equal number of B molecules are on the right?
If an equal number of A particles on both sides: No it doesn't increase noticebly. If there are an equal number of A and B particles on either side, then you can see the increased entropy that's due to the entropy of mixing.
4. Can heat flow reversibly from a hotter object to a cooler object that is 10 K cooler?
No, heat flow between objects of differing tempertatures is irreversible. Always.
5. Was there anything you had particular difficulty with in this reading? Is so, describe briefly. | 335 | 1,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-47 | latest | en | 0.928431 |
https://imperialcasinohotel.com/online-roulette-betting-strategy.html | 1,566,560,483,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318375.80/warc/CC-MAIN-20190823104239-20190823130239-00049.warc.gz | 505,022,024 | 7,498 | Roulette games have minimum bets, which will be posted on a placard at the table. Minimum bets work differently for inside bets (specific numbers) and outside bets (everything else). For outside bets, any bet you make has to be at least the table minimum. Inside bets can usually be as small as you like, as long as the total of all your inside bets is the table minimum. For example, with a table minimum of \$5, you could put \$5 on #14, or \$1 each on #14, #27, #8/9, #28/29/30, and #19/20/22/23. Remember that you can bet inside or outside if you like; there's no requirement to bet both on a given spin.
The purpose of roulette is to try to predict the outcome of where the white ball will land on each spin of the wheel. If we could only try to guess what number that was, the game would be a bit boring, as there would be no variety and it would get way too repetitive. Fortunately, the designers of the game have built in tons of betting options to make the game much more exciting, versatile, and downright fun. Let’s take a look at a few of the different categories of bets you can make.
But don’t get too comfortable here. If the \$20 bet loses, the player will double that to \$40. If that loses he goes up to \$80. If that loses, he goes up to \$160. If that loses, he puts up \$320. If that loses he goes to \$640. If that loses, he might be allowed to go to \$1280 but many casinos limit the maximum a player can wager. That \$1280 might be too much.
I would use a Martingale only on the even-money outside bets at roulette, the odd or even, high or low, red or black. These bets give the player 18 chances to win with 20 chances to lose on the American double-zero wheels and 18 chances to win with 19 chances to lose on the European Roulette (single-zero wheels). Obviously, if you can play the European wheel that is the preferred one as long as the betting ranges fit your bankroll.
Thanks. It has upset a lot of scammers who have gone to great lengths to discredit me and the reviews. As perhaps I should have expected it. There are many rubbish reviews about me that are made from competitors, so perhaps its not wise to listen to competitors about each other. So everyone can decide for themselves what they believe. Anyone can just buy a system in question and find out for themselves. You cant really rely on independently reviews alone as it is extremely common that competitors attack each other, usually using fake names on various sites.
Español: practicar una estrategia en la ruleta, Italiano: Giocare Strategicamente alla Roulette, Deutsch: Roulette Strategien, Français: mettre en œuvre une stratégie à la roulette, Русский: использовать системы игры в рулетку, Português: Praticar Estratégia de Roleta, Nederlands: Oefenen met roulettestrategieën, العربية: تعلم استراتيجيات لعبة الروليت
I would use a Martingale only on the even-money outside bets at roulette, the odd or even, high or low, red or black. These bets give the player 18 chances to win with 20 chances to lose on the American double-zero wheels and 18 chances to win with 19 chances to lose on the European Roulette (single-zero wheels). Obviously, if you can play the European wheel that is the preferred one as long as the betting ranges fit your bankroll.
It's important to understand that the outcome of the roulette wheel is truly random. If Black has come up for the last 10 spins in a row, the next spin is not more likely to be Red. Black and Red are still equally likely. There's an old saying, "The wheel has no memory." That means it doesn't know what it spun before, and even if it did, the wheel can't select what number comes up out of its own volition. There's more on this in my article Debunking the Gambler's Fallacy.
##### These are the bets we recommend for beginners who want to get more comfortable with roulette. (This does not mean they aren’t great bets for seasoned players, as well.) Instead of betting on specific numbers or groups of numbers, you are betting on what we have termed “the characteristics” of the number. This would include betting on the color of the number or on the evenness or oddness of the number. These bets always pay even money and are as simple as they sound. If you bet black and a black number rolls, you win. If you bet even and an even number rolls, you win. It’s that easy!
The odds of you winning will always be 50/50. So you have a 50% chance of LOSING \$1, and a 50% chance of WINNING \$0.50. You can’t just double bet size after losses, because then all you do is increase the amount you risk. Sure you may get lucky and win, but what happens if you lose? You’ll lose big. So there is no escaping the unfair payouts UNLESS you know which side of the coin is more likely to appear. Then you would be changing the odds of winning. And if you won much more often than 50% of the time, then the unfair payout wont matter as much.
A roulette strategy is any method that aims to win at roulette. In most cases it’s a set of mechanical rules that tell the player when and where to bet. There are more strategies to win roulette than any other casino game, but the vast majority of players consistently lose. This is partly because most roulette tips pages focus on casino promotion, rather than accurate tips.
# Here's another way to look at it: Let's say you bet \$10 on every number, one bet on each of the 38 spots. So you've just thrown down \$380 in bets. Only one of those numbers will win, and will pay 35 to 1, so you'll get back \$360 (the \$350 you won plus your original \$10 bet on that number). You bet a total of \$380 but you walked away with only \$360, so you lost \$20. That \$20 you lost represents the house edge of 5.26% (\$20 lost divided by the \$380 that you bet; \$20 ÷ \$380 = 5.26%).
7. Everything in roulette is long term, unless you have detailed data that accounts for why the ball lands where it does (like dominant diamond, rotor speed, ball bounce). You cannot possibly test a system properly from a few minutes or even weeks of play. Proper testing requires months, otherwise a loss or win can be plain good or bad luck. So for proper testing to be practical, you need at least 50,000 recorded spins from a real wheel. The only exception is if you have supporting information to back up results, like dominant diamond, rotor speed, ball bounce (so you can plainly see all factors contributing to where the ball lands).
##### The contribution of roulette towards this is very important. At one extreme you’ll find casinos which exclude roulette games entirely. Others weight the contribution to the play-through at a flat rate, for example, 10%. An alternative approach is to exclude bets which cover a certain amount of the wheel (for example 60% or more) and weight the mid-risk / low-risk bets.
Are there some dealers who can place the ball accurately into certain sections of the wheel? Here I will discuss if this is possible and what it would take for a dealer to be able to do such a thing. It isn’t an easy task. If you find a dealer with a signature how should you bet into it? Do other dealers think some dealers actually have this ability?
Officially, there are three variations of Roulette that exist, American, European and French Roulette. Players may bet on a single or a range of numbers, colors red or black, odd or even, or high (19–36) or low (1–18) numbers. The winning number is thus determined when the ball drops into one of 36 colored and numbered pockets on the wheel or a zero pocket (American roulette utilizes a double zero).
Cross-reference roulette system: Cross referencing is a type of analysis where all available data is considered, and used to detect usable patterns. What makes it special is the data cross-referenced to ensure accuracy. This enables the player to better find hidden patterns in spins, and in less time. Also it enables players to quickly adjust when conditions at the wheel change. The method of cross referencing is not exclusive to roulette, and can be applied to other casino games. But this particular roulette system is combined with other predictive methods that are exclusive to roulette.
##### Roulette Tips – Here is a collection of 8 roulette tips that should be helpful to new or intermediate players. You won’t find anything new or ground breaking here, but we truly believe that this is the best collection of tips that you’re going to find on the Internet. The reason for that is because we keep it real, we don’t make false claims about “winning a fortune” like other websites; we just give useful tips to help you understand the game.
As we’ve mentioned several times already, roulette is probably the easiest game in the casino to learn how to play. A roulette table will always consist of a wheel, a small white ball, and a table printed with all of your betting options. You will also have a dealer present live in the casino, or an electronic dealer if you’re playing online. Each game will consist of a round of betting, and then the dealer will spin the wheel, drop in the white ball, and wait to see where it lands. The object of the game is to try to predict with your bets what number the white ball is going to land on. After each spin of the wheel, the game resets completely and starts over with a new round of betting.
Online roulette is great for players who like the convenience and the ability to dictate how their experience goes. With online roulette, you get the ability to play at any time and from anywhere with an internet connection. You never have an issue getting a seat, and you don’t have to worry about reaching the other side of the betting felt, as you place all of your bets with the click of your mouse.
First, note that while tickets are billed at just \$59 on weekends, after the ticketing fee (\$7) and the \$10 upgrade to cross off two additional shows (something you'll want to do if you're a person who sees Broadway shows more often than just once in awhile) the ticket is closer to \$79, which is what most shows (excluding the hyped productions like Hamilton, Springsteen, Dear Evan Hansen, Book of Mormon, etc.) start at, anyway.
Most sites now require the player to sign various agreements before they can play casino games for real money. It is wise for the player to know what is expected of him and what rights or lack of rights that player has. Does an online roulette game need special techniques and strategies of play that differ from a regular casino’s roulette game? Which is the better game to play – an RNG game or a real game? This section will answer these questions.
Before the game starts, players are instructed to place their bets. This is the time that you get to choose what you would like to wager on for the next spin. We will go over the different bet types and options available to you in depth later. Bets are made by placing your chips onto the felt in the area designated for the bet you would like to make. | 2,486 | 10,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-35 | latest | en | 0.952719 |
https://jp.mathworks.com/matlabcentral/cody/problems/13-remove-all-the-consonants/solutions/1170176 | 1,580,083,865,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00147.warc.gz | 509,481,512 | 15,596 | Cody
# Problem 13. Remove all the consonants
Solution 1170176
Submitted on 24 Apr 2017 by Aaron T. Becker's Robot Swarm Lab
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
s1 = 'Jack and Jill went up the hill'; s2 = 'a a i e u e i'; assert(isequal(s2,refcn(s1)))
s2 = 'a a i e u e i'
2 Pass
s1 = 'I don''t want to work. I just want to bang on the drum all day.'; s2 = 'I o'' a o o. I u a o a o e u a a.'; assert(isequal(s2,refcn(s1)))
s2 = 'I o' a o o. I u a o a o e u a a.' | 210 | 602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-05 | latest | en | 0.708368 |
https://www.tldevtech.com/how-to-do-jigsaw-puzzles-like-an-expert/ | 1,716,559,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00376.warc.gz | 894,329,717 | 25,022 | # How to Do Jigsaw Puzzles Like an Expert
A jigsaw puzzle is one of the favorite pastimes of people of all ages. It requires attention to detail to solve the puzzle. Meanwhile, do you want to learn how to do jigsaw puzzle-like a pro? Then, this article is for you. The following tips can help you play jigsaw puzzle in the best possible way, which can offer you fun, entertainment, and satisfaction.
## Tips on doing Jigsaw puzzles
• Turn the picture side up of all the pieces
The first thing you must do is to turn the picture of each piece over, so you can see the picture side on the flat surface. It will help you to put the puzzle together easily.
• Divide the pieces into groups
As you turn the picture side over, you can sort the pieces into groups. For example, if you have a jigsaw with mountains and house, you can sort like this:
• Pieces with a house on it
• Pieces with a mountain on it
• Pieces with the sky on it (it can be sort into cloudy or blue)
Depending on the design of your puzzle, you can make A jigsaw or more groupings of the pieces.
• Assembling the border
Now that you have sorted the pieces into a group, you can assemble the border. Don’t worry if there are still missing pieces since it will be solved soon. With this, you will have the idea of the space you need to work out.
• Assemble through sorting colors, groups and patterns
You can now begin working on other pieces. You can start with the easy stuff to prevent you from giving up early. There are some instances when it is the right color, however, it could be incorrectly placed in the piece. So, you must pay attention to small details while solving your jigsaw puzzle.
• Pay attention to the shape of the pieces
One of the crucial factors to consider is the shape of the pieces. Usually, the jigsaw puzzle comes in 6 shapes, which start from 0 “knobs,” 4 “holes,” 0 holes, and 4 knobs, as well as all permutations in between. If you are a jigsaw puzzle lover, it will be easy for you to determine in just a glance if the piece will fit in where you want it to place.
You can divide the pieces into small piles according to their patterns, colors, and shapes. So, you can put together similar pieces perfectly.
Jigsaw puzzles are easy to solve if you spread it out on a flat surface where you can easily saw the picture of the pieces.
By following these tips, you can enjoy solving jigsaw puzzles like an expert. Solving jigsaw puzzles is a good way to spend time with your family. It can provide you with fun and entertainment that can make you feel relax. Additionally, it can also boost your mental abilities and concentration, which are essential in your daily life.
If you want to relax and take a break from your stressful routine, playing a jigsaw puzzle is the solution.
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### Basic Statistics for Business and Economics ch 7
What are the two families of continuous probability distributions? Uniform probability distribution and normal probability distribution.
Uniform Probability Distribution The simplest distribution for a continuous random variable. This distribution is rectangular in shape and is defined by minimum and maximum values.
Mean of the Uniform Distribution (mu) = (a + b)/2
Standard Deviation of the Uniform Distribution (sigma) = The square root of [(b-a)^2]/12
Uniform Distribution P(X) = 1/(b-a) [if x is between or equal to a and b]
Area = Height * Base
What are the characteristics of a normal probability distribution? It is bell-shaped and had a single peak at the center of the distribution. It is symmetrical about the mean. It falls off smoothly in either direction for the central values - it is asymptotic. Location of a normal distribution is determined by the mean.
Bell-Shaped The arithmetic mean, median, and mode are equal and located in the center of the distribution. The total are under the curve is 1.00. Half of the area under the normal curve is to the right of the center and the other half is to the left of it.
Symmetrical If the normal curve is cut vertically at the center value, the two halves will be mirror images.
Asymptotic The curve gets closer and closer to the X-axis but never actually touches it. The tails of the curve extend indefinitely in both directions.
The dispersion or spread of the distribution is determined by: The standard deviation (sigma).
Standard Normal Probability Distribution One member of the family of normal distributions that can be used to determine the probabilities for all normal distributions. It is unique because it has a mean of 0 and a standard deviation of 1.
z Value The signed distance between a selected value, designated X, and the mean (mu), divided by the standard deviation (sigma).
Empirical Rule For symmetrical bell-shaped frequency distribution about 68% of the observations will lie within plus and minus 1 standard deviation of the mean; 95% of the observations will lie within 2 standard deviations, and 99.7% within 3 standard deviations.
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1 Appendi A Review of Vectos This appendi is a summa of the mathematical aspects of vectos used in electicit and magnetism. Fo a moe detailed intoduction to vectos, see Chapte 1. A.1 DESCRIBING THE 3D WORLD: VECTORS Phsical phenomena take place in the 3D wold aound us. In ode to be able to make quantitative pedictions and give detailed, quantitative eplanations, we need tools fo descibing pecisel the positions and velocities of objects in 3D, and the changes in position and velocit due to inteactions. These tools ae mathematical entities called 3D vectos. 3D Coodinates We will use a 3D coodinate sstem to specif positions in space and othe vecto quantities. Usuall we will oient the aes of the coodinate sstem as shown in Figue A.1: + ais to the ight, + ais upwad, and + ais coming out of the page, towad ou. This is a ight-handed coodinate sstem: if ou hold the thumb, fist, and second finges of ou ight hand pependicula to each othe, and align ou thumb with the ais and ou fist finge with the ais, ou second finge points along the ais. (In some math and phsics tetbook discussions of 3D coodinate sstems, the ais points out, the ais points to the ight, and the ais points up, but we will also use a 2D coodinate sstem with up, so it makes sense alwas to have the ais point up.) Basic Popeties of Vectos: Magnitude and Diection A vecto is a quantit that has a magnitude and a diection. Fo eample, the velocit of a baseball is a vecto quantit. The magnitude of the baseball s velocit is the speed of the baseball, fo eample 20 metes/second. The diection of the baseball s velocit is the diection of its motion at a paticula instant, fo eample up o to the ight o west o in the + diection. A smbol denoting a vecto is witten with an aow ove it: Position v is a vecto. A position in space can also be consideed to be a vecto, called a position vecto, pointing fom an oigin to that location. Figue A.2 shows a position vecto that might epesent ou final position if ou stated at the oigin and walked 4 metes along the ais, then 2 metes paallel to the ais, then climbed a ladde so Figue A.1 Right-handed 3D coodinate sstem. = 3 m = 2 m = 4 m Figue A.2 A position vecto = 4, 3, 2 m and its,, and components. 1
2 2 Review of Vectos ou wee 3 metes above the gound. You new position elative to the oigin is a vecto that can be witten like this: = 4, 3, 2 m component =4m component =3m component =2m In thee dimensions a vecto is a tiple of numbes,,. Quantities like the position of an object and the velocit of an object can be epesented as vectos: = 4 m = 3 m = 2 m Figue A.3 The aow epesents the vecto = 4, 3, 2 m, dawn with its tail at location 0, 0, 2. Figue A.4 The position vecto 3, 1, 0, dawn at the oigin, in the plane. The components of the vecto specif the displacement fom the tail to the tip. The ais, which is not shown, comes out of the page, towad ou. Components of a Vecto =,, (a position vecto) 1 = 3.2, 9.2, 66.3 m (a position vecto) v = v,v,v (a velocit vecto) v 1 = 22.3, 0.4, 19.5 m/s (a velocit vecto) Each of the numbes in the tiple is efeed to as a component of the vecto. The component of the vecto v is the numbe v. The component of the vecto v 1 = 22.3, 0.4, 19.5 m/s is 19.5 m/s. A component such as v is not a vecto, since it is onl one numbe. It is impotant to note that the component of a vecto specifies the diffeence between the coodinate of the tail of the vecto and the coodinate of the tip of the vecto. It does not give an infomation about the location of the tail of the vecto (compae Figue A.2 and Figue A.3). Dawing Vectos In Figue A.2 we epesented ou position vecto elative to the oigin gaphicall b an aow whose tail is at the oigin and whose aowhead is at ou position. The length of the aow epesents the distance fom the oigin, and the diection of the aow epesents the diection of the vecto, which is the diection of a diect path fom the initial position to the final position (the displacement ; b walking and climbing ou displaced ouself fom the oigin to ou final position). Since it is difficult to daw a 3D diagam on pape, when woking on pape ou will usuall be asked to daw vectos which all lie in a single plane. Figue A.4 shows an aow in the plane epesenting the vecto 3, 1, 0. Vectos and Scalas A quantit which is epesented b a single numbe is called a scala. A scala quantit does not have a diection. Eamples include the mass of an object, such as 5 kg, o the tempeatue, such as 20C. Vectos and scalas ae ve diffeent entities; a vecto can neve be equal to a scala, and a scala cannot be added to a vecto. Scalas can be positive o negative: m =50kg T = 20 C Although a component of a vecto such as v is not a vecto, it s not a scala eithe, despite being onl one numbe. An impotant popet of a tue scala is that its value doesn t change if we oient the coodinate aes diffeentl. Rotating the aes doesn t change an object s mass, o the tempeatue, but it does change what we mean b the component of the velocit since the ais now points in a diffeent diection.
3 A.1. DESCRIBING THE 3D WORLD: VECTORS 3 Magnitude of a Vecto In Figue A.5 we again show the vecto fom Figue A.2, showing ou displacement fom the oigin. Using a 3D etension of the Pthagoean theoem fo ight tiangles (Figue A.6), the net distance ou have moved fom the stating point is (4 m)2 +(3m) 2 +(2m) 2 = 29 m = 5.39 m = 3 m = 4 m We sa that the magnitude of the position vecto is = 2 m =5.39 m The magnitude of a vecto is witten eithe with absolute-value bas aound the vecto as, o simpl b witing the smbol fo the vecto without the little aow above it,. The magnitude of a vecto can be calculated b taking the squae oot of the sum of the squaes of its components (see Figue A.6). Figue A.5 A vecto epesenting a displacement fom the oigin. MAGNITUDE OF A VECTOR If the vecto =,, then = (a scala). ( ) + 2 The magnitude of a vecto is alwas a positive numbe. The magnitude of a vecto is a single numbe, not a tiple of numbes, and it is a scala, not a vecto. The magnitude of a vecto is a tue scala, because its value doesn t change if ou otate the coodinate aes. Rotating the aes changes the individual components, but the length of the aow epesenting the vecto doesn t change. Can a Vecto be Positive o Negative? QUESTION Conside the vecto v = , 0, m/s. Is this vecto positive? Negative? Zeo? ( ) Figue A.6 The magnitude of a vecto is the squae oot of the sum of the squaes of its components (3D vesion of the Pthagoean theoem). None of these desciptions is appopiate. The component of this vecto is positive, the component is eo, and the component is negative. Vectos aen t positive, o negative, o eo. Thei components can be positive o negative o eo, but these wods just don t mean anthing when used with the vecto as a whole. On the othe hand, the magnitude of a vecto such as v is alwas positive. Mathematical Opeations Involving Vectos Although the algeba of vectos is simila to the scala algeba with which ou ae ve familia, it is not identical. Thee ae some algebaic opeations that cannot be pefomed on vectos. Algebaic opeations that ae legal fo vectos include the following opeations, which we will discuss in this chapte: adding one vecto to anothe vecto: a + w subtacting one vecto fom anothe vecto: b d finding the magnitude of a vecto: finding a unit vecto (a vecto of magnitude 1): ˆ multipling (o dividing) a vecto b a scala: 3 v o w/2 finding the ate of change of a vecto: Δ /Δt o d /dt. In late chaptes we will also see that thee ae two moe was of combining two vectos:
4 4 Review of Vectos the vecto dot poduct, whose esult is a scala the vecto coss poduct, whose esult is a vecto p 3p 2p p 1 2 p Opeations that ae Not Legal fo Vectos Although vecto algeba is simila to the odina scala algeba ou have used up to now, thee ae cetain opeations that ae not legal (and not meaningful) fo vectos: A vecto cannot be set equal to a scala. A vecto cannot be added to o subtacted fom a scala. A vecto cannot occu in the denominato of an epession. (Although ou can t divide b a vecto, note that ou can legall divide b the magnitude of a vecto, which is a scala.) Multipling a Vecto b a Scala A vecto can be multiplied (o divided) b a scala. If a vecto is multiplied b a scala, each of the components of the vecto is multiplied b the scala: If =,, then a = a, a, a 3p 2p Figue A.7 Multipling a vecto b a scala changes the magnitude of the vecto. Multipling b a negative scala eveses the diection of the vecto. If v = v,v,v then v b = v b, v b, v b ( 1 ) 6, 20, 9 = 3, 10, Multiplication b a scala scales a vecto, keeping its diection the same but making its magnitude lage o smalle (Figue A.7). Multipling b a negative scala eveses the diection of a vecto. Magnitude of a Scala You ma wonde how to find the magnitude of a quantit like 3, which involves the poduct of a scala and a vecto. This epession can be factoed: 3 = 3 The magnitude of a scala is its absolute value, so: 3 = 3 = Diection of a Vecto: Unit Vectos One wa to descibe the diection of a vecto is b specifing a unit vecto. A unit vecto is a vecto of magnitude 1, pointing in some diection. A unit vecto is witten with a hat (caet) ove it instead of an aow. The unit vecto â is called a-hat. QUESTION Is the vecto 1, 1, 1 a unit vecto? The magnitude of 1, 1, 1 is =1.73, so this is not a unit vecto. The vecto 1/ 3, 1/ 3, 1/ 3 is a unit vecto, since its magnitude is 1: ( 1 ) 2 +( 1 ) 2 +( 1 ) 2 = Note that eve component of a unit vecto must be less than o equal to 1.
5 A.1. DESCRIBING THE 3D WORLD: VECTORS 5 In ou 3D Catesian coodinate sstem, thee ae thee special unit vectos, oiented along the thee aes. The ae called i-hat, j-hat, and k-hat, and the point along the,, and aes, espectivel (Figue A.8): î= 1, 0, 0 ĵ= 0, 1, 0 ˆk = 0, 0, 1 One wa to epess a vecto is in tems of these special unit vectos: 0.02, 1.7, 30.0 =0.02î+( 1.7)ĵ+30.0ˆk We will usuall use the,, fom athe than the îĵˆk fom in this book, because the familia,, notation, used in man calculus tetbooks, emphasies that a vecto is a single entit. Not all unit vectos point along an ais, as shown in Figue A.9. Fo eample, the vectos ĝ = , , and ˆF = 0.424, 0.566, ae both unit vectos, since the magnitude of each is equal to 1. Note that eve component of a unit vecto is less than o equal to 1. k Figue A.8 The unit vectos î, ĵ, ˆk. Calculating Unit Vectos An vecto ma be factoed into the poduct of a unit vecto in the diection of the vecto, multiplied b a scala equal to the magnitude of the vecto. v = 1.5, 1.5, 0Ò m/s w = w ŵ Fo eample, a vecto of magnitude 5, aligned with the ais, could be witten as: 0, 5, 0 =5 0, 1, 0 v = 2, 2, 0Ò 2 2 Figue A.9 The unit vecto ˆv has the same diection as the vecto v, but its magnitude is 1, and it has no phsical units. Theefoe, to find a unit vecto in the diection of a paticula vecto, we just divide the vecto b its magnitude: CALCULATING A UNIT VECTOR ˆ = ( ), ˆ = =,, ( ) ( ), ( ) EXAMPLE Unit Vecto If v = 22.3, 0.4, 19.5 m/s, then ˆv = v v = 22.3, 0.4, 19.5 m/s = 0.753, , ( 22.3)2 +(0.4) 2 +( 19.5) 2 m/s Remembe that to divide a vecto b a scala, ou divide each component of the vecto b the scala. The esult is a new vecto. Note also that a unit vecto has no phsical units (such as metes pe second), because the units in the numeato and denominato cancel.
6 6 Review of Vectos Equalit of Vectos EQUALITY OF VECTORS A vecto is equal to anothe vecto if and onl if all the components of the vectos ae equal. w = means that w = and w = and w = The magnitudes and diections of two equal vectos ae the same: w = and ŵ =ˆ EXAMPLE Equal Vectos = 4, 3, 2 = ( )=5.39 ˆ = 4, 3, = 0.742, 0.557, If w = w = 4, 3, 2 w =5.39 ŵ = 0.742, 0.557, B Vecto Addition A ADDING VECTORS The sum of two vectos is anothe vecto, obtained b adding the components of the vectos. B A = A,A,A B = B,B,B A A + B = (A + B ), (A + B ), (A + B ) EXAMPLE Adding Vectos A + B B 1, 2, 3 + 4, 5, 6 = 3, 7, 9 A Figue A.10 The pocedue fo adding two vectos gaphicall: daw vectos tip to tail. To add A + B gaphicall, move B so the tail of B is at the tip of A then daw a new aow stating at the tail of A and ending at the tip of B. Waning: Don t Add Magnitudes! The magnitude of a vecto is not in geneal equal to the sum of the magnitudes of the two oiginal vectos! Fo eample, the magnitude of the vecto 3, 0, 0 is 3, and the magnitude of the vecto 2, 0, 0 is 2, but the magnitude of the vecto ( 3, 0, 0 + 2, 0, 0 ) is 1, not 5! Adding Vectos Gaphicall: Tip to Tail The sum of two vectos has a geometic intepetation. In Figue A.10 ou fist walk along displacement vecto A, followed b walking along displacement vecto B. What is ou net displacement vecto C = A + B? The component C
7 A.1. DESCRIBING THE 3D WORLD: VECTORS 7 of ou net displacement is the sum of A and B. Similal, the component C of ou net displacement is the sum of A and B. GRAPHICAL ADDITION OF VECTORS To add two vectos A and B gaphicall (Figue A.10): Daw the fist vecto A Move the second vecto B (without otating it) so its tail is located at the tip of the fist vecto Daw a new vecto fom the tail of vecto A to the tip of vecto B Vecto Subtaction The diffeence of two vectos will be ve impotant in this and subsequent chaptes. To subtact one vecto fom anothe, we subtact the components of the second fom the components of the fist: A B = (A B ), (A B ), (A B ) 1, 2, 3 4, 5, 6 = 5, 3, 3 Subtacting Vectos gaphicall: Tail to Tail To subtact one vecto B fom anothe vecto A gaphicall: Daw the fist vecto A Move the second vecto B (without otating it) so its tail is located at the tail of the fist vecto Daw a new vecto fom the tip of vecto B to the tip of vecto A Note that ou can check this algebaicall and gaphicall. As shown in Figue A.11, since the tail of A B is located at the tip of B, then the vecto A should be the sum of B and A B, as indeed it is: B +( A B)= A B A A B Figue A.11 The pocedue fo subtacting vectos gaphicall: daw vectos tail to tail; daw new vecto fom tip of second vecto to tip of fist vecto. Commutativit and Associativit Vecto addition is commutative: A + B = B + A Vecto subtaction is not commutative: A B B A The associative popet holds fo vecto addition and subtaction: The Zeo Vecto ( A + B) C = A +( B C) It is convenient to have a compact notation fo a vecto whose components ae all eo. We will use the smbol 0 to denote a eo vecto, in ode to distinguish it fom a scala quantit that has the value 0. 0 = 0, 0, 0 Fo eample, the sum of two vectos B +( B)= 0.
8 8 Review of Vectos 6 m Change in a Quantit: The Geek Lette Δ Fequentl we will want to calculate the change in a quantit. Fo eample, we ma want to know the change in a moving object s position o the change in its velocit duing some time inteval. The Geek lette Δ (capital delta suggesting d fo diffeence ) is used to denote the change in a quantit (eithe a scala o a vecto). We use the subscipt i to denote an initial value of a quantit, and the subscipt f to denote the final value of a quantit. If a vecto i denotes the initial position of an object elative to the oigin (its position at the beginning of a time inteval), and f denotes the final position of the object, then m Figue A.12 Relative position vecto. A = 1 A θ A Sting Figue A.13 A unit vecto whose diection is at a known angle fom the + ais. θ θ Figue A.14 A 3D unit vecto and its angles to the,, and aes. θ Δ = f i Δ means change of o f i (displacement) Δt means change of t ot f t i (time inteval) The smbol Δ (delta) alwas means final minus initial, not initial minus final. Fo eample, when a child s height changes fom 1.1 mto1.2m, the change is Δ =+0.1m, a positive numbe. If ou bank account dopped fom \$150 to \$130, what was the change in ou balance? Δ(bank account)= 20 dollas. Relative Position Vectos Vecto subtaction is used to calculate elative position vectos, vectos which epesent the position of an object elative to anothe object. In Figue A.12 object 1 is at location 1 and object 2 is at location 2. We want the components of a vecto that points fom object 1 to object 2. This is the vecto obtained b subtaction: 2 elative to 1 = 2 1. Note that the fom is alwas final minus initial in these calculations. Unit Vectos and Angles Suppose a taut sting is at an angle θ to the + ais, and we need a unit vecto in the diection of the sting. Figue A.13 shows a unit vecto  pointing along the sting. What is the component of this unit vecto? Conside the tiangle whose base is A and whose hpotenuse is  =1. Fom the definition of the cosine of an angle we have this: cos θ = adjacent hpotenuse = A 1, so A =cosθ In Figue A.13 the angle θ is shown in the fist quadant (θ less than 90 ), but this woks fo lage angles as well. Fo eample, in Figue?? the angle fom the + ais to a unit vecto ˆB is in the second quadant (θ geate than 90 ) and cos θ is negative, which coesponds to B being negative. What is tue fo is also tue fo and. Figue A.14 shows a 3D unit vecto ˆ and indicates the angles between the unit vecto and the,, and aes. Evidentl we can wite Vecto in plane ˆ = cos θ, cos θ, cos θ θ = 90º θ θ Figue A.15 If a vecto lies in the plane, cos θ =sinθ. These thee cosines of the angles between a vecto (o unit vecto) and the coodinate aes ae called the diection cosines of the vecto. The cosine function is neve geate than 1, just as no component of a unit vecto can be geate than 1. A common special case is that of a unit vecto ling in the plane, with eo component (Figue A.15). In this case θ + θ =90, so that cos θ = cos(90 θ )=sinθ, so that ou can epess the cosine of θ as the sine of θ,
9 A.1. DESCRIBING THE 3D WORLD: VECTORS 9 which is often convenient. Howeve, in the geneal 3D case shown in Figue A.14 thee is no such simple elationship among the diection angles, no among thei cosines. FINDING A UNIT VECTOR FROM ANGLES To find a unit vecto if angles ae given: Redaw the vecto of inteest with its tail at the oigin, and detemine the angles between this vecto and the aes. Imagine the vecto 1, 0, 0, which lies on the + ais. θ is the angle though which ou would otate the vecto 1, 0, 0 until its diection matched that of ou vecto. θ is positive, and θ 180. θ is the angle though which ou would otate the vecto 0, 1, 0 until its diection matched that of ou vecto. θ is positive, and θ 180. θ is the angle though which ou would otate the vecto 0, 0, 1 until its diection matched that of ou vecto. θ is positive, and θ 180. EXAMPLE Fom Angle to Unit Vecto A ope ling in the plane, pointing up and to the ight, suppots a climbe at an angle of 20 to the vetical (Figue A.16). What is the unit vecto pointing up along the ope? 20 Figue A.16 A climbe suppoted b a ope. Solution Follow the pocedue given above fo finding a unit vecto fom angles. In Figue A.17 we edaw the vecto with its tail at the oigin, and we detemine the angles between the vecto and the aes. If we otate the unit vecto 1, 0, 0 fom along the + ais to the vecto of inteest, we see that we have to otate though an angle θ =70. To otate the unit vecto 0, 1, 0 fom along the + ais to the vecto of inteest, we have to otate though an angle of θ =20. The angle fom the + ais to ou vecto is θ =90. Theefoe the unit vecto that points along the ope is this: θ = 90º θ = 20º θ = 70º cos 70, cos 20, cos 90 = 0.342, 0.940, 0 Figue A.17 Redaw the vecto with its tail at the oigin. Identif the angles between the positive aes and the vecto. In this eample the vecto lies in the plane. FURTHER DISCUSSION You ma have noticed that the component of the unit vecto can also be calculated as sin 70 =0.940, and it is often useful to ecognie that a vecto component can be obtained using sine instead of cosine. Thee is howeve some advantage alwas to calculate in tems of diection cosines. This is a method that alwas woks, including in 3D, and which avoids having to decide whethe to use a sine o a cosine. Just use the cosine of the angle fom the elevant positive ais to the vecto. EXAMPLE Fom Unit Vecto to Angles A vecto points fom the oigin to the location 600, 0, 300 m. What is the angle that this vecto makes to the ais? To the ais? To the ais?
10 10 Review of Vectos Solution 600, 0, 300 ˆ = = 0.894, 0, ( 600)2 +(0) 2 + (300) 2 m But we also know that ˆ = cos θ, cos θ, cos θ, so cos θ = 0.894, and the accosine gives θ = Similal, cos θ =0, θ =90 (which checks; no component) cos θ =0.447,θ =63.4 θ = 63.4º θ = 153.4º Figue A.18 Look down on the plane. The diffeence in the two angles is 90, as it should be. FURTHER DISCUSSION Looking down on the plane in Figue A.18, ou can see that the diffeence between θ = and θ =63.4 is 90, as it should be. A.2 VECTOR MULTIPLICATION Vectos can be added and subtacted, and the can be multiplied b a scala. Two vectos can also be multiplied, but two diffeent kinds of vecto multiplication ae defined: the dot poduct and the coss poduct. In the pevious volume the dot poduct was intoduced in the contet of wok, and the coss poduct was intoduced in the contet of angula momentum. The Dot Poduct The dot poduct is an opeation involving two vectos. This is encounteed in the epession fo wok in Chapte 6: W = F Δ =(F Δ + F Δ + F Δ) If F = 3, 2, 4 N and Δ = 2, 0, 5 m, then F Δ = ((3 2) + ( 2 0) + (4 5)) N m = 14 N m The esult of a dot poduct opeation is a scala (like the quantit wok). Note that the dot poduct of a vecto with itself gives the squae of the magnitude of the vecto:,,,, =( 2,2,2 )= 2 The magnitude of the dot poduct can also be calculated as: F Δ = F Δ cos θ = F Δ = F Δ whee θ is the angle between the two vectos, placed tail to tail. In the VPthon pogamming language, dot(vecto1,vecto2) gives the dot poduct of two vectos. The Coss Poduct k The coss poduct is discussed in detail in Chapte 18 in the contet of the Biot- Savat law fo finding the magnetic field of moving chages. In the VPthon pogamming language, coss(vecto1,vecto2) gives the coss poduct of two vectos. It is possible to evaluate the coss poduct in tems of unit vectos along the thee aes (Figue A.19). Fist, note that î î=0, ĵ ĵ=0, and ˆk ˆk =0, since when we coss a vecto with itself the angle between the two vectos is eo, and sin 0 =0. Second, î ĵ = ˆk, since the angle is 90 and the ight-hand ule gives a esult in the + diection (out of the page; Figue A.19). On the othe hand, ĵ î= ˆk, because the ight-hand ule gives a esult in the diection (into the Figue A.19 Coss poducts of unit vectos.
11 A.3. SUMMARY 11 page). Similal, ĵ ˆk =î, ˆk ĵ= î, ˆk î=ĵ, and î ˆk = ĵ. Putting this all togethe, we obtain the following geneal esult: A B =(A B A B )î+(a B A B )ĵ+(a B A B )ˆk o A B = (A B A B ), (A B A B ), (A B A B ) This appoach to calculating a coss poduct is paticulal useful in compute calculations. Note the cclic natue of the subscipts:,,. Common Eos in Vecto Multiplication (1) A dot poduct of two vectos esults in a scala, not anothe vecto. (2) A coss poduct of two vectos esults in anothe vecto, not a scala. Technicall, although a component of a vecto is a single numbe, it is not a scala. If ou otate ou coodinate aes, the,, and components of a vecto change, but a tue scala such as m =5kg doesn t change. A.3 SUMMARY Vectos A3Dvecto is a quantit with magnitude and a diection, which can be epessed as a tiple,,. A vecto is indicated b an aow:. A scala is a single numbe. Legal mathematical opeations involving vectos include: adding one vecto to anothe vecto subtacting one vecto fom anothe vecto multipling o dividing a vecto b a scala finding the magnitude of a vecto taking the deivative of a vecto Opeations that ae not legal with vectos include: A vecto cannot be added to a scala A vecto cannot be set equal to a scala A vecto cannot appea in the denominato (ou can t divide b a vecto) The smbol Δ denotes subtaction The smbol Δ (delta) means change of : Δt=t f t i, Δ = f i. Δ alwas means final minus initial.
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Chapter 06 Data Mining Techniques
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E (entropy) : expected information based on the partitioning into subsets by A
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• Chapter 06 Data Mining Techniques
1. 1. Data Mining: Concepts and Techniques — Chapter 6 — Jiawei Han Department of Computer Science University of Illinois at Urbana-Champaign www.cs.uiuc.edu/~hanj ©2006 Jiawei Han and Micheline Kamber, All rights reserved January 20, 2014 Data Mining: Concepts and Techniques 1
2. 2. January 20, 2014 Data Mining: Concepts and Techniques 2
3. 3. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 3
4. 4. Classification vs. Prediction Classification predicts categorical class labels (discrete or nominal) classifies data (constructs a model) based on the training set and the values (class labels) in a classifying attribute and uses it in classifying new data Prediction models continuous-valued functions, i.e., predicts unknown or missing values Typical applications Credit approval Target marketing Medical diagnosis Fraud detection January 20, 2014 Data Mining: Concepts and Techniques 4
5. 5. Classification—A Two-Step Process Model construction: describing a set of predetermined classes Each tuple/sample is assumed to belong to a predefined class, as determined by the class label attribute The set of tuples used for model construction is training set The model is represented as classification rules, decision trees, or mathematical formulae Model usage: for classifying future or unknown objects Estimate accuracy of the model The known label of test sample is compared with the classified result from the model Accuracy rate is the percentage of test set samples that are correctly classified by the model Test set is independent of training set, otherwise over-fitting will occur If the accuracy is acceptable, use the model to classify data tuples whose class labels are not known January 20, 2014 Data Mining: Concepts and Techniques 5
6. 6. Process (1): Model Construction Classification Algorithms Training Data NAME Mike Mary Bill Jim Dave Anne RANK YEARS TENURED Assistant Prof 3 no Assistant Prof 7 yes Professor 2 yes Associate Prof 7 yes Assistant Prof 6 no Associate Prof 3 no January 20, 2014 Classifier (Model) IF rank = ‘professor’ OR years > 6 THEN tenured = ‘yes’ Data Mining: Concepts and Techniques 6
7. 7. Process (2): Using the Model in Prediction Classifier Testing Data Unseen Data (Jeff, Professor, 4) NAME Tom Merlisa George Joseph RANK YEARS TENURED Assistant Prof 2 no Associate Prof 7 no Professor 5 yes Assistant Prof 7 yes January 20, 2014 Tenured? Data Mining: Concepts and Techniques 7
8. 8. Supervised vs. Unsupervised Learning Supervised learning (classification) Supervision: The training data (observations, measurements, etc.) are accompanied by labels indicating the class of the observations New data is classified based on the training set Unsupervised learning (clustering) The class labels of training data is unknown Given a set of measurements, observations, etc. with the aim of establishing the existence of classes or clusters in the data January 20, 2014 Data Mining: Concepts and Techniques 8
9. 9. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 9
10. 10. Issues: Data Preparation Data cleaning Relevance analysis (feature selection) Preprocess data in order to reduce noise and handle missing values Remove the irrelevant or redundant attributes Data transformation Generalize and/or normalize data January 20, 2014 Data Mining: Concepts and Techniques 10
11. 11. Issues: Evaluating Classification Methods Accuracy classifier accuracy: predicting class label predictor accuracy: guessing value of predicted attributes Speed time to construct the model (training time) time to use the model (classification/prediction time) Robustness: handling noise and missing values Scalability: efficiency in disk-resident databases Interpretability understanding and insight provided by the model Other measures, e.g., goodness of rules, such as decision tree size or compactness of classification rules January 20, 2014 Data Mining: Concepts and Techniques 11
12. 12. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 12
13. 13. Decision Tree Induction: Training Dataset This follows an example of Quinlan’s ID3 (Playing Tennis) age <=30 <=30 31…40 >40 >40 >40 31…40 <=30 <=30 >40 <=30 31…40 31…40 >40 January 20, 2014 income student credit_rating high no fair high no excellent high no fair medium no fair low yes fair low yes excellent low yes excellent medium no fair low yes fair medium yes fair medium yes excellent medium no excellent high yes fair medium no excellent buys_computer no no yes yes yes no yes no yes yes yes yes yes no Data Mining: Concepts and Techniques 13
14. 14. Output: A Decision Tree for “ buys_computer” age? <=30 31..40 overcast student? no no January 20, 2014 >40 credit rating? yes yes yes excellent no fair yes Data Mining: Concepts and Techniques 14
15. 15. Algorithm for Decision Tree Induction Basic algorithm (a greedy algorithm) Tree is constructed in a top-down recursive divide-and-conquer manner At start, all the training examples are at the root Attributes are categorical (if continuous-valued, they are discretized in advance) Examples are partitioned recursively based on selected attributes Test attributes are selected on the basis of a heuristic or statistical measure (e.g., information gain) Conditions for stopping partitioning All samples for a given node belong to the same class There are no remaining attributes for further partitioning – majority voting is employed for classifying the leaf There are no samples left January 20, 2014 Data Mining: Concepts and Techniques 15
16. 16. Attribute Selection Measure: Information Gain (ID3/C4.5) Select the attribute with the highest information gain Let pi be the probability that an arbitrary tuple in D belongs to class Ci, estimated by |Ci, D|/|D| Expected information (entropy) needed to classify a tuple m in D: Info( D) = − p log ( p ) ∑ i =1 i 2 i Information needed (after using A to split D into v v |D | j partitions) to classify D: Info A ( D) = ∑ × I (D j ) j =1 | D | Information gained by branching on attribute A Gain(A) = Info(D) − Info A(D) January 20, 2014 Data Mining: Concepts and Techniques 16
17. 17. Attribute Selection: Information Gain Class P: buys_computer = “yes” Class N: buys_computer = “no” Info( D) = I (9,5) = − age <=30 31…40 >40 age <=30 <=30 31…40 >40 >40 >40 31…40 <=30 <=30 >40 <=30 31…40 31…40 >40 Infoage ( D) = 9 9 5 5 log 2 ( ) − log 2 ( ) =0.940 14 14 14 14 pi 2 4 3 ni I(pi, ni) 3 0.971 0 0 2 0.971 income student credit_rating high no fair high no excellent high no fair medium no fair low yes fair low yes excellent low yes excellent medium no fair low yes fair medium yes fair medium yes excellent medium no excellent high yes fair medium no excellent January 20, 2014 buys_computer no no yes yes yes no yes no yes yes yes yes yes no + 5 4 I (2,3) + I (4,0) 14 14 5 I (3,2) = 0.694 14 5 I (2,3) means “age <=30” has 5 14 out of 14 samples, with 2 yes’es and 3 no’s. Hence Gain(age) = Info( D) − Infoage ( D) = 0.246 Similarly, Gain(income) = 0.029 Gain( student ) = 0.151 Gain(credit _ rating ) = 0.048 Data Mining: Concepts and Techniques 17
18. 18. Computing Information-Gain for Continuous-Value Attributes Let attribute A be a continuous-valued attribute Must determine the best split point for A Sort the value A in increasing order Typically, the midpoint between each pair of adjacent values is considered as a possible split point (ai+ai+1)/2 is the midpoint between the values of ai and ai+1 The point with the minimum expected information requirement for A is selected as the split-point for A Split: D1 is the set of tuples in D satisfying A ≤ split-point, and D2 is the set of tuples in D satisfying A > split-point January 20, 2014 Data Mining: Concepts and Techniques 18
19. 19. Gain Ratio for Attribute Selection (C4.5) Information gain measure is biased towards attributes with a large number of values C4.5 (a successor of ID3) uses gain ratio to overcome the problem (normalization to information gain) v SplitInfo A ( D) = −∑ j =1 |D| × log 2 ( | Dj | |D| ) GainRatio(A) = Gain(A)/SplitInfo(A) Ex. | Dj | SplitInfo A ( D ) = − 4 4 6 6 4 4 × log 2 ( ) − × log 2 ( ) − × log 2 ( ) = 0.926 14 14 14 14 14 14 gain_ratio(income) = 0.029/0.926 = 0.031 The attribute with the maximum gain ratio is selected as the splitting attribute January 20, 2014 Data Mining: Concepts and Techniques 19
20. 20. Gini index (CART, IBM IntelligentMiner) If a data set D contains examples from n classes, gini index, gini(D) is defined as n gini( D) =1− ∑ p 2 j j =1 where pj is the relative frequency of class j in D If a data set D is split on A into two subsets D1 and D2, the gini index gini(D) is defined as |D | |D | gini A ( D) = Reduction in Impurity: 1 |D| gini( D1) + 2 |D| gini( D 2) ∆gini( A) = gini(D) − giniA ( D) The attribute provides the smallest ginisplit(D) (or the largest reduction in impurity) is chosen to split the node (need to enumerate all the possible splitting points for each attribute) January 20, 2014 Data Mining: Concepts and Techniques 20
21. 21. Gini index (CART, IBM IntelligentMiner) Ex. D has 9 tuples in buys_computer = “yes” and 5 in “no” 2 2 9 5 gini ( D) = 1 − − = 0.459 14 14 Suppose the attribute income partitions D into 10 in D1: {low, medium} 10 4 giniincome∈{low,medium} ( D ) = Gini ( D1 ) + Gini ( D1 ) and 4 in D2 14 14 but gini{medium,high} is 0.30 and thus the best since it is the lowest All attributes are assumed continuous-valued May need other tools, e.g., clustering, to get the possible split values Can be modified for categorical attributes January 20, 2014 Data Mining: Concepts and Techniques 21
22. 22. Comparing Attribute Selection Measures The three measures, in general, return good results but Information gain: Gain ratio: biased towards multivalued attributes tends to prefer unbalanced splits in which one partition is much smaller than the others Gini index: biased to multivalued attributes has difficulty when # of classes is large tends to favor tests that result in equal-sized partitions and purity in both partitions January 20, 2014 Data Mining: Concepts and Techniques 22
23. 23. Other Attribute Selection Measures CHAID: a popular decision tree algorithm, measure based on χ2 test for independence C-SEP: performs better than info. gain and gini index in certain cases G-statistics: has a close approximation to χ2 distribution MDL (Minimal Description Length) principle (i.e., the simplest solution is preferred): Multivariate splits (partition based on multiple variable combinations) The best tree as the one that requires the fewest # of bits to both (1) encode the tree, and (2) encode the exceptions to the tree CART: finds multivariate splits based on a linear comb. of attrs. Which attribute selection measure is the best? Most give good results, none is significantly superior than others January 20, 2014 Data Mining: Concepts and Techniques 23
24. 24. Overfitting and Tree Pruning Overfitting: An induced tree may overfit the training data Too many branches, some may reflect anomalies due to noise or outliers Poor accuracy for unseen samples Two approaches to avoid overfitting Prepruning: Halt tree construction early—do not split a node if this would result in the goodness measure falling below a threshold Difficult to choose an appropriate threshold Postpruning: Remove branches from a “fully grown” tree—get a sequence of progressively pruned trees Use a set of data different from the training data to decide which is the “best pruned tree” January 20, 2014 Data Mining: Concepts and Techniques 24
25. 25. Enhancements to Basic Decision Tree Induction Allow for continuous-valued attributes Dynamically define new discrete-valued attributes that partition the continuous attribute value into a discrete set of intervals Handle missing attribute values Assign the most common value of the attribute Assign probability to each of the possible values Attribute construction Create new attributes based on existing ones that are sparsely represented This reduces fragmentation, repetition, and replication January 20, 2014 Data Mining: Concepts and Techniques 25
26. 26. Classification in Large Databases Classification—a classical problem extensively studied by statisticians and machine learning researchers Scalability: Classifying data sets with millions of examples and hundreds of attributes with reasonable speed Why decision tree induction in data mining? relatively faster learning speed (than other classification methods) convertible to simple and easy to understand classification rules can use SQL queries for accessing databases comparable classification accuracy with other methods January 20, 2014 Data Mining: Concepts and Techniques 26
27. 27. Scalable Decision Tree Induction Methods SLIQ (EDBT’96 — Mehta et al.) Builds an index for each attribute and only class list and the current attribute list reside in memory SPRINT (VLDB’96 — J. Shafer et al.) Constructs an attribute list data structure PUBLIC (VLDB’98 — Rastogi & Shim) Integrates tree splitting and tree pruning: stop growing the tree earlier RainForest (VLDB’98 — Gehrke, Ramakrishnan & Ganti) Builds an AVC-list (attribute, value, class label) BOAT (PODS’99 — Gehrke, Ganti, Ramakrishnan & Loh) Uses bootstrapping to create several small samples January 20, 2014 Data Mining: Concepts and Techniques 27
28. 28. Scalability Framework for RainForest Separates the scalability aspects from the criteria that determine the quality of the tree Builds an AVC-list: AVC (Attribute, Value, Class_label) AVC-set (of an attribute X ) Projection of training dataset onto the attribute X and class label where counts of individual class label are aggregated AVC-group (of a node n ) Set of AVC-sets of all predictor attributes at the node n January 20, 2014 Data Mining: Concepts and Techniques 28
29. 29. Rainforest: Training Set and Its AVC Sets Training Examples age <=30 <=30 31…40 >40 >40 >40 31…40 <=30 <=30 >40 <=30 31…40 31…40 >40 AVC-set on Age income studentcredit_rating buys_computerAge Buy_Computer high no fair no yes no high no excellent no <=30 3 2 high no fair yes 31..40 4 0 medium no fair yes >40 3 2 low yes fair yes low yes excellent no low yes excellent yes AVC-set on Student medium no fair no low yes fair yes student Buy_Computer medium yes fair yes yes no medium yes excellent yes medium no excellent yes yes 6 1 high yes fair yes no 3 4 medium no excellent no January 20, 2014 AVC-set on income income Buy_Computer yes no high 2 2 medium 4 2 low 3 1 AVC-set on credit_rating Buy_Computer Credit rating yes no fair 6 2 excellent 3 3 Data Mining: Concepts and Techniques 29
30. 30. Data Cube-Based Decision-Tree Induction Integration of generalization with decision-tree induction (Kamber et al.’97) Classification at primitive concept levels E.g., precise temperature, humidity, outlook, etc. Low-level concepts, scattered classes, bushy classification-trees Semantic interpretation problems Cube-based multi-level classification Relevance analysis at multi-levels Information-gain analysis with dimension + level January 20, 2014 Data Mining: Concepts and Techniques 30
31. 31. BOAT (Bootstrapped Optimistic Algorithm for Tree Construction) Use a statistical technique called bootstrapping to create several smaller samples (subsets), each fits in memory Each subset is used to create a tree, resulting in several trees These trees are examined and used to construct a new tree T’ It turns out that T’ is very close to the tree that would be generated using the whole data set together Adv: requires only two scans of DB, an incremental alg. January 20, 2014 Data Mining: Concepts and Techniques 31
32. 32. Presentation of Classification Results January 20, 2014 Data Mining: Concepts and Techniques 32
33. 33. Visualization of a Decision Tree in SGI/MineSet 3.0 January 20, 2014 Data Mining: Concepts and Techniques 33
34. 34. Interactive Visual Mining by PerceptionBased Classification (PBC ) January 20, 2014 Data Mining: Concepts and Techniques 34
35. 35. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 35
36. 36. Bayesian Classification: Why? A statistical classifier: performs probabilistic prediction, i.e., predicts class membership probabilities Foundation: Based on Bayes’ Theorem. Performance: A simple Bayesian classifier, naïve Bayesian classifier, has comparable performance with decision tree and selected neural network classifiers Incremental: Each training example can incrementally increase/decrease the probability that a hypothesis is correct — prior knowledge can be combined with observed data Standard: Even when Bayesian methods are computationally intractable, they can provide a standard of optimal decision making against which other methods can be measured January 20, 2014 Data Mining: Concepts and Techniques 36
37. 37. Bayesian Theorem: Basics Let X be a data sample (“evidence”): class label is unknown Let H be a hypothesis that X belongs to class C Classification is to determine P(H|X), the probability that the hypothesis holds given the observed data sample X P(H) (prior probability), the initial probability E.g., X will buy computer, regardless of age, income, … P(X): probability that sample data is observed P(X|H) (posteriori probability), the probability of observing the sample X, given that the hypothesis holds E.g., Given that X will buy computer, the prob. that X is 31..40, medium income January 20, 2014 Data Mining: Concepts and Techniques 37
38. 38. Bayesian Theorem Given training data X , posteriori probability of a hypothesis H, P(H|X), follows the Bayes theorem P(H | X) = P(X | H )P(H ) P(X) Informally, this can be written as posteriori = likelihood x prior/evidence Predicts X belongs to C2 iff the probability P(Ci|X) is the highest among all the P(Ck|X) for all the k classes Practical difficulty: require initial knowledge of many probabilities, significant computational cost January 20, 2014 Data Mining: Concepts and Techniques 38
39. 39. Towards Naïve Bayesian Classifier Let D be a training set of tuples and their associated class labels, and each tuple is represented by an n-D attribute vector X = (x1, x2, …, xn) Suppose there are m classes C1, C2, …, Cm. Classification is to derive the maximum posteriori, i.e., the maximal P(Ci|X) This can be derived from Bayes’ theorem P(X | C )P(C ) i i P(C | X) = i P(X) Since P(X) is constant for all classes, only P(C | X) = P(X | C )P(C ) i i i needs to be maximized January 20, 2014 Data Mining: Concepts and Techniques 39
40. 40. Derivation of Naïve Bayes Classifier A simplified assumption: attributes are conditionally independent (i.e., no dependence relation between attributes): n P ( X | C i ) = ∏ P ( x | C i ) = P ( x | C i ) × P ( x | C i ) × ... × P( x | C i ) k 1 2 n k =1 This greatly reduces the computation cost: Only counts the class distribution If Ak is categorical, P(xk|Ci) is the # of tuples in Ci having value xk for Ak divided by |Ci, D| (# of tuples of Ci in D) If Ak is continous-valued, P(xk|Ci) is usually computed based on Gaussian distribution with a mean μ and ( x −µ ) − standard deviation σ 1 2σ 2 g ( x, µ, σ ) = and P(xk|Ci) is January 20, 2014 2πσ e 2 P ( X | C i ) = g ( xk , µ C i , σ C i ) Data Mining: Concepts and Techniques 40
41. 41. Naïve Bayesian Classifier: Training Dataset Class: C1:buys_computer = ‘yes’ C2:buys_computer = ‘no’ Data sample X = (age <=30, Income = medium, Student = yes Credit_rating = Fair) January 20, 2014 age <=30 <=30 31…40 >40 >40 >40 31…40 <=30 <=30 >40 <=30 31…40 31…40 >40 income student redit_rating c buys_compu high no fair no high no excellent no high no fair yes medium no fair yes low yes fair yes low yes excellent no low yes excellent yes medium no fair no low yes fair yes medium yes fair yes medium yes excellent yes medium no excellent yes high yes fair yes medium no excellent no Data Mining: Concepts and Techniques 41
43. 43. Avoiding the 0-Probability Problem Naïve Bayesian prediction requires each conditional prob. be nonzero. Otherwise, the predicted prob. will be zero n P ( X | C i ) = ∏P ( x k | C i ) k =1 Ex. Suppose a dataset with 1000 tuples, income=low (0), income= medium (990), and income = high (10), Use Laplacian correction (or Laplacian estimator) Adding 1 to each case Prob(income = low) = 1/1003 Prob(income = medium) = 991/1003 Prob(income = high) = 11/1003 The “corrected” prob. estimates are close to their “uncorrected” counterparts January 20, 2014 Data Mining: Concepts and Techniques 43
44. 44. Naïve Bayesian Classifier: Comments Advantages Easy to implement Good results obtained in most of the cases Disadvantages Assumption: class conditional independence, therefore loss of accuracy Practically, dependencies exist among variables E.g., hospitals: patients: Profile: age, family history, etc. Symptoms: fever, cough etc., Disease: lung cancer, diabetes, etc. Dependencies among these cannot be modeled by Naïve Bayesian Classifier How to deal with these dependencies? Bayesian Belief Networks January 20, 2014 Data Mining: Concepts and Techniques 44
45. 45. Bayesian Belief Networks Bayesian belief network allows a subset of the variables conditionally independent A graphical model of causal relationships Represents dependency among the variables Gives a specification of joint probability distribution Nodes: random variables Links: dependency Y X Z January 20, 2014 X and Y are the parents of Z, and Y is P the parent of P No dependency between Z and P Has no loops or cycles Data Mining: Concepts and Techniques 45
46. 46. Bayesian Belief Network: An Example Family History Smoker The conditional probability table (CPT) for variable LungCancer: (FH, S) (FH, ~S) (~FH, S) (~FH, ~S) LC LungCancer Emphysema 0.8 0.5 0.7 0.1 ~LC 0.2 0.5 0.3 0.9 CPT shows the conditional probability for each possible combination of its parents PositiveXRay Dyspnea Bayesian Belief Networks January 20, 2014 Derivation of the probability of a particular combination of values of X, from CPT: n P ( x1 ,..., xn ) = ∏ P ( x i | Parents (Y i )) i =1 Data Mining: Concepts and Techniques 46
47. 47. Training Bayesian Networks Several scenarios: Given both the network structure and all variables observable: learn only the CPTs Network structure known, some hidden variables: gradient descent (greedy hill-climbing) method, analogous to neural network learning Network structure unknown, all variables observable: search through the model space to reconstruct network topology Unknown structure, all hidden variables: No good algorithms known for this purpose Ref. D. Heckerman: Bayesian networks for data mining January 20, 2014 Data Mining: Concepts and Techniques 47
48. 48. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 48
49. 49. Using IF-THEN Rules for Classification Represent the knowledge in the form of IF-THEN rules R: IF age = youth AND student = yes THEN buys_computer = yes Rule antecedent/precondition vs. rule consequent Assessment of a rule: coverage and accuracy ncovers = # of tuples covered by R ncorrect = # of tuples correctly classified by R coverage(R) = ncovers /|D| /* D: training data set */ accuracy(R) = ncorrect / ncovers If more than one rule is triggered, need conflict resolution Size ordering: assign the highest priority to the triggering rules that has the “toughest” requirement (i.e., with the most attribute test) Class-based ordering: decreasing order of prevalence or misclassification cost per class Rule-based ordering (decision list): rules are organized into one long priority list, according to some measure Concepts and by experts January 20, 2014 Data Mining: of rule quality or Techniques 49
50. 50. Rule Extraction from a Decision Tree age? <=30 Rules are easier to understand than large trees student? One rule is created for each path from the root to no a leaf Each attribute-value pair along a path forms a conjunction: the leaf holds the class prediction 31..40 no >40 credit rating? yes yes excellent yes Rules are mutually exclusive and exhaustive no Example: Rule extraction from our buys_computer decision-tree IF age = young AND student = no THEN buys_computer = no IF age = young AND student = yes THEN buys_computer = yes IF age = mid-age THEN buys_computer = yes IF age = old AND credit_rating = excellent THEN buys_computer = yes IF age = young AND credit_rating = fair January 20, 2014 THEN buys_computer = no Data Mining: Concepts and Techniques 50 fair yes
51. 51. Rule Extraction from the Training Data Sequential covering algorithm: Extracts rules directly from training data Typical sequential covering algorithms: FOIL, AQ, CN2, RIPPER Rules are learned sequentially, each for a given class Ci will cover many tuples of Ci but none (or few) of the tuples of other classes Steps: Rules are learned one at a time Each time a rule is learned, the tuples covered by the rules are removed The process repeats on the remaining tuples unless termination condition, e.g., when no more training examples or when the quality of a rule returned is below a user-specified threshold Comp. w. decision-tree induction: learning a set of rules simultaneously January 20, 2014 Data Mining: Concepts and Techniques 51
52. 52. How to Learn-One-Rule? Star with the most general rule possible: condition = empty Adding new attributes by adopting a greedy depth-first strategy Picks the one that most improves the rule quality Rule-Quality measures: consider both coverage and accuracy Foil-gain (in FOIL & RIPPER): assesses info_gain by extending condition pos ' pos FOIL _ Gain = pos'×(log 2 − log 2 ) pos'+ neg ' pos + neg It favors rules that have high accuracy and cover many positive tuples Rule pruning based on an independent set of test tuples pos − neg FOIL _ Prune( R) = pos + neg Pos/neg are # of positive/negative tuples covered by R. If FOIL_Prune is higher for the pruned version of R, prune R January 20, 2014 Data Mining: Concepts and Techniques 52
53. 53. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 53
54. 54. Classification: A Mathematical Mapping Classification: predicts categorical class labels E.g., Personal homepage classification x = (x , x , x , …), y = +1 or –1 i 1 2 3 i x1 : # of a word “homepage” x2 : # of a word “welcome” Mathematically x ∈ X = ℜn, y ∈ Y = {+1, –1} We want a function f: X Y January 20, 2014 Data Mining: Concepts and Techniques 54
55. 55. Linear Classification x x x x x x x x x ooo o o o o January 20, 2014 x o o o o o o Binary Classification problem The data above the red line belongs to class ‘x’ The data below red line belongs to class ‘o’ Examples: SVM, Perceptron, Probabilistic Classifiers Data Mining: Concepts and Techniques 55
56. 56. Discriminative Classifiers Advantages prediction accuracy is generally high robust, works when training examples contain errors fast evaluation of the learned target function As compared to Bayesian methods – in general Bayesian networks are normally slow Criticism long training time difficult to understand the learned function (weights) Bayesian networks can be used easily for pattern discovery not easy to incorporate domain knowledge Easy in the form of priors on the data or distributions January 20, 2014 Data Mining: Concepts and Techniques 56
57. 57. Perceptron & Winnow • Vector: x, w x2 • Scalar: x, y, w Input: {(x1, y1), …} Output: classification function f(x) f(xi) > 0 for yi = +1 f(xi) < 0 for yi = -1 f(x) => wx + b = 0 or w1x1+w2x2+b = 0 • Perceptron: update W additively x1 January 20, 2014 • Winnow: update W multiplicatively Data Mining: Concepts and Techniques 57
58. 58. Classification by Backpropagation Backpropagation: A neural network learning algorithm Started by psychologists and neurobiologists to develop and test computational analogues of neurons A neural network: A set of connected input/output units where each connection has a weight associated with it During the learning phase, the network learns by adjusting the weights so as to be able to predict the correct class label of the input tuples Also referred to as connectionist learning due to the connections between units January 20, 2014 Data Mining: Concepts and Techniques 58
59. 59. Neural Network as a Classifier Weakness Long training time Require a number of parameters typically best determined empirically, e.g., the network topology or ``structure." Poor interpretability: Difficult to interpret the symbolic meaning behind the learned weights and of ``hidden units" in the network Strength High tolerance to noisy data Ability to classify untrained patterns Well-suited for continuous-valued inputs and outputs Successful on a wide array of real-world data Algorithms are inherently parallel Techniques have recently been developed for the extraction of rules from trained neural networks January 20, 2014 Data Mining: Concepts and Techniques 59
60. 60. A Neuron (= a perceptron) x0 w0 x1 w1 - µk xn ∑ f wn output y For Example Input weight vector x vector w weighted sum Activation function n y = sign(∑ wi xi + µ k ) i =0 The n-dimensional input vector x is mapped into variable y by means of the scalar product and a nonlinear function mapping January 20, 2014 Data Mining: Concepts and Techniques 60
61. 61. A Multi-Layer Feed-Forward Neural Network Output vector Output layer Err j = O j (1 − O j )∑ Errk w jk k θ j = θ j + (l) Err j wij = wij + (l ) Err j Oi Hidden layer Err j = O j (1 − O j )(T j − O j ) wij Input layer Oj = 1 −I j 1+ e I j = ∑ wij Oi + θ j i Input vector: X January 20, 2014 Data Mining: Concepts and Techniques 61
62. 62. How A Multi-Layer Neural Network Works? The inputs to the network correspond to the attributes measured for each training tuple Inputs are fed simultaneously into the units making up the input layer They are then weighted and fed simultaneously to a hidden layer The number of hidden layers is arbitrary, although usually only one The weighted outputs of the last hidden layer are input to units making up the output layer, which emits the network's prediction The network is feed-forward in that none of the weights cycles back to an input unit or to an output unit of a previous layer From a statistical point of view, networks perform nonlinear regression: Given enough hidden units and enough training samples, they can closely approximate any function January 20, 2014 Data Mining: Concepts and Techniques 62
63. 63. Defining a Network Topology First decide the network topology: # of units in the input layer, # of hidden layers (if > 1), # of units in each hidden layer, and # of units in the output layer Normalizing the input values for each attribute measured in the training tuples to [0.0—1.0] One input unit per domain value, each initialized to 0 Output, if for classification and more than two classes, one output unit per class is used Once a network has been trained and its accuracy is unacceptable, repeat the training process with a different network topology or a different set of initial weights January 20, 2014 Data Mining: Concepts and Techniques 63
64. 64. Backpropagation Iteratively process a set of training tuples & compare the network's prediction with the actual known target value For each training tuple, the weights are modified to minimize the mean squared error between the network's prediction and the actual target value Modifications are made in the “backwards” direction: from the output layer, through each hidden layer down to the first hidden layer, hence “backpropagation” Steps Initialize weights (to small random #s) and biases in the network Propagate the inputs forward (by applying activation function) Backpropagate the error (by updating weights and biases) Terminating condition (when error is very small, etc.) January 20, 2014 Data Mining: Concepts and Techniques 64
65. 65. Backpropagation and Interpretability Efficiency of backpropagation: Each epoch (one interation through the training set) takes O(|D| * w), with |D| tuples and w weights, but # of epochs can be exponential to n, the number of inputs, in the worst case Rule extraction from networks: network pruning Simplify the network structure by removing weighted links that have the least effect on the trained network Then perform link, unit, or activation value clustering The set of input and activation values are studied to derive rules describing the relationship between the input and hidden unit layers Sensitivity analysis: assess the impact that a given input variable has on a network output. The knowledge gained from this analysis can be represented in rules January 20, 2014 Data Mining: Concepts and Techniques 65
66. 66. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 66
67. 67. SVM—Support Vector Machines A new classification method for both linear and nonlinear data It uses a nonlinear mapping to transform the original training data into a higher dimension With the new dimension, it searches for the linear optimal separating hyperplane (i.e., “decision boundary”) With an appropriate nonlinear mapping to a sufficiently high dimension, data from two classes can always be separated by a hyperplane SVM finds this hyperplane using support vectors (“essential” training tuples) and margins (defined by the support vectors) January 20, 2014 Data Mining: Concepts and Techniques 67
68. 68. SVM—History and Applications Vapnik and colleagues (1992)—groundwork from Vapnik & Chervonenkis’ statistical learning theory in 1960s Features: training can be slow but accuracy is high owing to their ability to model complex nonlinear decision boundaries (margin maximization) Used both for classification and prediction Applications: handwritten digit recognition, object recognition, speaker identification, benchmarking time-series prediction tests January 20, 2014 Data Mining: Concepts and Techniques 68
69. 69. SVM—General Philosophy Small Margin Large Margin Support Vectors January 20, 2014 Data Mining: Concepts and Techniques 69
70. 70. SVM—Margins and Support Vectors January 20, 2014 Data Mining: Concepts and Techniques 70
71. 71. SVM—When Data Is Linearly Separable m Let data D be (X 1, y1), …, (X |D|, y|D|), where X i is the set of training tuples associated with the class labels yi There are infinite lines (hyperplanes) separating the two classes but we want to find the best one (the one that minimizes classification error on unseen data) SVM searches for the hyperplane with the largest margin, i.e., maximum marginal hyperplane (MMH) January 20, 2014 Data Mining: Concepts and Techniques 71
72. 72. SVM—Linearly Separable A separating hyperplane can be written as W ●X+b=0 where W={w1, w2, …, wn} is a weight vector and b a scalar (bias) For 2-D it can be written as w0 + w1 x1 + w2 x2 = 0 The hyperplane defining the sides of the margin: H1: w0 + w1 x1 + w2 x2 ≥ 1 for yi = +1, and H2: w0 + w1 x1 + w2 x2 ≤ – 1 for yi = –1 Any training tuples that fall on hyperplanes H1 or H2 (i.e., the sides defining the margin) are support vectors This becomes a constrained (convex) quadratic optimization problem: Quadratic objective function and linear constraints Quadratic Programming (QP) Lagrangian multipliers January 20, 2014 Data Mining: Concepts and Techniques 72
73. 73. Why Is SVM Effective on High Dimensional Data? The complexity of trained classifier is characterized by the # of support vectors rather than the dimensionality of the data The support vectors are the essential or critical training examples — they lie closest to the decision boundary (MMH) If all other training examples are removed and the training is repeated, the same separating hyperplane would be found The number of support vectors found can be used to compute an (upper) bound on the expected error rate of the SVM classifier, which is independent of the data dimensionality Thus, an SVM with a small number of support vectors can have good generalization, even when the dimensionality of the data is high January 20, 2014 Data Mining: Concepts and Techniques 73
74. 74. A 2 SVM—Linearly Inseparable Transform the original input data into a higher dimensional space Search for a linear separating hyperplane in the new space January 20, 2014 Data Mining: Concepts and Techniques 74 A 1
75. 75. SVM—Kernel functions Instead of computing the dot product on the transformed data tuples, it is mathematically equivalent to instead applying a kernel function K(X i , X j ) to the original data, i.e., K(X i , X j ) = Φ(X i ) Φ(X j ) Typical Kernel Functions SVM can also be used for classifying multiple (> 2) classes and for regression analysis (with additional user parameters) January 20, 2014 Data Mining: Concepts and Techniques 75
76. 76. Scaling SVM by Hierarchical Micro-Clustering SVM is not scalable to the number of data objects in terms of training time and memory usage “Classifying Large Datasets Using SVMs with Hierarchical Clusters Problem” by Hwanjo Yu, Jiong Yang, Jiawei Han, KDD’03 CB-SVM (Clustering-Based SVM) Given limited amount of system resources (e.g., memory), maximize the SVM performance in terms of accuracy and the training speed Use micro-clustering to effectively reduce the number of points to be considered At deriving support vectors, de-cluster micro-clusters near “candidate vector” to ensure high classification accuracy January 20, 2014 Data Mining: Concepts and Techniques 76
77. 77. CB-SVM: Clustering-Based SVM Training data sets may not even fit in memory Read the data set once (minimizing disk access) Construct a statistical summary of the data (i.e., hierarchical clusters) given a limited amount of memory The statistical summary maximizes the benefit of learning SVM The summary plays a role in indexing SVMs Essence of Micro-clustering (Hierarchical indexing structure) Use micro-cluster hierarchical indexing structure provide finer samples closer to the boundary and coarser samples farther from the boundary Selective de-clustering to ensure high accuracy January 20, 2014 Data Mining: Concepts and Techniques 77
78. 78. CF-Tree: Hierarchical Micro-cluster January 20, 2014 Data Mining: Concepts and Techniques 78
79. 79. CB-SVM Algorithm: Outline Construct two CF-trees from positive and negative data sets independently Need one scan of the data set Train an SVM from the centroids of the root entries De-cluster the entries near the boundary into the next level The children entries de-clustered from the parent entries are accumulated into the training set with the non-declustered parent entries Train an SVM again from the centroids of the entries in the training set Repeat until nothing is accumulated January 20, 2014 Data Mining: Concepts and Techniques 79
80. 80. Selective Declustering CF tree is a suitable base structure for selective declustering De-cluster only the cluster Ei such that Di – Ri < Ds, where Di is the distance from the boundary to the center point of Ei and Ri is the radius of Ei Decluster only the cluster whose subclusters have possibilities to be the support cluster of the boundary “Support cluster”: The cluster whose centroid is a support vector January 20, 2014 Data Mining: Concepts and Techniques 80
81. 81. Experiment on Synthetic Dataset January 20, 2014 Data Mining: Concepts and Techniques 81
82. 82. Experiment on a Large Data Set January 20, 2014 Data Mining: Concepts and Techniques 82
83. 83. SVM vs. Neural Network SVM Neural Network Relatively old Relatively new concept Nondeterministic Deterministic algorithm algorithm Nice Generalization Generalizes well but properties doesn’t have strong mathematical foundation Hard to learn – learned Can easily be learned in in batch mode using incremental fashion quadratic programming To learn complex techniques functions—use multilayer Using kernels can learn perceptron (not that trivial) very complex functions January 20, 2014 Data Mining: Concepts and Techniques 83
84. 84. SVM Related Links SVM Website http://www.kernel-machines.org/ Representative implementations LIBSVM: an efficient implementation of SVM, multi-class classifications, nu-SVM, one-class SVM, including also various interfaces with java, python, etc. SVM-light: simpler but performance is not better than LIBSVM, support only binary classification and only C language SVM-torch: another recent implementation also written in C. January 20, 2014 Data Mining: Concepts and Techniques 84
85. 85. SVM —Introduction Literature “Statistical Learning Theory” by Vapnik: extremely hard to understand, containing many errors too. C. J. C. Burges. A Tutorial on Support Vector Machines for Pattern Recognition. Knowledge Discovery and Data Mining, 2(2), 1998. Better than the Vapnik’s book, but still written too hard for introduction, and the examples are so not-intuitive The book “An Introduction to Support Vector Machines” by N. Cristianini and J. Shawe-Taylor Also written hard for introduction, but the explanation about the mercer’s theorem is better than above literatures The neural network book by Haykins January 20, 2014 Data Mining: Concepts and Techniques 85
86. 86. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 86
87. 87. Associative Classification Associative classification Association rules are generated and analyzed for use in classification Search for strong associations between frequent patterns (conjunctions of attribute-value pairs) and class labels Classification: Based on evaluating a set of rules in the form of P1 ^ p2 … ^ pl “Aclass = C” (conf, sup) Why effective? It explores highly confident associations among multiple attributes and may overcome some constraints introduced by decision-tree induction, which considers only one attribute at a time In many studies, associative classification has been found to be more accurate than some traditional classification methods, such as C4.5 January 20, 2014 Data Mining: Concepts and Techniques 87
88. 88. Typical Associative Classification Methods CBA (Classification By Association: Liu, Hsu & Ma, KDD’98) Mine association possible rules in the form of Build classifier: Organize rules according to decreasing precedence based on confidence and then support CMAR (Classification based on Multiple Association Rules: Li, Han, Pei, ICDM’01) Cond-set (a set of attribute-value pairs) class label Classification: Statistical analysis on multiple rules CPAR (Classification based on Predictive Association Rules: Yin & Han, SDM’03) Generation of predictive rules (FOIL-like analysis) High efficiency, accuracy similar to CMAR RCBT (Mining top-k covering rule groups for gene expression data, Cong et al. SIGMOD’05) Explore high-dimensional classification, using top-k rule groups Achieve high January 20, 2014 classification Mining: Conceptsrun-time efficiency Data accuracy and high and Techniques 88
89. 89. A Closer Look at CMAR CMAR (Classification based on Multiple Association Rules: Li, Han, Pei, ICDM’01) Efficiency: Uses an enhanced FP-tree that maintains the distribution of class labels among tuples satisfying each frequent itemset Rule pruning whenever a rule is inserted into the tree Given two rules, R and R , if the antecedent of R is more general 1 2 1 than that of R2 and conf(R1) ≥ conf(R2), then R2 is pruned Prunes rules for which the rule antecedent and class are not positively correlated, based on a χ2 test of statistical significance Classification based on generated/pruned rules If only one rule satisfies tuple X, assign the class label of the rule If a rule set S satisfies X, CMAR divides S into groups according to class labels uses a weighted χ2 measure to find the strongest group of rules, based on the statistical correlation of rules within a group assigns X January 20, 2014 the class label of the strongest group Techniques Data Mining: Concepts and 89
90. 90. Associative Classification May Achieve High Accuracy and Efficiency (Cong et al. SIGMOD05) January 20, 2014 Data Mining: Concepts and Techniques 90
91. 91. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 91
92. 92. Lazy vs. Eager Learning Lazy vs. eager learning Lazy learning (e.g., instance-based learning): Simply stores training data (or only minor processing) and waits until it is given a test tuple Eager learning (the above discussed methods): Given a set of training set, constructs a classification model before receiving new (e.g., test) data to classify Lazy: less time in training but more time in predicting Accuracy Lazy method effectively uses a richer hypothesis space since it uses many local linear functions to form its implicit global approximation to the target function Eager: must commit to a single hypothesis that covers the entire instance space January 20, 2014 Data Mining: Concepts and Techniques 92
93. 93. Lazy Learner: Instance-Based Methods Instance-based learning: Store training examples and delay the processing (“lazy evaluation”) until a new instance must be classified Typical approaches k-nearest neighbor approach Instances represented as points in a Euclidean space. Locally weighted regression Constructs local approximation Case-based reasoning Uses symbolic representations and knowledgebased inference January 20, 2014 Data Mining: Concepts and Techniques 93
94. 94. The k -Nearest Neighbor Algorithm All instances correspond to points in the n-D space The nearest neighbor are defined in terms of Euclidean distance, dist(X 1 , X 2 ) Target function could be discrete- or real- valued For discrete-valued, k-NN returns the most common value among the k training examples nearest to xq Vonoroi diagram: the decision surface induced by 1NN for a typical set of training examples _ _ _ + _ _ . + January 20, 2014 + xq . _ + . . . . Data Mining: Concepts and Techniques 94
95. 95. Discussion on the k -NN Algorithm k-NN for real-valued prediction for a given unknown tuple Returns the mean values of the k nearest neighbors Distance-weighted nearest neighbor algorithm Weight the contribution of each of the k neighbors according to their distance to the query xq w≡ 1 Give greater weight to closer neighbors d ( xq , x )2 i Robust to noisy data by averaging k-nearest neighbors Curse of dimensionality: distance between neighbors could be dominated by irrelevant attributes To overcome it, axes stretch or elimination of the least relevant attributes January 20, 2014 Data Mining: Concepts and Techniques 95
96. 96. Case-Based Reasoning (CBR) CBR: Uses a database of problem solutions to solve new problems Store symbolic description (tuples or cases)—not points in a Euclidean space Applications: Customer-service (product-related diagnosis), legal ruling Methodology Instances represented by rich symbolic descriptions (e.g., function graphs) Search for similar cases, multiple retrieved cases may be combined Tight coupling between case retrieval, knowledge-based reasoning, and problem solving Challenges Find a good similarity metric Indexing based on syntactic similarity measure, and when failure, backtracking, and adapting to additional cases January 20, 2014 Data Mining: Concepts and Techniques 96
97. 97. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 97
98. 98. Genetic Algorithms (GA) Genetic Algorithm: based on an analogy to biological evolution An initial population is created consisting of randomly generated rules E.g., if A1 and ¬A2 then C2 can be encoded as 100 Each rule is represented by a string of bits If an attribute has k > 2 values, k bits can be used Based on the notion of survival of the fittest, a new population is formed to consist of the fittest rules and their offsprings The fitness of a rule is represented by its classification accuracy on a set of training examples Offsprings are generated by crossover and mutation The process continues until a population P evolves when each rule in P satisfies a prespecified threshold Slow but easily parallelizable January 20, 2014 Data Mining: Concepts and Techniques 98
99. 99. Rough Set Approach Rough sets are used to approximately or “roughly” define equivalent classes A rough set for a given class C is approximated by two sets: a lower approximation (certain to be in C) and an upper approximation (cannot be described as not belonging to C) Finding the minimal subsets (reducts) of attributes for feature reduction is NP-hard but a discernibility matrix (which stores the differences between attribute values for each pair of data tuples) is used to reduce the computation intensity January 20, 2014 Data Mining: Concepts and Techniques 99
100. 100. Fuzzy Set Approaches Fuzzy logic uses truth values between 0.0 and 1.0 to represent the degree of membership (such as using fuzzy membership graph) Attribute values are converted to fuzzy values e.g., income is mapped into the discrete categories {low, medium, high} with fuzzy values calculated For a given new sample, more than one fuzzy value may apply Each applicable rule contributes a vote for membership in the categories Typically, the truth values for each predicted category are summed, and these sums are combined January 20, 2014 Data Mining: Concepts and Techniques 100
101. 101. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 101
102. 102. What Is Prediction? (Numerical) prediction is similar to classification construct a model use model to predict continuous or ordered value for a given input Prediction is different from classification Classification refers to predict categorical class label Prediction models continuous-valued functions Major method for prediction: regression model the relationship between one or more independent or predictor variables and a dependent or response variable Regression analysis Linear and multiple regression Non-linear regression Other regression methods: generalized linear model, Poisson regression, log-linear models, regression trees January 20, 2014 Data Mining: Concepts and Techniques 102
103. 103. Linear Regression Linear regression: involves a response variable y and a single predictor variable x y = w0 + w 1 x where w0 (y-intercept) and w1 (slope) are regression coefficients Method of least squares: estimates the best-fitting straight line | D| ∑( x − x )( yi − y ) w = 1 ∑( x − x ) i =1 i | D| i =1 2 w = y −w x 0 1 i Multiple linear regression: involves more than one predictor variable Training data is of the form (X 1 , y1), (X 2 , y2),…, (X |D| , y|D|) Ex. For 2-D data, we may have: y = w0 + w1 x1+ w2 x2 Solvable by extension of least square method or using SAS, SPlus January 20, 2014 Data Mining: Concepts and Techniques 103
104. 104. Nonlinear Regression Some nonlinear models can be modeled by a polynomial function A polynomial regression model can be transformed into linear regression model. For example, y = w0 + w1 x + w2 x2 + w3 x3 convertible to linear with new variables: x2 = x2, x3= x3 y = w0 + w1 x + w2 x2 + w3 x3 Other functions, such as power function, can also be transformed to linear model Some models are intractable nonlinear (e.g., sum of exponential terms) possible to obtain least square estimates through extensive January 20, 2014 calculation on moreConcepts and Techniques Data Mining: complex formulae 104
105. 105. Other Regression-Based Models Generalized linear model: Foundation on which linear regression can be applied to modeling categorical response variables Variance of y is a function of the mean value of y, not a constant Logistic regression: models the prob. of some event occurring as a linear function of a set of predictor variables Poisson regression: models the data that exhibit a Poisson distribution Log-linear models: (for categorical data) Approximate discrete multidimensional prob. distributions Also useful for data compression and smoothing Regression trees and model trees Trees to predict continuous values rather than class labels January 20, 2014 Data Mining: Concepts and Techniques 105
106. 106. Regression Trees and Model Trees Regression tree: proposed in CART system (Breiman et al. 1984) CART: Classification And Regression Trees Each leaf stores a continuous-valued prediction It is the average value of the predicted attribute for the training tuples that reach the leaf Model tree: proposed by Quinlan (1992) Each leaf holds a regression model—a multivariate linear equation for the predicted attribute A more general case than regression tree Regression and model trees tend to be more accurate than linear regression when the data are not represented well by a simple linear model January 20, 2014 Data Mining: Concepts and Techniques 106
107. 107. Predictive Modeling in Multidimensional Databases Predictive modeling: Predict data values or construct generalized linear models based on the database data One can only predict value ranges or category distributions Method outline: Minimal generalization Attribute relevance analysis Generalized linear model construction Prediction Determine the major factors which influence the prediction Data relevance analysis: uncertainty measurement, entropy analysis, expert judgement, etc. Multi-level prediction: drill-down and roll-up analysis January 20, 2014 Data Mining: Concepts and Techniques 107
108. 108. Prediction: Numerical Data January 20, 2014 Data Mining: Concepts and Techniques 108
109. 109. Prediction: Categorical Data January 20, 2014 Data Mining: Concepts and Techniques 109
110. 110. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 110
111. 111. Classifier Accuracy Measures C1 C2 C1 True positive False negative C2 False positive True negative classes total recognition(%) 6954 46 7000 99.34 buy_computer = no 412 2588 3000 86.27 total buy_computer = no buy_computer = yes buy_computer = yes 7366 2634 10000 95.52 Accuracy of a classifier M, acc(M): percentage of test set tuples that are correctly classified by the model M Error rate (misclassification rate) of M = 1 – acc(M) Given m classes, CM , an entry in a confusion matrix , indicates # i,j of tuples in class i that are labeled by the classifier as class j Alternative accuracy measures (e.g., for cancer diagnosis) sensitivity = t-pos/pos /* true positive recognition rate */ specificity = t-neg/neg /* true negative recognition rate */ precision = t-pos/(t-pos + f-pos) accuracy = sensitivity * pos/(pos + neg) + specificity * neg/(pos + neg) This model can also be used for cost-benefit analysis January 20, 2014 Data Mining: Concepts and Techniques 111
112. 112. Predictor Error Measures Measure predictor accuracy: measure how far off the predicted value is from the actual known value Loss function : measures the error betw. yi and the predicted value yi’ Absolute error: | yi – yi’| Squared error: (yi – yi’)2 d d Test error (generalization error): the average loss over the test set Mean absolute error: ∑| y i =1 i − yi ' | d Relative absolute error: i= 1 d i | ∑y i= 1 i =1 i − yi ' ) 2 d ∑ ( yi − yi ' ) 2 d d | ∑y Mean squared error: ∑(y i − yi ' | −y | Relative squared error: i =1 d ∑(y i =1 i − y)2 The mean squared-error exaggerates the presence of outliers Popularly use (square) root mean-square error, similarly, root relative squared error January 20, 2014 Data Mining: Concepts and Techniques 112
113. 113. Evaluating the Accuracy of a Classifier or Predictor (I) Holdout method Given data is randomly partitioned into two independent sets Training set (e.g., 2/3) for model construction Test set (e.g., 1/3) for accuracy estimation Random sampling: a variation of holdout Repeat holdout k times, accuracy = avg. of the accuracies obtained Cross-validation (k-fold, where k = 10 is most popular) Randomly partition the data into k mutually exclusive subsets, each approximately equal size At i-th iteration, use D as test set and others as training set i Leave-one-out: k folds where k = # of tuples, for small sized data Stratified cross-validation: folds are stratified so that class dist. in each fold is approx. the same as that in the initial data January 20, 2014 Data Mining: Concepts and Techniques 113
114. 114. Evaluating the Accuracy of a Classifier or Predictor (II) Bootstrap Works well with small data sets Samples the given training tuples uniformly with replacement i.e., each time a tuple is selected, it is equally likely to be selected again and re-added to the training set Several boostrap methods, and a common one is .632 boostrap Suppose we are given a data set of d tuples. The data set is sampled d times, with replacement, resulting in a training set of d samples. The data tuples that did not make it into the training set end up forming the test set. About 63.2% of the original data will end up in the bootstrap, and the remaining 36.8% will form the test set (since (1 – 1/d)d ≈ e-1 = 0.368) Repeat the sampling procedue k times, overall accuracy of the k model: acc( M ) = ∑ (0.632 × acc( M i ) test _ set +0.368 × acc( M i ) train _ set ) i =1 January 20, 2014 Data Mining: Concepts and Techniques 114
115. 115. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 115
116. 116. Ensemble Methods: Increasing the Accuracy Ensemble methods Use a combination of models to increase accuracy Combine a series of k learned models, M , M , …, M , 1 2 k with the aim of creating an improved model M* Popular ensemble methods Bagging: averaging the prediction over a collection of classifiers Boosting: weighted vote with a collection of classifiers Ensemble: combining a set of heterogeneous classifiers January 20, 2014 Data Mining: Concepts and Techniques 116
117. 117. Bagging: Boostrap Aggregation Analogy: Diagnosis based on multiple doctors’ majority vote Training Given a set D of d tuples, at each iteration i, a training set D of d i tuples is sampled with replacement from D (i.e., boostrap) A classifier model M is learned for each training set D i i Classification: classify an unknown sample X Each classifier M returns its class prediction i The bagged classifier M* counts the votes and assigns the class with the most votes to X Prediction: can be applied to the prediction of continuous values by taking the average value of each prediction for a given test tuple Accuracy Often significant better than a single classifier derived from D For noise data: not considerably worse, more robust Proved improved accuracy in prediction January 20, 2014 Data Mining: Concepts and Techniques 117
118. 118. Boosting Analogy: Consult several doctors, based on a combination of weighted diagnoses—weight assigned based on the previous diagnosis accuracy How boosting works? Weights are assigned to each training tuple A series of k classifiers is iteratively learned After a classifier Mi is learned, the weights are updated to allow the subsequent classifier, Mi+1, to pay more attention to the training tuples that were misclassified by Mi The final M* combines the votes of each individual classifier, where the weight of each classifier's vote is a function of its accuracy The boosting algorithm can be extended for the prediction of continuous values Comparing with bagging: boosting tends to achieve greater accuracy, January 20, 2014 Data Mining: Concepts and Techniques 118
119. 119. Adaboost (Freund and Schapire, 1997) Given a set of d class-labeled tuples, (X 1 , y1), …, (X d , yd) Initially, all the weights of tuples are set the same (1/d) Generate k classifiers in k rounds. At round i, Tuples from D are sampled (with replacement) to form a training set Di of the same size Each tuple’s chance of being selected is based on its weight A classification model Mi is derived from Di Its error rate is calculated using Di as a test set If a tuple is misclssified, its weight is increased, o.w. it is decreased Error rate: err(X j ) is the misclassification error of tuple X j . Classifier d Mi error rate is the sum of the weights of the misclassified tuples: error ( M i ) = ∑ w j × err ( X j ) j The weight of classifier Mi’s vote is January 20, 2014 log 1 − error ( M i ) error ( M i ) Data Mining: Concepts and Techniques 119
120. 120. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 120
121. 121. Model Selection: ROC Curves ROC (Receiver Operating Characteristics) curves: for visual comparison of classification models Originated from signal detection theory Shows the trade-off between the true positive rate and the false positive rate The area under the ROC curve is a measure of the accuracy of the model Rank the test tuples in decreasing order: the one that is most likely to belong to the positive class appears at the top of the list The closer to the diagonal line (i.e., the closer the area is to 0.5), the less accurate is the model January 20, 2014 Vertical axis represents the true positive rate Horizontal axis rep. the false positive rate The plot also shows a diagonal line A model with perfect accuracy will have an area of 1.0 Data Mining: Concepts and Techniques 121
122. 122. Chapter 6. Classification and Prediction What is classification? What is prediction? Support Vector Machines (SVM) Issues regarding classification Associative classification and prediction Lazy learners (or learning from Classification by decision tree your neighbors) induction Other classification methods Bayesian classification Prediction Rule-based classification Accuracy and error measures Classification by back Ensemble methods propagation Model selection January 20, 2014 Data Mining: Concepts and Techniques 122
123. 123. Summary (I) Classification and prediction are two forms of data analysis that can be used to extract models describing important data classes or to predict future data trends. Effective and scalable methods have been developed for decision trees induction, Naive Bayesian classification, Bayesian belief network, rule-based classifier, Backpropagation, Support Vector Machine (SVM), associative classification, nearest neighbor classifiers, and case-based reasoning, and other classification methods such as genetic algorithms, rough set and fuzzy set approaches. Linear, nonlinear, and generalized linear models of regression can be used for prediction. Many nonlinear problems can be converted to linear problems by performing transformations on the predictor variables. Regression trees and model trees are also used for January 20, 2014 Data Mining: Concepts and Techniques 123
124. 124. Summary (II) Stratified k-fold cross-validation is a recommended method for accuracy estimation. Bagging and boosting can be used to increase overall accuracy by learning and combining a series of individual models. Significance tests and ROC curves are useful for model selection There have been numerous comparisons of the different classification and prediction methods, and the matter remains a research topic No single method has been found to be superior over all others for all data sets Issues such as accuracy, training time, robustness, interpretability, and scalability must be considered and can involve trade-offs, further complicating the quest for an overall superior method January 20, 2014 Data Mining: Concepts and Techniques 124
125. 125. References (1) C. Apte and S. Weiss. Data mining with decision trees and decision rules . Future Generation Computer Systems, 13, 1997. C. M. Bishop, Neural Networks for Pattern Recognition . Oxford University Press, 1995. L. Breiman, J. Friedman, R. Olshen, and C. Stone. Classification and Regression Trees. Wadsworth International Group, 1984. C. J. C. Burges. A Tutorial on Support Vector Machines for Pattern Recognition. Data Mining and Knowledge Discovery, 2(2): 121-168, 1998. P. K. Chan and S. J. Stolfo. Learning arbiter and combiner trees from partitioned data for scaling machine learning . KDD'95. W. Cohen. Fast effective rule induction . ICML'95. G. Cong, K.-L. Tan, A. K. H. Tung, and X. Xu. Mining top-k covering rule groups for gene expression data. SIGMOD'05. A. J. Dobson. An Introduction to Generalized Linear Models . Chapman and Hall, 1990. G. Dong and J. Li. Efficient mining of emerging patterns: Discovering trends and differences. KDD'99. January 20, 2014 Data Mining: Concepts and Techniques 125
126. 126. References (2) R. O. Duda, P. E. Hart, and D. G. Stork. Pattern Classification, 2ed. John Wiley and Sons, 2001 U. M. Fayyad. Branching on attribute values in decision tree generation . AAAI’94. Y. Freund and R. E. Schapire. A decision-theoretic generalization of on-line learning and an application to boosting . J. Computer and System Sciences, 1997. J. Gehrke, R. Ramakrishnan, and V. Ganti. Rainforest: A framework for fast decision tree construction of large datasets . VLDB’98. J. Gehrke, V. Gant, R. Ramakrishnan, and W.-Y. Loh, BOAT -- Optimistic Decision Tree Construction. SIGMOD'99. T. Hastie, R. Tibshirani, and J. Friedman. The Elements of Statistical Learning: Data Mining, Inference, and Prediction. Springer-Verlag, 2001. D. Heckerman, D. Geiger, and D. M. Chickering. Learning Bayesian networks: The combination of knowledge and statistical data . Machine Learning, 1995. M. Kamber, L. Winstone, W. Gong, S. Cheng, and J. Han. Generalization and decision tree induction: Efficient classification in data mining . RIDE'97. B. Liu, W. Hsu, and Y. Ma. Integrating Classification and Association Rule . KDD'98. JanuaryJ.20, 2014 Pei, CMAR: Accurate and Efficient Classification Based on Data Mining: Concepts and Techniques 126 W. Li, Han, and J.
127. 127. References (3) T.-S. Lim, W.-Y. Loh, and Y.-S. Shih. A comparison of prediction accuracy, complexity, and training time of thirty-three old and new classification algorithms. Machine Learning, 2000. J. Magidson. The Chaid approach to segmentation modeling: Chi-squared automatic interaction detection. In R. P. Bagozzi, editor, Advanced Methods of Marketing Research, Blackwell Business, 1994. M. Mehta, R. Agrawal, and J. Rissanen. SLIQ : A fast scalable classifier for data mining. EDBT'96. T. M. Mitchell. Machine Learning. McGraw Hill, 1997. S. K. Murthy, Automatic Construction of Decision Trees from Data: A MultiDisciplinary Survey , Data Mining and Knowledge Discovery 2(4): 345-389, 1998 J. R. Quinlan. Induction of decision trees . Machine Learning, 1:81-106, 1986. J. R. Quinlan and R. M. Cameron-Jones. FOIL: A midterm report. ECML’93. J. R. Quinlan. C4.5: Programs for Machine Learning . Morgan Kaufmann, 1993. J. R. Quinlan. Bagging, boosting, and c4.5. AAAI'96. January 20, 2014 Data Mining: Concepts and Techniques 127
128. 128. References (4) R. Rastogi and K. Shim. Public: A decision tree classifier that integrates building and pruning. VLDB’98. J. Shafer, R. Agrawal, and M. Mehta. SPRINT : A scalable parallel classifier for data mining. VLDB’96. J. W. Shavlik and T. G. Dietterich. Readings in Machine Learning . Morgan Kaufmann, 1990. P. Tan, M. Steinbach, and V. Kumar. Introduction to Data Mining. Addison Wesley, 2005. S. M. Weiss and C. A. Kulikowski. Computer Systems that Learn: Classification and Prediction Methods from Statistics, Neural Nets, Machine Learning, and Expert Systems. Morgan Kaufman, 1991. S. M. Weiss and N. Indurkhya. Predictive Data Mining. Morgan Kaufmann, 1997. I. H. Witten and E. Frank. Data Mining: Practical Machine Learning Tools and Techniques, 2ed. Morgan Kaufmann, 2005. X. Yin and J. Han. CPAR: Classification based on predictive association rules . SDM'03 H. Yu, J. Yang, and J. Han. Classifying large data sets using SVM with hierarchical clusters . KDD'03. January 20, 2014 Data Mining: Concepts and Techniques 128
129. 129. January 20, 2014 Data Mining: Concepts and Techniques 129 | 19,678 | 76,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-22 | latest | en | 0.672166 |
http://experiment-ufa.ru/x=sqrt(1000) | 1,508,621,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00710.warc.gz | 108,504,419 | 6,427 | # x=sqrt(1000)
## Simple and best practice solution for x=sqrt(1000) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for x=sqrt(1000) equation:
Simplifying
x = sqrt(1000)
Reorder the terms for easier multiplication:
x = 1000qrst
Solving
x = 1000qrst
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Simplifying
x = 1000qrst`
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https://kndrck.co/posts/zk_pedersen_commitments/ | 1,558,958,235,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00012.warc.gz | 522,635,362 | 5,303 | # Pederson Commitments
Posted on 2019-01-20 by Kendrick Tan
# Background
While doing my research on bulletproofs, a common term kept popping up – Pederson Commitments. Only problem was that most of the articles I found online were very academic based, which contained lots of big words that I didn’t understand.
This article will attempt to explain the motivation behind pederson commitments, how they work, accompanied with simple examples.
# Pederson Commitments
## Motivation
Pederson Commitments, as the name suggests, is a type of commitment scheme. The goal of a commitment scheme is to allow a party to commit to a chosen value, without revealing that value, with the ability to reveal and verify the committed value later.
One example use case would be betting – You want to let others know you’ve placed your bet by committing to a value. This lets the other parties know that you’ve placed your bet, but not what you betted on. Should you win the bet, you can reveal your committed value and retrieve your prize.
## Introduction
Pederson Commitments, like any other commitment scheme needs to uphold two very important properties:
• Hiding - The value chosen by the party, should not be known by anyone else
• Binding - The only value that can verify the commited value is the inital chosen value.
### Hiding
Pederson Commitments contains two generators - $g, h$. If we only had one generator to hide the chosen value, an adversary could potentially look up tables of commonly committed values to figure out what value was committed.
As such, we include a second generator $h$ to add the extra randomness into the algorithm to fulfill the hiding property.
## Homomorphism
For Pederson Commitments to work, it needs homomorphism. What this means is that it allows us to preserve structure between two arbitrary structures, e.g:
$f(a + b) = f(a) + f(b)$
## Algorithm
1. Choose a large prime number $p$
p = 260978677425009836700364089744760003717
1. Generate secret value $s$ and generator $g$ to be between $(1, p - 1)$
s = random.randint(1, p - 1)
g = random.randint(1, p - 1)
1. Generate generator $h$ using $h = g^s \ mod \ p$
h = pow(g, s, p)
1. To commit a value $c = g^v h^r \ mod \ p$, generate random value $r$ between $(1, p - 1)$
v = 42
r = random.randint(1, p - 1)
commited_value = (pow(g, v, p) * pow(h, r)) % p
1. Sender sends $c$ to the receiver, and reveals $v, r$ to the receiver at some point in the future. Receiver checks to see if the received commited value and the newly calculated value matches. (Note: $g, h, p$ are known by the receiver in advance)
# received r and v
calculated_commitment = (pow(g, value, p) * pow(h, r)) % p
if calculated_commitment == commited_value:
print('Verified!')
## Proof That Pederson Commitments Are Homomorphic
Recall the formula to commit a value $C(A)$:
\begin{align} C(A) &= g^A \cdot h^{r_A} \ mod \ p \newline C(A + B) &= g^{A+B} \cdot h^{r_A + r_B} \ mod \ p \newline \newline C(A) + C(B) &= (g^A \cdot h^{r_A} + g^B \cdot h^{r_B}) \ mod \ p \newline &= g^{A+B} \cdot h^{r_A + r_B} \ mod \ p \newline \newline \therefore C(A) + C(B) &= C(A + B) \end{align}
# Example
import random
# Prime number
p = 260978677425009836700364089744760003717
# Generator
g = random.randint(1, p - 1)
class Verifier:
def __init__(self, p, g):
self.p = p
# Generator
self.g = g
# Secret number
self.s = random.randint(1, p - 1)
# Hiding Generator - order of q and subgroup of Z_p
self.h = pow(self.g, self.s, self.p)
"""
Multiplying values in the pedersen commitment is
similar to adding the values together before
committing them
Proof:
C(A) x C(B) = (g^A)(h^(r_A)) * (g^B)(h^(r_B)) mod p
= g^(A+B) * h^(r_A + r_B) mod p
= C(A+B)
"""
cm = 1
for c in commitments:
cm = cm * c
return cm % self.p
def verify(self, c, x, *r) -> bool:
r_sum = sum(r)
res = (pow(self.g, x, self.p) * pow(self.h, r_sum, self.p)) % self.p
if c == res:
return True
return False
class Prover:
def __init__(self, p, g, h):
self.p = p
self.g = g
self.h = h
def commit(self, value):
"""
C(x) = (g^x)*(h^r) mod p
where h = (g^s) mod p
"""
r = random.randint(1, self.p - 1)
# Commit message
c = (pow(self.g, value, self.p) * pow(self.h, r, self.p)) % self.p
return c, r
# Values we want to commit and prove later on
value1 = 50
value2 = 42
# Verifier and prover
verifier = Verifier(p, g)
prover = Prover(p, g, verifier.h)
# Commit message
c1, r1 = prover.commit(value1)
c2, r2 = prover.commit(value2)
# Verify result
result1 = verifier.verify(c1, value1, r1)
result2 = verifier.verify(c2, value2, r2)
if result1:
print('Verified commitment 1')
else:
print('Commitment 1 unverified')
if result2:
print('Verified commitment 2')
else:
print('Commitment 2 unverified')
# Prove homomorphic property
print('Homomorphic property not verified') | 1,406 | 4,802 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-22 | longest | en | 0.924918 |
http://www.openwetware.org/index.php?title=BISC220/S14:_Mod_1_Lab_4&diff=prev&oldid=701541 | 1,444,381,144,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737922497.60/warc/CC-MAIN-20151001221842-00236-ip-10-137-6-227.ec2.internal.warc.gz | 824,635,677 | 10,441 | # BISC220/S14: Mod 1 Lab 4
(Difference between revisions)
Revision as of 15:22, 4 June 2013 (view source) (→Determining the amount of β-Galactosidase in the Total Protein by β-Galactosidase Enzyme Activity Using A420 measurement of ONP production)← Previous diff Revision as of 15:23, 4 June 2013 (view source) (→ENZYME SPECIFIC ACTIVITY OR VELOCITY)Next diff → Line 61: Line 61: # Calculate specific activity from absorbance using the Beer-Lambert formula. The molar extinction co-efficient of ONP is 4800 M-1 cm-1 and the path length of the cuvette used is 1 cm. The concentrations and total protein content in each of your fractions were determined by the Bradford dye assay. # Calculate specific activity from absorbance using the Beer-Lambert formula. The molar extinction co-efficient of ONP is 4800 M-1 cm-1 and the path length of the cuvette used is 1 cm. The concentrations and total protein content in each of your fractions were determined by the Bradford dye assay. - == ENZYME SPECIFIC ACTIVITY OR VELOCITY == + == Enzyme Specific Activity (Velocity) == '''Data:'''
'''Data:'''
ONPG assay for enzyme specific activity:
ONPG assay for enzyme specific activity:
## Revision as of 15:23, 4 June 2013
Wellesley College BISC 220 Cellular Physiology
## Determination of the Specific Activity of β-Galactosidase
Specific Activity
The purification of an enzyme is an attempt to enrich the extract for the desired enzyme while eliminating other cellular components, notably other proteins. One measure of the success of a purification step can be obtained by assaying the activity of the desired enzyme at saturating substrate concentration relative to the total amount of protein present. Specific Activity of an enzyme (also sometimes referred to as maximum velocity, Vmax) is defined as the amount of product formed/unit time (enzyme activity) per milligram (mg) of protein. An enzyme’s specific activity can be employed to evaluate the relative purity of fractions obtained during the purification. In order to evaluate the success of your purification of β-galactosidase, you must measure both the amount of β-galactosidase activity and the total amount of protein (mg) present in the starting material and in your final purified product.
Specific Activity = amount of product formed/unit time/mg protein
β-galactosidase specific activity is often expressed as µmoles of product (ONP) formed per minute per mg of protein.
Specific Activity of β-galactosidase = µmolesONP/minute/mg protein
Once you know the specific activity of your crude extract and your purified fraction, you can proceed to calculate other values useful in determining the success of a purification step such as:
1. total activity = (specific activity) x (total mg protein in preparation)
2. % yield – the amount of protein of interest retained in the purified fraction
= (total activity of the purified fraction/total activity of the starting material (crude extract)) * 100
3. purification factor – the fold increase of protein of interest in the purified fraction compared to the crude extract
= (specific activity of the purified fraction/specific activity of the starting material (crude extract))
Table I shows the results obtained during a β-galactosidase purification by researchers, Wallenfels et al. (1959), working with β-galactosidase. The inverse relationship between total activity and specific activity is clear. The yield (12%) was reasonable and the enzyme was purified substantially (13.5 fold).
References:
Kagedal L (1998) “Immobilized Metal Ion Affinity Chromatography” In Protein Purification 2nd ed. (Janson J-C and Ryden L eds) Wiley-Liss, New York.
Porath J, Carlsson J, Olsson I, Belfrage G (1975) Metal chelate affinity chromatography, a new approach to protein fractionation. Nature 258: 598-599.
Porath J, Olin B (1983) Immobilized metal ion affinity adsoprtion and immobilized metal ion affinity chromatography of biomaterials. Serum protein affinities for gel-immobilized iron and nickel ions. Biochemistry 22: 1621-1630.
Wallenfels K, Zarnitz ML Laule G, Bender H, Keser M (1959) Biochem Z 331: 459.
## Determining the amount of β-Galactosidase in the Total Protein by β-Galactosidase Enzyme Activity Using A420 measurement of ONP production
Microsoft Word File: Media:Determining the amount of β gal by enzyme activity.doc
In-vivo, β-galactosidase cleaves lactose to yield galactose and glucose, but in vitro, the appearance of the products of galactose and glucose are difficult to monitor. The colorless compound ortho-nitro-phenyl-galactoside (ONPG) is substituted for lactose and yields upon hydrolysis (cleavage) by β-galactosidase a yellow compound, ortho-nitrophenol, (ONP) and galactose. The addition of concentrated sodium carbonate (Na2CO3) shifts the pH to a very basic pH 11, a condition which inactivates the enzyme. The amount of colored product (ONP), formed from the colorless substrate (ONPG), can be quantified using a spectrophotometer and then converted to a concentration of ONP using its molar extinction coefficient ( Enzyme Specific Activity or Velocity for a sample calculation).
The total protein assay you performed has allowed you to determine the protein concentration of your two fractions (the crude extract and the purified fraction) but you do not know how much of that protein is our enzyme of interest, beta-galactosidase. You will now perform a specific β-galactosidase activity assay on a series of dilutions of both the crude extract (CE) and of the purified fraction (PF) in order to determine the concentration of enzyme that will yield an absorbance in the most accurate range of measurement of the spectrophotometers (0.1-1.0 for the Hitachi spec). Our goal is to find a diluted form of β-galactosidase in the CE & in the PF that gives an absorbance reading of close to 0.5A in the specific activity assay.
Suggestions for appropriate dilutions based on previous experimentation are: CE: 1:25; 1:50; 1:100; 1:200. Since there is more concentrated beta-galactosidase in the PF, dilutions of: 1:50, 1:100; 1:200; and 1:400 should be tested.
Making a serial dilution of the CE and PF for the SA assay:
Since you have a limited volume of both crude extract and purified fractions and you will need a minimum of 100 µl for each assay, it is advisable to make a little more than you need of each dilution but not so much more that you waste your fractions. It is perfectly acceptable, and often preferable, to use part of a stronger concentration to make the next weaker one.
For example, let's assume you want to end up with 250 µl of each dilution. You could start by making 500 μL of the 1:25 dilution of the CE fraction from last week. After preparing 500 µl of a 1:25 dilution and mixing well you could use 250µl of that 1:25 dilution added to an equal volume of buffer to make 500µl of a 1:50 dilution. If you wanted to continue to dilute with buffer equal volumes of each sequentially weaker concentration, you would end up with 250µl of each of the concentrations desired: 1:25, 1:50, 1:100, 1:200. This is a serial dilution.
How would you go about making 250 μL of each of the PF dilutions you want 1:50, 1:100, 1:200 and 1:400? Please show your dilution strategy to your instructor before you proceed.
After your instructor has approved your dilution strategy and you have on ice all of your labeled microfuge tubes with each of the specified dilutions of the CE and PF in Z-buffer (60mM Na2HPO4, 60mM NaH2PO4, 1mM MgSO4, 0.27% beta-mercaptoethanol), you are now ready to start the assay.
Protocol for Assay of Specific Activity (at saturating substrate concentration)
1. Label a set of 8 glass tubes with the 4 dilutions of CE and the 4 dilutions of PF to be tested. For the CE, you will test: 1:25, 1:50, 1:100, 1:200. For the PF, test: 1:50. 1:100, 1:200, 1:400. Label tube 9 "BLANK" as a reagent blanks for the spectrophotometer.
2. Pipet 1.9ml of Z-buffer into the 8 glass test tubes prepared above. Pipet 2ml Z-buffer into the 2 reagent blank (tube 9).
3. Pipet 100μl of the appropriate enzyme dilution into the labeled tube containing Z-buffer prepared in #1. Mix well by vortexing. Put NO enzyme in the blanks!
4. Equilibrate all 10 tubes to 28C in a water bath. Five minutes should be sufficient.
5. Start the reaction by adding 400 µl (0.4 ml) of substrate (ONPG, 4mg/ml) to the first tube, vortex immediately, and quickly return the tube to the water bath. In order to insure that all reactions occur for exactly the same amount of time, add 400µl (0.4 ml) substrate at carefully timed intervals, such as every 20 or 30 seconds. What is the effective concentration of ONPG?
6. At exactly 5 minutes after adding ONPG to the first tube, start adding 1000µl (1 ml) of stop buffer, 1M Na2CO3 to each tube in the same order at the same time interval. What is the effective concentration of stop buffer? Mix well after each addition. Since you will be calculating specific activity of beta-galactosidase as µmoles of product formed (ONP)/ minute/mg of total protein, timing of the reaction is critical.
7. Pour some of each tube into a set of labeled cuvettes. Make sure the cuvettes are 2/3 to ¾ full. It doesn’t matter if each has exactly the same volume.
8. Read A420 in the spectrophotometer. Don’t forget to change the wavelength from 595nm to 420nm. Zero the instrument using the reagent blanks.
9. Calculate specific activity from absorbance using the Beer-Lambert formula. The molar extinction co-efficient of ONP is 4800 M-1 cm-1 and the path length of the cuvette used is 1 cm. The concentrations and total protein content in each of your fractions were determined by the Bradford dye assay.
## Enzyme Specific Activity (Velocity)
Data:
ONPG assay for enzyme specific activity:
• Let's assume you have used 0.1 ml of a 1:500 dilution of one of your samples
• The absorbance reading at 420 nm (A) of your reaction tube was 0.450
• The total volume in the reaction tube when the A420 reading was taken was 3.4 ml
• Assume the inside diameter (l) of the reaction tube or curvette was 1.0 cm
• Molar extinction coefficient of ONP (e) = 4800 M-1 cm-1
Calculations for the enzyme assay:
Concentration (in moles/liter) = A/(e)(l)
A = absorbance
e= molar extinction coefficient (M-1 cm-1)
l = pathlength (cm)
The concentration of ONP in moles/liter in your reaction tube is given by:
[ONP] = 0.450/(4800 M-1cm-1)(1.0 cm)
Our example: [ONP] = 0.94 x 10-4 moles/L
The amount of ONP that was produced per minute in each ml of the reaction mixture in your reaction tube is given by:
[ONP]/time = (0.94 x 10-4 moles/L)/5 minutes (our reaction time)
[ONP]/time = 1.9 x 10-5 moles/L/min
Expressed in different units (moles --> mmoles and liter --> ml):
[ONP]/time = 1.9 x 10-5 mmoles ONP/ml/min
Convert this to µmoles ONP/ml/min:
[ONP]/time = 0.019 µmoles ONP/ml/min
The ml here refers to the volume in the reaction tube. The total volume in the reaction tube at the time the reaction was measured was 3.4 ml. Therefore, the amount (µmoles) of ONP produced per minute in the total reaction mixture must have been:
µmoles of ONP/min = (3.4 ml total volume) * 0.019 µmoles/min/ml
total µmoles of ONP/min = 0.064 µmoles/min
The 3.4 ml of total reaction volume contained only 0.1 ml of enzyme. If the 0.1 ml enzyme volume came from a 1/500 dilution containing 2.5 mg/ml of total protein before it was diluted, then in 0.1 ml of this 1/500 dilution you would have 0.0005 mg of protein.
DO NOT USE THE VALUES 2.5mg/ml or 1/500 IN THE FORMULA BELOW, UNLESS THAT WAS YOUR CALCULATED PROTEIN CONCENTRATION AND THE DILUTION YOU USED.
(2.5 mg/ml) * (0.1 ml) * (1/500) = 0 .0005 mg protein
Knowing the amount of ONP produced per min and the protein concentration expressed in milligrams, you may calculate the activity or velocity of your sample. This could also be referred to as the specific activity (SA) because in this case, the substrate was saturating:
SPECIFIC ACTIVITY OR VELOCITY= (0.064 µmoles ONP/min)/0.0005 mg protein
SPECIFIC ACTIVITY = 128 µmoles ONP produced/min/mg protein | 3,190 | 12,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-40 | longest | en | 0.913266 |
https://codesource.io/blog/how-to-get-unique-values-from-a-python-list/ | 1,721,055,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00430.warc.gz | 149,014,280 | 39,304 | In this article, you will learn how to get unique values from a Python list.
Python is an Object-Oriented and interpreted programming language. In python, there are some fundamental topics that are used to store collections of Data. These are list, tuple, dictionary, set, etc. lists are considered as the data structure that stores a group of elements. You can create a python list by using square brackets [] and may store different types of elements. Sometimes, we need to get the unique values from a python list and there are several ways of performing this action. In this article, we are going to explore them and see how we can get unique values from a Python list
Approach One
We can get unique values from a python list by using for loop and the conditional statement. The for loop will traverse the elements of the existing list and the conditional if-statement check whether the value of the list is unique or not. If it is unique then simply append it to a new list and ignore the repeated values. See the below code example to understand it more clearly.
def make_unique_list(my_list):
get_unique_list = []
for x in my_list:
if x not in get_unique_list:
get_unique_list.append(x)
print(get_unique_list)
my_list = [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
print('Original list:')
print(my_list)
print('Unique list: ')
make_unique_list(my_list)
# Output:
# Original list:
# [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
# Unique list:
# [1, 3, 4, 5, 7, 10]
Approach Two
To get unique values from a Python list, we can use the set() property. We need to insert our list into a set. By default, the set property only takes the value once. Let’s say you insert one three times into a set but it only takes the one for the first time. By doing so, we can get the unique values from the set. Finally, we need to convert it to the list. See the below code example:
def make_unique_list(my_list):
make_list_set = set(my_list)
get_unique_list = (list(make_list_set))
print(get_unique_list)
my_list = [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
print('Original list:')
print(my_list)
print('Unique list: ')
make_unique_list(my_list)
# Output:
# Original list:
# [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
# Unique list:
# [1, 3, 4, 5, 7, 10]
Approach Three
We can also get the unique values from a python list with the help of NumPy. At first, we need to import it and this library provides a function named np.unique() We can use this function to get the unique values. But NumPy workers with the array that is similar to the list. We will convert our python list to a NumPy array. See the below code example:
import numpy as np
def make_unique_list(my_list):
convert_list = np.array(my_list)
print(np.unique(convert_list))
my_list = [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
print('Original list:')
print(my_list)
print('Unique list: ')
make_unique_list(my_list)
# Output:
# Original list:
# [1, 3, 3, 4, 5, 5, 7, 7, 7, 10]
# Unique list:
# [ 1 3 4 5 7 10]
This is all about getting unique values from a Python list. You may follow these approaches to get unique values from a Python list. | 861 | 3,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.732422 |
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Difference in BigDecimal behavior
```I have two pieces of code new BigDecimal("1.240472701") and new BigDecimal(1.240472701). Now if i use compareTo method of java on both the methods then i get that they are not equal.
When i printed the values using System.out.println() method of java. I get different results for both the values. For example
new BigDecimal("1.240472701") -> 1.240472701
new BigDecimal(1.240472701) -> 1.2404727010000000664291519569815136492252349853515625
So i want to understand what could be reason for this?
```
```Since user thegauravmahawar provided the answer from docs. Yes, it is because of Scaling in BigDecimal case.
So the values might seem equal to You but internally java uses Scaling while storing the value of BigDecimal type.
Reason: Scaling.
Improvement:
You could call setScale to the same thing on the numbers you're comparing:
like this
new BigDecimal ("7.773").setScale(2).equals(new BigDecimal("7.774").setScale (2))
This will save you from making any mistake.
```
```You can refer the Java doc of public BigDecimal(double val) for this:
public BigDecimal(double val)
Translates a double into a BigDecimal
which is the exact decimal representation of the double's binary
floating-point value. The scale of the returned BigDecimal is the
smallest value such that (10^scale × val) is an integer.
The results of this constructor can be somewhat unpredictable. One
might assume that writing new BigDecimal(0.1) in Java creates a
BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with
a scale of 1), but it is actually equal to
0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that
matter, as a binary fraction of any finite length). Thus, the value
that is being passed in to the constructor is not exactly equal to
0.1, appearances notwithstanding.
The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates
a BigDecimal which is exactly equal to 0.1, as one would expect.
Therefore, it is generally recommended that the String constructor be
used in preference to this one.
When a double must be used as a source
for a BigDecimal, note that this constructor provides an exact
conversion; it does not give the same result as converting the double
to a String using the Double.toString(double) method and then using
the BigDecimal(String) constructor. To get that result, use the static
valueOf(double) method.
```
```The string "1.240472701" is a textual representation of a decimal value. The BigDecimal code parses this and creates a BigDecimal with the exact value represented in the string.
But the double 1.240472701 is merely a (close) approximation of that exact decimal value. Double cannot represent all decimal values exactly, so the exact value stored in the double differs slightly. If you pass that to a BigDecimal, it takes that differing value and turns it into an exact BigDecimal. But the BigDecimal only has the inexact double to go by, it does not know the exact text representation. So it can only represent the value in the double, not the value of the source text.
In the first case:
String --> BigDecimal
Because BigDecimal is made to exactly represent decimal values, that conversion is exact.
In the second case:
1 2
Source code text --> double --> BigDecimal
In the second case, precision is lost in the first conversion (1). The second conversion (2) is exact, but the input -- the double -- is an inexact representation of the source code text 1.240472701 (in reality, it is 1.2404727010000000664291519569815136492252349853515625).
So: never initialize a BigDecimal with a double, if you can avoid it. Use a string instead.
That is why the first BigDecimal is exact and the second is not.```
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# Eric Miller
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A 0.50-kg piece of aluminum at $250\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is dropped into 1.0 kg of water at $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . After equilibrium is reached, what is the net entropy change of the system?
82 J/K
Suppose 20 g of ice at $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is added to 300 g of water at $60\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . What is the total change in entropy of the mixture after it reaches thermal equilibrium?
A heat engine operates between two temperatures such that the working substance of the engine absorbs 5000 J of heat from the high-temperature bath and rejects 3000 J to the low-temperature bath. The rest of the energy is converted into mechanical energy of the turbine. Find (a) the amount of work produced by the engine and (b) the efficiency of the engine.
a. 2000 J; b. $40\text{%}$
A thermal engine produces 4 MJ of electrical energy while operating between two thermal baths of different temperatures. The working substance of the engine rejects 5 MJ of heat to the cold temperature bath. What is the efficiency of the engine?
A coal power plant consumes 100,000 kg of coal per hour and produces 500 MW of power. If the heat of combustion of coal is 30 MJ/kg, what is the efficiency of the power plant?
$60\text{%}$
A Carnot engine operates in a Carnot cycle between a heat source at $550\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and a heat sink at $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$ Find the efficiency of the Carnot engine.
A Carnot engine working between two heat baths of temperatures 600 K and 273 K completes each cycle in 5 sec. In each cycle, the engine absorbs 10 kJ of heat. Find the power of the engine.
$64.4\text{%}$
A Carnot cycle working between $100\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and $30\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is used to drive a refrigerator between $-10\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and $30\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$ How much energy must the Carnot engine produce per second so that the refrigerator is able to discard 10 J of energy per second?
Challenge problems
(a) An infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that $dU=TdS-dW$ . (b) When heat is added to an ideal gas, its temperature and volume change from ${T}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{T}_{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{V}_{2}$ . Show that the entropy change of n moles of the gas is given by
$\text{Δ}S=n{C}_{v}\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{T}_{2}}{{T}_{1}}+nR\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{V}_{2}}{{V}_{1}}$ .
derive
Using the result of the preceding problem, show that for an ideal gas undergoing an adiabatic process, $T{V}^{\text{γ}-1}$ is constant.
With the help of the two preceding problems, show that $\text{Δ}S$ between states 1 and 2 of n moles an ideal gas is given by
$\text{Δ}S=n{C}_{p}\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{T}_{2}}{{T}_{1}}-nR\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{p}_{2}}{{p}_{1}}$ .
derive
A cylinder contains 500 g of helium at 120 atm and $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . The valve is leaky, and all the gas slowly escapes isothermally into the atmosphere. Use the results of the preceding problem to determine the resulting change in entropy of the universe.
A diatomic ideal gas is brought from an initial equilibrium state at ${p}_{1}=0.50\phantom{\rule{0.2em}{0ex}}\text{atm}$ and ${T}_{1}=300\phantom{\rule{0.2em}{0ex}}\text{K}$ to a final stage with ${p}_{2}=0.20\phantom{\rule{0.2em}{0ex}}\text{atm}$ and ${T}_{1}=500\phantom{\rule{0.2em}{0ex}}\text{K}.$ Use the results of the previous problem to determine the entropy change per mole of the gas.
18 J/K
The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle , which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle are as follows:
1. Isobaric intake stroke ( OA ). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure ${p}_{0}$ as the piston expands, increasing the volume of the cylinder from zero to ${V}_{A}$ .
2. Adiabatic compression stroke ( AB ). The temperature of the mixture rises as the piston compresses it adiabatically from a volume ${V}_{\text{A}}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{\text{B}}$ .
3. Ignition at constant volume ( BC ). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat ${Q}_{1}$ causes the pressure to increase from ${p}_{B}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{p}_{C}$ at the constant volume ${V}_{\text{B}}\left(={V}_{\text{C}}\right)$ .
4. Adiabatic expansion ( CD ). The heated mixture of gasoline and air expands against the piston, increasing the volume from ${V}_{C}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{D}$ . This is called the power stroke , as it is the part of the cycle that delivers most of the power to the crankshaft.
5. Constant-volume exhaust ( DA ). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at ${V}_{A}\left(={V}_{D}\right)$ . Most of the available energy is lost here, as represented by the heat exhaust ${Q}_{2}$ .
6. Isobaric compression ( AO ). The exhaust valve remains open, and the compression from ${V}_{A}$ to zero drives out the remaining combustion products.
(a) Using ( i ) $e=W\text{/}{Q}_{1}$ ; ( ii ) $W={Q}_{1}-{Q}_{2}$ ; and ( iii ) ${Q}_{1}=n{C}_{v}\left({T}_{C}-{T}_{B}\right)$ , ${Q}_{2}=n{C}_{v}\left({T}_{D}-{T}_{A}\right)$ , show that
$e=1-\frac{{T}_{D}-{T}_{A}}{{T}_{C}-{T}_{B}}$ .
(b) Use the fact that steps (ii) and (iv) are adiabatic to show that
$e=1-\frac{1}{{r}^{\gamma -1}}$ ,
where $r={V}_{A}\text{/}{V}_{B}$ . The quantity r is called the compression ratio of the engine.
(c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for $r=6$ and $\gamma =1.4$ (the value for air), $e=0.51$ , or an efficiency of $51\text{%}.$ Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about $25\text{%}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}30\text{%}$ .
An ideal diesel cycle is shown below. This cycle consists of five strokes. In this case, only air is drawn into the chamber during the intake stroke OA . The air is then compressed adiabatically from state A to state B , raising its temperature high enough so that when fuel is added during the power stroke BC , it ignites. After ignition ends at C , there is a further adiabatic power stroke CD . Finally, there is an exhaust at constant volume as the pressure drops from ${p}_{D}$ to ${p}_{A}$ , followed by a further exhaust when the piston compresses the chamber volume to zero.
(a) Use $W={Q}_{1}-{Q}_{2}$ , ${Q}_{1}=n{C}_{p}\left({T}_{C}-{T}_{B}\right)$ , and ${Q}_{2}=n{C}_{v}\left({T}_{D}-{T}_{A}\right)$ to show that $e=\frac{W}{{Q}_{1}}=1-\frac{{T}_{D}-{T}_{A}}{\gamma \left({T}_{C}-{T}_{B}\right)}$ .
(b) Use the fact that $A\to B$ and $C\to D$ are adiabatic to show that
$e=1-\frac{1}{\gamma }\phantom{\rule{0.2em}{0ex}}\frac{{\left(\frac{{V}_{C}}{{V}_{D}}\right)}^{\gamma }-{\left(\frac{{V}_{B}}{{V}_{A}}\right)}^{\gamma }}{\left(\frac{{V}_{C}}{{V}_{D}}\right)-\left(\frac{{V}_{B}}{{V}_{A}}\right)}$ .
(c) Since there is no preignition (remember, the chamber does not contain any fuel during the compression), the compression ratio can be larger than that for a gasoline engine. Typically, ${V}_{A}\text{/}{V}_{B}=15\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{V}_{D}\text{/}{V}_{C}=5$ . For these values and $\gamma =1.4,$ show that $\epsilon =0.56$ , or an efficiency of $56\text{%}$ . Diesel engines actually operate at an efficiency of about $30\text{%}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}35\text{%}$ compared with $25\text{%}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}30\text{%}$ for gasoline engines.
proof
Consider an ideal gas Joule cycle, also called the Brayton cycle, shown below. Find the formula for efficiency of the engine using this cycle in terms of ${P}_{1}$ , ${P}_{2}$ , and $\gamma$ .
Derive a formula for the coefficient of performance of a refrigerator using an ideal gas as a working substance operating in the cycle shown below in terms of the properties of the three states labeled 1, 2, and 3.
${K}_{\text{R}}=\frac{3\left({p}_{1}-{p}_{2}\right){V}_{1}}{5{p}_{2}{V}_{3}-3{p}_{1}{V}_{1}-{p}_{2}{V}_{1}}$
Two moles of nitrogen gas, with $\gamma =7\text{/}5$ for ideal diatomic gases, occupies a volume of ${10}^{-2}{\text{m}}^{3}$ in an insulated cylinder at temperature 300 K. The gas is adiabatically and reversibly compressed to a volume of 5 L. The piston of the cylinder is locked in its place, and the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heat from the compressed gas leaves the gas, and the temperature of the gas becomes 300 K again. The gas is then slowly expanded at the fixed temperature 300 K until the volume of the gas becomes ${10}^{-2}{\text{m}}^{3}$ , thus making a complete cycle for the gas. For the entire cycle, calculate (a) the work done by the gas, (b) the heat into or out of the gas, (c) the change in the internal energy of the gas, and (d) the change in entropy of the gas.
A Carnot refrigerator, working between $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and $30\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is used to cool a bucket of water containing ${10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$ of water at $30\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ to $5\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ in 2 hours. Find the total amount of work needed.
$W=110,000\phantom{\rule{0.2em}{0ex}}\text{J}$
relation between Celsius and Kelvin
Newton's second laws is call with
Really
Arzoodan
what is mean by thermodynamics
it is study about temperature and it's equilibrium
thiru
Its the study of heat and its relation with others kind of energy
Antonio
state caulombs law clearly
show mathematically that an electron has the greater speed than the proton when they attract each other
show mathematically that an electron has the greater speed than the proton when they attract each other
srikanta
@ezra & srikanta; for electrons: a=ke^2/(mr^2) and for protons: a=kp^2/(mr^2)
Sikandar
what is electrostatics
the study of charge at rest
Gulzar
@Hero; the study of charges at rest is the electrostatics
Sikandar
okay what is electrostatic?
Abd
charge at rest
Nawal
set of character...
Arzoodan
oky
Abd
Gauss law, electric fields, dipoles,...
Antonio
good
Abd
A proton initially at rest falls through a p.d of 25000V. what speed does it gain?
@Minister; use equation v= sq root(2×eV/m)
Sikandar
what is the reaction of heat on magnet
Magnetization decreases with increase in temperature. But in case of diamagnetic substance heat has no role on magnetization.
srikanta
what is a physical significant of electric dipole moment .
A dipole moment it's a mechanical electrical effect used in nature
Antonio
what is the uses of carbon brushes in generator
to minimize heat
constand
at what temperature is the degree Fahrenheit equal to degree Celsius
Celsius and Faharaneith are different, never equal
Antonio
find their liners express of n=a+b/T² ( plot graph n against T)
Radio Stations often advertis "instant news,,if that meens you can hear the news the instant the radio announcer speaks it is the claim true? what approximate time interval is required for a message to travel from Cairo to Aswan by radio waves (500km) (Assume the waves Casbe detected at this range )
what is growth and decay
Can someone please predict the trajectory of a point charge in a uniform electric field???? | 3,858 | 12,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 74, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-22 | longest | en | 0.839471 |
https://itstillruns.com/convert-cc-mph-8651577.html | 1,620,676,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00626.warc.gz | 342,883,967 | 28,214 | # How to Convert CC to MPH
by Bryan Stokes II
Jupiterimages/Photos.com/Getty Images
Although there is no direct conversion between CC, or cubic centimeters, and MPH, miles per hour, there us a relationship between these two units of measurement. In motorcycles, engine displacement is measured in cubic centimeters. This measurement indicates the amount of space that pistons travel through in the engine and can provide an indicator of engine power and, thus, speed.
### Identifying the fastest bike
#### Step 1
Identify the displacement of the motorcycle. This will typically be expressed as a three or four-digit number followed by the letters "cc." Higher displacement indicates larger amounts of available power and, thus, higher speed.
#### Step 2
Weigh the motorcycle or find out the total weight from a reference manual.
Divide the weight by the displacement to identify the ratio of weight to displacement. Displacement can be used as a rough analogue to power. The ratio of weight to power quantifies how much power is available to move each pound of the motorcycle and how quickly the bike will accelerate, as well as the top speed. A lower ratio indicates that the motorcycle will be faster than another. | 246 | 1,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-21 | longest | en | 0.922947 |
https://astronomy.stackexchange.com/questions/49329/calculate-how-many-days-left-for-synodic-period-from-particular-planet | 1,721,879,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00264.warc.gz | 93,446,780 | 38,264 | # Calculate how many days left for synodic period from particular planet
One of the many tools used in Astronomy are the formulas used to determine Orbital Motion. There are two basic forms of orbits:
• Sidereal Period
• Synodic Period
For Jupiter:
$$\mathrm{\frac{1}{P} = \frac{1}{E} - \frac{1}{S}}$$
where
P = sidereal period in both equations
S = synodic period in both equations
E = Earth's orbit in both equations
Synodic period in this case, for Jupiter is 398.88 days.
Query: Is there any formula to calculate how many days left for Jupiter to reach its next synodic period from Earth? or how many days left until Jupiter is in opposition?
• I'm not sure what you mean by "reach it's next synodic period" The synodic period is a length of time, it doesn't have a fixed start point. I suppose you could mean "how many days left until Jupiter is in opposition" (since oppositions occur once in each synodic period) Can you confirm that or explain what you do mean? Commented May 14, 2022 at 5:47
• okay. Opposition yea that's the point. How long does it takes to finish synodic period, or reaching opposition. Thanks for the information. Commented May 14, 2022 at 5:50
If you know the date of the last opposition, then the next one will be about 399 days later (ie a year, a month and a bit). That is just counting days. So if you know that there was an opposition on June 1st, then the next one will be 399 days later, on about July 4th of the following year.
But this assumes circular orbits and so is good for an estimate, but is not exact It's usually out by a few days. Properly calculating the time of opposition requires you to calculate the position of the planets following elliptical orbits (and perturbed by each other's gravity etc). Fortunately, you can get a table which lists the dates of opposition:
August 20, 2021: Aquarius
September 26, 2022: Pisces
November 1, 2023: Aries
December 6, 2024: Taurus
January 9, 2026: Gemini
February 10, 2027: Leo
March 13, 2028: Virgo
April 13, 2029: Virgo
May 14, 2030: Libra
June 16, 2031: Ophiuchus
July 20, 2032: Sagittarius
August 25, 2033: Back in Aquarius (source)
The calculation of these dates is ultimately based on observation. There is no way to calculate them purely from the knowledge of the synodic or sidereal period.
So, as of 14th May 2022, there are 135 days until the next opposition.
• Thanks. How about inner planets like Mercury, Venus? Commented May 14, 2022 at 8:11
• Well they dont have oppositions, but you could look up the dates of their conjunctions (ie when the Earth would be in conjunction viewed from Mercury. Commented May 14, 2022 at 10:45
• Thanks again. How about between Mercury-Venus, Venus-Mars, and Jupiter-Saturn? Without respecting to Sun. Commented Oct 24, 2022 at 6:08 | 749 | 2,786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.955896 |
http://math.libretexts.org/TextMaps/Calculus_Textmaps/Map%3A_Calculus_(Guichard)/05._Curve_Sketching/The_second_derivative_test | 1,484,603,134,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00427-ip-10-171-10-70.ec2.internal.warc.gz | 185,915,878 | 13,153 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# The second derivative test
The basis of the first derivative test is that if the derivative changes from positive to negative at a point at which the derivative is zero then there is a local maximum at the point, and similarly for a local minimum. If $$f'$$ changes from positive to negative it is decreasing; this means that the derivative of $$f'$$, $$f''$$, might be negative, and if in fact $$f''$$ is negative then $$f'$$ is definitely decreasing, so there is a local maximum at the point in question.
Note well that $$f'$$ might change from positive to negative while $$f''$$ is zero, in which case $$f''$$ gives us no information about the critical value. Similarly, if $$f'$$ changes from negative to positive there is a local minimum at the point, and $$f'$$ is increasing. If $$f''>0$$ at the point, this tells us that $$f'$$ is increasing, and so there is a local minimum.
Example 5.3.1
Consider again $$f(x)=\sin x + \cos x$$, with $$f'(x)=\cos x-\sin x$$ and $$f''(x)=-\sin x -\cos x$$. Since
$f''(\pi/4)=-\sqrt{2}/2-\sqrt2/2=-\sqrt2 < 0,$
we know there is a local maximum at $$\pi/4$$. Since
$f''(5\pi/4)=--\sqrt{2}/2--\sqrt2/2=\sqrt2>0,$
there is a local minimum at $$5\pi/4$$.
When it works, the second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails, and sometimes the second derivative is quite difficult to evaluate; in such cases we must fall back on one of the previous tests.
Example 5.3.2
Let $$f(x)=x^4$$. The derivatives are $$f'(x)=4x^3$$ and $$f''(x)=12x^2$$. Zero is the only critical value, but $$f''(0)=0$$, so the second derivative test tells us nothing. However, $$f(x)$$ is positive everywhere except at zero, so clearly $$f(x)$$ has a local minimum at zero. On the other hand, $$f(x)=-x^4$$ also has zero as its only critical value, and the second derivative is again zero, but $$-x^4$$ has a local maximum at zero.
## Exercises 5.3
Find all local maximum and minimum points by the second derivative test.
Ex 5.3.1$$y=x^2-x$$ (answer)
Ex 5.3.2$$y=2+3x-x^3$$ (answer)
Ex 5.3.3$$y=x^3-9x^2+24x$$ (answer)
Ex 5.3.4$$y=x^4-2x^2+3$$ (answer)
Ex 5.3.5$$y=3x^4-4x^3$$ (answer)
Ex 5.3.6$$y=(x^2-1)/x$$ (answer)
Ex 5.3.7$$y=3x^2-(1/x^2)$$ (answer)
Ex 5.3.9$$y = 4x+\sqrt{1-x}$$ (answer)
Ex 5.3.10$$y = (x+1)/\sqrt{5x^2 + 35}$$ (answer)
Ex 5.3.11$$y= x^5 - x$$ (answer)
Ex 5.3.12$$y = 6x +\sin 3x$$ (answer)
Ex 5.3.13$$y = x+ 1/x$$ (answer)
Ex 5.3.14$$y = x^2+ 1/x$$ (answer)
Ex 5.3.15$$y = (x+5)^{1/4}$$ (answer)
Ex 5.3.16$$y = \tan^2 x$$ (answer)
Ex 5.3.17$$y =\cos^2 x - \sin^2 x$$ (answer)
Ex 5.3.18$$y = \sin^3 x$$ (answer) | 1,091 | 3,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2017-04 | latest | en | 0.80285 |
http://www.slideserve.com/amethyst-gardner/entanglement-interpretation-of-black-hole-entropy-in-string-theory | 1,493,071,103,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119838.12/warc/CC-MAIN-20170423031159-00462-ip-10-145-167-34.ec2.internal.warc.gz | 680,414,713 | 19,854 | # Entanglement interpretation of black hole entropy in string theory - PowerPoint PPT Presentation
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Entanglement interpretation of black hole entropy in string theory. Amos Yarom. Ram Brustein. Martin Einhorn. What is entanglement entropy?. What does BH entropy mean?. BH Microstates Entanglement entropy Horizon states. How does it relate to BH entropy?.
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Entanglement interpretation of black hole entropy in string theory
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## Entanglement interpretation of black hole entropy in string theory
Amos Yarom.
Ram Brustein.
Martin Einhorn.
What is entanglement entropy?
### What does BH entropy mean?
• BH Microstates
• Entanglement entropy
• Horizon states
How does it relate to BH entropy?
How does string theory evaluate BH entropy?
How are these two methods relate to each other?
All |↓22↓|
elements
1
2
### Entanglement entropy
S=0
S1=Trace (r1lnr1)=ln2
S2=Trace (r2lnr2)=ln2
r0
### Black holes
f(r0)=0
Coordinate singularity
Space-time singularity
f(0)=-
r=0
t
r=r0
t
x
“Kruskal” extension
r=0
t
r=r0
x
x
### “Kruskal” extension
The vacuum state
r=0
t
r=r0
x
|0
Trin(y’ y’’
rout(y’1,y’’1) =
Exp[-SE] DfD2
f(x,0+)=y’(x)
f(x,0)=y(x)
f(x,0+)=y’(x)
f(x,0-)=y’’(x)
t
f(x,0-)=y’’(x)
out y’1 y’’1 Exp[-SE] Df
f(x,0+) = y’1(x)y2(x)
y’(x)
y’’(x)
f(x,0-) = y’’1(x)y2(x)
x
f(x,0+) = y’1(x)
f(x,0-) = y’’1(x)
Finding rout
Kabat & Strassler (1994), R. Brustein, M. Einhorn and A.Y. (2005)
t
out y’1 y’’1 Exp[-SE] Df
y’1(x)
x
y’’1(x)
f(x,0+) = y’1(x)
f(x,0-) = y’’1(x)
Finding rin
Kabat & Strassler (1994), R. Brustein, M. Einhorn and A.Y. (2005)
’| e-bH|’’
b=T-1=f ’(r0)/4p
### BTZ BH
t
x
BTZ BH
What is entanglement entropy?
What is entanglement entropy of BH’s
How does string theory evaluate BH entropy?
How are these two methods relate to each other?
Black hole entanglement entropy
S.P. de Alwis, N. Ohta, (1995)
?
### BH entropy in string theory
TBH
TFT
=
SBH
=
SFT(TBH)
Anti deSitter
+BH
CFT
What is entanglement entropy?
What is entanglement entropy of BH’s
How does string theory evaluate BH entropy?
How are these two methods relate to each other?
S/A
1/R
Free theory:
l 0
Semiclassical gravity:
R>>ls
S. S. Gubser, I. R. Klebanov, and A. W. Peet (1996)
, T>0
S=A/3
SBH=A/4
How to relate them?
?
?
### Dualities
R. Brustein, M. Einhorn and A.Y. (2005)
### Dualities
R. Brustein, M. Einhorn and A.Y. (2005)
Tracing
Tracing
Dualities
R. Brustein, M. Einhorn and A.Y. (2005)
=
t
q
r
### Explicit construction: BTZ BH
Maldacena and Strominger (1998), Marolf and Louko (1998), Maldacena (2003)
CFTCFT, T=0
CFT, T>0
|0
### Consequences
R. Brustein and A.Y. (2003)
Area scaling
### Area scaling of correlation functions
EE = V V E(x) E(y) ddx ddy
= V V FE(|x-y|) ddx ddy
= D(x) FE(x) dx
= D(x) 2g(x) dx
= - ∂x(D(x)/xd-1) xd-1 ∂xg(x) dx
Geometric term:
Operator dependent term
D(x)=V V d(xxy) ddx ddy
### Geometric term
D(x)= V V d(xxy) ddx ddy
D(x)= d(xr) ddr ddR
ddR V + Ax +O(x2)
d(xr) ddr xd-1 +O(xd)
D(x)=C1Vxd-1 ± C2 Axd + O(xd+1)
### Area scaling of correlation functions
EE = V V E(x) E(y) ddx ddy
= V1 V2 FE(|x-y|) ddx ddy
= D(x) FE(x) dx
= D(x) 2g(x) dx
= - ∂x(D(x)/xd-1) xd-1 ∂xg(x) dx
UV cuttoff at x~1/L
∂ x(D(x)/xd-1) 1/L
A
D(x)=C1Vxd-1 + C2 Axd + O(xd+1)
Consequences
R. Brustein M. Einhorn and A.Y. (in progress)
Non unitary evolution
Consequences
R. Brustein M. Einhorn and A.Y. (in progress)
### Summary
• BH entropy is a result of:
• Entanglement
• Microstates
• Counting of states using dual FT’s is consistent with entanglement entropy.
End
Srednicki (1993)
S1=S2 | 1,746 | 4,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-17 | longest | en | 0.652118 |
http://math.stackexchange.com/questions/393532/same-eigenvalues-different-eigenvectors-but-orthogonal | 1,469,599,416,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00039-ip-10-185-27-174.ec2.internal.warc.gz | 154,324,656 | 17,546 | # Same eigenvalues, different eigenvectors but orthogonal
I am using a two different computational libraries to calculate the eigenvectors and eigenvalues of a symmetric matrix. The results show that the eigenvalues calculated with both libraries are exactly the same, however, the eigenvectors differ. Nevertheless, both seem to be correct since their eigenvectors are orthogonal and the factorization is also correct. How can that be possible? I asked the developers and although they seemed confused, they said that "probably" has to do with the directions of the vectors.
If it is possible, is there a kind of margin error between those two different range of eigenvectors so I could compare them and make sure they are in range (they are right)?
-
The eigenvalues must be the same, not the eigenvectors. For instance, $I_2$ has $(1,0)$ and $(0,1)$ as natural basis of eigenvectors. But $(1,1)$ and $(1,-1)$ is also a basis of eigenvectors for $I_2$. – 1015 May 16 '13 at 13:40
Have you ruled out that the dimension of the eigenspace is more than 1? – nayrb May 16 '13 at 13:40
@vadim123 A dimension one space too has many possible bases. – 1015 May 16 '13 at 13:42
If the developers themselves don't know, it is unlikely that anybody else will know how they designed their algorithm. – 1015 May 16 '13 at 13:47
@Manolete Probably because they're using slightly different algorithms to compute them... Without looking at the algorithms closely, it's impossible to say where exactly the difference lies, though. It's also rather irrelevant, the important thing to take away from this is that the eigenvectors are not uniquely determined by the matrix. – fgp May 16 '13 at 13:56
If $v$ is an eigenvector for $\lambda$, then so is $\alpha v$ for every $\alpha$ in your field $F$. It is not alarming to have different eigenvectors for a single eigenvalue.
It is also not alarming to have linearly independent eigenvectors with the same eigenvalue. In fact, since your matrix is symmetric, it is diagonalizable, and so an eigenvalue of multiplicity $n$ must have $n$ linearly independent eigenvectors for that eigenvalue. | 521 | 2,124 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-30 | latest | en | 0.944526 |
https://www.education.com/resources/geometry/?page=52 | 1,620,460,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988850.21/warc/CC-MAIN-20210508061546-20210508091546-00100.warc.gz | 769,602,582 | 25,344 | # Geometry Resources
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### Geometry Resources
Geometry is a math subject that deals with shapes and their properties and spans many grades. Starting in preschool, students are introduced to 2D shapes. As they get older, students learn how to plot coordinates on a graph and work with 3D shapes and angles. Geometry can be a tough subject, but our resources will help your student become a master in no time!
### Geometry Basics
Need to brush up on your geometry skills? Geometry deals with the discovery of patterns and calculation of areas, volumes, lengths, and angles. Read about some common concepts below, then move on to our geometry resources!
The circle
In geometry, students are introduced to some new mathematical terms relating to circles. Pi, commonly denoted by the π symbol, is a mathematical constant and is usually approximated as 3.14159. The radius of a circle is the distance from the middle of the circle to any point on the circle, while diameter is two times the radius. Lastly, the circumference is the distance around the circle once. These terms can then be found in some important formulas for circles:
• Area = π x (radius2)
• Circumference = 2 x π x radius or Circumference = π x diameter
Polygons
Students are introduced to polygons as early as preschool. Polygons are defined as closed, flat (or two dimensional) shapes with at least three straight lines. To study some of these shapes in more depth, visit our resources on triangles (3 sides), quadrilaterals (4 sides), pentagons (5 sides), and hexagons (6 sides).
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While two dimensional shapes only have a width and height, three dimensional shapes add the third dimension of depth. To learn more about what these shapes are, check out our 3D shapes resources. With this third dimension, the properties of the shapes change. Three dimensional shapes have the properties of:
• Surface area: the total area of all surfaces, or the amount of area you would be able to paint over.
• Volume: the amount of space inside the shape. For example, the amount of water it could hold. For more information on this subject, see our resources on volume.
Geometry covers many topics, and your child is sure to enjoy learning more about all of these math topics as they go through our resources! | 728 | 3,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-21 | latest | en | 0.909774 |
http://forums.informationbuilders.com/eve/forums/a/tpc/f/7971057331/m/3327043776?r=8567041486 | 1,669,711,531,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00665.warc.gz | 20,762,286 | 13,438 | Focal Point Focal Point Forums WebFOCUS/FOCUS Forum on Focal Point [CASE OPENED] Maximum Limit of Alphanumeric Define/Compute Fields
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[CASE OPENED] Maximum Limit of Alphanumeric Define/Compute Fields
Master
posted July 20, 2015 03:53 AM
As per documentation(WF8), maximum limit for alpha fields is 4095 characters.
But below sample code uses around 30,000 for field XYZ_1 and it still works fine in 8009.
Any one foresee any issues with beyond 4K characters?
```-* File test_max_alpha_define.fex
-SET &XYZ='12312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121211212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312122121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212312121231212121212121212121212121212121231212123121212121212121212121212121212123121212312121212121212121212121212121212312121231212121212121212121212121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-TYPE &XYZ.LENGTH
-SET &XYZ=&XYZ || 'ABC';
-TYPE XYZ=&XYZ
-TYPE &XYZ.LENGTH
SET WEBVIEWER = ON
SET WEBVIEWALLPG = OFF
SET WEBVIEWCLOSE = OFF
SET WEBVIEWTARG = OFF
SET WEBVIEWHELP = OFF
APP PATH IBISAMP
DEFINE FILE CAR
XYZ/I11 = &XYZ.LENGTH;
XYZ_1/A30600 = '&XYZ.EVAL';
END
TABLE FILE CAR
PRINT
CAR
COUNTRY
DEALER_COST
RETAIL_COST
XYZ
XYZ_1
ON TABLE HOLD FORMAT ALPHA
END
-RUN
TABLE FILE HOLD
PRINT
*
""
END
-RUN
```
Thanks,
Ram
This message has been edited. Last edited by: Ram Prasad E,
Posts: 542 | Location: Dearborn, MI | Registered: June 03, 2009 IP
Virtuoso
posted July 20, 2015 11:00 AM Hide Post
The 4095 limit is a storage limit for fixed format files, 3968 is for FOCUS file storage and 4096 for XFOCUS field storage.
There is a 32k (roughly) limit that is for SQL field actual/usage, so WF is built to handle this length of Alpha fields.
Alan.
WF 7.705/8.007
Posts: 1451 | Location: Portugal | Registered: February 07, 2007 IP
Master
posted July 20, 2015 01:21 PM Hide Post
Posts: 542 | Location: Dearborn, MI | Registered: June 03, 2009 IP
Master
posted July 29, 2015 12:23 AM Hide Post
I have opened a case with IBI for this.
-Ram
Posts: 542 | Location: Dearborn, MI | Registered: June 03, 2009 IP
Gold member
posted October 06, 2016 04:45 AM Hide Post
quote:
low sample code uses around 30,
Hi Ram,
Just reopening this.. What can be the maximum length that Alphanumeric can hold. i.e. A&listbox6.LENGTH
-SET &Phone_no = STRREP(&listbox6.LENGTH,&listbox6,02,'OR',1,',',&listbox6.LENGTH,'A&listbox6.LENGTH');
Thanks,
Ramya
WebFocus 7702
HTML
Posts: 73 | Registered: January 02, 2012 IP
Gold member
posted October 06, 2016 06:51 AM Hide Post
Hi Ram
The function is not happening if the value exceeds A4096
Thanks,
Ramya
WebFocus 7702
HTML
Posts: 73 | Registered: January 02, 2012 IP | 10,385 | 31,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.575675 |
juzzcurious.com | 1,624,043,592,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00128.warc.gz | 314,638,091 | 148,923 | # Language of Equations – What is E = mc2 ?
6 minutes
#### Equations of Physics
In our school days or college days we had many problems and most of them were not relationship problems but they were real mathematics and physics problems. Still many of us find difficulty when we are dealing with the equation and formulas. There are tons of equations in physics also; equations are quite normal for you if you’re a physicist, most probably these would have been the reason for many to hate physics in your school days. Literally equation will speak to us if we are ready to hear from that and I know it sounds maniac, you may ask how the equation E=mc2 ; and F=ma is going to speak to us.
#### Language of Equations
Language can explained as a medium or tool to share our thoughts with others and it make sense if the speaker and observer are familiar with that language. This is what happening when we are dealing with the equation, we can’t understand what they speak to us because we don’t understand it. There is a famous quote saying “Mathematics is the language of Physics” and most of the theory is initially proven by equation, not by experiment. Still we are sticking with equation from launching our rocket into space and landing our probe on the surface of moon, from the working of computer to the working of mobile phone and so many are there..! I hope now you understand the importance of equations.
#### Popular Equation
‘E=mc2 ’ is not the only famous equation but I noticed people using this equation often in many place like in a physics seminar, science exhibition, note-books and so on. At first I too didn’t understand this equation and later it was surprising me because the way Einstein derived this equation was impressive. This equation was the product of the Einstein’s theory of relativity; This equation of Einstein didn’t bought him Noble prize but photoelectric Theory. This equation is very popular in physics which doesn’t mean this is the greatest equation in physics. There are many equation like this which are even more greater in their way. For an example Schrodinger equation, which are super hard to understand itself but its role is a must in quantum physics.
“m=E/c^2” :
This “m=E/c^2 ” is the format of equation in the Einstein’s paper, later it is converted into “ ”. Let us get some glimpse about the world famous “E=mc2” equation. The reason behind the birth of this equation is law of conservation of energy and it was explained in our previous article – “Ocean of Light”, do check to get some idea about that law. So according to that law the energy is conserved; Energy is nothing but capacity to do a work. In this equation we can notice “E” is referred as Energy, “m” is referred as mass and “c” referred as velocity of light in air medium. In olden days people thought that energy is different from mass, in simple words they thought energy of an object does not depend on the weight of the particle.
This thought was broken by the Great equation of Einstein, this equation relates the energy of a particle with its mass of the particle.
Let us imagine you are going to a meat shop and you see a live ‘1 kg chicken’ there which seems enough for your family. The vendor weighs that chicken and gives it to the butcher to clean and cut it into small pieces. After all process the package reaches your hand and again you weighed the package, you can notice the 1 kg chicken is reduced to around 800 grams. Where the remain weight of the chicken is gone? Obviously you can’t argue with vendor for stealing your meat instead the loss of weight is due to the feathers and wastes of chicken. So the total weight of the package you got and the weight of the wastes cleaned by the butcher must be equal to 1 kg (Weight of live chicken).
Now let us see the same thing in a very small scale. Let us take a Hydrogen atom which is made up of two Sub-atomic particles namely Proton and Electron. Let us take mass of a proton as ‘A’ and mass of an electron as ‘B’, so the mass of a whole hydrogen atom must be ‘A+B’. But it is not correct in this case because there is a loss of mass like we saw in the case of chicken.
#### Who took that mass?
The answer for these question is given only by Einstein’s equation, the lost mass is converted into energy and that is the energy which is used to bind atom together, that energy is known as Binding Energy. This was the mind blowing solution for this problem.
#### What do we get?
This equation laid foundation for may useful things in the modern world and hence marks the birth of Nuclear Fusion. The Nuclear Fusion was very important in those days in order to produce energy for the Industrial revolution and Nuclear weapon for the World War. Even though Einstein is considered as a great mind in physics, it was his letter to the President of America which insisted them to built a Nuclear Weapon, which America dropped on Japan during the World war. This horrifying incident made the world to understand the enormous energy of the Nuclear Weapon and also made Einstein to regret for his letter. If you want to know more about nuclear physics, leave in the comment.
Even light is a form of energy and some amount of mass is transferred into light. For example when we turn on a flash light the mass inside the battery is converted into light energy and you may find a reduction in the mass of the flash light, but it is too small to identify the reduced mass in our world. Another great example is not only sun but all stars loss some of their mass and convert them into light energy. Don’t worry that the Sun will die because we still have 5 Million years ahead.
This Equation of Einstein relates the mass and energy under a constant of speed of light. There are contribution of many scientist to built this equation but those clues only helped Einstein to give birth to this equation. We should remember when Einstein published this paper he was a young man. To understand how the equation is constructed we need to get some knowledge in the theory of relativity. This is most famous equation of Einstein and Einstein worked a lot on this equation but it didn’t bring him Noble prize and made him famous. There are some gossips that during his time many didn’t understood the theory of relativity but the sad reality is that many times Legendary things do not get proper recognition.
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I am on the search for such information. | 23,769 | 105,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-25 | latest | en | 0.978383 |
https://www.univerkov.com/the-ball-was-thrown-vertically-upward-at-a-speed-of-4-m-s-in-0-8-s-it-will-cover-the-path/ | 1,642,477,621,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00535.warc.gz | 1,111,891,317 | 6,173 | # The ball was thrown vertically upward at a speed of 4 m / s. In 0.8 s, it will cover the path.
At the top of the ball trajectory V = 0:
V = V0 – gt = 0,
where V0 is the initial speed of the ball, V0 = 4 m / s;
g – acceleration due to gravity, g = 10 m / s2.
Ball lifting time: t1 = V0 / g = 4/10 = 0.4 s.
Maximum lifting height of the ball:
h1 = V0 t1 −g t1 ^ 2/2 = 4 × 0.4 – 10 × 0.4 ^ 2/2 = 0.8 m.
Ball drop time: t2 = t – t1 = 0.8 – 0.4 = 0.4 s.
Lowering the ball in time t2:
h2 = g t ^ 2 2/2 = 10 × 0.4 ^ 2/2 = 0.8 m.
Thus, in time t = 0.8 s the ball will cover the path:
s = h1 + h2 = 0.8 + 0.8 = 1.6 m.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 329 | 933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-05 | longest | en | 0.898113 |
http://www.math.utexas.edu/pipermail/maxima/2006/000383.html | 1,369,224,434,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701670866/warc/CC-MAIN-20130516105430-00089-ip-10-60-113-184.ec2.internal.warc.gz | 577,663,979 | 2,447 | FW: [Maxima] bug report, or am I doing something wrong?
Luke Sharkey 99LSharkey at ormskirk.lancs.sch.uk
Fri Feb 10 04:55:26 CST 2006
```So does nobody have any feedback for me on the log question, etc.?
-----Original Message-----
From: Luke Sharkey
Sent: 06 February 2006 11:01
To: 'maxima-admin at math.utexas.edu'
Subject: RE: [Maxima] bug report, or am I doing something wrong?
"Maxima's choice of branches for multi-valued functions."
So what can be done about it: when can we know whether the calculated
integral / differential is more likely to be corrector not? Am I just
going to have to plot it each time?
Also: (this caused me a *lot* of headache before I realised what was
going on).
Its about the log() function. I checked the documentation, and finally
realised that log() actually means the *natural logarithm*.
Firstly, how does one do log to the base 10 with maxima?
Secondly, on every calculator, maths program (Microsoft Excel, etc) and
textbook I have ever used, "log" has ALWAYS meant "log to the base 10",
and "ln()" log to the base e", unless it is specified in some way, e.g.
log2() meaning log to the base 2. I propose that the "log" function
should be changed in meaning from natural log to log to the base 10. I
personally think a new function should be created as ln() for natural
logarithms.
At the very least, every time the log function is used, some text should
appear below indicating that the such and such a calculation involving a
log has been "calculated with log to the base xyz" underneath the
output, so people like me don't use it assuming that "log" means "log to
the base 10".
Thirdly, programs like Excel use "LOG" in uppercase. Maxima should be
able to recognise when "LOG" is typed, or copied and pasted in from
Excel, that it should be recognised as "log": this should be automatic.
Thanks.
L Sharkey
________________________________________
From: maxima-admin at math.utexas.edu [mailto:maxima-admin at math.utexas.edu]
On Behalf Of Robert Dodier
Sent: 31 January 2006 15:45
To: Maxima List
Cc: Luke Sharkey
Subject: [Maxima] Fwd: bug report, or am I doing something wrong?
A quick look suggests the results below stem from Maxima's
choice of branches for multi-valued functions.
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``` | 651 | 2,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | latest | en | 0.915589 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/84/1/p/a/ | 1,723,664,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00301.warc.gz | 668,214,513 | 57,167 | # Properties
Label 84.1.p.a Level $84$ Weight $1$ Character orbit 84.p Analytic conductor $0.042$ Analytic rank $0$ Dimension $2$ Projective image $D_{3}$ CM discriminant -3 Inner twists $4$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [84,1,Mod(53,84)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(84, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([0, 3, 4]))
N = Newforms(chi, 1, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("84.53");
S:= CuspForms(chi, 1);
N := Newforms(S);
Level: $$N$$ $$=$$ $$84 = 2^{2} \cdot 3 \cdot 7$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 84.p (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$0.0419214610612$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\zeta_{6})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{2} - x + 1$$ x^2 - x + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image: $$D_{3}$$ Projective field: Galois closure of 3.1.588.1 Artin image: $C_3\times S_3$ Artin field: Galois closure of 6.0.21168.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
The $$q$$-expansion and trace form are shown below.
$$f(q)$$ $$=$$ $$q + \zeta_{6}^{2} q^{3} - \zeta_{6} q^{7} - \zeta_{6} q^{9} +O(q^{10})$$ q + z^2 * q^3 - z * q^7 - z * q^9 $$q + \zeta_{6}^{2} q^{3} - \zeta_{6} q^{7} - \zeta_{6} q^{9} - q^{13} + \zeta_{6} q^{19} + q^{21} + \zeta_{6}^{2} q^{25} + q^{27} - \zeta_{6}^{2} q^{31} + \zeta_{6} q^{37} - \zeta_{6}^{2} q^{39} - q^{43} + \zeta_{6}^{2} q^{49} - q^{57} - 2 \zeta_{6} q^{61} + \zeta_{6}^{2} q^{63} - \zeta_{6}^{2} q^{67} - \zeta_{6}^{2} q^{73} - \zeta_{6} q^{75} + \zeta_{6} q^{79} + \zeta_{6}^{2} q^{81} + \zeta_{6} q^{91} + \zeta_{6} q^{93} + 2 q^{97} +O(q^{100})$$ q + z^2 * q^3 - z * q^7 - z * q^9 - q^13 + z * q^19 + q^21 + z^2 * q^25 + q^27 - z^2 * q^31 + z * q^37 - z^2 * q^39 - q^43 + z^2 * q^49 - q^57 - 2*z * q^61 + z^2 * q^63 - z^2 * q^67 - z^2 * q^73 - z * q^75 + z * q^79 + z^2 * q^81 + z * q^91 + z * q^93 + 2 * q^97 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - q^{3} - q^{7} - q^{9}+O(q^{10})$$ 2 * q - q^3 - q^7 - q^9 $$2 q - q^{3} - q^{7} - q^{9} - 2 q^{13} + q^{19} + 2 q^{21} - q^{25} + 2 q^{27} + q^{31} + q^{37} + q^{39} - 2 q^{43} - q^{49} - 2 q^{57} - 2 q^{61} - q^{63} + q^{67} + q^{73} - q^{75} + q^{79} - q^{81} + q^{91} + q^{93} + 4 q^{97}+O(q^{100})$$ 2 * q - q^3 - q^7 - q^9 - 2 * q^13 + q^19 + 2 * q^21 - q^25 + 2 * q^27 + q^31 + q^37 + q^39 - 2 * q^43 - q^49 - 2 * q^57 - 2 * q^61 - q^63 + q^67 + q^73 - q^75 + q^79 - q^81 + q^91 + q^93 + 4 * q^97
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/84\mathbb{Z}\right)^\times$$.
$$n$$ $$29$$ $$43$$ $$73$$ $$\chi(n)$$ $$-1$$ $$1$$ $$-\zeta_{6}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
53.1
0.5 + 0.866025i 0.5 − 0.866025i
0 −0.500000 + 0.866025i 0 0 0 −0.500000 0.866025i 0 −0.500000 0.866025i 0
65.1 0 −0.500000 0.866025i 0 0 0 −0.500000 + 0.866025i 0 −0.500000 + 0.866025i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
7.c even 3 1 inner
21.h odd 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 84.1.p.a 2
3.b odd 2 1 CM 84.1.p.a 2
4.b odd 2 1 336.1.bn.a 2
5.b even 2 1 2100.1.bn.c 2
5.c odd 4 2 2100.1.bh.a 4
7.b odd 2 1 588.1.p.a 2
7.c even 3 1 inner 84.1.p.a 2
7.c even 3 1 588.1.c.b 1
7.d odd 6 1 588.1.c.a 1
7.d odd 6 1 588.1.p.a 2
8.b even 2 1 1344.1.bn.b 2
8.d odd 2 1 1344.1.bn.a 2
9.c even 3 1 2268.1.m.a 2
9.c even 3 1 2268.1.bh.b 2
9.d odd 6 1 2268.1.m.a 2
9.d odd 6 1 2268.1.bh.b 2
12.b even 2 1 336.1.bn.a 2
15.d odd 2 1 2100.1.bn.c 2
15.e even 4 2 2100.1.bh.a 4
21.c even 2 1 588.1.p.a 2
21.g even 6 1 588.1.c.a 1
21.g even 6 1 588.1.p.a 2
21.h odd 6 1 inner 84.1.p.a 2
21.h odd 6 1 588.1.c.b 1
24.f even 2 1 1344.1.bn.a 2
24.h odd 2 1 1344.1.bn.b 2
28.d even 2 1 2352.1.bn.a 2
28.f even 6 1 2352.1.d.b 1
28.f even 6 1 2352.1.bn.a 2
28.g odd 6 1 336.1.bn.a 2
28.g odd 6 1 2352.1.d.a 1
35.j even 6 1 2100.1.bn.c 2
35.l odd 12 2 2100.1.bh.a 4
56.k odd 6 1 1344.1.bn.a 2
56.p even 6 1 1344.1.bn.b 2
63.g even 3 1 2268.1.bh.b 2
63.h even 3 1 2268.1.m.a 2
63.j odd 6 1 2268.1.m.a 2
63.n odd 6 1 2268.1.bh.b 2
84.h odd 2 1 2352.1.bn.a 2
84.j odd 6 1 2352.1.d.b 1
84.j odd 6 1 2352.1.bn.a 2
84.n even 6 1 336.1.bn.a 2
84.n even 6 1 2352.1.d.a 1
105.o odd 6 1 2100.1.bn.c 2
105.x even 12 2 2100.1.bh.a 4
168.s odd 6 1 1344.1.bn.b 2
168.v even 6 1 1344.1.bn.a 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
84.1.p.a 2 1.a even 1 1 trivial
84.1.p.a 2 3.b odd 2 1 CM
84.1.p.a 2 7.c even 3 1 inner
84.1.p.a 2 21.h odd 6 1 inner
336.1.bn.a 2 4.b odd 2 1
336.1.bn.a 2 12.b even 2 1
336.1.bn.a 2 28.g odd 6 1
336.1.bn.a 2 84.n even 6 1
588.1.c.a 1 7.d odd 6 1
588.1.c.a 1 21.g even 6 1
588.1.c.b 1 7.c even 3 1
588.1.c.b 1 21.h odd 6 1
588.1.p.a 2 7.b odd 2 1
588.1.p.a 2 7.d odd 6 1
588.1.p.a 2 21.c even 2 1
588.1.p.a 2 21.g even 6 1
1344.1.bn.a 2 8.d odd 2 1
1344.1.bn.a 2 24.f even 2 1
1344.1.bn.a 2 56.k odd 6 1
1344.1.bn.a 2 168.v even 6 1
1344.1.bn.b 2 8.b even 2 1
1344.1.bn.b 2 24.h odd 2 1
1344.1.bn.b 2 56.p even 6 1
1344.1.bn.b 2 168.s odd 6 1
2100.1.bh.a 4 5.c odd 4 2
2100.1.bh.a 4 15.e even 4 2
2100.1.bh.a 4 35.l odd 12 2
2100.1.bh.a 4 105.x even 12 2
2100.1.bn.c 2 5.b even 2 1
2100.1.bn.c 2 15.d odd 2 1
2100.1.bn.c 2 35.j even 6 1
2100.1.bn.c 2 105.o odd 6 1
2268.1.m.a 2 9.c even 3 1
2268.1.m.a 2 9.d odd 6 1
2268.1.m.a 2 63.h even 3 1
2268.1.m.a 2 63.j odd 6 1
2268.1.bh.b 2 9.c even 3 1
2268.1.bh.b 2 9.d odd 6 1
2268.1.bh.b 2 63.g even 3 1
2268.1.bh.b 2 63.n odd 6 1
2352.1.d.a 1 28.g odd 6 1
2352.1.d.a 1 84.n even 6 1
2352.1.d.b 1 28.f even 6 1
2352.1.d.b 1 84.j odd 6 1
2352.1.bn.a 2 28.d even 2 1
2352.1.bn.a 2 28.f even 6 1
2352.1.bn.a 2 84.h odd 2 1
2352.1.bn.a 2 84.j odd 6 1
## Hecke kernels
This newform subspace is the entire newspace $$S_{1}^{\mathrm{new}}(84, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2} + T + 1$$
$5$ $$T^{2}$$
$7$ $$T^{2} + T + 1$$
$11$ $$T^{2}$$
$13$ $$(T + 1)^{2}$$
$17$ $$T^{2}$$
$19$ $$T^{2} - T + 1$$
$23$ $$T^{2}$$
$29$ $$T^{2}$$
$31$ $$T^{2} - T + 1$$
$37$ $$T^{2} - T + 1$$
$41$ $$T^{2}$$
$43$ $$(T + 1)^{2}$$
$47$ $$T^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2}$$
$61$ $$T^{2} + 2T + 4$$
$67$ $$T^{2} - T + 1$$
$71$ $$T^{2}$$
$73$ $$T^{2} - T + 1$$
$79$ $$T^{2} - T + 1$$
$83$ $$T^{2}$$
$89$ $$T^{2}$$
$97$ $$(T - 2)^{2}$$ | 3,860 | 7,525 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.490724 |
https://rdrr.io/github/rossellhayes/nombre/man/ratio.html | 1,680,311,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00623.warc.gz | 536,698,625 | 7,539 | # ratio: Convert numbers to ratio character vectors (two to one, one... In rossellhayes/nombre: Number Names
ratio R Documentation
## Convert numbers to ratio character vectors (two to one, one in three, five out of ten)
### Description
Convert numbers to ratio character vectors (two to one, one in three, five out of ten)
### Usage
ratio(x, sep = "in", max_n = Inf, negative = "negative", ...)
nom_ratio(x, sep = "in", max_n = Inf, negative = "negative", ...)
### Arguments
x A numeric vector sep A character vector separating components of the ratio. Defaults to "in". max_n A numeric vector. When the absolute value of x is greater than max_n, x remains numeric instead of being converted to words. If max_n is negative, no xs will be converted to words. (This can be useful when max_n is passed by another function.) Defaults to Inf, which converts all xs to words. negative A character vector to append to negative numbers. Defaults to "negative". ... Arguments passed on to fracture::frac_mat denomIf denom is not NULL, all fractions will have a denominator of denom. This will ignore all other arguments that affect the denominator. base_10If TRUE, all denominators will be a power of 10. common_denomIf TRUE, all fractions will have the same denominator. If the least common denominator is greater than max_denom, max_denom is used. max_denomAll denominators will be less than or equal to max_denom. If base_10 is TRUE, the maximum denominator will be the largest power of 10 less than max_denom. A max_denom greater than the inverse square root of machine double epsilon will produce a warning because floating point rounding errors can occur when denominators grow too large.
### Details
x is converted to a fraction by fracture::frac_mat().
### Value
A character vector of the same length as x
Other number names: adverbial(), cardinal(), collective(), denominator(), numerator(), ordinal()
### Examples
paste0("Our team is outnumbered ", nom_ratio(10), ".")
paste0("The chances of winning are ", nom_ratio(1/1000000, sep = "in"), ".")
nom_ratio(c(1, 10, 100))
nom_ratio(c(0, 0.5, 1.5))
nom_ratio(c(0, 0.125, 0.625, 1), sep = "out of", common_denom = TRUE)
nom_ratio(5 / 10, sep = "in", base_10 = TRUE)
nom_ratio(6 / 25, sep = "in")
nom_ratio(6 / 25, sep = "out of", max_denom = 10)
rossellhayes/nombre documentation built on June 2, 2022, 10:22 a.m. | 610 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.784064 |
https://www.wordunscrambler.net/unscramble-gtioltc | 1,638,540,378,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00358.warc.gz | 1,120,640,667 | 12,737 | # Unscramble GTIOLTC
GTIOLTC unscrambles into 49 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT!
### 2 letter words made by unscrambling GTIOLTC
#### Unscrambled 9 2 Letter Words
Above are the words made by unscrambling GTIOLTC (CGILOTT). To further help you, here are a few lists related to/with the letters GTIOLTC
### The Value of GTIOLTC In Word Scramble Games
The letters GTIOLTC are worth 10 points in Scrabble
The letters GTIOLTC are worth 13 in points Words With Friends
• G = 3 points in WWF & 2 points in Scrabble
• T = 1 points in WWF & 1 points in Scrabble
• I = 1 points in WWF & 1 points in Scrabble
• O = 1 points in WWF & 1 points in Scrabble
• L = 2 points in WWF & 1 points in Scrabble
• T = 1 points in WWF & 1 points in Scrabble
• C = 4 points in WWF & 3 points in Scrabble
## What Does GTIOLTC Mean... If you Unscramble it?
### Possible Definitions of GTIOLTC
If we unscramble these letters, GTIOLTC, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
### Glottic
• Alt. of Glottidean
## Permutations of GTIOLTC
According to our other word scramble maker, GTIOLTC can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
#### Definition of Permutation
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the letters gtioltc scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time GTIOLTC comes up in a word scramble game.
We stopped it at 49, but there are so many ways to scramble GTIOLTC!
### Scramble Words
Unscramble these letters to make words...
scrambled using word scrambler... | 531 | 1,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.899855 |
https://demo.formulasearchengine.com/wiki/Kolmogorov_structure_function | 1,579,801,711,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00548.warc.gz | 400,152,297 | 29,770 | # Kolmogorov structure function
In 1973 Kolmogorov proposed a non-probabilistic approach to statistics and model selection. Let each data be a finite binary string and models be finite sets of binary strings. Consider model classes consisting of models of given maximal Kolmogorov complexity. The Kolmogorov structure function of an individual data string expresses the relation between the complexity level constraint on a model class and the least log-cardinality of a model in the class containing the data. The structure function determines all stochastic properties of the individual data string: for every constrained model class it determines the individual best-fitting model in the class irrespective of whether the true model is in the model class considered or not. In the classical case we talk about a set of data with a probability distribution, and the properties are those of the expectations. In contrast, here we deal with individual data strings and the properties of the individual string focussed on. In this setting, a property holds with certainty rather than with high probability as in the classical case. The Kolmogorov structure function precisely quantify the goodness-of-fit of an individual model with respect to individual data.
The Kolmogorov structure function is used in the algorithmic information theory, also known as the theory of Kolmogorov complexity, for describing the structure of a string by use of models of increasing complexity.
## Kolmogorov's definition
Kolmogorov (left) talks on the structure function (see drawing on the blackboard) in (Tallinn, 1973).
The structure function was originally proposed by Kolmogorov in 1973 at a Soviet Information Theory symposium in Tallinn, but these results were not published [1] p. 182. But the results were announced in [2] in 1974, the only written record by Kolmogorov himself. One of his last scientific statements is (translated from the original Russian by L.A. Levin):
"To each constructive object corresponds a function ${\displaystyle \Phi _{x}(k)}$ of a
natural number k---the log of minimal cardinality of x-containing sets that allow definitions of complexity at most k. If the element x itself allows a simple definition, then the function ${\displaystyle \Phi }$ drops to 0 even for small k. Lacking such definition, the element is "random" in a negative sense. But it is positively "probabilistically random" only when function ${\displaystyle \Phi }$ having taken the value ${\displaystyle \Phi _{0}}$ at a relatively small ${\displaystyle k=k_{0}}$, then changes approximately as ${\displaystyle \Phi (k)=\Phi _{0}-(k-k_{0})}$.
[Kolmogorov, in the announcement cited above]
## Contemporary definition
It is discussed in.[1] It is extensively studied in [3] where also the main properties are resolved. The Kolmogorov structure function can be written as
${\displaystyle h_{x}(\alpha )=\min _{S}\{\log |S|:x\in S,K(S)\leq \alpha \}}$
where x is a binary string of length n with ${\displaystyle x\in S}$ where S is a contemplated model (set of n-length strings) for x, ${\displaystyle K(S)}$ is the Kolmogorov complexity of S and ${\displaystyle \alpha }$ is a nonnegative integer value bounding the complexity of the contemplated S's. Clearly, this function is nonincreasing and reaches ${\displaystyle \log |\{x\}|=0}$ for ${\displaystyle \alpha =K(x)+c}$ where c is the required number of bits to change x into ${\displaystyle \{x\}}$ and ${\displaystyle K(x)}$ is the Kolmogorov complexity of x.
### The algorithmic sufficient statistic
We define a set S containing x such that
${\displaystyle K(S)+K(x|S)=K(x)+O(1)}$.
The function ${\displaystyle h_{x}(\alpha )}$ never decreases more than a fixed independent constant below the diagonal called sufficiency line L defined by
${\displaystyle L(\alpha )+\alpha =K(x)}$.
It is approached to within a constant distance by the graph of ${\displaystyle h_{x}}$ for certain arguments (for instance, for ${\displaystyle \alpha =K(x)+c}$). For these ${\displaystyle \alpha }$'s we have ${\displaystyle \alpha +h_{x}(\alpha )=K(x)+O(1)}$ and the associated model S (witness for ${\displaystyle h_{x}(\alpha )}$) is called an optimal set for $x$, and its description of ${\displaystyle K(S)\leq \alpha }$ bits is therefore an algorithmic sufficient statistic. We write algorithmic' for Kolmogorov complexity' by convention. The main properties of an algorithmic sufficient statistic are the following: If S is an algorithmic sufficient statistic for x, then
${\displaystyle K(S)+\log |S|=K(x)+O(1)}$.
That is, the two-part description of x using the model S and as data-to-model code the index of x in the enumeration of S in ${\displaystyle \log |S|}$ bits, is as concise as the shortest one-part code of x in ${\displaystyle K(x)}$ bits. This can be easily seen as follows:
${\displaystyle K(x)\leq K(x,S)+O(1)\leq K(S)+K(x|S)+O(1)\leq K(S)+\log |S|+O(1)\leq K(x)+O(1)}$,
using straightforward inequalities and the sufficiency property, we find that ${\displaystyle K(x|S)=\log |S|+O(1)}$. (For example, given ${\displaystyle S\ni x}$, we can describe x self-delimitingly (you can determine its end) in ${\displaystyle \log |S|+O(1)}$ bits.) Therefore, the randomness deficiency ${\displaystyle \log |S|-K(x|S)}$ of x in S is a constant, which means that x is a typical (random) element of S. However, there can be models S containing x that are not sufficient statistics. An algorithmic sufficient statistic S for x has the additional property, apart from being a model of best fit, that ${\displaystyle K(x,S)=K(x)+O(1)}$ and therefore by the Kolmogorov complexity symmetry of information (the information about x in S is about the same as the information about S in x) we have ${\displaystyle K(S|x^{*})=O(1)}$: the algorithmic sufficient statistic S is a model of best fit that is almost completely determined by x. (${\displaystyle x^{*}}$ is a shortest program for x.) The algorithmic sufficient statistic associated with the least such ${\displaystyle \alpha }$ is called the algorithmic minimal sufficient statistic.
With respect to the picture: The MDL structure function ${\displaystyle \lambda _{x}(\alpha )}$ is explained below. The Goodness-of-fit structure function ${\displaystyle \beta _{x}(\alpha )}$ is the least randomness deficiency (see above) of any model ${\displaystyle S\ni x}$ for x such that ${\displaystyle K(S)\leq \alpha }$. This structure function gives the goodness-of-fit of a model S (containing x) for the string x. When it is low the model fits well, and when it is high the model doesn't fit well. If ${\displaystyle \beta _{x}(\alpha )=0}$ for some ${\displaystyle \alpha }$ then there is a typical model ${\displaystyle S\ni x}$ for x such that ${\displaystyle K(S)\leq \alpha }$ and x is typical (random) for S. That is, S is the best-fitting model for x. For more details see [1] and especially [3] and.[4]
### Selection of properties
Within the constraints that the graph goes down at an angle of at least 45 degrees, that it starts at n and ends approximately at ${\displaystyle K(x)}$, every graph (up to a ${\displaystyle O(\log n)}$ additive term in argument and value) is realized by the structure function of some data x and vice versa. Where the graph hits the diagonal first the argument (complexity) is that of the minimum sufficient statistic. It is incomputable to determine this place. See.[3]
### Main property
It is proved that at each level ${\displaystyle \alpha }$ of complexity the structure function allows us to select the best model S for the individual string x within a strip of ${\displaystyle O(\log n)}$ with certainty, not with great probability.[3]
## The MDL variant
The Minimum description length (MDL) function: The length of the minimal two-part code for x consisting of the model cost K(S) and the length of the index of x in S, in the model class of sets of given maximal Kolmogorov complexity ${\displaystyle \alpha }$, the complexity of S upper bounded by ${\displaystyle \alpha }$, is given by the MDL function or constrained MDL estimator:
${\displaystyle \lambda _{x}(\alpha )=\min _{S}\{\Lambda (S):S\ni x,\;K(S)\leq \alpha \},}$
where ${\displaystyle \Lambda (S)=\log |S|+K(S)\geq K(x)-O(1)}$ is the total length of two-part code of x with help of model S.
### Main property
It is proved that at each level ${\displaystyle \alpha }$ of complexity the structure function allows us to select the best model S for the individual string x within a strip of ${\displaystyle O(\log n)}$ with certainty, not with great probability.[3]
### Application in statistics
The mathematics developed above were taken as the foundation of MDL by its inventor Jorma Rissanen.[5]
## Probability models and the Kolmogorov structure function
For every computable probability distribution P it can be proved [6] that ${\displaystyle -\log P(x)=\log |S|+O(\log n)}$. For example, if P is the uniform distribution on the set S of strings of length n, then each ${\displaystyle x\in S}$ has probability ${\displaystyle P(x)=1/|S|}$. In the general case of computable probability mass functions we incur a logarithmic additive error term. Kolmogorov's structure function becomes
${\displaystyle h'_{x}(\alpha )=\min _{P}\{-\log P(x):P(x)>0,K(P)\leq \alpha \}}$
where x is a binary string of length n with ${\displaystyle -\log P(x)>0}$ where P is a contemplated model (computable probability of n-length strings) for x, ${\displaystyle K(P)}$ is the Kolmogorov complexity of P and ${\displaystyle \alpha }$ is an integer value bounding the complexity of the contemplated P's. Clearly, this function is nonincreasing and reaches ${\displaystyle \log |\{x\}|=0}$ for ${\displaystyle \alpha =K(x)+c}$ where c is the required number of bits to change x into ${\displaystyle \{x\}}$ and ${\displaystyle K(x)}$ is the Kolmogorov complexity of x. Then ${\displaystyle h'_{x}(\alpha )=h_{x}(\alpha )+O(\log n)}$. For every complexity level ${\displaystyle \alpha }$ the function ${\displaystyle h'_{x}(\alpha )}$ is the Kolmogorov complexity version of the maximum likelihood (ML).
### Main property
It is proved that at each level ${\displaystyle \alpha }$ of complexity the structure function allows us to select the best model S for the individual string x within a strip of ${\displaystyle O(\log n)}$ with certainty, not with great probability.[3]
## The MDL variant and probability models
The MDL function: The length of the minimal two-part code for x consisting of the model cost K(P) and the length of ${\displaystyle -\log P(x)}$, in the model class of computable probability mass functions of given maximal Kolmogorov complexity ${\displaystyle \alpha }$, the complexity of P upper bounded by ${\displaystyle \alpha }$, is given by the MDL function or constrained MDL estimator:
${\displaystyle \lambda '_{x}(\alpha )=\min _{P}\{\Lambda (P):P(x)>0,\;K(P)\leq \alpha \},}$
where ${\displaystyle \Lambda (P)=-\log P(x)+K(P)\geq K(x)-O(1)}$ is the total length of two-part code of x with help of model P.
### Main property
It is proved that at each level ${\displaystyle \alpha }$ of complexity the MDL function allows us to select the best model P for the individual string x within a strip of ${\displaystyle O(\log n)}$ with certainty, not with great probability.[3]
## Extension to rate distortion and denoising
It turns out that the approach can be extended to a theory of rate distortion of individual finite sequences and denoising of individual finite sequences [7] using Kolmogorov complexity. Experiments using real compressor programs have been carried out with success.[8] Here the assumption is that for natural data the Kolmogorov complexity is not far from the length of a compressed version using a good compressor.
## References
1. {{#invoke:citation/CS1|citation |CitationClass=book }}
2. Abstract of a talk for the Moscow Mathematical Society in Uspekhi Mat. Nauk Volume 29, Issue 4(178) in the Communications of the Moscow Mathematical Society page 155 (in the Russian edition, not translated into English)
3. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
4. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
5. {{#invoke:citation/CS1|citation |CitationClass=book }}
6. A.Kh. Shen, The concept of (α, β)-stochasticity in the Kolmogorov sense, and its properties, Soviet Math. Dokl., 28:1(1983), 295--299
7. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
8. {{#invoke:Citation/CS1|citation |CitationClass=journal }}
## Other literature
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }}
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }}
• {{#invoke:citation/CS1|citation
|CitationClass=book }}, Especially pp. 401–431 about the Kolmogorov structure function, and pp. 613–629 about rate distortion and denoising of individual sequences.
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }}
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }}
• {{#invoke:Citation/CS1|citation
|CitationClass=journal }} | 3,363 | 13,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 76, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | longest | en | 0.904137 |
https://housemetric.co.uk/house-price-analysis/BN24-5/Stone-Cross | 1,716,502,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00328.warc.gz | 253,897,148 | 15,056 | # House prices in BN24 5 (Stone Cross)
This article reveals price per square metre data and various charts to help you understand current housing market in Stone Cross (BN24 5).
## Defining 'BN24 5'
This analysis is limited to properties whose postcode starts with "BN24 5", this is also called the postcode sector. It is shown in red on the map above. There are no official names for postcode sectors so I've just labelled it Stone Cross.
You can click on the map labels to change to a neighbouring sector, or you can enter a different postcode sector (e.g. CM23 4) below.
FYI, a postcode sector is the full postcode without the last two letters.
## Price per square metre
Knowing the average house price in Stone Cross is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property:
£ per sqm = price ÷ internal area
E.g. Elfran, Dittons Road, Stone Cross, sold for £320,000 on Feb-2024. Given the internal area of 96 square metres, the price per sqm is £3,333.
England & Wales have been officially metric since 1965. However house price per square foot is prefered by some estate agents and those of sufficiently advanced age ;-) They may want to convert square meters on this page to square feet.
The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £4,800-£5,000, £5,000-£5,200, £5,200-£5,400 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 158 sales in Stone Cross (BN24 5) that took place in the last two years.
##### Distribution of £ per sqm for Stone Cross
Distribution of £ per sqm house prices in Stone Cross
You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc.
The spread of prices will give you a feel of the typical range to expect in Stone Cross (BN24 5). Of the 158 transactions, half were sold for between £3,630 and £4,480 per square metre. The median, or 'middle', price per square metre in 'BN24 5' is £4,100. Notably, only 25% of properties that sold recently were valued at more than £4,480 sqm. For anything to be valued more than this means it has to be more desireable than the clear majority of homes.
## Price map for Stone Cross
Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Stone Cross. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values.
House price heatmap for Stone Cross
## Comparison with neighbouring postcode sectors
The table below shows how 'BN24 5' compares to neighbouring postcode sectors.
Postcode sector Lower quartile Middle quartile Upper quartile
BN24 5 Stone Cross £3,630 sqm £4,100 sqm £4,480 sqm
BN24 6 Pevensey Bay £3,310 sqm £4,350 sqm £5,330 sqm
## Will Stone Cross house prices drop in 2024?
I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the chart below for BN24 5 (Stone Cross) compared with both the wider area BN24 and inflation (CPIH from the Office of National Statistics). The dashed trend lines in the chart show the average over time.
##### Historic price per square metre in Stone Cross,Stone Cross
House price trends for Stone Cross
For the most recent sales activity, rather than a summarized average, it is better to see the underlying data. This is shown in the chart below, where blue dots represent individual sales, click on them to see details. If there is an obvious trend you should be able to spot it here amid the noise from outliers.
##### Most recent BN24 5 sales
Recent trends for Stone Cross
Data from Land Registry comes in gradually over time. I update it every month but it takes about 5 months for the majority of sales for Stone Cross to be recorded. Disclaimer: I do not verify and cannot guarantee the accuracy of any data shown. Outliers exist in the data, typically these are where the EPC registry records the internal area incorrectly, sometimes although very rarely the Land Registry price paid data can be wrong. The data provided throughout this website about Stone Cross and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information.
## Street level data
Street Avg size Avg £sqm Recent sales
Rattle Road, Stone Cross, BN24 5D 125 sqm £3,815 25
Mallow Drive, Stone Cross, BN24 5G 90 sqm £3,839 16
Wood Sage Way, Stone Cross, BN24 5F 90 sqm £3,836 16
High Street, Stone Cross, BN24 5L 117 sqm £3,021 15
St Johns Drive, Stone Cross, BN24 5H 94 sqm £3,804 12
Beechfield Close, Stone Cross, BN24 5F 80 sqm £4,490 12
Springfield Close, Stone Cross, BN24 5J 82 sqm £4,087 12
Medina Drive, Stone Cross, BN24 5E 76 sqm £4,311 11
## Raw data
Our analysis of Stone Cross is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea. | 1,364 | 5,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.922065 |
https://www.numerade.com/questions/determine-whether-each-integral-is-convergent-or-divergent-evaluate-those-that-are-convergent-disp-8/ | 1,632,705,130,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00278.warc.gz | 934,825,415 | 36,319 | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
WZ
# Determine whether each integral is convergent or divergent. Evaluate those that are convergent.$\displaystyle \int_{-\infty}^\infty (y^3 - 3y^2)\ dy$
## $-\infty$, integral is divergent
#### Topics
Integration Techniques
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### Video Transcript
problem is determine whether age into grotesque wouldn't order and weren't evaluated. Those that are converted but this improper, integral a definition we can write. This's Iko too integral from negative Infinity to zero I just Q minus three times while square. Why us? Integral from zero to infinity function y two threes. Power minus three times. Why square? Why on the way computer The first into girl thiss improper Integral By definition, this's the coach is the limit some number A does too Negative entity from integral from a to zero. It's a function wide too. Three minus three times Why square device? Look at this definite Integral This is Echo two one fourth. Why two? Those power minus three. I'm sorry. Oh yes. Three times one hour, three terms. I too agree from a too zero on DH planning zero and a tooth. This function this is equal to zero minus won the war floor, eh? To force minus and two. Great! This's the code too. A two three sp Our times wass one Force, eh? And alas, what? This is what goes to negative infinity and to cube goes to make to infinity on DH This part goes to hospital in vanity. Negative. Infinity times positive Infinity, Uh, is it goes to negative profanity. So this Integral and Steadward
WZ
#### Topics
Integration Techniques
Lectures
Join Bootcamp | 408 | 1,738 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-39 | longest | en | 0.887647 |
https://oeis.org/A091612 | 1,638,589,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00071.warc.gz | 455,375,875 | 4,123 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A091612 Column 1 of triangle A091603. 2
1, -1, -2, 0, 1, 3, 4, 3, 3, 0, 0, -3, -4, -7, -9, -9, -9, -9, -9, -4, -5, -4, -3, 2, 3, 5, 6, 11, 11, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 11, 10, 13, 13, 10, 11, 10, 10, 3, 4, 0, 1, -2, -2, -3, -3, -9, -8, -7, -7, -19, -19, -19, -19, -19, -20, -21, -21, -21, -21, -27 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS G. C. Greubel, Table of n, a(n) for n = 1..500 MATHEMATICA b[n_, i_, k_]:= b[n, i, k]= If[n==0, 1, If[i>n, 0, Sum[b[n-i*j, i+1, Min[k, Quotient[n-i*j, i+1]]], {j, 0, k}]]]; t[n_, k_]:= t[n, k]= If[k>n, 0, b[n, 1, k] - b[n, 1, k-1]]; (* t = A091602 *) M := With[{p = 110}, Table[t[n, k], {n, p}, {k, p}]]; T := Inverse[M]; (* T = A091603 *) Table[T[[n, 1]], {n, 100}] (* G. C. Greubel, Nov 27 2021 *) CROSSREFS Cf. A091603. Sequence in context: A004542 A207331 A134405 * A253672 A213861 A108458 Adjacent sequences: A091609 A091610 A091611 * A091613 A091614 A091615 KEYWORD sign,changed AUTHOR Christian G. Bower, Jan 23 2004 STATUS approved
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Last modified December 3 22:32 EST 2021. Contains 349468 sequences. (Running on oeis4.) | 724 | 1,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.695046 |
https://www.fool.ca/2020/05/04/heres-what-makes-a-tsx-stock-cheap/ | 1,653,225,817,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00602.warc.gz | 886,882,016 | 67,783 | # Here’s What Makes a TSX Stock Cheap
Deciding if a TSX stock is cheap goes beyond the share price. Numerous factors go into forming the valuation of a business.
No matter what it is that someone’s buying, one thing we all need to know before we can make a purchase is the price of the object. Whether that’s a burger from a restaurant, a new car from the dealership, or a stock on the TSX, it’s almost impossible for a buyer to decide if they want something unless they know the price.
What are we referring to when we reference the price of a TSX stock?
Often beginner investors may think that the price of one individual share of the company has an effect on the valuation of the business as a whole. For example, a company with a stock price of \$50 a share is more expensive than a stock that has a price of \$10 a share.
While investors look at the price of the share to determine a company’s value, it isn’t the price alone that makes up the valuation of the business.
## How price per share is determined
The price of a TSX stock is calculated by taking the price of one share and multiplying it by the number of shares outstanding for the company. This is why looking at just the stock price on its own is irrelevant.
A company with one billion shares outstanding and a stock price of \$50 a share would have a total market value of \$50 billion.
However, a business with only one million shares outstanding but a stock price of \$100 (double that of \$50) would have a total market value of just \$100 million — substantially less than \$ 50 billion.
So, the price of the TSX stock is just the total market value divided by the number of shares.
While this can be complicated, we can make it easier by calculating per-share figures. An example of the most common figure is earnings per share.
## What price do we talk about when we say a TSX stock is cheap?
When we say a business is cheap, we are referring to a business’s value. There are several ways to calculate the value of a business.
Most investors like to use an earnings ratio to determine how much a business is worth.
So, in general, a TSX stock trading for 10 times the amount of net income it earned last year is cheaper than a stock that is trading for a price 20 times the net income it earned in the same year.
Like everything else in investing, there are, however, different ways to calculate the value of a business. Price will always be given to you; it’s the value of the business you have to consider to see if an investment is worth making.
## Understanding prices and values for TSX stocks
So far, when talking about the prices for a TSX stock, I have referenced the market price, or market cap, of the business.
While there is nothing wrong with using this price, it doesn’t tell the full story. In this day and age, almost every publicly traded stock has some type of debt.
Investors need to factor in this debt when considering the value of the stock. This is called the enterprise value; it represents the total cost of a company.
For example, If you saw a house that was worth \$1 million and selling for that price, you may buy it.
However, if that same house was still worth \$1 million, but there was \$750,000 in debt that came with the house as well, would you still pay \$1 million for it?
This is what enterprise value measures. When you buy a stock with a tonne of debt, you have to calculate what effect that has on the valuation of the business.
A TSX stock such as Hydro One, for example, has a market cap of \$14.8 billion but an enterprise value of \$19.9 billion.
## Bottom line
Understanding how a stock is priced and the total value you have to pay for a company is crucial. It’s a prerequisite to making any high-quality, long-term investment. If you don’t know what you’re paying to own a TSX stock, you can’t accurately calculate the value.
So, make sure to do your due diligence to find the cheapest stocks possible.
This article represents the opinion of the writer, who may disagree with the “official” recommendation position of a Motley Fool premium service or advisor. We’re Motley! Questioning an investing thesis — even one of our own — helps us all think critically about investing and make decisions that help us become smarter, happier, and richer, so we sometimes publish articles that may not be in line with recommendations, rankings or other content.
Fool contributor Daniel Da Costa has no position in any of the stocks mentioned.
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The pandemic bubble has burst, and interest rate reality strikes. Make the most of this market with five investing principles. | 1,193 | 5,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.954071 |
https://www.doubtnut.com/question-answer/vertices-of-a-parallelogram-taken-in-order-are-a-2-14-b-10-1-c-123-and-d-the-distance-between-the-pa-38183937 | 1,624,076,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00037.warc.gz | 679,222,207 | 66,078 | Class 12
MATHS
Different Products Of Vectors And Their Geometrical Applications
# Vertices of a parallelogram taken in order are A, ( 2,-1,4) , B (1,0,-1) , C ( 1,2,3) and D. <br> The distance between the parallel lines AB and CD is
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Text Solution
sqrt63sqrt(6//5)2sqrt23
c
Solution :
Let point D be (a_(1),a_(2),a_(3)) <br> a_(1)+1=3or a_(1)=2 <br> a_(2) +0= 1 l or a_(2)=1 <br> a_(3)-1 = 7 or a_(3)=8 <br> a_(3)-1=7 or a_(3)=8 <br> vecd=||vec((AB))xxvec((AD))|/|vec(AB)|| <br> vec(AB)= -hati+hatj-5hatk <br> vec(AD)=0hati+2hatj=4hatk<br> vec(AB)xx vec(AD)=|{:(hati,hatj,hatk),(-1,1,-5),(0,2,4):}| <br> 14hati+4hatj-2hatk <br> 2(7hati+2hatj-hatk)<br>Rightarrow d=2sqrt2 <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_V_3DG_C02_E01_227_S01.png" width="80%">
Image Solution
Find answer in image to clear your doubt instantly:
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5:11 | 707 | 1,576 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-25 | latest | en | 0.492252 |
https://techscience.com/cmc/v67n1/41220 | 1,620,729,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991982.8/warc/CC-MAIN-20210511092245-20210511122245-00618.warc.gz | 588,918,327 | 7,144 | Vol.67, No.1, 2021, pp.1253-1267, doi:10.32604/cmc.2021.014781
OPEN ACCESS
ARTICLE
High Order Block Method for Third Order ODEs
• A. I. Asnor1, S. A. M. Yatim1, Z. B. Ibrahim2, N. Zainuddin3
1 School of Distance Education, Universiti Sains Malaysia, USM Penang, 11800, Malaysia
2 Department of Mathematics, Faculty of Science, Universiti Putra Malaysia, Serdang, Selangor Darul Ehsan, 43400, Malaysia
3 Department of Fundamental and Applied Sciences, Universiti Teknologi PETRONAS, Seri Iskandar, Perak Darul Ridzuan, 32610, Malaysia
Received 16 October 2020; Accepted 28 November 2020; Issue published 12 January 2021
Abstract
Many initial value problems are difficult to be solved using ordinary, explicit step-by-step methods because most of these problems are considered stiff. Certain implicit methods, however, are capable of solving stiff ordinary differential equations (ODEs) usually found in most applied problems. This study aims to develop a new numerical method, namely the high order variable step variable order block backward differentiation formula (VSVO-HOBBDF) for the main purpose of approximating the solutions of third order ODEs. The computational work of the VSVO-HOBBDF method was carried out using the strategy of varying the step size and order in a single code. The order of the proposed method was then discussed in detail. The advancement of this strategy is intended to enhance the efficiency of the proposed method to approximate solutions effectively. In order to confirm the efficiency of the VSVO-HOBBDF method over the two ODE solvers in MATLAB, particularly ode15s and ode23s, a numerical experiment was conducted on a set of stiff problems. The numerical results prove that for this particular set of problem, the use of the proposed method is more efficient than the comparable methods. VSVO-HOBBDF method is thus recommended as a reliable alternative solver for the third order ODEs.
Keywords
Block method; stiff ODEs; third order; variable step variable order | 481 | 2,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-21 | latest | en | 0.891793 |
https://ladywiththepants.com/865ef3da8c4852105f14d30293a908f2.html | 1,653,658,152,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00026.warc.gz | 429,505,284 | 2,998 | # What is volume in math?
## What is volume in math?
In math, volume is the amount of space in a certain 3D object. For instance, a fish tank has 3 feet in length, 1 foot in width and two feet in height. To find the volume, you multiply length times width times height, which is 3x1x2, which equals six. So the volume of the fish tank is 6 cubic feet.
## Why is volume important in the real world?
Whether you're measuring out ingredients for a recipe, filling up a car's gas tank or just adding detergent to the washing machine, math and volume come are used often in daily life. From measuring liquids to assessing drinking amounts, volume is necessary.
## What are three facts about volume?
VolumeThe length is the longest distance between the object's extremities.The width (or breadth) refers to the size of the object in a direction perpendicular to its length.The height (or depth) stands for the size of that object in the direction perpendicular to both the length and the width.
## How is density used in everyday life?
Everyday Density Examples Wood generally floats on water because it is less dense than water. Rocks, generally being denser than water, usually sink. Helium balloons rise because helium is less dense than the surrounding air. Over time, the helium escapes the balloon and is replaced by air, causing it to sink.
## What is volume in the real world?
In a 3-dimensional shape, volume describes the amount of space that the shape encloses.
## Why is volume important to engineering?
In the engineering world, we refer to this as volume. Volume is a useful tool for engineers in any discipline. Civil, environmental and architectural engineers might consider the volume of people that need to occupy a new office building they are designing, or water that is contained in a new reservoir design.
## Why is density important?
Density is an intensive property, meaning that it is a property that is the same no matter how much of a substance is present. Density is an important concept because it allows us to determine what substances will float and what substances will sink when placed in a liquid.
Can you get a banned Xbox unbanned?
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## Does Jenna come back to life?
But how will they bring Mystic Falls' dearly departed denizen back to life? As first reported by our friends at Zap2it, Jenna won't be revived via the CW saga's signature witchery or otherworldly means, but rather through a series of flashbacks that examine Elena's life before her parents' untimely death.
## What episode does Jeremy come back?
Jeremy Gilbert does return to Mystic Falls during The Vampire Diaries season 6 finale to say goodbye to Elena after she succumbs to Kai's sleeping spell. He makes another brief cameo during the season 8 finale, helping Caroline and Alaric get the Salvatore School for the Young and Gifted off the ground. | 625 | 2,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-21 | latest | en | 0.94456 |
https://www.enotes.com/homework-help/5x4-529301 | 1,480,944,114,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541696.67/warc/CC-MAIN-20161202170901-00478-ip-10-31-129-80.ec2.internal.warc.gz | 927,518,850 | 10,959 | # 5x4
freemihai | College Teacher | (Level 2) Adjunct Educator
Posted on
If we are going to perform multiplication operation, then we need to remember that multiplication means repeated addition.
For example, 5x4 (read five times four) is equal to 20. So, all we have to do is to recall that multiplication is telling us how many groups of five numbers to add together. In this case, four groups of five numbers are added together.
5x4 = 4 + 4 + 4 + 4 + 4 = 20
Or we can think in this way: 5 groups of four numbers are added together.
4x5 = 5+5+5+5 = 20 | 160 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-50 | longest | en | 0.961006 |
https://www.mathspage.com/prime-factorizations/solved/prime-factorization-of-49 | 1,675,163,048,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00789.warc.gz | 887,446,052 | 3,264 | USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# What is the Prime Factorization Of 49?
• Equcation for number 49 factorization is: 7 * 7
• It is determined that the prime factors of number 49 are: 7
## Is 49 A Prime Number?
• No the number 49 is not a prime number.
• Forty-nine is a composite number. Because 49 has more divisors than 1 and itself.
## How To Calculate Prime Number Factorization
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What is Prime Number Factorization?
• In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
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### Re^3: I can afford to spend so much time at the Monastery because:
by wolfger (Deacon)
on Feb 09, 2009 at 20:20 UTC ( #742567=note: print w/replies, xml ) Need Help??
1. Albert Einstein clearly explained time, via special relativity.
2. As Benjamin Franklin said "Time is Money"
3. 1 Tim 6:10 clearly states that "The love of money is the root of all evil"(actually, the greek can actually be read "...the root of all sorts of evil")
4. Therefore, "Einstein is the root of all evil" QED (you can get him off the hook a little bit by using the latter tranlation of 1 Tim 6:10)
I hope you code better than you do logical proofs...
1. That Einstein explained time does not make him "time" itself.
2. I will accept the equivalency though it is unproven.
3. That "love of money" is the root of something does not make money itself the root of something.
4. If a = b and d (being "the love of b") is the root of c, that does not in any way show a relationship between e (the person who describes a) and d.
• Comment on Re^3: I can afford to spend so much time at the Monastery because:
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Re^4: I can afford to spend so much time at the Monastery because:
by Illuminatus (Curate) on Feb 09, 2009 at 23:45 UTC
Sorry. I didn't catch the reference.
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[erix]: (but then there's always a next (im)possible goal) Corion avoids running by starting on time [choroba]: my phone showed me 1 hour 10 km which was my goal, it changed a bit after uploading to the website [erix]: choroba I believe you ;) [1nickt]: Corion I assume you are talking about Retry? Attempt doesn;t offer callbacks and has a much simpler interface. (Also written by someone I trust...)
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Results (184 votes). Check out past polls. | 562 | 2,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | longest | en | 0.930769 |
https://www.thejavaprogrammer.com/evaluate-postfix-expression-in-java/ | 1,726,815,330,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00344.warc.gz | 940,101,735 | 23,081 | Evaluate Postfix Expression in Java
In this article, we will look at how to evaluate a given Postfix expression to a result value. This is a problem related to Stack Data Structure. We will look at the algorithm with a step-by-step example along with the implementation in code. Let us first have a quick look at Postfix Expressions.
So, Postfix Expressions are the mathematical expressions where for each pair of Operator and Operands, the operators come after operands. Hence, the evaluation order is from left to right. Here operators are written after operands.
For example, XY+ is a Postfix Expression and its equivalent Infix is X+Y. Compilers generally use Postfix Notations to evaluate a given expression with ease without multiple scanning. It becomes easier to evaluate a given expression due to the order of operators and operands.
Now, Consider the Postfix Expression: 8 2 3 * + 7 / 1 –
The Result after evaluation of this expression will be 1. At first, 2*3 = 6, its sum with 8 gives 14. 14 is then divided by 7 to give 2. After subtracting 1 with 2 we get 1 which is our result.
Note: The Postfix expression will be given as a String Input.
Postfix Evaluation Algorithm
We will use a Stack, the Stack will contain the Operands and the resultant value of each pair of operands for every operator. At the end of the algorithm, the Postfix stack will contain the resultant value obtained from the total expression.
• We traverse the whole String, for each character we encounter we have basically two cases to consider:
• If Character is an Operand: If the current character is an operand/Integer we push it into Stack.
• If Character is an Operator: If the current character is an operator we pop two operands from the Postfix stack evaluate the result with the respective operator and push the result back to Postfix Stack.
• We continue the above two steps for each such character. In the end, the Postfix stack will have only one value i.e. the final answer.
Example
Now, let’s understand the algorithm with a step-by-step example. We consider the same Postfix Expression: 8 2 3 * + 7 / 1 – .
Step 1:
We traverse the string (ignoring spaces) throughout its length from start(0) to End(Length – 1 = 9). At index = 0, we get 8 which is Integer/Operand so we push it into Postfix Stack. Similarly, at index = 1 and index = 2, we get 2 and 3 respectively which are also Operands so we push them into the stack. After these operations the Postfix Stack now looks like this:
Step 2:
We continue traversing and at index = 3, we get ‘ * ‘ an operator. This falls under Case 2 so we pop two items/Operands from the stack i.e. 3 and 2 and compute 3 * 2 = 6 and push the result back to stack. The Postfix stack now is:
Step 3:
Next at index = 4, we get ‘ + ‘ again an operator. So we again pop the two items compute the result 8 + 6 = 14 and push it back to stack again for future evaluation. Now at index = 5, we get 7 which is an operand so we push it into the stack. The stack now looks like this:
Step 4:
Moving again to the right at index = 6 we get, ‘ / ‘ an operator so we again pop the items 7 and 14 and compute the result 14 / 7 = 2, then push it to the stack. At index = 7, we get, 1 an integer so we simply push it into the stack.
Step 5:
Now, at index = 8, we get the ‘ – ‘ operator so we then pop the remaining elements from the stack and compute their result as 2 – 1 =1 and push it back to stack again. Finally, we reach the end of the string, and the Postfix Stack will have the only value which is the result of the total evaluation:
Note: We evaluate each pair of popped operands with the operator in reverse order to avoid errors in the result.
Implementation in Java
```import java.util.*;
public class PostfixEvaluation
{
static boolean isOperator(char ch)
{
if(ch == '+' || ch == '-' || ch == '*' || ch == '/')
return true;
return false;
}
static void evaluatePostfix(String exp)
{
Stack<Integer> postFix = new Stack<>(); // Create postfix stack
int n = exp.length();
for(int i=0;i<n;i++)
{
if(isOperator(exp.charAt(i)))
{
// pop top 2 operands.
int op1 = postFix.pop();
int op2 = postFix.pop();
// evaluate in reverse order i.e. op2 operator op1.
switch(exp.charAt(i))
{
case '+': postFix.push(op2 + op1);
break;
case '-': postFix.push(op2 - op1);
break;
case '*': postFix.push(op2 * op1);
break;
case '/': postFix.push(op2 / op1);
break;
}
}
// Current Char is Operand simple push into stack
else
{
// convert to integer
int operand = exp.charAt(i) - '0';
postFix.push(operand);
}
}
// Stack at End will contain result.
System.out.println(postFix.pop());
}
public static void main(String args[])
{
String exp = "823*+7/1-";
System.out.print("The PostFix Evaluation for the Given Expression "+exp+" is: ");
evaluatePostfix(exp);
}
}```
Output:
`The PostFix Evaluation for the Given Expression 823*+7/1- is: 1`
Time Complexity: We do a single traversal of the Given Postfix Expression throughout its length, so the Time Complexity is O(n), n is the length of Expression.
Space Complexity: We use a Stack which will at the most store all the Operands until we get an Operator so space complexity is O(T), T is the total number of operands in expression.
So that’s all about the article you can try out different examples with the algorithm discussed above and execute the code for a clear idea. | 1,326 | 5,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.887569 |
http://www.mineshgovenjidesigns.com/4hsuy/converse%2C-inverse-contrapositive-examples-6206d5 | 1,628,094,557,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00228.warc.gz | 79,128,810 | 8,603 | Example. Use the statement: Any two points are collinear. The converse would be If B then A or B -> A I told you it was easy. With the inverse, we negated the propositions. EXAMPLES Direct statement: If you drink Pepsi, then you are happy. The Converse, Inverse and Contrapositive. The mini-lesson targeted the fascinating concept of converse statement. To get the contrapositive, we must switch AND negate the propositions. 1) p → q 2) t → ~ w 3) ~ m → p 4) ~ p → ~ q. Definition of the Converse, contrapositive, and inverse of an implication Let P and Q be propositional variables. of p →q ¬q → ¬ p is the . In summary, the original statement is logically equivalent to the contrapositive, and the converse statement is logically equivalent to the inverse. The converse of the conditional statement is “If Q then P.” The contrapositive of the conditional statement is “If not Q then not P.” The inverse of the conditional statement is “If not P then not Q.” We will see how these statements work with an example. Example: Find the converse, inverse, and contrapositiveof “It raining is … So, by the law of contrapositive, the inverse and the converse also have the same truth value. p →q. * The converse … q →p is the . Note: As in the example, the contrapositive of any true proposition is also true. Examples Example 1. True/False Converse: _____True / … we can form new conditional statements . The inverse of a statement is taking the negation of each variable so with the last example we would have the original statement of: If A then B or A -> B The inverse would be If ~A then ~B or ~A -> ~B Contrapositive Improve your math knowledge with free questions in "Converses, inverses, and contrapositives" and thousands of other math skills. of p →q. of p →q ¬ p → ¬ q is the . Inverse. Index Prev Up Next Try these examples out: In 1 – 4, write the inverse of the statement in symbolic form. converse of proposition contrapositive of proposition Contents For the proposition P Q, the proposition Q P is called its converse, and the proposition Q P is called its contrapositive.. For example for the proposition "If it rains, then I get wet", At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! Fancy. Finally, we move on to the contrapositive. This packet will cover "if-then" statements, p and q notation, and conditional statements including contrapositive, inverse, converse, and biconditional. Example 7: Write four related conditional statements (if-then form, the converse, the inverse, and the contrapositive) of “ Guitar players are musicians.” Decide whether each statement is true or false. The following contrapositive statement is logically equivalent to the original if-then statement: "If I do not help you with your homework, then you will not do the dishes." See also. 16) If today is Monday, then it is not the weekend. So we can also write the inverse of p → q as ~p → ~q. The logical converse and inverse of the same conditional statement are logically equivalent to each other. Contents. Choose from 124 different sets of converse contrapositive flashcards on Quizlet. In this case, unlike the last example, the inverse of the statement is true. Justifying each question with examples. Now you can easily find the converse, inverse, and contrapositive of any conditional statement you are given! This picture shows the relationship between a conditional and its inverse, converse, and contrapositive. Contrapositive Examples. If it snows today, I will ski tomorrow. Let’s end this video with an example for you to process how to analyze a statement to write the converse, inverse, and contrapositive statements. Definition of the Converse, Contrapositive, and Inverse of an Implication. The inverse is "If a polygon is not a quadrilateral, then it does not have four sides." Writ the converse, inverse and contrapositive of each of the following statements :
(1) A family becomes literate if the women in it are literate. If-Then Form … Hope you enjoyed learning! Start studying Converse, Inverse, Contrapositive. Start studying Converse, Inverse, Contrapositive. How converse and inverse is not equivalent to conditional statement? Learn converse contrapositive with free interactive flashcards. Converse and Contrapositive Subjects to be Learned. I will be in the talent show only if I can do the same comedy routine I … With the converse, we switched the propositions. inverse. Finally, if you negate everything and flip p and q (taking the inverse of the converse, if you're fond of wordplay) then you get the contrapositive. About Cuemath. Converse: If you are happy, then you drink Pepsi. Previous: The Definition of the Contrapositive. Mathematical Excursions (MindTap Course List) Write the a. converse, b. inverse, and c. contrapositive of the given statement. The Contrapositive Okay, enough with the warm-up, now it's time to get really weird. Converse, Inverse and contrapositive of an implication.Hello friends, Welcome to my channel mathstips4u.In my last video we have seen logical equivalence and some of their examples.In this video we are going to learn converse, inverse and contrapositive of an implication and some of their examples.If p → q is an implication, then there arises following three implications1. Converse, Inverse, and Contrapositive of a Conditional Statement What we want to achieve in this lesson is to be familiar with the fundamental rules on how to convert or rewrite a conditional statement into its converse, inverse, and contrapositive. Examples Example 1 . A positive integer is a prime only if it has no divisors other than 1 and itself. A statement and the inverse contrapositive from cognitive and didactical points of view. Okay, allow me to begin with the first question. I come to class whenever there is going to be a quiz. converse. 1. Counter example: _____ For 16 – 17, write the converse, inverse, and contrapositive statements for each conditional statement. Elementary Foundations: An introduction to topics in discrete mathematics Jeremy Sylvestre. You are probably familiar with at least a little Canadian geography, and know that British Columbia (BC) is a geographical area within Canada. For any given proposition p q, p is also known as the premise or hypothesis and q as the conclusion.For any such a proposition, we have furthermore q p is the converse of p q; p q is the inverse of p q; q p is the contrapositive of p q. In 5 – 8, write the converse of the statement in symbolic form. | 1,477 | 6,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-31 | latest | en | 0.865363 |
https://math.stackexchange.com/questions/3251079/is-multiplicity-of-root-of-polynomial-meaningful-in-any-way | 1,566,090,282,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00479.warc.gz | 568,616,368 | 28,720 | Is multiplicity of root of polynomial meaningful in any way? [closed]
Is multiplicity of root of polynomial meaningful in any way?
I encounter this problem, when I find roots of some polynomial and there are fewer roots found than the order of the polynomial. Which means that some are repeated.
But I wonder if it's necessary to know that "these roots are repeated"? Does it serve any practical purpose?
closed as unclear what you're asking by Lord Shark the Unknown, Xander Henderson, user1551, Leucippus, CesareoJun 5 at 11:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• What is a "practical purpose"? – Lord Shark the Unknown Jun 4 at 19:05
• Is there a specific context you are thinking in? For example, the multiplicities of the eigenvalues (which are roots of a matrix's characteristic polynomial) are important in determining whether a given matrix has an eigenbasis. – paulinho Jun 4 at 19:06
• @paulinho Well just in general. I was finding roots of a 5th order complex polynomial today and I noticed that my factorization showed only 4. Thus I was thinking, do I need to factor it more to discover multiplicities. – mavavilj Jun 4 at 19:07
• Yes, in a big way. Roots of multiplicity greater than one often cause numerical issues. – copper.hat Jun 4 at 19:12
Yes, there are lots of situations where it's important:
1. The degree of the polynomial is the number of roots (over the complex field). So if you don't include multiplicities, you can end up with a polynomial of lower degree. Polynomials are defined up to a constant scaling by their roots (that is, if you take two polynomials that are the same except that one is the other times a constant, then the two will have the same roots), so if you don't include multiplicities, you don't get the same polynomial.
2. The multiplicity of the root is one more than the number of derivatives with a root. For instance, if the multiplicity is 3, then the first 2 derivatives will have a root there.
3. If the root has odd multiplicity, the graph will intersect the x-axis. If it is even, it will be tangent.
4. When you're working with the characteristic polynomial of a matrix, multiple roots can indicate that the eigenspace for that eigenvalue is multidimensional. If that's too advanced for you, the simplified version is that you can have things associated with each root, and when you have different things associated with the same root, that root tends to be multiple. Conversely, if a root is multiple, that may mean that you have several of the things associated with the root.
5. There are wide variety of more esoteric applications; as just one example, elliptic curve math requires distinguishing between single and double roots.
In addition to the given solution, when using a numerical method such as Newton-Raphson, it is important to take the number of repeated roots in consideration to get a better speed. If you know that there are repeated roots you'd use the modified version of the algorithm. See for example newton-raphson-method-for-double-roots.
Also, if you are constructing the polynomial from a graph, it is important to recognize repeated roots so that you could create the correct polynomial function, for example $$(x-1)$$ is a different function from $$(x-1)^2$$.
(I don't know if this counts as an answer to the question, but it's too long for a comment.)
If a polynomial $$p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0$$ has a root $$w$$ of multiplicity $$m$$, and $$C$$ is a closed disc with centre $$w$$ containing no other roots of $$p$$, then there is a number $$\delta > 0$$ such that any polynomial $$p^*(z) = z^n + a^*_{n-1}z^{n-1} + \cdots + a^*_1z + a^*_0$$ has exactly $$m$$ roots, counted with multiplicity, in the disc $$C$$, so long as $$|a^*_i - a_i| < \delta \quad (i = 0, 1, \ldots, n-1).$$ Conversely, if $$p(z) = (z - w)^mq(z)$$, where $$q(w) \ne 0$$, then, for any $$\delta > 0$$, there is a disc $$C$$ with centre $$w$$ such that the coefficients of any polynomial of the form $$p^*(z) = (z - w_1)\cdots(z - w_m)q(z)$$ differ from the corresponding coefficients of $$p(z)$$ by less than $$\delta$$, so long as $$w_1, \ldots, w_m \in C.$$ So multiple roots behave like distinct roots with small changes in the polynomial. | 1,145 | 4,538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.948446 |
https://www.esaral.com/q/a-jar-of-height-h-is-filled-with-a-transparent-liquid-93619 | 1,726,084,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00325.warc.gz | 712,212,418 | 11,554 | # A jar of height h is filled with a transparent liquid
Question:
A jar of height h is filled with a transparent liquid of refractive index µ. At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.
Solution:
tan ic d/2/h
ic = d/2h
d = 2h tan ic
d = 2h ×1/√μ2 – 1 | 113 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.811326 |
https://number.academy/117010 | 1,660,686,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00464.warc.gz | 401,574,477 | 11,984 | # Number 117010
Number 117,010 spell 🔊, write in words: one hundred and seventeen thousand and ten . Ordinal number 117010th is said 🔊 and write: one hundred and seventeen thousand and tenth. Color #117010. The meaning of number 117010 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 117010. What is 117010 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 117010.
## What is 117,010 in other units
The decimal (Arabic) number 117010 converted to a Roman number is (C)(X)(V)MMX. Roman and decimal number conversions.
#### Weight conversion
117010 kilograms (kg) = 257960.2 pounds (lbs)
117010 pounds (lbs) = 53075.4 kilograms (kg)
#### Length conversion
117010 kilometers (km) equals to 72707 miles (mi).
117010 miles (mi) equals to 188310 kilometers (km).
117010 meters (m) equals to 383887 feet (ft).
117010 feet (ft) equals 35666 meters (m).
117010 centimeters (cm) equals to 46066.9 inches (in).
117010 inches (in) equals to 297205.4 centimeters (cm).
#### Temperature conversion
117010° Fahrenheit (°F) equals to 64987.8° Celsius (°C)
117010° Celsius (°C) equals to 210650° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
117010 seconds equals to 1 day, 8 hours, 30 minutes, 10 seconds
117010 minutes equals to 2 months, 3 weeks, 4 days, 6 hours, 10 minutes
### Codes and images of the number 117010
Number 117010 morse code: .---- .---- --... ----- .---- -----
Sign language for number 117010:
Number 117010 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 117010
### Multiplications
#### Multiplication table of 117010
117010 multiplied by two equals 234020 (117010 x 2 = 234020).
117010 multiplied by three equals 351030 (117010 x 3 = 351030).
117010 multiplied by four equals 468040 (117010 x 4 = 468040).
117010 multiplied by five equals 585050 (117010 x 5 = 585050).
117010 multiplied by six equals 702060 (117010 x 6 = 702060).
117010 multiplied by seven equals 819070 (117010 x 7 = 819070).
117010 multiplied by eight equals 936080 (117010 x 8 = 936080).
117010 multiplied by nine equals 1053090 (117010 x 9 = 1053090).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 117010
Half of 117010 is 58505 (117010 / 2 = 58505).
One third of 117010 is 39003,3333 (117010 / 3 = 39003,3333 = 39003 1/3).
One quarter of 117010 is 29252,5 (117010 / 4 = 29252,5 = 29252 1/2).
One fifth of 117010 is 23402 (117010 / 5 = 23402).
One sixth of 117010 is 19501,6667 (117010 / 6 = 19501,6667 = 19501 2/3).
One seventh of 117010 is 16715,7143 (117010 / 7 = 16715,7143 = 16715 5/7).
One eighth of 117010 is 14626,25 (117010 / 8 = 14626,25 = 14626 1/4).
One ninth of 117010 is 13001,1111 (117010 / 9 = 13001,1111 = 13001 1/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
117010
#### Is Prime?
The number 117010 is not a prime number. The closest prime numbers are 116993, 117017.
#### Factorization and factors (dividers)
The prime factors of 117010 are 2 * 5 * 11701
The factors of 117010 are 1 , 2 , 5 , 10 , 11701 , 23402 , 58505 , 117010
Total factors 8.
Sum of factors 210636 (93626).
#### Powers
The second power of 1170102 is 13.691.340.100.
The third power of 1170103 is 1.602.023.705.101.000.
#### Roots
The square root √117010 is 342,067245.
The cube root of 3117010 is 48,911126.
#### Logarithms
The natural logarithm of No. ln 117010 = loge 117010 = 11,670015.
The logarithm to base 10 of No. log10 117010 = 5,068223.
The Napierian logarithm of No. log1/e 117010 = -11,670015.
### Trigonometric functions
The cosine of 117010 is -0,188053.
The sine of 117010 is -0,982159.
The tangent of 117010 is 5,22278.
### Properties of the number 117010
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 117010 in Computer Science
Code typeCode value
PIN 117010 It's recommendable to use 117010 as a password or PIN.
117010 Number of bytes114.3KB
CSS Color
#117010 hexadecimal to red, green and blue (RGB) (17, 112, 16)
Unix timeUnix time 117010 is equal to Friday Jan. 2, 1970, 8:30:10 a.m. GMT
IPv4, IPv6Number 117010 internet address in dotted format v4 0.1.201.18, v6 ::1:c912
117010 Decimal = 11100100100010010 Binary
117010 Decimal = 12221111201 Ternary
117010 Decimal = 344422 Octal
117010 Decimal = 1C912 Hexadecimal (0x1c912 hex)
117010 BASE64MTE3MDEw
117010 MD517ce80c3401c238baa2295720648d971
117010 SHA1e3e1ac8c057875c02eda0c6866c39c6143de0640
117010 SHA2248dcf93edc5dbe14626c286747398b527c803fa963bfabcfe2692c36c
117010 SHA256fa7ae4ed4137446159fa63903cd3b588192c0961d73e151e7107b9eb72c8ab34
117010 SHA384d2fb2006e8b7cec43485c77dc7227ba7118a50377efe1a3d26aa44950236b922f2e8efe9d8ac811f648f4bf41ab03b95
More SHA codes related to the number 117010 ...
If you know something interesting about the 117010 number that you did not find on this page, do not hesitate to write us here.
## Numerology 117010
### Character frequency in number 117010
Character (importance) frequency for numerology.
Character: Frequency: 1 3 7 1 0 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 117010, the numbers 1+1+7+0+1+0 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## Interesting facts about the number 117010
### Asteroids
• (117010) 2004 JG1 is asteroid number 117010. It was discovered by CSS, Catalina Sky Survey from Catalina Station, Mount Bigelow on 5/10/2004.
## № 117,010 in other languages
How to say or write the number one hundred and seventeen thousand and ten in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 117.010) ciento diecisiete mil diez German: 🔊 (Anzahl 117.010) einhundertsiebzehntausendzehn French: 🔊 (nombre 117 010) cent dix-sept mille dix Portuguese: 🔊 (número 117 010) cento e dezessete mil e dez Chinese: 🔊 (数 117 010) 十一万七千零十 Arabian: 🔊 (عدد 117,010) مائة و سبعة عشر ألفاً و عشرة Czech: 🔊 (číslo 117 010) sto sedmnáct tisíc deset Korean: 🔊 (번호 117,010) 십일만 칠천십 Danish: 🔊 (nummer 117 010) ethundrede og syttentusindti Dutch: 🔊 (nummer 117 010) honderdzeventienduizendtien Japanese: 🔊 (数 117,010) 十一万七千十 Indonesian: 🔊 (jumlah 117.010) seratus tujuh belas ribu sepuluh Italian: 🔊 (numero 117 010) centodiciassettemiladieci Norwegian: 🔊 (nummer 117 010) en hundre og sytten tusen og ti Polish: 🔊 (liczba 117 010) sto siedemnaście tysięcy dziesięć Russian: 🔊 (номер 117 010) сто семнадцать тысяч десять Turkish: 🔊 (numara 117,010) yüzonyedibinon Thai: 🔊 (จำนวน 117 010) หนึ่งแสนหนึ่งหมื่นเจ็ดพันสิบ Ukrainian: 🔊 (номер 117 010) сто сiмнадцять тисяч десять Vietnamese: 🔊 (con số 117.010) một trăm mười bảy nghìn lẻ mười Other languages ...
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## Comment
If you know something interesting about the number 117010 or any natural number (positive integer) please write us here or on facebook. | 2,452 | 7,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-33 | latest | en | 0.672962 |
https://algebra-calculators.com/access-answers-to-ncert-class-7-maths-chapter-15-visualising-solid-shapes-exercise-15-3/ | 1,675,643,957,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00020.warc.gz | 109,104,651 | 17,907 | ## Access Answers to NCERT Class 7 Maths Chapter 15 – Visualising Solid Shapes Exercise 15.3
### Access Answers to NCERT Class 7 Maths Chapter 15 – Visualising Solid Shapes Exercise 15.3
1. What cross-sections do you get when you give a
(i) vertical cut (ii) horizontal cut
to the following solids?
(a) A brick
Solution:-
The cross-section of a brick when it is cut into vertically is as shown in the figure below,
The cross-section of a brick when it is cut into horizontally is as shown in the figure below,
(b) A round apple
Solution:-
The cross-section of a round apple when it is cut into vertically is as shown in the figure below,
The cross-section of a round apple when it is cut into horizontally is as shown in the figure below,
(c) A die
Solution:-
The cross-section of a die when it is cut into vertically is as shown in the figure below,
The cross-section of a die when it is cut into horizontally is as shown in the figure below,
(d) A circular pipe
Solution:-
The cross-section of a circular pipe when it is cut into vertically is as shown in the figure below,
The cross-section of a circular pipe when it is cut into horizontally is as shown in the figure below,
(e) An ice cream cone
Solution:-
The cross-section of an ice cream when it is cut into vertically is as shown in the figure below,
The cross-section of an ice cream when it is cut into horizontally is as shown in the figure below, | 328 | 1,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | longest | en | 0.941658 |
http://www.business-analysis-made-easy.com/practical-business-math.html | 1,529,645,198,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864354.27/warc/CC-MAIN-20180622045658-20180622065658-00426.warc.gz | 379,125,818 | 8,125 | # Calculating Markup
by Uma
(Delaware, USA)
Please explain more about markup. Preferably with examples. I don't understand how to calculate the markup using price and margin.
Dee Reavis
Okay, here is an example: You can buy a widget for \$15. You have no trouble selling it for \$20. Your margin or gross profit is \$5, which is the difference between the sales price and the purchase cost. Cost markup is \$5/\$15 = .33 or 33%. Sales price markup is \$5/\$20 = .25 or 25%. If you are still confused by this just use the Margin and Markup Calculator.
## What Business Ratios Are Important To Calculate?
by Dora
(Gilbert, Arizona)
I want to know what ratios a Business Analyst should compute to find the businesses productivity and ROI.
Feb 11, 2010 Key Business Ratios by: Dee Reavis Here are some key business ratios that can help to evaluate the health of a company.Current RatioThis ratio is the current assets divided by the current liabilities. It gives an indication of whether a company is likely able to pay its debts.Quick RatioWhen you total a companies cash, marketable securities and its accounts receivable and divide the total by its current liabilities you get the current ratio. This is another measure of how well a company can pay its debts. If this ratio is at least 1 the company is in good shape. Even higher is betterDebt To EquityDebt is a 2 edged sword. When business is booming, debt makes a company grow faster. That is called leverage. When business is bad, big chunks to the company's capital can go to paying debt. The dept to equity ratio gives a measure of where that a company is at in that leverage to risk spectrum.Sales To InventoryThis one measure how fast inventory is turning over. If this ratio is 4, then it means that sales equals the amount invested in inventory times 4.Profit MarginProfit margin is the ratio of the profit received for a product divided by the revenue from that product expressed as a percentage. A company has less risk if its profit margin is high.
## Gross Margin Mark Up
by Donnah
(Ocala, FL)
How much should I mark up a product (course of study) if I want to achieve a 90% gross margin?
### Comments for Gross Margin Mark Up
Mar 24, 2010 Gross Margin Mark Up by: Dee Reavis The formula for calculation is:`Gross Margin Percentage = (Revenue-Cost of Product)/RevenueSo to achieve 90% GMP then:(Revenue - Cost of Product) = .9 x Revenue or.1 x Revenue = Cost of ProductRevenue = 10 x Cost of ProductThe markup would have to be 900%.`
## Margin Example
by Olu's Margriet
(Ila Orangun Osun State,nigerial)
Explain and show how margin works.
Apr 23, 2010 Margin Example by: Dee Reavis There are 2 ways that you might want to use margin in a business: gross margin and profit margin.1. Gross Margin: Gross margin is the difference between the sales price of an item you are selling and its cost to you.Suppose you are making a dress to sell. The cost for materials and labor to make the dress is \$15.00. You sell the dress for \$25.00. The gross margin is \$10.00. In terms of percent it is ((25-15)/25)x100 or 40%.2. Profit Margin: Profit margin is similar, but usually applied on broader level. It is net income divided by net sales.If your clothes making business has an income after costs of \$15,000 for total sales of \$50,000, then your profit margin is (15,000/50,000)x100 or 30%.
## Calculate Percentages
by DAVID
(VIRGINIA)
75% OF WHAT NUMBER IS 96?
This is easily calculated by dividing .75 into 96.
96/.75=128
## Calculate Percentage
Let's say that I'm taking a test, and the test has 24 questions. I end up getting 5 wrong. What percentage of the test did I get right?
This is how to calculate percentage:
(24-5)/24=.79 or 79%
Custom Search | 903 | 3,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-26 | latest | en | 0.94865 |
http://math.stackexchange.com/questions/95395/the-taylor-series-of-int-0x-operatornamesinct-dt | 1,469,322,952,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00318-ip-10-185-27-174.ec2.internal.warc.gz | 164,180,995 | 17,958 | # The Taylor series of $\int_0^x \operatorname{sinc}(t) dt$
I tried to find what is the Taylor series of the function $$\int_0^x \frac{\sin(t)}{t}dt .$$
Any suggestions?
-
Recall the series for $\sin\,t$, divide out $t$ for each term, and integrate each term. – J. M. Dec 31 '11 at 12:27
Pls try to format your math, it's not difficult to learn (right-click->"show source" over the equation to see) – leonbloy Dec 31 '11 at 13:11
@J.M. I think you should put your comment as an answer. – a.r. Dec 31 '11 at 16:27
Yes, JM this one belongs to you. I shall defer. – ncmathsadist Dec 31 '11 at 19:08
Sometimes "sinc" means this one $\mathrm{sinc}(x) = \sin(\pi x)/(\pi x)$. – GEdgar Jan 9 '13 at 14:28
To settle this:
We have the Maclaurin expansion
$$\sin\,t=t\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right)$$
Upon obtaining the expansion of $\dfrac{\sin\,t}{t}$ from this, integrate each term of this series expansion, using the formula
$$\int_0^x t^k \mathrm dt=\frac{x^{k+1}}{k+1}$$
-
I just want to extend J.M.'s solution to a full solution of the exercise:
$$\int_0^x \frac{\sin(t)}{t} \,dt$$
Is with Maclaurin expansion ($\sin\,t=t\left(1-t^2/3!+t^4/5!-t^6/7!+\cdots\right)$) equals to
$$\int_0^x \frac{t}{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = \int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt$$
We do now integrate this series to:
$$\int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots$$
Above we did use the flowing fact: $$\int_0^x t^k \, dt=\left.\frac{t^{k+1}}{k+1}\right|_0^x = \frac{x^{k+1}}{k+1}-\frac{0^{k+1}}{k+1} = \frac{x^{k+1}}{k+1}$$
We can now write the result as a series: $$x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) * (2n+1)!} = Si(x)$$
And see that we get the Sine Integral.
- | 857 | 1,971 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-30 | latest | en | 0.704517 |
https://www.thefreedictionary.com/orthogonal+matrix | 1,653,443,223,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00227.warc.gz | 1,157,621,544 | 11,669 | # orthogonal matrix
Also found in: Encyclopedia, Wikipedia.
## orthogonal matrix
n
(Mathematics) maths a matrix that is the inverse of its transpose so that any two rows or any two columns are orthogonal vectors. Compare symmetric matrix
Collins English Dictionary – Complete and Unabridged, 12th Edition 2014 © HarperCollins Publishers 1991, 1994, 1998, 2000, 2003, 2006, 2007, 2009, 2011, 2014
Mentioned in ?
References in periodicals archive ?
The optimal precoding and decoding matrices can be chosen from an arbitrary orthogonal matrix O [member of] [C.sup.MxM] as follows:
X is the orthogonal matrix composed of the feature vectors corresponding to the eigenvalues.
Every possible rotation R (a 3 x 3 special split orthogonal matrix) can be constructed from either one of the two related split quaternions q = [q.sub.0]1 + [q.sub.1]i + [q.sub.2]j + [q.sub.3]k or -q = -[q.sub.0]1 - [q.sub.1]i - [q.sub.2]i - [q.sub.3]k using the transformation law [8]:
and [mathematical expression not reproducible] is an orthonormal matrix, [[bar.V].sub.0] [member of] [R.sup.rxr] is an orthogonal matrix, and [[bar.[summation]].sub.0] [member of] [R.sup.rxr] is a diagonal matrix with nonnegative elements.
it is not difficult to see that Q is just the orthogonal matrix generated by applying one QR iteration from [B.sup.T]B to [B.sup.T]B with zero shift [8].
Taking into account the definition of an orthogonal matrix, the result will also remain an orthogonal matrix.
Given a symmetric positive semidefinite matrix W [greater than or equal to] 0, rank (W) [less than or equal to] r, r >:, there exists an orthogonal matrix U such that
[DELTA](H) [greater than or equal to] 1, and only when H is an orthogonal matrix, [DELTA](H) = 1.
This technique is based on the theorem from linear algebra which says that a rectangular matrix A can be given as the product of three matrices, as follows: (i) an orthogonal matrix UA, (ii) a diagonal matrix [summation]A, (iii) the transpose of the orthogonal matrix VA.
Here H represents an orthogonal matrix. This form normalized to have total mass unity is represented by (dH).
Ultrasound-assisted extraction (UAE) of pectic polysaccharide from oriental tobacco leaves was studied by orthogonal matrix method (L9(3)4).
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